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Suppose $n$ is prime and $x \in Z$ satisfies $x^2 \equiv 1 \mod n.$ Prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$. Suppose $n$ is prime and $x \in Z$ satisfies $x^2 \equiv 1 \mod n.$ Prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$.
So far I have done the following proof, but I am unsure how to complete it:
We will begin by explaining why $x^2 \equiv 1 \mod n$ iff $n|(x^2 - 1)$. Then use this to prove that $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$.
If $x^2 \equiv 1 \mbox{ mod } n$, then $x^2 = nk +1 $, for some $k \in Z$. Therefore $nk = x^2 -1$, and $n|(x^2 -1)$.
On the other hand, if $n|(x^2 -1)$, then $nk = x^2 -1$ for some $k \in Z$, and then we can say $x^2 = nk +1$, and thus $x^2 \equiv 1 \mod n$.
From here, however I am not sure how to demonstrate that this shows $x \equiv 1 \mod n$ or $x \equiv -1 \mod n$.
| You can factor $(x^2-1)=(x+1)(x-1)$. Since $n$ is prime, $n\mid (x^2-1)$ implies $n\mid x+1$ or $n\mid x-1$. So $x\equiv 1$ mod $n$ or $x\equiv -1$ mod $n$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove $\sum_p p^{-2} < \frac{1}{2}$? I am trying to prove $\sum_p p^{-2} < \frac{1}{2}$, where $p$ ranges over all primes. I think this should be doable by elementary methods but a proof evades me.
Questions already asked here (eg. What is the value of $\sum_{p\le x} 1/p^2$? and Rate of convergence of series of squared prime reciprocals) deal with the exact value of the above sum, and so require some non-elementary math.
| If you know $\displaystyle \sum_{n=1}^\infty \frac{1}{n^2} = \frac{\pi^2}{6}$ then you could simply say $$ \displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} $$ $$\lt \frac{\pi^2}{6} - \frac{1}{1^2}- \frac{1}{4^2}- \frac{1}{6^2}- \frac{1}{8^2}- \frac{1}{9^2}- \frac{1}{10^2}- \frac{1}{12^2}- \frac{1}{14^2}- \frac{1}{15^2}- \frac{1}{16^2} $$ $$ \approx 0.49629 $$ $$ \lt \frac12.$$
Alternatively if you do not know that, instead use $\displaystyle \frac{1}{k^2} \le \int_{x=k-1}^k \frac{1}{x^2}\, dx = \frac{1}{k-1} - \frac{1}{k}$ so $\displaystyle \sum_{n=k}^\infty \frac{1}{n^2} \le \int_{x=k-1}^\infty \frac{1}{x^2}\, dx = \frac{1}{k-1}$ and you can say: $$ \displaystyle \sum_{p \text{ prime}} \frac{1}{p^2} \lt \frac{1}{2^2}+ \frac{1}{3^2}+ \frac{1}{5^2}+ \frac{1}{7^2}+ \frac{1}{11^2}+ \frac{1}{13^2}+ \frac{1}{17-1} \approx 0.4982 \lt \frac12.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$\sin(4x) = -2\sin(2x)$ solutions in $[0,2\pi)$? My textbook gives the following answer:
$\sin(4x) + 2\sin(2x) = 0$
$2\sin(2x)\cos(2x) + 2\sin(2x) = 0$
$2\sin2x (\cos(2x) + 1) = 0$
$2\sin 2x = 0$
$\sin2x = 0$
$2x = πk$
$x = kπ/2$
In the interval $[0,2π)$ you have the solutions $0,π/2,π$ and $3π/2$. The book then shows the other solutions from $(\cos(2x) + 1) = 0.$
Here's my question: why is it $2x = πk$ and not $2x = 2πk$?
I'm just in high school so I probably won't understand if there's a really complicated explanation.
| Use the following identities
$$\sin(2 x) \equiv 2 \sin(x) \cos(x) $$
$$\cos(2 x) \equiv 2 \cos^2(x)-1 $$
$$\sin(4 x) \equiv \sin(2 x) \cos(2 x) = 8 \sin(x) \cos^3(x) - 4 \sin(x)\cos(x) $$
$$\sin(4 x) +2 \sin(2 x) = 8 \sin(x) \cos^3(x) - 4 \sin(x)\cos(x) + 4 \sin(x) \cos(x) $$
which simplifies to
$$ 8 \sin(x) \cos^3(x) =0 $$
with solution $x = k \frac{\pi}{2}$ since either $\sin(x)$ has to be zero or $\cos(x)$ has to be zero.
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve the following linear congruences For any integers $a, b$, let $N_{a,b}$ denote the number of
positive integers $x<1000$ satisfying $x\equiv a\pmod{27}$ and $x\equiv b\pmod{37}$. Then find $N_{a,b}$.
progress: $x\equiv a\pmod{27}$ and $x\equiv b\pmod{37}$ implies
$x=27m+a$ and $x=37n+b$ for some integers $m,n$. For common sosultion, $27m+a=37n+b\implies 27m-37n=b-a$. Now $\gcd(27,37)=1$ sow there exists integers $u,v$ such that $27u+37v=1$.
Is it possible to solve using this relation ?
Thank u in advance
| The numbers $u$ and $v$ can be found with the extended euclidean algorithm
$37=27\times 1+10$
$27=2\times 10+7$
$10=7\times 1+3$
$7=3\times 2+1$
Giving $1=7-3\times 2=7-(10-7)\times 2=3\times 7-2\times 10=3(27-2\times 10)-2\times 10=3\times 27-8\times 10=3\times 27-8\times (37-27)=11\times 27-8\times 37$
This gives the inverses : $27^{-1}=11 \pmod{37}$ and $37^{-1}=-8=19 \pmod {27}$
Now, the number $u=19\times 37\times a+11\times 27\times b$
satisfies $u\equiv a\pmod {27}$ and $u\equiv b\pmod{37}$. Taking $u$ modulo $27\times 37=999$ gives the smallest solution.
| {
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Comparing $n^n$ and $n!^2$ I tried to prove that if $n>2$ then $(n!)^2>n^n$ but did not managed. That is the trick to compare those as both grows rapidly? Induction seems hard: $((n+1)!)^2=(n+1)^2(n!)^2>(n+1)^2n^n$ but why $(n+1)^2n^n>(n+1)^{n+1}$? I also noted that $n^n=e^{n\ln n}$ and $n!^2=e^{2\ln n!}=e^{2\sum_{i=1}^n \ln i}$ but got stuck.
| Suppose
$(n!)^2 > n^n$.
Then
$((n+1)!)^2
=(n!)^2(n+1)^2
>n^n(n+1)^2
$
so we need
$n^n(n+1)^2
\ge (n+1)^{n+1}
$
which is the same as
$n^n
\ge (n+1)^{n-1}
$
or,
dividing by
$n^{n-1}$,
$n
\ge (1+1/n)^{n-1}
$.
Multiplying by
$1+1/n$,
this becomes
$n+1
\ge (1+1/n)^n
$.
This is true because
$(1+1/n)^n
< e
$,
as has been shown many times here.
Here is one of the easier:
Using the binomial theorem,
$\begin{array}\\
(1+1/n)^n
&=\sum_{k=0}^n \binom{n}{k}(1/n)^k\\
&=\sum_{k=0}^n \frac{n!}{k!(n-k)!}(1/n)^k\\
&=\sum_{k=0}^n \frac{\prod_{j=0}^{k-1}(n-j)}{k!n^k}\\
&=\sum_{k=0}^n \frac{\prod_{j=0}^{k-1}(1-j/k)}{k!}\\
&<\sum_{k=0}^n \frac{1}{k!}\\
&<\sum_{k=0}^{\infty} \frac{1}{k!}\\
&= e
\end{array}
$
Actually,
all we need is
a bound on
$\sum_{k=0}^{\infty} \frac{1}{k!}
$.
An easy one
is gotten from
$k! \ge 2\cdot 2^{k-2}$
for $k \ge 2$,
easily proved by induction.
Then
$\begin{array}\\
\sum_{k=0}^{\infty} \frac{1}{k!}
&=1+1+\sum_{k=2}^{\infty} \frac{1}{k!}\\
&\lt 2+\sum_{k=2}^{\infty} \frac{1}{2\cdot 2^{k-2}}\\
&\lt 2+\frac1{2}\sum_{k=0}^{\infty} \frac{1}{2^k}\\
&= 2+1\\
&= 3
\end{array}
$
| {
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"timestamp": "2023-03-29T00:00:00",
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Zero divided by zero must be equal to zero What is wrong with the following argument (if you don't involve ring theory)?
Proposition 1: $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}$ is not equal to $0$
$\frac{0}{0}$ is not equal to $0 \Rightarrow \frac{0}{0} = x$ , some $x$ not equal to $0$ $\Rightarrow$ $2(\frac{0}{0}) = 2x$ $\Rightarrow$ $\frac{2\cdot 0}{0} = 2x$ $\Rightarrow$ $\frac{0}{0} = 2x$ $\Rightarrow$ $x = 2x$ $\Rightarrow$ $ x = 0$ $\Rightarrow$[because $x$ is not equal to $0$]$\Rightarrow$ contradiction
Therefore, it is not the case that $\frac{0}{0}$ is not equal to $0$
Therefore, $\frac{0}{0} = 0$.
Q.E.D.
Update (2015-12-01) after your answers:
Proposition 2: $\frac{0}{0}$ is not a real number
Proof [Update (2015-12-07): Part 1 of this argument is not valid, as pointed out in the comments below]:
Suppose that $\frac{0}{0}= x$, where $x$ is a real number.
Then, either $x = 0$ or $x$ is not equal to $0$.
1) Suppose $x = 0$, that is $\frac{0}{0} = 0$
Then, $1 = 0 + 1 = \frac{0}{0} + \frac{1}{1} = \frac{0 \cdot 1}{0 \cdot 1} + \frac{1 \cdot 0}{1 \cdot 0} = \frac{0 \cdot 1 + 1 \cdot 0}{0 \cdot 1} = \frac{0 + 0}{0} = \frac{0}{0} = 0 $
Contradiction
Therefore, it is not the case that $x = 0$.
2) Suppose that $x$ is not equal to $0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow$ contradiction
Therefore, it is not the case that $x$ is a real number that is not equal to $0$.
Therefore, $\frac{0}{0}$ is not a real number.
Q.E.D.
Update (2015-12-02)
If you accept the (almost) usual definition, that for all real numbers $a$, $b$ and $c$, we have $\frac{a}{b}=c$ iff $ a=cb $, then I think the following should be enough to exclude $\frac{0}{0}$ from the real numbers.
Proposition 3: $\frac{0}{0}$ is not a real number
Proof: Suppose that $\frac{0}{0} = x$, where $x$ is a real number.
$\frac{0}{0}=x \Leftrightarrow x \cdot 0 = 0 = (x + 1) \cdot 0 \Leftrightarrow \frac{0}{0}=x+1$
$ \therefore x = x + 1 \Leftrightarrow 0 = 1 \Leftrightarrow \bot$
Q.E.D.
Update (2015-12-07):
How about the following improvement of Proposition 1 (it should be combined with a new definition of division and fraction, accounting for the $\frac{0}{0}$-case)?
Proposition 4: Suppose $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, and that the rule $a \cdot \frac{b}{c} = \frac{a \cdot b}{c}$ holds for all real numbers $a$, $b$ and $c$.
Then, $\frac{0}{0} = 0$
Proof: Suppose that $\frac{0}{0}=x$, where $x \ne 0$.
$x = \frac{0}{0} \Rightarrow 2x = 2 \cdot \frac{0}{0} = \frac{2 \cdot 0}{0} = \frac{0}{0} = x \Rightarrow x = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
Suggested definition of division of real numbers:
If $b \ne 0$, then
$\frac{a}{b}=c$ iff $a=bc$
If $a=0$ and $b=0$, then
$\frac{a}{b}=0$
If $a \ne 0$ and $b=0$, then $\frac{a}{b}$ is undefined.
A somewhat more minimalistic version:
Proposition 5. If $\frac{0}{0}$ is defined, so that $\frac{0}{0} \in \mathbb{R}$, then $\frac{0}{0}=0$.
Proof: Suppose $\frac{0}{0} \in \mathbb{R}$ and that $\frac{0}{0}=a \ne 0$.
$a = \frac{0}{0} = \frac{2 \cdot 0}{0} = 2a \Rightarrow a = 0 \Rightarrow \bot$
$\therefore \frac{0}{0}=0$
Q.E.D.
| Try to cancel 0/0 with different numbers each time you will get different solutions for 0/0 . Suppose 1×0/1×0 you get solution 1. Suppose 2×0/0×1 you get solution 2 . Suppose n×0/0×1 you get solution n. ∵0/0 has infinitely many solutions 0/0 is not defined.
| {
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Find the area of the region enclosed by the inner loop of the curve $r = 4 + 8 \sin \theta$ The loop is generated for $\theta \in \left[\frac76\pi, \frac{11}{6}\pi\right]$
(this is from setting $r = 0$).
So, $$A = \int_{\frac76\pi}^{\frac{11}{6}\pi} \frac12(4 + 8 \sin \theta)^2 \,d\theta=\\
= \int_{\frac76\pi}^{\frac{11}{6}\pi} 8 + 32 \sin \theta + 32 \sin^2\theta\, d\theta =\\
= \int_{\frac76\pi}^{\frac{11}{6}\pi} 8 + 32 \sin \theta + 16(1 - \cos(2\theta))\, d\theta=\\
= \int_{7\pi/6}^{11\pi/6} 24 + 32 \sin \theta - 16 \cos(2\theta)\, d\theta =\\
= \bigg[24\theta - 32 \cos \theta - 8 \sin(2\theta)\bigg]_{\frac76\pi}^{\frac{11}{6}\pi} =\\
= (44\pi - 16\sqrt{3} + 4\sqrt{3}) - (28\pi + 16\sqrt{3} - 4\sqrt{3}) =\\
= 16\pi - 24\sqrt{3}$$
Is my answer right?
| Yes.
Another way of doing this would be:
$$\int_{7\pi/6}^{11\pi/6} \int_{0}^{4 + 8\sin \theta} rdrd\theta$$
$$ \ldots = 16\pi - 24\sqrt{3}$$
If you fancy multiple integrals.
| {
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"timestamp": "2023-03-29T00:00:00",
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why the limit of this f(x) when x approach infinity is equal to infinity? why the limit of this f(x) when x approach infinity is equal to infinity?:
$$ \lim_{x\to \infty} \frac{x^2 + 4x + 5}{x-1}$$
$$ \lim_{x\to \infty} \frac{1 + \frac{4}{x} + \frac{5}{x^2}}{\frac{1}{x}-\frac{1}{x^2}} = \infty $$
I know that all the fraction with X be denominator is equal to Zero but is that mean at the end the $ \lim_{x\to \infty} = \frac{1}{0}$ why it is equal to $ \infty $ ?
| $$\lim_{x\to\infty}\left(\frac{x^2+4x+5}{x-1}\right)=$$
The leading term in the denominator of $\frac{x^2+4x+5}{x-1}$ is $x$.
So divide the numerator and denominator by this:
$$\lim_{x\to\infty}\left(\frac{x+4+\frac{5}{x}}{1-\frac{1}{x}}\right)=$$
The expressions $\frac{5}{x}$ and $\frac{1}{x}$ both tend to zero as $x$ approaches $\infty$:
$$\lim_{x\to\infty}\left(\frac{x+4+0}{1-0}\right)=$$
$$\lim_{x\to\infty}\left(\frac{x+4}{1}\right)=$$
$$\lim_{x\to\infty}\left(x+4\right)=$$
$$\lim_{x\to\infty}x=\infty$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Help me evaluate this infinite sum I have the following problem:
For any positive integer n, let $\langle n \rangle$ denote the integer nearest to $\sqrt n$.
(a) Given a positive integer $k$, describe all positive integers $n$ such that $\langle n \rangle = k$.
(b) Show that $$\sum_{n=1}^\infty{\frac{2^{\langle n \rangle}+2^{-\langle n \rangle}}{2^n}}=3$$
My progress: The first one is rather easy. As $$\left( k-\frac{1}{2} \right) < \sqrt n < \left( k+\frac{1}{2} \right) \implies \left( k-\frac{1}{2} \right)^2 < n < \left( k+\frac{1}{2} \right)^2 \implies \left( k^2-k+1 \right) \leq n \leq \left( k^2+k \right)$$
Actually, there would be $2k$ such integers.
But, I have no idea how to approach the second problem. Please give me some hints.
| The idea is to rewrite the sum as a double sum by observing that $$\langle m^2 + k \rangle = m$$ for $k \in \{-m+1, \ldots, m\}$. Therefore, $$\begin{align*} S &= \sum_{n=1}^\infty \frac{2^{\langle n \rangle} + 2^{-\langle n \rangle}}{2^n} \\ &= \sum_{m=1}^\infty \sum_{k=-m+1}^{m} \frac{2^m + 2^{-m}}{2^{m^2+k}} \\ &= \sum_{m=1}^\infty \frac{2^m + 2^{-m}}{2^{m^2}} \sum_{k=-m+1}^m \frac{1}{2^k} \\ &= \sum_{m=1}^\infty \frac{2^m + 2^{-m}}{2^{m^2}} \left(2^m - 2^{-m}\right) \\ &= \sum_{m=1}^\infty \frac{2^{2m} - 2^{-2m}}{2^{m^2}} \\ &= \sum_{m=1}^\infty 2^{-m(m-2)} - 2^{-m(m+2)} \\ &= \sum_{m=1}^\infty 2^{-m(m-2)} - \sum_{k=3}^\infty 2^{-(k-2)k} \\ &= \sum_{m=1}^2 2^{-m(m-2)} = 2^1 + 2^0 = 3.\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $ c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ prove that it is Isosceles Triangle
In a $\triangle ABC\;,$ If $\displaystyle c(a+b)\cos \frac{B}{2}=b(a+c)\cos \frac{C}{2}\;,$ Then how can we prove that $\triangle ABC$
is an Isoceles $\triangle.$
$\bf{My\; Try::}$ Using $\displaystyle \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}=k\;,$ We get
$$k\sin C\left[k\sin A+k\sin B\right]\cdot \cos \frac{B}{2} = k\sin B\left[k\sin A+k\sin C\right]\cdot \cos \frac{C}{2} $$
So we get $$\sin C\left[\sin \left(\frac{A+B}{2}\right)\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\sin \left(\frac{A+C}{2}\right)\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
Now Using $A+B+C=\pi\;,$ We get $\displaystyle \frac{A+B}{2}=\frac{\pi}{2}-\frac{C}{2}$ and $\displaystyle \frac{A+C}{2}=\frac{\pi}{2}-\frac{B}{2}$
So we get $$\sin C\left[\cos \frac{C}{2}\cdot \cos \left(\frac{A-B}{2}\right)\right]\cos \frac{B}{2}=\sin B\left[\cos \frac{B}{2}\cdot \cos \left(\frac{A-C}{2}\right)\right]\cos \frac{C}{2}$$
So we get $$\sin C\cdot \cos \left(\frac{A-B}{2}\right)=\sin B\cdot \cos \left(\frac{A-C}{2}\right)$$
Now if we put $B=C\;,$ Then these two are equal.
My question is how can we prove it.
Help me, Thanks
| we have $$\frac{cos\left(\frac{B}{2}\right)}{cos\left(\frac{C}{2}\right)}=\frac{ab+bc}{ac+bc}$$
Applying Componendo and Dividendo we get
$$\frac{cos\left(\frac{B}{2}\right)-cos\left(\frac{C}{2}\right)}{cos\left(\frac{B}{2}\right)+cos\left(\frac{C}{2}\right)}=\frac{a(b-c)}{ab+ac+bc}$$ $\implies$
$$-tan\left(\frac{B+C}{2}\right)tan\left(\frac{B-C}{2}\right)=\frac{a(b-c)}{ab+ac+bc} \tag{1}$$
Now $$tan\left(\frac{B+C}{2}\right)=cot\left(\frac{A}{2}\right)$$
and by Napier's Rule $$tan\left(\frac{B-C}{2}\right)=\frac{b-c}{b+c}cot\left(\frac{A}{2}\right)$$
so $(1)$ becomes
$$\frac{-(b-c)cot^2\left(\frac{A}{2}\right)}{b+c}=\frac{a(b-c)}{ab+ac+bc} $$ $\implies$
$$(b-c) \times \left(\frac{a}{ab+bc+ac}+\frac{cot^2\left(\frac{A}{2}\right)}{b+c}\right)=0 $$ which is possible only if $$b=c$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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On the complete solution to $x^2+y^2=z^k$ for odd $k$? While trying to answer this question, I was looking at a computer output of solutions to $x^2+y^2 = z^k$ for odd $k$ and noticed certain patterns. For example, for $k=5$ we have $x,y,z$,
$$10, 55, 5\\25, 50, 5\\38, 41, 5\\117, 598, 13\\122, 597, 13\\338, 507, 13\\799, 884, 17$$
Question: Is it true that all integer solutions to $x^2+y^2=z^k$ for odd $k>1$ are given by just two formulas, namely,
*
*Primitives $\gcd(x,y)=1$:
$$A^2+B^2 = (a^2+b^2)^k\tag1$$
where $A,B$ is the expansion of $(a+bi)^k = A+Bi$. Example, $$(a+bi)^3 =(a^3 - 3 a b^2)+ (3 a^2 b - b^3)i$$ hence, $$(a^3 - 3 a b^2)^2+ (3 a^2 b - b^3)^2 = (a^2+b^2)^3$$ and so on for other $k$.
*Non-primitives $\gcd(x,y)\neq1$:
$$a^2(a^2+b^2)^{k-1}+b^2(a^2+b^2)^{k-1} = (a^2+b^2)^k\tag2$$
where, for both Forms $1$ and $2$, we use some rational $a,b$?
Example. The first three solutions for $k=5$ above use:
$$a,b = 2/5,\;11/5\quad \text{Form 2}$$
$$a,b = 1,\;2\quad \text{Form 2}$$
$$a,b = -2,\;1\quad \text{Form 1}$$
| Good news: the answer is Yes. Moreover, all integer solutions are given by just one formula - the second one. It covers both primitive and non-primitive solutions. That is, for every solution $(x,y,z)$ to $x^2+y^2=z^k$ there exist rational numbers $a,b$ such that $x=a(a^2+b^2)^{(k-1)/2}$, $y=b(a^2+b^2)^{(k-1)/2}$ and $z=a^2+b^2$.
Bad news: This formula is quite useless. The reason is that you use rational parameters to describe integer solutions, and the question for which rational parameters the corresponding values of $x,y,z$ are integers is as difficult as solving the equation.
Now the proof. Take $a=\frac{x}{z^{(k-1)/2}}$, $b=\frac{y}{z^{(k-1)/2}}$. Then $a^2+b^2=(x^2+y^2)/z^{k-1} = z^k/z^{k-1}=z$, $a(a^2+b^2)^{(k-1)/2}=\frac{x}{z^{(k-1)/2}}z^{(k-1)/2}=x$ and $b(a^2+b^2)^{(k-1)/2}=y$.
| {
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Finding a convergent majorant series I have a series
$$
\sum_{n=1}^\infty \left( \frac{1}{n^3} \cos(nt) - \frac{1}{(2n+1)^2} \sin(nt) \right)
$$
and I have to find a majorant series to this series.
