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computing an integral with sines and cosines Hi i'm trying to do this definite integral $$∫_{0}^{π }\frac{\sin ^3xdx}{\left(R^2+Z^2-2ZR\cos x\right)^{3/2}}=∫_{0}^{π}\frac{\sin xdx}{\left(R^2+Z^2-2RZ\cos x\right)}-∫_{0}^{π}\frac{\cos ^2x\sin xdx}{\left(R^2+Z^2-2RZ\cos x\right)}$$
And then i have to see what happens when $ Z<R$
I do the substitution: $u=R^2+Z^2-2RZ\cos x,du=2ZRsenx,\cos ^2x=\left(\frac{R^2+z^2-u}{2ZR}\right)^2$
$$\frac{1}{2ZR}∫\frac{du}{u^{3/2}}-\frac{1}{\left(2ZR\right)^2}\left(∫\frac{\left(R^2+Z^2\right)^2du}{u^{3/2}}-∫\frac{2\left(R^2+Z^2\right)u}{u^{3/2}}+∫\frac{u^2du}{u^{3/2}}\right)$$
$$\frac{-u^{-1/2}}{ZR}+\frac{\left(R^2+Z^2\right)^2u^{-1/2}}{2\left(ZR\right)^2}+\frac{\left(R^2+Z^2\right)u^{1/2}}{\left(RZ\right)^2}-\frac{2u^{3/2}}{3}=$$$$\frac{-(R^2+Z^2-2RZcosx)^{-1/2}}{ZR}+\frac{(R^2+Z^2)^2(R^2+Z^2-2RZcosx)^{-1/2}}{2(ZR)^2}+\frac{(R^2+Z^2)(R^2+Z^2-2RZcosx)^{1/2}}{(RZ)^2}-\frac{2(R^2+Z^2-2RZcosx)^{3/2}}{3} \big|_0^\pi
$$
then
$$\frac{1}{\left(ZR\right)^2}(\frac{1}{2}\left(\frac{1}{|R+Z|}-\frac{1}{|R-Z|}\right)\left(\left(R^2+Z^2\right)^2-2ZR\right)+\left(R^2+Z^2\right)\left(|R+Z|-|R-Z|\right)-\frac{1}{6}\left(R+Z\right)^3+\frac{1}{6}\left(|R-Z|\left(R-Z\right)^2\right)$$
when $Z<R$
I get $$\frac{2}{3R^2}\frac{3R-2ZR^2-4Z^3}{R^2-Z^2}$$
but it has to be $\frac{4}{3R^3}$
i've looked at it too much i dont know where im doing wrong.please help
| I know how to do it now, a friend helped me, but it's a different way of solving it:
$$∫_{0}^{π}\frac{\sin ^3xdx}{\left(R^2+Z^2-2ZR\cos x\right)^{3/2}}$$
integrating by parts:
$$u=sen^2x, du=2senx\cos x$$$$dv=\frac{senxdx}{\left(R^2+Z^2-2ZR\cos x\right)^{3/2}},v=\frac{-\left(R^2+Z^2-2ZR\cos x\right)^{-1/2}}{ZR}$$$$\frac{sen^2x}{-\left(R^2+Z^2-2ZR\cos x\right)^{1/2}ZR}\big|_{0}^{π}+\frac{2}{ZR}∫\frac{senx\cos xdx}{\left(R^2+Z^2-2ZR\cos x\right)^{1/2}}$$
the first part becomes $0$ and again with parts
$$dv=\left(R^2+Z^2-2ZR\cos x\right)^{-1/2},v=\frac{-\left(R^2+Z^2-2ZR\cos x\right)^{1/2}}{ZR}$$
$$\frac{-2}{ZR}\left(\cos x\frac{\left(R^2+Z^2-2ZR\cos x\right)^{1/2}}{-ZR}+∫\frac{\left(R^2+Z^2-2ZR\cos x\right)^{1/2}du}{ZR}\right)$$$$\frac{-2}{\left(ZR\right)^2}\left(|R+Z|+|R-Z|\right)+\frac{4}{3\left(ZR\right)^3}\left(|R+Z|^3-|R-Z||R-Z|^2\right)$$
and when $Z<R$
$$\frac{-2}{\left(ZR\right)^2}\left(2R\right)+\frac{2}{3\left(ZR\right)^3}\left(R^3+3R^2Z+3RZ^2+Z^3-R^3+3R^2Z-3RZ^2+Z^3\right)=\frac{-4}{ZR^2}+\frac{4R^2Z}{3Z^3R^3}+\frac{4}{3R^3}=\frac{4}{3R^3}$$
However i don't understand why it doesn't work with the way I did it!
| {
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Inverse trigonometric function problem I recently came across a problem:
If $\mathbb{sin^{-1}x+sin^{-1}y+sin^{-1}z=\pi}$ then prove that $\mathbb{x^4+y^4+z^4+4x^2y^2z^2=2(x^2y^2+y^2z^2+x^2z^2)}$
Now, I proceeded like this:
Let, $\mathbb{sin^{-1}x=A, sin^{-1}y=B, sin^{-1}z=C}$ Then, $\mathbb{x=sinA, y=sinB, z=sinC}$ and $\mathbb{A+B+C=\pi}$
If I put the values of x, y and z in the left hand side and the right hand side of the identity, both sides should be equal. But I cannot proceed anymore. Can anyone help me solve the problem?
| My idea is to express this based on a triangle $\triangle ABC$, where $x=\sin A$, $y = \sin B$, $z=\sin C$, and let $a,b,c$ be the lengths of $BC, AC, AB$ respectively. From Heron's law, $S:=$ the area of $\triangle ABC$ can be written as $$S=\frac{1}{4}\sqrt{(a+b+c)(a+b-c)(a-b+c)(-a+b+c)}$$
and hence, $$a^4+b^4+c^4+16S^2=2\left(a^2b^2+a^2c^2+b^2c^2\right)$$
which can be proved by plugging $S$ in the left hand side and cancelling out terms.
Next, we know that in a triangle, $\ S = \frac{1}{2}bc\sin A$ and $\sin A = \frac{a}{2R}$. Hence, the left hand side can be written as $$a^4+b^4+c^4+16S^2=a^4+b^4+c^4+4b^2c^2\sin^2A=a^4+b^4+c^4+\frac{a^2b^2c^2}{R^2}$$
which gives us
$$a^4+b^4+c^4+\frac{a^2b^2c^2}{R^2} = 2\left(a^2b^2+a^2c^2+b^2c^2\right)$$
Next, we divide both sides by $16R^4$ to get:
$$\left(\frac{a}{2R}\right)^4+\left(\frac{b}{2R}\right)^4+\left(\frac{c}{2R}\right)^4+4\cdot\left(\frac{a}{2R}\right)^2\left(\frac{b}{2R}\right)^2\left(\frac{c}{2R}\right)^2 \\= 2\left(\left(\frac{a}{2R}\right)^2\left(\frac{b}{2R}\right)^2+\left(\frac{a}{2R}\right)^2\left(\frac{c}{2R}\right)^2+\left(\frac{b}{2R}\right)^2\left(\frac{c}{2R}\right)^2\right)$$
and use $x=\sin A = \frac{a}{2R}$, $y=\sin B = \frac{b}{2R}$, $z=\sin C = \frac{c}{2R}$ into the equation to get
$$x^4+y^4+z^4+4x^2y^2z^2=2\left(x^2y^2+x^2z^2+y^2z^2\right)$$
Note: The range of $\sin^{-1}$ is $\left[-\frac{\pi}{2},\frac{\pi}{2}\right]$, so with the condition $\sin^{-1}x+\sin^{-1}y+\sin^{-1}z=\pi$, the only case where we can't use a triangle to represent the problem is when one of the $\sin^{-1}x = 0$. In this case, the left hand side and right hand side both equal $2\cdot \left(\frac{\pi}{2}\right)^4$, and equality is maintained.
| {
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Prove that $\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2} \text{ when } abc=1$ So I have a possible proof but I'm not certain it's right:
$\text{As } \frac{a_1}{x_1} + \frac{a_2}{x_2} + \frac{a_3}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}\text{, we obtain } \frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{9}{a^3(b+c) + b^3(a+c) + c^3(a+b)}$
$\begin{align}
a^3(b+c) + b^3(a+c) + c^3(a+b) & = a^2(\frac{1}{b} + \frac{1}{c}) + b^2(\frac{1}{a} + \frac{1}{c}) c^2(\frac{1}{a} + \frac{1}{b}) \\
& \geq \frac{4a^2}{b+c} + \frac{4b^2}{a+c} + \frac{4c^2}{a+b} \text{ by the same inequality as above} \\
& \geq 4\frac{(a+b+c)^2}{2(a+b+c)} \text{ (same reason)} \\
& \geq 2(a+b+c)
\end{align}$
Now, we have:
$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{9}{2(a+b+c)}$
I'm not sure if this is a logical progression to make though. Because if $a \geq b$ then $\frac{1}{a} \leq \frac{1}{b}$, so is this not true? I feel like it must be as I can get the desired result, but I'm confused about it.
However, if this is true it is then sufficient to prove that $a+b+c \geq 3$.
$a+b+c = a+b + \frac{1}{ab} \geq 2\sqrt{ab} + \frac{1}{ab}$
It can be proved with calculus that $\forall x\ge 0,f(x) = 2\sqrt{x} + \frac{1}{x} \geq 3$. Thus the inequality is true, and the proof is finished.
Can someone please explain either why my proof is true (namely what I am confused about) or why this false proof yields the right result?
| I don't think generally $\frac{a_1}{x_1} + \frac{a_2}{x_2} + \frac{a_3}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$($a_i=10$, $x_i=1$ is counterexample), it should have been$$\frac{a_1^2}{x_1} + \frac{a_2^2}{x_2} + \frac{a_3^2}{x_3} \geq \frac{(a_1+a_2+a_3)^2}{x_1+x_2+x_3}$$by Cauchy-Schwartz inequality.
And if $x<y$ then $1/x>1/y$, therefore you should calculate the upper bound of $a^3(b+c) + b^3(a+c) + c^3(a+b)$, not the lower bound.
Also, it is true that $a+b+c \ge 3$ and you can prove it much simply with AM-GM inequality: $a+b+c \ge 3\sqrt[3]{abc}=3$. However, it only proves that $\frac{9}{2(a+b+c)}\le \frac{3}{2}$, which is not what you want.
| {
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How can I get general solution $y(x)$ in next differential equation : $(1+ xy)dx + (1-xy)dy = 0$? I used mathematica, and got solution, but that solution was too dirty and including lots of hard functions.
Is there any way to get solution for this equation -
$$(1+ xy)dx + (1-xy)dy = 0$$
| Approach $1$:
$(1+xy)~dx+(1-xy)~dy=0$
$(xy-1)\dfrac{dy}{dx}=xy+1$
$\left(y-\dfrac{1}{x}\right)\dfrac{dy}{dx}=y+\dfrac{1}{x}$
Let $u=y-\dfrac{1}{x}$ ,
Then $y=u+\dfrac{1}{x}$
$\dfrac{dy}{dx}=\dfrac{du}{dx}-\dfrac{1}{x^2}$
$\therefore u\left(\dfrac{du}{dx}-\dfrac{1}{x^2}\right)=u+\dfrac{1}{x}+\dfrac{1}{x}$
$u\dfrac{du}{dx}=\left(1+\dfrac{1}{x^2}\right)u+\dfrac{2}{x}$
This belongs to an Abel equation of the second kind.
Approach $2$:
$(1+xy)~dx+(1-xy)~dy=0$
$(xy+1)\dfrac{dx}{dy}=xy-1$
$\left(x+\dfrac{1}{y}\right)\dfrac{dx}{dy}=x-\dfrac{1}{y}$
Let $u=x+\dfrac{1}{y}$ ,
Then $x=u-\dfrac{1}{y}$
$\dfrac{dx}{dy}=\dfrac{du}{dy}+\dfrac{1}{y^2}$
$\therefore u\left(\dfrac{du}{dy}+\dfrac{1}{y^2}\right)=u-\dfrac{1}{y}-\dfrac{1}{y}$
$u\dfrac{du}{dy}=\left(1-\dfrac{1}{y^2}\right)u-\dfrac{2}{y}$
This belongs to an Abel equation of the second kind.
| {
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Is this the proper use of U-Substitution? I am part of an R&D role at my job, and I have been given the task to model the magnetic field of a bowl shaped permanent magnet. So this isn't part of a textbook and I have no idea if the answer I am getting is actually correct for now, therefore I am more interested to make sure my procedure is correct of using the U-Substitution. I am essentially trying to find the total sum of the the distances of the surface of a bowl to a point "P" somewhere on the +y axis.
This is my integral:
$$
r = \int_0^H \sqrt{a^2+ (x+h)^2} \space\space dh
$$
These are my steps:
let $u =x +h$ ;$\frac{du}{dh}=1$
$$
r =\int_0^H \sqrt{a^2+ u^2} \space \space du
$$
let $t = a^2+u^2 $ ; $\frac{dt}{du}=2u$
$$
r =\frac{1}{2u} \int_0^H \sqrt{t} \space \space dt
$$
Solving and plugging back in:
$$
r =\frac{1}{2u} \left(\frac{2}{3} t ^ {\frac{3}{2}}\right)
$$
Plugging all back in:
$$
\frac{1}{2(x+h)} \left(\frac{2}{3}[a^2+(x+h)^2]^\frac{3}{2} \right)
$$
Simplifying...
$$
r = \frac{(a^2 + [x+h]^2)^\frac{3}{2}}{2(x+h)}
$$
There you have it? Are these steps correct?
Thank you very much for helping out.
| I know this question has been resolved but here's another way using integration by parts and an integration formula which, in my experience, has been pretty useful to know. I'm posting this because even though it's a "longer" route, I do think it's good to see alternative approaches. And to me it feels more straightforward or natural, especially since I have no experience using hyperbolic trig substitutions.
$$
r = \int_0^H \sqrt{a^2+ (x+h)^2} \, dh
$$
First substitute $t = x+h$, so then $dt = dh$ and the new limits of integration are $x$ and $x+H$.
$$
r = \int_x^{x+H} \sqrt{a^2+ t^2} \, dt
$$
Now integrate by parts. Let $u = \sqrt{a^2+t^2}$, so then $dv = dt$, $v = t$, and $du = \dfrac t{\sqrt{a^2+t^2}} \, dt$ and we have
\begin{align*}
r &= \int_x^{x+H} \sqrt{a^2+ t^2} \, dt\\[0.3cm]
&= t\sqrt{a^2+t^2}\bigg|_{t=x}^{t=x+H} - \int_x^{x+H} \frac{t^2}{\sqrt{a^2+t^2}} \, dt\\[0.3cm]
&= (x+H)\sqrt{a^2+(x+H)^2} - x\sqrt{a^2+x^2} - \int_x^{x+H} \frac{\color{red}{a^2} +t^2\color{red}{-a^2}}{\sqrt{a^2+t^2}} \, dt\\[0.3cm]
&= (x+H)\sqrt{a^2+(x+H)^2} - x\sqrt{a^2+x^2} - \int_x^{x+H} \frac{a^2 +t^2}{\sqrt{a^2+t^2}} \, dt + \int_x^{x+H} \frac{a^2}{\sqrt{a^2+t^2}} \, dt\\[0.3cm]
&= (x+H)\sqrt{a^2+(x+H)^2} - x\sqrt{a^2+x^2} - \underbrace{\int_x^{x+H} \sqrt{a^2+t^2} \, dt}_{\text{This is $r$}} + a^2 \ln(\sqrt{a^2+t^2}+t)\bigg|_{t=x}^{t=x+H}\\[0.3cm]
r &= (x+H)\sqrt{a^2+(x+H)^2} - x\sqrt{a^2+x^2} - r + a^2 \ln(\sqrt{a^2+(x+H)^2}+x+H) - a^2\ln(\sqrt{a^2+x^2}+x)
\end{align*}
Now add $r$ to both sides and divide by 2.
| {
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Minimum values of coefficients of a quadratic Given a function $f(x)= ax^{2} + bx + c$ where $a<b$ and $f(x)\geq{0}$ for all real values of x. Then how would one find the minimum value of the relation between coefficients of the give quadratic. For ex,
How would one find the min value of $\frac{a+b+c}{b-a}$.
my work so far
I concluded that $\frac{a+b+c}{b-a}$ is the same as $\frac{f(1)}{b-a}$ and as per the given conditions $b^2 -4ac\leq{0}$ and I tried finding some triplets of $a,b, c$ and find the minimum value by observation but had no luck.
All help is greatly appreciated
| Let $f(x)=ax^2+bx+c$ where
$$\forall x \in \mathbb R, f(x) \ge 0 \qquad
\text{and} \qquad a < b \tag A$$
$$\text{Find $\min \dfrac{a+b+c}{b-a}$} \tag{B}$$
If $a=0$, then the problem is ill-defined. When $a < 0$, then $\displaystyle \lim_{x \to \infty}f(x) \to -\infty$. So $a > 0$.
Since $f(0) = c$, then we must have $ c \ge 0$, but we can do better than that. Clearly $f(x)$ is a parabola above or on the $x$-axis. So, holding $a$ and $b$ fixed, the value of $c$ that will minimize $\dfrac{a+b+c}{b-a}$ is the value, $c = -\frac 14b^2$, that puts the vertex of $f(x)$ on the $x$-axis. Combining that with $c \ge 0$, we get $c=0$. Since we require $a < b$, we let $b = aN$ where
$N > 0$.
We get $f(x) = a(x^2 + Nx + 0)$ and we want to minimize
\begin{align}
\dfrac{a+b+c}{b-a}
&= \dfrac{a+aN+0}{aN-a} \\
&= \dfrac{1+N}{N-1} \\
&= 1 + \dfrac{2}{N-1}
\end{align}
which approaches $1$ as $N$ approaches infinity.
For example, $f(x) = x^2 + Nx + 0$, where $N > 1$ satisfies condition (A) and
$\dfrac{a+b+c}{b-a} = \dfrac{N+1}{N-1} \to 1$ as $N \to \infty$.
| {
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Help differentiating $\frac{(x-1)^2(x+2)^2}{(x+1)^2}$ I need help differentiating this expression below. I know you can use a mix of the product rule and quotient rule, but that is tedious and long. Is there a shorter method?
$$\dfrac{(x-1)^2(x+2)^2}{(x+1)^2}$$
| Another approach:
Let $$y=\frac{(x-1)^2(x+2)^2}{(x+1)^2}$$
Then,
\begin{align*}
\ln y&=2\ln (x-1)+2\ln(x+2)-2\ln(x+1)\\
\qquad \frac{y'}{y}&=\frac2{x-1}+\frac2{x+2}-\frac2{x+1}\\
\qquad y'&=\left(\frac2{x-1}+\frac2{x+2}-\frac2{x+1}\right)\frac{(x-1)^2(x+2)^2}{(x+1)^2}\\
\end{align*}
Although we have suppose that $x-1,\;x+2$ and $x+1$ are positive the obtained derivative holds whenever $x\neq \pm1,-2.$
Generally speaking, whenever you want to differentiate some sort of fraction with degrees $>1$, it is usually best to take the $\ln$ of both sides, and then this will clean up your original function a lot.
| {
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Are there any positive integers such that $\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1$?
Are there any positive integers $x,y,z$ such that $$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} = 1?$$ Prove/Disprove.
I've plugged in random positive integers for $x,y,z$ and I have not been able to get the equation to equal $1$.
| By the AM-GM inequality, one has
$$\frac{x}{y} + \frac{y}{z} + \frac{z}{x} \ge 3\sqrt[3]{\frac{x}{y} \cdot\frac{y}{z} \cdot\frac{z}{x} }=3>1 $$
and hence it is impossible.
| {
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Evaluation of $\sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$ Find the following sum
$$S= \sum _{n=1}^{\infty} \tan^{-1} \frac{2}{n^2+n+4}$$
I am not able to make it telescopic series. Could someone help me with this?
| The $k$th partial sum is
\begin{align*}
\sum_{n = 1}^k (\tan^{-1} \sqrt{n} - \tan^{-1} \sqrt{n + 1}) &= (\tan^{-1} 1 - \tan^{-1} \sqrt{2}) + (\tan^{-1} \sqrt{2} - \tan^{-1} \sqrt{3}) \\
&\quad + (\tan^{-1} \sqrt{3} - \tan^{-1} \sqrt{4}) + \dots + (\tan^{-1} \sqrt{k} - \tan^{-1} \sqrt{k + 1}) \\
&= \tan^{-1} 1 - \tan^{-1} \sqrt{k + 1} \\
&= \frac{\pi}{4} - \tan^{-1} \sqrt{k + 1}.
\end{align*}As $k$ goes to infinity, $\tan^{-1} \sqrt{k + 1}$ approaches $\frac{\pi}{2},$ so the limit of the sum as $n$ goes to infinity is $\frac{\pi}{4} - \frac{\pi}{2} = \boxed{-\frac{\pi}{4}}.$
| {
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Minimum degree rational equation with root $a+\sqrt{b}+\sqrt{c}+\sqrt{d}$. What is the minimum degree of an equation with rational coefficients that has a root $x=a+\sqrt{b}+\sqrt{c}+\sqrt{d}$ with $a,b,c,d$ primes numbers?
I know how to find an equation of second degree that has root $a+\sqrt{b}$
$$
x=a+\sqrt{b} \quad \rightarrow \quad (x-a)^2=b
$$
and a $4-$degree equation that has root $a+\sqrt{b}+\sqrt{c}$
$$
x=a+\sqrt{b}+\sqrt{c} \quad \rightarrow \quad (x-a)^2=(\sqrt{b}+\sqrt{c})^2 \quad \rightarrow \quad \left[(x-a)^2-b-c \right]^2=4bc
$$
But it seems that this simple method cannot be used for a root with more than two surds. There is it some other method?
| Let $x=a+ \sqrt{b}+ \sqrt{c}+ \sqrt{d}$ square this and move some terms around
\begin{eqnarray*}
(x-a)^2+b-c-d = -2(x-a)\sqrt{b}+ 2\sqrt{c} \sqrt{d}
\end{eqnarray*}
Square it again and move some more terms around
\begin{eqnarray*}
((x-a)^2+b-c-d)^2 -4b(x-a)^2-4cd = -8(x-a)\sqrt{b}\sqrt{c} \sqrt{d}
\end{eqnarray*}
Squaring one final time & we have
\begin{eqnarray*}
(((x-a)^2+b-c-d)^2 -4b(x-a)^2-4cd)^2 = 64(x-a)^2bcd
\end{eqnarray*}
So the equation that this quantity satisfies an equation of degree $\color{red}{8}$ as expected.
| {
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prove this by binomial theorem Prove that by binomial theorem :
$\dbinom{6}{0}+\dbinom{6}{2}+\dbinom{6}{4}+\dbinom{6}{6}=\dbinom{6}{1}+\dbinom{6}{3}+\dbinom{6}{5}$
I started to prove it by moving the right side to the left :
$\dbinom{6}{0}+\dbinom{6}{2}+\dbinom{6}{4}+\dbinom{6}{6}-\dbinom{6}{1}-\dbinom{6}{3}-\dbinom{6}{5}=0$
it should be done by the binomial coefficient ...?....
how ??
thanks in advance
| HINT:
$$0=(1-1)^n=\sum_{r=0}^n\binom nr1^{n-r}(-1)^r=\sum_{r=0}^{2r\le n}\binom n{2r}-\sum_{r=0}^{2r+1\le n}\binom n{2r+1}$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\sum_{n=1}^{\infty}\frac{n-1}{2}\frac{e^{-10}10^n}{n!}$ I need to evaluate $\sum_{n=1}^{\infty}\frac{n-1}{2}\frac{e^{-10}10^n}{n!}$ to solve a statistic problem. Mathematica gives an answer of $\frac{9}{2}+\frac{1}{2e^{10}}$. How can I evaluate explicitly without aid of computer?
| Observe
\begin{align}
\sum_{n=1}^{\infty}\frac{n-1}{2}\frac{e^{-10}10^n}{n!}=&\ \frac{e^{-10}}{2}\left(10\sum^\infty_{n=1} n\frac{10^{n-1}}{n!}-\sum^\infty_{n=1} \frac{10^n}{n!} \right)\\
=& \frac{e^{-10}}{2}\left(10\sum^\infty_{n=1} \frac{10^{n-1}}{(n-1)!}-\sum^\infty_{n=1} \frac{10^n}{n!} \right)\\
=& \frac{e^{-10}}{2}\left(10\sum^\infty_{n=0} \frac{10^n}{n!}-\sum^\infty_{n=0} \frac{10^n}{n!}+1 \right)\\
=&\ \frac{e^{-10}}{2}\left(10e^{10}-e^{10}+1 \right).\\
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2237882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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$\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx$ Integrate $$\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx$$
What I first did is I tried to complete the square for the numerator
$\sqrt{4(x^2-2x+3/4)} = \sqrt{4(x^2-2x+1-1/4)} = \sqrt{4(x-1)^2 - 1}$
Now I did trig-substitution:
$(x-1) = \frac{1}{2}\sec(\theta)$
$dx = \frac{1}{2}\sec(\theta)\tan(\theta)d\theta$
$$=\int \frac{\tan^2(\theta)\sec(\theta)}{\frac{1}{2}\sec(\theta) }d\theta = 2\tan(\theta) - 2\theta + C = 2\frac{\sqrt{4x^2-8x+3}}{2} - 2\sec^{-1}\left(\frac{x-1}{2}\right) + C$$
| If you substitute $u=\sqrt{4x^2-8x+3}$ you obtain $du=\dfrac{8x-8}{2\sqrt{4x^2-8x+3}}dx$. So :
$$\int \frac{\sqrt{4x^2-8x+3}}{x-1}dx={\displaystyle\int}\dfrac{u^2}{u^2+1}\,du$$
Perform long division and you can finish easily.
| {
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Exact solution to nonconvex optimization problem I would like to minimize $v+w+x+y+z$ subject to the following:
$$\frac{v+w}{x+y+z}=\frac{y}{z}=\frac{w}{x+y}$$
where $v,w,x,y,z\ge 1$
I tried entering this problem in WolframAlpha:
Minimize[{v + w + x + y + z, (v+w)/(x+y+z) == y/z && y/z== w/(x+y)&& v >= 1 && x >= 1 && y >= 1 && w >= 1 && z >= 1}, {v, w, x, y, z}]
It found a global min at $(v,w,x,y,z)=(1,\sqrt{2},1,1,\sqrt{2})$. What techniques might WolframAlpha be using to find this solution?
| *
*Simplifying the constraints. Set
$$
t=\frac{x+y+z}{v+w}=\frac{z}{y}=\frac{x+y}{w},
$$
then
$$
\left\{
\begin{array}{rcl}
x+y+z&=&t(v+w),\\
z&=&ty,\\
x+y&=&tw
\end{array}\right.
