Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Simplifying the derivative of $x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$ $$x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}$$
So I get:
$$-x^{\frac{2}{3}} \cdot \frac{1}{3} (6-x) ^{\frac{-2}{3}} + (6-x) ^{\frac{1}{3}} \cdot \frac{2}{3} x ^ {\frac{-1}{3}}$$
How does one go about simplifying this?
I guess I can pull out common terms like this:
$$\frac{1}{3} x ^{-\frac{1}{3}} (6-x)^{\frac{-2}{3}} ( -x + (6-x) \cdot 2)$$
Is that right?
|
Alternatively:
$$\frac{d}{dx}\left(x^{\frac{2}{3}} \cdot (6-x)^{\frac{1}{3}}\right)=\frac{d}{dx}\left((6x^2-x^3)^{\frac13}\right)=\frac{1}{3}(6x^2-x^3)^{-\frac{2}{3}}(12x-3x^2)=\\
\frac{4x-x^2}{(6x^2-x^3)^{\frac{2}{3}}}=\frac{4x-x^2}{x^{\frac{4}{3}}(6-x)^{\frac23}}=\frac{4-x}{x^{\frac{1}{3}}(6-x)^{\frac23}}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2768051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
}
|
Show that the given linear system has this specific solution if $\det \neq 0$
Show that the linear system $\begin{pmatrix} a_{11} & a_{12}\\
a_{21} & a_{22} \end{pmatrix} \begin{pmatrix} x_1\\ x_2
\end{pmatrix}= \begin{pmatrix} b_1\\ b_2 \end{pmatrix} \,\,\,\,\,\,
\text{ with }\,\,\,\, A= \begin{pmatrix} a_{11} & a_{12}\\ a_{21} &
a_{22} \end{pmatrix}$
has the unique solution $\begin{pmatrix} x_1\\ x_2 \end{pmatrix}$
with $x_1 = \frac{\det \begin{pmatrix} b_1 & a_{12}\\ b_2 & a_{22}
\end{pmatrix}}{\det A} \,\,\,\,\,\,; \,\,\,\,\, x_2 = \frac{\det
\begin{pmatrix} a_{11} & b_1\\ a_{21} & b_2 \end{pmatrix}}{\det A}$,
if $\det A \neq 0.$
I have an idea and I think it's almost correct but something is missing or wrong..
Let $\vec{x} = \begin{pmatrix}
x_1\\
x_2
\end{pmatrix}$ and let $\vec{b} = \begin{pmatrix}
b_1\\
b_2
\end{pmatrix}$. For the linear system $A\vec{x} = \vec{b}$ there exists the unique solution $\begin{pmatrix}
x_1\\
x_2
\end{pmatrix}$, if $\det A \neq 0$ because we can use the inverse of matrix $A$:
$$A\vec{x}=\vec{b} \Leftrightarrow A^{-1}A\vec{x} = A^{-1} \vec{b} \Leftrightarrow A^{-1} \vec{b} = \vec{x}$$
Know that $A^{-1} = \frac{1}{\det A} \begin{bmatrix}
a_{22} & -a_{12}\\
-a_{21} & a_{11}
\end{bmatrix}$
Then we have $A^{-1} \vec{b} = \vec{x} \Leftrightarrow \frac{1}{\det A} \begin{bmatrix}
a_{22} & -a_{12}\\
-a_{21} & a_{11}
\end{bmatrix}
\begin{bmatrix}
b_1\\
b_2
\end{bmatrix} =
\begin{bmatrix}
x_1\\
x_2
\end{bmatrix}$
But how is this supposed to be equal to the solution given in the task?
I did a mistake somewhere? :(
|
You are almost there; simply write out what the two determinants
$$\det \begin{pmatrix} b_1 & a_{12}\\ b_2 & a_{22}
\end{pmatrix}\qquad\text{ and }\qquad\det
\begin{pmatrix} a_{11} & b_1\\ a_{21} & b_2 \end{pmatrix}$$
are. You will find that these are precisely the coefficients of the vector
$$\begin{bmatrix}
a_{22} & -a_{12}\\
-a_{21} & a_{11}
\end{bmatrix}
\begin{bmatrix}
b_1\\
b_2
\end{bmatrix}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2771179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
Minimum value of $\dfrac{a+b+c}{b-a}$
$f(x)= ax^2 +bx +c ~ ~~(a<b)$ and $f(x)\ge 0~ \forall x \in \mathbb R$ .
Find the minimum value of $\dfrac{a+b+c}{b-a}$
Attempt:
$b^2 \le 4ac$
$f(1) = a+b+c$
$f(0) = c$
$f(-1) = a-b+c$
$a>0$
and $c>0$
I am unable to utilize these things to find the minimum value of the expression $\equiv \dfrac{a+b+\frac {b^2}{4a}}{b-a}$
The answer given is $3$.
|
Alt. hint: $\;f(-2) = 4a - 2 b + c \ge 0 \iff 2a + c \ge 2(b-a)\,$, then:
$$
\dfrac{a+b+c}{b-a} = 1 + \frac{2a+c}{b-a} \;\ge\; 1 + 2 = 3
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2771575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Calculate $\sum\limits_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately Is it possible to calculate $\sum_{k=0}^{20}(-1)^k\binom{k+2}2$ without calculating each term separately?
The original question was find the number of solutions to $2x+y+z=20$ which I calculated to be the coefficient of $x^{20}$ in $(1+x^2+x^4\dots)(1+x+x^2\dots)^2$ which simplified to the term above.
I know $\sum_{k=0}^{20}\binom{k+2}{2}=\binom{23}3$ but the $(-1)^k$ is ruining things.
|
Using the identity from Pascal's Triangle, we get
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{k+a}{b}
&=\sum_{k=0}^n(-1)^k\left[\binom{k+a+1}{b}-\binom{k+a}{b-1}\right]\\
&=\sum_{k=1}^{n+1}(-1)^{k-1}\left[\binom{k+a}{b}-\binom{k+a-1}{b-1}\right]\\
&=\frac12\left(\binom{a}{b}+(-1)^n\binom{n+a+1}{b}-\sum_{k=1}^{n+1}(-1)^{k-1}\binom{k+a-1}{b-1}\right)\\
&=\frac12\left(\binom{a}{b}+(-1)^n\binom{n+a+1}{b}-\sum_{k=0}^n(-1)^k\binom{k+a}{b-1}\right)\tag1
\end{align}
$$
Multiplying by $(-2)^b$, we get
$$
\begin{align}
(-2)^b\sum_{k=0}^n(-1)^k\binom{k+a}{b}
&=\frac12(-2)^b\left[\binom{a}{b}+(-1)^n\binom{n+a+1}{b}\right]\\
&+(-2)^{b-1}\sum_{k=0}^n(-1)^k\binom{k+a}{b-1}\\
&=\frac12\sum_{j=1}^b(-2)^j\binom{a}{j}+\frac{(-1)^n}2\sum_{j=1}^b(-2)^j\binom{n+a+1}{j}\\
&+[n\equiv0\bmod2]\\
&=\bbox[5px,border:2px solid #C0A000]{\frac12\sum_{j=0}^b(-2)^j\binom{a}{j}+\frac{(-1)^n}2\sum_{j=0}^b(-2)^j\binom{n+a+1}{j}}\tag2
\end{align}
$$
Setting $a=b=2$ yields
$$
\begin{align}
\sum_{k=0}^n(-1)^k\binom{k+2}{2}
&=\frac18\left[\binom{2}{0}-2\binom{2}{1}+4\binom{2}{2}\right]\\
&+\frac{(-1)^n}8\left[\binom{n+3}{0}-2\binom{n+3}{1}+4\binom{n+3}{2}\right]\\[6pt]
&=\frac18+\frac{(-1)^n}8\left(2n^2+8n+7\right)\tag3
\end{align}
$$
Setting $n=20$ yields
$$
\begin{align}
\sum_{k=0}^{20}(-1)^k\binom{k+2}{2}
&=\frac18+\frac{(-1)^{20}}8\left(2\cdot20^2+8\cdot20+7\right)\\[6pt]
&=121\tag4
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2772397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
}
|
How to find all rational solutions of $\ x^2 + 3y^2 = 7 $? I knew that for $ x^2 + y^2 = 1$ the x and y can be expressed by introducing one more variable where $\ m=y/(x+1) $, then $\ x= 2m/(1+m^2) $ and $\ y= (1-m^2)/(1+m^2) $. What about $\ x^2 + 3y^2 = 7 $, should I divide the equation by 7 in order to get the 1 at the right hand side ? Then how to deal with the $\ 3y^2 $ ? Thank you!
|
There are only six solutions which are also algebraic integers, and they can each be found from whichever one you discover first.
For instance, $$(2 + \sqrt{-3}) \left( \frac{-1 + \sqrt{-3}}{2} \right) = \frac{-5 + \sqrt{-3}}{2}$$ and $$\frac{-5 + \sqrt{-3}}{2} \left( \frac{-1 + \sqrt{-3}}{2} \right) = \frac{1 - 3 \sqrt{-3}}{2}.$$
There are infinitely many solutions, but for each viable denominator $d$ there are only two or four solutions. For example, from the very strategically withheld solution $$\frac{10}{31} + \frac{47 \sqrt{-3}}{31}$$ (which was inexplicably deleted from this page earlier today) we can obtain $$\omega \left( \frac{10}{31} + \frac{47 \sqrt{-3}}{31} \right) = \frac{151 + 37 \sqrt{-3}}{62}$$ and verify that that number does have a norm of $7$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2773097",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 5,
"answer_id": 1
}
|
A unit square has its corner chopped off to form a regular polygon with eight sides. What is the area of the polygon?
A unit square has its corner chopped off to form a regular polygon with eight sides. What is the area of the polygon? Source: ISI BMATH UGA 2017 paper
A regular polygon with 8 sides can be divided into eight congruent triangles .I tried to find the area of a triangle in the following method. An angle of a triangle is 360/8=45..then I draw a perpendicular bisector of the angle which is height of the triangle and I found it to be 1/2 as the square is of unit length .Now I got a right angled triangle from which I wanted to find the length of the base of the triangle but I couldn't do so.
|
An image makes everything simpler to understand:
Because the square side length is 1,
$$b + a + b = 1$$
Solving for $b$ we get
$$b = \frac{1 - a}{2} \label{NA1}\tag{1}$$
Squaring this (noting that we are limited to positive $a$ and $b$, i.e. that $a \gt 0$ and $b \gt 0$) we get
$$b^2 = \frac{1 - 2 a + a^2}{4} \label{NA2}\tag{2}$$
In the corners, we need $b$ such that the hypotenuse is $a$. Pythagorean theorem says
$$b^2 + b^2 = a^2$$
solving for $b$ we get
$$b^2 = \frac{a^2}{2} \label{NA3}\tag{3}$$
Combining $\eqref{NA2}$ and $\eqref{NA3}$ we get
$$b^2 = \frac{a^2 - 2 a + 1}{4} = \frac{a^2}{2}$$
i.e.
$$2 a^2 = a^2 - 2 a + 1$$
which simplifies to
$$a^2 + 2 a - 1 = 0$$
and noting that the first two terms are from squaring $a$, to
$$a^2 + 2 a + 1 - 2 = (a + 1)^2 - 2 = 0$$
Moving the last term ($-2$) to the other side, we get
$$(a + 1)^2 = 2$$
so taking a square root on both sides, and remembering that we need $a \gt 0$, we get
$$a = \sqrt{2} - 1$$
Substituting this to $\eqref{NA1}$ we can solve $b$,
$$b = \frac{1 - a}{2} = \frac{1 - \sqrt{2} + 1}{2} = \frac{2 - \sqrt{2}}{2} = 1 - \sqrt{\frac{1}{2}}$$
Note also that
$$b^2 = \left(1 - \sqrt{\frac{1}{2}}\right)^2 = 1 - 2 \sqrt{\frac{1}{2}} + \frac{1}{2} = \frac{3}{2} - \sqrt{2}$$
The corner triangles total area is the same as the area of two $b$-sided squares. So, the area of the octagon is the area of the unit square (1) minus the area of two $b$-sided squares:
$$A_{octagon} = 1 - 2 b^2 = 1 - 2 (\frac{3}{2} - \sqrt{2}) = 1 - 3 + 2 \sqrt{2} = 2 \sqrt{2} - 2 \approx 0.828$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2775575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
If $a_1 = 3$ and $a_{n+1}=a_n^2-2$, then $\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{1}{a_1 a_2 a_3}+\cdots = \frac{3-\sqrt{5}}{2}$ Let $a_1=3$，and $a_{n+1}=a_n^2-2$ (for $n=1, 2, \cdots$). Show that
$$\frac{1}{a_1}+\frac{1}{a_1 a_2}+\frac{1}{a_1a_2a_3}+\cdots+\frac{1}{a_1a_2 \cdots a_n}+\cdots=\frac{3-\sqrt{5}}{2}.$$
My thought
In fact, it's natural to find the general term expression of $a_n$. Indeed, $a_n$ could be expressed explicitly as follows $$a_n=x^{2^{n-1}}+\frac{1}{x^{2^{n-1}}}~,$$ where $x$ is any of the two roots of the equation $x^2-3x+1=0$, i.e. $x=\dfrac{3 \pm \sqrt{5}}{2},$ and $n=1,2,\cdots.$
It's simple to prove that by mathematical induction. Let $n=1$, then $$x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}=x+\dfrac{1}{x}=\dfrac{x^2+1}{x}=\dfrac{3x}{x}=3=a_1.$$Assume that $a_k=x^{2^{k-1}}+\dfrac{1}{x^{2^{k-1}}},$ then $$a_{k+1}=a_k^2-2=\left(x^{2^{k-1}}+\dfrac{1}{x^{2^{k-1}}}\right)^2-2=x^k+\frac{1}{x^k}+2-2=x^k+\frac{1}{k}.$$ Thus, the general term formula above of $a_n$ holds for any $n=1,2,\cdots.$ From this, we have have $$a_1=\frac{1+x^2}{x},a_2=\frac{1+x^4}{x^2},\cdots，a_n=\frac{1+x^{2^n}}{x^{2^{n-1}}}.$$ But can those offer some help to solve the problem？Besides,the general term expression is what someone told me, but he didn't tell me how to get this. Who can show me?
Another thought
Let $b_n=\dfrac{a_n-\sqrt{a_n^2-4}}{2}$, where $a_n>2$. Then \begin{align*}b_{n+1}&=\dfrac{a_{n+1}-\sqrt{a_{n+1}^2-4}}{2}=\dfrac{a_{n}^2-2-\sqrt{(a_{n}^2-2)^2-4}}{2}=\dfrac{a_{n}(a_n-\sqrt{a_n^2-4})-2}{2}=a_nb_{n}-1.&\end{align*}
Hence, $$b_n=\frac{1}{a_n}+\frac{b_{n+1}}{a_n}.$$Therefore,$$b_1=\frac{1}{a_1}+\frac{b_2}{a_1}=\frac{1}{a_1}+\frac{1}{a_1}\left(\frac{1}{a_2}+\frac{b_3}{a_2}\right)=\frac{1}{a_1}+\frac{1}{a_1a_2}+\frac{b_3}{a_1a_2}=\cdots$$
Expand like this repeatedly till infinitely. Then we have $$b_1=\dfrac{1}{a_1}+\dfrac{1}{a_1a_2}+\dfrac{1}{a_1 a_2 a_3}+\cdots.$$ It follows that $$\dfrac{1}{a_1}+\dfrac{1}{a_1a_2}+\dfrac{1}{a_1 a_2 a_3}+\cdots=b_1=\dfrac{3-\sqrt{5}}{2}.$$
Please correct me if I'm wrong!
|
Notice that $a_1=3>2$. Suppose that $a_k>2$, then $a_{k+1}=a_k^2-2>2^2-2=2$. By mathematical induction, we have $a_n>2,$ further, $a_n^2>a_n+2$. Hence, $a_{n+1}=a_n^2-2>(a_n+2)-2=a_n.$ This shows that $a_n$ is an increasing integer sequence. Therefore, $a_n \to +\infty$ as $n \to \infty$.
Thus, we have $$\lim_{n \to \infty}\frac{\dfrac{1}{a_1 a_2 \cdots a_{n+1}}}{\dfrac{1}{a_1 a_2 \cdots a_{n}}}=\lim_{n \to \infty}\dfrac{1}{a_{n+1}}=0.$$ According to the ratio test of positive series, we have that $\sum\limits_{n=1}^{\infty}\dfrac{1}{a_1a_2\cdots a_n}$ is convergent. Hence, let it be $\dfrac{a_1-\lambda}{2},$ where $\lambda$ is an unknown real number we need to find.
It's simple to find that $$\frac{a_2-\lambda a_1}{2}=\frac{1}{a_2}+\frac{1}{a_2a_3}+\cdots$$$$\frac{a_3-\lambda a_1a_2}{2}=\frac{1}{a_3}+\frac{1}{a_3a_4}+\cdots$$$$\vdots$$$$\frac{a_n-\lambda a_1a_2\cdots a_{n-1}}{2}=\frac{1}{a_n}+\frac{1}{a_na_{n+1}}+\cdots$$
Since $$0<\frac{1}{a_n}+\frac{1}{a_na_{n+1}}+\cdots <\frac{1}{a_n}+\frac{1}{a_n^2}+\cdots=\frac{1}{a_n-1}，$$ hence $$0<\frac{a_n-\lambda a_1a_2\cdots a_{n-1}}{2}<\frac{1}{a_n-1}.$$
According to the Squeeze Theorem,$$\lim_{n \to \infty}\frac{a_n-\lambda a_1a_2\cdots a_{n-1}}{2}=0.$$ Thus, $$\lambda=\lim_{n \to \infty}\frac{a_n}{a_1a_2\cdots a_{n-1}}.$$
Moreover, from $a_{n+1}=a_n^2-2$, we have $$a_n^2-4=a_{n-1}^2(a_{n-2}^2-4)=\cdots=a_1^2a_2^2\cdots a_{n-1}^2(a_1^2-4).$$
Thus, $$\left(\frac{a_n}{a_1a_2\cdots a_{n-1}}\right)^2=\frac{4}{a_1^2a_2^2\cdots a_{n-1}^2}+(a_1^2-4).$$
Taking the limits as $n \to \infty$ of the both sides, we have $$ \lambda^2=0+5=5.$$ Then,$$\lambda=\sqrt{5}.$$ As a result, $$\frac{1}{a_1}+\frac{1}{a_1 a_2}+\frac{1}{a_1a_2a_3}+\cdots+\frac{1}{a_1a_2 \cdots a_n}+\cdots=\frac{3-\sqrt{5}}{2}.$$ This is exactly what we want to prove.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2776422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Prove that $ a^n + b^n + c^n = d^n + e^n + f^n $ by induction If $a,b,c,d,e,f$ are six real numbers such that:
$$ a + b + c = d + e + f $$
$$ a^2 + b^2 + c^2 = d^2 + e^2 + f^2 $$
$$ a^3 + b^3 + c^3 = d^3 + e^3 + f^3 $$
Prove by mathematical induction that:
$$ a^n + b^n + c^n = d^n + e^n + f^n $$
I tried solving this question by correlating to
$$ a^k + b^k = (a + b)(a^{k-1} + b^{k-1}) - ab(a^{k-2} + b^{k-2}) $$
The problem is that terms with $abc$ do not come in common while expanding. Could you please help me solve this question?
This question originates from the level III excercise of SK Goyal's book of Algebra for JEE Mains and Advanced.
|
Consider the polynomial $p(x)=x^3-sx^2+ux-v$ where $s=a+b+c=d+e+f$, $u=ab+bc+ca=\frac 12\left((a+b+c)^2-a^2+b^2+c^2\right)=de+ef+fd$ and $v=abc=\frac 13\left((a^3+b^3+c^3)-s(a^2+b^2+c^2)+u(a+b+c)\right)=def$
Then $p(a)=p(b)=p(c)=p(d)=p(e)=p(f)=0$ and you can use $$a^rp(a)+b^rp(b)+c^rp(c)=0$$ to obtain an expression for the sum $a^{r+3}+b^{r+3}+c^{r+3}$ in terms of sums of lower powers and the common constants $s,u,v$. This can be used for the inductive step.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2778849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
How to evaluate $\int_{0}^{1}\arccos\sqrt{1-x}\ln(x-x^2)\ln\left(\frac{1}{x}-1\right)dx?$ How may one demonstrate this integral is equal to this result?
$$\int_{0}^{1}\arccos\sqrt{1-x}\ln(x-x^2)\ln\left(\frac{1}{x}-1\right)\mathrm dx=\pi\zeta(2)-2\pi[1-\ln^2(2)]$$
Where do I start from?
|
This answer will primarily show how Waiting most likely got his answer. To start off, let's first define $I$ to be the integral under question, and using the basic logarithmic rules we all learned in first grade, $I$ can be transformed as
$$\begin{align*}I & =\int\limits_0^1dx\,\arccos\sqrt{1-x}\log^2(1-x)-\int\limits_0^1dx\,\arccos\sqrt{1-x}\log^2x\\ & =\int\limits_0^1dx\,\arccos\sqrt x\log^2x-\int\limits_0^1dx\,\arccos\sqrt x\log^2(1-x)\tag1\end{align*}$$
From ($1$), make the substitution $x\mapsto\sqrt x$ to get rid of the radical in the $\arccos(\cdot)$ function. Therefore, ($1$) turns into the form
$$\begin{align*}I & =8\int\limits_0^1dx\, x\arccos x\log^2x-2\int\limits_0^1dx\, x\arccos x\log^2(1-x^2)\tag2\end{align*}$$
Next, substitute $x\mapsto\cos x$ to get
$$I=8\int\limits_0^{\pi/2}dx\, x\sin x\cos x\log^2\cos x-8\int\limits_0^{\pi/2}dx\, x\sin x\cos x\log^2\sin x\tag3$$
Now, integration by parts, as difficult as it may seem, can be used to reduce the integral into something the beta function can easily manage. First, denote $I_1$ and $I_2$ as the first and second integral respectively of ($3$). Letting $u=x$ and integrating by parts on $I_1$ gives
$$\begin{align*}I_1 & =\frac {\pi}2+4\int\limits_0^{\pi/2}dx\,\cos^2x\log^2\cos x-4\int\limits_0^{\pi/2}dx\,\cos^2x\log\cos x\\ & =\frac {\pi}2+4\int\limits_0^{\pi/2}dx\,\sin^2\log^2\sin x-4\int\limits_0^{\pi/2}dx\,\sin^2x\log\sin x\\I_2 & =\frac {\pi}2-4\int\limits_0^{\pi/2}dx\,\sin^2x\log^2\sin x+4\int\limits_0^{\pi/2}dx\,\sin^2 x\log\sin x\end{align*}$$
Note that in the second line, we exploited the identity $\cos\left(\frac {\pi}2-x\right)=\sin x$ and used the limit identity for integrals. Putting everything together in ($3$) leaves
$$\begin{align*}I & =8\int\limits_0^{\pi/2}dx\,\sin^2\log^2\sin x-8\int\limits_0^{\pi/2}dx\,\sin^2x\log\sin x\\ & =8\frac {\partial^2}{\partial a^2}\int\limits_0^{\pi/2}dx\,\sin^{a+2}x-8\frac {\partial}{\partial b}\int\limits_0^{\pi/2}dx\,\sin^{b+2}x\end{align*}$$
Using equation ($14$) from here gives the answer as
$$\begin{align*}I & =4\sqrt{\pi}\frac {\partial^2}{\partial a^2}\frac {\Gamma\left(\frac a2+\frac 32\right)}{\Gamma\left(\frac a2+2\right)}-4\sqrt{\pi}\frac {\partial}{\partial b}\frac {\Gamma\left(\frac b2+\frac 32\right)}{\Gamma\left(\frac b2+2\right)}\\ & =\frac {\pi}6\biggr[\pi^2+6\log 2\left(\log 4-2\right)-6\biggr]-\pi+2\pi\log 2\end{align*}$$
A little simplifying reveals the answer
$$\int\limits_0^1dx\,\arccos\sqrt{1-x}\log(x-x^2)\log\left(\frac 1x-1\right)\color{blue}{=\frac {\pi^3}6+2\pi\log^22-2\pi}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2780457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
}
|
Find the Limit of $\prod_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{(1+\frac{1}{n+x})^{n+x}}$ Consider the following productions
$$\prod_{n=1}^{\infty}\frac{1+\frac{1}{n}}{1+\frac{1}{n+x}}$$
and
$$\prod_{n=1}^{\infty}\frac{(1+\frac{1}{n})^n}{(1+\frac{1}{n+x})^{n+x}}$$
I know that the above are convergent.
Can anyone find the limit of these products?
|
For the first one, rewrite the product:
\begin{equation*}
\begin{split}
\prod_{n=1}^{\infty} \frac{(n+1)(n+x)}{n(n+x+1)}& = \frac{(1+1)(1+x)}{1(1+x+1)} \times\frac{(2+1)(2+x)}{2(2+x+1)} \times \frac{(3+1)(3+x)}{3(3+x+1)}\times ...\\
& = \frac{2(1+x)}{1(2+x)} \times\frac{3(2+x)}{2(3+x)} \times \frac{4(3+x)}{3(4+x)}\times ...\\
& = (1+x).
\end{split}
\end{equation*}
For the second one, rewrite the product as
\begin{equation*}
\begin{split}
\prod_{n=1}^{\infty} \frac{(n+1)^n(n+x)^{n+x}}{n^n(n+x+1)^{n+x}} & = \frac{(1+1)^1(1+x)^{1+x}}{1^1(1+x+1)^{1+x}} \times \frac{(2+1)^2(2+x)^{2+x}}{2^2(2+x+1)^{2+x}}\times \frac{(3+1)^3(3+x)^{3+x}}{3^3(3+x+1)^{3+x}}\times ...\\
& = \frac{2(1+x)^{1+x}}{(2+x)^{1+x}} \times \frac{3^2(2+x)^{2+x}}{2^2(3+x)^{2+x}}\times \frac{4^3(3+x)^{3+x}}{3^3(4+x)^{3+x}}\times ...\\
& = (1+x)^{1+x} \times \frac{(2+x)}{2}\times \frac{(3+x)}{3}\times ...\\
& = (1+x)^{1+x}\prod_{n=2}^{\infty}\frac{n+x}{n}.
