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Proving $\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}$ I've been going through some of my notes when I found the following inequality for $a,b,c>0$ and $abc=1$:
$$
\begin{equation*}
\sqrt{2}(a+b+c) \geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}
\end{equation*}
$$
This was what I attempted, but that yielded no result whatsoever:
$$
\begin{align}
\sqrt{2}(a+b+c) &\geq \sqrt{1+a^2} + \sqrt{1+b^2} + \sqrt{1+c^2}\\
&\geq \sqrt{abc+a²} + \sqrt{abc+b²} + \sqrt{abc+c²}\\
&\geq \sqrt{a(bc+a)}+\sqrt{b(ac+b)}+\sqrt{c(ab+c)}\\
&\geq \sqrt{a\left(\frac{1}{a}+a\right)}+\sqrt{b\left(\frac{1}{b}+b\right)}+\sqrt{c\left(\frac{1}{c}+c\right)}
\end{align}
$$
With this we then have
$$
\begin{align}
\sqrt{1+a²} &= \sqrt{a\left(\frac{1}{a}+a\right)}\\
&=\sqrt{a\left(\frac{1}{a}+\frac{a²}{a}\right)}\\
&=\sqrt{a\left(\frac{1+a²}{a}\right)}\\
&=\sqrt{1+a²}
\end{align}
$$
This yields nothing but frustration and gets me back to step 0
|
here is another way which is not use derivative:
first we need to know: $\sqrt{\dfrac{x+y+z}{3}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}}{3}$ (1)
to prove it, we have $\sqrt{\dfrac{x+y}{2}} \ge \dfrac{\sqrt{x}+\sqrt{y}}{2} \implies \sqrt{\dfrac{x+y+z+t}{4}} \ge \dfrac{\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{t}}{4}$
let $t= \dfrac{x+y+z}{3}$ we have (1)
squre both sides:
$2(a+b+c)^2 \ge 3+(a^2+b^2+c^2)+2\sum {\sqrt{1+a^2+b^2+a^2b^2}} \iff \sum a^2+4\sum ab -3 \ge 2\sum {\sqrt{1+a^2+b^2+a^2b^2}} \iff \sum a^2+2\sum ab -3 \ge 2\sqrt{3(3+2\sum a^2+\sum a^2b^2)} \iff (\sum a^2-3)^2+8(\sum a^2-3)(\sum ab)+16(\sum ab)^2 \ge 36+24\sum a^2+12 \sum a^2b^2 $
note $(\sum ab)^2-\sum a^2b^2=2abc(a+b+c)=2(a+b+c) \iff (\sum a^2-3)^2+8(\sum a^2)(\sum ab-3) +4((\sum ab)^2-6\sum ab+9)+24(a+b+c)-72 \ge 0$
$abc=1$, so it is trivial $a+b+c \ge 3, ab+bc+ac \ge 3$ so the last one is true.
|
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|
Trigonometric equation $\sec(3\theta/2) = -2$ - brain dead Find $\theta$ with $\sec(3\theta/2)=-2$ on the interval $[0, 2\pi]$. I started off with $\cos(3 \theta/2)=-1/2$, thus $3\theta/2 = 2\pi/3$, but I don't know what to do afterwards, the answer should be a huge list of $\theta$s, which I cannot seem to get.
|
$$\sec\left(\frac{3\theta}{2}\right)=-2 \implies\cos\left(\frac{3\theta}{2}\right)=-\frac{1}{2} \implies \cos\left(\frac{3\theta}{2}\right)=\cos\left(\frac{2\pi}{3}\right)$$
Thus, we have a general solution as follows$$\frac{3\theta}{2}=2n\pi \pm \frac{2\pi}{3} \implies \theta=\frac{2}{3}\left(2n\pi\pm\frac{2\pi}{3}\right)=\frac{4\pi}{9}\left(3n\pm1 \right)$$Where, $n$ is any integer. All the values of $\theta$ in the given interval $[0, 2\pi]$ can be easily determined by setting different positive values of the integer $n=0, 1, 2, ..$ & selecting the correct values under given constraints, we get following three values of $\theta$ $$\theta=\frac{4\pi}{9}, \frac{8\pi}{9}, \frac{16\pi}{9}$$
|
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|
A set of numbers Problem: Let $E(x)$ be the number defined by the following expression
\begin{equation*}
E(x)=\sqrt[3]\frac{x^3-3x+(x^2-1)\sqrt{x^2-4}}{2}+\sqrt[3]\frac{x^3-3x-(x^2-1)\sqrt{x^2-4}}{2}
\end{equation*}
where $x$ is a real number and
$\sqrt[3]{Z}$ denotes the real cubic root of the real number $Z$.
Determine the set
$ \{E(n); n\in \mathbb{N}$ and $n > 1\}$
|
$$E(x)^3=(Y+Z)^3=Y^3+3YZE(x)+Z^3\\
=x^3-3x+3E(x)\sqrt[3]{\frac{x^6-6x^4+9x^2-x^6+6x^4-9x^2+4}4}$$
|
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|
Using Eulers formula I am trying to figure out how
\begin{equation*}
e^{i(-1+i\sqrt{3})}=e^{-\sqrt{3}} (cos(1)-i sin(1))??
\end{equation*}
I know that Euler's formula states that
\begin{equation*}
e^{ix} = \cos(x) + i \sin(x)
\end{equation*}
but surely that would mean in this case $x=-1+i\sqrt{3}$? Can someone please elaborate. Many thanks.
|
As Shayan says,
$e^{i(-1 + i \sqrt{3})} = e^{-\sqrt{3} - i} = e^{-\sqrt{3}} e^{-i} = e^{-\sqrt{3}}(\cos(1) - i\sin(1)); \tag{1}$
it is also true that
$e^{i(-1 + i \sqrt{3})} = \cos(-1 + i \sqrt{3}) + i\sin(-1 + i \sqrt{3}), \tag{2}$
since for any $z \in \Bbb C$,
$e^{iz} = \cos z + i \sin z; \tag{3}$
in (3), $e^{iz}$, $\cos z$, and $\sin z$ are understood to be complex analytic (holomorphic) functions; (3) may be verified for $z \in \Bbb C$ via the power series expansions of these three entire functions about $0$:
$e^{iz} = \sum_0^\infty \dfrac{(iz)^n}{n!} = 1 + iz + i^2 \dfrac{z^2}{2} + i^3 \dfrac{z^3}{6} + \ldots = 1 + iz - \dfrac{z^2}{2} - i\dfrac{z^3}{6} + \ldots$
$= (1 - \dfrac{z^2}{2} + \ldots) + i(z - \dfrac{z^3}{6} + \ldots) = \cos z + i \sin z, \tag{4}$
etc. etc. etc.
|
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|
What mistake have I made when trying to evaluate the limit $\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b}$? Suppose $a$ and $b$ are positive constants.
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = ?$$
What I did first:
I rearranged $\sqrt{n+a} \sqrt{n+b} = n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}}$ and so: $$\lim \limits _ {n \to \infty} n - n \sqrt{1+ \frac{a}{n}} \sqrt{1+ \frac{b}{n}} = 0$$
Because both $\frac{a}{n}$ and $\frac{b}{n}$ tend to $0$.
What would give a correct answer:
Plotting the function $$f(x) = x - \sqrt{x+a} \sqrt{x+b}$$
Clearly indicates that it has an asymptote in $- \frac{a+b}{2}$. This result can be obtained multiplying the numerator and the denominator by $n + \sqrt{n+a} \sqrt{n+b}$:
$$\lim \limits _ {n \to \infty}n - \sqrt{n+a} \sqrt{n+b} = $$
$$-\lim \limits _{n \to \infty} \frac {n(a+b)}{n + \sqrt{n+a} \sqrt{n+b}} - \lim \limits _{n \to \infty} \frac {ab}{n + \sqrt{n+a} \sqrt{n+b}}$$
The second limit is clearly $0$ and the first one gives the correct answer (dividing the numerator and denominator by $n$).
Why the first way I tried is wrong? I might have done something silly but I cannot find it.
|
Your initial approach was fine. Just use the expansion $\sqrt{1+\frac{a}{n}}=1+\frac{a}{2n}+O\left(\frac{1}{n^2}\right)$ to show that
$$\begin{align}
\sqrt{1+\frac{a}{n}}\sqrt{1+\frac{b}{n}}&=\left(1+\frac{a}{2n}+O\left(\frac{1}{n^2}\right)\right)\left(1+\frac{b}{2n}+O\left(\frac{1}{n^2}\right)\right)\\\\\
&=1+\frac{a+b}{2n}+O\left(\frac{1}{n^2}\right)
\end{align}$$
Finally we have
$$\begin{align}
\lim_{n\to \infty}\left(n-\sqrt{n+a}\sqrt{n+b}\right)&=\lim_{n\to \infty}\,n\,\left( 1-1-\frac{a+b}{2n}+O\left(\frac{1}{n^2}\right)\right)\\\\
&=-\frac{a+b}{2}\end{align}$$
|
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|
Complex number $\frac{z}{z+1}=2+3i$ Given that $\frac{z}{z+1}=2+3i$, find the complex number $z$, giving your answer in the form of $x+yi$.
Can someone give me some hints for solving this question? Thanks
|
$$\frac{z}{z+1}=2+3i\Longleftrightarrow$$
$$\left(\frac{z}{z+1}\right)(z+1)=(2+3i)(z+1)\Longleftrightarrow$$
$$\frac{z(z+1)}{z+1}=(2+3i)z+(2+3i)\Longleftrightarrow$$
$$z=(2+3i)z+(2+3i)\Longleftrightarrow$$
$$z-(2+3i)=((2+3i)z+(2+3i))-(2+3i)\Longleftrightarrow$$
$$(-1-3i)z=2+3i\Longleftrightarrow$$
$$\frac{(-1-3i)}{-1-3i}z=\frac{2+3i}{-1-3i}\Longleftrightarrow$$
$$z=-\frac{11}{10}+\frac{3}{10}i$$
|
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|
Determining the max area of a trapezoid with no known sides A trapezoid is drawn inside a semi circle cross section with the upper base length, being the length of the circle diameter, $d$, and a lower base, $x$, touching the lower sides of the semi circle.
What is the maximum length of the lower base, $x$, of the trapezoid if its area is to be a maximum?
What is the maximum are of the trapezoidal cross section in terms of $d$?
So far I have determined from $A=(\frac{x+d}2)h$, that $h$ is equal to $\sqrt{(\frac{d}2)^2 - (\frac{x}2)^2}$
And therefore $A$ in terms of $x$ & $d$.
$$A=(\frac{x+d}2)(\sqrt{(\frac{d}2)^2 - (\frac{x}2)^2}$$
$$A=(\frac{x+d}2)(\sqrt{\frac{d^2}4 - \frac{x^2}4})$$
$$A=(\frac{x+d}2)(\frac{\sqrt{d^2 - x^2}}2)$$
$$A= \frac14 (x+d) (\sqrt{d^2 - x^2})$$
this is where I got stuck... help?
|
Answer:
I really do not know what the problem is when you got the steps absolutely right.
If differentiation is the problem:
$$\frac{dA}{dx} = \sqrt{d^2-x^2} + (x+d) \frac{-2x}{2\sqrt{d^2-x^2}}$$
$$\frac{dA}{dx} = 0$$
Assuming two functions $U = x+d$ and $V = \sqrt{d^2-x^2}$
$$\frac{dA}{dx} = V\frac{dU}{dx}+U\frac{dV}{dx}$$
$\frac{dU}{dx} = 1$ and $\frac{dV}{dx} = \frac{-2x}{2\sqrt{d^2-x^2}}$
$$\frac{dA}{dx} = \sqrt{d^2-x^2}+(x+d)\frac{-2x}{2\sqrt{d^2-x^2}} = d^2-x^2 -x(x+d) $$ $$= d^2 - 2x^2-xd = 2x^2 +xd-d^2=0$$ This a quadratic with x as the variable and two roots of x in terms of d is what follows.
the two roots of a quadratic$ax^2+bx+c = 0$ is as follows:
$$ x= \frac{-b\pm\sqrt{b^2-4ac}}{2a}$$.
In our case $a = 2$, $b = d$ and $c = -d^2$
\begin{align}
x &= \frac{-d\pm\sqrt{d^2 + 8d^2}}{4}\\
&= \frac{-d\pm\sqrt{9d^2}}{4}\\ &= \frac{-d\pm3d}{4}\\
&= \frac{d}{2} \text{ or } \frac{-2d}{2}\text{ (rej.)}\\
\end{align}.
Now substitute the value of d in the area and find A in terms of d.
Goodluck
Thanks
Satish
|
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|
How to evaluate $\lim\limits_{n\to\infty}\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$ I don't know how to transform the expression $\frac{1}{n}((x+\frac{a}{n})^2+(x+\frac{2a}{n})^2+...+(x+\frac{(n-1)a}{n})^2)$
The solution, after transformation is
$\frac{n-1}{n}x^2+2\frac{1+...+(n-1)}{n^2}ax+\frac{1^2+...+(n-1)^2}{n^3}a^2$
Thanks for replies.
|
Hint:
$$1 + \cdots + (n - 1) = \frac{n(n-1)}{2} = \frac{n^2}{2} - \frac{n}{2}$$
and
$$
1^2 + 2^2 + \cdots + (n - 1)^2 = \frac{(n-1)n(2n-1)}{6} = \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}$$
|
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|
integral $\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$ I want to compute this integral
$$\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt$$
where $0<b \leq a$.
I have this results
$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2sin^2(t)+b^2cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$
But I don't know how to prove this equality.
Which can help me,
Thanks for all.
|
Although there is already an accepted answer, I put this answer here, since someone might have use of the "tool" to work with the expression like this before integrating.
The case $a=b$ is trivial, so we assume $a>b$. We write the numerator $\cos 2t$ in the following way:
$$
\cos 2t=\cos^2t-\sin^2t=\alpha\bigl(b^2\cos^2t+a^2\sin^2t\bigr)+(1-\alpha b^2)\cos^2t-(1+\alpha a^2)\sin^2t,
$$
and we'll soon see how to choose $\alpha$ in a good way, where good means that we should get something that we can integrate. Let us first use the trig-one on the $\cos^2 t$ part, to get
$$
\cos 2t = \alpha\bigl(b^2\cos^2t+a^2\sin^2t\bigr)+(1-\alpha b^2)-\bigl[(1-\alpha b^2)+(1+\alpha a^2)\bigr]\sin^2 t.
$$
Now, we choose $\alpha$ so that the constant in brackets is zero, i.e.
$$
(1-\alpha b^2)+(1+\alpha a^2)=0 \iff \alpha=\frac{2}{b^2-a^2}.
$$
This gives
$$
\cos 2t = \frac{2}{b^2-a^2}(b^2\cos^2t+a^2\sin^2t)-\frac{a^2+b^2}{b^2-a^2}.
$$
Now comes the fun part. We can write the integrand
$$
\begin{aligned}
\frac{\cos 2t}{b^2\cos^2t+a^2\sin^2t}&=\frac{2}{b^2-a^2}-\frac{a^2+b^2}{b^2-a^2}\frac{1}{b^2\cos^2t+a^2\sin^2t}\\
&=\frac{2}{b^2-a^2}-\frac{a^2+b^2}{b^2-a^2}\frac{1}{b^2\cot^2t+a^2}\frac{1}{\sin^2t}.
\end{aligned}
$$
This is easy to integrate,
$$
\int \frac{\cos 2t}{b^2\cos^2t+a^2\sin^2t}\,dt = \frac{2t}{b^2-a^2}-\frac{a^2+b^2}{b^2-a^2} \frac{1}{ab}\text{arccot}\,\Bigl(\frac{b\cot t}{a}\Bigr).
$$
I'm sure you can insert the limits and get the result from this. I get
$$
\int_0^\pi \frac{\cos 2t}{b^2\cos^2t+a^2\sin^2t}\,dt=\frac{(a-b)\pi}{ab(a+b)}.
$$
|
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|
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
|
Here is my favorite one. Suppose for the sake of contradiction that $\sqrt{2} = \frac{a}b$ for integers $a,b$. Then $2b^2 = a^2$. Let $p$ be an odd prime that is not congruent to $\pm 1 \pmod 8$. Then by quadratic reciprocity,
$$\left( \frac{2b^2}p \right) = \left( \frac{2}p \right) = -1 \ne \left( \frac{a^2}p \right) = 1.$$
|
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|
Find probability of exactly one $6$ in first ten rolls of die, given two $6$s in twenty rolls I am trying to calculate the probability that, when rolling a fair die twenty times, I roll exactly one $6$ in the first ten rolls, given that I roll two $6$s in the twenty rolls.
My thoughts
Let $A = \{\text {Exactly one 6 in first ten rolls of a die} \}$ and $B = \{\text {Exactly two 6s in twenty rolls of a die} \}.$
Then I want to find
$$P[A\mid B] = \frac{P[A \cap B]}{P[B]}.$$
By the binomial distribution formula, we get that
$$P[B] = {20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}.$$
Furthermore I think that $P[A \cap B]$ is equal to the probability of rolling exactly one $6$ in ten rolls and then rolling exactly one $6$ in another set of ten rolls. That is,
$$P[A \cap B] = \left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2.$$
Am I correct in thinking this?
If so, then it follows that the required probability is
$$P[A \mid B] = \frac{\left[{10 \choose 1} \cdot \left(\frac{1}{6}\right)^1 \cdot \left(\frac{5}{6}\right)^9\right]^2}{{20 \choose 2} \cdot \left(\frac{1}{6}\right)^2 \cdot \left(\frac{5}{6}\right)^{18}},$$
which, I know, can be simplified further!
|
I think you are over thinking this. We know we get exatly two sixes in twenty rolls how many ways can that happen? Consider a roll to be 6 or not 6 we don't
care what number it is otherwise.
One of the sixes arived in any of the twenty rolls and the other in an of the nineteen remaining rolls and since a 6 is a 6 we divide by two because the order does not matter.
There are thus $\dfrac{20 \cdot 19}{2} = 190$ ways we can get exactly 2 6's in 20 rolls.
In how many ways can we have exactly one 6 in 10 rolls? Well it can be any of the 10 rolls, and it must be not 6 in the other 9 so there are 10 ways to get exactly one 6 in ten rolls, and 10 ways to get the second 6 in the last 10 rolls making
The answer is simply $\dfrac{10 \cdot 10}{190} = \dfrac{100}{190} = \dfrac{10}{19}$.
|
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How to find integrals of the form $\frac 1{(a^2+x^2)^n}$ How will we find the integrals of the form $\frac 1{(a^2+x^2)^n}$?
For example if we find the integral of $\frac{1}{(x^2+1)^3}$, I cannot see any substitution here. Moreover, if we do integration by parts, it becomes more and more messy. I have heard somewhere about reduction formula, so please can someone explain the derivation of it and why does it work?
|
Let your integral be $I_n$. Then we use integral by part:
Let $\frac{1}{(a^2+x^2)^{n}} = u$, $du = \frac{-2nx}{(a^2+x^2)^{n+1}}dx$
$dx = dv, v = x$
$I_n = \frac{x}{(a^2+x^2)^{n}} + \int \frac{2nx^2}{(a^2+x^2)^{n+1}}dx $
The integral on the RHS can be written as: $\int \frac{2n(x^2+a^2)}{(a^2+x^2)^{n+1}}dx - \int \frac{2na^2}{(a^2+x^2)^{n+1}}dx = 2nI_n-2na^2I_{n+1} $
Then: $I_n = \frac{x}{(a^2+x^2)^{n}} + 2nI_n-2na^2I_{n+1}$
Or: $I_{n+1} = \frac{x}{2na^2(a^2+x^2)^{n}} + \frac{2n-1}{2na^2}I_n$
From this recursive formula, you can calculate $I_n$ (It's gonna be very lengthy, so I won't write it explicitly here)
|
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|
Integer partitioning Suppose we have an integer $n$. I we want to partition the integer in the form of $2$ and $3$ only; i.e., $10$ can be partitioned in the form $2+2+2+2+2$ and $2+2+3+3$.
So, given an integer, how to calculate the total number of ways of doing such partitions and how many $2$'s and $3$'s are there in each of the partitions?
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The number of partitions of $n$ into parts in the set $\{2,3\}$ is the coefficient of $x^n$ in
$$\left(1+x^2+x^4+x^6+\ldots\right)\left(1+x^3+x^6+x^9+\ldots\right)=\frac1{(1-x^2)(1-x^3)}\;.$$
For $n=10$, for instance, we can ignore powers of $x$ higher than $10$, so we need only consider
$$(1+x^2+x^4+x^6+x^8+x^{10})(1+x^3+x^6+x^9)\;,$$
and by inspection the $x^{10}$ term is
$$x^4\cdot x^6+x^{10}\cdot x^0=2x^{10}\;:$$
the two partitions of $10$ that you listed in the question are the only ones.
However, we can do better. According to Wikipedia, the number of partitions of $n$ into parts of sizes $1,2$, and $3$ is the integer nearest to $\frac1{12}(n+3)^2$, i.e.,
$$\left\lfloor\frac{(n+3)^2}{12}+\frac12\right\rfloor\;;$$
call this $f(n)$, and let $g(n)$ be the number of partitions of $n$ into parts of size $2$ and $3$. Each partition of $n$ with parts in $\{1,2,3\}$ is a partition of some $k\le n$ into parts in $\{2,3\}$ together with some number of unit parts, so
$$f(n)=\sum_{k=2}^ng(k)\;,$$
and
$$g(n)=f(n)-f(n-1)=\left\lfloor\frac{(n+3)^2}{12}+\frac12\right\rfloor-\left\lfloor\frac{(n+2)^2}{12}+\frac12\right\rfloor\;.$$
As a quick check, for $n=10$ this becomes
$$\left\lfloor\frac{(13)^2}{12}+\frac12\right\rfloor-\left\lfloor\frac{(12)^2}{12}+\frac12\right\rfloor=14-12=2\;.$$
|
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|
The value of $\int^{\pi/2}_0 \frac{\log(1+x\sin^2\theta)}{\sin^2\theta}d\theta$ Problem :
Find the value of $\int^{\pi/2}_0 \frac{\log(1+x\sin^2\theta)}{\sin^2\theta}d\theta$
Now how to use Leibnitz's rule : $\frac{d}{dx}(\int^{\psi(x)}_{\phi(x)} f(t)dt) = \frac{d}{dx}\{\psi(x)\} f(\psi(x)) -\frac{d}{dx}\{\phi(x)\} f(\phi (x)\}$ here
please guide will be of great help thanks as there is $x$ and $\theta$ both
|
Let $I(x)$ be the integral
$$I(x)=\int_0^{\pi/2}\frac{\log(1+x\sin^2 \theta)}{\sin^2\theta}d\theta$$
and assume that $x>0$ is real-valued.
Now, taking a derivative with respect to $x$ gives
$$I'(x)=\int_0^{\pi/2}\frac{1}{1+x\sin^2\theta}d\theta=\frac{\pi}{2\sqrt{1+x}}$$
where this latter integral was evaluated using contour integration.
Next, integration of $I'(x)$ gives
$$I(x)=\pi\sqrt{1+x}+C$$
where $C$ is found by noting $I(0)=0$. Thus, $C=-\pi$.
