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I'm looking for several ways to prove that $\int_{0}^{\infty }\sin(x)x^mdx=\cos(\frac{\pi m}{2})\Gamma (m+1)$ I'm looking for several ways to prove that $$\int_{0}^{\infty }\sin(x)x^mdx=\cos\left(\frac{\pi m}{2}\right)\Gamma (m+1)$$
for $-2< Re(m)< 0$
| Here is a fake proof: Let $x = \sqrt{t}$. Then
$$ \int_{0}^{\infty} x^{m} \sin x \, dx = \frac{1}{2} \int_{0}^{\infty} t^{m/2} \frac{\sin \sqrt{t}}{\sqrt{t}} \, dt. $$
Since
$$ \frac{\sin \sqrt{t}}{\sqrt{t}} = \sum_{n=0}^{\infty} \frac{\phi(n)}{n!} (-t)^{n} \quad \text{for} \quad \phi(n) = \frac{n!}{(2n+1)!} = \frac{\Gamma(n+1)}{\Gamma(2n+2)}, $$
Ramanujan's master theorem gives
\begin{align*}
\int_{0}^{\infty} t^{m/2} \frac{\sin \sqrt{t}}{\sqrt{t}} \, dt
&= \Gamma\left(\frac{m}{2}+1\right)\phi\left(-\frac{m}{2}-1\right) \\
&= \frac{\Gamma\left(\frac{m}{2}+1\right)\Gamma\left(-\frac{m}{2}\right)}{\Gamma\left(-m\right)}
= \frac{\Gamma\left(\frac{m}{2}+1\right)\Gamma\left(-\frac{m}{2}\right)}{\Gamma(m+1)\Gamma\left(-m\right)}\Gamma(m+1)\\
&= \frac{\sin \pi m}{\sin \left( \frac{\pi m}{2} \right)} \Gamma(m+1)
= 2\cos\left(\frac{\pi m}{2} \right) \Gamma(m+1).
\end{align*}
Therefore we have
$$ \int_{0}^{\infty} x^{m} \sin x \, dx =\cos\left(\frac{\pi m}{2} \right) \Gamma(m+1). $$
And here is a rigorous proof: By integration by parts, for $0 < a < b < \infty$,
\begin{align*}
\int_{a}^{b} x^{m} \sin x \, dx
&= \left[ x^{m} (1 - \cos x) \right]_{a}^{b} - \int_{a}^{b} m x^{m-1} (1 - \cos x) \, dx.
\end{align*}
Since $ \Re m \in (-2, 0)$, $x^{m} (1 - \cos x)$ tends to $0$ either as $x \to 0^{-}$ or $x \to \infty$. This proves
$$ \lim_{\substack{a &\to& 0 \\ b &\to& \infty}} \left[ x^{m} (1 - \cos x) \right]_{a}^{b} = 0. $$
Also, by noting that $0 \leq 1 - \cos x \leq 1 \wedge x^2 $, we have
$$ \left| m x^{m-1} (1 - \cos x) \right| \leq |m| x^{\Re m-1} (1 \wedge x^{2}). $$
Since the RHS is integrable, the same is true for the LHS $m x^{m-1} (1 - \cos x)$. Thus the integral
$$ \int_{0}^{\infty} x^{m} \sin x \, dx $$
exists as (at least) in improper sense, and we have
\begin{align*}
\int_{0}^{\infty} x^{m} \sin x \, dx
&= - \int_{0}^{\infty} m x^{m-1} (1 - \cos x) \, dx \\
&= - \frac{m}{\Gamma(1-m)} \int_{0}^{\infty} \frac{\Gamma(1-m)}{x^{1-m}} (1 - \cos x) \, dx \\
&= - \frac{m}{\Gamma(1-m)} \int_{0}^{\infty} \left( \int_{0}^{\infty} t^{-m}e^{-xt} \, dt \right) (1 - \cos x) \, dx.
\end{align*}
But since
\begin{align*}
\int_{0}^{\infty} \int_{0}^{\infty} \left| t^{-m}e^{-xt} (1 - \cos x) \right| \, dtdx
&\leq \int_{0}^{\infty} \int_{0}^{\infty} t^{-\Re m}e^{-xt} (1 - \cos x) \, dtdx \\
&= \int_{0}^{\infty} \frac{\Gamma(1-\Re m)}{x^{1-\Re m}} (1 - \cos x) \, dtdx
< \infty,
\end{align*}
Fubini's theorem shows that
\begin{align*}
\int_{0}^{\infty} x^{m} \sin x \, dx
&= - \frac{m}{\Gamma(1-m)} \int_{0}^{\infty} \int_{0}^{\infty} t^{-m}e^{-xt} (1 - \cos x) \, dxdt \\
&= - \frac{m}{\Gamma(1-m)} \int_{0}^{\infty} \frac{t^{-m-1}}{t^2 + 1} \, dt.
\end{align*}
Applying the beta function identity and the Euler reflection formula for the Gamma function, it follows that
\begin{align*}
\int_{0}^{\infty} x^{m} \sin x \, dx
&= \frac{\pi m}{2\Gamma(1-m)} \csc \left( \frac{\pi m}{2} \right)
= \cos \left( \frac{\pi m}{2} \right) \Gamma(m+1)
\end{align*}
as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Frullani version of the classic $\log \left( 1 + 2\alpha \cos px + \alpha^2\right)$ integral. I read in a paper about Frullani integrals the following claim
$$
\begin{align*}
I & :=\int_0^\infty \frac{1}{x}\log\left(\frac{1 + 2\alpha \cos px + \alpha^2}{1 + 2\alpha \cos qx + \alpha^2}\right)\,\mathrm{d}x
=
\left\{
\begin{array}{lrr}
\log \frac{p}{q} \log\left( 1 + \alpha^2\right) & \text{if} & \alpha^2>1\\
\log \frac{p}{q} \log\left( 1 + \frac{1}{\alpha}\right) & \text{if} & \alpha^2<1
\end{array}
\right.
\end{align*}
$$
Where the author points at the dangers of using this method. The problem
lies in the fact that the function $f(n,r) := \log(1+2n \cos rx + n^2)$ does not have a limit
as $x\to \infty$. And hence the Frullani identity can not be used.
A very similar integral $f(n,-r)$ has been evaluated several times before.
See for an example A question in Complex Analysis $\int_0^{2\pi}\log(1-2r\cos x +r^2)\,dx$
and can also be solved by differentiating under the integral sign . Even $f(n,-r)^2$ has been evaluated.
My question is: Can any of this be used to prove the claim above?
| Untilize the Frullani-like integral
$\int_0^\infty \frac{\cos px -\cos qx}xdx
=\ln\frac qp$
to obtain
\begin{align}
I_{a^2<1}= &\int_{0}^{\infty} \frac{1}{x} \, \ln \frac{1+2a\cos px+a^{2}}{1+2 a\cos qx + a^{2}} \ dx\\
=&\> 2 \int_{0}^{\infty}
\sum_{k=1}^\infty \frac{(-a)^{k+1}}{k}\frac{\cos(k px)-\cos(k qx)}x \ dx\\
= &\>2\ln \frac qp \sum_{k=1}^\infty \frac{(-a)^{k+1}}{k}
= 2\ln \frac qp \ \ln(1+a)
\\
\\
I_{a^2>1}=&\int_{0}^{\infty} \frac{1}{x} \, \ln \frac{1+\frac2a\cos px+\frac1{a^2}}{1+\frac2a\cos qx+\frac1{a^2}} \ dx
= 2\ln \frac qp \ \ln(1+\frac1a)
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Residue Computation How can we compute the residue of the function $\dfrac{(1+zx)^{k+1/3}}{z^2(z-a)} $ at $z=0$?? where $x$ is some fixed real number and $k$ is some fixed positive integer? I have tried computing laurent series of each but it gets too complicated to do.
| The residue of $\frac{f(z)}{z^2}$ at $z = 0$, where $f$ is holomorphic near $0$ is $f'(0)$, or, which sometimes is easier to find, the coefficient of $z$ in the Taylor expansion of $f$. Here we have
$$f(z) = \frac{(1+zx)^c}{(z-a)},$$
which is not totally trivial to differentiate (but not too hard either). We can develop the numerator into a binomial series
$$(1+xz)^c = \sum_{\nu=0}^\infty \binom{c}{\nu}(xz)^\nu = 1 + c\cdot xz + \frac{c(c-1)}{2}x^2z^2 + \dotsb$$
and the other factor into a geometric series
$$\frac{1}{z-a} = \frac{(-1)}{a} \frac{1}{1- \frac{z}{a}} = \frac{(-1)}{a} \sum_{\kappa=0}^\infty \left(\frac{z}{a}\right)^\kappa = -\frac{1}{a} - \frac{z}{a^2} - \frac{z^2}{a^3} - \dotsb$$
Now we need the coefficient of $z^1$ in the product of the two series,
$$-\left(\frac{1}{a} + \frac{z}{a^2} + \dotsb\right)\left(1 + cxz + \dotsb\right) = -\frac1a - \left(\frac{1}{a^2} + \frac{cx}{a}\right)z + \dotsb$$
| {
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For what $x\in[0,2\pi]$ is $\sin x < \cos 2x$ What's the set of all solutions to the inequality $\sin x < \cos 2x$ for $x \in [0, 2\pi]$? I know the answer is $[0, \frac{\pi}{6}) \cup (\frac{5\pi}{6}, \frac{3\pi}{2}) \cup (\frac{3\pi}{2}, 2\pi]$, but I'm not quite sure how to get there.
This is what I have so far: $\\
\sin x - \cos 2x < 0\\
\sin x - \cos^2 x + \sin^2 x < 0\\
2\sin^2 x + \sin x - 1 < 0\\
-1 < \sin x < \frac{1}{2}$
Any help will be much appreciated.
| You reached the right inequality. A slightly different way of finishing goes as follows.
We are interested in where $(2\sin x-1)(\sin x+1)\lt 0$.
We find the places where $(2\sin x-1)(\sin x+1)=0$. This happens at $x=\frac{\pi}{6}$, $x=\frac{5\pi}{6}$, and $x=\frac{3\pi}{2}$.
These $3$ points divide the interval $[0,2\pi]$ into $4$ pieces (intervals). On each of these pieces, our function does not change sign. So in each piece, we can choose a convenient test point to find the sign of our function on the whole piece.
For example, on the piece $[0,\frac{\pi}{6})$ choose the test point $x=0$. Result: our function is $\lt 0$ at the test point, and therefore $\lt 0$ on the whole interval $[0,\frac{\pi}{6})$.
On the piece $(\frac{\pi}{6}, \frac{5\pi}{6})$, choose the test point $x=\frac{\pi}{2}$. There our function is positive. So we do not use this interval.
Two more pieces to go.
| {
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Proof by induction that the sum of terms is integer I'm having some trouble in order to solve this induction proof.
Proof that $\forall{n} \in \mathbb{N}$ the number $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n$ is an integer.
I've tried proving this by induction, but I've not succeed so far. What I did was:
Let's assume that $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$. So, by induction,
$n=1$
$\frac{1}{5} + \frac{1}{3} + \frac{7}{15} = 1$ and $1 \in \mathbb{Z}$.
Inductive step
I want to prove that if $\frac{1}{5}n^5+\frac{1}{3}n^3 + \frac{7}{15}n = k$, with $k \in \mathbb{Z}$ then $\frac{1}{5}(n+1)^5+\frac{1}{3}(n+1)^3 + \frac{7}{15}(n+1) = l$, with $l \in \mathbb{Z}$
And then I'm stuck. I've tried developing every binomial but, for example, for $\frac{1}{5}(n+1)^5$ I get a $\frac{1}{5}n^4$ that is not in the inductive hypothesis, so therefore, I cannot get rid of it and it's not an integer, so I cannot conclude that $l \in \mathbb{Z}$.
Any help or ideas from where I can follow? Thanks in advance!
| You want to try to write this expression in terms of the previous one. So, start by expanding out the terms:
$$
\begin{align*}
\frac{1}{5}(n+1)^5&=\frac{1}{5}\sum_{k=0}^{5}\binom{5}{k}n^k\\
&=\frac{1}{5}(n^5+5n^4+10n^3+10n^2+5n+1)\\
&=\frac{1}{5}n^5+n^4+2n^3+2n^2+n+\frac{1}{5}
\end{align*}
$$
Do the same thing for $(n+1)^3$ and $(n+1)$. Now, you can pick out the terms corresponding to the inductive hypothesis; all that's left is to show that the other remaining terms ALSO give you an integer.
| {
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"timestamp": "2023-03-29T00:00:00",
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Successive Differentiation and division I am working on successive differentiation. I have ran into some confusion and would like some help with the process of differentiating when dealing with division. Here is the problem that sparked my intentions to post here: $$y = \frac{x^2 + a}{x + a}$$ I am reading Calculus Made Easy by Silvanus P. Thompson and unfortunately there is no example on how I should go about this. I can find $\frac{dy}{dx}$ but can't seem to find $\frac{d^2 y}{dx^2}$ for this exercise. I find $\frac{dy}{dx}$ by doing this: $$y = \frac{(x + a)(2x) - (x^2 + a)(1)}{(x + a)^2}$$ to get $y = \frac{x^2 + 2xa - a}{(x + a)^2}$. This is where I am stuck. I have attempted to simply perform the same operation on this result to differentiate again but it isn't adding up. Any help on this is much appreciated!
| $$ \dfrac{dy}{dx} = \frac{(x + a)(2x) - (x^2 + a)(1)}{(x + a)^2} =\dfrac{2x^2 + 2ax - x^2 - a}{(x+a)^2} = \frac{x^2 + 2xa - a}{(x + a)^2} $$
So far so good.
Now we need to differentiate again, using the quotient rule, as you did when finding $\dfrac{dy}{dx}$. It gets messy-looking, but in this case, simplifies relatively nicely in the end:
$$\dfrac{d^2y}{dx^2} = \dfrac{(x+a)^2(2x + 2a) - (x^2 + 2xa - a)[2(x+a)(1)]}{((x+ a)^2)^2}$$
We can factor the numerator, expand, and then simplify:
$$\begin{align}\dfrac{d^2y}{dx^2} & = \dfrac{2(x+a)[(x+a)^2 - (x^2 + 2xa - a)]}{(x+ a)^4} \\ \\
& = \dfrac{2[x^2 + 2ax + a^2 - x^2 - 2ax + a]}{(x+ a)^3}\\ \\
& =\dfrac{2( a^2 + a)}{(x + a)^3} \end{align}$$
We could have gotten clever by rewriting $\dfrac{dy}{dx}$:
We can rewrite $\dfrac{dy}{dx}$ thusly:
$$\begin{align} \dfrac{dy}{dx} = \frac{x^2 + 2xa - a}{(x + a)^2} & = \dfrac{x^2 + 2xa + a^2 -a -a^2}{(x+ a)^2} \\ \\
& = \dfrac{(x+a)^2 - a(a + 1)}{(x + a)^2} \\ \\
& = 1 - \dfrac{a(a + 1)}{(x + a)^2}\end{align}$$
The result of differentiating the result (using essentially, the power rule) would be the same as given above.
| {
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If $p =\frac{4\sin\theta \cos\theta}{\sin\theta +\cos\theta}$ Find the value of $\frac{p+2\sin\theta}{p-2\sin\theta}$ Problem :
If $\displaystyle p =\frac{4\sin\theta\cos\theta}{\sin\theta +\cos\theta}$, find the value of $\displaystyle \frac{p+2\sin\theta}{p-2\sin\theta} + \frac{p+2\cos\theta}{p-2\cos\theta}$.
Please help how to proceed in such problem..Thanks..
| HINT:
So, we have $$\frac p{2\sin\theta}=\frac{2\cos\theta}{\sin\theta+\cos\theta}$$
Apply Componendo and dividendo to get $$\frac{p+2\sin\theta}{p-2\sin\theta}=\frac{2\cos\theta+\sin\theta+\cos\theta}{2\cos\theta-(\sin\theta+\cos\theta)}=\frac{\sin\theta+3\cos\theta}{\cos\theta-\sin\theta}$$
Similarly, $$\frac p{2\cos\theta}=\frac{2\sin\theta}{\sin\theta+\cos\theta}\implies \frac{p+2\cos\theta}{p-2\cos\theta}=...=\frac{3\sin\theta+\cos\theta}{\sin\theta-\cos\theta}=-\frac{3\sin\theta+\cos\theta}{\cos\theta-\sin\theta}$$
Alternatively, $$\frac{p+2\sin\theta}{p-2\sin\theta}=\frac{\frac p{2\sin\theta}-1}{\frac p{2\sin\theta}+1}=\frac{\frac{2\cos\theta}{\sin\theta+\cos\theta}+1}{\frac{2\cos\theta}{\sin\theta+\cos\theta}-1}=\frac{2\cos\theta+(\sin\theta+\cos\theta)}{2\cos\theta-(\sin\theta+\cos\theta)}$$
$$\implies \frac{p+2\sin\theta}{p-2\sin\theta}=\frac{\sin\theta+3\cos\theta}{\cos\theta-\sin\theta}$$
Similarly for $$\frac{p+2\cos\theta}{p-2\cos\theta}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve problems involving roots. $\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$ How to solve problems involving roots. If we square them they may go to fourth degree.There must be some technique to solve this.
$$\sqrt{(x+3)-4\sqrt{x-1}} + \sqrt{(x+8)-6\sqrt{x-1}} =1$$
| Put $y=\sqrt{x-1}$, getting $x=y^2+1$; thus the equation becomes
$$
\sqrt{y^2-4y+4}+\sqrt{y^2-6y+9}=1
$$
which should ring a bell.
It becomes $|y-2|+|y-3|=1$ that can be treated without resorting to squaring; divide it into cases:
If $y<2$, the equation becomes $2-y+3-y=1$, or $2y=4$, that means $y=2$, absurd.
If $2\le y\le 3$, the equation becomes $y-2+3-y=1$, an identity.
If $y>3$, the equation becomes $y-2+y-3=1$, or $2y=6$ and $y=3$, absurd.
Therefore the solutions are all the numbers $x$ such that $2\le\sqrt{x-1}\le3$, that is $4\le x-1\le 9$ or $5\le x\le 10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/449300",
"timestamp": "2023-03-29T00:00:00",
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How to find the sum of the series $1+\frac{1}{2}+ \frac{1}{3}+\frac{1}{4}+\dots+\frac{1}{n}$? How to find the sum of the following series?
$$1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \dots + \frac{1}{n}$$
This is a harmonic progression. So, is the following formula correct?
$\frac{(number ~of ~terms)^2}{sum~ of~ all ~the~ denominators}$
$\Rightarrow $ if $\frac{1}{A} + \frac{1}{B} +\frac{1}{C}$ are in H.P.
Therefore the sum of the series can be written as :
$\Rightarrow \frac{(3)^3}{(A+B+C)}$
Is this correct? Please suggest.
| There is a new formula, that I've created. It can be written in a few different ways, below are two of them:
$\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{2n}+\pi\int_{0}^{1} (1-u)\cot{\pi u}\left(1-\cos{2\pi n u}\right)\,du$
$\sum_{k=1}^{n}\frac{1}{k}=\frac{1}{2n}+\frac{\pi}{2}\int_{0}^{1} (1-u)\cot{\frac{\pi u}{2}}\left(1-\cos{\pi n u}\right)\,du$
There is also a generalization that goes beyond the harmonic numbers.
For a harmonic progression where $a$ and $b$ are integers:
$\sum _{k=1}^n \frac{1}{a k+b}=-\frac{1}{2b}+\frac{1}{2(a n+b)}+2\pi\int_0^1 (1-u)\sin[\pi a n u]\sin[(a n+2b)\pi u]\cot[\pi a u]\,du$
(Notice the formula at the top is a particular case where $a=1$, $b=0$.) For odd powers, the general formula is:
\begin{multline}
\sum_{j=1}^{n}\frac{1}{(a j+b)^{2k+1}}=-\frac{1}{2b^{2k+1}}+\frac{1}{2(a n+b)^{2k+1}}+(-1)^{k}(2\pi)^{2k+1}\\ \int_{0}^{1}\sum_{j=0}^{k}\frac{B_{2k-2j}\left(2-2^{2k-2j}\right)}{(2k-2j)!(2j+1)!}(1-u)^{2j+1}\sin{[\pi a nu]}\sin{[(a n+2b)\pi u]}\cot{[\pi au]}\,du
\end{multline}
And for a harmonic progression where $a$ and $b$ are complex:
\begin{multline}\nonumber
\sum_{j=1}^{n}\frac{1}{(ai j+b)^k} =-\frac{1}{2b^k}+\frac{1}{2(ai n+b)^k}+ e^{-2\pi b/a}\left(\frac{2\pi}{a}\right)^k\\ \int_{0}^{1}\sum_{j=1}^{k}\frac{\phi\left(e^{-2\pi b/a},-j+1,0\right)(1-u)^{k-j}}{(j-1)!(k-j)!}e^{2\pi bu/a}\left(\frac{\sin{2\pi n u}}{2}+i \frac{1-\cos{2\pi n u}}{2}\right)\cot{\pi u}\,du
\end{multline}
The proofs are on papers that I've posted to the arXiv.
| {
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Order of $b$ when $ba=ab^3$ and $\operatorname{ord}(a)=4$.
Let $G$ be a non-abelian group and let $a,b \in G$ such that
*
*$\operatorname{ord}(a) = 4$, and
*$ba=ab^3$
Given that $\operatorname{ord}(b)$ is an odd prime number, compute its value.
So, $\operatorname{ord}(b)$ may be $3,5,7,11,13,...$ but I have no idea how I can compute it precisely.
I will be grateful for any help.
| You can find out a multiple of the order of $b$ by shifting $a^4 = e$ past $b$, one $a$ after the other:
$$\begin{align}
b &= b\cdot e = b(a^4) = (ba)(a^3) = a(b^3a)a^2\\
&= a (b^2ab^3)a^2 = a(bab^6)a^2 = a^2 (b^9) a^2\\
&= a^3(b^{27}a) = a^4 b^{81}\\
&= b^{81}
\end{align}$$
So we have $b^{81} = b \iff b^{80} = e$, hence the order of $b$ divides $80 = 2^4\cdot 5$. The order of $b$ is an odd prime dividing $80$, hence $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplify a radical expression for a limit I'm supposed to find the slope of the tangent of the function $$D(p)=\frac{20}{\sqrt{p-1}}$$ at $p=5$.
Using the definition of slope of a tangent at a point, I get
$$\lim_{h \to 0}\frac{\frac{20}{\sqrt{(5+h)-1}}-10}{h}$$
Which, using "regular" (high school) algebra I simplified to $$\frac{10(2-\sqrt{h+4})}{h\sqrt{h+4}}.$$
Rationalizing the numerator or denominator doesn't help in terms of removing $h$ as a factor in the denominator of the slope. How do you find the limit algebraically in this case (or can you)?
| Multiplying by the conjugate, we obtain:
$$ \begin{align*}
\dfrac{10(2-\sqrt{h+4})}{h\sqrt{h+4}}
&=\dfrac{10(2-\sqrt{h+4})}{h\sqrt{h+4}} \cdot \dfrac{2+\sqrt{h+4}}{2+\sqrt{h+4}}\\
&=\dfrac{10(4-(h+4))}{h\sqrt{h+4}(2+\sqrt{h+4})}\\
&=\dfrac{-10h}{h\sqrt{h+4}(2+\sqrt{h+4})}\\
&=\dfrac{-10}{\sqrt{h+4}(2+\sqrt{h+4})}\\
\end{align*} $$
So taking the limit as $h\to0$ yields:
$$
\dfrac{-10}{\sqrt{0+4}(2+\sqrt{0+4})}=\dfrac{-10}{2(2+2)}=\dfrac{-5}{4}
$$
| {
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} |
Approximation of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $ I would like to find an approximation when $ n \rightarrow\infty$ of $ \frac{n!}{(n-2x)!}(n-1)^{-2x} $. Using Stirling formula, I obtain $$e^{\frac{-4x^2+x}{n}}. $$ The result doesn't seem right!