\end{split}
\end{equation*}
The product is equal to zero for $x<0$, it is equal to $1$ for $x=0$, and for $x>0$ the product is divergent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2787307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
}
|
Intersection between $y=\frac{x^2+1}{x+1}$ and $y=x+2$ on $y$-axis I stumbled upon this problem: Considering the function $f:(-1;\infty)\rightarrow R$, $$f(x)=\frac{x^2+1}{x+1}$$ Find the sum of the values on positive $y$-axis where the tangent of the function is perpendicular to the line $y=x+2$
After letting $(0,a)$ be the point on positive $y$-axis, with the equation of the tangent $y-{{y}_{0}}=f'({{x}_{0}})(x-{{x}_{0}})$ gives slope to be $-1$ So the tangent is $y-a=-1(x-0)=-x$
Is this correct? Can you help me furthermore with this problem?
|
$f(x)=\frac{x^2+1}{x+1}$
$\implies f'(x) = \frac{2x(x+1)-x^2-1}{(x+1)^2}$
$\implies f'(x) = \frac{2x^2+2x-x^2-1}{(x+1)^2}$
$\implies f'(x) = \frac{x^2-1+2x}{(x+1)^2}$
set this to $-1$ as tangent of $f(x)$ is perpendicular to $y=x+2$
$\frac{x^2-1+2x}{(x+1)^2}=-1$
$x^2-1+2x=-(x^2+1+2x)$
$2x^2+4x =0$
$2x(x+2)=0$
$\therefore x=0,-2$
So at $x=0,-2$ the tangents drawn will be perpendicular to the line $y-x+2$
On the OY axis lies $x=0,y=1$, So at the point $(0,1)$ the tangent of the function $f(x)$ is perpendicular to $y=x+2$
EDIT 1:
graph for reference
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2787738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$
Proving by definition that $\lim_{(x,y) \to (1,2)}\frac{3x-4y}{x+y}=-\frac{5}{3}$
Take $\epsilon>0$, I want to find $\delta>0$ such that:
$$\lVert (x-1,y-2)\rVert <\delta \Rightarrow \left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert<\epsilon$$
So I started by adding both fractions and obtained:
$$\left\lvert \frac{3x-4y}{x+y}+\frac{5}{3}\right\rvert=\frac{7}{3}\left\lvert\frac{2x-y}{x+y}\right\rvert=\frac{7}{3}\left\lvert\frac{2x-y-2+2}{x+y}\right\rvert=\frac{7}{3}\left\lvert\frac{2(x-1)-(y-2)}{x+y}\right\rvert$$
Now, I have $\lvert x-1\rvert\leq\sqrt{(x-1)^2+(y-2)^2}<\delta$ and $\lvert y-2\rvert\leq\sqrt{(x-1)^2+(y-2)^2}<\delta$
However, im not being able to bound $\frac{1}{\lvert x+y\rvert}$
Am I on the correct track? Any suggestions? Thank you very much.
|
$\lim_{(x,y) \to (1,2)}\dfrac{3x-4y}{x+y}
=-\dfrac{5}{3}
$
I like to let
variables go to zero,
so let
$x = 1+u, y=2+v$.
Then
$\begin{array}\\
\dfrac{3x-4y}{x+y}+\dfrac{5}{3}
&=\dfrac{3(1+u)-4(2+v)}{1+u+2+v}+\dfrac{5}{3}\\
&=\dfrac{3u-4v-5}{3+u+v}+\dfrac{5}{3}
\qquad\text{You can see here that the limit will be -5/3}\\
&=\dfrac{3(3u-4v-5)+5(3+u+v)}{3(3+u+v)}\\
&=\dfrac{14u-7v}{3(3+u+v)}
\qquad\text{and the constant term cancels out}\\
&=\dfrac{14u-7v}{9+3(u+v)}\\
\end{array}
$
From this we see that
if $u$ and $v$ are small
the the numerator is small
and the denominator
is not small,
being around $9$.
To make this rigorous,
if $|u|, |v| < c$
where
(to get a lower bound
on the denominator)
$0 < c < \frac12$,
then
$|14u-7v|
\lt 21c
$ and
$|9+3(u+v)|
\ge 9-3|u+v|
\ge 9-3(2c)
\gt 6
$
so
$|\dfrac{14u-7v}{9+3(u+v)}|
\le \dfrac{21c}{6}
=\dfrac{7c}{2}
$.
Therefore,
to make
$|\dfrac{14u-7v}{9+3(u+v)}|
\lt d$,
choose
$\dfrac{7c}{2}
\lt d$
or
$c \lt \dfrac{2d}{7}
$.
Finally,
to go back to
the original problem,
if
$|x-1| < \dfrac{2d}{7}$
and
$|y-2| < \dfrac{2d}{7}$
where
$\dfrac{2d}{7} < \frac12$
then
$|\dfrac{3x-4y}{x+y}+\dfrac{5}{3}|
\lt d
$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2787946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Prove that $\sqrt{11}-1$ is irrational by contradiction I am working on an assignment in discrete structures and I am blocked trying to prove that $\sqrt{11}-1$ is an irrational number using proof by contradiction and prime factorization.
I am perfectly fine doing it with only $\sqrt{11}$, but I am completely thrown off by the $-1$ when it comes to the prime factorization part.
My current solution looks like this :
$$ \sqrt {11} -1 = \frac {a}{b}$$
$$ \sqrt {11} = \frac {a}{b} + 1$$
$$ \sqrt {11} = \frac {a+b}{b}$$
$$ 11 = \left(\frac {a+b}{b}\right)^2$$
$$ 11 = \frac {(a+b)^2}{b^2}$$
$$ 11 = \frac {a^2 + 2ab + b^2}{b^2}$$
$$ 11 b^2 = a^2 + 2ab +b ^2$$
$$ 10b^2 = a^2 + 2ab $$
At that point, is it acceptable to conclude that a² is a multiple of 11 even though we have a trailing $2ab$?
The required method is then to conclude using prime factorization that $a = 11k$ and replace all that in the formula above to also prove $b$, however, I am again stuck with the ending $2ab$.
Would it instead be correct to prove that $\sqrt{11}$ is rational using the usual method and that, by extension, $\sqrt{11} - 1$ is also rational?
Thank you
|
No, you cannot conclude that $a^2$ is a multiple of $11$. You can instead rewrite
$$
a^2=10b^2-2ab=2(5b^2-ab)
$$
so $2\mid a$. Write $a=2c$, with $c$ integer. Then
$$
4c^2=2(5b^2-2bc)
$$
or $5b^2=2(c^2+bc)$. Since $2\nmid 5$, we conclude $2\mid b$.
This is a contradiction to $a$ and $b$ being coprime.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2789537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
}
|
Indefinite integral of $\sqrt{x^2-x}$ i was trying to compute the indefinite integral:
$$
\int\sqrt{x^2-x}dx
$$
but i got stuck: after a few (unsuccessful) attempts for some $u$-substitution, i tried integration by parts:
$$
\int\sqrt{x^2-x} \ dx=\int(x)'\sqrt{x^2-x} \ dx= \\ x\sqrt{x^2-x}-\int x(\sqrt{x^2-x})'dx= \\
=x\sqrt{x^2-x}-\frac{1}{2}\int x\frac{2x-1}{\sqrt{x^2-x}}dx= \\
=x\sqrt{x^2-x}-\int \frac{x^2}{\sqrt{x^2-x}}dx+\frac{1}{2}\int \frac{x}{\sqrt{x^2-x}}dx=...
$$
and now what? Can anybody help?
|
Integrate by parts instead as follows
\begin{align}\int\sqrt{x^2-x}\ dx
= &\int\frac{\sqrt{x^2-x}}{2(x-\frac12)}d[(x-\frac12)^2]\\
= &\ \frac12(x-\frac12) \sqrt{x^2-x}-\frac18\int \frac 1{ \sqrt{x^2-x}}dx \\
=& \ \frac12(x-\frac12) \sqrt{x^2-x}
-\frac18\tanh^{-1}\frac{\sqrt{x^2-x}}{x-\frac12}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2790831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
}
|
Matrix Exponentiation $A^{15}$ A is a 2x2 matrix. Let it satisfy $A^2 = A - I$, where $I = \begin{bmatrix} 1 & 0\\ 0 & 1\end{bmatrix}$. I want to find $A^{15}$:
My Approach:
I isolated $A$:
$A^2-A=-I$
$A(A-I)=-I$
From that, I just tried to solve the system that would generate:
$\begin{bmatrix}a & b \\ c & d\end{bmatrix}\begin{bmatrix}a-1 & b \\ c & d-1\end{bmatrix} = \begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$
Turns out this doesn't help me get closer to the answer as it leads to:
$\begin{bmatrix}a(a-1)+bc & b(d-1)+ab\\c\left(a-1\right)+dc &d(d-1)+bc\end{bmatrix} = \begin{bmatrix}-1 & 0 \\ 0 & -1\end{bmatrix}$
Is that the correct approach, or else, which property am I missing in order to solve this problem?
|
We can show immediately that $A^3=-I$, so $A^{15}=(A^3)^5=(-I)^5=-I$, for instance:
$$
\begin{aligned}
A^3+I
&=(A+I)\underbrace{(A^2-A+I)}_{=0}\\
&=0\ .
\\&\qquad\qquad\text{Alternatively:}
\\
A^{15}+I
&=(A^3+I)(A^{12}-A^9+A^6-A^3+I)\\
&=(A+I)\underbrace{(A^2-A+I)}_{=0}(A^{12}-A^9+A^6-A^3+I)\\
&=0\ .
\end{aligned}
$$
Or one can compute the eigenvalues $\lambda1,\lambda_2$, of $A$, see that $\lambda_1^3=\lambda_2^3=-1$, are the eigenvalues of $A^3$, so $A^3=-I$, etc.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2793674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
}
|
Finding the Jordan canonical form of A and Choose the correct option Let $$ A = \begin{pmatrix}
0&0&0&-4 \\ 1&0&0&0 \\ 0&1&0&5 \\ 0&0&1&0
\end{pmatrix}$$
Then a Jordan canonical form of A is
Choose the correct option
$a) \begin{pmatrix}
-1&0&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$b) \begin{pmatrix}
-1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$c) \begin{pmatrix}
1&1&0&0 \\ 0&1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
$d) \begin{pmatrix}
-1&1&0&0 \\ 0&-1&0&0 \\ 0&0&2&0 \\ 0&0&0&-2
\end{pmatrix}$
My attempt : I know that Determinant of A = product of eigenvalues of A, as option c and d is not correct because Here Determinant of A = 4
that is $ \det A = -(-4) \begin{pmatrix}1 & 0 &0\\0& 1 & 0\\ 0&0&1\end{pmatrix}$
I'm in confusion about option a) and b).......how can I find the Jordan canonical form of A ?
PLiz help me.
Any hints/solution will be appreciated.
Thanks in advance
|
The matrix is diagonalizable, since it is a $4\times4$ matrix with $4$ distinct eigenvalues. Therefore, the correct option is a).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2796632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
}
|
Generating Function for $(4,9,16,25,36,.....)$ I have a sequence $(4,9,16,25,36,...)$ it is being generated by $a_n=(n+1)^2)$
I have found the generating function for $n^2$ here:
Proving the generating function of $n^2$.
I know I can shift a sequence:
$(1,4,9,16,25,....)$ to the right $(0,1,4,9,25,...)$ via $x^1*A(x)$ with $ A(x)$ a generating function. Does this also work with a left shift? Is there a definition? $n$ was a natural numbers. Maybe something else?
|
We consider a power series $A(x)$ with constant term $a_0=0$.
\begin{align*}
A(x)=\sum_{j=j_0}^\infty a_jx^j
\end{align*}
A power series $B(x)=\sum_{j=j_0}^\infty a_{j+1}x^j$ can be derived via
\begin{align*}
A(x)&=\sum_{j=j_0}^\infty a_j x^j=\sum_{j=j_0-1}^\infty a_{j+1} x^{j+1}\\
&=x\sum_{j=j_0}^\infty a_{j+1}x^j+a_{j_0}x^{j_0}\\
&=xB(x)+a_{j_0}x^{j_0}
\end{align*}
We conclude the left-shifted power series $B(x)$ has the representation
\begin{align*}
\color{blue}{B(x)=\frac{1}{x}A(x)-a_{j_0}x^{j_0-1}}
\end{align*}
In the current situation we have
\begin{align*}
A(x)&=\frac{x(1+x)}{(1-x)^3}\\
&=x+4x^2+9x^3+16x^4+\cdots\\\\
\color{blue}{B(x)}&=\frac{1}{x}A(x)-1\\
&=\frac{1+x}{(1-x)^3}-1\\
&=\frac{x(x^2-3x+4)}{(1-x)^3}\\
&\,\,\color{blue}{=4x+9x^2+16x^3+\cdots}
\end{align*}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2798231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
}
|
Prove using calculus or otherwise the following inequality Prove that$$ \frac {2}{\pi -2}\ \le\ \frac {\sin x-x\cos x}{x-\sin x}\ <\ 2$$where $$x\in\left(0,\frac{\pi}{2}\right]$$
My Attempt:$$ f'(x)=\frac{ x\sin x(x-\sin x)-(1-\cos x)(\sin x-x\cos x)}{(x-\sin x)^2}=\frac{ x^2\sin x-(1-\cos x)x-\sin x(1-\cos x)}{(x-\sin x)^2}$$
After this not able to get breakthrough.
|
I did by using brute force
$$f'(x)=\frac{\sin x\left(x-\frac{1-\cos x}{2\sin x}(1+\sqrt{5+4\cos x})\right)\left(x-\frac{1-\cos x}{2\sin x}(1-\sqrt{5+4\cos x})\right)}{(x-\sin x)^2}=\frac{\sin x\left(x-\tan \frac{x}{2}(1+\sqrt{5+4\cos x})\right)\left(x-\tan \frac{x}{2}(1-\sqrt{5+4\cos x})\right)}{(x-\sin x)^2}$$
Let $$g(x)=x-\tan \frac{x}{2}\left(1+\sqrt{5+4\cos x}\right)$$
$$g'(x)=1-\frac{1}{2}\sec^2\frac{x}{2}\left(1+\sqrt{5+4\cos x}\right)-\tan \frac{x}{2}.\frac{-4\sin x}{2\sqrt{5+4\cos x}}$$
$$=1-\frac{1+\sqrt{5+4\cos x}}{1+\cos x}+\frac{2(1-\cos x)}{\sqrt{5+4\cos x}}$$
$$=\frac{\cos x\left(\sqrt{5+4\cos x}-4\right)-2\cos^2x-3}{(1+\cos x)\sqrt{5+4\cos x}}<0$$
$g(x)$ is a decreasing function. So,if $x>0$,then $g(x)<g(0)$. i.e $g(x)<0$
Similarly, Let $$h(x)=x-\tan \frac{x}{2}\left(1-\sqrt{5+4\cos x}\right)$$
$$h'(x)=1-\frac{1}{2}\sec^2\frac{x}{2}\left(1-\sqrt{5+4\cos x}\right)-\tan \frac{x}{2}.(-1)\frac{-4\sin x}{2\sqrt{5+4\cos x}}$$
$$=1-\frac{1-\sqrt{5+4\cos x}}{1+\cos x}-\frac{2(1-\cos x)}{\sqrt{5+4\cos x}}$$
$$=\frac{\cos x\left(\sqrt{5+4\cos x}+4\right)+2\cos^2x+8}{(1+\cos x)\sqrt{5+4\cos x}}>0$$
$h(x)$ is a increasing function. So,if $x>0$,then $h(x)>h(0)$. i.e $h(x)>0$
Thus it can be concluded that $$f'(x)<0$$ and we are done.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2800023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Rearrangement inequality and minimal value of $\frac{\sin^3x}{\cos x} +\frac{\cos^3x}{\sin x}$ For $x \in \left(0, \dfrac{\pi}{2}\right)$, is the minimum value of $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x} = 1$? So considering ($\dfrac{1}{\cos x}$, $\dfrac{1}{\sin x}$) and ($\sin^3x$, $\cos^3x$), is it right to use the rearrangement inequality and conclude that $\dfrac{\sin^3x}{\cos x} +\dfrac{\cos^3x}{\sin x}$ is more than or equal to $\dfrac{\sin^3x}{\sin x} +\dfrac{\cos^3x}{\cos x}$ which is equal to $1$? Thanks.
|
Put $a = \sin x, b = \cos x \implies a^2+b^2=1\implies \dfrac{a^2}{b}+\dfrac{b^2}{a} > \dfrac{a^2}{1} + \dfrac{b^2}{1} = a^2+b^2 = 1$ . In fact this is a different inequality, but still looks nice though. To return to the question above, observe that $\dfrac{a}{b}+\dfrac{b}{a} \ge 2$ by AM-GM inequality. Thus $(a^2+b^2)\left(\dfrac{a}{b}+\dfrac{b}{a}\right) \ge 2\implies \dfrac{a^3}{b} + \dfrac{b^3}{a} + 2ab \ge 2\implies \dfrac{a^3}{b}+\dfrac{b^3}{a} \ge 2 -2ab \ge 2 - (a^2+b^2) = 2-1 = 1$ by AM-GM inequality again.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2800162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
}
|
Proving there exists $k$ such that $p(n)p(n+1)=p(k)$ Exact Question
Let $p(x)$ be monic quadratic polynomial over $\mathbb{Z}$. Show that for any integer $n$, there exists an integer $k$ such that $p(n)p(n+1)=p(k)$
Expanding the polynomial just creates a jumbled mess. Is there any intuitive way for this? Any hints appreciated
|
Let $p(x)=x^2+bx+c$ due to monicity (if this is even a word). Then $$\begin{align}p(n)p(n+1)&=(n^2+bn+c)((n+1)^2+b(n+1)+c)\\&=(n^2+bn+c)(n^2+(2+b)n+1+b+c)\\&=(n^2+bn+c)^2+(n^2+bn+c)(2n+1+b)\\&=(n^4+2bn^3+(b^2+2c)n^2+2bcn+c^2)+(n^3+(1+3b)n^2+(b^2+b+c)n+c+bc)\\&=n^4+(1+2b)n^3+(1+b^2+3b+2c)n^2+(b^2+b+2bc+c)n+(bc+c+c^2)\end{align}$$ Now if $p(n)p(n+1)=p(k)$ then $$n^4+(1+2b)n^3+(1+b^2+3b+2c)n^2+(b^2+b+2bc+c)n+(bc+c^2)=k^2+bk\tag{1}$$ for some $k$.
Firstly, to get $bc+c^2$, we must have that the coefficient independent of $n$ in $k$ is $c$ because then $k^2+bk$ 'ends' in $bc+c^2$.
Next, since $p(n)p(n+1)$ is also monic, so must $k$. Hence we currently have $$k=n^2+\text{something}\cdot n+c$$
Finally, substituting this into $(1)$ yields $$\boxed{k=n^2+(1+b)n+c}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2800287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Evaluate $\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}$. Problem
Evaluate $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}.$$
Solution
Notice that $$\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}=\lim\limits_{x \to \infty}x \cdot \left[\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}\right].$$
According to Taylor's Formula $(1+x)^{\alpha}=1+\dfrac{\alpha}{1!}x+\mathcal{O}(x)$, we have $$\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=1+\frac{2}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right)$$and $$\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=1-\frac{2}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right).$$Therefore,$$\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}=\frac{4}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right).$$As a result,
\begin{align*}
\lim\limits_{x \to \infty}\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2}-1}}&=\lim\limits_{x \to \infty}x \cdot \left[\left(1+\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}-\left(1-\frac{\sqrt{2}}{x}\right)^{\sqrt{2}}\right]\\
&=\lim\limits_{x \to \infty}x \cdot \left[
\frac{4}{x}+\mathcal{O}\left(\frac{\sqrt{2}}{x}\right)\right]\\
&=4.
\end{align*}
Hope to see another solution. Thanks!
|
Let $f(x) = x^{\sqrt{2}}$. Let's accept for the moment(actually, Yves' answer justifies this statement) that
$$f'(x) \sim \frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{2\sqrt{2}}$$
, reminiscent of the definition of the derivative.
Now, it then follows that
$$\lim_{x \to \infty} \frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{x^{\sqrt{2} - 1}} = \lim_{x \to \infty} \left(\frac{(x+\sqrt{2})^{\sqrt{2}}-(x-\sqrt{2})^{\sqrt{2}}}{2\sqrt{2}} \cdot \frac{2\sqrt{2}}{x^{\sqrt{2}-1}}\right) = \lim_{x \to \infty} \left(f'(x)\cdot \frac{2\sqrt{2}}{x^{\sqrt{2}-1}}\right)$$
But $f'(x) = \sqrt{2}x^{\sqrt{2}-1}$, so the above simplifies to
$$\lim_{x \to \infty}\left(\sqrt{2}x^{\sqrt{2} -1} \cdot \frac{2\sqrt{2}}{x^{\sqrt{2}-1}}\right) = \lim_{x\to\infty} 4 = 4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2800617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Solving recurrence relation of this form by iteration for closed form The recurrence relation is of the form:
$$ G(n) = 5G(n-1) +\frac{4^n}{4^2},\quad G(1)=3$$
My answer always differs from that on wolfram Alpha where is my mistake??
My steps are
$ G(n) = 5^2G(n-2) +\frac{5\cdot 4^n}{4^3}+\frac{4^n}{4^2} $
$G(n)= 5^3G(n-3) +\frac{5^2\cdot 4^n}{4^4}+\frac{5\cdot 4^n}{4^3}+\frac{4^n}{4^2} $
and so on after $n-1$ steps of the process we reach $G(1)$ and
$G(n)= \frac{5^n}{5}G(1) +4^n(\frac{5^n}{5^2\cdot 4^n\cdot 4}+\dots+\frac{5}{4^3}+\frac{1}{4^2}) $
then I use the sum of geometric series and sub 3 for $G(1)$. I think the mistake is some where in the above.
|
You have extra $4$ here: $+4^n(\frac{5^n}{5^2\cdot 4^n\cdot \color{red}{4}}+$.
Preserving the degrees of $4$:
$$\begin{align}G(n) = &5G(n-1) +\frac{4^n}{4^2}=5\left(5G(n-2) +\frac{4^{n-1}}{4^2}\right)+\frac{4^n}{4^2}=\\&5^2G(n-2) +\frac{5\cdot 4^{n-1}}{4^2}+\frac{4^n}{4^2}=5^2\left(5G(n-3) +\frac{4^{n-2}}{4^2}\right)+\frac{5\cdot 4^{n-1}}{4^2}+\frac{4^n}{4^2}=\\
&5^3G(n-3) +\frac{5^2\cdot 4^{n-2}}{4^2}+\frac{5\cdot 4^{n-1}}{4^2}+\frac{4^n}{4^2}=\cdots \\
&5^{n-1}G(1)+\frac{5^{n-2}\cdot 4^2}{4^2}+\frac{5^{n-3}\cdot 4^3}{4^2}+\cdots +\frac{5\cdot 4^{n-1}}{4^2}+\frac{4^n}{4^2}=\\
&3\cdot 5^{n-1}+\frac{5^{n-2}\cdot 4^2}{4^2}\left(1+\frac{4}{5}+\left(\frac45\right)^2+\cdots +\left(\frac45\right)^{n-2}\right)=\\
&3\cdot 5^{n-1}+5^{n-2}\cdot \frac{1\cdot \left(1-\left(\frac45\right)^{n-1}\right)}{1-\frac45}=\\
&3\cdot 5^{n-1}+5^{n-1}-4^{n-1}=\\
&4\cdot 5^{n-1}+4^{n-1}.\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2801807",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Solve the equation $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$
Solve $3^{\log_4(x)+\frac{1}{2}}+3^{\log_4(x)-\frac{1}{2}}=\sqrt{x}$.
I am able to reduce the LHS to $\sqrt{x}=3^{\log_4(x)} \cdot \dfrac{4}{3}$. Squaring both sides do not seem to lead to a result. Do you know how to proceed?
|
$$
3^{log_4x+\frac{1}{2}}+3^{log_4x-\frac{1}{2}}=\sqrt{x}\\
(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) 3^{log_4x}=\sqrt{x}\\
$$
Write $3 = 4^{\log_4 3}$:
$$
(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) 4^{\log_4 (3) \cdot log_4x}=\sqrt{x}\\
$$
Take log to base 4:
$$
\log_4(3^{\frac{1}{2}}+3^{-\frac{1}{2}}) +\log_4 (3) \cdot \log_4x=\frac12 \log_4 x\\
$$
So
$$
\log_4(x) = \frac{\log_4(3^{\frac{1}{2}}+3^{-\frac{1}{2}})}{\frac12 -\log_4 (3) }
$$
and
$$
x = 4^{\frac{\log_4(3^{\frac{1}{2}}+3^{-\frac{1}{2}})}{\frac12 -\log_4 (3) }}
$$
or, simplified further,
$$
x = 4^{\frac{2\log_4(3^{\frac{1}{2}}+3^{-\frac{1}{2}})}{1-2\log_4 (3) }}\\
= 4^{\frac{\log_4((3^{\frac{1}{2}}+3^{-\frac{1}{2}})^2)}{1-2\log_4 (3) }} =
4^{\frac{\log_4(3 + 2 + \frac13)}{1-2\log_4 (3) }} \\
=
4^{\frac{\log_4(\frac{16}{3})}{1-2\log_4 (3) }} =(\frac{16}{3})^ {\frac{1}{1-2\log_4 (3) }}
$$
You may also want to write it with the $\exp$ function or find some other convenient way of expressing it.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2806518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
}
|
Triple Integrals in Spherical Coordinates, problem with boundaries Calculate
$$\iiint \frac{1}{{x^2+y^2+z^2}}dA$$
Where
$x^2+y^2+(z-2)^2\le1$
I've used spherical coordinates, like this:
$x=\rho\sin\phi\cos\theta$; $y=\rho\sin\phi\sin\theta$; $z=\rho\cos\phi$ and $J=\rho^2\sin\phi$
but then I am having a rough time with the boundaries.
I am stuck at-
$$\rho^2\sin^2\phi\cos^2\theta + \rho^2\sin^2\phi\sin^2\theta +(\rho \cos\phi-2)^2 \le 1$$
$$ \rho^2-4\rho\cos\phi \ +4\le1$$
and I don't know where to go from here.