The convergent majorant series I was supposed to find is
$$
\sum_{n=1}^\infty \left( \frac{1}{n^3} + \frac{1}{(2n+1)^2} \right).
$$
however, I don't exactly get why.
I thought that $\frac{1}{n^3}$ is also a convergent majorant series.
I know that the first term $\frac{1}{n^3} \cos(nt)$ is always less than or equal to $\frac{1}{n^3}$ and subtracting the second term $\frac{1}{(2n+1)^2} \sin(nt)$ will only make it smaller, so why doesn't this inequality hold:
$$
|f_n(t)| = |\frac{1}{n^3} \cos(nt) - \frac{1}{(2n+1)^2} \sin(nt)| \leq \frac{1}{n^3}
$$
| Hint. You may use
$$
|a+b|\leq |a|+|b|,\qquad a,b \in \mathbb{R},
$$ giving, for $n=1,2,3 \cdots,$
$$
\left| \frac{1}{n^3} \cos(nt) - \frac{1}{(2n+1)^2} \sin(nt) \right|\leq \left| \frac{1}{n^3} \cos(nt)\right|+\left| - \frac{1}{(2n+1)^2} \sin(nt)\right|\leq \frac{1}{n^3} + \frac{1}{(2n+1)^2}
$$ since $|\cos (nt)|\leq 1$ and $|\sin (nt)|\leq 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Hidden Markov Models and Viterbi Algorithm: Fair and Biased Die So following is the problem that I am trying to solve using Viterbi algorithm and HMM:
Before attempting to write a program, I want to do this problem by hand for the first 3 observations($651$). Based on the question, I understand that:
$P(i | Fair) = \frac1{6} , 1\leq i \leq 6 $
$P( 6 | Biased ) = \frac5{10} = \frac1{2} $
$P( i | Biased ) = \frac1{10} , 1\leq i \leq 5 $
and the transition matrix is
$Fair$ $Biased$
$\begin{bmatrix}
0.95 & 0.05 \\
0.1 & 0.9 \\
\end{bmatrix}$
but where should I go from here ?
EDIT: Assume that initially either fair or biased is equally likely($\frac1{2}$) from a "fictitious" state $O$.
I managed to compute the first 4 highest probabilities: $1 , 0.25 , \frac9{400}, \frac{81}{40000}$ corresponding to $O , B, B, B$
Is this right ?
| Your figures look like you're doing it right. I get the following, skipping the first fictitious state, which is probably not necessary.
Iteration $1$:
\begin{align}
\nu_1(F) &= P(X_1\mid F)P(F) = P(6\mid F)P(F) = \dfrac{1}{6}\dfrac{1}{2} = \dfrac{1}{12} \\
& \\
\nu_1(B) &= P(X_1\mid B)P(B) = P(6\mid B)P(B) = \dfrac{1}{2}\dfrac{1}{2} = \dfrac{1}{4} \\
\end{align}
Iteration $2$:
\begin{align}
\nu_2(F) &= P(5\mid F)\max\{\nu_1(F)P(F\mid F),\; \nu_1(B)P(F\mid B)\} = \dfrac{1}{6}\max\{\dfrac{1}{12}\cdot 0.95,\;\dfrac{1}{4}\cdot 0.1\} \\
&= \dfrac{1}{6}\dfrac{1}{12}\cdot 0.95 = \dfrac{19}{1440} \qquad\qquad [FF]
& \\
\nu_2(B) &= P(5\mid B)\max\{\nu_1(F)P(B\mid F),\; \nu_1(B)P(B\mid B)\} = \dfrac{1}{10}\max\{\dfrac{1}{12}\cdot 0.05,\;\dfrac{1}{4}\cdot 0.9\} \\
&= \dfrac{1}{10}\dfrac{1}{4}\cdot 0.9 = \dfrac{9}{400} \qquad\qquad [BB]
\end{align}
Iteration $3$:
\begin{align}
\nu_3(F) &= P(1\mid F)\max\{\nu_2(F)P(F\mid F),\; \nu_2(B)P(F\mid B)\} = \dfrac{1}{6}\max\{\dfrac{19}{1440}\cdot 0.95,\;\dfrac{9}{400}\cdot 0.1\} \\
&= \dfrac{1}{6}\dfrac{19}{1440}\cdot 0.95 = 0.002089... \qquad\qquad [FFF]
& \\
\nu_3(B) &= P(1\mid B)\max\{\nu_2(F)P(B\mid F),\; \nu_2(B)P(B\mid B)\} = \dfrac{1}{10}\max\{\dfrac{19}{1440}\cdot 0.05,\;\dfrac{9}{400}\cdot 0.9\} \\
&= \dfrac{1}{10}\dfrac{9}{400}\cdot 0.9 = \dfrac{81}{40000} = 0.002025 \qquad\qquad [BBB]
\end{align}
So I get $[FFF]$ very slightly in front after $3$ iterations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If x is real , the numbers $ 5^{1+x} + 5^{1-x} $ , a/2 , $ 25^x + 25^{-x} $ form an A.P. then to which interval must ' a ' belong to? A) [ 1 , 5 ]
B) [ 2 , 5 ]
C) [ 5 , 12 ]
D) [ 12 , $ \infty $ ]
My progress so far :
Let $ 5^{x} = y $
Since they are in AP so ,
=> $ 5y $ + $ 5\over y $ + $ y^2 $ + $ 1\over (y^2) $ = a
=> $ {5y^3 + 5y + y^4 + 1 }\over (y)^2 $ = a
As you can see , It becomes a mess . How do I solve it with a better approach .
| As they are in AP, we have
$$
a = 5^{1+x} + 5^{1-x} + 25^{x} + 25^{-x}.
$$
So say $5^x = t$, then,
$$
a = 5t + \frac{5}{t} + t^2 + \frac{1}{t^2}.
$$
Now
$$
5t+\frac{5}{t} \geq 10
\qquad\text{and}\qquad
t^2+\frac{1}{t^2} \geq 2
$$
by the AM-GM inequality.
So $a \ge 12$, are the values that $a$ can take.
Hope that helps.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $a_{n+1}=\frac {a_n^2+5} {a_{n-1}}$ then $a_{n+1}=Sa_n+Ta_{n-1}$ for some $S,T\in \Bbb Z$. Question
Let $$a_{n+1}:=\frac {a_n^2+5} {a_{n-1}},\, a_0=2,a_1=3$$
Prove that there exists integers $S,T$ such that $a_{n+1}=Sa_n+Ta_{n-1}$.
Attempt
I calculated the first few values of $a_n$: $a_2=7,a_3=18, a_4=47$ so I'd have the system of diophantine equations:
$$
7=3S+2T\\
18=7S+3T\\
47=18S+7T
$$
Now: it seems that all of the $a_i$ are pairwise coprime, so these equations should always have solutions, but how could I check that the intersection of all the solutions is not $\emptyset$?
| Assume that $a_n$ can be expressed as $a_{n+1}=Sa_n+Ta_{n-1}$.
Then $a_n$ is expressed as
$$a_n=A\alpha^n+B\beta^n$$
Then
$$a_{n+1}a_{n-1}=(A\alpha^{n+1}+B\beta^{n+1})(A\alpha^{n-1}+B\beta^{n-1})$$
$$=A^2\alpha^{2n}+AB\alpha^{n-1}\beta^{n+1}+AB\alpha^{n+1}\beta^{n-1}+B^2\beta^{2n}$$
And
$$a_n^2+5=A^2\alpha^{2n}+2AB\alpha^n\beta^n+B^2\beta^{2n}+5$$
Therefore,
$$AB(\alpha^{n-1}\beta^{n+1}+\alpha^{n+1}\beta^{n-1})=2AB\alpha^n\beta^n+5$$
should hold for all $n$.
This is possible only when $\alpha\beta=1$ and if $\alpha\beta=1$,
$$AB(\alpha^2+\beta^2)=2AB+5$$
$$AB({\alpha-\beta})^2=5\tag{1}$$
Using initial conditions
$$a_0=A+B$$
$$a_1=A\alpha+B\beta$$
Solving linear equations,
$$A=\frac{a_1-a_0\beta}{\alpha-\beta}$$
$$B=\frac{a_0\alpha-a_1}{\alpha-\beta}$$
Applying to (1),
$$(a_0\beta-a_1)(a_0\alpha-a_1)+5=0$$
$$a_0^2-a_0a_1(\alpha+\beta)+a_1^2+5=0$$
$$\therefore \alpha+\beta=\frac{a_0^2+a_1^2+5}{a_0a_1}$$
As $\alpha$ and $\beta$ are roots of $x^2-Sx-T$,
$$S=\alpha+\beta=\frac{a_0^2+a_1^2+5}{a_0a_1}$$
$$T=-\alpha\beta=-1$$
(*) In our case, $a_0=2$ and $a_1=3$ gives us $\alpha+\beta=3$ that leads to lulu's solution.
| {
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"url": "https://math.stackexchange.com/questions/1566162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Find the equation of tangent line to the given curve
Find the equation of tangent line to $(x^2+y^2)^{3/2} = 2xy$ at the point $\left( \frac{1}{\sqrt{2}}, \frac{1}{\sqrt{2}} \right)$
I just can't figure out how to take the derivative of this function!
| You can use the implicit function theorem, which states that (in particular)
$$F(x,y)=0, \quad \text{and} \quad F_y(x,y) \neq 0 \quad \Rightarrow \quad \frac{dy}{dx}(x,y)=-\frac{F_x(x,y)}{F_y(x,y)}.$$
First define
$$ F(x,y) \equiv (x^2+y^2)^{3/2}-2xy $$
then calculate
\begin{align}
F_x(x,y) &= \frac{3}{2}(x^2+y^2)^{1/2}2x-2y \\[2ex]
F_y(x,y) &= \frac{3}{2}(x^2+y^2)^{1/2}2y-2x
\end{align}
and it follows that
$$ \frac{dy}{dx}(x,y)= -\frac{\frac{3}{2}(x^2+y^2)^{1/2}2x-2y}{\frac{3}{2}(x^2+y^2)^{1/2}2y-2x}.$$
Finally, substituting the desired coordinates we get
$$ \frac{dy}{dx}\left(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}}\right)= -1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to take an integral using half angle trigonometric substitution. So i have this question which is asking to take the integral using a predefined trigonometric substitution which is $$u=\tan\frac{x}{2}$$
and the integral equation is $$\int\frac{\sin x\ dx}{(6\cos x-2)(3-2\sin x)}$$ How would i go on about this problem? Because to begin with i do not know how i would even use the given substitution method. Any help is appreciated thank you.
| This substitution is used for integrals involving only trigonometric expressions. This method is very useful as it transforms the trigonometric integral into just rational integral. You should know how to write $\sin x, \cos x, \tan x$ in terms of $\tan \frac{x}{2}$
(try proving)
$\sin x=\dfrac{2\tan \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
$\cos x=\dfrac{1-\tan^2 \frac{x}{2}}{1+\tan^2 \frac{x}{2}}$
$\tan x=\dfrac{2\tan \frac{x}{2}}{1-\tan^2 \frac{x}{2}}$
To find $dx$ in terms of $du$, differentiate both sides.
$u = \tan \dfrac{x}{2}$
$du = \sec^2 \dfrac{x}{2} \dfrac{1}{2} dx = (1+\tan^2 \dfrac{x}{2})\dfrac{1}{2} dx=(1+u^2)\dfrac{1}{2}dx \Rightarrow dx=\dfrac{2}{1+u^2}du$
For this example, The integral is converted to,
$$\int\dfrac{\sin x}{(6cosx-2)(3-2sinx)}dx=\int\dfrac{\dfrac{2u}{1+u^2}}{(6\dfrac{1-u^2}{1+u^2}-2 )(3-2\dfrac{2u}{1+u^2})}\dfrac{2}{1+u^2}du$$
$$=\int\dfrac{4u}{(1+u^2)(6-6u^2-2-2u^2)(3+3u^2-4u)}du$$
$$=\int\dfrac{-4u}{(1+u^2)(8u^2-4)(3u^2-4u+3)}du$$
$$=\int\dfrac{3u^2-4u+3-(3u^2+3)}{(1+u^2)(8u^2-4)(3u^2-4u+3)}du$$
$$=\int\dfrac{du}{(1+u^2)(8u^2-4)}-\int\dfrac{3\ du}{(8u^2-4)(3u^2-4u+3)}$$
and integrate using partial fractions and substitute again $u=\tan \dfrac{x}{2}$.
Look how nicely the trigonometric integral(which made no sense) is transformed to a rational integral(which can be solved at least by brute force).
Try more examples.
(
You will get to realize that some problems looks like can be solved using half angle substitutions but really can be solved without substitution. Example
$$\int\dfrac{\sin x}{(\cos x -2)(2\cos x +3)}dx$$
)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $abc$.
The product of two $3$-digit numbers with digits $abc$, and $cba$ is $396396$, where $a > c$. Find the value of $abc$.
In order to solve this, should I just find the prime factorization of $396396$ and then find the two $3$-digit factors?
| Hint:
Let $[xyz]$ denote a thre-digit number. Then you want $a,b,c$ such that $[abc]\cdot[cba]=396396$, with $a>c$.
Write $[abc]=100\cdot a+100\cdot b+1\cdot c$ and $[cba]=100\cdot c+10\cdot b+1\cdot a$, then
$$[abc]\cdot[cba]=(100\cdot a+100\cdot b+1\cdot c)\cdot(100\cdot c+10\cdot b+1\cdot a)$$
$$=10\ 000\cdot ac+1000\cdot(ab+bc)+100\cdot(a^2+b^2+c^2)+10\cdot(ab+bc)+ac$$
and this should equal $396396$. Can you continue from here?
Edit:
Listing these terms as follow:
*
*$10\ 000$: $ac$
*$1000$: $\ \ \ ab+bc$
*$100$: $\ \ \ \ \ a^2+b^2+c^2$
*$10$: $\ \ \ \ \ \ ab+bc$
*$1$: $\ \ \ \ \ \ \ \ ac$
You can see some patterns, e.g. the terms $ac$ and $ab+bc$ occur twice. Now compare this with the structure of $396396$ and consider the prime factorization of $396396$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show $\sin(\frac{\pi}{3})=\frac{1}{2}\sqrt{3}$ I have to show that
$$\sin\left(\frac{\pi}{3}\right)=\frac{1}{2}\sqrt{3}$$
and
$$\cos\left(\frac{\pi}{3}\right)=\frac{1}{2}$$
Should I use the exponential function?
| Note that:
$sin(3\theta)=3sin(\theta)-4sin^{3}(\theta)$
Let $\theta =\frac{\pi}{3}$ and let $sin(\frac{\pi}{3})=x$
Then:
$0=x(3-4x^{2})$.
Hence either $x=0$ or $x^{2}=\frac{3}{4}$
But since $sin(0)=0$ and the $sine$ function has a period of $2\pi$ then we must conclude that $sin(\frac{\pi}{3})=\frac{\sqrt3}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1567824",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help me with the result of this determinant.. $$
D =
\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
2 & 1 & 1 & \dots & 1 & 0 \\
3 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
n-1 & 1 & 0 & \dots & 0 & 0 \\
n & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
=n*1*(-1)^\frac{n(n-1)}{2}
$$
I don't quite understand the solution of this determinant. I do understand that if we use Laplace expansion along the last row we get
$$
D = n*
\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}
$$
But how does the remaining determinant euqal: $1*(-1)^\frac{n(n-1)}{2}$?
Edit:
$$
\begin{vmatrix}
0 & 0 & 1 \\
0 & 1 & 0 \\
1 & 0 & 0 \\
\end{vmatrix}
=(-1)^{4}
\begin{vmatrix}
0 & 1 \\
1 & 0 \\
\end{vmatrix}
=(-1)^{4+3+2}
$$
I thought it should go: $(-1)^{3+2+1}$ or is the power actually the sum of row and column coordinates?
| To compute
$$\begin{vmatrix}
1 & 1 & 1 & \dots & 1 & 1 \\
1 & 1 & 1 & \dots & 1 & 0 \\
1 & 1 & 1 & \dots & 0 & 0 \\
\vdots & \vdots & \vdots &\ddots & \vdots & \vdots \\
1 & 1 & 0 & \dots & 0 & 0 \\
1 & 0 & 0 & \dots & 0 & 0 \\
\end{vmatrix}$$
You can proceed like that: If we swap the $k$-th row with the $n-k$-th row of the matrix for $k=1,2,\ldots$ then, at some point, we will get a lower triangular matrix with only $1$ on its main diagonal (so its determinant is $1$). Now, each swap corresponds to an elementary operation of the Gauss-Jordan method and thus changes the sign of the determinant (i.e. multiplies it by $-1$). So count the number of swaps to get the result.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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(Perhaps) An Easy Combinatorics Problem: (Perhaps not so easy...)
Still have some difficulty with problems like this.
Suppose I have Box with various buttons inside. For the sake of example, let the box contain: $$[ 4 \textrm{ Red}, 5 \textrm{ Blue}, 7 \textrm{ Green} ]$$ How many distinct tuples of, say, 5 buttons can I draw from the box? In the coordinates $(r,b,g)$ I can enumerate them:
$$\{(4,0,1),(4,1,0),(3,2,0),(3,1,1),(3,0,2),(2,3,0),(2,2,1),(2,1,2),(2,0,3),(1,4,0),(1,3,1),(1,2,2),(1,1,3),(1,0,4),(0,5,0),(0,4,1),(0,3,2),(0,2,3),(0,1,4),(0,0,5)\}$$
And I count twenty.
How do I do this in general?
Given a box (multiset) of buttons $B$,
$$B = \{B_1,B_2,\cdots,B_n\}$$
composed of $n$ different styles with respective sizes $|B_j|$ how many distinct $n$-tuples can I get by drawing $k$ buttons from such a box?
From the Selected Answer:
In the general case with my notation, we want the coefficient of $x^k$ in the expansion
$$P(x) = \Pi_{j=1}^n \frac{1-x^{|B_j|}}{1-x} $$
for the number of distinct $k$-tuples selected from the box.
In the cases that I'm dealing with, I think I'll have a big $n$ but maybe a smallish $k$. It appears that writing out the whole inclusion-exclusion form is a bit of a bear. But I get the general way to solve this now.
Thanks!
| The generating function approach is to write:
$$(1+x+x^2+x^3+x^4)(1+x+\cdots + x^5)(1+x+\cdots + x^7) = \frac{(1-x^5)(1-x^6)(1-x^8)}{(1-x)^3}$$
The coefficient of $x^n$ is the number of ways of taking $n$ buttons out.
Now, $$\frac{1}{(1-x)^3} = \sum_{k=0}^\infty \binom{k+2}{2}x^k$$
And: $$(1-x^5)(1-x^6)(1-x^8)=1-x^5-x^6-x^8+x^{11}+x^{13}+x^{14}-x^{19}$$
So we see the coefficient of $x^n$ is:
$$\binom{n+2}{2} - \binom{n-3}{2} - \binom{n-4}{2} - \binom{n-6}{2} + \binom{n-9}{2}+\binom{n-11}{2} +\binom{n-12}{2} - \binom{n-17}{2}$$
Where the values of the binomial $\binom{M}{2}$ is taken to be zero when $M<2$.
This is the same value you'd get with an "inclusion/exclusion" argument.
In general, if your box has $R,G,B$ buttons of each color, the formula would be:
$$\binom{n+2}{2} - \binom{n+2-R}{2} - \binom{n+2-G}{2} - \binom{n+2-B}{2} + \binom{n+2-R-G}{2}+\binom{n+2-R-B}{2} +\binom{n+2-G-B}{2} - \binom{n+2-R-G-B}{2}$$
The value $2$ gets replaced by $k-1$ and the formula gets worse with $k$ buttons, but the above approach works.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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What are the valors of $4x_1-x_1^2+x_3^2$? $x_1$ and $x_2$ $x_3$ real numbers such that
$x_1<x_2<x_3$ are solutions of the equation :
$x^3-3x^2+(a+2)x-a=0$ where a is real
What are the valors of $4x_1-x_1^2+x_3^2$
After factorizing i get :
$(x-1)(x(x-2)+a)=0$
The problem is that i don't know relation of solutions like in seconde degree of equation ( viet relation of solution )
Can someone give hint !
| From what you get $(x−1)(x(x−2)+a)=0$, we can find the other two roots, which satisfy $$x^2-2x+a=(x-1)^2-(1-a)=0$$ Since all roots are real, hence we have $a\leq 1$ ($a\neq1$ since the roots are distinct) then $x_1=1-\sqrt{1-a},x_3=1+\sqrt{1+a}$.
Hence, $$4x_1−x^2_1+x^2_3=4(1-\sqrt{1-a})-(1-\sqrt{1-a})^2+(1+\sqrt{1-a})^2$$
$$=4(1-\sqrt{1-a})+2\cdot(2\sqrt{1-a})=4$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find all positive integers $(a,b,c,n)$ such that $2^n=a!+b!+c!$ I have found the solutions by a little calculation $(2,3,5,7)$ and $(2,3,4,5)$. But I don't know if there's any other solutions or not?
| Let $a \le b \le c$.
If $m \ge 2$ m! is even so if $a = 1$ (odd), $b = 1$ and $c! = 2^n - 2 = 2(2^{n-1} - 1)$ so $c = 2; n=1$ or $c = 3; n = 2$. Two answers so far. (1,1,2,1)(1,1,3,2)
If $a > 1$ then $a! + b! + c! = a!(1 + b!/a! + c!/a!) = 2^n$ If $a \ge 3$ $3|a!$ and so $3|2^n$ which is impossible so $a = 2$.
So $1 + b!/2 + c!/2 = 2^{n-1}$ $n$ can't be $1$. so $b!/2$ must be odd so $b= 3$. So $c!/2 = 2^{n -1} - 4= 4(2^{n -3} - 1)$ so $c$ = 4 and $n = 5$. Or 3*5...c = $2^{n-3}-1).$ Possibly, $c=5$ and $15 = $ $2^{n-3} - 1$; $n = 7$. So two more answers $(2,3,4,5)$ and $(2,3,5,7)$.
Any more will require $3*5...c = 2^{n-3}-1)$ which is odd so $c<6$. So those $4$ are the only solutions.
| {
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How to find all pairs $(x,y)$ of integers such that $y^2 = x(x+1)(x+2)$? Here $y^2$ is divisible by $12.$ And satisfying all those conditions I think $y=0$ is the only solution. But I can't show it mathematically.
| Assume $y \ne 0$.
If $m | y^2$ $m>2$ then $m$ divides exactly 1 of $x, x + 1, x + 2$ so $m^2$ divides exactly 1 of $x, x+1, x +2$ so each $x, x+1, x + 2 = 2^nm^2$ for some (maybe 0) power of two and some $m$ odd.
Suppose $m, n$ odd and $n^2 = m^2 + 2$.
Then $(n + m)(n - m) =2$. But this has no integer solution.
So $x$ and $x+2$ aren't both odd. So $x$ and $x+2$ are both even. $2$ divides one of them and $4$ divides the other. So $8|y^2$ so $16|y^2$ and $8$ divides the other. So we have:
$8*4^l*m^2 = 2*k^2 \pm 2$ for odd $m,k$ so $4^{l+1}m^2 = k^2 \pm 1$ which means $2^{l+1}m = \pm \sqrt{k^2 \pm 1}$ but that can only happen if one of $x$ or $x+2$ is 0.
