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How to solve equation: $ \frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0 $ How to solve this equation?
$$
\frac{x+1}{9x^3-4x} + \frac{3x}{27x^3 - 12x + 8} - \frac{1}{(3x-2)^2}=0
$$
I try
$$
\frac{81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16}{x(3x-2)^2(3x+2)(27x^3 - 12x + 8)}=0
$$
And then
$$
81x^5 - 81x^4 - 90 x^3 + 36 x^2 + 16x -16 =0
$$
here $( x \neq 0; x \neq \frac{2}{3}; x \neq - \frac{2}{3}; 27x^3 - 12x + 8 \neq 0 )$.
Is there some simple (and different) way to solve it?
|
This graph helps to show what's happening.
It's not an answer but a companion to Claude's, in that it helps illustrate the turning points he identified and shows the fact that there's one solution. Note the scales are different, I had to stretch the Y axis.
|
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|
Factoring question involving 4 integers. For all non-negative integer values of $a$, $b$, $c$, $d$ given that $ac+bd+bc+ad=42$ and $c^2-d^2=12$, then determine all possible values of $a+b+c+d$.
This was a question on one of my previous tests and I am not sure how to do it.
|
$c^2 - d^2 = 12 \iff (c+d)(c-d)=12$. The only integers $c,d$ satysfying this equality are $c=4, d=2$. Putting these values into $ac + bd + bc + ad = 42$ we get that $7a+7b = 42 \iff a+b=6$. That means that there are 5 quadruples (a,b,c,d), namely (1,5,4,2), (2,4,4,2), (3,3,4,2), (4,2,4,2), (5,1,4,2).
|
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|
Ordered triples of n-powerful integers
Let’s say that an ordered triple of positive integers (a, b, c) is n-powerful
if:
*
*$a \le b \le c$,
*$gcd(a, b, c) = 1$ and
*$a^n + b^n + c^n$ is divisible by $a + b + c$.
For example, $(1, 2, 2)$ is 5-powerful.
a) Determine all ordered triples (if any) which are n-powerful for all $n \ge 1$.
b) Determine all ordered triples (if any) which are
2004-powerful and 2005-powerful, but not 2007-powerful.
This question is taken from the 2005 Canada National Olympiad.
I have established some results, but these are not of (much) direct relevance to the questions asked.
Let:
*
*$P^n$ be the set of all n-powerful triples
*$s=a+b+c$
*$\mathbb{P}$ be the set of all primes
*$p$ be an odd prime
By the binomial theorem,
$$\begin{align}
a^n+b^n+c^n&=a^n+b^n+\Big(s-(a+b)\Big)^n \\
&\equiv a^n+b^n+(-1)^n(a+b)^n\pmod{s} \tag{1}
\end{align}$$
Part (a)
Possibly the simplest solution is $\boxed{n=s\in\mathbb{P}_{>2}}$. Any triple satisfying this and the gcd condition is n-powerful, e.g. $(1,4,6)\in P^{11}$.
Suppose we have $4\mid s,n\text{ even}$
Then $a^n\equiv 0,1\pmod{4},b^n\equiv 0,1\pmod{4},\text{ and }(a+b)^n\equiv 0,1\pmod{4}$ so for n-power (1) requires that $a,b$ are both even, so that $gcd(a,b,c)>1$. So then $(a,b,c)$ is not n-powerful.
Therefore $$\boxed{n\text{ even}\implies 4\nmid s}$$
For $n \ge 3,1\le a\le b$ we have
$$\begin{align}
(a+b)^n &\ge a^n+b^n + na^{n-1}b + nab^{n-1}\\
&\ge a^n+b^n+6ab \\
&> a^n+b^n+(a+2b)
\end{align}$$
For $n=2$, the RHS (1) exceeds $(a+2b)$.
The requirement $1\le a \le b \le c$ implies only that $1\le a \le b \land s \ge a+2b$. So as long as $a,b$ are chosen such $1\le a \le b$ and $gcd(a,b)=1$ then we can find at least one $s$ satisfying (1) with $s\ge a+2b$, giving the triple $(a,b,s-a-b)$.
Suppose now that $n=ks$ with $k$ a positive integer and $s$ an odd prime.
Then we have by Fermat's Little Theorem:
$$\begin{align}
a^n \equiv (a^s)^k \equiv a^k\pmod{s} \tag{2}\\
b^n\equiv b^k\pmod{s} \tag{3}
\end{align}$$
In addition, since $s$ is prime, we have
$$\begin{align}
(a+b)^n=\Big((a+b)^s\Big)^k&\equiv(a^s+b^s)^k\pmod{s}&\left(s\mid\binom{s}{j}\text{ for }1\le j\le s-1\right) \\
&\equiv(a+b)^k\pmod{s}&(\text{by Fermat's Little})
\end{align}$$
Trivially, $$(-1)^n=(-1)^k$$
So then (1) reduces to:
$$a^n+b^n+c^n\equiv a^k+b^k+(-1)^k(a+b)^k\pmod{s} \tag{4}$$
which is of the same form as (1).
This gives a recipe for generating higher-power n-powerful triples from lower-power ones. For example,
*
*$1^2+2^2+4^2 \equiv 0\pmod7 \mapsto 1^{14}+2^{14}+4^{14}\equiv1+4+16\equiv0\pmod7$
*i.e. $(1,2,4)\in P^2 \mapsto (1,2,4)\in P^{14}$
I cannot see how to take these ideas further.
Part (b)
If $(a,b,c)$ is both 2004-powerful and 2005-powerful then we require simultaneously that:
$$\begin{align}
a^{2004}+b^{2004}+(a+b)^{2004}&\equiv0\pmod{s} \\
a^{2005}+b^{2005}-(a+b)^{2005}&\equiv0\pmod{s}
\end{align}$$
We can clear the $(a+b)^n$ term by multiplying the first equation through by $(a+b)$ and adding to the second equation to get:
$$(a+b)(a^{2004}+b^{2004})+a^{2005}+b^{2005}\equiv0\pmod{s}$$
but this seems to lead nowhere.
|
Let $P_i = a^i + b^i + c^i$
Hint for part 2: Apply Newton's Identities.
Show that $P_{2007} = A P_{2004} + B P _{2005} + C (a+b+c) $ where $A,B,C$ are constants to be determined.
Not sure about part 1 as yet, but there are solution sets of $(1,1,1)$ and $(1,1,4)$.
|
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|
Finding $a$ yielding minimum value for quadratic root expression $(x_1+2x_2)(x_2+2x_1)$ The problem is:
We have the expression $(x_1+2x_2)(x_2+2x_1)$, where $x_1$ and $x_2$ are the roots of $$f(x)=x^2+ax+a+\frac{1}{5}$$
Find the value(s) of $a$ yielding the least possible value for this expression.
My solution is: for readability's sake, $x_1=t, x_2=z$, then
$$(x_1+2x_2)(x_2+2x_1)=(t+2z)(z+2t)= .. =2(t+z)^2-tz$$
Using Vieta's formulas, we get
$$2(t+z)^2-tz=2\left(\frac{-B}{A}\right)^2-\frac{C}{A}=2a^2-a-0.2$$
From here, the minimum point is the apex of the parabola:
$$a_0=\frac{-B}{2A}=\frac{1}{4}$$
Yet the textbook's answer is $$-\frac{1}{4}$$
Is there a mistake in my calculations?
|
Given $x_1$ and $x_2$ are the roots of $\displaystyle f(x)=x^2+ax+a+\frac{1}{5} = 0$
So $\displaystyle x_{1}+x_{2} = -a$ and $\displaystyle x_{1}\cdot x_{2} = a+\frac{1}{5}$
Now $$\displaystyle (x_1+2x_2)(x_2+2x_1) = x_{1}\cdot x_{2}+2\left[x_{1}^2+x_{2}^2\right]+4x_{1}\cdot x_{2} = 5x_{1}x_{2}+2\left[\left(x_{1}+x_{2}\right)^2-2x_{1}\cdot x_{2}\right]$$
So Let $$\displaystyle f(a) = -5\left(a+\frac{1}{5}\right)+2\left[a^2-2\left(a+\frac{1}{5}\right)\right] = 2a^2+a+\frac{1}{5}$$
So $$\displaystyle f(a) = 2\left[a^2+\frac{a}{2}+\frac{1}{10}\right] = 2\left[a^2+\frac{a}{2}+\frac{1}{4^2}-\frac{1}{4^2}+\frac{1}{10}\right] = 2\left[\left(a+\frac{1}{4}\right)^2+\frac{3}{80}\right]$$
Thus, the minimum of the fuunction $f(a)$ occurs when $\displaystyle \left(a+\frac{1}{4}\right)^2 = 0\Rightarrow a= -\frac{1}{4}.$
|
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|
Find $\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx$ $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\frac{u}{v}$$ where $u$ and $v$ are in their lowest form. Find the value of $\dfrac{1000u}{v}$
$$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^4-2x^2+1}}dx=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt{2x^2(x^2-1)+1}}dx$$ I put $x^2-1=t$ but no benefit. Please guide me.
|
Notice, we have $$\int_{1}^{2}\frac{x^2-1}{x^3\sqrt {2x^4-2x^2+1}}dx$$
$$=\int_{1}^{2}\frac{x^2-1}{x^3\sqrt {2\left(x^2-\frac{1}{2}\right)^2-\frac{1}{2}+1}}dx$$
$$=\frac{1}{2}\int_{1}^{2}\frac{2x(x^2-1)}{x^4\sqrt {2\left(x^2-\frac{1}{2}\right)^2+\frac{1}{2}}}dx$$ Now, $$x^2-\frac{1}{2}=t\implies 2xdx=dt$$
$$=\frac{1}{2}\int_{1/2}^{7/2}\frac{\left(t-\frac{1}{2}\right)dt}{\left(t-\frac{1}{2}\right)^2\sqrt {2t^2+\frac{1}{2}}}dx$$
$$=\frac{1}{2}\int_{1/2}^{7/2}\frac{dt}{\left(t-\frac{1}{2}\right)\sqrt {2t^2+\frac{1}{2}}}dx$$
I hope you can take it from here.
|
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|
Where does this sequence $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.... converge? The given sequence is $\sqrt{7}$,$\sqrt{7+ \sqrt{7}}$,$\sqrt{7+\sqrt{7+\sqrt{7}}}$,.....and so on.
the sequence is increasing so to converge must be bounded above.Now looks like they would not exceed 7. The given options are
*
*${1+\sqrt{33}}\over{2}$
*${1+\sqrt{32}}\over{2}$
*${1+\sqrt{30}}\over{2}$
*${1+\sqrt{29}}\over{2}$
How to proceed now.
Thanks for any help.
|
Trick: Let $X = \sqrt{ 7 + \sqrt{ 7 + ... } } $. We have $X = \sqrt{ 7 + X } $ and so $X^2 = 7 + X $. Now you solve the quadratic equation.
|
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|
Integrating $\frac{1}{(x^4 -1)^2}$ How to solve the the following integral? $$\int{\frac{1}{(x^4 -1)^2}}\, dx$$
|
HINT: Integrate by parts
$$\int\dfrac{dx}{(x^4-1)^2}=\dfrac1{x^3}\int\dfrac{x^3}{(x^4-1)^2}-\int\left(\dfrac{d(1/x^3)}{dx}\cdot\int\dfrac{x^3}{(x^4-1)^2}\right)dx$$
$$=-\dfrac1{x^3}\cdot\dfrac1{4(x^4-1)}-\dfrac34\int\dfrac{dx}{x^4(x^4-1)}$$
Now $\dfrac1{x^4(x^4-1)}=\dfrac{x^4-(x^4-1)}{x^4(x^4-1)}=\dfrac1{x^4-1}-\dfrac1{x^4}$
Again, $\dfrac1{x^4-1}=\dfrac12\cdot\dfrac{x^2+1-(x^2-1)}{(x^2+1)(x^2-1)}=?$
|
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|
$y=\tan^{-1}\frac{1}{x^2+x+1}+\tan^{-1}\frac{1}{x^2+3x+3}+\tan^{-1}\frac{1}{x^2+5x+7}+\tan^{-1}\frac{1}{x^2+7x+13}......$to n terms. Prove that if $y=\tan^{-1}\frac{1}{x^2+x+1}+\tan^{-1}\frac{1}{x^2+3x+3}+\tan^{-1}\frac{1}{x^2+5x+7}+\tan^{-1}\frac{1}{x^2+7x+13}......$to n terms.Then $\frac{dy}{dx}=\frac{1}{1+(x+n)^2}-\frac{1}{1+x^2}$
I could simplify only first term of $y$,not rest.
$\tan^{-1}\frac{1}{x^2+x+1}=\tan^{-1}\frac{(x+1)-x}{1+x(x+1)}=\tan^{-1}x+1-\tan^{-1}x$
and i could not judge $n$th term in $y$.Please help me.
|
We can write $$\displaystyle \tan^{-1}\left(\frac{1}{x^2+x+1}\right) = \tan^{-1}\left(\frac{1}{1+(x+1)\cdot x}\right) = \tan^{-1}\left(x+1\right)-\tan^{-1}(x)$$.
Similarly $$\displaystyle \tan^{-1}\left(\frac{1}{x^2+3x+3}\right) = \tan^{-1}\left(\frac{1}{1+(x+2)\cdot (x+1)}\right) = \tan^{-1}\left(x+2\right)-\tan^{-1}(x+1)$$.
Similarly $$\displaystyle \tan^{-1}\left(\frac{1}{x^2+5x+7}\right) = \tan^{-1}\left(\frac{1}{1+(x+3)\cdot (x+2)}\right) = \tan^{-1}\left(x+3\right)-\tan^{-1}(x+2)$$.
Similarly for $\bf{n^{th}}$ term
$$\displaystyle \tan^{-1}\left(\frac{1}{x^2+x(2n-1)+n^2-n+1}\right) = \tan^{-1}\left(\frac{1}{1+(x+n)\cdot (x+n-1)}\right) = \tan^{-1}\left(x+n\right)-\tan^{-1}(x+n-1)$$.
Now Adding all these, we get $$\displaystyle y = \tan^{-1}(x+n)-\tan^{-1}(x)$$
And Differentitae both side w . r . to $x\;,$ We get $$\displaystyle \frac{dy}{dx} = \frac{1}{1+(x+n)^2}-\frac{1}{1+x^2}.$$
|
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|
Prove the trigonometric identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$ While solving an equation i came up with the identity $\cos(x) + \sin(x)\tan(\frac{x}{2}) = 1$.
Prove whether this is really true or not. I can add that $$\tan\left(\frac{x}{2}\right) = \sqrt{\frac{1-\cos x}{1+\cos x}}$$
|
The identity is true.
\begin{align*}
\cos x + \sin x \tan \frac{x}{2} & =(\cos^2\frac{x}{2}-\sin^2\frac{x}{2})+(2\sin \frac{x}{2}\cos \frac{x}{2})(\frac{\sin \frac{x}{2}}{\cos\frac{x}{2}})\\
& = \cos^2\frac{x}{2}+\sin^2 \frac{x}{2}\\
& = 1
\end{align*}
|
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|
Does the elliptic function $\operatorname{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\big|\frac{1}{2}\right)$ have a closed form? Given the complete elliptic integral of the first kind $K(k)$ for the modulus $k$,
can the elliptic function $$\text{cn}\left(\frac{2}{3}K\left(\frac{1}{2}\right)\bigg|\frac{1}{2}\right)$$ be expressed in closed form?
|
To explain how an algebraic equation for $Z:=\operatorname{cn} \left(\frac{2K}{3}|m\right)$ can be obtained, notice that
*
*The usual parity properties imply that
$$\operatorname{cn} \left(\frac{4K}{3}\biggl|\,m\right)=\operatorname{cn} \left(2K-\frac{2K}{3}\biggl|\,m\right)=-\operatorname{cn} \left(-\frac{2K}{3}\biggl|\,m\right)=-\operatorname{cn} \left(\frac{2K}{3}\biggl|\,m\right)=-Z.$$
*On the other hand the doubling formula implies that
$$\operatorname{cn} \left(\frac{4K}{3}\biggl|\,m\right)=\frac{1-2\operatorname{sn}^2 \left(\frac{2K}{3}|m\right)+m\operatorname{sn}^4 \left(\frac{2K}{3}|m\right)}{1-m\operatorname{sn}^4 \left(\frac{2K}{3}|m\right)}=\frac{2Z^2-1+m\left(1-Z^2\right)^2}{1-m\left(1-Z^2\right)^2}.$$
Thus $Z$ satisfies the equation
$$\frac{2Z^2-1+m\left(1-Z^2\right)^2}{1-m\left(1-Z^2\right)^2}=-Z.$$
Although naively it is of $5$th order, it has an obvious root $Z=-1$, so we can deduce from it a $4$th order equation
$$\boxed{\quad m\left(Z^2-1\right)\left(Z-1\right)^2-2Z+1=0\quad}$$
In particular, the relevant solution for $m=k^2=\frac12$ is given by
$$Z=\frac{1-\sqrt2\cdot\sqrt[4]{3}+\sqrt3}{2}\approx 0.435421.$$
|
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If $ a + b + c \mid a^2 + b^2 + c^2$ then $ a + b + c \mid a^n + b^n + c^n$ for infinitely many $n$
Let $ a,b,c$ positive integer such that $ a + b + c \mid a^2 + b^2 + c^2$.
Show that $ a + b + c \mid a^n + b^n + c^n$ for infinitely many positive integer $ n$.
(problem composed by Laurentiu Panaitopol)
So far no idea.
|
If $a,b,c,n\in\Bbb Z_{\ge 1}$, $a+b+c\mid a^2+b^2+c^2$, then $$a+b+c\mid a^n+b^n+c^n$$
is true when $n\nmid 3$, but not necessarily when $n\mid 3$.
$$x^2+y^2+z^2+2(xy+yz+zx)=(x+y+z)^2$$
$$\implies x+y+z\mid 2(xy+yz+zx)$$
$$\implies x+y+z\mid (x^k+y^k+z^k)(xy+yz+zx)$$
for all $k\ge 1$ (to see why, check cases when $x+y+z$ is even and when it's odd).
$$x^{n+3}+y^{n+3}+z^{n+3}=(x^{n+2}+y^{n+2}+z^{n+2})(x+y+z)$$
$$-(x^{n+1}+y^{n+1}+z^{n+1})(xy+yz+zx)+(x^n+y^n+z^n)xyz$$
for all $n\ge 1$. We know $$x+y+z\mid (x^{n+2}+y^{n+2}+z^{n+2})(x+y+z)$$
$$-(x^{n+1}+y^{n+1}+z^{n+1})(xy+yz+zx)$$
Now let $(x,y,z)=(x_1,y_1,z_1)=(1,3,9)$. $$x_1+y_1+z_1\nmid x_1^3+y_1^3+z_1^3$$ $$x_1+y_1+z_1\nmid \left(x_1^3+y_1^3+z_1^3\right)x_1y_1z_1$$ $$\implies x_1+y_1+z_1\nmid x_1^6+y_1^6+z_1^6$$
Since $x_1+y_1+z_1$ is coprime to $x_1,y_1,z_1$, we get $$x_1+y_1+z_1\nmid (x_1^6+y_1^6+z_1^6)x_1y_1z_1,$$
and so $x_1+y_1+z_1\nmid x_1^9+y_1^9+z_1^9$, etc.
Therefore $x+y+z$ cannot generally (for all $x,y,z\in\mathbb Z_{\ge 1}$) divide $x^{3m}+y^{3m}+z^{3m}$ for any given $m\ge 1$.
However, we easily get $x+y+z$ always divides $x^n+y^n+z^n$ for $n$ not divisible by $3$,
because $x+y+z\mid (x+y+z)xyz$ and $x+y+z\mid \left(x^2+y^2+z^2\right)xyz$,
because $x+y+z\mid x^2+y^2+z^2$ (given), so $x+y+z\mid x^4+y^4+z^4, x^5+y^5+z^5$,
so $x+y+z\mid \left(x^4+y^4+z^3\right)xyz, \left(x^5+y^5+z^5\right)xyz$,
so $x+y+z\mid x^7+y^7+z^7, x^8+y^8+z^8$, etc.
|
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How to find $ab+cd$ given that $a^2+b^2=c^2+d^2=1$ and $ac+bd=0$? It is given that $a^2+b^2=c^2+d^2=1 $
And it is also given that $ac+bd=0$
What then is the value of $ab+cd$ ?
|
A solution by Sumit Ray
$ac=-bd$
$\frac{a}{b} = -\frac{d}{c} = k$
$a=bk \text{ and }d=-ck$
$a^2+b^2=1\implies b^2 = \frac{1}{k^2+1}\implies c^2 = \frac{1}{k^2+1}$
Thus $b^2 - c^2 =0$
Now \begin{align*}ab+cd &= b^2\cdot k-c^2\cdot k\\&= k(b^2-c^2)
= 0\end{align*}
|
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How to find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$ from the following equation? I have a question about polynomial.
Given a polynomial:
$$x^4-7x^3+3x^2-21x+1=0$$
Given too that the roots of this polynomial are $a, b, c,$ and $d$.
Find the value of $(a+b+c)(a+b+d)(a+c+d)(b+c+d)$?
My attempt:
It seems I need to apply Vieta formula to find the relationship between its roots. From there, I can get:
$$a+b+c+d = 7$$
$$ab+ac+ad+bc+bd+cd= 3$$
$$abc+abd+acd+bcd= 21$$
$$abcd= 1$$
Then,
$$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$$
$$= (a²+b²+ab+3)(c²+d²+cd+3)$$
$$= 3(a^2+b^2+c^2+d^2)+abcd+(ac)^2+(ad)^2+(bc)^2+(bd)^2+a^2cd+b^2cd+abc^2+abd^2$$
From there, I don't have any idea how to go further.
Can somebody help me to explain to solve this equation?
Thanks.
|
Based on Vieta's formula and Brian and GAVD answers:
From Vieta formula, we got:
$a+b+c+d=7$
$ab+ac+ad+bc+bd+cd=3$
$abc+abd+acd+bcd=21$
$abcd=1$
Then:
$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$
$= (a+b+c+d-d)(a+b+c+d-c)(a+b+c+d-b)(a+b+c+d-a)$
$= (7-d)(7-c)(7-b)(7-a)$
Simplify and we got:
$(a+b+c)(a+b+d)(a+c+d)(b+c+d)$
$=(7−a)(7−b)(7−c)(7−d)$
$=7^4−7^3(a+b+c+d)+7^2(ab+ac+ad+bc+bd+cd)\\ \ \ −7(abc+abd+bcd+acd)+abcd$
Substitute all values and we got:
$=7^4-7^3(7)+7^2(3)-7(21)+1$
$=7(21)-7(21)+1$
$=1$
|
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|
Show that the function is continuous To show that the function $f: \mathbb{R}^2 \rightarrow\mathbb{R}$ with $f=\left\{\begin{matrix}
\frac{x^3-y^3}{x^2+y^2} & , (x,y) \neq (0,0)\\
0 & , (x,y)=(0,0)
\end{matrix}\right.$ is continuous on $(0,0)$ we have to show that $|f(x,y)-f(x_0,y_0)| \leq L ||(x,y)-(x_0,y_0)||$ so we have to show that $\left |\frac{x^3-y^3}{x^2+y^2}\right | \leq L \sqrt{x^2+y^2}$.