Thank you
|
By letting $z=2+w$ the problem boils down to computing
$$\iiint_{x^2+y^2+w^2\leq 1}\frac{d\mu}{x^2+y^2+w^2+4w+4} $$
or, by setting $w=\rho\cos\theta,y=\rho\sin\theta\sin\varphi,x=\rho\sin\theta\cos\varphi$,
$$ 2\pi\int_{0}^{1}\int_{0}^{\pi}\frac{\rho^2\sin\theta}{\rho^2+4\rho\cos\theta+4}\,d\theta\,d\rho.$$
Let us focus on the inner integral:
$$ \int_{0}^{\pi}\frac{\rho^2\sin\theta}{\rho^2+4\rho\cos\theta+4}\,d\theta = \frac{\rho}{2}\,\log\left(\frac{2+\rho}{2-\rho}\right) $$
by the tangent half-angle substitution. By integration by parts, the final outcome is $2\pi-\frac{3\pi}{2}\log(3)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2806764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Why does the circle intersect the line?
Is there an easy geometric way to prove that the circle is tangent to the line $\overleftrightarrow {CD}$, where $C=(3,-6)$, $D=(6,-2)$, and $B=(6,0)$? I can do this by using calculus but I think there has to be a nicer/shorter solution.
Thanks in advance.
Edit: I found that $CE$ has distance $3$, so maybe one could use the theorem of Thales to obtain a right angle in the triangle $(0,-6), E,(6,-6)$ and use this?
|
Yes, you set the equation for the circle equal to the equation of the line and solve for x. If there is only one solution, the line is tangent and if two solutions the line cuts through the circle.
$\sqrt{6^2 - x^2} = \frac{4}{3}x - 10$
$36 - x^2 = \frac{16}{9}x^2 - \frac{80}{3}x + 100$
$\frac{25}{9}x^2 - \frac{80}{3}x + 64 = 0$
$(\frac{5}{3}x - 8)^2 = 0$
$\frac{5}{3}x = 8$
$x = 4.8$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2806957",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 11,
"answer_id": 5
}
|
Mistake in solving this simple inequality? Suppose have the the inequality, where $x,y$ are variables,
$$
\frac{x+y}{2}\leq \frac{\alpha}{\beta+\gamma}(x+y)
$$
Clearly, this is true if $\frac12 <\frac{\alpha}{\beta+\gamma}$, but if I try to solve it by adding/subtracting I get
$$
x(1-\frac{2\alpha}{\beta+\gamma})\leq y(\frac{2\alpha}{\beta+\gamma} -1) = -y(1-\frac{2\alpha}{\beta+\gamma})
$$
which gives, (assuming I am not dividing by a negative number, if are then flip the inequality sign )
$$
x\leq -y
$$
which is very different from $\frac12 <\frac{\alpha}{\beta+\gamma}$?
Why do these two methods give very different results (or are they same and I am overlooking something?)
|
Method 2:
$x(1-\frac{2\alpha}{\beta+\gamma})\leq y(\frac{2\alpha}{\beta+\gamma} -1) = -y(1-\frac{2\alpha}{\beta+\gamma})$
Case 1: $1 = \frac {2\alpha}{\beta + \gamma}$
Then $0 = 0$ and nothing can be determined.
Case 2: $1 > \frac {2\alpha}{\beta + \gamma}$
Then $x\le -y$
Case 3: $1 < \frac {2\alpha}{\beta + \gamma}$
Then $x \ge -y$.
Method 1:
$\frac{x+y}{2}\leq \frac{\alpha}{\beta+\gamma}(x+y)$
$(x+y) = \leq \frac{2\alpha}{\beta+\gamma}(x+y)$
Case 1: $1 = \frac {2\alpha}{\beta + \gamma}$
The $x + y = x+y$ and nothing can be determined.
Case 2: $1 > \frac {2\alpha}{\beta + \gamma}$
The if $x+y > 0$ we would have $x+1 > \frac {2\alpha}{\beta + \gamma}(x+y)$ but that is a contradiction.
So $x + y \le 0$ and $x \le -y$.
Case 3: $1 < \frac {2\alpha}{\beta + \gamma}$
If $x + y < 0$ we would have $x+1 > \frac {2\alpha}{\beta + \gamma}(x+y)$ but that is a contradiction.
So we have $x + y \ge 0$ and $x \ge -y$.
...
so both methods come to same conclusion.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2808147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
System of differential equations verification I have a two questions about these types of equations:
*
*Is on checking if I had solve one case correctly
*A simpler way to solve/represent the solutions of it by using Jordan forms I think
So I want to solve this system
$x' = -23x -8y$
$y' = 60x +21y$
I first find eigen values with
$
det
\begin{bmatrix}
-23-\lambda & -8 \\
60 & 21-\lambda
\end{bmatrix} =0 \therefore \lambda_1 =1, \lambda_2=-3
$
I then find the eigen vectors $v$(with $\lambda_1 $) and $u$(with $\lambda_2 $) using the each eigen value by solving
$
\begin{pmatrix}
-24 & -8 \\
60 & 20
\end{pmatrix}\begin{pmatrix}
v_1 \\
v_2
\end{pmatrix} =\begin{pmatrix}
0 \\
0
\end{pmatrix} \therefore 3v_1=-v_2
$
$
\begin{pmatrix}
-20 & -8 \\
60 & 24
\end{pmatrix}\begin{pmatrix}
u_1 \\
u_2
\end{pmatrix} =\begin{pmatrix}
0 \\
0
\end{pmatrix} \therefore u_2=\frac{-5u_1}{2}
$
then we choose $v_1 = 1$ and $u_1=2$, arbitrarily:
$
\begin{pmatrix}
v_1 \\
v_2
\end{pmatrix} =\begin{pmatrix}
1 \\
-3
\end{pmatrix}
$
$
\begin{pmatrix}
u_1 \\
u_2
\end{pmatrix} =\begin{pmatrix}
2 \\
-5
\end{pmatrix}
$
then the general solutions are $ y(t)=Ce^{Dt}C^{-1}y_0 $ :
$
y(t)= \begin{pmatrix}
1&2 \\
-3&-5
\end{pmatrix} \begin{pmatrix}
e^{\lambda_1 t}&0 \\
0&e^{\lambda_2 t}
\end{pmatrix} \begin{pmatrix}
-5&-2 \\
3&1
\end{pmatrix} \begin{pmatrix}
x_0\\
y_0
\end{pmatrix}, \lambda_1 = 1, \lambda_2 =-3
$
How might I then find a specific solution given initial conditions?
In class the teacher showed a simpler way to represent the solutions with an example below, how may I apply that to the this question(above)?
$x' =-3x+y$
$y' =-3y$
$
\begin{pmatrix}
x' \\
y'
\end{pmatrix} = \begin{pmatrix}
-3&1 \\
0&-3
\end{pmatrix}\begin{pmatrix}
x \\
y
\end{pmatrix}
$
and then they said something about $\begin{pmatrix}
-3&1 \\
0&-3
\end{pmatrix}$ being in Jordan form(for 2 by 2 matrix) and so $\lambda =-3, k_{dimension-of- matrix}=2 $ then:
$
\begin{pmatrix}
x \\
y
\end{pmatrix} = e^{-3t}\begin{pmatrix}
1&t \\
0&1
\end{pmatrix}\begin{pmatrix}
x_0 \\
y_0
\end{pmatrix}
$
|
$$y=c_1e^{\lambda _1t }v_1 +c_2 e^{\lambda _2t }v_2=$$
$$ c_1e^{t } \begin{pmatrix}
1 \\
-3
\end{pmatrix} +c_2 e^{2t } \begin{pmatrix}
2 \\
-5 \end{pmatrix} $$
You can find $c_1$ and $c_2$ using a given initial value.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2810804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Finding the value of $a+2b+3c$ where $a,b,c$ are roots of a cubic equation. Let $a$, $b$, and $c$ be positive real numbers with $a<b<c$ such that $a+b+c=12$, $a^2+b^2+c^2=50$, and $a^3+b^3+c^3=216$. Find $a+2b+3c$.
I solved it like this:
$$2(ab+bc+ca)=(a+b+c)^2-(a^2+b^2+c^2)$$$$ab+bc+ca=47$$
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$$$abc=60$$
S0 $a,b,c$ are solutions of $x^3-12x^2+47x-60=0$
Placing $x=y+4$$$y^3+12y^2+48y+64-12y^2-96y-192+47y+188-60=0$$$$y^3-y=0$$
So $y=-1,0,1$ and $x=3,4,5$.
Thus $a=3;b=4;c=5$ and $a+2b+3c=26$
But is there any other way of solving it without actually making a cubic equation and then solving it to get the values of $a,b,c$
|
Make the change: $a=x+4, b=y+4,c=z+4$. Then the equations will be:
$$\begin{cases} x+y+z=0 \\
(x+4)^2+(y+4)^2+(z+4)^2=50 \\
(x+4)^3+(y+4)^3+(z+4)^3=216 \end{cases} \Rightarrow
\begin{cases} x+y+z=0 \\
x^2+y^2+z^2=2 \\
x^3+y^3+z^3=0 \end{cases}$$
Express $x+y=-z$. Square it and subtract the second. Cube it and subtract the third.
$$\begin{cases}xy=-1 \\
3xy(x+y)=0\end{cases} \Rightarrow (x,y,z)=(-1,1,0); (1,-1,0).$$
Note: All $6$ possible solutions will be the permutations of $(-1,1,0)$.
Since $a<b<c$, from the first we get:
$$(a,b,c)=(3,4,5).$$
Note: All $6$ possible solutions without constraint $a<b<c$ will be the permutations of $(3,4,5)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2810911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Given $f(x,y)=5+2x+4y+x^2+y^2+(x^2y^4)^\frac15$, show that it is differentiable at $(0,0)$. I was given the function:
$f(x,y)=5+2x+4y+x^2+y^2+(x^2y^4)^\frac15$
I need to show it is differentiable at $(0,0)$.
I started using the method of differentials and infinitesimal functions:
$\Delta f=f(0+\Delta x,0+\Delta y)-f(0,0)$
$\Delta f=2\Delta x + 4 \Delta y + (\Delta x)^2 + (\Delta y)^2 + ((\Delta x)^2(\Delta y)^4)^\frac15$
Now I cannot see a way how I can obtain infinitesimal functions* $\alpha$ and $\beta$ such that:
$\Delta f=2\Delta x + 4 \Delta y + \alpha\Delta x + \beta \Delta y$, where $\alpha,\beta=o(\rho)$
(Where $\rho^2 = x^2+y^2$, and $o$ is the Landau little-$o$ notation defined as $\alpha=o(\rho)\iff\lim \alpha/\rho=0$).
I managed to get 3 "infinitesimal functions": $\alpha=\Delta x$, $\beta=\Delta y$, and $\gamma=(\Delta x^2\Delta y^{-1})^\frac15$ but I am having trouble showing $\gamma=o(\rho)$ because frankly it doesn't seem to be.
Any hints? If this is not possible, are there any other methods that one would recommend?
*My prof defines infinitesimal functions as: $\alpha(x)$ is infinitesimal at $a$ if $\lim_{x\rightarrow a}\alpha(x)=0$
|
Another approach using the definition of differentiability:
To use the horrible limit
$$\lim_{(x,y)\to(0,0)}{\dfrac{f(x,y)-[f(0,0)+f'_x(0,0)(x-0)+f'_y(0,0)(y-0)]}{\sqrt{{(x-0)}^2+{(y-0)}^2}}}\tag 1$$
we need first to calculate the partial derivatives at $(0,0)$ using the fact that $f$ is $C^1(\text{Dom}(f))$. Let's begin.
Remember that $f(x,y)=5+2x+4y+x^2+y^2+\sqrt[5]{x^2y^4}$.
Finding the domain of the function
$\text{Dom}(f)=\mathbb R^2$ because is a sum of continuous.
Finding partial derivatives
$$\begin{array}{llc}
f'((0,0);(a,b))&&= \\
&\displaystyle\lim_{h\to 0}{\dfrac{f(ah,bh)-f(0,0)}{h}}&= \\
&\displaystyle\lim_{h\to 0}{\dfrac{\left(5+2ah+4bh+a^2h^2+b^2h^2+\sqrt[5]{a^2h^2b^4h^4}\right)-5}{h}}&= \\
&\displaystyle\lim_{h\to 0}{\dfrac{h\left(2a+4b+a^2h+b^2h+\frac{\sqrt[5]{a^2b^4h^6}}{h}\right)}{h}}&= \\
&\displaystyle\lim_{h\to 0}{\left(2a+4b+a^2h+b^2h+\sqrt[5]{a^2b^4h}\right)}&= \\
&2a+4b.
\end{array}$$
Thus the directional derivatives are continuous for all $(x,y)\in\mathbb R^2$, in particular for $(x,y)=(0,0)$, thus $f(x,y)\in C^1\left(\mathbb R^2\right)$ so
$$\begin{matrix}
f'((0,0);(1,0))&=&f'_x(0,0)&=&2\cdot 1+4\cdot 0&=&2, \\
f'((0,0);(0,1))&=&f'_y(0,0)&=&2\cdot 0+4\cdot 1&=&4.
\end{matrix}$$
Proving the differentiability
From $(1)$
$$\begin{array}{lcc}
\displaystyle\lim_{(x,y)\to(0,0)}{\dfrac{5+2x+4y+x^2+y^2+\sqrt[5]{x^2y^4}-[5+2(x-0)+4(y-0)]}{\sqrt{x^2+y^2}}} &=& \\
\displaystyle\lim_{(x,y)\to(0,0)}{\dfrac{x^2+y^2+\sqrt[5]{x^2y^4}}{\sqrt{x^2+y^2}}}&\underbrace =_{(x,\,y)\;=\;(\rho\cos\theta,\,\rho\sin\theta)}&\\
\displaystyle{\lim_{\rho\to 0^+}}_{\theta\in[0,2\pi)}{\dfrac{\rho^2\cos^2\theta+\rho^2\sin^2\theta+\sqrt[5]{\rho^2\cos^2\theta\cdot\rho^4\sin^4\theta}}{\sqrt{\rho^2\cos^2\theta+\rho^2\sin^2\theta}}}&=&\\
\displaystyle{\lim_{\rho\to 0^+}}_{\theta\in[0,2\pi)}{\dfrac{\rho^2+\rho\sqrt[5]{p\cos^2\theta\cdot\sin^4\theta}}{\rho}}&=&\\
\displaystyle{\lim_{\rho\to 0^+}}_{\theta\in[0,2\pi)}{\left(\rho+\sqrt[5]{p\cos^2\theta\cdot\sin^4\theta}\right)}&=&\\
0,&&
\end{array}$$
hence
$$\boxed{f(x,y)\quad\text{is differentiable at}\quad(0,0)}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2812656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
If $\lambda_{n} = \int_{0}^{1} \frac{dt}{(1+t)^n}$ , then value of $\lim_{n \to \infty} \lambda_{n}^{1/n}$?
Let $$\lambda_n = \int_0^1\frac{dt}{(1+t)^n} \quad \forall n \in \Bbb{N}.$$
To evaluate the integral,
Let $1+t = u$ then $dt =du$, when $t =0$ we get $u=1$ and when $t = 1$ we get $u = 2$; so that $\lambda_{n} = \int_{1}^{2} \frac{du}{u^n}$ we get after solving $\lambda_{n} = \frac{2^{1-n} - 1}{1-n}$ for $n \neq 1$.
For $n=1$, $\lambda_{1} = \int_{0}^{1}\frac{dt}{1+t} = ln2$ and hence I think $\lambda_{n}$ exists for all $n \in \Bbb{N}$.
In order to prove that the sequence $\lambda_{n}$ is bounded, I thought if I would prove that if the sequeences are converging then it will be bounded. So checking for convergence of $\lambda_{n}$,
that is $\lim_{n \rightarrow \infty}\lambda_{n} = \lim_{n \rightarrow \infty} \frac{2^{1-n}-1}{1-n}$ , so as $n \rightarrow \infty, \lambda_{n} \rightarrow 0$ and hence as the sequence is convergent we get that the sequence of $\lambda_{n} $ are bounded. Is this correct?.
Next I have to check for the correctness of $\lim_{n \rightarrow \infty} \lambda_{n}^{\frac{1}{n}} = 1$.
$\lim_{n \rightarrow \infty} \lambda_{n}^{\frac{1}{n}} = \lim_{n \rightarrow \infty} (\frac{2^{1-n}}{1-n} - \frac{1}{1-n})^{\frac{1}{n}}$, it is coming $0^0$ indeterminate form.
So let $L =\lim_{n \rightarrow \infty}(\frac{2^{1-n}}{1-n} - \frac{1}{1-n})^{\frac{1}{n}}$.
$ln(L) = \lim_{n \rightarrow \infty}(\frac{1}{n} ln(\frac{2^{1-n}}{1-n} - \frac{1}{1-n}))$.
So next I think I would apply $\lim f.g = \lim f . \lim g$, so that I would be getting $ln(L) = \lim_{n \rightarrow \infty} \frac{1}{n} . \lim_{n \rightarrow \infty} ln(\frac{2^{1-n}}{1-n} - \frac{1}{1-n}) = \lim_{n \rightarrow \infty} \frac{1}{n} . ln (\lim_{n \rightarrow \infty} (\frac{2^{1-n}}{1-n} - \frac{1}{1-n}))$, now I am thinking how to evaluate the limit $ln (\lim_{n \rightarrow \infty} (\frac{2^{1-n}}{1-n} - \frac{1}{1-n}))$ as its of the form $ln(0)$?.
|
$$\lim_{n \to \infty} \frac{1}{n} \log \frac{1-2^{-(n-1)}}{n-1} = \lim_{n \to \infty} \frac{1}{n} \log(1-2^{-(n-1)}) - \lim_{n \to \infty} \frac{1}{n} \log(n-1) = 0,$$
so $\lambda_n^{1/n} \to 1$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2814183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Finding The Zeros Of $\frac{z^2\sin z}{\cos z -1}$ $$f(z)=\frac{z^2\sin z}{\cos z -1}$$
$$\frac{z^2\sin z}{\cos z -1}=0\iff z^2\sin z=0$$ so $z=\pi k$ and $z=0$ are zeros, to find the order we must derive $\frac{z^2\sin z}{\cos z -1}$?
|
Note that the denominator $\cos z - 1$ has zeroes at $z = 2m\pi$, so not all zeroes of $\sin z$ are zeroes of $f(z)$. To show this, apply L'Hopital's rule
$$ \lim_{z\to 2m\pi \ne 0} \frac{z^2\sin z}{\cos z - 1} = \lim_{z\to 2m\pi \ne 0}\frac{2z\sin z + z^2\cos z}{-\sin z} = \text{DNE} $$
Hence, the zeroes of $\sin z$ where $z=n\pi = 2m\pi$ are actually poles. This leaves the odd multiples, $z = (2m+1)\pi$, which are indeed zeroes of $f(z)$.
$z = 0$ is also a zero, which you can show by applying L'Hopital once more
To simplify computations, we rewrite
$$ f(z) = \frac{z^2\sin z}{\cos z - 1} = -\frac{2z^2 \sin(\frac{z}{2})\cos(\frac{z}{2})}{2\sin^2(\frac{z}{2})} = -z^2\cot\left(\frac{z}{2}\right) $$
which confirms $z = (2m+1)\pi$ as the zeroes of $\cot(\frac{z}{2})$. Futhermore
$$ f'(z) = -2z\cot\left(\frac{z}{2}\right) + \frac{z^2}{2}\csc^2\left(\frac{z}{2}\right) $$
You can check that
$$ \lim_{z\to 0} f'(z) = -4\lim_{z\to0} \frac{\frac{z}{2}}{\sin\left(\frac{z}{2}\right)}\cos\left(\frac{z}{2}\right) + 2 \lim_{z\to 0} \frac{\left(\frac{z}{2}\right)^2}{\sin^2\left(\frac{z}{2}\right)} = -2 $$
and
$$ f'\big((2m+1)\pi\big) = \frac{(2m+1)^2\pi^2}{2} $$
Proving that these are all first-order zeroes
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2814928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Proof for $\sum_{k=2}^\infty \frac{1}{k^4-1}= \frac{7}{8}-\frac{\pi}{4}\coth(\pi)$ How can this identity be derived?
I have been searching the internet but I have no clue where to find a proof for this identity. Any help is highly appreciated.
$$\sum_{k=2}^\infty \frac{1}{k^4-1}= \frac{7}{8}-\frac{\pi}{4}\coth(\pi)$$
|
Using the residue theorem, we have
$$\begin{align}
\oint_{|z|=N+1/2}\frac{\cot(\pi z)}{(z^2+1)}\,dz&=2\pi i \sum\text{Res}\left(\frac{\cot(\pi z)}{z^2+1}\right)\\\\
&=2\pi i \left(\frac{2\cot(\pi i)}{2i}+\sum_{n=-N}^N \frac{1}{\pi(n^2+1)}\right)\tag1
\end{align}$$
As $N\to \infty$, the integral on the left-hand side of $(1)$ vanishes. Hence, we have
$$\sum_{n=-\infty}^\infty \frac{1}{n^2+1}=\pi \coth(\pi)$$
which after exploiting symmetry yields
$$\sum_{k=2}^\infty \frac{1}{n^2+1}=\frac{\pi\coth(\pi)}{2}-1\tag 2$$
Finally, using $\frac{1}{k^4-1}=\frac12\left(\frac1{k^2-1}-\frac{1}{k^2+1}\right)$ along with $(2)$ and the value of the telescoping series $\sum_{k=2}^\infty \frac{1}{k^2-1}$ yields the coveted result.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2817313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
How to solve $\lim_{x\to1}=\frac{x^2+x-2}{1-\sqrt{x}}$? let $f(x)=\dfrac{x^2+x-2}{1-\sqrt{x}}$
How do I solve this limit?
$$\lim_{x\to1}f(x)$$
I can replace the function with its content
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}$$
Then rationalizing the denominator
$$\lim_{x\to1}\dfrac{x^2+x-2}{1-\sqrt{x}}\cdot\dfrac{1+\sqrt{x}}{1+\sqrt{x}}$$
With $(a + b)(a - b) = a^2 - b^2$, I can remove the irrational denominator.
$$\lim_{x\to1}\dfrac{(x^2+x-2)(1+\sqrt{x})}{1-x}$$
I the multiply the two parenthesis
$$\lim_{x\to1}\dfrac{\sqrt{x} \cdot x^2+\sqrt{x}\cdot x- \sqrt{x} \cdot 2 + x^2 + x - 2}{1-x}$$
I'm not sure where to continue to solve this limit.
|
$$\lim_{x\to 1}\frac{x^2+x-2}{1-\sqrt{x}}=\lim_{y\to 1}\frac{y^4+y^2-2}{1-y}=-\lim_{y\to 1}\frac{y^4+y^2-2}{y-1}$$
And that last limit is, by definition, the derivative $\frac{d}{dy}(y^4+y^2)=4y^3+2y$ at $y=1.$
So the result is $-6.$
Or you can just factor:
$$\frac{y^4+y^2-2}{y-1}=\frac{y^4-1}{y-1}+\frac{y^2-1}{y-1}=1+y+y^2+y^3 + 1+y.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2818925",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
How did they get the term $(n+1)^3$ in the step of inductive proof which says $\sum_{k=1}^{n+1} k^3=\sum_{k=1}^n k^3 + (n+1)^3$? I'm struggling to understand on what was done in this inductive step. How did they get the $(n+1)^3$ term?
Proof Solution
|
$\sum\limits_{k=1}^{\color{red}{n+1}} k^3 = 1^3 + 2^3 + 3^3 + ....... + n^3 + \color{blue}{(n+1)^3}=$
$[1^3+2^3 + 3^3 + ...... + n^3] + \color{blue}{(n+1)^3} =$
$\sum\limits_{k=1}^{\color{red}{n}} k^3 + \color{blue}{(n+1)^3}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2819183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
}
|
Factoring $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$ I am trying to factor the expressions $(x+y+z)^3 xyz - (xz+xy+yz)^3$ and $(x-a)^3(b-c)^3+(x-b)^3(c-a)^3+(x-c)^3(a-b)^3$.
I am rather stuck though. Is there a general method for going about this? I always find myself having to guess which is not so useful here. I notice in the first one that if $x=y=z$ then the polynomial is zero. I'm not sure how useful this is though in the case of three variables.
|
Brute-forcing the first factorization by expanding and collecting the powers of $\,x\,$:
$$
\begin{align}
(x+y+z)^3 xyz - (xz+xy+yz)^3 &= yz \,x^4 - (y^3+z^3) \,x^3 + yz(y^3+z^3)\,x - y^3 z^3 \\
&=yz(x^4-y^2z^2)-(y^3+z^3)x(x^2-yz) \\
&= yz(x^2-yz)(x^2+yz)-(y^3+z^3)x(x^2-yz) \\
&= (x^2-yz)(\ldots)
\end{align}
$$
Therefore $\,x^2-yz\,$ is a factor, and by symmetry so are $\,y^2-xz\,$ and $\,z^2-xy\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2819615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Evaluating a complex integral with the goniometric function I was evaluating a this complex integral via gamma function:
$\int_0^\infty \sin (x^p) \,dx$ $\;$for $p \gt 1$,
so I expressed it as an imaginary part of $\int_0^\infty \exp(-ix^p) \, dx$ $\;$for $p \gt 1$
*
*The formula of the gamma function is $\Gamma (z) = \int_0^\infty x^{z-1} e^{-x} \, dx $
I used the substitution $-y^{1/p}=xi$, $\;$ $\;$ $dx= \frac 1 p y^{\frac{1}{p}-1}i \, dy$ $\;$ $\;$and $\;$ $\;$ $\frac {1}{p} = \alpha$
Then $\int_0^\infty \alpha i y^{\alpha-1}e^{-y} \, dx = \alpha i \Gamma (\alpha) = \ i \frac {1}{p} \Gamma (\frac {1}{p})$
The solution according to my textbook is $\ \frac {1}{p} \Gamma (\frac {1}{p}) \sin (\frac {\pi}{2p})$
But I think $\sin (\frac {\pi}{2p})$ is right if I have ${i}^p$, but I got just $i$.
My solution is then $\ \frac {1}{p} \Gamma (\frac {1}{p}) \sin (\frac {\pi}{2}) =\frac {1}{p} \Gamma (\frac {1}{p})$.
Did I miss something important?
