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How can I attack $\int_{-\infty}^\infty \frac{x}{[(x-vt)^2+a^2]^2} dx$? I want to evaluate
$$\int_{-\infty}^\infty \frac{x}{[(x-vt)^2+a^2]^2} dx$$
I know that
$$\int \frac{1}{(\xi^2+a^2)^2} = \frac{\xi}{2a^2(\xi^2+a^2)}+\frac{1}{2a^3}\arctan\frac{\xi}{a}$$
Using integration by parts and choosing $a:= a$, $\xi := x-vt$ and thus
$$u = x \Rightarrow u' = 1 $$
$$g' = \frac{1}{[(x-vt)^2+a^2]^2} \Rightarrow g = \frac{x-vt}{2a^2((x-vt)^2+a^2)}+\frac{1}{2a^3}\arctan\frac{x-vt}{a}$$
Therefore I get
$$\int_{-\infty}^\infty \frac{x}{[(x-vt)^2+a^2]^2} dx = \left. x\cdot \frac{x-vt}{2a^2((x-vt)^2+a^2)}+\frac{1}{2a^3}\arctan\frac{x-vt}{a}\right\vert_{-\infty}^\infty - \int_{-\infty}^\infty \frac{x-vt}{2a^2((x-vt)^2+a^2)}+\frac{1}{2a^3}\arctan\frac{x-vt}{a} dx$$
Which isn't really helpful especially because
$$\left.x\cdot \frac{x-vt}{2a^2((x-vt)^2+a^2)}+\frac{1}{2a^3}\arctan\frac{x-vt}{a}\right\vert_{-\infty}^\infty$$
Makes me divide $\frac{\infty}{\infty}$ in the frist term ...
So how can I do this better?
Thank you very much for your help.
FunkyPeanut
| Start by substituting $\xi=x-vt$ to get
$$\eqalign{\int_{-\infty}^\infty \frac{x}{((x-vt)^2+a^2)^2} dx
&=\int_{-\infty}^\infty \frac{\xi+vt}{(\xi^2+a^2)^2} dx\cr
&=\int_{-\infty}^\infty \frac{\xi}{(\xi^2+a^2)^2} dx
+vt\int_{-\infty}^\infty \frac{1}{(\xi^2+a^2)^2} dx\ .\cr}$$
Now the first integral is zero because the integrand is odd (but see below), and you already know how to do the second, giving
$$vt\biggl[\frac{\xi}{2a^2(\xi^2+a^2)}+\frac{1}{2a^3}\arctan\frac{\xi}{a}\biggr]_{-\infty}^\infty\ .$$
Now if $\xi\to\pm\infty$, the first term tends to $0$, because it has $\xi$ in the numerator but $\xi^2$ in the denominator. Also, the $\arctan$ term tends to $\pm\pi/2$, provided that $a$ is positive. So it all comes out to
$$\int_{-\infty}^\infty \frac{x}{((x-vt)^2+a^2)^2} dx
=\frac{vt\pi}{2a^3}\ .$$
Note that if $f$ is an odd function,
$$\int_{-\infty}^\infty f(\xi)\,dx=0\ ,$$
provided that $\int_0^\infty f(\xi)\,dx$ converges. This is true in your problem because the integrand is effectively $1/\xi^3$ for large $\xi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/774142",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A closed form for the infinite series $\sum_{n=1}^\infty (-1)^{n+1}\arctan \left( \frac 1 n \right)$ It is known that $$\sum_{n=1}^{\infty} \arctan \left(\frac{1}{n^{2}} \right) = \frac{\pi}{4}-\tan^{-1}\left(\frac{\tanh(\frac{\pi}{\sqrt{2}})}{\tan(\frac{\pi}{\sqrt{2}})}\right). $$
Can we also find a closed form for the value of $$\sum_{n=1}^{\infty} (-1)^{n+1} \arctan \left(\frac{1}{n} \right)? $$
Unlike the other infinite series, this infinite series only converges conditionally.
| Let $ \displaystyle S(a) = \sum_{n=1}^{\infty}(-1)^{n-1} \arctan \left(\frac{a}{n} \right)$.
Since $ \displaystyle \sum_{n=1}^{\infty} (-1)^{n-1} \frac{n}{n^{2}+a^{2}}$ converges uniformly on $\mathbb{R}$,
$$ \begin{align} S'(a) &= \sum_{k=1}^{\infty} (-1)^{n-1} \frac{n}{a^{2}+n^{2}} \\ &= \frac{1}{2} \sum_{n=1}^{\infty} (-1)^{n-1} \left(\frac{1}{n-ia} + \frac{1}{n+ia} \right) \\ &= \frac{1}{2} \sum_{n=0}^{\infty} (-1)^{n} \left(\frac{1}{n+1-ia}+ \frac{1}{n+1+ia} \right). \end{align}$$
Then using the fact $$ \sum_{k=0}^{\infty} \frac{(-1)^{k}}{z+k} = \frac{1}{2} \Big[\psi\left( \frac{z+1}{2}\right) - \psi \left(\frac{z}{2} \right) \Big] $$ (where $\psi(z)$ is the digamma function), we have
$$ S'(a) = \frac{1}{4} \left[\psi \left(1- \frac{ia}{2} \right) - \psi\left(\frac{1}{2}- \frac{ia}{2} \right) + \psi\left(1+ \frac{ia}{2} \right) -\psi\left(\frac{1}{2}- \frac{ia}{2} \right)\right].$$
Integrating back we find
$$ S(a) = \frac{i}{2} \left[\log \Gamma \left(1-\frac{ia}{2} \right) - \log \Gamma \left(\frac{1}{2}-\frac{ia}{2} \right) - \log \Gamma \left(1+\frac{ia}{2} \right) + \log \Gamma \left(\frac{1}{2}+\frac{ia}{2} \right)\right] + C.$$
But since $S(0) = 0$, the constant of integration is $0$.
Therefore,
$$ \begin{align} S(1) &= \sum_{n=1}^{\infty}(-1)^{n-1} \arctan \left(\frac{1}{n} \right) \\ &= \frac{i}{2} \left[\log \Gamma \left(1-\frac{i}{2} \right) - \log \Gamma \left(\frac{1}{2}-\frac{i}{2} \right) - \log \Gamma \left(1+\frac{i}{2} \right) + \log \Gamma \left(\frac{1}{2}+\frac{i}{2} \right)\right] \\ &\approx 0.5066709032. \end{align}$$
The answer can be put in the same form as the answer given by achille hui by using the Schwarz reflection principle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/776182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 3
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no. of positive integral solutions of ||x - 1| - 2| + x = 3 What are the no. of positive integral solutions of ||x - 1| - 2| + x = 3 ?
My effort
||x - 1| - 2| = 3 - x
|x - 1| - 2 = 3 - x OR |x - 1| - 2 = x - 3
|x - 1| = 5 - x OR |x - 1| = x - 1
x - 1 = 5 - x OR x - 1 = x - 5 OR x - 1 $\geq$ 0
2x = 6 OR x $\geq$ 1
Therefore, x $\geq$ 1
But the any value of x greater than 1 except 3 does not satisfy the equation. Where have I gone wrong?
| You should start with expanding the innermost absolute value:
$$
\vert \vert x-1 \vert -2 \vert = \begin{cases} \vert x-3 \vert & x \geqslant 1 \\ \vert -1-x \vert & x < 1 \end{cases} = \begin{cases} x-3 & x \geqslant 1, x > 3 \\ 3-x & x \geqslant 1, x \leqslant 3 \\ 1+x & x < 1, x \geqslant -1 \\ -1-x & x < 1, x < -1 \end{cases} = \begin{cases} x-3 & x > 3 \\ 3-x & 1 \leqslant x \leqslant 3\\ 1+x & -1 \leqslant x < 1 \\ -1-x & x < -1 \end{cases}
$$
You now seek solutions for each of these branches being equal to $3-x$ within the appropriate domain, giving you three integral solutions $x=1, x=2, x=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/778419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation of tangent perpendicular to line I've got the homework question which I cannot solve.
Find the equation of the tangents to $4x^2+y^2=72$ that are
perpendicular to the line $2y+x+3=0$.
What I have done so far:
I have found the gradient of the line which is $m_1 = -\frac12$.
Which means that the equation perpendicular to the tangent of $4x^2+y^2=72$ is $m_2 = 2$, since $m_1m_2 = -1$.
I've found the derivative of $4x^2+y^2=72$ which is $\dfrac{dy}{dx} = -\dfrac{4x}{y}$.
The next thing I have done is set $m_2=\dfrac{dy}{dx}$.
Now what?
| Let the point of tangency be the point $(x,y)$ on our curve. Your calculation shows that $\frac{-4x}{y}=2$, that is, $y=-2x$.
Also, we have $4x^2+y^2=72$. Substituting $-2x$ for $y$ in this equation, we get $8x^2=72$, and therefore $x=\pm 3$. That gives the two points $(3,-6)$ and $(-3,6)$.
The line with slope $2$ through $(3,-6)$ has equation $y-(-6)=2(x-3)$, which simplifies to $y=2x-12$. The other line has equation $y=2x+12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/778779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find PV $\int_0^\infty \frac{\log \cos^2 \alpha x}{\beta^2-x^2} \, \mathrm dx=\alpha \pi$ $$
I:=PV\int_0^\infty \frac{\log\left(\cos^2\left(\alpha x\right)\right)}{\beta^2-x^2} \, \mathrm dx=\alpha \pi,\qquad \alpha>0,\ \beta\in \mathbb{R}.$$
I am trying to solve this integral, I edited and added in Principle value to clarify the convergence issue that the community pointed out. I tried to use $2\cos^2(\alpha x)=1+\cos 2\alpha x\,$ and obtained
$$
I=-\log 2 \int_0^\infty \frac{\mathrm dx}{\beta^2-x^2}+\int_0^\infty \frac{\log (1+\cos 2 \alpha x)}{\beta^2-x^2}\mathrm dx,
$$
simplifying
$$
I=\frac{ \pi \log 2 }{2\beta}+\int_0^\infty \frac{\log (1+\cos 2 \alpha x)}{\beta^2-x^2}\mathrm dx
$$
but stuck here. Note the result of the integral is independent of the parameter $\beta$. Thank you
Also for $\alpha=1$, is there a geometrical interpretation of this integral and why it is $\pi$?
Note this integral
$$
\int_0^\infty \frac{\log \sin^2 \alpha x}{\beta^2-x^2} \,\mathrm dx=\alpha \pi-\frac{\pi^2}{2\beta},\qquad \alpha>0,\beta>0
$$
is also FASCINATING, note the constraint $\beta>0$ for this one. I am not looking for a solution to this too obviously on the same post, it is just to interest people with another friendly integral.
| We make use of the identity
$$ \sum_{n=-\infty}^{\infty} \frac{1}{a^{2} - (x + n\pi)^{2}} = \frac{\cot(x+a) - \cot(x-a)}{2a}, \quad a > 0 \text{ and } x \in \Bbb{R}. $$
Then for $\alpha, \beta > 0$ it follows that
\begin{align*}
I := \mathrm{PV}\int_{0}^{\infty} \frac{\log\cos^{2}(\alpha x)}{\beta^{2} - x^{2}}
&= \frac{1}{2} \mathrm{PV} \int_{-\infty}^{\infty} \frac{\log\cos^{2}(\alpha x)}{\beta^{2} - x^{2}} \, dx \\
&= \frac{\alpha}{2} \mathrm{PV} \int_{-\infty}^{\infty} \frac{\log\cos^{2}x}{(\alpha\beta)^{2} - x^{2}} \, dx \\
&= \frac{\alpha}{2} \sum_{n=-\infty}^{\infty} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \frac{\log\cos^{2}x}{(\alpha\beta)^{2} - (x+n\pi)^{2}} \, dx \\
&= \frac{\alpha}{2} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \left( \sum_{n=-\infty}^{\infty} \frac{1}{(\alpha\beta)^{2} - (x+n\pi)^{2}} \right) \log\cos^{2}x \, dx \\
&= \frac{1}{4\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\cos^{2}x \, dx,
\end{align*}
where interchanging the order of integration and summation is justified by Tonelli's theorem applied to the summation over large indices $n$. Then
\begin{align*}
I
&= \frac{1}{4\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\cos^{2}x \, dx \\
&= \frac{1}{2\beta} \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} (\cot(x+\alpha\beta) - \cot(x-\alpha\beta)) \log\left|2\cos x\right| \, dx \tag{1}
\end{align*}
Here, we exploited the following identity to derive (1).
$$ \mathrm{PV} \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} \cot(x+a) \, dx = 0 \quad \forall a \in \Bbb{R}. $$
Now with the substitution $z = e^{2ix}$ and $\omega = e^{2i\alpha\beta}$, it follows that
\begin{align*}
I
&= \frac{1}{2\beta} \Re \mathrm{PV} \int_{|z|=1} \left( \frac{\bar{\omega}}{z - \bar{\omega}} - \frac{\omega}{z - \omega} \right) \log(1 + z) \, \frac{dz}{z}. \tag{2}
\end{align*}
Now consider the following unit circular contour $C$ with two $\epsilon$-indents $\gamma_{\omega,\epsilon}$ and $\gamma_{\bar{\omega},\epsilon}$.
Then the integrand of (2)
$$ f(z) = \left( \frac{\bar{\omega}}{z - \bar{\omega}} - \frac{\omega}{z - \omega} \right) \frac{\log(1 + z)}{z} $$
is holomorphic inside $C$ (since the only possible singularity at $z = 0$ is removable) and has only logarithmic singularity at $z = -1$. So we have
$$ \oint_{C} f(z) \, dz = 0. $$
This shows that
\begin{align*}
I
&= \frac{1}{2\beta} \Re \lim_{\epsilon \downarrow 0} \left( \int_{-\gamma_{\omega,\epsilon}} f(z) \, dz + \int_{-\gamma_{\bar{\omega},\epsilon}} f(z) \, dz \right) \\
&= \frac{1}{2\beta} \Re \left( \pi i \mathrm{Res}_{z=\omega} f(z) + \pi i \mathrm{Res}_{z=\bar{\omega}} f(z) \right) \\
&= \frac{1}{2\beta} \Re \left( - \pi i \log(1 + \omega) + \pi i \log(1 + \bar{\omega}) \right) \\
&= \frac{\pi}{\beta} \arg(1 + \omega)
= \frac{\pi}{\beta} \arctan(\tan (\alpha \beta)).
\end{align*}
In particular, if $\alpha\beta < \frac{\pi}{2}$ then we have
$$ I = \pi \alpha. $$
But due to the periodicity of $\arg$ function, this function draws a scaled saw-tooth function for $\alpha > 0$. Of course, $I$ is an even function of both $\alpha$ and $\beta$, so the final result is obtained by even extension of this saw-tooth function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
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Prove or disprove inequality $\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}$. Let $a$, $b$ and $c$ be real numbers greater than $0$. Prove inequality $$\displaystyle{\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\le\frac{a^4+b^4+c^4}{2abc}}.$$
| $2abc\cdot \dfrac{a^2}{b+c} = 2a^3\cdot \dfrac{bc}{b+c} \leq 2a^3\cdot \dfrac{b+c}{4} = \dfrac{a^3(b+c)}{2}$. Thus:
$2abc\cdot \dfrac{a^2}{b+c} + 2abc\cdot \dfrac{b^2}{c+a} + 2abc\cdot \dfrac{c^2}{a+b} \leq \dfrac{(a^3b + ab^3) + (a^3c + ac^3) + (b^3c + bc^3)}{2} \leq \dfrac{(a^4 + b^4) + (b^4 + c^4) + (c^4 + a^4)}{2} = a^4 + b^4 + c^4$, and this is true because:
$x^3y + xy^3 \leq x^4 + y^4 \iff (x^3 - y^3)(x - y) \geq 0$ is true $\forall x,y \geq 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How prove the constant term of $\left(1+x+\frac{1}{x}\right)^p\equiv1\pmod {p^2}$ if $p>3$ is odd prime number,show that:
the constant term of $$\left(1+x+\dfrac{1}{x}\right)^p\equiv1\pmod {p^2}$$
My try: since
$$(1+x+\dfrac{1}{x})^p=\sum_{k=0}^{p}\binom{p}{k}\left(x+\dfrac{1}{x}\right)^k=\sum_{k=0}^{p}\binom{p}{k}\sum_{j=0}^{k}\binom{k}{j}x^{k-2j}$$
so when $k=2j$,then the term is constant.
But how prove this constant $\equiv 1\pmod {p^2}$?
| For $p=2$, $(1+x+\frac{1}{x})^2=x^2+2x+3+\frac{2}{x}+\frac{1}{x^2}$.
$3\equiv 1\ (mod\ 4)$ ???
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/781750",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$l^2+m^2=n^2$ $\implies$ $lm$ is always a multiple of 3 when $l,m,n,$ are positive integers. Let $l,m,n$ be any three positive integers such that $l^2+m^2=n^2$
Then prove that $lm$ is always a multiple of 3.
| We show that $3$ must divde $lm$, by showing that $3$ divides $l$ or $3$ divides $m$.
Any integer $x$ is either divisible by $3$, or is congruent to $1$ or $-1$ modulo $3$. And iff $x\equiv \pm 1\pmod{3}$, then $x^2\equiv 1 \pmod{3}$.
Thus if neither $l$ nor $m$ is divisible by $3$, then $l^2+m^2\equiv 2\pmod{3}$. It follows that $l^2+m^2$ cannot be a perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $|f(x)| \le \frac{3}{2}$ when $f(x)=ax^2+bx+c$ Suppose $f(x) = ax^2+bx+c$ where $x \in [-1,1]$. If $f(-1),f(0),f(1)\in [-1,1]$ show that $|f(x)| \le \frac{3}{2}$ $\forall x \in [-1,1]$.
This is how I tried:
$f(0)=c$
$f(1)=a+b+c$
$f(-1)=a-b+c$
Putting $f(0)=c$ we get $f(1)-f(0)=a+b$, $f(-1)-f(0)=a-b$. Solving for $a$ and $b$ respectively we get $$a=\frac{f(1)+f(-1)-2f(0)}{2}$$ and $$b=\frac{f(1)-f(-1)}{2}$$ Then
$$|f(x)|=|(\frac{f(1)}{2})(x^2+x)+(\frac{f(-1)}{2})(x^2-x)+(f(0))(1-x^2)|$$
$$\le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|$$. Now individually the maximum of $|x^2+x|$ happens at $x=1$ , $|x^2-x|$ at $x=-1$ and $|1-x^2|$ at $x=0$. So All I get $|f(x)| \le 3$. But I need $\frac{3}{2}$.
Thanks for the help!!
| As mentioned above, this has already been answered in Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. with the better bound $5/4$. But it seems that your
proof is almost complete and slightly shorter than (or at least different from) the answers given there.
You already showed that
$$
|f(x)| \le |\frac{x^2+x}{2}|+|\frac{x^2-x}{2}|+|1-x^2|
$$
and for $0 \le x \le 1$ the right-hand side is
$$
\frac{x^2+x}{2} -\frac{x^2-x}{2}+1-x^2
= x + 1 - x^2 = \frac 54 - (x - \frac 12)^2 \le \frac 54 < \frac 32 \quad .
$$
Similarly for $-1 \le x \le 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/783410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Help on an integral I can't figure out how to compute the following integral:
$$\int_{-\infty}^{+\infty}\frac{\mathrm{d}z}{\sqrt{(x^2+y^2+z^2)^3}}$$
I think I should do some substitution, but I didn't figure it out. Can you please give me a hint?
According to Mathematica the result should be
$$\left[\frac{z}{(x^2+y^2)\sqrt{x^2+y^2+z^2}}\right]_{-\infty}^{+\infty}=\frac{2}{x^2+y^2}$$
Thank you very much for your effort.
| Let $z=\sqrt{x^2+y^2}\tan\theta$, then $dz=\sqrt{x^2+y^2}\sec^2\theta\ d\theta$.
$$
\begin{align}
\int_{-\infty}^{+\infty}\frac{dz}{\sqrt{(x^2+y^2+z^2)^3}}&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sqrt{x^2+y^2}\sec^2\theta\ d\theta}{\sqrt{((x^2+y^2)+(x^2+y^2)\tan^2\theta)^3}}\\
&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sqrt{x^2+y^2}\sec^2\theta}{\sqrt{(x^2+y^2)^3}\sqrt{(1+\tan^2\theta)^3}}d\theta\\
&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\sec^2\theta\ }{\sqrt{(x^2+y^2)^2}\sec^3\theta}d\theta\\
&=\int_{\Large-\frac\pi2}^{\Large\frac\pi2}\frac{\cos\theta\ }{x^2+y^2}d\theta\\
&=\left.\frac{1}{x^2+y^2}\sin\theta\right|_{\Large-\frac\pi2}^{\Large\frac\pi2}\\
&=\frac{2}{x^2+y^2}.
\end{align}
$$
Since $\tan\theta=\dfrac{z}{\sqrt{x^2+y^2}}$, then $\sin\theta=\dfrac{z}{\sqrt{x^2+y^2+z^2}}$. Hence
$$
\left.\frac{1}{x^2+y^2}\sin\theta\right|_{\Large-\frac\pi2}^{\Large\frac\pi2}=\left.\frac{1}{x^2+y^2}\cdot\dfrac{z}{\sqrt{x^2+y^2+z^2}}\right|_{-\infty}^{\infty}=\frac{2}{x^2+y^2}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve: $\tan2x=1$ Are there any errors in my work? Thanks in advance! (Sorry for the bad format. I'm still new to this)
$\tan2x=1$
$\frac{2\tan x}{1-\tan^2x}=1$
$2\tan x=1-\tan^2x$
$0=1-\tan^2x-2\tan x$
$0 =-\tan^2x-2\tan x +1$
$0=\tan^2x+2\tan x-1$
$\frac{-(2)\sqrt{2^2-4(1)(-1)}}{2(1)}$
$x=0.4142, x=2.4142$
$\tan^{-1}(0.4142)$
$x=22.5, x=202.5$
| Your solution is not very rigorous after $0 = tan^2x + 2tanx - 1$. After that line, the
equation should be $tanx = \frac{-2 \pm \sqrt{2^2 - 4(1)(-1)}}{2(1)}$.
Also, after that, it's $tanx = \sqrt{2} - 1 \approx 0.4142$ or $tanx = \sqrt{2} + 1 \approx 2.4142$.
Then, $x = tan^{-1}(0.4142) \approx 22.5, 202.5$
| {
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"timestamp": "2023-03-29T00:00:00",
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Deriving $\sum\limits_{i=1}^n \frac{1}{i^2} = \frac{π^2}{6}-ψ^{(1)}(n+1)$. How is the partial sums formula of the $$\sum\limits_{i=1}^n \frac{1}{i^2} = \frac{π^2}{6}-ψ^{(1)}(n+1)$$
derived?
| It can be shown with the Hurwitz Zeta function (see especially formulas (1) and (2))
$$\sum\limits_{i=1}^n \frac{1}{i^2}
= \sum\limits_{i=1}^{\infty} \frac{1}{i^2} - \sum\limits_{i=n+1}^{\infty} \frac{1}{i^2} \\
= \sum\limits_{i=1}^{\infty} \frac{1}{i^2} - \sum\limits_{i=0}^{\infty} \frac{1}{(i+n+1)^2} \\
= \zeta(2)-\zeta(2,n+1)\\
= \frac{\pi^2}{6}-ψ^{(1)}(n+1)$$
and the well known value $\zeta(2) = \frac{\pi^2}{6}$ for the Riemann Zeta function.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/784778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Trigonometry problem cosine identity Let $\cos^6\theta = a_6\cos6\theta+a_5\cos5\theta+a_4\cos4\theta+a_3\cos3\theta+a_2\cos2\theta+a_1\cos\theta+a_0$. Then $a_0$ is
(A) $0$ (B) $\frac{1}{32}$ (C) $\frac{15}{32}$ (D) $\frac{10}{32}$
Any hints on how to approach this?
| Using \begin{align}&2\cos^2 x=\cos 2x +1, \\&4\cos^3x=\cos 3x+3\cos x, \\ \text{and}&2\cos a \cos b= \cos(a+b)+\cos(a-b),\end{align}
we obtain,
\begin{align}&\cos^6 x=(\cos^3 x)^2\\
\\=&\left(\dfrac{\cos 3x+3\cos x}{4}\right)^2\\
\\=&\dfrac1{16}(\cos^2 3x + 9\cos^2 x+6\cos x\cos 3x)\\
\\=&\dfrac{1}{16}\left(\dfrac{\cos 6x+1}{2}+9\dfrac{\cos 2x +1}2+3\cos 4x +3\cos 2x\right).\end{align}
It is seen that the constant term is $\dfrac{5}{16}$ or $\dfrac{10}{32}$.