Finally, we have
$$I(x)=\pi(\sqrt{1+x}-1)$$
NOTE:
Here, we will evaluate $\int_0^{\pi/2}\frac{1}{1+x\sin^2\theta}d\theta$ for $x<0$ using contour integration. Note that in general, we have
$$\begin{align}
\int_0^{\pi/2}\frac{1}{1+x\sin^2\theta}d\theta & = \int_0^{\pi/2}\frac{1}{1+x\left(\frac{1-\cos 2\theta}{2}\right)}d\theta \tag 1 \\\\
&=-\frac{1}{2x}\int_0^{2\pi}\frac{1}{\cos \theta-a}d\theta \tag 2
\end{align}$$
where $a=\frac{2+x}{x}=1+\frac2x$. In going from $(2)$ to $(3)$ we exploited the periodicity of the cosine function and effected the substitution $2x \to x$.
Next, we move to the complex plane by letting $z=e^{i\theta}$ so that $dz=ie^{i\theta}dz$ and the integration is over the closed-contour unit circle$C$, for which $|z|=1$. Then, we have
$$\begin{align}
-\frac{1}{2x}\int_0^{2\pi}\frac{1}{\cos \theta-a}d\theta &=-\frac{1}{ix}\oint_C \frac{dz}{z^2-2az+1}\\\\
&=-\frac{2\pi}{x}\text{Res}\left(\frac{1}{(z-a+\sqrt{a^2-1})(z-a-\sqrt{a^2-1})}\right)\\\\
&=-\frac{2\pi}{x}\text{Res}\left(\frac{1}{(z-\frac{x+2}{x}-\frac{2\sqrt{x+1}}{|x|})(z-\frac{x+2}{x}+\frac{2\sqrt{x+1}}{|x|})}\right)
\end{align}$$
There are three cases we need to examine.
Case 1: $x>0$
For $x>0$, the only pole is at $z=\frac{x+2-2\sqrt{x+1}}{x}$. The residue is thus $-\frac{x}{4\sqrt{x+1}}$ and we have
$$I'(x)=\frac{\pi}{2\sqrt{x+1}}$$
Case 2:
For $<-1x<0$, the only pole is at $z=\frac{x+2}{x}+2\frac{\sqrt{x+1}}{|x|}=\frac{x+2-2\sqrt{x+1}}{x}$. The residue is $-frac{x}{4\sqrt{x+1}} and we have
$$I'(x)=\frac{\pi}{2\sqrt{x+1}}$$
Case 3:
For $x<-1$, we note that $|\frac{x+2}{x}+2\frac{\sqrt{x+1}}{|x|}|=1$ and the poles are complex conjugates residing on the unit circle. Thus, the integral diverges.
We can intepret, however, a Cauchy Principal value of the integral. Here, we exclude the poles by modifying the contour $C$ with a semi-circle deformation around each pole. The two contributions from integration around the deformations are equal in magnitude and of opposite sign. Thus, we have
$$I'(x)=0$$
|
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|
How to integrate $\int_{-1}^1 \tan^{-1}\left(\frac{1}{\sqrt{1-x^2}}\right )\,dx$? Evaluate
$$\int_{-1}^1 \tan^{-1}\bigg (\dfrac{1}{\sqrt{1-x^2}}\bigg ) dx $$
Could somebody please help integrate this without using Differentiation under the Integral Sign?
|
First, integrate by parts to reduce the problem to calculating
$$\int \frac{x^2}{\sqrt{1-x^2}(2-x^2)}\,dx.$$
now split into two more manageable terms
$$\int \frac{2}{\sqrt{1-x^2}(2-x^2)}\,dx + \int \frac{x^2-2}{\sqrt{1-x^2}(2-x^2)}\,dx.$$
The left term is the only tricky one. Substitute $x= \sin u$ to get rid of the square root
$$\int \frac{2}{2-\sin^2 u}\,du = \int \frac{2}{2\cos^2 u+\sin^2 u}\,du = \int \frac{2\sec^2 u}{2+\tan^2 u}\,du$$
and finally substitute $v = \frac{\tan u}{\sqrt{2}}$.
|
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|
Limit of $\dfrac{(1+4^x)}{(1+3^x)}$? I don't remember how to find the limit in this case. I take $x$ towards $+\infty$.
$\lim\limits_{x\to \infty} \dfrac{1+4^x}{1+3^x}$
I do not know where to start. I would instinctively say that $1$ can't be right because $4^x$ goes faster than $3^x$ and thus one would move towards infinity, but this is apparently not the case....
|
$\lim\limits_{x\to \infty} \dfrac{1+4^x}{1+3^x}$
$=\lim\limits_{x\rightarrow \infty}\frac{1}{1+3^x}$+
$\lim\limits_{x\rightarrow \infty}\frac{4^x}{1+3^x}$
since $\lim\limits_{x\rightarrow \infty}\ (1+3^x)=\infty$
$\lim\limits_{x\rightarrow \infty}\frac{1}{1+3^x}=0$
The leadind term in the denominator of $\frac{4^x}{1+3^x}$ is $3^x.$ Divide the numerator and denomirator by this:
$\lim\limits_{x\rightarrow \infty}\frac{(\frac{4}{3})^x}{1+3^{-x}}$
The expression $3^{-x}$ tends to zero as $x$ approches $\infty$
$=\lim\limits_{x\rightarrow \infty}(\frac{4}{3})^x$
$=\left(\frac{4}{3}\right)^{\lim_{x\rightarrow \infty}x}=\left(\frac{4}{3}\right)^{\infty}=\color{red}\infty$
|
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|
Integrate $ \sin x /(1 + A \sin x)$ over the range $0$,$2 \pi$ for $A=0.2$ Wolfram Alpha indicates the following solution form:-
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
=
(1/A)\left( x - \frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}} + constant\right)^{2\pi}_0
$$
My thinking is that substituting for $x$ by $2 \pi$ and by $0$ the corresponding values of $\tan(x/2)$ are the same, namely $\tan(\pi) = \tan(0) = 0$. It follows therefore that the fractional term:-
$$
\frac{2 \tan^{-1} \left( \frac{A + \tan{(x/2)}}{\sqrt{(1-A^2)}}\right)}{\sqrt{(1-A^2)}}
$$
has the same value ( let us call it $Q$ ) for $x=2\pi$ and $x=0.$ Therefore the definite integral becomes
$$
\int_0^{2\pi} \frac{\sin x}{1 + A \sin x} dx
=
(1/A)\left( (2\pi - Q + constant) - (0 - Q + constant) \right)
= 2\pi/A.
$$
For $A = 0.2$ we derive the result
$$
\int_0^{2\pi} \frac{\sin x}{1 + 0.2 \sin x} dx
= 10 \pi.
$$
However Wolfram Alpha gives the answer as $-0.64782$ (approx.) which is reasonable looking at the graph.
So what is wrong with my reasoning? ( I realize that the $\tan$ function is not continuous over the range $0,2\pi$ but I don't know what that implies for the analysis of the definite integral ).
EDIT
I have accepted @abel's alternative method for solving the integral.
I have posted a related question here which asks why the approach which I followed is incorrect.
|
$$\int_0^{2\pi}\frac{\sin x}{1+a\sin x }\, dx = \int_0^{2\pi}\left(\frac 1 a-\frac{1}{a(1+a\sin x)}\right)\, dx=\frac{2\pi}a-\frac1a \int_0^{2\pi}\frac{1}{1+a\sin x }\, dx =\frac{2\pi}a-\frac Ja $$
now consider
$$\begin{align}J &= \int_0^{2\pi}\frac{1}{1+a\sin x }\, dx\\
&=\int_0^{\pi}\frac{1}{1+a\sin x }\, dx + \int_\pi^{2\pi}\frac{1}{1+a\sin x }\, dx \\
&=\int_0^{\pi}\frac{1}{1+a\sin x }\, dx + \int_0^\pi\frac{1}{1-a\sin x }\, dx \\
&=2\int_0^{\pi/2}\frac{1}{1+a\sin x }\, dx + 2\int_0^{\pi/2}\frac{1}{1-a\sin x }\, dx \\
&= 2f(a) + 2f(-a)\tag 1\end{align}$$
where $$f(a) = \int_0^{\pi/2}\frac{1}{1+a\sin x }\, dx,\quad u = \tan(x/2), x = 2\tan^{-1}u, dx = \frac{2du}{1+u^2} $$ therefore
$$\begin{align}f(a) &=2\int_0^1\frac{du}{1+2au+u^2} \\
&= 2\int_0^1\frac{du}{(u+a)^2 + 1- a^2}\\&=\frac{2}{\sqrt{1-a^2}} \tan^{-1}\left(\frac{u+a}{\sqrt{1-a^2}}\right)\big|_0^1\\
&=\frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\frac{1+a}{\sqrt{1-a^2}}\right) - \tan^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)\right)\\
&=\frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\sqrt{\frac{1+a}{1-a}}\right) - \tan^{-1}\left(\frac{a}{\sqrt{1-a^2}}\right)\right)\\
f(a) + f(-a) &= \frac{2}{\sqrt{1-a^2}}\left(\tan^{-1}\left(\sqrt{\frac{1+a}{1-a}}\right) + \tan^{-1}\left(\sqrt{\frac{1-a}{1+a}}\right)\right)\\
&=\frac{\pi}{\sqrt{1-a^2}}\end{align}$$
finally, $$\int_0^{2\pi}\frac{\sin x}{1+a\sin x }\, dx = \frac{2\pi}a-\frac{2\pi}{a\sqrt{1-a^2}}=\frac{2\pi(\sqrt{1-a^2} - 1)}{a\sqrt{1-a^2}}$$
|
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|
Derivative of Function with Rational Exponents $f(x)= \sqrt[3]{2x^3-5x^2+x}$ I have a question following:
$$f(x)=\sqrt[3]{2x^3-5x^2+x}$$
Here's what I did,
$$f(x)=\sqrt[3]{2x^3-5x^2+x}
\\ = (2x^3-5x^2+x)^{3\over2}
\\\\f'(x) = {3\over 2}(2x^3-5x^2+x)^{3\over2}(6x^2-10x+1)$$
Did I do this correctly?? Because I have different answer on the answer page. Can I reduce or factor any?? Or was there any mistakes?
Thanks
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$$
d/dx \ (2x^3-5x^2+x)^{(\frac{1}{3})}
$$
let $f(x)$ be $2x^3-5x^2+x$ and $g(x)$ be $x^{(\frac{1}{3})}$
$$
d/dx \ g(f(x)) = g'(f(x))*f'(x)
$$
so
$$
d/dx \ (2x^3-5x^2+x)^{(\frac{1}{3})} = \frac{1}{3}(2x^3-5x^2+x)^{\frac{-2}{3}}*(6x^2-10x+1)
$$
which can be simplified further if need to.
|
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|
$x^2-y^2=196$, can we find the value of $x^2+y^2$? $x$ and $y$ are positive integers.
If $x^2-y^2=196$, can we know what the value of $x^2+y^2$ is?
Can anyone explain this to me? Thanks in advance.
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$$
x^2-y^2=(x-y)(x+y)=14^2
$$
Since $(x+y)-(x-y)=2y$ either both $x+y$ and $x-y$ need to be even or both need to be odd. Since their product is even, both need to be even. There are two possible factorizations: $14\cdot14$ and $2\cdot98$. $14\cdot14$ implies $y=0$, so we want $x-y=2$ and $x+y=98$. That is, $x=50$ and $y=48$. Therefore,
$$
x^2+y^2=4804
$$
|
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|
if three integer such diophantine equation How find $x+y+z$
following Diophantine equation
$$xy^2+yz^2+zx^2=x^2y+y^2z+z^2x+x+y+z$$
ie:$(x-y)(y-z)(z-x)=x+y+z$
where $x,y,z$ are integers.
can find $x+y+z$
I tried some values and got some near equalities when $x,y,z$ at least two are equal, $x+y+z=0$,for other case, who have an idea to discuss...plz. Thanks in advance.
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Suppose $x\le y\le z$ and put $y=x+a=z-b$ where $a,b\ge 0$
Then you will find $$-a\cdot b\cdot-(a+b)=ab(a+b)=3y+a-b$$ so that $$3y= ab(a+b)+b-a$$
Then you can search for values of $a,b$ which give an integer multiple of $3$ on the right-hand side. So if $a=b=3$ you get $3y=54$ so that $y=18$.
Note that if $a$ and $b$ are not both multiples of $3$ then one of $ab(a+b)$ and $a-b$ is a multiple of $3$ and the other is not.
So put $a=3c, b=3d$ and obtain the general solution $$y=9cd(c+d)+d-c, x=y-3c, z=y+3d$$
|
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|
Problem with sine in a right triangle Given a triangle $ABC$ with angles a,b & c, prove that if $\sin^2(a) + \sin^2(b) + \sin^2(c) = 2$ then the triangle is right angled (has an angle of $90^o$).
If I assume the triangle is right angled and have AB, AC and BC as sides, with BC being the base, then I can say that statement is true, since $\sin^2(a)$ would be 1 where $a=90^o$.
Also, since it's a right angled triangle we can say that $\sin^2(b) = AC^2/BC^2$ and $\sin^2(c) = AB^2/BC^2$. The sum of $\sin^2(b) + \sin^2(c)$ would be equal to $(AC^2+AB^2)/BC^2$ which is 1 (pythagoras theorem). Hence the sum of squares of sines of all 3 angles is indeed 2.
But how do I prove this? As in, I'm only given the starting equation, how do I get from there to a right angled triangle?
|
Notice, in $\Delta ABC$ , we have $a+b+c=180^o$
Given that
$$\sin^2 a+\sin^2 b+\sin^2 c=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (180^o-(a+b))=2$$ $$\implies \sin^2 a+\sin^2 b+\sin^2 (a+b)=2$$
$$\implies (\sin a\cos b+\cos a\sin b)^2=2-\sin^2 a-\sin^2 b$$ $$\implies \sin^2 a\cos^2 b+\cos^2 a\sin^2 b+2\sin a\sin b\cos a\cos b=\cos^2 a+\cos^2 b$$
$$\implies -(1-\sin^2 a)\cos^2 b-(1-\sin^2 b)\cos^2 a+2\sin a\sin b\cos a\cos b=0$$ $$\implies -\cos^2 a\cos^2 b-\cos^2 a\cos^2 b+2\sin a\sin b\cos a\cos b=0$$
$$\implies -2\cos a\cos b(\cos a\cos b-\sin a\sin b)=0$$ $$\implies \cos a\cos b\cos(a+b)=0$$ $$ \color{blue}{\text{if } \quad \cos a=0 \implies a=90^o}$$ $$ \color{blue}{\text{if } \quad \cos b=0 \implies b=90^o}$$ $$ \color{blue}{\text{if } \quad \cos (a+b)=0 \implies a+b=90^o \iff c=90^o}$$
We find that above three cases show that $\color{blue}{\Delta ABC}$ is $\color{blue}{\text{right angled}}$.
|
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|
Please help to find term's coefficient in the following example I trying find the number of all solutions in the following:
$ x_1 + x_2 + x_3 + x_4 + x_5 = 24 $
where:
2 of variables are natural odd numbers
3 of variables are natural even numbers
none of variables are equal to $0$ or $1$
(all the variables are $>= 2$)
So I've made generating functions according to the limitations as follow:
$ (x^3 + x^5 + x^7 + ... + x^{21} + x^{23})^2 $
$ (x^2 + x^4 + x^6 + ... + x^{22} + x^{24})^3 $
Now I need to find the coefficient of $ x^{24} $
I've tried to solve it but get stuck, here is what I've done:
$ x^6(1 + x^2 + x^4 + ... + x^{18} + x^{20})^2 $
$ x^6(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $
$ x^{12}(1 + x^2 + x^4 + ... + x^{18} + x^{20})^2(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $
Now we actually looking for coefficient of $ x^{24-12}=x^{12} $ in:
$ (1 + x^2 + x^4 + ... + x^{18} + x^{20})^2(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $
$ (\frac{1 - x^{21}}{1 - x})^2(\frac{1 - x^{23}}{1 - x})^3 $
Here I'm stuck. How to proceed in order to find coefficient of $ x^{12} $?
Regards.
PS: Solution was given but just out of curiosity, how to solve it using generating-functions?
|
Let's take it from the point where we ask for the coefficient of $x^{12}$ in:
$$ (1 + x^2 + x^4 + ... + x^{18} + x^{20})^2(1 + x^2 + x^4 + ... + x^{20} + x^{22})^3 $$
[Note, however, that a factor of $\binom{5}{2}=10$ is already left out of the calculation at this point, which would account for varying positions of the two odd summands $x_i$. See the comments on Ashtay's Answer for more details.]
Replace $x^2$ with $y$ and ask for the coefficient of $y^6$:
$$ (1 + y + y^2 + \ldots + y^9 + y^{10})^2 (1 + y + y^2 + \ldots + y^{10} + y^{11})^3 $$
Of course we can remove all polynomial in $y$ terms with exponents greater than $6$ since these cannot contribute to the coefficient of $y^6$, then notice the polynomial factors are now equal:
$$ (1 + y + y^2 + \ldots + y^6)^5 $$
We can continue to explicitly work out the coefficient of $y^6$ by repeated squarings (and removing terms beyond $y^6$):
$$ (1 + y + \ldots + y^6)^2 \; \to \; (1 + 2y + 3y^2 + 4y^3 + 5y^4 + 6y^5 + 7y^6) $$
$$ (1 + 2y + 3y^2 + \ldots + 7y^6)^2 \; \to \; (1 + 4y + 10y^2 + 20y^3 + 35y^4 + 56y^5 + 84y^6) $$
Multiplication by $(1+y+y^2+\ldots + y^6)$ of the last result gives the desired term $210y^6$. It follows that $210$ was the original coefficient of $x^{24}$ you wanted (if my arithmetic is right!).
|
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|
Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is I got this task two days ago, quite difficult for me, since I have not done applications of Vieta's formulas and Bezout's Theorem for a while. If can someone solve this and add exactly how I am supposed to use these two theorem's on this task, I would be thankful.
Given $x^2 + 4x + 6$ as factor of $x^4 + ax^2 + b$, then $a + b$ is equal to?
|
Let $r_1,r_2$ be the roots of $x^2+4x+6$. By Vieta's $r_1+r_2 = -4$ and $r_1r_2 = 6$.
Since $x^2+4x+6$ is a factor of $x^4+ax^2+b$, we have that $r_1,r_2$ are also roots of $x^4+ax^2+b$.
Then, since $x^4+ax^2+b$ is an even polynomial, $-r_1,-r_2$ are also roots of $x^4+ax^2+b$.
It is easy to see that $r_1,r_2,-r_1,-r_2$ are distinct, so we've found all four roots of $x^4+ax^2+b$.
Therefore, $x^4+ax^2+b$ $= (x-r_1)(x-r_2)(x+r_1)(x+r_2)$ $= \left[(x^2-(r_1+r_2)x+r_1r_2\right]\left[(x^2+(r_1+r_2)x+r_1r_2\right]$ $= (x^2+4x+6)(x^2-4x+6)$ $= x^4 - 4x^2 + 36$.
Hence, $a = -4$, $b = 36$, and thus, $a+b = 32$.
Note: I'm not sure how Bezout's Theorem is needed here.
EDIT: Little Bezout's Theorem is also known as the polynomial remainder theorem, which was used above.
|
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|
Is it a composite number? How do I prove $19\cdot8^n+17$ is a composite number?
Or is that number just a prime?
So I tried to find a divisor in the cases $ n = 2k $ and $ n = 2k + 1 $. But I had no success.
Do you have any ideas?
|
Three cases:
*
*$n$ is even. Then $$19\cdot 8^n+17\equiv 1\cdot (-1)^n+17\equiv 1+17\equiv 0\pmod{\! 3}$$
*$n=4k+1$ for some $k\in\Bbb Z_{\ge 0}$. Then $$19\cdot 8^{4k+1}+17\equiv 6\cdot \left(8^2\right)^{2k}\cdot
8+4\equiv 6\cdot (-1)^{2k}\cdot 8+4$$
$$\equiv 48+4\equiv 52\equiv 0\pmod{\! 13}$$
*
*$n=4k+3$ for some $k\in\Bbb Z_{\ge 0}$. Then $$19\cdot 8^{4k+3}+17\equiv (-1)\cdot \left(8^2\right)^{2k}\cdot 8^3+17\equiv (-1)\cdot (-1)^{2k}\cdot 8^3+17$$
$$\equiv -(-2)^3+17\equiv 8+17\equiv 25\equiv 0\pmod{\! 5}$$
|
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|
Solving a complex number inequality involving absolute values. Here is the relevant paragraph (from "Complex numbers from A to Z" by Titu Andreescu and Dorin Andrica) :
Original question : How does $\left | 1+z \right |=t$ imply $\left | 1-z+z^2 \right |=\sqrt{\left | 7-2t^2 \right |}$?
(I checked for $z=i$ , it seems it is wrong ...)
EDIT: It seems it is indeed wrong. So , how can I prove the inequality?(perhaps even the lower and upper bounds need to be changed)
|
I propose the following. As $|z|=1$ we have $z^{-1}=\bar{z}$. Then, defining $x=Re(z)$, you get :
$$|1-z+z^2|=|z| \times |z+z^{-1}-1| = |z+\bar{z}-1|=|2x-1|$$
Moreover, $|1+z|^2=(1+z)(1+\bar{z})=1+|z|^2+z+\bar{z})=2(1+x)$. Hence :
$$|1+z|+|1-z+z^2|=\sqrt{2(1+x)}+|2x-1|:=f(x)$$
For $x\in[-1,\frac{1}{2}]$, $f(x)=1+\sqrt{2(x+1)}-2x$, hence $f'(x)=\left(\frac{1}{\sqrt{2(x+1)}}-2 \right)$. This is positive from $x=-1$ to $x=-\frac{7}{8}$ and then negative till $x=\frac{1}{2}$.
Then, for $x\in[\frac{1}{2},1]$, $f(x)=\sqrt{2(x+1)}+2x-1$, so $f$ is increasing.
Thus, we have $min(f(-1),f(\frac{1}{2})) \leq f(x) \leq max\left(f(-\frac{7}{8}),f(1)\right)$. Yet $f(-1)=3$, $f(1)=3$.
Yet, $f(-1)=f(1)=3$, $f(\frac{1}{2})=\sqrt{3}$ and $f(-\frac{7}{8})=\frac{13}{4}$.
So $\boxed{\sqrt{3} \leq |1+z|+|1-z+z^2| \leq \frac{13}{4}}$, which is not at all the wanted results. ^^ I'm going to do some numeric tests.
Edit : so, after a few test with mathematica, if we take $z=\frac{1+i\sqrt{3}}{2}$ (corresponding to my minimum $x=\frac{1}{2}$), we have indeed the value $\sqrt{3}$, which is smaller than $\sqrt{7/2}$, so I guess there is a problem in the exercise.
|
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|
Divisibility of a polynomial by another polynomial I have this question:
Find all numbers $n\geq 1$ for which the polynomial $x^{n+1}+x^n+1$ is divisible by $x^2-x+1$. How do I even begin?
So far I have that $x^{n+1}+x^n+1 = x^{n-1}(x^2-x+1)+2x^n-x^{n-1}+1,$ and so the problem is equivalent to finding $n$ such that $2x^n-x^{n-1}+1$ is divisible by $x^2-x+1.$
A solution that I found goes as follows (but I don't understand it!):
Assume that $x^n+1=(x^m-x+1)Q(x).$ The polynomial $x^m-x+1$ has a real root in the interval $(0,1)$ but $x^n+1$ has no positive real roots. So, no such pairs $m,n$ exist.
Any help?
|
Let $\omega = e^{\frac{i\pi}{3}}$, then $\omega^2 - \omega + 1 = 0$.