I have done the following:
$$\left |\frac{x^3-y^3}{x^2+y^2}\right |=\frac{|x-y||x^2+xy+y^2|}{x^2+y^2}\leq \dfrac{(|x-y|)(x^2+y^2+|xy|)}{x^2+y^2} \leq 2\dfrac{(|x-y|)(x^2+y^2)}{x^2+y^2} = 2(|x-y|)\leq 2(|x|+|y|)\leq 2\sqrt{(|x|+|y|)^2}=2\sqrt{|x|^2+|y|^2+2|xy|} \leq 2\sqrt{|x|^2+|y|^2+|x|^2+|y|^2}=2\sqrt{2(|x|^2+|y|^2)}=2\sqrt{2}\sqrt{|x|^2+|y|^2}=2\sqrt{2}\sqrt{x^2+y^2}=2\sqrt{2}||(x,y)||$$
Is this correct?
|
Note that$$\frac{x^3}{x^2 + y^2} = x\cdot\frac{x^2}{x^2 + y^2}$$ In absolute value, this is $\le |x|.$ As $(x,y)\to (0,0),|x| \to 0.$ Same thing for $y^3/(x^2 + y^2).$ So $f(x,y)$ is the difference of two functions that both $\to 0$ as $(x,y) \to (0,0);$ hence $\lim_{(x,y)\to (0,0)}f(x,y) = 0.$ This shows $f$ is continuous at $(0,0).$
|
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|
Trouble understanding inequality proved using AM-GM inequality I am studying this proof from Secrets in Inequalities Vol 1 using the AM-GM inequality to prove this question from the 1998 IMO Shortlist. However, I'm lost on the very first line of the solution.
Let $x,y,z$ be positive real numbers such that $xyz =1$. Prove that
$$
\frac{x^3}{(1+y)(1+z)} + \frac{y^3}{(1+z)(1+x)} + \frac{z^3}{(1+x)(1+y)} \ge \frac{3}{4}
$$
Using AM-GM in the following form:
$\frac{x^3}{(1+y)(1+z)} + \frac{1+y}{8} + \frac{1+z}{8} \ge \frac{3x}{4}$.
Now, this is where I get lost. Where was this expression derived from?
We conclude that
$\sum_{cyc}\frac{x^3}{(1+y)(1+z)} + \frac{1}{4}\sum_{cyc}(1+x) \ge \sum_{cyc}\frac{3x}{4} \rightarrow \sum_{cyc}\frac{x^3}{(1+y)(1+z)} \ge \frac{1}{4}\sum_{cyc}(2x-1) \ge \frac{3}{4}$
The equality holds for $x=y=z=1$
|
Let me try to cover the motivation for writing something like
$$\frac{x^3}{(1+y)(1+z)} + \frac{1+y}8 + \frac{1+z}8$$
for doing AM-GM, seemingly out of the blue.
We obviously start with the first term. Now clearly, to get rid of the denominator, it would be great to consider the terms $(1+y)$ and $(1+z)$. As there is $x^3$ in the numerator, we can use these as separate terms to do a three term AM-GM, to get a simple $x$ term after the AM-GM. An equally good alternative would be to use the single term $(1+y)(1+z)$ and do a two term AM-GM. So far so good.
Now note that AM-GM will have equality iff all terms are equal. However we know equality is when $x=y=z=1$. We have to maintain this equality possibility in each step taken in the proof, otherwise it cannot hold finally. This leads immediately to having the coefficient $\frac18$ attached to the second and third terms, so that the value of all three terms are the same, viz. $\frac14$ when $x=y=z=1$. Now you have the right set of terms to do the AM-GM!
|
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|
Does the limit $\lim\limits_{x\to0}\left(\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}\right)$ exist? Does the limit: $$\lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$ exist?
|
Given $$\displaystyle \lim\limits_{x\to0}\frac{1}{x\tan^{-1}x}-\frac{1}{x^2}$$
Now Let $$\tan^{-1}(x) = y\Rightarrow x=\tan y\;,$$ Then when $x\rightarrow 0,$ Then $y=\tan^{-1}(0)\rightarrow 0$
So limit convert into $$\displaystyle \lim_{y\rightarrow 0}\frac{1}{y\cdot \tan y}-\frac{1}{\tan^2 y} = \lim_{y\rightarrow 0}\frac{\tan y-y}{y\tan^2 y}$$
Now Using $\bf{D-L\; Hopital \; Rule}$
So $$\displaystyle \lim_{y\rightarrow 0}\frac{\sec^2 y-1}{y\cdot 2\tan y\cdot \sec^2 y+\tan^2 y} = \lim_{y\rightarrow 0}\frac{1-\cos^2 y}{2y\tan y+\sin^2 y} = \lim_{y\rightarrow 0}\frac{\sin^2 y}{2y\tan y+\sin^2 y}$$
So limit $$\displaystyle \lim_{y\rightarrow 0}\frac{1}{\frac{2y\tan y}{\sin^2 y}+1} =\lim_{y\rightarrow 0}\frac{1}{\left(\frac{y}{\sin y}\cdot {2}{\cos y}+1\right)}= \lim_{y\rightarrow 0}\frac{1}{2+1} =\frac{1}{3}$$
above we used $$\displaystyle \lim_{y\rightarrow 0}\frac{\sin y}{y} = 1$$
|
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|
Proving that $\frac{1}{2}<\frac{2}{3}<\frac{3}{4}<$...$<\frac{n-1}{n}$ In an attempt to find a pattern, I did this:
Let a,b,c,d be non-zero consecutive numbers. Then we have:
$a=a$
$b=a+1$
$c=a+2$
$d=a+3$
This implies:
$\frac{a}{b}=\frac{a}{a+1}$
$\frac{b}{c}=\frac{a+1}{a+2}$
$\frac{c}{d}=\frac{a+2}{a+3}$
I don't know how that helps. I'm greatly seeking your help. Thank you very much.
|
If $n>1$ is an integer we have
\begin{align*}
n-1&<n\\
\implies \frac{1}{n-1}&>\frac{1}{n}\\
\iff 1-\frac{1}{n-1}&<1-\frac{1}{n}\\
\iff \frac{n-2}{n-1}&<\frac{n-1}{n}
\end{align*}
|
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|
transformation of uniformly distributed random variable f(x)=1/2pi into Y=cosx Let $X$ be a uniformly distributed function over $[-\pi;\pi]$. That is
$
f(x)=\left\{\begin{matrix}
\frac{1}{2 \pi} & -\pi\leq x\leq \pi \\
0 & otherwise
\end{matrix}\right.\\
$
Find the probability density function of $Y = \cos X$.
I figured out how to derive the final result using the change of variable method, but don't quite get why it needs to be multiplied by $2$.
I cannot get the result using the other cdf method at all.
Would you please help and clarify the steps of the both methods (variable change and cdf) for this type of the problem?
$$f(x)=y=cos(x) $$
$$f^{-1} (y) = \arccos(y)$$
$$f^{-1} (y)'= 1 / \sqrt{1-y^2}$$
$$fy(y)= \mathbf 2 \cdot \left(\frac{1}{2\pi}\right) \cdot \left|-1/\sqrt{1-y^2}\right|
= \frac 1\pi \cdot \sqrt{1-y^2}$$
using cdf method
\begin{align}
Fy(y) &= P(Y \leq y) = P(\cos x \leq y) = P(x \leq -\arccos y) + P(x \geq \arccos y) \\ \\
&= \int_{-\pi}^{-\arccos(y)} \frac{1}{2 \pi} \; dx + \int_{\arccos(y)}^{\pi} \frac{1}{2 \pi} dx
= \left[\frac{x}{2 \pi}\right]_{-\pi}^{-\arccos(y)}
+ \left[\frac{x}{2 \pi}\right]_{\arccos(y)}^{\pi} \\ \\
&= \frac{-\arccos(y)}{2 \pi} - \frac{-\pi}{2 \pi} + \frac{\pi}{2 \pi} -\frac{\arccos(y)}{2 \pi} \\ \\
&= \frac{-\arccos(y)}{2 \pi} + \frac{\pi}{2 \pi} - \frac{arccos(y)}{2 \pi} + \frac{\pi}{2 \pi} = 1- \frac{\arccos(y)}{ \pi} \\ \\
f(y) &= F'(y) = \left(1- \frac{\arccos(y)}{ \pi}\right)' = \frac{1}{\pi \sqrt{1-y^2}}
\end{align}
|
We deal with your question about why we multiply by $2$. We are interested in finding the cumulative distribution function $F(y)$ of $Y$. The only interesting part is when $-1\lt y\lt 1$, because $F(y)=1$ if $y\ge 1$ and $F(y)=0$ when $y\le -1$.
It is geometrically perhaps a little easier to find $G(y)=\Pr(Y\gt y)$. Then the probability that $Y\le y$ is $1-G(y)$. For $y$ between $-1$ and $1$, $Y\gt y)$ when $-\arccos y\lt X\lt \arccos y$. This interval has length $2\arccos y$, and therefore for $y$ between $-1$ and $1$ we have $F(y)=1-\frac{2\arccos y}{2\pi}$.
If we deal with $F(y)$ directly, note that there are two intervals (of equal length) in which $\cos x\le y$.
|
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|
If in a triangle $ABC$,$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A$,then prove that triangle will be equilateral triangle If in a triangle $ABC$ we have
$$1=2\cos A\cos B\cos C+\cos A\cos B+\cos B\cos C+\cos C\cos A\ ,$$
then the triangle will be equilateral triangle.
I tried but except few steps,could not prove it. Putting $A=B=C=\frac{\pi}{3}$ makes both sides equal. How should i prove it?Please help me.
|
As $A+B+C=\pi,\cos(A+B)=\cdots=-\cos C$
and using Prove that $\cos (A + B)\cos (A - B) = {\cos ^2}A - {\sin ^2}B$,
$$2\cos A\cos B\cos C=\cos C[\cos(A-B)+\cos(A+B)]$$
$$=-\cos(A+B)\cos(A-B)-\cos^2C$$
$$=-(\cos^2A-\sin^2B)-\cos^2C$$
$$=1-(\cos^2A+\cos^2B+\cos^2C)$$
So, the equation reduces to $$\sum(\cos A-\cos B)^2=0$$
Can you take it from here?
|
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|
simultaneous equation help Can someone help me with to solve this system of equations ?
$$
\left\{
\begin{array}{c}
y=x+1 \\
y^2+2x^2=2
\end{array}
\right.
$$
|
$$\begin{cases} y=x+1 \\ { y }^{ 2 }+2{ x }^{ 2 }=2 \end{cases}\Rightarrow \begin{cases} y=x+1 \\ { \left( x+1 \right) }^{ 2 }+2{ x }^{ 2 }=2 \end{cases}\Rightarrow 3{ x }^{ 2 }+2x-1=0\quad \Rightarrow x=\frac { -1\pm 2 }{ 3 } ,y=1+\frac { -1\pm 2 }{ 3 } \\ $$
|
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|
Solve equation $m^2=n^5-4$ in $\mathbb{N}$ I have an exercise to solve an equation like in the title.
My goal until now is that both $m$ and $n$ are odd, but then I can not continue. Can you help me?
Thanks a lot!
|
Consider $m^2+4 (\bmod 11):$
$m=0 (\bmod 11) \Rightarrow m^2+4 = 4 (\bmod 11),$
$m=1,10 (\bmod 11) \Rightarrow m^2+4 = 5 (\bmod 11),$
$m=2, 9 (\bmod 11) \Rightarrow m^2 +4= 8 (\bmod 11),$
$m=3, 8 (\bmod 11) \Rightarrow m^2+4 = 2 (\bmod 11),$
$m=4, 7 (\bmod 11) \Rightarrow m^2+4 = 9 (\bmod 11),$
$m=5, 6 (\bmod 11) \Rightarrow m^2+4 = 7 (\bmod 11).$
But $n^5 (\bmod 11)$:
$n=0 (\bmod 11) \Rightarrow n^5=0 (\bmod 11),$
$n=1,3,4,5,9 (\bmod 11) \Rightarrow n^5=1 (\bmod 11),$
$n=2,6,7,8,10 (\bmod 11) \Rightarrow n^5=10 (\bmod 11).$
So, $n^5\ne m^2+4 (\bmod 11)$, and $n^5\ne m^2+4$, if $n,m\in \mathbb{N}.$
|
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|
Get number of additions that lead to a specific sum with given summands Suppose you have $n$ variables. Each of these variables (e.g. $a$) have their own interval between $0$ and $a_i$ (whole numbers). The only rule is that all these variables have to add up to a given value $s$.
An example:
$n = 3$, so the three variables are $a$, $b$ and $c$
$a_i = 2$, so $0\le$ $a$ $\le2$
$b_i = 3$, so $0\le$ $b$ $\le3$
$c_i = 4$, so $0\le$ $c$ $\le4$
$s = 6$, so $a+b+c=6$
The 9 unique possible solutions for these given variables are the following (the order of the variables is still important, so solution 7 and solution 9 are not the same):
$a=0$ $b=2$ $c=4$
$a=0$ $b=3$ $c=3$
$a=1$ $b=1$ $c=4$
$a=1$ $b=2$ $c=3$
$a=1$ $b=3$ $c=2$
$a=2$ $b=0$ $c=4$
$a=2$ $b=1$ $c=3$
$a=2$ $b=2$ $c=2$
$a=2$ $b=3$ $c=1$
What is the correct formula or program to only get the number of unique, possible solutions for any value of $n$, $s$ and the other variables, which would be 9 in this case?
|
Let me work out for this particular case. You should be able to put it into a formula.
If you can't, I'll see later.
We can use stars and bars with inclusion-exclusion.
Without any restrictions, there are ${6+3-1\choose 3-1} = {8\choose 2}= 28$ solutions.
To subtract cases that violate, (say) the $a_i$ restriction, pre-place $(2+1)=3$ in the $a_i$ compartment, and so on, then for combos of two, and combos of three, if possible, so
$${8\choose2} - \left[{5\choose 2} + {4\choose 2} + {3\choose 2}\right] = 9$$
Here it was not possible to violate constraints for combos of 2 or 3, but suppose we raise s to 16, then we would have
$${18\choose2} - \left[{15\choose 2} + {14\choose 2} + {13\choose 2}\right] + \left[{11\choose 2} + {10\choose 2} + {9\choose 2}\right] - \left[{6\choose 2}\right]$$
|
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|
How to get formula for sums of powers? Assuming I have Bernoulli numbers:
$B = [\frac{1}{1},\frac{1}{2},\frac{1}{6},\frac{0}{1},-\frac{1}{30}, \frac{0}{1}, \frac{1}{42}, ...]$
How can I get the coefficients of the sums of powers formulas?
For example the sum of squares is $(1/3)n^3 + (1/2)n^2 + 1/6n$
|
Stiring Numbers of the Second Kind is also good for building sum of powers formula.
$x^n = \sum _{k=0} ^n S(n,k) (x)_k$
Example, for sum of x^2 formula, "integrate" the falling factorial form.
$$x^2 = (1)(x)_1 + (1)(x)_2$$
$$\sum_{k=0}^{n-1} x^2
= \frac{(n)_2}{2}+\frac{(n)_3}{3}
= \frac{n^3}{3} - \frac{n^2}{2} + \frac{n}{6}$$
|
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|
Computing matrices to a power of $6$ Compute
$\begin{pmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{pmatrix}^6.$
How would I solve this question. I found out that it's square would be $\begin{pmatrix} 2 & -2\sqrt{3} \\ 2\sqrt{3} & 2 \end{pmatrix}.$ What next? Is there any easier way?
|
Set
$J = \begin{bmatrix} 0 & -1 \\ 1 & 0 \end{bmatrix}; \tag{1}$
we seek $A^6$, where
$A = \begin{bmatrix} \sqrt{3} & -1 \\ 1 & \sqrt{3} \end{bmatrix} = \sqrt{3} I + J; \tag{2}$
we note that we may write $A$ as
$A = 2(\dfrac{\sqrt{3}}{2} I + \dfrac{1}{2}J); \tag{3}$
noting further that
$\dfrac{\sqrt{3}}{2} = \cos \dfrac{\pi}{6}, \tag{4}$
$\dfrac{1}{2} = \sin \dfrac{\pi}{6}, \tag{5}$
we see we may write
$A = 2(\cos \dfrac{\pi}{6} I + \sin \dfrac{\pi}{6} J); \tag{6}$
next, we observe that
$J^2 = - I; \tag{7}$
from (7) we may deduce that matrices of the general form $(\cos \theta) I + (\sin \theta) J$ satisfy a variant of de Moivre's formula;
$((\cos \theta) I + (\sin \theta) J)^n = (\cos n\theta) I + (\sin n\theta) J; \tag{8}$
(8) is, like its scalar equivalent, easy to prove by induction; since the proof is virtually identical to that for the scalar version, we won't repeat it here, but see
https://en.m.wikipedia.org/wiki/De_Moivre%27s_formula. With the aid of (7), we compute
$A^6 = 2^6(\cos \dfrac{\pi}{6} I + \sin \dfrac{\pi}{6} J)^6 = 64(\cos \pi I + \sin \pi J) = -64I. \tag{9}$
|
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|
Partial fraction expansion with quadratic factors in the denominator Question: expand in partial fractions:
$$\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64} .$$
I factored the denominator as $(x-2)^2 (x^2+2x+4)^2$.
With a denominator like $(x-1)(x-2)^2$ I know it will be:
$\frac A {x-1} + \frac B {x-2} + \frac C {(x-2)^2}$ (first of all I don't get why that is?).
But in this exercise, will $\frac {x^5+x^4+3x^3-8x^2+28x+48} {x^6-16x^3+64}$ be equal to $\frac A {x-2} + \frac B {(x-2)^2} + \frac C {x^2+2x+4} + \frac D {(x^2+2x+4)^2}$?
Thanks in advance.
|
When the denominator contains factors like $(ax^2 + bx + c)^p$ with $b^2 - 4ac < 0$ (such as $x^2 + 2x + 4$ in your case), then the fractions that appear in the expansions are
$$\frac {A_1 x + B_1} {ax^2 + bx + c} + \frac {A_2 x + B_2} {(ax^2 + bx + c)^2} + \dots + \frac {A_p x + B_p} {(ax^2 + bx + c)^p} .$$
In your case, the expansion is
$$ \frac A {x-2} + \frac B {(x-2)^2} + \frac {Cx + D} {x^2 + 2x + 4} + \frac {Ex + F} {(x^2 + 2x + 4)^2} .$$
|
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|
Integrate $\int\frac{\left(1+x\right)\sin x}{\left(x^{2}+2x\right)\cos^{2}x-\left(1+x\right)\sin2x}dx$ (Q) $\displaystyle \int\frac{\left(1+x\right)\sin x}{\left(x^{2}+2x\right)\cos^{2}x-\left(1+x\right)\sin2x}dx$
Tried a lot to expand denominator and reduce it to bring its derivative on top , but all in vain .
|
Let $$\displaystyle I = \int\frac{(1+x)\sin x}{(x^2+2x)\cos^2 x-(1+x)\sin 2x}dx$$
$$\displaystyle I = \int\frac{(1+x)\sin x}{(x^2+2x+1)\cos^2 x-(1+x)\sin 2x-\cos^2 x}dx$$
$$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x\right]^2-2(x+2)\sin x\cdot \cos x-(1-\sin^2 x)}dx$$
So $$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x\right]^2-2(x+1)\sin x\cdot \cos x+\sin^2 x-1}dx$$
$$\displaystyle I = \int\frac{(1+x)\sin x}{\left[(x+1)\cos x-\sin x\right]^2-1^2}dx$$
Now Let $(x+1)\cos x-\sin x = t\;,$ Then $(x+1)\sin xdx = -dt$
So Integral $$\displaystyle I = -\int\frac{1}{t^2-1}dt = -\frac{1}{2}\int\left[\frac{1}{t-1}-\frac{1}{t+1}\right]dt$$
So we get $$\displaystyle I = \frac{1}{2}\left[\ln|t+1|-\ln|t-1|\right]+\mathcal{C} = \frac{1}{2}\ln\left|\frac{t+1}{t-1}\right|+\mathcal{C}$$
So we get $$\displaystyle I = \frac{1}{2}\ln \left|\frac{(x+1)\cos x-\sin x+1}{(x+1)\cos x-\sin x-1}\right|+\mathcal{C}$$
|
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Evaluating $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$ Without L'Hopital or Calculus? What is: $\lim _{x\to 1}\left(\frac{\left(2x^2-1\right)^{\frac{1}{3}}-x^{\frac{1}{2}}}{x-1}\right)$?
Thanks in advance
Much appreciated!
|
$$
\begin{align}
\lim_{x\to1}\frac{\left(2x^2-1\right)^{\frac13}-x^{\frac12}}{x-1}
&=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-(x+1)^{\frac12}}{x}\tag{1}\\
&=\lim_{x\to0}\frac{\left(2x^2+4x+1\right)^{\frac13}-1}{x}-\lim_{x\to0}\frac{(x+1)^{\frac12}-1}{x}\tag{2}\\
&=\lim_{x\to0}\frac{2x^2+4x}{x\left(\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1\right)}\tag{3}\\
&-\lim_{x\to0}\frac{x}{x\left((x+1)^{\frac12}+1\right)}\tag{4}\\
&=\lim_{x\to0}\frac{2x+4}{\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1}\tag{5}\\
&-\lim_{x\to0}\frac{1}{(x+1)^{\frac12}+1}\tag{6}\\
&=\frac43-\frac12\tag{7}\\[3pt]
&=\frac56\tag{8}
\end{align}
$$
Explanation:
$(1)$: substitute $x\mapsto x+1$
$(2)$: subtract and add $1$ to the numerator so that we can split the limit into two
$(3)$: multiply and divide by $\left(2x^2+4x+1\right)^{\frac23}+\left(2x^2+4x+1\right)^{\frac13}+1$
$(4)$: multiply and divide by $(x+1)^{\frac12}+1$
$(5)$: cancel $x$ in numerator and denominator
$(6)$: cancel $x$ in numerator and denominator
$(7)$: evaluate $(5)$ and $(6)$ at $x=0$
$(8)$: simplify
|
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|
Evaluating $\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right)$ I'm trying to evaluate the limit
$$\lim _{x\to 1} \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} .$$
I used an online limit calculator to find the result, which gives
$$\lim _{x\to 1} \frac{x^{\frac{1}{6}}+1}{2\left(\sqrt[3]{x}+x^{\frac{1}{6}}+1\right)}.$$
Then, plugging the value $1$ for $x$, you get $\frac{1}{3}$.
I don't see how did they reach that conclusion. This is how I tried to tackle it:
$$\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} = \frac{\sqrt[3]{x}-1}{2\sqrt{x}-2} \cdot \frac{\sqrt[3]{x}+1}{\sqrt[3]{x}+1},$$
which then yields
$$\frac{x-1}{(2\sqrt{x}-2)(\sqrt[3]{x}+1)},$$
and that becomes
$$\frac{x-1}{2\cdot(\sqrt{x}-1)\cdot(\sqrt[3]x+1)}.$$
That's
$$\frac{x-1}{2\cdot(\sqrt[6]{x}+\sqrt{x}-\sqrt[3]{x}-1)},$$
and this will still evaluate to $\frac{0}{0}$.
How did they solve this, exactly?
|
Hint:
$$\lim _{x\to 1}\left(\frac{\sqrt[3]{x}-1}{2\sqrt{x}-2}\right) = \frac{1}{2}\lim _{x\to 1}\left(\left(\frac{\sqrt[3]{x}-1}{\sqrt{x}-1}\right)\left(\frac{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}{\sqrt[3]{x^2} + \sqrt[3]{x} + 1}\right)\left(\frac{\sqrt{x} + 1}{\sqrt{x} + 1}\right)\right)$$
Now use the following equation to simplify:
$$(\sqrt[3]{x}-1)(\sqrt[3]{x^2} + \sqrt[3]{x} + 1) = (\sqrt{x}-1)(\sqrt{x}+1) = x - 1.$$
|
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|
Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$ Solve for $x^{11} + \frac{1}{x^{11}}$ if $x^3 + \frac{1}{x^3} = 18.$
I'm familiar with variants of this problem, especially those where the exponent in the given is a factor of the exponent in what we want to solve for, but this one stumped me.
|
If $x^3 + \frac{1}{x^3} = 18$ then squaring gives $x^6 + 2 + x^{-6} = 18^2$ so $x^6 + x^{-6} = 322$. Squaring again gives $x^{12} + x^{-12} = 322^2 - 2 = 103682$
Then $(x + x^{-1})(x^{11} + x^{-11}) = (x^{12} + x^{-12}) + (x^{10} + x^{-10})$.