(This of course, makes use of the question's assumption that the number $a_0$ is unique.)
| {
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} |
Find minimum value of the expression x^2 +y^2 subject to conditions Find the values of $x,y$ for which $x^2 + y^2$ takes the minimum value where $(x+5)^2 +(y-12)^2 =14$.
Tried Cauchy-Schwarz and AM - GM , unable to do.
| Another way is triangle inequality (essentially think of the triangle between the origin, the centre of the circle and any point on the circle):
$$\sqrt{x^2+y^2} +\sqrt{(x+5)^2+(y-12)^2} \ge \sqrt{(-5)^2+(12)^2} \implies x^2+y^2 \ge \left(13 - \sqrt{14}\right)^2$$
| {
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Maximium value of $(b-a)\Big(\dfrac 34-\dfrac{a+b}2-\dfrac{a^2+ab+b^2}3\Big)$ For $b>a$ what is the maximum possible value of $(b-a)\Big(\dfrac 34-\dfrac{a+b}2-\dfrac{a^2+ab+b^2}3\Big)$ ?
| For a fixed $a$, the expression is
$$\frac34(b-a)-\frac12(b^2-a^2)-\frac13(b^3-a^3)$$
which is a third degree polynomial.
Differentianting respect to $b$ gives:
$$\frac 34-b-b^2$$ whose roots are $-3/2$ and $1/2$.
The maximum is reached when $b=1/2$. If $a>1/2$, there is no maximum.
| {
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the series $\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})}$ converges uniformly in $[-a,a]$ I have to show that the series $\displaystyle{\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}\sin(1+\frac{x}{n})}$ converges uniformly in $[-a,a], a>0$.
$$$$
That's what I have tried:
$w_n=\frac{(-1)^n}{\sqrt{n}} \sin1 \cos(\frac{x}{n})+\frac{(-1)^n}{\sqrt{n}} \cos1 \sin(\frac{x}{n})=\frac{(-1)^n}{\sqrt{n}} \sin1 (\cos(\frac{x}{n})+1-1)+\frac{(-1)^n}{\sqrt{n}} \cos1 \sin(\frac{x}{n})= \frac{(-1)^n}{\sqrt{n}} \sin1 (\cos(\frac{x}{n})-1)+\frac{(-1)^n}{\sqrt{n}} \sin1 +\frac{(-1)^n}{\sqrt{n}} \cos1 \sin(\frac{x}{n})=\sin1 \cdot g_n(x)+ \cos1 \cdot f_n(x)+\frac{(-1)^n}{\sqrt{n}} \sin1 $ .
We set $d_n(x)=w_n-\frac{(-1)^n}{\sqrt{n}} \sin1=\sin1 \cdot g_n(x)+ \cos1 \cdot f_n(x)$
$|g_n(x)|=|\frac{(-1)^n}{\sqrt{n}}(\cos(\frac{x}{n})-1)|=\frac{1}{\sqrt{n}} (\cos(\frac{x}{n})-1) (\cos(\frac{x}{n})+1)=\frac{1}{\sqrt{n}}(\sin^2(\frac{x}{n})) \leq \frac{1}{\sqrt{n}} \cdot \frac{x^2}{n^2} \leq \frac{1}{\sqrt{n}} \cdot \frac{a^2}{n^2}=\frac{a^2}{n^{\frac{5}{2}}}$
$\sum_{n=1}^{\infty} \frac{a^2}{n^{\frac{5}{2}}}< +\infty$.
$|f_n(x)|=|\frac{(-1)^n}{\sqrt{n}} \sin(\frac{x}{n})| \leq \frac{1}{\sqrt{n}} \frac{x}{n} \leq \frac{a}{n^{\frac{3}{2}}}$
$\sum_{n=1}^{\infty} \frac{a}{n^{\frac{3}{2}}}< +\infty$.
Also,from Dirichlet's test,we know that $\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$ converges,let $b:=\sum_{n=1}^{\infty} \frac{(-1)^n}{\sqrt{n}}$
Therefore,we have $d_n(x) \leq \sin1 \frac{a^2}{n^{\frac{5}{2}}}+ \cos1 \frac{a}{n^{\frac{3}{2}}}$.
$\sum_{n=1}^{\infty} \sin1 \frac{a^2}{n^{\frac{5}{2}}}+ \cos1 \frac{a}{n^{\frac{3}{2}}}$ converges,as $\sum_{n=1}^{\infty} \frac{a^2}{n^{\frac{5}{2}}}$ converges and $\sum_{n=1}^{\infty} \frac{a}{n^{\frac{3}{2}}}$ converges too.
So, $\sum_{n=1}^{\infty} d_n(x)$ converges uniformly to a $s(x)$.
$\sum_{n=1}^{\infty} d_n(x)=\sum_{n=1}^{\infty} (w_n-\frac{(-1)^n}{\sqrt{n}} \sin1)= \sum_{n=1}^{\infty} w_n-b \overset{uniformly}{=}s(x) \Rightarrow \sum_{n=1}^{\infty} w_n=b+s(x):=q(x):[-a,a] $
So,the series converges uniformly to $q(x)$.
Could you tell me if it is right?
| Hint You may use Dirichlet's criterion, since $\sin (xn^{-1})$ is eventually decreasing in $[0,M]$ and $1-\cos (xn^{-1})$ is eventually decreasing in $[0,M]$ too, and both go uniformly to zero, and $$\sum_{n\geqslant 1} (-1)^{n}n^{-1/2}$$ converges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solving an integral using Laplace transform and inverse Laplace transform I want to solve this integral equation using Laplace:
$$ Y(t) + 3{\int\limits_0^t Y(t)}\operatorname d\!t = 2cos(2t)$$
if $$ \mathcal{L}\{Y(t)\} = f(s)$$
then, $$ f(s) + 3 \frac{f(s)}{s} = \frac{2s}{s^2+4} $$
doing some operations I obtain
$$ f(s) = \frac{2s^2}{s^3+3s^2+4s+12} $$
and using Ruffini in the denominator
$$ f(s) = \frac{2s^2}{(s+3)(s^2+4)} $$
How can I calculate the inverse Laplace transform of the following function to get the solution?
| Building on my comment, try the following:
$$f(s)=\frac{2s^2}{(s+3)(s^2+4)} = \frac{As+B}{s^2+4} + \frac{C}{s+3}$$
So $As(s+3) + B(s+3) + C(s^2+4) = 2s^2$, whence\
\begin{align*}
A+C &= 2\\
3A + B &= 0\\
3B+4C &= 0
\end{align*}
We can shortcut the solution a bit, as follows:
Plug in $s = -3$, we get: $13C = 18$, whence $C = \frac{18}{13}$.
Plug in $s = 0$, we get $3B+4C = 0$, so $B = -\frac{24}{13}$.
Then using the first equation, we get $A = \frac{8}{13}$.
So we have
\begin{align*}
f(s) &= \frac{\frac{8}{13}s}{s^2+4} - \frac{\frac{24}{13}}{s^2+4} + \frac{\frac{18}{13}}{s+3}\\
&= \frac{1}{13}\left(8\frac{s}{s^2+4} - 12\frac{2}{s^2+4} + \frac{18}{s+3}\right)
\end{align*}
And at this point I can refer you to a table of Laplace transforms (which implicitly gives the termwise inverse Laplace transform).
$$f(t) = \frac{1}{13}\left(8\cos(2t) - 12\sin(2t) + 18e^{-3t}\right)$$
| {
"language": "en",
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Integral determination I am trying to figure out this integral:
$\int \frac{x}{(x^2+4)^6}dx$
Substitution:
$t = x^2+4$
$dt = 2xdx => dx=\frac{dt}{2x}$
Then:
$\int \frac{x}{(x^2+4)^6}dx = \int \frac{x}{t^6}\frac{dt}{2x} = \frac{1}{2} \frac{1}{t^6} + C = $ but how to continue now?
| $$\int \frac{xdx}{(x^2+4)^6}=\frac12 \int \frac{dt}{t^6}=\int t^{-6}dt=\frac{1}{2}\cdot \frac{t^{-5}}{-5}+c=\frac{-1}{10(x^2+4)^5}+c$$
| {
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How to finish this integration? I'm working with the integral below, but not sure how to finish it...
$$\int \frac{3x^3}{\sqrt[3]{x^4+1}}\,dx = \int \frac{3x^3}{\sqrt[3]{A}}\cdot \frac{dA}{4x^3} = \frac{3}{4} \int \frac{dA}{\sqrt[3]{A}} = \frac{3}{4}\cdot\quad???$$
where $A=x^4+1$ and so $dA=4x^3\,dx$
| Assuming I interpreted your very low quality picture correctly...
You got a good start on the problem. That's the right substitution. Rewrite...
$$\frac{3}{4} \int \frac{dA}{\sqrt[3]{A}} = \frac{3}{4} \int A^{-1/3}\,dA$$
...and use the backwards power rule...
$$ = \frac{3}{4} \frac{A^{2/3}}{2/3} +C = \cdots$$
| {
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Find norm of operator I have a linear functional $$A: L_2[0,2] \to \mathbb R, Ax = \int_0^2(t^2+2)x(t)dt$$
I need to find $C$, trying to measure $C$ and $||Ax||$ to find it, but how can I do it in this problem?
| Here I'll play the role of Robin to Batman, and emulate my mentor's work in a more elementary manner, as befits a protoge of the master!
First of all, note that $y(t) = t^2 +2 \in L^2[0, 2]$; since the inner product $\langle u, v \rangle$ on $L^2[0, 2]$ is given by
$\langle u, v \rangle = \int_0^2 u(t)v(t) dt, \tag{1}$
and thus
$\langle t^2 + 2, t^2 + 2 \rangle = \int_0^2 (t^2 + 2)^2 dt = \int_0^2 (t^4 + 4t^2 + 4)^2 dt$
$= (\dfrac{t^5}{5} + \dfrac{4t^3}{3} + 4t \vert_0^2 = \dfrac{32}{5} + \dfrac{32}{3} + 8 = \dfrac{96}{15} + \dfrac{160}{15} + \dfrac{120}{15} = \dfrac{376}{15}, \tag{2}$
so that
$\Vert t^2 + 2 \Vert^2 = \langle t^2 + 2, t^2 + 2 \rangle = \dfrac{376}{15}, \tag{3}$
or
$\Vert t^2 + 2 \Vert = \sqrt{\dfrac{376}{15}}; \tag{4}$
since $\Vert t^2 + 2 \Vert$ exists (and is finite!), we have indeed have that $t^2 + 2 \in L_2[0, 2]$; thus we may apply the Cauchy-Schwarz inequality to $t^2 + 2$ and $x(t)$ to conclude that
$\vert \langle t^2 + 2, x(t) \rangle \vert \le \Vert t^2 + 2 \Vert \Vert x(t) \Vert; \tag{5}$
since $\langle t^2 + 2, x(t) \rangle$ is in fact the functional $Ax$, we see that (5) immediately shows that $Ax$ is bounded by $\Vert t^2 + 2 \Vert$;
$\vert A(x) \vert \le \Vert t^2 + 2 \Vert \Vert x(t) \Vert = \sqrt{\dfrac{376}{15}} \Vert x(t) \Vert, \tag{6}$
i.e.,
$\Vert A \Vert \le \sqrt{\dfrac{376}{15}}. \tag{7}$
Now taking
$x(t) = t^2 + 2 \tag{8}$
(2)-(4) show that
$\vert A(t^2 + 2) \vert = \Vert t^2 + 2 \Vert^2 = \sqrt{\dfrac{376}{15}} \Vert t^2 + 2 \Vert; \tag{9}$
since $\vert Ax(t) \vert$ in fact takes the value $\sqrt{376/15}\Vert x(t) \Vert$ for $x(t) = t^2 + 2$, we have that indeed
$\Vert A \Vert = \sqrt{\dfrac{376}{15}}. \tag{10}$
Hope this helps! Cheerio,
and as always,
Fiat Lux!!!
| {
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"timestamp": "2023-03-29T00:00:00",
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Finding determinant for matrix using upper triangle method Here is an example of a matrix, and I'm trying to evaluate its determinant:
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
2 & 5 & -2 & 9 \\
3 & 7 & 0 & 1 \\
\end{pmatrix}
$$
When applying first row operation i get:
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & -1 & -6 & 7 \\
0 & -2 & -6 & -2 \\
\end{pmatrix}
$$
Now, if I continue doing row operations until i get the upper triangle, the determinant will be 14 (which is said to be the correct one).
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & 0 & 2 & 3 \\
0 & 0 & 0 & -7 \\
\end{pmatrix}
$$
However, if I instead apply this certain operation, R4 --> (1/-2)R4...
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & -1 & -6 & 7 \\
0 & 1 & 3 & 1 \\
\end{pmatrix}
$$
...and then carry on with operations, I get a different final answer: The determinant will be 7 in this case!
$$
\begin{pmatrix}
1 & 3 & 2 & 1 \\
0 & 1 & 4 & -4 \\
0 & 0 & -1 & 5 \\
0 & 0 & 0 & -7 \\
\end{pmatrix}
$$
Could someone explain that to me - is this operation illegal?
R4 --> (1/-2)R4
How so? Because i always tend to use it, just to simply things a little.
| Different row-operations affect the determinant of the matrix differently. Adding a multiple of one row to another will not change the determinant. However, multiplying a row by some factor will lead to the determinant being multiplied by the same factor.
So, since you multiplied $R_4$ by the factor $- \frac 12$, the resulting determinant will be $- \frac 12$ times what the determinant of the original matrix was.
| {
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Algebraic Solving Contest Problem The problem is as follows
If $x^2+x-1=0$, compute all possible values of $\frac{x^2}{x^4-1}$
This was a no-calculator 10 min for 2 problem format contest.
I started by using quadratic formula, but the answers I got were too ugly to be plugged into the equation before time ran out. I tried several algebraic manipulations, but after losing time on the other problem, i found no quick answer to this. I presume there is one step i'm missing to make that leap. Can this be done easily?
| Let $a, b$ be the roots of $x^2+x-1=0$. Then $a+b = -1, ab = -1$. For simplicity, consider now the reciprocals of the values we wish to find, i.e. $\dfrac{a^4-1}{a^2} = k_1$ and $\dfrac{b^4-1}{b^2} = k_2$, then
$$k_1 + k_2 = a^2+b^2-\left(\frac1{a^2}+\frac1{b^2} \right), \quad k_1 k_2 = \frac{(a^4-1)(b^4-1)}{(ab)^2}$$
Now $a^2+b^2 = (a+b)^2-2ab = 3$, so $k_1 + k_2 = 3-3=0$.
Also $(a^4-1)(b^4-1) = 1+(ab)^4-(a^4+b^4) = 1+1-(3^2-2) = -5$, so $k_1 k_2 = -5$.
So $k_1, k_2$ are roots of $y^2-5=0 \implies \{k_1, k_2\} = \{\sqrt5, -\sqrt5\}$. Take reciprocals now.
| {
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"timestamp": "2023-03-29T00:00:00",
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Extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange Multipliers
Find the extrema of $x+y+z$ subject to $x^2 - y^2 = 1$ and $2x + z = 1$ using Lagrange multipliers.
So I set it up:
$$
1 = 2x\lambda_1 + 2\lambda_2 \\
1 = -2y\lambda_1 \\
1 = \lambda_2
$$
Plug in for $\lambda_2$:
$$
1 = 2x\lambda_1 + 2 \\
1 = -2y\lambda_1 \\
$$
So we work with:
$$
1 = 2x\lambda_1 + 2 \\
1 = -2y\lambda_1 \\
1 = x^2 - y^2 \\
1 = 2x + z
$$
After some algebra I got $x = y$ as a solution but that's impossible because of the constraint $1 = x^2 - y^2$. What am I missing?
| Consider the constraints: you can simplify them in
$$ y=\pm\sqrt{y^2+1}, \qquad z=\mp 2\sqrt{y^2+1}+1$$
for all $y\in\mathbb{R}$.
In this way you can convert the constrained optimization in two unconstrained optimizations: substitute the two solutions in the objective function and you will find
$$- \sqrt{y^2+1}+y+1, \qquad \sqrt{y^2+1}+y+1.$$
These two real functions are monotone increasing so they don't possess any stationary point.
| {
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probability with replacement A box contains 4 red, 3 black and 2 white cubes. A cube is randomly drawn and has its color noted. The cube is then replaced, together with 2 more of the same color. A second cube is then drawn. Find the probability that
1. The second cube is black
2. the second cube is not white
| Divide into cases. With probability $\frac{4}{9}$ a red is drawn. If that happens, the probability of a black next is $\frac{3}{11}$.
With probability $\frac{3}{9}$ a black is drawn. If that happens, the probability of a black next is $\frac{5}{11}$.
With probability $\frac{2}{9}$ a white is drawn. If that happens, the probability of a black next is $\frac{3}{11}$.
Thus the probability that the second ball drawn is black is $\frac{4}{9}\cdot\frac{3}{11}+\frac{3}{9}\cdot\frac{5}{11}+\frac{2}{9}\cdot\frac{3}{11}$.
We could have saved some work on this part of the question by noting that the only thing that is relevant is whether the first ball drawn is black or not. If the first ball drawn is not black, the probability of a black next is $\frac{3}{11}$. If the first ball drawn is black, the probability of a ball next is $\frac{5}{11}$. That gives answer $\frac{6}{9}\cdot \frac{3}{11}+\frac{3}{9}\cdot\frac{5}{11}$.
The second problem is done in basically the same way, and is left to you.
| {
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Proof for $\sin(x) > x - \frac{x^3}{3!}$ They are asking me to prove $$\sin(x) > x - \frac{x^3}{3!},\; \text{for} \, x \, \in \, \mathbb{R}_{+}^{*}.$$ I didn't understand how to approach this kind of problem so here is how I tried:
$\sin(x) + x -\frac{x^3}{6} > 0 \\$
then I computed the derivative of that function to determine the critical points. So:
$\left(\sin(x) + x -\frac{x^3}{6}\right)' = \cos(x) -1 + \frac{x^2}{2} \\ $
The critical points:
$\cos(x) -1 + \frac{x^2}{2} = 0 \\ $
It seems that x = 0 is a critical point.
Since $\left(\cos(x) -1 + \frac{x^2}{2}\right)' = -\sin(x) + x \\ $
and $-\sin(0) + 0 = 0 \\$
The function has no local minima and maxima. Since the derivative of the function is positive, the function is strictly increasing so the lowest value is f(0).
Since f(0) = 0 and 0 > 0 I proved that $ \sin(x) + x -\frac{x^3}{6} > 0$. I'm not sure if this solution is right. And, in general, how do you tackle this kind of problems?
| Take a decreasing sequence of positive real numbers $a_n$ such that $a_n\to 0$.
Now, consider the sequence $b_k=\sum_{n=1}^k (-1)^{n-1}a_n$. The alternating series criterion guarantee us that it converges to some $b$.
Note that $b_1=a_1$, $b_2=b_1-a_2\in(0,b_1)$, $b_3=b_2+a_3\in(b_2,b_1)$, etc. So the limit $b$ is lesser that the terms $b_{2k+1}$ and greater than $b_{2k}$.
Then, if $x<\sqrt 6$,
$$\sin x=\sum_{n=1}^\infty(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}>\sum_{n=1}^2(-1)^{n-1} \frac{x^{2n-1}}{(2n-1)!}=x-\frac{x^3}{3!}$$
If $x\geq\sqrt 6$, the function $f(x)=x-x^3/6$ is decreasing and $f(\sqrt 6)=0$, so $f(x)<0$ for $x>\sqrt 6$. Since $\sin x>0$ for $0<x<\pi$, we have that $\sin x>f(x)$ for $0<x<\pi$. (Note that $\sqrt 6<\pi$).
Last, for $x\geq \pi$, $\sin x\geq -1$ and $f(x)<f(\pi)<f(3)=3-4.5<-1$.
| {
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Integral $I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}$ Hi how can we prove this integral below?
$$
I:=\int_0^1 \frac{\log^2 x}{x^2-x+1}\mathrm dx=\frac{10\pi^3}{81 \sqrt 3}
$$
I tried to use
$$
I=\int_0^1 \frac{\log^2x}{1-x(1-x)}\mathrm dx
$$
and now tried changing variables to $y=x(1-x)$ in order to write
$$
I\propto \int_0^1 \sum_{n=0}^\infty y^n
$$
however I do not know how to manipulate the $\log^2 x$ term when doing this procedure when doing this substitution. If we can do this the integral would be trivial from here.
Complex methods are okay also, if you want to use this method we have complex roots at $x=(-1)^{1/3}$. But what contour can we use suitable for the $\log^2x $ term?
Thanks
| Setting $x\mapsto\frac1x$, we obtain
$$
\int_0^1\frac{\ln^2x}{x^2-x+1}\ dx
=\frac12\int_0^\infty\frac{\ln^2x}{x^2-x+1}\ dx.\tag1
$$
Note that $1+x^3=(1+x)(x^2-x+1)$, hence
$$
\frac12\int_0^\infty\frac{\ln^2x}{x^2-x+1}\ dx=\frac12\int_0^\infty\left[\frac{\ln^2x}{1+x^3}+\frac{x\ln^2x}{1+x^3}\right]\ dx.\tag2
$$
Now consider beta function
$$
\int_0^\infty\dfrac{x^{\large \alpha-1}}{1+x^{\beta}}\ dx=\frac{\pi}{\beta\sin\left(\frac{\alpha\pi}{\beta}\right)}\quad;\quad\text{for}\ 0<\alpha<\beta.\tag3
$$
Differentiating $(3)$ with respect to $\alpha$ twice yields
$$
\int_0^\infty\dfrac{x^{\alpha-1}\ln^2x}{1+x^{\beta}}\ dx=\frac{\pi^3\left[\cos\left(\frac{2\alpha\pi}{\beta}\right)+3\right]}{2\beta^3\sin^3\left(\frac{\alpha\pi}{\beta}\right)}.\tag4
$$
Evaluating $(2)$ using $(4)$, we obtain
\begin{align}
\frac12\int_0^\infty\left[\frac{\ln^2x}{1+x^3}+\frac{x\ln^2x}{1+x^3}\right]\ dx&=\frac{\pi^3}{4\cdot3^3}\left[\frac{\cos\left(\frac{2\pi}{3}\right)+3}{\sin^3\left(\frac{\pi}{3}\right)}+\frac{\cos\left(\frac{4\pi}{3}\right)+3}{\sin^3\left(\frac{2\pi}{3}\right)}\right]\\
\int_0^1\frac{\ln^2x}{x^2-x+1}\ dx&=\large\color{blue}{\frac{10\pi^3}{81\sqrt3}}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/803389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 8,
"answer_id": 1
} |
Calculus limits please demonstrate question? I have been trouble doing some more difficult limits
I have these two examples from a teacher and think if i can just be shown the working and answer i should be fine from there out. Thanks for your help.