So one of $x, x+1, x+1$ is $0$ and $y^2 = 0$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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${\int\int\int}_B dxdydz$ where $B$ is the region delimited by $x²+y²+z² = 4$ and $x²+y²=3z$ Take the following integral over the specified region:
${\int\int\int}_B dxdydz$ where $B$ is the region delimited by $x²+y²+z² = 4$ and $x²+y²=3z$
(i'm answering my own question because I was writing it and then found out my error, so I didn't want to erase all my work)
| This is the intersection betweet the paraboloid $z = \frac{x²+y²}{3}$ and the sphere $x²+y²+z²=4$. Their intersection forms a circle with radius $\sqrt{3}$. Therefore we just need to find the height of the region by integrating with respect to $z$, from the paraboloid to the sphere:
$$\int_{\frac{x²+y²}{3}}^{\sqrt{4-x²-y²}}dz = \sqrt(4-x²-y²)-\frac{x²+y²}{3}$$
then we need to integrate this height all over the circle with radius $\sqrt{3}$, which is our region $B_1$:
$${\int\int}_{B_1}\sqrt(4-x²-y²)-\frac{x²+y²}{3}dxdy$$
So to make this easier, I'm gonna integrate in polar coordinates by doing the substitution $x = p\cos(t), y= p\sin(t)$ where $t$ goes from $0$ to $2\pi$ because our circle is centered at the origin, and $p$ goes from $0$ to $\sqrt{3}$. The jacobian determinant of the transformation is $p$, so our integral becomes:
$$\int_0^{2\pi}\int_0^{\sqrt{3}}\left(\sqrt{4-p²}-\frac{1}{3}p²\right)p \ dp \ dt$$
$$\int_0^{2\pi}\int_0^{\sqrt{3}}\left(\sqrt{4-p²}\ p-\frac{1}{3}p^3\right) \ dp \ dt = $$
$$-\frac{1}{2}\int_0^{2\pi}\int_0^{\sqrt{3}}\sqrt{4-p²}\ (-2p) \ dp \ dt -
\int_0^{2\pi}\int_0^{\sqrt{3}}\frac{1}{3}p^3 \ dp \ dt = $$
$$-\frac{1}{2}\int_0^{2\pi}\int_{4}^{1}\sqrt{u} \ du - \int_0^{2\pi}\frac{\sqrt{3}^4}{12}dt = $$
$$\frac{1}{2}\int_0^{2\pi}\int_{1}^{4}\sqrt{u} \ du - \int_0^{2\pi}\frac{9}{12}dt = $$
$$\frac{1}{2}\int_0^{2\pi}\frac{2}{3}u^{3/2} \ du - \frac{3}{4}\int_0^{2\pi}dt = $$
$$\frac{1}{3}\int_0^{2\pi}(4)^{3/2}-1^{3/2} \ du - \frac{3}{4}2\pi = $$
$$\frac{7}{3}2\pi - \frac{3}{4}2\pi = \frac{19}{12}2\pi = \frac{19\pi}{6}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Can you simplify this term? $$X=\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}$$
| $$\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}=
\frac{r^2}{r^2}\frac{\frac{c}{r^2}+\frac{1-c}{(1+r)^{T+1}}}{\frac{c}{r}+\frac{1-c}{(1+r)^T}-1}=
\frac{{c}+r^2\frac{1-c}{(1+r)^{T+1}}}{r{c}+r^2\frac{1-c}{(1+r)^T}-r^2}=\\
\frac{(1+r)^{T+1}}{(1+r)^{T+1}}\frac{{c}+r^2\frac{1-c}{(1+r)^{T+1}}}{r{c}+r^2\frac{1-c}{(1+r)^T}-r^2}=
\frac{{c}(1+r)^{T+1}+r^2({1-c})}{r{c}(1+r)^{T+1}+r^2({1-c}){(1+r)}-r^2(1+r)^{T+1}}=
\frac{{c}(1+r)^{T+1}+r^2({1-c})}{r\left({c}-r\right)(1+r)^{T+1}+r^2({1-c}){(1+r)}}=
\frac{{c}(1+r)^{T+1}+r^2({1-c})}{\left(({c}-r)(1+r)^{T}+r({1-c})\right){(1+r)r}}
$$
$$%=\frac{c-r^2 (c-1) (r+1)^{-T-1}}{r(c -r)-r^2(c-1) (r+1)^{-T}}$$
would this help?
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit of $\lim_{x\to 0} (\frac{1+\tan x}{1+\sin x})^{\csc^3x}$ I failed to find the limit of:lim(x->0) $(\frac{1+tan(x)}{1+sinx})^{\frac{1}{sin^3(x)}}$?
as X approches 0
How do I find the answer for this?
Thanks in advance. the answer supposed to be sqr(e). but my answer was 1.
Can anyone please help me find my mistake?
I DID:
$lim_{x \to 0} (\frac{1+tan(x)}{1+sin(x)})^{\frac{1}{sin^3(x)}}$
$lim_{x \to 0} (\frac{((1+tan(x))^{1/sin(x)}}{((1+ sin(x))^{1/sin(x)}})^{1/sin^2(x)}$
now I look inside:
$lim_{x \to 0} ((1+tan(x))^{1/sin(x)}$ is e
$lim_{x \to 0} ((1+sin(x))^{1/sin(x)}$ is also e
so we get:
$lim_{x \to 0} (\frac{e}{e})^{\frac{1}{sin^2(x)}}$
$lim_{x \to 0} (1)^{\frac{1}{sin^2(x)}}$ = 1
| $$\\ \lim _{ x\rightarrow 0 }{ { \left( \frac { 1+tan(x) }{ 1+sin\left( x \right) } \right) }^{ \frac { 1 }{ sin^{ 3 }(x) } } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { tan(x)-\sin { \left( x \right) } }{ 1+sin\left( x \right) } \right) }^{ \frac { 1 }{ sin^{ 3 }(x) } } } =\lim _{ x\rightarrow 0 }{ { \left( 1+\frac { 1 }{ \frac { 1+sin\left( x \right) }{ tan(x)-\sin { \left( x \right) } } } \right) }^{ \frac { 1+sin\left( x \right) }{ tan(x)-\sin { \left( x \right) } } \left( \frac { tan(x)-\sin { \left( x \right) } }{ 1+sin\left( x \right) } \frac { 1 }{ sin^{ 3 }(x) } \right) } } =\\ =\lim _{ x\rightarrow 0 }{ { \left[ { \left( 1+\frac { 1 }{ \frac { 1+sin\left( x \right) }{ tan(x)-\sin { \left( x \right) } } } \right) }^{ \frac { 1+sin\left( x \right) }{ tan(x)-\sin { \left( x \right) } } } \right] }^{ \frac { tan(x)-\sin { \left( x \right) } }{ 1+sin\left( x \right) } \frac { 1 }{ sin^{ 3 }(x) } } }$$
if we simply this expression: $$\frac { tan(x)-\sin { \left( x \right) } }{ 1+sin\left( x \right) } \frac { 1 }{ sin^{ 3 }(x) } $$
$$\frac { tan(x)-\sin { \left( x \right) } }{ 1+sin\left( x \right) } \frac { 1 }{ sin^{ 3 }(x) } =\frac { \sin { \left( x \right) } \left( \frac { 1 }{ \cos { \left( x \right) } } -1 \right) }{ 1+sin\left( x \right) } \frac { 1 }{ sin^{ 3 }(x) } =\frac { 1-\cos { \left( x \right) } }{ \cos { \left( x \right) \left( 1+sin\left( x \right) \right) } } \frac { 1 }{ \sin ^{ 2 }{ \left( x \right) } } =\frac { 1-\cos { \left( x \right) } }{ \cos { \left( x \right) \left( 1+sin\left( x \right) \right) } } \frac { 1 }{ 1-\cos ^{ 2 }{ \left( x \right) } } =\\ \frac { 1-\cos { \left( x \right) } }{ \cos { \left( x \right) \left( 1+sin\left( x \right) \right) } } \frac { 1 }{ \left( 1-\cos { \left( x \right) } \right) \left( 1+\cos { \left( x \right) } \right) } =\frac { 1 }{ \cos { \left( x \right) \left( 1+sin\left( x \right) \right) \left( 1+\cos { \left( x \right) } \right) } } $$ so
$$ ={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { tan(x)-\sin { \left( x \right) } }{ 1+sin\left( x \right) } \frac { 1 }{ sin^{ 3 }(x) } } }={ e }^{ \lim _{ x\rightarrow 0 }{ \frac { 1 }{ \cos { x } \left( 1+sin\left( x \right) \right) } \frac { 1 }{ 1+{ \cos { x } } } } }={ e }^{ \frac { 1 }{ 2 } }$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding Symmetry Group $S_3$ in a function I was considering functions $f: \Bbb{C} \rightarrow \Bbb{C}$ and I defined the following instrument (I call it the Symmetry Group of a function)
$$ \text{Sym}(f) = \left< m(x)|f(m(x))=f(x) \right> $$
An intuitive example is to consider $\text{Sym}(e^x)$ and observe that
$$m(x) = x + 2i \pi $$
has the property that
$$ e^{m(x)} = e^{x+2i\pi}=e^x e^{2i \pi} = e^x $$
And the group generated by $m(x)$ under composition is the set of functions
$$ x + 2i\pi k, k \in \Bbb{Z}$$
Which is isomorphic to $\Bbb{Z}$ under function composition. So one can then say that $$\text{Sym}(e^x) \cong \Bbb{Z}$$
What I was curious about was if there are any elementary functions such that
$$ \text{Sym}(g(x)) \cong S_3$$
In attempt to build one I considered
$$ g(x) = x^{\frac{-1 + i \sqrt{3}}{2}} + x^{\left( {\frac{-1 + i \sqrt{3}}{2}}\right)^2} + x + \frac{1}{x} +x^{-\frac{-1 + i \sqrt{3}}{2}}+ x^{-\left({\frac{-1 + i \sqrt{3}}{2}}\right)^2} $$
G has as a generator for its symmetries the functions $L_1 = \frac{1}{x}$ and $L_2 = x^{\frac{-1 + i \sqrt{3}}{2}}$
Which can be observed as
$$G(L_1) = G(L_2) = G(x)$$
But the problem is that $L_1(L_2) = L_2(L_1)$ so clearly this isn't a generating set for $S_3$. It's not obvious at this point, how to go about making a function that has $S_3$ as its underlying symmetry group
Some Examples:
$\Bbb{Z}_2$ can be realized as $\text{Sym}\left(x + \frac{1}{x}\right)$ as this function is invariant under the substitutions $x \rightarrow x$ and $x\rightarrow \frac{1}{x}$
The proof arises from the following: Suppose we wish to find all transformations $T$ $x$ such that
$$ x + \frac{1}{x} = T(x) + \frac{1}{T(x)}$$
We can the derive that
$$ T(x)^2 - \left(x+ \frac{1}{x}\right)T(x) + 1 = 0$$
Which yields that
$$ T(x) = \frac{x + \frac{1}{x} \pm \sqrt{x^2+2+\frac{2}{x^2}-4}}{2}$$
simplifying to
$$ T(x) = \frac{x + \frac{1}{x} \pm (x-\frac{1}{x})}{2}$$
and that gives $T(x) = x, T(x) = \frac{1}{x}$ observe the these transformations form a group of order $2$ so they must be isomorphic to $\Bbb{Z}_2$
And in general I conjecture that:
$\Bbb{Z}_n$ can be realized as $$\text{Sym} \left( x + x^{\sqrt[n]{1}_1} + x^{\sqrt[n]{1}_2} + ... x^{\sqrt[n]{1}_{n-1}}\right)$$
| There's a little ambiguity as to the type of functions you are considering. For example, you seem ok with allowing the function to have some singularities at zero as your $z+1/z$ example indicates. The group $S_3$ acts naturally on $\mathbf{C} \cup \{\infty\}$ via the following rational functions:
$$\Sigma = \left\{z, \ 1/z, \ 1-z,\frac{1}{1-z}, \ 1 - \frac{1}{z}, \ \frac{z}{z-1}\right\}$$
In particular, if $h(z)$ is any function $h: \mathbf{C} \cup \{\infty\} \rightarrow \mathbf{C} \cup \{\infty\}$ then
$$f(z) = h(z) + h(1/z) + h(1-z) + h\left(\frac{1}{1-z}\right) + h\left(1-\frac{1}{z}\right) + h\left(\frac{z}{z-1}\right)$$
will be invariant under $\Sigma$. It could be the case that $f(z)$ is invariant under more symmetries, of course. For example, if $h(z) = z$, then $f(z) = 3$.
On the other hand, if $h(z) = z^2 + c$ for any constant $c$, then $f(z)$ is a non-trivial rational function. Moreover, one finds that (in this case)
$$f(x) - f(y) = \frac{2(x-y)(x+y-1)(xy - 1)(1-x+xy)(1-y+xy)(-x-y+xy)}{(x-1)^2 x^2 (y-1)^2 y^2}.$$
Assuming that $y \in \mathrm{Sym}(f)$, the numerator is zero, and so (under very weak continuity hypotheses) one of the six factors in the numerator are zero, leading to $y \in \Sigma$. So it seems that $f(z)$ is a suitable function in your case. A particularly nice choice of constant $c$ is $c = -7/4$, in which case $f(2) = 0$, and so
$$f(z) = f(z) - f(2) = \frac{(z-2)^2 (z+1)^2 (2z - 1)^2}{2 z^2 (z-1)^2}.$$
In this case, the square-root of this function is invariant under the even elements of $\Sigma = S_3$ and sent to its negative under the odd elements.
A slightly more general nice family (but no longer a square) is given (for a parameter $t$) by
$$f(x) = \frac{2(x-t)(x+t-1)(xt - 1)(1-x+xt)(1-t+xt)(-x-t+xt)}{(x-1)^2 x^2 (t-1)^2 t^2}.$$
I might as well add a complete list of such examples coming from polynomials. Suppose that $f(x)$ is a polynomial, and $y \in \mathrm{Sym}(f)$. Then we must have $f(x) - f(y) = 0$. But $f(x) - f(y)$ is a rational function in $y$, and so has a finite number of algebraic solutions. If we insist that our functions are entire functions on $\mathbf{C} \cup \{\infty\}$, then this forces $y$ to be a rational function (other algebraic functions will not be single valued), and (by degree considerations) a function of the form:
$$y = \frac{a x + b}{c x + d}.$$
The choice of constants is only well defined up to scaling. This gives an injective map:
$$\mathrm{Sym}(f) \rightarrow \mathrm{PGL}_2(\mathbf{C}).$$
The finite subgroups of the right hand side are well known, and so, in particular, we deduce:
Claim: Let $f$ be a rational function. Then $\mathrm{Sym}(f)$ is either cyclic, dihedral, or one of the exceptional groups $A_4$, $S_4$, and $A_5$.
Cyclic examples are easy to construct. Let $f(x) = x^n$, and then $\mathrm{Sym}(f)$ consists of $y = \zeta x$ for an $n$th root of unity $\zeta$. This corresponds to the map:
$$a \in \mathbf{Z}/n \mathbf{Z} \mapsto \left( \begin{matrix} \zeta^a & 0
\\ 0 & 1 \end{matrix} \right) \in \mathrm{PGL}_2(\mathbf{C}).$$
Naturally one can also take $f(x) = h(x^n)$ for a generic rational function $h(x)$.
Note that other examples (such as $f(x) = x + x^{-1}$) can be obtained from
these examples by suitable change of variables, namely, because
$$\left( \begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix} \right)
\left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right)
\left( \begin{matrix} 1 & - 1 \\ 1 & 1 \end{matrix} \right)^{-1} =
\left( \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right),$$
and we find that
$$h(x) = x + \frac{1}{x}, \qquad h\left(\frac{x-1}{x+1}\right) = g(x^2), \qquad
g(x) = 2 \cdot \frac{x+1}{x-1}.$$
Note that the dihedral representation of $D_{2n}$ inside $\mathrm{PGL}_2(\mathbf{C})$ is
given by the image of $\mathbf{Z}/n \mathbf{Z}$ together with the matrix
$$\left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right) ,$$
Hence we can write down the examples
$$f(x) = h\left(x^n + \frac{1}{x^n}\right),$$
for a generic function $h$ (taking $h(x) = x$ will do). Here $\mathrm{Sym}(f)$ is generated by $x \mapsto \zeta x$ and $x \mapsto 1/x$.
One can construct the other examples in a similar manner. For fun, I computed an example with $\mathrm{Sym}(f) = A_4$. The group $A_4$ has (several) projective representations
$$A_4 \rightarrow \mathrm{PGL}_2(\mathbf{C})$$
realized by $2$-dimensional representations of the Schur cover $\mathrm{SL}_2(\mathbf{F}_3)$.
One such example maps the non-trivial elements of the
Klein $4$-subgroup $K$ to
$$K \setminus \{e\} = \left\{\left( \begin{matrix} 0 & 1 \\ 1 & 0 \end{matrix} \right), \left( \begin{matrix} -1 & 0 \\ 0 & 1 \end{matrix} \right), \left( \begin{matrix} 0 & -1 \\ -1 & 0 \end{matrix} \right)\right\},$$
and this group is normalized by the order three (in $\mathrm{PGL}_2$) element
$$\left( \begin{matrix}i & -i \\ 1 & 1 \end{matrix} \right)$$
Writing down the corresponding $12$ elements of $A_4$ and letting
$$f(z) = \sum_{A_4 \subset \mathrm{PGL}_2(\mathbf{C})} h(\gamma z),$$
doing a calculation
as above with $h(z) = z^2 + c$, one finds that
$$\begin{aligned}
f(x) - f(y) = & \ 2(x - y)(x + y)(-1 + xy)(1 + xy)(-i - ix - y + xy)(i + ix - y + xy)
(-i - x - iy + xy)\\
\times & \
\frac{(i + x - iy + xy)(i - x + iy + xy)(-i + x + iy + xy)
(i - ix + y + xy)(-i + ix + y + xy)}{(-1 + x)^2x^2(-i + x)^2(i + x)^2
(1 + x)^2(-1 + y)^2y^2(-i + y)^2(i + y)^2(1 + y)^2} \end{aligned}
$$
Since translating $f(x)$ preserves the symmetry group, one can (for example) choose $f(x)$ to vanish at $x = y$ for any fixed $y$, and then $f(x)=f(x) - f(y)$ as above. For example, if $y = 2$, then
$$450 f(x) = \frac{(-2 + x) (2 + x) (-1 + 2 x) (1 + 2 x) (9 + x^2) (5 - 6 x + 5 x^2) (5 + 6 x + 5 x^2) (1 + 9 x^2)}{(-1 + x)^2 x^2 (1 +
x)^2 (1 + x^2)^2}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Show that $\sqrt{x^2 +1 }$ is uniformly continuous. $x \in \mathbb{R}$ Let $x,y \in \mathbb{R} $ such that;
If $|x| \leq \delta$ then $|y| = |x| +(|y|-|x|) \leq |x| +||y|-|x|| \leq |x| + |y-x| < 2 \delta$
$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| = \sqrt{y^2 +1 } - \sqrt{x^2 +1 } \leq \sqrt{y^2 +1 } - 1 \leq y^2 +1 -1 < 2\delta := \epsilon $$
Now, this is the part I am having trouble with
if $|x| > \delta$
$$|\sqrt{x^2 +1 } - \sqrt{y^2 +1 }| \leq |\frac{x^2 - y^2}{\sqrt{x^2 +1 } + \sqrt{y^2 +1 }}| \color{red}{\leq \frac{(x+y)^2}{2\sqrt{\delta^2 + 1}}}...$$
I don't think the red term makes much sense. As the $2xy$ term could very well be negative.
Could someone suggest a next step please.
| Usually, before trying to attack this kind of problems with the definition of uniform continuity, it is much easier and much faster to try to prove that the functions involved are Lipschitz-continuous.
In your case,
$$\left| \sqrt{x^2 + 1} - \sqrt {y^2 + 1} \right| = \left| \frac {x^2 - y^2} {\sqrt{x^2 + 1} + \sqrt {y^2 + 1}} \right| = \left| \frac {x + y} {\sqrt{x^2 + 1} + \sqrt {y^2 + 1}} \right| \cdot |x - y| = \left| \frac {x + y} {\sqrt{x^2 + 1} + \sqrt {y^2 + 1}} \right| \cdot |x - y| \le \left( \frac {|x| + |y|} {\sqrt{x^2 + 1} + \sqrt {y^2 + 1}} \right) \cdot |x - y| \le 1 \cdot |x-y| ,$$
so $\sqrt {x^2 + 1}$ is Lipschitz-continuous of Lipschitz constant $1$, whence uniform continuity follows immediately (remember that one characterization of uniform continuity is: $f$ is uniformly continuous on $I$ if and only if for $|x_n - y_n| \to 0$, we have $|f(x_n) - f(y_n)| \to 0$ for $(x_n)_n, (y_n)_n \subset I$).
| {
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why $ 1 - \cos^2x = \sin^2x $? I'm trying to prove this result $$\lim_{x\to 0} \frac{1 - \cos(x)}{x} = 0$$ In this process I have come across an identity $1-\cos^2x=\sin^2x$. Why should this hold ? Here are a few steps of my working:
\begin{array}\\
\lim_{x\to 0} \dfrac{1 - \cos(x)}{x}\\ = \lim_{x\to 0} \left[\dfrac{1 - \cos(x)}{x} \times \dfrac{1 + \cos(x)}{1 + \cos(x)}\right] \\
=\lim_{x\to 0} \left[\dfrac{1 - \cos^2(x)}{x(1+\cos(x))}\right] \\
=\lim_{x\to 0} \left[\dfrac{\sin^2(x)}{x(1+\cos(x))}\right]
\end{array}
| It's a Pythagorean identity and comes from $$\sin^2 x + \cos ^2 x = 1$$
Just subtract $\cos ^2 x$ from both sides and you have your answer. Now, as to where $\sin^2 x + \cos ^2 x = 1$ comes from:
Let's say you have a right triangle with legs $a$ and $b$. By the Pythagorean theorem, the hypotenuse is $$\sqrt {a^2 + b^2}$$
Next, $\sin x$ is defined as $\dfrac{opposite}{hypotenuse}$, and $\cos x$ is defined as $\dfrac{adjacent}{hypotenuse}$
So in your triangle you have $\sin x = \dfrac{a}{\sqrt {a^2 + b^2}}$ and $\cos x = \dfrac{b}{\sqrt {a^2 + b^2}}$
$\sin^2 x = \dfrac{a^2}{a^2 + b^2}$
$\cos^2 x = \dfrac{b^2}{a^2 + b^2}$
$\sin^2 x + \cos^2 x = \dfrac{a^2}{a^2 + b^2} + \dfrac{b^2}{a^2 + b^2} = 1$
| {
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"timestamp": "2023-03-29T00:00:00",
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Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$ Find $\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}$
Using Taylor series:
$$\ln(1+4^x)=\frac{2\cdot 4^x-4^{2x}}{2}+O(4^{2x}),\ln(1+3^x)=\frac{2\cdot 3^x-3^{2x}}{2}+O(3^{2x})\Rightarrow$$
$$\lim\limits_{x\to \infty}\frac{\ln(1+4^x)}{\ln(1+3^x)}=\lim\limits_{x\to \infty}\frac{2\cdot 4^x-4^{2x}}{2\cdot 3^x-3^{2x}}=\infty$$
The limit should be $0$. Could someone point out what is wrong?
| Since it is of the form $\frac{\infty}{\infty}$ L'Hopital's rule gives \begin{align}\lim_{x\to \infty}\frac{(\ln{(1+4^x)})'}{(\ln{(1+3^x)})'}&=\frac{\ln 4}{\ln 3}\cdot\lim_{x\to \infty}\frac{4^x(1+3^x)}{3^x(1+4^x)}=\\&=\frac{\ln4}{\ln3}\cdot\lim_{x\to \infty}\frac{12^x\left(\frac1{3^x}+1\right)}{12^x\left(\frac{1}{4^x}+1\right)}=\frac{\ln4}{\ln3}\cdot\lim_{x\to \infty}\frac{\frac1{3^x}+1}{\frac{1}{4^x}+1}=\frac{\ln4}{\ln3}\cdot \frac11=\frac{\ln4}{\ln3}\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583582",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$ If $\arcsin x+\arcsin y+\arcsin z=\pi$,then prove that $(x,y,z>0)$
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=2xyz$
$\arcsin x+\arcsin y+\arcsin z=\pi$,
$\arcsin x+\arcsin y=\pi-\arcsin z$
$\arcsin(x\sqrt{1-y^2}+y\sqrt{1-x^2})=\pi-\arcsin z$
$x\sqrt{1-y^2}+y\sqrt{1-x^2}=z$
Similarly,$y\sqrt{1-z^2}+z\sqrt{1-y^2}=x$
Similarly,$x\sqrt{1-z^2}+z\sqrt{1-x^2}=y$
Adding the three equations,we get
$x\sqrt{1-x^2}+y\sqrt{1-y^2}+z\sqrt{1-z^2}=\frac{x+y+z}{2}$
I am stuck here,please help me.Thanks.
| Hint:
Write
$$x = \sin X \qquad y = \sin Y \qquad z = \sin Z$$
Where $X + Y + Z = \pi$ (meaning that those are the angles of a triangle).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1585113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the volume between two paraboloids Find the volume of the solid enclosed by the paraboloids $z = 1-x^2-y^2$ and $z = -1 + (x-1)^2 + y^2$.