\quad\Rightarrow\quad y=v,
$$
and the problem becomes
$$
\min\, (1+t)(y+w)\quad\text{subject to }
\left\{
\begin{array}{rcl}
1&\le&ty,\\
1+y&\le&tw,\\
1&\le&y,\\
1&\le&w.
\end{array}\right.\quad\Leftrightarrow\quad
\left\{
\begin{array}{rcl}
1-ty&\le&0,\\
1+y-tw&\le&0,\\
1-y&\le&0,\\
1-w&\le&0.
\end{array}\right.\tag{*}
$$
*Applying the KKT necessary condition.
$$
\left\{
\begin{array}{rcl}
1+t-\mu_1t+\mu_2-\mu_3&=&0,\\
1+t-\mu_2t-\mu_4&=&0,\\
y+w-\mu_1y-\mu_2w&=&0,\\
\mu_1(1-ty)&=&0,\\
\mu_2(1+y-tw)&=&0,\\
\mu_3(1-y)&=&0,\\
\mu_4(1-w)&=&0,\\
\mu_k\ge 0\text{ and }(*)
\end{array}
\right.\quad\Leftrightarrow\quad
\left\{
\begin{array}{rcl}
1+(1-\mu_1)t+\mu_2-\mu_3&=&0,\quad(1)\\
1+(1-\mu_2)t-\mu_4&=&0,\quad(2)\\
(1-\mu_1)y+(1-\mu_2)w&=&0,\quad(3)\\
\mu_1(1-ty)&=&0,\quad(4)\\
\mu_2(1+y-tw)&=&0,\quad(5)\\
\mu_3(1-y)&=&0,\quad(6)\\
\mu_4(1-w)&=&0,\quad(7)\\
\mu_k\ge 0\text{ and } (*)&&\ \ \ \ \quad(8)
\end{array}
\right.
$$
*Solution. Equation (3) together with feasibility (*) leaves only 2 possibilites:
Case 1: $1-\mu_1\ge 0$, $1-\mu_2\le 0$ or
Case 2: $1-\mu_1\le 0$, $1-\mu_2\ge 0$.
Consider those cases.
Case 1. We have $\mu_2>0$ and from (1) $\mu_3>0$, hence, from (5) and (6) it follows that $$y=1,\quad tw=1+y=2.$$
- Assume $\mu_1=0$. Then from (3) we have $(\mu_2-1)w=1$ and from (2)
$\mu_4=1-t/w$.
If $\mu_4=0$ then $t=w=\sqrt{2}$ and the objective value is $\color{red}{(1+\sqrt{2})^2}$.
If $\mu_4>0$ then from (7) $w=1$, thus $t=2$ and the objective value is $\color{red}{6}$.
- Assume $\mu_1>0$. Then from (4) we get $1=ty=t$, thus $w=2$ and the objective value is $\color{red}{6}$.
Case 2. We have $\mu_1>0$ and from (2) $\mu_4>0$, hence, from (4) and (7) it follows that
$$
w=1,\quad ty=1.
$$
- Assume $\mu_2=0$. Then from (3) we get $(\mu_1-1)y=1$ and from (1) $\mu_3=1-t/y$.
If $\mu_3=0$ then $t=y=1$ and the objective value is $\color{red}{4}$.
If $\mu_3>0$ then from (6) $y=1$, hence, $t=1$ and the objective value is $\color{red}{4}$.
- Assume $\mu_2>0$. Then from (5) $1+y=tw=t$. It gives $1=ty=(1+y)y$ $\Rightarrow$ $y=\frac{1}{1+y}\le \frac12$. It is not possible as it contradicts $y\ge 1$.
Comparing the red objective values, it is clear that the smallest is for $y=1$, $t=w=\sqrt{2}$. Finally, $x=tw-y=1$, $z=ty=\sqrt{2}$ and $v=y=1$.
| {
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How to show that $e^x\le x + e^{x^2}$ for any real $x $
Show that $e^x\le x + e^{x^2}$ for any real $x$.
I have trouble proving this inequality for $x \in (0, 1)$. Any hint?
| Let $f(x)=x+e^{x^2}-e^x$, $f(0)=0$
$f'(x)=1+2xe^{x^2}-e^x$, $f'(0)=0$
$f''(x)=2e^{x^2}+4x^2e^{x^2}-e^x=e^x(2e^{x^2-x}-1)+4x^2e^{x^2}$
$e^{x^2-x}=e^{(x-\frac{1}{2})^2-\frac{1}{4}}\ge e^{-\frac{1}{4}}>\frac{1}{2}$, so $f''(x)>0$ for $x\in(0,1)$, done.
Another try:
By Taylor expansion about $x=0$, $e^x=1+x+\frac{x^2}{2}+R_3(x)$
$R_3(x)=\frac{e^c}{3!}x^3\le \frac{e}{6}x^3$
$x+e^{x^2}=x+(1+x^2+\cdots)>1+x+x^2$
$\frac{1}{2}>\frac{e}{6}\ge \frac{e}{6}x \Rightarrow \frac{1}{2}x^2>\frac{e}{6}x^3 \Rightarrow 1+x+x^2>1+x+\frac{x^2}{2}+R_3(x)\Rightarrow x+e^{x^2}>e^x$
| {
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"source": "stackexchange",
"question_score": "2",
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} |
Prove this inequality $m-(mx^2-x+m)e^{-x}x\ge 0,\forall x\in (0,1+\frac{1}{m}]$ Let $m>0$ show that
$$m-(mx^2-x+m)e^{-x}x\ge 0,\forall x\in (0,1+\dfrac{1}{m}]$$
It suffices to prove that
$$m(1-x^3e^{-x}-xe^{-x})+x\ge 0,x\in(0,1+\frac{1}{m}]$$
Idea $1$:,if we Let $f(m)=m(1-x^3e^{-x}-xe^{-x})+x$ it is clear $f(0)=x>0$,but why $f(m)\ge 0,\forall m>0$
Idea 2: if we let $$g(x)=m-(mx^2-x+m)e^{-x}x$$ Consider
$$\begin{align*}g'(x)&=-(2mx-1)e^{-x}x+(mx^2-x+m)e^{-x}x-(mx^2-x+m)e^{-x}\\
&=e^{-x}(mx^3-x^2+mx-2mx^2+x-mx^2+x-m)\\
&=e^{-x}(mx^3-(3m+1)x^2+(m+2)x-m)\end{align*}$$ following is hard to prove it
so maybe this inequality have other methods to solve it?
| Since $mx\le m+1$,$$m-(mx^2-x+m)e^{-x}x\ge m-((m+1)x-x+m)e^{-x}x=m(1-e^{-x}x(x+1))$$Therefore, we only need to prove $1-e^{-x}x(x+1) \ge 0$ for positive $x$. Here, we know that $e^{-x}\le 1/(1+x+\frac{1}{2}x^2+\frac{1}{6}x^3)$. Appplying this, we get$$1-e^{-x}x(x+1)\ge1-\frac{6x^2+6x}{x^3+3x^2+6x+6}$$
Now it's left to prive $x^3+3x^2+6x+6\ge 6x^2+6x$, or $x^3-3x^2+6\ge0$ for positive $x$.
$x^3-3x^2+6=(x-2)^2(x+1)+2>0$, so the inequality is proved.
The equality does not hold, because the inequality $x^3-3x^2+6>0$ is strict.
| {
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"url": "https://math.stackexchange.com/questions/2244753",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
Prove $f(x) = \sqrt x$ is uniformly continuous on $[0,\infty)$ I have seen a proof in $\sqrt x$ is uniformly continuous
Below shows an alternative proof. Please correct me if im wrong.
Proof:
For any given $\varepsilon >0$,
Let $\delta_1 = \frac{\varepsilon}{2}$, $\forall x,y \in [1,\infty)$ with $|x-y|<\delta_1$
Since $|\sqrt x + \sqrt y|\geq2$
$$|\sqrt x - \sqrt y| = \frac{|x-y|}{|\sqrt x+\sqrt y|} < |x-y| < \delta_1 = \frac{\varepsilon}{2}$$
Hence, $\sqrt x$ is uniformly continuous on $[1,\infty)$.
$\sqrt x$ is continuous on [0,1] , so $\sqrt x$ is uniformly continuous on [0,1].
So, there exist $\delta_2 > 0$ such that $\forall x,y \in [0,1], |x-y|<\delta_2$, $|\sqrt x -\sqrt y| <\frac{\varepsilon}{2}$
Let $\delta = \min{(\delta_1,\delta_2)}$
$\forall x,y \in [0,\infty)$ with $|x-y|<\delta$,
Case 1: $x,y \in [0,1]$
Proven above as $|x-y| < \frac{\varepsilon}{2} < \varepsilon$
Case 2: $x,y \in [1,\infty)$
Proven above as $|x-y| < \frac{\varepsilon}{2} < \varepsilon$
Case 3: $x \in [0,1] , y \in [1,\infty]$
$$|\sqrt x-\sqrt y| = |\sqrt x -1+1-\sqrt y| \leq |\sqrt x-1| + |\sqrt y -1| < \frac{\varepsilon}{2} + \frac{\varepsilon}{2}= \varepsilon$$
by applying case 1 and case 2.
| Let $0 < \epsilon < 1$ and $x,y \in [0,\infty)$ with $|x-y| < \epsilon^2$.
Consider $|\sqrt{x} - \sqrt{y}| = ||\sqrt{x}| - |\sqrt{y}|| \leq |\sqrt{x} + \sqrt{y}|$.
So, $|\sqrt{x} - \sqrt{y}|^2 = |\sqrt{x} - \sqrt{y}|\cdot |\sqrt{x} - \sqrt{y}| \leq |\sqrt{x} - \sqrt{y}|\cdot |\sqrt{x} + \sqrt{y}| = |x-y|$, that is, $|\sqrt{x} - \sqrt{y}| \leq |x-y|^{\frac{1}{2}}$ which implies that $\sqrt{\centerdot}$ is Hölder-continuous.
Finally, $|\sqrt{x} - \sqrt{y}| \leq |x-y|^{\frac{1}{2}} < \sqrt{\epsilon^2} = \epsilon$.
Q.E.D.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the exact value of $\sin(\pi+\alpha)+\cos(\frac{3\pi}{2}+\alpha)+\tan(-\frac{\pi}{2}+\alpha)$ Given that $\tan(\alpha) = -3$ find the exact value.
I tried:
$$\sin(\pi+\alpha)+\cos(\frac{3\pi}{2}+\alpha)+\tan(-\frac{\pi}{2}+\alpha) = \\
(\sin(\pi)\cos(\alpha)+\cos(\pi)\sin(\alpha))+(\cos(\frac{3\pi}{2})\cos(\alpha)-\sin(\frac{3\pi}{2})\sin(\alpha))+\tan(-\frac{\pi}{2}+\alpha) = \\
-\sin(\alpha)+\sin(\alpha)+\frac{\tan(-\frac{\pi}{2})+\tan(\alpha)}{1-\tan(-\frac{\pi}{2})\tan(\alpha)} = \\
\frac{\tan(-\frac{\pi}{2})+\tan(\alpha)}{1-\tan(-\frac{\pi}{2})\tan(\alpha)} = \\
\frac{-\tan(\frac{\pi}{2})-3}{1-(-\tan(\frac{\pi}{2})(-3)} = \\
\frac{-\tan(\frac{\pi}{2})-3}{1-3\tan(\frac{\pi}{2})} = \\
???$$
What do I do next?
| You are right, the first two terms cancel out. However, the third term cannot be evaluated using the known formula, since both numerator and denominator are infinite.
Instead, consider the following chain of statements:
$$
\tan \left(-\frac \pi 2 + \alpha \right) = - \tan\left(\frac \pi 2 - \alpha\right) = -\cot \alpha = \frac{-1}{\tan \alpha} = \frac{-1}{3}
$$
| {
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"source": "stackexchange",
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Prove that if $x$ is odd then $x^2 + (x + 2)^2$ is divisible by $2$ but not for $4$ This is what I tried:
$$x^2 + x^2 + 4x + 4=2x^2 + 4(x + 1),$$
so it's divisible by $2$, since this expression is a sum of a multiple of $2$ and a multiple of $4$.
Therefore, for the expression not to be a multiple of $4$, $2x^2$ can't be multiple of $4$, that's what I can't prove.
Brazilian student, sorry for my English.
| HINT:
$$x^2+(x+2)^2=\{(x+2)-x\}^2+2x(x+2)$$
As $x$ is odd $\iff x+2$ is odd, so will be $x(x+2)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to determine the outward normal for spehrical and cylindrical coordinates? Suppose I have a spherical surface $x^2+y^2+z^2=1$, and I correspondingly parametrize it as $r(\phi, \theta)=<\sin\phi\cos\theta,\ \sin\phi\sin\theta,\ \cos\phi>$. To find out the direction of normal, we compute $r_\phi\times r_\theta$ or $r_\theta\times r_\phi$, and they are in opposite direction. How do we know which is outward normal and how do we prove it?
Same situation in cylindrical coordinates. Suppose I have $x^2+y^2=1$ and I correspondingly parametrize it as $r(z,\theta)=<\cos\theta, \sin\theta, z>$. To find out the direction of normal, we compute $r_z\times r_\theta$ or $r_\theta\times r_z$. How do we know which is outward normal?
| You should be able to determine which direction a vector has relative to the surface by checking the angle between
the vector in question and the vector going from the origin to the surface point.
For a point on the spherical surface, the vector from the origin is
$$
\mathbf{r} =
\begin{pmatrix}
\sin \phi \cos \theta \\
\sin \phi \sin \theta \\
\cos \phi
\end{pmatrix}
$$
Calculating $\mathbf{r}_\phi$ and $\mathbf{r}_\theta$:
$$
\mathbf{r}_\phi =
\begin{pmatrix}
\cos \phi \cos \theta \\
\cos \phi \sin \theta \\
- \sin \phi
\end{pmatrix}
%
\qquad
%
\mathbf{r}_\theta =
\begin{pmatrix}
- \sin \phi \sin \theta \\
\sin \phi \cos \theta \\
0
\end{pmatrix}
$$
Define the vector $\mathbf{v}$ as the cross product $\mathbf{r}_\phi \times \mathbf{r}_\theta$:
$$
\mathbf{v} =
\mathbf{r}_\phi \times \mathbf{r}_\theta =
\begin{pmatrix}
\sin^2 \phi \cos \theta \\
\sin^2 \phi \sin \theta \\
\sin \phi \cos \phi
\end{pmatrix}
$$
Finding the magnitude, we obtain $|\mathbf{v}| = \sin \phi$ so $\mathbf{v}$ makes sense as a normal vector as long as $0 < \phi < \pi / 2$.
Now consider the angle $\alpha$ between $\mathbf{r}$ and $\mathbf{v}$, we can
find its cosine using the scalar product:
$$
|\mathbf{v}| \cos \alpha =
\mathbf{r} \cdot \mathbf{v}
=
\sin^3 \phi \cos^2 \theta + \sin^3 \phi \sin^2 \theta + \sin \phi \cos^2 \phi
$$
which simplifies to
$$
|\mathbf{v}| \cos \alpha =
\sin \phi
\quad \Rightarrow \cos \alpha = 1
\quad \Rightarrow \alpha = 0
$$
So, with the assumption that $0 < \phi < \pi$, we have shown that $\mathbf{v} =
\mathbf{r}_\phi \times \mathbf{r}_\theta$ points in the same direction as $\mathbf{r}$.
I.e. that $\mathbf{v}$ is an outward normal.
A similar method can be applied using cylindrical coordinates.
| {
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Show that $0 \le a^{\frac{1}{2}} \le b^{\frac{1}{2}}$
Show that if $0 \le a \le b$, then $0 \le a^{\frac{1}{2}} \le b^{\frac{1}{2}}$ in a $C^*$-algebra $A$.
Assume that $A$ is unital.
Suppose that $a$ and $b$ are invertible. Then $$0 \le a \le b \implies 0 \le a^{\frac{-1}{2}}aa^{\frac{-1}{2}} \le a^{\frac{-1}{2}}ba^{\frac{-1}{2}} \implies 0 \le e \le a^{\frac{-1}{2}}ba^{\frac{-1}{2}} \implies 0 \le a^{\frac{1}{2}}b^{-1}a^{\frac{1}{2}} \le e$$
Thus we have $$0 \le a^{\frac{1}{2}}b^{\frac{-1}{2}}b^{\frac{-1}{2}}a^{\frac{1}{2}} \le e \implies 0 \le \left(a^{\frac{1}{2}}b^{\frac{-1}{2}}\right)\left(a^{\frac{1}{2}}b^{\frac{-1}{2}}\right)^* \le e$$
Hence, $\|a^{\frac{1}{2}}b^{\frac{-1}{2}}\|\le 1$ which implies that $r(a^{\frac{1}{2}}b^{\frac{-1}{2}}) \le 1$
But $$\sigma(a^{\frac{1}{2}}b^{\frac{-1}{2}}) \cup \{0\}=\sigma(b^{\frac{-1}{4}}a^{\frac{1}{2}}b^{\frac{-1}{4}}) \cup \{0\}$$ gives that $r(b^{\frac{-1}{4}}a^{\frac{1}{2}}b^{\frac{-1}{4}}) \le 1$. Let $n=b^{\frac{-1}{4}}a^{\frac{1}{2}}b^{\frac{-1}{4}}$. Then $n$ is normal. Let $B=C^*(1,n)$. Then $B$ is an abelian $C^*$-algebra. Moreover $$\sigma_A(1-n)=\sigma_B(1-n)=\{\tau(1-n):\tau \in \Omega(B)\}=\{\tau(1)-\tau(n): \tau \in \Omega(B)\}=\{1-\tau(n): \tau \in \Omega(B)\} \subset \mathbb{R}^+$$
Hence $0 \le n \le 1$ which is same as $$b^{\frac{-1}{4}}a^{\frac{1}{2}}b^{\frac{-1}{4}} \le 1$$ implying $a^{\frac{1}{2}} \le b^{\frac{1}{2}}$.
How do I adapt this proof for any general $a$ and $b$? Since $1-b \le 1-a$, dividing through out by $2||1-a||$ we will have invertible elements. Then I can do as above.
Thanks for the help!
| You need to note that if $0 \leq x \leq y$ then for $\epsilon > 0$ that
$0 \leq x+ \epsilon 1 \leq y + \epsilon 1$. Now, for $\epsilon >0$, $-\epsilon \notin \sigma(x),\sigma(y)$ so we have $x+ \epsilon 1$ and $y + \epsilon 1$ are both invertible. Hence, from your work above we have that $(x+\epsilon 1)^{1/2} \leq (y+\epsilon 1)^{1/2}$.
It is not hard to see that
$(x + \epsilon 1)^{1/2} \to_{|| \cdot ||} x$ and $(y+ \epsilon 1)^{1/2} \to_{|| \cdot ||} y$.
Hence, the case you have already proved generalizes the result after noting these facts.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\lim_{z\to 0} \frac{zRe(z)}{\overline{z}}$
$$\lim_{z\to 0} \frac{z\cdot Re(z)}{\overline{z}}$$
$$\lim_{z\to 0} \frac{z\cdot Re(z)}{\overline{z}}=\lim_{(x,y)\to (0,0)} \frac{(x+yi)\cdot x}{x-yi}=\lim_{(x,y)\to (0,0)} \frac{x^2+yxi}{x-yi}\cdot\frac{x+yi}{x+yi}=\lim_{(x,y)\to (0,0)}\frac{x^3+yx^2i+y^2xi-y^2x}{x^2+y^2}=\lim_{(x,y)\to (0,0)}\frac{x^3-y^2x}{x^2+y^2}+i\frac{yx^2+y^2x}{x^2+y^2}$$
How should I continue?
| Using your notation:
$$\frac{x(x+iy)}{x-iy}=\frac{x^3-xy^2}{x^2+y^2}+\frac{2x^2y}{x^2+y^2}i$$
and using polar coordinates (why can we? Detail the following):
$$\begin{cases}\cfrac{x^3-xy^2}{x^2+y^2}=r\left(\cos^3\theta-\cos\theta\sin^2\theta\right)\xrightarrow[r\to0]{}0\\{}\\
\cfrac{2x^2y}{x^2+y^2}=r\left(\cos^2\theta\sin\theta\right)\xrightarrow[r\to0]{}0\end{cases}\;\;\implies\frac{ z\,\text{Re}\,z}{\overline z}\xrightarrow[z\to0]{}0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Solving an equation containing cube roots: $\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1$ I'm trying to figure out a way to solve this equation:
$$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1.$$
I tried to cube both sides, but I ended up with an equation looking like this:
$$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1}-\sqrt[3]{5x+7})=-6.$$
At this point I'm out of stuff to do. Any help would be appreciated. Thanks in advance.
| Following up on OP's "almost-there" attempt (and correcting the $-1\color{red}{2}\,$ in the second equation):
$\sqrt[3]{5x+7}-\sqrt[3]{5x-12}=1 \tag{1}$
i tried to cube both sides, but i ended up with an equation looking like this:
$\sqrt[3]{(5x-12)(5x+7)}(\sqrt[3]{5x-1\color{red}{2}}-\sqrt[3]{5x+7})=-6 $
The rightmost factor is $-1$ per the original equation $(1)\,$, which leaves:
$$-\sqrt[3]{(5x-12)(5x+7)}=-6 \tag{2}$$
Then the numbers $\,\sqrt[3]{5x+7}\,$ and $\,-\sqrt[3]{5x-12}\,$ have sum $\,1\,$ per $(1)\,$, and product $\,-6\,$ per $(2)\,$, so they are the roots of the quadratic $\,t^2-t-6=0 \iff t \in \{-2, 3\}\,$:
*
*$\sqrt[3]{5x+7} = -2 \iff 5x +7 = -8 \iff x = -3$
*$\sqrt[3]{5x+7} = 3 \iff 5x + 7 = 27 \iff x = 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2248632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Please verify my power series solution to the differential equation $y''-xy'+y=0$. Please verify my answer to the following differential equation:
$$y''-xy'+y=0$$
Let $y = {\sum_{n=0}^\infty}C_nx^n$, then $y' = {\sum_{n=1}^\infty}nC_nx^{n-1}$ and $y''={\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}$
Substituting this to the equation we get
$${\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}-x{\sum_{n=1}^\infty}nC_nx^{n-1}+{\sum_{n=0}^\infty}C_nx^n = 0$$
$${\sum_{n=2}^\infty}n(n-1)C_nx^{n-2}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=0}^\infty}C_nx^n = 0$$
Getting the $x^n$ term on all the terms
$${\sum_{n=0}^\infty}(n+2)(n+1)C_{n+2}x^{n}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=0}^\infty}C_nx^n = 0$$
Getting the $0$th term from the first and the third summations we get
$$2C_2+C_0 + {\sum_{n=1}^\infty}(n+2)(n+1)C_{n+2}x^{n}-{\sum_{n=1}^\infty}nC_nx^n+{\sum_{n=1}^\infty}C_nx^n = 0$$
Factoring $x^n$ we get
$$2C_2+C_0 + {\sum_{n=0}^\infty}[(n+2)(n+1)C_{n+2}-nC_n+C_n]x^n= 0$$
i.$$2C_2+C_0 = 0 => C_2 = \frac{-C_0}{2}$$
ii.$$(n+2)(n+1)C_{n+2}-nC_n+C_n = 0$$
Therefore solving ii. for $C_{n+2}$
$$C_{n+2}=\frac{(n-1)C_n}{(n+2)(n+1)}, n=0,1,2,3,...$$
If $n = 0$,
$$C_2 = \frac{-C_0}{2!}$$
If $n=1$,
$$C_3 = 0$$
If $n=2$,
$$C_4 = \frac{C_2}{3*4} = \frac{-C_0}{4!}$$
If $n=3$,
$$C_5 = \frac{2C_3}{4*5}=0$$
If $n=4$,
$$C_6 = \frac{3C_4}{5*6} = \frac{-C_0}{6!}$$
Upon seeing the pattern we realize that if $n=2m$ then
$$C_{2m} = \frac{-C_0}{2m!}$$
And if $n=2m+1$ then
$$C_{2m+1} = 0$$
So the final answer would be
$$y = {\sum_{n=0}^\infty}C_nx^n => {\sum_{m=0}^\infty}\frac{-C_0*x^{2m}}{2m!}$$
| Your equation is a second order linear equation so it should be a two-dimensional space of solutions (that is, solutions that depend on two free parameters) while your final answer depends only on one free parameter $C_0$ so this means you did something wrong.