Below is how I derive my approximation. I use mainly Stirling Approximation and $e^x =(1+\frac{x}{n})^n $.
$$\frac{n!}{(n-1)^{2 x} (n-2 x)!}\approx \frac{\left(\sqrt{2 \pi } n^{n+\frac{1}{2}} e^{-n}\right) (n-1)^{-2 x}}{\sqrt{2 \pi } e^{2 x-n} (n-2 x)^{-2 f+n+\frac{1}{2}}}\approx \frac{n^{n+\frac{1}{2}} e^{-2 x} n^{-2 x} \left(1-\frac{1}{n}\right)^{-2 x}}{(n-2 x)^{n-2 x+\frac{1}{2}}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} \left(\left(1-\frac{1}{n}\right)^n\right)^{-\frac{2 x}{n}}\approx e^{-2 x} \left(\frac{n}{n-2 x}\right)^{n-2 x+\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{-n+2 x-\frac{1}{2}} \left(\frac{n-2 x}{n}\right)^n e^{\frac{2 x}{n}}\approx \left(\frac{n-2 x}{n}\right)^{2 x-\frac{1}{2}} e^{\frac{2 x}{n}}\approx \left(\left(1-\frac{2 x}{n}\right)^n\right)^{\frac{2 x}{n}-\frac{1}{2 n}} e^{\frac{2 x}{n}}\approx \left(e^{-2 x}\right)^{\frac{2 x}{n}-\frac{1}{2 n}}=e^{\frac{x-4 x^2}{n}}$$
I would appreciate your input!
| We will first approximate $$f(n,k)=\frac{n!}{(n-k)!}n^{-k} = \prod_{i=0}^{k-1} \left(1-\frac{i}{n}\right)$$
Actually, we will approximate the logarithm of this value.
$$\begin{align}\log f(n,k) &= \sum_{i=0}^{k-1} \log\left(1-\frac{i}{n}\right)\\&= -\sum_{i=0}^{k-1}\sum_{j=1}^\infty \frac{1}{j}\left(\frac{i}{n}\right)^j\\&=-\sum_j \frac{1}{jn^{j}}\sum_{i=0}^{k-1} i^j
\end{align}$$
So $$\log f(n,k) = -\frac{k(k-1)}{2n} -\frac{k(k-1)(2k-1)}{12n^2} + O(n^{-3})$$
Also $$\log\left(\left(\frac{n-1}{n}\right)^{-k}\right) = k\left(\frac{1}{n}+\frac{1}{2n^2}+O(n^{-3})\right)=\frac{k}{n}+\frac{k}{2n^2} + O(n^{-3})$$
Adding,we get that $$\log\left(\frac{n!}{(n-k)!}(n-1)^{-k}\right)=\frac{2k-k(k-1)}{2n}+O(n^{-2}) = \frac{3k-k^2}{2n}+O(n^{-2})$$
Putting $k=2x$ we get:
$$\frac{n!}{(n-2x)!}(n-1)^{-2x} = \exp\left(\frac{3x-2x^2}{n}+O(n^{-2})\right)$$
| {
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If $a,b,c$ are in Geometric Progression and$ a-b,c-a,b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to? If $a,b,c$ are in Geometric Progression and $a-b$,$c-a$,$b-c$ are in Harmonic Progression, then the value of $a+4b+c$ is equal to?
The statements I could get are
1)$b^2=ac$
2)$(c-a)^2=2(b-a)(b-c)$
I get that $(a+4b+c)^2=24ac+6ab+6bc$.
how to proceed and get to the answer??Options given are
a)1
b)0
3)2abc
d)b^2+ac
| The condition for a harmonic progression is that
$$ \frac{1}{a-b} + \frac{1}{b-c} = \frac{ 2}{ c-a}. $$
Clearing denominators, we get
$$ a^2 -2b^2 + c^2 + 2ab + 2 bc - 4ac = 0. $$
This gives us
$$(a+b+c)^2 = a^2 + b^2 +c^2 + 2ab + 2bc + 2ac = 0 + 3b^2 + 6ac =9b^2.$$
Hence, either $ a+b+c = 3b$ or $a+b+c = -3b$.
Now, in the first case, since $a+c = 2b$ and $ac = b^2$, hence we must have $a=b=c$. This means that $a-b, c-a, b-c$ is not a harmonic progression.
Thus, we must have $a+b+c = - 3b$, or that $a+4b+c = 0$.
| {
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A natural number equation For what values of $n$ the equation $x^2 - (2n+1)xy + y^2 + x = 0$ has no solution in natural numbers ? (for $n=1$ it has a trivial solution).
| Letting $m=2n+1$ just for brevity, we see that:
$$(x-y)^2 = x((m-2)y-1)\\
(x+y)^2 = x((m+2)y-1)$$
Now, note that if $p|x$, then $p|x+y$ so $p|y$ and therefore $p\not\mid (m\pm2)y-1$.
Therefore, $x,(m-2)y-1$ are relatively prime, so must be perfect squares. Similarly, $(m+2)y-1$ must be a perfect square.
In the original equation, we can substitute $x=x_0^2$ and $y=x_0y_0$ (necessarily, $x_0|y$) and dividing out the common factors to get:
Substituting, this means we need to solve:
$$x_0^2 -mx_0y_0 + y_0^2 + 1 = 0$$
This has an integer solution if and only if $D = (my_0)^2 - 4(y_0^2+1) = (m^2-4)y_0^2 - 4$ is a perfect square, which means if and only if we have solution to:
$$U^2 - (m^2-4)V^2 = -4$$
Mike Bennett below in comments asserts this does not have a solution except when $m=\pm 3$, which coincides with $n=-2,1$.
This is not hard to reduce this to the equation: $$W^2-(m^2-4)Z^2=-1$$
The continued fraction expansion of $\sqrt{m^2-4}$, when $m>3$ is odd, can be written as:
$$\sqrt{m^2-4} = \left[m-1,\overline{1,\frac{m-3}{2},2,\frac{m-3}{2},1,2(m-1)},\dots\right]$$
Since the length of the repetition is even, there is no solution to $W^2-(m^2-4)Z^2=-1$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this : $ \prod^{\infty}_{n=2}(1-\frac{1}{n^2}) $ How to solve this :
$$ \prod^{\infty}_{n=2}(1-\frac{1}{n^2}) $$
We can write it as follows :
$$ \prod^{\infty}_{n=2}\frac{(n-1)(n+1)}{n^2} $$
How can we proceed further please guide.. thanks..
| HINT:
$$\text{If }T_r=\frac{(r-1)(r+1)}{r^2},$$
$$T_{r+1}=\frac{r(r+2)}{(r+1)^2}\text{ and } T_{r-1}=\frac{(r-2)(r)}{(r-1)^2}$$
Clearly, the denominator of any $n$th term will be cancelled by the numerators of the previous & the next term for $n\ge3$
Putting $n=2,3,\cdots,n-2,n-1,n$
$$\prod_{2\le r\le n}\frac{(1\cdot3)(2\cdot4)(3\cdot5)\cdots(n-3)(n-1)(n-2)(n)(n-1)(n+1)}{(2\cdot2)(3\cdot3)(4\cdot4)\cdots(n-2)(n-2)(n-1)(n-1)\cdot n\cdot n}=\frac{n+1}{2n}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $\cos^4 \theta −\sin^4 \theta = x$. Find $\cos^6 \theta − \sin^6 \theta $ in terms of $x$. Given $\cos^4 \theta −\sin^4 \theta = x$ , I've to find the value of $\cos^6 \theta − \sin^6 \theta $ .
Here is what I did:
$\cos^4 \theta −\sin^4 \theta = x$.
($\cos^2 \theta −\sin^2 \theta)(\cos^2 \theta +\sin^2 \theta) = x$
Thus
($\cos^2 \theta −\sin^2 \theta)=x$ ,
so $\cos 2\theta=x$ .
Now $x^3=(\cos^2 \theta −\sin^2 \theta)^3=\cos^6 \theta-\sin^6 \theta +3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $
So if I can find the value of $3 \sin^4 \theta \cos^2 \theta -3 \sin^2 \theta \cos^4 \theta $ in terms of $x$ , the question is solved. But how to do that ?
| It would be easier if you decompose $\cos^6 \theta -\sin^6 \theta $ into $(\cos^2 \theta -\sin^2 \theta)(\cos^4 \theta +\cos^2 \theta \sin^2 \theta+\sin^4 \theta)$
| {
"language": "en",
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"source": "stackexchange",
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For $a$, $b$, $c$, $d$ the sides of a quadrilateral, show $ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$. (A generalization of IMO 1983 problem 6)
Let $a$, $b$, $c$, and $d$ be the lengths of the sides of a quadrilateral. Show that
$$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0 \tag{$\star$}$$
Background: The well known 1983 IMO Problem 6 is the following:
IMO 1983 #6. Let $a$, $b$ and $c$ be the lengths of the sides of a triangle.
Prove that $$a^{2}b(a - b) + b^{2}c(b - c) +c^{2}a(c - a)\ge 0. $$
See: here.
A lot of people have discussed this problem. So this problem (IMO 1983) has a lot of nice methods.
Now I found for quadrilaterals a similar inequality. Are there some nice methods for inequality $(\star)$?
| use the ptolemy inequality !
$ab^2(b-c)+bc^2(c-d)+cd^2(d-a)+da^2(a-b)\ge 0$$\Longrightarrow$
$2(ab^3+bc^3+cd^3+da^3)\ge$$b^2\cdot{a}(b+c)+c^2\cdot{b}(c+d)+d^2\cdot{c}(a+d)+a^2\cdot{d}(a+b)$$\Longrightarrow$
$2(ab^3+bc^3+cd^3+da^3)\ge$$b^2\cdot{a}l_{1}+c^2\cdot{b}l_{2}+d^2\cdot{c}l_{1}+a^2\cdot{d}l_{2}$$\Longrightarrow$
$8abcd\ge$$b^2\cdot{a}l_{1}+c^2\cdot{b}l_{2}+d^2\cdot{c}l_{1}+a^2\cdot{d}l_{2}$$\Longrightarrow$
$8\ge$$bl_{1}/cd+cl_{2}/ad+dl_{1}/ab+al_{2}/bc$
if $bl_{1}/cd,cl_{2}/ad,dl_{1}/ab,al_{2}/bc\ge2$
then, $l_{1}l_{2}abcd\ge bc\cdot(cd)^2+ad\cdot(ab)^2+ab\cdot(bc)^2+cd\cdot(ad)^2$
$\Longrightarrow$$ac+bd\ge 4\sqrt{abcd}$$\Longrightarrow$$l_{1}l_{2}\ge ac+bd$
which contradict with the ptolemy inequality !
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/466271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "47",
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Generalized Euler sum $\sum_{n=1}^\infty \frac{H_n}{n^q}$ I found the following formula
$$\sum_{n=1}^\infty \frac{H_n}{n^q}= \left(1+\frac{q}{2} \right)\zeta(q+1)-\frac{1}{2}\sum_{k=1}^{q-2}\zeta(k+1)\zeta(q-k)$$
and it is cited that Euler proved the formula above , but how ?
Do there exist other proofs ?
Can we have a general formula for the alternating form
$$\sum_{n=1}^\infty (-1)^{n+1}\frac{H_n}{n^q}$$
| $$
\begin{align}
&\sum_{j=0}^k\zeta(k+2-j)\zeta(j+2)\\
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\sum_{j=0}^k\frac1{m^{k+2-j}n^{j+2}}\tag{1}\\
&=(k+1)\zeta(k+4)
+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{m^2n^2}
\frac{\frac1{m^{k+1}}-\frac1{n^{k+1}}}{\frac1m-\frac1n}\tag{2}\\
&=(k+1)\zeta(k+4)
+\sum_{\substack{m,n=1\\m\ne n}}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{3}\\
&=(k+1)\zeta(k+4)
+2\sum_{m=1}^\infty\sum_{n=m+1}^\infty\frac1{nm^{k+2}(n-m)}-\frac1{mn^{k+2}(n-m)}\tag{4}\\
&=(k+1)\zeta(k+4)
+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{(n+m)m^{k+2}n}-\frac1{m(n+m)^{k+2}n}\tag{5}\\
&=(k+1)\zeta(k+4)\\
&+2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^{k+3}n}-\frac1{(m+n)m^{k+3}}\\
&-2\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}\tag{6}\\
&=(k+1)\zeta(k+4)
+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}
-4\sum_{n=1}^\infty\sum_{m=1}^\infty\frac1{n(n+m)^{k+3}}\tag{7}\\
&=(k+1)\zeta(k+4)
+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}
-4\sum_{n=1}^\infty\sum_{m=n+1}^\infty\frac1{nm^{k+3}}\tag{8}\\
&=(k+1)\zeta(k+4)
+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}
-4\sum_{n=1}^\infty\sum_{m=n}^\infty\frac1{nm^{k+3}}+4\zeta(k+4)\tag{9}\\
&=(k+5)\zeta(k+4)
+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}
-4\sum_{m=1}^\infty\sum_{n=1}^m\frac1{nm^{k+3}}\tag{10}\\
&=(k+5)\zeta(k+4)
+2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}
-4\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{11}\\
&=(k+5)\zeta(k+4)
-2\sum_{m=1}^\infty\frac{H_m}{m^{k+3}}\tag{12}
\end{align}
$$
Letting $q=k+3$ and reindexing $j\mapsto j-1$ yields
$$
\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)
=(q+2)\zeta(q+1)-2\sum_{m=1}^\infty\frac{H_m}{m^q}\tag{13}
$$
and finally
$$
\sum_{m=1}^\infty\frac{H_m}{m^q}
=\frac{q+2}{2}\zeta(q+1)-\frac12\sum_{j=1}^{q-2}\zeta(q-j)\zeta(j+1)\tag{14}
$$
Explanation
$\hphantom{0}(1)$ expand $\zeta$
$\hphantom{0}(2)$ pull out the terms for $m=n$ and use the formula for finite geometric sums on the rest
$\hphantom{0}(3)$ simplify terms
$\hphantom{0}(4)$ utilize the symmetry of $\frac1{nm^{k+2}(n-m)}+\frac1{mn^{k+2}(m-n)}$
$\hphantom{0}(5)$ $n\mapsto n+m$ and change the order of summation
$\hphantom{0}(6)$ $\frac1{mn}=\frac1{m(m+n)}+\frac1{n(m+n)}$
$\hphantom{0}(7)$ $H_m=\sum_{n=1}^\infty\frac1n-\frac1{n+m}$ and use the symmetry of $\frac1{m(n+m)^{k+3}}+\frac1{n(n+m)^{k+3}}$
$\hphantom{0}(8)$ $m\mapsto m-n$
$\hphantom{0}(9)$ subtract and add the terms for $m=n$
$(10)$ combine $\zeta(k+4)$ and change the order of summation
$(11)$ $H_m=\sum_{n=1}^m\frac1n$
$(12)$ combine sums
| {
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"source": "stackexchange",
"question_score": "79",
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How can I solve $\cos^2 x + \sin x +1 = 0$? The solution set of the equation $$\cos^2 x + \sin x +1 = 0$$ is? I haven't studied trigonometry, I'm kinda lost on this issue ...
| We can use the identity $\sin^2 x + \cos^2 x = 1$.
$$\begin{align} \cos^2 x + \sin x + 1 & = 0 \\ \\ (1 - \sin^2 x) + \sin x + 1 & = 0 \\ \\ \sin^2 x - \sin x - 2 & = 0\end{align}$$
Now, you have a quadratic in $\sin x$. For example, set $t = \sin x$, and you'll have a quadratic in $t$, find the roots, then find the corresponding solutions to each root.
Putting $\sin x = t,$ we obtain $$t^2 - t - 2 = (t-2)(t+1) = 0$$
Roots: $t = 2,\;\;t = -1$.
Since $t = \sin x$, that means that
*
*$t = \sin x = 2$ (impossible, since $-1 \leq \sin x \leq 1, \;\forall x \in \mathbb R$), or
*$t = \sin x = -1 \implies x = 3\pi/2 + 2k\pi$, where $k$ is any integer.
| {
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Prove that $|z| \le |z-1|+1$ and $|z-1| \le |z|+1$ Prove that for any $ z \in \mathbb{C}$:
*
*$|z| \le |z-1|+1$
*$|z-1| \le |z|+1$
I take any $z \in \mathbb{C}$ such that $z=a+bi$ for some $a,b \in \mathbb{R}$. I received:
*
*$\sqrt{a^2 +b^2} \le \sqrt{a^2+b^2 -2a+1}+1$
*$\sqrt{a^2+b^2 -2a+1}\le \sqrt{a^2+b^2}+1$
But I don't know how can I prove that these inequality are always true.
| We do it the ugly way, to show it can be done. But the right way is to use the Triangle Inequality. For example, we want to show that
$$\sqrt{a^2+b^2} \le \sqrt{(a-1)^2+b^2}+1.\tag{1}$$
(Note that we did not expand yet. Why bother, unless it is useful?)
Since both sides are non-negative, Inequality (1) is equivalent to the following inequality, obtained by squaring both sides:
$$a^2+b^2\le (a-1)^2+b^2+2\sqrt{(a-1)^2 +b^2}+1.\tag{2}$$
This inequality is equivalent to
$$2a-2\le 2\sqrt{(a-1)^2+b^2}.$$
(Now we did expand one of the $(a-1)^2$.)
But $b^2\ge 0$, so $2\sqrt{(a-1)^2+b^2}\le 2\sqrt{(a-1)^2}=2|a-1|$. Thus we only need to show that
$$2(a-1)\le 2|a-1|.$$
This is obvious.
A similar manipulation takes care of the second inequality.
| {
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How to factor $2x^2-x-3$? I know its:
$$(x+1)(2x-3)$$
But how do you come to that conclusion?
| Solution 1:
We usually do some tests by replacing x=0; x=1 and x=-1 in the equation (usual roots, you can take it as habit) here we can notice that -1 is a solution for the equation (root) so :
$$2x^2 - x - 3 =(x+1)\times 2 \times Q$$ and $$Q=ax+b;$$
Variables a and b can be extracted easily for this example (sometimes we can even do an euclidean division)
Result is :
$$ 2x^2 - x - 3 =(x+1)\times2\times(x-3)=2(x+1)(x-3)$$
Solution 2:
The equation is 2nd degree one so either it has 1 double root, 2 roots or no solution in R
(1) $$2x^2 - x - 3= ax^2 +bx+c=0$$ so a=2, b=-1 and c=-3
We calculate $$Delta=b^2-(4ac)=1-(4\times2\times(-3))=25 >0$$
We have two roots for equation 1 so we can write : $$2x^2 - x - 3 = a(x-x1)(x-x2)$$
$$x1=\frac{-b - \sqrt{Delta}}{2a} = \frac{(1-5)}{4}=-1;$$
$$x2=\frac{-b + \sqrt{Delta}}{2a} = \frac{(1+5)}{4}=3;$$
So
$$2x^2 - x - 3 = a(x-x1)(x-x2) = 2(x+1)(x-3)$$
| {
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Solving for $\tan \theta$ given $\sin \theta/2$ QUESTION:
I'm having a hard time figuring this problem out. I've looked through my lectures and cannot find a problem that relates to this one. I do have my identities pulled up in front of me. I'm unsure where to start though. Can someone give me a kick in the right direction. Also could someone give me some rep points so I can format my questions nicer next time. Thanks
PROBLEM:
Find $\tan \theta$ if $\sin(\theta/2) = 3/5$
| Since we're given $\sin\frac{\theta}{2}=\frac{3}{5}$, we can use the Half-Angle Formula for Sine to write $\tan\theta$ as a function of $\sin\frac{\theta}{2}$.
Now normally the Half-Angle Formula for Sine is $\sin\frac{\theta}{2}=\pm\sqrt{\frac{1-\cos\theta}{2}}$, with the $\pm$ being dependent on which quadrant the angle $\frac{\theta}{2}$ lies within. However, for sake of simplicity, we'll just skip to starting with the equation $\left|\sin\frac{\theta}{2}\right|=\sqrt{\frac{1-\cos\theta}{2}}$ instead.
$\left|\sin\frac{\theta}{2}\right|=\sqrt{\frac{1-\cos\theta}{2}}$
$\left(\left|\sin\frac{\theta}{2}\right|\right)^2=\left(\sqrt{\frac{1-\cos\theta}{2}}\right)^2$
$\sin^2\left(\frac{\theta}{2}\right)=\frac{1-\cos\theta}{2}$
$2\sin^2\left(\frac{\theta}{2}\right)=1-\cos\theta$
$\cos\theta=1-2\sin^2\left(\frac{\theta}{2}\right)$
$\sec\theta=\frac{1}{1-2\sin^2\left(\frac{\theta}{2}\right)}$
$\sec^2\theta=\frac{1}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}$
$\sec^2\theta-1=\frac{1}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}-1$
$\tan^2\theta=\frac{1}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}-\frac{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}$
$\tan^2\theta=\frac{1-\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}{\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}$
And in the end we get
$\tan\theta=\pm\frac{\sqrt{1-\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}}{1-2\sin^2\left(\frac{\theta}{2}\right)}$
Note that the $\pm$ is still dependent on quadrant.
From here, we can simply substitute in for $\sin\frac{\theta}{2}$ with $\frac{3}{5}$.
$\tan\theta=\pm\frac{\sqrt{1-\left[1-2\sin^2\left(\frac{\theta}{2}\right)\right]^2}}{1-2\sin^2\left(\frac{\theta}{2}\right)}$
$\tan\theta=\pm\frac{\sqrt{1-\left[1-2\left(\frac{3}{5}\right)^2\right]^2}}{1-2\left(\frac{3}{5}\right)^2}$
$\tan\theta=\pm\frac{\sqrt{674}}{7}$
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Problem finding $\int\sqrt[3]{\frac{x+1}{x-1}}dx$ Given the integral $\int\sqrt[3]{\frac{x+1}{x-1}}dx$, I can't find its value. I can't one correct substitution. I tried $t=\sqrt[3]{x+1}$ but then $dt\cdot {3\sqrt[3]{x+1}^2}=dx$ but then I'm stuck with integrating $\frac{3t^3}{\sqrt[3]{t^3-2}}$.Almost the same when substituing $t=\sqrt[3]{x-1}$. I also tried multiplying and diving by $\sqrt[3]{(\frac {x+1}{x-1})^{2/3}}$ but again 3rd root ruins everything. Which substitution will solve the integral?
| Hint: Put $$\frac{x+1}{x-1}=u^3$$ Then, $$dx=\frac{3u^2(u^3-1)-3u^2(u^3+1)}{(u^3-1)^2}du=\frac{-6u^2}{(u^3-1)^2}du$$the integral becomes $$-\int \frac{6u^3}{(u^3-1)^2}du $$Then use partial fraction expansion.
| {
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Congruence Equation $3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$ How to solve the following congruence equation?
$$3n^3+12n^2+13n+2\equiv0,\pmod{2\times3\times5}$$
If $t_n$ be the $n$th triangular number, then
$$t_1^2+t_2^2+...+t_n^2=\frac{t_n(3n^3+12n^2+13n+2)}{30}$$
so if we solve this congruence equation we find the values of $n$ that $t_n$ divides $t_1^2+t_2^2+...+t_n^2$.
| Work first modulo $2$. We can rewrite the congruence as $n^3+n\equiv 0\pmod{2}$. Note that this always holds.