If there is a $n$ such that $x^2 - x + 1 $ divides $P_n(x) = x^{n+1} + x^n + 1$, then
$$ 0 = P_n(\omega) = \omega^n(\omega+1)+1.$$
$$ \omega^n(\omega+1) = -1.$$
$\omega+1 = \sqrt{3}e^{\frac{i\pi}{6}}$ is a vector of length $\sqrt{3}$. After a rotation it becomes $-1$, a vector of length $1$.
This is absurd.
|
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|
needs solution of the equation ${(2+{3}^{1/2}})^{x/2}$ + ${(2-{3}^{1/2}})^{x/2}$=$2^x$ $$\left(2+{3}^{1/2}\right)^{x/2} + \left(2-{3}^{1/2}\right)^{x/2} = 2^x.$$
Clearly $x = 2$ is a solution. i need others if there is any. Please help.
|
square this equation we have
$$(2+\sqrt{3})^x+(2-\sqrt{3})^x+2=4^x$$
so
$$\left(\dfrac{2+\sqrt{3}}{4}\right)^x+\left(\dfrac{2-\sqrt{3}}{4}\right)^x+\dfrac{2}{4^x}=1$$
It is clear
$$f(x)=\left(\dfrac{2+\sqrt{3}}{4}\right)^x+\left(\dfrac{2-\sqrt{3}}{4}\right)^x+\dfrac{2}{4^x}$$ is decreasing,because use $y=a^x,0<a<1$ is decreasing
and Note
$$f(2)=1$$
so
$$x=2$$
or consider
$$f(x)=\left(\dfrac{\sqrt{2+\sqrt{3}}}{2}\right)^x+\left(\dfrac{\sqrt{2-\sqrt{3}}}{2}\right)^x$$
|
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|
Two different trigonometric identities giving two different solutions Using two different sum-difference trigonometric identities gives two different results in a task where the choice of identity seemed unimportant. The task goes as following:
Given $\cos 2x =-\frac {63}{65} $ , $\cos y=\frac {7} {\sqrt{130}}$, under the condition $0<x,y<\frac {\pi}{2}$, calculate $x+y$.
The first couple of steps are the same: finding $\sin$ and $\cos$ values for both $x$ and $y$.
From $\cos 2x$ we have:
\begin{align*}
\cos 2x &=-\frac {63}{65} \\
\cos2x &= \cos^2x-\sin^2x = \cos^2x - (1-\cos^2x)=2\cos^2x -1 \\
2\cos^2x -1 &= -\frac {63}{65} \\
2\cos^2x &=\frac {-63+65}{65} \\
\cos^2x &=\frac {1}{65} \\
\cos x &=\frac {1} {\sqrt{65}}
\end{align*}
(taking only the positive value of $\cos x$ because $\cos x$ is always positive under the given domain)
\begin{align*}
\sin^2x &=1-\frac {1}{65} \\
\sin^2x &=\frac {64}{65} \\
\sin x &=\frac {8} {\sqrt{65}}
\end{align*}
(again, only positive value)
From $\cos y$ we have:
\begin{align*}
\cos y &=\frac {7} {\sqrt{130}} \\
\cos^2y &=\frac {49} {130} \\
\sin^2y &=1-\frac {49} {130} \\
\sin^2y &=\frac {81} {130} \\
\sin y &=\frac {9} {\sqrt{130}}
\end{align*}
Now that we've gathered necessary information, we proceed to calculate value of some trigonometric function of $x+y$, hoping we will get some basic angle:
sin(x+y):
\begin{align*}
\sin(x+y) &=\sin x \cos y + \sin y \cos x =\frac {8} {\sqrt{65}} \frac {7} {\sqrt{130}} + \frac {9} {\sqrt{130}}\frac {1} {\sqrt{65}} \\
\sin(x+y) &=\frac {65} {\sqrt{65}\sqrt{130}} \\
\sin(x+y) &=\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {\pi}{4}+2k{\pi}$ OR $x+y =\frac {3\pi}{4}+2k{\pi}$
Since $x$ and $y$ are in the first quadrant, their sum must lie in first or second quadrant.
Solutions are:
$x+y= \{\frac {\pi}{4}, \frac {3\pi}{4} \}$
cos(x+y):
\begin{align*}
\cos(x+y) &= \cos x \cos y - \sin x \sin y =\frac {1} {\sqrt{65}} \frac {7} {\sqrt{130}} - \frac {8} {\sqrt{65}}\frac {9} {\sqrt{130}} \\
\cos(x+y) &=-\frac {65} {\sqrt{65}\sqrt{130}} \\
\cos(x+y) &=-\frac {\sqrt{2}}{2}
\end{align*}
Thus,
$x+y =\frac {3\pi}{4}+2k{\pi} $ OR $x+y =\frac {5\pi}{4}+2k{\pi}$
Now we can only have one solution: $x+y=\{\frac {3\pi}{4}\}$
Similar happens with $\cos(x-y)$.
My question is: why do these two formulas give two different solutions? General insight would be great, since I found a lot of examples with similar problems. Thank you in advance.
|
To know which quadrant $x+y$ is in, you need to know the signs of both $\sin(x+y)$ and $\cos(x+y)$ In your case, $\sin(x+y) \gt 0$ and $\cos(x+y) \lt 0$. This puts $x+y$ in the second quadrant. So the solution is the one angle $x+y = \dfrac{3 \pi}4$.
|
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|
Partial fraction of $\frac 1{x^6+1}$ Can someone please help me find the partial fraction of $$1\over{x^6+1}$$
?
I know the general method of how to find the partial fraction of functions but this seems a special case to me..
|
For starters, write $x^6+1 = (x^2+1)(x^4-x^2+1)$. If you are allowing complex roots you can rewrite $x^2+1 = (x-i)(x+i)$. Then you can continue to factor $x^4-x^2+1$.
The way I came up with this initial factorization is by realizing that $\pm i$ is a root of $x^6+1$, which means $(x-i)$ and $(x+i)$ divide $x^6+1$, which means so does $(x-i)(x+i) = x^2+1$. Then I used polynomial division to divide $x^6+1$ by $x^2+1$ which left me with the quantity $x^4-x^2+1$. Hence, $x^6+1 = (x^2+1)(x^4-x^2+1)$. You can find the roots of $x^4-x^2+1$ by treating it as a quadratic in terms of $x^2$ (that is, make the substitution $x^2 = y$ and solving the quadratic $y^2-y+1=0$)
|
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|
Proving that $\frac{1}{a^2}+\frac{1}{b^2} \geq \frac{8}{(a+b)^2}$ for $a,b>0$ I found something that I'm not quite sure about when trying to prove this inequality.
I've proven that
$$\dfrac{1}{a}+\dfrac{1}{b}\geq \dfrac{4}{a+b}$$
already. My idea now is to replace $a$ with $a^2$ and $b^2$, so we now have
$$\dfrac{1}{a^2}+\dfrac{1}{b^2}\geq \dfrac{4}{a^2+b^2}$$
So to prove the required result, we just need to show that
$$\dfrac{4}{a^2+b^2} \geq \dfrac{8}{(a+b)^2}$$
or equivalently
$$(a+b)^2 \geq 2(a^2+b^2)$$
But this inequality cannot be true since the pair $(a,b)=(1,2)$ doesn't work. If anything, the reverse is always true!
What have I done wrong here?
|
Use Holder
$$\left(\dfrac{1}{a^2}+\dfrac{1}{b^2}\right)(a+b)(a+b)\ge (1+1)^3=8$$
|
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|
Compute definite integral Question: Compute
$$\int_0^1 \frac{\sqrt{x-x^2}}{x+2}dx.$$
Attempt: I've tried various substitutions with no success. Looked for a possible contour integration by converting this into a rational function of $\sin\theta$ and $\cos \theta$.
|
Letting $x=u^2, dx=2udu$ gives $\displaystyle2\int_0^1\frac{u^2\sqrt{1-u^2}}{u^2+2}du$; then letting $u=\sin\theta, du=\cos\theta d\theta$ gives
$\displaystyle2\int_0^{\pi/2}\frac{\sin^2\theta\cos^2\theta}{\sin^2\theta+2}d\theta=2\int_0^{\pi/2}\frac{\tan^2\theta\sec^2\theta}{(\sec^4\theta)(3\tan^2\theta+2)}d\theta=2\int_0^{\infty}\frac{t^2}{(t^2+1)^2(3t^2+2)}dt$ [$t=\tan\theta$]
$\displaystyle=2\int_0^{\infty}\left(\frac{2}{t^2+1}+\frac{1}{(t^2+1)^2}-\frac{6}{3t^2+2}\right)du=2\left[\frac{5}{2}\tan^{-1}t+\frac{t}{2(t^2+1)}-\sqrt{6}\tan^{-1}\frac{\sqrt{6}t}{2}\right]_0^{\infty}$
$\displaystyle=5\cdot\frac{\pi}{2}-2\sqrt{6}\cdot\frac{\pi}{2}=\big(5-2\sqrt{6}\big)\frac{\pi}{2}$.
|
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|
Prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$? I have to prove that $2^{15}-1$ is divided by $11\cdot31\cdot61$.
I have proven using congruencies that $2^{15}-1$ is divided by $31$. However we have
$$2^5\equiv 10 \mod{11}$$
$$2^{15}\equiv 10^3=1000\equiv 10 \pmod {11}$$
Therefore
$$2^{15}-1\equiv 9 \pmod{11}.$$
So it is impossible to prove!!
|
You are correct: the factorization you are given is wrong.
Since $N=2^{15}-1=(2^5)^3-1=(2^5-1)((2^5)^2+2^5+1)$ you get
$$
2^{15}-1=31\cdot(1024+32+1)=31\cdot 1057
$$
However, you can also use
$$
2^{15}-1=(2^3)^5-1=(2^3-1)((2^3)^4+(2^3)^2+2^3+1)
$$
so $N$ is also divisible by $7$. Divide $1057$ by $7$ to get
$$
N=31\cdot 7\cdot 151
$$
(and $151$ is prime).
|
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|
Prove this inequality $\frac{1}{1+a}+\frac{2}{1+a+b}<\sqrt{\frac{1}{a}+\frac{1}{b}}$
Let $a,b>0$ show that
$$\dfrac{1}{1+a}+\dfrac{2}{1+a+b}<\sqrt{\dfrac{1}{a}+\dfrac{1}{b}}$$
It suffices to show that
$$\dfrac{(3a+b+3)^2}{((1+a)(1+a+b))^2}<\dfrac{a+b}{ab}$$
or
$$(a+b)[(1+a)(1+a+b)]^2>ab(3a+b+3)^2$$
this idea can't solve it to me,are we aware of an elementary way of proving that? Thanks in advance.
|
Use Cauchy-Schwarz inequality we have
$$\left(\dfrac{1}{1+a}+\dfrac{2}{1+a+b}\right)^2\le\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}\right)$$
so suffices to show that
$$\dfrac{a}{(1+a)^2}+\dfrac{4b}{(1+a+b)^2}<1$$
since
$$\dfrac{4b}{(1+a+b)^2}<\dfrac{4b}{4(1+a)b}=\dfrac{1}{1+a}$$
it suffices to show that
$$\dfrac{a}{(1+a)^2}+\dfrac{1}{1+a}<1$$
$$\Longleftrightarrow a+1+a<(1+a)^2\Longleftrightarrow a^2>0$$ it is clear
|
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|
$(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$? The question given is
Show that $(x+y+z)^3-(y+z-x)^3-(z+x-y)^3-(x+y-z)^3=24xyz$.
What I tried is suppose $a=(y+z-x),\ b=(z+x-y)$ and $c=(x+y-z)$ and then noted that $a+b+c=x+y+z$. So the question statement reduced to $(a+b+c)^3-(a^3+b^3+c^3)$. Then I tried to invoke the identity $(a^3+b^3+c^3-3abc)=(a+b+c)(a^2+b^2+c^2-ab-bc-ac)$ by adding and subtracting $3abc$ in the question statement. After doing all this when I substituted back the values of $a,b$ and $c$, I ended up with the initial question statement.
Any hints will be appreciated.
|
Put $z = 0$, the LHS vanishes. So $z$ must be a factor. By symmetry $xyz$ must be a factor. As LHS is of third degree, the only other factor must be a constant, try $x=y=z=1$ to get that...
|
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|
$\operatorname{lcm}(2x, 3y-x, 3y+x)$ Problem
Find $\textrm{lcm} (2x, 3y-x, 3y+x)$, where $y > x > 0$ and $\gcd(x,y) = 1$.
Attempt
I noticed after some numerical calculation that the answer seems to depend on the parity of $x$ and $y$. However, I am basically at a loss of how to approach this. Any help would be appreciated.
|
You need to distinguish a few cases.
First note that if $3 \nmid x$, then $\gcd(3y - x, 3y + x)$ is either $1$ or $2$. Indeed, otherwise either a prime $p \neq 3$ or $4$ would divide both $6y$ and $2x$, but $x$ and $y$ are coprime by hypothesis. Furthermore, that $\gcd$ is $2$ if and only if both $x$ and $y$ are odd. Since no prime that divides $x$ can divide $3y-x$ or $3y+x$, again because $\gcd(x,y) = 1$, if $3 \nmid x$ we have
$$
\text{lcm}(2x, 3y-x, 3y+x) =
\begin{cases}
2x(3y-x)(3y+x) & \text{if } 2 \mid xy \\
\frac{1}{2}x(3y-x)(3y+x) & \text{otherwise}
\end{cases}
$$
Now suppose that $x = 3z$ for some $z \in \Bbb{Z}$. Since $\gcd(x,y) = 1$ no prime that divides $z$ can divide $y-z$ or $y+z$. Furthermore, just as above $\gcd(y-z,y+z)$ is $2$ if both $x$ and $y$ are odd, and it is $1$ otherwise. Thus if $3 \mid x$ we have
$$
\text{lcm}(2x, 3y-x, 3y+x) =
\begin{cases}
2x(y-z)(y+z) & \text{if } 2 \mid xy \\
\frac{1}{2}x(y-z)(y+z) & \text{otherwise}
\end{cases}
$$
|
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|
Hints on solving $y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=0$
$$y''-\frac{x}{x-1}y'+\frac{1}{x-1}y=0$$
Is there any simple method to solve this equation?
I need hints please $\color{red}{not}$ a full answer
|
Applying the idea in my comment:
$$0 = y'' - \frac{x-1+1}{x-1}y' + \frac{1}{x-1}y = y'' -\frac{1}{x-1}y' - y' + \frac{1}{x-1}y.$$
Multiplying by $\dfrac{1}{x-1}$, this becomes
$$ 0 = \frac{1}{x-1}y'' - \frac{1}{(x-1)^2}y' - \frac{1}{x-1}y' + \frac{1}{(x-1)^2}y.$$
We can realize the first two terms together as $\dfrac{d}{dx}\left(\dfrac{1}{x-1}y'\right)$ and the second two terms can similarly be realized as $-\dfrac{d}{dx}\left(\dfrac{1}{x-1}y\right).$ (This is just the method of integrating factors in disguise.) Putting these two together gives
$$ 0 = \frac{d}{dx}\left(\frac{1}{x-1}y'\right) - \frac{d}{dx}\left(\frac{1}{x-1}y\right) = \frac{d}{dx}\left(\frac{1}{x-1}(y'-y)\right).$$
That is to say that
$$ C = \frac{1}{x-1}(y'-y).$$
This can be solved by integrating factors as well. This method has the benefit of not doing a perhaps strange change of function but it does require a clever form of $1$.
|
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|
Is it true that $\sin x > \frac x{\sqrt {x^2+1}} , \forall x \in (0, \frac {\pi}2)$? Is it true that $$\sin x > \dfrac x{\sqrt {x^2+1}} , \forall x \in \left(0, \dfrac {\pi}2\right)$$ (I tried differentiating , but it's not coming , please help)
|
Square both sides (assume $x\in(0,\pi/2)$):
$$\sin x>\frac{x}{\sqrt{x^2+1}}\iff 1-\cos^2 x>1-\frac{1}{x^2+1}$$
$$\iff \cos ^2 x<\frac{1}{x^2+1}=\frac{\sin^2 x+\cos^2 x}{x^2+1}$$
$$\iff x^2\cos^2 x+\cos^2 x<\sin^2 x + \cos^2 x$$
$$\iff \tan^2 x> x^2\iff \tan x>x,$$
which is true ($(\tan x-x)'>0$ for $x\in(0,\pi/2)$ and $\tan(0)-0=0$).
|
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Can we obtain $f(y+x)=y+f(x)$ from $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$?
$\mathbb Z^+$ is the set of positive integers. Find all functions $f:\mathbb{Z}^+\rightarrow \mathbb{Z}^+$ such that
$$f(m^2+f(n))=f(m)^2+n\quad(\clubsuit)$$
Let $P(x,y)$ be the assertion: $f(x^2+f(y))=f(x)^2+y \; \forall x,y \in \mathbb{Z}^+.$
$P(x,x)$ gives us $f(x^2+f(x))=f(x)^2+x$.
$P(x,x^2+f(x))$ gives us $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$.
Can we obtain $f(y+x)=y+f(x)$ from $f(x^2+f(x)^2+x)=f(x)^2+x^2+f(x)$ ?
|
Here is an incomplete idea to actually solve the quoted problem (as opposed to your specific question about $f(x+y)=y+f(x)$). This idea might go nowhere. Define $a$ to be $f(1)$. It's easy to show using induction that if $a=1$, then $f$ is the identity function.
If $a=2$, we can arrive at a contradiction:
$$
\begin{align}
f(1^2+f(1))&=f(1)^2+1\\
f(1+2)&=4+1\\
f(3)&=5\\
\implies f(1^2+f(3))&=f(1)^2+3\\
f(1+5)&=4+3\\
f(6)&=7\\
\implies f(1^2+f(6))&=f(1)^2+6\\
f(1+7)&=4+6\\
f(8)&=10\\
\implies f(1^2+f(8))&=f(1)^2+8\\
f(1+10)&=4+8\\
f(11)&=12
\end{align}
$$
But
$$
\begin{align}
f(3^2+f(1))&=f(3)^2+1\\
f(9+2)&=25+1\\
f(11)&=26
\end{align}
$$
This contradiction rules out $f(1)$ being $2$. Perhaps you can find a pattern to generalize and show that $f(1)$ must be $1$, which would prove that $f$ must be the identity. (I had no luck finding such a pattern though.)
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1350094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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|
Evaluating a function at a point where $x =$ matrix. Given $A=\left(
\begin{array} {lcr}
1 & -1\\
2 & 3
\end{array}
\right)$
and $f(x)=x^2-3x+3$ calculate $f(A)$.
I tried to consider the constant $3$ as $3$ times the identity matrix ($3I$) but the answer is wrong. Appreciate any ideas.
|
We find that
$$
A^2 = \pmatrix{
-1 & -4\\
8 & 7}, \quad
-3A =
\pmatrix{
-3&3\\
-6&-9}, \quad
3I = \pmatrix{
3&0\\0&3
}
$$
Adding these together, we find
$$
f(A) = \pmatrix{
-1&-1\\
2&1
}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1350275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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|
Basis for a eigenspace (multiple choice problem) The following (multiple choice) problem is from a test review.
For the given matrix $A$, find a basis for the corresponding eigenspace for the given eigenvalue.
$$A = \begin{bmatrix}1 & 6 & 6 \\ 6 & 1 & -6 \\ -6 & 6 & 13\end{bmatrix};\quad \lambda = 7.$$
The four given options are
A) $\left\{ \begin{bmatrix}1 \\ 0 \\ 1 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix} \right\}$
B) $\left\{ \begin{bmatrix}0 \\ 1 \\ -1 \end{bmatrix} \right\}$
C) $\left\{ \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix} \right\}$
D) $\left\{ \begin{bmatrix}1 \\ 0 \\ -1 \end{bmatrix}, \begin{bmatrix}0 \\ 1 \\ 1 \end{bmatrix} \right\}$
The correct answer is A.
My answer is not among the answer choices. What am I doing wrong?
First, reduce $(A-\lambda I)$:
$$\begin{align}
A - \lambda I &=
\begin{bmatrix}1 & 6 & 6 \\ 6 & 1 & -6 \\ -6 & 6 & 13\end{bmatrix} - \begin{bmatrix}7 & 0 & 0 \\ 0 & 7 & 0 \\ 0 & 0 & 7\end{bmatrix} \\
&= \begin{bmatrix}-6 & 6 & 6 \\ 6 & -6 & -6 \\ -6 & 6 & 6\end{bmatrix} \\
&\sim \begin{bmatrix}-1 & 1 & 1 \\ 1 & -1 & -1 \\ -1 & 1 & 1\end{bmatrix} \\
&\sim \begin{bmatrix}1 & -1 & -1 \\ 0 & 0 & 0 \\ 0 & 0 & 0\end{bmatrix}
\end{align}$$
Parametric form solving for $(A-\lambda I) = \vec 0$:
$$\begin{align}
x_1 &= x_2 + x_3 \\
x_2 &= x_2 \\
x_3 &= x_3
\end{align}$$
has solutions $x_2(1,1,0) + x_3(1,0,1)$ for all $x_2 , x_3$.
So a basis for the null-space of $(A-\lambda I)$ consists of the vectors $\begin{bmatrix}1 \\ 1 \\ 0\end{bmatrix}$ and $\begin{bmatrix}1 \\0 \\ 1\end{bmatrix}$.
So, am I wrong or is the solution wrong?
|
As @GTonyJacobs points out in the comments, you are correct but the key is also correct. Vector spaces can have many different bases!
Note that you have proven that $A-7I$ has rank one so the rank-nullity theorem implies that the nullspace of $A-7I$ has dimension two. This allows us to immediately eliminate choices (B) and (C).
Now, note that
\begin{align*}
A\begin{bmatrix}1\\0\\1\end{bmatrix}
&= \begin{bmatrix}7\\0\\7\end{bmatrix}
= 7\begin{bmatrix}1\\0\\1\end{bmatrix}
&
A\begin{bmatrix}0\\1\\-1\end{bmatrix}
&=\begin{bmatrix}0\\7\\-7\end{bmatrix}
=7\begin{bmatrix}0\\1\\-1\end{bmatrix}
\end{align*}
This proves that choice (A) is correct.
Also, note that choice (D) is incorrect since
$$
A\begin{bmatrix}1\\0\\-1\end{bmatrix}
=\begin{bmatrix}-5\\12\\-19\end{bmatrix}
\neq7\begin{bmatrix}1\\0\\-1\end{bmatrix}
$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1350733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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|
How to find the generating function from recurrence relation $t_n=(1+c~q^{n-1})~p~t_{n-1}+a +b+nbq, ~~n\ge 2$? How to find the generating function $T(z)=\sum_{n=0}^{\infty} t_n~z^n$ from the recurrence relation $$t_n=(1+c~q^{n-1})~p~t_{n-1}+a +b+nbq,\qquad n\ge 2.$$ given that $$t_1=b+(1+c~p)(a~q^{-1}+b),$$ and $$t_0=\frac{a}{q}+b,$$ where $a, b, c$ are constants and $p+q=1.$
|
Consider $t_{n+1} = (1+c~q^{n})~p~t_{n} + a + b + (n+1)bq$ for which
\begin{align}
\sum_{n=0}^{\infty} t_{n+1} \, x^{n} &= p \, T(x) + c \, p \, T(q x) + \frac{a + b}{1-x} + b q \, \partial_{x} \left( \sum_{n=0}^{\infty} x^{n} \right) \\
\frac{T(x) - t_{0}}{x} &= p \, T(x) + c \, p \, T(q x) + \frac{a + b}{1-x} + b q \, \partial_{x} \frac{1}{1-x}
\end{align}
or
\begin{align}
c p x \, T(q x) + (p x -1) T(x) + \frac{(a + b) x}{1- x} + \frac{b q \, x}{(1-x)^{2}} + t_{0} = 0
\end{align}
where $T(x) = \sum_{n=0}^{\infty} t_{n} \, x^{n}$.