Now to me it seems clear that since we have
$$
(x+ x^{-1})^3 = (x^3 + x^{-3}) + 3(x + x^{-1}),
$$
if we perform the substitution $t = x+ x^{-1}$ as suggested by @Remember_Me, then we get $t^3 - 3t - 18 = 0$. Now, doing the rational root test yields the root $t=3$, and then we can solve for the other two roots, but all choices will give the same result for $x^{11} + x^{-11}$, so we may as well take $x+ x^{-1} = 3$.
Then (squaring each time) we have $x^2 + x^{-2} = 7$, $x^{4} + x^{-4} = 47$ and $x^{8} + x^{-8} = 2207$ and so $(x^8 + x^{-8})(x^2 + x^{-2}) = (x^{10} + x^{-10}) + (x^6 + x^{-6}).$ Thus
$$
x^{10} + x^{-10} = 2207 \times 7 - 322 = 15127,
$$
and so finally we get (from the above equation)
$$
3(x^{11} + x^{-11}) = 103682 + 15127 = 118809,
$$
so $x^11 + x^{-11} = 39603$, and we avoided...most...of the messy calculation!
|
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|
Show that $g(x)=\sqrt{x^{2}+4}$ is continuous at $x=1$ Let $g(x)=\sqrt{x^{2}+4}$ from $\mathbb{R}\rightarrow \mathbb{R}$. I want to show that $g$ is continuous at $x=1$.
I have to show that for any $\epsilon>0$, there exists a $\delta>0$ such that $|x-1|< \delta \implies |g(x)-g(1)|<\epsilon$.
So I do some rough work first and start with $$|g(x)-g(1)|$$ $$=|\sqrt{x^{2}+4} -\sqrt{5}|$$ $$\leq|\sqrt{x^{2}+4}| + |\sqrt{5}|$$.
I am stuck on what to do after. The plan is to choose the right $\delta$.
|
As suggested in a comment of imranfat, multiply by the conjugate, i.e.
\begin{align*}
\sqrt{x^2+4}-\sqrt{5}=\frac{(\sqrt{x^2+4}-\sqrt{5})(\sqrt{x^2+4}+\sqrt{5})}{\sqrt{x^2+4}+\sqrt{5}}=\frac{x^2-1}{\sqrt{x^2+4}+\sqrt{5}}=\frac{(x-1)(x+1)}{\sqrt{x^2+4}+\sqrt{5}}.
\end{align*}
Now, if $|x-1|<\delta$, then $|x+1|=|x-1+2|\leq|x-1|+2\leq\delta+2$. Since $\sqrt{x^2+4}+\sqrt{5}\geq\sqrt{5}\geq1$, it follows that
\begin{align*}
|\sqrt{x^2+4}-\sqrt{5}|=\frac{|x-1||x+1|}{\sqrt{x^2+4}+\sqrt{5}}\leq\delta(\delta+2).
\end{align*}
Can you go on from here?
|
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|
What is row reduced echelon form? How to row reduce this matrix? I'm not being able to grasp the concept of row reduced echelon form. Please, explain how to row reduce one of the the following matrices.
$A =
\begin{bmatrix}
1&3&4&5\\3&9&12&9\\1&3&4&1
\end{bmatrix}$
$B=
\begin{bmatrix}
1&2&1&2\\0&1&0&1\\-1&2&0&3
\end{bmatrix}$
|
In order to obtain the reduced row echelon form (rref) of a matrix, we apply some row operations. According to this article, for the first case we have:
$\begin{array}{l} \begin{bmatrix} 1 & 3 & 4 & 5\\ 3 & 9 & 12 & 9\\ 1 & 3 & 4 & 1 \end{bmatrix}\overset{R_2:=3R_1 - R_2}{\to}\begin{bmatrix} 1 & 3 & 4&5\\0&0&0&6 \\ 1&3&4&1 \end{bmatrix}\overset{R_3:=R_1 - R_3}{\to}\begin{bmatrix} 1&3&4&5\\0&0&0&6\\0&0&0&4 \end{bmatrix}\overset{R_2:=\frac 16R_2}{\to}\begin{bmatrix}1&3&4&5\\ 0&0&0&1\\0&0&0&4\\\end{bmatrix}\\\overset{R_3:=4R_2-R_3}{\to}
\begin{bmatrix}1&3&4&5\\0&0&0&1\\0&0&0&0\end{bmatrix}\overset{R_1:=R_1 - 5R_2}{\to}\begin{bmatrix}1&3&4&0\\ 0&0&0&1\\0&0&0&0 \end{bmatrix}.
\end{array}$
The last matrix satisfies all the conditions of the reduced row echelon form of a matrix.
|
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|
Put $(7+5\sqrt{2})^{\frac{1}{3}}$ in the form $x+y(\sqrt{2})$ I said, let:
$(7+5\sqrt{2})^{\frac{1}{3}}=((x+y\sqrt{2})^{3})^{\frac{1}{3}}$
Therefore,
$(7+5\sqrt{2})=(x+y\sqrt{2})^{3}$
Hence,
$(7+5\sqrt{2})=x^{3}+3x^{2}y(\sqrt{2})+3xy^{2}(\sqrt{2})^{2}+y^{3}
(\sqrt{2})^3$
However, from here how do I go? Anyone have any ideas?
Thanks a bunch in advance.
|
Remembering that $\sqrt{2}^2=2$ we get, from where you left off.
$$
7+5\sqrt{2} = x^3+3x^2y(\sqrt{2})+3xy^2(2)+y^3(2)(\sqrt{2})\\
= x^3+6xy^2+3x^2y(\sqrt{2}) +2y^3 (\sqrt{2})\\
= x^3+6xy^2+(3x^2y+2y^3)(\sqrt{2})
$$
So, we see that $7=x^3+6xy^2$ and $5=3x^2y+2y^3$. Which still looks like a 3 pipe problem untill we stare at the equations for a bit and see that $x=y=1$ is a solution.So $\sqrt[3]{7+5\sqrt{2}}=1+\sqrt{2}$ Which we can verify by calculating $(1+\sqrt{2})^3$.
|
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|
How to solve this limit 2 variables $\lim_{(x,y) \to (4,1)} \frac{y \sqrt x - 2y - \sqrt x + 2}{4 - x + x \sqrt y - 4 \sqrt y}$ Please anybody can help me solve this?
$$\lim \frac{y \sqrt x - 2y - \sqrt x + 2}{4 - x + x \sqrt y - 4 \sqrt y}$$
with $(x,y) \to (4,1)$
Thank you!
|
$$\frac{y\sqrt x-2y-\sqrt x+2}{4-x+x\sqrt y-4 \sqrt y}=\frac{\sqrt x (y-1)-2(y-1)}{\sqrt y(x-4)-(x-4)}=\frac{(y-1)(\sqrt x-2)}{(x-4)(\sqrt y-1)}=\frac{\sqrt y+1}{\sqrt x+2}$$
You can show that the function $f(x,y)=\frac{\sqrt y+1}{\sqrt x+2}$ is continuous at the point $(x,y)=(4,1)$ :
$$|f(x,y)-f(4,1)|=|\frac{\sqrt y+1}{\sqrt x+2}-\frac{1}{2}|=|\frac{2\sqrt y+2-\sqrt x-2}{2(\sqrt x+2)}|\leq \frac{2|\sqrt y-1|+|2-\sqrt x|}{2(\sqrt x+2)}<\epsilon$$ for $\|(x,y)-(4,1)\|_{l^2}$ small enough.
So the limit really exists and it is $f(4,1)=\frac{1}{2}$
|
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|
What are the $\sin$ and $\cos$ of discrete calculus? I'm getting acquainted with Discrete Calculus, and I really like taking functions that arise in traditional calculus and finding what their counterparts in discrete-land are.
For example, if we define our difference operator
$$\Delta f(n) = f(n + 1) - f(n)$$
(the analogue of the familiar derivative operator $D = \frac{d}{dx}$) then we can ask: what is the $e^x$ of this operator? That is, what is the function $g$ such that $\Delta g(n) = g(n)$? The answer is $2^x$, since:
$$\Delta 2^x = 2^{x+1} - 2^x = 2(2^x) - 2^x = 2^x$$
Likewise, we can ask what the $g(x) = x^n$ of discrete calculus is -- it turns out to be
$$g(x) = \overbrace{x(x - 1)(x - 2) \dots (x - (n - 1))}^{n \text{ factors}}$$
My question is: is there a $\sin$ and $\cos$ of discrete calculus? By which I mean, are there any functions $f$ such that $\Delta^2 f = -f$?
|
Note that if $f(n)$ is a discrete sinusoid, then
$$
f(n+1)=f(n)+\Delta f(n)
$$
and
$$
\Delta f(n+1) = \Delta f(n) + \Delta^2 f(n) = \Delta f(n) - f(n),
$$
or
$$
\left(\begin{matrix}f(n+1) \\ \Delta f(n+1)\end{matrix}\right)=\left(\begin{matrix}1 & 1 \\ -1 & 1 \end{matrix}\right)\left(\begin{matrix}f(n)\\ \Delta f(n)\end{matrix}\right).
$$
Also,
$$
\left(\begin{matrix}1 & 1 \\ -1 & 1 \end{matrix}\right)^2 = \left(\begin{matrix}1 & 1 \\ -1 & 1 \end{matrix}\right)\left(\begin{matrix}1 & 1 \\ -1 & 1 \end{matrix}\right)=\left(\begin{matrix}0 & 2 \\ -2 & 0 \end{matrix}\right),
$$
$$
\left(\begin{matrix}1 & 1 \\ -1 & 1 \end{matrix}\right)^3 = \left(\begin{matrix}1 & 1 \\ -1 & 1 \end{matrix}\right)\left(\begin{matrix}0 & 2 \\ -2 & 0 \end{matrix}\right)=\left(\begin{matrix}-2 & 2 \\ -2 & -2 \end{matrix}\right),
$$
$$
\left(\begin{matrix}1 & 1 \\ -1 & 1 \end{matrix}\right)^4 = \left(\begin{matrix}1 & 1 \\ -1 & 1 \end{matrix}\right)\left(\begin{matrix}-2 & 2 \\ -2 & -2 \end{matrix}\right)=\left(\begin{matrix}-4 & 0 \\ 0 & -4 \end{matrix}\right)=-4\hat{I}.
$$
So the discrete $\cos$, which starts with $f(0)=1$ and $\Delta f(0)=0$, is
$$
\left( 1, 1, 0, -2, -4, -4, 0, 8, 16, \ldots\right),
$$
and the discrete $\sin$, which starts with $f(0)=0$ and $\Delta f(0)=1$, is
$$
\left(0, 1, 2, 2, 0, -4, -8, -8, 0, \ldots\right).
$$
The fact that these grow as $2^{n/2}$, rather than being periodic, is a consequence of your non-symmetric choice of $\Delta^2$. If you define $\Delta^2 f(n)=f(n-1) - 2f(n) + f(n+1)$, say, you will find functions closer to what you might have expected.
|
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|
Maclaurin expansion of $y=\frac{1+x+x^2}{1-x+x^2}$ to $x^4$
Maclaurin expansion of $$\displaystyle y=\frac{1+x+x^2}{1-x+x^2}\,\,\text{to } x^4$$
I have tried by using Maclaurin expansion of $\frac1{1-x}=1+x+x^2+\cdots +x^n+o(x^n)$, but it seems not lead me to anything.
|
Hint: Expand the fraction with $(x+1)$
$$y=\frac{(1+x+x^2)(x+1)}{(x+1)(1-x+x^2)}=\frac{(1+x+x^2)(x+1)}{1+x^3}$$
Now use the geometric series you wrote down for $$\frac{1}{1-(-x^3)}$$
$$y=(1+x+x^2)(x+1)(1+(-x^3)+O(x^6))$$
|
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|
Proving a progression inequality Prove that $$1+ \frac{1}{2^3} + \cdot \cdot \cdot + \frac{1}{n^3} < \frac{5}{4} $$
I got no idea of how to approach this problem.
|
Since:
$$\begin{eqnarray*} \frac{1}{n^3} &=& \frac{1}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)+\frac{1}{2}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\frac{2n+1}{2n^3(n+1)^3}\\&\leq&\frac{1}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)+\frac{1}{2}\left(\frac{1}{n^3}-\frac{1}{(n+1)^3}\right)+\frac{1}{4}\left(\frac{1}{n^4}-\frac{1}{(n+1)^4}\right)\tag{1}\end{eqnarray*}$$
$$ \zeta(3) \leq \frac{1}{2}+\frac{1}{2}+\frac{1}{4}=\frac{5}{4} $$
follows by creative telescoping. With an extra step we may also prove $\frac{6}{5}\leq\zeta(3)\leq\frac{5}{4}$.
More tight approximations (both upper and lower bounds) can be derived from the identity:
$$ \zeta(3) = \sum_{n\geq 1}\frac{5(-1)^{n+1}}{2n^3\binom{2n}{n}}\tag{2}$$
that was crucial in Apery's proof of the irrationality of $\zeta(3)$.
|
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|
How to prove $(6666\ldots66)^2 + 8888\ldots88 = 4444\ldots44$ (with $n$ 6s and 8s, $2n$ 4s) How to prove that, if $n$ is a positive integer, then
$$
(\underbrace{666 \ldots 6}_{n \text{ copies of } 6})^2 +
\underbrace{888 \ldots 8}_{n \text{ copies of } 8} =
\underbrace{444 \ldots 4}_{2n \text{ copies of } 4}?
$$
|
Given $$\underbrace{(666666666666\ldots)^2}_{n~\text{times}}+\underbrace{(888888888888\ldots)}_{n~\text{times}}$$
Now we can write $$\underbrace{666666666666\ldots}_{n~\text{times}} = (6+6\cdot 10+6\cdot 10^2+\cdots+6\cdot 10^{n-1})$$
$$\displaystyle =6\left[\frac{10^n-1}{10-1}\right] = \frac{2}{3}\left[10^n-1\right]$$
Similarly we can write $$\underbrace{88888888888\ldots}_{n~\text{times}} = (8+8\cdot 10+8\cdot 10^2+\cdots+8\cdot 10^{n-1})$$
$$\displaystyle =8\left[\frac{10^n-1}{10-1}\right] = \frac{8}{9}\left[10^n-1\right]$$
So we get $$\displaystyle \left\{\frac{2}{3}\left[10^n-1\right]\right\}^2+\frac{8}{9}\left[10^n-1\right] = \frac{4}{9}(10^n-1)^2+\frac{8}{9}(10^n-1)$$
So we get $$\displaystyle = \frac{4}{9}(10^n-1)\cdot \left[10^n-1+2\right] = \frac{4}{9}(10^{2n}-1)$$
So we can write $$\displaystyle \frac{4}{9}(10^{2n}-1) = \underbrace{(444444444444\ldots)}_{2n~\text{times}}$$
|
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|
Discrete mathematics - Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$. Find all integer solutions of the equation $a^2+ b^2 + c^2=a^2 b^2$.
This is one of the questions we presented in one session to contest preparation PUTNAM. It turns out that I can't get from the problem. Could someone just give me a hint? (Please, don't give me the answer. Simply, an argument that can help me advance in the problem or theorem might suffice.)
|
The equation is equivalent to
$$
c^2+1=(a^2-1)(b^2-1)
$$
Since the left side is either $1$ or $2$ mod $4$, and the right side is either $0$ or $1$ mod $4$, they must both be $1$ mod $4$, which means that $a,b,c$ must be even.
Since any prime which is $3$ mod $4$ is a Gaussian prime, no prime which is $3$ mod $4$ can divide $c^2+1=(c+i)(c-i)$. Thus, the only primes which can divide $c^2+1$ must be $1$ mod $4$.
Since $a^2-1$ and $b^2-1$ are both $3$ mod $4$, they must have a prime factor which is $3$ mod $4$ and hence cannot divide the left hand side (except if they are both $-1$). Thus, the only solution is when $a,b,c=0$.
|
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|
Given area of square $= 9+6\sqrt{2}$ Without calculator show its length in form of $(\sqrt{ c}+\sqrt{ d})$ $\sqrt{9+6\sqrt{2}}$ to find length
But how do I express the above in the form of $\sqrt{c} + \sqrt{d}$.
|
Let the side of the square be $(\sqrt c+\sqrt d)$ then its area is given as $$(\sqrt c+\sqrt d)^2=9+6\sqrt 2\tag 1$$
$$c+d+2\sqrt{cd}=9+6\sqrt 2$$ by comparing the corresponding rational & irrational parts, we get $$c+d=9\tag 2$$
$$2\sqrt{cd}=6\sqrt 2\tag 3$$
Now, we know $$(\sqrt c-\sqrt d)^2=(\sqrt c+\sqrt d)^2-2\sqrt{cd}$$$$=(9+6\sqrt 2)-2(6\sqrt 2)$$$$(\sqrt c-\sqrt d)^2=9-6\sqrt 2\tag 4$$ multiplying (1) & (4) we get $$(\sqrt c+\sqrt d)^2(\sqrt c-\sqrt d)^2=(9+6\sqrt 2)(9-6\sqrt 2)$$
$$(c-d)^2=9\implies c-d=\pm 3$$ 1. taking positive sign, we get $c-d=3\tag 5$ solving (2) & (5), we get $$\color{red}{c=6, \ d=3}$$
2. taking negative sign, we get $c-d=-3\tag 6$ solving (2) & (6), we get $$\color{red}{c=3, \ d=6}$$ Both the above cases give the same result. Hence, the side of the square is $$\color{red}{\sqrt c+\sqrt d}=\color{blue}{\sqrt 6+\sqrt 3}$$
or $$\color{red}{\sqrt c+\sqrt d}=\color{blue}{\sqrt 3+\sqrt 6}$$
|
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|
Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{ c + 1} −\sqrt{c}$ for all real $c \geq 1$. Prove that $\sqrt{ c} − \sqrt{c − 1} \geq \sqrt{c + 1} −\sqrt{c}$ for all real $c \geq 1$.
Can anyone provide some form of guidance? So far all I have been able to think of is writing $c$ as $x^2$ for some $x$, or eliminating the radical on one side...
|
Consider $$B = \sqrt c - \sqrt{c-1},\quad D = \sqrt {c+1} - \sqrt{c}$$
By mean value theorem, there exist $$b\in (c-1, c),\quad d\in (c, c+1)$$ that satisfy
$$\frac1{2\sqrt b} = B,\quad \frac1{2\sqrt d} = D$$
and since $b\le d$, $$\begin{align*}
\dfrac1{2\sqrt b} &\ge \dfrac1{2\sqrt d}\\
\sqrt c - \sqrt{c-1} &\ge \sqrt {c+1} - \sqrt{c}
\end{align*}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
How to show this limit exists I know the limit is 0, but how would I prove the limit? $$\lim_{(x,y)\to(0,0)} {\frac{x^3y^2}{x^4+y^6}}$$
|
As
\begin{align*}
x^4+y^6 = \frac{1}{2} x^4 + \frac{1}{2} x^4 + y^6 \geq \frac{3}{\sqrt[3]{4}}x^{8/3}y^2,
\end{align*}
then
\begin{align*}
\Big|\frac{x^3y^2}{x^4+y^6}\Big| \leq \frac{\sqrt[3]{4}}{3}|x|^{1/3} \rightarrow 0.
\end{align*}
|
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|
Why are the derivatives of these two equations different? I am trying to implicitly differentiate the following: $\frac {x}{x-y}\ =\ y^2-1 $
I originally multiplied the whole equation by $(x - y)$ and the result of implicitly differentiating the resulting equation ($\ x=xy^2-x-y^3+y $ ) was $\frac {dy}{dx}\ = \frac {2-y^2}{2xy-3y^2+1} $ but then afterwards, when I tried again and differentiated the original expression, $\frac {x}{x-y}\ =\ y^2-1 $, (without manipulating it) I ended up with the correct answer of $\frac {-y}{2y(x-y)^2-x} $.
My question is why does simply changing the form of the equation you're trying to implicitly differentiate (like I have done above) change what you get as $\frac {dy}{dx} $ ?
|
Contrary to what was said in the comments, you lose no information by your manipulation. In the original equation, you knew $x\ne y$. You don't lose anything by multiplying by $x-y$. You have correctly differentiated both expressions, so we have:
$$
\frac{dy}{dx}=\frac{2-y^2}{2xy-3y^2+1}, \text{ and} \\ \frac{dy}{dx}=\frac{-y}{2y(x-y)^2-x}
$$
We would like to reconcile these two expressions. To do this, we use the equation we started with: $\frac{x}{x-y}=y^2-1$. There are many ways to see that the two expressions are equivalent (on the curve defined by $\frac{x}{x-y}=y^2-1$). One way is to note first that:
$$
\begin{split} \frac{dy}{dx}=\frac{-y}{2y(x-y)^2-x}=\frac{dy}{dx}&=\frac{-y}{(x-y)\left(2y(x-y)-\frac{x}{x-y}\right)} \\ &=\frac{\frac{-y}{x-y}}{2y(x-y)-\frac{x}{x-y}} \\ &=\frac{\frac{-y}{x-y}}{2xy-2y^2-\frac{x}{x-y}} \\ &=\frac{\frac{-y}{x-y}}{2xy-2y^2-(y^2-1)}, \text{ since }\frac{x}{x-y}=y^2-1 \\ &= \frac{\frac{-y}{x-y}}{2xy-3y^2+1} \end{split}
$$
As you can see we are almost at the second expression now. The final part has already been done in the comments. I will repeat it here for completeness. Taking the other expression we have:
$$
\begin{split} \frac{2-y^2}{2xy-3y^2+1} &=\frac{2-\frac{2x-y}{x-y}}{2xy-3y^2+1}, \text{ since }y^2=1+\frac{x}{x-y} \\ &=\frac{\frac{2x-2y-2x+y}{x-y}}{2xy-3y^2+1} \\ &=\frac{\frac{-y}{x-y}}{2xy-3y^2+1}\end{split}
$$
which is what we had above. So both expressions are equivalent. In general you can manipulate your equation before you implicitly differentiate. You do need to be careful if you divide by something that may be $0$. Multiplying by $0$ is fine, always.
|
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|
Prove that the probability that $x+y\leq 1,$ given that $x^2+y^2\geq \frac{1}{4}$ is $\frac{8-\pi}{16-\pi}$. Let $2$ positive real numbers $x,y$ satisfy $x\leq 1,y\leq 1$ are chosen at random.Prove that the probability that $x+y\leq 1,$ given that $x^2+y^2\geq \frac{1}{4}$ is $\frac{8-\pi}{16-\pi}$.