Now work modulo $3$. The congruence can be rewritten as $n+2\equiv 0\pmod{3}$. This holds precisely if $n\equiv 1\pmod{3}$.
Finally, work modulo $5$. The congruence can be rewritten as $3n^3-3n^2+3n-3\equiv 0\pmod{5}$, and then as $(n-1)(n^2+1)\equiv 0\pmod{5}$. This has the solutions $n\equiv 1\pmod{5}$, $n\equiv 2\pmod{5}$ and $n\equiv 3\pmod{5}$.
So the conditions come down to $n\equiv 1\pmod{3}$ and $n\equiv 1$, $2$, or $3$ modulo $5$.
The solutions are therefore $n\equiv 1$, $n\equiv 7$ and $n\equiv 13\pmod{15}$.
| {
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Differentiating w. r. t. $x$
Differentiate $$ \text{arccot} \frac{1-x}{1+x} $$ with respect to $x$
After putting $x= \cos \theta$, I got $$\text{arccot} \left(\tan^2 \frac{\theta}{2}\right)$$
Then how do I reach the answer?
Thank you in advance
| Method $1:$
HINT:
Putting $x=\tan \theta \implies \theta=\arctan x$
$$\frac{1-x}{1+x}=\frac{1-\tan\theta}{1+\tan\theta}=\frac{\tan\frac\pi4-\tan\theta}{1+\tan\theta\tan \frac\pi4}\text{ as } \tan\frac\pi4=1$$
$$=\tan\left(\frac\pi4-\theta\right)=\cot\{\frac\pi2-\left(\frac\pi4-\theta\right)\}=\cot\left(\frac\pi4+\theta\right)=\cot\left(\frac\pi4+\arctan x\right)$$
Then, $$ \text{arccot} \left(\frac{1-x}{1+x}\right) =?$$
Method $2:$
Let $$ y= \text{arccot} \frac{1-x}{1+x}\implies \cot y=\frac{1-x}{1+x} $$
Applying Componendo and dividendo, $$x=\frac{1-\cot y}{1+\cot y}=\frac{\tan y-1}{1+\tan y} (\text{ multiplying the numerator & the denominator by}\tan y)$$
$$\implies x=\frac{\tan y-\tan\frac\pi4}{1+\tan\frac\pi4\tan y}=\tan\left(y-\frac\pi4\right)\text{ using }\tan(A-B)=\frac{\tan A-\tan B}{1+\tan A\tan B}$$
Differentiating wrt $y,$
$$ \frac{dx}{dy}=\frac{d \tan\left(y-\frac\pi4\right)}{d(y-\frac\pi4)}\cdot\frac {d(y-\frac\pi4)}{dy}=\sec^2\left(y-\frac\pi4\right)=1+\tan^2\left(y-\frac\pi4\right)=1+x^2$$
Now use this
| {
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Factoring $x^5 + x + 1$
Factor. $\displaystyle x^5 + x + 1$.
I'm trying my hand on these types of expressions. In the book, it mentions that for these types you should rewrite the $x$ term such as $2x-x$. But I don't know how to make any use of this technique.
What is this type of factoring called by the way?
| As Gerry suggests, we can add and subtract $\;+ x^2 - x^2 = 0 $ without changing the expression:
$$\begin{align} x^5 + x + 1 & = (x^5-x^2)+(x^2+x+1)\\ \\ & = x^2(x^3 - 1) + (x^2 + x + 1) \\ \\ & = x^2(x-1)\color{blue}{\bf (x^2 + x + 1)} + \color{blue}{\bf (x^2 + x + 1)}\end{align}$$
Now factor out the common factor...which gives us $$\Big(x^2(x-1) + 1\Big)(x^2 + x + 1) = (x^3 - x^2 + 1)(x^2 + x + 1)$$
| {
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Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$ Given that $T=\frac{x-iy}{x+iy}$, where $x, y, T \in\Bbb R$, show that $$\frac{1+T^2}{2T}=\frac{x^2-y^2}{x^2+y^2}$$
I have got rid of $i$ in the denominator, leaving $$\frac{x^2-y^2-2ixy}{x^2+y^2}$$
but now I am stuck. What are the next steps please?
| It may be easier to note that our expression is equal to
$$\frac{1}{2}\left(\frac{1}{T}+T\right).$$
Substituting, we get
$$\frac{1}{2}\left(\frac{x+iy}{x-iy}+\frac{x-iy}{x+iy} \right).$$
Bring to the common denominator $x^2+y^2$. The top is then $(x+iy)^2+(x-iy)^2=2x^2-2y^2$, and we are finished.
| {
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Partial Fraction decomposition when denominator is in $x^2 + a$ form Why isn't the expansion of $$ \frac{s^3 - 2s^2 + 16s - 2}{(s^2+1)(s^2+16)} $$
in the form of
$$ \frac{As+B}{s^2+1} + \frac{Cs+D}{s^2+16} $$ since (as I recall) the denominator is of square power and you should decompose it (in the numerator) until the s diminishes.
Apparently wolframalpha is saying the correct form should be
$$ \frac{A}{s^2+1} + \frac{B}{s^2+16} $$
Please clarify.
Thank you in advance.
| There is a fast way, which is seldom seen on a calculus textbook. Polynomials $s^2 + 1$ and $s^2 + 16$ are coprime and squarefree, so we can solve it in a manner similar to linear denominators.
The numerator on $s^2 + 1$ is the unique representative of
$$\frac {s^3 - 2s^2 + 16s - 2}{s^2 + 16} + \left\langle s^2 + 1 \right\rangle$$
whose degree less than $\deg \left( s^2 + 1 \right) = 2$. Let's denote this as
$$\left\lceil s^2 + 1 \right\rceil \frac {s^3 - 2s^2 + 16s - 2}{s^2 + 16}.$$
To extract multiples of $s^2 + 1$, we can simply set $s^2 = -1$ and let the problem solves itself.
$$\left\lceil s^2 + 1 \right\rceil \frac {s^3 - 2s^2 + 16s - 2}{s^2 + 16}
= \frac {-\left(s - 2\right) + 16s - 2}{15} = s.$$
Similarly,
$$\left\lceil s^2 + 16 \right\rceil \frac {s^3 - 2s^2 + 16s - 2}{s^2 + 1}
= \frac {-16\left(s - 2\right) + 16s - 2}{-15} = -2.$$
Both numerators are found.
$$\frac {s^3 - 2s^2 + 16s - 2}{\left(s^2 + 1\right) \left(s^2 + 16\right)} = \frac{s}{s^2+1} - \frac{2}{s^2+16}.$$
This method was published in Guoce Xin's A Fast Algorithm for Partial Fraction Decompositions (2004). It is covered in the last section in the article.
| {
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inverse laplace transform - with symbolic variables Transform: $$ F(s) = \frac{2s^2 + (a-6b)s + a^2 - 4ab}{(s^2-a^2)(s-2b)} $$
My steps:
$$ F(s) = \frac{2s^2 + (a-6b)s + a^2 - 4ab}{(s+a)(s-a)(s-2b)} $$
$$ = \frac{A}{s+a} + \frac{B}{s-a} + \frac{C}{s-2b} + K $$
$$ K = 0$$
$$A = F(s) * (s+a) $$ at s = -a
$$ A = \frac{2a^2 + (a-6b)(-a) +a^2 - 4ab}{4ab+2a^2} $$
$$ A = \frac{a+b}{2b+a}$$
I problems like this (that I've seen) at the step above the fraction would reduce into just a number. In this case, it doesn't and its surprising because I've never seen it happened before. Is there something I am doing wrong here?
| The routine way is to use the way we get in partial fractions and you correctly noted that. If we set:
$$ \frac{2s^2 + (a-6b)s + a^2 - 4ab}{(s^2-a^2)(s-2b)}=F= \frac{A}{s+a} + \frac{B}{s-a} + \frac{C}{s-2b}
$$ then by doing boring :-) handy calculations we can find $A,B$ and $C$. We have then:
$$\frac{-Asa+2Aab+As^2-2Asb+Bsa-2Bab+Bs^2-2Bsb+Cs^2-Ca^2}{(-s+2b)(-s^2+a^2)}$$ Now if we put $s=-a$ in the numerators, we get $A=\frac{a+b}{a+2b}$ and this is what you already got. With the similar approach $s=+a$ for $B$ and $s=2b$ for $C$, we get:
$$B=\frac{2a-5b}{a-2b},~~C=\frac{4b^2+2ab-a^2}{a^2-4b^2}$$
| {
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Finding the simplest form How can we simplify the expression below:
For any real numbers $a$ and $b$,
$
\frac{\frac{\frac{a+b}{2}+ b}{3}+ b} {4}+ b \cdots $
This is same as asking what is $$\frac{a+ b\sum_{i=1}^ni!}{n!}$$ as $n$ approaches to $+ \infty$?
Thanks.
| Let's rewrite your sum like this:
$\dfrac{a+ b\sum_{i=1}^ni!}{n!}=\dfrac{a+ b \left( 1!+2!+ \cdots + \left( n-2\right)! + \left( n-1\right)! +n! \right)}{n!}$
$=\dfrac{a}{n!}+b \left( \dfrac{1!}{n!} +\dfrac{2!}{n!} + \cdots \dfrac{ \left( n-2\right)!}{n!} + \dfrac{\left( n-1\right)!}{n!} + \dfrac{n!}{n!} \right)$
$=\dfrac{a}{n!}+b \left( \dfrac{1}{n!} +\dfrac{2}{n!} + \cdots \dfrac{ 1 }{n \left( n-1 \right)} + \dfrac{1}{n} + 1 \right)$
Take the limit as $n$ approches $+ \infty$ and you get $b$.
| {
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Sum of the series $1+\frac{1\cdot 3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+\cdots$ Decide if the sum of the series
$$1+\frac{1\cdot3}{6}+\frac{1\cdot3\cdot5}{6\cdot8}+ \cdots$$
is: (i) $\infty$, (ii) $1$, (iii) $2$, (iv) $4$.
| I believe the general term after the first is
$$\frac{(2n+1)! \cdot 4 \cdot 2}{2^n \cdot n! \cdot 2^{n+2} \cdot (n+2)! } \ , $$
with $ \ n \ \ge \ 1 . $
The series then ought to be convergent. The series is clearly larger than
$$1 \ + \ \sum_{n=0}^{\infty} \frac{1}{2} \cdot \left( \ \frac{5}{8} \ \right)^n \ , $$
so the sum is bigger than 2, but not infinite. That leaves 4 among the choices.
| {
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Help on finding the math behind the 68/95/99 rule for Normal distribution I also see on the section on the Central Limit Theorem that the normal distribution is used and gives the graph of the 68/95/99 rule. However, I haven't seen the actual math used to prove this or how it gets computed. Can anyone give me a link or explain/show how the math behind the statement is done?
| If you don't want to just use numerical integration: for large $t$ the standard normal CDF has the asymptotic series (which can be established by integration by parts)
$$
\Phi(t) \approx 1 + \frac{\exp(-t^2/2)}{\sqrt{2\pi}} \left(- \frac{1}{t} + \frac{1}{t^3} - \frac{3}{t^5} + \frac{3 \cdot 5}{t^7} - \frac{3 \cdot 5 \cdot 7}{t^9} + \ldots\right) $$
and thus
$$P(-t \le X \le t) = 2 \Phi(t) - 1 = 1 + 2 \frac{\exp(-t^2/2)}{\sqrt{2\pi}} \left(- \frac{1}{t} + \frac{1}{t^3} - \frac{3}{t^5} + \frac{3 \cdot 5}{t^7} - \frac{3 \cdot 5 \cdot 7}{t^9} + \ldots\right)$$
This is only an asymptotic series, which does not converge for any $t$. But
if $t$ is large you can get good approximations using appropriate partial sums: the value lies between any two adjacent partial sums.
Thus $P(-2 \le X \le 2)$ lies between $1 + 2 \dfrac{\exp(-2)}{\sqrt{2\pi}} \left(-\dfrac{1}{2} + \dfrac{1}{8}\right) \approx 0.9595$ and
$1 + 2 \dfrac{\exp(-2)}{\sqrt{2\pi}} \left(-\dfrac{1}{2} + \dfrac{1}{8} - \dfrac{3}{32}\right) \approx 0.9494$ (the actual value is $.9544997360$ to $10$ decimal places). Similarly (taking $5$ or $6$ terms of the sum) $P(-3 \le X \le 3)$ lies between approximately $.99728$ and $.99733$ (the actual value is $.9973002039$).
| {
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How to prove :$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$ For sequence $u_n$ satisfing : $$\begin{cases} u_1=\sqrt{2015}\\ u_{n+1}=u_n^2-2\end{cases}$$
How to prove : $$\lim_{n\to+\infty}\left(\dfrac{u_{n+1}}{u_1.u_2...u_n}\right)^2=2011$$
| The recurrence relation may be expressed as
$$
{u_{n}}^2 = \frac{{u_{n+1}}^2 - 4}{{u_{n}}^2 - 4}.
$$
Which allows telescopic cancellation,
$$
\begin{align*}
\left ({u_{1}}{u_{2}}\cdots{u_{n}}\right )^2 &=
\left (\frac{\color{#036}{{u_{2}}^{2} - 4}}{{u_{1}}^2 - 4} \right )
\left ( \frac{\color{#063}{{u_{3}}^{2} - 4}}{\color{#036}{{u_{2}}^2 - 4}} \right )
\cdots
\left ( \frac{{u_{n+1}}^{2} - 4}{\color{#630}{{u_{n}}^2 - 4}} \right )\\[10pt]
&= \frac{{u_{n+1}}^2 - 4}{{u_{1}}^2 - 4}.
\end{align*}
$$
Hence,
$$
\begin{align*}
\lim_{n\to\infty} \left ( \frac{u_{n+1}}{{u_{1}}{u_{2}}\cdots{u_{n}}} \right )^2
&= \lim_{n\to\infty} {u_{n+1}}^2 \cdot \frac{{u_{1}}^2 - 4}{{u_{n+1}}^2 - 4}\\
&= \lim_{n\to\infty} \left ( {u_{1}}^2 - 4\right ) \cdot \frac{1}{1 - 4/{u_{n+1}}^2}\\
&= {u_{1}}^2 - 4 &\text{since $\lim_{n\to\infty} 1/u_{n+1} = 0$}.
\end{align*}
$$
Substituting $u_1 = \sqrt{2015}$ yields the final result
$$\lim_{n\to\infty} \left ( \frac{u_{n+1}}{{u_{1}}{u_{2}}\cdots{u_{n}}} \right )^2 = 2011.$$
| {
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Help with minimization problem help me, if $x$ and $y$ are real such that $3x-4y = 12$, determine the minimum value of $z = x ^ 2 + y ^ 2$?$$$$I thought of $$3x-4y = 12\Longrightarrow x=4\frac{y+3}{3}\\z = x ^ 2 + y ^ 2\Longrightarrow z = \left(4\frac{y+3}{3}\right) ^ 2 + y ^ 2$$ and then?
| To answer your question, "and then?", with details, this is a way the calculation you began could continue, and this is without calculus:
$z=(4\frac{y+3}{3})^2 + y^2 = 16\frac{(y+3)^2}{9} + \frac{9y^2}{9} = \frac{16(y^2+6y+9)+9y^2}{9} = \frac{25}{9}y^2 + \frac{32}{3}y + 16$.
Center of a parabola is at $y = \frac{-b}{2a} = \frac{-32/3}{50/9} = -\frac{48}{25}$.
Finally, $x=4\frac{y+3}{3} = \frac{4(27/25)}{3} =\frac{36}{25}$.
The method described by Brian Scott is a bit easier, I think.
| {
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Prove that $1^3 + 2^3 + 3^3 +\cdots+ n^3 = \frac14n^4 + \frac12n^3 + \frac14n^2$ I have to prove that this is true using mathematical induction.
I have this:
for every $n \in \mathbb N$: $1^3 + 2^3 + 3^3 + ... + n^3 = \frac 14n^4 + \frac 12n^3 + \frac 14n^2$
for $n = 1: 1^3 = 1/4 + 1/2 + 1/4$, hence $P(1)$ is true.
Let $N \in \mathbb N$ be given and assume that $P(N)$ is true, that is $$1^3 + 2^3 + 3^3 + ... + N^3 = \frac 14N^4 + \frac 12N^3 + \frac 14N^2$$
For n = $N$ + 1:
And now what? I just couldn't solve it.
| Perhaps the easiest way to go about this involves noticing that $$\frac 14N^4 + \frac 12N^3 + \frac 14N^2=\frac {N^2(N+1)^2}4$$
If you add $(N+1)^3$ to each side of $P(N)$ you get$$1^3+2^3+\dots N^3+(N+1)^3=\frac {N^2(N+1)^2}4+(N+1)^3$$
The left-hand side is what you need for $P(N+1)$ and the task is to show that the right-hand side can also be put into the correct form. You can extract an obvious common factor and put everything over a common denominator and see what happens.
If you don't happen to notice the convenient form for the right-hand side, you can still work with $\frac 14N^4 + \frac 12N^3 + \frac 14N^2$ to get the result.
| {
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Sum $ \sum\limits_{k=1}^{n} (-1)^k \frac{2k+3}{k(k+1)} $ I have the following Sum
$$ \sum\limits_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)} $$
and I need to calculate the sums value by creating the partial sums.
I started by checking if $$\sum\limits_{n=1}^{\infty} \left| (-1)^n \frac{2n+3}{n(n+1)} \right|$$ converges.
i tried to check for convergence with the 2 criterias a) decreasing and b) zero sequence but yes then i tried to transform the equotation to
$$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n(n+1)} $$
the first fraction is "ok" - the 2nd one i did partial fraction decomposition and finally got $$ \sum\limits_{n=1}^{\infty} \frac{2}{n+1} + \frac{1}{n} - \frac{1}{n+1} $$
i then tried to see a pattern by find out the first sequences but im not sure if i'm on the right track.
| i think this is an easy answer
$$ \sum_{n=1}^{\infty} (-1)^n \frac{2n+3}{n(n+1)} $$
$$ \frac{2n+3}{n(n+1)} = \frac{3(n+1) - n}{n(n+1)} = \frac{3}{n} - \frac{1}{n+1} $$
$$ \sum_{n=1}^{\infty} \left(\frac{3(-1)^n}{n} - \frac{(-1)^n}{n+1} \right) $$
$$ = \int_0^1 3\sum_{n=1}^{\infty} (-1)^n x^{n-1} - \sum_{n=1}^{\infty} (-1)^n x^n \ dx $$
$$ = \int_0^1 -\frac{3}{1+x} + \frac{x}{x+1} \ dx $$
$$ =\int_0^1 \frac{x+1 - 4}{x+1} \ dx = \int_0^1 1 - \frac{4
}{x+1} \ dx = \left[x - 4\ln(x+1) \right]_0^1 = 1 - 4\ln 2 $$
i think you know $$ \sum_{n=1}^{\infty} x^n = \frac{x}{1-x} $$
| {
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Math inequality proof If $a, b$ are positive real numbers and $a + b = 1$, prove that
$$
\left(a +\frac{1}{a}\right)^2 + \left(b +\frac{1}{b}\right)^2 \geq \frac{25}{2}
$$
Thank you.
| yes,use $Cuachy-Schwarz$ inequality,we have
$$\left[\left(a+\dfrac{1}{a}\right)^2+\left(b+\dfrac{1}{b}\right)^2\right][1+1]\ge\left(a+b+\dfrac{1}{a}+\dfrac{1}{b}\right)^2=\left(1+\dfrac{1}{a}+\dfrac{1}{b}\right)^2$$
use $Cauchy-Schwarz $ inequality,we have
$$\left(\dfrac{1}{a}+\dfrac{1}{b}\right)(a+b)\ge(1+1)^2$$
| {
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Proving that a sequence is between certain values at certain n I'm given that $a_1=1$, and for every $n \gt1, a_{n+1} = a_n + \frac{1}{a_{n}}$. I need to prove that $20 < a_{200} < 24$. I tried finding a limit at infinity setting both limits to $L$ ( for $a_n$ and $a_{n+1}$ but that gave me nothing useful. All I figured out was that this sequence is increasing..
| I've decided to submit my answer.
It's a little different from Oleg567's,
but the results are the same.
I first try to estimate the growth of $a_n$.
From the desired bounds,
it looks like it might be of order
$\sqrt{n}$.
Suppose
$a_n
=c\sqrt{n}
$.
Then
$c\sqrt{n}+1/(c\sqrt{n})
=c\sqrt{n+1}
=c\sqrt{n}\sqrt{1+1/n}
\approx c\sqrt{n}(1+1/(2n))
= c\sqrt{n}+c/(2\sqrt{n})
$
or
$1/(c\sqrt{n})
\approx c/(2\sqrt{n})
$
or
$1/c \approx c/2$
or
$c \approx \sqrt{2}$.
It therefore looks like
$a_n \approx \sqrt{2n}$.
Suppose $a_n > \sqrt{2n}$.
Then
$a_{n+1}
> \sqrt{2n}+1/\sqrt{2n}
$.
To show
$a_{n+1} > \sqrt{2(n+1)}$,
we need
$\sqrt{2n}+1/\sqrt{2n}
> \sqrt{2(n+1)}$.
Squaring both sides,
this is
$2n+2+1/(2n)
> 2(n+1)
$
which is true.
Setting $n=200$,
$a_{200} > \sqrt{400} = 20$.
Suppose
$a_n \le \sqrt{(2+k)n}$ for some $k$.
Then, since
$a_n^2 > 2n$,
$a_{n+1}^2
=a_n^2+2+1/a_n^2
< (2+k)n+2+1/(2n)
$
and for this to be
$\le (2+k)(n+1)$
we need
$(2+k)n+2+1/(2n)
\le (2+k)(n+1)$
or
$2+1/(2n) \le 2+k$
or
$1/(2n)\le k$.
This will certainly be true for
large enough $n$.
We have
$a_2 = 2$, $a_3 = 5/2$,
and $a_4 = 5/2+2/5 = 29/10$.
For $n=3$,
we want
$25/4 \le 3(2+k)=6+3k$
or
$1/4 \le 3k$
or $k \ge 1/12$.