Using triple integrals, it is known that $V = \iiint_R \mathrm dx\,\mathrm dy\,\mathrm dz$, and I will have to change variables. But I can't just say that $r^2 = x^2 + y^2$, because the second paraboloid has a "$(x-1)^2$" term. How do I change variables without making a mess?
| Since $z=1-x^2-y^2$ is a downward paraboloids and $z=-1+(x-1)^2+y^2$ is upward , the limits for $z$ are:
$$
-1+(x-1)^2+y^2 \le z \le 1-x^2-y^2
$$
The projection on the $xy$ plane of the curve of intersection between the two paraboloids is the circumference:
$$
x^2+y^2-x-\frac{1}{2}=0
$$
so the limits for $y$ are:
$$-\sqrt{\frac{1}{2}+x-x^2} \le y \le \sqrt{\frac{1}{2}+x-x^2}
$$
that has real solution if $x$ in the limits
$$
\frac{1-\sqrt{3}}{2}\le x \le \frac{1+\sqrt{3}}{2}
$$
so the volume is given by:
$$
\int_{\frac{1-\sqrt{3}}{2}}^{\frac{1+\sqrt{3}}{2}} \int_{-\sqrt{\frac{1}{2}+x-x^2}}^{\sqrt{\frac{1}{2}+x-x^2}} \int_{[-1+(x-1)^2+y^2]}^{[1-x^2-y^2]} dzdydx
$$
Since the problem has not an axis of symmetry, it is unlikely that the use of cylindrical coordinates gives a simpler integration.
| {
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"url": "https://math.stackexchange.com/questions/1585444",
"timestamp": "2023-03-29T00:00:00",
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Compute $(235432_7 \cdot 2551_7) \pmod{311_7} = N_7 = ?$ This is for my assembly language class. I am finding different answers. My answer was $15_7$. But a friend got 2824. Can someone please explain the correct way to do it if $15_7$ is wrong?
$$(235432_7 \cdot 2551_7) \pmod{311_7} = N_7 = ?$$
| $235432$ in base 7 is $2+3\cdot 7 + 4\cdot 7^2+5\cdot 7^3+3\cdot 7^4+2\cdot 7^5 = 42751$ in base 10
$2551$ in base 7 is $1+5\cdot 7+5\cdot 7^2+2\cdot 7^3 = 967$ in base 10
$311$ in base 7 is $1+1\cdot 7 + 3\cdot 7^2 = 155$ in base 10
$42751\cdot 967\pmod{155}\equiv 12$
$12$ in decimal is $5+1\cdot 7$ is $15$ in base 7
Note, if the answer was expected to be in base 7, then your friends answer of $2824$ doesn't even make sense since $8$ is not a valid digit in base 7.
| {
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"timestamp": "2023-03-29T00:00:00",
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How is this limit being solved? I can't grasp it I am going over limits for my finals as I notice this example in my schoolbook discribing limits of the undefined form $0\over0$ in the shape of an irrational fraction.
$$\lim\limits_{x \to 1} {\sqrt[3]{x}-1\over\sqrt{2x-1}-1} = {0\over0} = \lim\limits_{x \to 1} {(\sqrt[3]{x}-1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)(\sqrt{2x-1}+1)\over(\sqrt{2x-1}-1)(\sqrt{2x-1}+1)(\sqrt[3]{x^2}+\sqrt[3]{x}+1)}$$
I really don't get what is going on here. I know that you are supposed to multiply both the nominator and the denominator by the added value of either. But it appears that here it has been mutliplied by both? And where does this term $(\sqrt[3]{x^2}+\sqrt[3]{x}+1)$ come from?
There is no worked out example, except for the solution after this which is $1\over3$ but thats it.
Could anyone please explain this to me? I would be really gratefull!
Also sorry if I didn't use the correct terms here and there, not a native english speaker.
Thanks in advance.
Cheers, Michiel
| $${\sqrt[3]{x}-1\over\sqrt{2x-1}-1} = \frac{\sqrt[3]{x}-1}{\sqrt{2x-1}-1}\cdot \frac{(\sqrt{2x-1}+1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}{(\sqrt{2x-1}+1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$
$$=\frac{\left((\sqrt[3]{x}-1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)\right)(\sqrt{2x-1}+1)}{\left((\sqrt{2x-1})^2-1^2\right)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$
$$=\frac{(x-1)(\sqrt{2x-1}+1)}{2(x-1)\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}=\frac{\sqrt{2x-1}+1}{2\left(\sqrt[3]{x^2}+\sqrt[3]{x}+1\right)}$$
$$\stackrel{x\to 1}\to \frac{\sqrt{2(1)-1}+1}{2\left(\sqrt[3]{1^2}+\sqrt[3]{1}+1\right)}=\frac{1}{3}$$
| {
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"url": "https://math.stackexchange.com/questions/1586052",
"timestamp": "2023-03-29T00:00:00",
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Solve the equation $\frac{1}{\sin^{2k}(x)}+\frac{1}{\cos^{2k}(x)} = 8$
Solve the equation $\dfrac{1}{\sin^{2k}(x)}+\dfrac{1}{\cos^{2k}(x)} = 8$ where $k$ is an integer and $x$ is a real number.
Attempt
We have that $\dfrac{1}{\sin^{2k}(x)}+\dfrac{1}{\cos^{2k}(x)} = \dfrac{\sin^{2k}(x)+\cos^{2k}(x)}{\cos^{2k}(x)\sin^{2k}(x)} = 8 \implies \sin^{2k}(x)+\cos^{2k}(x) - 8\cos^{2k}(x)\sin^{2k}(x) = 0.$ I get stuck here and am not sure how to proceed.
| We have
$$8=\dfrac1{\sin^{2k}(x)}+\dfrac1{\cos^{2k}(x)} \geq \dfrac{2}{\left\vert\sin^k(x)\cos^k(x)\right\vert} = \dfrac{2^{k+1}}{\left\vert\sin^k(2x)\right\vert} \geq 2^{k+1}$$
Hence, we have $k+1 \leq 3 \implies k \leq 2$.
If $k=0$, we have $\dfrac1{\sin^{2k}(x)}+\dfrac1{\cos^{2k}(x)}=2$.
If $k < 0$, we have $$\sin^{-2k}(x) + \cos^{-2k}(x) \leq \sin^2(x) + \cos^2(x) = 1$$
Hence, the only options are $k=1$ and $k=2$.
If $k=2$, we see that
$$8 = \dfrac1{\sin^4(x)} + \dfrac1{\cos^4(x)} \geq \dfrac8{\sin^2(2x)} \implies \sin^2(2x) \geq 1 \implies \sin^2(2x) = 1$$
This implies that
$$2x = n\pi + \dfrac{\pi}2 \implies x = \dfrac{n\pi}2 + \dfrac{\pi}4$$
When $k=1$, we have that
$$\dfrac1{\sin^2(x)} + \dfrac1{\cos^2(x)} = \dfrac1{\sin^2(x)\cos^2(x)} = \dfrac4{\sin^2(2x)} = 8$$
Hence,
$$\sin^2(2x) = \dfrac12 \implies 2x = n\pi \pm \dfrac{\pi}4 \implies x = \dfrac{n\pi}2 \pm \dfrac{\pi}8$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$? How to compute the area of that portion of the conical surface $x^2+y^2=z^2$ which lies between the two planes $z=0$ and $x+2z=3$ ? I can't even figure out what the integrand will be ( should it be $\sqrt{z^2-x^2}$ ?
) and not even the limits . Please help . thanks in advance
|
In order to calculate the surface of the cone $\mathcal{C}$
\begin{align*}
\mathcal{C}:x^2+y^2=z^2
\end{align*}
between the planes
\begin{align*}
z=0\qquad\text{ and }\qquad
z=\frac{3-x}{2}\tag{1}
\end{align*}
we consider a parameter representation $\Phi(t,\varphi)$ of $\mathcal{C}$
\begin{align*}
\Phi(t,\varphi)
=\begin{pmatrix} x\\ y\\ z\end{pmatrix}
=\begin{pmatrix} t\cos \varphi \\ t \sin \varphi\\ t \end{pmatrix}\qquad\qquad
\begin{matrix}0\leq \varphi \leq 2\pi\\0\leq t \leq \frac{3}{2+\cos\varphi}\end{matrix}\tag{2}
\end{align*}
Comment:
*
*The apex of the cone is $(0,0,0)$ and at height $z=t$ the cone admits a representation by a circle with radius $t$ and polar coordinates:
$(t\cos \varphi,t\sin \varphi)$.
*The limits of the parameter representation $t$ in (2) are due to the planes in (1)
\begin{align*}
0&\leq t\leq \frac{3-x}{2}\qquad\text{and}\qquad x=t\cos\varphi
\end{align*}
The lateral surface $S_{lat}(\mathcal{C})$
The area $S_{lat}(\mathcal{C})$ of the lateral surface of $\mathcal{C}$ is
\begin{align*}
S_{lat}(\mathcal{C})&=\iint_\mathcal{C}\left\|\frac{\partial \Phi}{\partial \varphi}\times\frac{\partial\Phi}{\partial t}\right\|dS\\
&=\int_{0}^{2\pi}\int_{0}^{\frac{3}{2+\cos\varphi}}
\left\|\begin{pmatrix}-t\sin\varphi\\t\cos\varphi\\0\end{pmatrix}\times\begin{pmatrix}\cos\varphi\\\sin\varphi\\1\end{pmatrix}\right\|\,dt\,d\varphi\\
&=\int_{0}^{2\pi}\int_{0}^{\frac{3}{2+\cos\varphi}}
\left\|\begin{pmatrix}t\cos\varphi\\t\sin\varphi\\-t\end{pmatrix}\right\|\,dt\,d\varphi\\
&=\int_{0}^{2\pi}\int_{0}^{\frac{3}{2+\cos\varphi}}
t\sqrt{2}\,dt\,d\varphi\\
&=\frac{9\sqrt{2}}{2}\int_{0}^{2\pi}\frac{1}{(2+\cos\varphi)^2}
\,d\varphi\tag{3}\\
&=2\sqrt{6}\pi
\end{align*}
The integral (3) was calculated with the help of WolframAlpha.
The top surface $S_{top}(\mathcal{C})$
The area of the top surface $S_{top}(\mathcal{C})$ is the intersection of the cone $\mathcal{C}$ with the plane $z=\frac{3-x}{2}$. Its projection on the $xy$-plane is the ellipse $\mathcal{E}:x^2+y^2=\left(\frac{3-x}{2}\right)^2$ and after normalisation we obtain
\begin{align*}
\mathcal{E}: \frac{1}{4}(x+1)^2+\frac{1}{3}y^2&=1
\end{align*}
Let $\mathcal{D}$ denote the region within the ellipse $\mathcal{E}$ which has area $A(\mathcal{D})=2\sqrt{3}\pi$.
The area of the intersection of the plane $z=f(x,y)=\frac{3-x}{2}$ with the cone $\mathcal{C}$ is given by
\begin{align*}
S_{top}(\mathcal{C})&=\iint_\mathcal{D}\sqrt{\left(\frac{\partial f}{\partial x}\right)^2+\left(\frac{\partial f}{\partial y}\right)^2+1}dA\\
&=\iint_\mathcal{D}\sqrt{\left(-\frac{1}{2}\right)^2+\left(0\right)^2+1}dA\\
&=\frac{\sqrt{5}}{2}\iint_\mathcal{D}dA\\
&=\frac{\sqrt{5}}{2}2\sqrt{3}\pi\\
&=\sqrt{15}\pi
\end{align*}
$$ $$
We conclude: The area of the surface of the cone between the planes in (1) is
\begin{align*}
S_{lat}(\mathcal{C})+S_{top}(\mathcal{C})&=(2\sqrt{6}+\sqrt{15})\pi
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove this inequality $\sum \cos{A}\ge\frac{1}{4}(3+\sum\cos{(A-B)})$ Prove that in any triangle $ABC$ the following inequality holds
$$\cos{A}+\cos{B}+\cos{C}\ge\dfrac{1}{4}(3+\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$
And I have gotten
$$8(\cos{A}+\cos{B}+\cos{C})\ge 6+2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})$$
$$2(\cos{(A-B)}+\cos{(B-C)}+\cos{(C-A)})+3=(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$
$$\Longleftrightarrow 8\sum_{cyc}\cos{A}\ge 3+(\sum_{cyc}\cos{A})^2+(\sum_{cyc}\sin{A})^2$$
then Any hints, ideas? Thanks in advance.
| use
$$\sum\cos{A}=\dfrac{R+r}{R},\sum\cos{A}\cos{B}=\dfrac{s^2+r^2-4R^2}{4R^2},\sum\sin{A}\sin{B}=\dfrac{s^2+4Rr+r^2}{4R^2}$$
$$\Longleftrightarrow \dfrac{4R+4r}{R}\ge 3+\dfrac{s^2+r^2-4R^2}{4R^2}+\dfrac{s^2+4Rr+r^2}{4R^2}$$
$$\Longleftrightarrow 4R^2+6Rr\ge s^2+r^2$$
use Gerrentsen inequality
$$s^2\le 4R^2+4Rr+3r^2$$
we only prove following
$$4R^2+6Rr\ge 4R^2+4Rr+4r^2$$
it equal to Euler inequality
$$R\ge 2r$$
| {
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"timestamp": "2023-03-29T00:00:00",
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$2^m+3^n$ is a square Determine all pairs ($m,n$) of positive integers such that $2^m+3^n$ is a square.
One such pair is ($4,2$) and I think that there are no other solutions. Am I correct? If I am can somebody please give me a hint to prove it?
| If $m=1$ , then we have $2+3^n$, which cannot be a square because the residue modulo $3$ is $2$.
If $m>1$, then $n$ cannot be odd because the residue of $2^m+3^n$ modulo $4$ would be $3$. So, $2^m$ and $3^n$ must both be squares. So, we need a pythagorean triple $(u,v,w)$ , where $u$ is a power of $3$ and $v$ is a power of $2$.
The general solution is $u=m^2-n^2$ , $v=2mn$ , so $mn$ must be a power of $2$, so $m$ and $n$ must both be a power of $2$, while $(m-n)(m+n)$ must be a power of $3$. If $m>1$ and $n>1$, then $(m-n)(m+n)$ is even. $m=1$ is also impossible, so we must have $n=1$. So, $(2^k-1)(2^k+1)$ must be a powe of $3$
But $2^k+1$ is never a power of $3$ for $k>3$ because of the proven catalan-conjecture. If $k=2$, then $2^k+1=5$, if $k=3$, then $2^k+1=9=3^2$, but $2^k-1=7$. So, we must have $k=1$ and therefore $m=2$ and thereofore $u=3$ , $v=4$, corresponding with the only solution $2^4+3^2=4^2+3^2=5^2$
| {
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Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$
Prove that given $a,b,c > 0$, it is possible to construct a triangle with sides of length $a,b,c$ if and only if $pa^2+qb^2 > pqc^2$ for any $p,q$ with $p+q = 1$.
Should I prove this using the triangle inequality? Also should I split the proof up into two parts for the if and only if?
| $\Rightarrow$ We will consider triangle $OAB$ wher $A,\ B$ are
vectors on $\mathbb{R}^2$ Define $$a=|A|,\ b=|B|,\ c=|A-B|$$ Then
$$ |A|^2+|B|^2-2A\cdot B=|A-B|^2 $$
Then $pA+qB,\ p+q=1$ is in a line passing through $A,\ B$. Then
\begin{align} |pA+qB|^2&=p^2a^2+q^2b^2+2pqA\cdot B \\&=
p^2a^2+q^2b^2+pq (a^2+b^2-c^2)
\\& =pa^2+ qb^2 -pqc^2 \ \ast\end{align}
Since $a,\ b>0$, so $|pA+qB|>0$ Hence "only if" is proved
$\Leftarrow$ In $\ast$, $$pa^2+ qb^2 -pqc^2=(pa)^2+ (qb)^2-2(pa)(pb)
C $$ where $$ C= - \frac{a^2+b^2-c^2}{2ab }$$
If $C>1$ then $c>a+b $ So for $pa=qb$ we have $pa^2+qb^2
> pqc^2>pq(a+b)^2\Rightarrow (pa-qb)^2>0 $ It is a contradiction
If $C<-1$ then $|a-b|>c$ If $a>b$ then $a>b+c$ so that for $ p=-t,\
t>0,\ q=1+t$ we have $ (b-tc)^2>0$ Let $t=\frac{b}{c}$ Hence it is a
contradiction
So $|C|\leq 1$ So we have $\theta$ s.t. $\cos\ \theta=-C$ If we have
triangle of sides $a,\ b,\ c'$ where $\theta$ is an angle between
sides of length $a,\ b$, then $\cos\ \theta=(a^2+b^2-(c')^2)/2ab$
That is $c=c'$ This complete the proof
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve an irrational equation? I want to solve this equation
$$2 (x-2) \sqrt{5-x^2}+(x+1)\sqrt{5+x^2} = 7 x-5.$$
I tried
The given equation equavalent to
$$2 (x-2) (\sqrt{5-x^2}-2)+(x+1)(\sqrt{5+x^2}- 3)=0$$
or
$$(x-2)(x+1)\left [\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2}\right ] = 0.$$
I see that, the equation
$$\dfrac{x+2}{\sqrt{5+x^2} + 3} - \dfrac{2(x-1)}{\sqrt{5-x^2} + 2} = 0$$
has unique solution $x = 2$, but I can not solve. How can I solve this equation or solve the given equation with another way?
| First of all note that the RHS of your original equation can be written as $4(x-2)+3(x+1)$. Now transfer the terms on either side of the equation obtaining $$2(x-2)\big[\sqrt {5-x^2}-2\big] = (x+1)\big[3-\sqrt{5+x^2}\big].$$ The LHS vanishes for $\pm 1$ and $2$. The RHS vanishes for $-1$ and $\pm2$. Two of the root are therefore $-1$ and $2$.
EDIT: Since the RHS remains positive and the LHS remains negative from $-1$ to $2$, there are no further roots between these two. In the range $-\sqrt 5$ to -1, the LHS is more than RHS and in the range $2$ to $\sqrt 5$, LHS is less than RHS
| {
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Range of function $ f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ Where $x>0$
Find the range of the function $\displaystyle f(x) = x\sqrt{x}+\frac{1}{x\sqrt{x}}-4\left(x+\frac{1}{x}\right),$ where $x>0$
$\bf{My\; Try::}$ Let $\sqrt{x}=t\;,$ Then $\displaystyle f(t) = t^3+\frac{1}{t^3}-4\left(t^2+\frac{1}{t^2}\right)\;,$
Now After Simplification, We get $\displaystyle f(t) = \left(t+\frac{1}{t}\right)^3-3\left(t+\frac{1}{t}\right)-4\left[\left(t+\frac{1}{t}\right)^2-2\right]$
Now Put $\displaystyle t+\frac{1}{t} = u\;,$ Then $\displaystyle \sqrt{x}+\frac{1}{\sqrt{x}} = u\;,$ So we get $u\geq 2$ (Using $\bf{A.M\geq G.M}$)
And our function convert into $\displaystyle f(u) = u^3-4u^2-3u+8\;,$ Where $u\geq 2$
Now Using Second Derivative Test, $f'(u) = 3u^2-8u-3$ and $f''(u) = 6u-8$
So for Max. and Min., We put $\displaystyle f'(u)=0\Rightarrow u=3$ and $f''(3)=10>0$
So $u=3$ is a point of Minimum.
So $f(2)=8-4(4)-3(2)+8 = -6$ and $f(3) = -10$
and Graph is Like this
So Range is $$\displaystyle \left[-10,\infty \right)$$
My question is can we solve it any other way, Like using Inequality
If yes, Then plz explain here
Thanks
| We have
$$\color{blue}{x\sqrt{x} + \dfrac1{x\sqrt{x}}-4\left(x+\dfrac1x\right) = \underbrace{\dfrac{\left(1+\sqrt{x}\right)^2\left(x-3\sqrt{x}+1\right)^2}{x^{3/2}}}_{\text{Is non-negative}}-10}$$
Hence, the minimum is $-10$
| {
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Calculation of $\max$ and $\min$ value of $f(x) = \frac{x(x^2-1)}{x^4-x^2+1}.$
Calculation of $\max$ and $\min$ value of $$f(x) = \frac{x(x^2-1)}{x^4-x^2+1}$$
My try: We can write $$f(x) = \frac{\left(x-\frac{1}{x}\right)}{\left(x^2+\frac{1}{x^2}\right)-1} = \frac{\left(x-\frac{1}{x}\right)}{\left(x-\frac{1}{x}\right)^2+1}$$
Now put $\left(x-\frac{1}{x}\right)=t,x\ne0$. Then we get $$f(t) = \frac{t}{t^2+1} = \frac{1}{2}\left(\frac{2t}{1+t^2}\right)$$
Now put $t=\tan \theta$. Then $$f(\theta) = \frac{1}{2}\frac{2\tan \theta}{1+\tan^2 \theta} = \frac{1}{2}\sin 2\theta$$
So we get $$-\frac{1}{2}\leq f(\theta)\leq \frac{1}{2}\Rightarrow f(\theta)\in \left[-\frac{1}{2}\;,\frac{1}{2}\right]$$
My question is: Is my solution right? If not, then how can we solve it?
| The denominator 'factorises', giving us this:
$$f(x) = {x(x^2-1) \over (x^2-1)^2 + x^2}
$$
We can establish this is less than $1/2$ fairly easily:
$${uv \over {u^2+v^2}} \le {1 \over 2}
\iff u^2 + v^2 - 2uv \ge 0
\iff (u-v)^2 \ge 0
$$
($u,v$ not both 0, but this holds for us.)