Since the $x+y\leq 1$ and $x^2+y^2\geq \frac{1}{4}$,so the required probability is area outside circle of radius $\frac{1}{2}$,centered at $(0,0)$ and inside the line segment whose $x-$intercept and $y-$intercept are 1 each in the first quadrant.so probability$=\frac{1}{2}-\frac{\pi}{16}=\frac{8-\pi}{16}$
But my answer does not match the given answer.Somewhere i must have faltered,Can you please guide me to the right method?
|
For the conditional probability you need to divide by the "given" area.
in this case $ Area(x^2+y^2 \ge \frac 14)=1-\frac{\pi}{16}$
|
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|
How to Solve for $x$ in the equation $\sin(2x) = -\frac{1}{2}$ within the interval of $[0, 2\pi]$? So far, I only have the solutions $11\pi/12$, and $7\pi/12$ (I think there are a total of $4$). I think this has something to do with trig identities, and should be able to be solved without using a calculator. However, I have only been able to get the above solutions using my calculator.
|
If $$\sin(2x) = -\frac{1}{2}$$ then
\begin{align*}
2x & = \arcsin\left(-\frac{1}{2}\right) + 2n\pi & 2x & = \pi - \arcsin\left(-\frac{1}{2}\right) + 2m\pi\\
2x & = -\frac{\pi}{6} + 2n\pi & 2x & = \pi - \left(-\frac{\pi}{6}\right) + 2m\pi\\
x & = -\frac{\pi}{12} + n\pi & 2x & = \frac{7\pi}{6} + 2m\pi\\
& & x & = \frac{7\pi}{12} + m\pi
\end{align*}
where $m, n \in \mathbb{Z}$. The requirement that $x \in [0, 2\pi]$ implies $n = 1, 2$ and $m = 0, 1$.
|
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|
How do I transform, with long division, this polynomial into the answer given? I need to transform:
$ \frac {x^5}{(x-2)(x+2)(x^2+4)} $
into
$ \frac{-2x}{x^2+4} + x + \frac{1}{x-2} + \frac{1}{x+2} $
How can I solve it?
Thanks.
|
Let me try. We have $$\frac{x^5}{(x-2)(x+2)(x^2+4)} = \frac{x^5}{x^4-16}= x + \frac{16x}{x^4-16} = x + 2x\left(\frac{1}{x^2-4} - \frac{1}{x^2+4}\right) = x -\frac{2x}{x^2+4} +\frac{2x}{(x+2)(x-2)} = x -\frac{2x}{x^2+4} + \frac{1}{x-2} + \frac{1}{x+2}.$$
|
{
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|
Calculating $\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$ Without L'Hopital,
$$\lim_{x\to 0}\frac{x^2+x\cdot \sin x}{-1+\cos x}$$
That's
$$\frac{x^2+x\cdot \sin x}{-1+\left(1-2\sin^2\frac{x}{2}\right)} = \frac{x^2+x\cdot \sin x}{-2\sin^2\frac{x}{2}}$$
Let's split this
$$\frac{x\cdot x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}} + \frac{x\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}}$$
On the left one, it would be great to have $\frac{x}{2} \cdot \frac{x}{2}$. To do so, I will multiply and divide by $\frac{1}{4}$:
$$\frac{\frac{x}{2}\cdot \frac{x}{2}}{\frac{1}{4}\cdot-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}} + \frac{x\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}}$$
Then we use the identity $\frac{x}{\sin x} = 1$:
$$-2 + \frac{x\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}}$$
On the right side, I'd like to have $\frac{x}{2}$ there, so I will multiply and divide by $\frac{1}{2}$:
$$-2 + \frac{\frac{x}{2}\cdot \sin x}{-2\cdot\sin\frac{x}{2}\cdot \sin\frac{x}{2}\cdot\frac{1}{2}}$$
We use the identity $\frac{x}{\sin x} = 1$:
$$-2 + \frac{\sin x}{-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2}}$$
Hmm... we could perform the addition I suppose:
$$\frac{\sin x - 2(-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2})}{-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2}}$$
Simplify:
$$\frac{\sin x + (4\cdot-2\sin\frac{x}{2}\cdot-2\frac{1}{2})}{-2\cdot\sin\frac{x}{2}\cdot\frac{1}{2}} = \frac{\sin x +8\sin\frac{x}{2}}{\sin\frac{x}{2}}$$
Then
$$\frac{\sin x}{\sin\frac{x}{2}} + \frac{8\sin\frac{x}{2}}{\sin\frac{x}{2}} = \frac{\sin x}{\sin\frac{x}{2}} + 8$$
I'm close! The answer should be $-4$, but I don't know what to do with
$$\frac{\sin x}{\sin\frac{x}{2}} + 8$$
So basically I need to know what to do with $\frac{\sin x}{\sin\frac{x}{2}}$, but I also included my whole procedure just in case I've been doing it wrong all along.
|
Notice, $$\lim_{x\to 0 }\frac{x^2+x\sin x}{-1+\cos x}$$
$$=\lim_{x\to 0 }\frac{x^2+x\sin x}{-1+1-2\sin ^2\frac{x}{2}}$$
$$=-\frac{1}{2}\lim_{x\to 0}\frac{x\sin x+x^2}{\sin ^2\frac{x}{2}}$$
$$=-\frac{1}{2}\lim_{x\to 0}\frac{x^2\left(\frac{\sin x}{x}+1\right)}{\sin ^2\frac{x}{2}}$$
$$=-2\lim_{x\to 0}\frac{\frac{\sin x}{x}+1}{\left(\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2}$$
$$=-2\left(\frac{1+1}{1^2}\right)=-2(2)=\color{red}{-4}$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
About Integration $\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx$ What i want to prove is following integral
\begin{align}
\int_{-\infty}^{\infty} e^{-x^2} \cos(x^2) dx=\frac{1}{2} \sqrt{\left(1+\sqrt{2}\right) \pi }
\end{align}
can you give some explicit method to obtain this result?
|
Let your integral be labelled
$$I_1=\int_{-\infty}^{\infty} e^{-x^2}\cdot \cos(x^2)\space dx,$$
and a second integral
$$I_2=\int_{-\infty}^{\infty} e^{-x^2}\cdot\sin(x^2)\space dx.$$
It follows that
$$I_1-i\cdot I_2=\int_{-\infty}^{\infty} e^{-x^2}\cdot(\cos(x^2)-i\cdot \sin(x^2))\space dx.$$
Apply Euler's formula in complex analysis:
$$I_1-i\cdot I_2=\int_{-\infty}^{\infty} e^{-x^2}\cdot e^{-ix^2} dx=\int_{-\infty}^{\infty} e^{-(1+i)x^2}dx.$$
Let $x=(1+i)^{-1/2}\ t$ such that $dx=(1+i)^{-1/2}\ dt$, where $t\in(-\infty,\infty)$:
$$I_1-i\cdot I_2=(1+i)^{-1/2}\cdot\int_{-\infty}^{\infty} e^{-t^2}dt.$$
Evaluate the Gaussian integral:
$$I_1-i\cdot I_2=(1+i)^{-1/2}\cdot\sqrt{\pi}.$$
Rewrite the expression by making use of general properties of the exponential function and logarithms:
$$I_1-i\cdot I_2=e^{\ln((1+i)^{-1/2})}\cdot\sqrt{\pi}=e^{-1/2\ln(1+i)}\cdot\sqrt{\pi}.$$
The complex number $1+i$ can be converted from cartesian notation to polar notation; $1+i=\sqrt{2}\space e^{i \pi/4}$. Taking the natural logarithm on both sides gives $\ln(1+i)=\ln(\sqrt{2}\space e^{i \pi/4})=\ln(\sqrt{2})+i \pi/4$. Therefore,
$$I_1-i\cdot I_2=e^{-1/2(\ln(\sqrt{2})+i \pi/4)}\cdot\sqrt{\pi} = e^{\ln(\frac{1}{\sqrt{\sqrt{2}}})}\cdot e^{-i \pi/8}\cdot\sqrt{\pi}=\frac{1}{\sqrt{\sqrt{2}}}\cdot e^{-i \pi/8}\cdot \sqrt{\pi}.$$
Apply Euler's formula in complex analysis:
$$I_1-i\cdot I_2=\frac{1}{\sqrt{\sqrt{2}}}\cdot (\cos(\pi/8)-i\cdot \sin(\pi/8))\cdot \sqrt{\pi}.$$
Rewrite $\cos(\pi/8)$ by making use of the half-angle formula for the cosine function, $\cos(x)=2\cos^2(x/2)-1$:
$$\cos(\pi/4)=2\cos^2(\pi/8)-1\rightarrow \cos^2(\pi/8)=\frac{1+\cos(\pi/4)}{2}=\frac{1+\sqrt{2}}{2\sqrt{2}}=\frac{2+\sqrt{2}}{4}.$$
Then, $\cos(\pi/8)=\frac{\sqrt{2+\sqrt{2}}}{2}$.
Rewrite $\sin(\pi/8)$ by making use of the half-angle formula for the sine function, $\cos(x)=1-2\sin^2(x/2)$:
$$\cos(\pi/4)=1-2\sin^2(\pi/8)\rightarrow \sin^2(\pi/8)=\frac{1-\cos(\pi/4)}{2}=\frac{1-\sqrt{2}}{2\sqrt{2}}=\frac{2-\sqrt{2}}{4}.$$
Then, $\sin(\pi/8)=\frac{\sqrt{2-\sqrt{2}}}{2}$.
Substitution into the expression gives
$$I_1-i\cdot I_2=\frac{1}{\sqrt{\sqrt{2}}}\cdot (\frac{\sqrt{2+\sqrt{2}}}{2}-i\cdot \frac{\sqrt{2-\sqrt{2}}}{2})\cdot \sqrt{\pi}= (\frac{\sqrt{\sqrt{2}+1}}{2}-i\cdot \frac{\sqrt{\sqrt{2}-1}}{2})\cdot \sqrt{\pi}.$$
After expanding the terms it follows that
$$I_1=\frac{\sqrt{\sqrt{2}+1}}{2}\cdot \sqrt{\pi},\space I_2=\frac{\sqrt{\sqrt{2}-1}}{2}\cdot \sqrt{\pi}.$$
The integral you wanted to evaluate is $I_1$.
|
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"url": "https://math.stackexchange.com/questions/1458055",
"timestamp": "2023-03-29T00:00:00",
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|
How do I rationalize higher index roots? Often I've found exercises like
$$\frac{2-\sqrt[6]{3x+64}}{5x}$$
Where I need the rationalize. But I am not sure how to do it with indexes greater than $2$. How do I rationalize higher index roots? (the above one is just an example).
That is, perform
$$\frac{2-\sqrt[6]{3x+64}}{5x} \cdot \frac{2+\sqrt[6]{3x+64}}{2+\sqrt[6]{3x+64}}$$
|
(See my comments on original for links that explain this in detail... effectively, if we have a sum or difference involving an $n^{th}$ root we take either the sum of difference of $n^{th}$ roots and evaluate, using the given terms as the first factor and then plug in values for the remaining factor... it's messy, but it does work)
$$a^6 - b^6 = (a-b)(a^5 + a^4b + a^3b^2 + a^2b^3 + ab^4 + b^5 )$$
We now set $a = 2$ and $b =\sqrt[6]{3x+64}$
$$\frac{2-(3x+64)^{1/6}}{5x} * \frac{(3x+64)^{5/6}+2 (3 x+64)^{2/3}+4 \sqrt{3 x+64}+8 (3 x+64)^{1/3}+16 (3 x+64)^{1/6}+32}{(3x+64)^{5/6}+2 (3 x+64)^{2/3}+4 \sqrt{3 x+64}+8 (3 x+64)^{1/3}+16 (3 x+64)^{1/6}+32}$$
This yields
$$\frac{-3x}{5x[(3x+64)^{5/6}+2 (3 x+64)^{2/3}+4 \sqrt{3 x+64}+8 (3 x+64)^{1/3}+16 (3 x+64)^{1/6}+32]}$$
I hope you're happy... all that TeXing :P
|
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|
Are these boundaries true? I want to find the double integral
$$\iint xy \, dx\,dy $$
over the region
bounded by the positive $y$- axis, the line $y=\sqrt3 \ x$ and the circle
$x^2+y^2=4$.
My solution is that $x=0$ to $x=(1/\sqrt3) y$, and $y=0$ to $y=2$.
Is that correct?
|
My solution is that $x=0$ to $x=(1/\sqrt3) y$, and $y=0$ to $y=2$.
If you express it in that kind of language, it would be best to be explicit that you're putting the integral with respect to $y$ on the outside.
This would be
$$
\int_0^2 \left( \int_0^{y/\sqrt 3} xy \, dx \right) \, dy
$$
and that is not correct. It ignores the equation $x^2+y^2=4$.
If $x^2+y^2 = 4$ and $x\sqrt3 = y$, then $3x^2 + x^2 = 4$ then $x= \pm 1$. Since you say the "positive" $y$-axis, we need $x=+1$.
The equation $y^2+x^2=4$, is equivalent to $y = \pm\sqrt{4-x^2}$, and conjoining that with $y\ge 0$ we have $y = \sqrt{4-x^2}$.
So we can say $x$ goes from $0$ up to $1/\sqrt3$, and then for any fixed value of $x$, the other variable $y$ goes from $x\sqrt3$ up to $\sqrt{4-x^2}$, so we have
$$
\int_0^1 \left( \int_{x\sqrt3}^{\sqrt{4-x^2}} xy\, dy \right) \, dx
$$
(If there is any question about which of the two bounds on the inside integral is bigger, draw a picture.) Details below.
If you want to do it in the other order, then it's more complicated. You'd have
$$
\int_0^1 \left( \int_0^{y/\sqrt3} xy \, dx \right) \, dy + \int_1^2 \left( \int_0^{\sqrt{4-y^2}} xy \, dx \right) \, dy.
$$
In order to actually evaluate the integral, I would not do it like this; I would use polar coordinates.
Details: We have $\displaystyle \int_0^1 \left( \int_{x\sqrt3}^{\sqrt{4-x^2}} xy\, dy \right) \, dx$. The inside integral is $$\int_{x\sqrt3}^{\sqrt{4-x^2}} xy \, dy = \left[ \frac{xy^2} 2 \right]_{y:=x\sqrt3}^{y:=\sqrt{4-x^2}}$$ $$= \frac{x(4-x^2)} 2 - \frac{x\cdot 3x^2} 2 = 2x - 2x^3.$$ Then the outside integral becomes
$$ \int_0^1 \left( 2x - 2x^3 \right) \, dx = \left[ x^2 - \frac{x^4} 2 \right]_0^1 = \frac 1 2$$
But (as I said above) this is not how I would normally do this. I would use polar coordinates.
|
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|
Evaluating $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$ Determine the value of $x^4 + \frac{1}{x^4}$ given that $x^2 - 3x + 1 = 0$. I've tried forcing in a difference of squares, looked for various difference of $n$s or sum of odd powers that I could equate this to, but have yet to find a solution.
|
Here is an algebraic approach.
$x^2 - 3x + 1 = 0$ implies $\dfrac1x = 3-x$ and so $x^4 + \dfrac{1}{x^4}=x^ 4+(3-x)^4$.
The extended Euclidean algorithm for $\gcd(x^ 4+(3-x)^4,x^2 - 3x + 1)$ gives
$$
47 = 1\cdot(x^ 4+(3-x)^4)+ (-2 x^2 + 6 x - 34)\cdot(x^2 - 3x + 1)
$$
and so $x^4 + \dfrac{1}{x^4}=47$.
|
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|
Two apparently different evaluations of $\int \frac{x-1}{9x^2-18x+17}dx$
Evaluate the indefinite integral $$\int \frac{x-1}{9x^2-18x+17} \, dx .$$
This is an exercise from a book I'm studying. It gives the answer as:
$$\ln(9x^2 -18x+17)^\frac{1}{18} +C .$$
This is an easy integral. You just notice that the numerator is the derivative of the denominator.
But I didn't notice exactly that at first, so I solved it in a slightly different way. I did:
$$\int \frac{x-1}{9x^2-18x+17}\, dx=\frac{1}{18}\int \frac{x-1}{\frac{x^2}{2}-x+\frac{17}{18}}\, dx$$ so $$z=\frac{x^2}{2}-x+\frac{17}{18}, \quad \frac{dz}{dx}=x-1 $$
I find the answer in the usual way as: $$\frac{1}{18} \int\frac{1}{z} dz= \frac{1}{18}\ln\left(\frac{x^2}{2}-x+\frac{17}{18}\right)+C , $$
which is (I believe) different from the answer the book gives, because the arguments of $ln$ are different. What is the problem here? Where am I wrong?
|
$$\int \frac{x-1}{9x^2-18x+17}dx=\dfrac{1}{18}\log(9x^{2}-18x+17)+Cst$$
indeed,
*
*Substitute $u=9x^2-18x+17$ and $du=(18x-18)dx$
\begin{align}
\int \frac{x-1}{9x^2-18x+17}dx&=\dfrac{1}{18}\int \dfrac{1}{u}du\\
&=\dfrac{1}{18}\log(u)+Cst
\end{align}
*
*Substitute back for $u=9x^2-18x+17$:
\begin{align}
\int \frac{x-1}{9x^2-18x+17}dx&=\dfrac{1}{18}\log(u)+Cst\\
&=\dfrac{1}{18}\log(9x^2-18x+17 )+Cst
\end{align}
|
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|
How to evaluate the limit of $\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$ when $x\to1$ without using L'Hospital's rule?
How do I evaluate this without using L'Hospital's rule:$$\lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}\ ?$$
Note : I used L'Hospital's Rule I find $\frac{-4}{3}$ but in wolfram alpha is $0$
|
$$\displaystyle \lim_{x \to 1}\frac {2\cos({x-1})-2}{({x²}-2\sqrt{x}+1 )}$$
$$=-2\cdot\dfrac1{\lim_{x\to1}\{1+\cos(x-1)\}}\cdot\left(\lim_{x\to1}\dfrac{\sin(x-1)}{x-1}\right)^2\cdot\lim_{x\to1}\dfrac{(x-1)^2}{x^2-2\sqrt x+1}$$
Set $\sqrt x-1=y\implies x=(y+1)^2$ in
$$\lim_{x\to1}\dfrac{(x-1)^2}{x^2-2\sqrt x+1}$$
to get $$\lim_{y\to0}\dfrac{\{(1+y)^2-1\}^2}{(1+y)^4-2(y+1)+1} =\lim_{y\to0}\dfrac{y^2(2+y)^2}{2y+O(y^2)}=\cdots=0$$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Solve $y\frac{dy}{dx}+x=\sqrt{x^2+y^2}$ using line integrals I'm aware of the fact that the ODE
\begin{align}y\frac{dy}{dx}+x=\sqrt{x^2+y^2},\end{align}
can be solved using substitution methods. I've worked out the solution quite easily by setting $u=\sqrt{x^2+y^2}$ and everything works out. But what if I rewrite the equation as
\begin{align}\frac{dy}{dx}+\left(\frac{y}{x}\right)^{-1}&=\sqrt{\frac{x^2+y^2}{y^2}}\\\implies \frac{dy}{dx}+\left(\frac{y}{x}\right)^{-1} &=\sqrt{\left(\frac{y}{x}\right)^{-2}+1}.\end{align}
Now let $u=\displaystyle\frac{y}{x}\implies \displaystyle\frac{dy}{dx}=\frac{d\left(ux\right)}{dx}$, obtaining
\begin{align}\frac{d\left(ux\right)}{dx}&=\sqrt{\frac{1}{u^2}+1}-\frac{1}{u}\\\implies \int\left(\sqrt{\frac{1}{u^2}+1}-\frac{1}{u}\right)^{-1}\:d\left(ux\right)&=\int\:dx\\\implies \int\frac{u\:dx+x\:du}{\sqrt{\frac{1}{u^2}+1}-\frac{1}{u}}&=x+C_0.\end{align}
Assuming I haven't messed anything up (I feel like I've made a mistake here), how do I evaluate the integral on the LHS? The RHS looks familiar, however, because the answers I've found have all taken the form $y=\sqrt{\left(x+C\right)^2-x^2}$.
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Another simplest way :
From your given equation we have ,
$$2(y\,dy+x\,dx)=2\sqrt{x^2+y^2}\,dx$$
$$\implies \frac{d(x^2+y^2)}{\sqrt{x^2+y^2}}=2\,dx$$
$$\implies 2\sqrt{x^2+y^2}=2x+2C \implies \sqrt{x^2+y^2}=x+C$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1462703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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|
How does $\tan^{-1}(x-\sqrt{1+x^2})=\frac{1}{2}\tan^{-1}x+C$ directly? I'm teaching baby calculus recitation this semester, and I meet a problem to calculate the derivative of
$$y=\tan^{-1}(x-\sqrt{1+x^2})$$
Just apply the chain rule and after some preliminary algebra, I find
$$\frac{dy}{dx}=\frac{1}{2(1+x^2)}$$
What surprises me is that the result implies
$$y=\frac{1}{2}\tan^{-1}x+C$$
Can anyone tell me how to see that directly?
|
First, you have $\arctan (\tan y)=y$ for $y\in]-\pi/2, \pi/2[$.
And for all $y\neq \pi/4+k\pi/2$,
$$\tan (2y)=\frac{2\tan y}{1-\tan^2 y}$$
Thus, for all $z\in]-1,1[$, $y=\arctan z$ is in $]-\pi/4,\pi/4[$, and you can apply the above:
$$\tan(2\arctan z)=\frac{2z}{1-z^2}$$
And since $2\arctan z \in ]-\pi/2,\pi/2[$, you can also apply the first identity to get
$$\arctan \tan (2\arctan z)=2\arctan z=\arctan \left(\frac{2z}{1-z^2}\right)$$
Bear in mind that this is proved only for $z\in]-1,1[$, and it actually is wrong outside this interval.
Now, for $x>0$, $x-\sqrt{1+x^2}<0$ since $1+x^2>x^2$ implies $\sqrt{1+x^2}>x$.
But it's also $>-1$, since
$$(1+x)^2=x^2+1+2x>x^2+1$$
And taking square root, $1+x>\sqrt{1+x^2}$
Thus, for $x>0$, $-1<x-\sqrt{1+x^2}<0$.
It's thus a good candidate to apply the formula above with arctangent, which is true for $|z|<1$:
That is, for $x>0$,
$$2\arctan (x-\sqrt{1+x^2})=\arctan \left(\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}\right)$$
And
$$\frac{2(x-\sqrt{1+x^2})}{1-(x-\sqrt{1+x^2})^2}=\frac{2(x-\sqrt{1+x^2})}{-2x(x-\sqrt{1+x^2})}=-\frac{1}{x}$$
Thus, still for $x>0$,
$$2\arctan (x-\sqrt{1+x^2})=-\arctan\frac{1}{x}$$
And since $\arctan x+\arctan \frac1x=\pi/2$ for $x>0$, you have
$$2\arctan (x-\sqrt{1+x^2})=\arctan (x) -\pi/2$$
Now, we have to prove the same for $x<0$.
Then, $x-\sqrt{1+x^2}<-1$, so we can't apply our formula with arctangent.
However, its inverse is in $]-1,0[$, and for $x<0$,
$$\arctan (x-\sqrt{1+x^2})+\arctan \left(\frac{1}{x-\sqrt{1+x^2}}\right)=-\pi/2$$
$$\arctan (x-\sqrt{1+x^2})-\arctan (x+\sqrt{1+x^2})=-\pi/2$$
$$\arctan (x-\sqrt{1+x^2})=\arctan (x+\sqrt{1+x^2})-\pi/2$$
Then, still for $x<0$, $x+\sqrt{1+x^2} \in ]0,1[$ and
$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{1-(x+\sqrt{1+x^2})^2}$$
$$2\arctan (x+\sqrt{1+x^2})=\arctan \frac{2(x+\sqrt{1+x^2})}{-2x(x+\sqrt{1+x^2})}=-\arctan \frac1x$$
And $\arctan x+\arctan \frac1x=-\pi/2$ (remember that $x<0$), hence
$$2\arctan (x+\sqrt{1+x^2})=\arctan (x)+\pi/2$$
And finally
$$2\arctan (x-\sqrt{1+x^2})=2\arctan (x+\sqrt{1+x^2})-\pi=\arctan (x)-\pi/2$$
This formula is now proved for $x>0$ and for $x<0$. It's also true for $x=0$, since then it amounts to $2\arctan (-1)=-\pi/2$.