$(a)\; \lim\limits_{x\to-2}\dfrac{3x^2+8x+4}{2x^2+x-6} \\ \; \\ (b)\;\lim\limits_{x\to\infty}\dfrac{3x^4-11x^3+42x}{-15x^4+3x^2 +101}$
| $$\lim\limits_{x\to-2}\dfrac{3x^2+8x+4}{2x^2+x-6} = \lim_{x\to -2} \dfrac{(3x+2)(x+2)}{(2x -3)(x+2)} = \lim_{x\to -2}\frac{3x+2}{2x-3} = \frac{-4}{-7}= \frac 47 $$
For the second limit, divide numerator and denominator by $x^4$, and evaluate.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ How to prove the following
$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$
I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are not best ideas to approach the problem.
Any other ideas ?
| $\displaystyle J=\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$
Perform the change of variable $y=\sqrt{\dfrac{1-x}{1+x}}$,
$\displaystyle J=-4\int^1_0 \dfrac{\log(x)}{1-x^2}dx=-4\int_0^1 \left( \log x\times\sum_{n=0}^{+\infty}x^{2n}\right) dx=-4\sum_{n=0}^{+\infty}\left(\int_0^1 x^{2n}\log x dx\right)$
$\displaystyle J=4\sum_{n=0}^{+\infty} \dfrac{1}{(2n+1)^2}=4\left(\zeta(2)-\sum_{n=1}^{+\infty}\dfrac{1}{(2n)^2}\right)=4\times\left(1-\dfrac{1}{4}\right)\times\zeta(2)=4\times\dfrac{3}{4}\times\dfrac{\pi^2}{6}=\dfrac{\pi^2}{2}$
Note that:
0) For $s>1$, $\displaystyle \zeta(s)=\sum_{n=1}^{+\infty} \dfrac{1}{n^s}$
1) $\zeta(2)=\dfrac{\pi^2}{6}$
2) For $n\geq 0$, $\displaystyle \int_0^1 x^n\log x dx=\left[\dfrac{x^{n+1}}{n+1}\times\log x\right]_0^1-\int_0^1 \left(\dfrac{x^{n+1}}{n+1}\times\dfrac{1}{x}\right)dx=-\left[\dfrac{x^n}{n+1}\right]_0^1=-\dfrac{1}{(n+1)^2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 10,
"answer_id": 4
} |
Line of intersection of planes How to find the line of intersection of the planes
$x·\begin{pmatrix}
1\\
2\\
3\\
\end{pmatrix}=0
$
and $x=\lambda_1\begin{pmatrix}
2\\
1\\
2\\
\end{pmatrix}+\lambda_2 \begin{pmatrix}
1\\
0\\
-1\\
\end{pmatrix}$. All I can figure is x is orthagonal to $\begin{pmatrix}
1\\
2\\
3\\
\end{pmatrix}$ and $x_1+2x_2+3x_3=0$, and then I'm stuck on solving those two equations to find a line of intersection.
| So each vector in the second plane looks like $\begin{bmatrix}2\lambda_1+\lambda_2\\ \lambda_1\\ 2\lambda_1-\lambda_2\end{bmatrix}$, and to be in the first plane this would have to be perpendicular to $[1,2,3]^\top$. This yields $2\lambda_1+\lambda_2+2\lambda_1+6\lambda_1-3\lambda_2=0$, hence $10\lambda_1-2\lambda_2=0$ and $\lambda_2=5\lambda_1$.
This describes the line $\lambda_1\begin{bmatrix}7\\ 1\\ -3\end{bmatrix}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Limit of $\frac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}$ as $x \rightarrow 0$ Find $\lim_{x \to 0} \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}$
I came across this limit a long time ago and could easily obtain a straightforward solution by finding the asymptotic expansion. But since the limit turns out to be nice despite the messy coefficients, I'm just curious if there is some reason other than just coincidental coefficients.
| I suppose and think that this is really related to the coincidental coefficients. We can show it building for each piece a Taylor expansion built at $x=0$. As results, we have
$$\tan ^{-1}\left(\sin ^{-1}(x)\right)=x-\frac{x^3}{6}+\frac{13 x^5}{120}-\frac{173 x^7}{5040}+O\left(x^{9}\right)$$
$$\sin ^{-1}\left(\tan ^{-1}(x)\right)=x-\frac{x^3}{6}+\frac{13 x^5}{120}-\frac{341 x^7}{5040}+O\left(x^{9}\right)$$
$$\tan (\sin (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{107 x^7}{5040}+O\left(x^{9}\right)$$
$$\sin (\tan (x))=x+\frac{x^3}{6}-\frac{x^5}{40}-\frac{55 x^7}{1008}+O\left(x^{9}\right)$$ As a result $$ \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))} \approx 1$$ In my opinion, it is quite important to notice for the terms appearing in numerator and denominator the identity of the first three terms (this is why a fourth term had to be introduced in order to avoid a $\frac{0}{0}$ situation).
Pushing the expansion to much higher orders would give
$$ \dfrac{\tan^{-1}(\sin^{-1}(x))-\sin^{-1}(\tan^{-1}(x))}{\tan(\sin(x))-\sin(\tan(x))}=1-\frac{5 x^2}{3}+\frac{3937 x^4}{1890}-\frac{24779 x^6}{11907}+O\left(x^8\right)$$ If we modify the problem to $$ \dfrac{\tan^{-1}(\sin^{-1}(a x))-\sin^{-1}(\tan^{-1}(a x))}{\tan(\sin(b x))-\sin(\tan(b x))}$$ the result of the expansion would be $$\frac{a^7}{b^7}-\frac{5 x^2 \left(13 a^9+29 a^7 b^2\right)}{126
b^7}+O\left(x^4\right)$$ For a completely general formulation such as $$ \dfrac{\tan^{-1}(\sin^{-1}(a x))-\sin^{-1}(\tan^{-1}(b x))}{\tan(\sin(c x))-\sin(\tan(d x))}$$ ($a,b,c,d$ being different) the result of the expansion would be $$\frac{a-b}{c-d}-\frac{x^2 \left((a-b) \left(a^2+a b+b^2+c^2+c d+d^2\right)\right)}{6
(c-d)}+O\left(x^4\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/809632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How come the function and the inverse of the function are the same? What is the inverse of the function:
$$f(x)=\frac{x+2}{5x-1}$$
?
Answer:
$$f^{-1}(x)=\frac{x+2}{5x-1}$$
Can one of you explain how the inverse is the same exact thing as the original equation?
| Let $f(x)=y$. So we have $y=\frac{x+2}{5x-1}$. Now "swap" the variables so we have
$$x = \frac{y'+2}{5y'-1}$$
where $y' = f^{-1}(x)$ (to distinguish $y'$ from $y$). Now let's solve for $y'$.
\begin{align}
x &= \frac{y'+2}{5y'-1} \\
x(5y'-1) &= y'+2 & \text{multiply both sides by } 5y'-1 \\
5xy' - x&=y'+2 & \text{distribute $x$ on the left side} \\
5xy'-y'-x&=2 & \text{move all $y'$ to left side} \\
5xy'-y'&=x+2 & \text{and add $x$ to right side} \\
(5x-1)y'&=x+2 & \text{factor out $y'$ on left side} \\
y'&=\frac{x+2}{5x-1} & \text{divide both sides by $5x-1$}
\end{align}
Thus, $f^{-1}(x)=y'=\frac{x+2}{5x-1}=y=f(x)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/810394",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How can I find the integral? I want to find the integral $$\int_{0}^{1}{\sqrt[3]{2x^3-3x^2-x+1}}\,\mathrm{d}x.$$
I tried, $$\int_{0}^{1}{\sqrt[3]{(-1 + 2 x) (-1 - x + x^2)}}\,\mathrm{d}x.$$
Put $t =-1 - x + x^2$, then $\mathrm{d} t = (2x + 1)\mathrm{d}x.$ And now, I can not find the integral.
| Note that $2x^3-3x^2-x+1=2(x-\frac{1}{2})^3-\frac{5}{2}(x-\frac{1}{2})$. Hence the integrand is symmetric about $x=\frac{1}{2}$, and since you are integrating with limits also symmetric about $\frac{1}{2}$, the value of the integral is just $0$.
$$\int_{0}^{1}{\sqrt[3]{2x^3-3x^2-x+1}}\,\mathrm{d}x=0$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find sum of $\frac{1}{\sin\theta\cdot \sin2\theta} + \frac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \sin (n+1)\theta}$ $$\sum_{k=1}^n \frac{1}{\sin k\theta \sin (k+1)\theta} = \dfrac{1}{\sin\theta\cdot \sin2\theta} + \dfrac{1}{\sin2\theta\cdot \sin3\theta} + \cdots + \frac{1}{\sin n \theta \sin (n+1)\theta}$$ up to $n$ terms.
I tried but in vain
| Here is a hint: the sum telescopes, with some careful manipulation:
$$\sin \theta = \sin((k+1)\theta - k\theta) = \sin(k+1)\theta \cos k\theta - \sin k\theta \cos (k+1)\theta.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/813413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Help with limit equation The question is:
$$\lim_{x \to + \infty} \frac{3x^4 - 11x^3 + 42x}{-15x^4 + 3x^2 +101}$$
So far my answer:
$$\lim_{x \to + \infty} \frac{x^4}{x^4} \frac{3 - \frac{11x^3}{x^4}+\frac{42x}{x^4}}{-15 + \frac{3x^2}{x^4}+ \frac{101}{x^4}}$$
cancel the $x^4$ gives us:
$$\lim_{x \to + \infty} \frac{3 - \frac{11x^3}{x^4}+\frac{42x}{x^4}}{-15 + \frac{3x^2}{x^4}+ \frac{101}{x^4}}$$
Now as x becomes larger (as $x \to +\infty$) $\frac{1}{x^4}$ goes to $0$
this gives:
$$\lim_{x \to + \infty} \frac{3 - 11\cdot 0 + 42 \cdot 0}{-15 + 3\cdot 0 + 101 \cdot 0}$$
$$=- \frac{3}{15}$$
I have been following my text book so this is how I got to where I'm at I was just wondering if I have gone about this the right way?
| Yes. It is right. When you have ratios of polynomials, however, there is an easy way to analyze $$\lim_{x \to +\infty} \frac{p(x)}{q(x)}$$
It is $+\infty$ or $-\infty$ if the degree of $p(x)$ is greater than the degree of $q(x)$, and the sign depends on the sign of the ratio of $p(x)$ and $q(x)$'s dominant coefficients. It is zero is the degree of $p(x)$ is less than the degree of $q(x)$. If both degrees are equal, the limit is the ratio of the dominant coefficients. Your exercise is an example of the last case.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/816607",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from,
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
| Note that the denominator of $\frac{(x^2 + x-6)}{x-2}$ and
$\frac{(x+3)(x-2)}{x-2}$ is the same, so it remains to show that the numerators are the same; i.e. that $x^2+x-6\equiv(x+3)(x-2)$.
Now, $\color{green}{(x+3)(x-2)} \equiv x^2\underbrace{-2x+3x}_{\equiv \ +x}-6 \equiv \color{green}{x^2+x-6},$ as required.
To expand brackets, use the distributive property (i.e. the fact that multiplication is distributive over addition).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/817424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue)
$$
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$ $$
1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$
I ask
$$
f\left( z \right) = - \frac{2}{{\left( {4z + 9} \right)\left( {4z + 7}\right)}}
$$
is to :
$$\sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)\left( {4n + 7}\right)}}} = \left( {\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] + \mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}}\right] } \right)$$
I found:
\begin{array}{l}
\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{9}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{9}{4}} \right)\left( {4z + 7} \right)}}} \right] \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{4z + 7}}} \right] = \frac{1}{4}\left[ {\frac{{ - \pi }}{{ - 2}}} \right] = \frac{\pi }{8} \\
\mathop {\lim }\limits_{z \to - \frac{7}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)\left( {4z + 7} \right)}}} \right] = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\left( {z + \frac{7}{4}} \right)\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{7}{4}} \right)\left( {4z + 9} \right)}}} \right] \\
\quad \quad \quad \quad \quad \quad \quad \quad \quad = \frac{1}{4}\mathop {\lim }\limits_{z \to - \frac{9}{4}} \left[ {\frac{{\pi \cot \left( {\pi z} \right)}}{{\left( {4z + 9} \right)}}} \right] = \frac{\pi }{8} \\
\end{array}
\begin{array}{l}
\sum\limits_{n = - \infty }^\infty {\frac{2}{{\left( {4n + 9} \right)4n + 7}}} = - \frac{\pi }{4} = - \left( {\frac{\pi }{8} + \frac{\pi }{8}} \right) \\
\Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = 1 - \frac{\pi }{8} = \frac{{7 - \pi }}{8} \\
\end{array}
I have a question for the result
$$\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = - \frac{1}{5} \Rightarrow s = 1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} = \frac{4}{5} \ne \frac{{7 - \pi }}{8}$$
thank you in advance
| $1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - \frac{1}{{23}} + \frac{1}{{25}} - .... = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 1}}} - \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 7}}} $
\begin{array}{l}
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - \frac{1}{{23}} + \frac{1}{{25}} - .... = \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 1}}} - \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 7}}} \\
f\left( z \right) = \frac{1}{{8z + 1}} - \frac{1}{{8z + 7}} = \frac{6}{{\left( {8n + 1} \right)\left( {8n + 7} \right)}} \\
\frac{1}{8}\mathop {\lim }\limits_{z \to - \frac{1}{8}} \left( {z + \frac{1}{8}} \right)\frac{{6\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{1}{8}} \right)\left( {8z + 7} \right)}} = \frac{\pi }{8}\cot \left( { - \frac{\pi }{8}} \right) = - \frac{{\pi \left( {1 + \sqrt 2 } \right)}}{8} \\
\frac{1}{8}\mathop {\lim }\limits_{z \to - \frac{1}{8}} \left( {z + \frac{7}{8}} \right)\frac{{6\pi \cot \left( {\pi z} \right)}}{{\left( {z + \frac{7}{8}} \right)\left( {8z + 1} \right)}} = - \frac{\pi }{8}\cot \left( { - \frac{{7\pi }}{8}} \right) = - \frac{{\pi \left( {1 + \sqrt 2 } \right)}}{8} \\
\sum\limits_{ - \infty }^{ + \infty } {\frac{6}{{\left( {8n + 1} \right)\left( {8n + 7} \right)}}} = - \left( {Residu\left( {\pi f\left( z \right)\cot \left( {\pi z} \right); - \frac{1}{8}} \right) + Re sidu\left( {\pi f\left( z \right)\cot \left( {\pi z} \right); - \frac{7}{8}} \right)} \right) \\
\Rightarrow \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 1}}} - \sum\limits_{n = 0}^{ + \infty } {\frac{1}{{8n + 7}}} = \frac{{\pi \left( {1 + \sqrt 2 } \right)}}{8} \\
\end{array}
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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limits of function without using L'Hopital's Rule $\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x+ 1 - x}} = 1$ Good morning.
I want to show that without L'Hopital's rule :
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - \ln x}}{{x\ln x + 1 - x}} = 1$
I did the steps
$
\begin{array}{l}
\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 + \ln \left( x \right)}}{{x\ln \left( x \right) - x + 1}} = \mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} \\
\ln \left( {y + 1} \right) = 1 - \frac{{y^2 }}{2} + o\left( {y^2 } \right);and\quad \mathop {\lim }\limits_{y \to 0} o\left( {y^2 } \right) = 0 \\
\Rightarrow \left( {y + 1} \right)\ln \left( {y + 1} \right) = 1 + y - \frac{{y^2 }}{2} + o\left( {y^2 } \right) \\
\mathop {\lim }\limits_{y \to 0} \frac{{y + \ln \left( {y + 1} \right)}}{{\left( {y + 1} \right)\ln \left( {y + 1} \right) - y}} = \frac{{1 + y - \frac{{y^2 }}{2}}}{{1 + y - \frac{{y^2 }}{2}}} = 1 \\
\end{array}
$
help me what you please
| by use of Taylor series when ${x \to 1}$ you'll get $\ln x=(x-1)-\frac{{(x - 1)^2}}{2x^2}$
$\mathop {\lim }\limits_{x \to 1} \frac{{x - 1 - (x - 1)+\frac{{(x - 1)^2}}{2x^2}}}{{{x(x - 1)}-\frac{{(x - 1)^2}}{2x} + 1 - x}}$
after factoring $(x-1)$ you'll have
$\mathop {\lim }\limits_{x \to 1} \frac{{\frac{{(x - 1)}}{2x^2}}}{x-\frac{{(x - 1)}}{2x}-1}$
factor another $(x-1)$
$\mathop {\lim }\limits_{x \to 1} \frac{{\frac{1}{2x^2}}}{1-\frac{1}{2x}}=1$
here's the graph to ensure.
https://www.google.com/webhp?sourceid=chrome-instant&ion=1&espv=2&ie=UTF-8#q=y%3D(x-1-ln(x))%2F(x*ln(x)%2B1-x)
| {
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"url": "https://math.stackexchange.com/questions/818908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Factoring in the derivative of a rational function Given that
$$
f(x) = \frac{x}{1+x^2}
$$
I have to find
$$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$
\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} =
\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} =
\frac{x+xa^2-a-ax^2}{(1+x^2)(1+a^2)(x-a)}
$$
Now, is it possible to factor $x+xa^2-a-ax^2$? I can't seem to find a way, as for simplifying the whole thing. Is there any rule I can use, and I'm unable to see?
| Factor:
$$x+xa^2−a−ax^2=x-a-ax^2+xa^2=(x-a)-ax(x-a)=(1-ax)(x-a)$$
Thus, $$\frac{f(x)-f(a)}{x-a}=\frac{\frac{(1-ax)(x-a)}{(1+x^2)(1+a^2)}}{(x-a)}=\boxed{\frac{1-ax}{(1+x^2)(1+a^2)}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
} |
This double integral $$ \int_0^1\int_0^1x^3y^2\sqrt{1+x^2+y^2}\hspace{1mm}dxdy$$
We have to compute this up to 4 decimal places
| If you just need an approximation of this integral, you could use numerical methods. Matlab or Wolfram's Double Integral Calculator, for example, can do it just fine :
$$ \int_{0}^{1} \int_{0}^{1} x^{3}y^{2}\sqrt{1+x^{2}+y^{2}} \, \mathrm{d}x \mathrm{d}y \simeq 0.125065 $$
But if you want an exact computation, this might be harder.
To begin with, I wouldn't use polar coordinates here because you integrate on the unit square $[0,1] \times [0,1]$ and the parametrization of this unit square in polar coordinates is not straightforward.
Let
$$ I = \int_{0}^{1} \int_{0}^{1} x^{3}y^{2} \sqrt{1+x^{2}+y^{2}} \, \mathrm{d} x \mathrm{d}y.$$
Using Fubini's theorem, you can write :
$$I = \int_{0}^{1} y^{2} \Bigg( \int_{0}^{1} x^{3} \sqrt{1+y^{2}+x^{2}} \, \mathrm{d}x \Bigg) \, \mathrm{d}y. \tag{1}$$
Now, we have to compute an integral of this form :
$$ J=\int_{0}^{1} x^{3} \sqrt{a+x^{2}} \, \mathrm{d}x $$
with $a \in \mathbb{R}^{+}$. This can be done using two changes of variable :
$$
\begin{align*}
J &= {} \int_{0}^{1} x^{3} \sqrt{a+x^{2}} \, \mathrm{d}x \\[2mm]
& \mathop{=} \limits_{u=x^2} \frac{1}{2} \int_{0}^{1} u \sqrt{a+u} \, \mathrm{d}u \\[2mm]
& = \frac{1}{2} \int_{a}^{a+1} (v-a)\sqrt{v} \, \mathrm{d}v \\[2mm]
& = \frac{1}{5} (a+1)^{\frac{5}{2}} + \frac{2}{15}a^{\frac{5}{2}}-\frac{a}{3}(a+1)^{\frac{3}{2}} \\
\end{align*}
$$
As a consequence, $(1)$ becomes :
$$ I = \int_{0}^{1} y^{2} \Big[ \frac{1}{5}(2+y^{2})^{\frac{5}{2}} + \frac{2}{15}(1+y^{2})^{\frac{5}{2}} - \frac{1}{3}(1+y^{2})(2+y^{2})^{\frac{3}{2}} \Big] \, \mathrm{d}y. $$
At this point, we have to compute the three following integrals :
$$I_{1} = \int y^{2}(1+y^{2})^{\frac{5}{2}} \, \mathrm{d}y, \quad I_{2} = \int y^{2}(1+y^{2})^{\frac{3}{2}} \, \mathrm{d}y \quad \mathrm{and} \quad I_{3} =\int (1+y^{2})^{\frac{3}{2}} \, \mathrm{d}y.$$
One method to compute these integrals (it might not be the easiest way) would be to consider the change of variable $y = \sinh(t)$. Even though, the computation is not completely straightforward. If you want, I can give more explanations about the computation of these integrals. Wolfram gives the following results :
$$
\begin{align*}
I_{1} & = {} \frac{1}{8} \Big( 19\sqrt{3} - 5 \sinh^{-1}\big(\frac{1}{\sqrt{2}} \big) \Big) \\
I_{2} & = \frac{1}{384} \big( 317\sqrt{2} - 15 \sinh^{-1}(1) \big) \\
I_{3} & = \frac{1}{8} \Big( 11\sqrt{3} - \sinh^{-1}\big( \frac{1}{\sqrt{2}} \big) \Big)
\end{align*}
$$
In the end :
$$I = \frac{317\sqrt{2} - 15 \sinh^{-1}(1)}{2880} + \frac{1}{40}\Bigg(19\sqrt{3}-5\sinh^{-1} \Big( \frac{1}{\sqrt{2}} \Big) \Bigg)+\frac{1}{24}\Bigg( \sinh^{-1}\Bigg( \frac{1}{\sqrt{2}} \Big) - 11\sqrt{3} \Bigg).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How prove this Rāmā ujan Aiya kār identity $\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}=3$ show that
$$\sqrt{1+2\sqrt{1+3\sqrt{1+4\sqrt{1+5\sqrt{1+\cdots}}}}}=3$$
I know
$$3=\sqrt{1+8}=\sqrt{1+2\sqrt{16}}=\sqrt{1+2\sqrt{1+15}}=\sqrt{1+2\sqrt{1+3\sqrt{1+4\cdot6}}}=\cdots$$
| For your question answer is given in this
http://www.isibang.ac.in/~sury/ramanujanday.pdf
| {
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$\sum\limits_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}$ isn't divisible by 5 I have no idea
Prove that for any $n$ natural number this sum $$\sum\limits_{k=0}^{n}\binom{2n+1}{2k+1}2^{3k}$$ isn't divisible by $5$.