Since we also want
$k \ge 1/(2n) = 1/6$,
we can use $k = 1/6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/489099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Prove that if $a_{n+1} = a_n^2$, the last $n$ digits of $a_{n+1}$ are the same as the last $n$ digits of $a_n$. I have been working on this problem for a while. I know that I have to prove it using induction, but I'm unsure of the next step. The formula for the terms is: $a_{n+1} = 5^{2n}$ with $a_1 = 5$. The sequence is 5, 25, 625, etc. This is what I have thus far:
$a_{n+1} = a_n^2 = 5^{2n}$
$5^{2(n+1)} - 5^{2n} \equiv 0 \pmod {10^{n+1}}$
$5^{2n}5^2 - 5^{2n} \equiv 0 \pmod {10^{n+1}}$
$5^{2n}(24) \equiv 0 \pmod {10^{n+1}}$
Any hints? Thanks
| The correct formula for $a_{n+1}$ in terms of $a_1$ is
$$
a_{n+1}=a_1^{2^n}\tag{1}
$$
You have $a_1=5$, so what you want to show is that
$$
5^{2^n}\equiv 5^{2^{n-1}}\pmod{10^n}\tag{2}
$$
which is equivalent to
$$
10^n\mid(5^{2^{n-1}}-1)5^{2^{n-1}}\tag{3}
$$
Now,
$$
\begin{align}
5^{2^{n-1}}-1
&=(5^{2^{n-2}}+1)(5^{2^{n-2}}-1)\\
&=(5^{2^{n-2}}+1)(5^{2^{n-3}}+1)(5^{2^{n-3}}-1)\\
&\hphantom{=\ }\vdots\\
&=\underbrace{(5^{2^{n-2}}+1)(5^{2^{n-3}}+1)(5^{2^{n-4}}+1)\dots(5^{2^0}+1)(5^{2^0}-1)}_{\text{$n$ terms, all of which are even, so divisible by $2^n$}}\tag{4}
\end{align}
$$
Therefore,
$$
2^n\mid5^{2^{n-1}}-1\tag{5}
$$
Furthermore, $2^{n-1}\ge n$ for $n\ge1$, so
$$
5^n\mid5^{2^{n-1}}\tag{6}
$$
$(5)$ and $(6)$ show that $(3)$ is true, which is equivalent to $(2)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/490025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Domain and Range of f(x) Find the domain and range of Function?
$$F(x) = \frac{1}{\sqrt{25-x^2}}$$
Domanin f(x) has real value , if
$$25-x^2\geqslant0$$
$$-x^2\geqslant-25$$
$$x^2\le25$$
$$-5\le x\le5$$
$$D_f= [-5,5]$$
let y= f(x)
$$y = \frac{1}{\sqrt{25-x^2}}$$
then what should i do to find range?
| As noted above, the domain is $(-5,5)$.
Another way to find the range is to find the values of $y$ for which the equation
$\displaystyle y=\frac{1}{\sqrt{25-x^2}}$ has a solution for x:
1) Since $\displaystyle\frac{1}{\sqrt{25-x^2}}>0,$ we have that $y>0$.
2) $\displaystyle y^2=\frac{1}{25-x^2} \implies 25-x^2=\frac{1}{y^2}\implies x^2=25-\frac{1}{y^2}=\frac{25y^2-1}{y^2}=\frac{(5y+1)(5y-1)}{y^2}$,
so
$\;\;\;\;\displaystyle x^2\ge0\implies \frac{(5y+1)(5y-1)}{y^2}\ge0 \implies y\le -1/5$ or $y\ge1/5$.
Since $y>0$ and $y\le-1/5$ or $y\ge1/5$, $\;\;\;\;y\ge1/5\;\;\;;$ so the range is $[1/5,\infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/491905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\int \frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}} \mathrm dx$ I am having real trouble evaluating this indefinite integral
$$\int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx$$
Need I mention that I already tried WolframAlpha with little success? It returned a complicated expression full of many $\arctan$ terms.
Here is what I have already tried:
*
*Seperating denominator as $(1-\sqrt{2}\sin{x}\cos{x})(1-\sqrt{2}\sin{x}\cos{x})$ and applying partial fractions.
*Multiplying and dividing by $\sec^4{x}$
Please provide any help. It would be appreciated if you give a hint rather than the whole solution.
| $$\sin^4x+\cos^4x=(\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x=1-\frac{\sin^22x}2$$
If we set $\int(\sin x+\cos x)dx=\sin x-\cos x,\sin2x=1-u^2$
$$\implies \int{\frac{\sin{x}+\cos{x}}{\sin^4{x}+\cos^4{x}}}dx=2\int\frac{du}{2-(1-u^2)^2}=2\int\frac{du}{(\sqrt2+1-u^2)(\sqrt2-1+u^2)}$$
$$=\frac2{2\sqrt2}\int\frac{(\sqrt2+1-u^2)+(\sqrt2-1+u^2)}{(\sqrt2+1-u^2)(\sqrt2-1+u^2)}du$$
$$=\frac1{\sqrt2}\left(\int\frac{du}{\sqrt2-1+u^2}+\int\frac{du}{\sqrt2+1-u^2}\right)$$
Now apply $\displaystyle\int\frac{dy}{a^2+y^2}=\frac1a\arctan \frac ya+K$
and $\displaystyle\int\frac{dy}{a^2-y^2}=\frac1{2a}\ln\big|\frac{a+y}{a-y}\big|+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
Finding $a, b\in\mathbb{R}$ so that $\lim_{x\to \infty} \sqrt{ax^2 + bx} - \sqrt{x^2 + x + 1} = 1$ Find the values of real constants $a$ and $b$ such that
$$\lim_{x\to \infty} \sqrt{ax^2 + bx} - \sqrt{x^2 + x + 1} = 1$$
If I obtained $a = 1$ and $b=0$, is that the correct answer?
Or is $b$ actually some other value? If so, then why?
| Hint: $$\dfrac{ax^2 + bx - x^2 - x - 1}{\sqrt{ax^2 + bx} + \sqrt{x^2 + x + 1}} = \dfrac{(a-1)x + (b-1) - \frac{1}{x}}{\sqrt{a + \frac{b}{x}} + \sqrt{1 + \frac{1}{x} + \frac{1}{x^2}}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/493758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that: $\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$ Let $x,y\in \mathbb{R}$ know that $4x^2-9y^2=36$
Prove that:
$$\left | 2x-y-4 \right |\geq 4\sqrt{2}+4$$
| $36 + 9y^2 = 4x^2$ Suggests we could use $y = 2 \tan A$ to get $x = \pm 3\sec A$. Given the objective, it is sufficient to consider $x = 3 \sec A, ~~ A \in [0, \frac{\pi}{2})$
Then $(2x - y - 4 )^2 = 4(3 \sec A - \tan A - 2)^2 $
It is easily checked that the minimum occurs when $3 \tan A = \sec A$ or when $\sin A = \frac{1}{3}$. So the minimum value is $4(3 \sec A - \tan A - 2)^2 = 4(3\frac{3}{2\sqrt{2}} - \frac{1}{2\sqrt{2}}-2)^2 = \left(4\sqrt{2}-4\right)^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/495377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding a set of representatives in a certain array made from a multiset Let $M$ be an $n \times n$ array whose entries are from the multiset consisting of a single one, three twos, five threes,..., and $2n-1$ $n$'s, with the additional condition that, for $1\leq k\leq n$, no row or column have more than $k$ $k$'s.
Is it the case that one can always find $r_1, r_2,\dots,r_n$, all different, such that row $r_1$ contains a $1$, row $r_2$ contains a $2,\dots,$ and row $r_n$ contains an $n$, or one can find $c_1, c_2,\dots,c_n$, all different, such that column $c_1$ contains a $1$, column $c_2$ contains a $2,\dots,$ column cn contains an $n$?
| It turns out that it is not the case that one can always find a set of distinct representatives for matrices as described in the question. Here is an example of a 9x9 matrix where no set can be chosen:
$$\begin{pmatrix}x & x & x & 6 & 5 & 5 & 5 & 5 & 5\\x & x & x & 5 & 6 & 6 & 6 & 6 & 5\\x & x & x & 6 & 6 & 6 & 6 & 6 & 6\\x & x & x & 7 & 7 & 7 & 7 & 7 & 7\\x & x & x & 7 & 7 & 7 & 7 & 7 & 7\\x & x & x & 7 & 8 & 8 & 8 & 8 & 8\\9 & 8 & 8 & 8 & 8 & 8 & 8 & 8 & 8\\8 & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 8\\9 & 9 & 9 & 9 & 9 & 9 & 9 & 9 & 9\end{pmatrix}$$
The $x$'s may be any appropriate arrangement of the proper number of $1$'s, $2$'s, $3$'s, $4$'s, and $5$'s.
The bottom three rows contain only two different numbers: $8$ and $9$. So its not possible to choose three of them containing three different numbers.
Similarly, the rightmost six columns contain only five different numbers, so its not possible to choose six of them containing six different numbers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/497445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factor $(x+y)^7-(x^7+y^7)$ I encountered the following problem while preparing for upcoming math contests.
Factor $(x+y)^7-(x^7+y^7)$.
I got zero for $(x+y)^7-(x^7+y^7)$, however, the solutions said it's
$$7xy(x+y)(x^2+xy+y^2)(x^2+xy+y^2)$$
Can someone explain how this is possible?
| ( Motivation for this (late) answer is to demistify the derivation of the square factor, and show that it reduces to the routine factorization of an associated quadratic. )
Using that $\,x^7 + 1 = (x+1)(x^6 - x^5 + x^4 - x^3 + x^2 - x + 1)\,$:
$$
(x+1)^7 - x^7 - 1 = (x+1)\left((x+1)^6 - x^6 + x^5 - x^4 + x^3 - x^2 + x - 1\right) \tag{1}
$$
The second factor is a palindromic polynomial, which suggests the substitution $\,u = x + \dfrac{1}{x}\,$. Then $\,x^2 + \dfrac{1}{x^2} = u^2 - 2\,$, $\,x^3 + \dfrac{1}{x^3} = u^3 - 3u\,$, and:
$$
\begin{align}
& (x+1)^6 - x^6 + x^5 - x^4 + x^3 - x^2 + x - 1
\\ =\; &(x^2 + 2x + 1)^3 - x^6 + x^5 - x^4 + x^3 - x^2 + x - 1
\\ =\; &x^3\left(\left(x+\frac{1}{x}+2\right)^3 - \left(x^3 + \frac{1}{x^3} \right)+\left(x^2 + \frac{1}{x^2}\right)-\left(x + \frac{1}{x}\right)+1\right)
\\ =\; &x^3 \left((u+2)^3 - (u^3 - 3u) + (u^2 - 2) - u + 1\right)
\\ =\; &x^3 \left(7 u^2 + 14 u + 7\right)
\\ =\; &7x^3(u+1)^2
\\ =\; &7x(ux+x)^2
\\ =\; &7x(x^2 + x + 1)^2 \tag{2}
\end{align}
$$
Piecing together $\,(1)\,$ and $\,(2)\,$:
$$
(x+1)^7 - x^7 - 1 = 7x (x+1) \left(x^2+x+1\right)^2
$$
Substituting $\,x \mapsto \dfrac{x}{y}\,$ and multiplying by $\,y^7\,$ gives the identity in OP's question.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/498584",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 2
} |
Logic for decomposing a permutation into different products composed of transpositions I know that any permutation cycle can be decomposed into transpositions as follows:
$(a_1,a_2...,a_n) = (a_1,a_{n-1})...(a_1,a_2)$
But in my book there is an example of the following form
$(1,2,3,4,5) = (5,4)(5,2)(2,1)(2,5)(2,3)(1,3)$
I verified that it holds true. But I cant seem to figure out how the author came up with that. Also, What is the generic algorithm to produce all decompositions of permutation in terms of transpositions.
| In his question and comments the OP keeps asking HOW did the author of his book come up with
$\tag a \sigma = (1,2,3,4,5) = (5,4)(5,2)(2,1)(2,5)(2,3)(1,3)$
But we have the transposition algebra rules:
\begin{align}(1) \quad (ab)(ab)&=1_{Id}\\
(2) \quad (ab)(cd)&=(cd)(ab)\\
(3) \quad (ab)(bc)&=(ca)(ab)\\
(4) \quad (ab)(bc)&=(bc)(ca)\end{align}
The first identity (1) can be used to expand a product of $n$ transpositions into another one with length $n+2$ - you can always employ this 'silly padding'. The other manipulations (2) thru (4) can be used to 'disguise' this 'silly padding'.
Observe that given a product of transpositions we can, step by step, 'move' any fixed transposition to a new position in the composition product, but other transpositions might have to be modified along the way.
Since $2$ occurs $4$ times in $(5,4)(5,2)(2,1)(2,5)(2,3)(1,3)$ we write
$\quad (1,2,3,4,5) = (3,4,5,1,2) = (2, 1) (2, 5) (2, 4) (2, 3) =$
$\quad \quad (5 ,4) (5 ,4) (2, 1) (2, 5) (2, 4) (2, 3) (1,3) (1,3) =$
$\quad \quad (5 ,4) (5 ,4) (2, 5) (5, 1) (2, 4) (1, 2) (2,3) (1,3) =$
$\quad \quad (5 ,4) (5 ,2) (2, 4) (5, 1) (2, 4) (1, 2) (2,3) (1,3) =$
$\quad \quad (5 ,4) (5 ,2) (5, 1) (1, 2) (2,3) (1,3) =$
$\quad \quad (5 ,4) (5 ,2) (2, 1) (2, 5) (2,3) (1,3) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 1
} |
lim calculus problem with infinity $$\lim_{n→∞} \left( \left(\frac{n+3}{n+5}\right)^{n+4} + \sqrt[2n]{3n}\right)$$
I tried replacing n+5 with u and simplifying but the answer seems off. Help would be greatly appreciated.
| We need to do some rewriting:
$$\begin{align}
\left(\frac{n+3}{n+5}\right)^{n+4} &= \left(\frac{n+5-2}{n+5}\right)^{n+4} \\
&= \left(1-\frac{2}{n+5}\right)^{n+5-1} \\
&= \underbrace{ \left(1-\frac{2}{n+5}\right)^{n+5} }_{\to e^{-2}}\ \underbrace{ \left(1-\frac{2} {n+5}\right)^{-1} }_{\to 1}
\end{align}
$$
So the left term goes to $e^{-2}$. The right side can be written as $(\sqrt[n]{3n})^{\frac12}$. The $n$-th root of any polynomial in $n$ goes to $1$, so this goes to $1$. The overall limit is $e^{-2}-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/499916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A closed form for $\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt{x}\right)\ e^{-x\sqrt{2}}}{x}dx$ Let $\operatorname{erfi}(x)$ be the imaginary error function
$$\operatorname{erfi}(x)=\frac{2}{\sqrt{\pi}}\int_0^xe^{z^2}dz.$$
Consider the integral
$$I=\int_0^\infty\frac{\sin(x)\ \operatorname{erfi}\left(\sqrt{x}\right)\ e^{-x\sqrt{2}}}{x}dx.$$
Its numeric value is approximately $0.625773669454426\dots$
Is it possible to express $I$ in a closed form using only elementary functions, integers and constants $\pi$, $e$?
| My strategy here is to use Parseval's equality to express the integral in a simpler form. This requires a strategic splitting of the integrand into Fourier transforms.
Begin by writing
$$\text{erfi}({\sqrt{x}})=\frac{2}{\sqrt{\pi}} \sqrt{x} \int_0^1 dt \, e^{x t^2}$$
and consider the following Fourier Transform:
$$\int_{-\infty}^{\infty} dx \, \theta(x) \, \text{erfi}(\sqrt{x}) \, e^{-\sqrt{2} x} \, e^{i k x}$$
where $\theta(x)$ is the Heaviside step function, which is $1$ when $x \gt 0$ and $0$ when $x \lt 0$. Using a change in the order of integration, we may evaluate this Fourier transform in exact form:
$$\begin{align}\int_{0}^{\infty} dx \, \text{erfi}(\sqrt{x}) \, e^{-\sqrt{2} x} \, e^{i k x} &=\frac{2}{\sqrt{\pi}} \int_0^1 dt \, \int_{0}^{\infty} dx \, \sqrt{x} e^{x t^2} \, e^{-\sqrt{2} x} \, e^{i k x}\\ &= \frac{2}{\sqrt{\pi}} \int_0^1 dt \, \int_{0}^{\infty} dx \, \sqrt{x} e^{-(\sqrt{2}-t^2-i k) x}\\ &= \int_0^1 \frac{dt}{(\sqrt{2}-i k - t^2)^{3/2}} \\ &=\frac{1}{\sqrt{2}-i k} \frac{1}{\sqrt{\sqrt{2}-1-i k}}\end{align} $$
Note that the third line comes from the integral
$$\int_0^{\infty} dx \, \sqrt{x} e^{-a x} = \frac{\sqrt{\pi}}{2 a^{3/2}}$$
The result in the fourth line may be obtained using a trig substitution in the integral in the third line; the only trick is pretending that $\sqrt{2}-i k$ may be set to some $b^2$ parameter, and then proceeding with the usual trig substitution.
Now, the rest of the original integrand is $\sin{x}/x$, which Fourier transform is simply $\pi$ when $|k| \lt 1$ and $0$ otherwise. We may then invoke Parseval's equality, which states that, for functions $f$ and $g$ and their respective Fourier transforms $F$ and $G$, we have
$$\int_{-\infty}^{\infty} dx \, f(x) g(x) = \frac{1}{2 \pi} \int_{-\infty}^{\infty} dk \, F(k) G(k)$$
Here,
$$f(x) = \text{erfi}(\sqrt{x}) \, e^{-\sqrt{2} x}\, \theta(x) $$
$$g(x) = \frac{\sin{x}}{x}$$
$$F(k) = \frac{1}{\sqrt{2}-i k} \frac{1}{\sqrt{\sqrt{2}-1-i k}}$$
$$G(k) = \begin{cases}\pi & |k| \lt 1 \\ 0 & k \gt 1\end{cases}$$
Thus, we have reduced the integral to the evaluation of the following:
$$\frac12 \int_{-1}^1 \frac{dk}{\sqrt{2}-i k} \frac{1}{\sqrt{\sqrt{2}-1-i k}}$$
Now sub $v^2=\sqrt{2}-1-i k$ and get the following integral
$$-i \int_{\sqrt{\sqrt{2}-1-i}}^{\sqrt{\sqrt{2}-1+i}} \frac{dv}{1+v^2}$$
which evaluates to
$$-i \left [\arctan{\sqrt{\sqrt{2}-1+i}} - \arctan{\sqrt{\sqrt{2}-1-i}} \right ] $$
which is equal to
$$\tanh ^{-1}\left(\frac{\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2
\sqrt{2}}\right)}}{1+\sqrt{4-2 \sqrt{2}}}\right) $$
or
$$\frac12 \log{\left [\frac{1+\sqrt{4-2 \sqrt{2}}+\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2
\sqrt{2}}\right)}}{1+\sqrt{4-2 \sqrt{2}}-\sqrt{2 \left(1-\sqrt{2}+\sqrt{4-2
\sqrt{2}}\right)}}\right ]} \approx 0.625774$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "25",
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How much caffeine is left in your system? The half-life of caffeine in your system is 6 hours. If you drink 100mg of caffeine every day, what percentage is left in your system after a week?
If you drink 100mg caffeine every day for a month (30d), how long does it take for the level of caffeine in your system to drop below 1mg?
| The answer to both depends on when (and how, since one rarely drinks the coffee instantaneously) the coffee is consumed during the day.
For the purposes of this problem, however, I will assume that all the coffee is consumed at exactly $t_{\text{coffee}}$ every day and that we are looking for the quantity remaining at the end of the $7$th day.
At the end of a day, the contribution from that day is $100 ( \frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}}$, after $n$ days, this contribution decreases to
$100 ( \frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}} ( \frac{1}{2^{4} } )^n$ (since $4 = \frac{24}{6}$). Summing up, and dividing by $700$ to get the ratio gives:
$$\frac{100}{700} (\frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}}
\left( ( \frac{1}{2^{4} } )^6 + ... + ( \frac{1}{2^{4} } )^1 + 1\right) =
\frac{1}{7} (\frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}}
\left( \frac{1-(\frac{1}{2^{4}})^7 }{1-\frac{1}{2^{4} }} \right)$$
So, doing the computation shows that the percentage left varies between $0.95 \% -15.24 \%$ (taking $t_{\text{coffee}} \in [0,24)$). Note that this is the percentage of the total caffeine consumed during the week, which is a bit misleading, since if you took the average over $n$ days, and let $n \to \infty$, then the percentage would decrease to zero.
The other computation follows similar lines (but is less misleading):
I am assuming that you drink coffee every day for 30 days and then go cold turkey.
Then you want to find the minimum $n$ (whole days) such that
$$100 (\frac{1}{2} )^{ \frac{24-t_{\text{coffee}}}{6}}
\left( ( \frac{1}{2^{4} } )^{29} + ... + ( \frac{1}{2^{4} } )^1 + 1\right) (\frac{1}{2^4} )^{ n} \le 1$$
A quick calculation shows that the answer is $n=1$ or $n=2$ (depending on $t_{\text{coffee}}$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/501435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$? Why is $\arctan\frac{x+y}{1-xy} = \arctan x +\arctan y$?
It is said that this is derived from trigonometry, but I couldn't find why this is the case.
| Here's a proof that has essentially nothing to do with trigonometry:
Hold $y$ constant and differentiate the function $$f(x) = \arctan{\frac{x + y}{1 - xy}} - \arctan{x} - \arctan{y}$$ to find that
\begin{align}f'(x) &= \frac{1}{1 + \left(\frac{x + y}{1 - xy}\right)^2} \cdot \left(\frac{(1 - xy) - (x + y)(-y)}{(1 - xy)^2}\right) - \frac{1}{1 + x^2} \\
&= \frac{(1 - xy)^2}{(1 - xy)^2 + (x + y)^2} \cdot \left(\frac{1 - xy + xy + y^2}{(1 - xy)^2}\right) - \frac{1}{1 + x^2} \\
&= \frac{1 + y^2}{1 - 2xy + x^2y^2 + x^2 + 2xy + y^2} - \frac{1}{1 + x^2} \\
&= \frac{1 + y^2}{1 + x^2 + y^2 + x^2y^2} - \frac{1}{1 + x^2} \\
&= \frac{1 + y^2}{(1 + y^2) + x^2 (1 + y^2)} - \frac{1}{1 + x^2} \\
&= \frac{1}{1 + x^2} - \frac{1}{1 + x^2} = 0
\end{align}
So $f$ is a constant. Letting $x = 0$, we find that $f(0) = 0$ and the identity follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/502189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Laplace transformation of $ \frac{\sin^2{t}}{t^2} $? I tried convolution theorem also tried this process..
$$Laplace(\frac{f(t)}{t}) = \int_s^\infty F(u) du $$
So,
$$Laplace(\frac{\sin{t}}{t} * \frac{\sin{t}}{t}) = \int_s^\infty F(u) du * \int_s^\infty F(u) du $$
where F(u) = Laplace(f(t))
I know this look's funny as there's no rule like the last equation.