Indeed, your equations don't determine what is $C_1$ which actually means that $C_1$ can be arbitrary. In addition, you made a mistake in deducing the general pattern. For example,
$$ C_6 = \frac{3}{6 \cdot 5} C_4 = -\frac{3}{6 \cdot 5} \frac{1}{4!} C_0 = -\frac{3C_0}{6!} \neq -\frac{C_0}{6!}.$$
In fact, we have
$$ C_{2m} = \frac{(2m-3)C_{2m-2}}{(2m)(2m-1)} = \frac{(2m-3)(2m-5)C_{2m-4}}{(2m)(2m-1)(2m-2)(2m-3)} = \dots = -\frac{(2m - 3)(2m - 5) \dots 1}{(2m)!} C_0 = - \frac{(2m - 3)(2m - 5) \dots 1}{(2m)(2m-1)(2m-2)(2m-3)(2m-4) \dots 1} C_0 = -\frac{C_0}{(2m-1)2^{m}m!}$$
and the general solution is given by
$$ y(x) = C_1 \cdot x - C_0 \sum_{m=0}^{\infty} \frac{x^{2m}}{(2m-1)2^m m!}. $$
| {
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"url": "https://math.stackexchange.com/questions/2254025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proof that consecutive terms of a sequence are coprime I've been dealing with the following problem for a while:
Let $a_n$ be a sequence in $\mathbb{N}$, such that $a_{0} =7, a_{1} =9, a_{n} =5 \cdot a_{n-1}-2 \cdot a_{n-2}$. Prove that $a_n$ and $a_{n+1}$ are coprime.
I've tried coming up with a general formula for $a_n$ but couldn't. I know that the gcd between $a_n$ and $a_{n+1}$ should be 1, but how should I prove it? I guess that I could say that there's a prime that divide's both of them and get a contradiction...
So if p divides $a_n$ and $a_{n+1}$ it must divide their sum:
$5(a_n + a_{n-1}) \equiv 2 (a_{n-1} + a_{n-2}) \mod {p}$
$p$ can't divide both 5 and 2 at the same time unless it's 1. But what can I say about the rest? Is this the best way to go?
Thanks
| Just solving the recurrence:
Let $a_n$ be a sequence in $\mathbb{N}$, such that $a_{0} =7, a_{1} =9, a_{n} =5 \cdot a_{n-1}-2 \cdot a_{n-2}$.
This is a homogeneous linear recurrence relation with constant coefficionts, so the usual algorithm works.
The characteristic polynomial is
$$
p(t) = t^2 - 5t + 2
$$
Going for the roots
$$
0 = (t - 5/2)^2 + 2 - 25/4 = (t - 5/2)^2 - 17/4 \iff \\
t = \frac{5 \pm\sqrt{17}}{2}
$$
This gives the general solution
$$
a_n =
k_1 \left( \frac{5 + \sqrt{17}}{2} \right)^n +
k_2 \left( \frac{5 - \sqrt{17}}{2} \right)^n
$$
Applying the initial values gives
$$
a_0 = k_1 + k_2 = 7 \\
a_1 = k_1 \frac{5 + \sqrt{17}}{2} +
k_2 \frac{5 - \sqrt{17}}{2} = 9
$$
and
$$
k_1 = \frac{7 - \sqrt{17}}{2} \\
k_2 = \frac{7 + \sqrt{17}}{2} \\
$$
and we get
$$
a_n =
\frac{7 - \sqrt{17}}{2} \left( \frac{5 + \sqrt{17}}{2} \right)^n +
\frac{7 + \sqrt{17}}{2} \left( \frac{5 - \sqrt{17}}{2} \right)^n
$$
No idea yet if that form is of any help.
| {
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"url": "https://math.stackexchange.com/questions/2254765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Formula for $\left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots$ I was doing an exercise about the integral $$ I_{n} = \int_{0}^{1} (1-x^{2})^{n}\, \mathrm{d}x $$ and I tried two approaches. On the one hand, by substituting $x = \sin u $ I found that $$ I_{n} = \frac{2}{3} × \frac{4}{5} × \cdots × \frac{2n}{2n+1} = \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots\left(1-\frac{1}{2n+1}\right).$$
On the other hand, the binomial expansion formula gave me $$I_{n} = \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{2k+1}.$$
So I find that $$ \sum_{k=0}^{n}\binom{n}{k}\frac{(-1)^{k}}{2k+1} = \left(1-\frac{1}{3}\right)\left(1-\frac{1}{5}\right)\cdots\left(1-\frac{1}{2n+1}\right),$$
but I want to understand this identity without integration. Is there a direct way to prove this? Perhaps by induction or generating polynomials or something.
I tried induction, and found that proving the identity is equivalent to proving $$ \sum_{k=0}^{n} \binom{n}{k}(-1)^{k} \left( \frac{1}{2n+3} - \frac{1}{2k+3} \right) = 0 $$
but here I get stuck.
| An expression
$$\sum_{k=0}^n (-1)^k\binom{n}k f(a+k)$$
indicates an $n$-fold differencing operation. Here you have $f(x)=1/(2x+1)$
and $a=0$.
Define
$$\Delta f(x)=f(x)-f(x+1),$$
$$\Delta^2 f(x)=\Delta f(x)-\Delta f(x+1)=f(x)-2f(x+1)+f(x+2)$$
etc. Then
$$\Delta^n f(x)=\sum_{k=0}^n (-1)^k\binom{n}k f(x+k).$$
In our example $f(x)=1/(2x+1)$. So
$$\Delta f(x)=\frac1{2x+1}-\frac1{2x+3}=\frac2{(2x+1)(2x+3)}$$
$$\Delta^2 f(x)=\frac2{(2x+1)(2x+3)}-\frac2{(2x+3)(2x+5)}
=\frac{2\times 4}{(2x+1)(2x+3)(2x+5)}$$
etc. By induction
$$\Delta^n f(x)=\frac{2\times4\times\cdots\times (2n)}
{(2x+1)(2x+3)\cdots(2x+2n+1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2254933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
In triangle $ABC$ find angle $\angle BAC$ given that...
In triangle $ABC$, $D$ is a point on $BC$ such that $CD=2BD$.Also we know that $\angle B=45$ and $\angle ADC=60$.Find angle $\angle BAC$.
I tried to find two similar triangles but failed! I heared the aimed solution method uses similar triangles...
|
Let $E$ be the projection of $A$ on $BC$ and let us assume $AE=\sqrt{3}$.
We have $ED=1$ and $EB=\sqrt{3}$, from which $DB=\sqrt{3}-1$ and $CE=2\sqrt{3}-3$.
By the Pythagorean theorem, $AB=\sqrt{6}$ and $AC=2\sqrt{6-3\sqrt{3}}=\sqrt{6}(\sqrt{3}-1)$.
By the cosine theorem:
$$\cos\widehat{BAC}=\frac{AB^2+AC^2-BC^2}{2\,AB\cdot AC} = \frac{6+12(2-\sqrt{3})-18(2-\sqrt{3})}{12(\sqrt{3}-1)}=\frac{1}{2}$$
hence $\widehat{BAC}=\color{red}{60^\circ}$ as wanted.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2255965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
Let $B\in M_{2}(\mathbb{R})$ and let $T:M_{2}(\mathbb{R})\rightarrow M_{2}(\mathbb{R})$ defined by $T(A)=BA$. Find the rank of $T$ Let $B\in M_{2}(\mathbb{R})$ and let $T:M_{2}(\mathbb{R})\rightarrow M_{2}(\mathbb{R})$ defined by $T(A)=BA$. Find the rank of $T$
$\text{Im}(T)=\left\{ w\in M_{2}(\mathbb{R}):T(v)=w, \forall v\in M_{2}(\mathbb{R})\right\} $
$\dim(\text{Im}(T))= \text{rank} (T)$
Now, Let
$B=\begin{pmatrix}b_{1} & b_{2}\\
b_{3} & b_{4}
\end{pmatrix}$
$A=\begin{pmatrix}a_{1} & a_{2}\\
a_{3} & a_{4}
\end{pmatrix}$
Then, $BA=\begin{pmatrix}b_{1}a_{1}+b_{2}a_{3} & b_{1}a_{2}+b_{2}a_{4}\\
b_{3}a_{1}+b_{4}a_{3} & b_{3}a_{2}+b_{4}a_{4}
\end{pmatrix}=\begin{pmatrix}\alpha_{1} & \alpha_{2}\\
\alpha_{3} & \alpha_{4}
\end{pmatrix}$ with $\alpha_{i}\in M_{2}(\mathbb{R})$ and $1\leq i\leq4$
$\Rightarrow\begin{cases}
b_{1}a_{1}+b_{2}a_{3}=\alpha_{1}\\
b_{1}a_{2}+b_{2}a_{4}=\alpha_{2}\\
b_{3}a_{1}+b_{4}a_{3}=\alpha_{3}\\
b_{3}a_{2}+b_{4}a_{4}=\alpha_{4}
\end{cases}$
In this step i'm stuck, what am I supposed to clear to get the rank of $T$?
| First we find the matrix $\ T$ in respect to standard basis of $\ M_{2x2} $
Let $\ B=\pmatrix{a&b\\c&d}$
$\ T(\pmatrix{1&0\\0&0})=\pmatrix{a&0\\c&0}$
$\ T(\pmatrix{0&1\\0&0})=\pmatrix{0&a\\0&c}$
$\ T(\pmatrix{0&0\\1&0})=\pmatrix{b&d\\0&0}$
$\ T(\pmatrix{0&0\\0&1})=\pmatrix{0&b\\0&d}$
Then to the respect to the standard basis we form a matrix $\ T$
$\ T=\pmatrix{a&0&b&0\\0&a&b&d\\c&0&0&0\\0&c&0&d}$ then in the respect to $\ a,b,c,d$ we find the $\ rankT$
For $\ a,b,c,d\neq0\ rankT=4$
Cases where they are $\ 0$ I leave up to you to calculate the rank of T
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2257714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the derivatives of $\frac{d}{dx}\int _2^{x^3}\sin \left(t^2\right)dt$ and $\frac{d}{dx}\int _{\tan x}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt$ I know that $\frac{d}{dx}\int _a^xf\left(t\right)dt=f\left(x\right)\:$ and $\frac{d}{dx}\int _a^{g\left(x\right)}f\left(t\right)dt=f\left(g\left(x\right)\right)\cdot g'\left(x\right)$
I think I know what to do on the first problem, but I'm stumped on the second one. $a=\tan \left(x\right)$ rather than some constant, so I feel like I did not do it correctly.
a. $\frac{d}{dx}\int _2^{x^3}\sin \left(t^2\right)dt$
$$=\sin \left(\left(x^3\right)^2\right)\cdot 3x^2$$
$$=\sin \left(x^6\right)\cdot 3x^2$$
b. $\frac{d}{dx}\int _{\tan x}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt$
$$=\sqrt[3]{1+\sqrt{x+2}^2}\cdot \frac{1}{2\sqrt{x+2}}$$
$$=\sqrt[3]{x+3}\cdot \frac{1}{2\sqrt{x+2}}$$
$$=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}$$
| The Fundamental Theorem of Calculus implies that $$\frac{d}{dx}\int_{g(x)}^{h(x)}f(t)dt = f(h(x))h'(x)-f(g(x))g'(x)$$
So, your first problem is correct only because $$\frac{d}{dx}\int_2^{x^3}\sin \left(t^2\right)dt=\sin((x^3)^2)\cdot3x^2-\sin(2^2)\cdot\underbrace{\frac{d}{dx}(2)}_{=0}$$
and this works when the lower bound is any constant (similar if upper bound is constant).
For the second one, do the same: $$\frac{d}{dx}\int_{\tan(x)}^{\sqrt{x+2}}\sqrt[3]{1+t^2}dt=\sqrt[3]{1+(\sqrt{x+2})^2}\frac{d}{dx}\left(\sqrt{x+2}\right)-\sqrt[3]{1+\tan^2(x)}\cdot\frac{d}{dx}(\tan(x))$$ $$=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}-\sqrt[3]{\sec^2(x)}\cdot\sec^2(x)=\frac{\sqrt[3]{x+3}}{2\sqrt{x+2}}-\sec^{8/3}(x)$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Simplifying expression $\sqrt{6+2\sqrt{5}} - \sqrt{6-2\sqrt{5}}.$ I'm stuck. Squaring it will change it's value. Is there any general method of simplifying expressions of the form $$\sqrt{a+b}-\sqrt{a-b} = c \ \ ?$$
| It's clearly positive. Its square is
$$a^2=6+2\sqrt 5+6-2\sqrt 5-2\sqrt{(6+2\sqrt5)(6-2\sqrt5)}
=12-2\sqrt{16}=4$$
so $a=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2258833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdot...\cdot\frac{2n-1}{2n}\le \frac{1}{\sqrt{3n+1}}$ We want to prove by induction that
$$\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdots\frac{2n-1}{2n}\le \frac{1}{\sqrt{3n+1}}$$ for all $n,k \in \mathbb{Z^+}$
For k=1 : $\frac1{2}\le\frac{1}{2}$ which it is true.
Assume the induction hypothesis $\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdots\frac{2k-1}{2k}\le \frac{1}{\sqrt{3k+1}}$
We need to prove $\frac1{2}\cdot\frac3{4}\cdot\frac5{6}\cdots\frac{2(k+1)-1}{2(k+1)}\le \frac{1}{\sqrt{3(k+1)+1}}$
We can take the induction hypothesis and multiply by $\frac{2(k+1)-1}{2(k+1)}$ on both sides.
Then, we need to prove that $\frac{1}{\sqrt{3k+1}}\cdot\frac{2(k+1)-1}{2(k+1)}\le \frac{1}{\sqrt{3(k+1)+1}}$
$\frac{(2k+1)^2}{4(3k+1)(k+1)^2}\le \frac{1}{(3k+4)} \Leftrightarrow$
$(3k+4)(2k+1)^2\le 4(3k+1)(k+1)^2 \Leftrightarrow$
$12k^3+28k^2+19k+4\le 12k^3+16k^2+20k+4 \Leftrightarrow$
$12k^2-k\le 0$ which is not true!
Where is the mistake??
| $$\begin{align}4(3n+1)(n+1)^2&= 4(3n+1)(n^2+2n+1) \\
&= 4 (3n^3+6n^2+3n+n^2 + 2n+1)\\
&= 12n^3 + \mathbf{28n^2} + 20n + 4\\
&\ne 12n^3 + 16n^2 + 20n + 4\end{align}$$
For the record, I just plugged the larger algebraic steps into a calculator with arbitrary numbers and saw there was arbitrary mismatch at that line.
I trust you can carry on from here.
| {
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"timestamp": "2023-03-29T00:00:00",
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Find the value of the Product $\prod_{k=1}^{\infty} \frac{2k(2k+2)}{(2k+1)^2}$ Find the value of the Product $$P=\prod_{k=1}^{\infty} \frac{2k(2k+2)}{(2k+1)^2}$$
we have
$$P=\prod_{k=1}^{\infty}\left(1-\frac{1}{(2k+1)^2}\right)$$ Taking $ln$ on both sides we get
$$\ln(P)=\sum_{k=1}^{\infty}\ln\left(1-\frac{1}{2k+1}\right)+\sum_{k=1}^{\infty}\ln\left(1+\frac{1}{2k+1}\right)$$
Is there any way to continue further?
| From the formula $$\prod_{k\geq0}\frac{\left(k+a\right)\left(k+b\right)}{\left(k+c\right)\left(k+d\right)}=\frac{\Gamma\left(c\right)\Gamma\left(d\right)}{\Gamma\left(a\right)\Gamma\left(b\right)},\,a+b=c+d$$ we have $$\prod_{k\geq1}\frac{2k\left(2k+2\right)}{\left(2k+1\right)^{2}}=\prod_{k\geq0}\frac{\left(k+1\right)\left(k+2\right)}{\left(k+3/2\right)^{2}}=\frac{\Gamma^{2}\left(3/2\right)}{\Gamma\left(1\right)\Gamma\left(2\right)}=\color{red}{\frac{\pi}{4}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2265797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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How to find the least real number $M$? Determine the least real number $M$ such that the inequality
$|ab(a^2-b^2)+bc(b^2-c^2)+ca(c^2-a^2)|\le$ $M(a^2+b^2+c^2)^2$
holds for all real numbers $a$, $b$ and $c$.
| By AM-GM
$$\frac{|\sum\limits_{cyc}ab(a^2-b^2)|}{(a^2+b^2+c^2)^2}=\frac{9|(a-b)(b-c)(c-a)(a+b+c)|}{((a-b)^2+(b-c)^2+(c-a)^2+(a+b+c)^2)^2}\leq$$
$$\leq\frac{9|\frac{(a-b+b-c)^2}{4}(c-a)(a+b+c)|}{\left(\frac{(a-b+b-c)^2}{2}+(c-a)^2+(a+b+c)^2\right)^2}=\frac{9|(c-a)^3(a+b+c)|}{\left(3(c-a)^2+2(a+b+c)^2\right)^2}\leq$$
$$\leq\frac{9|(c-a)^3(a+b+c)|}{\left(4\sqrt[4]{(c-a)^6\cdot2(a+b+c)^2}\right)^2}=\frac{9}{16\sqrt2}.$$
It's obvious that the equality occurs, which says that the answer is $\frac{9}{16\sqrt2}.$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
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How to find the sum of the series $(1) + (2 + 3) +(4 + 5 + 6) + \cdots$ to $n$ terms? How to find the sum of this series?
$$(1) + (2 + 3) +(4 + 5 + 6) + \cdots \text{ to }\, n \, \text{ terms }$$
Answer is given as:
$$\frac {1}{8} n (n + 1) (n^2 + n +2) $$
| The first element in all sum is $\frac{n(n-1)}{2}+1$.
Thus, the sum of all elements in $n$-th sum is $$\frac{n^2(n-1)}{2}+1+2+...+n=\frac{n^2(n-1)}{2}+\frac{n(n+1)}{2}=\frac{n}{2}(n^2+1).$$
Thus the given sum is
$$\sum_{k=1}^n\frac{k}{2}(k^2+1)=\frac{1}{2}\sum_{k=1}^n(k^3+k)=\frac{n^2(n+1)^2}{8}+\frac{n(n+1)}{4}=\frac{n(n+1)(n^2+n+2)}{8}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Ratio test involving factorials: $a_{n} = \frac{n-2}{(n+1)!}$; finding $\lim_{n \to \infty} |\frac{a_{n+1}}{a_n}|$ The correct answer is apparently $0$ however I end up with $1$. I do the following:
$$a_{n+1} = \frac{(n+1)-2}{((n+1)+1)!} = \frac{n-1}{(n+2)!}$$
$$
\begin{split}
\lim_{n \to \infty} \left|\frac{a_{n+1}}{a_n}\right|
&= \lim_{n \to \infty} \left|\frac{n-1}{(n+2)(n!)}
\times \frac{(n+1)(n!)}{n-2}\right| \\
&= \lim_{n \to \infty} \left|\frac{(n-1)(n+1)}{(n+2)(n-2)}\right| \\
&= \lim_{n \to \infty} \left|\frac{n^2-1}{n^2-2}\right| \\
&= \lim_{n \to \infty} \frac{n^2}{n^2} \\
&= 1
\end{split}
$$
I assumed as $n$ approaches $\infty$ the '$-1$' and '$-2$' become meaningless, however I'm pretty certain that is now incorrect unless I've made an error earlier on.
| It should be $$\begin{align*}\lim_{n \to \infty} \left|\frac{n-1}{(n+2)(n+1)n!} \cdot \frac{(n+1)n!}{n-2}\right| = \lim_{n\to \infty} \frac{n-1}{(n+2)(n-2)} = 0\end{align*}$$
since $$\begin{align*}(n+2)! &= (n+2)(n+1)\underbrace{\color{blue}{(n)(n-1)\cdots (2)(1)}}_{\color{blue}{n!}} \\ & = (n+2)(n+1)\color{blue}{n!}\end{align*}$$
In general, we have $(n+k)! = (n+k)(n+k-1)\cdots (n+1)n!$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Minimum value of $\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}$ Let $x,y,z>0$ and $x+y+z=xyz$. What is the minimum value of $$\frac{x}{y^2}+\frac{y}{z^2}+\frac{z}{x^2}?$$
In the case when $x=y=z$, the equation $x+y+z=xyz$ translates to $3x=x^3$, or $x=\sqrt{3}$. If $A$ denotes the quantity that we want to minimize, then $A=\sqrt{3}$ as well.
If we use the inequality of arithmetic and geometric means, we get
$$A\geq \frac{3}{\sqrt[3]{xyz}}=\frac{3}{\sqrt[3]{x+y+z}}.$$
| Let $x_1 = 1/x, y_1 = 1/y, z_1 = 1/z$. Then $x_1y_1 + x_1z_1 + y_1z_1 = 1$.
$$
\sum_{cyc} \frac{x}{y^2} \geq \sum_{cyc} \frac1x = x_1 + y_1 + z_1
$$
Due to rearrangement inequality $(x_1 + y_1 + z_1)^2 \geq 3(x_1y_1 + x_1z_1 + y_1z_1) = 3$. So
$$
x_1 + y_1 + z_1 \geq \sqrt{3}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2272279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
Primes dividing sum of binomial coefficients. Given that $p>3$ is a prime, we have $k=\lfloor \frac {2p}{3} \rfloor$ then prove or disprove that $\sum_{i=1}^k\binom{p}{i}\equiv 0\pmod{p^2}$.
Its easy to see how $p$ divides each of the binomial coefficients, but I tried using maximal powers of $p$ in each coefficient but was stuck can someone help.
| Ahmad showed the problem is equivalent to
*
*$H_{2k}-H_k \equiv 0 \pmod p$ for $p \equiv 1 \pmod 3$
*$H_{2k}-H_k+(2k+1)^{-1} \equiv 0 \pmod p$ for $p \equiv 2 \pmod 3$
1:
$$H_{2k}-H_k \equiv \sum_{i=1}^{2k} i^{-1} - \sum_{i=1}^{k} i^{-1} \equiv \sum_{i=k+1}^{2k} i^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=\frac{3}{2}k+1}^{2k} i^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=-\frac{3}{2}k-1}^{-2k} (-i)^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=3k+1-\frac{3}{2}k-1}^{3k+1-2k} (-i)^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=k+1}^{\frac{3}{2}k} (-i)^{-1} \equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + \sum_{i=k+1}^{\frac{3}{2}k} (-1)^{-1}(i)^{-1}\equiv
\sum_{i=k+1}^{\frac{3}{2}k} i^{-1} + -i^{-1}
\equiv
\sum_{i=k+1}^{\frac{3}{2}k} 0
\equiv
0
\pmod p$$
The idea is that for each inverse the opposite inverse is also in the sum and therefore the sum cancelled out. You need to show $k$ is even so no inverse has no companion. If $k$ is odd, $3k+1$ is divisible by $2$ and therefore not a prime.