By working out term by term it can be determined that
\begin{align}
t_{1} &= p(1+c) \, t_{0} + (bq + a + b) \\
t_{2} &= p^{2} \, (1+c)(1+cq) \, t_{0} + (bq + a + b)(p(1+cq) + 1) + bq \\
\cdots &= \cdots
\end{align}
where the general form is
\begin{align}
t_{n} = t_{0} \, p^{n} \, \prod_{k=1}^{n} \{ 1 + c q^{k-1} \} + (bq + a + b) \, \sum_{r=1}^{n} p^{n-r} \, \left(\prod_{k=r+1}^{n} \{ 1 + c q^{k-1} \}\right) + bq \, \sum_{r=2}^{n} a_{r} \, p^{n-r} \, \left(\prod_{k=r+1}^{n} \{1 + c q^{k-1} \} \right)
\end{align}
where the terms indicate that $a_{r}$ may be $r-1$ (based upon $n=1,2,3,4$).
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1352249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Finding subgroups of $\mathbb{Z}_{13}^*$
I need to find all nontrivial subgroups of $G:=\mathbb{Z}_{13}^*$ (with multiplication without zero)
My attempt:
$G$ is cyclic so the order of subgroup of $G$ must be $2,3,4,6$
Now to look for $g\in G$ such that $g^2=e,g^3=e,g^4=e,g^6=e$
$$\begin{align}
&12^1=12\mod 13\\
&12^2=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{12,1\}}$ is a subgroup of order $2$
$$\begin{align}
&3^1=3\mod 13\\
&3^2=9\mod 13\\
&3^3=1\mod 13\\
\end{align}$$
$$\begin{align}
&9^1=9\mod 13\\
&9^2=3\mod 13\\
&9^3=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{9,3,1\}}$ is a subgroup of order $3$
$$\begin{align}
&5^1=5\mod 13\\
&5^2=12\mod 13\\
&5^3=8\mod 13\\
&5^4=1\mod 13\\
\end{align}$$
$$\begin{align}
&8^1=8\mod 13\\
&8^2=12\mod 13\\
&8^3=5\mod 13\\
&8^4=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{1,5,12,8\}}$ is a subgroup of order $4$
$$\begin{align}
&4^1=4\mod 13\\
&4^2=3\mod 13\\
&4^3=12\mod 13\\
&4^4=9\mod 13\\
&4^5=10\mod 13\\
&4^6=1\mod 13\\
\end{align}$$
$$\begin{align}
&10^1=10\mod 13\\
&10^2=9\mod 13\\
&10^3=12\mod 13\\
&10^4=3\mod 13\\
&10^5=4\mod 13\\
&10^6=1\mod 13\\
\end{align}$$
$\Rightarrow \color{blue}{\{4,3,12,9,10,1\}}$ is a subgroup of order $6$
Is it correct? is there any easier method?
|
Your solution is right. As, I am noting that(may be i'm wrong), you are applying that "A cyclic subgroup of order $4$ must contain $2$ elements of order $4$ and $1$ element of order $2$, and you searching those elements and listing them. You can reduce your calculation by searching one element of each order, and then you can generate your required subgroups, e.g. $5$ is element of order $4$ so,
$$<5>=\{1,5,8,12\}$$ is subgroup of order $4$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1352348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
Can someone please check my work?: $\cos^2(x)=1-\sin(x)$ $$\begin{align}\cos^2(x)&=1-\sin(x)\\
1-\sin^2(x)&=1-\sin(x)\\
(1-\sin x)(1+\sin x)&= 1-\sin(x)
\end{align}$$
divide both sides by $1 - \sin(x)$
End up with $1 + \sin(x)$
The answer is supposed to be in radians between $0$ and $2 \pi$.
So I get $1+\sin(x)=0$
$$\sin(x)=-1 = -90\text{ degrees } = -\pi/2 \text{ or }3\pi/2$$
|
Hey there you are partially correct , one thing you missed in your solution
Follow my solution
$\cos^2x=1-\sin x$
$\Longrightarrow (1-\sin x)(1+\sin x)=1-\sin x$
From this equation we have
$1- \sin x=0$ and $1+ \sin x=0$
$\sin x = 1$ and $\sin x = -1$
$\sin x= \sin(\frac\pi2)$ and $\sin x= \sin(-\frac\pi2)$
We have if $\sin x= \sin\alpha$ then $x= n\pi+(-1)^n\alpha$
Use this result for both equations and find solution.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1353326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Determining $\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$
$$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$$
Attempt:
Simplification of the root factor:
$$\sqrt{x^4+2x^3-x^2+2x+1}=\frac{1}{x}\sqrt{\left(x^2+\frac{1}{x^2}\right)+2\left(x+\frac{1}{x}\right)-1}.$$
Arranging the rest of the factors as:
$$\frac{x-1}{x^2(x+1)}=\frac{x^2-1}{x^2(x+1)^2}$$
Now I did the following substitution: let $(x+\frac{1}{x})=t$, so $(1-\frac{1}{x^2})dx=dt$
Arranging the integral: $$I=\int \frac{\sqrt{t^2+2t-3}}{t+2}dt.$$
But what to do next?
I tried integration by parts for this but couldn't simplify my result.
|
$\bf{My\; Solution::}$ Given $$\displaystyle \int\frac{(x-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx = \int\frac{(x^2-1)\cdot \sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x^2+2x+1)}dx$$
Above we multiply both $\bf{N_{r}}$ and $\bf{D_{r}}$ by $(x+1).$
$$\displaystyle = \int\frac{\left(1-\frac {1}{x^2}\right)\cdot \sqrt{x^2\cdot \left(x^2+2x-1+\frac{2}{x}+\frac{1}{x^2}\right)}}{ \left(x+2+\frac{1}{x}\right)}dx$$
Now Let $ \displaystyle \left(x+\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1-\frac{1}{x^2}\right)dx = dt$
So Integral $$\displaystyle = \int\frac{\sqrt{t^2+2t-3}}{t+2}dt = \frac{t^2+2t-3}{(t+2)\sqrt{t^2+2t-3}}dt = \int\frac{t(t+2)-3}{(t+2)\sqrt{t^2+2t-3}}dt$$
So Integral $$\displaystyle = \underbrace{\int\frac{t}{\sqrt{t^2+2t-3}}dt}_{I} - \underbrace{\int\frac{3}{(t+2)}\cdot \frac{1}{\sqrt{t^2+2t-3}}dt}_{J}..........\color{\red}\checkmark.$$
So $$\displaystyle I = \int\frac{t}{\sqrt{t^2+2t-3}}dt = \int\frac{(t+1)-1}{\sqrt{(t-1)^2-2^2}} = \int\frac{(t-1)}{\sqrt{(t-1)^2-2^2}}-\int\frac{1}{\sqrt{(t-1)^2-2^2}}dt$$
Now Let $(t-1) = z\;\;,$ Then $dt = dz$
So $$\displaystyle I = \int\frac{z}{\sqrt{z^2-2^2}}dz-\int\frac{1}{\sqrt{z^2-2^2}}dz = \sqrt{z^2-4}-\ln \left|(t+1)+\sqrt{t^2+2t-3}\right|$$
Now $$\displaystyle J = 3\int\frac{1}{(t+2)\sqrt{t^2+2t-3}}dt = 3\int\frac{1}{(t+2)\sqrt{(t+2)^2-2(t+2)+1-4}}$$
Now Let $(t+2) = u\;,$ Then $dt = du$ and Integral $$\displaystyle = 3\int\frac{1}{u\sqrt{u^2-2u+1-4}}=3\int\frac{1}{u\sqrt{(u-1)^2-4}}du$$
Now $\displaystyle (u-1) = 2\sec \theta \;, $ Then $du= 2\sec \theta \cdot \tan \theta.$
Now after that You can Solve It.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inequality involving number of binary Lyndon words of length $n$ and $n+1$
Let $f(n)$ be the number of binary Lyndon words of length $n$. This sequence is given by OEIS entry A001037. Is it true that $2f(n) \ge f(n+1)$ for all positive $n$?
I have found a general formula to calculate $f(n)$:
$$
f(n) = \frac{1}{n} \sum_{d \mid n} \mu(n/d) 2^d.
$$
Symmetry also allows us to rewrite it another way:
$$
f(n) = \frac{1}{n} \sum_{d \mid n} \mu(d) 2^{n/d}.
$$
Here $\mu(x)$ is the Möbius function, and $n$ is a positive integer (the case $n=0$ can be handled separately). It is possible to substitute these identities into the inequality above and perform some simplifications, but it does not seem to go anywhere. Any help is appreciated.
EDIT: My original formula was incorrect, I forgot the factor of $1/n$ in front of the summation.
|
Step 1: For $n \in \mathbb{N}_+$,$$
\frac{2^n}{n} - \frac{2^{\frac{n}{2}}}{2} \leqslant f(n) \leqslant \frac{2^n}{n} + \frac{2^{\frac{n}{6}}}{6}. \tag{1}
$$
Proof: For the upper bound,$$
n f(n) = \sum_{d \mid n} μ\left( \frac{n}{d} \right) 2^d \leqslant 2^n + \sum_{\substack{d \mid n,\ d < n\\μ(n / d) = 1}} 2^d. \tag{2}
$$
If $d \mid n$, $d < n$ and $μ\left( \dfrac{n}{d} \right) = 1$, then $\dfrac{n}{d}$ has at least two distinct prime divisors, which implies $\dfrac{n}{d} \geqslant 6$, i.e. $d \leqslant \dfrac{n}{6}$. Thus $(2) \leqslant 2^n + \dfrac{n}{6} 2^{\frac{n}{6}} \Rightarrow f(n) \leqslant \dfrac{2^n}{n} + \dfrac{2^{\frac{n}{6}}}{6}$.
For the lower bound,$$
n f(n) = \sum_{d \mid n} μ\left( \frac{n}{d} \right) 2^d \geqslant 2^n - \sum_{\substack{d \mid n,\ d < n\\μ(n / d) = -1}} 2^d. \tag{3}
$$
If $d \mid n$, $d < n$ and $μ\left( \dfrac{n}{d} \right) = -1$, then $\dfrac{n}{d}$ has at least one prime divisor, which implies $\dfrac{n}{d} \geqslant 2$, i.e. $d \leqslant \dfrac{n}{2}$. Thus $(3) \geqslant 2^n - \dfrac{n}{2} 2^{\frac{n}{2}} \Rightarrow f(n) \geqslant \dfrac{2^n}{n} - \dfrac{2^{\frac{n}{2}}}{2}$. Therefore, (1) holds.
Step 2: $2f(n) \geqslant f(n + 1)$ holds for $n \geqslant 15$.
Proof: By (1. 1), it suffices to prove that $2 \left( \dfrac{2^n}{n} - \dfrac{2^{\frac{n}{2}}}{2} \right) \geqslant \dfrac{2^{n + 1}}{n + 1} + \dfrac{2^{\frac{n + 1}{6}}}{6}$. Because$$
2 \left( \frac{2^n}{n} - \frac{2^{\frac{n}{2}}}{2} \right) \geqslant \frac{2^{n + 1}}{n + 1} + \frac{2^{\frac{n + 1}{6}}}{6} \Longleftrightarrow \frac{2^{n + 1}}{n(n + 1)} \geqslant 2^{\frac{n}{2}} + \frac{2^{\frac{n + 1}{6}}}{6},
$$
and $2^{\frac{n + 1}{4}} \geqslant n + 1$ for $n \geqslant 15$, then$$
\frac{2^{n + 1}}{n(n + 1)} \geqslant 2^{\frac{n + 1}{2}} = 2^{\frac{n}{2}} + (\sqrt{2} - 1) 2^{\frac{n}{2}} \geqslant 2^{\frac{n}{2}} + \frac{2^{\frac{n}{2}}}{6} \geqslant 2^{\frac{n}{2}} + \frac{2^{\frac{n + 1}{6}}}{6}.
$$
Therefore, $2f(n) \geqslant f(n + 1)$ holds for $n \geqslant 15$.
Step 3: $2f(n) \geqslant f(n + 1)$ holds for $0 \leqslant n \leqslant 14$.
Proof: Looking up in A001037 verifies this.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Integral calculation: $\int_0^1 (x-\sqrt{x}+\sqrt[3]{x}-\sqrt[4]{x}+\cdots) \, dx$ $$K=\int_0^1 (x-\sqrt{x}+\sqrt[3]{x}-\sqrt[4]{x}+\cdots) \, dx$$
I'm looking for an exact solution for this integral. Thank you very much.
|
Given
$$
K = \int_0^1 ( x - \sqrt{x} + \sqrt[3]{x} - \sqrt[4]{x} + \cdots )
$$
Write this as
$$
K = - \int_0^1 \sum_{k=1}^\infty (-1)^k x^{1/k}
$$
Thus
$$
K = - \sum_{k=1}^\infty (-1)^k \int_0^1 x^{1/k} = \sum_{k=1}^\infty (-1)^k \left[ \frac{1}{1/k+1} x^{1/k+1} \right]_0^1
$$
So
$$
K = - \sum_{k=1}^\infty (-1)^k \frac{k}{1+k}
$$
We obtain
$$
K = \frac{1}{2} - \frac{2}{3} + \frac{3}{4} - \frac{4}{5} + \cdots
$$
or
$$
K = - \frac{1}{2 \cdot 3} - \frac{1}{4 \cdot 5} - \frac{1}{6 \cdot 7} - \cdots
$$
or
$$
K = - \frac{1}{6} - \frac{1}{20} - \frac{1}{42} - \cdots
$$
Which can be written as $\ln(2) - 1$, so
$$
\int_0^1 ( x - \sqrt{x} + \sqrt[3]{x} - \sqrt[4]{x} + \cdots ) = \ln(2) - 1
$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving $\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$ The problem is to prove that
$$\sin 20^\circ \, \sin 40^\circ \, \sin 60^\circ \, \sin 80^\circ =\frac{3}{16}$$
All my attempts were to get them in $\sin (2A)$ form after eliminating $\sin 60^\circ$ in both sides. Unfortunately, all these attempts were futile.
Any hints are welcomed.
|
We have:
$\sin20^{\circ} = \sin(30^\circ - 10^\circ)= \frac{1}{2}\cos10^\circ - \frac{\sqrt{3}}{2}\sin10^\circ$
$\sin40^{\circ} = \sin(30^\circ + 10^\circ)= \frac{1}{2}\cos10^\circ + \frac{\sqrt{3}}{2}\sin10^\circ$
$\sin 60^\circ = \frac{\sqrt{3}}{2}$
$\sin80^\circ =\cos 10^\circ$
Therefore,
$A = \sin20^\circ \sin40^\circ \sin60^\circ \sin80^\circ = \frac{\sqrt{3}}{2} \left((\cos10^\circ )^2 - \frac{3}{4} \right)\cos10^\circ$
Moreover :
$ \cos30^\circ = 4(\cos10^\circ )^3 - 3\cos 10^\circ$, from this cubic equation we find out the value of $cos10^\circ$, and then plug it to the expression of $A$ we will get the answer.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proving the integral series $\int _0^1\left(1-x^2\right)^n\,dx=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}$
We have the series $\left(I_n\right)_{n\ge 1\:}$ where $$I_n=\int _0^1\left(1-x^2\right)^n\,dx.$$
Prove that $$I_n=\frac{2}{3}\cdot \frac{4}{5}\cdot\ldots\cdot \frac{2n}{2n+1}.$$
I tried to integrate that function for $n=1,2,3$ to see whether there's a pattern (recurrence relation), but I just can't figure out how to write the pattern and how to prove that statement. Must I use induction?
|
Integrate by parts to obtain:
$$I_n=\int_0^1 \left(1-x^2\right)^n\,dx=\left(x\left(1-x^2\right)^n\right|_0^1+2n\int_0^1 x^2\left(1-x^2\right)^{n-1}\,dx$$
$$\Rightarrow I_n=2n\int_0^1 \left(x^2-1+1\right)\left(1-x^2\right)^{n-1}\,dx$$
$$\Rightarrow I_n=2nI_{n-1}-2nI_n \Rightarrow I_n=\frac{2n}{2n+1}I_{n-1}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Prove that $\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=3/2$
Prove that $$\cos^2(\theta) + \cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})=\frac{3}{2}$$
I thought of rewriting $$\cos^2(\theta +120^{\circ}) + \cos^2(\theta-120^{\circ})$$ as $$\cos^2(90^{\circ}+ (\theta +30^{\circ})) + \cos^2(90^{\circ}-(\theta-30^{\circ}))$$ However I don't seem to get anywhere with this.
Unfortunately I don't know how to solve this question. I would be really grateful for any help or suggestions. Many thanks in advance!
|
$\cos^2(A-120^\circ)+\cos^2(A+120^\circ)=1+\cos^2(A-120^\circ)-\sin^2(A+120^\circ)$
and Using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$, $\cos^2(A-120^\circ)-\sin^2(A+120^\circ)=1+\cos(2A)\cos(240^\circ)=?$
$\cos(240^\circ)=\cos(180^\circ+60^\circ)=-\cos60^\circ=?$
Now use $\cos2A=2\cos^2A-1$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Polynomial division remainder Given the polynomials $f,g,h \in \mathbb{R}\left[X\right]$ with $$f=(x-1)^n-x^n+1$$ $$g=x^2-3x+2$$ $$h=x^2-x$$ where $n\ge3$
Find the remainder of dividing the polynomial f to g. Prove that if $n$ is odd, $h$ divides $f$ with no remainder.
I noticed that the three polynomials have a common root in $x=1$ meaning I can write all three of them like $(x+1)*p(x)$, thus having no remainder. I don't know if this helps me though. How do I solve this?
|
Use the remainder theorem: for any polynomial $p(x)$, the remainder when dividing it by $x - a$ is given by $p(a)$.
Notice that $f(1) = 0$, so $x - 1$ divides evenly into $f(x)$. Therefore, for some $p(x)$, we may write,
$$f(x) = (x - 1)p(x).$$
Further, since $f(2) = p(2)$ is the common remainder when dividing both $p(x)$ and $f(x)$ by $2$, we can write,
$$p(x) = q(x)(x - 2) + p(2) = q(x)(x - 2) + f(2),$$
for some $q(x)$. Therefore,
$$f(x) = (x - 1)(q(x)(x - 2) + f(2)) = (x^2 - 3x + 2)q(x) + f(2)(x - 1).$$
When dividing $f(x)$ by $x^2 - 3x + 2$, we therefore get a remainder of,
$$f(2)(x - 1) = (2 - 2^n)(x - 1).$$
The second question is far more simple. Since $x^2 - x = x(x - 1)$, we just need to individually verify that $x$ and $x - 1$ divide evenly into $f(x)$ when $n$ is even. We do this by verifying $f(0) = f(1) = 0$. Notice that $f(1) = 0$ still, regardless of the value of $n$, but,
$$f(0) = (0 - 1)^n - 0^n + 1 = (-1)^n + 1 = 0,$$
precisely when $n$ is odd, and we are done.
|
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|
Integrate $\int \frac{(x+1)}{(x^2+2)^2}dx$ This is the question I want to ask
Integrate $$\int \frac{(x+1)}{(x^2+2)^2}\,dx.$$
I tried it using algebric manipulation
Integrate $x/(x^2+2)^2+1/(x^2+2)^2$.
Then the latter part would not be solved.
|
$\bf{My\; Solution::}$ Let $$\displaystyle I = \int\frac{1}{(x^2+2)^2}dx$$
Let $x=\sqrt{2}\tan \theta\;,$ Then $dx = \sqrt{2}\sec^2 \theta d\theta$
So Integral $$\displaystyle I = \int\frac{\sqrt{2}\sec^2 \theta }{4\sec^4 \theta }d\theta = \frac{1}{2\sqrt{2}}\int \cos ^2\theta d\theta = \frac{1}{4\sqrt{2}}\int (1+\cos 2 \theta )d\theta$$
So $$\displaystyle I = \frac{1}{4\sqrt{2}}\left(\theta +\frac{\sin 2\theta }{2}\right)+\mathcal{C} = \frac{1}{4\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{4\sqrt{2}}\cdot \frac{\sqrt{2}x}{x^2+2}+\mathcal{C}$$
So $$\displaystyle I = \int \frac{1}{(x^2+2)^2}dx = \frac{1}{4\sqrt{2}}\tan^{-1}\left(\frac{x}{\sqrt{2}}\right)+\frac{1}{4}\cdot \frac{x}{x^2+2}+\mathcal{C}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Conditional expectation on Gaussian random variables If we suppose that the two independent random variables $X \sim \mathcal{N}(0,\sigma^2_x)$ and $N \sim \mathcal{N}(0,\sigma^2_n)$ and that $S = X + N$, how would I work out the conditional expectation $E[X\mid S=s]$?
|
Since $X$ and $N$ are independent and normal, any combinations of $X$ and $N$ are normal. As any combination of $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ is also a combination of $X$ and $N$, that is, any combinations of $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ are normal, then $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ are jointly normal. Moreover, note that
\begin{align*}
E\bigg((X+N)\Big( \frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N\Big) \bigg) &= 0.
\end{align*}
Then $X+N$ and $\frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N$ are independent. Consequently,
\begin{align*}
E\big( (X+N) \mid S\big) &= S,\\
E\bigg( \Big( \frac{1}{\sigma_x^2}X - \frac{1}{\sigma_n^2}N\Big) \mid S\bigg) &= 0.
\end{align*}
Solving this system, we obtain that
\begin{align}
E(X\mid S) = \frac{\sigma_x^2}{\sigma_x^2+\sigma_n^2}S.
\end{align}
In other words,
\begin{align}
E(X\mid S=s) = \frac{\sigma_x^2}{\sigma_x^2+\sigma_n^2}s.