Please help...
| \begin{align*}
\mathcal L \left( \frac{\sin^2(x)}{x^2} \right ) &= \int_0^\infty e^{-sx} \frac{\sin^2(x)}{x^2} dx \\
&= \left[e^{-sx} \sin^2(x) \int \frac 1{x^2}dx \right ]_0^\infty - \int_0^\infty \left( 2 e^{-s x} \sin (x) \cos (x)-s e^{-s x} \sin ^2(x) \right ) \frac{(-1)}{x} dx \\
&= \int_0^\infty e^{-sx} \frac{\sin(2x)}{x} dx - s \int_0^\infty e^{-sx} \frac{\sin^2(x)}{x} dx \\ &= \arctan \left( \frac 2 s \right )- s\frac{1}{4} \left(\log \left(s^2+4\right)-2 \log (s)\right)
\end{align*}
To find $\displaystyle \int_0^\infty e^{-sx} \frac{\sin(2x)}{x} dx $
\begin{align*}
\int_0^\infty e^{-sx} \frac{\sin(2 x)}{x} dx &=
\int_0^\infty e^{- \frac s 2 (2x)} \frac{\sin(2x)}{2x} d(2x)\\
&= \int_0^\infty e^{- \frac s 2 u} \frac{\sin (u)}{u}du\\
&= \int_0^\infty e^{- \frac s 2 u} \sin (u) \int_0^\infty e^{- u t} dt du \\
&= \int_0^\infty \int_0^\infty e^{ - \left( \frac s 2 + t \right )u } \sin(u) du dt\\
&= \int_0^\infty \frac{4}{(4 + (s + 2 t)^2)} dt \\
&= \frac \pi 2 - \arctan \left( \frac s 2 \right ) \\
&= \arctan \left( \frac 2 s \right )
\end{align*}
The relation $\displaystyle \arctan(x) + \arctan \left( \frac 1 x\right) = \frac \pi 2$ has been used above
To find $\displaystyle \int_0^\infty e^{-sx} \frac{\sin^2(x)}{x} dx$ we proceed in the same way.
\begin{align*}
\int_0^\infty e^{-sx} \frac{\sin^2(x)}{x}dx &= \int_0^\infty \int_0^\infty e^{-(s+t)x} \sin^2(x)dx \\
&= \int_0^\infty \int_0^\infty e^{-(s+t)x} \frac{1 - \cos(2x)}{2}dx dt \\
&= \frac 1 4 \log \left( 1 + \frac 4 {s^2} \right )\\
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/504812",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find a closed form solution to a recurrence of the following form? I need to find the closed form solution to the following recurrence -:
$ T(n) = 8*T(n/2) + 0.25*n^2$ with $T(1) = 1$ and $n=2^j$ and this is what I have tried so far but just can't seem to get a pattern out for this recurrence. Letting 0.25 to behave as a variable x, I have,
$T(n) = T(n/2) + n^2*x$
$T(2) = 8 + 2^2 * x $
$T(4) = 8*(8 + 2^2 * x) + 4^2 * x $
$T(8) = 8*(8*(8 + 2^2 * x) + 4^2 * x) + 8^2 * x $
I extracted the highest power of 8 from the above equation for T(8), to check if I can find a pattern but couldn't how can I possibly solve this recurrence?
| You may be interested to know that the recurrence
$$ T(n) = 8\;T(\lfloor n/2 \rfloor) + 1/4 \; n^2 $$
where $T(0) = 0$ and $T(1) = 1/4$ actually has a closed form solution for all $n$ and not just for $n$ a power of two. (We will show how to solve this with $T(1) = 1$ at the end of this post.)
This closed form is obtained by unrolling the recursion according to the binary digits of $n.$ Let those digits be given by
$$n = \sum_{k=0}^{\lfloor \log_2 n \rfloor} d_k 2^k.$$
Then the closed form solution (exact for all $n$) is given by
$$ T(n) = \frac{1}{4} \sum_{k=0}^{\lfloor \log_2 n \rfloor} 8^k
\left(\sum_{j=k}^{\lfloor \log_2 n \rfloor} d_j 2^{j-k} \right)^2.$$
To get an upper bound consider the case of all one digits, which gives
$$ T(n) \le \frac{1}{4} \sum_{k=0}^{\lfloor \log_2 n \rfloor} 8^k
\left(\sum_{j=k}^{\lfloor \log_2 n \rfloor} 2^{j-k} \right)^2 =
\frac{20}{21} 8^{\lfloor \log_2 n \rfloor}
- 4^{\lfloor \log_2 n \rfloor}
+ \frac{1}{3} 2^{\lfloor \log_2 n \rfloor}
- \frac{1}{28}.$$
This bound is attained.
To get a lower bound consider the case of a one digit followed by a string of zeros, giving
$$ T(n) \ge \frac{1}{4} \sum_{k=0}^{\lfloor \log_2 n \rfloor} 8^k
\left( 2^{\lfloor \log_2 n \rfloor-k} \right)^2
= \frac{1}{2} 8^{\lfloor \log_2 n \rfloor}
- \frac{1}{4} 4^{\lfloor \log_2 n \rfloor}.$$
This bound too is attained.
Joining these two it now becomes evident that
$$T(n)\in \Theta\left(8^{\lfloor \log_2 n \rfloor}\right)
= \Theta\left(2^{3\lfloor \log_2 n \rfloor}\right)
= \Theta(n^3).$$
The above calculation has $T(1) = 1/4$ while the query asks for $T(1) = 1$. It is not difficult to see that the answer to the original query can be obtained by taking
$$ T(n) + \frac{3}{4} 8^{\lfloor \log_2 n \rfloor}.$$
This gives the following upper and lower bounds
$$ \frac{143}{84} 8^{\lfloor \log_2 n \rfloor}
- 4^{\lfloor \log_2 n \rfloor}
+ \frac{1}{3} 2^{\lfloor \log_2 n \rfloor}
- \frac{1}{28}
\quad\text{and}\quad
\frac{5}{4} 8^{\lfloor \log_2 n \rfloor}
- \frac{1}{4} 4^{\lfloor \log_2 n \rfloor}.$$
The asymptotic complexity is not affected.
There is more material on this method here.
| {
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"url": "https://math.stackexchange.com/questions/505004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0$ mod n iff $n \equiv \pm 1$ mod 6 The problem is as follows: Prove that $$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 \equiv 0 \, \text{mod n} \ \text{if and only if } n \equiv \pm 1 \, \text{mod} 6.$$
My idea is to of course rewrite the summation. We have $$1^2 + 2^2 + 3^2 + \cdots + (n-1)^2 = \frac{(n-1)n(2n-1)}{6}.$$
Now, in order for this to be an integer, must we have in particular that $6 \mid n-1 \Rightarrow n \equiv 1 \, \text{mod} 6$?
We also know that the sum $$1^2 + 2^2 + 3^2 + \ldots + (n-1)^2 + n^2 = \frac{n(n+1)(2n+1)}{6}.$$ Must we have in particular that then $6 \mid n+1 \Rightarrow n \equiv -1 \, \text{mod} 6$?
I think the statement holds if and only if the whole way so we are done with both ways?
| $n|\sum\limits_{k=1}^{n-1}k^2\iff n|\frac{n\cdot(n-1)\cdot(2n-1)}{6}$
$\iff 6|(n-1)\cdot(2n-1)$
$\impliedby n\equiv 1\pmod 6$
If $n\not\equiv 1 \pmod 6$, then since $(2n-1)$ is odd, $n\equiv 1\pmod 2\, \&\, 2n\equiv 1\pmod 3$
i.e., $n\equiv-1\pmod 2 \,\&\, n\equiv -1\pmod 3\implies n\equiv -1 \pmod 6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/506986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Laurent series of a trigonometric function Find the Laurent Serie(and residue) aroud $z_0=0$ of the function
$f(z) = \frac{1}{1-\cos z}$.
Progress:
It looks very trivial but it seems to get complicated so I'll only try with 3-4 terms:
We can use that $\cos z = 1- \frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...+\sum_{0}^{\infty}\frac{(-1)^k z^{2k}}{(2k)!}$ and hence
$f(z) = \frac{1}{1-\cos z} = \frac{1}{1-(1- \frac{z^2}{2!}+\frac{z^4}{4!}-\frac{z^6}{6!}+...+\sum_{0}^{\infty}\frac{(-1)^k z^{2k}}{(2k)!})} \Rightarrow f(z)= \frac{1}{\frac{z^2}{2!}-\frac{z^4}{4!}+\frac{z^6}{6!}}
$
This is not correct apparently so I tried to factor out $z^2$ and$\frac{z^2}{2!}$ but I was unable to proceed. I suspect that I'll arrive to a nestled sum.
| Factor out a $z^2/2!$ from the denominator to get
$$\begin{align}\frac{1}{1-\cos{z}} &= \frac{2}{z^2} \frac{1}{\displaystyle1-2 \sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{2 k}}{(2 (k+1))!}}\\ &= \frac{2}{z^2} + \frac{4}{z^2} \sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{2 k}}{(2 (k+1))!} + \frac{2}{z^2} \left ( \sum_{k=1}^{\infty} (-1)^{k+1}\frac{z^{2 k}}{(2 (k+1))!}\right )^2+\cdots \\ &\approx \frac{2}{z^2} +\frac{1}{6} +\cdots\end{align}$$
The residue at $z=0$ is the coefficient of $1/z$ and is thus zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/509889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
second order ODE. I have this second order ODE:
$$-au''+bu'=x^2$$
where $u=u(x)$, $a=\text{constant}>0$, $b=\text{constant}\ge0 $
and $$u(0)=1 ,u(1)=0$$
I tried to solve it and I get:
$$u(x)=D+Ce^{bx/a}$$
where $D,C$ are constant.
Is my answer correct?
How I can find $D$ and $C$?
I realy need help because there are 3 other quistions depending on this one.
Thanks.
| Hint
I'd guess particular solution as
\begin{align}
u_p' &= Ax^2+Bx+C \\
u_p'' &= 2Ax + B
\end{align}
Now, substitute it to the ODE given
$$
-2Aax - Ba + Abx^2 + Bbx + Cb = Abx^2 + (Bb - 2Aa)x + Cb - Ba = x^2
$$
from which you can find
\begin{align}
Ab &= 1 \\
Bb - 2Aa &= 0 \\
Cb - Ba &= 0
\end{align}
$$
A = \frac 1b; B = \frac {2a}{b^2}; C = \frac {2a^2}{b^3}
$$
So, particular solution
$$
u_p = \frac A3 x^3 + \frac B2 x + Cx = \frac 1{3b}x^3 + \frac a{b^2}x^2 + \frac {2a^2}{b^3} x
$$
Therefore, general solution
$$
u = C_1 + C_2 e^{\frac ba x} + \frac 1{3b}x^3 + \frac a{b^2}x^2 + \frac {2a^2}{b^3} x
$$
If you substitute BCs you get
\begin{align}
u(0) &= C_1 + C_2 = 1 \\
u(1) &= C_1 + C_2 e^{\frac ba} + \frac 1{3b} + \frac a{b^2} + \frac {2a^2}{b^3} = 0
\end{align}
After solving which you get your constants $C_1$ and $C_2$.
Update
As for the case with $b = 0$, that's different story. You need to solve
$$
-au'' = x^2
$$
which has a solution
$$
u = -\frac 1{12a}x^4 + C_1 x + C_2
$$
and you need to substitute BC to this one rather than the solution I provided above.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing a recursive sequence is Cauchy let $a_1=1$, $a_2=2$, and define $a_n=\frac{1}{2}(a_{n-1}+a_{n-2})$. How can I show that this sequence is Cauchy?
I began with $|a_n-a_{n-1}|=\frac{1}{2}(a_n-a_{n-1})$
which goes to $\frac{1}{2}(a_{n-1}+a_{n-2})-\frac{1}{2}(a_{n-2}+a_{n-3})$
I am have been toying with this for quite some time and it is just not clicking. Can somebody please help!
| You said yourself that
$|a_n-a_{n-1}| = \frac{1}{2}(a_{n-1}+a_{n-2}) - \frac{1}{2}(a_{n-2}+a_{n-3})$.
Cancelling the $a_{n-2}$ terms leaves $\frac{1}{2}(a_{n-1}-a_{n-3})$. Expanding $a_{n-1}$ gives
$\frac{1}{2}(a_{n-1}-a_{n-3}) = \frac{1}{2}\left(\frac{1}{2}(a_{n-2}+a_{n-3})-a_{n-3})\right) = \frac{1}{2}\left(\frac{1}{2}(a_{n-2}-a_{n-3}))\right)$.
This means $(a_n-a_{n-1})$ = $\frac{1}{4} (a_{n-2} - a_{n-3})$. If $a_1$ and $a_2$ are fixed then $a_n-a_{n-1}$ follows exponential decay and converges to $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/514083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to evaluate the summation $S_b$ This question is from my notebook, not hw or else, only exercise to understand better. I tried by myself. However, since my trail are too trivial, I dont need to write here. i am confused a bit. I want to learn how to solve such question types. I guess I need to use residue. I am grateful to help me thank you.
| we'll use the integral $$ \oint_{C_N} \frac{\pi\cot(\pi z)}{z^2 + b^2 } \ dz $$
where $C_N $ is the square contour of side $ 2N + 1 $
By residues theory :
$$ \oint_{C_N} \frac{\pi\cot(\pi z)}{z^2 + b^2 } \ dz = 2\pi i\left( \sum_{-N}^N \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , n \right) + \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , bi \right) + \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , -bi \right) \right)$$
the integral $ \to 0 $ as $ N \to \infty $
$$ \Rightarrow \sum_{-\infty }^{\infty} \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , n \right) + \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , bi \right) + \mathrm{Res} \left( \frac{\pi\cot(\pi z)}{z^2 + b^2 } , -bi \right) = 0 $$
$$ \Rightarrow \sum_{-\infty}^{-1} \frac{1}{n^2 + b^2} + \sum_{1}^{\infty} \frac{1}{n^2 + b^2 } + \frac{1}{b^2} + \lim_{z \to bi} \frac{\pi\cot(\pi z)}{z + bi} + \lim_{z \to -bi} \frac{\pi\cot(\pi z)}{z - bi} = 0$$
$$ \Rightarrow 2\sum_{n=1}^{\infty} \frac{1}{n^2 + b^2 } + \frac{1}{b^2} - \frac{\pi\coth(b\pi)}{b} = 0 $$
$$ \Rightarrow \sum_{n=1}^{\infty} \frac{1}{n^2 + b^2 } = \frac{\pi\coth(b\pi)}{2b} - \frac{1}{2b^2} $$
| {
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"url": "https://math.stackexchange.com/questions/515720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I prove by induction? For example if i wanted to prove:
$1^2 + \dots + n^2 = \frac {n(n + 1)(2n + 1)} {6}$
by induction.
I'm not sure where to start.
Thanks.
| Since $\frac{1(1 +1)(2+1)}{6} = 1$, then base case holds. Suppose result holds for $n$. Then
$$\sum_{i=1}^{n+1} i^2 = \sum_{i=1}^{n}i^2 + (n+1)^2 = \frac{n(n+1)(2n + 1)}{6} + (n+1)^2 $$
$$ \frac{n(n+1)(2n + 1)}{6} + (n+1)^2 = \frac{(n+1)[n(2n+1) + 6(n+1)}{6} = \frac{(n+1)(2n^2 + 7n + 1)}{6} =_{*} \frac{(n+1)(n+2)(2n + 3)}{6} = \frac{(n+1)((n+1) + 1)(2(n+1) + 1)}{6} $$
(*): Since $2n^2 + 7n + 1 = (n+2)(2n+3)$. The problem is solved by math induction.
| {
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"url": "https://math.stackexchange.com/questions/515966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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if $f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$ If $\displaystyle f(x) = x-\frac{1}{x}.$ Then no. of solution of the equation $f(f(f(x))) = 1$
$\underline{\bf{My\;\; Try}}::$ Given $\displaystyle f(x) = x-\frac{1}{x} = \frac{x^2-1}{x}.$ Now Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get.
$\displaystyle f(f(x)) = x-\frac{1}{x}-\frac{x}{x^2-1} = \frac{x^2-1}{x}-\frac{x}{x^2-1} = \frac{(x^2-1)^2-x^2}{x.(x^2-1)} = \frac{x^4-3x^2+1}{x.(x^2-1)}$
again Replace $\displaystyle x\rightarrow \frac{1}{x}\;,$ We Get
$\displaystyle f(f(f(x))) = \frac{\left(\frac{x^2-1}{x}\right)^4-3\left(\frac{x^2-1}{x}\right)^2+1}{\left(\frac{x^2-1}{x}\right).\left\{\left(\frac{x^2-1}{x}\right)^2-1\right\}}$
Now I did not understand how can I solve This $8^{th}$ Degree equation,
So Help Required,
Thanks
| Hint: plot the graph of $f$, and notice that:
*
*it steadily grows from $-\infty$ to $+\infty$ as $x$ grows grom $-\infty$ to $0$
*$x=0$ is a vertical asymptote
*it steadily grows from $-\infty$ to $+\infty$ as $x$ grows grom $0$ to $+\infty$.
Hence $x_1=f(x)$ can be considered as an independent variable on each interval $(-\infty,0)$ and $(0,+\infty)$. It follows that $f \circ f$ has three vertical asymptotes: $x=-1$ and $x=1$ (coming from $f(x)=0$) and $x=0$.
You repeat this argument and discover that $f \circ f\circ f$ has seven vertical asymptotes. Before the smallest asymptote and after the largest one, $f \circ f \circ f$ is steadily increasing from $-\infty$ to $+\infty$.
Since it is easily shown that $f \circ f \circ f$ is steadily increasing between any two consecutive asymptotes, you conclude that $f(f(f(x)))=1$ has exactly eight distinct solutions.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to factor and find zeros of $2x^6+5x^4-x^2$ In the equation $y=2x^6+5x^4-x^2$, how can I factor it in order to find the zeros or x-intercepts? This is what I've gotten to, but I can't see what to do next: $x^2(x^2(2x^2+5)-1)$
| $$y=2x^6+5x^4-x^2 = x^2(2x^4 + 5x^2 - 1)$$
For the factor $\;2x^4 + 5x^2 -1,\;$ put $t = x^2$:
$$2x^4 + 5x^2 - 1 = 2t^2 + 5t - 1$$
Suggestion: Use the quadratic formula to find the roots (also zeros): $t_1, t_2$, then back substitute, knowing $t = x^2 \implies x = \pm \sqrt t$. So for each zero $t_i$, there are two corresponding zeros, $x_j, x_k$ such that $x_{j, k} = \pm t_i$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$
Find the smallest natural number that leaves residues $5,4,3,$ and $2$ when divided respectively by the numbers $6,5,4,$ and $3$.
I tried
$$x\equiv5\pmod6\\x\equiv4\pmod5\\x\equiv3\pmod4\\x\equiv2\pmod3$$What $x$ value?
| HINT: Notice that your congruences are equivalent to the following ones:
$$\left\{\begin{align*}
x\equiv-1\pmod6\\
x\equiv-1\pmod5\\
x\equiv-1\pmod4\\
x\equiv-1\pmod3
\end{align*}\right.$$
In other words, $x+1$ is divisible by $6,5,4$, and $3$. What’s the smallest positive integer with that property?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Parametrize the intersection of 2 planes. Parametrize the intersection of $\frac {x^2} {3}+y^2+\frac {z^2} {10} = 1$ with $z=2$ (level curve) plane.
Here is what I did.
Plugged in $z=2$ into the plane $\frac {x^2} {3}+y^2+\frac 25=1$. I got $y^2=9-5x^2$ Then I substituted $y^2$ into the plane $\frac {x^2} {3}+9-5x^2+\frac25=1$ to solve for $x^2$. I got $x^2=1.8$ and then got $y=0$. How should I parametrize the intersection from here onward?
$\mathbf Error$ in my calculation, $y^2=\frac 35-\frac{x^2}{3}$
| You started out correctly, by plugging $z = 2$ into the equation and finding $\frac{x^2}{3} + y^2 + \frac{2}{5} = 1$, or
$$\frac{x^2}{3} + y^2 = \frac{3}{5}.$$
The next step is to put the equation in the standard form for an ellipse. (Recall that the standard form is $\frac{(x - h)^2}{a^2} + \frac{(y - k)^2}{b^2} = 1$.)
\begin{align}
\frac{5}{3} \cdot \frac{x^2}{3} + \frac{5}{3} \cdot y^2 &= \frac{5}{3} \cdot \frac{3}{5} \\
\frac{x^2}{\frac{9}{5}} + \frac{y^2}{\frac{3}{5}} &= 1
\end{align}
Here we have $h = k = 0$, $a = \sqrt{\frac{9}{5}} = \frac{3}{\sqrt{5}}$, and $b = \sqrt{\frac{3}{5}}$. The next step is to parametrize the ellipse, and recall that the parametrization for the $z$ coordinate is $z(t) = 2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How find the $(x^2_{1}+x_{1}x_{2}+x^2_{2})(x^2_{2}+x_{2}x_{3}+x^3_{3})(x^2_{3}+x_{3}x_{1}+x^2_{1})$ Let $x_1$, $x_2$, and $x_3$ be the roots of the equation
$$4x^3-6x^2+7x-8=0.$$
Find this value:
$$(x_1^2+x_1x_2+x_2^2)(x_2^2+x_2x_3+x_3^3)(x_3^2+x_3x_1+x_1^2)$$
My try:
$$x_1+x_2+x_3=\frac 3 2$$
$$x_1 x_2+x_2 x_3+x_1 x_3=\frac 7 4$$
$$x_1 x_2 x_3=2$$
Now I have
$$x_1^2+x_1x_2+x_2^2=(x_1+x_2)^2-x_1x_2=(\frac 3 2-x_3)^2-\frac 2 {x_3}.$$
But this is very ugly, and I think this problem should have a cleaner solution. Thanks.
| Notice
$$
x_1^2 + x_1x_2 + x_2^2 = \frac{x_1^3 - x_2^3}{x_1 - x_2}
= \frac{(6x_1^2 -7x_1 + 8)-(6x_2^2 -7x_2 + 8)}{4(x_1-x_2)}\\
= \frac{6(x_1+x_2)-7}{4} = \frac{6(\frac{3}{2} - x_3) - 7}{4}
= \frac{1-3x_3}{2} = \frac32(\frac13 - x_3)
$$
and similar expression holds for $x_2^2 + x_2x_3 + x_3^2$ and $x_3^2 + x_3x_1 + x_1^2$.
The expression we want is equal to
$$\begin{align}
& \left(\frac{3}{2}\right)^3(\frac13 - x_1)(\frac13 - x_2)(\frac13 - x_3)\\
= & \frac14 \left(\frac{3}{2}\right)^3\left[4\left(\frac13\right)^3 - 6\left(\frac13\right)^2 + 7\left(\frac13\right) - 8\right]\\
= & -\frac{167}{32}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/521225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequalities with absolute value question Solve: $ 5< \left\vert\dfrac{x+10}{x-10}\right\vert<6$
attempt at a solution:
Dividing into two:
$5<\left\vert \dfrac{x+10}{x-10}\right\vert $ And $\left\vert \dfrac{x+10}{x-10}\right\vert<6 $
For first we solve:
$ \dfrac{x+10}{x-10}<-5 $ or $ \dfrac{x+10}{x-10} >5 $ which yields $( -∞,15)$
For Second:
$ -6<\dfrac{x+10}{x-10} $ and $ \dfrac{x+10}{x-10} <6 $ which yields $(14, ∞)$
intersecting the two we get $(14,15)$
This solution was deemed wrong by the text book.