EDIT
I tried to calculate this integral for $p = 2$ and the textbook is right, but why?
|
You can first substitute $u=x^p$:
\begin{align}
I=\int_0^\infty \sin (x^p)\,dx&=\frac{1}{p}\int_0^\infty u^{\frac{1}{p}-1}\sin u \,du \\ \\
&=\frac{1}{p} \Im\int_0^\infty u^{\frac{1}{p}-1} e^{iu}\,du \\ \\
&=\frac{1}{p}\Gamma\left(\frac{1}{p}\right) \Im i^{1/p} \\ \\
&=\frac{1}{p}\Gamma\left(\frac{1}{p}\right) \Im e^{\frac{i\pi}{2p}} \\ \\
&=\Gamma\left(1+\frac{1}{p}\right)\sin\frac{\pi}{2p}
\end{align}
You can also evaluate the integral via a useful property of the Laplace Transform:
\begin{align}
\int_0^\infty f(x)\,g(x)\,dx=\int_0^\infty \mathcal{L}^{-1}\{f(x)\}(s)\mathcal{L}\{g(x)\}(s)\,ds
\end{align}
Then,
\begin{align}
\int_0^\infty \sin (x^p)\,dx&=\frac{1}{p}\int_0^\infty u^{\frac{1}{p}-1}\sin u \,du \\ \\
&=\frac{1}{p\,\Gamma\left(1-\frac{1}{p}\right)}\int_0^\infty \frac{s^{-\frac{1}{p}}}{s^2+1}\,ds \qquad s^2\mapsto u \\ \\
&=\frac{1}{2p\,\Gamma\left(1-\frac{1}{p}\right)} \int_0^\infty \frac{u^{-\frac{1}{2}(\frac{1}{p}+1)}}{1+u}\,du \\ \\
&=\frac{1}{2p\,\Gamma\left(1-\frac{1}{p}\right)} \mathcal{B}\left[\frac{1}{2}\left(1-\frac{1}{p}\right),\frac{1}{2}\left(1+\frac{1}{p}\right) \right] \\ \\
&= \frac{1}{2p\,\Gamma\left(1-\frac{1}{p}\right)} \Gamma\left(\frac{1}{2}\left(1-\frac{1}{p}\right)\right) \Gamma \left(\frac{1}{2}\left(1+\frac{1}{p}\right) \right) \\ \\
&=\frac{\pi}{2p\,\Gamma\left(1-\frac{1}{p}\right)}\sec\left(\frac{\pi}{2p} \right) \\ \\
&= \frac{\pi \sin \left(\frac{\pi}{2p} \right)}{p\,\sin \left(\frac{\pi}{p} \right)\,\Gamma\left(1-\frac{1}{p}\right)} \\ \\
&= \frac{1}{p} \Gamma \left(\frac{1}{p}\right) \sin \frac{\pi}{2p} \\ \\
&= \Gamma \left(\frac{1}{p}+1\right) \sin \frac{\pi}{2p}
\end{align}
Feel free to ask if you have any questions!
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2820135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
}
|
Does $4(x^3+2x^2+x)^3(3x^2+4x+1) = 4x^3(3x+1)(x+1)^7$? Please provide proof
The answer in the back of the book is different to both my calculations and also the online calculator I crosschecked my answer with...
The Question:
"Differentiate with respect to $x$:"
$
(x^3+2x^2+x)^4
$
My Answer: $4(x^3+2x^2+x)^3(3x^2+4x+1)$
The Book's Answer: $4x^3(3x+1)(x+1)^7$
|
You're both right.
Note that
$$
x^3+2x^2+x = x (x + 1)^2
$$
$$
3x^2+4x+1 = (3 x + 1) (x + 1)
$$
On the other hand, you might have started with
$$
(x^3+2x^2+x)^4 = x^4 (x + 1)^8
$$
whose derivative is
$$
4x^3 (x + 1)^8 + 8x^4 (x + 1)^7
= 4x^3 (x+1 +2x) (x + 1)^7
= 4x^3 (3x+1) (x + 1)^7
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2823614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
}
|
Solving $y'''-y=\frac{1}{3}\left(e^x+e^{wx}+e^{w^2x}\right)$ by power series I have the following differential equation
$$y'''-y=\frac{1}{3}\left(e^x+e^{wx}+e^{w^2x}\right)$$
where $w=e^{2i\pi/3}$
I am trying to solve by power series. So let $y(x)=\sum_{n=0}^\infty a_n\frac{x^n}{n!}$. Then $y'''(x)=\sum_{n=0}^\infty a_{n+3}\frac{x^n}{n!}$. Also, as a power series, the RHS is $\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!}$. Setting them equal gives
$$\sum_{n=0}^\infty a_{n+3}\frac{x^n}{n!}-\sum_{n=0}^\infty a_{n}\frac{x^n}{n!}=\sum_{n=0}^\infty \frac{x^{3n}}{(3n)!}$$
After expansion of the first 3 terms, we have that $a_0=1, a_1=a_2=0$. The remaining expansion therefore produces the following;
$$a_{3(n+1)}-a_{3n}=1$$
$$a_{3(n+1)+1}-a_{3n+1}=0$$
$$a_{3(n+1)+2}-a_{3n+2}=0$$
Now if $n=0,$ we have that $a_3-1=1\rightarrow a_3=2, a_4-0=0\rightarrow a_4=0, a_5-a_2=0\rightarrow a_2=0$.
Thus, we have that $a_{3k+1}=a_{3k+2}=0$ for all $k\in \mathbb{N}$. Continuing then we have $a_6-a_3=1\rightarrow a_6=1+2=3.$ Using an induction argument we thus have $a_{3n}=n+1$. Thus, the solution would be
$$y(x)=\sum_{n=0}^\infty(n+1)\frac{x^{3n}}{(3n)!}$$
Is this a correct approach?
|
Indeed, the term with $x^n$ in the right-hand side is
$$
\frac{x^n}{3n!}(1+w^n+w^{2n})
$$
since $w^3=1$, writing $n=3q+r$, we have
$$
1+w^n+w^{2n}=1+w^r+w^{2r}
$$
If $r=0$, this equals $3$; if $r=1$ or $r=2$ this equals $0$.
The left-hand side is
$$
\sum_{n=0}^{\infty}(a_{n+3}-a_n)\frac{x^n}{n!}
$$
so we must have $a_{n+3}-a_{n}=0$ when $n$ is not a multiple of $3$ and $a_{n+3}-a_n=1$ otherwise.
This does not allow you to conclude that $a_n=0$ when $n$ is not divisible by $3$, but just that $a_{3n+1}=a_1$ and $a_{3n+2}=a_2$. Similarly, $a_0$ cannot be determined.
If we set $b_n=a_{3n}$, then the recurrence is $b_{n+1}=b_n+1$, so $b_n=a_0+n$. If we set $a_0=a$, $a_1=b$ and $a_2=c$, the final result is
$$
a_n=\begin{cases}
a+(n/3) & \text{if $n\equiv0\pmod{3}$} \\[4px]
b & \text{if $n\equiv1\pmod{3}$} \\[4px]
c & \text{if $n\equiv2\pmod{3}$}
\end{cases}
$$
Since you're given a third order differential equation, you have to expect three constants.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2824940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Show that all eigenvalues of this pentadiagonal matrix are double degenerate I am trying to show in general that the following pentadiagonal matrix $\mathbf{M}$ has double degenerate eigenvalues,
\begin{equation}
\mathbf{M} =
\left[ \begin{array}{cccccccc}
4 & 0 & a & 0 & 0 & 0 & 0 & 0 \\
0 & 3 & 0 & a & 0 & 0 & 0 & 0\\
a & 0 & 2 & 0 & a & 0 & 0 & 0\\
0 & a & 0 & 1 & 0 & a & 0 & 0\\
0 & 0 & a & 0 & 1 & 0 & a & 0\\
0 & 0 & 0 & a & 0 & 2 & 0 & a\\
0 & 0 & 0 & 0 & a & 0 & 3 & 0\\
0 & 0 & 0 & 0 & 0 & a & 0 & 4
\end{array} \right]\!,
\end{equation}
where $a$ is just some real number. It doesn't particularly matter what the matrix elements are on the diagonal, so long as they are symmetric about the anti-diagonal the eigenvalues will still be double degenerate. Numerically, I find that a pentadiagonal matrix of this form and of arbitrary size always yields double degenerate eigenvalues.
In block form we can express $\mathbf{M}$ as
\begin{equation}
\mathbf{M} =
\left[ \begin{array}{cc}
\mathbf{A} & \mathbf{B} \\
\mathbf{J}^{-1}\mathbf{B}\mathbf{J} & \mathbf{J}^{-1}\mathbf{A}\mathbf{J} \\
\end{array} \right]\!,
\end{equation}
where
\begin{equation}
\mathbf{A}=\left[\begin{array}{cccc}
4 & 0 & a & 0 \\
0 & 3 & 0 & a \\
a & 0 & 2 & 0 \\
0 & a & 0 & 1
\end{array} \right],
%
\; \; \mathbf{B}=\left[ \begin{array}{cccc}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
a & 0 & 0 & 0 \\
0 & a & 0 & 0
\end{array} \right]
\end{equation}
and $\mathbf{J}$ is the exchange matrix, hence the top-left and bottom-right blocks are permutation-similar (as are the top-right and bottom-left blocks). However, this has not helped me show that all eigenvalues are double degenerate.
So far I have that the truncated 4x4 matrix
\begin{equation}
\mathbf{M}_{\text{trunc}}=\left[\begin{array}{cccc}
2 & 0 & a & 0 \\
0 & 1 & 0 & a \\
a & 0 & 1 & 0 \\
0 & a & 0 & 2
\end{array} \right]
\end{equation}
has a characteristic polynomial
\begin{equation}
p(\lambda)=(a^2+(1-\lambda)\lambda + 2(\lambda-1))^2
\end{equation}
which has roots which are evidently double degenerate.
Any help at all would be much appreciated!
|
I should have made this only a comment, sorry:
Mathematica easily grinds out the characteristic polynomial of $\bf M$, $$a^8-6 a^6 \lambda ^2+34 a^6 \lambda -46 a^6+11 a^4 \lambda ^4-122 a^4
\lambda ^3+497 a^4 \lambda ^2-882 a^4 \lambda +577 a^4-6 a^2 \lambda
^6+94 a^2 \lambda ^5-596 a^2 \lambda ^4+1950 a^2 \lambda ^3-3454 a^2
\lambda ^2+3116 a^2 \lambda -1104 a^2+\lambda ^8-20 \lambda ^7+170
\lambda ^6-800 \lambda ^5+2273 \lambda ^4-3980 \lambda ^3+4180 \lambda
^2-2400 \lambda +576\,,$$
which simplifies to $\left(a^4+a^2 ((17-3 \lambda ) \lambda -23)+(\lambda -4) (\lambda -3)
(\lambda -2) (\lambda -1)\right)^2\,.$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2826148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Solving $\cos(3x) = \cos(2x)$ I'm struggling with solving given trigonometric equation
$$\cos(3x) = \cos(2x)$$
Let's take a look at the trigonometric identities we can use:
$$\cos(2x) = 2\cos^2-1$$
and
$$\cos(3x) = 4\cos^3(x) -3\cos(x)$$
Plugging into the equation and we have that
$$4\cos^3(x) -3\cos(x) = 2\cos^2(x)-1$$
$$4\cos^3(x) -3\cos(x) - 2\cos^2(x)+1= 0$$
Recalling $t = \cos (x)$,
$$4t^3-2t^2-3t +1 = 0$$
Which is a cubic equation. Your sincerely helps will be appreciated.
Regards!
|
Calling
$$
\cos x = \frac{e^{ix}+e^{-ix}}{2}
$$
we have
$$
e^{3ix}+e^{-3i x} = e^{2ix}+e^{-2i x}
$$
or calling $z = e^{ix}$
$$
z^6+1 = z^5+z\to z^6-z^5-z+1 = (z-1)^2(z^4+z^3+z^2+z+1) = (z^5-1)(z-1) = 0
$$
so the solutions are obvious.
$$
x = \frac{2\pi}{5}k,\;\; \mbox{for}\;\; k=0,1,2,\cdots
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2826222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 2
}
|
Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$
Approximate integral: $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx.$
My attempt:
Let I = $\int_3^4 \frac{x}{\sqrt{(7x-10-x^2)^3}}dx$
$u=7-x\implies I=\frac 72 \int_3^4 \frac 1{\sqrt{(7x-10-x^2)^3}}dx$ Now I have to approximate this...
I got to this point too:
$u=3-x \implies I= \int_0^1 \frac {1}{\sqrt{(-x^2-x+2)^3}}dx.$
and: $\frac 1{-x^2-x+2}\leq1\implies \int_0^1 \frac{1}{\sqrt{(-x^2-x+2)^3}}dx\leq \int_0^1-\frac{1}{(x+2)(x-1)}dx$ and the right one does not converge...
I have these values:
$a) (\frac 9{10};\frac {19}{20})$
$b) (\frac {19}{20},1)$
$c) (1;\frac {21}{20})$
$d) (\frac {21}{20};\frac {11}{10})$
$e) (\frac {11}{10},\frac {23}{20})$
$f) (\frac {23}{20}; \frac 65)$.
|
$$ \int_{3}^{4}\frac{x}{\sqrt{(x-2)^3 (5-x)^3}}\stackrel{x\mapsto\frac{7}{2}+z}{=}\int_{-1/2}^{1/2}\frac{z+7/2}{\sqrt{\left(9/4-z^2\right)^3}}\,dz=\frac{7}{2}\int_{-1/2}^{1/2}\frac{dz}{\sqrt{(9/4-z^2)^3}} $$
since the integral of an odd integrable function over a symmetric interval (with respect to the origin) equals zero. By symmetry again, the RHS equals
$$ 7\int_{0}^{1/2}\frac{dz}{\sqrt{(9/4-z^2)^3}}=28\int_{0}^{1}\frac{du}{\sqrt{(9-u^2)^3}}=\frac{28}{9}\int_{0}^{1/3}\frac{dv}{(1-v^2)^{3/2}} $$
or
$$ \frac{28}{9}\int_{0}^{\arcsin(1/3)}\frac{d\theta}{\cos^2\theta}=\frac{28}{9}\tan\arcsin\frac{1}{3}=\color{blue}{\frac{7}{9}\sqrt{2}}\approx 1.09994. $$
This also has a nice geometric/probabilistic interpretation, since $\frac{\mathbb{1}_{(0,1)}(x)}{2\sqrt{x}}$ is the PDF of a squared $U(0,1)$ random variable and the given integral essentially is a convolution integral.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2827122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Floor value of sum whose general term satisfy recursive relation.
Consider the sequence $x_{n}$ given by $\displaystyle x_{1} = \frac{1}{3}$ and $x_{k+1}=x^2_{k}+x_{k}$
and Let $\displaystyle S = \frac{1}{x_{1}}+\frac{1}{x_{2}}+\cdots \cdots +\frac{1}{x_{2008}}$. Then $\lfloor S \rfloor $ is
Try: From $\displaystyle x_{k+1}=x^2_{k}+x_{k}=x_{k}(x_{k}+1)\Rightarrow \frac{1}{x_{k+1}}=\frac{1}{x_{k}}-\frac{1}{x_{k}+1}$
So $\displaystyle S=\sum^{2008}_{k=1}\frac{1}{x_{k+1}} = \sum^{2008}_{k=1}\bigg(\frac{1}{x_{k}}-\frac{1}{x_{k+1}}\bigg)=\frac{1}{a_{1}}-\frac{1}{a_{2009}} = 3-\frac{1}{a_{2009}}$
Could some help me how can i calculate value of $a_{2009},$ Thanks
|
Let $y_n=\frac1{x_n}$. Then
$$\tag1y_{n+1}=y_n-\frac1{\frac1{y_n}+1}=y_n-\frac{y_n}{y_n+1}=y_n\cdot\left(1-\frac1{y_n+1}\right).$$
Note that this means $0<y_{n+1}\le \frac a{a+b}y_n$ if $0<y_n\le \frac ab $.
Thus if $y_n\le \frac ab<1$ then $$\tag2\sum_{k=n}^\infty y_n\le y_n\sum_{k=0}^\infty\left(\frac a{a+b}\right)^{k}=\frac{a+b}{b}y_n\le \frac{(a+b)a}{b^2}. $$
We compute the first few terms
$$ \begin{align}y_1&=3\\
y_2&=3-\frac34=\frac{9}{4}\\
y_3&=\frac94-\frac{9}{9+4}=\frac{81}{52}\end{align}$$
Apparently, the exact fractions are beginning to become ugly. We note from $(1)$ that $\alpha<y_n<\beta$ implies $\alpha\cdot\left(1-\frac1{\alpha+1}\right) <x_{n+1}<\beta\cdot\left(1-\frac1{\beta+1}\right)$. Very moderate will turn out to be good enough. Say, if we have $\frac {r_n}{10}\le y_n\le \frac{s_n}{10}$ with integers $r_n,s_n$, then $\frac {r_{n+1}}{10}\le y_{n+1}=\frac{s_{n+1}}{10}$ if we let
$$\tag3r_{n+1}\le r_n-\left\lceil \frac{10r_n}{r_n+10}\right\rceil,\qquad s_{n+1}=s_n-\left\lfloor\frac{10s_n}{s_n+10}\right\rfloor.$$
So instead of computing complicated fractions or unreliable numerical approximations, we use $(3)$ to compute exact but manageable bounds, starting over with $r_1=s_1=30$ to find
$$\begin{array}tr_2=22&r_3=14&r_4=9&r_5=4\\
s_2=23&s_3=17&s_4=11&s_5=6\end{array} $$
From this we conclude
$$8=\frac{30+22+15+9+4}{10} \le y_1+\ldots+y_5<S$$
From $y_5\le \frac 6{10}$ and $(2)$, we find $$\sum_{n=5}^\infty y_n\le\frac{96}{100} $$
so that
$$ S=\sum_{n=1}^\infty y_n\le \frac{30+23+17+11}{10}+\frac{96}{100}=9.06.$$
This is a pity. With $8<S\le 9.06$ we cannot yet determine $\lfloor S\rfloor$ with certainty. But recall that we already computed $y_3=\frac{81}{52}$ and this is $<\frac{16}{10}$. Thus we can use $s_3=16$ instead of $s_3=17$ in our upper bound and this is enough to show $S<8.96$ (we could also use this to improve the values for $s_4$, $s_5$ and the tail sum, but we do not need that). We conclude
$$ \lfloor S\rfloor = 8.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2829257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
}
|
Bounds for an integral I am trying to show that $$\frac{1}{5} < \int_5^8 \frac{2x-7}{2x+5} dx <1$$ since for the integral $$5\le x \le 8 \rightarrow 15\le 2x+5 \le 21$$$$-\frac{12}{15}\le-\frac{12}{2x+5}\le -\frac{12}{21}\rightarrow \frac{3}{15}\le 1- \frac{12}{2x+5}\le \frac{3}{7}$$ By taking integral $$\frac{3}{5}\le \int_5^8 \left(1-\frac{12}{2x+5}\right)dx\le\frac{9}{7}$$ Whilee the first one could be reduced to what is needed, how can I deal with the right bound?
|
Tangent line to $y=\frac{2x-7}{2x+5}$ at $x=5$ and $x=7$ are: $y=\frac{8}{75}x-\frac13$ and $y=\frac{24}{361}x-\frac{35}{361}$, respectively.
Refer to the graph:
$\hspace{0.5cm}$
Hence:
$$\int_5^8\frac{2x-7}{2x+5}dx<\int_5^6 \left(\frac{8}{75}x-\frac13\right)dx+\int_6^8 \left(\frac{24}{361}x-\frac{35}{361}\right)dx\approx \\
0.2533333333+0.736842105=0.9901754383<1.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2829373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
Showing that $4b^2+4b = a^2+a$ has no non-zero integer solutions? The problem is
Show that $4b^2+4b = a^2+a$ has no integer solutions where none of $a, b$ are zero.
I have a solution but I think there must be some better ways:
My Solution:
$$4b^2+4b = a^2+a$$
$$(2b+a)(2b-a)+4b-a= 0$$
Now letting $x = 2b + a$ and $y = 2b-a$, we see that $x+y = 4b$. Substituting,
$$xy+\dfrac {x+y}{2}+y=0$$
$$2xy+x+3y=0$$
From this we see that $y|x$, so we can substitute $x = ky$ for some integer $k$
$2ky^2+ky+3y = 0$
$$k = \dfrac {3}{2y+1}$$
From here we get that $y \in \{-2, -1 , 1 \}$, and each of the cases can be checked individually.
|
A varied version of @Oscar Lanzi's answer:
Rewrite the equation
$$a^2+a-(4b^2+4b)=0$$
$$\Delta=1+4(4b^2+4b)=(4b+2)^2-3\overset{def}{\equiv} K^2-3$$
If $a$ is an integer, then there exists some integer $M$ such that $\Delta=M^2$.
$$K^2-3=M^2\iff K^2-M^2=3$$
So there will only be a set of $(K,M)$ or $(a,b)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2831439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
}
|
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$ If the circle $(x-a)^2+(y-b)^2=r^2$ and the line $y=mx+c$ do not meet:
Prove that $m^2(r^2-a^2)+2am(b-c)+2bc-b^2-c^2+r^2<0$
These are the steps I have thus taken (although they may be wrong/useless):
*
*Rearranged the circle equation for y:
$y=\sqrt{r^2-x^2+2ax-a^2}+b$
*Set the circle and line equations equal to each other as in simultaneous equations.
I was then planning on finding some discriminant and as we know the line and circle do not meet I could set the discriminant < 0 and show it is the same as the long inequality.
Any solutions/pointers appreciated as I cannot seem to work this problem out.
|
Rewrite the equation of the circle subbing in for $y=mx+c$:
\begin{align*}
(x-a)^2+(y-b)^2&=x^2-2ax+a^2+y^2-2by+b^2\\
&=x^2-2ax+a^2+(mx+c)^2-2b(mx+c)+b^2\\
&=x^2-2ax+a^2+m^2x^2+2mcx+c^2-2bmx-2bc+b^2\\
&=(m^2+1)x^2+2(m(c-b)-a)x+(a^2+b^2+c^2-2bc)=r^2\\
\end{align*}
Therefore solve
$$(m^2+1)x^2+2(m(c-b)-a)x+(a^2+b^2+c^2-2bc-r^2)=0$$
giving
$$x=\frac{-2(m(c-b)-a)\pm\sqrt{4(m(c-b)-a)^2-4(m^2+1)(a^2+b^2+c^2-2bc-r^2)}}{2(m^2+1)}$$
So we require
$$(m(c-b)-a)^2-(m^2+1)(a^2+b^2+c^2-2bc-r^2)<0$$
which simplifies as
$$ m^2(r^2- a^2 ) + 2 a m(b -c) + 2 bc-b^2 - c^2+r^2 <0$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2831778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
}
|
Finding critical points of $f(x,y)= \sin x+\sin y + \cos(x+y)$
Find the critical points of function$$
f(x,y)=\sin x + \sin y + \cos(x+y),$$
where $0<x<\dfrac{\pi}{2}$, $0<y<\dfrac{\pi}{2}$.
What I have done:
$$f_{x}=\cos(x)-\sin(x+y),\\
f_{y}=\cos(y)-\sin(x+y).$$
From $f_{x}=0$, $\cos(x)=\sin(x+y)$. From $f_{x}=0$, $\cos(y)=\sin(x+y)$. I do not know where to go from here.
My attemps:
$$\sin\left(\frac{\pi}{2}-x\right)=\sin(x+y)=\sin\left(\frac{\pi}{2}-y\right).$$
|
$$f_x = \cos x - \sin(x+y) = 0\\
f_y = \cos y - \sin(x+y) = 0$$
Subtracting one from the other we get $\cos x = \cos y$ and with the restrictions of $x,y$ to $(0,\frac {\pi}{2})$ we can say $x= y$ and $\cos x - \sin 2x = 0$
$$\cos x(1-\sin x) = 0\\x = \frac {\pi}{6}$$
$\frac {\pi}{2}$ would also solve the equation but the domain says strictly less than $\frac{\pi}{2}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2832009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
Find the length of AB in Triangle ABC In $\Delta ABC , m \angle A = 2 m \angle C$ , side $BC$ is 2 cm longer than side $AB$ . $AC = 5 $What is $AB$ ?
Well I thought you can use trigonometry or Complete Pythagoras theorem , but I don't really know how to apply it
|
Let $AB=x$ and $AD$ be bisector of $\Delta ABC$.
Thus, $$\Delta ABD\sim\Delta CBA,$$ which gives
$$\frac{BD}{x}=\frac{x}{x+2}$$ or
$$BD=\frac{x^2}{x+2},$$ which gives
$$DC=x+2-\frac{x^2}{x+2}=\frac{4x+4}{x+2}$$ and since
$$\frac{AB}{AC}=\frac{BD}{DC},$$ we obtain:
$$\frac{x}{5}=\frac{\frac{x^2}{x+2}}{\frac{4x+4}{x+2}},$$
which gives $x=4$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2833571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
Find absolute maximum and minimum values by parametrizing the boundaries $f(x,y) = 2\cos x + 3\sin y$ $; R= {(x , y): 0 \leq x \leq 2\pi \\\mbox{and}\\ 0 \leq y \leq \pi} $
I need to find the absolute maximum value and absolute minimum value in the region $R$, and I do have to parametrize the boundary pieces of $R$ to find critical points there.
I tried taking $(x,y) = (r\cos(\theta),r\sin(\theta))$ for $\theta \in [0,2\pi]$
and then I got $h(\theta) = 2\cos(r\cos(\theta))+3\sin(r\sin(\theta))$
After that $h'(\theta)=2r\sin(\theta)\cdot\sin(r\cos(\theta))+3r\cos(\theta)\cdot(\cos(r\sin(\theta))$.
I can't find values of $\theta$ for which $h'(\theta)=0$.