\end{align}
|
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|
Definite Integration with Trigonometric Substitution I'm working on a question that involves using trigonometric substitution on a definite integral that will later use u substitution but I am not sure how to go ahead with this.
$$\int_1^2\frac1{x^2\sqrt{4x^2+9}}dx$$
My first step was to use $\sqrt{a^2+x^2}$ as $x=a\tan\theta$ to get...
$$2x=3\tan\theta :x=\frac32\tan\theta$$
$$dx=\frac32\sec^2\theta$$
Substituting:
$$\int\frac{\frac32\sec^2\theta}{\frac94\tan^2\theta\sqrt{9\tan^2\theta+9}}$$
The problem here is how do I change the limit it goes to?
$$\frac43=\tan\theta$$
and
$$\frac23=\tan\theta$$
Following DR.MV's answer so far..
$$\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec^2\theta}{\tan^2\theta\sqrt{9\sec^2\theta}}d\theta$$
$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\sec\theta}{\tan^2\theta}d\theta$$
$$=\frac29\int_{\tan^-1(\frac23)}^{\tan^-1(\frac43)}\frac{\cos\theta}{\sin^2\theta}d\theta$$
Now $u=\sin\theta$ so $du=\cos\theta d\theta$
$$=\frac29\int_{?}^{?}\frac{1}{u^2}du$$
This is where I am stuck now...
|
There was a typo in the current post. After enforcing the substitution $2x=3\tan \theta$, the integral ought to read
$$\begin{align}I&=\int_{\arctan(2/3)}^{\arctan(4/3)}\frac{\frac32 \sec^2\theta}{\frac94 \tan^2\theta\sqrt{9\tan^2\theta+9}}d\theta\\\\
&=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\frac{ \sec^2\theta}{ \tan^2\theta\,\sec \theta}d\theta\\\\
&=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\frac{ \sec^2\theta}{ \tan^2\theta\,\sec \theta}d\theta\\\\
&=\frac29 \int_{\arctan(2/3)}^{\arctan(4/3)}\cot \theta \csc \theta d\theta\\\\
&=\frac29 \left.(-\csc \theta)\right|_{\arctan(2/3)}^{\arctan(4/3)}\\\\
&=\frac{\sqrt{13}}{9}-\frac{5}{18}\end{align}$$
NOTES:
Remark 1: When making a substitution of variables in a definite integral, the limits of integration change accordingly. In this example, the substitution was $x=\frac32 \tan \theta$. When $x=1$ at the lower limit, $\tan \theta =\frac23\implies \theta =\arctan(2/3)$. Similarly, when $x=2$ at the upper limit, $\tan \theta =\frac43\implies \theta =\arctan(4/3)$.
Remark 2:
To evaluate $\sin (\arctan(2/3))$, we recall that the arctangent is an angle whose tangent is $2/3$. A picture sometimes facilitates the analysis wherein we draw a right triangle with vertical side of length $2$ and horizontal side of length $3$ forming a right angle.
Note that the angle the hypotenuse makes with the horizontal side is $\arctan(2/3)$. Inasmcuh as the hypotenuse is of length $\sqrt{2^2+3^2}=\sqrt{13}$, we see $\sin(\arctan(2/3))=\frac{2}{\sqrt{13}}$ and thus $\csc (\arctan(2/3))=\frac{\sqrt{13}}{2}$.
|
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|
Prove the inequality $\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$ with the constraint $abc=1$ If $a,b,c$ are positive reals such that $abc=1$, then prove that $$\sqrt\frac{a}{a+8} + \sqrt\frac{b}{b+8} +\sqrt\frac{c}{c+8} \geq 1$$ I tried substituting $x/y,y/z,z/x$, but it didn't help(I got the reverse inequality). Need some stronger inequality. Thanks.
|
The required inequality is trivialized by the claim below. The equality case is when $a=b=c=1$.
Claim: If $a,b,c>0$ are such that $abc=1$, then $\displaystyle\sqrt{\frac{a}{a+8}}\geq \frac{a^{4/9}}{a^{4/9}+b^{4/9}+c^{4/9}}$. The equality holds if and only if $a=b=c=1$.
Proof: Note that the required inequality is equivalent to
$$\left(a^{4/9}+b^{4/9}+c^{4/9}\right)^2 \geq a^{-1/9}(a+8)\,,$$
which is also equivalent to
$$\left(b^{4/9}+c^{4/9}\right)\left(a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\right) \geq 8a^{-1/9}\,.$$
To prove the previous inequality, we invoke the AM-GM Inequality twice:
$$b^{4/9}+c^{4/9}\geq 2b^{2/9}c^{2/9}$$
and
$$a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\geq 4a^{1/9}a^{1/9}b^{1/9}c^{1/9}=4a^{2/9}b^{1/9}c^{1/9}\,.$$
Thus,
$$
\begin{align}
\left(b^{4/9}+c^{4/9}\right)\left(a^{4/9}+a^{4/9}+b^{4/9}+c^{4/9}\right) &\geq \left(2b^{2/9}c^{2/9}\right)\left(4a^{2/9}b^{1/9}c^{1/9}\right)
\\
&=8a^{2/9}b^{1/3}c^{1/3}=8a^{-1/9}\left(abc\right)^{1/3}=8a^{-1/9}\,,
\end{align}$$
which is what we want. By the equality condition of the AM-GM Inequality, the equality happens iff $a=b=c=1$.
P.S.: I just realized why this inequality looks so familiar. It is equivalent to IMO'2001#2 (http://imo.wolfram.com/problemset/IMO2001_solution2.html). Substitute $a$, $b$, and $c$ by $x^3$, $y^3$, and $z^3$, then homogenize the required inequality via the condition $xyz=1$, and you will see what I'm talking about.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Formulae for sequences Given that for $1^3 + 2^3 + 3^3 + \cdots + n^3 = \frac{n^2(n+1)^2}{4}$
deduce that $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3 = \frac{n^2(3n+1)(5n+3)}{4}$
So far:
the sequence $(n+1)^3 + (n+2)^3 +\cdots+ (2n)^3$ gives $2^3 + 3^3 + 4^3 +\cdots,$ when n=1.
The brackets in the formula for the second sequence are $2n$ and $4n+2$ bigger than $(n+1)$ in the original i.e. $(3n+1)(5n+3).$
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$$s_1=1^3+2^3+3^3+..n^3=(\frac{n(n+1)}{2})^2\\s_2=1^3+2^3+3^3+..n^3+(n+1)^3+...+(2n)^3=(\frac{2n(2n+1)}{2})^2\\s_2-s_1=(n+1)^3+...+(2n)^3=\\(\frac{2n(2n+1)}{2})^2-(\frac{n(n+1)}{2})^2=\\\frac{n^2}{4}(4(2n+1)^2-(n+1)^2)=\\\frac{n^2}{4}((4n+2+n+1)(4n+2-n-1)$$
|
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|
Finding $4$ variables using $3$. if I have:
$ x=\dfrac{a-.5b-.5c+.25d}{a+b+c+d}$
$ y=\dfrac{\dfrac{b\sqrt{3}}{2}+\dfrac{c\sqrt{3}}{2}+\dfrac{d\sqrt{3}}{4}}{a+b+c+d}$
$ z=a+b+c+2d $
Then how do I get back to:
$ a= $ , $ b= $ , $ c= $ , and $ d= $ ?
When there is a divide by zero error, $ x=0 $ and $ y=0 $.
When
$(a,b,c,d)=(e,e,0,0)$ and $e<>0$
then
$(x,y,z)=(.25,\dfrac{\sqrt{3}}{4},2e)$
but also when
$(a,b,c,d)=(0,0,0,e)$ and $e<>0$
then
$(x,y,z)=(.25,\dfrac{\sqrt{3}}{4},2e)$
again. So, everytime
($a=b$ and $c=0$)
then
$(a,b,0,0)=(0,0,0,a+b)$.
$x>=-.5$
$x\leq1$
$y\geq\dfrac{\sqrt{3}}{2}$
$y\leq-\dfrac{\sqrt{3}}{2}$
$z\geq0$
$z\leq765$ this could mathematically be up to $1020$ but will be always be $\leq 765$, because $a+b+2d \leq 510$.
$a\geq0$
$b\geq0$
$c\geq0$
$d\geq0$
$a\leq255$
$b\leq255$
$c\leq255$
$d\leq255$
$a$, $b$, $c$, and $d$ are all integers.
I can find $a$, $b$, and $c$, when I don't use $d$. but I would like to find what the highest $d$ value can be; while $a$, $b$, and $c$ are between or equal to $0$ and $255$.
When $d=0$, I get the formulas I want:
$a=\dfrac{z}{3}(2x+1)$
$b=\dfrac{z}{3}*(\sqrt{3}y-x+1)$
$c=-\dfrac{z}{3}(x+\sqrt{3}y-1)$
$d=0$
When $θ$ of $(x,y)$ from $(0,0)$ is $60°$ then only $d$ exists.
When $θ$ of $(x,y)$ from $(0,0)$ is between $-120°$ and $60°$ then only $a$,$c$, and $d$ exists.
When $θ$ of $(x,y)$ from $(0,0)$ is between $60°$ and $240°$ then only $b$,$c$, and $d$ exists.
When $θ$ of $(x,y)$ from $(0,0)$ is $240°$ (aka $-120°$) then only $c$ exists.
But I'd like the highest $d$ value; not lowest.
note: $atan(\dfrac{y}{x})$.
|
Hints: You can't get back $b$ and $c$ individually, but you can solve for $a$, $d$, and $e\equiv b+c$ as follows: transform your system into
\begin{align*}
(a+e+d)x&=a-0.5e+d\\
(a+e+d)y&=\frac{\sqrt{3}}{2}e+\frac{\sqrt{3}}{4}d\\
z&=a+e+2d
\end{align*}
which can be further written as
$$
\begin{pmatrix}
x-1& x+0.5&x-1\\
y&y-\sqrt{3}/2&y-\sqrt{3}/4\\
1&1&2
\end{pmatrix}
\times
\begin{pmatrix}
a\\ e\\d
\end{pmatrix}
=
\begin{pmatrix}
0\\ 0\\z
\end{pmatrix}\cdot
$$
Edit: call the $3\times 3$ matrix on the LHS above $A$. Then your answer is
$$
\begin{pmatrix}a \\ e \\ d\end{pmatrix}=A^{-1}\begin{pmatrix}
0\\ 0\\z
\end{pmatrix}\cdot
$$
In fact, if you use $A^{-1}=\text{adj}(A)/\det(A)$, you only need to compute the last column of $A^{-1}$.
|
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|
Matrix Equation: Reduced Row Echelon Form Given a matrix $A$ and its reduced row echelon form $R$, how can one find an invertible matrix $P$ such that $PA=R$?
I was given a $4\times4$ matrix $A$, I found $R$, and I tried to find $P$ by inspection. Since it was too difficult, I tried finding $P^{-1}$ by inspection because $A=P^{-1}R$ and succeeded (the greater number of zeros helps tremendously).
However, is there a more reliable method?
|
A matrix can be reduced with some sequence of three elementary row operations: swapping rows, multiplying a row by a constant, and adding one row to another. Luckily for us, each of these operations is linear, so each can be represented as a matrix multiplication. Then we just have to chain all of those matrix multiplications together.
Here's an example. Say we want to reduce the following matrix:
$$
A=
\left(\begin{array}{ccc}
0 & 1 & 5 \\
1 & 0 & 3 \\
3 & 0 & 9 \\
\end{array}\right)
$$
There are many ways we could get to the reduced row echelon form, but let's start by dividing the third row by three.
$$
P_1A =
\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \frac{1}{3}
\end{array} \right)
\left(\begin{array}{ccc}
0 & 1 & 5 \\
1 & 0 & 3 \\
3 & 0 & 9
\end{array}\right)
=
\left(\begin{array}{ccc}
0 & 1 & 5 \\
1 & 0 & 3 \\
1 & 0 & 3
\end{array}\right)
$$
Then let's subtract the second row from the third.
$$
P_2(P_1A)=
\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & -1 & 1
\end{array} \right)
\left(\begin{array}{ccc}
0 & 1 & 5 \\
1 & 0 & 3 \\
1 & 0 & 3
\end{array}\right)
=
\left(\begin{array}{ccc}
0 & 1 & 5 \\
1 & 0 & 3 \\
0 & 0 & 0
\end{array}\right)
$$
Finally, let's swap the first two rows.
$$
P_3(P_2P_1A)=
\left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array} \right)
\left(\begin{array}{ccc}
0 & 1 & 5 \\
1 & 0 & 3 \\
0 & 0 & 0
\end{array}\right)
=
\left(\begin{array}{ccc}
1 & 0 & 3 \\
0 & 1 & 5 \\
0 & 0 & 0
\end{array}\right)
$$
Great, we got to the row echelon form using three elementary operations. Notice that now we can find our permutation matrix by composing these operation matrices.
$$
P=P_3P_2P_1=
\left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & 0 & 1
\end{array} \right)
\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & -1 & 1
\end{array} \right)
\left( \begin{array}{ccc}
1 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & \frac{1}{3}
\end{array} \right)
=
\left( \begin{array}{ccc}
0 & 1 & 0 \\
1 & 0 & 0 \\
0 & -1 & \frac{1}{3}
\end{array} \right)
$$
|
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|
Mathematical Induction proof for a cubic equation. If $ x^3 = x +1$, prove by induction that $ x^{3n} = a_{n}x + b_n + \frac {c_n}{x}$, where $a_1=1, b_1=1, c_1=0$ and
$a_n = a_{n-1} + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,\dots $
For $n=1$ we have $x_3 = a_1x + b_1 + \frac {c_1}{x}$ = x + 1 + $\frac{0}{x}$ = x + 1, which is true.
Assume the case for $n=k$ is true, so $x^{3k} = a_kx + b_k + \frac{c_k}{x}$
So for $n = k+1$ (and using $a_n = a_{n-1} + b_{n-1}, b_n = a_{n-1} + b_{n-1} + c_{n-1}, c_n = a_{n-1} + c_{n-1}$ for $n=2,3,...), $ we have:
$ \begin{align}
x^{3(k+1)} &= (a_{k+1-1}x + b_{k+1-1}){x} + (a_{k+1-1} + b_{k+1-1} + c_{k+1-1}) +
( \frac{(a_{k+1-1} + c_{k+1-1})}{x}) \\
&= (a_k + b_k){x} + ( a_k + b_k + c_k) + \frac{(a_k + c_k)}{x} \\
&= a_{k+1}{x} + b_{k+1} + \frac{c_{k+1}}{x} \\\end{align} $
This is the same form as $ x^{3k} $ but for n=k+1.
Therefore, if the result is true for k, it is also true for (k+1).
Would you proceed like this or would you add the $(k+1)$th term and show that it is equal to the $k$th term with $n$ (or $k$) replaced by $k+1$?
|
Hint: Multiply $x^{3n} = a_{n}x + b_n + \frac {c_n}{x}$ on the left by $x^3$ and on the right by $x+1$. Regroup and deduce a formula for $a_{n+1}$, $b_{n+1}$, $c_{n+1}$ in terms of $a_{n}$, $b_{n}$, $c_{n}$.
|
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|
Number theory with binary quadratic I found this questions from past year maths competition in my country, I've tried any possible way to find it, but it is just way too hard.
Given $$ \frac {x^2-y^2+2y-1}{y^2-x^2+2x-1} = 2$$ find $x-y$
I'm not sure if given choices is right... (A)2 (B)3 (C)4 (D)5 (E)6
I've tried to move them $$x^2-y^2+2y-1 = 2y^2-2x^2+4x-2$$
$$x^2-y^2+2y-1 - 2y^2+2x^2-4x+2 = 0$$
$$3x^2-3y^2+2y-4x+1=0$$
$$(3x-1)(x-1)-(3y-1)(y+1)+1=0$$
I've stuck in here, not sure if I've found x and y, or not...
EDIT: I've move other questions to other posts, thanks for helping me identifying the questions category.
|
Recognize the squares of binomials in both the numerator and denominator to rewrite the equation as $$\begin{align}\frac{x^2-(y-1)^2}{y^2-(x-1)^2} &= 2, \end{align}$$ and thus, factoring, $$\require\cancel \begin{align}\frac{(x-y+1)\cancel{(x+y-1)}}{(y-x+1)\cancel{(y+x-1)}}&=2. \end{align}$$ Finally, let $t=x-y$ and multiply both sides by $1-t$ to have $$\begin{align} t+1&=2(1-t) \\ 3t&=1 \\ t&=\frac{1}{3}. \end{align}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Solve $10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$ How to solve the following equation?
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$
My attempt:
$$ 10x^4 - (7x^2+1)(x^2+x+1)=0$$
Thats all i can
Update
Tried to open brakets and simplify:
$$(7x^2+1)(x^2+x+1) = 7x^4+7x^3+7x^2+x^2+x+1=7x^4+7x^3+8x^2+1 $$
$$10x^4 - (7x^2+1)(x^2+x+1)= 3x^4-7x^3-8x^2-1=0 $$
|
$$10x^4-7x^2(x^2+x+1)+(x^2+x+1)^2=0$$
Set $t=x^2,z=x^2+x+1$.
$\Longrightarrow$
$$\begin{align}10t^2-7tz+z^2&=(2t-z)(5t-z)\\&=(2x^2-(x^2+x+1))(5x^2-(x^2+x+1))\\&=(x^2-x-1)(4x^2-x-1)\end{align}$$
$$\boxed{\color{red}{x_{1,2}=\frac{1}{2}\pm\frac{\sqrt5}{2},\;x_{3,4}=\frac{1}{8}\pm \frac{\sqrt{17}}{8}}}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
creative method to obtain range of newton function ?! I am searching for more proof that the range of $y=\frac{x}{x^2+1}$ is $ \frac{-1}{2}\leq y \leq \frac{+1}{2}$
these are my tries :
domain is $\mathbb{R}$
first : $$\quad{y=\frac{x}{x^2+1}\\yx^2+y=x \rightarrow x^2y-x+y=0 \overset{\Delta \geq 0 }{\rightarrow} 1-4y^2 \geq 0 \rightarrow \frac{-1}{2}\leq y \leq \frac{+1}{2} }$$
second :
$$ \quad{y=\frac{x}{x^2+1} \overset{x=tan \alpha }{\rightarrow} \space y=\frac{tan \alpha}{tan^2 \alpha+1}=\frac{tan \alpha}{\frac{1}{cos^2\alpha}}=sin \alpha cos \alpha =\\\frac{2}{2} sin \alpha cos \alpha =\frac{1}{2}sin 2\alpha\\ -1\leq sin 2\alpha \leq 1 \rightarrow -\frac{1}{2} \leq \frac{1}{2}sin 2\alpha \leq \frac{1}{2} \rightarrow \frac{-1}{2}\leq y \leq \frac{+1}{2}}$$ 3rd:
$$\quad{\frac{1}{y}=\frac{x^2+1}{x}=x+\frac{1}{x} \overset{|x+\frac{1}{x}|\geq 2}{\rightarrow} |\frac{1}{y}| \geq 2 \rightarrow |y| \leq \frac{1}{2} \rightarrow \frac{-1}{2}\leq y \leq \frac{+1}{2}}$$
4th :$$f(x)=\frac{x}{x^2+1} \\f'=\frac{1-x^2}{1+x^2}=0 \rightarrow x=\pm1 \\ \left\{\begin{matrix}
f(1) &=\frac{1}{2} \\
f(-1) & =-\frac{1}{2}\\
\lim_{x \rightarrow +\infty }\frac{x}{x^2+1} & =0\\
\lim_{x \rightarrow -\infty }\frac{x}{x^2+1} & =0
\end{matrix}\right. \overset{x \in \mathbb{R}}{\rightarrow} \space \frac{-1}{2}\leq y \leq \frac{+1}{2}$$
now : is there (creative ) 5th ,6th ,... method to find range of this function ?
|
By contradiction: assume $y > \frac{1}{2}$
$$\frac{x}{x^2 +1} > \frac{1}{2} \Rightarrow 0 > (x-1)^2$$
and $y<-\frac{1}{2}\Rightarrow 0 > (x+1)^2 $.
But we also see that $y(-1) = -\frac{1}{2}$ and $y(1) = \frac{1}{2}$.
|
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|
Determine if this series $ \sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}$ converges Determine if the following series converges:
$$
\sum_{n=0}^\infty\frac{n^6+13n^5+n+1}{n^7+13n^4+9n+2}.
$$
(http://i.stack.imgur.com/qWiuy.png)
I don't know how to start.
|
You want to use a comparison test. Remember with series that you can disregard a finite number of terms. When $n \geq 2$
$$n^7+13n^5+9n+2 \leq n^7 +13n^7 + 9n^7+n^2 = 24n^7,$$
which implies
$$ \frac{1}{n^7+13n^5+9n+2} \geq \frac{1}{n^7 +13n^7 + 9n^7+n^2} \geq \frac{1}{24n^7}.$$
For the numerator, you have
$$n^6 + 13n^5 + n + 1 \geq n^6.$$
Therefore
$$ \sum_{n=2}^\infty \frac{n^6 + 13n^5 + n + 1}{n^6 + 13n^5 + 9n + 2} \geq \frac{1}{24} \sum_{n=2}^\infty \frac{n^6}{n^7} = \frac{1}{24}\sum_{n=2}^\infty \frac{1}{n} =\infty, $$
which shows that the series diverges by comparison with the harmonic series.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Am I getting the right answer for the integral $I_n= \int_0^1 \frac{x^n}{\sqrt {x^3+1}}\, dx$?
Let $I_n= \int_0^1 \dfrac{x^n}{\sqrt {x^3+1}}\, dx$. Show that $(2n-1)I_n+2(n-2)I_{n-3}=2 \sqrt 2$ for all $n \ge 3$. Then compute $I_8$.
I get an answer for $I_8={{2 \sqrt 2} \over 135}(25-16 \sqrt 2)$, could somebody please check against my answer and see if I made a mistake.
|
Integrating by parts, we have
$$\begin{align}
I_n&=\int_0^1\frac{x^n}{(1+x^3)^{1/2}}dx\\\\
&=\left.\frac23 x^{n-2}(x^3+1)\right|_0^1-\frac23(n-2)\int_0^1x^{n-3}(x^3+1)^{1/2}dx \tag 1\\\\
&=\frac232^{1/2}-\frac23(n-2)\int_0^1\frac{x^{n-3}(x^3+1)}{(x^3+1)^{1/2}}\,dx \\\\
&=\frac232^{1/2}-\frac23 (n-2)I_n-\frac23(n-2)I_{n-3}\\\\
(2n-1)I_n+2(n-2)I_{n-3}&=2^{3/2} \tag 2\\\\
\end{align}$$
Using $(1)$, we can see that $I_2$ given by
$$I_2=\frac23(\sqrt{2}-1) \tag 3$$
Then, using $(2)$ and $(3)$ and iterating once to $I_5$
$$I_5=\frac29 (2-\sqrt{2})$$
and again to $I_8$ reveals
$$I_8=\frac{2}{45}(7\sqrt{2}-8)$$
|
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|
How to integrate $\int \frac{4}{x\sqrt{x^2-1}}dx$ In order to solve the following integral:
$$\int \frac{4}{x\sqrt{x^2-1}}dx$$
I tried different things such as getting $u = x^2 + 1$, $u=x^2$ but it seems that it does not work. I also tried moving $x$ to the radical and writing the denominator as $\sqrt{x^4-x^2}$ but it also didn't work. Its look reminds me of $$\int \frac {1}{\sqrt{a^2-x^2}}dx = \arcsin\left(\frac{x}{a}\right) +c$$ but there is an extra $x$ in the denominator that I can't get rid of.
|
We have, taking $u=\sqrt{x^{2}-1}
$, $$\int\frac{4}{x\sqrt{x^{2}-1}}dx=4\int\frac{1}{u^{2}+1}du.
$$ I think you can get from here. Expanding a little bit, note that $$ u=\sqrt{x^{2}-1}\Rightarrow du=\frac{x}{\sqrt{x^{2}-1}}dx
$$ and $$ x^{2}=u^{2}+1
$$ then $$4\int\frac{1}{x\sqrt{x^{2}-1}}dx=4\int\frac{1}{x^{2}}\frac{x}{\sqrt{x^{2}-1}}dx=4\int\frac{1}{u^{2}+1}du.
$$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Struggling with an inequality: $ \frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1 $ Prove that for every natural numbers, $m$ and $n$, this inequality holds:
$$
\frac{1}{\sqrt[n]{1+m}} + \frac{1}{\sqrt[m]{1+n}} \ge 1
$$
I tried to use Bernoulli's inequality, but I can't figure it out.
|
If we change a little this inequality we will have:
$$ \sqrt[m \cdot n]{(1 + n )^n} + \sqrt[m \cdot n]{(1 + m)^m} \geq \sqrt[m \cdot n]{(1+n)^n \cdot (1+m)^m}$$
From the first part we will have:
$$ \sqrt[m \cdot n]{(1 + n )^n} + \sqrt[m \cdot n]{(1 + m)^m} \geq 1 + \frac{n}{m} + 1 + \frac{m}{n} = \frac{(m+n)^2}{m \cdot n}$$
From tail we have inequality( using this fact Inequality of arithmetic and geometric means):
$$ \sqrt[m \cdot n]{(1+n)^n \cdot (1+m)^m} \leq \frac{ n \cdot (1+n) + m \cdot ( 1 + m)}{m \cdot n} = \frac{(n+m)^2 - 2\cdot m \cdot n + m + n }{m \cdot n}$$
As a result we have:
$ \frac{(n+m)^2 }{m \cdot n} \geq \frac{(n+m)^2 - 2\cdot m \cdot n + m + n }{m \cdot n}$
or
$$ 2\cdot m \cdot n \geq m + n $$ that is true, because $m$ and $n$ are natural.
|
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"timestamp": "2023-03-29T00:00:00",
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|
For which $x, y\in\mathbb{R ^+}$ do we have $|xy-\frac{1}{xy}|\le|x-\frac{1}{x}|+|y-\frac{1}{y}|$? I need to find all $x, y\in\mathbb{R^+}$ such that the following inequality holds.
$$\Big| xy-\dfrac{1}{xy}\Big|\le\Big|x-\dfrac{1}{x}\Big|+\Big|y-\dfrac{1}{y}\Big|$$
If I substitute $x=2$ and $y=3$ clearly that inequality fails.