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"language": "en",
"url": "https://math.stackexchange.com/questions/1466415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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|
Compute $\int\frac{1}{1+x^4}dx$ I was given the following hint and I can solve the problem by using the following equation. But I'm curious about how one can get the equation. Can someone give me some wikipedia links about it or hints to manipulate the equation?$$\frac{1}{1+x^4}=\frac{x-\sqrt{2}}{2\sqrt{2}(-x^2+\sqrt{2}x-1)}+\frac{x+\sqrt{2}}{2\sqrt{2}(x^2+\sqrt{2}x+1)}$$
BTW, are there any easier method to solve this integral? As seen on the answer, it seems not very possible. ;)
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To find this decomposition, we have to factorize $x^4 + 1$. As $x^4 + 1$ does not have real roots, it does not have real linear factors. This gives the ansatz
$$ x^4 + 1 = (x^2 + ax + b)(x^2 + cx + d) $$
We have
$$
(x^2 + ax + b)(x^2 + cx + d) = x^4 + (a+c)x^3 + (b + d + ac)x^2 + (bc + ad)x + bd $$
Comparing the coefficients with $x^4+ 1$, we have
\begin{align}
a + c &= 0\tag 1\\
b + d + ac &= 0\tag 2\\
bc + ad &= 0\tag 3\\
bd &= 1\tag 4
\end{align}
Now $a = -c$ from (1), this gives
\begin{align}
a &= -c\\
b + d - a^2 &= 0\tag 2\\
a(d - b) &= 0\tag 3\\
bd &= 1\tag 4
\end{align}
As $a = 0$ is impossible, as then (2) would give $b = -d$, which contradicts (4), we must have $a \ne 0$ and hence $d = b$ from (3), this leaves us with
\begin{align}
a &= -c\\
2b - a^2 &= 0\tag 2\\
d &= b\\
b^2 &= 1\tag 4
\end{align}
By (2), $b$ is positive, hence $b = 1$ by (4). This gives $d = 1$, and hence $a = \pm \sqrt 2$. Due to symmetry in $a$ and $c$, we may let $a = \sqrt 2$, giving
$$ x^4 + 1 = (x^2 + \sqrt 2 x + 1)(x^2 - \sqrt 2 x + 1) $$
Now we use partial fraction decomposition, starting with the ansatz
$$ \frac 1{x^4 + 1} = \frac{\alpha x + \beta}{x^2 + \sqrt 2 x + 1} + \frac{\gamma x + \delta}{x^2 - \sqrt 2 x + 1} $$
Clearing denominators gives
\begin{align*}
1 &= (\alpha x + \beta)(x^2 - \sqrt 2 x + 1) + (\gamma x + \delta)(x^2 + \sqrt 2 x + 1)\\
&= (\alpha + \gamma)x^3 + (\beta -\sqrt 2 \alpha + \delta + \sqrt 2 \gamma)x^2 + (\alpha + \gamma - \sqrt 2\beta + \sqrt 2 \delta)x + \beta + \delta
\end{align*}
Comparing coefficients gives
\begin{align*}
\alpha + \gamma &= 0\\
\beta + \delta + \sqrt 2(\gamma - \alpha) &= 0\\
\alpha + \gamma + \sqrt 2(\delta - \beta) &= 0\\
\beta + \delta &= 1
\end{align*}
From the first, the third equation and the forth equation we have $\beta = \delta = \frac 12$, the second one gives $\gamma - \alpha = \frac 1{\sqrt 2}$, as $\alpha = -\gamma$ by the first equation $\gamma = \frac 1{2\sqrt 2}$, $\alpha = - \frac 1{2\sqrt 2}$, giving
$$ \frac 1{x^4 + 1} = \frac{- x + \sqrt 2}{2\sqrt 2(x^2 + \sqrt 2 x + 1)} + \frac{x + \sqrt 2}{2\sqrt 2(x^2 - \sqrt 2 x + 1)}
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1470540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
Contest style inequality Can anyone help me with this inequality? For $a,b,c>0:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
My try:
I first tried inserting a simpler inequality in between the expressions but it feels like nothing simple fits. Next I noticed we can normalise: restricting $a+b+c=1$ it can be made to look like this:
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{a}{1-a}}+\sqrt{\frac{b}{1-b}}+\sqrt{\frac{c}{1-c}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}$$
Another idea is to let $x=a/b,y=b/c, z=c/a:$
$$\sqrt{\frac{2}{3}}\left(\sqrt{\frac{1}{x(y+1)}}+\sqrt{\frac{1}{y(z+1)}}+\sqrt{\frac{1}{z(x+1)}}\right)\leq \sqrt{x+y+z}$$
But I can't see where to go from here.
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I can obtain something a bit weaker. Since the square root is a concave fucntion, we have that
$$
1/3(\sqrt{A}+\sqrt{B}+\sqrt{C})\leq \sqrt{1/3 (A+B+C)}.
$$
This implies
$$
\frac{\sqrt{3}}{3}\left(\sqrt{A}+\sqrt{B}+\sqrt{C}\right)\leq \sqrt{A+B+C}.
$$
Now, take $A=a/(b+c)$, $B=b/(a+c)$ and $C=c/(b+a)$. Since $a,b,c>0$ we have $A<a/b$, $B<b/c$ and $C<c/a$. Then
$$
\frac{1}{\sqrt{3}}\left(\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{a+c}}+\sqrt{\frac{c}{b+a}}\right)\leq \sqrt{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}}.
$$
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"url": "https://math.stackexchange.com/questions/1471811",
"timestamp": "2023-03-29T00:00:00",
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|
Limit Infinity Minus Infinity form without using infinite series Expansions Evaluate $$L=\lim_{x \to 0} \frac{1}{\sin^2x}-\frac{1}{\sinh^2x}$$ If we take L.C.M and use LHopital's Rule it becomes quite Tedious. I tried in this way.
$$
\sinh x=\frac{e^x-e^{-x}}{2}
$$
Now
$$\lim_{x \to 0}\frac{e^x-e^{-x}}{x}=2 \implies \lim_{x \to 0}\left(\frac{e^x-e^{-x}}{x}\right)^{-2}=\frac{1}{4}
$$
Now
$$
\sinh x=\left(\frac{e^x-e^{-x}}{x}\right)\left(\frac{x}{2}\right)
$$
So
$$
\lim_{x \to 0}\frac{1}{\sinh^2x}=\lim_{x \to 0}\left(\frac{e^x-e^{-x}}{x}\right)^{-2}\:\lim_{x \to 0}\frac{4}{x^2}=\lim_{x \to 0}\frac{1}{x^2}
$$
Hence
$$
L=\lim_{x \to 0}\frac{1}{\sin^2x}-\frac{1}{x^2}=\lim_{x \to 0}\frac{(x-\sin x)(x+\sin x)}{(x\sin x)(x\sin x)}=\lim_{x \to 0}\frac{x-\sin x}{x\sin x}\:\lim_{x \to 0}\frac{x+\sin x}{x\sin x}$$
Individually if i evaluate above limits I am getting Zero. But answer is not correct. Can anyone spot out where I am going wrong
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For every $x$ with $0<|x|<\pi$ we have
$$
\frac{1}{\sin^2x}-\frac{1}{x^2}=\frac{x^2-\sin^2x}{x^2\sin^2x}=\frac{x-\sin x}{x^2\sin x}\cdot\frac{x+\sin x}{\sin x}
$$
Using l'Hopital's rule we have:
\begin{eqnarray}
\lim_{x\to0}\frac{x-\sin x}{x^2\sin x}&=&\lim_{x\to0}\frac{(x-\sin x)'}{(x^2\sin x)'}=\lim_{x\to0}\frac{1-\cos x}{2x\sin x+x^2\cos x}=\lim_{x\to0}\frac{(1-\cos x)'}{(2x\sin x+x^2\cos x)'}\\
&=&\lim_{x\to0}\frac{\sin x}{2\sin x+4x\cos x-x^2\sin x}=\lim_{x\to0}\frac{(\sin x)'}{(2\sin x+4x\cos x-x^2\sin x)'}\\
&=&\lim_{x\to0}\frac{\cos x}{2\cos x+4\cos x-4x\sin x-2x\sin x-x^2\cos x}=\frac{1}{2+4}=\frac16\\
\lim_{x\to0}\frac{x+\sin x}{\sin x}&=&\lim_{x\to0}\left(\frac{x}{\sin x}+1\right)=\lim_{x\to0}\left[\frac{(x)'}{(\sin x)'}+1\right]=\left(\frac{1}{\cos x}+1\right)=\frac11+1=2.
\end{eqnarray}
It follows that
$$\tag{1}
\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)=\left(\lim_{x\to0}\frac{x-\sin x}{x^2\sin x}\right)\cdot\left(\lim_{x\to0}\frac{x+\sin x}{\sin x}\right)=\frac16\cdot2=\frac13.
$$
Similarly, for every $x\ne 0$ we have
$$
\frac{1}{\sinh^2x}-\frac{1}{x^2}=\frac{x^2-\sinh^2x}{x^2\sinh^2x}=\frac{x-\sinh x}{x^2\sinh x}\cdot\frac{x+\sinh x}{\sinh x}
$$
Using l'Hopital's rule we have:
\begin{eqnarray}
\lim_{x\to0}\frac{x-\sinh x}{x^2\sinh x}&=&\lim_{x\to0}\frac{(x-\sinh x)'}{(x^2\sinh x)'}=\lim_{x\to0}\frac{1-\cosh x}{2x\sinh x+x^2\cosh x}=\lim_{x\to0}\frac{(1-\cosh x)'}{(2x\sinh x+x^2\cosh x)'}\\
&=&\lim_{x\to0}\frac{-\sinh x}{2\sinh x+4x\cosh x+x^2\sinh x}=\lim_{x\to0}\frac{(\sinh x)'}{(2\sinh x+4x\cosh x-x^2\sinh x)'}\\
&=&\lim_{x\to0}\frac{-\cosh x}{2\cosh x+4\cosh x+4x\sinh x+2x\sinh x+x^2\cosh x}=\frac{-1}{2+4}=-\frac16\\
\lim_{x\to0}\frac{x+\sinh x}{\sinh x}&=&\lim_{x\to0}\left(\frac{x}{\sinh x}+1\right)=\lim_{x\to0}\left[\frac{(x)'}{(\sinh x)'}+1\right]=\left(\frac{1}{\cosh x}+1\right)=\frac11+1=2.
\end{eqnarray}
It follows that
$$\tag{2}
\lim_{x\to0}\left(\frac{1}{\sinh^2x}-\frac{1}{x^2}\right)=\left(\lim_{x\to0}\frac{x-\sinh x}{x^2\sinh x}\right)\cdot\left(\lim_{x\to0}\frac{x+\sinh x}{\sinh x}\right)=-\frac16\cdot2=-\frac13.
$$
Combining (1) and (2) we get:
\begin{eqnarray}
\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{\sinh^2x}\right)&=&\lim_{x\to0}\left[\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)-\left(\frac{1}{\sinh^2x}-\frac{1}{x^2}\right)\right]\\
&=&\lim_{x\to0}\left(\frac{1}{\sin^2x}-\frac{1}{x^2}\right)-\lim_{x\to0}\left(\frac{1}{\sinh^2x}-\frac{1}{x^2}\right)\\
&=&\frac13-\left(-\frac13\right)=\frac13+\frac13=\frac23.
\end{eqnarray}
|
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"timestamp": "2023-03-29T00:00:00",
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|
verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not. verify if $8x^2 - 2x - 3 \equiv 0 \pmod{75}$ has solutions or not.
I tried to write the left term in a form $y^2 \equiv a \pmod{75}$, where $a \in \mathbb{Z}$ so then I can use quadratic repriocity laws to solve this problem. But i get a fraction:
$8x^2 - 2x = 2((2x - \frac{1}{2})^{2} - \frac{1}{4}) $. So the previous statement is equivalent to proving:
$(2x - \frac{1}{2})^2 \equiv 3\frac{1}{2} \cdot \frac{1}{2} \mod 75$. which obviously does not help me further.
Any tips hints on this problem?
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Multiply both sides by $8$:
$$(8x-1)^2\equiv 25\pmod{75}$$
So $5\mid 8x-1$. I.e. $3x\equiv 1\equiv 6\pmod{5}\stackrel{:3}\iff x\equiv 2\pmod{5}$. Assuming this,
$$\iff \left(\frac{8x-1}{5}\right)^2\equiv 1\pmod{3}$$
$$\iff \frac{8x-1}{5}\equiv \pm 1\pmod{3}$$
$$\iff 8x-1\equiv \pm 5\equiv \pm 2\pmod{3}$$
$$\iff \left(2x\equiv 0\pmod{3}\, \text{ or }\, 2x\equiv 2\pmod{3}\right)$$
$$\iff \left(x\equiv 0\pmod{3}\, \text{ or }\, x\equiv 1\pmod{3}\right)$$
By Chinese Remainder Theorem we'll surely have a solution. In this case, $x\equiv 12\pmod{15},\, x\equiv 7\pmod{15}$ are all the solutions.
|
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|
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
Find two numbers, $x,y$ whose sum is $35$ and $x^2y^5$ is maximum.
My answer:
$$x+y=35$$
$x^2y^5$ is maximum
$$y=35-x$$
$$\frac{d}{dx} x^2(35-x)^5$$
Which rule to apply here after? I reached:
$$(35-x)^4(-5x^2+(35-x)2x)=0$$
Either $(35-4x)^4 =0$ or $x^2-7=0$
|
$$x+y=35$$
$$y=35-x$$
Then plug this into the function you want to maximize to create a single variable function. So we want to maximize $$f(x)=x^2(35-x)^5.$$
To find all local extremes, we need to take the derivative, set it equal to $0$, and solve for $x$. Using the product rule, we get
$$f'(x)=2x\cdot (35-x)^5+x^2\cdot 5(35-x)^4\cdot (-1)$$
Now set it equal to $0$ and we can factor out a $x(35-x)^4$:
$$0=2x\cdot (35-x)^5+x^2\cdot 5(35-x)^4\cdot (-1)$$
$$0=x(35-x)^4\Big[2(35-x)-5x\Big]$$
Now we have 3 different factors. We can set each one equal to 0 and solve separately.
Factor 1:
$$x=0$$
Factor 2:
$$(35-x)^4=0$$
$$35-x=0$$
$$x=35$$
Factor 3:
$$2(35-x)-5x=0$$
$$70-2x-5x=0$$
$$70-7x=0$$
$$70=7x$$
$$x=10$$
Now we need to plug each of these local extrema back into $f(x)$ to see which is the absolute max.
$$f(0)=(0)^2(35-0)^5=0$$
$$f(35)=(35)^2(35-35)^5=0$$
$$f(10)=(10)^2(35-10)^5=976,562,500$$
Clearly $x=10$ gives the maximum. Now we just need to figure out what $y$ is when $x=10$.
$$y=35-x$$
$$y=35-10$$
$$y=25$$
So, $x=10$ and $y=25$ are the numbers whose sum is 35 and maximize $x^2y^5$.
Here's a link to a site with a little more detail to explain a similar problem. The function you derive in this other example is easier than the one we had to do here, but the concept is the same.
https://jakesmathlessons.com/derivatives/solution-find-two-numbers-whose-sum-is-23-and-whose-product-is-a-maximum/
|
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"timestamp": "2023-03-29T00:00:00",
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|
Proof verification - Sum formulas in trig We have $a: \mathbb{R}^2 \to \mathbb{R}^2$ represent rotation about $(0,0)$ over an angle $\alpha$. We know that $a$ is a linear map and we know that $a$ corresponds to the matrix $A = \begin{pmatrix}
\cos(\alpha) & -\sin(\alpha) \\
\sin(\alpha) & \cos(\alpha) \\
\end{pmatrix}$. Rotation over an angle $\alpha + \beta$ would thus be given by the matrix $
\begin{pmatrix}
\cos(\alpha + \beta) & -\sin(\alpha + \beta) \\
\sin(\alpha + \beta) & \cos(\alpha + \beta) \\
\end{pmatrix}$. By the additivity property of a linear map, we know that $$
\begin{pmatrix}
\cos(\alpha + \beta) & -\sin(\alpha + \beta) \\
\sin(\alpha + \beta) & \cos(\alpha + \beta) \\
\end{pmatrix} = \begin{pmatrix} \cos(\alpha ) & -\sin(\alpha) \\
\sin(\alpha ) & \cos(\alpha ) \\
\end{pmatrix} \begin{pmatrix} \cos(\beta) & -\sin(\beta) \\
\sin(\beta) & \cos(\beta) \\
\end{pmatrix} $$
$$ = \begin{pmatrix} \cos(\alpha ) \cos(\beta) - \sin(\alpha)\sin(\beta) & -\cos(\alpha)\sin(\beta) - \sin(\alpha)\cos(\beta) \\
\sin(\alpha)\cos(\beta) + \cos(\alpha)\sin(\beta) & -\sin(\alpha)\sin(\beta) + \cos(\alpha)\cos(\beta) \\
\end{pmatrix} $$
QED.
Is this proof correct? My only gripe with it at the moment is explaining why $$
\begin{pmatrix}
\cos(\alpha + \beta) & -\sin(\alpha + \beta) \\
\sin(\alpha + \beta) & \cos(\alpha + \beta) \\
\end{pmatrix} = \begin{pmatrix} \cos(\alpha ) & -\sin(\alpha) \\
\sin(\alpha ) & \cos(\alpha ) \\
\end{pmatrix} \begin{pmatrix} \cos(\beta) & -\sin(\beta) \\
\sin(\beta) & \cos(\beta) \\
\end{pmatrix} $$
instead of $$
\begin{pmatrix}
\cos(\alpha + \beta) & -\sin(\alpha + \beta) \\
\sin(\alpha + \beta) & \cos(\alpha + \beta) \\
\end{pmatrix} = \begin{pmatrix} \cos(\alpha ) & -\sin(\alpha) \\
\sin(\alpha ) & \cos(\alpha ) \\
\end{pmatrix} + \begin{pmatrix} \cos(\beta) & -\sin(\beta) \\
\sin(\beta) & \cos(\beta) \\
\end{pmatrix} $$
since additivity is $f(x+y) = f(x) + f(y)$. Intuitively, I completely understand it, but I don't know how to explain it mathematically, or which mathematical property of a linear map explains it.
|
Rotating by $\alpha + \beta$ degrees can be thought of as first rotating by $\beta$ degrees, then rotating by $\alpha$ degrees. In other words, if $R_\theta:\Bbb R^2 \to \Bbb R^2$ is the rotation by angle $\theta$, then
$$
R_\beta (R_\alpha(x)) = R_{\alpha + \beta}(x)
$$
That is, the combined rotation is the composition of two linear maps, not their sum. In order to find the linear map of a composition, one "multiplies" the corresponding matrices.
|
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|
Show (with epsilón - delta) that $\lim_{x -> \infty}(\sqrt{n^2+n}-n)=\frac{1}{2}$ I need show that $\forall \epsilon > 0$ exist $M \in \mathbb{N}$ such that $|\sqrt{n^2+n}-n-\frac{1}{2}| < \epsilon$ for $n \geq M$.
I tried considering $|\frac{(\sqrt{n^2+n}-n)\sqrt{n^2+n}+n)}{\sqrt{n^2+n}+n)}-\frac{1}{2}|=|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}+n}|<|\frac{n-\sqrt{n^2+n}}{2\sqrt{n^2+n}}|$
I need help because now I don't know how abound more...
Thanks.
|
Consider $\sqrt{n^2+n}-n-\frac{1}{2}$. Multiply top and missing bottom by
$\sqrt{n^2+n}+n+\frac{1}{2}$. We get
$\frac{-1/4}{\sqrt{n^2+n}+n+\frac{1}{2}}$.
The absolute value of this is less than $\frac{1}{8n}$. Now finding an appropriate $M$ is straightforward.
|
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|
Polynomial Multiplication in GF(256) I would like to compute the following:
10100011 * 01100011 in GF(256) using the AES irreducible polynomial.
So first we get the polynomials:
10100011 = x^8 + x^6 + x + 1
01100011 = x^7 + x^6 + x^2 + 1
Multiplying these out, I got the result:
x^56 + x^48 + x^42 + x^36 + x^16 + x^12 + x^3 + x^2 + x + 1
So when I try to reduce this using the AES irreducible polynomial, x^8 + x^4 + x^3 + x + 1, I am getting the result:
x^7 + x^6 with a remainder of x^42 + x^36 + x^28 + x^24 +x^21 + x^18 + x^16 + x^12 + x^8 + x^6 + x^3 + x^2 + x + 1
This cannot be correct, but I'm having trouble finding where I made a mistake. I have a feeling my multiplication of the two polynomials is incorrect.
|
The exponentiation law is:
x^a * x^b = x^(a+b)
|
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|
Equation - what first? I have this equation:
$$
(x+y)(x^x + y^y) = 2009.
$$
I must designate all pairs of integers satisfying the equation.
What first? I tried multiply brackets , but to no avail
|
hint: $2009=7^2\times 41$ so keeping in mind that $x+y\lt x^x+y^y$ we're looking for integers $x$ and $y$ such that
$$(x+y)(x^x+y^y)=7^2\times 41$$
This means finding integers such that $x+y=41$ and $x^x+y^y=49$ or $x+y=7$ and $x^x+y^y=7\times 41$
|
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|
Answer to some trigonometry series I have two series as bellow
$$\frac{2}{N}\sum_{n=0}^{N-1}\cos^2(\frac{2\pi n(l-q)}{N}) \quad l\neq q \quad(1)$$
$$ \sum_{q=0}^{N-1}cos(\frac{2\pi nq}{N})cos(\frac{2\pi rq}{N}) \quad n\neq r \quad (2)$$
Where $l$, $q$, $r$, $n$ and $N$ are integers. I have computed the two series in MATLAB. The first one is equal to $1$ and the second one is zero. But I cannot prove this analytically. Can anyone help me understand why (1) equals $1$ and (2) equals zero?
Thanks in advance
|
For Reference, I am putting two formulas I've used in the solution of these questions :
(Sum of sines and cosines with angle are in A.P.)
$\displaystyle \sum_{k=0}^{n} \sin(k\beta) = \dfrac{\sin \dfrac{n\beta}{2}}{\sin \dfrac{\beta}{2}} \cdot \sin \left [ (n+1)\dfrac{\beta}{2} \right]$
$\displaystyle \sum_{k=0}^{n} \cos(k\beta) = \dfrac{\sin \dfrac{n\beta}{2}}{\sin \dfrac{\beta}{2}} \cdot \cos \left [ (n+1)\dfrac{\beta}{2} \right]$
1st Question - Partial Solution::
$\displaystyle \begin{aligned} \dfrac{2}{N}\sum_{n=0}^{N-1}\cos^2 \left (\dfrac{2\pi n(l-q)}{N} \right) &= \dfrac{2}{N} \sum_{n=0}^{N-1} \left [ \dfrac{1 + \cos \dfrac{4 \pi n(l-q)}{N}}{2} \right] \\ \\ &= \dfrac{2}{N} \left [\dfrac{N-1}{2} + \dfrac {\displaystyle \sum_{n=0}^{N-1} \cos \dfrac{4 \pi n(l-q)}{N}}{2} \right] \\ \\ &= \dfrac{2}{N} \left [\dfrac{N-1}{2} + \dfrac{\sin \dfrac{2 \pi (l-q)(N-1)}{N}}{2\sin \dfrac{2 \pi (l-q)}{N}} \cdot \cos \dfrac{2 \pi (l-q)N}{N} \right] \\ \\ &= \dfrac{2}{N} \left [\dfrac{N-1}{2} + \dfrac{1}{2} \right] =1 .\end{aligned} $
(try to understand the last step here. )
2nd Question -- Hint::
$\begin{aligned} \sum_{q=0}^{N-1}cos(\frac{2\pi nq}{N})cos(\frac{2\pi rq}{N}) &= \dfrac{1}{2}\sum_{q=0}^{N-1} 2cos(\frac{2\pi nq}{N})cos(\frac{2\pi rq}{N}) \\ &= \dfrac{1}{2}\sum_{q=0}^{N-1} \sin \left (\frac{2\pi (n+r)q}{N} \right) + \sin \left(\frac{2\pi (n-r)q}{N} \right)\end{aligned}$.
|
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|
Find the sum from the system of equations
If $x,y, z$ satisfy: $$x + y = z^2 + 1, y + z = x^2 + 1, x + z = y^2 + 1 $$ Find the value of $2x +3y + 4z$.