How should I proceed from here?
|
Making the parameterizations
$$
x = \pi(1+\sin (u))\\
y = \frac{\pi}{2}(1+\sin (v))
$$
we have
$$
f(u,v) = 3 \cos \left(\frac{1}{2} \pi \sin (v)\right)-2 \cos (\pi \sin (u))
$$
so the stationary points are the solutions for
$$
\left\{
\begin{array}{rcl}
2 \pi \cos (u) \sin (\pi \sin (u))=0 \\
-\frac{3}{2} \pi \cos (v) \sin \left(\frac{1}{2} \pi \sin (v)\right)=0 \\
\end{array}
\right.
$$
giving the solutions
$$
\left[
\begin{array}{ccc}
x & y & f \\
\pi & -\frac{\pi }{2} & -2 \\
\pi & \frac{\pi }{2} & -2 \\
\pi & 0 & 1 \\
0 & -\frac{\pi }{2} & 2 \\
0 & \frac{\pi }{2} & 2 \\
2 \pi & -\frac{\pi }{2} & 2 \\
2 \pi & \frac{\pi }{2} & 2 \\
0 & 0 & 5 \\
2 \pi & 0 & 5 \\
\end{array}
\right]
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2836318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Size of partial quotients in continued fraction expansion of $\sqrt{n}$ It is a well-known result that the partial fraction expansion of $\sqrt{n}$ for some non-square natural number $n$ is periodic of the form
$$[a_0, \overline{a_1, a_2, \ldots, a_{l-1}, 2a_0}]$$
where $a_0 = \lfloor\sqrt{n}\rfloor$ and $a_1, \ldots, a_{l-1}$ is palindromic.
However, here is something that I could not find anywhere explicitly: Is anything more known about the size of the $a_1, \ldots, a_{l-1}$? Are they always ${}\leq a_0$? Looking at the first $10^5$ or so examples seems to suggest so.
If not that, are they at least ${}<2a_0$?
|
It is more convenient to look at this in terms of
$$\omega = \sqrt{n} + \lfloor \sqrt{n}\rfloor = \bigl[\overline{q_0,q_1,\dotsc,q_{l-1}}\bigr]\,.$$
It is clear that we have $q_0 = 2a_0$ and $q_k = a_k$ for $k \geqslant 1$, since $\omega$ differs from $\sqrt{n}$ exactly by the integer $a_0$.
Proposition: The partial quotients of the simple continued fraction expansion of $\omega$ satisfy $q_k \leqslant a_0$ for $k \not\equiv 0 \pmod{l}$, where $l$ is the minimal period of the continued fraction expansion. Additionally, $q_k = a_0$ can happen at most once per period, in the middle.
Proof: We can write the complete quotients of $\omega$ in the form
$$\omega_k = \frac{\sqrt{n} + b_k}{c_k}$$
with integers $b_k, c_k$ satisfying
*
*$0 < b_k < \sqrt{n}$,
*$0 < c_k < \sqrt{n} + b_k$, and
*$c_k \mid (n - b_k^2)$.
This is proved by induction. The base case $k = 0$ is immediate, for $$\omega_0 = \omega = \frac{\sqrt{n} + \lfloor \sqrt{n}\rfloor}{1}\,,$$
i.e. $b_0 = \lfloor \sqrt{n}\rfloor$ and $c_0 = 1$. For the induction step, we note that $b_{k+1} = q_{k}c_{k} - b_{k}$ and $c_{k+1} = \frac{n - b_{k+1}^2}{c_{k}}$. Since $b_{k+1} \equiv -b_k \pmod{c_k}$, we have $$(n - b_{k+1}^2) \equiv (n - b_{k}^2) \equiv 0 \pmod{c_{k}}$$ by the induction hypothesis, so $c_{k+1}$ is an integer (that $b_{k+1}$ is an integer is obvious). And by the continued fraction algorithm
\begin{align}
\omega_{k+1} &= \frac{1}{\omega_k - q_k} \\
&= \biggl(\frac{\sqrt{n} + b_k}{c_k} - q_k\biggr)^{-1} \\
&= \biggl(\frac{\sqrt{n} - (q_kc_k - b_k)}{c_k}\biggr)^{-1} \\
&= \frac{c_k}{\sqrt{n} - b_{k+1}} \\
&= \frac{c_k(\sqrt{n} + b_{k+1})}{n - b_{k+1}^2} \\
&= \frac{\sqrt{n} + b_{k+1}}{c_{k+1}}\,.
\end{align}
Thus it is immediate that 3. holds for $k+1$ if it holds for $k$, and 2. follows from 1. since we always have $\omega_{k+1} > 1$. To see that 1. holds, note that $b_{k+1} < \sqrt{n}$ is equivalent to $\omega_k - q_k = \omega_k - \lfloor \omega_k\rfloor > 0$, which is true because $\omega_k$ is irrational. And
\begin{align}
&& 0 &< b_{k+1} \\
&\iff& 0 &< q_k c_k - b_k \\
&\iff& \frac{b_k}{c_k} &< q_k\,.
\end{align}
If $b_k < c_k$, then this is true because $q_k = \lfloor \omega_k\rfloor \geqslant 1$. And if $c_k \leqslant b_k$, then $c_k < \sqrt{n}$ and thus
$$q_k = \biggl\lfloor \frac{\sqrt{n} + b_k}{c_k}\biggr\rfloor \geqslant \biggl\lfloor \frac{c_k + b_k}{c_k}\biggr\rfloor = 1 + \biggl\lfloor \frac{b_k}{c_k}\biggr\rfloor > \frac{b_k}{c_k}\,.$$
Having established $b_k \leqslant \lfloor\sqrt{n}\rfloor = a_0$ and $c_k > 0$ for all $k$, we see that
$$q_k = \biggl\lfloor \frac{\sqrt{n} + b_k}{c_k}\biggr\rfloor = \biggl\lfloor \frac{a_0 + b_k}{c_k}\biggr\rfloor \leqslant \biggl\lfloor \frac{2a_0}{c_k}\biggr\rfloor\,.$$
Thus $q_k \leqslant a_0$ unless $c_k = 1$. But if $c_k = 1$, then clearly $q_k = a_0 + b_k$, $b_{k+1} = a_0 = b_1$ and $c_{k+1} = n - a_0^2 = c_1$, i.e. $\omega_{k+1} = \omega_1$, which means $k$ is a multiple of the minimal period $l$.
Additionally, we see that $q_r = a_0$ if and only if $c_r = 2$ and $b_r = a_0$. Then we have $b_{r+1} = a_0 = b_r$ and
$$c_rc_{r+1} = n - b_{r+1}^2 = n - b_r^2 = c_{r-1}c_r\,,$$
whence $c_{r+1} = c_{r-1}$. From this one finds
$$b_{r+m} = b_{r+1-m} \qquad\text{and}\qquad c_{r+m} = c_{r-m}$$
as well as $q_{r+m} = q_{r-m}$ for $0 \leqslant m \leqslant r$ using the recursion $b_{k+1} = q_kc_k - b_k$ and $c_kc_{k+1} = n - b_{k+1}^2$. Thus if $q_r = a_0$, then $c_{2r} = c_0 = 1$ and $2r$ is a multiple of $l$. So $q_k = a_0$ can happen at most once in a period, and if it happens the period has even length and $q_{l/2} = a_0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2838367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
}
|
expected sum of the numbers that appear on two dice, each biased so that a 3 comes up twice as often as each other number? This question is already posted here,but i want to check my approach.
Question
What is the expected sum of the numbers that appear on
two dice, each biased so that a $3$ comes up twice as often
as each other number?
My Approach
Let the probability of getting number other than $3$ is $p$
so
$$\frac{1}{p}+\frac{1}{p}+\frac{2}{p}+\frac{1}{p}+\frac{1}{p}+\frac{1}{p}=1 \Rightarrow p=\frac{1}{7}$$
Proability of getting $3=\frac{2}{7}$
and rest other $=\frac{1}{7}$
let $E(X_1)$ be the expectation of getting sum on rolling $1$ dice.
$E(X_1)=1 \times \frac{1}{7}+2 \times \frac{1}{7}+3 \times \frac{2}{7}+4 \times \frac{1}{7}+5 \times \frac{1}{7}+6 \times \frac{1}{7}=\frac{24}{7}$
Now
expected sum of the numbers that appear on two dice
$$E(X_1 +X_2)=E(X_1)+E(X_2)=\frac{24}{7}+\frac{24}{7}=\frac{48}{7}\approx 6.86$$
Is my approach correct?
|
This is the same as two seven sided dice with two 3s.
Expected value $E= 2\cdot \frac{1+2+3+3+4+5+6}{7} = 6.8571$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2839494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
}
|
If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ If the minimum and the maximum values of $y = (x^2-3x+c)/(x^2+3x+c)$ are $7$ and $1/7$ , then the value of $c$ is?
I cross multiplied the equation and tried to find it's discriminant, but I don't think it gets me anywhere. A little hint would be appreciated!
|
Solving $$y=\frac{x^2-3x+c}{x^2+3x+c}$$ and $$y=7$$
we put $$\frac{x^2-3x+c}{x^2+3x+c}=7$$
to get $$x^2-4x+c=0$$ which should have exactly one root.
So putting $Discriminant=0$
we get $c=4$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2842481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
}
|
$a+b+c=1.$ Show that $\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq \frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Assume $a,b,c>0$ and $a+b+c=1.$ Show that
$\frac{1}{1-a}+\frac{1}{1-b}+\frac{1}{1-c} \geq
\frac{2}{1+a}+\frac{2}{1+b}+\frac{2}{1+c}$
Here's what I tried:
$\frac{a+b+c}{b+c}+\frac{a+b+c}{a+b}+\frac{a+b+c}{a+b} \geq
\frac{1+a+b+c}{1+a}+\frac{1+a+b+c}{1+b}+\frac{1+a+b+c}{1+c}$
$\rightarrow \frac{a}{b+c}+\frac{c}{a+b}+\frac{b}{a+b} \geq
\frac{b+c}{1+a}+\frac{a+c}{1+b}+\frac{a+b}{1+c}$
Let $A=b+c, B=a+c,$ and $ C=a+b,$ so $A+B+C=2.$
$\rightarrow \frac{1-A}{A}+\frac{1-B}{B}+\frac{1-C}{C} \geq \frac{A}{2-A}+\frac{B}{2-B}+\frac{C}{2-C}.$
Or $\frac{1}{A}+\frac{1}{B}+\frac{1}{C} - 3 \geq \frac{A}{2-A}+\frac{B}{2-B}+\frac{C}{2-C}.$
I am stuck here, thought about using AM-GM-HM to get rid of the reciprocal on the RHS but it doesn't work if applied directly.
|
Just another way, note for $x\in(0,1)$,
$$f(x)=\frac1{1-x}-\frac2{1+x}-\frac98(3x-1)=\frac{(3x-1)^2(3x+1)}{8(1-x^2)}\geqslant0$$
while the inequality is just $f(a)+f(b)+f(c)\geqslant 0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2844178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Probabilities of selection with replacement. Say you have a bag with 4 balls, 2 of which are red and 2 are blue. You select 2 balls randomly from the bag and you want to know the probability of selecting exactly 1 red ball.
Without replacement, there are two simple methods:
*
*$\binom{2}{1}\frac{2}{4}\frac{2}{3} = \frac{2}{3}$
*$\frac{ \binom{2}{1}\binom{2}{1}} {\binom{4}{2}} = \frac{2}{3}$
With replacement we can do this:
*$\binom{2}{1}\frac{2}{4}\frac{2}{4} = \frac{1}{2}$
My question: Is there a way I can find the probability of exactly 1 red ball with replacement using ONLY combinations in the numerator and denominators, like in example 2?
I tried using the $\binom{r+n-1}{r}$ formula:
$\frac{ \binom{2+1-1}{1}\binom{2+1-1}{1}} {\binom{4+2-1}{2}} = \frac{ \binom{2}{1}\binom{2}{1}} {\binom{5}{2}}$ but I got $\frac{2}{5}$ instead of the correct answer which is $\frac{1}{2}$. I'm curious to know.
|
With replacement we just have independent trials with success probability $\frac{2}{4} = \frac{1}{2}$. We do two trials, one of which must be a success (red) and one a failure. We have $\binom{2}{1}$ ways of choosing the trial with a success and all sequences with one sucess and one failure have probability $\frac12 \frac12$ by the product rule.
So generalising for $r$ reds out of $n$ trials in a bag with $k$ reds and $m-k$ blues (so $m$ balls total), we have $$P(r \text{ reds}) = \binom{n}{r}(\frac{k}{m})^r(\frac{m-k}{m})^{n-r}= \frac{\binom{n}{r}k^r(m-k)^{n-r}}{m^n}$$
while the corresponding probability without replacement equals, provided $m \le n, r \le k, n-r \le m-k$:
$$P(r \text{ reds}) = \frac{\binom{k}{r}\binom{m-k}{n-r}}{\binom{m}{n}}= $$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2844456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Complex Partial Fraction Decomposition The question I need help with is:
Prove that
$$\sum_{k=0}^{6}\frac{1-z^{2}}{1-2z\cos\left(\frac{2k\pi}{7}\right)+z^{2}}=\frac{7(z^{7}+1)}{1-z^{7}}$$
I have already tried brute forcing this by combining the LHS into a single fraction. While this worked, it is an extremely long proof.
I was wondering if there is a more elegant approach that uses partial fractions. I tried decomposing each term in the sum of the LHS into $$-1+\frac{B}{z-\omega^{k}}+\frac{C}{z-\omega^{-k}}$$but this gave me very complicated expressions for constants B and C so I gave up.
|
(Edit: the following was posted before the OP added "I have already tried brute forcing this ...".)
I don't see the elegant solution offhand, but the problem can certainly be brute-forced as sketched below. Note that the first term of the sum (excluding the $\,\,1-z^2$ factor ) is $\,\dfrac{1}{(z-1)^2}\,$ then the rest of terms are pairwise equal since $\,\cos \left(2k\pi/7\right) = \cos\left(2(7-k)\pi/7)\right)\,$. Therefore the sum of those remaining $\,6\,$ terms is twice the sum of the first $\,3\,$, which works out to:
$$
\begin{align}
& \frac{1}{z^2-(\omega+\omega^6)z+1}+\frac{1}{z^2-(\omega^2+\omega^5)z+1}+\frac{1}{z^2-(\omega^3+\omega^4)z+1} \\[15px]
=\;\; &{\frac{(z^2-(\omega^2+\omega^5)z+1)(z^2-(\omega^3+\omega^4)z+1)\\+(z^2-(\omega+\omega^6)z+1)(z^2-(\omega^3+\omega^4)z+1)\\+(z^2-(\omega+\omega^6)z+1)(z^2-(\omega^2+\omega^5)z+1)}{(z-\omega)(z-\omega^6)\cdot(z-\omega^2)(z-\omega^5) \cdot (z-\omega^3)(z-\omega^4)}}
\end{align}
$$
The denominator of the latter fraction is $\,\dfrac{z^7-1}{z-1}\,$, and the numerator eventually evaluates to:
$$
3z^4 -2(\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega)z^3 \\+(\omega^{11}+\omega^{10}+2\omega^9+2\omega^8+2\omega^6+2\omega^5+\omega^4+\omega^3+6)z^2 \\-2(\omega^6+\omega^5+\omega^4+\omega^3+\omega^2+\omega)z +3
$$
Using that $\,\omega^7=1\,$ and $\,1+\omega+\omega^2+\omega^3+\omega^4+\omega^5+\omega^6=0\,$ the above simplifies to:
$$
3z^4 +2z^3 +4z^2 +2z +3
$$
Then the problem reduces to verifying the algebraic identity:
$$
(1-z^2)\left(\frac{1}{(z-1)^2} + 2\cdot\frac{3z^4+2z^3+4z^2+2z+3}{\dfrac{z^7-1}{z-1}}\right) = \dfrac{7(z^7+1)}{1-z^7}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2846658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
}
|
Show that the system has a limit cycle. The full question reads:
Show that the system with $\dot{x} = x-y-(x^2 + \frac{3}{2} y^2)x$ and $\dot{y} = x+y-(x^2 + \frac{1}{2} y^2)y$ has a limit cycle.
I wanted to check the correctness of my method. My thought was to find the equilibrium points (If I am correct, the only one is $(0,0)$, then find the stability of that point. Then I attempted to show that if we go far enough out in the domain, that the flow of the direction field is going in the opposite direction of the direction field directly surrounding the equilibrium point. Then we can use Poincare-Bendixon and be done. Is this correct, or should I handle it differently? Thanks!
|
Another approach is to proceed using the Poincare-Bendixon theorem.
Our aim then is to show that the system contains only one unstable equilibrium point, and that every trajectory of the system remains in a closed bounded subset.
From your original problem, let
$$
f_1(x,y) = x - y - x^3 - \frac{3}{2}y^2{x} \nonumber \\
f_2(x,y) = x + y - x^2{y} - \frac{1}{2}y^3 \nonumber \\
$$
Then
$$
\dot{x} = f_1(x,y) \nonumber \\
\dot{y} =f_2(x,y) \nonumber \\
$$
The equilibrium point is $x=0$ and $y =0$
The Jacobian is
\begin{align}
J &=
\begin{bmatrix}
\frac{\partial{f_1}}{\partial x} & \frac{\partial{f_1}}{\partial y} \\
\frac{\partial{f_2}}{\partial y} & \frac{\partial{f_2}}{\partial y} \\
\end{bmatrix} \nonumber
_{\substack{x=0 \\ y=0}} \\
&=
\begin{bmatrix}
1 & -1 \\
1 & 1 \\
\end{bmatrix}
\end{align}
The eigenvalues are $1+i$ and $1-i$.
Hence the system is unable (unstable node or focus), which is the first criterion of the Poincare-Bendixon theorem.
Secondly, let us choose a closed bounded subset given by
$$
V(x,y)=x^2 + y^2 \le c \nonumber
$$
where c is a positive constant.
Therefore the chosed subset $V(x,y)$ is a circle of radius $\sqrt c$.
We only need to show that there exists a finte value of $c$ for which the vector field of $f(x,y)$ never leaves the set enclosed by $c$.
This is identical to the statement:
$$
\nabla {V(x,y)} \cdot f(x,y) \le 0 \nonumber
$$
Now,
$$
\nabla {V(x,y)} \cdot f(x,y) = \frac{\partial V}{\partial x}f_1(x,y) + \frac{\partial V}{\partial y}f_2(x,y) \nonumber
$$
Therefore
$$
\frac{\partial V}{\partial x}f_1(x,y) + \frac{\partial V}{\partial y}f_2(x,y) \nonumber
$$
$$
= 2x\left[x-y-\left(x^2+\frac{3}{2}y^2\right)x\right] + 2y\left[x+y-\left(x^2+\frac{1}{2}y^2\right)y\right] \nonumber \\
= 2{\left(x^2+y^2\right)} -2x^2{\left(x^2+y^2+\frac{1}{2}y^2\right)}-2y^2{\left(x^2+y^2-\frac{1}{2}y^2\right)} \\
= 2{\left(x^2+y^2\right)} - 2\left(x^2+y^2\right)^2 + y^4-x^2y^2
$$
But
$$
y^4-x^2y^2 \le \left(x^2+y^2\right)^2
$$
Therefore
$$
2{\left(x^2+y^2\right)} - 2\left(x^2+y^2\right)^2 + y^4-x^2y^2 \le 2{\left(x^2+y^2\right)} - 2\left(x^2+y^2\right)^2 +\left(x^2+y^2\right)^2
= 2{\left(x^2+y^2\right)} - \left(x^2+y^2\right)^2
$$
This implies that
$$
\frac{\partial V}{\partial x}f_1(x,y) + \frac{\partial V}{\partial y}f_2(x,y) \le 2{\left(x^2+y^2\right)} - \left(x^2+y^2\right)^2 = 2c - c^2 = (2-c)c \nonumber
$$
Finally, if we choose $c=2$, we are guaranteed that
$$
\frac{\partial V}{\partial x}f_1(x,y) + \frac{\partial V}{\partial y}f_2(x,y) \le 0 \nonumber
$$
which satisfies the second criterion of the Poincare-Bendixon theorem.
Therefore, it can be concluded that a limit cycle exists.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2847203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Finding the residue Finding the fourth derivative in order to get residue seems me very complicated, is there another way? $$Res\left( z=i,\frac { { e }^{ iz } }{ { \left( { z }^{ 2 }+1 \right) }^{ 5 } } \right) =\lim _{ z\rightarrow i }{ \frac { 1 }{ 4! } \frac { { d }^{ 4 } }{ { d }{ z }^{ 4 } } \left( { \left( z-i \right) }^{ 5 }\frac { { e }^{ iz } }{ { \left( z-i \right) }^{ 5 }{ \left( z+i \right) }^{ 5 } } \right) } $$
|
Since the residuum is the coefficient of $(z-i)^{-1}$ in the Laurent-series expansion of $\frac{e^{iz}}{(z^2+1)^5}$ at $z=i$, we can expand the function and extract the coefficient.
We obtain
\begin{align*}
\color{blue}{\mathrm{Res}}&\color{blue}{\left( z=i,\frac { { e }^{ iz } }{ { \left( { z }^{ 2 }+1 \right) }^{ 5 } } \right)}\\
&=\mathrm{Res}\left(t=0,\frac{e^{i(t+i)}}{t^5(t+2i)^5}\right)\tag{1}\\
&=[t^{-1}]\frac{e^{i(t+i)}}{t^5(t+2i)^5}\tag{2}\\
&=[t^4]\frac{e^{i(t+i)}}{(t+2i)^5}\tag{3}\\
&=\frac{1}{(2i)^5}[t^4]e^{i(t+i)}\sum_{j=0}^\infty\binom{-5}{j}\left(\frac{t}{2i}\right)^j\tag{4}\\
&=\frac{1}{32ie}[t^4]e^{it}\sum_{j=0}^{\infty}\binom{j+4}{j}\left(-\frac{t}{2i}\right)^j\tag{5}\\
&=\frac{1}{32ie}\left(\frac{i^0}{0!}\binom{8}{4}\left(-\frac{1}{2i}\right)^4+\frac{i^1}{1!}\binom{7}{3}\left(-\frac{1}{2i}\right)^3
+\frac{i^2}{2!}\binom{6}{2}\left(-\frac{1}{2i}\right)^2\right.\\
&\qquad\qquad\left.+\frac{i^3}{3!}\binom{5}{3}\left(-\frac{1}{2i}\right)^1+\frac{i^4}{4!}\binom{4}{4}\left(-\frac{1}{2i}\right)^0\right)\tag{6}\\
&=\frac{1}{32ie}\left(\frac{35}{8}+\frac{35}{8}+\frac{15}{8}+\frac{5}{12}+\frac{1}{24}\right)\\
&\,\,\color{blue}{=-\frac{133i}{384e}}
\end{align*}
Comment:
*
*In (1) we shift the residuum to $0$ by setting $t=z-i$.
*In (2) we use the coefficient of $[t^n]$ operator to denote the coefficient of $t^n$ in a series.
*In (3) we apply the rule $[t^{p-q}]A(t)=[t^p]t^qA(t)$.
*In (4) we factor out $(2i)^5$ and use the binomial series expansion.
*In (5) we use the binomial identity $\binom{-p}{q}=\binom{p}{q}(-1)^q$.
*In (6) we select the coefficient of $t^4$, by recalling $e^{it}=\sum_{k=0}^\infty\frac{(it)^k}{k!}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2847781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Factorization Manipulation Let $x$ and $y$ be real numbers. Consider $t=x^2+10y^2-6xy-4y+13$. So what is the $t$ as smallest number? The solution is $9$.
My trying:
$$t=x^2+10y^2-6xy-4y+13$$
$$=x^2+9+10y^2-6xy-4y+4$$
$$=(x+3)^2-3x+2y(5y-2)-6xy+4$$.
So let $x=-3$ and $y=0$. From this I got $t=13$. My answer is false. Can you help?
|
Use Gauß' method to write a quadratic form as a linear combination of squares of linear forms:
\begin{align}
t&=x^2+10y^2-6xy-4y+13 =(x-3y)^2 -9y^2+10y^2-4y+13 \\
&=(x-3y)^2+y^2-4y+13 =(x-3y)^2+(y-2)^2-4+13\\
&=(x-3y)^2+(y-2)^2+9.
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2849669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
}
|
Minimum value of $\frac{b+1}{a+b-2}$
If $a^2 + b^2= 1 $ and $u$ is the minimum value of the $\dfrac{b+1}{a+b-2}$, then find the value of $u^2$.
Attempt:
Then I tried this way: Let $a= bk$ for some real $k$.
Then I got $f(b)$ in terms of b and k which is minmum when $b = \dfrac{2-k}{2(k+1)}$ ... then again I got an equation in $k$ which didn't simplify.
Please suggest an efficient way to solve it.
|
A bit geometry;
1)$x^2+y^2 = 1$, a circle, centre $(0,0)$, $r=1$.
2) Minimum of $C$:
$C:=\dfrac{y+1}{x+y-2}$
(Note: $x+y-2 \not =0$).
$C(x+y-2) = y+1$, or
$Cx +y(C-1) -(2C+1)= 0$, a straight line.
The line touches or intersects the circle 1)
if the distance line-to-origin $\le 1$ (radius).
Distance to $(0,0):$
$d =\dfrac{|2C+1|}{\sqrt{C^2+(C-1)^2}} \le 1.$
$(2C+1)^2 \le C^2 + (C-1)^2;$
$4C^2 +4C +1 \le 2C^2 -2C+1;$
$2C(C+3) \le 0.$
Hence: $-3 \le C \le 0$.
Minimum at $C =-3$.
Used: Line to point distance formula:
http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2852399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
}
|
Can we conclude that two of the variables must be $0$?
Assuming $$a^2+b^2+c^2=1$$ and $$a^3+b^3+c^3=1$$ for real numbers $a,b,c$, can we conclude that two of the numbers $a,b,c$ must be $0$ ?
I wonder whether mathworld's result that only the triples $(1,0,0)$ , $(0,1,0)$ , $(0,0,1)$ satisfy the given equation-system , is actually true.
Looking at $(a+b+c)^3$ and $(a+b+c)^2$ , using
\begin{align}
&(a+b+c)^3= (a+b+c)^2(a+b+c)=\\
&(1+2(ab+ac+bc))(a+b+c)= \\
&2(a+b)(a+c)(b+c)+a+b+c+2abc
\end{align}
and eliminating $(a+b)(a+c)(b+c)$, with $S:=a+b+c$ , I finally got $$(S-1)^2(S+2)=6abc$$
I guess this is not enough to show the above result (if it is true at all).