How can I attempt this question? Any hint?
Thank you.
|
Since both members are positive, we can square and get this equivalent $$x^2y^2+\frac{1}{x^2y^2}-2\le x^2+\frac{1}{x^2}-2+y^2+\frac{1}{y^2}-2+2|x-\frac{1}{x}||y-\frac{1}{y}|$$
Then gettting all terms but the last one to the left, factorizing and multipying by $x^2y^2$ we get
$$ (x^2y^2+1)(x^2-1)(y^2-1) \le 2x^2y^2|x-\frac{1}{x}||y-\frac{1}{y}| $$
From this we know that all $x,y$ satisfying $(x^2-1)(y^2-1) \le 0 $ are solution. For other values, we can square again and cancel repeted (positive) factors, we get $$(x^2y^2+1)^2 \le 4x^2y^2$$
And from this it is easy to deduce that
$$(x^2y^2-1)^2 \le 0$$
what is to say, $(xy)^2=1$
I hope this is a little helpful!
New edit: actually, if $ x^2= \frac{1}{y^2}$ then it is also true that $(x^2-1)(y^2-1) \le 0 $. So $x,y$ solve the inequality if and only if they satisfy this last condition, which means that $|x| \le 1 $ and $|y| \ge 1 $ or viceversa
|
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|
How to integrate ${x^3}/(x^2+1)^{3/2}$? How to integrate $$\frac{x^3}{(x^2+1)^{3/2}}\ \text{?}$$
I tried substituting $x^2+1$ as t, but it's not working
|
$$\int\frac{x^3}{(x^2+1)^{3/2}}dx$$
$u:=x^2,du=2xdx$
$$=\frac{1}{2}\int\frac{u}{(u+1)^{3/2}}du$$
$s:=u+1,ds=du$
$$=\frac{1}{2}\int\frac{s-1}{s^{3/2}}ds$$
$$=\sqrt s+\frac{1}{\sqrt s}$$
$s=u+1,u=x^2$
$$=\sqrt{x^2+1}+\frac{1}{\sqrt{x^2+1}}+C$$
$$\boxed{\color{blue}{=\frac{x^2+2}{\sqrt{x^2+1}}+C}}$$
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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|
Find all values that solve the equation For which values a, the equation
$$ a\sin{x}+(a+1)\sin^2{\frac{x}{2}} + (a-1)\cos^2{\frac{x}{2}} =1 $$
has a solution?
My idea: I think it's possible to factorize equation or reduce equation to the form like: $a(\sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}) =1 $
Let's go:
$$ 2a\sin{\frac{x}{2}}\cos{\frac{x}{2}} + asin^2{\frac{x}{2}} + sin^2{\frac{x}{2}} + a\cos^2{\frac{x}{2}} - \cos^2{\frac{x}{2}} = 1$$
$$ a\left(sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 = 1 - \sin^2{\frac{x}{2}} + \cos^2{\frac{x}{2}}$$
$$ a\left(\sin{\frac{x}{2}}+\cos{\frac{x}{2}}\right)^2 =2\cos^2{\frac{x}{2}}$$
I can't finish...
|
HINT....The equation simplifies to $ a\sin x +a- \cos x=1$ can you take it from there?
|
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|
Find the equation of the circle. Find the equation of the circle whose radius is $5$ which touches the circle $x^2 + y^2 - 2x -4y - 20 = 0$ externally at the point $(5,5)$
|
The circle: $x^2+y^2-2x-4y-20=0$ has center $(1, 2)$ & a radius $=\sqrt{(-1)^2+(-2)^2-(-20)}=5$ & the unknown circle has a radius $5$ Hence the point $(5, 5)$ is the mid point of line joining their centers
Let the center of unknown circle be $(a, b)$ then the point $(5, 5)$ is mid point of lines joining the centers $(a, b)$ & $(1, 2)$ hence we have $$\left(\frac{a+1}{2}, \frac{b+2}{2}\right)\equiv(5, 5)$$ by comparing the corresponding coordinates we get $$\frac{a+1}{2}=5\implies a=9$$ $$\frac{b+2}{2}=5\implies b=8$$ Hence the equation of the circle having center $(9, 8)$ & a radius $5$ is given as $$(x-9)^2+(y-8)^2=5^2=25$$
$$\color{blue}{(x-9)^2+(y-8)^2=25}$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
The value of the definite integral The value of the definite integral $\displaystyle\int\limits_0^\infty \frac{\ln x}{x^2+4} \, dx$ is
(A) $\dfrac{\pi \ln3}{2}$ (B) $\dfrac{\pi \ln2}{3}$ (C) $\dfrac{\pi \ln2}{4}$ (D) $\dfrac{\pi \ln4}{3}$
I tried using integration by parts,
\begin{align}
& \int_0^\infty \frac{\ln x}{x^2+4}dx = \ln x\int_0^\infty \frac{1}{x^2+4}dx-\int_0^\infty \left(\frac{d}{dx}\ln x\int_0^\infty \frac{1}{x^2+4}\right) \, dx \\[10pt]
= {} & \left[\ln x \frac{1}{2}\tan^{-1}\frac{x}{2}\right]-\int_0^\infty \frac{1}{x}\frac{1}{2}\tan^{-1}\frac{x}{2} \, dx
\end{align}
I could not move ahead. Can someone help me to get final answer?
|
Using the substitution $\displaystyle \theta=\frac{1}{2}\arctan\left(\frac{x}{2}\right)$ so $\displaystyle \text{d}\theta=\frac{1}{x^2+4} \text{d}x$ and $\displaystyle x=2\tan(2\theta)$ we get $$\int_{0}^{\infty}\frac{\ln(x)}{x^2+4}\text{d}x=\int_{0}^{\frac{\pi}{4}}\ln\left(2\tan(2\theta)\right)\text{d}\theta= \\ =\int_{0}^{\frac{\pi}{4}}\ln(2)\text{d}\theta+\int_{0}^{\frac{\pi}{4}}\ln\sin(2\theta)\text{d}\theta-\int_{0}^{\frac{\pi}{4}}\ln\cos(2\theta)\text{d}\theta $$
Last two integrals cancel each other, so we get $$\int_{0}^{\infty}\frac{\ln(x)}{x^2+4}\text{d}x=\int_{0}^{\frac{\pi}{4}}\ln(2)\text{d}\theta=\frac{\pi \ln(2)}{4}$$
Of course, we can generalize and compute $\displaystyle \int_{0}^{\infty}\frac{\ln(x)}{x^2+k^2}\text{d}x$.
Using the substitution $\displaystyle \theta=\frac{1}{k}\arctan\left(\frac{x}{k}\right)$ we get $\displaystyle \text{d}\theta=\frac{1}{x^2+k^2}\text{d}x \ , \ x=k\tan(k\theta)$, hence $$\int_{0}^{\infty}\frac{\ln(x)}{x^2+k^2}\text{d}x=\int_{0}^{\Large \frac{\pi}{2k}}\ln(k\tan(k\theta))\text{d}\theta= \\ =\int_{0}^{\Large \frac{\pi}{2k}}\ln(k)\text{d}\theta+\int_{0}^{\Large \frac{\pi}{2k}}\ln(\sin(k\theta))\text{d}\theta-\int_{0}^{\Large \frac{\pi}{2k}}\ln(\cos(k\theta))\text{d}\theta$$
Now, we prove that $\displaystyle I=\int_{0}^{\Large \frac{\pi}{2k}}\ln(\sin(k\theta))\text{d}\theta-\int_{0}^{\Large \frac{\pi}{2k}}\ln(\cos(k\theta))\text{d}\theta=0$.
For the first integral we use the substitution $\displaystyle u=\frac{\pi}{2}-k\theta$ and for the second integral we use the substitution $\displaystyle u=k\theta$, hence $$I=-\frac{1}{k}\int_{\Large \frac{\pi}{2}}^{0}\ln(\cos(u))\text{d}u-\frac{1}{k}\int_{0}^{\Large \frac{\pi}{2}}\ln(\cos(u))\text{du}=0$$
Eventually, $\displaystyle \boxed{\color{#2E2EFE}{\int_{0}^{\infty}\frac{\ln(x)}{x^2+k^2}\text{d}x=\int_{0}^{\Large \frac{\pi}{2k}}\ln(k)\text{d}\theta =\frac{\pi}{2k}\ln(k)}}$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Matrix $\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$ to a large power Compute
$\begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}^{99}$
What is the easier way to do this other than multiplying the entire thing out?
Thanks
|
Following the comment of Matt Samuel:
$$
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix} =
\begin{pmatrix}
1 & 2 \\
0 & 1
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}^3
=
\begin{pmatrix}
1 & 3 \\
0 & 1
\end{pmatrix}
$$
$$
\begin{pmatrix}
1 & 1 \\
0 & 1
\end{pmatrix}
\begin{pmatrix}
1 & n \\
0 & 1
\end{pmatrix}
=
\begin{pmatrix}
1 & n+1 \\
0 & 1
\end{pmatrix}
$$
|
{
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"timestamp": "2023-03-29T00:00:00",
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|
External bisectors of the angles of ABC triangle form a triangle $A_1B_1C_1$ and so on If the external bisectors of the angles of the triangle ABC form a triangle $A_1B_1C_1$,if the external bisectors of the angles of the triangle $A_1B_1C_1$ form a triangle $A_2B_2C_2$,and so on,show that the angle $A_n$ of the $n$th derived triangle is $\frac{\pi}{3}+(-\frac{1}{2})^n(A-\frac{\pi}{3})$,and that the triangles tend to become equilateral.
My attempt:The angle $A_1$ of triangle $A_1B_1C_1$ is $\frac{\pi}{2}-\frac{A}{2}$,angle $A_2$ of triangle $A_2B_2C_2$ is $\frac{A}{4}-\frac{3\pi}{4}$ but my answer is nowhere look like resembling final answer.Is my approach correct.If not,what is the correct way to solve this question?Can someone guide me?
|
You are on the right track, you just need to continue. You have a correct formula for $A_1$, so format it differently.
$$\begin{align}
A_1 &= \frac{\pi}2-\frac A2 \quad\text{(your formula)}\\
&= \frac{\pi}3+\frac{\pi}6+\left(-\frac 12\right)A \\
&= \frac{\pi}3+\left(-\frac 12\right)\left(A-\frac{\pi}3\right) \\
&= \frac{\pi}3+\left(-\frac 12\right)^1\left(A-\frac{\pi}3\right) \\
\end{align}$$
Your formula for $A_1$ from $A$ also gives $A_{n+1}$ from $A_n$, so combining your formula with my derivation above and using induction we get
$$\begin{align}
A_{n+1} &= \frac{\pi}3+\left(-\frac 12\right)\left(A_n-\frac{\pi}3\right) \\
&= \frac{\pi}3+\left(-\frac 12\right)\left[\frac{\pi}3+
\left(-\frac 12\right)^n\left(A-\frac{\pi}3\right)-\frac{\pi}3\right] \\
&= \frac{\pi}3+\left(-\frac 12\right)\left[
\left(-\frac 12\right)^n\left(A-\frac{\pi}3\right)\right] \\
&= \frac{\pi}3+\left(-\frac 12\right)^{n+1}\left(A-\frac{\pi}3\right) \\
\end{align}$$
which finishes the proof by induction.
So the given formula is correct. As we let $n\to\infty$ we see that $\left(-\frac 12\right)^n\to 0$ and thus $A_n\to\frac{\pi}3$. We can do the same to $B$ and $C$ to get $B_n\to\frac{\pi}3$ and $C_n\to\frac{\pi}3$, so all three angles approach $60°$, and the (increasingly large) triangles $\triangle A_nB_nC_n$ approach equilateral.
Here is a quick proof of your formula $A_1 = \frac{\pi}2-\frac A2$, though I used Greek letters for angles in this diagram.
This should be self-explanatory, and the final value for $A_1= \frac{\pi}2-\frac A2$ comes directly from $A_1=\frac{\beta+\gamma}2$ and $\alpha+\beta+\gamma=\pi$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1382062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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|
How many divisors of the combination of numbers?
Find the number of positive integers that are divisors of at least one of $A=10^{10}, B=15^7, C=18^{11}$
Instead of the PIE formula, I would like to use intuition.
$10^{10}$ has $121$ divisors, and $15^7$ has $64$ divisors, and $18^{11}$ has $276$ divisors.
Number of divisors total with no restriction is: $461$. $A,B \to $ there are $5^{0} \to 5^{7} = 8$ divisors. $B, C \to$ there are: $3^{0} \to 3^{7} = 8$ divisors. $A, C \to$ there are $2^{0} \to 2^{10} = 11$ divisors.
So far: $461 - 27 = 434$. I took out: $15$ divisors from $B$, $19$ from $A$, and $19$ from $C$. So in total:
$$434 + 53 = 487$$
But this isnt right.
|
$10^{10}$ has $121$ divisors, and $15^7$ has $64$ divisors, and $18^{11}$ has $276$ divisors.
$A,B \to $ there are $5^{0} \to 5^{7} = 8$ divisors. $B, C \to$ there are: $3^{0} \to 3^{7} = 8$ divisors. $A, C \to$ there are $2^{0} \to 2^{10} = 11$ divisors.
These are correct. Now $1$ is the only divisor of the three numbers $A,B,C$. So, the number of positive integers that are divisors of at least one of $A,B,C$ is given by
$$(121+64+276)-(8+8+11)+1=435.$$
You may want to see the inclusion–exclusion principle.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1382185",
"timestamp": "2023-03-29T00:00:00",
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|
$A+B+C=2149$, Find $A$ In the following form of odd numbers
If the numbers
taken from the form where $A+B+C=2149$
Find $A$
any help will be appreciate it, thanks.
|
Let's write the numbers as $a_k:=2k-1$, starting at $k=1$. Then, $a_1=1$, $a_2=3$, $a_3=5$, and so on. When $A=a_k$ is in row $n$ (the first row is row $0$), then $B=a_{k+n+1}$ and $C=a_{k+n+2}=B+2$. Hence, we have
\begin{align*}
&& 2149 &= a_k + 2a_{2k+n+1} +2 \\
&\Rightarrow& 2147 &= 2k-1 + 2(2(k+n+1)-1) = 6k + 4n + 1
\end{align*}
Note that the $n$-th row (starting at $n=0$) of the triangle starts with $a_{k_n}$, where
$$k_n = 1+\cdots+n = \frac{n(n+1)}2.$$
Therefore, $k=k_n+q$ where $0\le q\le n$. Substituting, we get
\begin{align*}
&& 2147 &= 3n(n+1) + 10q + 4n + 1 \\
&\Rightarrow & 0 &= 3 n^2 + 7 n + 10 q -2146
\end{align*}
Observe that the linear term $10q$ does not affect the position of the zeros of this quadratic polynomial much. Hence, I just plugged in $q=n/2$ and this polynomial has a zero around $25$. So, $n=25$ seems a good guess.
Indeed, plugging in $n=25$ into our earlier equation gives
\begin{align*}
&& 2147 &= 6*k + 4*25 + 1 \\
&\Rightarrow & 2046 &= 6*k \\
&\Rightarrow & 341 &= k
\end{align*}
So we have $A=a_{341}=681$ in row $n=25$ and $B=2*(341+25+1)-1=733$, $C=735$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1384690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Minus sign in logarithm of integral's solution I want to solve the following integral:
$
\int \frac{dp}{a(1-p)u-bp}
$
where $a$, $b$ and $u$ are some constants.
After integration I get:
$
p = -\frac{\log(-apu+au-bp)}{au+b} + C.
$
According to WolframAlpha it is equivalent to (citing precisely: "Which is equivalent for restricted $p$, $a$, $b$ and $u$ values to"):
$
p = -\frac{log(a(p-1)u+bp)}{au+b} + C,
$
or transforming it a bit, it is equivalent to:
$
p = -\frac{\log(apu-au+bp)}{au+b} + C.
$
So for some reasons a minus is "removed" from the logarithm. Could you please explain to me why?
|
\begin{align}
I &= \int \frac{dp}{a(1-p)u - b p} = \frac{1}{au} \, \int \frac{dp}{1 - \left( 1 + \frac{b}{au}\right) p } \\
&= \frac{1}{au+b} \, \int \frac{dx}{1-x} \hspace{10mm} \text{where} \quad x = \left(1+\frac{b}{au}\right) \, p \\
&= - \frac{1}{au + b} \, \int \frac{dx}{x-1} \\
&= - \frac{1}{au+b} \, \ln(x-1) \\
&= - \frac{\ln(au\, (1-p) + bp) - \ln(au)}{au+b}.
\end{align}
|
{
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx$ Find the maximum value of $72\int\limits_{0}^{y}\sqrt{x^4+(y-y^2)^2}dx $ for $y\in[0,1].$
I tried to differentiate the given function by using DUIS leibnitz rule but the calculations are messy and I tried to solve directly by integrating it but that also is not working.Can someone please help me in solving this question?
|
As observed by @MichaelGaluza, by Leibniz's rule, it suffices to show
$$
\sqrt{y^{4}+\left(y-y^{2}\right)^{2}}+\int_{0}^{y}\frac{\left(1-2y\right)\left(y-y^{2}\right)}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\geq0.
$$
Note that $y-y^{2}=y\left(1-y\right)>0$ for $y\in\left(0,1\right)$.
Now, for $y\leq\frac{1}{2}$, we have $\left(1-2y\right)\left(y-y^{2}\right)\geq0$,
so that the claim is trivial.
Thus, we can assume $y\in\left(\frac{1}{2},1\right)$. In this case,
we have $\left(1-2y\right)\left(y-y^{2}\right)<0$. Furthermore,
$$
\int_{0}^{y}\frac{1}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\leq\int_{0}^{y}\frac{1}{\sqrt{\left(y-y^{2}\right)^{2}}}\,{\rm d}y=\frac{y}{y-y^{2}}=\frac{1}{1-y}.
$$
Multiplying by $\left(1-2y\right)\left(y-y^{2}\right)<0$ yields
$$
\int_{0}^{y}\frac{\left(1-2y\right)\left(y-y^{2}\right)}{\sqrt{x^{4}+\left(y-y^{2}\right)^{2}}}\,{\rm d}x\geq\frac{\left(1-2y\right)\left(y-y^{2}\right)}{1-y}=\left(1-2y\right)y,
$$
so that it suffices to show
\begin{align*}
& \sqrt{y^{4}+\left(y-y^{2}\right)^{2}}\geq\left(2y-1\right)y\\
\left(\text{since }\left(2y-1\right)y\geq0\text{ because }y>\frac{1}{2}\right)\Longleftrightarrow & y^{4}+\left(y-y^{2}\right)^{2}\geq\left(2y-1\right)^{2}y^{2}\\
\Longleftrightarrow & y^{4}+y^{2}-2y^{3}+y^{4}=y^{4}+\left(y-y^{2}\right)^{2}\geq\left(4y^{2}-4y+1\right)y^{2}=4y^{4}-4y^{3}+y^{2}\\
\Longleftrightarrow & 0\geq2y^{4}-2y^{3}=2y^{3}\left(y-1\right)\\
\left(\text{since }y>0\right)\Longleftrightarrow & y-1\leq0\\
\Longleftrightarrow & y\leq1,
\end{align*}
which is true. This proves that the derivative is indeed nonnegative,
so that the maximum is attained at $y=1$.
|
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"url": "https://math.stackexchange.com/questions/1386453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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|
Prove that $f(x,y,z)$ is reducible if and only if $a,b,c,d$ is a geometric progression.
Let $a,b,c,d$ be real numbers not all $0$, and let $f(x,y,z)$ be the polynomial in three variables defined by
$$f(x,y,z) = axyz + b(xy + yz + zx) + c(x + y + z) + d.$$
Prove that $f(x,y,z)$ is reducible if and only if $a,b,c,d$ is a geometric progression.
What I have tried...
Let $g$ is symmetric We may assume that $g = x + y + z + k$ for some constant $k.$
And I'm stuck here!!, any help will be appreciated.
|
We have:
$$
\begin{align*}
f(x,y,z) & = axyz + b(xy + yz + zx) + c(x + y + z) + d.\\
& = axyz + b(xy) + b(yz + zx) + c(x+y)+ cz + d\\
& = xy(az+b) + (x+y)(bz+c) + (cz+d)\\
\end {align*}
$$
The only way we can make it reduce further is if $(az+b), (bz+c)$ and $(cz+d)$ are related geometrically, i.e., $(bz+c)$ and $(cz+d)$ are of the type $m\cdot(az+b)$.
Let the geometric progression be $a, b=at, c=at^2, d=at^3$.
Then $(bz+c) = at(z) + at^2 = t(az+at) = t(az+b)$. Similarly, $(cz+d) = t^2(az+b)$.