This gives us (by getting $x + y + z$ that)
$z^2 + z + 1 = x^2 + x + 1 = y^2 + y + 1 \implies z^2 + z = x^2 + x = y^2 + y$.
Using the first and last, I also got:
$2x + y + z = z^2 + y^2 + 2$
But I cannot get much farther!
EDIT:
$x -z = z^2 - x^2 = (z-x)(z+x) \implies z + x = -1$.
$y - x = x^2 = y^2 = (x-y)(x+y) \implies x + y = -1$
Thus, $z - y = 0$ and $z = y$.
$2x + 7y$ is to be found then, and $x = -1 -y $ so: $2(-1 - y) - 7y = -2 - 9y$ is to be found.
Any ideas here?
|
As you've already noted, we can set $z^2 +z +1 = x^2 +x +1 = y^2 +x +1= \lambda^2$. We can do this because $x^2 +x+1>0 $ , $\forall x \in \mathbb{R}$.
Recall the law of cosines and observe that $z^2 +z +1=\lambda^2$ can be rewritten as $\lambda^2=z^2+ 1^2 -2 \cdot z \cdot 1 \cdot \cos 120^\circ $, which is the above law for a triangle with sides of length $z$ and $\lambda$ and $1$.
Applying this method to the other two equations we get that $x,y,z$ are the cevians of an equilateral triangle with sides of length $\lambda$ and since we've already established that one those is $1$, they're all equal to $1$, i.e $x=y=z=1$, the result then follows as mentioned above. For a visualization see this figure.
|
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|
Algebraic values of sine at sevenths of the circle At the end of a calculation it turned out that I wanted to know the value of
$$\sin(2\pi/7) + \sin(4\pi/7) - \sin(6\pi/7).$$
Since I knew the answer I was supposed to get, I was able to work out that the the above equals $\sqrt{7}/2$, and I can confirm this numerically. How would I prove this? I suspect I want to use the seventh roots of unity in some way but I am not sure how to proceed.
|
Using Prosthaphaeresis Formulas,
$$\sin2x+\sin4x-\sin6x=2\sin2x\cos2x-(2\sin2x\cos4x)$$
$$=2\sin2x(\cos2x-\cos4x)=4\sin2x\sin3x\sin x$$
Now from this, $\sin(2n+1)x=(2n+1)\sin x+\cdots+2^{2n}(-1)^n\sin^{2n+1}x$
If $\sin(2n+1)x=0,(2n+1)x=m\pi$ where $m$ is any integer
$x=\dfrac{m\pi}{2n+1}$ where $m\equiv0,\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$
So, the roots of $$2^{2n}(-1)^nt^{2n+1}x+\cdots+(2n+1)t=0$$ are $\sin\dfrac{m\pi}{2n+1}$ where $m\equiv0,\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$
So, the roots of $$2^{2n}(-1)^nt^{2n}x+\cdots+2n+1=0$$ are $\sin\dfrac{m\pi}{2n+1}$ where $m\equiv\pm1,\pm2,\cdots,\pm n\pmod{2n+1}$
$\implies\prod_{r=-n}^n\sin\dfrac{m\pi}{2n+1}=\dfrac{2n+1}{2^{2n}(-1)^n}$
Now as $\sin(-y)=-\sin y,$
$\implies\prod_{r=1}^n(-1)^n\sin^2\dfrac{m\pi}{2n+1}=\dfrac{2n+1}{2^{2n}(-1)^n}$
As $0<\dfrac{m\pi}{2n+1}<\dfrac\pi2$ for $1\le m\le n,$
$\implies\prod_{r=1}^n\sin\dfrac{m\pi}{2n+1}=\dfrac{\sqrt{2n+1}}{2^n}$
Here $n=3$
|
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|
$\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of n such that l is non zero finite real number. Find l Problem :
$\lim\limits_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$ value of $n$ such that $l$ is non zero finite real number. Find value of $l$.
My approach :
$$\lim_{x \to 0} \frac{x^n}{\cos(\sin x) -\cos x}=l$$
$$ \Rightarrow \lim_{x \to 0} \frac{x^n}{-2\sin(\frac{\sin x+x}{2})\sin(\frac{\sin x-x}{2})} =l $$
Please suggest how to move further , will be of great help thanks.
|
In order to simplify matters we are going to look at the reciprocal of the given expression. The denominator can be developed into a series as follows: From
$$\cos x=1-{1\over2}x^2+{1\over24}x^4+?x^6$$
and
$$\eqalign{\cos(\sin x)
&=1-{1\over2}x^2\left(1-{1\over 6}x^2+?x^4\right)^2+{1\over24}x^4(1+?x^2)^4+?x^6\cr
&=1-{1\over2}x^2+\left({1\over6}+{1\over24}\right)x^4+?x^6\cr}$$
it follows that
$$\cos(\sin x)-\cos x={1\over6}x^4+?x^6\ .$$
This implies
$$\lim_{x\to0}{\cos(\sin x)-\cos x\over x^4}={1\over6}\ ,$$
respectively:
$$\lim_{x\to0}{x^4\over \cos(\sin x)-\cos x}={6}\ ,$$
and $n=4$ is the only exponent leading to a limit $\ne0, \infty$.
|
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|
Factoring a polynomial (multivariable)
Factor $ (a - b)^3 + (b - c)^3 + (c-a)^3$ by SYMMETRY.
Okay, this is the problem. Let $f(a) = (a - b)^3 + (b-c)^3 + (c-a)^3$ obviously, if you let $a = b$ then, $f(b) = 0$, thus $(a - b)$ is a factor of $f(a)$. Then someone said :
If $(a - b)$ is a factor then $(b - c)$ and $(a-c)$ must be factors as well by symmetry.
But $f(a, b, c)$ is not symmetric, actually, $f(b, a, c) = -f(a, b, c)$ it is an alternating polynomial, so what is up with the solution?
|
The symmetry to use is not to change $f(a,b,c)$ to $f(b,a,c)$.
Instead, change $f(a,b,c)$ to $f(b,c,a)$ or $f(c,a,b)$.
This is the same as replacing $(a - b)^3 + (b - c)^3 + (c - a)^3$ by either
$(b - c)^3 + (c - a)^3 + (a - b)^3$ or $(c - a)^3 + (a - b)^3 + (b - c)^3$,
which clearly makes no change in the value.
So the same reasoning that showed $(a-b)$ is a factor
must work for $(b-c)$ and $(c-a)$ as well.
And if $(c-a)$ is a factor, $(a - c)$ is a factor.
|
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|
Problem related to integration I tried using partial fractions but I am not sure whether it is the right approach or not. I need help with this problem
Evaluation of Integral $$\int\frac{3x^4-1}{(x^4+x+1)^2}dx $$
|
Let $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = \int\frac{3x^4-1}{x^2\left(x^3+1+x^{-1}\right)^2}dx = \int\frac{3x^2-x^{-2}}{(x^3+1+x^{-1})^2}dx$$
Now Put $x^3+1+x^{-1} = t\;,$ Then $(3x^2-x^{-2})dx = dt$
So Integral $$\displaystyle I = \int\frac{1}{t^2} dt = -\frac{1}{t}+\mathcal{C} = -\frac{1}{x^3+1+x^{-1}}+\mathcal{C}$$
So we get $$\displaystyle I = \int\frac{3x^4-1}{(x^4+x+1)^2}dx = -\left[\frac{x}{x^4+x+1}\right]+\mathcal{C}$$
|
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|
Fourier sine series for $x^3$ It is asked to find the Fourier Sine Series for $x^3$ given that
$$\frac{x^2}{2} = \frac{l^2}{6} + \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} \cos\left(\frac{n \pi x}{l} \right)$$
integrating term by term. (This result was found in another exercise). As suggested, I integrated:
$$\int \frac{x^2}{2} dx = \frac{x^3}{6} = \frac{l^2x}{6} + \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} \int \cos\left(\frac{n \pi x}{l} \right)dx + C$$
$$ \Rightarrow \frac{x^3}{6} = \frac{l^2x}{6}+ \frac{2l^2}{\pi^2} \sum_{n=1}^\infty (-1)^n \frac{1}{n^2} \frac{l}{n \pi} \sin\left(\frac{n \pi x}{l} \right) + C $$
$$\Rightarrow x^3 = l^2x + \frac{12l^3}{\pi^3}\sum_{n=1}^\infty (-1)^n \frac{1}{n^3} \sin\left(\frac{n \pi x}{l} \right) + C$$
It looks like that wolfram gives a different answer. I don't know if the problem is the constant $C$ or if I made something wrong. Please, follow the book approach, don't try to calculate the coefficient of Fourier sine series.
Thanks!
|
1) You can simply conclude that $C=0$ as the equation must hold when $x=0$.
2) The answer of wolfram is different from you in appearance. The reason is that in the answer by wolfram, $x$ is also written in sin Fourier series and then combined with other terms in your equation.
|
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|
How to prove that the following system of equations has only one solution? $
\begin{cases}
(x - 1)^2 + (y + 1)^2 = 25 \\
(x + 5)^2 + (y + 9)^2 = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
I have to solve this system of equations. After substituting $y = -\frac{3}{4}x - \frac{13}{2}$ into $(x - 1)^2 + (y + 1)^2 = 25$, I got that $(x + 2)^2 = 0$, so $x = -2$, so $y = -\frac{3}{4} \cdot (-2) - \frac{13}{2} = -5$.
How to prove that this is the only solution?
|
The first two are equations of circles. Two circles may have no point in common, or only one tangency point, or two points of intersection or all points if they're actually the same circle.
It's easy to check this is not the first nor the last case. Now, subtract the first equation from the second one:
$
\begin{cases}
(x^2 - 2x + 1) + (y^2 + 2y + 1) = 25 \\
(x^2 + 10x + 25) + (y^2 + 18y + 81) = 25 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
$
\begin{cases}
x^2 - 2x + 1 + y^2 + 2y + 1 = 25 \\
12x + 24 + 16y + 80 = 0 \\
y = -\frac{3}{4}x - \frac{13}{2}
\end{cases}
$
Now the second and third equation make a system of linear equations, so they may have only zero, one or infinity of satisfying points – but certainly not two. Q.E.D.
|
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|
Given that $\pi=2\int_{-1}^1\sqrt{1-x^2}dx$ use the properties of the integral to compute $\int_{-2}^{2}(x-3)\sqrt{4-x^2}dx$ in terms of $\pi$. This is part c of the exercise set 2.4 in Apostol's Calculus 2nd Ed. We are only supposed to use the properties of the integral such as: linearity, additivity with respect to the interval of integration, invariance under translation, expansion or contraction of the interval of integration, or any consequences of them. A hint would be helpful.
|
By linearity
$$\int_{-2}^{2} (x-3) \sqrt{4-x^2}dx = \int_{-2}^{2} x \sqrt{4-x^2}dx - 3 \int_{-2}^{2} \sqrt{4-x^2}dx$$
But $\int_{-b}^b f(x) dx = 0$ for odd $f(x)$ (Optional Exercise 25), and $f(x) = x \sqrt{4-x^2}$ is odd.
\begin{align}
\int_{-2}^{2} (x-3) \sqrt{4-x^2}dx &= - 3 \int_{-2}^{2} \sqrt{4-x^2}dx\\
&= -6 \int_{-1}^{1} \sqrt{4-(2x)^2}dx \qquad \mbox{Contraction, k = 1/2, Thm 1.19}\\
&= -12 \int_{-1}^{1} \sqrt{1-x^2}dx \qquad \mbox{Homogeneous property Thm 1.3}\\
&= -6 \pi
\end{align}
|
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|
Deriving Frobenius number for two denominations? I would like to know if the following question has an intelligent solution:
Determine the largest bet that cannot be made using chips of $7$ and $9$ dollars.
After not being able to solve it I found a solution online which writes out all combinations of $7$ and $9$ up to $90$ and then notes that we can produce all numbers after $47$ so the largest bet is $47$.
Then after asking this question here on the site I got pointed to the formula for the Frobenius number on Wikipedia: $g(a,b) = ab - a- b$. But Wikipedia does not explain this formula.
How to derive this formula? Why is the Frobenius number for two coins
$ab - a - b$ where $a,b$ are the denominations of the coins?
|
A very nice explanation is given at Cut the knot.
You want the number for $p, q$ with $\gcd(p, q) = 1$ (otherwise it makes no sense). Consider the $q$ sequences:
$\begin{align}
&0 + 0, 0 + q, 0 + 2 q, \dotsc \\
&p + 0, p + q, p + 2 q, \dotsc \\
&\vdots \\
&(q - 1) p + 0, (q - 1) p + q, \dotsc
\end{align}$
They have no elements in common. Now take the series:
$\begin{align}
\sum_{k \ge 0} z^{r p + k q}
&= z^{r p} \sum_{k \ge 0} z^{k q} \\
&= \frac{z^{r p}}{1 - z^q}
\end{align}$
Add them all up:
$\begin{align}
\sum_{0 \le r \le q - 1} \frac{z^{r p}}{1 - z^q}
&= \frac{1}{1 - z^q} \sum_{0 \le r \le q - 1} z^{r p} \\
&= \frac{1}{1 - z^q} \frac{1 - z^{q p}}{1 - z^p} \\
&= \frac{1 - z^{p q}}{(1 - z^p) (1 - z^q)}
\end{align}$
The coefficients of this are all 0 (the number isn't representable) or 1 (the number is representable). We get the series with 1 for non-representable ones by:
$\begin{align}
\frac{1}{1 - z} - \frac{1 - z^{p q}}{(1 - z^p) (1 - z^q)}
&= \frac{(1 - z^p) (1 - z^q) - (1 - z) (1 - z^{p q})}
{(1 - z) (1 - z^p) (1 - z^q)}
\end{align}$
This is a polynomial (it is easy to see that large enough numbers are all representable). Its degree is the last non-representable number, and that is just the degree of the numerator less the degree of the denominator:
$\begin{align}
(1 + p q) - (p + q + 1)
= p q - p - q
\end{align}$
As the coefficients are all 1, we can also get the number of non-representable ones as:
$\begin{align}
\lim_{z \to 1} \frac{(1 - z^p) (1 - z^q) - (1 - z) (1 - z^{p q})}
{(1 - z) (1 - z^p) (1 - z^q)}
\end{align}$
Applying l'Hôpital thrice gives:
$\begin{align}
\frac{−3 p q (p q − p − q + 1)}{- 6 p q}
= \frac{p q - p - q + 1}{2}
\end{align}$
|
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|
How to simplify $\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$ to $\sqrt[3]{3}$ $$\frac{1}{\sqrt[3]{3}-1} - \frac{2}{\sqrt[3]{3}+1}$$
I have simplified above to:
$$\frac{3-\sqrt[3]{3}}{(\sqrt[3]{3}+1)(\sqrt[3]{3}-1)}$$
What is equal to:
$$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}$$
WolframAlpha says this can be shown as $\sqrt[3]{3}$, but I can't find out how to do this.
|
$$\frac{3-\sqrt[3]{3}}{\sqrt[3]{9}-1}=\frac{(3^{1/3})^3-3^{1/3}}{3^{2/3}-1}=\frac{3^{1/3}(3^{2/3}-1)}{3^{2/3}-1}=3^{1/3}(=\sqrt[3]{3})$$
|
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Simplify $(-2\sin(t)-2\sin(2t))^2+(2\cos(t)-2\cos(2t))^2$? In calculating the length of a deltoid one gets the following string of trigonometric functions:
$$(-2\sin(t)-2\sin(2t))^2+(2\cos(t)-2\cos(2t))^2$$
$$=4\big(\sin^2(2t)+2\sin(t)\sin(2t)+\cos^2(2t)-2\cos(t)\cos(2t)+\sin^2(t)+\cos^2(t)\big)$$
$$(\sin^2x+\cos^2x=1)$$
$$=4\big(1+2\sin(t)\sin(2t)-2\cos(t)\cos(2t)+1\big)$$
Question is, how to simplify this further?
|
Notice,
$$4(1+2\sin t\sin 2t-2\cos t\cos 2t+1)$$
$$=8(1+\sin t\sin 2t-\cos t\cos 2t)$$
$$=8(1-(\cos t\cos 2t-\sin t\sin 2t)$$
using $\color{red}{\cos A\cos B-\sin A\sin B=\cos (A+B)}$,
$$=8(1-\cos(t+2t))$$ $$=8(1-\cos 3t)$$
using $\color{red}{\cos A=1-2\sin^2\frac{A}{2}}$,
$$=8\left(1-\left(1-2\sin^2 \frac{3t}{2}\right)\right)$$
$$=8\left(1-1+2\sin^2 \frac{3t}{2}\right)$$
$$=8\left(2\sin^2 \frac{3t}{2}\right)=\color{blue}{16\sin^2 \frac{3t}{2}}$$
|
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"language": "en",
"url": "https://math.stackexchange.com/questions/1491073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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|
How to prove $\frac xy + \frac yx \ge 2$ I am practicing some homework and I'm stumped.
The question asks you to prove that
$x \in Z^+, y \in Z^+$
$\frac xy + \frac yx \ge 2$
So I started by proving that this is true when x and y have the same parity, but I'm not sure how to proceed when x and y have opposite partiy
This is my proof so far for opposite parity
$x,y \in Z^+ $ $|$ $x \gt 0,$ $y \gt 0$. Let x be even $(x=2a, $ $ a \in Z^+)$ and y be odd $(y=2b+1, $ $b \in Z^+)$. Then,
$\frac xy + \frac yx \ge 2$
$\frac {2a}{2b+1} + \frac {2b+1}{2a} \ge 2$
$\frac {2b^2 + 4a^2 + 4a + 1}{2b(2a+1)} \ge 2$
$4b^2 + 4a^2 +4a + 1 \ge 4b(2a+1)$
$4b^2 + 4a^2 + 4a +1 \ge 8ab + 4b$
$4b^2 - 4b + (2a + 1)^2 \ge 8ab$
$(2b-1)^2 + 1 + (2a+1)^2 \ge 8ab$
I feel like this is the not the correct way to go about proving it, but I can't think of a better way to do it. Does anyone have any suggestions? Just hints please, not a solution.
|
Clearly $(x-y)^2 \geq 0$
So $x^2-2xy+y^2 \geq 0 \Rightarrow x^2+y^2 \geq 2xy $
Since $x $ and $y$ are positive integers , $x \neq 0 $ and $y \neq 0$.
Thus we can divide by $xy$.
So we have $\frac{x^2+y^2}{xy} \geq \frac {2xy}{xy} \Rightarrow \frac {x}{y}+\frac{y}{x} \geq 2$.
|
{
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|
What would be the fastest way of solving the following inequality $\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$ What would be the fastest way of solving the following inequality:
$\frac{(x+1)}{(x-1)(x+2)}>\frac{(x)}{(x-1)(x-3)}$
|
Clearly
$\frac{x+1}{\left(x-1\right)\left(x+2\right)}-\frac{x}{\left(x-1\right)\left(x-3\right)}>0$
$\Rightarrow \frac{-4x-3}{\left(x+2\right)\left(x-1\right)\left(x-3\right)}>0$
Then draw a straight line and mark $x$ values that make the fractors zero. That is $-2 ,-\frac{3}{4} ,1,3$.
Then check the validity for some $x$ value less than $-2$.
Say $x=-3$.
For $x=-3$ we have that inequality does not hold.
So LHS is negative for $x=-3$
Then clearly $x \in \:\left(-2,-\frac{3}{4}\right)\cup \left(1,\:3\right)$
|
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|
Given integer roots of $x^2+mx-n=0$ and $x^2-mx+n=0$. Show $6 \mid n$. Suppose that $m$ and $n$ are integers such that both the quadratic equations $x^2+mx-n=0$ and $x^2-mx+n=0$ have integer roots. Prove that $n$ is divisible by $6$.
I figured out the roots of the equations through quadratic formula-for the first one it is $\frac{-m \pm \sqrt{m^2-4n}}{2}$ and for the second one it was $\frac{m \pm \sqrt{m^2+4n}}{2}$ .
Then I proved that the discriminant has the same parity as that of $m$ through congruences. Therefore, it was clear that if $m^2-4n$ and $m^2+4n$ were perfect squares,then the roots would be integers. I started with $m^2-4n$ and equated it with $a^2$, where $a$ is some integer. Then I rearranged the terms and factorized things to get $(m-a)(m+a)=4n$. This shows that at least one of them should be even but if one is even then the other should also be even. This shows that $m$ and $a$ are of the same parity.
I don't know what to do next. I want some help to prove that $n$ is even. I could figure out how to prove that $n$ is divisible by 3 once it is proved that $n$ is even.
|
Claim 1:
Suppose both $m$ and $n$ were odd, then both $x^2+mx-n=0$ and $x^2-mx+n=0$ cannot have any integer solutions.
Reason:
From the first equation we have $n=x(x+m)$. Thus if $x$ is even, then $n$ must be even and if $x$ is odd then $x+m$ is even, either way $n$ would be even which contradicts the assumption that $n$ is odd. Likewise we can do the second equation.
Claim 2: $n$ must be even.
Reason:
Let $A$ be the integer root of $x^2+mx-n=0$ and $B$ be the integer root of $x^2-mx+n=0$ . Suppose $n$ was odd. In which case both $A$ and $B$ must be odd (since the product of roots will be $\pm n$). By claim 1, $m$ must be even (otherwise no integer roots).
Moreover
\begin{align*}
A^2+mA-n & = 0\\
B^2-mB+n &=0
\end{align*}
This gives us $2n=(A+B)(A-B+m)$. The right hand side is divisible by $4$ so the left hand side must be divisible by $4$ as well. But with $n$ odd that is not possible.
Thus $n$ must be even.
|
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|
Solving linear system equation How would I solve this linear system equation?
$$\begin{cases}
2w+x-y=4\\
3z-x=6\\
-2y-x+9z+4w=7
\end{cases}$$
First I arranged them so I could make the metric and then I was stuck and I don't realy know how to continue, please help me.
|
Write your system into normal form:
\begin{cases}
x-y+2w=4\\
x-3z=-6\\
x+2y-9z-4w=-7
\end{cases}
Now, depending on the tools you have available, there are several possibilities. The most efficient is Gaussian elimination: the matrix of the system is
\begin{align}
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
1 & 0 & -3 & 0 & -6 \\
1 & 2 & -9 & -4 & -7
\end{array}\right]
&\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
0 & 1 & -3 & -2 & -10 \\
1 & 2 & -9 & -4 & -7
\end{array}\right] && R_2\gets R_2-R_1 \\[6px]
&\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
0 & 1 & -3 & -2 & -10 \\
0 & 3 & -9 & -6 & -11
\end{array}\right] && R_3\gets R_3-R_1 \\[6px]
&\to
\left[\begin{array}{cccc|c}
1 & -1 & 0 & 2 & 4 \\
0 & 1 & -3 & -2 & -10 \\
0 & 0 & 0 & 0 & 19
\end{array}\right] && R_3\gets R_3-3R_2 \\[6px]
\end{align}
which shows the system has no solution.