2:
Same idea just with odd number of terms.
$$\begin{align}
H_{2k}-H_k+(2k+1)^{-1} \equiv \\ &
\equiv
(2k)^{-1}+(2k-1)^{-1}+\cdots+(1)^{-1} - ((k)^{-1}+(k-1)^{-1}+\cdots+(1)^{-1}) + (2k+1)^{-1} \\ &
\equiv
(2k)^{-1}+(2k-1)^{-1}+\cdots+(k+1)^{-1} +(2k+1)^{-1} \\ &
\equiv
(2k)^{-1}+(2k-1)^{-1}+\cdots+(k+1)^{-1} + (2k+1)^{-1} \\ &
\equiv
-(k+2)^{-1}+-(k+3)^{-1}+\cdots+(k+1)^{-1} + -(k+1)^{-1} \\ &
\equiv
0
\pmod p \end{align}$$
Again we need to show $k$ is always odd. If $k$ is even $3k+2$ is divisible by $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
show that $11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $
Let $n$ be postive integer,show that
$$11\nmid n^6-2n^5+4n^4-8n^3+16n^2-32n+64 $$
I have use ugly metods to solve consider this case,$n\equiv 0,\pm 1,\pm 2,\pm 3,\pm 4,\pm 5\pmod {11}$,have other more simple methods it?
| Suppose $11\mid n^6-2n^5+4n^4-8n^3+16n^2-32n+64$. Then $n^7 + 2^7 = 0 (\mod 11)$ therefore $n^7 + 7 = 0 (\mod 11)$ equivalent $n^7= 4 (\mod 11)$. Using Fermat's little theorem, we get $1 = 4n^3(\mod 11) $ equivalent $3 = n^3(\mod 11)$
From $n^7= 4 (\mod 11)$ we get $9n = 4 (\mod 11) $ therefore $n=9(\mod 11) $. Now $n^6-2n^5+4n^4-8n^3+16n^2-32n+64 (\mod 11)$ for $n=-2$ becomes $7 \cdot2^6 (\mod 11)$ which is $\ne 0 \mod 11$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2274569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to integrate $\int 2xe^{x^2-y^2}\cos(2xy)- 2ye^{x^2-y^2}\sin(2xy) \ \mathrm dy$? Find the value of:
$$\int 2xe^{x^2-y^2}\cos(2xy)- 2ye^{x^2-y^2}\sin(2xy) \ \mathrm dy$$
i have tried using integration by parts but it doesn't seem to work. How would I go about this integral ?
| $$\int \left(2xe^{x^2-y^2}\cos 2xy - 2ye^{x^2-y^2}\sin 2xy\right) \, dy = \int 2xe^{x^2-y^2} \cos 2xy \, dy - \int 2ye^{x^2-y^2} \sin 2xy \, dy $$
Let's integrate that first one on the RHS by parts. We'll take $u = e^{x^2-y^2}$. So then $du = -2ye^{x^2-y^2} \, dy$, $dv = 2x \cos 2xy \, dy$, and $v = \sin 2xy$. (It may not be clear at first, but note that $v$ can easily be obtained from $dv$ with a simple substitution integral. Substitute $t = 2xy$ into $\int 2x \cos 2xy \, dy$ to get $v = \sin 2xy$.)
With integration by parts, we have
\begin{align*}
\int 2xe^{x^2-y^2} \cos2xy \, dy &= e^{x^2-y^2}\sin 2xy - \int \sin (2xy) \cdot \left(-2y e^{x^2-y^2}\right) \, dy\\[0.3cm]
&= e^{x^2-y^2}\sin 2xy + \int 2y e^{x^2-y^2} \sin 2xy \, dy\\[0.3cm]
\end{align*}
Now use this with the very first line of my post to get a great cancellation, giving you the desired result immediately.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2279500",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $\sqrt{3+\sqrt{3+x}}=x$
Solve
$$\sqrt{3+\sqrt{3+x}}=x$$
My try:
$$\sqrt{3+\sqrt{3+x}}=x \\
3+\sqrt{3+x}=x^2\\\sqrt{3+x}=x^2-3\\3+x=(x^2-3)^2$$
$$x^4-6x^2+9=x+3\\x^4-6x^2-x+6=0$$
Now ?
| We have to factorize $x^4 - 6x^2 - x + 6 = 0.$
We would observe that the values of x would be $1,-2,\frac{1±\sqrt{13}}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2280929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Inverse Z transform of $\frac {-81}{z^4(z-3)(z-1)}$ The question given like this:
Find the convolution of $3^nu(-n+3)*u(n-2)$
Attempt:
I tried to solve it through z transform method
Let $Y(z)=X1(z)X2(z)$
Now $Z$ transform of $x1(n)=3^nu(-n+3)$ will be $\frac{-81z^{-3}}{z-3}$
Z transform of $u(n-2)=\frac{1}{z(z-1)}$
now $Y(z)=\frac {-81}{z^4(z-3)(z-1)}$
I tried to take inverse z transform of this function by residue theorem like this
$Residue at z=3,will be \frac{-3^{n-1}}{2}$.
$Residue at z=1,will be \frac{-81}{2}$
$Residue at z=0,will be -40$
BUT the solution given like this
$$
y(n)=
\begin{cases}
\frac{81}{2},n>=5\\
\frac{3^{n-1}}{2},n<5\\
\end{cases}
$$
Now What mistake i am doing,any other method other than Residue throem will be helpful.
Thanks
| $$Y(z) = \frac{-81}{z^4(z-3)(z-1)}$$
$$X(z) = z^4Y(z) =\frac{-81}{(z-3)(z-1)}$$
$$X(z) = \frac{-81}{(z-3)(z-1)} = \frac{A}{z-1} + \frac{B}{z-3} $$
$$A = \frac{81}{2}; B=\frac{-81}{2}$$
$$x(n) = \frac{81}{2}[1-3^{n-1}]u(n-1)$$
$$ X(z) = z^4Y(z)$$
$$x(n-4) = y(n)$$
$$y(n) = \frac{81}{2}[1-3^{n-5}]u(n-5)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n=\frac{4^p+1}{5}$ is a strong 2-pseudoprime for all $p>5$ I want to prove that $n=\frac{4^p+1}{5}$ is a strong 2-pseudoprime for all $p>5$ where $p$ is a prime.
i.e. $n$ is a composite number which passes the Miller-Rabin test.
It is easy to check that it is indeed a composite no.
We write $n-1=2^st$ where $t$ is an odd number. $n$ passes the test if either
$$a^t\equiv1 \space(mod \space n)$$ or
$$a^{{2^i}t}\equiv-1 \space(mod \space n)$$ for some $i<s$.
Then $n-1=\frac{4^p+1}{5}-1= \space ... \space=2^2(\frac{4^{p-1}\space-1}{5})$so $s=2$ and $t=\frac{4^{p-1}\space-1}{5}$
So is $$2^{\frac{4^{p-1}\space-1}{5}}\equiv1\space(mod \space \frac{4^p+1}{5})$$
or
$$2^{2^1(\frac{4^{p-1}\space-1}{5})}\equiv-1\space(mod \space \frac{4^p+1}{5})$$?
I checked the results for the first few primes (Yup, WolframAlpha did help).
It turns out that it is always the latter case which holds
My guess is that:
if $p\equiv3 \space (mod \space 4)$ then $2^{\frac{4^{p-1}\space-1}{5}}\equiv2^p\space(mod \space \frac{4^p+1}{5})$
if $p\equiv1 \space (mod \space 4)$ then $2^{\frac{4^{p-1}\space-1}{5}}\equiv-2^p\space(mod \space \frac{4^p+1}{5})$.
Squaring both sides and the result follows. Could anyone show why my guess is correct (or not?)?
| Just want to fill in the gap that $n$ is indeed a composite number
since $p$ is an odd prime, $(p+1)/2$ is a positive integer.
$4^p+1=(2^p+1)^2-2^{p+1}=(2^p+1)^2-2^{2((p+1)/2)}=(2^p+2^{(p+1)/2}+1)(2^p-2^{(p+1)/2}+1)$
We show that $5$ divides one of $(2^p+2^{(p+1)/2}+1)$ and $(2^p-2^{(p+1)/2}+1)$. What's left from the dividend is $>1$ and so multiplied by that not divisible by $5$ we get a composite no.
There are 4 cases:
-when $p=4k+1$ and $k$ is odd
-when $p=4k+1$ and $k$ is even
-when $p=4k+3$ and $k$ is odd
-when $p=4k+3$ and $k$ is even...
for example, when $p=4k+3$ and $k$ is odd
$2^p+2^{(p+1)/2}+1=2^{4k+3}+2^{2k+2}+1=8(16^k)+4(4^k)+1\equiv3(1^k)-(-1)^k+1\equiv0\space(mod\space5)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2281956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
PreCalc Roots of Unity Let $\omega$ be a complex number such that $\omega^5 = 1$ and $\omega \neq 1$. Find
$$\frac{\omega}{1 + \omega^2} + \frac{\omega^2}{1 + \omega^4} + \frac{\omega^3}{1 + \omega} + \frac{\omega^4}{1 + \omega^3}$$
So far, I have simplified that down to $\frac{-2}{(1+\omega)^2}$, but I don't know how to continue. Thank you!
| We need $$\sum_{r=1}^4\dfrac{x^r}{1+x^{2r}}$$ where $x^5=1$
Let $y=\dfrac x{1+x^2}=\dfrac1{x+\dfrac1x}$
$$\implies\dfrac1{y^5}=\left(x+\dfrac1x\right)^5=x^5+\dfrac1{x^5}+\binom51\left(x^3+\dfrac1{x^3}\right)+\binom52\left(x+\dfrac1x\right)$$
Now as $x^3+\dfrac1{x^3}=\left(x+\dfrac1x\right)^3-3\left(x+\dfrac1x\right)=\dfrac1{y^3}-\dfrac3y$
$$\implies\dfrac1{y^5}=1+\dfrac11+5\left(\dfrac1{y^3}-\dfrac3y\right)+\dfrac{10}y$$
As $y\ne0,$ multiply both sides by $y^5$ to get $$2y^5-5y^4+\cdots=0$$
By Vieta's formula, $$\sum_{r=1}^5\dfrac{x^r}{1+x^{2r}}=\dfrac52$$
But $r=5\implies\dfrac{x^r}{1+x^{2r}}=\dfrac{x^5}{1+(x^5)^2}=\dfrac1{1+1}$ as $x^5=1$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2282765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Conditional probability for bivariate random variables question: $$ f_{(X,Y)(x,y)}=
\begin{cases} 8xy, &0\le y\le x \le 1\\0.&\text{elsewhere} \end{cases} $$
Find $ P\left( X\le \frac{1}{2}|Y\ge \frac{1}{4}\right) $
What is wrong with my solution:
$$
\frac{\int_{\frac{1}{4}}^{\frac{1}{2}}\int_{\frac{1}{4}}^x8xydydx}{\int_{\frac{1}{4}}^{1}4y-4y^3dy}
$$
seeing that $ 4y-4y^3 $is the marginal pdf of Y
| We note that the probability density function is:
$$ f_{X,Y}(x,y) = \begin{cases} 8xy & 0 \leq y \leq x \leq 1 \\ 0 & \text{else} \end{cases} $$
We can visualize the three regions of interest for this problem,
*
*$\color{darkblue}{\text{The support of the distribution, the set } S \text{ of points } (x,y) \text{ such that } 0 \leq y \leq x \leq 1}$
*$\color{goldenrod}{\text{The set } R_1 \text{ of points } (x,y) \text{ such that } x \leq \frac{1}{2}}$
*$\color{green}{\text{The set } R_2 \text{ of points } (x,y) \text{ such that } y \geq \frac{1}{4}}$
as follows:
The conditional probability can be expressed as:
$$ \mathbb{P} \left[X \leq \frac{1}{2} \,\,\bigg\rvert\,\, Y \geq \frac{1}{4} \right] = \frac{\mathbb{P} \left[X \leq \frac{1}{2} , Y \geq \frac{1}{4} \right]}{\mathbb{P} \left[Y \geq \frac{1}{4} \right]} $$
The numerator probability is calculated by integrating the probability density function over the area where the event is true, which is $R_1 \cap R_2$ (and, since we don't care about zero probability mass, this region's intersection with the support $S$). We use the colored diagram to see what this region is and what the limits of integration are. We see that:
\begin{align*}
\mathbb{P} \left[X \leq \frac{1}{2} , Y \geq \frac{1}{4} \right] &= \iint_{R_1 \cap R_2} f_{X,Y}(x,y) \, dA \\
&= \iint_{R_1 \cap R_2 \cap S} f_{X,Y}(x,y) \, dA \\
&= \int_\frac{1}{4}^\frac{1}{2} \int_\frac{1}{4}^x 8xy \, dy \, dx \\
&= \int_\frac{1}{4}^\frac{1}{2} \left[ 4xy^2 \right]_{y = \frac{1}{4}}^{y=x} \, dx \\
&= \int_\frac{1}{4}^\frac{1}{2} \left(4x^3 - \frac{1}{4}x \right) \, dx \\
&= \left[x^4 - \frac{x^2}{8} \right]_\frac{1}{4}^\frac{1}{2} \\
&= \frac{1}{16} - \frac{1}{32} -\frac{1}{256} + \frac{1}{16\cdot8} \\
&= \frac{16-8-1+2}{256} \\
&= \frac{9}{256} \\
\end{align*}
The denominator probability is calculated similarly on $R_2$ (and its intersection with the support $S$, since its where the probability mass is). We use the colored diagram to see what this region is and what the limits of integration are. We see that:
\begin{align*}
\mathbb{P} \left[Y \geq \frac{1}{4} \right] &= \iint_{R_2} f_{X,Y}(x,y) \, dA \\
&= \iint_{R_2 \cap S} f_{X,Y}(x,y) \, dA \\
&= \int_\frac{1}{4}^1 \int_\frac{1}{4}^x 8xy \, dy \, dx \\
&= \int_\frac{1}{4}^1 \left(4x^3 - \frac{1}{4}x \right) \, dx \\
&= \left[x^4 - \frac{x^2}{8} \right]_\frac{1}{4}^1 \\
&= 1 - \frac{1}{8} - \frac{1}{256} + \frac{2}{256} \\
&= \frac{7}{8} + \frac{1}{256} \\
&= \frac{225}{256}
\end{align*}
Putting the results together, we see that the conditional probability we are looking for is:
$$ \mathbb{P} \left[X \leq \frac{1}{2} \,\,\bigg\rvert\,\, Y \geq \frac{1}{4} \right] = \frac{9}{225} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2284526",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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In a quadrilateral $ABCD, $ it is given that $AB$ is parallel to $CD$...
In a quadrilateral $ABCD, $ it is given that $AB$ is parallel to $CD$ and the diagonals are perpendicular to each other. Show that
(i)$AD\cdot BC\ge AB\cdot CD$
(ii)$AD+BC\ge AB+CD$
I have no idea how to proceed.
Any help will be appreciated.
| Let $O$ be the point where $AC$ and $BD$ meets. Let $AO = a, BO = b, CO = c, DO = d$. Then, we know that $$\frac{a}{c}=\frac{b}{d}:=t\implies a = tc, \ \ b =td$$ for some $t$ because $AC\parallel BD$. The first inequality is equivalent to showing $$(a^2+d^2)(b^2+c^2)\geq(a^2+b^2)(c^2+d^2)$$ which is true because $$Left-Right = (a^2-c^2)(b^2-d^2) = c^2d^2(t^2-1)^2\geq 0\implies Left\geq Right$$
The second inequality is equivalent to showing $$\sqrt{a^2+d^2} + \sqrt{b^2+c^2}\geq\sqrt{a^2+b^2}+\sqrt{c^2+d^2}$$
Taking the square on both sides, it is equivalent to showing $$\sqrt{(a^2+d^2)(b^2+c^2)}\geq\sqrt{(a^2+b^2)(c^2+d^2)}$$
which is the same as the first inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2287569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $w_1 = 3\cdot cis(\frac{4\pi}{3})$ and $w_2=1/3 \cdot cis(\alpha)$. Determine $\alpha$ so that $w_1\cdot w_2$ is a real number I tried:
$$3 \operatorname*{cis}(\frac{4\pi}{3})\cdot \frac{1}{3} \operatorname*{cis}(\alpha) = \operatorname*{cis}(\frac{4\pi}{3}+\alpha) = \cos(\frac{4\pi}{3}+\alpha) + \sin(\frac{4\pi}{3}+\alpha)i$$
Since this has to be a real number I did:
$$\sin(\frac{4\pi}{3}+\alpha) = 0 \Leftrightarrow \frac{4\pi}{3}+\alpha = 2k\pi \lor \frac{4\pi}{3}+\alpha = \pi+2k\pi \Leftrightarrow \\
\alpha = 2k\pi-\frac{4\pi}{3} \lor \alpha = \pi+2k\pi-\frac{4\pi}{3}$$
But my book states the solution is $\alpha = \frac{2\pi}{3}+k\pi$. What went wrong?
By the way I know theres a shorter way of representing the solution and it goes something like $\alpha = \text{something }+(-1)^k\arcsin(\text{number})$, anyone knows it?
| Note that $$\sin { x } =0\quad \Rightarrow x=k\pi ,\quad k\in \mathbb Z $$ so $$\\ \sin { \left( \frac { 4\pi }{ 3 } +\alpha \right) =0\quad \Rightarrow \frac { 4\pi }{ 3 } +\alpha =k\pi ,\quad \Rightarrow } \alpha =-\frac { 4\pi }{ 3 } +k\pi ,\quad or\quad \alpha =\frac { 2\pi }{ 3 } +k\pi $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2288182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability that the sum of $k$ dice is $n$ The probability of two dice summing to k is simple enough, make a diagram of the throws,
$$
\begin{array}{c}
&|&1&2&3&4&5&6\\\hline
1&|&2&3&4&5&6&7\\
2&|&3&4&5&6&7&8\\
3&|&4&5&6&7&8&9\\
4&|&5&6&7&8&9&\color{green}{10}\\
5&|&6&7&8&9&\color{green}{10}&11\\
6&|&7&8&9&\color{green}{10}&11&12\\
\end{array}
$$
and the probability of a sum is just the length of the corresponding diagonal divided by $36$. For instance, $\mathrm{P}(\mathrm{sum} = 10) = \frac{3}{36} = \frac{1}{12}$. However, it's not as easy to draw a diagram for three dice or more, so how would a general formula look?
| Following the notation and steps of this previous answer: let $n$ dice of $D$ sides each one numbered from $1$ to $D$, then we can represent the throw of these dice as
$$\begin{align}f(x)&=(x^1+x^2+\ldots+x^D)^n=\left(x\sum_{k=0}^{D-1} x^k\right)^n=\left(x\frac{1-x^D}{1-x}\right)^n\\&=x^n\sum_{k=0}^n(-1)^k\binom{n}{k}x^{kD}\sum_{j=0}^\infty\binom{n+j-1}{n-1}x^j\tag{1}\end{align}$$
Now we want to find a formula for the coefficient of the monomial $x^S$ in the expansion of $f$, denoted by $[x^S]f(x)$. This coefficient represent the different ways that the sides of the $n$ thrown dice add up to $S$. From $(1)$ we can see that $S$ will have the form $S=n+kD+j$, thus
$$\bbox[5px,border:2px solid gold]{[x^S]f(x)=\sum_{k=0}^M(-1)^k\binom{n}{k}\binom{S-kD-1}{n-1},\quad M:=\left\lfloor \frac{S-n}{D}\right\rfloor}\tag2$$
The details of the algebraic manipulations to achieve the formula stated in $\rm (2)$ are in the answer previously linked. I dont know if this formula can be simplified further in an useful way.
Finally note that because the dice are fair then each side of each dice have the same probability to show up, so
$$
P(\operatorname{sum}=S)=\frac{[x^S]f(x)}{D^n}\tag3
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2290090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Characteristic Polynomials of a Matrix in $\mathbb F_2^3$ Let $K = \mathbb F_2$ and $V= K^3$, $\phi \in End(V)$
How many possible characteristic Polynomials exist?
My attempt:
Since its a 3x3 - Matrix and the diagonals are either $0$ or $1$ i get $8 $ different polynomials:
$ a_3(x-1)^3 +a_0\qquad ; \qquad a_3(x-1)^3 +a_1(x-1) +a_0$
$a_3x(x-1)^2 +a_0 \qquad ; \qquad a_3(x-1)^2x + a_1(x-1) +a_1x +d$
$ax^2(x-1)+b \qquad ; \qquad a_3x^2(x-1) + a_1x +a_1(x-1) +d$
$ax^3 +b \qquad ; \qquad a_3x^3 + a_1x$
Is that correct?
| As each characteristic polynomial in this case is a monic polynomial of degree $3$, there can be at most $8$ polynomials, which can be found by looking for all different combinations of coefficients of $x^2$, $x^1$ and $x^0$. Each coefficient must be in $\mathbb{F}_2$, thus we have:
$$
\begin{eqnarray*}
x^3+0x^2+0x+0 &=& x^3\\
x^3+0x^2+0x+1 &=& x^3+1\\
x^3+0x^2+1x+0 &=& x^3+x\\
x^3+0x^2+1x+1 &=& x^3+x+1\\
x^3+1x^2+0x+0 &=& x^3+x^2\\
x^3+1x^2+0x+1 &=& x^3+x^2+1\\
x^3+1x^2+1x+0 &=& x^3+x^2+x\\
x^3+1x^2+1x+1 &=& x^3+x^2+x+1
\end{eqnarray*}
$$
In order to show that there are exactly 8 polynomials, we now need to show that each of those polynomials can actually occur as a characteristic polynomial of a $3\times 3$-Matrix over $\mathbb{F}_2$. This is trivial, we only need to take the companion matrices of the polynomials.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2291068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Computing $\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} $
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} \space \text{ converges by the alternating series test, but what is its value?} $$
$1 - \sin(x) \space \text{ is an entire function} $, so by Weierstrass Factorization Theorem, it can be written as a product of its zeros, which occur at $x = \frac{(4n+1)\pi}{2} $ and $\frac{-(4n+3) \pi}{2}$
Equating this to its Taylor series, we get
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x)(1 + \frac{2}{3\pi}x)(1 - \frac{2}{5\pi}x)(1 + \frac{2}{7\pi}x)... $$
$$1 - x + O(x^3) = (1 - \frac{2}{\pi}x(1-\frac13)+O(x^2))(1 - \frac{2}{\pi}x(\frac15-\frac17)+O(x^2))... $$
$$1 - x + O(x^3) = 1 - \frac{2}{\pi}x(1-\frac13+\frac15-\frac17+...)+O(x^2)$$
$$x - O(x^3) = \frac{2}{\pi}x \sum_{n=0}^\infty \frac{(-1)^n}{2n+1}-O(x^2)$$
Equating $x$ terms,
$$\sum_{n=0}^\infty \frac{(-1)^n}{2n+1} = \frac{\pi}{2}$$
However, I seem to be off by a factor of $2$ (actual answer is $\frac{\pi}{4} $). Does anyone see where this went wrong?
| This is the value of the Dirichlet beta function at the input $s=1$, which is defined as the series
$$\beta(s)=\sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)^s}$$
There is an integral representation related with the Gamma Function and the Dirichlet beta function.
$$\beta(s)\Gamma(s)=\int_0^\infty \frac{x^{s-1}e^{-x}}{1+e^{-2x}}\, \mathrm dx$$
Sketch of the proof:\begin{align}\mathcal{I}=\int_0^\infty \frac{x^{s-1}e^{-x}}{1+e^{-2x}}\, \mathrm dx&=\int_0^\infty x^{s-1}e^{-x}\bigg(\sum_{n=0}^\infty(-e^{-2x})^k\bigg)\\&=\sum_{n=0}^\infty (-1)^k\int_0^\infty x^{s-1}e^{-(2k+1)x}\, \mathrm dx\\&=\sum_{n=0}^\infty \frac{(-1)^k}{(2k+1)^s}\Gamma(s)\\&=\beta(s)\Gamma(s)\end{align}
In particular, we would find
\begin{align}\sum_{n=0}^\infty \frac{(-1)^n}{2n+1}&=\beta(1)\Gamma(1)\\&=\int_0^\infty \frac{e^{-x}}{1+e^{-2x}}\, \mathrm dx\\&=\int_0^\infty \frac{e^{x}}{e^{2x}+1}\ \mathrm dx\\&=\int_0^\infty\frac{1}{1+t^2}\, \mathrm dt\, \text{, via substituting $t=e^x$}\\&=\frac{\pi}{2}\end{align}
Hence proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2292606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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If $ { a\over a+1} + { b\over b+1 } + { c\over c+1 } = 1 $ prove $ abc \le 1/8 $ If $ a,b,c $ are positive real numbers and $ { a\over a+1} + {b\over b+1} + {c\over c+1} = 1 $ then prove that $ abc \le {1\over 8} $
Could I get some help with this?