$\begin{array}{l}
\left( {1 + x} \right)^{2n + 1} - \left( {1 - x} \right)^{2n + 1} = \sum\limits_{k = 0}^{2n + 1} {\left( {\begin{array}{*{20}c}
{2n + 1} \\
k \\
\end{array}} \right)} x^k - \sum\limits_{k = 0}^{2n + 1} {\left( {\begin{array}{*{20}c}
{2n + 1} \\
k \\
\end{array}} \right)} \left( { - x} \right)^k = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c}
{2n + 1} \\
{2k + 1} \\
\end{array}} \right)} x^{1 + k} \\
x = 2 \Rightarrow 3^{2n + 1} + 1 = \sum\limits_{k = 0}^{2n + 1} {\left( {\begin{array}{*{20}c}
{2n + 1} \\
k \\
\end{array}} \right)} 2^k - \sum\limits_{k = 0}^{2n + 1} {\left( {\begin{array}{*{20}c}
{2n + 1} \\
k \\
\end{array}} \right)} \left( { - 2} \right)^k \\
3^{2n + 1} + 1 = \sum\limits_{k = 0}^n {\left( {\begin{array}{*{20}c}
{2n + 1} \\
{2k + 1} \\
\end{array}} \right)} 2^{k + 1} \\
\end{array}$
$
\begin{array}{l}
\left\{ \begin{array}{l}
3^0 \equiv 1\left[ 5 \right] \\
3^1 \equiv 3\left[ 5 \right] \\
3^2 \equiv 4\left[ 5 \right] \\
3^3 \equiv 2\left[ 5 \right] \\
3^4 \equiv 1\left[ 5 \right] \\
\end{array} \right. \Rightarrow \forall m \in\mathbb N:\left\{ \begin{array}{l}
3^{4m} \equiv 1\left[ 5 \right] \\
3^{4m + 1} \equiv 3\left[ 5 \right] \\
3^{4m + 2} \equiv 4\left[ 5 \right] \\
3^{4m + 3} \equiv 2\left[ 5 \right] \\
\end{array} \right. \\
3^{2\left( {2m} \right) + 1} + 1 \equiv 2\left[ 5 \right]and3^{2\left( {2m + 1} \right) + 1} + 1 \equiv 3\left[ 5 \right] \\
\end{array}
$
thank you in advance
| We will use $\mathbb{Z}_5[\sqrt2]$
$$
\begin{align}
a_n
&=\sum_{k=0}^n\binom{2n+1}{2k+1}2^{3k}\\
&=\frac1{4\sqrt2}\left(\sum_{k=0}^{2n+1}\binom{2n+1}{k}\sqrt8^{\,k}-\sum_{k=0}^{2n+1}\binom{2n+1}{k}(-\sqrt8)^k\right)\\[3pt]
&=\frac{\left(1+2\sqrt2\right)^{2n+1}-\left(1-2\sqrt2\right)^{2n+1}}{4\sqrt2}\\[9pt]
&\equiv3(1+\sqrt2)(-1+\sqrt2)^n+3(1-\sqrt2)(-1-\sqrt2)^n\pmod5\tag{1}
\end{align}
$$
$-1+\sqrt2$ and $-1-\sqrt2$ are roots of $x^2-3x-1\equiv0\pmod5$, so we have
$$
a_n\equiv3a_{n-1}+a_{n-2}\pmod5\tag{2}
$$
where $a_0=1$ and $a_1=1$. Therefore, $a_n$ mod $5$ is
$$
1,1,4,3,3,2,4,4,1,2,2,3,1,1,\dots\tag{3}
$$
which repeats, since $(2)$ is a second order linear recurrence. Therefore, none of the $a_n$ are $0$ mod $5$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Proving that $\sqrt[3] {2} ,\sqrt[3] {4},1$ are linearly independent over rationals I was trying to prove that $\sqrt[3] {2} ,\sqrt[3] {4}$ and $1$ are linearly independent using elementary knowledge of rational numbers. I also saw this which was in a way close to the question I was thinking about. But I could not come up with any proof using simple arguments. So if someone could give a simple proof, it would be great.
My try:
$a \sqrt[3] {2}+b\sqrt[3] {4}+c=0$ Then taking $c$ to the other side cubing on both sides we get $2a^3+4b^3+6ab(a+b)=-c^3$. I could not proceed further from here.
Apart from the above question i was also wondering how one would prove that $\sqrt{2},\sqrt{3},\sqrt{5},\sqrt{7},\sqrt{11},\sqrt{13}$ are linearly independent. Here assuming $a\sqrt{2}+b\sqrt{3}+c\sqrt{5}+...=0$ and solving seems to get complicated. So how does one solve problems of this type?
| First show that $1$ and $\sqrt[3]2$ are linearly independent. (This should be relatively easy.)
Then in order for $1$, $\sqrt[3]2$ and $\sqrt[3]4$ to be linearly dependent we must have
$$\sqrt[3]{4}=a+b\sqrt[3]2$$
for some $a,b\in\mathbb Q$.
(Since $\sqrt[3]{4}$ is a linear combination of $1$ and $\sqrt[3]2$.)
If we multiply the above equation by $\sqrt[3]2$, we get
\begin{gather*}
8=a\sqrt[3]2+b\sqrt[3]4=a\sqrt[3]2+b(a+b\sqrt[3]2)=(a+b^2)\sqrt[3]2+ab\\
8-ab=(a+b^2)\sqrt[3]2
\end{gather*}
Since $1$ and $\sqrt[3]2$ are linearly independent, we get $2-ab=a+b^2=0$.
So we get
\begin{align*}
ab&=8\\
b^2&=-a
\end{align*}
which yields $b^2=-\frac2b$ and $b^3=-8$. The only rational solution is $b=-2$. This would mean that $a=-b^2=-4$. So in the equation $\sqrt[3]{4}=a+b\sqrt[3]2$ we get that the LHS is positive and the RHS is negative, a contradiction.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Urn with balls, distribution of random variable From an urn containing $6$ balls numerated $1,\ldots,6$ we randomly choose one, then again and stop only when we picked the ball with number $1$ on it. Let $X$ be the greatest number that appeared on balls we already pulled out. What's the distribution of $X\ $? And $\mathbb{E}X=\ ?$
| Edit: Later on, we describe a fast solution. But for reasons of nostalgia, we keep our first slow way.
The slow way: We assume that the balls are removed one at a time and not replaced. Then the probabilities can be found with a careful examination of cases.
With probability $\frac{1}{6}$ we have $X=1$.
We have $X=2$ precisely if we got the sequence $2,1$. This has probability $\frac{1}{6\cdot 5}$.
We have $X=3$ in several ways: the sequence $3,1$, the sequence $2,3,1$ and the sequence $3,2,1$. The combined probability is $\frac{1}{6\cdot 5}+\frac{2}{6\cdot 5\cdot 4}$.
For the number of ways in which $X=4$, note that this happens with the sequences
$4,1$, or $4,2,1$, or $2,4,1$, or $3,4,1$ or $4,3,1$, or
$x,y,z,1$, where $x,y,z$ is one of the $6$ permutations of $2,3,4$. The probability is $\frac{1}{6\cdot 5}+\frac{4}{6\cdot 5\cdot 4}+ \frac{6}{6\cdot 5\cdot 4\cdot 3}$.
And so on. Actually, we are almost finished, since once we find $\Pr(X=5)$, we know $\Pr(X=8)$, since the probabilities add to $1$. It may still be a good idea to do an independent computation of $\Pr(X=6)$, as a check.
Once we have the probability distribution of $X$, finding $E(X)$ is mechanical.
A much faster way: The computations can be streamlined. Let us find $\Pr(X\le 5)$. This is the probability that $6$ comes after $1$. By symmetry this is $\frac{1}{2}$. So $\Pr(X=6)=\Pr(X\le 6)-\Pr(X\le 5)=\frac{1}{2}$.
To find $\Pr(X=5)$, we calculate $\Pr(X\le 5)-\Pr(X\le 4)$. The probability that $X$ is $\le 4$ is the probability $1$ comes before $5$ and $6$, which is $\frac{1}{3}$. So $\Pr(X=5)=\frac{1}{6}$.
To find $\Pr(X= 4)$, we find $\Pr(X\le 4)-\Pr(X\le 3)$. The probability that $X$ is $\le 3$ is the probability that $1$ comes before $4$, $5$, and $6$. This is $\frac{1}{4}$, so $\Pr(X=4)=\frac{1}{12}$.
Continue, it's almost over.
The idea works smoothly for balls numbered $1$ to $n$.
Remarks: $1.$ The "fast" method can be adapted to sampling with replacement.
We calculate, as a sample, the probability that $X$ is $\le 4$. This can happen in various ways: First is $6$, first is between $2$ and $4$ and the second is a $1$, first $2$ are between $2$ and $4$ and the third is a $1$, and so on. The probability is given by
$$\Pr(X\le 4)=\frac{1}{6}\left(1+\frac{3}{6}+\frac{3^2}{6^2}+\cdots \right).$$
Thus $\Pr(X\le 4)=\frac{1}{3}$. Similarly, we can calculate $\Pr(X\le 3)$, and then by subtraction $\Pr(X=4)$. If we compare with the numbers obtained in the no replacement case, we will notice something interesting.
$2.$ There is a fancier, but ultimately faster way of finding $E(X)$, the method of indicator random variables. This can be done without finding the distribution of $X$. But since we were asked to find that distribution, we might as well use it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/829439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Three Circles Meeting at One Point We have three triples of points on the plane, that is, $X=\{x_1, x_2, x_3\}$, $Y=\{y_1, y_2, y_3 \}$, and $Z=\{z_1, z_2, z_3\}$, where $x_i, y_i, z_i$ are points on the plane. I was wondering if there is a simple (or not-so-simple) algebraic relation satisfied by the coordinates of the 9 points if we know that three circles determined by $X, Y, Z$ pass through a single point.
| Let $\triangle ABC$ have edge-lengths $a=|BC|$, $b=|CA|$, $c=|AB|$. Suppose circles of radius $u$, $v$, $w$ about respective points $A$, $B$, $C$ meet at a common point $P$.
Coordinatizing, we can write $A=(0,0)$, $B=(c,0)$, $C=(b\cos A, b\sin A)$, $P=(p,q)$, so that
$$\begin{align}
u^2 &= p^2 + q^2 &(1)\\
v^2 &= (p-c)^2 + q^2 = p^2 + q^2 - 2 p c + c^2 &(2)\\
w^2 &= (p-b \cos A)^2 + (q-b\sin A)^2= p^2 + q^2-2b p \cos A - 2 b q \sin A + b^2 &(3)
\end{align}$$
Use $(1)$ to eliminate $p^2+q^2$ from $(2)$ and $(3)$:
$$\begin{align}
u^2 - v^2 + c^2 &= 2 p c &(4)\\
u^2 - w^2 + b^2 &= 2 b p \cos A + 2 b q \sin A &(5)
\end{align}$$
Then, solve the system $(4)$, $(5)$ for $p$ and $q$. Substituting this solution into $(1)$, replacing $\sin^2 A$ with $1-\cos^2 A$, and then writing $\cos A = \frac{1}{2bc}(-a^2+b^2+c^2)$, gives this polynomial condition for the concurrence of the three circles:
$$\begin{align}
0 &= a^2 b^2 c^2 + a^2 u^4 + b^2 v^4 + c^2 w^4 \\
&\quad- \left(\;a^2 u^2 + v^2 w^2\;\right)\left(-a^2 + b^2 + c^2\;\right) \\
&\quad- \left(\;b^2 v^2 + w^2 u^2\;\right)\left(\phantom{-}a^2 - b^2 + c^2\;\right) \\
&\quad- \left(\;c^2 w^2 + u^2 v^2\;\right)\left(\phantom{-}a^2 + b^2 - c^2\;\right)
&(\star)\end{align}$$
To answer your question, you "only" need to determine centers $A$, $B$, $C$, distances, $a$, $b$, $c$, and radii $u$, $v$, $w$ from your point-sets $X$, $Y$, $Z$, and then make appropriate substitutions into $(\star)$. Doing this symbolically is straightforward, if tedious ---see, for instance, the MathWorld "Circumcircle" article--- and the result will be an enormous polynomial. As a practical matter, it's probably best to compute the various elements numerically, and then substitute.
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\dfrac{\sigma_1(n)}{n} = \sigma_{-1}(n)$ where $\sigma_x(n)$ is the sum of the $x$th powers of the positive divisors of $n$. I computed $\dfrac{\sigma_1(n)}{n}$ and $\sigma_{-1}(n)$ on a good hundred values of $n$, and they seem to always match.
For example:
$\dfrac{\sigma_1(6)}{6} = \dfrac{1 + 2 + 3 + 6}{6} = \dfrac{12}{6} = 2$
$\sigma_{-1}(6) = \dfrac{1}{1} + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{6} = \dfrac{6}{6} + \dfrac{3}{6} + \dfrac{2}{6} + \dfrac{1}{6} = \dfrac{12}{6} = 2$
Humm... Now that I'm actually writing all of this down, I think I have an idea how to prove this...
For every positive integer $n$, all the $m$ divisors of $n$ can be grouped into $\dfrac{m}{2}$ pairs {$d_k$ ; $d_{m+1-k}$} such that $d_k \times d_{m+1-k} = n$ with $1 \leq k \leq m$
$d_k = \dfrac{n}{d_{m+1-k}}$
$\dfrac{\sigma_1(n)}{n} = \dfrac{d_1}{n} + \dfrac{d_2}{n} + \dfrac{d_3}{n} +$ $... + \dfrac{d_{m-2}}{n} + \dfrac{d_{m-1}}{n} + \dfrac{d_m}{n} $
$\sigma_{-1}(n) = \dfrac{1}{d_1} + \dfrac{1}{d_2} + \dfrac{1}{d_3} +$ $... + \dfrac{1}{d_{m-2}} + \dfrac{1}{d_{m-1}} + \dfrac{1}{d_m} + $
$\sigma_{-1}(n) = \dfrac{d_m}{n} + \dfrac{d_{m-1}}{n} + \dfrac{d_{m-2}}{n} +$ $... + \dfrac{d_3}{n} + \dfrac{d_2}{n} + \dfrac{d_1}{n}$
$\sigma_{-1}(n) = \dfrac{\sigma_1(n)}{n}$
And voila!
I'm not totally sure it's perfectly rigorous but it seems to be correct.
If $n$ is a square there's $d_{\frac{m+1}{2}}$ that isn't in a pair but we have: $d_{\frac{m+1}{2}}^2 = n$ so $d_{\frac{m+1}{2}} = \dfrac{n}{d_{\frac{m+1}{2}}}$
Well, how about you try to find some different proofs for my equality then? Maybe someone can find an even simpler and more elegant proof.
.
Hallowed is 120.
$\dfrac{\sigma_1(120)}{120} = \dfrac{360}{120} = 3$
$\sigma_{-1}(120) = 3$
| $\sigma_k(p^a)=\dfrac{1-p^{k(a+1)}}{1-p^k}=p^{ka}\dfrac{p^{-k(a+1)}-1}{p^{-k}-1}=p^{ka}\sigma_{-k}(p^a)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{\sin\theta}{1-\cos\theta} \equiv \csc\theta + \cot\theta$ This must be proved using elementary trigonometric identities.
I have not been able to come to any point which seems useful enough to include in this post.
| Multiply the LHS by $1+\cos\theta$ yields
\begin{align}
\frac{\sin\theta}{1-\cos\theta}\cdot\frac{1+\cos\theta}{1+\cos\theta}&=\frac{\sin\theta(1+\cos\theta)}{1-\cos^2\theta}\\
&=\frac{\sin\theta(1+\cos\theta)}{\sin^2\theta}\qquad;\qquad\color{red}{\cos^2\theta+\sin^2\theta=1}\\
&=\frac{1+\cos\theta}{\sin\theta}\\
&=\frac{1}{\sin\theta}+\frac{\cos\theta}{\sin\theta}\\
&=\color{blue}{\csc\theta+\cot\theta}.\qquad\qquad\blacksquare
\end{align}
| {
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Give a delta-epsilon proof that $\lim_{x \to \frac{1}{2}} \frac{1}{x} = 2$, is my answer right? One of my first $\delta$-$\epsilon$ proofs, but I don't feel that I'm heading in the right direction. I want to proof that:
$$
\lim_{x \to \frac{1}{2}} \frac{1}{x} = 2
$$
By the limit definition, For every $\epsilon > 0$, I must find a $\delta$, such that for every $x \in \operatorname{Dom}(f)$:
$$
\left| x - \frac{1}{2}\! \right| < \delta \Rightarrow \left|\! \frac{1}{x} - 2\right| < \epsilon
$$
I assume $f(x) = \frac{1}{x}$ is a function $f : \mathbb{R}\backslash \{0\} \rightarrow \mathbb{R}$, since $\frac{1}{x}$ is a reciprocal.
Now,
$$ \begin{align}
\left|\!\frac{1}{x} - 2\right| &= \left|\!\frac{1}{x} - 2\right|\\
&= \frac{|1-2x|}{|x|} \\
&= \frac{|\!\frac{1}{2}-x|}{2|x|} \end{align}
$$
Since $|x-y| = |y-x|$ and therefore,
$$
< \frac{\delta}{2|x|} \\
$$
I recognise that I have to take $\delta<\frac{1}{2}$ to prevent dividing by zero. But finding an estimate in terms of $\delta$ for $\frac{1}{2|x|}$ seems difficult. My attempt:
$$
2|x| = 2\left(x - \frac{1}{2} + \frac{1}{2}\right) < 2\left|x-\frac{1}{2}\!\right|+1 = 2\delta+1
$$
Concluding,
$$
\left|\!\frac{1}{x} - 2\right| < \frac{\delta}{2\delta+1} < \epsilon
$$
And therefore,
$$
\delta < \min\left(\frac{1}{2}, \frac{1}{\frac{1}{\epsilon} - 2}\right)
$$
Is this conceptual proof right? I have the feeling that I'm missing something.
| The one equation should be
\begin{equation}
\frac{|1-2x|}{|x|} = \frac{2|\frac{1}{2}-x|}{|x|}\ .
\end{equation}
So you have that
\begin{equation}
d(x,\frac{1}{2}) < \delta \Rightarrow d(\frac{1}{x},2) < \frac{2 \delta}{|x|}\ .
\end{equation}
Then, the second problem I see is that you take $2|x| < 2\delta +1$ and conclude that
\begin{equation}
\frac{\delta}{2|x|} < \frac{\delta}{2\delta +1}
\end{equation}
The last inequality has to be $>$ since you devide by something larger.
| {
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"timestamp": "2023-03-29T00:00:00",
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Why all composite numbers have this property? Define $f(n)=\sum\limits_{A \in S} f_{1}(n,A),\ n>2,\ n \in \mathbb{Z}$, where $S$ is the power set of $\{\frac{1}{2},\cdots ,\frac{1}{n-1}\}$.
Define $\ f_1(n,\varnothing)=1,\ f_{1}(n,A)=(-1)^{\#A-2n\Sigma(A)}$, where $\#A$ is the size of the set $A$ and $\Sigma(A)$ is the sum of the elements of A.
Let's take $n = 5$, for example:
$$
\begin{align*}
S = \{\\
& \varnothing,\\
& \{\frac{1}{2}\},\ \{\frac{1}{3}\},\ \{\frac{1}{4}\},\\
& \{\frac{1}{2},\frac{1}{3}\},\ \{\frac{1}{2},\frac{1}{4}\},\ \{\frac{1}{3},\frac{1}{4}\},\\
& \{\frac{1}{2},\frac{1}{3},\frac{1}{4}\}\\
\}
\end{align*}
$$
$f(5)=1+(-1)^{1-2 \cdot 5 \cdot (\frac{1}{2})}+ \cdots +(-1)^{2-2 \cdot 5 \cdot (\frac{1}{2}+\frac{1}{3})} + \cdots + (-1)^{3-2 \cdot 5 \cdot (\frac{1}{2}+\frac{1}{3}+\frac{1}{4})}=4.7320508075688767+1.2679491924311326i$
My question is: Why for every composite number $n$ is the real part of $f(n)$ approximately $0$, while prime numbers do not have this property? Examples:
$$
\begin{align*}
f(4) & = 1.887379141862766e-15-1.1102230246251565e-15i\\
f(22) & = -8.325340417059124e-12-7.568612403474617e-1i
\end{align*}
$$
Source: http://mymathforum.com/number-theory/43341-prime-prime.html
P.S. Python code for $f(n),f_{1}(n,A)$:
import itertools
def f_1(n,A):
return (-1) ** (len(A) - 2 * n * (sum(A)))
def f(n):
l = [itertools.combinations([1/x for x in range(2,n)], x) for x in range(1,n-1)]
return round(sum([f_1(n,y) for x in l for y in x]).real) + 1
print(f(4))
# Output: 0
| Because $f_1(n,A) = (-1)^{\#A-2n\sum A} = \prod_{x\in A} (-1)^{1-2nx}$, the sum over the power set of any set $B$ factors:
$$
\sum_{A\subset B} f_1(n,A) = \prod_{x\in B} \big( 1 + (-1)^{1-2nx} \big).
$$
In particular,
$$
f(n) = \prod_{k=2}^{n-1} \big( 1 + (-1)^{1-2n/k} \big).
$$
If $n$ is composite, then there exists an integer $2\le k\le n-1$ such that $k$ divides $n$; for such a $k$, we have $1 + (-1)^{1-2n/k} = 1+(-1)^{\text{some odd integer}}=0$. Therefore $f(n)$ is exactly equal to $0$ when $n$ is composite.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $
Show that
$$ 7\mid x \text{ and } 7\mid y \Longleftrightarrow 7\mid x^2+y^2 $$
Indeed,
First let's show
$7\mid x \text{ and } 7\mid y \Longrightarrow 7\mid x^2+y^2 $
we've $7\mid x \implies 7\mid x^2$ the same for $7\mid y \implies 7\mid y^2$ then
$ 7\mid x^2+y^2 $
*
*Am i right and can we write $a\mid x \implies a\mid x^P ,\ \forall p\in \mathbb{N}^*$
Now let's show
$7\mid x^2+y^2 \Longrightarrow 7\mid x \text{ and } 7\mid y$
$7\mid x^2+y^2 \Longleftrightarrow x^2+y^2=0 \pmod 7 $
for
\begin{array}{|c|c|c|c|c|} \hline
x& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline
x^2& 0 & 1 & 4& 2 & 2 & 4 & 1 &\pmod 7\\ \hline
y& 0 & 1 & 2& 3 & 4 & 5 & 6 \\ \hline
y^2& 0 & 1 & 4& 2 & 2 & 4 & 1 & \pmod 7 \\ \hline
\end{array}
which means we have one possibility that $x=y= 0 \pmod 7 $
*
*Am I right and are there other ways?
| Your example is just a special case of this theorem,
Theorem
If $p\equiv 3\pmod 4$ Then whenever $p\mid x^2+y^2$, we have $p\mid x$ and $p\mid y$.
Proof
Assume that $p\mid x^2+y^2$ with $p\not\mid x,y$ then
$$x^2+y^2\equiv 0\pmod p$$
$$x^2\equiv -y^2\pmod p$$
Which means $$x^{2\frac{p-1}{2}}\equiv (-1)^{\frac{p-1}{2}}y^{2\frac{p-1}{2}} \pmod p$$
$$x^{p-1}\equiv - y^{\frac{p-1}{2}}\pmod p$$
That’s because $(p-1)/2$ is odd, but let’s see what we’ve got, using Fermat’s little theorem $$1\equiv -1\pmod p$$
Hence $p=2$ but that’s clearly a contradiction. $\quad \quad \quad \quad \quad \quad\square$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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help: isosceles triangle circumscribing a circle of radius r Please help me show that: the equilateral triangle of altitute $3r$ is the isosceles triangle of least area circumscribing a circle of radius $r$.
Iassumed the following:
base = $2a$
height = $h$
radius of circle = $r$
Area = $\frac{1}{2}(2a)h = ah$
$tan(2\theta) = \frac{h}{a}$ and $tan\theta = \frac{r}{a} $
Ii tried using double angle identities to represent h in terms of a and \theta but I have failed miserably.
This is what I ended up with:
$\displaystyle{%
h={2r\cos^{2}\left(\theta\right) \over 2\cos^{2}\left(\theta\right) - 1}}$
I know i need to differentiate the area and set to zero, but I can't get it. please direct me how to solve this problem. Thank you;
| This is a nice example of a problem where the "right" choice for the independent variable makes a solution reasonably tractable, whereas other choices will lead to rather rough going. I tried a straight-ahead Cartesian approach (which led to a quintic equation) and the trigonometric approach you attempted (which gave a derivative function running up to fourth powers of sine which was going to be difficult to solve -- though it could be shown that one would obtain the expected angle for an equilateral triangle).