Another possibility is to get $x$ from the second equation: $x=3z-6$. Substitute in the third equation to get
$$
3z-6+2y-9z-4w=-7
$$
or
$$
2y-6z-4w=-1
$$
Substitute also in the first equation to find $3z-6-y+2w=4$, so you get
$$
y=3z+2w-10
$$
and now back in the other equation
$$
6z+4w-6z-4w-20=-1
$$
or
$$
-20=-1
$$
that is false.
|
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|
Olympiad problem about finding minimum value with $x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2$ Let $x,y,z$ be positive real numbers such that $x^2y^2+y^2z^2+z^2x^2\ge x^2y^2z^2$.
Find the minimum value of
$$\frac{x^2y^2} {z^3(x^2+y^2)}+\frac {y^2z^2} {x^3(y^2+z^2)}+\frac {z^2x^2} {y^3(z^2+x^2)}$$
I'm pretty sure that the answer would be $\frac {\sqrt {3}} {2}$, when all parameters are $\sqrt {3}$. But I couldn't prove it after some hours of thinking. So can anyone help me?
Any help would be welcome.
Thanks:D.
|
If $x=y=z=\sqrt3$ then we get a value $\frac{\sqrt3}{2}$.
We'll prove that it's a minimal value, for which it's enough to prove that
$$\sum_{cyc}\frac{y^2z^2}{x^3(y^2+z^2)}\geq\frac{\sqrt3}{2}\cdot\sqrt{\frac{x^2y^2+x^2z^2+y^2z^2}{x^2y^2z^2}}$$ or
$$\sum_{cyc}\frac{y^3z^3}{x^2(y^2+z^2)}\geq\frac{\sqrt3}{2}\cdot\sqrt{x^2y^2+x^2z^2+y^2z^2}.$$
Let $xy=c$, $xz=b$ and $yz=a$ and since our inequality is homogeneous,
we can assume that $a^2+b^2+c^2=3$.
Thus, we need to prove that
$$\sum_{cyc}\frac{a^3}{3-a^2}\geq\frac{3}{2}$$ or
$$\sum_{cyc}\left(\frac{a^3}{3-a^2}-\frac{1}{2}\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)(2a^2+3a+3)}{3-a^2}\geq0$$ or
$$\sum_{cyc}\left(\frac{(a-1)(2a^2+3a+3)}{3-a^2}-2(a^2-1)\right)\geq0$$ or
$$\sum_{cyc}\frac{(a-1)^2(2a^2+6a+3)}{3-a^2}\geq0.$$
Done!
|
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|
How to prove this inequality? $\sum_{k=2}^{n-1}k\log k < \frac{1}{2} n^2\log n-\frac{1}{8}n^2$? How can I prove this inequality?
$$\sum_{k=2}^{n-1}k\log k < \frac{1}{2} n^2\log n-\frac{1}{8}n^2.$$
|
Summation by parts is enough. We have:
$$ \sum_{k=1}^{n-1} k\log k = \frac{n^2-n}{2}\cdot\log n-\sum_{k=1}^{n-2}\frac{k^2-k}{2}\cdot\log\left(1+\frac{1}{k}\right)$$
but $k\mapsto k\cdot \log\left(1+\frac{1}{k}\right)$ is an increasing function, hence:
$$ \sum_{k=2}^{n-1} k\log k \leq \frac{n^2-n}{2}\cdot \log n- \sum_{k=2}^{n-2}(k-1)\log\left(\frac{3}{2}\right) $$
leads to the stronger bound:
$$ \sum_{k=2}^{n-1} k\log k \leq \frac{n^2-n}{2}\cdot \log n - \frac{(n-2)(n-3)}{2}\cdot\log\left(\frac{3}{2}\right) $$
for any $n\geq 4$.
|
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|
How many ways are there to select an ordered pair of numbers from $1$ to $7$ so that the sum is even? How many ways are there to select an ordered pair of numbers from $1$ to $7$ so that the sum is even?
Soln: The way I tried approaching this problem was, I made a grid of $7 \times 7$ squares. So $49$ ordered pairs in total. Also the sum of two numbers is even only if the numbers selected are either both even or both odd. So, how do I proceed further?
Thank you.
|
Your table is
$\begin{array}[ht]{|p{2cm}|||p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|p{0.5cm}|} \hline \text{sum } & 1 &2 &3 &4 &5 &6 & 7\\ \hline \hline \hline 1 &\color{red}2 &3 &\color{red}4 &5 &\color{red}6 &7&\color{red}8 \\ \hline 2 & 3 &\color{red}4 &5 &\color{red}6 &7&\color{red}8 &9 \\ \hline 3&\color{red}4 &5 &\color{red}6 &7 &\color{red}8&9&\color{red}{10} \\ \hline 4 &5 &\color{red}6 &7&\color{red}8&9&\color{red}{10}&11 \\ \hline 5 &\color{red}6 &7&\color{red}8&9&\color{red}{10}&11&\color{red}{12} \\ \hline 6 &7&\color{red}8&9&\color{red}{10}&11&\color{red}{12}&13 \\ \hline 7&\color{red}8&9&\color{red}{10}&11&\color{red}{12}&13&\color{red}{14} \\ \hline \end{array}$
The red marked numbers are the even sums. Now you can sum up the up-right diagonals of the red marked numbers: $1+3+5+7+5+3+1=25$
Additional you can calculate the probability of selecting an even sum.
The number of possible outcomes is $7\cdot 7=49$
Therefore the probability to select an ordered pair of numbers from
1 to 7 so that the sum is even is $\frac{25}{49}\approx 51\%$
|
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|
Let $p$ and $q = 2p + 1$ be odd primes. Show that $(−1) ^{\frac{p−1}{ 2}} 2$ is a primitive root modulo q. Let $p$ and $q = 2p + 1$ be odd primes. Show that $(−1) ^{\frac{p−1}{ 2}} 2$ is a primitive root modulo q.
We see that our number must have order $q-1$ to be a primitive root. If we check the divisors which are $1,2,p,2p$ and see that $\left((−1) ^{\frac{p−1}{ 2}} 2\right)^\frac{q-1}{d} \equiv 1 \pmod{q}$ we are finished.
Obviously $1$ and $2$ fail so we need to check $p$.
$p = \frac{q-1}{2}$, so we see $\left((−1) ^{\frac{p−1}{ 2}} 2\right)^\frac{q-1}{d} \equiv \left((−1) ^{\frac{p−1}{ 2}} 2\right)^2 \equiv 4 \pmod{q}$. So this fails as well. The only one to check is $2p$ but that is equal to $q-1$ so we are finished.
Kees
|
I.e. we want to prove $\left((-1)^{\frac{p-1}{2}}2\right)^{d}\not\equiv 1\pmod{q}$ for all $d\in\{1,2,p\}$.
If $(-1)^{\frac{p-1}{2}}2\equiv 1\pmod{q}$, then $\pm 2\equiv 1\pmod{q}$, so $q=3$, which is not of the form $2p+1$.
If $\left((-1)^{\frac{p-1}{2}}2\right)^2\equiv 1\pmod{q}$, then $4\equiv 1\pmod{q}$, i.e. $q=3$, which is not of the form $2p+1$.
If $\left((-1)^{\frac{p-1}{2}}2\right)^p\equiv 1\pmod{q}$, then we check two cases:
$1)\ \ $ $p\equiv 1\pmod{4}$. Then $\left((-1)^{\frac{p-1}{2}}2\right)^p\equiv 2^p\pmod{q}$, so $2^p\equiv 1\pmod{q}$. By Euler's criterion $2^p\equiv 2^{\frac{q-1}{2}}\equiv \left(\frac{2}{q}\right)\pmod{q}$, so we must have $\left(\frac{2}{q}\right)=1$, i.e. by Quadratic Reciprocity $q\equiv \pm 1\pmod{8}$. However, also $q=2p+1\equiv 2\cdot 1+1\equiv 3\pmod{4}$, so $q\equiv -1\pmod{8}$, so $-1\equiv 2p+1\pmod{8}$, i.e. $p\equiv 3\pmod{4}$, which contradicts $p\equiv 1\pmod{4}$.
$2)\ \ $ $p\equiv 3\pmod{4}$. Then $\left((-1)^{\frac{p-1}{2}}2\right)^p\equiv -2^p\pmod{q}$, so $-2^p\equiv 1\pmod{q}$. By Euler's criterion $2^p\equiv 2^{\frac{q-1}{2}}\equiv \left(\frac{2}{q}\right)\pmod{q}$, so we must have $\left(\frac{2}{q}\right)=-1$, i.e. by Quadratic Reciprocity $q\equiv \pm 3\pmod{8}$. However, also $q=2p+1\equiv 2\cdot 3+1\equiv 3\pmod{4}$, so $q\equiv 3\pmod{8}$, so $3\equiv 2p+1\pmod{8}$, i.e. $p\equiv 1\pmod{4}$, which contradicts $p\equiv 3\pmod{4}$.
|
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|
Show that $\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$ for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$. Show that $$\left| \sqrt2-\frac{h}{k} \right| \geq \frac{1}{4k^2},$$
for any $k \in \mathbb{N}$ and $h \in \mathbb{Z}$.
I tried many different ways to expand left side and estimate it but always got stuck at some point.
|
I will show that
$|\sqrt{n}-\frac{x}{y}|
>\frac1{(\sqrt{n}+\sqrt{n+1})y^2}
$.
For $n=2$,
$\sqrt{2}+\sqrt{3}
< 3.15
$,
so
$|\sqrt{2}-\frac{x}{y}|
>\frac1{3.15 y^2}
$.
I also show that if
$|\sqrt{n}-\frac{x}{y}|
<\frac1{(2\sqrt{n}+\epsilon)y^2}
$.
then
$y
<\sqrt{ \frac1{2\epsilon\sqrt{n}}}
$.
In general,
if $n$ is not a perfect square,
then
$|x^2-ny^2| \ge 1
$
for integers $x$ and $y$.
Therefore,
if $z = \sqrt{n}$,
$\begin{align*}
1
&\le |x^2-ny^2|\\
&=|x-zy||x+zy|\\
\text{so}\\
|x-zy|
&\ge \frac1{|x+zy|}\\
&= \frac1{x+zy}\\
\text{or}\\
|z-\frac{x}{y}|
&\ge \frac1{y}\frac1{x+zy}\\
&\ge \frac1{y^2}\frac1{z+x/y}\\
\end{align*}
$
If
$|z-\frac{x}{y}|
< \frac1{cy^2}
$,
then
$\frac{x}{y}
< z+\frac1{cy^2}
$
and
$\frac1{z+x/y}
\le y^2|z-\frac{x}{y}|
<\frac1{c}
$
so that
$c
<z+x/y
<z+z+\frac1{cy^2}
=2z+\frac1{cy^2}
$.
Therefore
$c^2-2zc
< \frac1{y^2}
\le 1
$
or
$c^2-2zc+z^2
< z^2+1
$
or
$(c-z)^2
< n+1
$
or
$c < z+\sqrt{n+1}
=\sqrt{n}+\sqrt{n+1}
$.
Therefore
$|\sqrt{n}-\frac{x}{y}|
>\frac1{(\sqrt{n}+\sqrt{n+1})y^2}
$.
For $n=2$,
$c <
\sqrt{2}+\sqrt{3}
< 3.15
$.
(added later)
Note that,
since
$c^2-2zc
< \frac1{y^2}
$,
$c < z+\sqrt{z^2+\frac1{y^2}}
=z+z\sqrt{1+\frac1{z^2y^2}}
<z+z(1+\frac1{2z^2y^2})
=2z+\frac1{2zy^2})
$.
Therefore,
if $c > 2z$,
$\frac1{2zy^2}
> c-2z
$
or
$2zy^2
< \frac1{c-2z}
$
or
$y
<\sqrt{ \frac1{2z(c-2z)}}
$.
|
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|
Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plane $x+y+z=1$
Find the highest and lowest points on the ellipse of intersection of the cylinder $x^2+y^2 = 1$ and the plain $x+y+z=1$
Hi i was doing this question but i'm not sure i was right. Does this make sense?
Let
$$f(x) = x^2+y^2 = 1$$
$$f(y) = x+y+z=1, \quad \text{our constraint}$$
$$ \nabla f(x) = \lambda \nabla g(x) $$
$f_x = 2x = \lambda$ ..... (1)
$f_y = 2y = \lambda$ ...... (2)
$f_z = 0 = \lambda$
(1) = (2)
$2x = 2y = \lambda = 0 $
$x=y$
Sub into f(x)
$$f(x) = x^2 + x^2 = 1$$ $$f(x) = 2x^2 = 1$$ $$f(x) = x = \pm \frac{1}{\sqrt{2}} $$ $$ \therefore \, y = \pm \frac{1}{\sqrt{2}}$$
Points: $\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}
\right), \quad \left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}} \right)$
Sub these points into g
$$g\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, z \right) = \frac{2}{\sqrt{2}} + z = 10 $$
$$ z = 1 - \frac{2}{\sqrt{2}} $$
$$g\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, z \right) = -\frac{2}{\sqrt{2}} + z = 10 $$
$$ z = 1 + \frac{2}{\sqrt{2}} $$
So highest and lowest points are....
$$\left(\frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}}, 1 - \frac{2}{\sqrt{2}} \right) \quad lowest$$
$$\left(-\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 1 + \frac{2}{\sqrt{2}} \right) \quad highest$$
|
No, this is not correct. It looks like you assumed that the first function described in the problem was the objective function (that which is to be optimized) and the second function is the constraint function.
But if you read the problem, you see that both $x^2 + y^2 = 1$ and $x+y+z=1$ are constraints. So you need a Lagrange multiplier approach with two multipliers.
But wait! If these are both constraints, what is the objective function? Read it again. What words in the problem say “this is an optimization problem.” Optimal means superlative, and the superlative words are highest and lowest.
What function would have its maximum value achieved at the highest point of a set? The height function $f(x,y,z) = z$.
Here is a fleshed-out solution. The Lagrange equation is
$$
\nabla (z) = \lambda \nabla (x^2 + y^2) + \mu \nabla (x+y+z)
$$
In coordinates,
\begin{align}
0 &= \lambda(2x) + \mu \tag{1}\\
0 &= \lambda(2y) + \mu \tag{2}\\
1 &= \mu \tag{3}
\end{align}
Then we have the equations of constraint
\begin{align}
x^2 + y^2 = 1 \tag{4} \\
x + y + z = 1 \tag{5}
\end{align}
Substituting (3) into (1) and (2) gives
\begin{align*}
x =y= -\frac{1}{2\lambda}
\end{align*}
Then by (4),
$$
\left(-\frac{1}{2\lambda}\right)^2 + \left(-\frac{1}{2\lambda}\right)^2 = 1
\implies \lambda^2 = \frac{1}{2} \implies \lambda = \pm \frac{1}{\sqrt{2}}
$$
If $\lambda = \frac{1}{\sqrt{2}}$, then $x = y = -\frac{1}{\sqrt{2}}$ and by (5) $z=1+\sqrt{2}$. If $\lambda = - \frac{1}{2}$ then $x=y=\frac{1}{\sqrt{2}}$ and $z=1-\sqrt{2}$. So the highest point is $\left(-\frac{\sqrt{2}}{2},-\frac{\sqrt{2}}{2},1+\sqrt{2}\right)$ and the lowest point is $\left(\frac{\sqrt{2}}{2},\frac{\sqrt{2}}{2},1-\sqrt{2}\right)$
|
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"timestamp": "2023-03-29T00:00:00",
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|
Equation of a circle in matrix form I have an equation $ \left( x-3 \right)^{2}+\left( y-3 \right)^{2}=9 $, and am trying to apply a matrix rotation of 180 degrees to it, however, I am having difficulty transferring the equation of the circle into matrix form so to complete the transformation.
Thanks
|
The equation $(x-3)^2+(y-3)^2=9$ is can be described using matrix and vector $[x \ y \ z]^{\text{T}}$ as follows. Then $z=1$.
$$
\begin{bmatrix}
x & y & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & -3 & \\
0 & 1 & -3 & \\
-3 & -3 & 9 &
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}
=0
$$
The rotation matrix is
$$
\begin{bmatrix}
\cos{\theta} & - \sin{\theta} & 0 & \\
\sin{\theta} & \cos{\theta} & 0 & \\
0 & 0 & 1 &
\end{bmatrix}
$$
$[x' \ y' \ z']^{\text{T}}$ is the transformed vector by the rotation matrix. When you want to translate 180 degree as $\theta = \pi$, the equation is decided below.
$$
\begin{bmatrix}
x' & y' & 1
\end{bmatrix}
\begin{bmatrix}
1 & 0 & -3 & \\
0 & 1 & -3 & \\
-3 & -3 & 9 &
\end{bmatrix}
\begin{bmatrix}
x' \\
y' \\
1
\end{bmatrix}
=0
$$
$$
\begin{bmatrix}
x & y & 1
\end{bmatrix}
\begin{bmatrix}
-1 & 0 & 0 & \\
0 & -1 & 0& \\
0 & 0 & 1 &
\end{bmatrix}
\begin{bmatrix}
1 & 0 & -3 & \\
0 & 1 & -3 & \\
-3 & -3 & 9 &
\end{bmatrix}
\begin{bmatrix}
-1 & 0 & 0 & \\
0 & -1 & 0& \\
0 & 0 & 1 &
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}
=0
$$
$$
\begin{bmatrix}
x & y & 1
\end{bmatrix}
\begin{bmatrix}
\ 1 & \ 0 & \ 3 & \\
\ 0 & \ 1 & \ 3 & \\
\ 3 & \ 3 & \ 9 &
\end{bmatrix}
\begin{bmatrix}
x \\
y \\
1
\end{bmatrix}
=0
$$
Therefore, the equation can be expressed $(x+3)^2+(y+3)^2=9$ eventually.
|
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|
Calculate area by using double integral Area $D$ confined by lines:
$x=\sqrt{4-{y}^{2}},\: y=\sqrt{3x}, \: x\geq 0$
Need to calculate $I = \iint\limits_D \, dx\,dy$
My steps:
*
*Draw $D$:
*Set boundaries:
$I = {I}_{1} + {I}_{2}$
${I}_{2} = \frac{1}{4}\left(\pi {r}^{2} \right)=\pi$
${I}_{1} = \int_{0}^{\sqrt{3}} dy \int_{\frac{1}{3}{y}^{2}}^{\sqrt{4-{y}^{2}}}dx = \int_{0}^{\sqrt{3}}\sqrt{4-{y}^{2}}-\frac{1}{3}{y}^{2} \: dy = \left[2 \arcsin\frac{y}{2} + \frac{y}{2}\sqrt{4-{y}^{2}} \right]_{0}^{\sqrt{3}} - \frac{1}{3}\left[\frac{{y}^{3}}{3} \right]_{0}^{\sqrt{3}} = \frac{\sqrt{3}}{6}+\frac{2\pi }{3}$
So $I = \frac{\sqrt{3}}{6}+\frac{2\pi }{3} + \pi = \frac{5\pi }{3} + \frac{\sqrt{3}}{6}$
I have check ${I}_{1}$ via WolframAlpha.
But answer should be $\frac{2\pi - \sqrt{3}}{6}$
|
You selected the wrong area. You want the bit bound by all 3 lines mentioned.
You can get that by doing $2\pi$ take away your answer. Or recalculate the easier integral.
|
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|
Why two Inequalities are true $A$ is a subset of real numbers.
Consider the set $A$ of all real
numbers $x$ such that $x^2 \le 2$.
This set is nonempty and bounded from above, for example by 2. Call the $\sup(A),\; y$. Then $y^2 = 2$, because the other possibilities
$y^2 \lt 2$ and $y^2 \gt 2$ both lead to a contradiction.
Assume that $y^2 \gt 2$, so there exists $z \gt 0$ such that $y^2 − 4z \gt 2$.
Then:
$$ (y − z)^2 = y^2 − 2zy + z^2 > y^2 − 4z > 2 \tag{1} $$
Therefore $y − z$ is also an upper bound of $A$ and $y$ cannot be the least upper bound.
If $y^2 \lt 2$, so there exists $c \gt 0$ such that $y^2 + 5c \lt 2$ and $c^2 \lt c$.
Then:
$$ (y + c)^2 = y^2 + 2cy + c^2 < y^2 + 5c < 2. \tag{2} $$
Therefore also $y + c \in A$ and $y$ cannot be the least upper bound.
I apologize for this basic question. The inequalities labelled 1 and 2 appear to be true to me, but I can't see why they are true logically. I don't see the "then" logical link.
|
In (1), the equality part is easy and the second inequality is just an assumption. So, you only need to prove that $y^2-2zy+z^2 > y^2-4z$ which is equivalent to $z^2-2zy+4z > 0$ and, since $z>0$, this is equivalent to $z-2y+4 > 0$, or, in other words, $z > 2y-4$. So, you only need to prove $z > 2y-4$. Now, since $y$ is upper-bounded by $2$, we have $y < 2$ which is equivalent to $2y - 4 < 0$ and since $z > 0$, we have $z > 2y-4$ as required.
Similarly for (2), inequality $y^2+2cy+c^2 < y^2+5c$ is equivalent to $c^2 + 2cy - 5c < 0$ which, since $c > 0$ is equivalent to $c + 2y - 5 < 0$. This holds because $c < 1$ (since $c^2 < c$ and $c > 0$) and $y < 2$ ($y$ is upper-bounded by $2$). Therefore, $c + 2y < 1 + 2 * 2 = 5$.
|
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|
Forgot my basic math... So a younger (college student) asked me for my help to solve a basic math question, and to my surprise, I've forgot some basic math rules, rendering me unable to answer the problem.
According to online math-generators the answer should'nt be what I'm getting.. And I am unable to follow the auto-generated steps... I hope this forum can be of help, as I actually really do like maths, (but obviously have been neglecting it)
The problem is a reduction problem, and is as follows:
$\left(\frac{1}{2}\right) \times\left(\frac{2}{2a}\right)+\frac{\left(\frac{2a}{6}\right)}{\left(\frac{6}{5}\right)}$
What I do is this:
$$\left(\frac{1}{2}\right) \times\left(\frac{2}{2a}\right)+\frac{\left(\frac{2a}{6}\right)}{\left(\frac{6}{5}\right)} $$
$$=\left(\frac{2}{6a}\right)+\frac{\left(\frac{2a}{6}\right)}{\left(\frac{6}{5}\right)} $$
$$=\left(\frac{1}{3}\right)+\frac{\left(\frac{1a}{3}\right)}{\left(\frac{6}{5}\right)} $$ (I fear this may be wrong?)
$$=\frac{\left(\frac{4}{4a}\right)}{\left(\frac{6}{5}\right)} $$
$$=\left(\frac{4\times5}{4\times6a}\right) $$
$$=\frac{20}{24a} $$
$$=\frac{5}{6a}$$
|
Small mistake from step 1 to step 2. Notice that:
$$ \left(\frac{1}{2}\right) \times \left(\frac{2}{2a}\right) = \left( \frac{1\times 2}{2\times 2a} \right) = \left(\frac{2}{4a} \right) = \left( \frac{1}{2a} \right).$$
You will then end up with
$$ \left( \frac{1}{2a} \right) + \frac{\left(\frac{2a}{6}\right)}{\left(\frac{6}{5}\right)}.$$
You can simplify $\frac{2a}{6}$ to $\frac{a}{3}$, leaving you with
$$ \left( \frac{1}{2a} \right) + \frac{\left(\frac{a}{3}\right)}{\left(\frac{6}{5}\right)}.$$
From here, simplify $ \frac{\left(\frac{a}{3}\right)}{\left(\frac{6}{5}\right)}$ using the fact that
$$ \frac{\left(\frac{A}{B}\right)}{\left(\frac{C}{D}\right)} = \left(\frac{A}{B}\right) \times \left(\frac{D}{C}\right).$$
Once you do that, find a common denominator to add the fractions.
|
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|
Applying Euler's substitution to evaluate the integral? $$\int \frac{dx}{\sqrt{x^2+a^2}} $$
let $$\sqrt{x^2+a^2}=x+t $$
$$a^2=2xt+t^2 $$
$$x=\frac{1}{2}\Big( \frac{a^2}{t}-t\Big) $$
$$dx=-\frac{1}{2}\Big(\frac{a^2}{t^2}+1\Big)dt$$
hence $$x+t=\frac{a^2+t^2}{2t} $$
this gives $$\int \frac{dx}{\sqrt{x^2+a^2}}=-\ln({\sqrt{x^2+a^2}-x}) $$
Which is wrong. What mistake I have made?
|
Hint
You are not wrong and you did a good job but, may be, you just forgot the integration constant at the end of the calculation.