What I have tried-
Making the denominators equal and adding to get another equation of the form $ {N\over M}= 1 $ then multiplying by $ M $ both the sides and simplifying to get $ 2abc + ab + bc + ac = 1 $ , from which we have,
$ abc \lt 1/2 $
But which obviously isn't enough to prove that $ abc \le 1/8 $
I also tried using Jensen's inequality which states that
$ f(E(X)) \ge E(f(X)) $ when $ f $ is a concave function and $ f(E(X)) \le E(f(X)) $ when $ f $ is a convex function, and where $ E(X) $ denotes the expectation of X ,
but that didn't really help.
| By AM-GM:$$\frac{1}{a+1}=1-\frac{a}{a+1}=\frac{b}{b+1}+\frac{c}{c+1}\geq2\sqrt{\frac{bc}{(b+1)(c+1)}}.$$
By the permutation symmetry, the inequality is invariant with respect to the cyclic permutation of $(a,b,c)$.
Thus, $$\frac{1}{(a+1)(b+1)(c+1)}\geq\frac{8abc}{(a+1)(b+1)(c+1)}$$
and we are done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294748",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
How to find $\lim\limits_{x \to 8} \frac{\frac{1}{\sqrt{x +1}} - \frac 13}{x-8}$ I am trying to find the limit as $x\to 8$ of the following function. What follows is the function and then the work I've done on it.
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}$$
\begin{align}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} &= \frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8} \times \frac{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}} \\\\
& = \frac{\frac{1}{x+1}-\frac{1}{9}}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac{8-x}{(x-8)\left(\frac{1}{\sqrt{x +1}} + \frac{1}{3}\right)}\\\\
& = \frac {-1}{\frac{1}{\sqrt{x +1}} + \frac{1}{3}}\end{align}
At this point I try direct substitution and get:
$$ = \frac{-1}{\frac{2}{3}}$$
This is not the answer. Could someone please help me figure out where I've gone wrong?
| Let $\sqrt{x+1}-3=h\implies x=(3+h)^2-1$
$$ \lim_{x\to 8}\frac{\frac{1}{\sqrt{x +1}} - \frac{1}{3}} {x-8}=\lim_{h\to0^+}\dfrac{\dfrac1{h+3}-\dfrac13}{(3+h)^2-9}=\lim_{h\to0^+}\dfrac{3-(h+3)}{3h(6+h)(h+3)}=?$$
Cancel out $h$ safely as $h\ne0$ as $h\to0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2294991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$ then $x+\dfrac{1}{x}=?$
let :
$$x = \frac{{\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18} }}{{\sqrt 2 + \sqrt 3 + \sqrt 4 }}$$
then :
$$x+\dfrac{1}{x}=?$$
My try :
$$\sqrt 3 + \sqrt 6 + \sqrt {16} + \sqrt {18}=\sqrt3+\sqrt 3\times\sqrt2+4+3\sqrt2 \\ \sqrt3(1+\sqrt2+3)+4$$
Now what ?
| (Too long for a comment.) For a more systematic, albeit grossly overkill in this case, approach:
*
*with $\,u=\sqrt{2}\,$ and $\,v=\sqrt{3}\,$:
$$
x = \frac{{\sqrt 3 + \sqrt 6 + 4 + 3 \sqrt {2} }}{{\sqrt 2 + \sqrt 3 + 2 }} = \frac{4+ 3u + v + uv}{2+u+v}
$$
*
*eliminate $u$ by Resultant[x (2 + u + v) - (4 + 3 u + v + u v), u^2 - 2, u]:
$$
v^2 x^2 - 2 v^2 x - v^2 + 4 v x^2 - 8 v x - 4 v + 2 x^2 - 4 x - 2
$$
*
*then $v$ by Resultant[v^2 x^2 - 2 v^2 x - v^2 + 4 v x^2 - 8 v x - 4 v + 2 x^2 - 4 x - 2, v^2 - 3, v]:
$$
-23 x^4 + 92 x^3 - 46 x^2 - 92 x - 23 = -23 (x^2 - 2 x - 1)^2
$$
*
*then let $y = x + 1/x \iff x^2-xy+1=0\,$, and Resultant[x^2 - 2 x - 1, x^2 - x y + 1, x]:
$$
8 - y^2
$$
Therefore $x+1/x=y=\pm \sqrt{8}\,$ and, since $\,x\,$ is known to be positive, $\,x+1/x=+\sqrt{8}=2\sqrt{2}\,$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2295387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Find $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$ $\lim_{n\to\infty}\sin^4x+\frac{1}{4}\sin^42x+.....+\frac{1}{4^n}\sin^42^nx$
I feel that this question will be solved by telescoping series but i cannot express it as telescoping series.
$\sin^4x=(\frac{1-\cos2x}{2})^2=\frac{1+\cos^22x-2\cos2x}{4}$
I am not able to solve it further.
The options are $(A)\sin^4x,(B)\sin^2x,(C)\cos^2x$(D)does not exist
| Consider : $\displaystyle \frac{\sin^{4}(2^{i}x)}{4^i} + \frac{\sin^{4}(2^{i+1}x)}{4^{i+1}} = \frac{1}{4^{i}}\big(\frac{1}{8}\cos(4\cdot2^{i}x) -\frac{1}{2}\cos(2\cdot2^{i}) + \frac{3}{8} + \frac{1}{32}\cos(2\cdot2^{i+1})-\frac{1}{8}\cos(2\cdot2^{i+1}) + \frac{3}{8}\big)$
Now you could subtract two terms in neighbour parts.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2297465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Prove by induction that $ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$
Prove by induction that
$$ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$ with $n \geq 1$
*
*Testing n=1:
$$\sin(x)= \frac {1-\cos(2x)}{2\sin(x)}$$
$$2\sin^2(x)=1-\cos(2x)$$
$$2\sin^2(x)=1-[\cos^2(x)-\sin^2(x)]$$
$$\sin^2(x)=1-\cos^2(x)$$
$$\sin^2(x)+\cos^2(x) =1$$
It shows that n=1 yields a true identity (Pythagorean identity)
*Let's assume that $P_n$ is true:
$$\ sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$
*Let's consider adding $\sin(2n+1)x$ to $P_n$:
$$ \sin(x) +\sin(3x) +...+ \sin (2n-1)x+ \sin(2n+1)x= \frac{1-\cos(2nx)}{2\sin x} + \sin(2n+1)x$$
Considering the RHS:
$$\frac{1-\cos(2nx)}{2\sin x} + \sin (2n+1)x$$
$$\frac{1-\cos(2nx)+ \sin(2n+1)x \cdot 2\sin x}{2 \sin x}$$
$$\frac{1-\cos(2nx)+ 2[\sin(2n+1)x \cdot sin x]}{2sinx}$$
$$\frac{1-\cos(2nx)+ 2 \cdot \frac{1}{2} [\cos[(2n+1)x-x]-\cos[(2n+1)x+x]]}{2\sin x}$$
$$\frac{1-\cos(2nx)+ \cos(2nx)-\cos(2n+2)x}{2\sin x}$$
$$\frac{1-\cos(2n+2)x}{2\sin x}$$
It follows that
$$\sin(x) +\sin(3x) +...+ \sin [(2(n+1)-1)x]= \frac{1-\cos(2(n+1)x)}{2\sin x}$$
Therefore,
$$ \sin(x) +\sin(3x) +...+ \sin [(2n-1)x]= \frac{1-\cos(2nx)}{2\sin x}$$ is true
Any input is much appreciated.
| Also, you can multiply both sides by $2\sin x$ and apply telescoping:
$$2\sin x\sin x +2\sin 3x\sin x +…+ 2\sin (2n-1)x\sin x=$$
$$[\cos0-\require{cancel}\cancel{\cos2x}]+[\cancel{\cos2x}-\cancel{\cos4x}]+\cdots+[\cancel{\cos(2n-2)x}-\cos2nx]=$$
$$1-\cos2nx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2298751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Complex Numbers and the Triangle Inequality I'm working on the following problem for my introductory complex variables course.
By factoring $z^4-4z^2+3$ in two quadratic factors and using inequality derived from the triangle inequality, show that if $z$ lies on the circle $|z|=2$, then $$
\bigg|\frac{1}{z^4-4z^2+3}\bigg| \ge\frac{1}{19} $$
I'm not really sure how to attack this problem, I've tried multiple methods but can't seem to get anywhere with them. My first attempt was factoring $z^4-4z^2+3$ into $z^2(z+2)(z-2)+3$ to try and use it with the triangle inequality, but am not really sure how to implement this into the triangle inequality.
A general hint towards solving problems similar to this will suffice. I'm not looking for an exact answer, but any help will be appreciated.
Thanks!
| Definitely a typo.
$z^4 - 4z^2 + 3 = (z^2 -1)(z^2 - 3)$. If $|z| = 2$ then $|z^2| = 4$.
$|z^2 - 1| \le |z^2| + 1 = 5; |z^2 -3| \le |z^2| + 3 = 7$ so
$|z^4 - 4z^2 + 3| = |z^2 -1||z^2 -3|\le 5*7 = 35$
And $|\frac 1 {z^4 - 4z^2 + 3}| \ge \frac 1{35}$
Equality holds if $|z^2 - 1| = |z^2| +1$ and $|z^2 - 3| = |z^2| +3$ which happens if $z^2 = -4$ or if $z = \pm 2i$.
So $|\frac 1 {z^4 - 4z^2 + 3}|_{z = \pm 2i} = \frac 1{35} < \frac 1{19}$
On the other hand if $|z^2| = 4$ then $|z^2 - 1| \ge |z^2| - 1 = 3$ and $|z^2 - 3| > |z^2| - 3 = 1$ (with equality holding if $z = \pm 2$) and so $ \frac 1{35} \le |\frac 1 {z^4 - 4z^2 + 3}| \le \frac 13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2299176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Complex: $z^2 = 1+2i$ (find real- and imaginary part of $z$) these are my toughs:
$$z^2 = 1 + 2i \Longrightarrow (x+yi)(x+yi) = 1 + 2i$$
so: $x^2-y^2 = 1$ and $2xy = 2$
then i got that $x = 1/y$ but i cant continue to find the real- and imaginary part of z anymore. Appriciated any help
| You have $x^2 - y^2 = 1$ and $x = \frac{1}{y}$. Substitute the latter into the former to get $$\frac{1}{y^2} - y^2 = 1 \implies 1 - y^4 = y^2 \implies y^4 + y^2 - 1 =0.$$
You can solve the quadratic $u^2 + u -1 = 0$ giving solution $u = \frac{1}{2}(-1 \pm \sqrt{5})$. So you need to solve $y^2 = \frac{1}{2}(-1 + \sqrt{5})$ (since $y$ is real) which is now straightforward.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Cannot understand trig manipultion $a$ is a radius in a origin centered circle.
Any point on a circle circumference has coordinates $(a\ \cos\theta, a \ \sin\theta)$.
I do not understand the last transformation in this expression $x = \sqrt{a^2 - y^2} = \sqrt{a^2 - a^2 \sin^2\theta } = a \cos\theta $
How do we get $ \sqrt{a^2 - a^2 \sin^2\theta } = a \cos\theta $ ???
Note: this is just a part of calc lecture in MIT ocw
| $$ \sqrt{ a^2 - a^2 \sin^2 \theta } $$
$$ = \sqrt{ a^2 ( 1 - \sin^2 \theta ) } $$
$$ = a \sqrt{ \cos^2 \theta } $$
$$ = a \cos \theta $$
We can simplify $1-\sin^2\theta$, because of the pythagorean identity $\sin^2 \theta + \cos^2 \theta = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2300951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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An integral and series to prove that $\log(5)>\frac{8}{5}$ A relationship between $\log(5) \approx 1.6094$ and $\dfrac{3}{2}=1.5$ can be justified by the harmonic approximation $$\log(5) \approx H_2=1+\frac{1}{2}=\frac{3}{2}$$
that can be obtained by regrouping Lehmer's logarithm
$$\log(5) = \sum_{k=0}^\infty \left(\frac{1}{5k+1}+\frac{1}{5k+2}+\frac{1}{5k+3}+\frac{1}{5k+4}-\frac{4}{5k+5}\right)$$
symmetrically around the negative term
$$\log(5)=\frac{3}{2}+\sum_{k=1}^\infty \left( \frac{1}{5k-2}+\frac{1}{5k-1}-\frac{4}{5k}+\frac{1}{5k+1}+\frac{1}{5k+2} \right)$$
The corresponding integral is
$$\log(5)-\frac{3}{2}=\int_0^1 \frac{x^2(1-x)(1+3x+x^2)}{1+x+x^2+x^3+x^4}\:dx$$
(answer https://math.stackexchange.com/a/1656356/134791 by Olivier Oloa)
This is a direct proof that $\log(5)>\dfrac{3}{2}$ because the integrand is non-negative in $(0,1)$.
However, $\dfrac{8}{5}=1.6$ would be a closer approximation using small numbers, so the natural question is:
What are Dalzell-type integral and series for $\log(5)-\dfrac{8}{5}$?
| Here is an initial solution.
Take two truncations of the series
$$\sum_{k=1}^\infty \left(\frac{1}{5k-2}+\frac{1}{5k-1}-\frac{4}{5k}+\frac{1}{5k+1}+\frac{1}{5k+2}\right)$$
that evaluate to $\log(5)-r_1$ and $\log(5)-r_2$ such that
$$r_1<\frac{8}{5}<r_2$$
and build a linear combination of the truncations
$$\alpha r_1+(1-\alpha)r_2=\frac{8}{5}$$
with $0<\alpha<1$.
Because of linearity, the resulting series will have positive terms by construction (when combined properly) and the corresponding integral has positive integrand.
$$\log(5)=\frac{8}{5}+\frac{1}{287} \int_0^1 \frac{x^7(1-x)(1+3x+x^2)(89+198x^{5})}{1+x+x^2+x^3+x^4} dx$$
An acceptable solution should have smaller degree in the numerator.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2301252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $\lim_{n\to\infty}b_n=\frac{\sqrt{b^2-a^2}}{\arccos\frac{a}{b}}$ If $a$ and $b$ are positive real numbers such that $a<b$ and if
$$a_1=\frac{a+b}{2}, b_1=\sqrt{(a_1b)},..., a_n=\frac{a_{n-1}+b_{n-1}}{2},b_n=\sqrt{a_nb_{n-1}},$$ then show that
$$\lim_{n\to\infty}b_n=\frac{\sqrt{b^2-a^2}}{\arccos\frac{a}{b}}.$$
I tried to calculate explicitly the first few terms $a_2,b_2$ etc but the terms got too complicated really quickly and I couldn't spot any pattern.
| If we set $a=b\cos\theta$, we can show by induction $$b_n=b\prod^n_{k=1}\cos\frac{\theta}{2^k},\quad a_n=b_n\cos\frac{\theta}{2^n},$$ using $$\frac{1+\cos\frac{\theta}{2^k}}{2}=\cos^2\frac{\theta}{2^{k+1}}.$$
But $$\prod^n_{k=1}\cos\frac{\theta}{2^k}=\frac{\sin\theta}{\theta}\,\frac{\frac{\theta}{2^n}}{\sin\frac{\theta}{2^n}},$$ as we can show by repeated application of the identity $\sin2\alpha=2\sin\alpha\cos\alpha$, so the limit of the product as $n\rightarrow\infty$ is $\sin\theta/\theta$. Since $$b\sin\theta=b\sqrt{1-\cos^2\theta}=\sqrt{b^2-a^2},$$ the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2301340",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove that $\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$
Prove that $$\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$$
My attempt,
Let an equation $x^{2n+1}-1=0$, which has roots $$\cos \frac{2\pi}{2n+1}+i \sin \frac{2\pi}{2n+1}, \cos \frac{4\pi}{2n+1}+i \sin \frac{4\pi}{2n+1},...,\cos \frac{4n\pi}{2n+1}+i \sin \frac{4n\pi}{2n+1}$$
$$(1+\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+...+\cos \frac{4n\pi}{2n+1})+(\sin \frac{2\pi}{2n+1}+\sin \frac{4\pi}{2n+1}+...+\sin \frac{4n\pi}{2n+1})i=0$$
$$\cos \frac{2\pi}{2n+1}+ \cos \frac{4\pi}{2n+1}+...+ \cos \frac{4n\pi}{2n+1}=-1$$
Since $$\cos \frac{2\pi}{2n+1}=\cos \frac{4n\pi}{2n+1}$$
$$\cos \frac{4\pi}{2n+1}=\cos \frac{(4n-2)\pi}{2n+1}$$
So, $$2(\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+...+ \cos \frac{2n\pi}{2n+1})=-1$$
which proves that $$\cos \frac{2\pi}{2n+1}+\cos \frac{4\pi}{2n+1}+\cos \frac{6\pi}{2n+1}+...+\cos \frac{2n\pi}{2n+1}=\frac{-1}{2}$$
My question:
Is my attempt with using complex number too tedious and long? Is there another way to solve this question?
| $$\sum_{k=1}^n\cos\frac{2k\pi}{2n+1}=\sum_{k=1}^n\frac{2\sin\frac{\pi}{2n+1}\cos\frac{2k\pi}{2n+1}}{2\sin\frac{\pi}{2n+1}}=\frac{\sum\limits_{k=1}^n\left(\sin\frac{(2k+1)\pi}{2n+1}-\sin\frac{(2k-1)\pi}{2n+1}\right)}{2\sin\frac{\pi}{2n+1}}=$$
$$=\frac{\sin\frac{(2n+1)\pi}{2n+1}-\sin\frac{\pi}{2n+1}}{2\sin\frac{\pi}{2n+1}}=-\frac{1}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2303799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Solving $\tan(a+b)\tan(b+c)\tan(a+c)=1$ I have a simple question :
What are the condition on $a,b,c$ to have :
$$\tan(a+b)\tan(b+c)\tan(a+c)=1$$
I think we can use the Huilier formula but I am not sure how to solve it.
Thanks
Edit:
With :
$\tan(a)\tan(b)<1$,
$\tan(a)\tan(c)<1$, and
$\tan(b)\tan(c)<1$.
My second idea would be to build a cyclic quadrilateral (convex in your case) with this formulas :
$\tan(\frac{\alpha}{2})\tan(\frac{\beta}{2})=\tan(\frac{\gamma}{2})\tan(\frac{\delta}{2})=1$
With $\alpha+\beta+\gamma+\delta=\pi$
On the other hand an other idea would be to use the semiperimeter :
For a cyclic quadrilateral with successive sides $a, b, c, d$ semiperimeter $s$, and angle A between sides $a$ and $d$,we have
$\tan(\frac{A}{2})=\sqrt{\frac{(s-a)(s-d)}{(s-b)(s-c)}}$
| You are looking for the solutions to
$$ \bbox[lightyellow] {
\left\{ \matrix{
\tan \left( {a + b} \right)\tan \left( {a + c} \right)\tan \left( {b + c} \right) = 1 \hfill \cr
\tan \left( a \right)\tan \left( b \right),\;\tan \left( a \right)\tan \left( c \right),\;\tan \left( b \right)\tan \left( c \right) < 1 \hfill \cr} \right.
} \tag{1}$$
The equation is symmetric in $a,b,c$, and the solutions will
be centered around the symmetric solution
$$ \bbox[lightyellow] {
a = b = c = \pi /8
} $$
This hints to try and make the substitution
$$ \bbox[lightyellow] {
\left\{ \matrix{
a + b = \pi /4 + \gamma \hfill \cr
b + c = \pi /4 + \alpha \hfill \cr
a + c = \pi /4 + \beta \hfill \cr} \right.\quad \Leftrightarrow \quad \left\{ \matrix{
a = \pi /8 + \left( { - \alpha + \beta + \gamma } \right)/2 \hfill \cr
b = \pi /8 + \left( {\alpha - \beta + \gamma } \right)/2 \hfill \cr
c = \pi /8 + \left( {\alpha + \beta - \gamma } \right)/2 \hfill \cr} \right.
} \tag{2}$$
Making this change of variables we get, for $tan(a+b)$
$$ \bbox[lightyellow] {
\tan \left( {a + b} \right) = \tan \left( {\pi /4 + \gamma } \right) = {{1 + \tan \gamma } \over {1 - \tan \gamma }}
} \tag{3.a}$$
while for $tan(a)tan(b)$ we get
$$ \bbox[lightyellow] {
\eqalign{
& \tan a\tan b = \tan \left( {\pi /8 + \gamma /2 - \left( {\alpha - \beta } \right)/2} \right)\tan \left( {\pi /8 + \gamma /2 + \left( {\alpha - \beta } \right)/2} \right) = \cr
& = {{\tan \left( {\pi /8 + \gamma /2} \right) - \tan \left( {\left( {\alpha - \beta } \right)/2} \right)} \over {1 + \tan \left( {\pi /8 + \gamma /2} \right)\tan \left( {\left( {\alpha - \beta } \right)/2} \right)}}{{\tan \left( {\pi /8 + \gamma /2} \right) + \tan \left( {\left( {\alpha - \beta } \right)/2} \right)} \over {1 - \tan \left( {\pi /8 + \gamma /2} \right)\tan \left( {\left( {\alpha - \beta } \right)/2} \right)}} = \cr
& = {{\tan ^{\,2} \left( {\pi /8 + \gamma /2} \right) - \tan ^{\,2} \left( {\left( {\alpha - \beta } \right)/2} \right)} \over {1 - \tan ^{\,2} \left( {\pi /8 + \gamma /2} \right)\tan ^{\,2} \left( {\left( {\alpha - \beta } \right)/2} \right)}} \cr}
} $$
and imposing the condition that $tan(a)tan(b)<1$ we get
$$ \bbox[lightyellow] {
\eqalign{
& \tan a\tan b < 1\quad \to \; \cr
& \to \;\;\tan ^{\,2} \left( {\pi /8 + \gamma /2} \right) - \tan ^{\,2} \left( {\left( {\alpha - \beta } \right)/2} \right) < 1 - \tan ^{\,2} \left( {\pi /8 + \gamma /2} \right)\tan ^{\,2} \left( {\left( {\alpha - \beta } \right)/2} \right)\;\; \to \cr
& \to \;\;\tan ^{\,2} \left( {\pi /8 + \gamma /2} \right)\left( {1 + \tan ^{\,2} \left( {\left( {\alpha - \beta } \right)/2} \right)} \right) < 1 + \tan ^{\,2} \left( {\left( {\alpha - \beta } \right)/2} \right)\;\; \to \cr
& \to \;\;\tan ^{\,2} \left( {\pi /8 + \gamma /2} \right) < 1\;\; \to \cr
& \to \;\; - 3\pi /4 < \gamma < \pi /4\quad \to \quad \tan \gamma < 1 \cr}
} \tag{3.b}$$
and analogously for the other combinations.