Instead we will work with the slope $ \ m \ $ of the sides of the isosceles triangle. If we center the circle of radius $ \ r \ $ on the origin, the "base" will lie along the line $ \ y \ = \ -r \ $ ; we will work with one vertex at $ \ ( \ a, \ -r \ ) \ $ , continuing your notation. The apex of this triangle is then located at $ \ ( \ 0, \ h \ - \ r \ ) \ $ and the area of the triangle is $ \ A \ = \ ah \ $ . The slope of the marked side is then $ \ m \ = \ -\frac{h}{a} \ $ .
We need some sort of constraint for this optimization: that is provided by the fact that the circumscribed triangle must be tangent to the circle on each of its sides. We will call the tangent point on the marked side $ \ ( \ X, \ Y \ ) \ $ . The equation of the line along which this side lies is $ \ y \ = \ ( \ h \ - \ r \ ) \ - \ \frac{h}{a} x \ $ .
Since the radius of the circle which extends to the tangent point is perpendicular to the tangent line, the equation of the line for this radius is $ \ y \ = \ -\frac{1}{m} x \ $ . Its intersection with the tangent line is thus given by
$$ \ -\frac{1}{m} X \ = \ ( \ h \ - \ r \ ) \ + \ m X \ \ \Rightarrow \ \ X \ = \ -\frac{m}{m^2 \ + \ 1} \ ( \ h \ - \ r \ ) $$
$$ \Rightarrow \ \ Y \ = \ \frac{1}{m^2 \ + \ 1} \ ( \ h \ - \ r \ ) \ \ . $$
Now we apply the constraint: the tangent point lies on the circle, so
$$ X^2 \ + \ Y^2 \ = \ r^2 \ \ \Rightarrow \ \ \left[ \ \left(-\frac{m}{m^2 \ + \ 1}\right)^2 \ + \ \left(\frac{1}{m^2 \ + \ 1}\right)^2 \ \right] \ ( \ h \ - \ r \ )^2 \ = \ r^2 $$
$$ \Rightarrow \ \ \left(\frac{1}{m^2 \ + \ 1}\right) \ ( \ h^2 \ - \ 2rh \ + \ r^2 ) \ = \ r^2 \ \ . $$
If, for the moment, we call $ \ \mu \ = \ \frac{1}{m^2 \ + \ 1} \ $ , then we can solve the quadratic equation $ \ \mu \ h^2 \ - \ 2 \mu r \ h \ + \ ( \ \mu \ - \ 1 \ ) r^2 \ = \ 0 \ $ to obtain
$$ h \ = \ \frac{2 \mu r \ \pm \ \sqrt{4 \mu^2 r^2 \ - \ 4 \mu \ ( \mu \ - \ 1 \ ) \ r^2}}{2 \mu} \ = \ r \ \left( 1 \ \pm \ \frac{1}{\sqrt{\mu}} \right) \ \ ; $$
since it is plainly the case that $ \ h \ > \ r \ $ , we discard the "negative solution" to produce $ \ h \ = \ r \ ( \ 1 \ + \ \sqrt{m^2 \ + \ 1} \ ) \ $ .
We have made our way through the first thicket and prepare to enter the second one. The area function is now found from
$$ m \ = \ -\frac{h}{a} \ \ \Rightarrow \ \ a \ = \ -\frac{h}{m} $$
$$ \Rightarrow \ \ A \ = \ ah \ = \ -\frac{h^2}{m} \ = \ - r^2 \ \left( \frac{m^2 \ + \ 2 \ + \ 2\sqrt{m^2 \ + \ 1}}{m} \right) \ \ . $$
It is this latter function for which we wish to locate the "critical value" for $ \ m \ $ ; hence,
$$ \frac{dA}{dm} \ = \ - r^2 \ \left( 1 \ - \ \frac{2}{m^2} \ - \ \frac{2}{m^2 \ \sqrt{m^2 \ + \ 1}} \right) $$
$$ = \ - r^2 \ \left( \frac{m^2 \ \sqrt{m^2 \ + \ 1} \ - \ 2 \sqrt{m^2 \ + \ 1} \ - \ 2} {m^2 \ \sqrt{m^2 \ + \ 1}} \right) \ = \ 0 \ \ . $$
Since the denominator cannot be zero (as $ \ m \ $ is non-zero), we only need to consider when the numerator is zero:
$$ m^2 \ \sqrt{m^2 \ + \ 1} \ - \ 2 \sqrt{m^2 \ + \ 1} \ - \ 2 \ = \ 0 \ \ \Rightarrow \ \ (m^2 \ - \ 2 ) \ \sqrt{m^2 \ + \ 1} \ = \ 2 $$
$$ \Rightarrow \ \ m^2 \ + \ 1 \ = \ \frac{4}{(m^2 \ - \ 2)^2} \ \ \Rightarrow \ \ m^6 \ - \ 3 \ m^4 \ = \ m^4 \ ( \ m^2 \ - \ 3 \ ) \ = \ 0 \ \ . $$
[squaring both sides of the equation and simplifying in this last line]
We have already rejected zero as a possible solution, so we have $ \ m \ = \ \pm \sqrt{3} \ $ . But these are the slopes of the sides of an equilateral triangle above the (horizontal) base; this then is our circumscribed isosceles triangle of minimal area.
| {
"language": "en",
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Meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$. I know that, for $|x|\leq 1$, $e^x$ can be bounded as follows:
\begin{equation*}
1+x \leq e^x \leq 1+x+x^2
\end{equation*}
Likewise, I want some meaningful lower-bound of $\sqrt{a^2+b}-a$ when $a \gg b > 0$.
The first thing that comes to my mind is $\sqrt{a^2}-\sqrt{b} < \sqrt{a^2+b}$, but plugging this in ends up with a non-sense lower-bound of $-\sqrt{b}$ even though the target number is positive.
\begin{equation*}
\big(\sqrt{a^2}-\sqrt{b} \big) - a < \sqrt{a^2+b}-a
\end{equation*}
How can I obtain some positive lower-bound?
| Maybe..
$$
\sqrt{1+x} = 1 + \frac{1}{2} x - \frac{1}{8}x^2 + \frac{1}{16}x^3 + O(x^4).
$$
So, for small enough $x$,
$$
\sqrt{1+x} > 1 + \frac{1}{2}x - \frac{1}{8}x^2
$$
hence
$$
\sqrt{a^2 + b} = a \sqrt{1 + (b/a^2)} > a\Big(1 + \frac{b}{2a^2} -\frac{b^2}{8a^4} \Big)
$$
when $a$ is much larger than $b$. So, in this case,
$$
\sqrt{a^2+b} - a > \frac{b}{2a} -\frac{b^2}{8a^3}
$$
| {
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If $4k^3+6k^2+3k+l+1$ and $4l^3+6l^2+3l+k+1$ are powers of two, how to conclude $k=1, l=2$ It is given that
$$4k^3+6k^2+3k+l+1=2^m$$
and
$$4l^3+6l^2+3l+k+1=2^n$$
where $k,l$ are integers such that $1\leq k\leq l$.
How do we conclude that the only solution is $k=1$, $l=2$?
I tried subtracting the two equations to get:
$$2(l-k)(2l^2+2lk+2k^2+3l+3k+1)=2^m(2^{n-m}-1)$$
But I am unable to proceed further.
Thanks for any help!
Updates:
An observation is that $k$ and $l$ have opposite parity.
Because if both $k$ and $l$ are even (or both odd), then $4k^3+6k^2+3k+l+1\equiv 1 \pmod 2$, which is a contradiction.
| The following argument has a lot of cases, but at least works (I think) to show the only solution is $2,1$.
Let $p(x,y)=4x^3+6x^2+3x+y+1.$ [I changed notation; $1 \le k \le l$ will be here $1 \le y \le x$], and then one wants each of $p(x,y)$ and $p(y,x)$ to be powers of $2$. Since the variables are each $\ge 1$ these powers of $2$ are at least $4+6+3+1+1=15,$ i.e. each is at least the fourth power of two.
Mod $2$, $p(x,y)=x+y+1$ so that the latter must be even, in particular $x\neq y$ so that $x>y.$ Now let $p(x,y)=2^b,\ p(y,x)=2^a$ and since $x>y$ implies $p(x,y)>p(y,x)$ we have $2^b>2^a$. So we have $b>a\ge 4.$ Our first step is to consider the sum $S=p(x,y)+p(y,x)$ which is $2^a+2^b=2^a(2^c+1)$ where $c=b-a\ge 1.$ we have
$$S=2(x+y+1)(2(x^2-xy+y^2)+x+y+1). \tag{1}$$
Here since $x+y+1$ is even it is coprime to $2^c+1$ and so divides $2^a$, making it a power of $2,$ say $2^r$. And here $r \ge 2$ since $x+y+1\ge 3.$ In this notation $S=2^{r+2}(x^2-xy+y^2+2^{r-1}).$ Since $2^{r-1}$ is still even, and $x^2-xy+y^2$ is odd because $x,y$ have opposite parity, we now see that the $2$-power in $S$, namely $2^a,$ must match the two power in the last expression for $S$, namely $2^{r+2}.$ This gives $r+2=a.$
We have arrived at $x+y+1=2^{a-2}.$ We apply this to $p(y,x)=2^a.$ We can write
$$p(y,x)=4y^3+6y^2+2y+(x+y+1)=4y^3+6y^2+2y+2^{a-2},$$
and so $g(y) \equiv 4y^3+6y^2+2y=2^a-2^{a-2}=3\cdot 2^{a-2}.$ This suggests looking at the congruence classes for $y$ mod $3$, dividing $g(y)$ through by $3$, and the result is to be $2^{a-2},$ a power of $2$ which is at least $4$.
If $y=3t$ then $g/3=2t(18t^2+9t+1).$ Here $t=1$ leads to the second factor as $28$, so that cannot divide a power of $2$. Otherwise $t$ is even, making the second factor odd. So $y=3t$ is impossible.
If $y=3t-1$ then $g/3=2t(3t-1)(6t-1)$ has the odd factor $6t-1$ so $y=3t-1$ is not possible.
The remaining case is $y=3t+1$ in which $g/3=2(3t+1)(6t^2+7t+2).$ So in this case (because of the last factor $6t^2+7t+2,$) we need $t$ even, and if it is not zero then the factor $3t+1$ is odd and greater than $1$, not possible.
The conclusion so far of this is that it must be that we're in the $y=3t+1$ with $t=0$, that is $y=1$ is forced.
Now returning to $x+y+1=2^r$ where $r=a-2$ and filling in the known $y=1$ we have $x=2^r-2=2(2^{r-1}-1).$ The least $r$ here being $2$, this means that either $x=2$ or else $x$ is of the form $2(4t-1).$ However if the latter occurs then when $p(x,1)$ is computed one gets
$$p(2(4t-1),1)=4(512t^3-288t^2+54t-3),$$
which is not a power of $2$ since the factor in parentheses is odd.
Thus we have $x=2,y=1$ forced by the assumptions, as desired.
[Thanks to yoyostein (OP) for catching a sign difference on the $xy$ term of the quadratic involved in the factorization of $p(x,y)+p(y,x),$ which is now $x^2-xy+y^2$ as it should be.]
| {
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What am I supposed to do here now? $\tan(\pi/8) = \sqrt{2} -1$ complex analysis
Find $\sqrt{1+i}$, and hence show $\tan(\pi/8) = \sqrt{2}-1$
Okay so I know that $\sqrt{1+i} = 2^{1/4}e^{i\pi/8}$ and I know $\sin x = \frac{e^{ix} - e^{-ix}}{2i}$ and $\cos x = \frac{e^{ix} + e^{-ix}}{2}$
If i directly substitute those definitions of sine and cosine into $$\tan(x) = \frac{\sin(x)}{\cos(x)}$$
then I am going to end up with a complex number in the form of $x + iy$. My key says
$$\tan \pi/8 = \sqrt{\frac{\sqrt{2}-1}{\sqrt{2}+1}} = \sqrt{2}-1.$$
Did they just use $\Re(\tan \pi/ 8) = \frac{\Re (\sin )}{\Re \cos}$?
| If you let $$\sqrt{1+i}=x+iy$$, then $$(x^2-y^2)+i2xy=1+i$$. then $$x^2-y^2=1$$ and $$2xy=1$$. Hence $$(x^2+y^2)^2=(x^2-y^2)^2+4x^2y^2=2$$. Hence $$x^2+y^2=\sqrt{2}$$. Now adding this two terms $$x^2=\frac{1+\sqrt{2}}{2}$$ and $$y^2=\frac{\sqrt{2}-1}{2}$$. So $$\sqrt{1+i} =\sqrt{\frac{1+\sqrt{2}}{2}}+i\sqrt{\frac{\sqrt{2}-1}{2}}$$
You already have $$\sqrt{1+i} = 2^{1/4}e^{i\pi/8}=2^{\frac{1}{4}}\left(\cos \frac{2\pi}{8}+ i\sin \frac{2 \pi}{8}\right)$$
Comparing both these $$\cos \frac{ \pi}{8}=\frac{1}{2^{\frac{1}{4}}}\sqrt{\frac{\sqrt{2}+1}{2}}$$ and $$\sin \frac{ \pi}{8}=\frac{1}{2^{\frac{1}{4}}}\sqrt{\frac{\sqrt{2}-1}{2}}$$
Then you have your answer
| {
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Please check my solution of $\int \sin^6(x)\cos^3(x) dx$ $$\int \sin^6(x)\cos^3(x) dx = \int \sin^6(x)(1-\sin^2(x))\cos(x)dx$$
$$\int \sin^6(x)\cos(x)dx - \int\sin^8x\cos xdx$$
Now, $\cos xdx = d(\sin x)$
$$\int u^6du - \int u^8du = \frac{1}{7}u^7 - \frac{1}{9}u^9 + C$$
$$\frac{1}{7}\sin^7(x) - \frac{1}{9}\sin^9(x) + C$$
However, WolframAlpha says it's:
Can anyone tell if those expressions are equal?
| Your expression is correct. Using the double-angle identity $\cos 2x=1-2\sin^2 x$, you can verify that the more awkward Alpha version is also correct. For $7(1-2\sin^2 x)+11=18-14\sin^2 x$ and $\frac{18}{126}=\frac{1}{7}$ and $\frac{14}{126}=\frac{1}{9}$.
Remark: Fairly often, with trigonometric functions, verification can be more complicated. As a simple example, suppose one calculation gives $2\cos^2 x+C$ as the integral, and another calculation gives $\cos 2x+C$. The functions $\cos 2x$ and $2\cos^2 x$ are not the same, but they differ by a constant, so if one is valid, then so is the other.
| {
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how to derive the fact that the integral of $1/\sin^2(x) = -\cot (x)$ I know how that the integral of $\dfrac{1}{\sin^2(x)} = -\cot (x)$, but how does derive this fact? Can you use half-angle formula to do this integral?
| We can use the tangent half-angle substitution to get rid of any trigonometric expression during the integration.
Let $\sin x=\dfrac{2t}{1+t^2}$. Then $\cos x=\dfrac{1-t^2}{1+t^2}$ and $dx=\dfrac{2dt}{1+t^2}$, and we have
$$\int\frac{1}{\sin^2x}dx=\int\frac{(1+t^2)^2}{4t^2}\frac{2}{1+t^2}dt=\int\frac{1+t^2}{2t^2}dt=\int\frac{1}{2}+\frac{1}{2t^2}dt=\frac{t}{2}-\frac{1}{2t}.$$
Now
$$\frac{t}{2}-\frac{1}{2t}=\frac{t^2-1}{2t}=\frac{-\frac{1-t^2}{1+t^2}}{\frac{2t}{1+t^2}}=\frac{-\cos x}{\sin x}=-\cot x.$$
| {
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If $a+\frac{1}{b}, b+\frac{1}{c}, c+\frac{1}{a}\in\mathbb{Z}$, find $a+b+c$. Let $a,b,c$ be positive rational numbers such that
$a+\frac{1}{b}, b+\frac{1}{c}, c+\frac{1}{a}$
are all integers.
Find all the possible values of $a+b+c$.
it would be too complicate to solve by quadratic equation(the discriminant is 6 degree polynomial of 3 variables...).
Either, no thought at all…
| Ok I worked it out
$$a+b+c=3, \frac{7}{2} ,\frac{25}{6}, \frac{23}{6}$$
Assume $$\frac{x}{y}+\frac{z}{w}=\frac{xw+yz}{yw}$$
is an integer with both fractions in reduced form,
then $y|w$ and $w|y$ so $y=w$ (they are positive).
This means that
$a,b,c$ must be of the form
$$\frac{p}{q}, \frac{q}{r}, \frac{r}{p}$$ respectively.
where $p,q,r$ are relatively prime in pairs.
Let us assume that $p\leq q \leq r$.
If $r=1$ then all are one and $a+b+c=3$.
So assume $r>1$ then $r|p+q$ and $p+q \leq 2r$ so either
$p+q=r$ or $p+q=2r$.
Taking the first possiblity,
we see that $p|2q$ and $q|2p$ now since they are relatively prime, $p=1,2$ and $q=1,2$
If $p=q=1$ we have
$$1, \frac{1}{2}, 2$$ and
$$a+b+c=\frac{7}{2}$$
If $p=1, q=2$ we have
$$\frac{1}{2}, \frac{2}{3}, 3$$ and
$$a+b+c=\frac{25}{6}$$
If $p=q=2$ then we get a previous case.
Now assume that
$p+q=2r$,
then $2p|3q+p$ and $2q|3p+q$ so $p=1,3$ and $q=1,3$
If $p=q=1$ then $r=1$ and we have previous case.
If $p=1$, $q=3$ then $r=2$
and
$$\frac{1}{3}, \frac{3}{2}, 2$$
and $$a+b+c =\frac{23}{6}$$
If $p=q=3$ we have a previous case.
This gives the values stated at the beginning.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluation of $ \lim_{x\rightarrow \infty}\left\{2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right\}$
Evaluate the limit
$$
\lim_{x\rightarrow \infty}\left(2x-\left(\sqrt[3]{x^3+x^2+1}+\sqrt[3]{x^3-x^2+1}\right)\right)
$$
My Attempt:
To simplify notation, let $A = \left(\sqrt[3]{x^3+x^2+1}\right)$ and $B = \left(\sqrt[3]{x^3-x^2+1}\right)$. Now
$$
\begin{align}
2x^2 &= A^3-B^3\\
x &= \sqrt{\frac{A^3-B^3}{2}}
\end{align}
$$
So the limit becomes
$$\lim_{x\rightarrow \infty}\left(\sqrt{\frac{A^3-B^3}{2}}-A-B\right)$$
How can I complete the solution from this point?
| Hint
Since $x$ goes to infinity, let us write $$\sqrt[3]{x^3+x^2+1}=x\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}}$$ and let us define $y=\frac{1}{x}+\frac{1}{x^2}$. Now, look at the Taylor expansion of $$\sqrt[3]{1+y}=1+\frac{y}{3}-\frac{y^2}{9}+O\left(y^3\right)$$ and replace $y$ by its definition and expand to get finally $$\sqrt[3]{x^3+x^2+1}=x \Big(1+\frac{1}{3 x}-\frac{1}{9 x^2}+...\Big)\simeq x-\frac{1}{9 x}+\frac{1}{3}$$ Similarly, the same method should give $$\sqrt[3]{x^3-x^2+1} \simeq x-\frac{1}{9 x}-\frac{1}{3}$$
I am sure that you can take from here.
More intuitively, you could have noticed that, since $x$ goes to $\infty$,$$\sqrt[3]{x^3 \pm x^2+1}=x\sqrt[3]{1 \pm\frac{1}{x}+\frac{1}{x^2}}$$ behaves just as $x$ and then the limit of the expression is $0$. However, this approach allows you to show "how" the expression goes to $0$ (by positive or negative values).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/851849",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 2
} |
For the polynomial For the polynomial, -2 is a zero. $h(x)= x^3+8x^2+14x+4$. Express $h(x)$ as a product of linear factors.
Can someone please explain and help me solve?
| Since its a cubic equation, you are looking at $$(x+A).(x+B).(x+C)=x^3+8x^2+14x+4$$
Obviously, you have to multiply and work hard towards solving this and getting the answer.
Alternatively, what you can do is represent the same thing as
$$x^3+8x^2+12x+2x+4=0$$
Or, $$x(x^2+8x+12)+2(x+2)$$
Or, $$x(x+6)(x+2)+2(x+2)$$
Or, $$(x+2)(x(x+6)+2)$$
Or, $$(x+2)(x^2+6x+2)$$
Solving the second equation for x will give $$x= -3 +\sqrt{7}$$
and $$x=-3-\sqrt{7}$$
and $$x=-2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852355",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
Basic induction proof methods so we're looking to prove $P(n)$ that
$$1^2+2^3+\cdots+n^3 = (n(n+1)/2)^2$$
I know the basis step for $p(1)$ holds.
We're going to assume $P(k)$
$$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$
And we're looking to prove $P(k+1)$
What I've discerned from the internet is that I should be looking to add the next term, $k+1$, to both sides so...
$$1^3+2^3+\cdots+k^3 + (k+1)^3=(k(k+1)/2)^2 + (k+1)^3$$
now I saw some nonsense since that we assumed $p(k)$ we can use it as a definition in our proof, specifically on the left hand side
so since
$$1^3+2^3+\cdots+k^3=(k(k+1)/2)^2$$
then
$$(k(k+1)/2)^2 + (k+1)^3 = (k(k+1)/2)^2 + (k+1)^3$$
and we have our proof
OK so far thats wrong
so far ive figured this.
$$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+1)+1)/2)^2$$
Then
$$1^3+2^3+\cdots+k^3 + (k+1)^3=((k+1)((k+2)/2)^2$$
using the definition
$$(k(k+1)/2)^2 + (k+1)^3 = ((k+1)((k+2)/2)^2$$
$$(k^2+k/2)^2 + (k^2+2k+1)(k+1) = (k^2+3k+2/2)^2$$
$$(k^4+k^2/4)+(k^2+2k^2+k+k^2+2k+1)= (k^4+9k^2+4/4)$$
Where should I go from here? It doesn't possibly look like these could equate, I'll keep going though
| Assuming $P(k)$, you add $(k+1)^3$ on both sides of
$$
1^3 + 2^3 + \ldots + k^3 = (k(k+1)/2)^2
$$
to get
\begin{align}
1^3 + 2^3 + \ldots + k^3 + (k+1)^3
& = (k(k+1)/2)^2 + (k+1)^3 \\
& = \frac 14 k^2(k+1)^2 + (k+1)^3 \\
& = \frac 14\left(k^4 + 2k^3 + k^2 + 4k^3 + 12k^2 + 12k + 4\right) \\
& = \frac 14\left(k^4 + 6k^3 + 13k^2 + 12k + 4\right) \\
& = \frac 14(k+1)^2(k+2)^2 \\
& = \left((k+1)(k+2)/2\right)^2.
\end{align}
This statement is $P(k+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/852689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
functions satisfying $f(x - y) = f(x) f(y) - f(a - x) f(a + y)$ and $f(0)=1$ A real valued function $f$ satisfies the functional equation
$$f(x - y) = f(x) f(y) - f(a - x) f(a + y) \tag 1 \label 1$$
Where $a$ is a given constant and $f(0) = 1$. Prove that $f(2a - x) = -f(x)$, and find all functions which satisfy the given functional equation.
My Try:
Put $x=y=0$ in equation \eqref{1}.
$\implies f(0)=f(0)^2-f(a)\cdot f(a)\implies f(a)^2=0\implies f(a)=0$.
Now Put $x=a$ and $y=x-a$.