And equality is obtained when $u=v$, i.e. $x=x^2-1$, and this has a positive solution (which is in fact the golden ratio).
I think a sketch is instructive. Note that:
*
*The denominator never vanishes (consider it as a quadratic in $x^2$)
*as $x \rightarrow \infty$, $f(x) \rightarrow 0$
*$f(x)$ is an odd function
These along with the extremal value work we've just done should be enough for a sketch.
| {
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Inequality, Cauchy Schwarz and Schur For $a,b, c>0$, prove that $$\frac{a^3}{a^3+b^3+abc}+\frac{b^3}{b^3+c^3+abc}+\frac{c^3}{c^3+a^3+abc}\geq 1$$
I tried the following $$\sum_{cyc}\frac{a^3}{a^3+b^3+abc}\cdot \sum_{cyc}a^3+b^3+abc\geq (a^{3/2}+b^{3/2}+c^{3/2})^2$$
where cyc stands for cyclic sum of $a,b,c$, then we could show that $$(a^{3/2}+b^{3/2}+c^{3/2})^2\geq 2(a^3+b^3+c^3)+3abc$$ to prove our original inequality. This looks similar to Schur but with inequality sign reversed. How would you proceed? Any ideas/hint?
| Let $x=\dfrac{b}{a},y=\dfrac{c}{b},z=\dfrac{a}{c},\Longrightarrow xyz=1$
then
$$\dfrac{a^3}{a^3+b^3+abc}=\dfrac{1}{1+x^3+\frac{x}{z}}=\dfrac{1}{xyz+x^3+x^2y}=\dfrac{xyz}{xyz+x^3+x^2y}=\dfrac{yz}{yz+x^2+xy}$$
and use Cauchy-Schwarz inequality we have
$$\left(\sum_{cyc}\dfrac{yz}{yz+x^2+xy}\right)\sum_{cyc}yz(yz+x^2+xy)\ge (xy+yz+xz)^2$$
This is clear
In fact
$$(xy+yz+xz)^2=yz(yz+x^2+xy)+zx(zx+y^2+yz)+xy(xy+z^2+zx)$$
so
$$\sum_{cyc}\dfrac{yz}{yz+x^2+xy}\ge 1$$
| {
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Prove for every odd integer $a$ that $(a^2 + 3)(a^2 + 7) = 32b$ for some integer $b$. I've gotten this far:
$a$ is odd, so $a = 2k + 1$ for some integer $k$.
Then $(a^2 + 3).(a^2 + 7) = [(2k + 1)^2 + 3] [(2k + 1)^2 + 7]$
$= (4k^2 + 4k + 4) (4k^2 + 4k + 8) $
$=16k^4 + 16k^3 + 32k^2 + 16k^3 + 16k^2 + 32k + 16k^2 + 16k + 32$
$=16k^4 + 32k^3 + 64k^2 + 48k + 32$
But this isn't a multiple of 32, at most I could say $(a^2 + 3)(a^2 + 7) = 16b$ for some integer $b$
| Rewrite your first step as
$$4(k^2 + k + 1) \cdot 4(k^2 + k + 2) = 16 (k^2 + k + 1)(k^2 + k + 2)$$
Now notice that $k^2 + k + 1$ and $k^2 + k + 2$ have opposite parities....
| {
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Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$ Please help me with this integral:
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}\,dx .$$
I know I must solve it by substitution, but I don't know how exactly.
|
$$\int{\frac{x^2 + 4}{x^2 + 6x +10}}dx$$
Hints:
$$(1)\int\bigg({1-\frac{6x+6}{x^2 + 6x +10}}\bigg)dx$$
$(2)$rewrite the integrand
${\frac{x+1}{x^2 + 6x +10}}$ as $\frac{2x+6}{2(x^2+6x+10)}-\frac{2}{x^2 + 6x +10}$
$(3)$ for inegrand $\frac{2x+6}{2(x^2+6x+10)}$ substitute $u=x^2+6x+10$ and $du=(2x+6)dx$
| {
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Evaluate $\int \frac{x^2}{x^2 -6x + 10}\,dx$ Evaluate $$\int \frac{x^2}{x^2 -6x + 10} \, dx$$
I'd love to get a hint how to get rid of that nominator, or make it somehow simpler.
Before posting this, I've looked into: Solve integral $\int{\frac{x^2 + 4}{x^2 + 6x +10}dx}$
And I've not understood how they simplied the nominator. I know that it has to match $2x-6$ somehow. but the way they put $(x-6)$ and multipled the integral and have suddenly in the nominator $x+1$ does not make sense to me.
| Hint : $$\int\frac{x^2}{x^2-6x+10}dx=\int1+\frac{6x-10}{x^2-6x+10}dx$$
$\dfrac{d}{dx} \ln{(x^2-6x+10)}=\frac{2x-6}{x^2-6x+10}$ , therefore make
$$\frac{6x-10}{x^2-6x+10}=\frac{3(2x-6)}{x^2-6x+10}+\frac{8}{x^2-6x+10}=3\frac{2x-6}{x^2-6x+10}+8\frac{1}{(x-3)^2+1}$$
$$I=x+3\ln(x^2-6x+10)+8\arctan{(x-3)}+constant$$
| {
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If a and b are non-negative real numbers then demonstrate inequality If $a$ and $b$ are non-negative real numbers then show that
$$(3a +\frac{4}{a+1}+\frac{8}{\sqrt{2(1+b^2)}})\cdot(3b +\frac{4}{b+1}+\frac{8}{\sqrt{2(1+a^2)}})\geq81.$$
Inequality is less common. To demonstrate the usual instruments (means inequality, ...) did not help. Does anyone have any idea?
| Well... Notice that by the AM-GM inequality:
$$ 3a+\frac{4}{a+1}=\left(a+1\right)+\frac{2\left(a^2+1\right)}{a+1}+1 \ge 2\sqrt{2\left(1+a^2\right)}+1 $$
Then using the Cauchy–Bunyakovsky–Schwarz inequality, we get
$$ LHS \ge \left(2\sqrt{2\left(1+a^2\right)}+1+\frac{8}{\sqrt{2\left(1+b^2\right)}}\right)\left(\frac{8}{\sqrt{2\left(1+a^2\right)}}+1+2\sqrt{2\left(1+b^2\right)}\right) \ge 81 $$
as desired. The equality occurs when $a=b=1$.
| {
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Finding the limit of the sequence $T_1=0,T_2=1, T_n=\frac{T_{n-1}+T_{n-2}}{2}$ Given that $T_1=0$, $T_2=1$ and $T_n=\frac{T_{n-1}+T_{n-2}}{2}$, show that the sequence converges to $\frac{2}{3}$.
| I am quite a fan of the following solution: We fist show that the sequence converges for arbitrary two starting values and then derive the limit indirectly from its dependence on the starting values.
Let $a,b \in \mathbb{R}$ and define the sequence $(c_n)_{n \in \mathbb{N}}$ by $c_0 = a$, $c_1 = b$ and $c_{n+2} = (c_n + c_{n+1})/2$ for all $n \in \mathbb{N}$. Then for all $n \geq 1$ we have
$$
| c_{n+1} - c_{n} |
= \left| \frac{c_n + c_{n-1}}{2} - c_n \right|
= \frac{|c_n-c_{n-1}|}{2}
$$
and thus
$$
| c_{n+1} - c_{n} |
= \frac{|c_n-c_{n-1}|}{2}
= \frac{|c_{n-1}-c_{n-2}|}{4}
= \dotsb
= \frac{|c_1-c_0|}{2^n}
= \frac{|b-a|}{2^n}.
$$
It follows that for all $m \geq n \geq 1$ we have
$$
|c_m - c_n|
\leq \sum_{k=n}^{m-1} |c_{k+1}-c_k|
= \sum_{k=n}^{m-1} \frac{|b-a|}{2^k}
\leq \sum_{k=n}^\infty \frac{|b-a|}{2^k}
= \frac{|b-a|}{2^{n-1}}.
$$
Thus the sequence $(c_n)_{n \in \mathbb{N}}$ is a Cauchy sequence and thus converges.
We can now calculate the limit in a rather nice way: For all $a,b \in \mathbb{R}$ let $\Phi(a,b) \in \mathbb{R}$ denote the limit of the above sequence with start values $a$ and $b$. We want to show that $\Phi(0,1) = \frac{2}{3}$.
For this notice that from the properties of convergent series and the special form of our sequence it directly follows that for all $a,b,c \in \mathbb{R}$ we have
$$
\Phi(a+c,b+c) = \Phi(a,b) +c, \quad
\Phi(ca,cb)= c\, \Phi(a,b), \quad
\Phi(a,b) = \Phi\left(b, \frac{a+b}{2}\right)
$$
From this we can now follow that
\begin{align*}
\Phi(0,1)
= \frac{1}{2} \Phi(0,2)
= \frac{1}{2} \Phi(2,1)
= \frac{1}{2} \Phi(0,-1) + 1
= -\frac{1}{2} \Phi(0,1) + 1.
\end{align*}
Comparing the left and right hand side of the above equality yields $\Phi(0,1) = \frac{2}{3}$.
From this we find even more generally that for all $a,b \in \mathbb{R}$ we have
$$
\Phi(a,b)
= \Phi(0,b-a) + a
= (b-a) \Phi(0,1) + a
= \frac{2}{3}(b-a) + a
= \frac{2b+a}{3}.
$$
| {
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A Locus in the complex plane I am having trouble understanding exactly what the following set represents:
The set of all complex numbers $z\neq 5$ such that $\frac{\lvert z -3\rvert }{\lvert z-5\rvert}=\frac{\sqrt2}{2}$
I have a feeling that it represents a circle of some sort, but I do not know how to prove it. I have tried expanding both sides with $z=x+yi$ but did not succeed.
| We can rewrite this as
$$|z-3|=\frac{\sqrt 2}2|z-5|$$
which can be interpreted geometrically as the locus of all points that has a distance from the point $3+0i$ that is $\frac{\sqrt 2}2$ times the distance from the point $5+0i$. From geometry we know that is a circle.
If you can solve this geometrically, you can find the nearest point to $3+0i$ and the farthest point from $3+0i$. These are points on the real axis and are easy to find. These points are the diameter of the desired circle, so find the center and radius and you are done.
Your approach of using $z=x+yi$ should also succeed. Use the equation I used, replace the absolute values with square roots of sums of squares, and square both sides of the equation. When you simplify you will get the equation of a circle.
$$|x+yi-3|=\frac{\sqrt 2}2|x+yi-5|$$
$$|(x-3)+yi|=\frac{\sqrt 2}2|(x-5)+yi|$$
$$\sqrt{(x-3)^2+y^2}=\frac{\sqrt 2}2\sqrt{(x-5)^2+y^2}$$
$$(x-3)^2+y^2=\frac 12\left[(x-5)^2+y^2\right]$$
$$2(x-3)^2+2y^2=(x-5)^2+y^2$$
$$2x^2-12x+18+2y^2=x^2-10x+25+y^2$$
$$x^2-2x+y^2=7$$
$$x^2-2x+1+y^2=7+1$$
$$(x-1)^2+y^2=8$$
$$(x-1)^2+y^2=(2\sqrt 2)^2$$
| {
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A trigonometric identities with the ratio of four terms like $1+(\frac{\tan x}{\sin y})^2$ Prove:
$$\frac{1+\left(\frac{\tan x}{\sin y}\right)^2}{1+\left(\frac{\tan x}{\sin z}\right)^2}=\frac{1+\left(\frac{\sin x}{\tan y}\right)^2}{1+\left(\frac{\sin x}{\tan z}\right)^2}$$
I started by opening the brackets and squaring but did not get the required answer.
| $$\frac{1+\left(\dfrac{\tan x}{\sin y}\right)^2}{1+\left(\dfrac{\tan x}{\sin z}\right)^2}=\frac{1+\left(\dfrac{\sin x}{\tan y}\right)^2}{1+\left(\dfrac{\sin x}{\tan z}\right)^2}$$
$$\iff\frac{1+\left(\dfrac{\tan x}{\sin y}\right)^2}{1+\left(\dfrac{\sin x}{\tan y}\right)^2}=\frac{1+\left(\dfrac{\tan x}{\sin z}\right)^2}{1+\left(\dfrac{\sin x}{\tan z}\right)^2}$$
So, if we can prove that $\dfrac{1+\left(\dfrac{\tan x}{\sin A}\right)^2}{1+\left(\dfrac{\sin x}{\tan A}\right)^2}$ is independent of $A$, we are done.
Method $\#1:$
$\dfrac{1+\left(\dfrac{\tan x}{\sin A}\right)^2}{1+\left(\dfrac{\sin x}{\tan A}\right)^2}=\dfrac{1+\tan^2x\csc^2A}{1+\sin^2x\cot^2A}$
$=\dfrac{\cos^2A+\sin^2x(1+\cot^2A)}{\cos^2x(1+\sin^2x\cot^2A)}=\sec^2x$ which is clearly independent of $A$
Method $\#2:$
$\dfrac{1+\left(\dfrac{\tan x}{\sin A}\right)^2}{1+\left(\dfrac{\sin x}{\tan A}\right)^2}=\dfrac{(\sin^2A\cos^2x+\sin^2x)\sin^2A}{\sin^2A\cos^2x(\sin^2A+\cos^2A\sin^2x)}=\sec^2x$ as $\sin^2A\cos^2x+\sin^2x=\sin^2A(1-\sin^2x)+\sin^2x=\sin^2A+\sin^2x(1-\sin^2A)=?$
| {
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Integrate the square root of the ratio of two quadratic polynomials $$\int \sqrt{\frac{x^2+x-1}{x^2-1}} dx$$
I have been trying to find this integral for a while and I just can't. Does it even have a closed form?
| Mathematica gives:
$\frac{\sqrt{\frac{x^2+x-1}{x^2-1}} \left((x+1)
\left(x^2+x-1\right)+\frac{\sqrt{\frac{x+1}{1-x}} \sqrt{\frac{2 x+\sqrt{5}+1}{1-x}}
(x-1)^2 \left(\sqrt{\frac{1+\sqrt{5}}{x-1}+2 \left(2+\sqrt{5}\right)} \left(\left(14-6
\sqrt{5}\right) F\left(\sin ^{-1}\left(\sqrt{-2+\sqrt{5}}
\sqrt{\frac{x+1}{1-x}}\right)|-9-4 \sqrt{5}\right)+\left(5 \sqrt{5}-11\right)
E\left(\sin ^{-1}\left(\sqrt{-2+\sqrt{5}} \sqrt{\frac{x+1}{1-x}}\right)|-9-4
\sqrt{5}\right)+2 \left(\sqrt{5}-3\right) \Pi \left(-2-\sqrt{5};\sin
^{-1}\left(\sqrt{-2+\sqrt{5}} \sqrt{\frac{x+1}{1-x}}\right)|-9-4
\sqrt{5}\right)\right)-2 \left(\sqrt{5}-1\right) \sqrt{\frac{\sqrt{5} x+x-2}{x-1}}
\left(F\left(\sin ^{-1}\left(\sqrt{-2+\sqrt{5}} \sqrt{\frac{x+1}{1-x}}\right)|-9-4
\sqrt{5}\right)-2 \Pi \left(-2-\sqrt{5};\sin ^{-1}\left(\sqrt{-2+\sqrt{5}}
\sqrt{\frac{x+1}{1-x}}\right)|-9-4 \sqrt{5}\right)\right)\right)}{2
\left(\sqrt{5}-3\right)}\right)}{x^2+x-1}$
| {
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Is $|9x-1|^3$ differentiable at $1/9$? I think it is, but I'm stuck at showing it. I know that it is equal to:
$$|9x - 1| \cdot |9x - 1|^2.$$
$|9x - 1|$ is not differentiable at $1/9$, and I am pretty sure that $|9x - 1| ^2$ is differentiable there.
| We can look separately at the limits of $\frac{f(x)-f(1/9)}{x-1/9}$ (for $x\neq 1/9$) on both sides of $1/9$: the goal is to see if they exist, and (if so) if they are the same. If this is the case, then $f$ is differentiable at $1/9$, and $f^\prime(1/9)$ equals this common limit.
See below for more details. The hidden parts can be revealed by placing your mouse over them.
Start by factorizing $(9x-1)^3$ to make a factor $x-\frac{1}{9}$ appear:
$$(9x-1)^3 = 9\left(x-\frac{1}{9}\right)(9x-1)^2$$
*
*For $x > \frac{1}{9}$, $\lvert 9x-1\rvert^3 = (9x-1)^3$, so that
$$\frac{f(x) - f(\frac{1}{9})}{x-\frac{1}{9}} = \frac{(9x-1)^3-0}{x-\frac{1}{9}} = 9(9x-1)^2 \xrightarrow[x\to1/9^+]{} 0$$
*For $x < \frac{1}{9}$, $\lvert 9x-1\rvert^3 = (1-9x)^3 = -(9x-1)^3$, so that
$$\frac{f(x) - f(\frac{1}{9})}{x-\frac{1}{9}} = -\frac{(9x-1)^3-0}{x-\frac{1}{9}} = -9(9x-1)^2 \xrightarrow[x\to1/9^-]{} 0$$
Therefore, the limits on both sides exist and coincide: $f^\prime(1/9)=\lim_{x\to1/9}\frac{f(x) - f(\frac{1}{9})}{x-\frac{1}{9}}$ thus exists.
| {
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Show that $\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}+\frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}+\frac{c^2}{\sqrt{(1+c^3)(1+a^3)}} \geq \frac{4}{3}$
For positive real numbers $a,b,c$ with $abc = 8$ prove that
$$
\frac{a^2}{\sqrt{(1+a^3)(1+b^3)}}
+ \frac{b^2}{\sqrt{(1+b^3)(1+c^3)}}
+ \frac{c^2}{\sqrt{(1+c^3)(1+a^3)}}
\geq \frac{4}{3}.
$$
Can we prove this by Cauchy-Schwarz or Jensen's inequality? If not how?
| By AM-GM $\sqrt{a^3+1}\leq\frac{a+1+a^2-a+1}{2}=\frac{a^2+2}{2}$.
Hence, it remains to prove that $\sum\limits_{cyc}\frac{a^2}{(a^2+2)(b^2+2)}\geq\frac{1}{3}$, which is
$\sum\limits_{cyc}(a^2b^2+2a^2)\geq72$, which is AM-GM again.
| {
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If $a^2 + b^2 = 1$, show there is $t$ such that $a = \frac{1 - t^2}{1 + t^2}$ and $b = \frac{2t}{1 + t^2}$ My question is how we can prove the following:
If $a^2+b^2=1$, then there is $t$ such that $$a=\frac{1-t^2}{1+t^2} \quad \text{and} \quad b=\frac{2t}{1+t^2}$$
| Since $x^2 + y^2 = 1$ draws a circle on the Euclidean plane, we have $(x, y) = (\cos \theta , \sin \theta)$ for some $\theta$. Now let $\tan (\theta/2) = t$, then $\tan \theta = \dfrac{2t}{1-t^2}$, $\cos\theta = \dfrac{1-t^2}{1+t^2}$, $\sin\theta = \dfrac{2t}{1+t^2}$ by trigonometry.
Another way to see this :
Let $(0, u)$ be a point on $y$-axis, and consider a line joining $(0, u)$ and $(1, 0)$. This line and the circle meet twice: once at $(1, 0)$, and by calculation the another point is at $\left(\dfrac{ u^2 - 1}{u^2 + 1}, \dfrac{2u}{ u^2 + 1}\right)$. Consider $t = 1/u$ (with $u\ne0$) then you have the wanted parametrization. You can recover the case of $t = 0$ and $t = \infty$ easily.
| {
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Geometric progression in an inequality Problem: Show that if $a>0$ and $n>3$ is an integer then $$\frac{1+a+a^2 \cdots +a^n}{a^2+a^3+ \cdots a^{n-2}} \geq \frac{n+1}{n-3}$$
I am unable to prove the above the inequality.
I used the geometric progression summation formula to reduce it to proving $\frac{a^{n+1}-1}{a^2(a^{n-3}-1)} \geq \frac{n+1}{n-3}$.
Also writing it as $$\frac{1+a+a^2 \cdots +a^n}{n+1} \geq \frac{a^2+a^3+ \cdots a^{n-2}}{n-3}$$
seems to suggest that some results on mean-inequalities can be used but I can't figure out what that is.
| In fact, that identity is useful. In details, by AM-GM inequality, $$ \left(n-3\right)\left(1+a^n\right) \ge 2\left(a^2+\cdots+a^{n-2}\right) $$ and $$ \left(n-3\right)\left(a+a^{n-1}\right) \ge 2\left(a^2+\cdots+a^{n-2}\right) $$ Thus, $$ LHS = \frac{1+a+a^{n-1}+a^{n}}{a^2+\cdots+a^{n-2}}+1 \ge \frac{4}{n-3}+1 =RHS $$
| {
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Finding the no. of non-negative integral solutions to $x+y+2z=33$.
Finding the no. of non negative integral solutions to $x+y+2z=33$.
I tried many pure combinatorial approaches (I don't like making individual cases, too long). But they went in vain. I was just pondering, when randomly, a thought came to my mind.
$$x+y+2z=33$$
$$x+2y+z=33$$
$$2x+y+z=33$$
All would have the same number of solutions (integral). So if we add them up, then also, the solutions would remain the same.
$$4x+4y+4z=99$$
$$x+y+z=24.75$$
Taking the floor of the right side (random), we get $x+y+z=24$.
Then, the number of solutions would be ${26\choose2}$. But it is incorrect. So, what can I employ.
| Clearly the number of non-negative integral solutions to the equations $x + y + 2z = 33$ is same as the number of similar solutions to $i + j + 2k = 33$. This is obviously equal to the coefficient of $x^{33}$ in the expansion $$(1 + x + x^{2} + \cdots + x^{i} + \cdots)(1 + x + x^{2} + \cdots + x^{j} + \cdots)(1 + x^{2} + \cdots + x^{2k} + \cdots)$$ which is the same as the coefficient of $x^{33}$ in $$\frac{1}{(1 - x)^{2}}\cdot\frac{1}{1 - x^{2}} = (1 + 2x + 3x^{2} + 4x^{3} + \cdots)(1 + x^{2} + x^{4} + x^{6} + \cdots)$$ and this coefficient is given by $$2 + 4 + 6 + \cdots + 32 + 34 = ((2 + 34)/2) \cdot 17 = 306$$
| {
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"timestamp": "2023-03-29T00:00:00",
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inequality with three variables and condition If $a$,$b$ and $c$ positive real numbers such that $a+b+c=1$, prove $\dfrac{b^2}{a+b^2}+\dfrac{c^2}{b+c^2}+\dfrac{a^2}{c+a^2} \geqslant \dfrac{3}{4}$. I have tried several methods to solve this,but can't get any result. Any idea?
| By C-S and Vasc we obtain $\sum\limits_{cyc}\frac{a^2}{a^2+c}=\sum\limits_{cyc}\frac{a^4}{a^4+a^2c(a+b+c)}\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^3c+a^2b^2+a^2bc)}\geq$
$\geq\frac{(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(a^4+a^2b^2+a^2bc)+\frac{(a^2+b^2+c^2)^2}{3}}=\frac{3(a^2+b^2+c^2)^2}{\sum\limits_{cyc}(4a^4+5a^2b^2+3a^2bc)}$.