So, the original equation is transformed into
$$ \bbox[lightyellow] {
\eqalign{
& \left\{ \matrix{
\tan \left( {a + b} \right)\tan \left( {a + c} \right)\tan \left( {b + c} \right) = 1 \hfill \cr
\tan a\tan b,\;\tan a\tan c,\;\tan b\tan c < 1 \hfill \cr} \right.\quad \to \cr
& \to \quad \left\{ \matrix{
a + b = \pi /4 + \gamma \hfill \cr
b + c = \pi /4 + \alpha \hfill \cr
a + c = \pi /4 + \beta \hfill \cr
\left( {{{1 + \tan \alpha } \over {1 - \tan \alpha }}} \right)\left( {{{1 + \tan \beta } \over {1 - \tan \beta }}} \right)\left( {{{1 + \tan \gamma } \over {1 - \tan \gamma }}} \right) = 1 \hfill \cr
- 3\pi /4 < \alpha ,\;\beta ,\;\gamma < \pi /4 \hfill \cr} \right.\quad \to \cr}
} $$
$$ \bbox[lightyellow] {
\to \quad \left\{ \matrix{
a = \pi /8 + \left( { - \alpha + \beta + \gamma } \right)/2 \hfill \cr
b = \pi /8 + \left( {\alpha - \beta + \gamma } \right)/2 \hfill \cr
c = \pi /8 + \left( {\alpha + \beta - \gamma } \right)/2 \hfill \cr
0 = \tan \alpha + \tan \beta + \tan \gamma + \tan \alpha \tan \beta \tan \gamma \hfill \cr
\tan \alpha ,\;\tan \beta ,\;\tan \gamma < 1 \hfill \cr} \right.
} \tag{4}$$
Conclusion
Equation (4) tells us that the solutions to (1) are reconducible to the roots of the symmetric polynomial
$$ \bbox[lightyellow] {
P(x,y,z) = x + y + z + xyz = x\left( {1 + yz} \right) + y + z
} $$
which can be easily found by expressing one variable in terms of the other two,
and which are rendered graphically by this plot.
Note the similarity of the solution to the above with the addition formula for $\tanh$
$$ \bbox[lightyellow] {
x = - {{y + z} \over {1 + yz}}\quad \leftrightarrow \quad - \tanh \left( {c + d} \right) = - {{\tanh c + \tanh d} \over {1 + \tanh c\tanh d}}
} $$
which would deserve to be further expanded.
Otherwise, dividing by one of the variables taken to be non-null (e.g., $z$),
the solutions are reconducible to the points on the hyperbola
$$ \bbox[lightyellow] {
h(\xi ,\eta ;z) = z^{\,2} \xi \eta + \xi + \eta + 1 = 0
} $$
and we can use the well developed apparatus of conic sections to analyze the behaviour
of the solutions (existence, axes, canonical form, etc.) and to find appropriated parametric
representations.
example :
In equation (4), choosing $\tan \alpha = 0,\,\;\tan \beta = 1/2$ we get $\tan \gamma = - 1/2 $, and then
$$ \bbox[lightyellow] {
\eqalign{
& \tan \alpha = 0,\,\;\tan \beta = 1/2,\;\tan \gamma = - 1/2 \cr
& \quad \quad \Downarrow \cr
& \alpha = 0,\,\;\beta = \arctan \left( {1/2} \right),\;\gamma = - \arctan \left( {1/2} \right) \cr
& \quad \quad \Downarrow \cr
& \left\{ \matrix{
a = \pi /8 \hfill \cr
b = \pi /8 - \arctan \left( {1/2} \right) \hfill \cr
c = \pi /8 + \arctan \left( {1/2} \right) \hfill \cr} \right. \cr
& \quad \quad \Downarrow \cr
& \tan \left( {a + b} \right)\tan \left( {a + c} \right)\tan \left( {b + c} \right) = \cr
& = \tan \left( {\pi /4 - \arctan \left( {1/2} \right)} \right)\tan \left( {\pi /4 + \arctan \left( {1/2} \right)} \right)\tan \left( {\pi /4} \right) = \cr
& = {{1 - 1/2} \over {1 + 1/2}}{{1 + 1/2} \over {1 - 1/2}} = 1 \cr}
} $$
addendum
The similarity with the addition formula for $\tanh$ provides in fact
an interesting parametric equation for $a,b,c$ in terms of $u,v$, which
for $-\infty<u,v<\infty$ always satisfies the original system (1)
$$ \bbox[lightyellow] {
\left\{ \matrix{
a = \pi /8 + \left( { - \alpha + \beta + \gamma } \right)/2 \hfill \cr
b = \pi /8 + \left( {\alpha - \beta + \gamma } \right)/2 \hfill \cr
c = \pi /8 + \left( {\alpha + \beta - \gamma } \right)/2 \hfill \cr
\alpha = \arctan \left( {\tanh u} \right) \hfill \cr
\beta = \arctan \left( {\tanh v} \right) \hfill \cr
\gamma = \arctan \left( {\tanh \left( { - u - v} \right)} \right) = - \arctan \left( {\tanh \left( {u + v} \right)} \right) \hfill \cr} \right.
} \tag{5}$$
In fact, we have
$$ \bbox[lightyellow] {
\left. {\left\{ \matrix{
0 \equiv \tan \alpha + \tan \beta + \tan \gamma + \tan \alpha \tan \beta \tan \gamma \; \hfill \cr
\tan \alpha ,\;\tan \beta ,\;\tan \gamma < 1\; \to \;\left\{ \matrix{
\tanh u,\;\tanh v < 1 \hfill \cr
- 1 < \tanh \left( {u + v} \right) \hfill \cr} \right. \hfill \cr} \right.\quad } \right|\; - \infty < u,v < \infty
} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2304173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Solving Pell's equation $x^2-5y^2=\pm4$ using elementary methods.
Solve Pell's equation $x^2-5y^2=\pm4$.
This equation arises when I tried to prove that the units of $\mathbb{Z}[\varphi]$, where $\varphi=\frac{1+\sqrt{5}}{2}$ is the golden ratio, are of the form $\pm\varphi^n$. I found that $x+\varphi y$ is a unit iff $x^2+xy-y^2=\pm1$, i.e. $(2x+y)^2-5y^2=\pm4$. Yet I am unable to solve this equation. I saw here a solution using algebraic number theory, but I am interested in how to solve this equation using elementary methods, without using results from algebraic number theory. Thanks in advance!
| For $x^2-5y^2=-4$ we have a base solution at $(x,y) = (1,1)$ and the unit solution for $x^2-5y^2=\pm1$ is $(x,y) = (2,1)$. So you can generate infinitely many solutions with:
\begin{equation*} x^2-5y^2 = (1-\sqrt{5})(1+\sqrt{5})(2-\sqrt{5})^n(2+\sqrt{5})^n = (-4)(-1)^n \end{equation*}
With $n=1$ that gives $(1+\sqrt{5})(2-\sqrt{5}) = -3 +\sqrt{5}, (1-\sqrt{5})(2+\sqrt{5}) = -3 -\sqrt{5}$ and we thus have $3^2-5(1^2) = 4$. We also have another base solution for $x^2-5y^2=4$ with $(x,y) = (\pm2, 0)$ which generates another family of solutions:
\begin{equation*} x^2-5y^2 = 4(2-\sqrt{5})^n(2+\sqrt{5})^n = (4)(-1)^n \end{equation*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
problem : conditions of similar matrices Could you please check my solution?
Q. State all conditions that make A and B similar.
$$A=
\begin{pmatrix}
0 & 4 \\
a & 4 \\
\end{pmatrix}
$$
$$B=
\begin{pmatrix}
2 & b \\
0 & c \\
\end{pmatrix}
$$
My solution :
Since similar matrices have the same eigenvalues, trA=4=trB. so C=2.
DetB=4=DetA so a=-1
Since both the sum and the product of the two eigenvalues are 4, the eigenvalues of A and B are 2,2 (multiplicity 2)
$$B-2I=
\begin{pmatrix}
0 & b \\
0 & 0 \\
\end{pmatrix}
$$
$$(B-2I)v=
\begin{pmatrix}
0 & b \\
0 & 0 \\
\end{pmatrix}\begin{pmatrix}
x \\
y \\
\end{pmatrix}=\begin{pmatrix}
by \\
0\\
\end{pmatrix}=\begin{pmatrix}
0 \\
0 \\
\end{pmatrix}
$$, where $$v=\begin{pmatrix}
x \\
y \\
\end{pmatrix}$$ is an eigenvector of B
The value of $by$ must always be 0. So b is 0.
So the answer is a=-1 and b=0 and c=2
| Your arguments for $c=2$ and $a=-1$ look convincing, but $b=0$ can't be right. That would lead to $B=2I$, which is not similar to anything except itself.
Your mistake is that you're assuming that the geometric multiplicity of the eigenvalue $2$ is $2$ (such that any $(^x_y)$ would be an eigenvector) whereas this is actually one of the cases where the geometric multiplicity is smaller than the algebraic multiplicity that you've correctly found to be $2$.
Actually $b$ can take any nonzero value, but specifically can't be $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2307986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to solve this limit without using L'Hopital's Rule: $\lim_{x\to0^-}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}$?
Find:
$$\lim_{x\to0^-}\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2}\cdot$$
I tried to simplify the expression, but I kept getting stuck. I also tried to make a substitution $u=\frac{1}{x}$, but I got stuck again. Please give me hint or explain what to do. Thanks.
| $$\frac{\sqrt{x^4\cos^2{x}+2x^2\sin^2{2x}-x^4}}{x^2} = \frac{\sqrt{x^4\cos^2{x}+8x^2 \sin^2x \cos^2x-x^4}}{x^2}=
\frac{\sqrt{x^4(\cos^2{x}-1)+8x^2 \sin^2x \cos^2x}}{x^2}=
\frac{\sqrt{-x^4\sin^2 x+8x^2 \sin^2x \cos^2x}}{x^2} =
\frac{\sqrt{x^2\sin^2 x(8 \cos^2x - x^2)}}{x^2} =
\frac{\sqrt{\sin^2 x(8 \cos^2x - x^2)}}{x} =
\sqrt{ \left(\frac{\sin x}{x}\right)^2 (8 \cos^2x - x^2)}
$$
Now, take the limit
$$\lim_{x\rightarrow 0^{-}}\sqrt{ \left(\frac{\sin x}{x}\right)^2 (8 \cos^2x - x^2)} = \lim_{x\rightarrow 0^{-}}\sqrt{ \left(\frac{\sin x}{x}\right)^2} \lim_{x\rightarrow 0^{-}}\sqrt{(8 \cos^2x - x^2)} = 1 \times \sqrt{8} = 2\sqrt{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2308860",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
What is the remainder when dividing $11^{(345^{678})}$ by 13? So basically I figured out that this problem comes down to the following:
We want to find the answer to $11^{(345^{678})} \ \text{mod} \ 13$.
Then, because $11$ and $13$ are coprime, we know that $11^{\phi(13)} = 1 \
\text{mod} \ 13$. Because 13 is prime, we know that $\phi(13) = 12$. So
$11^{12x} = 1 \ \text{mod} \ 13 $ for any integer x.
Then we need to find how many times 12 fits into $345^{678}$, so $345^{678} \ \text{mod} \ 12 $. However, $345$ and $12$ are not coprime, so we can't use Euler's theorem. This is the part where I don't actually know what to do anymore.
I tried the following: 345 mod 12 = 9 mod 12, so $345^{678} \ \text{mod} \ 12 = 9^{678} \ \text{mod} \ 12$.
We know that $9^2 \ \text{mod} \ 12 = 81 \ \text{mod} \ 12 = 9 \ \text{mod} \ 12$. And we know that $9^4 \ \text{mod} \ 12 = 9^2 \cdot 9^2 \ \text{mod} \ 12 = 9 \ \text{mod} \ 12$. So $9^2$, $9^4$, $9^8$, $9^{16}$, $9^{32}$, $9^{64}$,$9^{128}$, $9^{256}$ and $9^{512}$ are all 9 in mod 12.
So then, $9^{678} \ \text{mod} \ 12 = 9^{512} \cdot 9^{128} \cdot 9^{32} \cdot 9^{4} \cdot 9^2 \ \text{mod} \ 12 = 9^5 \ \text{mod} \ 12 = 9 \ \text{mod} \ 12 $.
Then, going back to the original question, we know that we need to calculate $11^{9} \ \text{mod} \ 13 = 11^8 \cdot 11 \ \text{mod} \ 13$.
Using the same trick as with the 9, I get that $11^{9} \ \text{mod} \ 13 = 8 \ \text{mod} \ 13$.
Is this correct? If not, what am I doing wrong? Is there a better way to attack such problems?
Thanks for reading,
K.
| Your answer is correct, bur finding the value of $345^{678}\bmod 12$ is too long:
as you observe, it is $9^{678}\bmod 12$. You missed that since $9^2\equiv9\mod13$, we also have (easy induction) $9^n\equiv 9\mod13$ for all $n\ge1$, so
$$11^{345^{678}}\equiv (-2)^9=(-2)^6(-2)^3\equiv (-1)\cdot(-8)=8.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2309183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proof of a logarithmic equation If \begin{align}\log_{16}{15} &= a\\
\log_{12}{18} &= b\\
\log_{25}{24} &= c\end{align}
then prove that $$c=\frac{5-b}{2(8a - 4ab -2b +1)}$$
My attempt: I tried to prove it by applying the standard logarithmic formulas such as $$\log_{a}{b} = \log_{c}{b}\times \log_{a}{c} $$ but the denominator is becoming too complex to perform an LCM of all the terms. Can there be a simpler way to prove this?
Any help is appreciated.
| Let $\log_52=x$ and $\log_53=y$.
Hence, $$a=\frac{\log_515}{\log_516}=\frac{1+y}{4x},$$
which gives $y=4ax-1$ and
$$b=\frac{\log_518}{\log_512}=\frac{x+2y}{2x+y},$$ which gives
$$(2b-1)x=(2-b)y$$ or
$$y=\frac{(2b-1)x}{2-b}$$ and we obtain
$$x\left(4a-\frac{2b-1}{2-b}\right)=1$$ or
$$x=\frac{2-b}{8a-4ab-2b+1},$$ which gives
$$y=\frac{2b-1}{8a-4ab-2b+1}.$$
Id est,
$$c=\frac{1}{2}(3x+y)=\frac{3(2-b)+2b-1}{2(8a-4ab-2b+1)}=\frac{5-b}{2(8a-4ab-2b+1)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2311820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Find $S = \frac{a}{b+c}+\frac{b}{c+a} + \frac{c}{a+b}$ if values of $a+b+c$ and $\frac1{a+b}+\frac1{b+c}+\frac1{a+c}$ are given I just stumbled upon a contest question from last year's city olympiad math contest:
Question: For the real numbers $a,b,c$ such that: $a+b+c = 6, \dfrac{1}{a+b}+\dfrac{1}{b+c} + \dfrac{1}{c+a} = \dfrac{47}{60}$, find the value of $S = \dfrac{a}{b+c}+\dfrac{b}{c+a} + \dfrac{c}{a+b}$.
Since I just saw it from an online forum "elsewhere", I thought I'd want to hear from other more skilled and experienced MSE members about your tactics and approaches to the solution of this interesting question.
| Even if the Joshua's answer is probably the most elegant solution, let me offer you an alternative way of solving it, that is through simplification due to arbitrary values.
You have three unknowns and only two equations. This allows you to get rid of one variable, for example, c = 0. (Note that only one can be zero).
Hence, the problem can be rewritten as:
1) $a + b = 6$
2) $\dfrac{1}{a + b} + \dfrac{1}{b} + \dfrac{1}{a} = \dfrac{1}{a + b} + \dfrac{a + b}{ab} = \dfrac{47}{60}$
$S = \dfrac{a}{b} + \dfrac{b}{a} = \dfrac{a^2 + b^2}{ab} = \dfrac{(a + b)^2 - 2ab}{ab}$.
At this point, you can clearly either find both the values of $a$ and $b$ (annoying) or trick a little more.
Substituting $x = a + b$ and $y = ab$, we have the perfectly equivalent problem
1) $x = 6$
2) $\dfrac{1}{x} + \dfrac{x}{y} = \dfrac{47}{60}$
$S = \dfrac{x^2 - 2y}{y}$.
Substituting the first equation to the second, we get $\dfrac{6}{y} = \dfrac{47}{60} - \dfrac{1}{6} = \dfrac{37}{60}$ and $y = \dfrac{10\cdot6^2}{37}$.
Finally, $S = \dfrac{x^2 - 2y}{y} = \left(6^2-2\cdot\dfrac{10\cdot6^2}{37}\right)\dfrac{37}{10\cdot6^2} = \left(1-\dfrac{20}{37}\right)\dfrac{37}{10} = \dfrac{37-20}{37} \cdot \dfrac{37}{10} = \dfrac{17}{10}$.
Answer $S = \dfrac{17}{10}$.
| {
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"answer_count": 6,
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} |
Find : $ \lim_{x\to 0} \frac{1+\sin x-\cos x+\log(1-x))}{x^3}$
Using squeeze theorem find the following limit $$\lim_{x\to 0} \frac{1+\sin x-\cos x+\log(1-x))}{x^3}$$
My approach :
I may only use the squeeze theorem. Please guide how to use this to solve the below question
Suppose we have inequality :
$h(x) \leq f(x) \leq g(x) $
$ \lim_{x\to c} h(x) \leq \lim_{x\to c} f(x) \leq \lim_{x\to c} g(x) $
$ \lim_{x\to c} h(x) \leq L \leq \lim_{x\to c} g(x) $
| For small positive $x$,
$$0\le\cos x\le 1$$ and by definite integration in $[0,x]$,
$$0\le\sin x\le x.$$
Integrating again,
$$0\le 1-\cos x\le\frac{x^2}2$$
and
$$0\le x-\sin x\le\frac{x^3}6.$$
and
$$0\le\frac{x^2}2-1+\cos x\le\frac{x^4}{24}.$$
Similarly, from
$$6\le\frac6{(1-x)^4}\le12$$
integrating four times
$$-\frac{x^4}2\le\ln(1-x)+x+\frac{x^2}2+\frac{x^3}3\le-\frac{x^4}4.$$
By combining these results, we have for the numerator
$$-\frac{x^3}2-ax^4\le n(x)\le-\frac{x^3}2+bx^4.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2314448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\sum_{n=1}^\infty \frac{1}{2^n-1}$ What is the exact value of$$\sum_{n=1}^\infty \frac{1}{2^n-1}$$
if it exists?
| Hint.
$$
\frac{1}{2^n-1}=\frac{1}{2^n}\cdot\frac{1}{1-2^{-n}}=\frac{1}{2^n}\sum_{k=0}^\infty \frac{1}{2^{kn}}=\sum_{k=1}^\infty \frac{1}{2^{kn}}.
$$
Hence
$$
\sum_{n=1}^\infty\frac{1}{2^n-1}=\sum_{n=1}^\infty\sum_{k=1}^\infty\frac{1}{2^{kn}}=\sum_{m=1}^\infty\frac{f(m)}{2^m},
$$
where $f(m)$ is the number of divisors of $m$.
| {
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"question_score": "1",
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prove that $ x-\frac{1}{6}x^3<\sin(x)
prove that :
there exists a deleted neighborhood of $x=0$
such that :
$$
x-\frac{1}{6}x^3<\sin(x)<x-\frac{1}{6}x^3+\frac{1}{120}x^5
$$
MyTry:
let:$$f(x):=\sin x-x+\dfrac{1}{6}x^3$$
And :
$$g(x):=\sin x-x+\dfrac{1}{6}x^3-\dfrac{1}{125}x^5$$
Now what ?
| You have\begin{align*}\sin x&=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}-\frac{x^9}{9!}+\frac{x^{11}}{11!}-\cdots\\&=x-\frac{x^3}6+\frac{x^5}{5!}\left(1-\frac{x^2}{6\times7}\right)+\frac{x^9}{9!}\left(1-\frac{x^2}{10\times11}\right)+\cdots\\&>x-\frac{x^3}6\end{align*}if $0<x<\sqrt{42}$. A similar argument proves that $\displaystyle\sin x<x-\frac{x^3}6+\frac{x^5}{120}$ in some interval $(0,\varepsilon)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2316231",
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"source": "stackexchange",
"question_score": "2",
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Find value of this Triple summation Find the value of $$S=\sum_{i=0}^{\infty}\sum_{j=0}^{\infty}\sum_{k=0}^{\infty}\frac{1}{3^{i+j+k}}$$ where $i \ne j \ne k$
i have tried in this way:
First fix $j$ and $k$, then we get
$$S_1=\sum_{\substack{i=0 \\ i\neq j,k}}^{\infty}\frac{1}{3^{i+j+k}}$$ where $j \ne k$
So we get $$S_1=\left(\frac{3}{2} \times\frac{1}{3^{j+k}}\right)-\frac{1}{3^{2j+k}}-\frac{1}{3^{j+2k}}$$
So now
$$S_2=\sum_{\substack{j=0 \\ j\neq k}}^{\infty}\left(\left(\frac{3}{2} \times\frac{1}{3^{j+k}}\right)-\frac{1}{3^{2j+k}}-\frac{1}{3^{j+2k}}\right)$$
So $$S_2=\left(\frac{9}{4} \times \frac{1}{3^k}\right)-\left(\frac{9}{8} \times \frac{1}{3^k}\right)-\left(\frac{3}{2}\times \frac{1}{9^k}\right)-\left(\frac{3}{2}\times \frac{1}{3^{2k}}\right)+\left(\frac{2}{3^{3k}}\right)$$
Finally
$$S=\sum_{k=0}^{\infty}\left(\frac{9}{4} \times \frac{1}{3^k}\right)-\left(\frac{9}{8} \times \frac{1}{3^k}\right)-\left(\frac{3}{2}\times \frac{1}{9^k}\right)+\sum_{k=0}^{\infty}\frac{2}{3^{3k}}-\sum_{k=0}^{\infty}\frac{\frac{3}{2}}{3^{2k}}$$
we get
$$S=\frac{27}{16}-\frac{27}{16}+\frac{54}{26}-\frac{27}{16}$$ hence
$$S=\frac{81}{208}$$
is there any other approach like using integration etc
| You have
$$S=S_0-S_1-S_2+S_3$$
where $S_0$ is the sum over all triples $(i,j,k)$,
$S_1$ is the sum over triples $(i,j,k)$ with $i=j$,
$S_2$ is the sum over triples $(i,j,k)$ with $j=k$
and
$S_3$ is the sum over triples $(i,j,k)$ with $i=j=k$.
This is basically inclusion-exclusion.
Then
$$S_0=\left(\frac32\right)^3=\frac{27}8,$$
$$S_1=S_2=\sum_{i=0}^\infty\frac1{9^i}\sum_{k=0}^\infty\frac1{3^k}
=\frac{27}{16}$$
and
$$S_3=\sum_{i=0}^\infty\frac1{27^i}=\frac{27}{26}$$
etc.
ADDED IN EDIT
If you actually want the sum over $(i,j,k)$ with $i$, $j$, $k$
distinct, rather than with just $i\ne j\ne k$, then inclusion-exclusion
will give the answer $S_0-3S_1+2S_3$ instead.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Ramanujan and his problem Following was proposed by Ramanujan:
$ \sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}=1+4\sin(10^o)$
Working on this I got the radical on the left equal to $(1+2\sqrt{2})$ implying that $\sin(10^o)=1/\sqrt{2}$
How is this possible? What is wrong here?
| Let $x=\sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}$ and $ y=\sqrt{11+2\sqrt{11-2\sqrt{11+2\sqrt{11-\cdots}}}}$ then
\begin{eqnarray*}
x=\sqrt{11-2y} \\
y=\sqrt{11+2x}
\end{eqnarray*}
so $x^2=11-2y$ & $y^2=11+2x$ ... after a little algebra ...
\begin{eqnarray*}
x^4-22x^2-8x+77=0 \\
(x^2+2x-7)(x^2+2x-11)=0
\end{eqnarray*}
So we have the possible solutions $x=-1 \pm 2 \sqrt{2}$ & $x=1 \pm 2 \sqrt{3}$.
One can verify that
\begin{eqnarray*}
\sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11+\cdots}}}}=\color{red}{2 \sqrt{2}-1}.
\end{eqnarray*}
If the $+$ and $-$'s alternate then the above value is correct ... Ramanujan actually does the $(-,+,-)$ repeating every $3$ times ... then the result is
\begin{eqnarray*}
\sqrt{11-2\sqrt{11+2\sqrt{11-2\sqrt{11\color{red}{-}\cdots}}}}=1+ \sin(10^o).