$\implies f(2a-x)=f(a)\cdot f(x-a)-f(0)\cdot f(x) = -f(x)$.
My question is how can I find all function which satisfy the given functional equation.
Help me.
Thanks.
| Partial progress:
Let $P(x,y)$ be the property that $f(x-y) = f(x)f(y)-f(a-x)f(a+y)$.
Then, $P(0,0)$ gives $f(0) = f(0)f(0)-f(a)f(a)$ which yields $f(a) = 0$.
Also, $P(a,x)$ gives $f(a-x) = f(a)f(x)-f(0)f(a+x)$ which yields $f(a-x)=-f(a+x)$.
Then, $P(x,x)$ gives $f(0) = f(x)f(x)-f(a-x)f(a+x)$ which yields $f(x)^2 + f(x+a)^2 = 1$ (*).
Using (*), we have $f(x)^2+f(x+a)^2 = 1$ and $f(x+a)^2+f(x+2a)^2 = 1$ for all $x \in \mathbb{R}$.
Hence, $f(x+2a) = \pm f(x)$ for all $x \in \mathbb{R}$. Therefore, $f(x+4a) = f(x)$ for all $x \in \mathbb{R}$.
It's pretty clear that $f(x) = \cos\dfrac{\pi x}{2a}$ is a solution. Now, can we show that there aren't any more?
| {
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"url": "https://math.stackexchange.com/questions/852861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
Proving an expression is composite I am trying to prove that $ n^4 + 4^n $ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.
This problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.
I would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1.
| Some interesting factorizations of a polynomial of type $x^4+\text{const}$:
$$ x^4+4=(x^2+2x+2)(x^2-2x-2) \tag{1}$$
$$ x^4+1=(x^2+\sqrt[]{2}x+1)(x^2-\sqrt[]{2}x+1) \tag{2}$$
So one can ask, how to select the coefficients $a,b,c,d$ in
$$(x^2+ax+b)(x^2+cx+d) \tag{3}$$
such that all coefficients of the resulting polynomial are zero except the constant term and the coefficient of the 4th power. The latter is $1$.
If we expand $(3)$ we get
$$x^4+(c+a)x^3+(d+a c+b)x^2+(a d+b c)x+b d$$
And the coefficients disappear, if
$$
\begin{eqnarray}
c+a &=& 0 \\
d+ ac +b &=& 0 \\
ad+bc &=& 0
\end{eqnarray}
$$
When solving for $b,c,d$ we get
$$
\begin{eqnarray}
c &=& -a \\
b &=& \frac{a^2}{2} \\
d &=& \frac{a^2}{2}
\end{eqnarray}
$$
and therefore
$$x^4+\frac{a^4}{4} = (x^2+a x+\frac{a^2}{2})(x^2-ax+\frac{a^2}{2})$$
For $a=2$ this gives $(1)$, $a=\sqrt[]{2}$ this gives $82)$ .
Substituting $a=2^{t+1}$ we get
$$x^4+4^{2t+1} = (x^2+2\cdot 2^t x+2^{2t+1})(x^2-2\cdot 2^tx+2^{2t+1})$$
Substituting $x=n=2t+1$ gives the required result for odd $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/853615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
I'm missing the right substitute $\sqrt3\cos x=1-\sin x$ Please show me how to solve the following equation for $x$.
I've tried multiple substitutes but can't seem to find the right one.
$$\sqrt3\cos x=1-\sin x$$
| You don't have to guess the right substitution. Every equation of the form
$$
a\cos x+b\sin x = c
$$
can be managed with the following procedure.
Divide both sides by $\sqrt{a^2+b^2}$, which is different from $0$ unless $a=b=0$, which would make it trivial.
Now we can observe that there is a unique $\varphi$ such that
$$
\frac{a}{\sqrt{a^2+b^2}}=\sin\varphi,\quad \frac{b}{\sqrt{a^2+b^2}}=\cos\varphi,
$$
for
$$
\biggl(\frac{a}{\sqrt{a^2+b^2}}\biggr)^2+
\biggl(\frac{b}{\sqrt{a^2+b^2}}\biggr)^2=1.
$$
Thus the equation becomes
$$
\sin\varphi\cos x+\cos\varphi\sin x=\frac{c}{\sqrt{a^2+b^2}}
$$
or
$$
\sin(x+\varphi)=\frac{c}{\sqrt{a^2+b^2}}
$$
which is an elementary equation for the sine.
In your case $a=\sqrt{3}$, $b=1$, so $a^2+b^2=4$ and
$$
\sin\varphi=\frac{a}{\sqrt{a^2+b^2}}=\frac{\sqrt{3}}{2},\quad
\cos\varphi=\frac{b}{\sqrt{a^2+b^2}}=\frac{1}{2}
$$
so $\varphi=\pi/3$. Since $c=1$, the equation becomes
$$
\sin\biggl(x+\frac{\pi}{3}\biggr)=\frac{1}{2}=\sin\frac{\pi}{6}
$$
that has the solutions
$$
x+\frac{\pi}{3}=\frac{\pi}{6}+2k\pi,\qquad
x+\frac{\pi}{3}=\pi-\frac{\pi}{6}+2k\pi
$$
or, in another form
$$
x=-\frac{\pi}{6}+2k\pi,\qquad x=\frac{\pi}{2}+2k\pi.
$$
An alternative method is to set $X=\cos x$, $Y=\sin x$ and to solve the system
\begin{cases}
aX+bY=c\\[1ex]
X^2+Y^2=1
\end{cases}
which in this case is
\begin{cases}
\sqrt{3}\,X+Y=1\\[1ex]
X^2+Y^2=1
\end{cases}
Solve with respect to $Y$ the linear equation: $Y=1-\sqrt{3}\,X$ and substitute:
$$
X^2+(1-\sqrt{3}\,X)^2=1
$$
that becomes
$$
4X^2-2\sqrt{3}\,X=0
$$
that factors as
$$
X=0\qquad\text{or}\qquad 2X-\sqrt{3}=0
$$
that become
$$
\begin{cases}
\cos x=0\\[2ex]
\sin x=1
\end{cases}
\qquad\text{or}\qquad
\begin{cases}
\cos x=\frac{\sqrt{3}}{2}\\[2ex]
\sin x=-\frac{1}{2}
\end{cases}
$$
that give the same solutions as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/854592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
} |
what is a smart way to find $\int \frac{\arctan\left(x\right)}{x^{2}}\,{\rm d}x$ I tried integration by parts, which gets very lengthy due to partial fractions.
Is there an alternative
| Compute using the series expansion:
\begin{align*}
\int \frac{\tan^{-1} x}{x^2} dx &= \int \frac 1 {x^2} \left(x - \frac{x^3}{3} + \frac{x^5}{5} - \cdots\right) dx \\
&= \int \frac 1 x - \frac{x}{3} + \frac{x^3}{5} - \cdots dx \\
&= \ln x - \frac{x^2}{2 \cdot 3} + \frac{x^4}{4 \cdot 5} - \frac{x^6}{6 \cdot 7} + \cdots
\end{align*}
Now using the identity
$$\frac 1 {n(n + 1)} = \frac 1 n - \frac 1 {n + 1}$$
we can rewrite the remaining series as
\begin{align*}
\frac{x^2}{2 \cdot 3} - \frac{x^4}{4 \cdot 5} + \frac{x^6}{6 \cdot 7} - \cdots &= \left(\frac{x^2}{2} - \frac{x^4}{4} + \frac{x^6}{6} - \cdots\right) -\left(\frac{x^2}{3} - \frac{x^4}{5} + \frac{x^6}{7}\right)
\end{align*}
These are both (related to) standard series for well-known functions; the second one is
$$1-\frac {\tan^{-1} x} x$$
and the first is $\frac 1 2\ln(x^2 + 1)$, using the fact that $$\ln(1 + t) = t - \frac {t^2}2 + \frac{t^3}3 - \frac{t^4}4 + \cdots$$
Putting all this together gives the result
$$\boxed{\displaystyle\int \frac{\tan^{-1} x}{x^2} dx = -\frac 1 2 \ln(1 + x^2) + \ln x - \frac{\tan^{-1} x}{x} + C}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/854997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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If $K = \frac{2}{1}\times \frac{4}{3}\times \cdots \times \frac{100}{99}.$ Then value of $\lfloor K \rfloor$
Let $K = \frac{2}{1}\times \frac{4}{3}\times \frac{6}{5}\times \frac{8}{7}\times \cdots \times \frac{100}{99}.$ Then what is the value of $\lfloor K \rfloor$, where $\lfloor x \rfloor$ is the floor function?
My Attempt:
By factoring out powers of $2$, we can write
$$
\begin{align}
K &= 2^{50}\times \left(\frac{1}
{1}\times \frac{2}{3}\times \frac{3}{5}\times \frac{4}{7}\times \frac{5}{9}\times\cdots\times \frac{49}{97}\times \frac{50}{99}\right)\\
&= 2^{50}\cdot 2^{25}\times \left(\frac{1\cdot 3 \cdot 5\cdots49}{1\cdot 3 \cdot 5\cdots 49}\right)\times \left(\frac{1}{51\cdot 53\cdot 55\cdots99}\right)\\
&= \frac{2^{75}}{51\cdot 53\cdot 55\cdots99}
\end{align}
$$
How can I solve for $K$ from here?
| Note that $K = \dfrac{2 \cdot 2 \cdot 4 \cdot 4 \cdots 100 \cdot 100}{1 \cdot 2 \cdot 3 \cdot 4 \cdots 99 \cdot 100} = \dfrac{2^{100}(50!)^2}{100!} = \dfrac{2^{100}}{\dbinom{100}{50}}$
It can be shown that the Central Binomial Coefficent satisfies:
$\left(1-\dfrac{1}{8n}\right)\dfrac{2^{2n}}{\sqrt{\pi n}} \le \dbinom{2n}{n} \le \dfrac{2^{2n}}{\sqrt{\pi n}}\left(1-\dfrac{1}{9n}\right)$
for all $n \ge 1$.
Thus, $12.56105 \approx \dfrac{450}{449}\sqrt{50\pi} \le \dfrac{2^{100}}{\dbinom{100}{50}} \le \dfrac{400}{399}\sqrt{50\pi} \approx 12.56456$
Therefore, $\lfloor K \rfloor = 12$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/855990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Evaluate the sum ${n \choose 1} + 3{n \choose 3} +5{n \choose 5} + 7{n \choose 7}...$ in closed form How do I evaluate the sum:$${n \choose 1} + 3{n \choose 3} +5{n \choose 5} + 7{n \choose 7} ...$$in closed form?
I don't really know how to start and approach this question. Any help is greatly appreciated.
| Starting with
\begin{align}
(1+t)^{n} = \sum_{k=0}^{n} \binom{n}{k} t^{k}
\end{align}
it is seen that
\begin{align}
\frac{1}{2} \left[ (1+t)^{n} - (1-t)^{n} \right] = \sum_{k=1}^{[(n+1)/2]} \binom{n}{2k-1} t^{2k-1}.
\end{align}
Now differentiating both sides leads to
\begin{align}
\sum_{k=1}^{[(n+1)/2]} (2k-1) \binom{n}{2k-1} t^{2k-2} = \frac{n}{2} \left[ (1+t)^{n-1} + (1-t)^{n-1} \right]
\end{align}
and upon setting $t=1$ the desired result is
\begin{align}
\sum_{k=1}^{[(n+1)/2]} (2k-1) \binom{n}{2k-1} = 2^{n-2} \ n.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/856418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Proof Verification for Discrete Math Class Prove that $n^2$ is even iff $n$ is even.
I proved it like this:
Case I: $n$ is even
1) $n = 2a$ $(a\in Z)$
2) $n^2 = 4a^2 = 2(2a^2)$
3) $2a^2 = K$ $(K \in Z)$
4) $n^2 = 2K$
Case II: $n$ is odd
1) $n = 2a + 1$ $(a\in Z)$
2)$n^2 = 4a^2 + 4a +1 = 2(2a^2 + 2a) + 1$
3) $2a^2 + 2a = K$ $K\in Z$
4) $n^2 = 2K+1$
My book does it slightly differently:
Is my proof still correct?
| You are attempting to prove $A \iff B$.
Your book proves $A \Rightarrow B$ and $B \Rightarrow A$.
You prove $A \Rightarrow B$ and $\lnot A \Rightarrow \lnot B$. However, $\lnot A \Rightarrow \lnot B$ is logically equivalent to $B \Rightarrow A$.
All of these things can be checked with truth tables. For example:
$\begin{array} {|cc|cc|}
A & B & B \Rightarrow A & \lnot A \Rightarrow \lnot B \\ \hline
\text{T} & \text{T} & \text{T} & \text{T} \\ \hline
\text{T} & \text{F} & \text{T} & \text{T} \\ \hline
\text{F} & \text{T} & \text{F} & \text{F} \\ \hline
\text{F} & \text{F} & \text{T} & \text{T} \\ \hline
\end{array}$
$\begin{array} {|cc|cc|c|c|}
A & B & A \Rightarrow B & B \Rightarrow \lnot A & (A \Rightarrow B) \text{ and } (B \Rightarrow A) & A \text{ iff } B\\ \hline
\text{T} & \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \hline
\text{T} & \text{F} & \text{F} & \text{T} & \text{F} & \text{F} \\ \hline
\text{F} & \text{T} & \text{T} & \text{F} & \text{F} & \text{F} \\ \hline
\text{F} & \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \hline
\end{array}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/856682",
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"source": "stackexchange",
"question_score": "2",
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How to find orthogonal projection of vector on a subspace? Well, I have this subspace: $V = \operatorname{span}\left\{ \begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix},\begin{pmatrix}1 \\3 \\4\end{pmatrix}\right\}$
and the vector $v = \begin{pmatrix}9 \\0 \\0\end{pmatrix}$
How can I find the orthogonal projection of $v$ on $V$?
This is what I did so far:
\begin{align}&P_v(v)=\langle v,v_1\rangle v_1+\langle v,v_2\rangle v_2 =\\=& \left\langle\begin{pmatrix}9 \\0 \\0\end{pmatrix},\begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix}\right\rangle\begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix}+\left<\begin{pmatrix}9 \\0 \\0\end{pmatrix},\begin{pmatrix}1 \\3 \\4\end{pmatrix}\right>\begin{pmatrix}1 \\3 \\4\end{pmatrix} = \begin{pmatrix}10 \\29 \\38\end{pmatrix}\end{align}
Is this the right method to compute this?
| Hint:
1) Compute an orthonormal base $v_1,v_2$ of $V$ using Gram-Schmidt
2) Consider the projector $p_V(x) = \sum_{i=1}^2 \langle v_i,x\rangle v_i$
3) Compute $p_V(v)$
So what you did is wrong because $\begin{pmatrix}\frac{1}{3} \\\frac{2}{3} \\\frac{2}{3}\end{pmatrix}$,$\begin{pmatrix}1 \\3 \\4\end{pmatrix}$ are not orthonormal vectors.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/857942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Calculate $\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$ I have homework questions to calculate infinity sum, and when I write it into wolfram, it knows to calculate partial sum...
So... How can I calculate this:
$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)}$$
| OK, I answer the question with the hint:
$$\sum_{k=1}^n \frac 1 {(k+1)(k+2)} = \sum_{k=1}^n \left(\frac 1 {k+1} - \frac 1 {k+2}\right) = \\ = \left( \frac 1 2 - \frac 1 3 \right) + \left( \frac 1 3 - \frac 1 4 \right) + \left( \frac 1 4 - \frac 1 5 \right) + \ldots + \left( \frac 1 {n+1} - \frac 1 {n+2} \right) = \\ = \frac 1 2 - \frac 1 {n+2}$$
(For my homework: $\lim_{n\to\infty} \frac 1 2 - \frac 1 {n+2} = \frac 1 2$)
Thanks!
| {
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"timestamp": "2023-03-29T00:00:00",
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"answer_id": 4
} |
Test for convergence $\sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1})$ Test for convergence $$\sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1})$$
So far I attempted to use the ratio test, but I'm stuck on what to do after.
$$\dfrac{\dfrac{\sqrt{n+2}-\sqrt{n}}{n+1}}{\dfrac{\sqrt{n+1}-\sqrt{n-1}}{n}} = \dfrac{n(\sqrt{n+2}-\sqrt{n})}{(n+1)(\sqrt{n+1}-\sqrt{n-1})}$$
| First note that
$$
\frac{\sqrt{n+1}+\sqrt{n-1}}{\sqrt n} > \frac{\sqrt{n+1}}{\sqrt n} > 1.
$$
Therefore
$$
\sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1}) < \sum_{n=1}^\infty \frac{1}{n}(\sqrt{n+1}-\sqrt{n-1}) \frac{\sqrt{n+1}+\sqrt{n-1}}{\sqrt n}= \sum_{n=1}^\infty \frac{2}{n^{3/2}}
$$
which converges; hence the original sum converges by the comparison test.
The intuition, if it helps, is that
\begin{multline*}
\sqrt{n+1}-\sqrt{n-1} = \sqrt{n(1+O(\tfrac1n))} - \sqrt{n(1+O(\tfrac1n))} \\
= \sqrt n(1+O(\tfrac1n)) - \sqrt n(1+O(\tfrac1n)) = O(\tfrac1{\sqrt n}).
\end{multline*}
and so we want to compare the given sum with $\sum \frac1{n\sqrt n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How do you find the following limit as x approaches infinity? $\lim_{x\to \infty} \sqrt{x^2+9} - \sqrt{x^2-2}$
I have tried multiplying by the conjugate but the square roots are throwing me off and I'm not sure what to do next. How do you solve this?
| $$\sqrt{x^2+9}-\sqrt{x^2-2}=\frac{(\sqrt{x^2+9}-\sqrt{x^2-2})(\sqrt{x^2+9}+\sqrt{x^2-2})}{\sqrt{x^2+9}+\sqrt{x^2-2}}=\frac{x^2+9-x^2+2}{\sqrt{x^2+9}+\sqrt{x^2-2}}=\frac{11}{\sqrt{x^2+9}+\sqrt{x^2-2}}$$
$$\lim_{x \to \infty} \sqrt{x^2+9}-\sqrt{x^2-2}=\lim_{x \to \infty } \frac{11}{\sqrt{x^2+9}+\sqrt{x^2-2}}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/861613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Use a proof by cases to show that $\lfloor n/2 \rfloor$ * $\lceil n/2 \rceil$ = $\lfloor \frac{n^2}{4} \rfloor$ for all integers $n$. Question
Use a proof by cases to show that $\lfloor n/2 \rfloor$ * $\lceil n/2 \rceil$ = $\lfloor \frac{n^2}{4} \rfloor$ for all integers $n$.
My Attempt:
I can only think of two cases,
*
*$n/2 \in \mathbb{Z}$
*$n/2 \notin \mathbb{Z}$
First case is straightforward:
$\lfloor n/2 \rfloor = \lceil n/2 \rceil = n/2$,
$\frac{n}{2}*\frac{n}{2} = \frac{n^2}{4}$
Second case troubled me,
$\lceil n/2 \rceil = \lfloor n/2 \rfloor + 1\\
\lceil n/2 \rceil = \lfloor n/2 + 1\rfloor$
$n/2 - 1 \leq \lfloor n/2 \rfloor < n/2\\
n/2 \leq \lfloor n/2 + 1 \rfloor < n/2 + 1$
I multiply both inequalities,
$\frac {n^2 - 2n}{4} \leq \lfloor n/2 \rfloor * \lfloor n/2 + 1 \rfloor < \frac{n^2 + 2n}{4}$
I need to prove that $\lfloor n/2 \rfloor * \lfloor n/2 + 1 \rfloor$ should be at least $n^2 /4$ and less than $n^2 /4 + 1$, this ensures that if I floor that, it will be $n^2/4$, but I'm lost.
My second attempt, (I didn't think the top have anywhere to go). This time I used some epsilon $\epsilon \in (0, 1)$,
$\lfloor n/2 \rfloor = n/2 - \epsilon\\
\lceil n/2 \rceil = n/2 + 1 - \epsilon$
$\lfloor n/2 \rfloor * \lfloor n/2 + 1 \rfloor = (n/2 - \epsilon)*(n/2 + 1 - \epsilon)\\
= n^2/4 + n/2 - n*\epsilon/2 - n*\epsilon/2 - \epsilon + \epsilon ^ 2\\
= n^2/4 + n/2 - 2n\epsilon/2 + 2\epsilon^2/2\\
= n^2/4 + \frac{n-2n\epsilon - 2\epsilon + 2\epsilon^2}{2}$
The problem now is I need to prove that $\frac{n-2n\epsilon - 2\epsilon + 2\epsilon^2}{2}$ is between 0 and 1. I don't really think this one is the solution is either so I gave up.
| Why make it so complicated:
Case 1: n is even. Let n = 2k.
$\lceil n/2 \rceil \times \lfloor n/2 \rfloor = k * k = \lfloor n^2/4 \rfloor$
Case 2: n is odd. Let n = 2k + 1.
$\lceil n/2 \rceil \times \lfloor n/2 \rfloor = k (k + 1) = \lfloor n^2/4 \rfloor$ as $n^2/4 = (4k^2 + 4k + 1)/4 = k(k+1) + 1/4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/864140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
I need help finding the critical values of this function. So $h(t)=t^{\frac{3}{4}}-7t^{\frac{1}{4}}$. So I need to set $h'(t)=0$. So for $h'(t)$ the fattest I've gotten to simplifying os $h'(t)=\frac{3}{4 \sqrt[4]{t}}-\frac{7}{4\sqrt[4]{t^3}}$ and that is as farthest as I can simplify. So i'm having a had time having $h'=0$ So could someone show me how to properly simplify and find the critical values step by step I would immensely appreciate it. In the most recent material we have covered in class ir seems that my biggest struggle comes from simplifying completely. Thanks in advance for the help.
| $h(t)=t^{\frac{3}{4}}-7t^{\frac{1}{4}}$
$\Rightarrow$ $0=h'(t)=\frac{3}{4}t^{-\frac{1}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=\frac{3}{4}t^{-\frac{3}{4}+\frac{2}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=\frac{3}{4}t^{-\frac{3}{4}}t^{\frac{2}{4}}-\frac{7}{4}t^{-\frac{3}{4}}=t^{-\frac{3}{4}}(\frac{3}{4}t^{\frac{2}{4}}-\frac{7}{4})$
Thus $t=(\frac{7}{3})^2$ Since one factor have to be $0$ and $t^{-\frac{3}{4}}$ is impossible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/865552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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} |
Simplify [1/(x-1) + 1/(x²-1)] / [x-2/(x+1)] Simplify: $$\frac{\frac{1}{x-1} + \frac{1}{x^2-1}}{x-\frac 2{x + 1}}$$
This is what I did.
Step 1: I expanded $x^2-1$ into: $(x-1)(x+1)$. And got: $\frac{x+1}{(x-1)(x+1)} + \frac{1}{(x-1)(x+1)}$
Step 2: I calculated it into: $\frac{x+2}{(x-1)(x+1)}$
Step 3: I multiplied $x-\frac{2}{x+1}$ by $(x-1)$ as following and I think this part might be wrong:
*
*$x(x-1) = x^2-x$. Times $x+1$ cause that's the denominator =
*$x^3+x^2-x^2-x = x^3-x$.
*After this I added the $+ 2$
*$\frac{x^3-x+2}{(x-1)(x+1)}$
Step 4: I canceled out the denominator $(x-1)(x+1)$ on both sides.