Is there any other way of solving this? Is there a mistake in this method of solution?
| Given that $10$ isn't even in the domain of the given function, your solution to the first inequality cannot be correct. Without actually seeing the details, though, it's hard to say what went wrong.
As an alternative, note that $\frac{x+10}{x-10}>0$ for $x>10$ and for $x<-10,$ and $\frac{x+10}{x-10}<0$ for $-10<x<10$. It's far simpler, then, to proceed casewise.
If $x<-10$, the following are equivalent: $$5<\left|\frac{x+10}{x-10}\right|<6\\5<\frac{x+10}{x-10}<6\\5(x-10)>x+10>6(x-10)\\5x-50>x+10>6x-60,$$ but $5x-50>x+10$ yields $x>15,$ so we can rule out all $x<-10$ as solutions to the inequality.
If $x>10,$ the following are equivalent: $$5<\left|\frac{x+10}{x-10}\right|<6\\5<\frac{x+10}{x-10}<6\\5(x-10)<x+10<6(x-10)\\5x-50<x+10<6x-60.$$ $5x-50<x+10$ yields $x<15,$ while $x+10<6x-60$ yields $14<x,$ so all $x$ with $14<x<15$ will be solutions to the inequality.
If $-10<x<10,$ the following are equivalent: $$5<\left|\frac{x+10}{x-10}\right|<6\\-5>\frac{x+10}{x-10}>-6\\-5(x-10)<x+10<-6(x-10)\\-5x+50<x+10<-6x+60.$$ $-5x+50<x+10$ yields $\frac{20}3<x,$ while $x+10<-6x+60$ yields $x<\frac{50}7,$ so all $x$ with $\frac{20}3<x<\frac{50}7$ will be solutions to the inequality.
Hence, the solution set is $$\left(\frac{20}3,\frac{50}7\right)\cup(14,15).$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of factors of a big number How to find the number of factors of $884466000$ without using a calculator?
| It's not too difficult to factor, since it's divisible by rather high powers of $2$ and $5$; right off the bat, we see that
$$884466000 = 1000 \cdot 884466 = 1000 \cdot 2 \cdot 442233$$
Now $442233$ has digit sum $4 + 4 + 2 + 2 + 3 + 3 = 18$, so it's divisible by $9$, giving
$$442233 = 49137 \cdot 9 = 16379 \cdot 27$$
Looking back at the original form of the number, it's also clearly divisible by $11$, since $$442233 = 11 \cdot 40203$$
So summarizing what we've got so far,
$$884466000 = 2^4 5^3 3^3 11^1 \cdot 1489$$
Now finally, $\sqrt{1489} < 40$, and checking the primes between $13$ and $37$, we find no more divisors. Hence $1489$ is prime.
Thus, every factor of $884466000$ is of the form
$$2^a 3^b 5^c 11^d 1489^e$$
with $0 \le a \le 4$, $0 \le b, c, \le 3$, and $0 \le d, e\le 1$, for a total of $5 \cdot 4^2 \cdot 2^2$ divisors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluating $\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}dx$ I am trying to prove that
\begin{equation}
\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x = \frac{\pi^{3}}{16} - 3\mathrm{G}\log\left(2\right)
\tag{1}
\end{equation}
where $\mathrm{G}$ is Catalan's Constant.
I was able to express it in terms of Euler Sums but it does not seem to be of any use.
\begin{align}
&\int_{0}^{1}\frac{\log\left(x\right)
\log\left(\,{1 - x^{4}}\,\right)}{1 + x^{2}}
\,\mathrm{d}x
\\[3mm] = &\
\frac{1}{16}\sum_{n = 1}^{\infty}
\frac{\psi_{1}\left(1/4 + n\right) -
\psi_{1}\left(3/4 + n\right)}{n} \tag{2}
\end{align}
Here $\psi_{n}\left(z\right)$ denotes the polygamma function.
Can you help me solve this problem $?$.
| I tried substitutions and the differentiation w.r.t a paramater trick like the other posters. Another partial result, or a trail of breadcrumbs to follow, is the following. We try a series expansion,
$$
\frac{\log\left(1-x^4\right)}{1+x^2} = \displaystyle \sum_{k=1}^{\infty} x^{4k}\left(x^{2} -1\right)H_k,
$$
where $H_k$ are the Harmonic numbers. Then
\begin{align}
\int_0^1 \frac{\log x \log \left(1-x^4 \right)}{1+x^2}\ \mathrm{d}x &=\displaystyle \sum_{k=1}^{\infty}\, H_k\int_0^1 x^{4k}\left(x^{2} -1\right)\log x \ \mathrm{d}x \\
&=\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+1)^2}-\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+3)^2}.
\end{align}
These sums look very similar to the ones evaluated in this post, in which they are transformed into alternating sums. Using the same techniques, or perhaps working back from the answers, we can hopefully show that
$$
\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+1)^2} = -G\left(\frac{\pi}{4}+\frac{\log 8}{2} \right) +\frac{7}{4}\zeta(3) +\frac{\pi^3}{32} - \frac{\pi^2}{16}\log 8,
$$
$$
\displaystyle \sum_{k=1}^{\infty} \, \frac{H_k}{(4k+3)^2} = -G\left(\frac{\pi}{4}-\frac{\log 8}{2} \right) +\frac{7}{4}\zeta(3) -\frac{\pi^3}{32} - \frac{\pi^2}{16}\log 8,
$$
Subtracting the second from the first gives us
$$
\frac{\pi^3}{16}-G\log 8.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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} |
If x and y are different integers , and if $2005 +x =y^2 ; 2005+y =x^2 $ then find xy... Problem :
If $2005 +x =y^2 ; 2005+y =x^2$ then find xy...
My approach :
Let $2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii) $
Now from (i) we get :
$ y = \sqrt{x + 2005}$
Now putting this value of y in (ii) we get :
$ \Rightarrow 2005 +\sqrt{x + 2005} =x^2$
$ \Rightarrow \sqrt{x + 2005} =x^2 -2005 $
Now squaring both sides we get :
$\Rightarrow (\sqrt{x + 2005})^2 =(x^2 - 2005)^2$
Is there any other way I can solve this problem please suggest... thanks.
| $2005 +x =y^2 .....(i) ; 2005+y =x^2 ......(ii)$
$(i)-(ii)\Leftrightarrow (x-y).(x+y+1)=0$
$\Rightarrow x=y ; x+y=-1$
$+x=y$, now we solve $x,y$ then $xy=...$
$+x+y=-1 \Rightarrow xy=-2004$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/524729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Reflection across a line?
The linear transformation matrix for a reflection across the line $y = mx$ is:
$$\frac{1}{1 + m^2}\begin{pmatrix}1-m^2&2m\\2m&m^2-1\end{pmatrix} $$
My professor gave us the formula above with no explanation why it works. I am completely new to linear algebra so I have absolutely no idea how to go about deriving the formula. Could someone explain to me how the formula is derived? Thanks
| You can have (far) more elegant derivations of the matrix when you have some theory available. The low-tech way using barely more than matrix multiplication would be:
The line $y = mx$ is parametrised by $t \cdot \begin{pmatrix}1\\m\end{pmatrix}$. The line orthogonal to it is parametrised by $r \cdot \begin{pmatrix}-m\\1\end{pmatrix}$. The line $y = mx$ shall be fixed, the line orthogonal to it shall be reflected, so you want a matrix $R$ with
$$R \begin{pmatrix}1 & -m\\ m & 1\end{pmatrix} = \begin{pmatrix}1 & m\\ m & -1\end{pmatrix},$$
and that means
$$\begin{align}
R &= \begin{pmatrix}1 & m\\m&-1\end{pmatrix} \begin{pmatrix}1&-m\\m&1\end{pmatrix}^{-1}\\
& = \begin{pmatrix}1&m\\m&-1\end{pmatrix}\cdot \frac{1}{1+m^2}\begin{pmatrix}1&m\\-m&1\end{pmatrix}\\
&= \frac{1}{1+m^2} \begin{pmatrix}1 - m^2 & 2m\\2m &m^2-1\end{pmatrix}.
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/525082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "45",
"answer_count": 6,
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} |
How to find the determinant of this matrix I have the following matrix:
$
\begin{bmatrix}
a & 1 & 1 & 1 \\
1 & a & 1 & 1 \\
1 & 1 & a & 1 \\
1 & 1 & 1 & a \\
\end{bmatrix}
$
My approach is to rref the matrix so that i can find the determinant by multiplying along the diagonals.
I attempted to do an rref and ended up with
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1-a & a-1 & 0 & 0 \\
0 & 1-a & a-1 & 1 \\
0 & 0 & 1-a & a-1 \\
\end{bmatrix}
$
I then factored out (1-a) to get this
$
\begin{bmatrix}
1 & a & 1 & 1 \\
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 1 \\
0 & 0 & 1 & -1 \\
\end{bmatrix}
$
which will then make my determinant a multipile of $(1-a)^3$:
$\det(A) = (1-a)^3x$
But now here I don't know what to do. I have a feeling that my approach is wrong. Any help or guide please?
| at lin3 col 4 is 0 not 1
det$\begin{bmatrix}
1 & a & 1 & 1 \\
1 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 1 & -1 \\
\end{bmatrix}$
~det$\begin{bmatrix}
a+3 & a & 1 & 1 \\
0 & -1 & 0 & 0 \\
0 & 1 & -1 & 0 \\
0 & 0 & 1 & -1 \\
\end{bmatrix}$
etc.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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How can I prove by induction that $9^k - 5^k$ is divisible by 4? Recently had this on a discrete math test, which sadly I think I failed. But the question asked:
Prove that $9^k - 5^k$ is divisible by $4$.
Using the only approach I learned in the class, I substituted $n = k$, and tried to prove for $k+1$ like this:
$$9^{k+1} - 5^{k+1},$$
which just factors to $9 \cdot 9^k - 5 \cdot 5^k$.
But I cannot factor out $9^k - 5^k$, so I'm totally stuck.
| $$\begin{align} 9\cdot 9^k - 5\cdot 5^k & = (4 + 5)\cdot 9^k - 5\cdot 5^k \\ \\ & = 4\cdot 9^k + 5 \cdot 9^k - 5\cdot 5^k \\ \\ & = 4\cdot 9^k + 5(9^k - 5^k)\\ \\ & \quad \text{ use inductive hypothesis}\quad\cdots\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/528709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Pascal's formula and Binomial Coefficients Prove Pascal's formula by substituting the values of the binomials coefficient as given in the equation
$$\binom nk= \frac{n!}{k!\,(n-k)!} = \frac{n(n-1)\cdots(n-k+1)}{k(k-1)\cdots1}.$$
I guess I just don't know what values I'm supposed to be substituting in.
| $$\binom nk=\frac{n!}{k!(n-k)!}=\frac{n!}{(n-k)!(n-(n-k))!}=\binom{n}{n-k}....[1]$$
$$\binom nk=\frac{n(n-1)!}{k(k-1)!(n-k)!}=\frac{ n}{k}\frac{(n-1)!}{(k-1)!((n-1)-(k-1))!}=\frac{n}{k}\binom {n-1}{k-1}...[2]$$
from [1] and [2] we have
$$\binom {n}{k}=\binom {n}{n-k}=\frac {n}{n-k}\binom {n-1}{n-k-1}=\frac{n-k+k}{n-k} \binom {n-1}{(n-1)-(n-k-1)}=$$
$$=\left(1+\frac{ k}{n-k}\right) \binom {n-1}{k}=\binom {n-1}{k}+\frac{ k}{n-k}\binom {n-1}{k}=$$
$$=\binom {n-1}{k}+\frac{ k}{n-k}\frac{n-1}{k}\binom {n-2}{k-1}=\binom {n-1}{k}+\frac{n-1}{n-k}\binom {n-2}{(n-2)-(k-1)}=$$
$$=\binom {n-1}{k}+\frac{n-1}{n-k}\binom {n-2}{n-k-1}=\binom {n-1}{k}+\binom {n-1}{n-k}=$$
$$=\binom {n-1}{k}+\binom {n-1}{(n-1)-(n-k)}=\binom {n-1}{k}+\binom {n-1}{k-1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Property on glide reflections I need to prove that the conjugate of a glide reflection is a glide reflection.
What I have tried: Let $m: X= \begin{pmatrix} x_1\\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} \cos \phi & \sin \phi \\ \sin \phi & -\cos \phi \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix}+\begin{pmatrix} a_1\\ a_2 \end{pmatrix}$
and $n:X= \begin{pmatrix} x_1\\ x_2 \end{pmatrix} \mapsto \begin{pmatrix} \cos \psi & \sin \psi \\ \sin \psi & -\cos \psi \end{pmatrix} \begin{pmatrix} x_1\\ x_2 \end{pmatrix}+\begin{pmatrix} b_1\\ b_2 \end{pmatrix}$
I wanted to calculate $n \circ m \circ n^{-1}$ and show that it has the form of a glide reflection but what is $n^{-1}$ ?
| Hint: The involved matrix, let's call her $M_{\psi}$, is invertible so simply solve $y=M_{\psi}x+b$ for $x$.
| {
"language": "en",
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If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$? All is in the title:
If $k^2-1$ is divisible by $8$, how can we show that $k^4-1$ is divisible by $16$?
I can't conclude from the fact that $k^2 - 1$ is divisible by $8$, that then $k^4-1$ is divisible by $16$.
| $\displaystyle (2a+1)^2=4a^2+4a+1=8\frac{a(a+1)}2+1=8b+1$ where $b$ is some integer
Now, $(2a+1)^4=(8b+1)^2=64b^2+16b+1\equiv1\pmod{16}$
This way we can show $(2a+1)^{2^n}\equiv1\pmod{2^{n+2}}$ for integer $n\ge1$
This is generalized as Carmichael function $$\lambda(2^{n+2})=2^n\text{ for }n\ge1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/532070",
"timestamp": "2023-03-29T00:00:00",
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Find the value of $\lim_{n\rightarrow\infty}\frac{a_n}{1\cdot 2}+\frac{a_{n-1}}{2\cdot 3}+\ldots +\frac{a_1}{n(n+1)}$ $\displaystyle \lim_{n\rightarrow\infty}\frac{a_n}{1\cdot 2}+\frac{a_{n-1}}{2\cdot 3}+\ldots +\frac{a_1}{n(n+1)}$ if $\lim_{n\rightarrow\infty}a_n=a$
Note that $\displaystyle\sum_{k=1}^n\frac{a_{n-k+1}}{n^2}\le S_n\le \sum_{k=1}^n\frac{a_{n-k+1}}{k^2}$
Can I bound $S_n$ by $a$ instead of $a_{n-k+1}$ in the numerator instead?
| We will show that the limit is $a$.
Let $\varepsilon>0$. First, $(a_n)$ is bounded, so $|a_n|<M$ for some $M>0$, for all $n\in\mathbb N$. There exists $n_0\in\mathbb N$ such that, for all $n\geq n_0$, $|a_n-a|<\frac{\varepsilon}{6}$. Also, there exists $n_1\in\mathbb N$ with $n_1>n_0$ such that, if $n\geq n_1$, then $\sum_{k=n_1}^{\infty}\frac{1}{k(k+1)}<\frac{\varepsilon}{6M}$. Finally, there exists $n_2\in\mathbb N$ such that, if $n\geq n_2$, we have that $\frac{1}{n+1}<\frac{\varepsilon}{3M}$. Then, if $n\geq n_0,2n_1,n_2$,$$\left|\frac{a_n}{1\cdot2}+\frac{a_{n-1}}{2\cdot3}+\dots+\frac{a_1}{n(n+1)}-a\right|=$$$$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_1-a}{n(n+1)}-a+\frac{a}{1\cdot2}+\frac{a}{2\cdot3}+\dots+\frac{a}{n(n+1)}\right|\leq$$$$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_1-a}{n(n+1)}\right|+\left|\frac{a}{1\cdot2}+\frac{a}{2\cdot3}+\dots+\frac{a}{n(n+1)}-a\right|.$$ The term in the second absolute value is equal to $$\left|a-\frac{a}{2}+\frac{a}{2}-\frac{a}{3}+\dots+\frac{a}{n}-\frac{a}{n+1}-a\right|=\left|-\frac{a}{n+1}\right|\leq\frac{M}{n+1}<\frac{\varepsilon}{3},$$ since $n\geq n_2$. For the first term, you have that $$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_1-a}{n(n+1)}\right|\leq$$$$\left|\frac{a_n-a}{1\cdot2}+\frac{a_{n-1}-a}{2\cdot3}+\dots+\frac{a_{n_1}-a}{(n-n_1+1)(n-n_1+2)}\right|+$$$$\left|\frac{a_{n_1-1}-a}{(n-n_1+2)(n-n_1+3)}+\dots+\frac{a_1-a}{n(n+1)}\right|.$$ Since $n_1>n_0$, the first term here is bounded by $$\left|\frac{\varepsilon/6}{1\cdot2}+\dots+\frac{\varepsilon/6}{(n-n_1+1)(n-n_1+2)}\right|<\frac{2\varepsilon}{6}=\frac{\varepsilon}{3},$$ and the second term is bounded by $$2M\sum_{k=n-n_1+2}^\infty\frac{1}{k(k+1)}<\frac{2M\varepsilon}{6M}=\frac{\varepsilon}{3},$$ since $n-n_1+2\geq n_1$. After adding those inequalities, we get that this limit is equal to $a$.
| {
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Express $\sin\frac{\pi}{8}$ and $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$ I've been trying with no success expressing this functions.
a) $\sin\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$
b) $\cos\frac{\pi}{8}$ with $\cos\frac{\pi}{4}$
I've tried formula of the double angle ($\sin$ and $\cos$) and the ecuation $\cos^2X+\sin^2X=1$.
Can anyone point what to use or where to start?
I appreciate it, thanks.
UPDATE
I think this is achived with the half formula as njguliyev and nbubis said.
$$\sin\frac{\pi}{8} = \sin(\frac{1}{2} * \frac{\pi}{4}) = \pm\sqrt\frac{1-\cos\frac{\pi}{4}}{2}$$
$$\cos\frac{\pi}{8} = \cos(\frac{1}{2} * \frac{\pi}{4}) = \pm\sqrt\frac{1+\cos\frac{\pi}{4}}{2}$$
| I mean just use the trig identities: $$\cos \frac{ PI }{4} = 1 - 2 \sin^2 \frac{ PI}{8} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given n = 4k. Prove that $\sum_{i=0}^{2k}\binom{n}{2i}(-1)^i=2^{2k}(-1)^k$.
Given $n = 4k$, where $k$ is a non-negative integer. Prove that $\sum_{i=0}^{2k}\binom{n}{2i}(-1)^i=2^{2k}(-1)^k$.
I have no idea how to prove this equation.
| Let $i=\sqrt{-1}$, and note that $(1+i)^4=(1-i)^4=-4$. Also, for $j \ge 0$, we have
$$\frac{1+(-1)^j}{2} = \begin{cases}1 &\text{if $2|j$}\\0 &\text{otherwise}\end{cases}.$$
Then
\begin{align}
\sum_{j\ge 0}\binom{4k}{2j}(-1)^j
&= \sum_{j\ge 0}\binom{4k}{2j}i^{2j} \\
&= \sum_{j\ge 0}\frac{1+(-1)^j}{2}\binom{4k}{j}i^{j} \\
&= \frac{1}{2}\sum_{j\ge 0}\binom{4k}{j}i^{j} + \frac{1}{2}\sum_{j\ge 0}\binom{4k}{j}(-i)^{j} \\
&= \frac{1}{2}(1+i)^{4k} + \frac{1}{2}(1-i)^{4k} \\
&= \frac{1}{2}(-4)^k + \frac{1}{2}(-4)^k \\
&= (-4)^k \\
&= 2^{2k}(-1)^k
\end{align}
| {
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"url": "https://math.stackexchange.com/questions/533978",
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"source": "stackexchange",
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How find the value of $\cos{x}+\cos{y}$ let
$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}+\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=1$$
Find the value $$cos{x}+\cos{y}=?$$
this following My ugly solution: let
$$\tan{\dfrac{x}{2}}=a,\tan{\dfrac{y}{2}}=b$$
then
$$\cos{x}=\dfrac{1-a^2}{1+b^2},\cos{y}=\dfrac{1-b^2}{1+b^2}$$
then
$$\dfrac{\cos{x}\cos{\dfrac{y}{2}}}{\cos{(x-\dfrac{y}{2})}}=\dfrac{1}{1+\tan{x}\tan{\dfrac{y}{2}}}=\dfrac{1-a^2}{1-a^2+2ab}$$
and
$$\dfrac{\cos{y}\cos{\dfrac{x}{2}}}{\cos{(y-\dfrac{x}{2})}}=\dfrac{1-b^2}{1-b^2+2ab}$$
so
$$\dfrac{1-a^2}{1-a^2+2ab}+\dfrac{1-b^2}{1-b^2+2ab}=1$$
then
$$\Longrightarrow (1-a^2)(1-b^2+2ab)+(1-b^2)(1-a^2+2ab)=(1-a^2+2ab)(1-b^2+2ab)$$
then
$$(1-b^2)(1-a^2+2ab)=2ab(1-b^2+2ab)$$
$$a^2+b^2=1-3a^2b^2$$
so
$$\cos{x}+\cos{y}=\dfrac{2-2a^2b^2}{1+a^2+b^2+a^2b^2}=1$$
My question: this problem have nice methods? Thank you,because My methods is very ugly.Thank you
| Multiplying through out the given Equation with 2
$$ \frac{2 Cos(x)Cos(\frac{y}{2})}{Cos(x-\frac{y}{2})} + \frac{2 Cos(y)Cos(\frac{x}{2})}{Cos(y-\frac{x}{2})}=2$$ $\implies$
$$\frac{Cos(x+\frac{y}{2})+Cos(x-\frac{y}{2})}{Cos(x-\frac{y}{2})}+\frac{Cos(y+\frac{x}{2})+Cos(y-\frac{x}{2})}{Cos(y-\frac{x}{2})}=2$$ $\implies$
$$\frac{Cos(x+\frac{y}{2})}{Cos(x-\frac{y}{2})}+\frac{Cos(y+\frac{x}{2})}{Cos(y-\frac{x}{2})}=0$$ $\implies$
$$2Cos(x+\frac{y}{2})Cos(y-\frac{x}{2})+2Cos(y+\frac{x}{2})Cos(x-\frac{y}{2})=0$$ $\implies$
$$Cos(\frac{x+3y}{2})+Cos(\frac{3x-y}{2})+Cos(\frac{3x+y}{2})+Cos(\frac{x-3y}{2})=0 $$ $\implies$
$$Cos(\frac{x}{2})Cos(\frac{3y}{2})+Cos(\frac{3x}{2})Cos(\frac{y}{2})=0$$ $\implies$
$$Cos(\frac{x}{2})Cos(\frac{y}{2}) \left(4Cos^2(\frac{x}{2})+4Cos^2(\frac{y}{2})-6\right)=0$$ $\implies$
$$ 4Cos^2(\frac{x}{2})+4Cos^2(\frac{y}{2})-6=0$$ $\implies$
$$ 2\left(1+Cos(x)+1+Cos(y)\right)=6$$ $\implies$
$$Cos(x)+Cos(y)=1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/534062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving inequality with non-negative reals Let $a_1 , a_2 , ... , a_n$ be $n$ non-negative real numbers all less than $1$ and satisfying
let $$a=\sqrt {\frac1{n}\sum_{i=1}^na^2_i}≥\dfrac{\sqrt{3}}{3}$$ then how do we prove that $$\frac{a_1}{1-a^2_1} + \frac{a_2}{1-a_2^2}+....+\frac{a_n}{1-a_n^2}≥\frac{na}{1-a^2}$$
| let $a^2_{i}=x_{i},1\le i\le n$,then
if $x_{1},x_{2},\cdots,x_{n}<1$, are non-negative real numbers,such
$$\dfrac{x_{1}+x_{2}+\cdots+x_{n}}{n}=r\ge\dfrac{1}{3}$$
then we have
$$\dfrac{\sqrt{x_{1}}}{1-x_{1}}+\dfrac{\sqrt{x_{2}}}{1-x_{2}}+\cdots+\dfrac{\sqrt{x_{n}}}{1-x_{n}}\ge\dfrac{n\sqrt{r}}{1-r}$$
Proof
Apply RCF-Theorem to the function
$$f(u)=\dfrac{\sqrt{u}}{1-u},0\le u\le 1$$
then
$$f'(u)=\dfrac{3u^2+6u-1}{4u\sqrt{u}(1-u)^2}$$
it follows that $f$ is convex on $[\dfrac{2}{\sqrt{3}}-1,1)$,since $s=\dfrac{1}{3}>\dfrac{2}{\sqrt{3}}-1$,the function $f$ is convex on $[s,1)$
By RCF-Theorem,
it suffices to show that $g(x)\le g(y)$ for $0\le x\le s\le y<1$ and $x+(n-1)y=ns$
where
$$g(t)=\dfrac{f(t)-f(s)}{t-s}$$
for convenience,let $$a=\sqrt{x},b=\sqrt{y},c=\sqrt{s}$$
we have
$$g(t^2)=\dfrac{f(t^2)-f(c^2)}{t^2-c^2}=\dfrac{1+ct}{(1-c^2)(1-t^2)(t+c)}$$
and
$$g(x)-g(y)=g(a^2)-g(b^2)=(a^2-b^2)\dfrac{a^2+b^2+c(a+b)+c^2-1+ab(1+c^2)+abc(a+b)}{(1-c^2)(1-a^2)(1-b^2)(a+c)(b+c)}$$
since
\begin{align*}
&a^2+b^2+c(a+b)+c^2-1+ab(1+c^2)+abc(a+b)\\
&\ge a^2+b^2+c(a+b)+c^2-1\\
&\ge a^2+b^2+c\sqrt{a^2+b^2}+c^2-1
\end{align*}
it is enought to show that
$$x+y+\sqrt{s(x+y)}+s-1\ge 0$$
Indeed,we have
$$x+y=\dfrac{ns+(n-2)s}{n-1}\ge\dfrac{ns}{n-1}$$
and therefore
$$x+y+\sqrt{s(x+y)}+s-1\ge\left(\dfrac{n}{n-1}+\sqrt{\dfrac{n}{n-1}+1}\right)s-1\ge 3s-1=0$$
Equality occurs only for $x_{1}=x_{2}=\cdots=x_{n}=r$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $x^{20} +x^{15}+ x^{10}+x^5+1$ irreducible in $\mathbb{Q}[x]$? Is $x^{20} +x^{15}+ x^{10}+x^5+1$ irreducible in $\mathbb{Q}[x]$?