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2317265",
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} |
Evaluating $\sum\limits_{k=0}^\infty n^{2^k}\prod\limits_{m=0}^k\frac1{n^{2^m}+1}$ I found this particular series in a book which asks to evaluate:
$$S=\lim_{k \to \infty}\frac{n}{n+1}+\frac{n^2}{(n+1)(n^2+1)}+\frac{n^4}{(n+1)(n^2+1)(n^4+1)}+...+\frac{n^{2^k}}{(n+1)(n^2+1)...(n^{2^k}+1)}$$
which I believe is equal to
$${\lim_{k \to \infty}\sum_{j=0}^k\frac{n^{2^j}}{\prod_{m=0}^j(n^{2^m}+1)}}$$
So, I did it like this:
$$\frac{n}{n+1}+\frac{n^2}{(n+1)(n^2+1)}+\frac{n^4}{(n+1)(n^2+1)(n^4+1)}+...+\frac{n^{2^k}}{(n+1)(n^2+1)...(n^{2^k}+1)}$$
$$=\frac{(n+1)-1}{n+1}+\frac{(n^2+1)-1}{(n+1)(n^2+1)}+\frac{(n^4+1)-1}{(n+1)(n^2+1)(n^4+1)}+...+\frac{(n^{2^k}+1)-1}{(n+1)(n^2+1)...(n^{2^k}+1)}$$
Now breaking this gives
$$1-\frac{1}{n+1}+\frac{1}{n+1}-\frac{1}{(n+1)(n^2+1)}+\frac{1}{(n+1)(n^2+1)}-\frac{1}{(n+1)(n^2+1)(n^4+1)}...+\frac{1}{(n+1)(n^2+1)...(n^{2^{k-1}}+1)}-\frac{1}{(n+1)(n^2+1)...(n^{2^{k-1}}+1)(n^{2^{k}+1})}$$
which telescopes to
$$S=1-\frac{1}{(n+1)(n^2+1)...(n^{2^{k-1}}+1)(n^{2^{k}+1})}$$
which gives $$\lim_{k \to \infty}S=1$$
Is this correct? If not, Could somebody help me in the right direction?
| Indeed, what you did shows that, for every $n\geqslant1$, $$S=1$$ However, for every $0\leqslant n<1$, $$S=1-\frac1{T(n)}$$ where $$T(n)=\prod_{k=0}^\infty(1+n^{2^k})$$ Now, for every $K$, $$\prod_{k=0}^K(1+n^{2^k})=\sum_{i=0}^{2^{K+1}-1}n^i=\frac{1-n^{2^{K+1}}}{1-n}$$ hence, for every $0\leqslant n<1$, $$T(n)=\frac1{1-n}$$ and finally, for every nonnegative $n$, $$S=\min\{n,1\}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318687",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
If $x=2\sin\frac{3\pi}{10}$, find the value of the expression. If $x=2\sin\frac{3\pi}{10}$, find the value of the expression
$$
W =2x^5+x^4-4x^3-4x^2-x+7
$$
After a bit of thought, I got :
$$
x = \frac{\sqrt{5}+1}{2}
$$
But I don't think my teacher expected me to substitute the value of $x$ into the expression. Is there anything I'm missing ?
| Use $x^2-x-1=0$. It must help:
$$2x^5+x^4-4x^3-4x^2+x+7=(x^2-x-1)(2x^3+3x^2+x)+7=7.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2318809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solving algebraic inequality I been struggling trying to solve this inequality by hand
$$x+\sqrt{x+3} < 0$$
how should I proceed?
| Let $f(x)=x+\sqrt{x+3}$.
Thus, $f$ is an increasing function.
Also, $$f\left(\frac{1-\sqrt{13}}{2}\right)=\frac{1-\sqrt{13}}{2}+\sqrt{\frac{1-\sqrt{13}}{2}+3}=$$
$$=\frac{1-\sqrt{13}}{2}+\sqrt{\frac{14-2\sqrt{13}}{4}}=\frac{1-\sqrt{13}}{2}+\frac{\sqrt{13}-1}{2}=0.$$
But the domain gives $x\geq-3$, which gives the answer $\left[-3,\frac{1-\sqrt{13}}{2}\right)$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\int_1^{\infty} {\frac{\ln{x}}{(x-1)(2x-1)}\,dx}$ Evaluate:
$$\int_1^{\infty} {\left(\frac{\ln{x}}{\left(x-1\right)\left(2x-1\right)}\,dx\right)}$$
It turns out that the value of the integral is exactly:
$$\frac{1}{12}\left(\pi^2+6\ln^2{2}\right)$$
as found by WolframAlpha, but Wolfram gives no indication of how it arrives at this curious result.
How would one solve the integral analytically? A first step might be the $u$-sub, $u=\frac{1}{x}$, which gives:
$$\int_1^{\infty} {\left(\frac{\ln{x}}{\left(x-1\right)\left(2x-1\right)}\,dx\right)}=-\int_0^1 {\left(\frac{\ln{x}}{\left(1-x\right)\left(2-x\right)}\,dx\right)}$$
Thanks!
| This is related to the dilogarithm function $\text{Li}_2(x)$ and the special value $$\text{Li}_2(\frac{1}{2}) = \frac{\pi^2}{12}-\frac{\ln^2 2}{2}$$
The dilogarithm has integral representation:
$$\text{Li}_2(z) = \frac{\pi^2}{6}+\int_0^{1-z} \frac{\ln t}{1-t} dt$$
For your question, we have
$$\begin{aligned}
\int_0^1 \frac{\ln{x}}{\left(1-x\right)\left(2-x\right)}dx &= \int_0^1 \frac{\ln x}{1-x}dx - \int_0^1\frac{\ln x}{2-x}dx \\
&= \int_0^1 \frac{\ln x}{1-x}dx - 2\int_0^{1/2}\frac{\ln 2x}{2-2x}dx \\
&= \int_0^1 \frac{\ln x}{1-x}dx-(\ln 2) \int_0^{1/2}\frac{1}{1-x}dx-\int_0^{1/2}\frac{\ln x}{1-x}dx \\
&= -\frac{\pi^2}{6}-\ln^2 2 -[\text{Li}_2(\frac{1}{2})-\frac{\pi^2}{6}] \\
&= -\frac{\pi^2}{12}-\frac{\ln^2 2}{2}
\end{aligned}$$
| {
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"url": "https://math.stackexchange.com/questions/2323365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
prove that $\frac{a_1+a_2+\dots+a_m}{m}\ge \frac{n+1}{2}$ Let $m$ and $n$ be positive integers. Let $a_1, a_2,\dots, a_m$ be distinct elements of $\{1, 2, \ldots, n\}$ such that whenever $a_i + a_j \le n$ for some $i,j, 1 \le i \le j \le m$, there exists $k, 1 \le k \le m$, with $a_i + a_j = a_k$. Prove that
$$\frac{a_1+a_2+\dots+a_m}{m}\ge \frac{n+1}{2}$$
Trial: I am thinking using AM$\ge$GM or $1+2+\dots+n= \frac{n(n+1)}{2}$. But unable to fully solve the problem.
| Using Induction
If $m=1$ then $a_1\ge \dfrac{n+1}2$
Assume for $m=m$ we have $\dfrac{a_1+a_2+\dots+a_m}{m}\ge \dfrac{n+1}{2}$
Now try to prove the inequality for $m=m+1$
We have
\begin{align*}
\frac{a_1+a_2+\dots+a_m+a_{m+1}}{m+1}&\ge \frac{\dfrac{m(n+1)}2+a_{m+1}}{m+1}\\
&\ge\frac{m}{m+1}. \frac{n+1}{2}+\frac{a_{m+1}}{m+1}\\
&\ge\frac{m}{m+1}. \frac{n+1}{2}+\frac{1}{m+1}\cdot\frac{n+1}{2} =\frac{n+1}2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2323620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is my mistake in this integral: $\int \tan^5(x) dx$? I'm doing an exercise from Stewart's Calculus textbook in which I have to evaluate the following integral:
$\int \tan^5(x) dx$
I start by rewriting the integral this way:
$\int \tan^5(x) dx = \int \frac{\sin^5(x)}{\cos^5(x)} dx = \int \frac{\sin^4(x)}{\cos^5(x)} \sin(x) dx$
And here I make a substitution:
$u = \cos(x)$
$-du = \sin(x)dx$
So the integral with the substitution becomes:
$\int \frac{(1-u^2)^2}{u^5} (-1) du$
$\int \frac{(1-2u^2+u^4)}{u^5} (-1) du$
$\int \frac{(2u^2-u^4-1)}{u^5} du$
$\int \frac{(2u^2)}{u^5} - \frac{(u^4)}{u^5} - \frac{1}{u^5} du = 2 \frac{u^{-2}}{(-2)} - \ln\vert u \vert - \frac{u^{-4}}{(-4)} + constant$
$= \frac{-1}{u^2} - \ln\vert u \vert + \frac{1}{4u^4} + constant $
$= \frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} + constant $
This antiderivative seems to be correct, because if I differentiate it I get:
$\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} + constant] $
$-(-2)\cos^{-3}(x)(-\sin(x))-\frac{1}{\cos(x)}(-\sin(x))+\frac{1}{4}(-4)\cos^{-5}(x)(-\sin(x))$
$ \frac{-2\sin(x)}{\cos^3(x)} +\tan(x)+\frac{\sin(x)}{\cos^5(x)}$
$\tan(x)[1-\frac{2}{\cos^2(x)} + \frac{1}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-2\cos^2(x)+1}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-\cos^2(x)+1-\cos^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^4(x)-\cos^2(x)+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^2(x)(\cos^2(x)-1)+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\cos^2(x)(-\sin^2(x))+\sin^2(x)}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^2(x)(1-\cos^2(x))}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^2(x)(\sin^2(x))}{\cos^4(x)}]$
$\tan(x)[\frac{\sin^4(x)}{\cos^4(x)}] = \tan^5(x)$
However, when I graph $\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} ] $ and $\tan^5(x)$ at a first glance they look like they are the same, but when I start to zoom in I notice that these curves are actually not equal:
The magenta curve is $\tan^5(x)$ and the black curve is $\frac{d}{dx} [\frac{-1}{\cos^2(x)} - \ln\vert \cos(x) \vert + \frac{1}{4\cos^4(x)} ] $
The answer that Stewart provides at the end of the book is the following:
$ \int \tan^5(x) dx = \frac{1}{4} \sec^4(x)-\tan^2(x)+\ln\vert\sec(x)\vert + C$
Which is confusing me even more, because if I try to differentiate his answer I get:
$\frac{d}{dx}[\frac{1}{4} \sec^4(x)-\tan^2(x)+\ln\vert\sec(x)\vert + C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)+\ln\vert1/\cos(x)\vert + C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)+\ln(1) - \ln\vert\cos(x)\vert+ C]$
$\frac{d}{dx}[\frac{1}{4} \cos^{-4}(x)-\tan^2(x)- \ln\vert\cos(x)\vert+ C]$
$= \frac{1}{4}(-4)\cos^{-5}(x)-2\tan(x)\sec^2(x)-\frac{1}{\cos(x)}(-\sin(x))$
$= \frac{-1}{\cos^5(x)}-2\tan(x)\sec^2(x)+\tan(x)$
$= \frac{-1}{\cos^5(x)}+\tan(x)(1-2\sec^2(x))$
$= \frac{-1}{\cos^5(x)}+\tan(x)(1-\sec^2(x) -\sec^2(x))$
$= \frac{-1}{\cos^5(x)}+\tan(x)(-\tan^2(x) -\sec^2(x))$
$= \frac{-1}{\cos^5(x)}-\tan^3(x)-\tan(x)\sec^2(x)$
$= \frac{-1}{\cos^5(x)}-\frac{\sin^3(x)}{\cos^3(x)}-\frac{\sin(x)}{\cos^3(x)}$
$=\frac{-1-\cos^2(x)\sin^3(x)-\cos^2(x)\sin(x)}{\cos^5(x)}$
$=\frac{-1-(1-\sin^2(x))\sin^3(x)-(1-\sin^2(x))\sin(x)}{\cos^5(x)}$
$=\frac{-1-(\sin^3(x)-\sin^5(x))-(\sin(x)-\sin^3(x))}{\cos^5(x)}$
$=\frac{-1-\sin^3(x)+\sin^5(x)-\sin(x)+\sin^3(x)}{\cos^5(x)}$
$=\frac{-1-\sin(x)+\sin^5(x)}{\cos^5(x)}$
$=\tan^5(x)-\frac{(1+\sin(x))}{\cos^5(x)}$
If I plot the derivative of Stewart's answer along the derivative of my answer and the original function $\tan^5(x)$ this is how the the graph looks like:
*
*Magenta curve: $\tan^5(x)$
*Black curve: $\frac{d}{dx}[\text{my answer}]$
*Blue curve: $\frac{d}{dx}[\text{Stewart's answer}]$
What's going on? Is my answer correct? If it isn't, what am I doing wrong? And what about Stewart's answer? Am I making a mistake when I differentiate it or is it correct?
Thanks for your help!
| If you're uncertain about a function being the correct antiderivative, you can always differentiate.
Alternatively, the difference between two antiderivatives should be constant. Taking the difference between yours and Stewart’s:
$$
\left(
\frac{-1}{\cos^2(x)} - \ln\lvert \cos(x) \rvert + \frac{1}{4\cos^4(x)} \right)
-
\left(
\frac{1}{4} \sec^4(x)-\tan^2(x)+\ln\lvert\sec(x)\rvert
\right)
$$
we see that the terms $-\ln\lvert\cos(x)\rvert$ and $-\ln\lvert\sec(x)\rvert$ cancel out, because $\sec(x)=(\cos(x))^{-1}$; for the same reason, $\frac{1}{4\cos^4(x)}$ and $-\frac{1}{4}\sec^4(x)$ cancel out and we remain with
$$
-\frac{1}{\cos^2(x)}+\tan^2(x)=
-\frac{1}{\cos^2(x)}+\frac{\sin^2(x)}{\cos^2x}=
-\frac{1-\sin^2(x)}{\cos^2(x)}=-1
$$
Since the difference is constant, either both are antiderivatives of $\tan^5(x)$ or neither is.
Thus the two functions you get by setting the constant to $0$ differ by $-1$, which accounts for the graph you show.
You can check more quickly that your function is an antiderivative by recalling that $\frac{1}{\cos^2(x)}=1+\tan^2(x)$, so your function is
$$
-1-\tan^2(x)-\ln\lvert\cos(x)\rvert+\frac{1}{4}(1+2\tan^2(x)+\tan^4(x))
$$
and the derivative is therefore
\begin{align}
-2&\,\tan(x)(1+\tan^2(x))+\tan(x)+\frac{1}{4}(4\tan(x)+4\tan^3(x))(1+\tan^2(x))\\
&=-2\tan(x)-2\tan^3(x)+\tan(x)+\tan(x)(1+2\tan^2(x)+\tan^4(x))\\
&=\tan^5(x)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
limit of a 2-dimensional function I know that $\frac{2xy}{x^2+y^2}$ is not steady at $(0,0)$. But I don't understand why this criteria $\left|\frac{2xy}{x^2+y^2}-0\right|\leq|2xy|$ which converges to $0$ as $x,y\to 0$ doesn't work here, which would show the contrary. When can I use this criteria in order to show that a $2$-dimensional is steady at $(0,0)$?
| (1). The definition of continuity of $f(x,y)$ at $(0,0)$ is that for every $e>0$ there exists $d>0$ such that whenever $|x|<d$ and $|y|<d$ we have $|f(x,y)-f(0,0)|<e.$ Or equivalently, "whenever $|x|+|y|<d$...." or "whenever $x^2+y^2<d^2$....".
(2). No matter what value you give to $f(0,0),$ if $f(x,y)=2xy/(x^2+y^2)$ when $(x,y)\neq (0,0),$ the function $f$ will be discontinuous at $(0,0).$ For if $y=x\ne 0$ then $f(x,y)=1.$ But if $y=2x\ne 0$ then $f(x,y)=4/5.$ So if $0\ne |x|<d/2$ then $f(x,x)=1$ and $f(x,2x)=4/5.$
(3). If $xy\ne 0$ the inequality $|\frac {2xy}{x^2+y^2}-0|\leq |2xy|$ is equivalent to $x^2+y^2\geq 1.$ To test for continuity of $f$ at $(0,0)$ you have to examine the behavior of $f(x,y)$ as $x^2+y^2$ tends to $0,$ not for $x^2+y^2\geq 1.$
(4). You may have meant to write $|\frac {2xy}{x^2+y^2}|\leq 1,$ which is true when $(x,y)\neq (0,0)$, but knowing only that a function takes values only in $[-1,1]$ tells us nothing nothing about its continuity.
(5). However we can use this inequality to prove the continuity of (for example) $g(x,y)=\pi \sqrt {|xy^{3/2}|}\;$ at $(0,0).$ We have $g(xy)=\pi \sqrt {|xy|}\; \sqrt {|y|}.$
So $|\pi \sqrt {|xy^{3/2}|}\;\leq$ $ \pi \sqrt {\frac {x^2+y^2}{2} }\;\sqrt |y|\leq$ $ \pi \sqrt {\frac {x^2+y^2}{2}}\;\sqrt {x^2+y^2}=$ $\frac {\pi}{\sqrt 2}(x^2+y^2)^{3/2}.$
So with $e>0$ and $d= \left(\frac {e\sqrt 2}{\pi} \right)^{2/3},$ if $x^2+y^2<d^2$ then $|g(x,y)|<e.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2324370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Finding all pairs $(a,b)$ of positive integers such that $a^2+nab+b^2$ is a perfect square. When $n=2$, the question is trivial. Is there a general method to find all such pairs for $n\ge{3}$ and $n\in{\mathbb{N}}$?
| Generally speaking, this equation has a lot of formulas for the solution. Because it is symmetrical.
Write the formula can someone come in handy. the equation:
$$Y^2+aXY+X^2=Z^2$$
Has a solution:
$$X=as^2-2ps$$
$$Y=p^2-s^2$$
$$Z=p^2-aps+s^2$$
more:
$$X=(4a+3a^2)s^2-2(2+a)ps-p^2$$
$$Y=(a^3-8a-8)s^2+2(a^2-2)ps+ap^2$$
$$Z=(2a^3+a^2-8a-8)s^2+2(a^2-2)ps-p^2$$
more:
$$X=(a+4)p^2-2ps$$
$$Y=3p^2-4ps+s^2$$
$$Z=(2a+5)p^2-(a+4)ps+s^2$$
more:
$$X=8s^2-4ps$$
$$Y=p^2-(4-2a)ps+a(a-4)s^2$$
$$Z=-p^2+4ps+(a^2-8)s^2$$
In the equation: $$X^2+aXY+bY^2=Z^2$$ there is always a solution and one of them is quite simple.
$$X=s^2-bp^2$$
$$Y=ap^2+2ps$$
$$Z=bp^2+aps+s^2$$
$p,s$ - integers asked us.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Three different result for the same indefinite integral I was killing time solving some indefinite integrals, when I found this one:
\begin{equation}
\int\frac{1}{(x+1)\sqrt{x^2+2x}}\ \mathrm{d}x
\tag{1}\label{integral}
\end{equation}
Not a particularly difficult one, I'll post my solution here:
\begin{equation}
\begin{split}
\int\frac{1}{(x+1)\sqrt{x^2+2x}}\ \mathrm{d}x = \int\frac{1}{(x+1)\sqrt{(x+1)^2-1}}\ \mathrm{d}x
\end{split}
\tag{2}\label{calculus}
\end{equation}
by substituting $t = x+1$, d$t = \mathrm{d}x$, and then $s = \sqrt{t^2-1} \rightarrow \mathrm{d}s = \frac{t}{\sqrt{t^2-1}}\mathrm{d}t$
\begin{equation}
\begin{split}
= \int\frac{1}{t\sqrt{t^2-1}}\ \mathrm{d}t &= \int\frac{1}{(x+1)\sqrt{(x+1)^2-1}}\ \mathrm{d}x =\\
&= \int\frac{\mathrm{d}s}{1+s^2} =\\
&= \arctan{s} + \mathrm{cost}\\
&= \arctan{\sqrt{t^2-1}} + \mathrm{cost}\\
&= \arctan{\sqrt{\left(x+1\right)^2-1}} + \mathrm{cost}
\end{split}
\tag{3}\label{calculus2}
\end{equation}
I then derived (I recommend, if you wanna check my results to use this online derivative calculator which actually shows steps...) my solution finding the starting function:
$$\frac{\mathrm{d}}{\mathrm{d}x}(\arctan{\sqrt{(x+1)^2-1}} + \mathrm{cost}) = \frac{1}{(x+1)\sqrt{x^2+2x}}$$
The graph (plotted with Grapher from Mac Os X), with $\color{red}{\text{function}}$ and $\color{blue}{\text{integral}}$ :
Now this integral comes from the exercise book Problems in Mathematical Analysis by Boris Demidovich and it's the number 1271, and even if I'm pretty sure this mine is the correct solution I lost quite time to understand the proposed solution of the book, which, if you don't have it and you can't check for yourself, is
1271. $\qquad-\arcsin(\frac{1}{1+x})$
deriving this function you'll find:
\begin{equation}
\begin{split}
\frac{\mathrm{d}}{\mathrm{d}x}&\left[-\arcsin(\frac{1}{1+x})\right] =\\
&= \dfrac{1}{\left(x+1\right)^2\sqrt{1-\frac{1}{\left(x+1\right)^2}}}\\
&= \dfrac{1}{\frac{\left(x+1\right)^2}{\sqrt{(x+1)^2}}\sqrt{\left(x+1\right)^2-1}}\\
&= \dfrac{1}{\frac{\left(x+1\right)^2}{|x+1|}\sqrt{\left(x+1\right)^2-1}}\\
&= \dfrac{1}{{\left(x+1\right)}\mathrm{sgn}(x+1)\sqrt{\left(x+1\right)^2-1}}\\
\end{split}
\tag{4}\label{calculus3}
\end{equation}
Plotting the result will give you an idea about the mistakte he could have done, in cyan the $\color{cyan}{Demidovich's~~primitive}$:
So I was pretty sure I was right and he was not, so I tried integrate the \eqref{integral} with Mathematica, with another unsatisfying outcome:
\begin{equation}
\int\frac{1}{(x+1)\sqrt{x^2+2x}}\ \mathrm{d}x = \frac{\sqrt{2}\sqrt{x}\sqrt{x+2}\arctan(\sqrt{\frac{x}{x+2}})}{\sqrt{x(x+2)}}
\end{equation}
This solution, in $\color{orange}{orange}$, is almost like mine, even if it does not comprehend the negative values of the function...
Also, if I try to derive (I've done it with calculator, as it's quite long to do for yourself...) you get
$$\left(\frac{\sqrt{2}\sqrt{x}\sqrt{x+2}\arctan(\sqrt{\frac{x}{x+2}})}{\sqrt{x(x+2)}}\right)' = $$
$$-\dfrac{\left(\sqrt{x+2}\left(-x^\frac{7}{2}-3x^\frac{5}{2}-2x^\frac{3}{2}\right)+\left(x+2\right)^\frac{3}{2}\left(x^\frac{5}{2}+x^\frac{3}{2}\right)\right)\arctan\left(\frac{\sqrt{x}}{\sqrt{x+2}}\right)-x^3-4x^2-4x}{\left(x\left(x+2\right)\right)^\frac{3}{2}\left(x^2+3x+2\right)}$$
Now, since I've found that in my range of definition $x^2-2x>0$ the primitive function that i found, derived, gives me the starting function, this should tell me that I am right and others (computer, this case) have bugs or they encounter problems deriving such a function. So which of the three solution $\color{blue}{mine}$, $\color{orange}{Mathematica's}$ or $\color{cyan}{Demidovich's}$ is the correct one? Why are them wrong, if they are? Does it depend on calculator's bug or it's my problem?
Thanks for attention.
| The expression provided by Mathematica simplifies in the case where $x < -2$ or $x > 0$:
FullSimplify[Integrate[1/((x + 1) Sqrt[x^2 + 2 x]), x], x^2 + 2 x > 0]
gives $$2 \tan^{-1} \sqrt{\frac{x}{x+2}}.$$ This is equivalent to your antiderivative on $x > 0$, and differs from your antiderivative by a constant factor on $x < -2$. Therefore, they are equivalent. What is clearly wrong is the "Demidovich" solution for $x < -2$: the antiderivative cannot be increasing when $x$ is negative.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2325631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\frac{1}{9} + \frac{1}{99} + \frac{1}{999} + \frac{1}{9999} + \cdots ={} $? Sum the following:
\begin{align}
S &= \frac{1}{9} + \frac{1}{99} + \frac{1}{999} + \frac{1}{9999} + \cdots\\[0.1in]
&= \sum_{n=1}^{\infty} \frac{1}{{10}^n - 1}
\end{align}
It's fairly straightforward to show that this sum converges:
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{{10}^n - 1}\\
&= \frac{1}{9} + \sum_{n=2}^{\infty} \frac{1}{{10}^n - 1}\\
&< \frac{1}{9} + \sum_{n=2}^{\infty} \frac{1}{{10}^n - 10}\\
&= \frac{1}{9} + \frac{1}{10}\sum_{n=2}^{\infty} \frac{1}{{10}^{n-1} - 1}\\
&= \frac{1}{9} + \frac{1}{10}S\, ,
\end{align}
which leads to $S < 10/81 = 0.\overline{123456790}$.