Step 5: And I'm left with: $\frac{x+2}{x^3-x+2}$
Step 6: Removed $(x+2)$ from both sides I got my UN-correct answer: $\frac{1}{x^3}$
Please help me. What am I doing wrong?
| First recall that
\[ \frac{a}{b} \pm \frac{c}{d} =\frac{ad \pm cb}{bd} \]
And
\[ \frac{\frac{a}{b}}{\frac{c}{d}} =\frac{ad}{bc} \]
Now just simplify, no fancy fractions needed:
\[
\frac{\frac{1}{x-1}+\frac{1}{x^2-1}}{x-\frac{2}{x+1}}
= \frac{\frac{(x^2-1)+(x-1)}{(x-1)(x^2-1)}}{\frac{x(x+1)-2}{x+1}}
= \frac{\frac{(x-1)(x+1)+(x-1)}{(x-1)^2(x+1)}}{\frac{(x+2)(x-1)}{x+1}}
= \frac{\frac{(x-1)(x+2)}{(x-1)^2(x+1)}}{\frac{(x+2)(x-1)}{x+1}}
= \frac{(x-1)(x+2)(x+1)}{(x-1)^3(x+1)(x+2)}
= \frac{1}{(x-1)^2}
\]
I attempted to be as clear as possible. If you'd like me to elaborate further, just let me know.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/866935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\lambda$ if $\int^{\infty}_0 \frac{\log(1+x^2)}{(1+x^2)}dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}dx$ Problem : If $\displaystyle\int^\infty_0 \frac{\log(1+x^2)}{(1+x^2)}\,dx = \lambda \int^1_0 \frac{\log(1+x)}{(1+x^2)}\,dx$ then find the value of $\lambda$.
I am not getting any clue how to proceed as if I put $(1+x^2)\,dx =t $ then its derivative is not available. Please suggest how to proceed in this. Thanks.
| From Evaluating $\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx$, you can obtain
$$\int_0^{\infty}\frac{\ln(x^2+1)}{x^2+1}dx=\pi\ln 2. $$
Define
$$I(\alpha)=\int_0^{1}\frac{\ln(\alpha x+1)}{x^2+1}dx. $$
Then
\begin{eqnarray*}
I'(\alpha)&=&\int_0^{1}\frac{x}{(\alpha x+1)(x^2+1)}dx=\int_0^1\left(\frac{x+\alpha}{(\alpha^2+1)(x^2+1)}-\frac{\alpha}{(\alpha^2+1)(x^2+1)}\right)dx.\\
&=&\frac{\pi\alpha+2\ln 2-4\ln(\alpha+1)}{4(\alpha^2+1)}.
\end{eqnarray*}
Hence
\begin{eqnarray*}
I(1)&=&\int_0^1\frac{\pi\alpha+2\ln 2-4\ln(\alpha+1)}{4(\alpha^2+1)}d\alpha\\
&=&\int_0^1\frac{\pi\alpha+2\ln 2}{4(\alpha^2+1)}d\alpha-I(1)
\end{eqnarray*}
and so
$$ I(1)=\frac{1}{2} \int_0^1\frac{\pi\alpha+2\ln 2}{4(\alpha^2+1)}d\alpha=\frac{1}{8}\pi\ln 2. $$
Thus $\lambda=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/867024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Number theory proofs regarding perfect squares How do you prove that $3n^2-1$ is never a perfect square
| Another approach. First establish that perfect squares are either $0$ or $1$ modulo $4$:
$$(2k)^2 = 4k^2 \equiv 0\pmod 4$$
$$(2k+1)^2 = 4k^2 + 4k + 1 \equiv 1 \pmod 4$$
If $n$ is even, $n = 2m$ and,
$$3n^2 - 1 = 12m^2 - 1 \equiv -1 \equiv 3 \pmod 4$$
If $n$ is odd, $n = 2m+1$ and,
$$3n^2 - 1 = 12m^2 + 12m + 3 - 1 \equiv 2 \pmod 4$$
In neither case is the condition for a perfect square modulo $4$ met.
Hence $3n^2 - 1$ is never a perfect square.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/867476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How do I reduce radian fractions? For example, I need to know $\sin (19π/12)$.
I need to use the subtraction formula. How do I get $(\text{what}) - (\text{what}) = 19π/12$? I am stuck at what are the radians
Do I divde it by something? What is the process?
| $2\pi$ radians represents a full circle turn so
*
*$\sin(2\pi+\theta)=\sin(\theta)$ and
*$\sin(2\pi-\theta)=\sin(-\theta)=-\sin(\theta)$
So $\sin \left(\frac{19}{12}\pi\right) = \sin \left(2\pi -\frac{5}{12}\pi\right) = \sin \left(-\frac{5}{12}\pi\right) = -\sin \left(\frac{5}{12}\pi\right)$
which since $\frac{5}{12}\pi=\frac{1}{6}\pi+\frac{1}{4}\pi$ makes $\sin \left(\frac{5}{12}\pi\right) = \sin \left(\frac{\pi}{6}+\frac{\pi}{4}\right) =\sin \left(\frac{\pi}{6}\right) \cos \left(\frac{\pi}{4}\right) + \cos \left(\frac{\pi}{6}\right) \sin \left(\frac{\pi}{4}\right) =\frac12 \sqrt{\frac12}+\frac{\sqrt{3}}{2} \sqrt{\frac12}$ and so gives $\sin \left(\frac{19}{12}\pi\right) = -\frac14 (\sqrt 2 + \sqrt 6) \approx -0.9659$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
What subject does this fall under (differential curves maybe?) I found a question where I don't even know where it comes from (except, vaguely, a Calc 2 class--but in my past Calc 2 class I never saw anything like this):
Find the angle of rotation needed to eliminate $xy$ from the equation $2x^2 +2\sqrt{3}xy+4y^2 +8x +8\sqrt{3}y=50$.
Can anyone tell me the basic method of doing this, or at least what topic this kind of problem falls under? Is this some kind of change of coordinates? Does that mean that you need to find two new parameters in an orthonormal basis such that the equation becomes ... linear? Or just a polynomial?
| Say that $Q\colon \mathbb R^2\rightarrow \mathbb R$ is the function that maps $(x,y)$ to $2x^2 +2\sqrt{3}xy+4y^2 +8x +8\sqrt{3}y$. Note that the set of points $(x,y)$ so that $Q(x,y)=50$ is an ellipse (picture).
Let $e_1$ be $(1,0)$ and $e_2$ be $(0,1)$, the vectors of the standard basis for $\mathbb R^2$.
Then we have:
$\begin{align*}
Q(xe_1 + ye_2) = 2x^2 +2\sqrt{3}xy+4y^2 +8x +8\sqrt{3}y &= 2\left[x^2 + \sqrt{3}xy + 2y^2\right]+8x+8\sqrt 3y\\
&= 2(x+\dfrac{\sqrt 3}{2}y)^2 + \dfrac{5}{2}y^2 + 8x + 8\sqrt{3}y.
\end{align*}$
Now we want to find a new basis $e_1', e_2'$ so that the expression of $Q$ in this basis
does not contain $xy$. Here, it suffices to change $x+\frac{\sqrt 3}{2}y$ to $x'$:
let $e_1' = e_1$ and $e_2' = -\frac{\sqrt 3}{2}e_1 + \frac{1}{2}e_2$.
Now, we have $Q(xe_1' + ye_2') = Q((x-\frac{\sqrt{3}}{2}y)e_1 + ye_2) = 2x^2 + \dfrac{5}{2}y^2 + 8x + 4\sqrt 3y$.
Basically, we tilted the $y$-axis so that it is parallel with one of the axis of the ellipse. Now, it simply remains to compute the angle between $e_2'$ and $e_2$, which you can do with a simple dot product ($e_2$ and $e_2'$ have norm $1$).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/871828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you get from this expression to the other? Briefly, I am following the solution to a problem, which I understand up to the point where this expression is reached:
$$u_x = \frac{vy + \sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)}}{1 + y^2}$$
The solution then states that this is equivalent to:
$$u_x = \frac{c^2 - v^2}{\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} - vy}$$
My question is, how do you get from one to the other? It is possibly obvious, but I just can't see it and it's driving me insane.
| \begin{align}
u_x =& \frac{c^2 - v^2}{\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} - vy}\\\\
=& \frac{\left(\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} + vy\right)(c^2 - v^2)}{\left(\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} - vy\right)\left(\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} + vy\right)}\\\\
=&\frac{\left(\sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)} + vy\right)(c^2 - v^2)}{(c^2 - v^2)(1 + y^2) }\\\\
=& \frac{vy + \sqrt{(vy)^2 + (c^2 - v^2)(1 + y^2)}}{1 + y^2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Logarithm Equality. $$\sqrt {\log_a(ax)^{\frac{1}{4}} + \log_x(ax)^{\frac{1}{4}}} + \sqrt {\log _a{(\frac{x}{a})^{\frac{1}{4}}} + \log_x (\frac{a}{x})^\frac{1}{4}} = a,$$ for $a>0$ and different than 1... I keep getting $a = 1$, but that cannot be. I use log identities to transform the above into $$\sqrt {\frac{1}{2} + \frac{2\ln | ax |}{4\ln | ax |}} + \sqrt {\frac{1}{4}\frac{2\ln | ax |}{\ln | ax |} - \frac{1}{2}} = a $$ which means $a = 1$. Maybe I am overlooking something, but I do not see what.
| $$\sqrt {
\log_a\left((ax)^\frac 14\right)
+ \log_x\left((ax)^\frac 14\right)
} + \sqrt {
\log_a\left(\left(\frac xa\right)^\frac 14\right)
+ \log_x\left(\left(\frac ax \right)^\frac 14\right)
} = a$$
$$\sqrt {
\frac 14\log_a\left(ax\right)
+ \frac 14\log_x\left(ax\right)
} + \sqrt {
\frac 14\log_a\left(\frac xa\right)
+ \frac 14\log_x\left(\frac ax\right)
} = a$$
$$\sqrt {
\log_a\left(ax\right)
+ \log_x\left(ax\right)
} + \sqrt {
\log_a\left(\frac xa\right)
+ \log_x\left(\frac ax\right)
} = 2a$$
$$\sqrt {
\log_a(a) + \log_a(x)
+ \log_x(a) + \log_x(x)
} + \sqrt {
\log_a(x) - \log_a(a)
+ \log_x(a) - \log_x(x)
} = 2a$$
$$\sqrt {
1 + \underbrace{\log_a(x) + \log_x(a)}_z + 1
} + \sqrt {
\log_a(x) - 1 + \log_x(a) - 1
} = 2a$$
$$\sqrt {
z + 2
} + \sqrt {
z - 2
} = 2a$$
$$z + 2 + 2\sqrt {z^2 - 4} + z - 2 = 4a^2$$
$$2\sqrt {z^2 - 4}= 4a^2 - 2z$$
$$4z^2 - 16 = 16a^4 - 16a^2z^2 + 4z^2$$
$$\frac{a^4 + 1}{a^2} = z^2$$
$$\frac{a^4 + 1}{a^2} = \left(\frac{\log(a)}{\log(x)} + \frac{\log(x)}{\log(a)}\right)^2$$
$$\frac{a^4 + 1}{a^2} = \frac{\log(a)^2}{\log(x)^2} + 2 + \frac{\log(x)^2}{\log(a)^2}$$
$$\frac{a^4 -2a^2 + 1}{a^2} = \frac{\log(a)^2}{\underbrace{\log(x)^2}_y} + \frac{\log(x)^2}{\underbrace{\log(a)^2}_w}$$
$$yw\left(\frac{a^2 - 1}{a}\right)^2 = w^2 + y^2$$
$$\text{...Etc}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show that $(a+b)^p\le 2^p (a^p+b^p)$ If I may ask, how can we derive that $$(a+b)^p\le 2^p (a^p+b^p)$$ where $a,b,p\ge 0$ is an integer?
| In fact we can make the stronger claim:
$$(a+b)^p\le 2^{p-1} (a^p+b^p)$$
if $a,b \ge 0$. As Mohammad Khosravi points out, this is equivalent to
$$(x+1)^p\le 2^{p-1} (x^p+1)$$
if $x \ge 0$. We prove this by induction. The case $p=1$ is easy: $x+1 \le x+1$. So suppose we have established that
$$(x+1)^{p-1}\le 2^{p-2} (x^{p-1}+1)$$
Then we have:
$$\begin{align}
(x+1)^p & = (x+1)(x+1)^{p-1}\\
& \le 2^{p-2}(x+1)(x^{p-1}+1)
\end{align}$$
Now we just have to show that $2^{p-2}(x+1)(x^{p-1}+1) \le 2^{p-1} (x^p+1)$, or equivalently that $2(x^p+1) - (x+1)(x^{p-1}+1) \ge 0 $:
$$\begin{align}
2(x^p+1) - (x+1)(x^{p-1}+1) & = x^p-x^{p-1}-x+1 \\
& = (x-1)(x^{p-1}-1) \\
& = (x-1)^2(x^{p-2} + x^{p-3} + \ldots + x + 1) \\
& \ge 0
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/872276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 1
} |
if $\frac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$,find $a_{n}$ Let
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\sum_{n=0}^{\infty}a_{n}x^n$$
Find the closed form $$a_{n}$$
since
$$(1-x^4)(1-x^3)(1-x^2)=(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)$$
so
$$\dfrac{1}{(1-x^4)(1-x^3)(1-x^2)}=\dfrac{1}{(1-x)^3(1+x+x^2+x^3)(1+x+x^2)(1+x)}$$
then I feel very ugly,can you someone have good partial fractions methods by hand?
because I take an hour to solve this problem.
ago I have solve $$x+2y+3z=n$$ the number of the positive integer solution $a_{n}$ I found
$$a_{n}=\left[\dfrac{(n+3)^2}{12}\right]$$
Thank you
| Incomplete answer (maybe it can be used):
If $\left|z\right|<1$ then $\frac{1}{1-z}=1+z+z^{2}+\cdots$
Applying this for $z=x^{4},x^{3},x^{2}$ you find:
$$\frac{1}{\left(1-x^{4}\right)\left(1-x^{3}\right)\left(1-x^{2}\right)}=\left(1+x^{4}+x^{8}+\cdots\right)\left(1+x^{3}+x^{6}+\cdots\right)\left(1+x^{2}+x^{4}+\cdots\right)$$
Then $a_{n}$ must equalize the cardinality of $\left\{ \left(p,q,r\right)\in\left\{ 0,1,2,\dots\right\} \mid4p+3q+2r=n\right\} $ (as Macavity comments).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/875792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
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Integral: $\int_0^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\,dx$ I am trying to solve the following by elementary methods:
$$\int_0^{\pi} \frac{x}{x^2+\ln^2(2\sin x)}\,dx$$
I wrote the integral as:
$$\Re\int_0^{\pi} \frac{dx}{x-i\ln(2\sin x)}$$
But I don't find this easier than the original integral. I have seen solutions which make use of complex analysis but I am interested in elementary approaches.
Any help is appreciated. Thanks!
| Here is an approach without using contour integration (Cauchy's theorem).
I've found the following result.
Theorem. Let $a$ be any real number.
Then
$$
\begin{align}
\displaystyle \int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 e^{a} \sin x)}\mathrm{d}x
& = \, \frac{2 \pi^2}{\pi^2+4a^2},\tag1\\\\
\int_{0}^{\pi} \frac{\ln(2 e^{a} \sin x)}{x^2+\ln^2(2 e^{a} \sin x)}\mathrm{d}x & = \frac{4\pi a}{\pi^2+4a^2}.\tag2
\end{align}
$$
from which, by putting $a=0$, you deduce the evaluation of the desired integral:
Example $1 (a)$. $$ \begin{align}
\int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 \sin x)}\:\mathrm{d}x
& = 2 \tag3
\end{align}
$$
and
Example $1 (b)$. $$ \begin{align}
\int_{0}^{\pi} \frac{\ln(2\sin x)}{x^2+\ln^2(2\sin x)}\mathrm{d}x& = 0. \\ \tag4
\end{align}
$$
Proof of the Theorem. We consider throughout the principal value of $\log(z)$ defined for $z\neq 0$ by
$$
\log(z)= \ln |z| + i \arg z, \, - \pi < \arg z \leq \pi,
$$
where $\ln$ is the base-$e$ logarithm $\ln e = 1$.
Let $a$ be any real number. Since $\displaystyle z \mapsto i \pi/2+a+\log(1+z) $ is analytic on $|z| < 1$ non-vanishing at the point $z=0$, one may obtain the following power series expansion
$$
\begin{align}
\displaystyle \frac{1}{i \pi/2+a+\log \left(1+z\right)} - \frac{1}{i \pi/2+a}
= \sum_{n=1}^{\infty} \displaystyle \frac{a_n(\alpha)}{n!} z^n,\tag5
\end{align}
$$ with
$$ \displaystyle a_1(\alpha) = -\alpha^2,
\quad a_2(\alpha) = \alpha^3+\alpha^2/2, \, ... ,
$$
$\displaystyle {a}_n(\cdot)$ being a polynomial of degree $n+1$ in $\displaystyle \alpha$,
where we have set $\displaystyle \alpha:=1/(i \pi/2+a)$.
Plugging $z=-e^{2ix}$ in $(3)$, for $0<x<\pi$,
noticing
$$
i x +\ln(2e^{a} \sin x) = \log\left(ie^{a}\left(1-e^{2ix} \right) \right),
$$
separating real and imaginary parts, gives the Fourier series expansions:
$$
\begin{align}
\displaystyle \frac{x}{x^2 +\ln^2(2e^{a}\sin x)}
& = \frac{2\pi}{\pi^2+4a^2}+ \sum_{n=1}^{\infty}(-1)^n \left( \frac{a_n^{-}(\alpha)}{n!} \cos (2nx)+\frac{a_n^{+}(\alpha)}{n!} \sin (2nx)\right)
\\ \frac{\ln(2e^{a}\sin x)}{x^2 +\ln^2(2e^{a}\sin x)}
& = \frac{4a}{\pi^2+4a^2}+ \sum_{n=1}^{\infty}(-1)^n \left( \frac{a_n^{+}(\alpha)}{n!} \cos (2nx)-\frac{a_n^{-}(\alpha)}{n!} \sin (2nx)\right)
\end{align}
$$ with $\displaystyle a_n^{+}(\alpha):= {\mathfrak{R}}a_n(\alpha)$ and $\displaystyle a_n^{-}(\alpha):= {\mathfrak{I}}a_n(\alpha)$.
Now, a termwise integration with respect to $x$ from 0 to $\pi$, justified by the convergence of the series $\sum_{n=1}^{\infty} \displaystyle \frac{a_n(\alpha)}{n!} $, leads to the Theorem due to
$$
\begin{align}
\displaystyle \int_{0}^{\pi} \cos(2nx)\:\mathrm{d}x = \int_{0}^{\pi} \sin(2nx)\:\mathrm{d}x = 0.
\end{align}
$$
One may observe that, by uniqueness of the Fourier coefficients, you have the following closed forms.
Example $2 (a)$.
$$
\begin{align}
\int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 \sin x)}\cos^{2} x \:\mathrm{d}x
& = 1, \tag6 \\
\int_{0}^{\pi} \frac{x}{x^2+\ln^2(2 \sin x)}\sin^{2} x\:\mathrm{d}x & = 1. \tag7
\end{align}
$$
$$
$$
Example $2 (b)$.
$$
\begin{align}
\int_{0}^{\pi} \frac{\ln(2 \sin x)}{x^2+\ln^2(2 \sin x)} \cos^{2} x \:\mathrm{d}x
& = -\frac{1}{\pi}, \tag8 \\
\int_{0}^{\pi} \frac{\ln(2 \sin x)}{x^2+\ln^2(2 \sin x)}\sin^{2} x\:\mathrm{d}x & = \frac{1}{\pi}. \tag9
\end{align}
$$
A general result does exist.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "24",
"answer_count": 1,
"answer_id": 0
} |
When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$... When the numerator of a fraction is increased by $4$, the fraction increases by $2/3$. What is the denominator of the fraction?
I tried,
Let the numerator of the fraction be $x$ and the denominator be $y$.
Accordingly, $$\frac{x+4}y=\frac xy+\frac 23$$
I am not able to find the second equation.
| In general you are correct: to solve for two unknowns, you would usually need two equations. But in this case you are lucky, and the one equation gives a solution for the one unknown you are asked to find.
$\dfrac{x+4}{y}=\dfrac{x}{y}+\dfrac{2}{3}$ so multiplying both sides by $3y$ gives $3x+12 = 3x+2y$ making the denominator $y=6$.
It gives no specific solution for $x$, for example: $\dfrac{5+4}{6}=\dfrac{5}{6}+\dfrac{2}{3}$ and $\dfrac{7+4}{6}=\dfrac{7}{6}+\dfrac{2}{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/880466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 2
} |
Solving recurrence relation: Product form Please help in finding the solution of this recursion.
$$f(n)=\frac{f(n-1) \cdot f(n-2)}{n},$$
where $ f(1)=1$ and $f(2)=2$.
| I believe the formula is, for $n\geq 3$,
$$f(n) = 2^{F_{n-1}}\prod_{k=3}^{n}k^{-F_{n-k+1}}$$
where $F_1,F_2,F_3,\ldots$ is the standard Fibonacci sequence whose first two terms are $1$.
You can see this by just writing out the first several terms. For example,
$$f(9) =\dfrac{2^{21}}{3^{13}4^8 5^5 6^3 7^2 8^1 9^1}
$$
Addendum (Inductive Proof): The base case ($n=3$) is trivial, since
$$f(3)=\dfrac{f(2)f(1)}3=\dfrac{2\cdot 1}3 = \frac23=2^1 3^{-1}=2^{F_2}3^{-F_1}$$
For the inductive step, suppose the formula holds for $3,\ldots,n$. We show it holds for $n+1$ as follows:
$$
f(n+1)=\frac{f(n)f(n-1)}{n+1}=(n+1)^{-1}\cdot \left(2^{F_{n-1}}\prod_{k=3}^{n}k^{-F_{n-k+1}}\right) \cdot \left(2^{F_{n-2}}\prod_{k=3}^{n-1}k^{-F_{n-k}}\right)
$$
$$
= 2^{F_{n-1}+F_{n-2}}(n+1)^{-F_2}\cdot \left(n^{-F_1}\prod_{k=3}^{n-1}k^{-F_{n-k+1}}\right)\cdot \left(\prod_{k=3}^{n-1}k^{-F_{n-k}}\right)
$$
$$
= 2^{F_n}(n+1)^{-F_2}n^{-F_1}\prod_{k=3}^{n-1}k^{-F_{n-k+1}-F_{n-k}}
= 2^{F_n} \prod_{k=3}^{n+1}k^{-F_{n-k+2}}
$$
$$
=2^{F_{(n+1)-1}} \prod_{k=3}^{n+1}k^{-F_{(n+1)-k+1}}
$$
as required. $\blacksquare$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Prove that $\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$
Given $a$, $b$ and $c$ are positive real numbers. Prove that:$$\sum \limits_{cyc}\frac {a}{(b+c)^2} \geq \frac {9}{4(a+b+c)}$$
Additional info: We can't use induction. We should mostly use Cauchy inequality. Other inequalities can be used rarely.
Things I have done so far: The inequality look is similar to Nesbitt's inequality.
We could re-write it as: $$\sum \limits_{cyc}\frac {a}{(b+c)^2}(2(a+b+c)) \geq \frac{9}{2}$$
Re-write it again:$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \frac{9}{2}$$
Cauchy appears:
$$\sum \limits_{cyc}\frac {a}{(b+c)^2}\sum \limits_{cyc}(b+c) \geq \left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2$$
So, if I prove $\left(\sum \limits_{cyc}\sqrt\frac{a}{b+c}\right)^2 \geq \frac {9}{2}$ then problem is solved.