Id est, it remains to prove that $4(a^2+b^2+c^2)^2\geq\sum\limits_{cyc}(4a^4+5a^2b^2+3a^2bc)$, which is $\sum\limits_{cyc}c^2(a-b)^2\geq0$. Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{b+2c}+\frac{b}{c+2a}+\frac{c}{a+2b} \geq 1$
For three positive real numbers $a,b,$ and $c$, prove that $$\dfrac{a}{b+2c}+\dfrac{b}{c+2a}+\dfrac{c}{a+2b} \geq 1.$$
Attempt
Rewritting we obtain $\dfrac{2 a^3+2 a^2 b-3 a^2 c-3 a b^2-3 a b c+2 a c^2+2 b^3+2 b^2 c-3 b c^2+2 c^3}{(a+2b)(2a+c)(b+2c)} \geq 0$. Then to I proveed to use rearrangement, AM-GM, etc. on the numerator?
| Very similar to Nesbitt's inequality. If we set $A=b+2c,B=c+2a,C=a+2b$, we have $ 4A+B-2C = 9c $ and so on, and the original inequality can be written as:
$$ \frac{4B+C-2A}{9A}+\frac{4C+A-2B}{9B}+\frac{4A+B-2C}{9C} \geq 1 $$
or:
$$ \frac{4B+C}{A}+\frac{4C+A}{B}+\frac{4A+B}{C} \geq 15 $$
that follows from combining $\frac{B}{A}+\frac{A}{B}\geq 2$ (consequence of the AM-GM inequality) with $\frac{B}{A}+\frac{C}{B}+\frac{A}{C}\geq 3$ (consequence of the AM-GM inequality again).
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \leq \frac{3}{2}$
Let $a,b,$ and $c$ be positive real numbers with $ab+bc+ca = 1$. Prove that $$\dfrac{a}{\sqrt{a^2+1}}+\dfrac{b}{\sqrt{b^2+1}}+\dfrac{c}{\sqrt{c^2+1}} \leq \dfrac{3}{2}$$
Attempt
The $ab+bc+ca = 1$ condition reminds of the rearrangement inequality. Thus, I would say that $a^2+b^2+c^2 \geq ab+bc+ca = 1$ then rewrite the given inequality as $4(a+b+c)^2 = 4(a^2+b^2+c^2) + 8(ab+bc+ca) = 4(a^2+b^2+c^2) \leq 9(a^2+1)(b^2+1)(c^2+1)$
I don't know what to do next.
| See my previous answer here: How prove this inequality $\sum\limits_{cyc}\sqrt{\frac{yz}{x^2+2016}}\le\frac{3}{2}$
For an elegant solution, apply the so-called Purkiss Principle.
| {
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How to compute $\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$? I have a problem with this limit, I don't know what method to use. I have no idea how to compute it. Is it possible to compute this limit with the McLaurin expansion? Can you explain the method and the steps used? Thanks. (I prefer to avoid to use L'Hospital's rule.)
$$\lim _{x\to 0}\frac{x\bigl(\sqrt{3e^x+e^{3x^2}}-2\bigr)}{4-(\cos x+1)^2}$$
| Let's use the elementary techniques to solve the simple limit
\begin{align}
L &= \lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{4 - (1 + \cos x)^{2}}\notag\\
&= \lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{4 - 4\cos^{4}(x/2)}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{1 - \cos^{4}(x/2)}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{1 - \cos(x/2)}{1 - \cos^{4}(x/2)}\cdot\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{1 - \cos(x/2)}\notag\\
&= \frac{1}{4}\lim_{t \to 1}\frac{t - 1}{t^{4} - 1}\cdot\lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{1 - \cos(x/2)}\text{ (putting }t = \cos(x/2))\notag\\
&= \frac{1}{16}\lim_{x \to 0}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{1 - \cos(x/2)}\notag\\
&= \frac{1}{16}\lim_{x \to 0}\frac{(x/2)^{2}}{1 - \cos(x/2)}\frac{x(\sqrt{3e^{x} + e^{3x^{2}}} - 2)}{(x/2)^{2}}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{\sqrt{3e^{x} + e^{3x^{2}}} - 2}{x}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{(3e^{x} + e^{3x^{2}}) - 4}{x(\sqrt{3e^{x} + e^{3x^{2}}} + 2)}\notag\\
&= \frac{1}{8}\lim_{x \to 0}\frac{(3e^{x} + e^{3x^{2}}) - 4}{x}\notag\\
&= \frac{1}{8}\left(3\lim_{x \to 0}\frac{e^{x} - 1}{x} + \lim_{x \to 0}\frac{e^{3x^{2}} - 1}{3x^{2}}\cdot 3x\right)\notag\\
&= \frac{1}{8}(3\cdot 1 + 1\cdot 3\cdot 0)\notag\\
&= \frac{3}{8}\notag
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Trigonometry Identity (Proof): $ \sin^4\theta +\cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $
Question: Prove that $$ \sin^4\theta +\cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$
What I have attempted (Usually I start of with the complex side)
So starting with the LHS
$$ \sin^4\theta +\cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$
$$ (\sin^2\theta)^2 + \cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$
$$ (1-\cos^2\theta)^2 + \cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$
$$ (1-\cos^2\theta)(1-\cos^2\theta) + \cos^4 \theta =1-2\sin^2 \theta \cos^2 \theta $$
$$ 1 - 2\cos^2\theta + \cos^4\theta + \cos^4\theta =1-2\sin^2 \theta \cos^2 \theta $$
$$ 1 - 2\cos^2\theta + 2\cos^4\theta =1-2\sin^2 \theta \cos^2 \theta $$
Now I am stuck... is my approach correct?
| Alternatively, you may just square the identity
$$
\cos^2\theta +\sin^2 \theta=1
$$
giving
$$
\cos^4\theta+2\cos^2\theta\sin^2\theta +\sin^4 \theta=1.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Modular arithmetic problem (mod $22$) $$\large29^{2013^{2014}} - 3^{2013^{2014}}\pmod{22}$$
I am practicing for my exam and I can solve almost all problem, but this type of problem is very hard to me. In this case, I have to compute this by modulo $22$.
| This number is obviously even so let's look at it modulo $11$ :
$$29^{2013^{2014}}-3^{2013^{2014}} \equiv 7^{2013^{2014}}-3^{2013^{2014}} \pmod{11}$$
Now look at the powers $7^x$ modulo $11$ and notice that : $$7^{10} \equiv 1 \pmod{11}$$ (this follows also from Fermat's little theorem )
So we need to look at $2013^{2014} \pmod{10}$ .
Use the same method $3^4 \equiv 1 \pmod{10}$ so :
$$2013^{2014} \equiv 3^{2014} \equiv 3^{2012} \cdot 3^2 \equiv 1 \cdot 9 \equiv 9 \pmod{10}$$
Putting them together :
$$7^{2013^{2014}} \equiv 7^9 \equiv 7^{-1} \equiv 8 \pmod{11}$$
We can proceed similarly for the other term because $3^{10} \equiv 1 \pmod{11}$ :
$$3^{2013^{2014}} \equiv 3^9 \equiv 3^{-1} \equiv 4 \pmod{11}$$
This means that :
$$29^{2013^{2014}}-3^{2013^{2014}} \equiv 8-4 \equiv 4 \pmod{11}$$
This number is even so modulo $22$ :
$$29^{2013^{2014}}-3^{2013^{2014}} \equiv 4 \pmod{22}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove if $2|x^{2} - 1$ then $8|x^{2} - 1$ I have seen this question posted before but my question is in the way I proved it. My books tells us to recall we have proven
if $2|x^{2} - 1$ then $4|x^{2} - 1$
Using this and the fact $x^{2} - 1 = (x+1)(x-1)$ and a previous proof in which we have shown if$a|b$ and $c|d$ then $ac|bd$ my proof is as follows. Assume $2|x^{2} - 1$ and $4|x^2 - 1$ then $2|(x-1)(x+1)$ and $4|(x+1)(x-1)$. Assume $2|x-1$ and $4|x+1$ then $(2)(4)|(x+1)(x-1) = 8|x^{2} - 1$. Would this proof be considered valid? I rewrote it and used some other known proofs to help me out but I dont know if by rewriting it, I have proven something else something not originally asked.
| If $2\vert x^2-1$, then $x$ must be odd, so write $x=2y+1, y\in\mathbb{Z}$.
Then $x^2-1=4y^2+4y=8\frac{y(y+1)}{2}$.
Note that at least one of $y,y+1$ is even, so the fraction on the right is always an integer.
Thus we have that $8\vert x^2-1$.
| {
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In $\triangle ABC,a,b,c$ are the sides of triangle satisfying $a^4+b^4+c^4-4abc+1=0$.Find $\frac{a^2+b^2+c^2}{S}$ In $\triangle ABC,a,b,c$ are the sides of triangle satisfying $a^4+b^4+c^4-4abc+1=0$
Find the value of $\frac{a^2+b^2+c^2}{S}$,where $S$ is area of the triangle $ABC$and find the value of $1+\frac{R}{r}$ where $R$ is the circumradius and $r$ is the inradius of the triangle $ABC$
My attempt:$a^4+b^4+c^4-4abc+1=0$
I expanded $(a+b+c)^4=a^4+4a^3b+4a^3c+6a^2b^2+6a^2c^2+12a^2bc+4ab^3+4ac^3+12ab^2c+12abc^2+4bc^3+4b^3c+6b^2c^2+b^4+c^4$
But this expression has got complicated and not seeming helpful and i do not know any other method to solve this question.
| HINT: use that $$\cos(A)+\cos(B)+\cos(C)=1+\frac{r}{R}$$ holds
| {
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Simplifying nested square roots ($\sqrt{6-4\sqrt{2}} + \sqrt{2}$) I guess I learned it many years ago at school, but I must have forgotten it. From a geometry puzzle I got to the solution
$\sqrt{6-4\sqrt{2}} + \sqrt{2}$
My calculator tells me that (within its precision) the result equals exactly 2, but I have no idea how to transform the calculation to symbolically get to that result.
(I can factor out one $\sqrt{2}$ from both terms, but that does not lead me anywhere, either)
| Start by trying to simplify $\sqrt{6-4\sqrt{2}}$. Let's assume there is some number $p+q\sqrt{2}$ for which
$$\sqrt{6-4\sqrt{2}} = p + q\sqrt{2}$$
Squaring both sides gives
$$6-4\sqrt{2} = (p + q\sqrt{2})^2 = p^2+2q^2 + 2pq\sqrt{2}$$
Comparing coefficients gives $6=p^2+2q^2$ and $-4=2pq$, i.e. $-2=pq$.
We need to solve $p^2+2q^2=6$ and $pq = -2$ simultaneously.
If $pq=-2$ then $q=-\frac{2}{p}$ and we can substitute this into $p^2+2q^2=6$. We get
\begin{eqnarray*}
p^2+2q^2 &=& 6 \\ \\
p^2 + 2\left(-\frac{2}{p}\right)^2 &=& 6 \\ \\
p^2 + \frac{8}{p^2} &=& 6 \\ \\
p^4+8 &=& 6p^2 \\ \\
p^4-6p^2+8 &=& 0 \\ \\
(p^2-2)(p^2-4) &=& 0
\end{eqnarray*}
Either $p^2-2=0$ or $p^2-4=0$, i.e. $p=\pm\sqrt{2}$ or $p=\pm 2$. The only valid solutions are $p = \pm2$ because we usually assume that $p$ and $q$ are rational numbers, i.e. fractions.
If $p=\pm 2$ then $pq=2$ gives $\pm 2q=-2$, and so $q=\mp 1$. Hence
$$p+q\sqrt{2} = \pm(2-\sqrt{2})$$
Recall that $\sqrt{6-4\sqrt{2}} = p + q\sqrt{2}$ and since, by definition, $\sqrt{6-4\sqrt{2}} \ge 0$ we conclude
$$\sqrt{6-4\sqrt{2}} = 2-\sqrt{2}$$
Finally:
$$\sqrt{6-4\sqrt{2}} \ \ {\color{red}{+\sqrt{2}}}= 2-\sqrt{2} \ \ {\color{red}{+\sqrt{2}}} = 2$$
| {
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Prove $\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b}{a+c}+\frac{b+c}{b+a}+\frac{c+a}{c+b}.$
Prove that for all positive real numbers $a,b,$ and $c$, we have $$\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}.$$
What I tried is saying $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} = \dfrac{a^2c+b^2a+c^2b}{abc} \geq \dfrac{3abc}{abc} = 3$. Then how can I use this to prove that $\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a} \geq \dfrac{a+b}{a+c}+\dfrac{b+c}{b+a}+\dfrac{c+a}{c+b}$?
| Assume $$\dfrac{a}{b}=x,\dfrac{b}{c}=y,\dfrac{c}{a}=z$$
So for instance
$$\dfrac{a+c}{b+c}=\dfrac{1+xy}{1+x}=x+\dfrac{1-x}{1+y}$$
And the problem would be transformed to:
$$\dfrac{x-1}{y+1}+\dfrac{y-1}{z+1}+\dfrac{z-1}{x+1}\ge0$$
$\equiv(x^2-1)(z+1)+(y^2-1)(x+1)+(z^2-1)(y+1)\ge0$
$\equiv \sum{x^2z}+\sum{x^2}\ge\sum{x}+3$
We have $xyz=1$, hence :$$\sum{x^2z}\ge3$$
And also $x+y+z\ge3$, so $$\sum{x^2}\ge\dfrac{(\sum{x})^2}{3}\ge\sum{x}$$
Problem solved
| {
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Prime factorization of Gaussian integers
I want to find $a, b\in\mathbb{Z}[i]$ such that $a(2+3i)+b(5+5i)=1$.
I don't know how to do this, but my first thought was to do something with the norm or otherwise factoring $5+5i=(2+i)(2-i)(1+i)$, but I don't see how to use this further.
The second question is for which $x\in\mathbb{Z}[i]$ we have $x=1\bmod (2-+3i)$ and $x=-1\bmod(5+5i)$? But I would like to try and solve this myself as soon as I have the first question.
Edit: I will attempt to solve this using the Euclidean algorithm as stated in the comments. I will be back later with more! (Although I would not mind if someone beat me to it) ;)
| Using the Extended Euclidean Algorithm as implemented in this answer, modified for Gaussian integers, we get
$$
\begin{array}{r}
&&2&-1+2i&1-i\\\hline
1&0&1&1-2i&2+3i\\
0&1&-2&-1+4i&-5-5i\\
5+5i&2+3i&1-i&1&0
\end{array}\tag{1}
$$
This says that
$$
(1-2i+(2+3i)k)\color{#C00000}{(5+5i)}+(-1+4i-(5+5i)k)\color{#C00000}{(2+3i)}=1\tag{2}
$$
where $k$ is a Gaussian integer. For example, $k=i$ gives the solution
$$
\underbrace{(-2)\color{#C00000}{(5+5i)}}_{-10-10i}+\underbrace{(4-i)\color{#C00000}{(2+3i)}}_{11+10i}=1\tag{3}
$$
Since $-10-10i=(-2)(5+5i)$, we have
$$
\begin{align}
-10-10i&\equiv1\pmod{2+3i}\\
-10-10i&\equiv0\pmod{5+5i}
\end{align}\tag{4}
$$
and since $11+10i=(4-i)(2+3i)$
$$
\begin{align}
11+10i&\equiv0\pmod{2+3i}\\
11+10i&\equiv1\pmod{5+5i}
\end{align}\tag{5}
$$
Adding $a$ times $(4)$ and $b$ times $(5)$ gives a solution to
$$
\begin{align}
x&\equiv a\pmod{2+3i}\\
x&\equiv b\pmod{5+5i}
\end{align}\tag{6}
$$
Applying $(6)$ to the particular question above
$$
\begin{align}
-21-20i&\equiv \phantom{-}1\pmod{2+3i}\\
-21-20i&\equiv -1\pmod{5+5i}
\end{align}\tag{7}
$$
Since the solution in $(7)$ is given mod $(2+3i)(5+5i)=-5+25i$, adding $(1-i)(-5+25i)=20+30i$, we also get the solution
$$
\begin{align}
-1+10i&\equiv \phantom{-}1\pmod{2+3i}\\
-1+10i&\equiv -1\pmod{5+5i}
\end{align}\tag{8}
$$
| {
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If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to If $(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$ then the expression $x^2+y^2-10x-6y+34$ CANNOT be equal to
$(A)\frac{1}{2}\hspace{1cm}(B)8\hspace{1cm}(C)2\hspace{1cm}(D)3$
$(x,y)$ satisfies $x^2+y^2-4x+2y+1=0$
$(x,y)$ satisfies $(x-2)^2+(y+1)^2=2^2$
The expression $x^2+y^2-10x-6y+34$ can be written as $(x-5)^2+(y-3)^2$
But i do not know how to further solve it.Please help.
| Two circles have at least one common point only if the distance between their centers is equal to or smaller than the sum of their radii.
All you need to do is rewrite equations as circles, extract the center, the radius of each circle and check.
$$(x-2)^2+(y+1)^2=4, a_{0}=2, b_{0}=-1, r_{0}=2$$
$$(x-5)^2+(y-3)^2=\frac{1}{2}, a_{1}=5, b_{1}=3, r_{1}=\frac{\sqrt{2}}{2}$$
$$(x-5)^2+(y-3)^2=8, a_{2}=5, b_{2}=3, r_{2}=2\sqrt{2}$$
$$(x-5)^2+(y-3)^2=2, a_{3}=5, b_{3}=3, r_{3}=\sqrt{2}$$
$$(x-5)^2+(y-3)^2=3, a_{4}=5, b_{4}=3, r_{4}=\sqrt{3}$$
For example you need to prove:
$$\sqrt{(5-2)^2+(3+1)^2}=5>2+\frac{\sqrt{2}}{2}$$
$$5>2+2\sqrt{2}$$
$$5>2+\sqrt{2}$$
$$5>2+\sqrt{3}$$
All of them pretty much obvious.
| {
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Real values of $x$ satisfying the equation $x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$
Real values of $x$ satisfying the equation $$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0$$
We can write it as $$512x^9+576x^6+216x^3-512x+219=0$$
I did not understand how can i factorise it.
Help me
| $\bf{I\; have \;Solved\; Like \;This \; Way}$
$$x^9+\frac{9}{8}x^6+\frac{27}{64}x^3-x+\frac{219}{512} =0\Rightarrow 512x^9+(9\cdot 64)x^6+(27\cdot 8)x^3-512x+219=0$$
So $$\underbrace{(8x^3)^3+3(8x^3)^2\cdot 3+3(3^2)\cdot 8x^3+3^3}-512x+219-3^3=0$$
So $$(8x^3+3)^3=512x-219\Rightarrow (8x^3+3)^3=512\left(x-\frac{192}{512}\right)=8^3\left(x-\frac{3}{8}\right)^{\frac{1}{3}}.$$
So $$8^3\left(x^3+\frac{3}{8}\right) = 8\left(x-\frac{3}{8}\right)^{\frac{1}{3}}\Rightarrow x^3+\frac{3}{8} = \left(x-\frac{3}{8}\right)^{\frac{1}{3}}$$
Now Let $\displaystyle f(x)=x^3+\frac{3}{8},$ Where $f:\mathbb{R}\rightarrow \mathbb{R}\;,$ Then $\displaystyle f^{-1}(x) = \left(x-\frac{3}{8}\right)^{\frac{1}{3}}\;,$ Where $f:\mathbb{R}\rightarrow \mathbb{R}\;$
So We have To solve $$f(x) = f^{-1}(x)$$
Now We now that $f(x)$ and $f^{-1}(x)$ is Symmetrical about $y=x$ line.
So $$f(x) = f^{-1}(x) =x$$
So $$x^3+\frac{3}{8}=x\Rightarrow 8x^3-8x+3=0$$
So $$(2x-1)\left[4x^2+2x-3\right]=0\Rightarrow x=\frac{1}{2}\;\;,x=\frac{-1\pm \sqrt{13}}{4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617737",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Substitution in a geometric series: $\sum_{n=0}^{N} x^n = \frac{x^{N+1} -1}{x-1}$ To find a formula for the sums of square of first n natural no, the following method is applied in generating functionology, we know
$$\sum_{n=0}^{N} x^n = \frac{x^{N+1} -1}{x-1}$$
So we apply the $\left\lbrace xD \right\rbrace^2$ operator on it and put $ x = 1$ to get the corresponding formula.
The $\left\lbrace xD \right\rbrace^2$ of right side I found as
$$\frac{(N+1)^2x^{N+1}}{x-1} - \frac{(N+1)x^{N+2}}{(x-1)^2} - \frac{(N+2)x^{N+2}-x}{(x-1)^2} + \frac{2x(x^{N+2} -x)}{(x-1)^3}$$ Here after substituting $x = 1$, however the denominator are all $0$ .How to eliminate it to get the correct formula i.e $ \frac{N (N+1) (2N +1)} {6}$
| Hint. I would set $$\epsilon:=x-1$$ then, as $\epsilon \to 0$, use the binomial theorem to get
$$
x^N=(1+\epsilon)^N=1+N\epsilon+\frac{1}{2} N(N-1)\:\epsilon^2+\frac{1}{6} N(N-1)(N-2)\:\epsilon^3+o(\epsilon^3). \tag1
$$ Inserting $(1)$ in your identity gives the sought result.
In fact, by applying $\left\lbrace (1+\epsilon)D_\epsilon \right\rbrace^2$ to
$$
\sum_{n=1}^{N} (1+\epsilon)^n=\frac{(1+\epsilon )^{N+1}-1}{\epsilon }, \quad \epsilon\neq0,\tag2
$$ you get, for $\epsilon\neq0$,
$$
\sum_{n=1}^{N} n^2\times (1+\epsilon)^n=-\frac{2+3 \epsilon +\epsilon ^2-(1+\epsilon )^{N+1} \left(2+\epsilon \left(3-2 (N+1)+N^2 \epsilon \right)\right)}{\epsilon ^3}.\tag3
$$ Then inserting $(1)$ into $(3)$ leads to
$$
\sum_{n=1}^{N} n^2\times (1+\epsilon)^n=\frac{\frac{N(N+1)(2N +1)}{6}\epsilon^3+o(\epsilon^3)}{\epsilon^3}=\frac{N(N+1)(2N +1)}{6}+o(1)
$$ and letting $\epsilon \to 0$ gives the announced result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Not getting the same solution when using the rule sin(x)\x=1 on a limit There is a rule in limits that when $x$ approaches zero:
$$\frac{\sin\left(x\right)}{x}=1$$
So I used this rule on the following exercise:
Evaluate
$$
\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin\left(2x\right)-\tan\left(2x\right)}
$$
I substituted $\sin(2x)$ with $2x$ by the following way:
$$\sin\left(2x\right)=\frac{\sin\left(2x\right)}{2x}\cdot 2x=1\cdot 2x=2x \Rightarrow \lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{2x-\tan\left(2x\right)}$$
But according to symbolab:
$$\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{2x-\tan\left(2x\right)}=-\frac{1}{16}$$
while
$$\lim _{x\to 0}\:\frac{x-\sin\left(x\right)}{\sin(2x)-\tan\left(2x\right)}=-\frac{1}{24}$$
Why am I getting this contradiction?