Now, back to our equation:
$$
\begin{align*}
f(x,y,z) & = xy(az+b) + (x+y)(bz+c) + (cz+d)\\
& = (az+b)[xy + t(x+y) + t^2]
\end {align*}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1387575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
If $\alpha$ and $\beta$ are the zeroes of $p(x) =x^2- px +q = 0$ Find $\alpha^2 + \beta^2$ and $\alpha^3 + \beta^3$.
|
For convenience, we reverse the sign of $q$, without loss of generality.
$$x^2-px-q$$ is the characteristic equation of the recurrence
$$x_{n+2}=px_{n+1}+qx_n,$$
that has the general solution
$$a\alpha^n+b\beta^n.$$
With $a=b=1$,
$$x_0=2,\\
x_1=\alpha+\beta=p,\\
x_2=\alpha^2+\beta^2=px_1+qx_0=p^2+2q,\\
x_3=\alpha^3+\beta^3=px_2+qx_1=p(p^2+2q)+pq=p^3+3pq,\\
x_4=\alpha^4+\beta^4=px_3+qx_2=p(p^3+3pq)+(p^2+2q)q=p^3+4p^2q+2q^2,\\
x_5=\alpha^5+\beta^5=px_4+qx_3=p(p^3+4p^2q+2q^2)+(p^3+3pq)q=p^4+5p^3q+5pq^2,\\\cdots
$$
The general formula is closely related to the development of $(p+q)^n$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1387742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
Show that $\sqrt{4+2\sqrt{3}}-\sqrt{3}$ is rational using the rational zeros theorem What I've done so far:
Let $$r = \sqrt{4+2\sqrt{3}}-\sqrt{3}.$$
Thus, $$r^2 = 2\sqrt{3}-2\sqrt{3}\sqrt{4+2\sqrt{3}}+7$$
and $$r^4=52\sqrt{3}-28\sqrt{3}\sqrt{4+2\sqrt{3}}-24\sqrt{4+2\sqrt{3}}+109.$$
I did this because in a similar example in class, we related $r^2$ and $r^4$ to find a polynomial such that $mr^4+nr^2 = 0$ for some integers $m,n$. However, I cannot find such relation here. Am I doing this right or is there another approach to these type of problems.
|
From $r = \sqrt{4+2\sqrt{3}}-\sqrt{3}$ we get $r+\sqrt{3}=\sqrt{4+2\sqrt{3}}$ and, squaring both sides,
$$
r^2+2r\sqrt{3}+3=4+2\sqrt{3}
$$
and so
$$
r^2-1=2(1-r)\sqrt{3}
$$
Square again:
$$
r^4-2r^2+1=12-24r+12r^2
$$
so
$$
r^4-14r^2+24r-11=0
$$
The rational root test only allows $1$, $-1$, $11$ and $-11$ as roots. Since $1$ is clearly a root we have
$$
(r-1)(r^3+r^2-13r+11)=0
$$
and $1$ is a root also of the second factor:
$$
(r-1)^2(r^2+2r-11)=0
$$
The roots of the second factor are
$$
-1+2\sqrt{3},\qquad -1-2\sqrt{3}
$$
Since $r>0$, we only have two possibilities: $r=1$ or $r=2\sqrt{3}-1$. The second possibility gives
$$
\sqrt{4+2\sqrt{3}}=3\sqrt{3}-1
$$
If we square this, we get
$$
4+2\sqrt{3}=28-6\sqrt{3}
$$
or
$$
24=8\sqrt{3}
$$
which is absurd. Thus we only remain with the possibility that $r=1$.
Easier: $4+2\sqrt{3}=3+2\sqrt{3}+1=(\sqrt{3}+1)^2$.
Alternatively, from $r^2-1=2(1-r)\sqrt{3}$ we deduce $r=1$ or
$$
r+1=-2\sqrt{3}
$$
that's absurd, because $r>0$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1388206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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|
Find the sum of binomial coefficients
Calculate the value of the sum
$$
\sum_{i = 1}^{100} i\binom{100}{i} = 1\binom{100}{1} +
2\binom{100}{2} +
3\binom{100}{3} +
\dotsb +
100\binom{100}{100}
$$
What I have tried:
$$\begin{align}
S &= 0\binom{100}{0}+1\binom{100}{1}+ \dotsb +99\binom{100}{99}+100\binom{100}{100} \\ \\
&=100\binom{100}{100}+99\binom{100}{99}+ \dotsb +1\binom{100}{1}+0\binom{100}{0}
\end{align}$$
and I'm stuck here, I don't know if it's true or not, any help will be appreciated.
|
HINT :
Using$$\binom{n}{i}=\binom{n}{n-i}$$
just add the two sums you write to get $$2S=100\binom{100}{0}+100\binom{100}{1}+\cdots+100\binom{100}{99}+100\binom{100}{100}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1388720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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|
Method of partial fractions when denumerator cannot be factorized? Suppose I'm given an expression: \begin{align*} P(x) = \frac{1+x}{1-2x-x^2}. \end{align*} The denumerator cannot be readily factorized, so I found the zeros, which are \begin{align*} \lambda_1 = -1 - \sqrt{2} \qquad \lambda_2 = -1 + \sqrt{2}. \end{align*}
I want to apply the method of partial fraction to $P(x)$ now. I wanted to write \begin{align*} P(x) = \frac{1+x}{1-2x-x^2} = \frac{A}{(x- \lambda_1)} + \frac{B}{(x- \lambda_2)} \end{align*} but this equality does not hold. Since $(x-\lambda_1) (x- \lambda_2) = x^2 + 2x -1 \neq 1-2x-x^2$. So what denumerators should I choose to go with $A$ and $B$?
|
In general when you have some factors in denumerator which is not factor-able you have to get its numerator as $ax+b$ and then equate the fractions and compute $a$ and $b$ and so on.
|
{
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"url": "https://math.stackexchange.com/questions/1390746",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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|
Find the general solution of $\cos(x)-\cos(2x)=\sin(3x)$ Problem:
Find the general solution of $$\cos(x)-\cos(2x)=\sin(3x)$$
I tried attempting this by using the formula$$\cos C-\cos D=-2\sin(\dfrac{C+D}{2})\sin(\dfrac{C-D}{2})$$
Thus, $$-2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)=\sin 3x$$
$$\Rightarrow -2\sin\left(\dfrac{x}{2}\right)\sin\left(\dfrac{3x}{2}\right)-\sin 3x=0$$
Unfortunately, I couldn't get further with this problem. Any help with this would be truly appreciated. Many thanks in advance!
|
*
*Expand the trig functions $$ \cos(x)+1-2 \cos^2(x) = 4 \sin(x) \cos^2(x)-\sin(x)$$
*Use the tangent half angle substitution $t=\tan(x/2)$, $\cos(x)=\frac{1-t^2}{1+t^2}$ and $\sin(x) = \frac{2 t}{1+t^2}$ $$ \frac{2 t^2 (3-t^2)}{(1+t^2)^2} = \frac{2 t (3 t^4-10 t^2+3)}{(1+t^2)^3}$$
*Collect terms $$\frac{2 t (t+1) (t^2-3) (t^2+2t-1)}{(1+t^2)^3} = 0$$
*Solve for $t$ $$\begin{align} t&=0\\t&=-1\\t&=\sqrt{3}\\t&=-\sqrt{3}\\t&=\sqrt{2}-1\\t&=-\sqrt{2}-1 \end{align}$$
*Solve for $x=2 \arctan(t)$ $$\begin{align}
x&=0\\
x&=-\frac{\pi}{2}\\
x&=\frac{2\pi}{3}\\
x&=-\frac{2\pi}{3}\\
x&=\frac{\pi}{4}\\
x&=-\frac{3\pi}{4}
\end{align}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1390849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
$\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ $\int\limits_{1/2}^{2}\frac{1}{x}\sin (x-\frac{1}{x})dx$ has value equal to
$(A)0\hspace{1cm}(B)\frac{3}{4}\hspace{1cm}(C)\frac{3}{4}\hspace{1cm}(D)2 $
I tried to solve this question by putting $x-\frac{1}{x}=t$ and limits have changed to $\frac{-3}{2}$ to $\frac{3}{2}$ but what to do with remaining $\frac{1}{x}$ and $dx$ in the integration.Please help....
|
$\bf{My\; Solution::}$ Let $$\displaystyle I = \int_{\frac{1}{2}}^{2}\frac{1}{x}\cdot \sin \left(x-\frac{1}{x}\right)dx$$
Now Let $\displaystyle\left(x-\frac{1}{x}\right) = t\;,$ Then $\displaystyle \left(1+\frac{1}{x^2}\right)dx = dt\Rightarrow \left(x+\frac{1}{x}\right)dx = xdt$
and Changing Limits
Now Using $$\displaystyle \left(x+\frac{1}{x}\right)^2 -\left(x-\frac{1}{x}\right)^2=4\Rightarrow \left(x-\frac{1}{x}\right)=\sqrt{\left(x-\frac{1}{x}\right)^2+4}=\sqrt{t^2+4}$$
So Integral $$\displaystyle I = \int_{-\frac{3}{2}}^{\frac{3}{2}}\sin t\cdot \frac{1}{\sqrt{t^2+4}}\cdot \frac{x}{x}dt = \int_{-\frac{3}{2}}^{\frac{3}{2}}\sin t\cdot \frac{1}{\sqrt{t^2+4}}dt$$
So we get $$\displaystyle I = \int_{-\frac{3}{2}}^{\frac{3}{2}}\underbrace{\frac{\sin t}{\sqrt{t^2+4}}}_{\bf{odd\; function}}dt = 0$$
Above we have used the formula $$\displaystyle \int_{-a}^{a}f(x)dx = 0\;,$$ If $f(x)$ is odd function.
|
{
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"url": "https://math.stackexchange.com/questions/1390926",
"timestamp": "2023-03-29T00:00:00",
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|
Let $(a, b, c)$ be a Pythagorean triple. Prove that $\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2$ is greater than 8 and never an integer.
Let $(a, b, c)$ be a Pythagorean triple, i.e. a triplet of positive integers with $a^2 + b^2 = c^2$.
a) Prove that $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > 8$$
b) Prove that there are no integer $n$ and Pythagorean
triple $(a, b, c)$ satisfying $$\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 = n$$
The foregoing question is from the 2005 Canada National Olympiad.
I need some help on part (b).
Part (a)
Examine the behaviour of the following function [that embodies the constraint $c^2=a^2+b^2$, $x$ can play the role of either $a$ or $b$] on $(0,c)$:
$$f(x) = \dfrac{c}{x} + \dfrac{c}{\sqrt{c^2-x^2}} \tag{1}$$
Now find the derivative: $$f'(x)=cx\left[\dfrac{1}{(c^2-x^2)^{3/2}}-\dfrac{1}{x^3}\right] \tag{2}$$
To find local extrema:
$$\begin{align}
f'(x)=0 &\implies x^3=(c^2-x^2)^{3/2} \\
&\implies x^6=(c^2-x^2)^3 &(\text{squaring}) \\
&\implies u^3=(k-u)^3 &(u=x^2,k=c^2) \\
&\implies 2u^3-3ku^2+3k^2u-k^3=0 \\
&\implies (2u-k)(u^2+-ku+k^2)=0 \tag{3}\\
\end{align}$$
So, solutions are: $$u=\dfrac{k}{2};u=\dfrac{k\pm\sqrt{k^2-4k^2}}{2}\notin\mathbb{R}$$
So take the only real solution as $$u=\dfrac{k}{2} \implies x^2=\dfrac{c^2}{2} \implies x=\dfrac{c}{\sqrt2}\quad(\text{since }x\in(0,c))$$
This is a local minima because $f$ is continuous on $(0,c)$ and $$\lim_{x\to0^+}{f(x)}=\lim_{x\to c^-}{f(x)}=+\infty$$
The minimum value is $$f\left(\dfrac{c}{\sqrt2}\right)=\sqrt2+\sqrt2=2\sqrt2 \tag{4}$$
This value is not achievable, because $\sqrt{2}$ is not a rational number. Hence,
$$[f(a)]^2 = \left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 > (2\sqrt2)^2 = 8 \tag{5}$$
Part (b)
$$\begin{align}
\left(\dfrac{c}{a}+\dfrac{c}{b}\right)^2 = n \implies c^2(a^2+b^2)=a^2b^2\cdot n \tag{6}
\end{align}$$
in a perhaps friendlier form.
I am not sure how to use this further.
|
If $(a,b,c) \in \mathbb{N}^3$ is a Pythagorean triple, then without loss of generality, we may assume that $a=\left(m^2-n^2\right)d$, $b=2mnd$, and $c=\left(m^2+n^2\right)d$ for some positive integers $m,n,d$ with $m>n$, $\gcd(m,n)=1$, and $m\not\equiv n\pmod{2}$.
We now have
$$t:=\frac{c}{a}+\frac{c}{b}=\frac{m^2+n^2}{m^2-n^2}+\frac{m^2+n^2}{2mn}=\frac{\left(m^2+n^2\right)\left(m^2-n^2+2mn\right)}{2mn\left(m^2-n^2\right)}\,.$$
Note that, since $t\in\mathbb{Q}$, we see that $t^2\in\mathbb{Z}$ iff $t\in\mathbb{Z}$. Now, $t\in\mathbb{Z}$ implies that $2mn$ divides $\left(m^2+n^2\right)\left(m^2-n^2+2mn\right)$. Since $\gcd(m,n)=1$ and $m\not\equiv n\pmod{2}$, we see that $\gcd\left(m^2+n^2,2mn\right)=1$. Therefore, $2mn$ must divide $m^2-n^2+2mn$, which means $2mn\mid m^2-n^2$. Again, we have $\gcd\left(2mn,m^2-n^2\right)=1$, which leads to a contradiction. Hence, $t\notin\mathbb{Z}$, so $t^2\notin\mathbb{Z}$.
Note that $t$ can get arbitrarily close to $2\sqrt{2}$. We can find $m$ and $n$ such that $\frac{m}{n}$ is arbitrarily close to $1+\sqrt{2}$. If $r:=\frac{m}{n}$, then
$$t=\frac{\left(r^2+1\right)\left(r^2-1+2r\right)}{2r\left(r^2-1\right)}\,,$$
so as $r\to 1+\sqrt{2}$, $t\to 2\sqrt{2}$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1391298",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "6",
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|
Help with a dominated convergence theorem problem The problem is to find an integrable function that bounds $f(x)={\dfrac{{{n^{\frac{3}{2}}}x}}{{1 + {n^2}{x^2}}}}$ on $[0,1]$ so we can calculate $\displaystyle\int_0^1 {\frac{{{n^{\frac{3}{2}}}x}}{{1 + {n^2}{x^2}}}dx} $ using dominated convergence theorem. I know by taking derivative with respect to $n$ we can find the maximum point $n=\frac{\sqrt 3}{x}$ which leads to $f(x)$ is bounded by $x^{-\frac{1}{2}}$, which is integrable on $[0,1]$, i.e. $|f(x)|\le x^{-\frac{1}{2}}$ on $[0,1]$.
My question is, is there nicer solution without taking derivatives? I have tried an hour using inequalities like $1+n^2x^2\ge 2nx$ or substituting $1$ in the denominator by $x$, or $x^\frac{1}{2}$, but all failed. I am so tired so I post this question here and hope someone can help.
I used derivative in my homework but my professor commented "try not using derivatives".
Thank you!
|
By AM-GM: $\dfrac{1+n^2x^2}{n^{3/2}x^{3/2}} = \dfrac{1}{(nx)^{3/2}} + (nx)^{1/2} = \dfrac{1}{(nx)^{3/2}} + \dfrac{(nx)^{1/2}}{3} + \dfrac{(nx)^{1/2}}{3} + \dfrac{(nx)^{1/2}}{3}$
$\ge 4 \sqrt[4]{\dfrac{1}{(nx)^{3/2}}\cdot \dfrac{(nx)^{1/2}}{3}\cdot\dfrac{(nx)^{1/2}}{3}\cdot\dfrac{(nx)^{1/2}}{3}}$ $= \dfrac{4}{3^{3/4}}$.
Hence, $\dfrac{n^{3/2}x^{3/2}}{1+n^2x^2} \le \dfrac{3^{3/4}}{4}$, and thus, $\dfrac{n^{3/2}x}{1+n^2x^2} \le \dfrac{3^{3/4}}{4x^{1/2}}$, which is integrable on $[0,1]$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1391644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Why I am getting different answer? I have just started learning single variable calculus. I'm confused in a problem from sometime. I didn't get why my answer is different from the book.
$$
\require{cancel}
\begin{align}
&\int\sin x \sin 2x \sin 3x\,dx\\
&=\int\sin x\;\,2\sin x\cos x \left(3\sin x - 4\sin^3 x\right)\,dx\\
&\qquad\text{Let }\sin x = t, \text{ then}\\
&\qquad\quad\cos x\, dx = dt\\
&=\int t\;2t\left(3t - 4 t^3\right)\,dt\\
&=\int 2t^2\left(3t - 4t^3\right)\,dt\\
&=\int\left(6t^3-8t^5\right)\,dt\\
&=6\int t^3\,dt - 8\int t^5\,dt\\
&=\cancel{6}\,3\frac{t^4}{\cancel{4}2}+c_1-\cancel{8}\,4\frac{t^6}{\cancel{6}3}+c_2\\
&=\frac32t^4-\frac43t^6+C\\
&=\frac32\sin^4x-\frac43\sin^6x+C
\end{align}
$$
The answer given in my book is
$$\displaystyle\frac{1}{4}\left[\frac{1}{6}\cos 6x - \frac{1}{4}\cos 4x - \frac{1}{2}\cos 2x\right] + C. \ $$
Where did I go wrong?
|
Both solutions are correct, they only differ by a constant. See also Wolfram Alpha.
This is similar to the following situation: Both $f(x) = \sin^2(x)$ and $g(x) = -\cos^2(x)$ are antiderivatives of $2\sin(x)\cos(x)$. Even if they look quite different, they only differ by a constant: $f(x) = 1 + g(x)$.
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1393098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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|
How to factor $4x^2 + 2x + 1$? I want to know how to factor $4x^2 + 2x + 1$? I found the roots using quadratic equation and got $-1 + \sqrt{-3}$ and $-1 - \sqrt{-3}$, so I thought the factors would be $(x - (-1 + \sqrt{-3}))$ and $(x - (-1 - \sqrt{-3}))$
However, according to MIT's course notes, the factors are $(1 - (-1 + \sqrt{-3})x)$ and $(1 - (-1 - \sqrt{-3})x)$
Course Notes (pg. 30)
Are these two expressions equivalent, or am I simply not factoring correctly? Thanks.
|
$$4x^2+2x+1$$
Finding roots by quadratic rule as follows
$$x=\frac{-2\pm\sqrt{2^2-4(4)(1)}}{2(4)}$$ $$=\frac{-2\pm2i\sqrt{3}}{8}$$
$$=-\frac{1}{4}\pm\frac{i\sqrt 3}{4}$$
$$x=-\frac{1}{4}+\frac{i\sqrt 3}{4}\ \vee\ x=-\frac{1}{4}-\frac{i\sqrt 3}{4}$$
Edit:
Now, we have the factors as follows $$4x^2+2x+1=4\left(x+\frac{1}{4}-\frac{i\sqrt 3}{4}\right)\left(x+\frac{1}{4}+\frac{i\sqrt 3}{4}\right)$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1393930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
nice classical nonhomogeneous inequality Let $a,b,c$ be positive reals and $abc=1$. Prove that $$10(a^4+b^4+c^4)+21\ge 17(a^3+b^3+c^3).$$
I have found a solution using MV and I'm wondering if there is a nice solution.
|
Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that
$$10(81u^4-108u^2v^2+18v^4+12uw^3)+21w^4-17(27u^3-27uv^2+3w^3)\geq0$$
or $f(u)\geq0$, where
$$f(u)=270u^4-360u^2v^2+60v^4+40uw^3-153u^3w+153uv^2w-10w^4.$$
But by Schur $w^3\geq4uv^2-3u^3$.
Thus,
$$f'(u)=1080u^3-720uv^2+40w^3-459u^2w+153v^2w\geq$$
$$\geq621u^3-720uv^2+99w^3=9(69u^3-80uv^2+11w^3)\geq$$
$$\geq9(69u^3-80uv^2+44uv^2-33u^3)=324u(u^2-v^2)\geq0,$$
which says that $f$ is an increasing function.
Thus, it's enough to prove our inequality for a minimal value of $u$,
which happens for equality case of two variables.
Let $b=a$ and $c=\frac{1}{a^2}$.
Id est, we need to prove that:
$$(a-1)^2(20a^{10}+6a^9-8a^8-22a^7-15a^6-8a^5-a^4+6a^3+13a^2+20a+1)\geq0,$$
which is true.
Done!
|
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"timestamp": "2023-03-29T00:00:00",
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|
If $x,y,z>0$ and $xyz=32,$ Then the minimum of $x^2+4xy+4y^2+4z^2$ is If $x,y,z$ are positive real no. and $xyz= 32\;,$ Then Minimum value of $$x^2+4xy+4y^2+4z^2$$ is
$\bf{My\; Try::}$ Here I have Used $\bf{A.M\geq G.M}$ Inequality
So $$\displaystyle \frac{x^2+4xy+4y^2+4z^2}{4}\geq \left(x^3\cdot y^3\cdot z^2\right)^{\frac{1}{4}}\;,$$ But I did not How can I solve it
Help me, Thanks
|
Applying the AM-GM is the right strategy, but you need to do it a bit differently.
$$x^2+4xy+4y^2+4z^2$$
$$= x^2+2xy+2xy+4y^2+2z^2+2z^2$$
$$\ge 6\sqrt[6]{x^2 \cdot 2xy \cdot 2xy \cdot 4y^2 \cdot 2z^2 \cdot 2z^2}$$
$$= 6\sqrt[6]{64x^4y^4z^4}$$
$$= 12(xyz)^{2/3}$$
$$= 12 \cdot 32^{2/3}$$
Equality holds when $x^2 = 2xy = 4y^2 = 2z^2$ and $xyz = 32$, which is attained when $(x,y,z) = (2^{13/6},2^{7/6},2^{10/6})$.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Inverse Trigonometric Function: Find the Exact Value of $\sin^{-1}\left(\sin\left(\frac{7\pi}{3}\right)\right)$ $$\arcsin\left(\sin\left(\frac{7\pi}{3}\right)\right)$$
I cannot use this formula, correct? $f(f^{-1}(x))=x$
The answer in the book is $\frac{\pi}{3}$
How do I approach solving a problem such as this?
The inverse sin function of $\sin\frac{7\pi}{3}$
Am I saying to myself there were 7 revolutions and $\frac{\pi}{3}$ a corresponding angle to $\frac{7\pi}{3}$
|
Let $\sin^{-1}\sin\dfrac{7\pi}3=x$ where $-\dfrac\pi2\le x\le\dfrac\pi2$
$$\implies\sin x=\sin\dfrac{7\pi}3$$
$\implies x=n\pi+(-1)^n\dfrac{7\pi}3$ where $n$ is any integer
If $n$ is even $=2m$(say), $x=2m\pi+\dfrac{7\pi}3=\dfrac{(6m+7)\pi}3\implies -\dfrac\pi2\le\dfrac{(6m+7)\pi}3\le\dfrac\pi2$
$\iff-3\le12m+14\le3\implies-2<-\dfrac{17}{12}\le m\le-\dfrac{11}{12}<0\implies m=-1$
Check for odd $n=2m+1$(say)
|
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"timestamp": "2023-03-29T00:00:00",
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|
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Find the number of real roots of $1+x/1!+x^2/2!+x^3/3! + \ldots + x^6/6! =0$.
Attempts so far:
Used Descartes signs stuff so possible number of real roots is $6,4,2,0$
tried differentiating the equation $4$ times and got an equation with no roots hence proving that above polynomial has $4$ real roots.
But using online calculators I get zero real roots. Where am I wrong?
|
We can compute the number of real roots using Sturm's Theorem.
$$
\begin{array}{rll}
\text{Sturm Chain}&+\infty&-\infty\\\hline
x^6+6x^5+30x^4+120x^3+360x^2+720x+720&+\infty&+\infty\\
6x^5+30x^4+120x^3+360x^2+720x+720&+\infty&-\infty\\
-5x^4-40x^3-180x^2-480x-600&-\infty&-\infty\\
-48x^3-432x^2-1728x-2880&-\infty&+\infty\\
45x^2+360x+900&+\infty&+\infty\\
384x+1920&+\infty&-\infty\\
-225&-225&-225
\end{array}
$$
There are $3$ changes of sign at $+\infty$ and $3$ changes of sign at $-\infty$. Thus, there are no real roots.
|
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"timestamp": "2023-03-29T00:00:00",
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|
Value of $\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}}$ Here is the question:
$$\frac{\sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\cdots+\sqrt{10+\sqrt{99}} }{\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\cdots+\sqrt{10-\sqrt{99}}} = \;?$$
(original image)
I think we need to simplify it writing it in summation sign as you can see here:
$$\frac{\sum\limits_{n=1}^{99} \sqrt{10 + \sqrt{n}}}{\sum\limits_{n=1}^{99} \sqrt{10 - \sqrt{n}}}$$
or in Wolfram Alpha input in comments.