Re-write in semi expanded form:$$2\left(\sum \limits_{cyc}\frac{a}{b+c}+2\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)}\right) \geq 9$$
We know that $\sum \limits_{cyc}\frac{a}{b+c} \geq \frac {3}{2}$.So$$4\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq 6$$
So the problem simplifies to proving this $$\sum \limits_{cyc}\sqrt\frac{ab}{(b+c)(c+a)} \geq \frac{3}{2}$$
And I'm stuck here.
| Without loss of generality, we assume that $a+b+c=1$. Then,
\begin{align*}
\frac{a}{(b+c)^2} + \frac{b}{(a+c)^2} + \frac{c}{(b+a)^2} &= \frac{a^2}{a(b+c)^2} + \frac{b^2}{b(a+c)^2} + \frac{c^2}{c(b+a)^2}\\
&\ge \left(\frac{a}{b+c} + \frac{b}{a+c}+\frac{c}{b+a} \right)^2\\
&=\left(\frac{1}{b+c} + \frac{1}{a+c}+\frac{1}{b+a} -3 \right)^2\\
&\ge \left(\frac{3}{2} -3 \right)^2\\
&=\frac{9}{4}.
\end{align*}
| {
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"answer_count": 6,
"answer_id": 5
} |
Prove $\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$
If $a$, $b$ and $c$ are positive real numbers, prove that:
$$\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2} \geq \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}$$
Additional info:We can use AM-GM and Cauchy inequalities mostly.We are not allowed to use induction.
Things I have tried so far:
Using Cauchy inequality I can write:$$\left(\frac{a^3}{b^2}+\frac{b^3}{c^2}+\frac{c^3}{a^2}\right)(a+b+c) \geq \left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)^2$$
but I can't continue this.I tried expanded form:$$\sum \limits_{cyc} \frac{a^5c^2}{a^2b^2c^2} \geq \sum \limits_{cyc} \frac{a^3c}{abc}$$
Which proceeds me to this Cauchy:$$\sum \limits_{cyc} \frac{a^5c^2}{abc}\sum \limits_{cyc}a(abc)\geq \left(\sum \limits_{cyc}a^3c\right)^2$$
I can't continue this one too.
The main Challenge is $3$ fraction on both sides which all of them have different denominator.and it seems like using Cauchy from first step won't leads to anything good.
| The next thing you can try is Cauchy again
$$\left(\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\right)(b+c+a)\geq (a+b+c)^2.$$
So
$$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge a+b+c.$$
Now you can eliminate $a+b+c$ from your first inequality.
| {
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
How to describe $G/U$? Let $G=SL_2(\mathbb{C})$ and let $U = \{\left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right): x \in \mathbb{C}\}$. We have an action of $U$ on $G$ by right multiplication. By definition, $G/U$ is the set of all $U$-orbits under this action. How to describe $G/U$ as a set?
Take $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in G$. The orbit of $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right)$ is the set $$ \{ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right)\left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right): x \in \mathbb{C} \}. $$
But I don't know how to classify orbits in $G/U$. Thank you very much.
Edit: $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in G$ and $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right) \in G$ are in the same orbit. Indeed, Suppose that
$\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right)$ = $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right)$. Then $x = (b'-b)/a$. We also have $cx+d=(cb'+ad-bc)/a=(cb'+1)/a$. Since $ad'-b'c=1$, $(cb'+1)/a=d'$. Therefore $x = (b'-b)/a$ is a solution of $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \left( \begin{matrix} 1 & x \\ 0 & 1 \end{matrix} \right)$ = $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right)$. Hence $\left( \begin{matrix} a & b \\ c & d \end{matrix} \right) \in G$ and $\left( \begin{matrix} a & b' \\ c & d' \end{matrix} \right) \in G$ are in the same orbit.
I think that $G/U = \{ \left( \begin{matrix} a & b \\ c & d \end{matrix} \right): a, c \in \mathbb{C}, b,d \text{ are fixed }\}$. Is this correct? Thank you very much.
| For any $a \ne 0,$ and $c \in \mathbb{C},$ the orbit of $\begin{pmatrix} a & 0 \\ c & a^{-1} \end{pmatrix}$ under $U$ gives you $$\Big\{\begin{pmatrix} a & ax \\ c & cx+a^{-1} \end{pmatrix}: \; x \in \mathbb{C}\Big\},$$ so these are all distinct orbits ($a$ and $c$ are uniquely determined by the left column) and run through all matrices in $SL_2(\mathbb{C})$ with nonzero upper-left element.
For any $c \ne 0,$ the orbit of $\begin{pmatrix} 0 & -c^{-1} \\ c & 0 \end{pmatrix}$ is $$\Big\{\begin{pmatrix} 0 & -c^{-1} \\ c & cx \end{pmatrix}: \; x \in \mathbb{C} \Big\},$$ so these are distinct orbits and run through all matrices in $SL_2(\mathbb{C})$ with left column $\begin{pmatrix} 0 \\ c \end{pmatrix}.$
In other words, the orbit is determined uniquely by the left column of the matrix
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
Simplifying from POS using boolean algeabra I have a boolean function, f expressed in the Product of Sum form.
$$f = (A+B+C)\cdot(A+B+ \overline C)\cdot(\overline A + \overline B + \overline C) $$
On simplification I get,
$$ f = ((A+B) + (C \cdot\overline C))\cdot (\overline A + \overline B + \overline C) $$
$$ = (A+B) \cdot (\overline A + \overline B + \overline C) $$
However, I do not know how to proceed after this step.
The answer is given as $ (A+B)\cdot(A+\overline C)$
Any advice on how to proceed ?
| Your answer is correct, while the given answer is incorrect (or there is a typo somewhere). Indeed, let $A = B = C = 1$. Then observe that:
\begin{align*}
f &= (A+B+C)\cdot(A+B+ \overline C)\cdot(\overline A + \overline B + \overline C) \\
&= (1 + 1 + 1) \cdot (1 + 1 + 0) \cdot (0 + 0 + 0) \\
&= 1 \cdot 1 \cdot 0 \\
&= 0
\end{align*}
while on the other hand, the proposed answer yields:
\begin{align*}
(A+B)\cdot(A+\overline C)
&= (1 + 1) \cdot (1 + 0) \\
&= 1 \cdot 1 \\
&= 1
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/884796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Recognizing the sequence 1/16, 1/8, 3/16, 1/4, 5/16, ...
What is the missing number?
$$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}, \ \ \ [?]$$
$$A. \frac{5}{4}\quad B. \frac{3}{4}\quad C. \frac{5}{8}\quad D. \frac{3}{8}$$
Spoiler: Answer is $D$, but I don't know why.
Thanks
| Another sequence-recognizing technique
is to look at the difference between consecutive terms.
In this case,
$\frac{1}{8}-\frac{1}{16}
= \frac{1}{16}
$,
$\frac{3}{16}-\frac{1}{8}
= \frac{1}{16}
$,
$\frac{1}{4}-\frac{3}{16}
= \frac{1}{16}
$,
and
$\frac{5}{16}-\frac{1}{4}
= \frac{1}{16}
$.
Since the difference between
consecutive terms is
$\frac{1}{16}$,
the next term should be
$\frac{5}{16}+\frac{1}{16}
=\frac{6}{16}
=\frac{3}{8}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/885137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
If $f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$, then what is the value of $f'(1)$
Find $f'(1)$ if $$f\left(x-\frac{2}{x}\right) = \sqrt{x-1}$$
My attempt at the question:
Let $(x-\dfrac{2}{x})$ be $g(x)$
Then $$f(g(x)) = \sqrt{x-1} $$
Differentiating with respect to x:
$$f'(g(x))\cdot g'(x) = \frac{1}{2\sqrt{x-1}} $$
Therefore
$$f'(g(x)) = \frac{1}{2(g'(x))\sqrt{x-1}} $$
Finding the value of $x$ for which $g(x) = 1$ : $ x=( -1) , x=2$
But as $x\neq (-1)$, as $\sqrt{x-1}$ becomes indeterminant, substitute x = 2.
we get: $$f'(1) = \frac13 $$
Which is not the correct answer. The correct answer is supposedly $1$. Need some help as to why my method is wrong.
| Let us assume ${\left(x-\frac{2}{x}\right) = 1, \text{the part inside}}$
It is found that $x = 2$ or $x = -1$
Taking the positive value of $x$
Therefore $f(1) = \sqrt {x-1}$ and
Thus $$f'(1) = \frac {1}{2}\cdot(x-1)^{\frac{1}{2}-1},$$ $ \text{1 for the value inside the function}$
$$\Rightarrow \frac{1}{2}\cdot(x-1)^{\frac{-1}{2}}$$
Putting the positive value $2$, we get
$f'(1) = \frac{1}{2}$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Recurrence of the form $2f(n) = f(n+1)+f(n-1)+3$ Can anyone suggest a shortcut to solving recurrences of the form, for example:
$2f(n) = f(n+1)+f(n-1)+3$, with $f(1)=f(-1)=0$
Sure, the homogenous solution can be solved by looking at the characteristic polynomial $r^2-2x+1$, so that in general a solution for the homogenous equation is of the form $f^h(n) = c_1+c_2n$. But how does one deal with the constant 3 in this case?
| we can use the generating function $F(x)=\sum_{k=1}^{\infty} f(k)x^k$
the recurrence relation may be written:
$$2f(n+1) = f(n+2)+f(n)+3$$
so multiplying by $x^{n+2}$ we have
$$2xf(n+1)x^{n+1} = f(n+2)x^{n+2}+x^2f(n)x^n+3x^{n+2}$$
and summing for $n=0$ to $\infty$
$$
2x(F(x)-f(0)) = F(x)-xf(1)-f(0) + x^2F(x) +3x^2(1-x)^{-1}
$$
since $f(1)=0$ and $f(0)=\frac32$ this gives (writing $F$ for $F(x)$)
$$
-(1-x)^2F= 3x - \frac32 +3x^2(1-x)^{-1}= \frac32(2x-1 +2x^2(1-x)^{-1})=\frac32(3x-1)(1-x)^{-1}
$$
so
$$
F(x)=-\frac32(3x-1)(1-x)^{-3}
$$thus for $n \ge 1$ we have $f(n)$, the coefficient of $x^n$ is $-\frac32 \left(3 \binom{n+1}2 - \binom{n+2}2 \right)$
which simplifies to
$$
f(n) = -\frac32 (n^2-1)
$$
| {
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"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 3
} |
Evaluating $ \int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx$ I am trying to evaluate the indefinite integral of
$$\int x \sqrt{\frac{1-x^2}{1+x^2}} \, dx.$$
The first thing I did was the substitution rule:
$u=1+x^2$, so that $\displaystyle x \, dx=\frac{du}2$ and $1-x^2=2-u$. The integral then transforms to
$$\int \sqrt{\frac{2-u}{u}} \, \frac{du}2$$
or $$\frac 12 \int \sqrt{\frac 2u - 1} \, du$$
I'm a bit stuck here. May I ask for help on how to proceed?
| $\bf{My\; Solution::}$ Let $\displaystyle I = \int x\sqrt{\frac{1-x^2}{1+x^2}}dx\;,$ Now Substitute $x^2=t\;,$ Then $\displaystyle xdx = \frac{dt}{2}$
So $\displaystyle I = \frac{1}{2}\int \sqrt{\frac{1-t}{1+t}}dt = \frac{1}{2}\int \frac{\left(1-t\right)}{\sqrt{1-t^2}}dt =\frac{1}{2}\int\frac{1}{\sqrt{1-t^2}}dt-\frac{1}{2}\int \frac{t}{\sqrt{1-t^2}}dt$
So $\displaystyle I = \frac{1}{2}\sin^{-1}(t)-\frac{1}{2}\sqrt{1-t^2}+\mathcal{C} = \frac{1}{2}\sin^{-1}(\sqrt{x})-\frac{1}{2}\sqrt{1-x}+\mathcal{C}$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 3
} |
Solving the logarithimic inequality $\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$ I tried solving the logarithmic inequality:
$$\log_2\frac{x}{2} + \frac{\log_2x^2}{\log_2\frac{2}{x} } \leq 1$$
several times but keeping getting wrong answers.
| Put $u=\log_2\left(\dfrac x2 \right)=\log_2x-1$
Note that $\log_2(x^2)=2(u+1)$, and $\log_2\left(\dfrac 2x\right)=-u$.
Hence inequality becomes
$$ \begin{align}
u-\dfrac {2(u+1)}u \leq1\\
\dfrac{u^2-3u-2}u \leq0\\
\dfrac{(u-\alpha)(u-\beta)}u \leq 0\\
\end{align}$$
where $\alpha=\dfrac {3+\sqrt{17}}2,\ \beta=\dfrac {3-\sqrt{17}}2\ $
$$
\begin{align}
u\leq\beta &,\ \quad 0\leq u\leq\alpha,\\\
\log_2\left(\frac x2\right)\leq \dfrac {3-\sqrt{17}}2 &,\quad \ 0\leq\log_2\left(\frac x2\right)\leq\dfrac{3+\sqrt{17}}2\\
\log_2x\leq \frac{5-\sqrt{17}}2&,\quad \ 1\leq \log_2 x\leq \frac{5+\sqrt{17}}2\\
\end{align}$$
As $x>0$,
$$0< x\leq2^{\frac{5-\sqrt{17}}2},\ \quad 2\leq x\leq2^{\frac{5+\sqrt{17}}2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Solving awkward quadratic equation to obtain "nice" solution. I would like to solve the following quadratic equation to get a "nice" analytic solution for $\rho$.
$\rho^2(r\sin\theta-2nr^2)+\rho(2nr^3-2r^2\sin\theta-2\sin\theta+2nr)-2nr^2+3r\sin\theta=0$
where $r=1-\cos\theta$ (I cannot see how this could be used to simplify the quadratic equation.)
I would hope that the solution to be found which be of a simple form as that would correspond nicely to the larger problem I am working on.
Also $\theta=\pi/n$
EDIT: I would hope that the solution $\displaystyle \rho=1-\frac{sin(\frac{\pi}{n})}{n}$ would be one of the solutions to the quadratic.
| It is not a particularly simple form.
Solution:
$\rho = (R \pm S)/Q$
Where
$Q = \textrm{ sin}( \frac{\pi}{2 n} )^3( 32 n - 16 \textrm{ cot}(\frac{\pi}{2 n}))$
$R = -12 \textrm{ cos}(\frac{\pi}{2 n}) + 6 \textrm{ cos}(\frac{3 \pi}{2 n} ) -
2 \textrm{ cos}( \frac{5 \pi}{2 n}) +
2 n \left(14 \textrm{ sin}( \frac{\pi}{2 n}) - 5 \textrm{ sin}( \frac{3\pi}{2 n}) +
\textrm{ sin}( \frac{5\pi}{2 n}) \right)$
$S = \sqrt{2} \sqrt{22 - 58 n^2 + 6 (1 + 17 n^2) \textrm{ cos}( \frac{\pi}{n} ) -
56 n^2 \textrm{ cos}(\frac{2\pi}{n}) + (9 + 11 n^2) \textrm{ cos}(\frac{3\pi}{n}) +
2 (-3 + n^2) \textrm{ cos}(\frac{4\pi}{n}) - (-1 + n^2) \textrm{ cos}(\frac{5\pi}{n}) -
20 n \textrm{ sin}( \frac{\pi}{n}) + 8 n \textrm{ sin}( \frac{2\pi}{n}) - 22 n \textrm{ sin}(\frac{3\pi}{n}) +
12 n \textrm{ sin}( \frac{4\pi}{n} ) - 2 n \textrm{ sin}( \frac{5\pi}{n})} $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/888335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then $|z|=$? If z is a complex number and $ \frac {z^2 + z+ 1} {z^2 -z +1}$ is purely real then find the value of $|z|$ .
I tried to put $ \frac {z^2 + z+ 1} {z^2 -z +1} =k $ then solve for $z$ and tried to find |z|, but it gets messy and I am stuck.
The answer given is |z|=1
| Proposition. $\frac{z^2+z+1}{z^2-z+1}$ is real if and only if $z\in\Bbb R$ or $|z|=1$.
I'll let you handle the $\Leftarrow$ direction. For $\Rightarrow$, write $k=\frac{z^2+z+1}{z^2-z+1}=\frac{(z+z^{-1})+1}{(z+z^{-1})-1}\implies z+z^{-1}=\frac{k+1}{k-1}$.
When is $z+z^{-1}$ real for nonreal $z$? (What's its imaginary part?)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/890723",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Proving or disproving inequality $ \frac{xy}{z} + \frac{yz}{x} + \frac{zx}{y} \ge x + y + z $
Given that $ x, y, z \in \mathbb{R}^{+}$, prove or disprove the
inequality
$$ \dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z $$
I have rearranged the above to:
$$ x^2y(y - z) + y^2z(z - x) + z^2x(x - y) \ge 0 \\
\text{and, } \dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} \ge \dfrac{1}{xy} + \dfrac{1}{xz} + \dfrac{1}{yz} $$
What now? I thought of making use of the arithmetic and geometric mean properties:
$$
\dfrac{x^2 + y^2 + z^2}{3} \ge \sqrt[3]{(xyz)^2} \\
\text{and, } \dfrac{x + y + z}{3} \ge \sqrt[3]{xyz}
$$
but I am not sure how, or whether that'd help me at all.
| Consider this inequality:
$$(a-b)^2 + (b-c)^2 + (c-a)^2 \ge 0$$
Expand and simplify the above expression to get:
$$a^2 + b^2 + c^2 \ge ab + bc + ca$$
Substitute $a = \dfrac{1}{x}, b = \dfrac{1}{y}$ and $c = \dfrac{1}{z}$:
$$\dfrac{1}{x^2} + \dfrac{1}{y^2} + \dfrac{1}{z^2} \ge \dfrac{1}{xy} + \dfrac{1}{yz} + \dfrac{1}{zx}$$
Multiply both sides with $xyz$ to get
$$\dfrac{xy}{z} + \dfrac{yz}{x} + \dfrac{zx}{y} \ge x + y + z$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/891997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Fermat: Prove $a^4-b^4=c^2$ impossible Prove by infinite descent that there do not exist integers $a,b,c$ pairwise coprime such that $a^4-b^4=c^2$.
| Lemma. If $a,b,c$ are coprime and $a^2+b^2=c^2$, then
$$ \{a,b\}=\{2pq,p^2-q^2\},\quad c=p^2+q^2 $$
with $p,q$ coprime and not both odd.
This follows from the fact that we are looking for rational points on the curve $x^2+y^2=1$, that allows the parametrization $\left(\frac{2t}{1-t^2},\frac{1-t^2}{1+t^2}\right)$. Assuming that $P=(x_0,y_0)$ is a rational point, we consider the lines through $P$ with a rational slope. They all must intersect the circle in a second point with rational coordinates, due to Viète's theorem.
The lemma hence gives two possibilities:
1. $a^2=p^2+q^2,\quad c=2pq,\quad b^2=p^2-q^2.$
2. $a^2=p^2+q^2,\quad c=p^2-q^2,\quad b^2=2pq.$
The infinite descent starts as soon as we notice that in 1. $p$ must be the sum of two squares, since $p^2=b^2+q^2$, while in 2. $p+q$ must be the sum of two squares since $a^2+b^2=(p+q)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/893544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\int\frac{dx}{x\sqrt{x^2-2}}$. The exercise is:
Calculate:$$\int\frac{dx}{x\sqrt{x^2-2}}$$
My first approach was:
Let $z:=\sqrt{x^2-2}$ then $dx = dz \frac{\sqrt{x^2-2}}{x}$ and $x^2=z^2+2$ $$\int\frac{dx}{x\sqrt{x^2-2}} = \int\frac{1}{x^2}dz = \int\frac{1}{z^2+2}dz = \frac{1}{2}\int\frac{1}{(\frac{z}{\sqrt{2}})^2 + 1}dz = \frac{1}{2} \arctan(\frac{z}{\sqrt{2}})$$
$$ = \frac{1}{2}\arctan(\frac{\sqrt{x^2-2}}{\sqrt{2}})$$
Yet when e.g. Maple derives the expression, it turns out my solution is false. I also cannot find the mistake, and I'd be glad if someone could point it out to me.
I am interested in your solutions, but in another approach I substituted $z:=\frac{\sqrt{2}}{x}$ and multiplied the integrand with $\frac{\sqrt{2}x}{\sqrt{2}x}$ and it worked out just fine.
| The mistake is that you are integrating over $z$,you can do a substitution $$u=\frac{z}{\sqrt{2}},dz=\sqrt{2}du\\\frac{1}{2}\int\frac{\sqrt{2}du}{u^2+1}=\frac{\sqrt{2}}{2}\arctan(u)=\frac{1}{\sqrt{2}}\arctan(\frac{\sqrt{x^2-2}}{\sqrt{2}})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/895602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Inequality involving a finite sum this is my first post here so pardon me if I make any mistakes.
I am required to prove the following, through mathematical induction or otherwise:
$$\frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$$
I tried using mathematical induction through:
$Let$ $P(n) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{n}} < 2{\sqrt{n}}$
$Since$ $P(1) = \frac{1}{\sqrt1} < 2{\sqrt{1}}, and$ $P(k) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}} < 2{\sqrt{k}},$
$P(k+1) = \frac{1}{\sqrt1} + \frac{1}{\sqrt2} + \frac{1}{\sqrt3} + ... + \frac{1}{\sqrt{k}}+ \frac{1}{\sqrt{k+1}} < 2{\sqrt{k+1}}$
Unfortunately, as I am quite new to induction, I couldn't really proceed from there. Additionally, I'm not sure how to express ${\sqrt{k+1}}$ in terms of ${\sqrt{k}}$ which would have helped me solve this question much more easily. I am also aware that this can be solved with Riemann's Sum (or at least I have seen it being solved in that way) but I do not remember nor quite understand it.
| $P(k+1)=P(k)+\dfrac{1}{\sqrt{k+1}}\leq 2\sqrt{k}+\dfrac{1}{\sqrt{k+1}}=\dfrac{2\sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}}$, in the other hand we have $2\sqrt{k}\sqrt{k+1}\leq k+k+1=2k+1$ (using $2ab\leq a^2+b^2$), now we get $2\sqrt{k}\sqrt{k+1}+1\leq 2(k+1)$, and finally $\dfrac{2\sqrt{k}\sqrt{k+1}+1}{\sqrt{k+1}}\leq 2\sqrt{k+1}$, so $P(k+1)\leq 2\sqrt{k+1}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Proof by Induction: $(1+x)^n \le 1+(2^n-1)x$ I have to prove the following by induction:
$$(1+x)^n \le 1+(2^n-1)x$$
for $n \ge 1$ and $0 \le x \le 1$.
I start by showing that it's true for $n=1$ and assume it is true for one $n$.
$$(1+x)^{n+1} = (1+x)^n(1+x)$$
by assumption:
$$\le (1+(2^n-1)x)(1+x)$$
$$= 1+(2^n-1)x+x+(2^n-1)x^2$$
That is what I was able to do on my own. I do not understand the next step of the solution (they are now using $x$ instead of $x^2$ at the end):
$$Because\,0 \le x \le 1$$
$$(1+x)^{n+1} \le 1 + (2^n-1)x+x+(2^n-1)x$$
$$=1+2(2^n-1)x+x$$
$$=1+(2^{n+1}-1)x$$
Because $0 \le x \le 1$ I assume that $(2^n-1)x^2 < (2^n-1)x$. So in my opinion, I just proved that $(1+x)^{n+1}$ is less than something greater than what I had to prove initially. Why is it enough to prove the original problem?
| I'm not exactly clear as to what your final sentence means. But, since $0\le x\le 1$, we know that $x^2\le x\cdot x\le 1\cdot x\le x$, so that
$$(1+x)^{n+1} \le 1 + (2^n-1)x + x + (2^n-1)x^2
\le 1 + (2^n-1)x + x + (2^n-1)x = 1+2(2^n-1)x + x = 1 + (2^{n+1}-1)x,$$
which is the statement for $n+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/896720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
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