This question is inspired by an exercise to determine the possible values of $a+b+c$ assuming the above equations, so this question could be a duplicate, but I am not sure whether it actually is.
|
Without loss of generality assume that $a\not=0$ and $b\not=0$ such that $a^2+b^2+c^2=1$ and $a^3+b^3+c^3=1$. Then $|a|<1$, $|b|<1$ and $|c|<1$ (otherwise $a^2+b^2+c^2>1$). Therefore
$$1=|a^3+b^3+c^3|\leq |a|^3+|b|^3+|c|^3<|a|^2+|b|^2+|c|^2=1$$
Contradiction.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2852504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
}
|
Minimizing in 3 variables
Find the least possible value of the fraction $\dfrac{a^2+b^2+c^2}{ab+bc}$, where $a,b,c > 0$.
My try:
$a^2+b^2+c^2 = (a+c)^2 - 2ac + b^2$,
$= (a+c)/b +b/(a+c) -2ac/b(a+c)$
AM > GM
$3\sqrt[3]{-2ac/b(a+c)}$
And I cant somehow move on.
|
Let $\dfrac{a^2+b^2+c^2}{ab+bc}=k>0$ as $a,b,c>0$
$\iff b^2-kb(a+c)+a^2+c^2=0$ which is a Quadratic Equation in $b$
As $b$ is real, the discriminant must be $\ge0$
i.e., $$k^2(a+c)^2-4(a^2+c^2)\ge0\iff k^2\ge\dfrac{4(a^2+c^2)}{(a+c)^2}$$
the equality occurs if $a=\dfrac{k(a+c)}2$
Now $2(a^2+c^2)-(a+c)^2=(a-c)^2\ge0\implies2(a^2+c^2)\ge(a+c)^2$
$\implies k^2\ge2$
the equality occurs if $a=c$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2853202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
}
|
prove this inequality $(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$ Let $x,y,z>0$,show that
$$(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3\le27 (xyz)^2$$
I have prove this inequality
$$(x+y-z)(y+z-x)(x+y-z)\le xyz$$
because it is three schur inequality
$$\Longleftrightarrow x^3+y^3+z^3+3xyz\ge xy(x+y)+yz(y+z)+zx(z+x)$$
how to solve this inequiality $xyz ⩾ (x+y-z)(y+z-x)(z+x-y)$
But I can't prove this inequality to prove $(1)$
|
We can assume that $x,y,z$ are sides of a triangle because otherwise the LHS is negative.
By Heron's formula we have
$$A^2 = s(s-x)(s-y)(s-z) = \frac1{16}(x+y+z)(x+y-z)(y+z-x)(x+z-y)$$
where $A$ is the area and $s$ is the semiperimeter.
Now recall this inequality:
$$4\sqrt{3}A \le \frac{9xyz}{x+y+z}$$
or $$A^2 \le \frac{27}{16} \frac{(xyz)^2}{(x+y+z)^2}$$
It is listed on wikipedia and for a proof see here.
We get
\begin{align}
(x+y-z)(y+z-x)(x+z-y)(x+y+z)^3 = 16{A^2}(x+y+z)^2 \le {27(xyz)^2}
\end{align}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2853415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
}
|
Find Principal Part Of $\frac{ze^{iz}}{(z^2+9)^2}$ Find the principal part of $\frac{ze^{iz}}{(z^2+9)^2}$ at $z_0=3i$
Can I say that $$\frac{ze^{iz}}{(z^2+9)^2}=\frac{\frac{ze^{iz}}{(z+3i)^2}}{(z-3i)^2}$$
And look at $g(z)=\frac{ze^{iz}}{(z+3i)^2}$?
|
Note that both parts $\frac{1}{(z-3i)^2}$ as well as $g(z)$ contribute to the principal part of the function.
We obtain
\begin{align*}
\frac{ze^{iz}}{(z^2+9)^2}&=\frac{1}{(z-3i)^2}\cdot\frac{ze^{iz}}{(z+3i)^2}\\
&=\frac{1}{(z-3i)^2}\cdot\frac{ze^{iz}}{(6i+z-3i)^2}\\
&=\frac{1}{(z-3i)^2}\cdot\frac{ze^{iz}}{(6i)^2\left(1+\frac{z-3i}{6i}\right)^2}\\
&=\frac{1}{(z-3i)^2}\cdot\frac{\left[3i+(z-3i)\right]e^{i(z-3i+3i)}}{-36}
\sum_{n=0}^\infty\binom{-2}{n}\left(\frac{z-3i}{6i}\right)^n\tag{1}\\
&=\left(-\frac{i}{12e^3(z-3i)^2}-\frac{1}{36e^3(z-3i)}\right)e^{i(z-3i)}\sum_{n=0}^\infty\binom{-2}{n}\left(\frac{z-3i}{6i}\right)^n\tag{2}\\
&=\left(-\frac{i}{12e^3(z-3i)^2}-\frac{1}{36e^3(z-3i)}\right)\\
&\qquad\qquad\cdot\left(1+i(z-3i)+\sum_{n=2}^\infty \frac{(i(z-3i))^n}{n!}\right)\\
&\qquad\qquad\cdot\left(1-\frac{1}{3i}(z-3i)+\sum_{n=2}^\infty\binom{-2}{n}\left(\frac{z-3i}{6i}\right)^n\right)\tag{3}\\
&=\color{blue}{-\frac{i}{12e^3(z-3i)^2}+\frac{1}{12e^3(z-3i)}}+\sum_{n=0}^\infty a_n(z-3i)^n\tag{4}
\end{align*}
Comment:
*
*In (1) we use the binomial series expansion and prepare the numerator for expansion at $z-3i$.
*In (2) we do some simplifications and separate the relevant terms with negative powers.
*In (3) we separate the constant and linear terms of the power series which contribute to the principal part.
*In (4) we multiply out and obtain the (blue colored) principal part.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2854580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
}
|
Proving the equality case in triangle inequality Background
When plotted on a real number line, it may be deduced that if
$$a,b,c \in \mathbb{R} $$
$$a < b < c$$
then
$$\left| {a - c} \right| = \left| {a - b} \right| + \left| {b - c} \right|$$
Problem
But the problem is with the proof. How can the above statement be proven true from the properties of order structure and the definition of absolute value?
|
To prove it you do cases
$|a-c| = \pm (a-c)$ and $|a-b| + |b- c| = \pm(a-b) + \pm (b-c)$.
consider the ways that $a,b,c$ may be ordered:
(your case is a subcase of case 6: so you can skip to the very last case if you want.)
1: Assume $a \ge b \ge c$ then
$|a-c| = a-c$ and $|a-b| + |b-c| = (a-b) + (b-c) = a-c = |a-c|$.
2: Assume $a \ge c \ge b$ then
$|a-c| = a-c$ and $|a-b| + |b-c| = (a-b) -(b-c) = a-2b + c$. Now $2b \le 2c$ so $-2b \ge -2c$ and $a-2b + c\ge a -2c +c = a-c = |a-c|$.
3: Assume $b \ge a \ge c$ then
$|a-c| = a-c$ and $|a-b| + |b-c| = -(a-b) +(b-c) = -a+2b - c$. Now $2b \ge 2a$ so $-a+2b - c \ge -a +2a -c = a-c = |a-c|$.
4: Assume $b \ge c \ge a$
$|a-c| = -(a-c) = c-a$ and $|a-b| + |b-c| = -(a-b) +(b-c) = -a+2b - c$. Now $2b \ge 2c$ so $-a+2b - c \ge -a +2c -c = -a+c = |a-c|$.
5: Assume $c \ge a \ge b$
$|a-c| = -(a-c) = c-a$ and $|a-b| + |b-c| = (a-b) -(b-c) = a-2b + c$. Now $2b \le 2a$ so $-2b \ge -2a$ and $a-2b + c\ge a -2a +c = -a+c = |a-c|$.
6: Assume $c \ge b \ge a$
$|a-c| = -(a-c) = c-a$ and $|a-b| + |b-c| = -(a-b) - (b-c) = (b-a) + (c-b) = -a + c = |a-c|$.
===
The "equality cases are 1) and 6) and they are very straightforward.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2855210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
}
|
Prove that $a^4+b^4+1\ge a+b$.
Prove that
$$a^4+b^4+1\ge a+b$$
for all real numbers $a,b$.
What I've tried:
1.I checked how AM-GM may help but doesn't look like it's useful here.
*I've tried:
$$(a^2+b^2)^2 -2(ab)^2+1 \ge a+b$$
But unfortunately, I can't find any way to continue this..
I'm sure that this isn't too hard, it's just that "I'm not seeing it", I would appreciate if clues are given first so I can answer this myself.
*This exercise is from the TAU entry exams.
|
Proof
Since $$\left(a^4-a^2+\frac{1}{4}\right)+\left(b^4-b^2+\frac{1}{4}\right)=\left(a^2-\frac{1}{2}\right)^2+\left(b^2-\frac{1}{2}\right)^2 \geq0,$$
hence $$a^4+b^4+1 \geq a^2+b^2+\frac{1}{2}.\tag1$$
Since $$\left(a^2-a+\frac{1}{4}\right)+\left(b^2-b+\frac{1}{4}\right)=\left(a-\frac{1}{2}\right)^2+\left(b-\frac{1}{2}\right)^2 \geq 0,$$
hence, $$a^2+b^2+\frac{1}{2} \geq a+b.\tag2$$
Combining $(1)$ and $(2)$, $$a^4+b^4+1 \geq a+b.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2857994",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 5,
"answer_id": 4
}
|
How to evaluate $\int\frac{1+x^4}{(1-x^4)^{3/2}}dx$? How do I start with evaluating this-
$$\int\frac{1+x^4}{(1-x^4)^{3/2}}dx$$
What should be my first attempt at this kind of a problem where-
*
*The denominator and numerator are of the same degree
*Denominator involves fractional exponent like $3/2$.
Note:I am proficient with all kinds of basic methods of evaluating integrals.
|
Using @Fimpellizieri 's brilliant observation on @Dahaka 's ingenious post:
\begin{equation}
\frac{d\,[f(x)]^a}{dx}=\frac{d\,f^a}{df}\frac{d\,f}{dx}=af^{a-1}\cdot f'=a\frac{f'}{f^{1-a}}\tag{1}
\end{equation}
we find the integral in the question belongs to a family that can be found from using $(1)$.
First consider $I=\int\frac{1+x^2}{(1-x^2)^{2}}dx$ and let $f(x)=x^{-1}-x$, $f'(x)=-x^{-2}-1=-1\cdot(x^{-2}+1)$, $a=-1$. Now by $(1)$
$$\frac{d\,[f(x)]^{-1}}{dx}=-1\cdot\frac{(-1\cdot(x^{-2}+1))}{(x^{-1}-x)^{1-(-1)}}
=\frac{(x^{-2}+1)}{(x^{-1}-x)^{2}}=\frac{1+x^2}{(1-x^2)^2}$$
and so
$$I=\int \frac{(1+x^{2})}{(1-x^{2})^{2}}dx=\int d\,([f(x)]^{-1})=[f(x)]^{-1}+c=(x^{-1}-x)^{-1}+c=\frac{x}{1-x^2}+c$$
For the next let $f(x)=x^{-2}-x^{2}$, $f'(x)=-2x^{-3}-2x=-2\cdot(x^{-3}+x)$, $a=-\frac{1}{2}$. Now by $(1)$
$$\frac{d\,[f(x)]^{-1/2}}{dx}=-\frac{1}{2}\cdot\frac{(-2\cdot(x^{-3}+x))}{(x^{-2}-x^{2})^{1-(-\frac{1}{2})}}
=\frac{(x^{-3}+x)}{(x^{-2}-x^{2})^{3/2}}=\frac{1+x^4}{(1-x^4)^{3/2}}$$
and so
$$I=\int \frac{1+x^4}{(1-x^4)^{3/2}}dx=\int d\,([f(x)]^{-1/2})=[f(x)]^{-1/2}+c=(x^{-2}-x^{2})^{-1/2}+c=\frac{x}{(1-x^4)^{1/2}}+c$$
For the next in the pattern we have
\begin{align*}
I&=\int\frac{1+x^6}{(1-x^6)^{4/3}}dx=\int\frac{(1+x^6)/x^4}{(1-x^6)^{4/3}/x^4}dx=\int\frac{x^{-4}+x^2}{(x^{-3}-x^3)^{4/3}}dx\\
&=\int d\left((x^{-3}-x^3)^{-1/3}\right)=(x^{-3}-x^3)^{-1/3}+c=\frac{x}{(1-x^6)^{1/3}}+c
\end{align*}
In general we have (with $f(x)=x^{-n}-x^{n}$, $f'(x)=-n\cdot(x^{-(n+1)}+x^{n-1})$, $a=-\frac{1}{n}$)
\begin{align*}
I&=\int\frac{1+x^{2n}}{(1-x^{2n})^{(n+1)/n}}dx=\int\frac{(1+x^{2n})/x^{(n+1)}}{(1-x^{2n})^{(n+1)/n}/x^{(n+1)}}dx=\int-\frac{1}{n}\cdot\frac{\left(-n\cdot(x^{-{(n+1)}}+x^{n-1})\right)}{(x^{-n}-x^{n})^{(n+1)/n}}dx\\
&=\int d\left((x^{-n}-x^n)^{-1/n}\right)=(x^{-n}-x^n)^{-1/n}+c=\frac{x}{(1-x^{2n})^{1/n}}+c
\end{align*}
giving the general result:
\begin{equation}
I=\int\frac{1+x^{2n}}{(1-x^{2n})^{(n+1)/n}}dx=\frac{x}{(1-x^{2n})^{1/n}}+c
\end{equation}
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2858747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 1
}
|
How to factorize $a(y) := xy^{3}+xy^{2}+(x+1)y+x \in GF(2)[x]_{x^{2}+x+1}[y]$ Could someone please help me to find irreducible factors of $a(y) := xy^{3}+xy^{2}+(x+1)y+x \in GF(2)[x]_{x^{2}+x+1}[y]$?
In $GF(2)[x]_{x^{2}+x+1}[y]$, we have $0,1,x,x+1$. So we use these in $a(y)$ one after the other:
$0:x⋅0^{3}+x⋅0^{2}+(x+1)⋅0+x=x$ => not a root
$1:x⋅1^{3}+x⋅1^{2}+(x+1)⋅1+x=x+x+x+1+x=4x+1=1$ => not a root
$x:x⋅x^{3}+x⋅x^{2}+(x+1)⋅x+x=x^{4}+x^{3}+x^{2}+x+x=x^{4}+x^{3}+x^{2}$
But I don't understand what should I do with $x^{4}+x^{3}+x^{2}$ now?
I can factorize it into $x^{2}⋅(x^{2}+x+1)$ but how can I understand that x is a root ot not?
Thank you for any help!
|
@hardmath, I'll try :-)
In $GF(2)[x]_{x^{2}+x+1}[y]$, we have $0,1,x,x+1$. So we use these in a(y) one after the other:
*
*$0:x⋅0^{3}+x⋅0^{2}+(x+1)⋅0+x=x$ => not a root.
*$1:x⋅1^{3}+x⋅1^{2}+(x+1)⋅1+x=x+x+x+1+x=4x+1=1$ => not a root.
*$x:x⋅x^{3}+x⋅x^{2}+(x+1)⋅x+x=x4+x3+x2+x+x=x4+x3+x2 =0$ => x is a root.
*$x+1:x⋅(x+1)^{3}+x⋅(x+1)^{2}+(x+1)⋅(x+1)+x=x+x+1+x+x=1$ => not a root.
Therefore, we know that $(y+x)$ is one of the factors.
$(xy^{3}+xy^{2}+(x+1)y+x):(y+x)=xy^{2}+y+1$
Again, we use $0,1,x,x+1$ in order to determine whether $xy^{2}+y+1$ is irreducible:
*
*$0:x⋅0^{2}+0+1=1$ => not a root.
*$1:x⋅1^{2}+1+1=x$ => not a root.
*$x:x⋅x^{2}+x+1=x^{3}+x+1=x$ => not a root.
*$x+1:x(x+1)^{2}+(x+1)+1=x(x^{2}+1)+x=x^{2}=1$ => not a root.
$xy^{2}+y+1$ has no roots, so it is irreducible over GF(2).
Therefore,
$xy^{3}+xy^{2}+(x+1)y+x=(y+x)⋅(xy^{2}+y+1)$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2859687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Remainder of a Polynomial when divided by a polynomial with nonreal roots What will be the remainder when $x^{2015}+x^{2016}$ is divided by $x^2+x+1$ without using the fact that $x^2+x+1$ has roots as non real cube roots of unity.
|
Hint: using that $a^3-1=(a-1)(a^2+a+1)$ and so $\,x^{3k}-1=(x-1)(x^2+x+1)(\ldots)\,$:
$$
\big(x^{2016}\color{red}{-1}\big)+\big(x^{2015}\color{red}{-x^2}\big)\color{red}{+1+x^2} \\
=\left(\left(x^3\right)^{672}-1\right)+x^2\left(\left(x^3\right)^{671}-1\right)+\big(x^2\color{blue}{+ x} + 1\big)\color{blue}{- x}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2862039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Logarithm equation
How to solve this logarithm equation?
$$\frac12\cdot[\log(x) + \log(2)) + \log[\sqrt{2x} + 1] = \log(6).$$
The answer is $2$.
I've tried to solve it, but I don't know how to proceed:
$\frac12\log(2x) + \log[\sqrt{2x} + 1] = \log(6)$
$\log([(2x)^\frac12] + \log[\sqrt{2x} + 1] = \log(6)$
$\log[\sqrt{2x}] + \log[\sqrt{2x} + 1] = \log(6)$
$\log(\sqrt{2x}\cdot [\sqrt{2x} + 1]) = \log(6)$
$\sqrt{2x}\cdot[\sqrt{2x} + 1] = 6$
$2x + \sqrt{2x} = 6$
$2x + \sqrt{2}\sqrt{x} - 6 = 0$
|
Square both sides of the equation $\sqrt{2x}=6-2x$ to obtain $2x=36-24x+4x^2$ which will give you this quadratic equation $4x^2-26x+36=0$. If we simplify the equation further by dividing by the common factor $2$ we obtain $2x^2-13x+18=0$. This should be easy to solve if you use the quadratic formula or by factorizing the left side and finding its roots.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2864900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Evaluating $\lim_{x\to0} \frac{\cos x - \cos 3x}{\sin 3x^2 - \sin x^2}$
$$ \lim_{x\to0}\frac{\cos x-\cos (3x)}{\sin (3x^2)-\sin (x^2)} $$
Is there a simple way of finding the limit?
I know the long one: rewrite it as
$$ -\lim_{x\to 0}\frac{\cos x-\cos(3x)}{\sin(3x^2)}\cdot\frac{1}{1-\dfrac{\sin(3x^2)}{\sin(x^2)}} $$
and then find both limits in separately applying L'Hospital's rule several times. The answer is $2$.
|
By standard limits
*
*$\frac{\sin x}x \to 1$
*$\frac{1-\cos x}{x^2} \to \frac12$
we have that
$$\frac{\cos x-\cos(3x)}{\sin(3x^2)-\sin(x^2)}=\frac{\frac{\cos x-1+1- \cos(3x)}{x^2}} {\frac{\sin(3x^2)-\sin(x^2)}{x^2}}=\frac{-\frac{1-\cos x}{x^2}+9\frac{1- \cos(3x)}{(3x)^2}} {3\frac{\sin(3x^2)}{3x^2}-\frac{\sin(x^2)}{x^2}}\to\frac{-\frac12+\frac92}{3-1}=2$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2867375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 9,
"answer_id": 6
}
|
Proving $\sum_{j=0}^n (-1)^j {n \choose j} F_{s+2n-2j} = F_{s+n} $, where $F_n$ is the $n$-th Fibonacci number
$$\sum_{j=0}^n (-1)^j {n \choose j} F_{s+2n-2j} = F_{s+n} $$
($F$ is Fibonacci number).
I have been trying to prove this by mathematical induction.
First I assume this is true for n.
If I could prove it works for n+1, then it is done.
$\sum_{j=0}^{n+1} (-1)^j {n+1 \choose j} F_{s+2n-2j+2} = F_{s+n+1} $
Also I could replace s with s+1 since it is valid for general constant s.
So I could get $\sum_{j=0}^n (-1)^j {n \choose j} F_{s+1+2n-2j} = F_{s+1+n} $
|
Here is a proof by induction over $n$. To recap, we want to show
$$F_{s+n} = \sum_{j=0}^n (-1)^j \binom{n}{j} F_{s+2n-2j} \tag{1}$$
The base case, $n=0$, is trivial. So suppose (1) holds for some value of $n$. We want show it holds for $n+1$. Since $s$ is arbitrary, we may substitute $s+1$ for $s$ in (1):
$$F_{s+n+1} = \sum_{j=0}^n (-1)^j \binom{n}{j} F_{s+1+2n-2j}$$
Next, we apply the Fibonacci identity $F_{n} = F_{n+1}-F_{n-1}$:
$$F_{s+n+1} = \sum_{j=0}^n (-1)^j \binom{n}{j} (F_{s+2+2n-2j}-F_{s+2n-2j})$$
So
$$\begin{align}
F_{s+n+1} &= \sum_{j=0}^n (-1)^j \binom{n}{j} F_{s+2+2n-2j} - \sum_{j=0}^n (-1)^j \binom{n}{j} F_{s+2n-2j} \\
&= F_{s+2+2n} + \sum_{j=1}^n (-1)^j \binom{n}{j} F_{s+2+2n-2j} - \sum_{j=0}^{n-1} (-1)^j \binom{n}{j} F_{s+2n-2j} - (-1)^n F_s \\
&= F_{s+2+2n} + \sum_{j=1}^n (-1)^j \binom{n}{j} F_{s+2+2n-2j} - \sum_{j=1}^{n} (-1)^{j-1} \binom{n}{j-1} F_{s+2+2n-2j} - (-1)^n F_s \tag{2}\\
&= F_{s+2+2n} + \sum_{j=1}^n (-1)^j \left[ \binom{n}{j} + \binom{n}{j-1} \right] F_{s+2+2n-2j} +(-1)^{n+1}F_s \\
&= F_{s+2+2n} + \sum_{j=1}^n (-1)^j \binom{n+1}{j} F_{s+2+2n-2j} +(-1)^{n+1}F_s \tag{3}\\
&= \sum_{j=0}^{n+1} (-1)^j \binom{n+1}{j} F_{s+2(n+1)-2j}
\end{align}$$
This completes the proof by induction.
Notes
(2) Shifting the index in the second sum
(3) Applying the recursive formula for binomial coefficients
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2868713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
}
|
Find the values of $x$ and $y$ that satisfies $\sin(x+y)=\sin x+\sin y$. I know that in general the following equality does not hold: $\sin(x+y)=\sin x + \sin y$. However, I have been looking for specific values of $x$ and $y$ that satisfies the given equation. This is what I have done so far:
$\sin(x+y)=\sin x\cos y + \sin y \cos x = \sin x + \sin y$. Then from this equation, I got
$\sin x(1-\cos y)+\sin y(1-\cos x)=0$.
There are two possibilities:
Case 1: $\sin x(1-\cos y)=0$ and $\sin y(1-\cos x)=0$. Solving equations for $x$ and $y$, we get that $x=y=2\pi k$ for some integer $k$.
Case 2: $\sin x(1-\cos y)=n$ and $\sin y(1-\cos x)=-n$ where $n$ is a nonzero real number. Then we have
$\sin x(1-\cos y)=\sin y(\cos x-1)$ Since $n\neq0$ then we can divide both sides of this equation by $\sin x \sin y)$ to obtain $\csc y - \cot y = \cot x - \csc x$. I'm stuck in here. Any suggestions and hints (not answers) will be welcomed...
|
use that $$\sin(x)+\sin(y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)$$ and $$\sin(x+y)=2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x+y}{2}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2871372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
}
|
compute $\frac{1}{2}(\frac{1}{3.5........(2n-1)} -\frac{1}{3.5........(2n+1)})$ compute the summation
$\sum_ {n=1}^{\infty} \frac{n}{3.5........(2n+1)}= ?$
My attempts : i take $a_n =\frac{n}{3.5........(2n+1)}$
Now =$\frac{1}{2}$ .$\frac{(2n+ 1) - 1}{3.5........(2n+1)}= \frac{1}{2}(\frac{1}{3.5........(2n-1)} -\frac{1}{3.5........(2n+1)})$
after that i can not able to proceed further,,,
pliz help me..
thanks u
|
Your work so far is good. Just keep going.
$$\sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{1}{2} \left(\frac{1}{1 \cdot 3 \cdot 5 \cdots (2n-1)} - \frac{1}{1 \cdot 3 \cdot 5 \cdots (2n+1)}\right)
= \frac{1}{2} \left(\frac{1}{1} - \frac{1}{1 \cdot 3} + \frac{1}{1\cdot 3} - \frac{1}{1 \cdot 3 \cdot 5} + \frac{1}{1 \cdot 3 \cdot 5} - \frac{1}{1 \cdot 3 \cdot 5 \cdot 7} + \cdots\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2871600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
If the roots of the equation $x^2 + 3x -1 = 0$ also satisfy the equation $x^4 + ax^2 + bx + c = 0$ find the value of $a + b + 4c + 100$
I tried really hard but the most I could get is the sum of the roots of the second equation is $3$
Please could someone solve this please !
It would mean the world to me
|
We have:
$$(x^2+3x-1)(x^2+px+q)=x^4+ax^2+bx+c$$
We can expand the left side:
$$x^4+x^3(3+p)+x^2(q+3p-1)+x(3q-p)-q=x^4+ax^2+bx+c$$
Because there is no $x^3$ element on the right side, we conclude
$$3+p=0$$
$$p=-3$$
We have then:
$$x^4+x^2(q-10)+x(3q+3)-q=x^4+ax^2+bx+c$$
thus:
$$\begin{cases}a=q-10 \\ b=3q+3 \\ c=-q\end{cases}$$
Subtracting these into $a+b+4c+100$ we obtain:
$$a+b+4c+100=q-10+3q+3-4q+100 = -7+100=93$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2873902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
}
|
Find the maximum of the value$(n^{n-1}-1)\sqrt[n]{a_{1}a_{2}\cdots a_{n}}+\sqrt[n]{\frac{a^n_{1}+a^n_{2}+\cdots+a^n_{n}}{n}}$
Let $n$ be give a positive integer, $a_{i}
\ge 0$,such that $a_{1}+a_{2}+\cdots+a_{n}=n$. Find the maximum value of
$$(n^{n-1}-1)\sqrt[n]{a_{1}a_{2}\cdots a_{n}}+\sqrt[n]{\dfrac{a^n_{1}+a^n_{2}+\cdots+a^n_{n}}{n}}$$
Using AM-GM we have
$$\sqrt[n]{a_{1}a_{2}\cdots a_{n}}\le\dfrac{a_{1}+a_{2}+\cdots+a_{n}}{n}=1$$
But other hand we have
$$a^n_{1}+a^n_{2}+\cdots+a^n_{n}\ge\dfrac{1}{n^{n-1}}(a_{1}+a_{2}+\cdots+a_{n})^n=\dfrac{1}{n^{n-1}}$$
|
dezdichado's answer is correct, I'll just write out all the computations.