More over if I susbsitute the following I do get the right answer
$$\tan\left(2x\right)=\frac{\sin\left(2x\right)}{\cos\left(2x\right)}=\frac{2x}{\cos\left(2x\right)}\Rightarrow \lim \:_{x\to \:0}\:\frac{x-\sin\left(x\right)}{2x-\frac{2x}{\cos\left(2x\right)}}=-\frac{1}{24}$$
If you want to test yourself: Symbolab with the excersice preloaded
| I think the limit notation has been made so confusing thanks to various educators (including book authors) who try to simplify it too much. When you say $$\lim_{x \to 0}\frac{\sin x}{x} = 1 \tag{1}$$ you don't mean $(\sin x)/x = 1$ when $x \to 0$. Hence it is just not possible to replace $\sin x$ by $x$. However whenever you see the expression $\lim_{x \to 0}\dfrac{\sin x}{x}$ you can replace it by $1$ (because of the equation $(1)$ above).
You can always replace a thing called $A$ by a thing called $B$ when you know that $A$ and $B$ are equal but if $A \neq B$ then you can't replace $A$ by $B$. Unfortunately people are so used to invalid replacements while calculating limits that they don't notice that it is wrong.
The way to solve your limit problem is as follows:
\begin{align}
L &= \lim_{x \to 0}\frac{x - \sin x}{\sin 2x - \tan 2x}\notag\\
&= \lim_{x \to 0}\dfrac{x - \sin x}{\sin 2x - \dfrac{\sin 2x}{\cos 2x}}\notag\\
&= \lim_{x \to 0}\dfrac{x - \sin x}{\sin 2x}\cdot\frac{\cos 2x}{\cos 2x - 1}\notag\\
&= \lim_{x \to 0}\dfrac{x - \sin x}{\dfrac{\sin 2x}{2x}\cdot 2x}\cdot\frac{\cos 2x}{\cos 2x - 1}\notag\\
&= \lim_{x \to 0}\dfrac{x - \sin x}{2x(\cos 2x - 1)}\cdot\dfrac{1}{\lim_\limits{x \to 0}\dfrac{\sin 2x}{2x}}\cdot \lim_{x \to 0}\cos 2x\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{x - \sin x}{x(\cos 2x - 1)}\cdot 1\cdot 1\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{\sin x - x}{2x\sin^{2}x}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\frac{x^{2}}{\sin^{2}x}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot\lim_{x \to 0}\frac{x}{\sin x}\cdot\lim_{x \to 0}\frac{x}{\sin x}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\cdot 1\cdot 1\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\sin x - x}{x^{3}}\notag\\
&= \frac{1}{4}\lim_{x \to 0}\frac{\cos x - 1}{3x^{2}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{12}\lim_{x \to 0}\frac{\cos^{2}x - 1}{x^{2}(1 + \cos x)}\notag\\
&= -\frac{1}{12}\lim_{x \to 0}\frac{\sin^{2}x}{x^{2}}\cdot\lim_{x \to 0}\frac{1}{1 + \cos x}\notag\\
&= -\frac{1}{12}\cdot 1\cdot\frac{1}{2}\notag\\
&= -\frac{1}{24}\notag
\end{align}
The above solution has been presented in a slightly more detailed fashion (than usually necessary for an exam) in order to show how the expression $\lim_{x \to 0}\dfrac{\sin x}{x}$ is replaced by $1$ (and how $\lim_{x \to 0}\cos x$ is replaced by $1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
Find the minimum value of $\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
Find the minimum value of
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta$
$a.)\ 1 \ \ \ \ \ \ \ \ \ \ \ \ b.)\ 3 \\
c.)\ 5 \ \ \ \ \ \ \ \ \ \ \ \ d.)\ 7 $
$\sin^{2} \theta +\cos^{2} \theta+\sec^{2} \theta+\csc^{2} \theta+\tan^{2} \theta+\cot^{2} \theta \\
=\sin^{2} \theta +\dfrac{1}{\sin^{2} \theta }+\cos^{2} \theta+\dfrac{1}{\cos^{2} \theta }+\tan^{2} \theta+\dfrac{1}{\tan^{2} \theta } \\
\color{blue}{\text{By using the AM-GM inequlity}} \\
\color{blue}{x+\dfrac{1}{x} \geq 2} \\
=2+2+2=6 $
Which is not in options.
But I am not sure if I can use that $ AM-GM$ inequality in this case.
I look for a short and simple way .
I have studied maths upnto $12$th grade .
| A nice way to simplify this problem, using nothing more than elementary trigonometry and algebra, is this:
As noted in another answer, the given expression simplifies to:
$$2\sec^2\theta +2\csc^2\theta - 1$$
Finding $\theta$ to minimize that expression is the same as finding $\theta$ to minimize
$$\begin{align}
\sec^2\theta+\csc^2\theta &= \frac{1}{\cos^2\theta}+\frac{1}{\sin^2\theta}\\
&=\frac{\sin^2\theta+\cos^2\theta}{\sin^2\theta\cos^2\theta}\\
&=\frac{1}{\sin^2\theta\cos^2\theta},
\end{align}$$
which is the same as finding $\theta$ to maximize
$$\sin^2\theta\cos^2\theta = (\sin\theta\cos\theta)^2,$$
which is the same as finding $\theta$ to maximize
$$|\sin\theta\cos\theta| = \frac12|\sin(2\theta)|.$$
This is clearly maximized when $\sin(2\theta)=\pm 1$, which happens when $\theta$ is $45^\circ$ away from a multiple of $90^\circ$. If you plug in $\theta=\frac{\pi}{4}$, you'll find that minimum value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 5
} |
How to calculate $\int_a^bx^2 dx$ using summation? So for this case, we divide it to $n$ partitions and so the width of each partition is $\frac{b-a}{n}$ and the height is $f(x)$.
\begin{align}
x_0&=a\\
x_1&=a+\frac{b-a}{n}\\
&\ldots\\
x_{i-1}&=a+(i-1)\frac{b-a}{n}\\
x_i&=a+i\frac{b-a}{n}
\end{align}
So I pick left point, which is $x_{i-1}$
I start with
\begin{align}
\sum\limits_{i=1}^{n} \frac{b-a}{n}f\left(a+\frac{(i-1)(b-a)}{n}\right)
&=\frac{b-a}{n}\sum\limits_{i=1}^{n} \left(a+\frac{(i-1)(b-a)}{n}\right)^2\\
&=\frac{b-a}{n}\left(na^2+ \frac{2a(b-a)}{n}\sum\limits_{i=1}^{n}(i-1)+\frac{(b-a)^2}{n^2} \sum\limits_{i=1}^{n}(i-1)^2\right)
\end{align}
Here I am stuck because I don't know what$ \sum\limits_{i=1}^{n}(i-1)^2$is (feel like it diverges). Could someone help?
| It is a well-known formula that
$$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$
So, in your case,
\begin{align*}
\sum_{i=1}^n (i-1)^2 &= 0^2 + 1^2 + \cdots + (n-1)^2 \\
&= 1^2 + \cdots + (n-1)^2 \\ \\
&= \frac{(n-1)n(2n-1)}{6}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Prove that $3(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}) \ge 10 + 8\cdot \frac{a^2+b^2+c^2}{ab+bc+ca}$
For the positive real numbers $a, b, c$ prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\cdot \dfrac{a^2+b^2+c^2}{ab+bc+ca}$$
I did the following:
$$\begin{split}\dfrac{a^2+b^2+c^2}{ab+bc+ca} & = \dfrac{(a+b+c)^2-2(ab+bc+ca)}{ab+bc+ca} \\ & = \dfrac{(a+b+c)^2}{ab+bc+ca} - 2 \\ & \le \dfrac{a}{b} + \dfrac{b}{c} + \dfrac{c}{a} -2\end{split}$$
I got the last one by applying Titu's Lemma
Thus I think it suffices to prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\bigg(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}-2\bigg) \\ \Longrightarrow 3a^2b+3b^2c+3c^2a+6abc \ge 5a^2c+5b^2a+5c^2b$$
But by rearrangement inequality this is not always true I think. The direction of the inequality should have been flipped.
| Note that we must prove that $$3\bigg(\dfrac{a+b}{c}+\dfrac{b+c}{a}+\dfrac{c+a}{b}\bigg) \ge 10 + 8\bigg(\dfrac{a^2+b^2+c^2}{ab+bc+ca}\bigg) \\ $$
Adding $6$ gives us
$$\dfrac{3(a+b)(b+c)(c+a)}{abc} \ge \dfrac{8(a+b+c)^2}{ab+bc+ca}$$
However, since $(a+b)(b+c)(c+a) \ge \dfrac{8}{9}(a+b+c)(ab+bc+ca)$, it is enough to prove that $(ab+bc+ca)^2 \ge 3abc(a+b+c)$, which follows from $(a+b+c)^2 \ge 3ab+3bc+3ca$. Our proof is done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Volume of Cone By Integration I am trying to find the volume of a cone by integrating it in spherical coordinates, but elementary geometry suggests that my approach is incorrect.
The specifications of the cone are $0\le R \le 5$, $0 \le \theta \le \frac{\pi}{3}$, and $0 \le \phi \le 2\pi$. My reasoning was as follows:
The volume element in spherical coordinates is given by: $$dV = R^{2}\sin{\theta}dRd\theta d\phi$$
Simply integrate this over the specified region to obtain the total volume:
\begin{align}
V & = \int_{0}^{2\pi}\int_{0}^{\frac{\pi}{3}}\int_{0}^{5}R^{2}\sin{\theta}dRd\theta d\phi = 2\pi\int_{0}^{\frac{\pi}{3}}\int_{0}^{5}R^{2}\sin{\theta}dRd\theta\\
& =2\pi\int_{0}^{\frac{\pi}{3}}[\frac{R^3}{3}]|_{0}^{5}\sin{\theta}d\theta\\
& =2\pi[\frac{125}{3}]\int_{0}^{\frac{\pi}{3}}\sin{\theta}d\theta = [\frac{125\pi}{3}]
\end{align}
But when I frame the problem in terms of elementary geometry, with the slant height $R = 5$ and using the formula $V = \frac{1}{3}\pi r^2 h$ where $r$ is the radius of the base and $h$ is the height of the cone, I obtain a different answer.
The base radius $r$ and the cone height $h$ should be related to the slant height (which is $5$) by simple trigonometry:
$r = 5 \sin\theta = 5 \sin(\frac{\pi}{3}) = \frac{5\sqrt{3}}{2}$
$h = 5 \cos\theta = 5 \cos(\frac{\pi}{3}) = \frac{5}{2}$
Placing this into the elementary geometry formula $V = \frac{1}{3}\pi r^2 h$ one obtains:
$$V = \frac{1}{3}\pi(\frac{5\sqrt{3}}{2})^{2}(\frac{5}{2})= \frac{1}{3}\pi(\frac{25*3}{4})(\frac{5}{2}) = \frac{125\pi}{8}$$
Which one is correct, if either of them, and why?
| Your problem is simple. The cone you described is an ice cream cone ie a cone with the a spherical cap since $R $ goes from 0 to 5, and it's volume cannot be given by $V=\frac{1}{3}\pi r^2 h $ (which is the volume of a cone without a cap), but by the integration you did.
Draw it and it will become more obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding Eigenvectors of a 3x3 Matrix (7.12-15) Please check my work in finding an eigenbasis (eigenvectors) for the following problem. Some of my solutions do not match answers in my differential equations text (Advanced Engineering Mathematics by Erwin Kreyszig, 1988, John Wiley & Sons).
For reference the following identity is given because some textbooks reverse the formula having $\lambda$ subtract the diagonal elements instead of subtracting $\lambda$ from the diagonal elements:
$$ det(A - \lambda I) = 0 $$
$$
A =
\begin{bmatrix}
3 & 1 & 4 \\
0 & 2 & 6 \\
0 & 0 & 5 \\
\end{bmatrix}
$$
By inspection the eigenvalues are the entries along the diagonal for this upper triangular matrix.
$$
\begin{align*}
\lambda_1 = 3 \qquad \lambda_2 = 2 \qquad \lambda_3 = 5
\end{align*}
$$
When $\lambda_1 = 3$ we have:
$$
A - 3I =
\begin{bmatrix}
3-3 & 1 & 4 \\
0 & 2-3 & 6 \\
0 & 0 & 5-3 \\
\end{bmatrix}
=
\begin{bmatrix}
0 & 1 & 4 \\
0 & -1 & 6 \\
0 & 0 & 2 \\
\end{bmatrix}
=
\begin{bmatrix}
0 & 0 & 0 \\
0 & 1 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
$$
\begin{align*}
x_1 = 1 \: (free \: variable) \qquad x_2 = 0 \qquad x_3 = 0 \\
\end{align*}
$$
$$
v_1 =
\begin{bmatrix}
1 \\
0 \\
0 \\
\end{bmatrix} \qquad (matches \: answer \: in \: text)
$$
When $\lambda_2 = 2$ we have:
$$
A - 2I =
\begin{bmatrix}
3-2 & 1 & 4 \\
0 & 2-2 & 6 \\
0 & 0 & 5-2 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 4 \\
0 & 0 & 6 \\
0 & 0 & 3 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 1 & 0 \\
0 & 0 & 0 \\
0 & 0 & 1 \\
\end{bmatrix}
$$
$$
\begin{align*}
x_1 = -x_2 \qquad x_2 = 1 \: (free \: variable) \qquad x_3 = 0 \\
\end{align*}
$$
$$
v_2 =
\begin{bmatrix}
-1 \\
1 \\
0 \\
\end{bmatrix} \qquad but \: answer \: in \: text \: is \qquad
\begin{bmatrix}
1 \\
-1 \\
0 \\
\end{bmatrix}
$$
What happened? Is it from a disagreement in what we should consider arbitrary or am I doing something fundamentally wrong?
When $\lambda_3 = 5$ we have:
$$
A - 5I =
\begin{bmatrix}
3-5 & 1 & 4 \\
0 & 2-5 & 6 \\
0 & 0 & 5-5 \\
\end{bmatrix}
=
\begin{bmatrix}
2 & 1 & 4 \\
0 & -3 & 6 \\
0 & 0 & 0 \\
\end{bmatrix}
=
\begin{bmatrix}
1 & 0 & -3 \\
0 & 1 & -2 \\
0 & 0 & 0 \\
\end{bmatrix}
$$
$$
\begin{align*}
x_1 = 3x_3 \qquad x_2 = 2x_3 \qquad x_3 = 1 \: (free \: variable) \\
\end{align*}
$$
$$
v_3 =
\begin{bmatrix}
3 \\
2 \\
1 \\
\end{bmatrix} \qquad (matches \: answer \: in \: text)
$$
| Eigenvectors are never unique. In particular, for the eigenvalue $2$ you can take, for example, $x_2=-1$ which gives you the answer in the book.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $\tan A+\tan B+\tan C=6$ and $\tan A\tan B=2 $ in $\triangle ABC$, then find the type of triangle.
In $\triangle ABC$, $\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2
$
Then the triangle is
$a.)\text{Right-angled isosceles} \\
b.) \text{Acute-angled isosceles}\\
\color{green}{c.)\text{Obtuse-angled}} \\
d.)\text{equilateral} $
$\ \ \ $
$\tan A+\tan B+\tan C=6 \\
\tan A\tan B=2 \\
A+B=180-C\\
\dfrac{\tan A+\tan B}{1-\tan A\tan B}=-\tan C\\
\tan A+\tan B=\tan C\\
\tan C=3 \\
\tan A\tan B=2\ \text{and} \ \tan A+\tan B=3 \\
\implies \tan A=2, \tan B=1\ \ \text{or}\ \ \tan A=1, \tan B=2
$
Now I am stucked, I look for a short and simple way.
I have studied maths upto $12$th grade .
Note:- Calculator is not allowed.
| WLOG, $C = \tan^{-1}(3) \approx 72^{\circ}, A = \tan^{-1}(1) = 45^{\circ}, B = \tan^{-1}(2) \approx 63^{\circ}$. This shows that $\triangle ABC$ is scalene !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Series expansion: $1/(1-x)^n$ What is the expansion for $(1-x)^{-n}$?
Could find only the expansion upto the power of $-3$. Is there some general formula?
| I realize this is an old thread, but I wanted to expand on the above answers on how to derive the formula for anyone else that might come along. Starting with the geometric series and taking successive derivatives:
$$
\begin{align}
\dfrac{1}{(1-x)} &= 1+x+x^2+x^3+x^4+x^5\dots+x^m+\dotsm\\
\dfrac{1}{(1-x)^2} &= 1+2x+3x^2+4x^3+5x^4\dots+mx^{m-1}+\dotsm\\
\dfrac{2\cdot 1}{(1-x)^3} &= 2+(3\cdot 2)x+(4\cdot 3)x^2+(5\cdot 4)x^3\dots+(m \cdot m-1)x^{m-2}+\dotsm\\
\vdots\\
\dfrac{(n-1)!}{(1-x)^n} &= \sum_{k=0}^\infty \dfrac{(k+n-1)!}{k!} x^k\\
\end{align}
$$
which can be simplified by dividing:
$$
\dfrac{1}{(1-x)^n} = \sum_{k=0}^\infty \dfrac{(k+n-1)!}{(n-1)!k!} x^k = \sum_{k=0}^\infty \binom{k+n-1}{n-1}x^k\\
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Integration by parts - hint I'm stuck on a passage on my textbook:
$$ \int \frac{1}{(1+t^2)^3} dt = \frac{t}{4(t^2+1)^2}+\frac{3}{4} \int \frac{1}{(t^2+1)^2} dt$$
I know that it should be easy but I just can't figure out what is the product of functions considered in this integration by parts..
can you help me?
thanks a lot
| $$\int \dfrac {1}{(t^2+1)^3}\,dt = \int \dfrac {(1+t^2) - t^2}{(t^2+1)^3}\, dt = \int \dfrac {1}{(t^2+1)^2}\,dt + \dfrac{1}{4}\int t\, d(t^2+1)^{-2}$$$$ = \int \dfrac {1}{(t^2+1)^2}\,dt + \dfrac{1}{4}\dfrac {t}{(t^2+1)^2} - \dfrac{1}{4}\int \dfrac{1}{(t^2+1)^2}\,dt $$$$ = \dfrac{1}{4}\dfrac {t}{(t^2+1)^2} + \dfrac{3}{4}\int \dfrac{1}{(t^2+1)^2}\,dt $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$? How do you solve B and C for $\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1}$ ?
$A = \left.\frac{s^2-s}{s^2+1} \right\vert_{s=-1} = \frac{1-(-1)}{1+1}=1$
| We have
$$
\frac{s-1}{s+1} \frac{s}{s^2+1} = \frac{A}{s+1} + \frac{Bs+C}{s^2+1} \tag1
$$
Multiplying $(1)$ by $s$ and making $s \to \infty$ gives
$$
1=A+B
$$ from which $\color{red}{B=0}$.
Making $s=0$ in $(1)$ gives
$$
0=A+C
$$ from which $\color{red}{C=-1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Proving $\frac{1}{\sqrt{1-x}} \le e^x$ on $[0,1/2]$. Is there a simple way to prove $$\frac{1}{\sqrt{1-x}} \le e^x$$ on $x \in [0,1/2]$?
Some of my observations from plots, etc.:
*
*Equality is attained at $x=0$ and near $x=0.8$.
*The derivative is positive at $x=0$, and zero just after $x=0.5$. [I don't know how to find this zero analytically.]
*I tried to work with Taylor series. I verified with plots that the following is true on $[0,1/2]$:
$$\frac{1}{\sqrt{1-x}} = 1 + \frac{x}{2} + \frac{3x^2}{8} + \frac{3/4}{(1-\xi)^{5/2}} x^3 \le 1 + \frac{x}{2} + \frac{3}{8} x^2 + \frac{5 \sqrt{2} x^3}{6} \le 1 + x + \frac{x^2}{2} + \frac{x^3}{6} \le e^x,$$
but proving the last inequality is a bit messy.
| For our interval, the inequality is equivalent to $1-x\ge e^{-2x}$. (We squared and flipped.)
This inequality can be proved using differential calculus. Let $f(x)=1-x-e^{-2x}$. Then $f'(x)=2e^{-2x}-1$. So $f(x)$ is increasing until $x=\frac{\ln 2}{2}\approx 0.34$ and then decreasing. Thus all we need to do is check its value at $x=1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
The order of pole of $\frac{1}{(2\cos z -2 +z^2)^2} $at $z=0$ What is the order of the pole of :
$$\frac{1}{(2\cos z -2 +z^2)^2}$$
at $z=0$
This is what I did :
$$\cos z = \frac{e^{iz}-e^{-iz}}{2} $$
then :
$$ \implies \frac{1}{(e^{iz}-e^{-iz}+z^2-2)^2} $$
How do I continue?
| Since
$$
\cos z = 1 - \frac12 z^2 + \frac1{4!}z^4 + \cdots
$$
we have
$$
(2\cos z - 2 + z^2)^2 = (\frac1{12}z^4 + \cdots)^2 = \frac{1}{144}z^8 + \cdots
$$
where $\cdots$ denote higher order terms. This shows that your function has an $8$-fold zero in the denominator, so the order of the pole is $8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How do I solve $\int4\cos^2(x) dx$? I have the basic idea of how to work out the integral of a trig function, but am having trouble in applying the concept. Would really appreciate it if someone could help me. Thanks!
| $\int 4 \cos^2(x) dx$
$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$
$=4\int\cos^2(x) dx$
$\mathrm{Use\:the\:following\:identity}:\quad \cos ^2\left(x\right)=\frac{1+\cos \left(2x\right)}{2}$
$=4\int \frac{1+\cos \left(2x\right)}{2}dx$
$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$
$=4\frac{1}{2}\int \:1+\cos \left(2x\right)dx$
$\mathrm{Apply\:Integral\:Substitution:}\:\int
> f\left(g\left(x\right)\right)\cdot g^{'}\left(x\right)dx=\int
> f\left(u\right)du,\:\quad u=g\left(x\right)$
$\mathrm{Substitute:}\:u=2x$
$\frac{du}{dx}=2$
$\quad \Rightarrow \:du=2dx$
$\Rightarrow \:dx=\frac{1}{2}du$
$=\int \left(1+\cos \left(u\right)\right)\frac{1}{2}du$
$=4\frac{1}{2}\int \left(1+\cos \left(u\right)\right)\frac{1}{2}du$
$\mathrm{Take\:the\:constant\:out}:\quad \int a\cdot f\left(x\right)dx=a\cdot \int f\left(x\right)dx$
$=4\frac{1}{2}\frac{1}{2}\int \:1+\cos \left(u\right)du$
$\mathrm{Apply\:the\:Sum\:Rule}:\quad \int f\left(x\right)\pm g\left(x\right)dx=\int f\left(x\right)dx\pm \int g\left(x\right)dx$
$=4\frac{1}{2}\frac{1}{2}\left(\int \:1du+\int \cos \left(u\right)du\right)$
$=4\frac{1}{2}\frac{1}{2}\left(u+\sin \left(u\right)\right)$
$\mathrm{Substitute\:back}\:u=2x$
$=4\frac{1}{2}\frac{1}{2}\left(2x+\sin \left(2x\right)\right)$
$\mathrm{Simplify}$
$=2x+\sin \left(2x\right)$
$Add\:a\:constant\:to\:the\:solution$
$=2x+\sin \left(2x\right)+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to evaluate $\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)$$
| Hint:
\begin{align}
\lim_{x\to\infty}\left(\frac{2x^3+x}{x^2+1}\sin\left(\frac{x+1}{4x^2+3}\right)\right)&=\left(\lim_{x\to\infty}\frac{(2x^3+x)(x+1)}{(x^2+1)(4x^2+3)}\right)\lim_{x\to\infty}\frac{\sin\left(\frac{x+1}{4x^2+3}\right)}{\frac{x+1}{4x^2+3}}\\
\end{align}
Now, by making $\color{blue}{t=\frac{x+1}{4x^2+3}}$, $t\to 0^+$ as $x\to \infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629634",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the value of : $\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$ I saw some resolutions here like $\sqrt{x+\sqrt{x+\sqrt{x}}}- \sqrt{x}$, but I couldn't get the point to find
$\lim_{x\to\infty}\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}$.