I think if $y=x^5$, then $P(y)=y^{4} +y^{3}+ y^{2}+y+1 =\displaystyle \frac{y^5-1}{y-1}$, $5$ is prime, then $P(y)$ is irreducible in $\mathbb{Q}[y]$ or $\mathbb{Q}[x^5]$ , how do I show that it is irreducible in $\mathbb{Q}[x]$?
| Let $f(x) = x^{20}+x^{15}+x^{10}+x^5+1 \in \mathbb{Z}[x]$, we have
$$\begin{array}{rrcl}
& (x^5-1)f(x) &=& x^{25} - 1\\
\implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1\\
\implies & ((x+1)^5-1)f(x+1) &=& (x+1)^{25} - 1 \pmod 5\\
\iff & ((x^5+1)-1)f(x+1) &=& (x^5+1)^5 - 1 \pmod 5\\
\iff & x^5 f(x+1) &=& (x^{25}+1) - 1 \pmod 5\\
\iff & f(x+1) &=& x^{20} \pmod 5
\end{array}$$
This implies aside from the leading term $x^{20}$, the coefficients of $x^{k}$ in $f(x+1)$ where $0 \le k < 20$ are all divisible by $5$. Notice the constant term in $f(x+1)$ is $f(1) = 5$ which
is not divisible by $5^2$. By Eisenstein's criterion,
$f(x+1)$ is irreducible over $\mathbb{Q}[x]$. As a corollary, so does $f(x)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Improper Integral $\int\limits_0^1\frac{\ln(x)}{x^2-1}\,dx$ How can I prove that?
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\frac{\pi^2}{8}$$
I know that
$$\int_0^1\frac{\ln(x)}{x^2-1}\,dx=\sum_{n=0}^{\infty}\int_0^1-x^{2n}\ln(x)\,dx=\sum_{n=0}^{\infty}\frac{1}{(2n+1)^2}=\frac{\pi^2}{8}$$
but I want another method.
| This comes from my answer here. Let
$$I = \int_0^{\infty} dx \frac{\log{x}}{x^2-1}$$
Note that the singularity at $x=1$ is removable in this integral and therefore we do not need to use a Cauchy principal value. We evaluate this integral by once again appealing to the residue theorem, but this time, we consider
$$\oint_{C'} dz \frac{\log^2{z}}{z^2-1}$$
where $C'$ is a keyhole contour with respect to the positive real axis. By integrating around this contour and noting that the integrand vanishes sufficiently fast as the radius of the circular section of $C'$ increases without bound, we get
$$\oint_{C'} dz \frac{\log^2{z}}{z^2-1} = -i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} + 4 \pi^2 \int_0^{\infty} dx \frac{1}{x^2-1}$$
This is equal to, by the residue theorem, $i 2 \pi$ times the sum of the residues of the poles of the integrand of the complex integral within $C'$. As the only pole is at $z=-1$, we see that
$$\begin{align}\oint_{C'} dz \frac{\log^2{z}}{z^2-1} &= i 2 \pi \frac{\log^2{(-1)}}{2 (-1)} \\ &= i 2 \pi \frac{\pi^2}{2}\end{align}$$
Now, the real part of the integral above is split into a Cauchy principal value and a piece indented about the singularity at $x=1$. The Cauchy principal value is zero:
$$\begin{align}PV \int_0^{\infty} dx \frac{1}{x^2-1} &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_{1+\epsilon}^{\infty} dx \frac{1}{x^2-1}\right]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} + \int_0^{1/(1+\epsilon)} \left (-\frac{dx}{x^2} \right ) \frac{1}{(1/x^2)-1} \right ]\\ &= \lim_{\epsilon \rightarrow 0} \left [\int_0^{1-\epsilon} dx \frac{1}{x^2-1} - \int_0^{1-\epsilon} \frac{dx}{x^2-1} \right ] \\ &= 0\end{align}$$
The indent in the contour, however, produces a contribution; let $x=1+\epsilon e^{i \phi}$ and $\phi \in [\pi,0]$:
$$4 \pi^2 i \epsilon \int_{-\pi}^0 d\phi \frac{e^{i \phi}}{2 \epsilon e^{i \phi}} = i \frac{\pi}{2} 4 \pi^2$$
so that
$$-i 4 \pi \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = i 2 \pi \frac{\pi^2}{2} - i \frac{\pi}{2} 4 \pi^2 = -i 2 \pi \frac{\pi^2}{2}$$
Therefore
$$I = \int_0^{\infty} dx \frac{\log{x}}{x^2-1} = \frac{\pi^2}{4}$$
But the sought-after integral is
$$\int_0^{1} dx \frac{\log{x}}{x^2-1} = \frac12 I = \frac{\pi^2}{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/537903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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How to solve system equation $x^3+y^3+z^3=x+y+z, x^2+y^2+z^2=xyz$ for $x,y,z\in\mathbb{R}$? How to solve system equation $\left\{\begin{matrix}x^3+y^3+z^3=x+y+z&\\x^2+y^2+z^2=xyz&\end{matrix}\right.$ , $x,y,z\in\mathbb{R}$ ?
| Express all of the terms in terms of the symmetric polynomials $s_1=x+y+z, s_2=xy+yz+zx$ and $s_3=xyz$.
You get $s_1^3-3s_1s_2+3s_3=x^3+y^3+z^3=x+y+z=s_1$ and $s_1^2-2s_2=x^2+y^2+z^2=xyz=s_3$.
Putting the second equation into the first gives $s_1^3-3s_1s_2+3s_1^2-6s_2=s_1$, which simplifies to $s_2=(s_1^3+3s_1^2-s1)/(3s_1+6)$ and $s_3=s_1^2-2s_2$.
Then the solutions to the original equations are the roots of the polynomial $x^3-s_1x^2+s_2x-s_3$ where $s_2$ and $s_3$ are given according to the previous paragraph.
However, for almost all values of $s_1$, 2 of these roots are complex, so these are not real solutions.
In order to get real solutions, you need the discriminant of $x^3-s_1x^2+s_2x-s_3$ to be non-negative. This simplifies to $s_1^2(s_1+2)(648+536s_1+426s_1^2+117s_1^3+12s_1^4-6s1^5-6s1^6+s1^7)$ being non-positive, and the only $s_1$'s for which this is satisfied are $s_1=0$ ($x,y,z=0$) and $-2.16821<=s_1<-2$.
Example non-0 solution are the roots to $100x^3+210x^2-2023x-4487$
| {
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How find this limit $\lim\limits_{x\to 0^{+}}\frac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$
Find the limit
$$\lim_{x\to 0^{+}}\dfrac{\sin{(\tan{x})}-\tan{(\sin{x})}}{x^7}$$
My attempt: Since
$$\sin{x}=x-\dfrac{1}{3!}x^3+\dfrac{1}{5!}x^5-\dfrac{1}{7!}x^7+o(x^7)$$
$$\tan{x}=x+\dfrac{1}{3}x^3+\dfrac{2}{15}x^5+\dfrac{1}{63}x^7+o(x^3)$$
So
$$\sin{(\tan{x})}=\tan{x}-\dfrac{1}{3!}(\tan{x})^3+\dfrac{1}{5!}(\tan{x})^5-\dfrac{1}{7!}(\tan{x})^7+o(x^7)$$
Though this method might solve, I think this problem has nicer methods. Thanks.
| For sure, L'Hopital's rule would be useful. But you could also use each Taylor expansion of $\sin(x)$ and $\tan(x)$ to expand $\sin (\tan (x))$ and $\tan (\sin (x))$. This would probably be tedious but it is doable (I made it).
Using what you wrote, you should arrive to $$\sin (\tan (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+O\left(x^8\right)$$ and $$\tan (\sin (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+O\left(x^8\right)$$ $$\sin (\tan (x))-\tan (\sin (x))=-\frac{x^7}{30}+O\left(x^8\right)$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Pre Calculus Synthetic Division two of the roots of the equation $2x^3-3x^2+px+q=0$ are $3$ and $-2$. Find the third root of the equation.
| Hint:
Let $f(x) = 2x^{3} - 3x^{2} + px + q$.
Since there are $3$ roots to a cubic equation, and $3$ and $-2$ are roots of $f(x)$, we can factor $f(x)$ as
$f(x) = (2x - k)(x-3)(x+2)$
where $k/2$ is the third root of $f(x)$.
Since $f(x) = 2x^{3} - 3x^{2} + px + q$, we can expand the previous factorization of $f(x)$ to get linear equations which will allow us be able to solve for $k$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculating a determinant How do I calculate the determinant of the following matrix:
$$\begin{matrix}
-1 & 1 & 1 &\cdots & 1 \\
1 & -1 &1 &\cdots &1 \\
1 & 1 & -1 &\cdots &1\\
\vdots & \vdots & \vdots& \ddots &\vdots \\
1&1&1&\cdots&-1\\ \end{matrix}$$
Thanks in advance!
| Recall the formula for the determinant of a matrix $A$
$$\det(A) = \sum_{\pi\in S_n} \sigma(\pi) \prod_{i=1}^n A_{i,\pi(i)}.$$
Now the sign $\sigma(\pi)$ of a permutation $\pi$ is given by
$$\sigma(\pi) = \prod_{c\in\pi} (-1)^{|c|-1}$$
where the product is over the cycles of the disjoint cycle composition of $n.$
We will thus be able to evaluate the determinant if we can construct a generating function of the set of permutations where we mark the length $|c|$ of every cycle with $|c|-1$ in one variable (call it $u$) and the number of fixed points with another (call it $v$, this will count the number of elements equal to $-1$ in the product $\prod_{i=1}^n A_{i,\pi(i)}$ because they lie on the main diagonal, i.e. correspond to fixed points).
This gives the combinatorial species
$$\mathcal{Q} = \mathfrak{P}(\mathcal{V}\mathfrak{C}_1(\mathcal{Z})
+ \mathcal{U}\mathfrak{C}_2(\mathcal{Z})
+ \mathcal{U}^2\mathfrak{C}_3(\mathcal{Z})
+ \mathcal{U}^3\mathfrak{C}_4(\mathcal{Z})
+\cdots).$$
which in turn produces the generating function
$$Q(z,u,v) =
\exp\left(vz + u\frac{z^2}{2} + u^2\frac{z^3}{3} + u^3\frac{z^4}{4}+\cdots\right).$$
This is
$$Q(z,u,v) = \exp\left(vz - z + \frac{z}{1} + u\frac{z^2}{2} + u^2\frac{z^3}{3} + u^3\frac{z^4}{4}+\cdots\right) \\
= \exp\left(vz - z + \frac{1}{u}\left(u\frac{z}{1} + u^2\frac{z^2}{2} + u^3\frac{z^3}{3} + u^3\frac{z^4}{4}+\cdots\right)\right)\\
= \exp\left(vz - z + \frac{1}{u}\log\frac{1}{1-uz}\right)
= \exp\left(vz - z \right)\left(\frac{1}{1-uz}\right)^{1/u}.$$
Now the term $q \times u^k v^m z^n / n!$ appearing in $Q(z,u,v)$ represents $q$ permutations that have $m$ fixed points and where $$\sum_{c\in\pi} (|c|-1) = k.$$ Furthermore for all these permutations we have that
$$\sigma(\pi)\prod_{i=1}^n A_{i,\pi(i)} = (-1)^k (-1)^m.$$
This implies that
$$\det(A) = n! [z^n] Q(z,-1,-1)
= n! [z^n] \exp\left(-z - z \right)\left(\frac{1}{1+z}\right)^{-1}\\
= n! [z^n] \exp(-2z) (1+z)
= n! ([z^n] \exp(-2z) + [z^{n-1}] \exp(-2z))
\\= n! \left(\frac{(-2)^n}{n!} + \frac{(-2)^{n-1}}{(n-1)!}\right)
= (-2)^{n-1} (-2 + n)
\\ = (n-2) (-2)^{n-1}.$$
| {
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Is this equation to prove that $aRb \iff a^2 - b^2 = 1$ is antisymmetric correct?
Over $\mathbb{R}$, $aRb \iff a^2 - b^2 = 1$.
I tried determining if it was antisymmetric. I seem to have done it, but while doing the equation, I stumbled upon a scenario that always made me doubt my decisions:
Have
$$aRb \land bRa$$
$$a^2 - b^2 = 1 \land b^2 - a^2 = 1$$
$$-a^2 + b^2 = -1 \land b^2 - a^2 = 1$$
$$-2a^2 + 2b^2 = 0$$
$$\color{#C00}{2(-a^2 + b^2) = 0}$$
$$\color{#C00}{-a^2 + b^2 = 0}$$
$$b^2 = a^2$$
$$b = a$$
I passed the $2$ to divide to the other side. The other side has a $0$, so the $2$ is essentially killed off with no apparent consequence.
Is this equation correct?
| The deduction you have does not work, as $a^2 = b^2$ does not imply $a = b$. It is actually much simpler. Assuming $aRb$ and $bRa$, you get (third line of equations) $-a^2 + b^2 = -1$ and $b^2 - a^2 = 1$. So $b^2 - a^2$ must be equal to $1$ and $-1$ at the same time, which is not possible.
| {
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Evaluating $\lim\limits_{n\to \infty}\left\{(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})\right\}^{\frac{1}{n}}$ Question is to evaluate :
$$\lim_{n\to \infty}\left\{ \left(1+\frac{1}{n}\right)\left(1+\frac{2}{n}\right)\dots\left(1+\frac{n}{n}\right)\right\}^{\frac{1}{n}}$$
I tried to do something like this but it is not getting better...
$(1+\frac{1}{n})(1+\frac{2}{n})\cdots(1+\frac{n}{n})=(\frac{n+1}{n})(\frac{n+2}{n})\cdots (\frac{n+n}{n})=\frac{1}{n^n}(n+1)(n+2)\dots(n+n)$
please give some hint to proceed further.
| Consider the log of $[(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}}$:
$$\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right).$$
The limit of this sum is equal to $\int_0^1\ln(1+x)dx$, so
$$
\begin{aligned}
\lim_{n\to\infty} [(1+\frac{1}{n})(1+\frac{2}{n})\dots(1+\frac{n}{n})]^{\frac{1}{n}}
&= \lim_{n\to\infty} \exp\left(\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right) \right)\\
&= \exp\left(\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^n \ln \left( 1+\frac{k}{n}\right) \right)\\
&= \exp \left( \int_0^1\ln(1+x)dx\right)
\end{aligned}$$
| {
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"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Infinite Series $\sum\limits_{n=1}^\infty\frac{(H_n)^2}{n^3}$ How to prove that
$$\sum_{n=1}^{\infty}\frac{(H_n)^2}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$
$H_n$ denotes the harmonic numbers.
| Since $$\sum_{n=1}^\infty\frac{H_{n-1}}{n}x^n=\frac12\ln^2(1-x)$$
divide both sides by $x$ then $\int_0^y$
$$\sum_{n=1}^\infty\frac{H_{n-1}}{n^2}y^n=\frac12\int_0^y\frac{\ln^2(1-x)}{x}dx$$
Next multiply both sides by $-\frac{\ln(1-y)}{y}$ then $\int_0^1$ and apply that $-\int_0^1 y^{n-1}\ln(1-y)dy=\frac{H_n}{n}$
$$\sum_{n=1}^\infty\frac{H_{n-1}H_n}{n^3}=\sum_{n=1}^\infty\frac{H_n^2}{n^3}-\sum_{n=1}^\infty\frac{H_n}{n^4}=-\frac12\int_0^1\int_0^y\frac{\ln^2(1-x)\ln(1-y)}{xy}dxdy$$
$$=-\frac12\int_0^1\frac{\ln^2(1-x)}{x}\left(\int_x^1\frac{\ln(1-y)}{y}dy\right)dx$$
$$=-\frac12\int_0^1\frac{\ln^2(1-x)}{x}\left(\text{Li}_2(x)-\zeta(2)\right)dx$$
$$=\zeta(2)\zeta(3)-\frac12\int_0^1\frac{\ln^2(1-x)\text{Li}_2(x)}{x}dx$$
$$=\zeta(2)\zeta(3)-\frac12\sum_{n=1}^\infty\frac1{n^2}\int_0^1 x^{n-1}\ln^2(1-x)dx$$
$$=\zeta(2)\zeta(3)-\frac12\sum_{n=1}^\infty\frac1{n^2}\left(\frac{H_n^2+H_n^{(2)}}{n}\right)$$
$$=\zeta(2)\zeta(3)-\frac12\sum_{n=1}^\infty\frac{H_n^2}{n^3}-\frac12\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}$$
Rearrange the terms
$$\sum_{n=1}^\infty\frac{H_n^2}{n^3}=\frac23\zeta(2)\zeta(3)+\frac23\sum_{n=1}^\infty\frac{H_n}{n^4}-\frac13\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=\frac{7}{2}\zeta(5)-\zeta(2)\zeta(3)$$
where we used $\sum_{n=1}^\infty\frac{H_n}{n^4}=3\zeta(5)-\zeta(2)\zeta(3)$ and $\sum_{n=1}^\infty\frac{H_n^{(2)}}{n^3}=3\zeta(2)\zeta(3)-\frac92\zeta(5)$
| {
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Solve: $\frac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$ Solve: $$\dfrac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$$
My try:
Conditions identify: $\left\{ \begin{array}{l} x-1>0\\1-\sqrt[3]{2-x}>0\end{array} \right.\Leftrightarrow \left\{ \begin{array}{l} x>1\\ \sqrt[3]{2-x}<1 \end{array} \right.\Leftrightarrow x>1$
$\dfrac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)\\\Leftrightarrow \log_{\frac{1}{2}}\sqrt{x-1}>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)\\\Leftrightarrow \sqrt{x-1}-1+\sqrt[3]{2-x}<0$
I don't know how to solve this Any equation: $\boxed{\sqrt{x-1}-1+\sqrt[3]{2-x}<0},$ please help me solve that into the result, please guide me, thanks.
| Ok, let me give you a hint.
$$\frac{1}{2}\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$$
so
$$\log_{\frac{1}{2}}\left(x-1\right)>2\log_{\frac{1}{2}}\left(1-\sqrt[3]{2-x}\right)$$
$$\log_{\frac{1}{2}}\left(x-1\right)>\log_{\frac{1}{2}}[\left(1-\sqrt[3]{2-x}\right)]^2$$
$$\left(x-1\right)<[\left(1-\sqrt[3]{2-x}\right)]^2$$
$$\left(x-1\right)<\left(1-2({2-x})^{\frac{1}{3}}+({2-x})^{\frac{2}{3}}\right)$$
$$0<\left((2-x)-2({2-x})^{\frac{1}{3}}+({2-x})^{\frac{2}{3}}\right)$$
$$0<\left((2-x)^{\frac{3}{3}}-2({2-x})^{\frac{1}{3}}+({2-x})^{\frac{2}{3}}\right)$$
| {
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How prove this inequality $\frac{x}{x+yz}+\frac{y}{y+zx}+\frac{z}{z+xy}\ge \frac{3}{2}$ let $x,y,z>0$,and such $x^n+y^n+z^n=3(n\ge 1),n\in N^*$,
show that:
$$\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy}\ge \dfrac{3}{2}$$
My try: if $n=1$ ,
since $x+y+z=3$,then
use Cauchy-Schwarz inequality
$$\left(\dfrac{x}{x+yz}+\dfrac{y}{y+zx}+\dfrac{z}{z+xy} \right)(x^2+y^2+z^2+3xyz)\ge (x+y+z)^2$$
then we only prove
$$\dfrac{9}{x^2+y^2+z^2+3xyz}\ge\dfrac{3}{2}$$
$$\Longleftrightarrow x^2+y^2+z^2+3xyz\le 6$$
Then I can't,and for $n$ how prove it?
| Using power means we have
$x+y+z\le 3\cdot {(\dfrac {x^n+y^n+z^n} {3})}^{\frac 1 n}=3$
Let $x\le y\le z$ then we have $z\ge 1$ so
$\dfrac x {x+yz}+\dfrac y {y+zx} \ge \dfrac 2 {z+1} \Leftrightarrow \\$
$\dfrac {(z-1)z(x-y)^2} {(x+yz)(y+zx)(z+1)} \ge 0$ is true
Hence it is sufficies to prove
$\dfrac 2 {z+1} +\dfrac z {z+xy} \ge \dfrac 3 2$
Since from AmGm inequality we have $xy\le (\dfrac {x+y} 2)^2 \le (\dfrac {3-z} 2)^2$ it is sufficies to prove
$\dfrac 2 {z+1} +\dfrac {4z} {4z+(3-z)^2} \ge \dfrac 3 2 \Leftrightarrow \\$
$\dfrac {(3-z)(z-1)^2} {(z+1)(z^2-2z+9)} \ge 0$
which is obvious since $x+y+z\le 3\Rightarrow z\le 3$
equality holds if and only if $x=y=z=1$ or $z\rightarrow 3,x=y\rightarrow 0$
second case of equality is valid only if $n=1$
| {
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"source": "stackexchange",
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Probability Problem on Divisibility of Sum by 3 From the 3-element subsets of $\{1, 2, 3, \ldots , 100\}$ (the set of the first 100 positive integers), a subset $(x, y, z)$ is picked randomly. What is the probability that $x + y + z$ is divisible by 3?