Clearly the maximum value is $n^{n-1}$.
Let $X = \frac{1}{n}\sum a_i^n$ and $Y=\prod a_i$. We would like to show that $\left(n^{n-1}-1\right)\sqrt[n]{Y}+\sqrt[n]{X}\le n^{n-1}$
Expanding and using AM-GM: $$n^n=\left(\sum a_i\right)^n=\sum a_i^n+\sum_{|\alpha|=n} \left(\begin{matrix}n\\ \alpha\end{matrix}\right)a^\alpha\ge nX+\left(n^n-n\right)Y$$ So $$n^{n-1}\ge X+\left(n^{n-1}-1\right)Y $$
Now applying Hölder's inequality gives us: $$\left[X+\left(n^{n-1}-1\right)Y\right]\left(1+n^{n-1}-1\right)^{n-1}\ge \left(\sqrt[n]{X}+\left(n^{n-1}-1\right)\sqrt[n]{Y}\right)^n$$
Simplifying that results in $$n^{n-1}n^{(n-1)^2}=\left(n^{n-1}\right)^n\ge \left(\sqrt[n]{X}+\left(n^{n-1}-1\right)\sqrt[n]{Y}\right)^n$$
Which completes the proof. So the maximum value is $n^{n-1}$ when all $a_i$ are equal to $1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2874956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Showing that $\int_{0}^{\infty} \frac{\cos(x)}{x^{2}+3}dx = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}$ via Contour Integration? In the text "Function Theory of One Complex Variable" Third Edition, I'm inquiring if my proof of $\text{Proposition (1)}$ is sound ?
$\text{Proposition (1)}$
$$\int_{0}^{\infty} \frac{\cos(x)}{x^{2}+3}dx = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}$$
$\text{Proof}$
Assume that $R>1$ define $\gamma_{R}$ such that,
$$\gamma_{R}^{1}(t) = t + i0 \, \, \text{if} \, \, -R \leq t \leq R$$
$$\gamma_{R}^{2}(t) = Re^{it} \, \text{if} \, \, \, \, \, \, 0\leq t \leq \pi.$$
$\, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, \, $
Consider our choice $f$ and that,
$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3} \, dz.$$
Clearly it's obvious that
$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3}dz = \sum_{\psi = 1,2} \oint_{\gamma_{R}^{\psi}}\frac{e^{iz}}{z^{2}+3}dz.$$
It's trivial that,
$$\oint_{\gamma_{R}^{1}}e^{iz}/({z^{2}+3})dx \rightarrow \int_{0}^{\infty} \frac{e^{ix}}{x^{2}+3}\operatorname{dx}$$
It's natural to claim that,
$$\Bigg| \lim_{R \rightarrow \infty}\oint_{ \gamma_{R}^{2}} \frac{e^{iz}}{z^{2}+3} dz \Bigg| \rightarrow 0. $$
Using the Estimation Lemma one can be relived that,
$$\bigg |\oint_{\gamma_{R}^{2}} \frac{e^{iz}}{z^{2}+3} dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|\frac{e^{iz}}{z^{2}+3}|\leq \pi R \cdot \frac{1}{R^{2} - 3} \rightarrow 0 \, \text{as} \, R \rightarrow \infty$$
Thus,
$$ \operatorname{Re}\int_{0}^{\infty} \frac{\cos(x)}{x^{2}+3}dx = \operatorname{Re} \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}} = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}.$$
However we need to consider that
$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3} \, dz = 2 \pi i \sum_{j=1,2} \operatorname{Ind_{\gamma}} \cdot \operatorname{Res_{f}(P_{j})}$$
It's easy to observe that,
$$\oint_{\gamma_{R}} \frac{e^{iz}}{z^{2}+3} \, dz = \big(2\pi i \cdot \frac{ie^{\sqrt{3}}}{2 \sqrt{3}}) = \frac{e^{-\sqrt{3}} \pi}{2 \sqrt{3}}$$
|
Thanks to @Batominovski's comments and insights I was able to note that there was errors in the original proof and a new one can be found below.
$\text{Proof}$
To proceed one must consider that,
$$ \int_0^\infty\,\frac{\cos(x)}{x^2+3}\,\text{d}x=\frac{1}{\sqrt{3}}\,\int_0^\infty\,\frac{\cos(\sqrt{3}t)}{t^2+1}\,\text{d}t=\frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}$$
$t:=\sqrt{3}x$
Assume that $R>1$ define $\gamma_{R}$ such that,
$$\gamma_{R}^{1}(t) = t + i0 \, \, \text{if} \, \, -R \leq t \leq R$$
$$\gamma_{R}^{2}(t) = Re^{it} \, \text{if} \, \, \, \, \, \, 0\leq t \leq \pi.$$
Consider our choice $f$ and that,
$$\frac{1}{\sqrt{3}}\,\oint_{\gamma_{R}}\,\frac{e^{(\sqrt{3}t)}}{z^2+1}\,\text{d}z $$
Clearly it's simple that
$$\frac{1}{\sqrt{3}} \oint_{\gamma_{R}}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz = \sum_{\psi = 1,2} \frac{1}{\sqrt{3}} \oint_{\gamma_{R}^{\psi}}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz.$$
Now it's imperative to claim that
$$\lim_{R\to\infty}\,\oint_{\gamma_R^1}\,\frac{\exp(\sqrt{3}z)}{z^2+3}\,\text{d}z= \frac{1}{\sqrt{3}} \,\int_{0}^{\infty}\, \, \frac{e^{(\sqrt{3}t)}}{z^2+1}\,\text{d}z = \frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}.$$
Now at this leg of our journey important to conjecture that,
$$\Bigg| \lim_{R \rightarrow \infty}\oint_{ \gamma_{R}^{2}}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz \Bigg| \rightarrow 0. $$
Thanks to Estimation Lemma we can be relived that,
$$\bigg |\oint_{\gamma_{R}^{2}} \frac{e^{(\sqrt{3}z)}}{z^2+3}dz \bigg | \leq \big\{\text{length}(\gamma_{R}^{2}) \big\} \cdot \sup_{\gamma_{R}^{2}}|\frac{e^{(\sqrt{3}z)}}{z^2+3}|\leq \pi R \cdot \frac{1}{R^{2} - 3} \rightarrow 0 \, \text{as} \, R \rightarrow \infty.$$
Thus, taking
$$\operatorname{Re} \bigg( \frac{1}{\sqrt{3}}\,\int_0^\infty\,\frac{\cos(\sqrt{3}t)}{t^2+1}\text{d}t \bigg) \, = \frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}$$
However we need to consider that
$$\frac{1}{\sqrt{3}} \oint_{\gamma_{R}} \frac{e^{(\sqrt{3}z)}}{z^2+1}\, dz = 2 \pi i \sum_{j=1,2} \operatorname{Ind_{\gamma}} \cdot \operatorname{Res_{f}(P_{j})} $$
After a trivial calculation we have that,
$$\frac{1}{\sqrt{3}} \oint_{\gamma_{R}} \frac{e^{(\sqrt{3}z)}}{z^2+1}\, dz = 2 \pi i + \operatorname{Res}\frac{e^{(\sqrt{3}z)}}{z^2+1}dz = \frac{\pi}{2\sqrt{3}}\,e^{(-\sqrt{3})}.$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2875240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Evaluate: $u =\int_0^\infty\frac{dx}{x^4 +7x^2+1}$ Evaluate:
$u =\displaystyle\int_0^\infty\dfrac{dx}{x^4 +7x^2+1}$
Attempt:
$$u = \int_0^\infty \dfrac{dx}{\left(x^2+ \left(\dfrac{7 - \sqrt {45}}{2}\right)\right)\left(x^2+ \left(\dfrac{7 + \sqrt {45}}{2}\right)\right)}$$
$$u = \int_0^\infty \dfrac{dx}{(x^2+a^2)(x^2+b^2)}$$
After partial fraction decomposition and simplifying I get:
$u = \dfrac{\pi}{2(a+b)}$
But answer is $\frac \pi 6$.
Where have I gone wrong?
|
$\newcommand{\I}{\mathfrak{I}}$$\newcommand{\dx}{\mathrm dx\,}$Let’s take the time to generalize this. Denote the generalized integral as$$\I=\int\limits_0^{\infty}\dx\frac 1{x^4+ax^2+b^2}$$Now factor out an $x^2$ from the denominator and complete the square to get
$$\I=\int\limits_0^{\infty}\frac {\dx}{x^2}\frac 1{\left(x-\frac bx\right)^2+a+2b}$$Make the transformation $x\mapsto\tfrac bx$ so that$$\I=\frac 1b\int\limits_0^{\infty}\dx\frac 1{\left(x-\frac bx\right)^2+a+2b}$$Therefore, it’s easy to see that
$$b\I=\int\limits_0^{\infty}\frac {\dx}{x^2}\frac b{\left(x-\frac bx\right)^2+a+2b}=\int\limits_0^{\infty}\dx\frac 1{\left(x-\frac bx\right)^2+a+2b}$$
Add the two integrals together to get
$$\begin{align*}\I & =\frac 1{2b}\int\limits_0^{\infty}\dx\left(1+\frac b{x^2}\right)\frac 1{\left(x-\frac bx\right)^2+a+2b}\\ & =\frac 1{2b}\int\limits_{-\infty}^{\infty}\frac {\dx}{x^2+a+2b}\\ & \color{blue}{=\frac {\pi}{2b\sqrt{a+2b}}}\end{align*}$$
Now set $a=7$ and $b=1$ to get the answer.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2879653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
}
|
Evaluate $\lim_{x \to \infty} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]$ $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]=?$
My Try :$[(x+2)\tan^{-1}(x+2) -x \tan^{-1}x] = x \tan^{-1} \frac {2}{1+2x+x^2} + 2. \tan^{-1}(x+2)$
$\underset{x \to \infty}{\lim} x \tan^{-1} \frac {2}{1+2x+x^2} =0$ [By manipulating L'hospital] and $\underset{x \to \infty}{\lim}2. \tan^{-1}(x+2) = \pi$
so $\underset{x \to \infty}{\lim} [(x+2)\tan^{-1}(x+2) -x \tan^{-1}x]= \pi$
Can anyone please correct me If I have gone wrong anywhere?
|
Your solution is fine.
Another possible way to do it would be using that, for $t>0$,
$$\tan^{-1}(t)+\tan ^{-1}\left(\frac{1}{t}\right)=\frac \pi 2$$ making
$$A=(x+2)\tan^{-1}(x+2) -x \tan^{-1}(x)$$ $$A=(x+2)\left(\frac \pi 2-\tan ^{-1}\left(\frac{1}{x+2}\right) \right)-x\left(\frac \pi 2-\tan ^{-1}\left(\frac{1}{x}\right) \right)$$
$$A=\pi+x\tan ^{-1}\left(\frac{1}{x}\right)-(x+2)\tan ^{-1}\left(\frac{1}{x+2}\right)$$ and use, for large $t$, the equivalent
$$t\tan ^{-1}\left(\frac{1}{t}\right)\sim 1$$
If you want to go beyond, you could use Taylor expansions and get
$$A=\pi -\frac{4}{3 x^3}+\frac{4}{ x^4}+O\left(\frac{1}{x^5}\right)$$
Using your pocket calculator, for $x=10$,
$$A=12 \tan ^{-1}(12)-10 \tan ^{-1}(10)\approx 3.14058$$ while the truncated expression gives
$$\pi -\frac{7}{7500}\approx 3.14066$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2880972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
}
|
Intersection of 3 planes along a line I have three planes:
\begin{align*}
\pi_1: x+y+z&=2\\
\pi_2: x+ay+2z&=3\\
\pi_3: x+a^2y+4z&=3+a
\end{align*}
I want to determine a such that the three planes intersect along a line. I do this by setting up the system of equations:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
x+ay+2z&=3\\
x+a^2y+4z&=3+a
\end{align*}
\end{cases}
$$
and solve for x, y and z. I subtract the first row from the second and third
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
(a-1)y+z&=1\\
(a^2-1)y+3z&=1+a.
\end{align*}
\end{cases}
$$
I subtract $(a+1)*\textrm{row}_1$ from the third row:
$$
\begin{cases}
\begin{align*}
x+y+z&=2\\
(a-1)y+z&=1\\
(2-a)z&=1+a.
\end{align*}
\end{cases}
$$
I think I am supposed to find an $a$ such that $0=0$ in the third row, but obviously there isn't any such $a$. Do I have the right idea about how to solve this? If so, where is my mistake?
|
You had the right idea.
Of yourse, in this case, there was only one parameter, and even that appeared in the form of simple expressions ($a$ and $a^2$). With more complicated problems, I suggest you start in a different way.
Compute the determinant of the matrix of coeeficients on the LHS:
$$\det\begin{pmatrix} 1 & 1 & 1 \\ 1 & a & 2 \\ 1 & a^2 & 4 \\ \end{pmatrix} = 4a+2+a^2-2a^2-4-a= -a^2+3a-2$$
This is zero iff $a=2$ or $a=1$. These are the only values to check: if the determinant is nonzero, then the system has a unique solution, so the planes intersect in a single point.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2882503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
}
|
Express $z = \dfrac{3i}{\sqrt{2-i}} +1$ in the form $a + bi$, where $a, b \in\Bbb R$.
Express $$z = \frac{3i}{\sqrt{2-i}} +1$$ in the form $a + bi$, where $a, b \in\Bbb R$.
I figure for this one I multiply by the conjugate of $\sqrt{2 +1}$? But I’m still struggle to achieve the form $a+bi$.
|
$\displaystyle z=\frac{3i}{\sqrt{2}-i+1}$
$\displaystyle =\frac{3i}{\sqrt{2}-i+1}\times\frac{\sqrt{2}+i+1}{\sqrt{2}+i+1}$
$\displaystyle =\frac{-3+(3+\sqrt{2})i}{(\sqrt{2}+1)^2+1}$
$\displaystyle =\frac{-3+(3+\sqrt{2})i}{4+2\sqrt{2}}$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2883855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
}
|
Find all value of integers n for which the expression is a perfect square.
Find all integers $n$ for which the expression $$n^4+6n^3+11n^2+3n+31$$ is a perfect square.
My approach :
First I simplified the expression and equated it to some square number $$(n^2+3n+1)^2 - 3(n-10) = k^2$$ Now if we try to remove the term "$3(n-10)$" then the equation will be satisfied. And it is done only be equating $n = 10$ . I don't know how to find other solution. Please help me in solving further.
|
Strategy: try to find two perfect squares such that the expression $f(n)=n^4+6n^3+11n^2+3n+31$ is strictly between them for all but finitely many $n$.
$(n^2+3n+1)^2$ is important to compute, as you also pointed out.
It is $n^4+6n^3+11n^2+6n+1$.
So our impression is that $f(n)$ is close to this, and that if $f(n)$ is a perfect square, it should be the square of $n^2+3n+1$.
We prove this for all but finitely many $n$. More precisely, we prove that for all but finitely many $n$, we have $(n^2+3n)^2< f(n)< (n^2+3n+2)^2$.
First part: $(n^2+3n)^2= n^4+6n^3+9n^2< n^4+6n^3+11n^2+3n+31 \Leftrightarrow 0< 2n^2+3n+31$. This holds for all $n\in \mathbb{Z}$ (the discriminant is negative).
Second part: $n^4+6n^3+11n^2+3n+31< (n^2+3n+2)^2 = n^4 + 6n^3 +13n^2 + 12n + 4 \Leftrightarrow$ $0< 2n^2+9n-27$.
This inequality fails iff $-6\leq n\leq 2$. So unfortunately, you have to check if $f(n)$ is a perfect square for these nine values. (Easy calculation.)
Once you are done with that, if $n$ is not among these nine values, then $f(n)$ is strictly between $(n^2+3n)^2$ and $(n^2+3n+2)^2$, so it has to be $(n^2+3n+1)^2$.
This leads to a unary equation that you have already solved.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2884100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
}
|
Taking square roots modulo $2^N$ I was trying to solve $y^2 - y \equiv 16 \pmod{512}$ by completing the square.
Here is my solution.
\begin{align}
y^2 - y &\equiv 16 \pmod{512} \\
4y^2 - 4y + 1 &\equiv 65 \pmod{512} \\
(2y-1)^2 &\equiv 65 \pmod{512} \\
2y - 1 &\equiv \pm 33 \pmod{512} &\text{Found by pointwise search.}\\
2y &\equiv 34, -32 \pmod{512} \\
y &\equiv 17, -16 \pmod{256} \\
y &\in \{17, 273, 240, 496\} \pmod{512} &\text{These values need to be verified.}\\
y &\in \{240, 273\}\pmod{512}
\end{align}
I had to solve $x^2 \equiv 65 \pmod{2^9}$ by a pointwise search. Is there any systematic method for solving equivalences of the form $x^2 \equiv a \pmod{2^N}$, or more generally $x^2 \equiv a \pmod{p^N}$ for a prime number $p$.
|
Can you solve it in $p$-adics?
For any $2$-adic integer $n$ we have the familiar Taylor/binomial-power series for $\sqrt{1+8n}$:
$\sqrt{1+8n}=1+(1/2)(8n)-...$
All omitted terms vanish $\bmod 512$ when $n$ is a multiple of $8$, thus with $n=8$
$\sqrt{65}\equiv 1+(1/2)(8×8) \bmod 512$
$\sqrt{65}\equiv 1+32\equiv 33 \bmod 512$
So one square root (the root closer to $1$ in $2$-adics) is $33 \bmod 512$.
For reference here are more terms in the binomial expansion used above:
$\sqrt{1+8n}=1+(2^2)n-(2^3)n^2+(2^5)n^3-(5×2^5)n^4+(7×2^7)n^5-(21×2^8)n^6+(33×2^{10})n^7-(429×2^9)n^8+...$
This is sufficiently many terms to obtain at least eleven bits in the square root ($\bmod 2048$) for any integer $n$ (not just multiples of $8$ as above); the exponent on $2$ is at least $11$ in all subsequent terms. The exponent on $2$ is always at least the number of terms rendered (e.g. $2^9$ in the ninth term), thus guaranteeing convergence in $2$-adics. Note also that for a conventional integer square the obtained square root is one greater than a multiple of $4$, which in conventional algebra could be the negative root (e.g. $\sqrt{9}=-3$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2885249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Is there a smarter way to differentiate the function $f(x) = \sin^{-1} \frac{2x}{1+x^2}$?
Given $f(x) = \sin^{-1} \frac{2x}{1+x^2}$,
Prove that $$f'(x) = \begin{cases}\phantom{-}\frac{2}{1+x^2},\,|x|<1 \\\\ -\frac{2}{1+x^2},\,|x|>1 \end{cases}$$
Obviously the standard approach would be to use the chain rule and simplify from there.
But I noticed that some of these expressions are familiar, specifically, from the tangent half-angle formulae:
If $x = \tan \frac \theta 2$, then $\sin \theta = \frac{2x}{1+x^2}$ and $\frac{d\theta}{dx} = \frac{2}{1+x^2}$.
So my question is: can this observation be used to construct a more elegant proof?
|
Implicit differentiation:
$$\sin(f(x)) = \frac{2x}{1 + x^2}$$
$$\cos(f(x)) f'(x) = \frac{2(1 - x^2)}{(1 + x^2)^2}$$
$$
f'(x) = \frac{2(1 - x^2)}{\cos(f(x))(1 + x^2)^2} =
\frac{2(1 - x^2)}{\sqrt{1 - \sin^2(f(x))}(1 + x^2)^2} =
\frac{2(1 - x^2)}{\sqrt{(1 - x^2)^2/(1 + x^2)^2}(1 + x^2)^2}\cdots
$$
and:
*
*$1 - x^2 < 0\hbox{ for }|x| > 1$;
*$1 - x^2 > 0\hbox{ for }|x| < 1$ .
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2886631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
Complex number question - spurious solutions...
$z_1 = 2 + 3i$ and $z_2 = 3 - 4i$
The complex number $z = x + iy$ is such that $\frac{z + z_1}{2z - z_2} = 1$.
Find the value of $x$ and the value of $y$.
Method 1:
$$z + z_1 = 2z - z_2 $$$$\Rightarrow x + iy + 2 + 3i = 2x + 2iy - 3 + 4i$$
Equating real and imaginary components, I obtain $x = 5$ and $y = -1$.
This is a single unique solution as expected.
Method 2:
$$\frac{x + iy + 2 + 3i}{2x + 2iy - 3 + 4i} = 1$$
$$\Rightarrow \frac{(x+2) + i(y+3)}{(2x-3) + i(2y + 4)} = 1$$ Multiplying top and bottom by the conjugate of the denominator gives: $$\frac{(x+2)(2x-3) + (y+3)(2y+4)}{(2x-3)^2 + (2y+4)^2} + i \frac{(y+3)(2x-3) - (x+2)(2y+4)}{(2x-3)^2 + (2y+4)^2} = 1$$
Straightaway, I can see that this will have two solutions for $x$ and $y$ when you equate components as one of the equations will be quadratic.
Working through the algebra gives $x= 5$ and $y=-1$ as above but also $x = \frac{3}{2}$ and $y = -2$. I have noticed that this additional solution gives a denominator of $0$ when substituted back into the problem, so I suppose the issue lies there somehow...
Question: I would like to be enlightened about the origins of this additional solution and why it appears in this 2nd method and not the 1st.
|
We have that
$$\frac{a}{b}=1\implies \frac{ab}{b^2}=1\implies ab=b^2\implies b=a\:\text{or}\: b=0.$$ But $b=0$ is not a solution of $\frac{a}{b}=1.$
For the same reason
$$\dfrac{1}{z}=2\implies \dfrac{\bar{z}}{z\bar{z}}=2\implies \bar{z}=2|z|^2.$$ $z=0$ is a solution of $\bar{z}=2|z|^2$ but not of $\dfrac{1}{z}=2.$
Conclusion: We have to be careful that we don't multiply and divide by a quantity that can be zero.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2887117",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
}
|
Find the maximum value of $a+b$ The question:
Find the maximum possible value of $a+b$ if $a$ and $b$ are different non-negative real numbers that fulfill
$$a+\sqrt{b} = b + \sqrt{a}$$
Without loss of generality let us assume that $a\gt b$. I rearrange the equation to get $$a - \sqrt{a} = b - \sqrt{b}$$
If $f(x)= x - \sqrt{x},$ then we are trying to solve $f(a)=f(b).$ Using some simple calculus I found that the turning point of $f(x)$ is $(\frac{1}{4}, -\frac{1}{4})$. Hence $0 \le b \le \frac{1}{4}$ and $\frac{1}{4} \le a \le 1$. From here, I have no idea how to proceed.
I used trial and error to find that when $a$ increases, the value of $a+b$ increases as well. Hence I hypothesise that $a+b$ is at a maximum when $a=1$ and $b=0$, which implies that $a+b=1$ is a maximum. Can anyone confirm this?
|
WLOG, let $b=ax,x\ge 0$. Then:
$$a+\sqrt{b} = b + \sqrt{a} \iff a+\sqrt{ax}=ax+\sqrt{a} \iff \\
\sqrt{a}(\sqrt{x}-1)=a(x-1) \stackrel{x\ne 1}{\iff} 1=\sqrt{a}(\sqrt{x}+1) \iff a=\frac{1}{(\sqrt{x}+1)^2}.$$
Hence:
$$a+b=a+ax=\frac{1}{(\sqrt{x}+1)^2}+\frac{x}{(\sqrt{x}+1)^2}=\frac{x+1}{(\sqrt{x}+1)^2}\le 1, x\ge 0.$$
Equality occurs for $x=0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2889732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 4
}
|
How can I show that these two polynomials are coprime? Let $n , m \in \mathbb{N}$ and let $k = \gcd(n , m)$. Then there exist $p , q \in \mathbb{N}$ coprime such that $n = p k$ and $m = q k$. And it is easy to see that
$$
\frac{X^n - 1}{X^k - 1} = \sum_{i = 0}^{p - 1} X^{k i} = f(X) \qquad \mbox{ and } \qquad \frac{X^m - 1}{X^k - 1} = \sum_{j = 0}^{q - 1} X^{k j} = g(X)\mbox{,}
$$
being $X$ an indeterminate. How can I show that $\gcd(f(X) , g(X)) = 1$?
|
The following is a step-by-step proof for $k=1$. The general case is proved in my other answer to the related question, and the details can be filled-in very similarly there.
It is given that $\,X^{p} = \left(X-1\right)f(X)+1\,$ and $X^{q} = \left(X-1\right)g(X)+1\,$.
Since $\,p,q\,$ are coprime, there exist integers $\,a,b\,$ such that $\,ap-bq=1\,$, and it can be assumed WLOG that they are positive i.e. $\,a,b \in \Bbb N\,$ (see here for example). Then $\,X^{ap}=X^{bq+1} = X \cdot X^{bq}\,$, and therefore:
$$
\begin{align}
0 &= \left(\left(X-1\right)f(X)+1\right)^{a} - X \cdot \left( \left(X-1\right)g(X)+1\right)^b \\[5px]
&= \Big(\underbrace{(X-1)^af^a(X)+ \binom{a}{1}(X-1)^{a-1}f^{a-1}(X)+\ldots+\binom{a}{a-1}(X-1)f(X)}_{u(X) \cdot f(X)}+1\Big) \\
&\quad - X \cdot \Big(\underbrace{(X-1)^bg^b(X)+ \binom{b}{1}(X-1)^{b-1}g^{b-1}(X)+\ldots+\binom{b}{b-1}(X-1)g(X)}_{v(X) \cdot g(X)}+1\Big) \\[5px]
&= u(X) \cdot f(X) - X \cdot v(X)\cdot g(X) + 1 - X
\end{align}
$$
Therefore $\,u(X) \cdot f(X) - X \cdot v(X)\cdot g(X)=X-1\,$.