I can compute it too! It's easy to write a script for this kind of question.
I need a way to solve it. How would you solve it on a piece of paper?
|
Let the numerator and the denominator
$$N= \sqrt{10+\sqrt{1}}+\sqrt{10+\sqrt{2}}+\ldots+\sqrt{10+\sqrt{99}}$$
$$D =\sqrt{10-\sqrt{1}}+\sqrt{10-\sqrt{2}}+\ldots+\sqrt{10-\sqrt{99}}$$
Apply the denesting formulas
$$\sqrt{a\pm\sqrt c} = \frac{1}{\sqrt2} \left( \sqrt{a+\sqrt{a^2-c}} \pm
\sqrt{a-\sqrt{a^2-c}} \right)$$
to get
$$\sqrt{10\pm\sqrt n} = \frac{1}{\sqrt2} \left( \sqrt{10+\sqrt{100-n}} \pm
\sqrt{10-\sqrt{100-n}} \right)$$
where $n=1,2,...99$. As a result,
$$N= \frac{1}{\sqrt2}(N+D), \>\>\>\>\>D= \frac{1}{\sqrt2}(N-D)$$
Take the ratio,
$$\frac ND=\frac{N+D}{N-D}$$
or,
$$\left(\frac ND\right)^2 - 2\frac ND -1 =0$$
Solve to obtain,
$$\frac ND = 1+\sqrt2$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Need help with an arithmetic sequence proving question It is given that $a_1, a_2, a_3, \ldots ,a_n$ are consecutive terms of an Arithmetic progression. I have to prove that
$$\sum_{k=2}^n (\sqrt{a_{(k-1)}} + \sqrt{a_k} )^{-1} = \frac{n-1}{\sqrt{a_1} + \sqrt{a_n} }$$
Using Mathematical induction I showed that it is true for $n = 2$.
Then, assuming that it works for $n = m$, I took the case wherein $n = m+1\ldots$
Using this the left hand side of the equation is :
$$\frac{m-1}{\sqrt{a_1} + \sqrt{a_m} } + (\sqrt{a_{m}} + \sqrt{a_{m+1}} )^{-1} $$
and the right hand side would be
$$\frac{m}{\sqrt{a_1} + \sqrt{a_{m+1}} }$$
How do I prove that the LHS = RHS??. I tried squaring both numerator and denominator, as well as using the properties of an arithmetic sequence but I havent been able to simplify the algebra.
|
Actually the induction is not such a mess:
$$
\frac{m-1}{\sqrt{a_1} + \sqrt{a_m} } + (\sqrt{a_{m}} + \sqrt{a_{m+1}} )^{-1}=\frac{(m-1)(\sqrt{a_m} -\sqrt{a_1})}{a_m - a_1 } + \frac{\sqrt{a_{m + 1}} - \sqrt{a_{m}} }{a_{m+1} - a_m}=\frac{(m-1)(\sqrt{a_m} -\sqrt{a_1})}{d(m - 1)}+ \frac{\sqrt{a_{m + 1}} - \sqrt{a_{m}} }{d}=\frac{(\sqrt{a_{m+1}} -\sqrt{a_1})}{d}=\frac{(a_{m+1} -{a_1})}{d(\sqrt{a_{m+1}} +\sqrt{a_1})}=\frac{m}{\sqrt{a_1} + \sqrt{a_{m+1}} }
$$
where d = common difference of the AP.
So LHS = RHS
|
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"timestamp": "2023-03-29T00:00:00",
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|
What is the least number of (fixed) parameters I can ask for, when calculating area of a triangle of unknown type? I need to calculate the area of a triangle, but I don't know, whether it is right angled, isoscele or equilateral.
What parameters do I need to calculate the area of a triangle of unknown type?
|
All you need is the lengths of each side of the triangle.
By Heron's Formula, we know for a triangle with sides $a,b,c$, we have
$$A=\sqrt{s(s-a)(s-b)(s-c)}\text{ ,where }s=\frac{a+b+c}2$$
Reference:
https://en.wikipedia.org/wiki/Heron%27s_formula
EDIT:
In response to suggestion by @Hurkyl , I now add the case of ASA and SsA.
(ASA)
With a known side $c$ and its neighbouring angles $\alpha,\beta$, we have:
$$A=\frac{c^2\sin\alpha\sin\beta}{2\sin{(\alpha+\beta)}}$$
Proof:
Let $\alpha,\beta,\theta$ be the corresponding angles of sides $a,b,c$ respectively, we have
$$\theta=\pi-\alpha-\beta\implies\sin\theta=\sin{(\pi-\alpha-\beta)}=\sin{(\alpha+\beta)}$$
By Sine Law,
$$\frac c{{\sin{(\pi-\alpha-\beta)}}}=\frac a{\sin\alpha}=\frac b{\sin\beta}$$
So $a=\frac{c\sin\alpha}{\sin{(\alpha+\beta)}}$, $b=\frac{c\sin\beta}{\sin{(\alpha+\beta)}}$
Therefore,
\begin{align}
A&=\frac12ab\sin\theta\\&=\frac12\cdot{\frac{c\sin\alpha}{\sin{(\alpha+\beta)}}}\cdot\frac{c\sin\beta}{\sin{(\alpha+\beta)}}\cdot\sin{(\alpha+\beta)}\\&=\frac{c^2\sin\alpha\sin\beta}{2\sin{(\alpha+\beta)}}
\end{align}
(SsA)
With 2 known sides $b,c$ and $\beta$, the corresponding angle of $b$, we have:
$$A=\frac12 c\sin\beta[\sqrt{1-(\frac{c\sin\beta}b)^2}+c\cos\beta]$$
Proof:
By Sine Law:
$$\frac b{\sin\beta}=\frac c{\sin\theta}$$
$$\sin\theta=\frac{c\sin\beta}b, \cos\theta=\sqrt{1-(\frac{c\sin\beta}b)^2}$$
By Cosine Law,
$$c^2=a^2+b^2-2ab\cos\theta$$
$$c^2-b^2+b^2\cos^2\theta=a^2-2ab\cos\theta+b^2\cos^2\theta$$
$$c^2-b^2\sin^2\theta=(a-b\cos\theta)^2$$
$$a=b\cos\theta+\sqrt{c^2-b^2\sin^2\theta}=b\cos\theta+c\cos\beta$$
So
\begin{align}
A&=\frac12ab\sin\theta\\&=\frac12(b\cos\theta+c\cos\beta)(b)\cdot\frac{c\sin\beta}b\\&=\frac12 c\sin\beta[\sqrt{1-(\frac{c\sin\beta}b)^2}+c\cos\beta]
\end{align}
|
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|
Angle bisector between two vectors, which are expressed by unit non-orthogonal vectors Given vectors $\,a=2m-2n\,$ and $\,b=3m+6n\,$, where $\ \left\lvert m \right \rvert =\left\lvert n \right \rvert =1\,$ and $\,\angle\left(m,n\right)=\dfrac{2\pi}{3},\,$ find vector of angle bisector of the angle $\angle\left(a,b\right)$, which intensity is $10\sqrt{3}$.
Advice on how to start?
EDIT: intensity=magnitude
|
First, compute angle $\theta := \angle (a,b)$ between $a$ and $b$ using cosine law.
$$
\left\langle a, b \right\rangle = \left\| a\right\| \left\| b\right\| \cos \theta
\implies
\cos \theta = \frac{\left\langle a, b \right\rangle}{\left\| a\right\| \left\| b\right\| },
$$
where $\left\langle \,\cdot\,, \cdot \,\right\rangle$ is inner product of vectors $a$ and $b$.
Note that
$$
\left\| a \right\| = \sqrt{\left\langle a, a \right\rangle }
= \sqrt{\left\langle 2m - 2n , 2m - 2n \right\rangle }
= 2 \sqrt{\left\langle m - n , m - n \right\rangle }
= 2 \sqrt{ \left\langle m,m \right\rangle - 2 \left\langle m, n \right\rangle + \left\langle m,m \right\rangle }
= 2\sqrt{\left\|m\right\|^2 - 2 \left\langle m, n \right\rangle + \left\|n\right\|^2}
$$
Since $\left\langle m, n \right\rangle = \left\|m\right\| \left\|n\right\|\cos\angle\left(m,n \right) = 1\cdot 1\cdot \cos \frac{2\pi}{3} = -\frac{1}{2}$,
we have
$$
\begin{aligned}
\left\| a \right\|
&= 2\sqrt{\left\|m\right\|^2 - 2 \left\langle m, n \right\rangle + \left\|n\right\|^2}
= 2\sqrt{1^2 + \frac{2}{2} + 1^2}
= 2\sqrt{3}
\\
\left\| b \right\|
&= \sqrt{\left\langle 3m+6n,3m+6n\right\rangle}
=3\sqrt{\left\|m\right\|^2 + 4\left\langle m,n\right\rangle + 4\left\|n\right\|^2}
=3\sqrt{1 - \frac{4}{2} + 4 }
= 3\sqrt{3}
\\
\left\langle a,b\right\rangle
&= \left\langle 2m-2n,3m+6n\right\rangle
= 6 \left\| m\right\| ^2 + 6 \left\langle m,n\right\rangle - 12 \left\| n\right\|^2
= 6 - \frac{6}{2} - 12 = -9
\end{aligned}
$$
And so
$$
\cos \theta
= \frac{\left\langle a, b \right\rangle}{\left\| a\right\| \left\| b\right\| }
= \frac{-9}{2\sqrt{3} \cdot 3\sqrt{3}}
= -\frac{1}{2}
$$
You are looking for a vector $v$ such that $\angle\left(a,v\right) = \theta/2 $ and $\left\| v\right\| = 10\sqrt 3$.
Assume $x,y$ are coefficients of decomposition of $v$ in terms of $m$ and $n$, i.e. $v = x m + y n$, then we write
$$
v = x m + yn, \quad
\left\| v\right\| = 10\sqrt 3, \quad
\cos \frac{\theta}{2} = \sqrt{\frac{1+ \cos \theta}{2}}
= \sqrt{\frac{1-\frac{1}{2}}{2}}
= \pm \frac{1}{2}
$$
Let us write out what we know about $v$:
$$
\begin{aligned}
\left\langle v,v\right\rangle
& = \left\| v \right\|^2 = 300
\\
\left\langle v,a\right\rangle
& = \left\| v \right\| \left\| a \right\| \cos \frac{\theta}{2}
= 10\sqrt{3}\cdot 2\sqrt{3}\cdot \frac{\pm 1}{2}= \pm 30
\\
\left\langle v,b\right\rangle
& = \left\| v \right\| \left\| b \right\| \cos \frac{\theta}{2}
= 10\sqrt{3}\cdot3\sqrt{3}\cdot\frac{\pm 1}{2}=\pm 45
\end{aligned}
$$
On the other hand,
$$
\begin{aligned}
30 = \left\langle v,a\right\rangle &= \left\langle xm+yn,2m-2n\right\rangle
= 2\left(x \left\| m\right\| ^2 + (y - x) \left\langle m,n\right\rangle - y \left\| n\right\|^2 \right) =
\\ &
= 2\left(x -\frac{y - x}{2} - y \right) = 3x-3y
\\
45 = \left\langle v,b\right\rangle &= \left\langle xm+yn,3m+6n\right\rangle
= 3\left(x \left\| m\right\| ^2 + (y + 2x) \left\langle m,n\right\rangle + 2 y \left\| n\right\|^2 \right)=
\\ &
= 3\left(x -\frac{y + 2 x}{2} + y \right) = \frac{9}{2}y
\end{aligned}
$$
Thus we have a linear system
$$
\begin{cases}
\displaystyle 3x - 3y = 30 \\
\displaystyle \dfrac{9}{2}y = 45
\end{cases}
\implies
\begin{cases}
\displaystyle x = 10 + y\\
\displaystyle y = 10
\end{cases}
\implies
\begin{cases}
\displaystyle x = 20\\
\displaystyle y = 10
\end{cases}
$$
Finally, we write the answer
$$
\bbox[5pt, border: 2pt solid #FF0000]{v = 20 m + 10 n}
$$
One can check that indeed $ \left\| v\right\| = 10\sqrt{3}$.
|
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"url": "https://math.stackexchange.com/questions/1400189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Inverse Laplacetransform of rational function with multiple pole I have to calculate the inverse Laplacetransorm of this function using Residue calculus
$$
\frac{s^4 + 6s^3 - 10s^2 + 1}{s^5}
$$
but I can't find any Residue rule that would solve this. Can you show me how to solve this without using partial fractions(I doubt that its even possible to do it using partial fractions).
I have used some very poorly described method from my book, that i never fully understood and never got the right results.
|
If $F(s)=\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)$ is the Laplace Transform of $f(t)$, then the Inverse Laplace Transform is given by
$$f(t)=\frac{1}{2\pi i}\int_{\sigma -i\infty}^{\sigma +i\infty}e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)\,ds \tag 1$$
where here $\sigma >0$.
For $t>0$, we will close the contour with an "infinite semi-circle" that enclosed the left-half plane, while for $t<0$ we will close the contour with an "infinite semi-circle" that encloses the right-half plane. Then, the integral $(1)$ is equal to $2\pi i$ times the residue of the integrand.
For $t>0$, we close the contour in the left-half plane for which $F(s)$ has a pole at $s=0$. We can find the residue by first expanding $e^{st}$ in its Taylor series
$$e^{st}=\sum_{n=0}^{\infty}\frac{(st)^n}{n!}=1+(st)+\frac12(st)^2+\frac16(st)^3+\frac{1}{24}(st)^4+\cdots \tag 2$$
and computing the coefficient on the term $s^{-1}$ of $e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)$. Proceeding, we find that
$$\text{Res}\left(e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right),s=0\right)=1+6t-5t^2+\frac{1}{24}t^4$$
Therefore, for $t>0$
$$f(t)=1+6t-5t^2+\frac{1}{24}t^4$$
For $t<0$, we close the contour in the right-half plane for which there are no singularities and find that $f(t)=0$ for $t<0$.
Putting it all together, we have
$$\bbox[5px,border:2px solid #C0A000]{f(t)=\left(1+6t-5t^2+\frac{1}{24}t^4\right)u(t)}$$
where $u(t)$ is the unit-step function.
NOTE:
In this note, we address a specific request regarding the calculation of the residue. To do this we rely on the expansion given by $(2)$.
Now, the residue is equal to the coefficient on the $\frac1s$ term of the product
$$\begin{align}
e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)&=\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)\sum_{n=0}^{\infty}\frac{(st)^n}{n!}\\\\
&=\frac1s\sum_{n=0}^{\infty}\frac{(st)^n}{n!} \tag{A}\\\\
&+\frac{6}{s^2}\sum_{n=0}^{\infty}\frac{(st)^n}{n!} \tag{B}\\\\
&-\frac{10}{s^3}\sum_{n=0}^{\infty}\frac{(st)^n}{n!}\tag{C}\\\\
&+\frac{1}{s^5}\sum_{n=0}^{\infty}\frac{(st)^n}{n!} \tag{D}
\end{align}$$
In $(A)$ the coefficient on the $s^{-1}$ term is $1$. This result comes from multiplying $\frac1s$ by the first term in the series, $\frac{(st)^0}{0!}=1$, where $0!$ is defined to be $1$. Thus, $$\frac1s \times \frac{(st)^0}{0!}=\frac1s$$
In $(B)$ the coefficient on the $s^{-1}$ term is $6t$. This result comes from multiplying $\frac{6}{s^2}$ by the second term in the series, $\frac{(st)^1}{1!}=st$. Thus, $$\frac{6}{s^2}\times \frac{(st)^1}{1!}=\frac{6t}{s}$$
In $(C)$ the coefficient on the $s^{-1}$ term is $-5t^2$. This result comes from multiplying $-\frac{10}{s^3}$ by the third term in the series, $\frac{(st)^2}{2!}=\frac12(st)^2$. Thus, $$-\frac{10}{s^3}\times \frac{(st)^2}{2!}=\frac{-5t^2}{s}$$
In $(D)$ the coefficient on the $s^{-1}$ term is $\frac{1}{24}t^4$. This result comes from multiplying $\frac{1}{s^5}$ by the fifth term in the series, $\frac{(st)^4}{4!}=\frac{1}{24}(st)^4$. Thus, $$\frac{1}{s^5}\times \frac{(st)^4}{4!}=\frac{\frac{1}{24}t^4}{s}$$
Adding the four contributions gives the expected result.
|
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"url": "https://math.stackexchange.com/questions/1400841",
"timestamp": "2023-03-29T00:00:00",
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|
Finding values of $a$ with which two equations are equivalent; getting rid of radical sign Two equations are given:
$$x^2+(a^2-5a+6)x=0$$
$$x^2+2(a-3)x+a^2-7a+12=0$$
We need to find the values of $a$ that will render them equivalent.
From the first equation,
$$x=-a^2+5a-6$$
From the second,
$$x=\frac{-2a+6\pm2\sqrt{a-3}}{2}$$
If the two equations are equivalent, then x=x
$$-a^2+5a-6=\frac{-2a+6\pm2\sqrt{a-3}}{2}$$
That is, we must have two roots, considering the $\pm$ sign. From this, I've gotten as far as (for the plus sign case):
$$2a^2-12a+18+2\sqrt{a-3}=0$$
For the minus case, there will be a minus before the term $a\sqrt{a-3}$. Basically these two equations will have the roots a=3 and a=4 respectively (from the answers section in my textbook).
But how does one get to that? How to get rid of the square root sign? Or is there an alternative way to solve the whole shebang?
|
Notice, we have $$x^2+(a^2-5a+6)x=0\tag 1$$
$$x^2+2(a-3)x+a^2-7a+12=0\tag 2$$
Now, since the coefficients of $x^2$ in both the equations is $1$ hence both the above equations will be equivalent to each other if their corresponding coefficients of $x$ & constant terms are equal hence, by comparison, we have the following cases $$\begin{cases} a^2-5a+6=2(a-3)\\ a^2-7a+12=0\end{cases}$$
or $$\begin{cases} a^2-7a+12=0\\ a^2-7a+12=0\end{cases}$$
or $$\begin{cases} (a-3)(a-4)=0\\ (a-3)(a-4)=0\end{cases}$$
Hence, we get $$a-3=0\iff a=3$$
or $$a-4=0\iff a=4$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1401086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
Prove the root is less than $2^n$
A polynomial $f(x)$ of degree $n$ such that coefficient of $x^k$ is $a_k$. Another constructed polynomial $g(x)$ of degree $n$ is present such that the coefficeint of $x^k$ is $\frac{a_k}{2^k-1}$. If $1$ and $2^{n+1}$ are roots of $g(x)$, show that $f(x)$ has a positive root less than $2^n$.
Personally a tough problem. Hints Only Please!
I got that: $f(x) = g(2x) - g(x)$
Also,
$f(x) = a_0 + \sum_{k=1}^{n} a_k x^k$ and $g(x) = \sum_{k=1}^{n} \frac{a_k x^k}{2^k - 1}$ and $h(x) = g(2x) = \sum_{k=1}^{n} \frac{2^k a_k x^k}{2^k - 1}$
$g(1) = \sum_{k=1}^{n} \frac{a_k}{2^k - 1} = 0$
$g(2^{n+1}) = \sum_{k=1}^{n} \frac{2^k a_k 2^{nk}}{2^k - 1} = 0$
Realize that: $g(2^{n + 1}) = g(2 \cdot 2^{n}) = h(2^n). $ Hence the root of $h(x) = g(2x) \implies x = 2^n$.
Since $h(1/2) = g(1) = 0$, it follows $x= \frac{1}{2}$ is a root for $h(x) = g(2x)$.
So I have:
$g(x) = 0 \implies x = \{1, 2^{n+1} \}$
$g(2x) = 0 \implies x = \{\frac{1}{2}, 2^{n}\}$.
Now suppose $x = 2^{n} + i$. Then,
$g(2x) = g(2^{n+1} + 2i) > g(2^{n+1}) = 0$.
$g(x) = g(2^n + i)$
But I'm not sure what to do next?
|
Assume that you meant that the coefficient of $x^k$ in $g(x)$ is $\frac{a_k}{2^k-1}$ for each $k=1,2,\ldots,n$ and that the constant term of $f(x)$ is $0$. If $f$ has no root in the interval $\left(1,2^{n}\right)$, then either it is strictly positive or strictly negative on this interval. Without loss of generality, assume that $f$ is strictly positive on $\left(1,2^n\right)$. Show that $g\left(2^{n+1}\right)>0$, which contradicts the condition that $g\left(2^{n+1}\right)=0$. Hence, $f$ must have a root in the interval $\left(1,2^n\right)$.
If the constant term of $f$ may be nonzero, then this problem is false. A counterexample is given by $f(x)=3x^2-5x+3$ and $g(x)=x^2-5x+4$.
|
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"url": "https://math.stackexchange.com/questions/1401440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Solve for $v$ - simplify as much as possible Solve for $v$. Simplify the answer.
$$-3 = -\frac{8}{v-1}$$
Here is what I tried:
$$-3 = \frac{-8}{v-1} $$
$$(-8) \cdot (-3) = \frac{-8}{v-1} \cdot (-8) $$
$$24 = v-1$$
$$25 = v$$
|
I'd probably start by taking a reciprocal of both sides:
$$
\begin{align}
-3&=-\frac{8}{v-1}\\
-\frac{1}{3}&=-\frac{v-1}{8}\\
(-8)-\frac{1}{3}&=-\frac{v-1}{8}(-8)\\
\frac{8}{3}&=v-1\\
\frac{8}{3}+1&=v\\
\frac{11}{3}&=v
\end{align}
$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1401865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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|
Is $(x^2 + 1) / (x^2-5x+6)$ divisible? I'm learning single variable calculus right now and at current about integration with partial fraction. I'm stuck in a problem from few hours given in my book. The question is to integrate $$\frac{x^2 + 1}{x^2-5x+6}.$$
I know it is improper rational function and to make it proper rational fraction we have to divide
$$\frac{x^2 + 1}{x^2-5x+6}$$
I'm trying from sometime but couldn't find the right solution.
Please help! Thank you in advance.
|
Notice, $$\frac{x^2+1}{x^2-5x+6}=\frac{(x^2-5x+6)+5(x-1)}{x^2-5x+6}$$
$$=1+5\frac{x-1}{x^2-5x+6}$$
$$\implies \frac{x-1}{x^2-5x+6}
=\frac{x-1}{(x-2)(x-3)}$$ $$=\frac{A}{x-2}+\frac{B}{x-3}$$
By comparing the corresponding coefficients, we get $A=-1$, $B=2$, hence
$$\frac{x-1}{x^2-5x+6}=-\frac{1}{x-2}+\frac{2}{x-3}$$ Hence, we have $$\frac{x^2+1}{x^2-5x+6}=1-\frac{5}{x-2}+\frac{10}{x-3}$$
|
{
"language": "en",
"url": "https://math.stackexchange.com/questions/1402824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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|
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.