I tried $\frac{1}{x}.(\sqrt{x+\sqrt{x+\sqrt{x}}})=\frac{\sqrt{x}}{x}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)=\frac{1}{\sqrt{x}}\left(\sqrt{1+\frac{\sqrt{x+\sqrt{x}}}{x}} \right)$ but now I got stuck. Could anyone help?
| Jonas produced the solution first, but I wanted to write up a solution for my own edification.
For $x>1$ we can write the following inequalities,
$$x < x+a\sqrt{x} < (1+a)x\qquad (a>0),$$
This allows us to come up with an upper and lower bound for the numerator,
$$\sqrt{x+\sqrt{x+\sqrt{x}}} < \sqrt{x+\sqrt{2x}}< \sqrt{(1+\sqrt{2})x} = \sqrt{x}\ \sqrt{(1+\sqrt{2})}$$
$$\sqrt{x+\sqrt{x+\sqrt{x}}} > \sqrt{x+\sqrt{x}} > \sqrt{x}$$
So our function, which we are taking the limit of, has the following bounds on it,
$$\frac{\sqrt{x}}{x} <\frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x}<\frac{\sqrt{x}}{x}\sqrt{1+\sqrt{2}}, $$
as $x\rightarrow \infty$ the function $\sqrt{x}/x$ goes to $0$. Our function is bounded above by something which goes to zero and below by something which goes to zero it must also go to zero; this is called the squeeze theorem.
$$\lim_{x\rightarrow \infty} \frac{\sqrt{x+\sqrt{x+\sqrt{x}}}}{x} = 0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1629846",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
How to evaluate $\lim\limits _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}}\right)^{5x+1}$? I have a problem with this limit, I have no idea how to compute it. Can you explain the method and the steps used(without L'Hopital if is possible)? Thanks
$$\lim _{x\to \infty }\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}}\right)^{5x+1}$$
I tried like that:
$$\lim _{x\to \infty }\left(e^{\left(5x+1\right)ln\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+2}}\right)}\right)$$
Then:
$$ln\left(\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+2}}\right)=\frac{\sqrt[3]{x^4+1}-\sqrt[6]{9x^8+2}}{x+\sqrt[6]{9x^8+2}}$$
$$=(5x+1)\frac{\sqrt[3]{x^4+1}-\sqrt[6]{9x^8+2}}{x+\sqrt[6]{9x^8+2}}$$
$$=\frac{5x^2\sqrt[3]x(1-\sqrt[3]{3})}{x+x\sqrt[3]x\sqrt[3]3}$$
Now I don't understand a thing:
the result of the latter fraction is $-\infty$ so $e^{-\infty}=\color{red}0$?
I'm sure I missed a few step, help me. Thanks
| $$\lim_{x\to\infty}\frac{x+\sqrt[3]{x^4+1}}{x+\sqrt[6]{9x^8+3}} =\lim_{x\to\infty}\dfrac{\dfrac1{x^{4/3}}+\left(1+\dfrac1{x^4}\right)^{1/3}}{\dfrac1{x^{4/3}}+\left(9+\dfrac3{x^8}\right)^{1/6}}=\dfrac1{9^{1/6}}$$
which is $<1$
Hence the given limit should converge to $0$
Keep in mind $$\lim_{n\to\infty}\left(1+\dfrac1n\right)^n=e$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
induction to prove the equation $3 + 9 + 15 + ... + (6n - 3) = 3n^2$ I have a series that I need to prove with induction. So far I have 2 approaches, though I'm not sure either are correct.
$$3 + 9 + 15 + ... + (6n - 3) = 3n^2$$
1st attempt:
\begin{align*}
& = (6n - 3) + 3n^2\\
& = 3n^2 + 6n - 3\\
& = (3n^2 + 5n - 4) + (n + 1)
\end{align*}
That seems way wrong lol ^^^
2nd attempt:
\begin{align*}
f(n) & = 3 + 9 + 15 + ... + (6n - 3)\\
f(n + 1) & = 6(n + 1) - 3\\
f(n + 1) & = 6(n - 3) + 6(n + 1) - 3\\
& = ?
\end{align*}
I don't know I feel like I'm headed in the wrong direction
I guess another attempt I have would be:
\begin{align*}
f(n) & = 3n^2\\
f(n+1) & = 3(n + 1)^2\\
& = 3(n^2 + 2n + 1)\\
& = 3n^2 + 6n + 3\\
& = f(n) + (6n + 3)
\end{align*}
| Without induction you can do it easily taking three common so $3(1+3+..(2n-1))$ now we know that sum of odd terms is a perfect square so we can write as $3n^2$ or sum of AP=$\frac{n}{2}(2+(n-1)2)=n^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
This sigma to binom? Can you please show me how to get from the left side to the right side?
$$\sum\limits_{k=0}^{20}\binom{50}{k}\binom{50}{20-k} = \binom{100}{20}$$
| $$(x+1)^{50}(1+x)^{50}=(1+x)^{100}$$
$$(\sum_{k=0}^{50}\binom{50}k x^k)\cdot(\sum_{k=0}^{50}\binom{50}kx^{50-k})=\sum_{k=0}^{50}\binom{100}r x^r$$
Consider the coefficients of $x^{20}$
$$\binom{100}{20}$$
$$=\binom{50}0\cdot\binom{50}{20}+\binom{50}1\cdot\binom{50}{19}+\cdots+
\binom{50}{19}\cdot\binom{50}1+\binom{50}{20}\cdot\binom{50}0$$
$$=\sum_{r=0}^{20}\binom{50}r\binom{50}{20-r}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1630679",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the sum $\sum _{ k=1 }^{ 100 }{ \frac { k\cdot k! }{ { 100 }^{ k } } } \binom{100}{k}$ Find the sum $$\sum _{ k=1 }^{ 100 }{ \frac { k\cdot k! }{ { 100 }^{ k } } } \binom{100}{k}$$
When I asked my teacher how can I solve this question he responded it is very hard, you can't solve it. I hope you can help me in solving and understanding the question.
| \begin{align}
\sum\limits_{k=1}^{100} \frac {k\cdot k!}{100^k} \frac{100!}{k!(100-k)!} &= \frac{100!}{100^{100}} \sum\limits_{k=1}^{100} \frac{k\cdot100^{100-k}}{(100-k)!}\\
&= \frac{100!}{100^{100}} \sum\limits_{k=0}^{99}\frac{(100-k)\cdot 100^k}{k!}\\ &=\frac{99!}{100^{99}} \sum\limits_{k=0}^{99}
\left( \frac {100^{k+1}}{k!} - \frac{100^k}{(k-1)!}\right)\\ &=
\frac{99!}{100^{99}}
\left(\frac{100^1}{0!}+
\sum\limits_{k=1}^{99} \frac {100^{k+1}}{k!} -
\sum\limits_{k=0}^{98} \frac{100^{k+1}}{k!}\right)\\&=
\frac{99!}{100^{99}}
\left(\frac{100^1}{0!}+
\frac{100^{100}}{99!} +
\sum\limits_{k=1}^{98} \frac {100^{k+1}}{k!} -
\sum\limits_{k=1}^{98} \frac{100^{k+1}}{k!}
-\frac{100^1}{0!}
\right)\\
&=
100
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1632928",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 0
} |
Prove that $L_n = \alpha^n +\beta^n$ for all integers $n\geq 0$ Let $\alpha =\left(\frac{1+\sqrt{5}}{2}\right)$ and $\beta = \left(\frac{1-\sqrt{5}}{2}\right)$. Prove that $L_n = \alpha^n +\beta^n$ for all integers $n\geq 0$ where $L_n$ denotes the Lucas numbers.
I managed to solve the base case:
$$L_2 = \left(\frac{1+\sqrt{5}}{2}\right)^2 + \left(\frac{1-\sqrt{5}}{2}\right)^2 = 3$$
I wasn't able to figure out how to proceed with the induction step and the rest of the proof. Any help is appreciated.
| The crucial fact is that $\alpha$ and $\beta$ are the roots of $x^2=x+1$ and so $\alpha+\beta=1$. This makes induction work easily:
$L_0 = 2 = 1+1 = \alpha^0+\beta^0$
$L_1 = 1 = \alpha+\beta = \alpha^1+\beta^1$
$L_{n+2}=L_{n+1}+L_n
= \alpha^{n+1}+\beta^{n+1}+\alpha^n+\beta^n
= \alpha^{n+1}+\alpha^n+\beta^{n+1}+\beta^n
$
$\qquad
= \alpha^n(\alpha+1)+ \beta^n(\beta+1)
= \alpha^n\alpha^2+ \beta^n\beta^2
= \alpha^{n+2}+\beta^{n+2}
$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding char polynomial in $Z_3$ $ K=Z_3 $
$ A \in K_{(4 \times 4)} $
$$A=
\begin{bmatrix}
a & -1 & -2 & -2 \\
0 & a-1 & -2 & 0 \\
-2 & 0 & a & 0 \\
-2 & -1 & 0 & a-2 \\
\end{bmatrix}
$$
I need to find $\det(A)$.
I came up with $\det(A)= (a)(a-1)(a)(a-2) - (a-2)(-2)(a-1)(-2)$
If I rewrite this I get:
$a^4 - 2a^3 - a^3 + 2a^2 - 4a^2 + 4a + 8a - 8 =
a^4 -2a^2 - 2$
My answer sheet gives me $\det(A)= x^4+1 =(x^2+x+2)(x^2+2x+2)$
What am I overlooking or maybe there is something wrong with my calculation?
| Keeping in mind the patterns $\left|\begin{array}{cccc}+&-&+&-\\-&+&-&+\\+&-&+&-\\-&+&-&+\end{array}\right|$ and $\left|\begin{array}{cccc}+&-&+\\-&+&-\\+&-&+\end{array}\right|$, expanding along the last column we get:
$$-(-2)\left|\begin{array}{cccc}0&a-1&-2\\-2&0&a\\-2&-1&0\end{array}\right|
+(a-2)\left|\begin{array}{cccc}a&-1&-2\\0&a-1&-2\\-2&0&a\end{array}\right|$$
At this point (or even earlier), it helps to work modulo $3$:
$$-1\left|\begin{array}{cccc}0&a-1&1\\1&0&a\\1&-1&0\end{array}\right|
+(a+1)\left|\begin{array}{cccc}a&-1&1\\0&a-1&1\\1&0&a\end{array}\right|\\
\begin{array}{l}=\underbrace{-\left[-1\underbrace{\left|\begin{array}{cccc}a-1&1\\-1&0\end{array}\right|}_{1}+1\underbrace{\left|\begin{array}{cccc}a-1&1\\0&a\end{array}\right|}_{a^2-a}~\right]}_{1-a^2+a}
+(a+1)\underbrace{\left[a\underbrace{\left|\begin{array}{cccc}a-1&1\\0&a\end{array}\right|}_{a^2-a}+1\underbrace{\left|\begin{array}{cccc}-1&1\\a-1&1\end{array}\right|}_{-1-a+1=-a}~\right]}_{a^3-a^2+a}\\
=1-a^2+a+(a+1)(a^3-a^2-a)\\
=1-a^2+a+a^4-a^3-a^2+a^3-a^2-a\\
=a^4-3a^2+1\\
=a^4+1\end{array}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637274",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Decompose $\frac{x^4 + 5}{x^5 + 6x^3}$ (partial fraction decomposition)
Write out the form of the partial fraction decomposition of the function. Do not determine the numerical values of the coefficients.
$$\frac{x^4 + 5}{x^5 + 6x^3}$$
So I factored the denominator to be $x^3(x^2+6)$ and here is the answer i got:
$\frac{Ax+B}{x^3}+\frac{Cx+D}{x^2+6}$
But this isn't the correct answer. Any help is appreciated!
Also have another one:
$\frac{5}{(x^2 − 16)^2}$
Here is what I got: $\frac{A}{x+4}+\frac{Bx+C}{\left(x+4\right)^2}+\frac{D}{x-4}+\frac{Ex+F}{\left(x-4\right)^2}$ which is also wrong... if anyone could help me that would be great
| The correct way is
$$\frac{x^4 + 5}{x^3(x^2 + 6)}=\frac{A}{x}+\frac{B}{x^2}+\frac{C}{x^3}+\frac{Dx+E}{x^2+6}$$
and
$$\frac{5}{(x^2 − 16)^2}=\frac{A}{x+4}+\frac{B}{\left(x+4\right)^2}+\frac{C}{x-4}+\frac{D}{\left(x-4\right)^2}$$
In general, if the denominator has degree $n$, then the numerator is given a degree $(n-1)$. This is because for any higher degree, one can divide and get a remainder where the numerator has a degree less than the denominator. Moreover, the numerator can have a degree less than $(n-1)$ if the coefficients of appropriate powers of $x$ are $0$.
When the denominator is a $k^{th}$ power of a polynomial in $x$, the numerator can be split into $k$ expressions each being some multiple of the polynomial in the denominator, which in turn can be split into $k$ different fractions. This gives rise to the form of expression used above.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Find all pairs of nonzero integers $(a,b)$ such that $(a^2+b)(a+b^2)=(a-b)^3$
Find all pairs of nonzero integers $(a,b)$ such that
$(a^2+b)(a+b^2)=(a-b)^3$
My effort
Rearranging the equation I have
\begin{array}
\space (a^2+b)(a+b^2)-(a-b)^3 &=0 \\
a^2(b^2+3b)+a(-3b^2+b)+2b^3 &=0 \\
\end{array}
Letting $a=x$, we have the polynomial $Q(x)$ such that $$Q(x)=x^2(b^2+3b)+x(-3b^2+b)+2b^3 $$
So I must have that the roots $a_1,a_2$ must be such that
\begin{array}
\space a_1+a_2=-\cfrac{-3b+1}{b+3} \\
a_1\cdot a_2=\cfrac{2b^2}{b+3} \\
\end{array}
I am stuck now,should I go by brute force and verify some values or
there's still something I can do here to simplify the problem ?
| $b=0$ is a solution for any $a$. We can then divide it out, looking for non-zero solutions. If we just feed it to the quadratic equation, we get $$a=\frac {3b-1 \pm \sqrt{(1-3b)^2-8(b+3)b^2}}{2(b+3)}$$ You must have $b \lt 0$ to have the discriminant positive. A quick search finds $(-1,-1), (9,-6), (8,-10), (9, -21)$ with no others by $b=-1000$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1638987",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
If $P(x) = (x^4+x^3-3x^2+4x-4)\cdot q(x) + (2x^3-5x^2+7x-3)$ find $P(2)$
If for the polynomial $P(x)$ is true that $$P(x) = (x^4+x^3-3x^2+4x-4)\cdot q(x) + (2x^3-5x^2+7x-3)$$ find $P(2)$
I assumed that the polynomial $(x^4+x^3-3x^2+4x-4)$ has $(x-2)$ as one of its factors, but it turns out that this is not true.
By the equation we got that $P(2)=16\cdot q(2) + 7$ but I don't know what I should do next.
| I give you what I think is the only chance that your problem has a meaning.
There are many possibilities for a such polynomial: you can choose for $g(x)$ any arbitrary polynomial and the resultant one $$P(x) = (x^4+x^3-3x^2+4x-4)\cdot g(x) + (2x^3-5x^2+7x-3)$$ satisfy in fact the property of your statement.
In particular you can do $P(x)=2x^3-5x^2+7x-3$ so the polynomial is the rest itself in whose case the answer is the constant polynomial $7$.
For this, you have written yourself the general answer, where $g(x)$ is an arbitrary polynomial (if $g$ has $x=2$ as a root then the rest is still $7$ of course).
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that the sequence of combinations contains an odd number of odd numbers
Let $n$ be an odd integer more than one. Prove that the sequence $$\binom{n}{1}, \binom{n}{2}, \ldots,\binom{n}{\frac{n-1}{2}}$$
contains an odd number of odd numbers.
I tried writing out the combination form as $$\frac{(2k+1)!}{(m!)((2k+1)-m)!}.$$ How do I use this to show that the sequence contains an odd number of odd numbers?
| $$
\overbrace{\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2}} + \overbrace{\binom n {\frac{n+1} 2} + \cdots + \binom n {n-1} + \binom n n} = 2^n.
$$
The two sums under the $\overbrace{\text{overbraces}}$ are equal; hence
$$
\binom n 0 + \binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2} = 2^{n-1}.
$$
Therefore
$$
\binom n 1 + \binom n 2 + \cdots + \binom{n}{\frac{n-1} 2} = 2^{n-1}-1 = \text{an odd number}.
$$
If a sum of finitely many terms is an odd number (as this one is) then the number of odd terms must be odd, since if it were even, then the sum would be even.
| {
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"timestamp": "2023-03-29T00:00:00",
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About Factorization I have some issues understanding factorization.
If I have the expression $x^{2}-x-7$ then (I was told like this) I can put this expression equal to zero and then find the solutions with the quadratic formula, so it gives me $x_{0,1}= 1 \pm 2\sqrt{2}$ then $$x^{2}-x-7 = (x-1-2\sqrt{2})(x-1+2\sqrt{2}).$$
That is correct I have checked it.
Now for the expression $3x^{2}-x-2$ if I do the same I have $x_{0} = 1$ and $x_1=\frac{-2}{3}$ so I would have $$3x^{2}-x-2 = (x-1)(x+\frac{2}{3})$$
but this is not correct since $(x-1)(x+\frac{2}{3}) = \frac{1}{3}(3x^{2}-x-2)$,
the correct factorization is $3x^{2}-x-2 = (3x+2)(x-1)$.
So I guess finding the roots of a quadratic expression is not sufficient for factorizing.
| The quadratic formula only gives the roots of the equation, which are unaltered by the constant factor.
Let $\ P(x)=3x^{2}-x-2 \ \text{ and } \ Q(x) = x^2 -\frac{x}{3}-\frac{2}{3}$
The roots of both the polynomials are the same, as
$$\begin{align}3x^{2}-x-2&=0 \\ \implies x^2 -\frac{x}{3}-\frac{2}{3}&=0\end{align}$$
but $P(x)$ and $Q(x)$ differ by a constant factor $3$, the leading coefficient of $P(x)$.
$$P(x) = 3\times Q(x)$$
| {
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How to solve $\int \frac{1}{1-y^2}$ with respect to $y$? I was solving an A Level paper when I came across this question. I tried substitution, but I'm not getting the answer with that. Would appreciate it if someone would help me.
| Step 1: Factorize the denominator
To compute the integral, you start of by using the third binomial theorem
$$\int \frac{1}{1-y^2}\ \mathrm{d}y = \int \frac{1}{1-y}\frac{1}{1+y}\ \mathrm{d}y$$
Step 2: Partial fractions
Use a technique called partial fractions to write the integrant as $\frac{A}{1-y}+\frac{B}{1+y}$. $A$ and $B$ are calculated as follows
$$\begin{array}[t]{lll}
\frac{A}{1-y}+\frac{B}{1+y} & = & \frac{1}{1-y}\frac{1}{1+y}\\
A(1+y)+B(1-y) & = & 1\\
1(A+B)+y(A-B) & = & 1\cdot 1 + y\cdot 0
\end{array}$$
Equating coefficients gives you
$$A+B=1,\ A-B = 0 \Rightarrow A = B = 0{,}5$$
Step 3: Substitution
The remaining integral
$$\int \left(\frac{0{,}5}{1-y}+\frac{0{,}5}{1+y}\right) \ \mathrm{d}y$$
can be solved straightforwardly unsing substitutions
$$u=y-1,\ s=y+1,\ \mathrm{d}u=\mathrm{d}s=\mathrm{d}y$$
and the fact that $\int \frac{1}{x}\ \mathrm{d}x = \ln(|x|)$.
$$\int \left(\frac{0{,}5}{1-y}+\frac{0{,}5}{1+y}\right) \ \mathrm{d}y = \int \frac{0{,}5}{-u}\ \mathrm{d}u+\int \frac{0{,}5}{s} \ \mathrm{d}s $$
$$=-0{,}5\cdot\ln(|u|)+0{,}5\cdot\ln(|s|)=\frac{-\ln(|y-1|)+\ln(|y+1|)}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1640202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integration of $\frac{dx}{(6x-4x^2)^{1/2}}$ and completing square. Integrate $$\frac{dx}{\sqrt{6x-4x^2}}$$
While completing the square of $6x-4x^2$, I want to know where did $9/16$ come from in the following after taking $4$ out as the common factor $$-4\bigg(x^2 -\frac{3x}{2} +\frac{9}{16}\bigg) +\frac{9}{4}$$. I want to know the steps of completing the square.
| $$-4x^2+6x = -4(x^2-\frac{3}{2}x)$$
$$ = -4\bigg(\big(x-\frac{3}{4}\big)^2-\frac{9}{16}\bigg) $$
$$ = -4\big(x-\frac{3}{4}\big)^2+\frac{9}{4}$$
Do you see it now?
| {
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How to simplify this equation to solve for m? It has been way too many years since high school. How can I simplify this equation to solve for m:
$\frac{x}{c+pm}=m$
I got to
$x = cm + pm^2$
and I don't know how to get any further. I wish this stuff stayed in my head over the last 20 years.
| Your equation
$$
x = c\, m + p\, m^2 \quad (1)
$$
is a quadratic equation of the form
$$
y = a \, x^2 + b \, x + c
$$
(where $x$ is the unknown, in your case $m$).
If $p \ne 0$, then we can divide both sides of equation $(1)$ by $p$ to get
$$
m^2 + \frac{c}{p} m = \frac{x}{p}
$$
we now try to align the left hand side for application of the binomic formula
$$
a^2 + 2 a b + b^2 = (a + b)^2 \quad (2)
$$
where $m$ is the $a$ and $c/p$ is $2 b$, so the $b^2$ is missing, we add this to both sides:
$$
m^2 + 2 m \frac{c}{2p} + \left(\frac{c}{2p}\right)^2 =
\frac{x}{p} + \left(\frac{c}{2p}\right)^2
$$
we can now apply equation $(2)$ and have a new left hand side:
$$
\left(m + \frac{c}{2p} \right)^2 = \frac{x}{p} + \left(\frac{c}{2p}\right)^2
$$
Taking the square root of both sides gives
$$
m + \frac{c}{2p}= \pm \sqrt{\frac{x}{p} + \left(\frac{c}{2p}\right)^2}
$$
and solving for $m$ gives:
$$
m = -\frac{c}{2p} \pm \sqrt{\frac{x}{p} + \left(\frac{c}{2p}\right)^2}
$$
The $\pm$ symbol indicates up to two solutions, one having the added square root, the other having the substracted square root.
To have solutions which are real numbers, the radicand
$$
\frac{x}{p} + \left(\frac{c}{2p}\right)^2
$$
must be non-negative, otherwise you get no real numbers as solutions, but complex numbers with non-zero imaginary components.
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