Let $\,h(X) = \gcd\left(f(X),g(X)\right)\,$, then $\,h(X)\,$ divides the LHS, so $\,h(X) \mid X-1\,$. But:
$$
\begin{align}
f(X) &= 1 + X + X^2 + \ldots+X^{p-1} \\
&= p + (X-1)+(X^2-1)+\ldots+(X^{p-1}-1) \\
&= p+\underbrace{(X-1)+(X-1)(X+1)+\ldots+(X-1)(X^{p-2}+X^{p-3}+\ldots+1)}_{(X-1) \cdot w(X)} \\
&= p + (X-1) \cdot w(X)
\end{align}
$$
Since $\,h(X) \mid f(X)\,$ and $\,h(X) \mid X-1\,$ it follows that $\,h(X) \mid p = f(X)-(X-1)\cdot w(X)\,$, so $\,h(X)\,$ is a constant polynomial, which completes the proof that $\,\gcd(f(X),g(X))=1\,$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2890374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
}
|
How can I factorize this: "$X^3 + X^2 + X - 3$" I am going to elementary school & I am living in one of those deprived areas of Africa.
I can solve mathematical questions like this:
$$X^3 + X^2 + X +1 = X^2(X+1)+(X+1) = (X+1)(X^2+1)$$
Or even
\begin{align}X^2 − 2X + X^2 - X + 1 &= (X^2 - 2X + 1) + (X^2 - X) \\
&= (X - 1)^2 + X(X - 1) \\
&= (X-1)(X-1+X) \\
&= (X - 1)(2X - 1)
\end{align}
But for a few months I have not been able to find a teacher around here who can factorize this:
$$X^3 + X^2 + X - 3$$
Do we have to solve it in this way?
$$X^3 + X^2 + X - 3 = X^2(X + 1) + X - 3$$
Or something else?
I'd appreciate your help with this.
|
Hint: $(x^3+x^2+x-3):(x-1)=x^2+2x+3$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2890772",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Solve the system $x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$
Find all real numbers $x,\ y,\ z$ that satisfy $$x+y=\sqrt{4z-1},\ x+z=\sqrt{4y-1},\ z+y=\sqrt{4x-1}$$
First natural move would be rewriting the system as:
$$x^{2}+y^{2}+2xy=4z-1$$
$$x^{2}+z^{2}+2xz=4y-1$$
$$z^{2}+y^{2}+2zy=4x-1$$
Thus,
$$2x^{2}+2y^{2}+2z^{2}+2xy+2yz+2zx-4x-4y-4z+3=0$$
And I tried to write the polynomial above as the sum of complete squares. Is this possible or do you have other ideas?
|
Necessarily $x,y,z\ge{\large{\frac{1}{4}}}$.
Without loss of generality, assume $x=\min(x,y,z)$.
Then $2x \le y+z$, hence
\begin{align*}
&(y+z)^2=4x-1\\[4pt]
\implies\;&(2x)^2\le 4x-1\\[4pt]
\implies\;&4x^2-4x+1\le 0\\[4pt]
\implies\;&(2x-1)^2\le 0\\[4pt]
\implies\;&(2x-1)^2=0\\[4pt]
\implies\;&x={\small{\frac{1}{2}}}\\[4pt]
\end{align*}
Then $y,z\ge {\large{\frac{1}{2}}}$, and $4x-1=1$, hence
\begin{align*}
&(y+z)^2=4x-1\\[4pt]
\implies\;&(y+z)^2=1\\[4pt]
\implies\;&y+z=1\\[4pt]
\implies\;&y=z={\small{\frac{1}{2}}}\\[4pt]
\end{align*}
Therefore $x=y=z={\large{\frac{1}{2}}}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2894140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
}
|
How does $\frac{x + \frac{1}{2}}{\frac{1}{x} + 2} = \frac{x}2$? How does simplifying $\dfrac{x+\frac12}{\frac1x+2}=\dfrac{x}2$
Plugging and chugging seems to prove this, but I don’t understand the algebra behind it.
How would you simplify $\dfrac{x+\frac12}{\frac1x+2}$ to get $\dfrac{x}2$?
I tried multiplying the expression by $x/x$, but that left me with $\frac{x^2+\frac{x}2}{1+2x}$
Is the fact that $\frac1x$ and $2$ are the reciprocals of $x$ and $\frac12$ respectively of any significance?
Thanks!
|
I'd start by multiplying by $2/2$, to make the expression easier to read.
$$\frac{x + 1/2}{1/x + 2} = \frac{2x + 1}{2/x + 4} = \frac{2x^2 + x}{4x + 2}$$
Factoring the top and bottom:
$$\frac{2x^2 + x}{4x + 2} = \frac{x(2x + 1)}{2(2x + 1)}$$
And cancel:
$$\frac{x(2x + 1)}{2(2x + 1)} = \frac{x}{2}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2894360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
}
|
Sum of roots of equation $x^4 - 2x^2 \sin^2(\displaystyle {\pi x}/2) +1 =0$ is My try:
$$x^4-2x^2\sin^2(\frac{\pi x}{2})+1=0\\x^4+1=2x^2\left (1-\cos^2\left(\frac{\pi x}{2}\right)\right )\\(x^2-1)^2=-2x^2\cos^2\left(\frac{\pi x}{2}\right)\\(x^2-1)^2+2x^2\cos^2\left(\frac{\pi x}{2}\right)=0\\x^2-1=0\,\text{and}\, 2x^2\cos^2\left(\frac{\pi x}{2}\right)=0$$
I am stuck , I am confused now what to do now
|
Notice that if $x_0$ is a root, then so is $-x_0$. Can you see what the sum is now?
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2895070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
}
|
Prove $\ \frac{z-1}{z+1} $ is imaginary no' iff $\ |z| = 1 $ Let $\ z \not = -1$ be a complex number. Prove $\ \frac{z-1}{z+1} $ is imaginary number iff $\ |z| = 1 $
Assuming $\ |z| = 1 \Rightarrow \sqrt{a^2+b^2} = 1 \Rightarrow a^2+b^2 = 1 $ and so
$$\ \frac{z-1}{z+1} = \frac{a+bi-1}{a+bi+1} = \frac{a-1+bi}{a+1+bi} \cdot \frac{a+1-bi}{a+1-bi} = \frac{(a^2+b^2)-1+2bi}{(a^2+b^2)+1 +2a} = \frac{bi}{1+a} $$
and therefore $\ \frac{z-1}{z+1}$ is imaginary
now let me assume $\ \frac{z-1}{z+1} $ is imaginary number, how could I conclude that $\ |z| =1 $ I really can't think of any direction..
Thanks
|
Assume $z\ne\pm1$. Then ${z-1\over z+1}\in i{\mathbb R}$ means that, viewed from $z$, the two points $\pm1$ are seen under a right angle. By Thales' theorem this is the case iff $z$ is lying on a circle with diameter $[{-1},1]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2898314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
}
|
Converting complex numbers into Cartesian Form 3 When calculating the real and imaginary parts of the complex number, do we take the angle as shown or the magnitude of it? I thought that we would just take the angle as shown, but apparently not, according to my textbook (unless its a typo):
$$z=2\sqrt{3}\operatorname{cis}\left(\frac{\pi }{3}\right)\:w=4\operatorname{cis}\left(\frac{-\pi }{6}\right)$$
Find $z + w$ in:
*
*Cartesian form
*Modulus-argument form
I worked out the complex number to be $3\sqrt{3}+i$:
$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:=\:2\sqrt{3}-2i\\
2\sqrt{3}\cos\left(\frac{\pi }{3}\right)+2\sqrt{3}i\:\sin\left(\frac{\pi }{3}\right)=\:\sqrt{3}\:+3i\\
\sqrt{3}\:+3i\:+\:2\sqrt{3}-2i\:=\:3\sqrt{3}+i$$
(Following from this I got the mod-arg form to be $[2\sqrt{7},0.19]$)
The textbook says pretty much the same thing, except
$$4\:\cos\left(-\frac{\pi }{6}\right)\:+\:4i\:\sin\left(-\frac{\pi }{6}\right)\:$$ is taken to be: $4\cos\left(\frac{\pi }{6}\right)+4i\sin\left(\frac{\pi }{6}\right)$, resulting in the complex number $=3\sqrt{3}+5i$.
|
We take the angle as shown, not just its absolute value. Therefore, $4\operatorname{cis}\left(\frac\pi6\right)\neq4\operatorname{cis}\left(-\frac\pi6\right)$. It looks as if you are right and your textbook is wrong.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2898424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
}
|
Unable to prove that $\sqrt{i} + \sqrt{-i}$ is a real number. I did this :-
$$
\sqrt{i} = \sqrt{\frac{1}{2}.2i} = \sqrt{\frac{1}{2}.(1 + 2i - 1)} = \sqrt{\frac{1}{2}.(1 + 2i + i^2)} = \sqrt{\frac{1}{2}.(1+i)^2} = \frac{1}{\sqrt{2}}(1+i) \\= \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ ---------
1} \\ \mbox{Also, }
\sqrt{-i} = \sqrt{-1.i} = \sqrt{-1}.\sqrt{i} = i\sqrt{i} = i(\frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}}) \mbox{ --------from 1} \\ = \frac{1}{\sqrt{2}}i - \frac{1}{\sqrt{2}} \\ = -\frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \mbox{ --------- 2} \\
\mbox{Adding 1 and 2, we get} \\ \sqrt{i} + \sqrt{-i} \\ = \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{2}} + i.\frac{1}{\sqrt{2}} \\= i.\frac{2}{\sqrt{2}} \\= i.\frac{\sqrt{2}.\sqrt{2}}{\sqrt{2}} \\= i\sqrt{2}
$$
This is a complex number, not a real number.
What am I doing wrong here? Is there another way to prove this ?
|
so to sum up. the question is incorrect. there are 4 solutions for this because root of any number has 2 solutions. in this question there are 2 real solutions and 2 solutions are on the imaginary axis
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2899061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Using $\epsilon$-$\delta$ approach, prove that $\lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3} $ How to prove that $$ \lim_{(x,y)\rightarrow (1,3)}\frac{x}{y}=\frac{1}{3}$$
By $\epsilon$-$\delta$ definition.
What i did is this:
Let $\epsilon$ be grater than zero; i need to find a $\delta(\epsilon)>0$ such that $|x/y-1/3|<\epsilon$ whenever $0<\sqrt{(x-1)^2+(y-3)^2}<\delta$.
Then, rewriting $$\bigg|\frac{x}{y}-\frac{1}{3}\bigg|=\bigg|\frac{3x-y}{3y}\bigg|= \bigg|\frac{3x-y+3-3}{3y}\bigg|=
\bigg|\frac{3(x-1)+(-y+3)}{3y}\bigg| $$
Now, by triangle inequality $$\bigg|\frac{3(x-1)+(-y+3)}{3y}\bigg|\leq \frac{3|x-1|+|y-3|}{3|y|}$$
since
$$|x-1|<\sqrt{(x-1)^2+(y-3)^2}<\delta \text{ and } |y-3|<\sqrt{(x-1)^2+(y-3)^2}<\delta$$
then
$$\frac{3|x-1|+|y-3|}{3|y|}<\frac{3\delta+\delta}{3|y|}=\frac{4\delta}{3|y|}$$
But i have that $|y-3|<\sqrt{(x-1)^2+(y-3)^2}<\delta$, then $y\in (3-\delta,\delta+3)$.
So i think the next step is to take $\delta =1/3$; if i do that i get $8/3<y$ that is $1/|y|<3/8$
then
$$\frac{4\delta}{3|y|}<\frac{3}{8}\frac{4\delta}{3}=\frac{\delta}{2}.$$
So, taking $\delta=\min\{1/3, 2\epsilon\}$, we have
$$\bigg|\frac{x}{y}-\frac{1}{3} \bigg|< \frac{3}{8}\frac{4\delta}{3} = \frac{\delta}{2}= \frac{1}{2}2\epsilon = \epsilon. $$
is this reasoning right?
|
You may also use this: if $$|x-1|<\epsilon\\|y-3|<\epsilon$$ we may bound $|\dfrac{x}{y}-\dfrac{1}{3}|$ sufficiently close to zero. To show that we have $$1-\epsilon<x<1+\epsilon\\3-\epsilon<y<3+\epsilon\\$$choosing $\epsilon$ small enough we obtain$$\dfrac{1-\epsilon}{3+\epsilon}<\dfrac{x}{y}<\dfrac{1+\epsilon}{3-\epsilon}$$or$$\dfrac{1}{3}-\dfrac{4\epsilon}{9+3\epsilon}=\dfrac{1-\epsilon}{3+\epsilon}<\dfrac{x}{y}<\dfrac{1+\epsilon}{3-\epsilon}=\dfrac{1}{3}+\dfrac{4\epsilon}{9-3\epsilon}$$therefore$$-\dfrac{1}{3}\epsilon<\dfrac{x}{y}-\dfrac{1}{3}<\dfrac{2}{3}\epsilon$$as long as $\epsilon<1$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2900047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
}
|
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$
If $x=\cos a + i \sin a$, $y=\cos b + i \sin b$, $z=\cos c +i \sin c$, and $x+y+z = xyz$, then show that $$\cos(a-b) + \cos(b-c) + \cos(c-a) + 1=0$$
Here's how I tried it
$$x+y+z=xyz $$
So, by De Moivre's Theorem,
$$(\cos a + \cos b + \cos c) + i(\sin a + \sin b + \sin c) = \cos(a+b+c) + i \sin(a+b+c)
$$
Equating real and imaginary parts,
$$\cos a + \cos b + \cos c= \cos(a+b+c)$$
and similarly for sine. Now,
$$(a-b) + (b-c) + (c-a) =0$$
What to do now? Please help. And please use De Moivre Theorem!
|
(This is a followup on Stefan Lafon's answer, too long for a comment.)
Using that $\,|x|=1 \iff \bar x = \dfrac{1}{x}\,$ and $\,\cos(a-b)= \operatorname{Re}\left(\dfrac{x}{y}\right)=\operatorname{Re}(x \bar y) = \operatorname{Re}(\bar x y)\,$, it follows that:
$$
\begin{align}
\cos(a-b) + \cos(b-c) + \cos(c-a) &= \cos(a-b) + \cos(a-c) + \cos(b-c) \\
&= \operatorname{Re}\left(x\bar y+ x \bar z + y \bar z\right) \\
&= \operatorname{Re}\left(x\left(\bar y + \bar z\right) + y \bar z\right) \\
&= \operatorname{Re}\left(x\left(\bar x \bar y \bar z - \bar x\right) + y \bar z\right) \\
&= \operatorname{Re}\left(\bar y \bar z + y\bar z -1 \right) \\
&= \operatorname{Re}\left(\left(y + \bar y\right) \bar z -1 \right) \\
&= -1 + 2 \operatorname{Re}(y) \operatorname{Re}(z)
\end{align}
$$
For the latter RHS to be $\,-1\,$, the second term must be zero, so one of $\,y,z\,$ must be a purely imaginary number of modulus $\,1\,$ i.e. $\,\pm i\,$. With some more legwork, it follows that the solution set of the given constraints is $\,\{(x,-x,z)\mid x=\pm i, |z| = 1\}\,$ or permutations thereof.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2900211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
Is $s(x)$ a linear combination of $a \cdot p(x) $and $b \cdot q(x)$ Let $p(x)=x^2+5x-3$ , $q(x)=4x^2+18x+4,$ and $s(x)=x^2+8x+2$.
I want to find two constants $a$ and $b$ such that $$s(x)=a \cdot p(x) +b \cdot q(x)$$
My attempt: $$ap(x)+bq(x) = ax^2+5ax-3a+4bx^2+18bx+4b = (a+4b)x^2+(5a+18b)x+4b-3a$$
I put this in the augmented matrix (the last column to the right is the coefficients of $s(x)$:
$\begin{bmatrix}1&4&1\\5&18&8\\-3&4&2\end{bmatrix}$.
I reduced this to echelon form $\begin{bmatrix}1&4&1\\0&2&-3\\0&-8&-1\end{bmatrix}$ and found out that the system is inconsistent, and has no solutions, meaning that there are no such $a$ and $b$ that satisify the given equation.
Is my solution correct?
|
Yes your way is fine, indeed with respect to the standard basis $\{1,x,x^2\}$ we have
*
*$v_p=(-3,5,1)$
*$v_q=(4,18,4)$
*$v_s=(1,8,1)$
and we need to solve
$$av_p+bv_q=v_s \iff \begin{bmatrix}-3&4\\5&18\\1&4\end{bmatrix}
\begin{bmatrix}a\\b\end{bmatrix}
=\begin{bmatrix}2\\8\\1\end{bmatrix}$$
and by the augmented matrix we obtain
$$\begin{bmatrix}-3&4&2\\5&18&8\\1&4&1\end{bmatrix}\to \begin{bmatrix}-3&4&2\\0&74&34\\0&16&5\end{bmatrix}$$
which has not solution.
As an alternative note that
$$\det\begin{bmatrix}-3&4&2\\5&18&8\\1&4&1\end{bmatrix}=-54+32+40-36+96-20=58 \neq 0$$
therefore the three vectors are linearly independent.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2900628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Calculate $\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$
Calculate
$$\lim_{x\to{0^+}}\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{\frac{x}{2}}}\cos{\frac{x}{2}}}$$
My Attempt:
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{\sin{x}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{2\sin{\frac{x}{2}}\cos{\frac{x}{2}}}}$$
$$\lim_{x\to{0^+}}\frac{\ln{(1-2\sin^2{\frac{x}{2}})}}{\frac{-2\sin^2{\frac{x}{2}}}{-2\sin^2{\frac{x}{2}}}}\cdot\frac{\frac{-2\sin^2{\frac{x}{4}}}{-2\sin^2{\frac{x}{4}}}}{\ln{(1-2\sin^2{\frac{x}{4}})}}\cdot\frac{\ln{\sin{\frac{x}{2}}}}{\ln{\sin{\frac{x}{2}}}+\ln{2}}$$
$$\frac{1}{4}\cdot 16\cdot 1$$
$$=4$$
Am I solving this correct?
|
Yes it looks fine, as a similar alternative we can use
$$\log_{\sin{x}}{\cos{x}}=\frac12\log_{\sin{x}}{\cos^2{x}}=\frac12\log_{\sin{x}}{(1-\sin^2{x})}=\frac12\frac{\log{(1-\sin^2{x})}}{\log {\sin{x}}}$$
$$\log_{\sin{(x/2)}}{\cos{(x/2)}}=\frac12\frac{\log{(1-\sin^2{(x/2)})}}{\log {\sin{(x/2)}}}$$
then
$$\frac{\log_{\sin{x}}{\cos{x}}}{\log_{\sin{(x/2)}}{\cos{(x/2)}}}
=\frac{\log{(1-\sin^2{x})}}{ {\log{(1-\sin^2{(x/2)})}}}
\frac{\log {\sin{(x/2)}}}{\log\sin{x}}=$$
$$=\frac{\log{(1-\sin^2{x})}}{-\sin^2 x}
\frac{-\sin^2 (x/2)}{ {\log{(1-\sin^2{(x/2)})}}}
\frac{\sin^2 x}{x^2}
\frac{4(x/2)^2}{\sin^2 (x/2)}
\frac{\log {\sin{(x/2)}}}{\log 2+\log\sin{(x/2)}+\log \cos (x/2)}$$$$\to 1\cdot 1\cdot 1 \cdot 4 \cdot 1=4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2901096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
}
|
Closed form for $t_n = 6t_{n-1}-9t_{n-2}$ where $t_0 = 5$ and $t_1 = 9$
Consider the sequence defined by
$$
\begin{cases}
t_0=5\\
t_1=9\\
t_n=6t_{n-1}-9t_{n-2} & \text{if }n\ge 2
\end{cases}
.$$
Find a closed form for $t_n$.
Your response should be a formula in terms of $n$, and should not contain terms such as $t_n,$ $t_{n-1},$ and so on. Do not include $``t_n=\text{''}$ in your response.
I tried forming a sequence by taking some values for $n$ and finding $t_n$. Once that was done, I moved on to find some pattern between $n$ and $t_n$ but couldn't find any. Here's the sequence:
$$\begin{array}{c || c *5{| c}}
n & 0 & 1 & 2 & 3 & 4 \\
\hline
t_n & 5 & 9 & 9 & -27 & -243
\end{array}$$
What am I suppose to do?
|
Another widely used approach is using generating functions, i.e.
$$f(x)=\sum\limits_{n=0}\color{red}{t_n}x^n=
5+9x+\sum\limits_{n=2}t_nx^n=
5+9x+\sum\limits_{n=2}\left(6t_{n-1}-9t_{n-2}\right)x^n=\\
5+9x+6x\left(\sum\limits_{n=2}t_{n-1}x^{n-1}\right)-9x^2\left(\sum\limits_{n=2}t_{n-2}x^{n-2}\right)=\\
5+9x+6x\left(\sum\limits_{n=1}t_{n}x^{n}\right)-9x^2\left(\sum\limits_{n=0}t_{n}x^{n}\right)=
5+9x+6x\left(f(x)-5\right)-9x^2f(x)$$
or
$$f(x)=5+9x+6x\left(f(x)-5\right)-9x^2f(x) \iff \\
f(x)=\frac{5-21x}{1-6x+9x^2}=
\frac{5-21x}{(1-3x)^2}=
\frac{7}{1-3x}-\frac{2}{(1-3x)^2}=\\
7\left(\sum\limits_{n=0}3^nx^n\right)-2\left(\sum\limits_{n=0}(n+1)3^nx^n\right)=\\
\sum\limits_{n=0}\color{red}{\left(7-2(n+1)\right)3^n}x^n$$
and
$$t_n=\left(5-2n\right)3^n$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2902485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
}
|
Marginal Pdf from joint pdf of discrete variable X,Y are random discrete variables with joint probability function of
$f(x,y) = \dfrac {1}{e^2\cdot y!\cdot (x-y)!} $ with $x \in\{ 0,1,2,...\}$ and $y\in\{0,1,2,...x\}$
Find marginals $f_X(x)$ , $f_Y(y)$.
I tried to compute firstly $f_Y(y)$ but i have a problem with the summation of the series:
$$f_Y(y) = \sum_{x=0}^\infty \dfrac {1}{e^2\cdot y!\cdot (x-y)!} = \frac {1}{e^2\cdot y!}\cdot \sum_{x=0}^\infty \frac {1}{(x-y)!} $$
How can I compute $f_Y(y)$ and $f_X(x)$ ?
Thank you very much!
|
The marginal $f_X(x)$ is
$$f_X(x) =\frac{1}{e^2} \sum\limits_y \frac{1}{y!(x-y)!} = \frac{1}{x!e^2} \sum\limits_y \frac{x!}{y!(x-y)!} = \frac{1}{x!e^2} \sum\limits_y \binom{x}{y}(1)^y(1)^{x- y} = \frac{1}{x!e^2} (1+1)^{x} $$
So
$$f_X(x) = \frac{1}{x!e^2} 2^x$$
Similarly
$$f_Y(y) =\frac{1}{y!e^2} \sum\limits_{x=0}^{\infty} \frac{1}{(x-y)!}=\frac{1}{y!e^2} \sum\limits_{x=y}^{\infty} \frac{1}{(x-y)!}$$
where the last equation is because $ y \leq x$. Let $k = x - y$, you'd get
$$f_Y(y) = \frac{1}{y!e^2}\sum\limits_{k=0}^{\infty} \frac{1}{k!} =\frac{1}{y!e^2}\sum\limits_{k=0}^{\infty} \frac{1^k}{k!} = \frac{1}{y!e^2}e^1 = \frac{1}{ey!}$$
The sum of $\frac{1}{k!}$ becomes $e$ due to Taylor series.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2904533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
}
|
Find $\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$ Find $$\lim_{x\to \infty}\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)$$ , without using squeeze theorem.
I have done the solution as below using squeeze theorem ...
$$Let \left[\left({x \over x^2+1}+{x \over x^2+2}+\cdots +{x\over x^2+x}\right)\right]=f(x)\implies \\ \left({x \over x^2+x}+{x \over x^2+x}+\cdots +{x\over x^2+x}\right)\lt f(x)\lt \left({x \over x^2+1}+{x \over x^2+1}+\cdots +{x\over x^2+1}\right) \\ {x^2 \over x+x^2}\lt f(x) \lt {x^2\over 1+x^2}\\ \text{applying limit on both sides }\\ \implies\lim_{x\to \infty}{x^2 \over x+x^2}= \lim_{x\to \infty}{x^2\over 1+x^2}=1\\ \implies \lim_{x\to \infty}f(x)=1$$
Can we do this without squeeze theorem?
|
With a Riemannian sum:
$$\sum_{k=1}^n\frac{n}{n^2+k}=n\frac 1{n^2}\sum_{k=1}^n\frac1{1+\dfrac k{n^2}}$$ can be seen as a Riemaniann sum truncated to the $n$ first terms among $n^2$. Then
$$\lim_{n\to\infty}n\int_0^{1/n}\frac{dx}{1+x}=\lim_{n\to\infty}n\log\left(1+\frac1n\right)=1.$$
No quite rigorous, though (because the sum converges to the integral at the same time that we increase $n$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2905900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
$\sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
The number of natural number $n\leq 50$ such that
$\displaystyle \sqrt{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$ is a natural number, is
Try: Let $\displaystyle x=\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}$
So $\displaystyle x=\sqrt[3]{n+x}\Rightarrow x^3=n+x\Rightarrow x^3-x-n=0$
could some help me how to solve it, Thanks
|
With $$\displaystyle \sqrt[3]{n+\sqrt[3]{n+\sqrt[3]{n+\cdots \cdots \cdots }}}$$ We get $$x^3-x-n=0$$
Thus $$n=x^3-x=x(x-1)(x+1)$$
Therefore for any natural number $x>1$ we have a natural number $n$ satisfying $$n=x^3-x=x(x-1)(x+1)$$
For example with $x=5$ we get $n=120$ and for $x=10$ we get $n=990.$
Another way of saying that is for any three natural numbers, we can have their product as $n$ and the middle of the three natural numbers to be $x$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/2907515",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
}
|
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.