Just rewriting your last expression
$$A=-\log\left({\sqrt{x^2+a^2}-x}\right)=\log\left(\frac 1 {\sqrt{x^2+a^2}-x}\right)$$ $$A=\log\left(\frac 1 {\sqrt{x^2+a^2}-x}\times\frac{\sqrt{x^2+a^2}+x}{\sqrt{x^2+a^2}+x}\right)$$ $$A=\log\left(\frac{\sqrt{x^2+a^2}+x}{x^2+a^2-x^2}\right)=\log\left(\sqrt{x^2+a^2}+x\right)-2\log(a)$$ The last term will be absorbed in the integration constant. So $$\int \frac{dx}{\sqrt{x^2+a^2}}=-\log\left({\sqrt{x^2+a^2}-x}\right)+K_1=\log\left(\sqrt{x^2+a^2}+x\right)+K_2$$ The two expressions are the same.
|
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|
Show that the sequence is null? Show that $\frac{14n + 26}{49n^3 +14}$ is a null sequence.
So we need to show that: $\forall \epsilon > 0, \exists X \in \mathbb{R}, \forall n > X,|\frac{14n + 26}{49n^3 +14}| < \epsilon $.
$|\frac{14n + 26}{49n^3 +14}|= \frac{14n + 26}{49n^3 +14} < \frac{14n+26}{n}$ (since $n> 49n^3 +14 \Rightarrow \frac{1}{n} > \frac{1}{49n^3 +14} $)
So, $\frac{14n+26}{n} < \frac{14n+n}{n} $ for n $\ge $ 26
Hence, n > $\frac{15}{\epsilon}$
Taking X:= Max {$\frac{15}{\epsilon}$,26} for n > X we have $|\frac{14n + 26}{49n^3 +14}| < \epsilon $.
Is this the right working? I'm unsure about when i was reducing it to get $\frac{15}{\epsilon}$?
|
You have made the bound too coarse.
For all $n \geq 1$ we have
$$
\frac{14n + 26}{49n^{2}+14} < \frac{14n + 26}{n^{2}} = \frac{14}{n} + \frac{26}{n^{2}};
$$
given any $\varepsilon > 0$, we have $\frac{14}{n} < \frac{\varepsilon}{2}$ if $n > \frac{28}{\varepsilon}$ and we have $\frac{26}{n^{2}} < \frac{\varepsilon}{2}$ if $n > \sqrt{\frac{52}{\varepsilon}}$; but then $n > \max \{\frac{28}{\varepsilon}, \sqrt{\frac{52}{\varepsilon}} \}$ only if
$$
\frac{14n + 26}{49n^{2} + 14} < \varepsilon.
$$
|
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|
A good way to solve this trigonometric equation $$\sin x+\cos x=\frac{1}{2}$$
What is the value of $\tan x$? I tried using
$$\sin2 x=\frac{2\tan x}{1+\tan^2x}$$ and $$\cos2 x=\frac{1-\tan^2x}{1+\tan^2x}$$ but we get a quadratic for $\tan\left(\frac{x}{2}\right)$ . So any better approach would be much appreciated. Thanks!
|
Given $$\sin x+\cos x=\frac{1}{2}$$
squaring both the sides, $$(\sin x+\cos x)^2=\left(\frac{1}{2}\right)^2$$
$$\sin^2 x+\cos^2 x+2\sin x\cos x=\frac{1}{4}$$
$$1+\sin 2x=\frac{1}{4}$$$$\sin 2x=-\frac{3}{4}$$
$$\frac{2\tan x}{1+\tan ^2x }=-\frac{3}{4}$$
$$3\tan^2 x+8\tan x+3=0$$
let $\tan x=t$ $$3t^2+8t+3=0$$
I hope you can take it from here.
|
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|
Show that if $a\neq b$ then $a^3+a\neq b^3+b$ Show that if $a\neq b$ then $a^3+a\neq b^3+b$
We assume that $a^3+a=b^3+b$ to show that $a=b$
$$\begin{align}
a^3+a=b^3+b &\iff a^3-b^3=b-a\\
&\iff(a-b)(a^2+ab+b^2)=b-a\\
&\iff a^2+ab+b^2=-1
\end{align}$$
Im stuck here !
|
If you have calculus, you can conclude this is impossible as follows, since $lim_{x, y \to \infty}x^2+xy+y^2=\infty$, there is a point where the minima is achieved. This point has vanishing partial derivatives, thus $\partial_x(x^2+xy+y^2)=2x+y=0$ and likewise $2y+x=0$, so $x=-2y=4x$ so $x=0$ and likewise $y=0$, so $x=y=0$, and we have a contridiction since $0^2+0*0+0^2=0$.
|
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|
Integral $\sqrt{1+\frac{1}{4x}}$ $$\mathbf\int\sqrt{1+\frac{1}{4x}} \, dx$$
This integral came up while doing an arc length problem and out of curiosity I typed it into my TI 89 and got this output
$$\frac{-\ln(|x|)+2\ln(\sqrt\frac{4x+1}{x}-2)-2x\sqrt\frac{4x+1}{x}}{8}\ $$
How would one get this answer by integrating manually?
|
Set $u=\cfrac{1}{4x}, du=\cfrac{-1}{4x^2}$. Substitute:
$$\int \sqrt{1+u}(-4x^2)du$$
Since $4x=\cfrac{1}{u}, x=\cfrac{1}{4u}$,
$$=\int \cfrac{-1}{4u^2}\sqrt{1+u}du$$
Integrate by parts: $s=\sqrt{1+u}, ds=\cfrac{1}{2\sqrt{1+u}}du, dt=\cfrac{1}{u^2}du, t=\cfrac{-1}{u}$,
$$=\cfrac{-1}{4}\bigg[\cfrac{-\sqrt{1+u}}{u} - \int\cfrac{-1}{u}\cfrac{1}{2\sqrt{1+u}}du\bigg]$$
Substitute again, $r=\sqrt{u+1}, dr=\cfrac{1}{2\sqrt{u+1}}du=\cfrac{1}{2r}du \rightarrow 2rdr=du$
$$=\cfrac{-1}{4}\bigg[\cfrac{-\sqrt{1+u}}{u} - \int\cfrac{-1}{u}\cfrac{1}{2r}2rdr\bigg]$$
$$=\cfrac{-1}{4}\bigg[\cfrac{-\sqrt{1+u}}{u} - \int\cfrac{1}{1-r^2}dr\bigg]$$
Use the identity $\int \cfrac{1}{1-x^2} dx = arctanh(x)+C$.
$$=\cfrac{-1}{4}\bigg[\cfrac{-\sqrt{1+u}}{u} - arctanh(r) + C\bigg]$$
Substitute back in: $u=\cfrac{1}{4x}, r=\sqrt{u+1}$,
$$=\cfrac{1}{4}\bigg[4x\sqrt{1+\cfrac{1}{4x}} + arctanh\bigg(\sqrt{\cfrac{1}{4x}+1}\bigg)\bigg]+C$$
|
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|
Upper Triangularising a Matrix Kay and Wilson Linear Algebra I have been reading Linear Algebra by Richard Kay and Robert Wilson and am specifically looking at pages 156-158.
I understand the book's proof on page 156 of the proposition 10.10:
"Let $V= \mathbb{C}^n $ be the $n$ dimensional vector space over $\mathbb{C}$ and suppose $f$ is a linear transformation from $V$ to $V$. Then there is a basis $\{v_1, v_2,..., v_n\}$ of $V$ such that , with respect to this basis, $f$ is upper triangular."
I undertand this to be another way of saying for any square matrix $A$ over $\mathbb{C}$ there is an invertible matrix $P$ over $\mathbb{C}$ such that $P^{-1}AP$ is upper triangular.
However when actually put into practice I don't understand what is going on.
The second example 10.13 it gives on page 157 is $A= \begin{pmatrix} 3 & 0 & -1 \\ -1 & 4 & -3 \\-1 & 0 & 5 \end{pmatrix}$
We see $\begin{pmatrix} 0 \\ 1 \\0 \end{pmatrix}$ is an eigenvector with eigenvalue $4$.
Then $A-4I = \begin{pmatrix} -1 & 0 & -1 \\ -1 & 0 & -3 \\-1 & 0 & 1 \end{pmatrix}$
So the image of $A-4I$ has basis formed by $f_1=$ $\begin{pmatrix} -1 \\ -1 \\-1 \end{pmatrix}$ and $f_2=$ $\begin{pmatrix} -1 \\ -3 \\1 \end{pmatrix}$.
We extend this to a basis of the whole space by adjoining $f_3=$ $\begin{pmatrix} 1 \\ 0 \\0 \end{pmatrix}$ and so we have base change matrix
$P=$ $\begin{pmatrix} -1 & 1 & 1 \\ -1 & -3 & 0 \\-1 & 1 & 0 \end{pmatrix}$
On calculating we get $P^{-1}AP$= $\begin{pmatrix} 3 & 1 & 1 \\ -1 & 5 & 0 \\0 & 0 & 4 \end{pmatrix}$
So we have the zeroes we need on the bottom row.
Now we look at
$B=$ $\begin{pmatrix} 3 & 1 \\ -1 & 5\end{pmatrix}$
We obviously need to deal with this section of the $3$x$3$ matrix but I am now confused as to the way the book is going about it-
To continue B has eigenvalue $4$ and eigenvector $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
$B-4I =$ $\begin{pmatrix} -1 & 1 \\ -1 & 1\end{pmatrix}$ so a basis for the image is $\begin{pmatrix} 1 \\ 1 \end{pmatrix}$
We extend this to a basis of $\mathbb{R^2}$ by adding $\begin{pmatrix} 1 \\ 0 \end{pmatrix}$
Why have we done this for this $2$x$2$ matrix? This section of the $3$x$3$ was the problem but why have we treated this like it's $\mathbb{R^2}$ and also separately from the bigger $3$x$3$?
The book then says "Going back to $\mathbb{R^3}$ what we done is replace $f_1 ,f_2, f_3$ with $f_1+f_2, f_1, f_3$. I can't see how. The section from the submatrix B to here has lost me. But if I can understand this bit the rest of stuff which is continued on from here is nice and simple:
The base change matrix from $f_1 ,f_2, f_3$ to $f_1+f_2, f_1, f_3$ is $Q=$ $\begin{pmatrix} 1 & 1 & 0 \\ 1 & 0 & 0 \\0 & 0 & 1 \end{pmatrix}$ which is clear and $Q^{-1}P^{-1}APQ$ on calculating is
$\begin{pmatrix} 4 & -1 & 0 \\ 0 & 4 & 1 \\0 & 0 & 4 \end{pmatrix}$ which is upper triangular.
|
Let's see why the procedure work. Let's denote the usual standard basis of $\mathbb{R}^3$ by $\{e_1,e_2,e_3\}$. Then we have
\begin{align}
(A-4I)e_1&= f_1\\ (A-4I)e_3 &= f_2
\end{align}
So, $A(f_1) \in Range(A-4I) = Span\{ f_1, f_2\} $ and similarly $Af_2\in Range(A-4I) = Span\{ f_1, f_2\} $.
Now extend $Span\{ f_1, f_2\}$ to get a basis for $\mathbb{R}^3$ say $\{f_1,f_2,f_3\}$; Because of the Property mentioned above i.e. $Af_1, Af_2 \in Span \{f_1,f_2\}$, we will get, w.r.t this basis matrix of $A$ will be of the following form
\begin{align}
\begin{pmatrix}
b_{1,1} & b_{1,2} & c_1 \\
b_{2,1} & b_{2,2} & c_2\\
0 & 0 & d
\end{pmatrix}=\begin{pmatrix}
B & C \\
0 & d
\end{pmatrix}
\end{align}
In other word we have found a invertible $3 \times 3$ matrix $P$ (where $P$ is the matrix of change of basis in $\mathbb{R}^3$) so that
\begin{align}
P^{-1}AP &= \begin{pmatrix}
B & C \\
0 & d
\end{pmatrix}
\end{align}
Now we will do the same procedure for $B$. Now we will get a $2\times 2 $ invertible matrix $Q_1$ so that
\begin{align}
Q_1^{-1}BQ_1 &= \begin{pmatrix}
r & s \\
0 & t
\end{pmatrix}
\end{align}
Now define a $3\times 3$ invertible matrix $Q$ in following manner
\begin{align}
Q = \begin{pmatrix}
Q_1 & 0\\
0 & 1
\end{pmatrix}
\end{align} (Note that $Q$ become invertible).
So now we have
\begin{align}
Q^{-1}P^{-1}APQ &= \begin{pmatrix}
Q_1^{-1} & 0\\
0 & 1
\end{pmatrix} \begin{pmatrix}
B & C \\
0 & d
\end{pmatrix}\begin{pmatrix}
Q_1 & 0\\
0 & 1
\end{pmatrix}\\
&= \begin{pmatrix}
Q_1^{-1}BQ_1 & Q_1^{-1}C \\
0 & d
\end{pmatrix}\\
&= \begin{pmatrix}
r & s & *\\
0 & t & *\\
0 & 0 & d
\end{pmatrix}
\end{align}
Hence you have your required Upper triangular form.
In your case you already wrote $P$. Note that, in this case we have
\begin{align}
B &= \begin{pmatrix}
3 & -1\\
1 & 5
\end{pmatrix}
\end{align} So repeating the previous procedure for $B$ we will get
\begin{align}
Q_1 &= \begin{pmatrix}
1 & 1\\
1 & 0
\end{pmatrix}
\end{align} so that
\begin{align}
Q_1^{-1}B Q_1 &= \begin{pmatrix}
4 & -1\\
0 & 4
\end{pmatrix}
\end{align}
And hence we get whatever we wanted.
|
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|
Can someone explain the integration of $\sqrt{v²+\tfrac14}$ to me? I am currently trying to integrate this root:
$$\sqrt{v^2+\frac{1}{4}}$$
According to several integration calculators on the web it is:
$$\frac{\operatorname{arsinh}(2v)}{8} +\frac{v\sqrt{v^2+\tfrac{1}{4}}}{2}$$
However, I just can't get my head around it. I have absolutely no idea where that $\operatorname{arsinh}$ is coming from.
My take on this would have been:
$$f = \left(v^2+\frac{1}{4}\right)^{\frac{1}{2}}\implies F = \frac{2}{3}\cdot\frac{1}{2v}\left(v^2+\frac{1}{4}\right)^{\frac{3}{2}}$$
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Since $\sqrt{v^2+\frac{1}{4}}$ is of the form $\sqrt{x^2+a^2}$, we can do a trigonometric substitution using $\tan\theta$.
Let $v=\frac{1}{2}\tan u$, $dv=\frac{1}{2}\sec^2(u) du$
$$\int \sqrt {v^2+\frac{1}{4}}dv=\int \sqrt {\frac{1}{4}\tan^2 u+\frac{1}{4}}\cdot\frac{1}{2}\sec ^2 (u)du$$
Using the trig identity $1+\tan^2\theta=\sec ^2 \theta$ we have
$$=\frac{1}{4}\int \sec ^3 (u)du$$
Now we can use integration by parts.
$$ds=\sec ^2(u)du$$
$$s=\tan u$$
$$t=\sec u$$
$$dt=\sec (u) \tan (u) du$$
Then
$$\int \sec ^3 (u)du=\sec u \tan u-\int \sec (u) \tan^ 2(u)du$$
Using the same identity we can substitute again.
$$=\sec u \tan u - \int \sec u (\sec ^2 u -1)du$$
$$\int \sec^3(u)du=\sec u \tan u -\int \sec ^3 (u)du + \int sec (u) du$$
But $\int \sec ^3 (u)du$ is what we are trying to find.
$$2\int\sec ^3 (u)du = \sec u \tan u+\int \sec (u) du$$
$$=\sec u \tan u +\ln|\sec u + \tan u |$$
$$\frac{1}{4}\int \sec ^3 (u)du=\frac{1}{8}\sec u \tan u +\frac{1}{8}\ln |\sec u + \tan u|$$
Now we are almost ready to switch back to $v$. We defined $\tan u = 2v$ earlier. If we draw a representative right triangle we can figure out $\sec u$ as well.
From the triangle it is clear that $\sec u =\sqrt {4v^2+1}$.
Therefore
$$\int\sqrt{v^2+\frac{1}{4}}dv=\frac{1}{4}v\sqrt{4v^2+1}+\frac{1}{8}\ln |2v+\sqrt{4v^2+1}| + C$$
which you can switch to the hyperbolic if you like.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1519383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
}
|
Geometry question on square $ABCD$ is a square and $AB$ = 1. Equilateral triangles $AYB$ and $CXD$ are drawn such that $X$ and $Y$ are inside the square.How can I find the length of $XY$ ?
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You can use also this approach if you like,also if, as already said, you can really solve it by simmetry arguments.
Area of $\Delta DPR = \cfrac {DP ^2 \cdot \sin P \cdot \sin D}{2 \sin R}=\cfrac {1}{4} \cdot \cfrac{ \sin 90 \cdot \sin 30 }{2 \sin60} =\cfrac { \sqrt {3}}{24}$.
( ----> Area of any triangle $\Delta ABC$ given side $a$ and two angles is : $\cfrac{a^2 \sin B \sin C}{2 \sin A}$)
So we have that the area of $DROY' =\cfrac {1}{4} - \cfrac { \sqrt {3}}{24}$.
This is the equivalent to say that $DROY'=\cfrac {(DY'+RO) \cdot OY'}{2}= \cfrac {1}{4} - \cfrac { \sqrt {3} }{24}$, now since $DY=OY'=\cfrac {1}{2}$ , we find that $RO= \cfrac {3- \sqrt{3}}{6} $.
Now note that $OX=\sqrt{3} \cdot OR$ (30-60-90 triangle properties) ,so we find that $OX=\cfrac {3- \sqrt{3}}{6} \cdot \sqrt{3} =\cfrac {\sqrt {3}}{2} -\cfrac {1}{2} $.
Now since $XY=2OX$ we find $XY=\sqrt{3} -1$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1523689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
}
|
Calculate the limit $\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$ $$\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)}$$
My attempt
\begin{align}
\lim_{n\to\infty} \left(\frac{n^2}{n^2 + 1} \right)^{(n-1)/(n+1)} &= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-n+1)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{1 + (-2n)/(n+1)} \\
&= \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}
\end{align}
since $\left(\frac{n^2+1}{n^2} \right) \to 1$ when $n\to\infty$
$$\lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)\left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)} = \lim_{n\to\infty} \left(\frac{n^2+1}{n^2} \right)^{(-2n)/(n+1)}$$
Also I tried to make similar simplifications into the braсkets but nothing happens and no proof that limit $= 1$ or whatever.
And here is a rule for the task. This is a limit of a sequence so no usage of functional simplifications and derives are allowed. If you have really beautiful solution for the task then post it anyway.
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Hints:
*
*For $0 \lt k \lt 1$ and $0 \lt x \lt 1$ you have $x \lt x^k \lt 1$
*$ \displaystyle \lim_{n\to\infty} \bigg(\dfrac{n^2}{n^2 + 1} \bigg) = \lim_{n\to\infty} \bigg(1-\dfrac{1}{n^2 + 1} \bigg) $
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1529304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
}
|
Frobenius Method Indicial Equation I need to verify that the indicial equation only has one root.
$xy''+(1-x)y'+\frac{1}{2}y=0$
Attempt:
$y=\sum\limits_{m=0}^\infty {a_mx}^{m+r}$
$y'=\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r-1}$
$y'=\sum\limits_{m=0}^\infty {(m+r-1)}{(m+r)}{a_mx}^{m+r-2}$
Substitute:
$x\sum\limits_{m=0}^\infty {(m+r-1)}{(m+r)}{a_mx}^{m+r-2}+(1+x)\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r-1}+\frac{1}{2}\sum\limits_{m=0}^\infty {a_mx}^{m+r}=0$
Reduce:
$\sum\limits_{m=0}^\infty {(m+r-1)}{(m+r)}{a_mx}^{m+r-1}+\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r-1} -\sum\limits_{m=0}^\infty {(m+r)}{a_mx}^{m+r}+\frac{1}{2}\sum\limits_{m=0}^\infty {a_mx}^{m+r}=0$
$\sum\limits_{m=0}^\infty[(m+r-1)(m+r)+(m+r)]a_{m}x^{m+r-1}-\frac{1}{2}\sum\limits_{m=0}^\infty {a_mx}^{m+r}=0$
Shift Index:
$\sum\limits_{m=0}^\infty[(m+r-1)(m+r)+(m+r)]a_{m}x^{m+r-1}-\frac{1}{2}\sum\limits_{m=1}^\infty {a_{m-1}x}^{m+r-1}=0$
$[r^2]a_0x^{r-1}+\sum\limits_{m=1}^\infty([(m+r-1)(m+r)+(m+r)]a_m-\frac{1}{2}a_{m-1})x^{m+r-1}=0$
Therefore r is the double root of $0$, but I am not sure if this is correct.How would I go about verifying there is only one root?
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The indicial must be at $x=0$ in this case. Dividing by the coefficient of the highest order derivative gives
$$
y'' + \left(\frac{1}{x}+1\right)y'+\frac{1}{2x}y = 0.
$$
The indicial equation is found by keeping the $\frac{1}{x}y'$ terms and $\frac{1}{x^{2}}y$ terms:
$$
y''+\frac{1}{x}y' = 0.
$$
Set $y=x^{p}$. The conditions to have such solutions is that $p$ satisfies the indicial equation
$$
p(p-1)+p = 0, \\
p^{2}=0.
$$
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1530820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
}
|
Trigonometry equation, odd-function. So I have the following equation:
$\sin\left(x-\frac{\pi}{6}\right) + \cos\left(x+\frac{\pi}{4}\right)=0$
It should be solved using the fact that Sin is an odd function, I can not really get the gripp of how and what I need to do? Any sugestions?
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So i found the solution using the fact that Sin is odd, here is my solution:
$\sin\left (x-\frac{\pi}{6}\right ) + \cos\left (x+\frac{\pi}{4}\right ) = 0\Leftrightarrow \cos\left (x+\frac{\pi}{4}\right ) = -\sin\left (x-\frac{\pi}{6}\right )$
Since sin is an odd function, $-\sin\left (x-\frac{\pi}{6}\right ) = \sin\left (x-\frac{\pi}{6} + \pi\right )$
This the gives the following:
$\cos\left (x+\frac{\pi}{4}\right ) = \sin\left (x-\frac{\pi}{6} + \pi\right ))\Leftrightarrow \cos\left (x+\frac{\pi}{4}\right ) = \cos\left (\frac{\pi}{2} -\frac{5\pi}{6} -x\right )$
This happends when:
$x + \frac{\pi}{4} = \frac{-2\pi}{6}-x+n2\pi \Leftrightarrow x=\frac{-7\pi}{24}+n\pi$
Thanks alot for the help, special thanx to @BLAZE, you gave me the hints to find the solution.
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{
"language": "en",
"url": "https://math.stackexchange.com/questions/1531671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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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.