This is a math Olympiad problem. I would welcome a good solution.
| Using the Polya Enumeration Theorem and supposing we are choosing the three-element subset from $\{1,n\}$ we see that we have the following combinatorial class (not labelled):
$$\mathcal{Q} = \mathfrak{P}_3(\mathcal{Z}+\mathcal{Z}^2+\cdots+\mathcal{Z}^n).$$
Now the cycle index for the set construction is given by
$$Z(A_n)-Z(S_n) = Z(S_n)_{a_k=(-1)^{k-1} a_k}.$$
For $n=3$ we get the cycle index
$$ Z = \frac{1}{6} a_1^3 - \frac{1}{2} a_1 a_2 + \frac{1}{3} a_3.$$
The generating function that we substitute into $Z$ is
$$z+z^2+\cdots+z^n = z(1+z+\cdots+z^{n-1}) = z \frac{1-z^n}{1-z}.$$
Therefore the generating function of all subsets by total size (not only those whose sum is divisible by three) is
$$f(z) = \frac{1}{6} \left(z \frac{1-z^n}{1-z}\right)^3
- \frac{1}{2} z \frac{1-z^n}{1-z} z^2 \frac{1-z^{2n}}{1-z^2} +
\frac{1}{3} z^3 \frac{1-z^{3n}}{1-z^3}.$$
We need the value at $z=1$ of this generating function (excluding powers not divisible by three). Setting $w=e^{2\pi i/3}$ we see that the total count of the subsets with sum divisible by three is
$$\left.\frac{1}{3}(f(wz)+f(w^2z)+f(w^3z))\right|_{z=1}.$$
Now suppose that $$n\equiv r\bmod 3$$
where $r\in(1,2,3).$
Then we have that
$$f(wz)_{z=1} =
\frac{1}{6} \left(\sum_{q=1}^r w^q\right)^3
-\frac{1}{2} \left(\sum_{q=1}^r w^q\right) \left(\sum_{q=1}^r w^{2q}\right)
+\frac{1}{3}n$$
and that
$$f(w^2z)_{z=1} =
\frac{1}{6} \left(\sum_{q=1}^r w^{2q}\right)^3
-\frac{1}{2} \left(\sum_{q=1}^r w^{2q}\right) \left(\sum_{q=1}^r w^{4q}\right)
+\frac{1}{3}n$$
and
$$f(w^3z)_{z=1} = \frac{1}{6}n^3-\frac{1}{2}n^2+\frac{1}{3}n
= {n\choose 3}.$$
Doing the arithmetic we find that
$$f(wz)_{z=1} + f(w^2z)_{z=1} =
\frac{1}{3} \frac{w^{2r}-w^r}{w^2-w}
- \left( \frac{w^{2r}-w^r}{w^2-w}\right)^2
+\frac{2}{3} n.$$
This gives for the total count the value
$$\frac{1}{9} \frac{w^{2r}-w^r}{w^2-w}
- \frac{1}{3} \left( \frac{w^{2r}-w^r}{w^2-w}\right)^2
+ \frac{2}{9} n
+\frac{1}{3} {n\choose 3}$$
which can be simplified further to
$$\frac{1}{9} (r-3)(3r-2)
+ \frac{2}{9} n
+\frac{1}{3} {n\choose 3}$$
which produces the sequence
$$0, 0, 1, 2, 4, 8, 13, 20, 30, 42, 57, 76, 98, 124, 155, 190, 230, 276, 327, 384,\ldots$$
which is A061866 from the OEIS.
We thus obtain the following formula for the probability
$$\frac{1}{3}
+{n\choose 3}^{-1}
\left(\frac{1}{9} (r-3)(3r-2)
+ \frac{2}{9} n\right).$$
Substituting $n=100$ in the above we get the final result,
$$\frac{817}{2450}.$$
| {
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$x^{311} \equiv 662 \mod{713}$ We have to find $x$ such that $x^{311} \equiv 662 \mod{713}$
The example given in my notes has a few typos and the professor is unavailable.
| $$713=23 \cdot 31 $$
By Chinese Remainder Theorem, you need to solve
$$x^{311} \equiv 662 \pmod{23} \\
x^{311} \equiv 662 \pmod{31}$$
or equivalently
$$x^{311} \equiv 18 \pmod{23} \\
x^{311} \equiv 11 \pmod{31}$$
It follows that $x$ is invertible in both arithmetics, and then by Fermat Little Theorem we have
$$x^3\equiv 18 \pmod{23} \Rightarrow x^{-1} \equiv x^{21} \equiv 18^7 \pmod{23}\Rightarrow x \equiv (18^7)^{-1} \pmod{23}$$
and
$$x^{11} \equiv 11 \pmod{31} \Rightarrow x^{3} \equiv x^{33} \equiv 11^{3} \pmod{31}\,.$$
Now use the CRT to find $x \pmod{713}$.
Second solution
As 662 is invertible, it follows that $x$ is invertible and hence, by Euler Theorem we have
$$x^{660} \equiv 1 \pmod {713} \,.$$
now, find $a,b$ so that
$$a\cdot 311 +b \cdot 660=1$$
Then
$$x= x^{1}=x^{a\cdot 311 +b \cdot 660}=(x^{311})^a\cdot(x^{660})^b=662^a \pmod{713} \,.$$
| {
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"answer_count": 1,
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Show that $\frac{a+b}{2} \ge \sqrt{ab}$ for $0 \lt a \le b$ I have to prove that
$$\frac{a+b}{2} \ge \sqrt{ab} \quad \text{for} \quad 0 \lt a \le b$$
The main issue I am having is determining when the proof is complete (mind you, this is my first time). So I did the following steps:
$$\begin{align}
\frac{a+b}{2} &\ge \sqrt{ab} \\
\left(\frac{a+b}{2}\right)^2 &\ge \left(\sqrt{ab}\right)^2 \\
\frac{a^2+2ab+b^2}{4} &\ge ab \\
a^2+2ab+b^2 &\ge 4ab \\
a^2-2ab+b^2 &\ge 0\\
(a-b)^2 &\ge 0 \\
\end{align}$$
Now this is where I stopped because if I square root each side, I will be left with $a-b \ge 0$ or in other words, $a \ge b$ which doesn't make a whole lot of sense to me. So ultimately the question is: how do I know when I'm done? and is what I did above correct?
Thanks!
| Actually, you don't need to square both sides, since $\frac{a+b}{2} \ge \sqrt{ab} \Leftrightarrow \frac{a+b-2\sqrt{ab}}{2} \ge 0
\Leftrightarrow \frac{(\sqrt a - \sqrt b)^2}{2} \ge 0$. And for any real number x, $x^2 \ge 0$ is always true, so $\frac{(\sqrt a - \sqrt b)^2}{2} \ge 0$ is true.
As for your last equation, $(a-b)^2 \ge 0$, for any real number $a, b$, it is always a truth.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/557653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 4,
"answer_id": 3
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Find the limit of a fraction So far I have,
$$
\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{1^2+1}}}{x-1}=\lim_{x\to 1} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{2}}}{x-1}
$$
I have no idea how to keep going with this, every way I try I get stuck and can't do anything with it.
| More generally,
consider
$$\lim_{x\to a} \frac{\frac{x}{\sqrt{x^2+1}} - \frac{1}{\sqrt{a^2+1}}}{x-a}.$$
I will use just algebra.
If $x \ne a$,
$\begin{align}
\frac{\frac{x}{\sqrt{x^2+1}} - \frac{a}{\sqrt{a^2+1}}}{x-a}
&=\frac{x\sqrt{a^2+1}-a\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)}\\
&=\left(\frac{x\sqrt{a^2+1}-a\sqrt{x^2+1}}{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)}\right)
\left(\frac{x\sqrt{a^2+1}+a\sqrt{x^2+1}}{x\sqrt{a^2+1}+a\sqrt{x^2+1}}\right)\\
&=\frac{x^2(a^2+1)-a^2(x^2+1)}
{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
&=\frac{x^2a^2+x^2-a^2x^2-a^2}
{\sqrt{x^2+1}\sqrt{a^2+1}(x-a)(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
&=\frac{x^2-a^2}
{(x-a)\sqrt{x^2+1}\sqrt{a^2+1}(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
&=\frac{x+a}
{\sqrt{x^2+1}\sqrt{a^2+1}(x\sqrt{a^2+1}+a\sqrt{x^2+1})}\\
\end{align}
$
Letting $x \to a$,
this becomes
$\frac{2a}
{(a^2+1)(2a\sqrt{a^2+1})}
=\frac{1}
{(a^2+1)^{3/2}}
$.
As a derivative,
$\begin{align}
\left(\frac{x}{\sqrt{x^2+1}}\right)'
&=\frac{\sqrt{x^2+1}-x(1/2)(2x)(x^2+1)^{-1/2}}{x^2+1}\\
&=\frac{\sqrt{x^2+1}-x^2(x^2+1)^{-1/2}}{x^2+1}\\
&=\frac{(x^2+1)-x^2}{(x^2+1)^{3/2}}\\
&=\frac{1}{(x^2+1)^{3/2}}\\
\end{align}
$
which is comforting (and much easier).
Note that the "$1$"
in $x^2+1$ and $a^2+1$
can be any value -
it is just carried along
and,
if the expression is
$\frac{x}{\sqrt{x^2+b}}$,
the result is
$\frac{b}{(x^2+b)^{3/2}}$.
| {
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"source": "stackexchange",
"question_score": "1",
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A limit problem related to $\log \sec x$
If $$f(x) = \dfrac{{\displaystyle 3\int_{0}^{x}(1 + \sec t)\log\sec t\,dt}}{(\log\sec x)\{x + \log(\sec x + \tan x)\}}$$ then prove that $$\lim_{x \to {\pi/2}^{-}}f(x) = \frac{3}{2}$$ and $$\lim_{x \to 0}\frac{f(x) - 1}{x^{4}} = \frac{1}{420}$$
Looking at the integral sign in numerator I see that the best way to attack this problem is via L'Hospital Rule. But that requires to show that the integral diverges to $\infty$ as $x \to {\pi/2}^{-}$. Assuming that this is the case I solved the first limit by applying L'Hospital's rule twice. But for the second limit it seems hopeless to try L'Hospital because of denominator $x^{4}$ which might require 4 times its application.
Looking at the functions involved it does not look easy to apply Taylor's series expansions. I am not sure if there is any elegant solution for the second problem. Please let me know any hints or a solution to the second limit.
Update: I tried some simplification along with LHR for the second limit but still the final solution is eluding.
Let $a(x), b(x)$ be the numerator and denominator of $f(x)$. Clearly we can see that
\begin{align}
B &= \lim_{x \to 0}\frac{b(x)}{x^{3}}\notag\\
&= \lim_{x \to 0}\frac{\log\sec x\{x + \log(\sec x + \tan x)\}}{x^{3}}\notag\\
&= -\lim_{x \to 0}\frac{\log\cos x\{x + \log(1 + \sin x) - \log \cos x\}}{x^{3}}\notag\\
&= -\lim_{x \to 0}\frac{\log\cos x}{x^{2}}\cdot\frac{x + \log(1 + \sin x) - \log \cos x}{x}\notag\\
&= -\lim_{x \to 0}\frac{\log(1 + \cos x - 1)}{\cos x - 1}\cdot\frac{\cos x - 1}{x^{2}}\cdot\frac{x + \log(1 + \sin x) - \log \cos x}{x}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\frac{x + \log(1 + \sin x) - \log \cos x}{x}\notag\\
&= \frac{1}{2}\lim_{x \to 0}\left(1 + \frac{\log(1 + \sin x)}{\sin x}\cdot\frac{\sin x}{x} - \frac{\log (1 + \cos x - 1)}{\cos x - 1}\cdot x\cdot \frac{\cos x - 1}{x^{2}}\right)\notag\\
&= \frac{1}{2}\cdot 2 = 1\notag
\end{align}
Thus we can write
\begin{align}
L &= \lim_{x \to 0}\frac{f(x) - 1}{x^{4}}\notag\\
&= \lim_{x \to 0}\frac{a(x) - b(x)}{b(x)x^{4}}\notag\\
&= \lim_{x \to 0}\frac{a(x) - b(x)}{x^{7}}\cdot\frac{x^{3}}{b(x)}\notag\\
&= \lim_{x \to 0}\frac{a(x) - b(x)}{x^{7}}\notag\\
&= \frac{1}{7}\lim_{x \to 0}\frac{a'(x) - b'(x)}{x^{6}}\text{ (via L'Hospital's Rule)}\notag\\
&= \frac{1}{7}\lim_{x \to 0}\frac{3(1 + \sec x)\log\sec x - \tan x\{x + \log(\sec x + \tan x)\} -\log\sec x\{1 + \sec x\}}{x^{6}}\notag\\
&= \frac{1}{7}\lim_{x \to 0}\frac{2(1 + \sec x)\log\sec x - \tan x\{x + \log(\sec x + \tan x)\}}{x^{6}}\notag\\
&= \frac{1}{7}\lim_{x \to 0}\frac{2(1 + \cos x)\log\sec x - \sin x\{x + \log(\sec x + \tan x)\}}{x^{6}\cos x}\notag\\
&= \frac{1}{7}\lim_{x \to 0}\frac{2(1 + \cos x)\log\sec x - \sin x\{x + \log(\sec x + \tan x)\}}{x^{6}}\notag\\
\end{align}
I wonder what could be done to go further.
| Here is my work for the first limit: Let $f(x) = \frac{num(x)}{den(x)}$. We get $\infty/\infty$ so we can use L'Hopital:
\begin{align*}
\frac{num'(x)}{den'(x)}&=\frac{3(1+\sec(x))\log \sec x}{(\tan(x))(x + \log(\sec x + \tan x)) + (\log \sec (x)) (1 + \sec(x)) }\\
&= \frac{3(1+\sec(x))\log \sec x}{(\tan(x))[x + (\log \sec x) +\log(1+\sin(x))] + (\log \sec x)(1+\sec(x)) )}\\
&= \frac{3}{\left(\frac{\tan(x)}{1+\sec(x)}\right)\left(1 + \frac{x + \log(1 + \sin(x))}{\log \sec x}\right)+ 1}
\end{align*}
Taking a limit as $x\rightarrow \pi/2^-$ gives $3/2$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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How to show that $\log (\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2})= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$ If $\log (a +b +c) =\log a+\log b+\log c$ then show that $$\log \left(\frac{2a}{1-a^2}+\frac{2b}{1-b^2}+\frac{2c}{1-c^2}\right)= \log\frac{2a}{1-a^2}+ \log \frac{2b}{1-b^2}+ \log \frac{2c}{1-c^2}$$
Trial: If put $a=\frac{2a}{1-a^2}$ and the similar we are done. Can we do this in this way? Otherwise how to do. Please help.
| OK, here's another way to do it:
Let $a\circ b=\dfrac{a+b}{1-ab}$.
Prove that this operation is associative.
Show that $a\circ b\circ c = \dfrac{a+b+c-abc}{1-ab-ac-bc}$.
That implies that if $a+b+c=abc$, then $a\circ b\circ c=0$.
It's easy to show that $0\circ 0=0$, so we have
\begin{align}
0 & = 0\circ 0 = (a\circ b\circ c)\circ(a\circ b\circ c) \\[12pt]
& = (a\circ a)\circ(b\circ b)\circ(c\circ c) \qquad (\text{This is where associativity is used.}) \\[12pt]
& = \frac{2a}{1-a^2}\circ\frac{2b}{1-b^2} \circ\frac{2c}{1-c^2}.
\end{align}
So that implies the numerator $\dfrac{2a}{1-a^2}+\dfrac{2b}{1-b^2} +\dfrac{2c}{1-c^2} -\dfrac{2a}{1-a^2}\cdot\dfrac{2b}{1-b^2} \cdot\dfrac{2c}{1-c^2}$, is $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/569381",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Hard contest type trigonometry proof Suppose that real numbers $x, y, z$ satisfy:
$$\frac{\cos x + \cos y + \cos z}{\cos(x + y + z)}
=
\frac{\sin x + \sin y + \sin z}{\sin (x + y + z )}
= p$$
Then prove that:
$$\cos (x + y) + \cos (y + z ) + \cos (x + z) = p$$
I am not even getting where to start? Please help.
| Since
\begin{align*}
p \cos(x+y+z) &= \cos x + \cos y + \cos z \\
p \sin(x+y+z) &= \sin x + \sin y +\sin z
\end{align*}
\begin{align*}
p e^{i(x+y+z)} & = e^{ix} + e^{iy} + e^{iz}
\end{align*}
Multiplying throughout by $e^{-i(x+y+z)}$, we get
\begin{align*}
p = e^{-i(y+z)} + e^{-i(z+x)} + e^{-i(x+y)}
\end{align*}
Since $p$ is real, equating the real parts, we get
$$\cos(y+z) + \cos(z+x) + \cos(x+y) = p $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/571706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Evaluate $ \sum\limits_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}$ The question was: Evaluate, ${\textstyle {\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}}.$
And I go, since $\frac{n}{n^{4}+n^{2}+1}\sim\frac{1}{n^{3}}$ and we know that ${\displaystyle \sum_{n=}^{\infty}\frac{1}{n^{3}}}$ converges. so ${\displaystyle \sum_{n=1}^{\infty}\frac{n}{n^{4}+n^{2}+1}}$ is convergent as well.
But I find it hard to calculate the sum. can you give me some hints?
| Easiest and fastest solution:
Basically, try small values of $n$ and you'll see a pattern:
$\frac{1}{1^4+1^2+1}+\frac{2}{2^4+2^2+1}+\frac{3}{3^4+3^2+1}+\frac{4}{4^4+4^2+1}\dots$
$\implies \frac{1}{3}+\frac{2}{21}+\frac{3}{91}+\frac{4}{273}$
$\implies \frac{1}{2}\left(\frac{1}{1}-\frac{1}{3}\right)+\frac{1}{4}\left(\frac{2}{3}-\frac{2}{7}\right)+\frac{1}{6}\left(\frac{3}{7}-\frac{3}{13}\right)+\frac{1}{8}\left(\frac{4}{13}-\frac{4}{21}\right)+\dots$
$\implies \frac{1}{2}-\frac{1}{6}+\frac{1}{6}-\frac{1}{14}+\frac{1}{14}-\frac{1}{26}+\frac{1}{26}-\frac{1}{42}+\dots$
Notice that we can just cancel out every single term except for the original one, that is $\boxed{\frac{1}{2}}$ and we're done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/571973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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"answer_id": 5
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Show $f(x) = (x^4+x^2+1)/(x^3+1) $ is $O(x)$ How would I find the witnesses $C$ and $k$ such that $f(x)$ is $O(x)$?
What I tried was $$(x^4+x^2+1)/(x^3+1) ≤ (x^4+x^4+x^4)/(x^3+x^3) = (3/2)x $$
for values $x>1$. $C = 3/2, k = 1$
Is this right?
| Hint: The degree of the numerator is $4$, while the degree of the denominator is $3$. The degree of $O(x)$ is $4-3=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/572496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Interval of convergence homework $$ \sum_{n=1}^{\infty} \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} (x-1)^n $$
First, I start with the ratio test:
$$\lim_{n\to \infty} \left| \left(\frac{-1}{4}\right)^{n+1} \frac{(5(n+1))^{n+1}}{(n+1)!} (x-1)^{n+1} \div \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} (x-1)^n \right| =$$
$$=\frac{1}{4} \lim_{n\to \infty} \left| \frac{(5(n+1))^{n+1} (x-1)^{n+1} n!}{(n+1)!(5n)^n (x-1)^n}\right|= \left| \frac{x-1}{4}\right| \lim_{n\to \infty} \left| \frac{5(n+1)^{n+1}}{n^{n+1}} \right|=$$
$$= \left| \frac{5x-5}{4}\right|\lim_{n\to \infty} \left| \frac{(n+1)^{n+1}}{n^{n+1}} \right|=\left| \frac{5ex-5e}{4}\right|$$
For the series to converge:
$$-\frac{5}{4e} +1 < x < \frac{5}{4e}+1$$
Now I should check the endpoints:
$$ \sum_{n=1}^{\infty} \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} \left(-\frac{5}{4e}\right)^n$$
and $$\sum_{n=1}^{\infty} \left(\frac{-1}{4}\right)^n \frac{(5n)^n}{n!} \left(\frac{5}{4e}\right)^n$$
After checking my answers with wolfram, I think I've made a mistake somewhere. Can somebody help me out.
| You made a small algebra error in solving the inequality at the end, interchanging the $4$ and the $5$: you should have
$$1-\frac4{5e}<x<1+\frac4{5e}\;.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/578635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Integral of a function I have to solve a probability problem, and at a final step I arrive at the following integral:
$$\int_z^\infty xe^{-x^2/a^2}\displaystyle\frac{1}{\sqrt{x^2-z^2}}dx$$
I tried to integrate by parts, taking $f(x)=\displaystyle\frac{1}{\sqrt{x^2-z^2}}$ and
$g'(x)= xe^{-x^2/a^2}$, but this choice leads to a more complicated integral
$\int_z^\infty f'(x)g(x)dx$.
Is there an analytical method to solve this problem?
| $\int_z^\infty xe^{-\frac{x^2}{a^2}}\dfrac{1}{\sqrt{x^2-z^2}}dx$
$=\int_z^\infty\dfrac{e^{-\frac{x^2}{a^2}}}{2\sqrt{x^2-z^2}}d(x^2)$
$=\int_{z^2}^\infty\dfrac{e^{-\frac{x}{a^2}}}{2\sqrt{x-z^2}}dx$
$=e^{-\frac{z^2}{a^2}}\int_{z^2}^\infty e^{-\frac{x-z^2}{a^2}}~d(\sqrt{x-z^2})$
$=e^{-\frac{z^2}{a^2}}\int_0^\infty e^{-\frac{x^2}{a^2}}~dx$
$=\dfrac{ae^{-\frac{z^2}{a^2}}\sqrt\pi}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/579780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Fractions in Questions and Answers
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