Numerically (Mathematica), we find that this sum is approximately $S \approx 0.122324$.
This sum can also be written as
\begin{align}
S &= \sum_{n=1}^{\infty} \frac{1}{{10}^n - 1}\\
&= \sum_{n=1}^{\infty} \frac{1/{10}^n}{1 - 1/{10}^n}\\
&= \sum_{n=1}^{\infty}\sum_{m=1}^{\infty} \frac{1}{{10}^{nm}}
\end{align}
(The last expression above is itself quite amusing, since the coefficient of $1/{10}^k$ is the number of distinct ways of writing $k$ as a product of two positive integers.)
Analytically, Mathematica evaluates this sum as
\begin{equation}
S = \frac{\ln(10/9) - \psi_{1/10}(1)}{\ln(10)}\, ,
\end{equation}
where $\psi_q(z)$ is the Q-Polygamma function. This is not really a nice, tidy, closed-form solution. Now, I can accept that this sum may not have such a nice, tidy sum, but something about it feels like it should have one. (Non-rigorous, I know!) Furthermore, I'm aware that Mathematica is by no means infallible, especially when it comes to simplification of certain expressions.
So I'm wondering if a neater solution exists.
| The given series is an irrational number (probably a trascendental number, too) and for its numerical evaluation it is possible to exploit some tailor-made acceleration techniques (like the one outlined here for the base-$2$ analogue). Besides that, I am not aware of any nice closed form for $\sum_{n\geq 1}\frac{d(n)}{10^n}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326330",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 1,
"answer_id": 0
} |
How does my calculator calculate the roots for a cubic polynomial? For example, many calculators come equipped with the quadratic formula, and so you can give them $A,B,C$ coefficients, and they will show you out the solution using the formula
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}$$
My calculator however finds $x$ when we have $x^3$ too, and this time it asks for $A,B,C,D$.
I was wondering how exactly it did this? Is there a cubic power "formula"? My calculator even calculates irrational roots so I'm not sure how exactly it can do this without a formula.
| "Cardano's method" solves the "reduced" cubic $x^3+ bx= c$ where b and c are real numbers. The basic idea is:
For any numbers u and v, $(u+ v)^3= u^3+ 3u^2v+ 3uv^2+ v^3$ and $3(u+ v)= 3u+ 3v$ so that $(u+ v)^3- 3uv(u+ v)= u^3+ v^3$.
So, given any two numbers, u and v, taking b= -3uv and $c= u^3- v^3$ we can construct the cubic equation $x^3+ bx= c$ that has solution $x= u+ v$. Now, what about the other way? If we know b and c, can we solve $u^3- v^3= c$ and $-3uv= c$ for u and v and so find x?
Yes, of course! From $-3uv= b$ we have $v= -\frac{b}{3u}$ so that $u^3- v^3= c$ becomes $u^3-\frac{b^3}{3^3u^3}= c$ and, multiplying by $u^3$, $u^6- \left(\frac{b}{3}\right)^3= cu^3$ which we can write as $(u^3)^2- cu^3- \left(\frac{b}{3}\right)^3= 0$, a quadratic equation in $u^3$.
Using the quadratic formula $u= \frac{c\pm\sqrt{c^2+ 4\frac{b^3}{3^3}}}{2}= \frac{c}{2}\pm\sqrt{\left(\frac{c}{2}\right)^2+ \left(\frac{b}{3}\right)^3}$
Then $v= -\frac{b}{3u}$ gives v so we can find x= u+ v.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2326445",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find sum of series $\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...$ How to find sum of above series
$$\frac{1}{6} +\frac{5}{6\cdot12} +\frac{5\cdot8}{6\cdot12\cdot18} +\frac{5\cdot8\cdot11}{6\cdot12\cdot18\cdot24}+...$$
How to find sum of series I can find its convergence but not sum of series.
Can anyone explain?
| We want to compute:
$$ S = \sum_{n\geq 1}\frac{1}{6^n n!}\cdot\frac{1}{2}\prod_{k=1}^{n}(3k-1) = \sum_{n\geq 1}\frac{\Gamma\left(n+\frac{2}{3}\right)}{2^{n+1}\,\Gamma\left(\frac{2}{3}\right)\,\Gamma(n+1)}=\sum_{n\geq 1}\frac{B\left(n+\frac{2}{3},\frac{1}{3}\right)}{2^{n+1}\cdot\frac{2\pi}{\sqrt{3}}}$$
that by Euler's Beta function equals
$$ \frac{\sqrt{3}}{4\pi}\sum_{n\geq 1}\int_{0}^{1}\frac{1}{2^n} x^{n-1/3}(1-x)^{-2/3}\,dx =\frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{x^{2/3}}{(1-x)^{2/3}(2-x)}\,dx$$
or
$$ \frac{\sqrt{3}}{4\pi}\int_{0}^{1}\frac{(1-x)^{2/3}}{x^{2/3}(1+x)}\,dx =\frac{3\sqrt{3}}{4\pi}\int_{0}^{1}\frac{(1-x^3)^{2/3}}{1+x^3}\,dx=\color{red}{\frac{2^{2/3}-1}{2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Finding the sum of an alternating series I want to find the sum of $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2}$$ I know that this is equal to $\frac{\pi^2}{12}$ thus I was thinking this must just be a taylor series of some trigonometric function but after looking it up, I cannot seem to find one that satisfies this. Any suggestions are greatly appreciated.
From below the user states that $$\sum_{n=1}^{\infty}\frac{(-1)^{n+1}}{n^2} = \sum_{n=0}^{\infty}\left(\frac{1}{2n+1} \right)^2 -\sum_{n=1}^{\infty}\left(\frac{1}{2n}\right)^2 = \sum_{n=1}^{\infty}\frac{1}{n^2} - 2\sum_{n=1}^{\infty}\left(\frac{1}{2n}\right)^2 = \ldots = \frac{\pi^2}{12}$$
I want to know the details of the $\ldots$ part.
| The series can be written as
$$\sum _0 ^{\infty}\left(\frac {1}{2n+1}\right)^2-\sum _1 ^{\infty}\left(\frac {1}{2n}\right)^2=\sum _1 ^{\infty} \frac {1}{n^2}-2\sum _1 ^{\infty}\left(\frac {1}{2n}\right)^2$$
By adding and subtracting:
$\sum _1 ^{\infty} \left(\frac {1}{2n}\right)^2$
Edited part :-
$$\sum _0 ^{\infty}\left(\frac {1}{2n+1}\right)^2+\sum_1 ^{\infty}\left(\frac {1}{2n}\right)^2-2\sum_1 ^{\infty}\left(\frac{1}{2n}\right)^2$$
Now first two sums can be written as
$$\sum _1 ^{\infty} \frac {1}{n^2}$$ So the next two can be written as:
$$\frac {1}{2}\sum _1 ^{\infty} \frac {1}{n^2}$$
Now we know all the summations. Thus the answer is:
$$\frac {\pi ^2}{6}-\frac {\pi^2}{2.6}=\frac {\pi^2}{12} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2327443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
number of integer solutions for $2^x\cdot 3^y$ Can someone help me figure out how to calculate the number of integer solutions to the equation: $N\ge2^x3^y|x,y\in\Bbb Z+$
For example if $N$ is $30$ we have $2\cdot 3$, $2^2\cdot 3$, $2^3\cdot 3$, $2\cdot 3^2$ so $4$ solutions
I know the number of integer solutions to $N\ge2^x$ is just $\left \lfloor \ln{N}/\ln{2} \right \rfloor $
So I was thinking something along the lines of taking the sum of $\log_2\left(\frac{N}{3^i}\right)$ from $i=1..(\log_3(N))$
Any help is greatly appreciated
|
The problem $2^x 3^y \le 30$ can be solved graphically as follows.
Taking the logarithm of both sides and simplifying, we get
\begin{align}
2^x 3^y &\le 30 \\
(\ln 2)x + (\ln 3)y &\le \ln 30 \\
\end{align}
The line above is the line described by $(\ln 2)x + (\ln 3)y = \ln 30$.
The indicated lattice points are the $(x,y)$ solutions to $ 2^x 3^y \le 30$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2330700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Log with $\sqrt x$ base I'd like to know how this simplification happened:
$$\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)=\log _2\left(x-2\right)$$
$$
\begin{array}{l}
\color{red}{2 \log _{2} x+\log _{\frac{1}{2}}(1-\sqrt{x})=\frac{1}{2} \log _{\sqrt{2}}(x-2 \sqrt{x}+2) \quad } \color{blue}{0<x<1} \\
\Leftrightarrow 2 \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2) \\
\Leftrightarrow \log _{2} x-\log _{2}(1-\sqrt{x})=\log _{2}(x-2 \sqrt{x}+2)-\log _{2} x
\end{array}
$$
| Just do it:
$k \log_a b = \log_a b^k$ no matter what type of base $a$.
(Because $k \log_a b = m \implies \log_a b = m/k \implies a^{m/k} = b \implies a^m = b^k \implies m = \log_a b^k$.)
And $\log_a b = \log_{a^x} b^x = x \log_{a^x} b$ for any valid $x$.
( Because $\log_a b = m \implies a^m = b \implies (a^m)^x =(a^x)^m = b^x \implies \log_{a^x} b^x = x \log_{a^x} b$.)
So ...
$\frac{1}{2}\log _{\sqrt{2}}\left(x-2\right)=\log _2\left(x-2\right)$
So $\frac 12\log_{\sqrt 2}(x-2) = \frac 12 \log_2 (x-2)^2 = \frac 12*2*\log_2(x-2) = \log_2(x-2)$
Or if that is too slick (I always like to double check things work by fundamental definitions)...
Let $\log_{\sqrt 2} (x-2) = z$ then
$\sqrt{2}^z = (x-2)$ then
$(\sqrt{2}^z)^2 = (x - 2)^2$ then
$2^z = (x-2)^2$ then
$\log_2 (x-2)^2 = z$
Let $\log_2 (x-2) = w$
The $2^w = (x-2)$
$2^{2w} = (x-2)^2$
$z = \log_2 (x-2)^2 = 2w = 2 \log_2 (x-2)$
So $\frac 12 \log_{\sqrt{2}}(x-2) = \frac 12 z = \frac 12 2w = w = \log_2(x-2)$.
It all works. Learn and get comfortable with these identities.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2331604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Elementary proof of a cotangent inequality
Let $0<x<\pi/2$. Then
$$ \cot{x} > \frac{1}{x}+\frac{1}{x-\pi}. $$
(This is still true for $-\pi<x<0$, but the given range is the one I'm concerned about.)
Is there an elementary proof of this? The local inequality $\cot{x}<1/x$ is easy to prove since it is equivalent to $x<\tan{x}$, which even has a simple geometric proof.
One can massage the Mittag-Leffler formula
$$ \cot{x} = \frac{1}{x} + \sum_{n \neq 0} \frac{1}{x-n\pi} + \frac{1}{n\pi} $$
into
$$ \cot{x} = \frac{1}{x} + \frac{1}{x-\pi} + \sum_{n=1}^{\infty} \frac{1}{z- (n+1)\pi}+\frac{1}{z+\pi n}, $$
and the terms in the second sum are all positive in the range considered since
$$ \frac{1}{z- (n+1)\pi}+\frac{1}{z+\pi n} = \frac{\pi-2z}{n(n+1)\pi^2+z(\pi-z)}>0, $$
but this is rather heavyweight for such a simple-looking inequality.
An equivalent formulation is
$$ \tan{y} > \frac{1}{\pi/2-y}-\frac{1}{\pi/2+y} = \frac{8y}{\pi^2-4y^2} $$
for $0<y<\pi/2$, if one desires more symmetry.
| Note that on the interval $[-\pi,\pi]$ $$0 \leq \operatorname{sinc}(x)=\frac{\sin(x)}{x}=\frac{\sin(x/2)\cos(x/2)}{x/2} \leq \cos(x/2)$$ with equality only in $\{-\pi,0,\pi\}$. Then on $(0, \pi)$
$$\begin{eqnarray}
\operatorname{sinc}^2(x) + \operatorname{sinc}^2(x-\pi) &< &\cos^2(x/2) + \cos^2((x-\pi)/2)\\& =& \cos^2(x/2) + \sin^2(x/2)\\& =& 1 \end{eqnarray}.$$ Dividing both sides by $\sin^2(x)$ on the same interval and noting that $\sin^2(x)=\sin^2(x-\pi)$ $$\frac1{x^2}+\frac1{(x-\pi)^2}<\frac1{\sin^2(x)}.$$ Then for $x\in(0,\pi/2)$ $$\frac1x + \frac1{x-\pi} = \int_x^{\pi/2}\left(\frac1{t^2}+\frac1{(t-\pi)^2}\right)\mathrm{d}t < \int_x^{\pi/2}\frac{\mathrm{d}t}{\sin^2(t)} = \cot(x).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2333813",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Finding GCD of two elements over a quadratic extension of integers I'm asking for a step-by-step explanation on finding $GCD(1+\sqrt{13}, 5+2\sqrt{13})$ in $\mathbb{Z}\left[\frac{1+\sqrt{13}}{2}\right]$.
| Note that $\Bbb{Z}[\frac{1+\sqrt{13}}{2}]$ is Norm Euclidean (see this post and the references given there), so we can apply the Euclidean algorithm to find the GCD.
$$\begin{align}
5+2\sqrt{13}&= 2(1+\sqrt{13}) +3 \\
1+\sqrt{13}&= -1\cdot 3 +(4+\sqrt{13}) \\
3 &= (4-\sqrt{13})(4+\sqrt{13}) + 0
\end{align}$$
So we see $4+\sqrt{13}$ is the GCD (up to associates) and $\dfrac{1+\sqrt{13}}{4+\sqrt{13}}=-3+\sqrt{13}$ and $\dfrac{5+2\sqrt{13}}{4+\sqrt{13}}=-2+\sqrt{13}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2335169",
"timestamp": "2023-03-29T00:00:00",
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Volume from iterated integrals and two regions The prompt is to find the volume of the solid which is described the equations and is bounded.
$$x^2+y^2+z^2=9 $$
$$x^2-3x+y^2=0 $$
The first one is a sphere with radius 3, the shadow is on the y-x plane.
For the second on I tried using completing the squares.
$$x^2-3x + y^2 =0 $$
$$x^2-3x+ 1/25 + y^2 = 1/25 $$
$$(x-1/5)^2 + y^2 = 1/25 $$
i dont know how to procede now. I also tried...
$$x^2+y^2-3x = 0$$
$$r^2-3x = 0 $$
$$ r^2 = 3x$$
$$ r^2 = 3cos\theta$$
$$ r = \sqrt{3cos\theta} $$
$$ \int_0^3\int_0^{2\pi} \int_0^{\sqrt{3cos\theta}}x^2+y^2+z^9rdrd\theta dz$$
Please correct me if the method to get the radius, if its wrong? Im kinda new to calculus.
| the first is indeed a sphere of radius $3$ centered at the origin. But the second is a cylinder of radius $3/2$ centered at $(3/2,0,0)$ You have to fix the completion of squares:
$$x^2-3x+y^2=0\;;x^2-3x+9/4+y^2=9/4\;;(x-3/2)^2+y^2=9/4$$
But you try too to use cylindrical coordinates and it seems a better way. The equation for the cylinder is ok:
$r^2=3\cos\theta\;;r=\sqrt{3\cos\theta}$ (dropping the minus sign as it has to be $r\geq0$)
The equation for the sphere is $r^2+z^2=9$. Isolating $z$ to use for the integration limits, $z=\pm\sqrt{9-r^2}$
We are calculating the volume, so, we integrate only for the volume element $rdrd\theta dz$ (in fact, you tried to integrate the value of the square of the distance to the origing all over the volume; if we had to interpret that integral, it is the mean over the points of the region of their squared distance to the origin times the volume)
For the integration limits:
$-\pi/2\lt\theta\lt\pi/2\;;0\lt r\lt\sqrt{3\cos\theta}\;;-\sqrt{9-r^2}\lt z\lt\sqrt{9-r^2}$
$$V=\int_{-\pi/2}^{\pi/2}\int_0^\sqrt{3\cos\theta}\int_{-\sqrt{9-r^2}}^\sqrt{9-r^2}rdzdrd\theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2336777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to calculate $\lim_{\varepsilon\rightarrow 1}\int_0^\varepsilon \int_0^z \int_0^y\frac 1 {1-x^3} \, dx \, dy \, dz$?
How to calculate the integral:
$$\lim_{\varepsilon\rightarrow 1}\int_0^\varepsilon \int_0^z \int_0^y\frac 1 {1-x^3} \, dx \, dy \, dz\quad?$$
One solution is about infinite series, but I don't fully understand that solution. Any other approaches?
| \begin{align}
& \int_0^\varepsilon \left( \int_0^z \left( \int_0^y\frac 1 {1-x^3} \, dx\right) \, dy\right) \, dz \\[12pt]
= {} & \iiint\limits_{0 \,\le\,x\,\le\,y\,\le\,z\,\le \, \varepsilon } \frac 1 {1-x^3} \,d(x,y,z).
\end{align}
Since the innermost function depends on $(x,y,z)$ only through $x$, putting $\displaystyle \int \cdots\,dx$ on the outside will mean on the inside we're just integrating constant functions, so that might be simpler.
From the expression
$$
0 \le x\le y\le z\le\varepsilon \tag 1
$$
we get $0\le x\le \varepsilon,$ so we have
$$
\int_0^\varepsilon \cdots\,dx.
$$
Then from $(1)$ we get $x\le y \le\varepsilon,$ so we get
$$
\int_0^\varepsilon \left( \int_x^\varepsilon \cdots\,dy \right) dx.
$$
Then from $(1)$ we have $y\le z\le\varepsilon,$ so we have
$$
\int_0^\varepsilon \left( \int_x^\varepsilon \left( \int_y^\varepsilon \cdots \,dz \right) dy \right) dx.
$$
So we have
\begin{align}
& \int_0^\varepsilon \left( \int_x^\varepsilon \left( \int_y^\varepsilon \frac 1 {1-x^3} \,dz \right) dy \right) dx \\[10pt]
= {} & \int_0^\varepsilon \int_x^\varepsilon \frac{\varepsilon-y}{1-x^3}\,dy\,dx \\[10pt]
= {} & \frac 1 2 \int_0^\varepsilon \frac{(\varepsilon-x)^2}{1-x^3} \, dx
\end{align}
If $\varepsilon=1,$ then
\begin{align}
& \frac{(\varepsilon-x)^2}{1-x^3} = \frac{(1-x)(1+x)}{(1-x)(1+x+x^2)} \\[10pt]
= {} & \frac{1+x}{1+x+x^2} = \underbrace{\frac{1/2}{1+x+x^2}}_{\Large\text{complete the square, etc.}} + \underbrace{\frac{(1/2) + x}{1+x+x^2}}_{\Large\text{routine substitution}}
\end{align}
At this point I would ponder the question of why it was expressed as $\lim\limits_{\varepsilon\to1}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2337475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The set of values of $p$ for which the points of extremum of the function $f(x)=x^3-3px^2+3(p^2-1)x+1$ lie in the interval $(-2,4)$ The set of values of $p$ for which the points of extremum of the function $f(x)=x^3-3px^2+3(p^2-1)x+1$ lie in the interval $(-2,4)$ is
$(A)(-3,5)$
$(B)(-3,3)$
$(C)(-1,3)$
$(D)(-1,5)$
$$f'(x)=3x^2-6px+3(p^2-1)=3[(p-x)^2-1]$$
For the points of extremum to lie between $(-2,4)$
The vertex of the parabola $f'(x)=3x^2-6px+3(p^2-1)$ should lie between $(-2,4)$
The vertex of the parabola $f'(x)=3x^2-6px+3(p^2-1)$ is $2p$
$$-2<2p<4\implies -1<p<2$$
But the answer is $(-1,3)$
| The x-coordinate of vertex of the quadratic equation is $-\dfrac{b}{2a}$, not $-\dfrac{b}{a}$.
Here is where you went wrong.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2339536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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I have to find all positive integers possible for $x$ and $y$ that fit $2x+1$ is divisible by $y$ and $2y +1$ is divisible by $x$ I think I have to simultaneously solve by doing
$$2x + 1 = ay$$
$$2y + 1 = bx$$
But if I continue solving simultaneously, it kind of loops and I get nowhere.
| If $x=1$ you have $y \in \{1,3\}$. Otherwise without loss of generality $2\le x\le y$. Since $y\mid 2x+1$, then
$$
x\le y\le 2x+1.
$$
Since $y$ is a divisor of $2x+1$, it can be only $2x+1$ or $\frac{2x+1}{d}$ with $d\ge 3$ odd (because the numerator is odd). The second case is impossible since
$$
x \le y =\frac{2x+1}{d} \le \frac{2x+1}{3} \Leftrightarrow x\le 1.
$$
Hence $y=2x+1$. You miss only to verify that $x\mid 2y+1=4x+3$, which implies $x \in \{1,3\}$. The unique case to check is $x=3$ which leads to $y=7$. Therefore the unique solutions are
$$
\{(1,1),(3,1),(1,3), (3,7), (7,3)\}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2340399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
} |
Correct answer of an indefinite integral Find the value of
$$ \int{\frac{dx}{x\sqrt{1-x^3}}} $$
I assumed $x^3 = \sin^2\theta$ and found the solution as
$$\frac{2}{3} \log\left|\frac{1}{x\sqrt{x}} - \frac{\sqrt{1-x^3}}{x\sqrt{x}} \right| + c$$
but the solution is given as
$$\frac{1}{3} \log{\left|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\right|} + c$$
Any help to reach to this provided solution will be appreciated.
| Note that
$$
\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}=
\frac{(\sqrt{1-x^3}-1)^2}{(\sqrt{1-x^3}+1)(\sqrt{1-x^3}-1)}
=
-\frac{(\sqrt{1-x^3}-1)^2}{x^3}.
$$
Hence
$$
\frac{1}{3}\ln\biggl|\frac{\sqrt{1-x^3}-1}{\sqrt{1-x^3}+1}\biggr|
=
\frac{2}{3}\ln\biggl|\frac{1-\sqrt{1-x^3}}{x^{3/2}}\biggr|.
$$
Thus, your answer is also correct.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
Limit of trig function raised to trig function I need help on evaluating this limit:
$\lim_{x\to0}(\frac{\sin{x}}{x})^{\frac{1}{1-\cos{x}}}$. So far I have tried setting the limit to L and taking the ln of both sides. So $\ln{L}=\lim_{x\to0}(\frac{1}{1-\cos{x}}\ln{(\frac{\sin{x}}{x})})$. I then tried using L'Hopital on the indeterminate 0/0 to no luck. Any hints would be appreciated.
| If you are allowed to use series expansion, consider $$A=\left(\frac{\sin{(x)}}{x}\right)^{\frac{1}{1-\cos{(x)}}}\implies \log(A)=\frac{1}{1-\cos{(x)}}\log\left(\frac{\sin{(x)}}{x}\right)$$ and use $$\sin{(x)}=x-\frac{x^3}{6}+\frac{x^5}{120}+O\left(x^7\right)$$ $$\frac{\sin{(x)}}{x}=1-\frac{x^2}{6}+\frac{x^4}{120}+O\left(x^6\right)$$ $$\log\left(\frac{\sin{(x)}}{x}\right)=-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right)$$ $$\cos(x)=1-\frac{x^2}{2}+\frac{x^4}{24}+O\left(x^6\right)$$ $$1-\cos(x)=\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right)$$ $$\log(A)=\frac{-\frac{x^2}{6}-\frac{x^4}{180}+O\left(x^6\right) }{\frac{x^2}{2}-\frac{x^4}{24}+O\left(x^6\right) }=\frac {-\frac{1}{6}-\frac{x^2}{180}+O\left(x^4\right) }{\frac{1}{2}-\frac{x^2}{24}+O\left(x^4\right)}$$ which already shows the limit for $\log(A)$.
If you want to go further, use the long division (or Taylor series again) to get $$\log(A)=-\frac{1}{3}-\frac{7 x^2}{180}+O\left(x^4\right)$$ and $$A=e^{\log(A)}=e^{-\frac{1}{3}}\left(1-\frac{7 x^2}{180} \right)+O\left(x^4\right)$$ which shows the limit and how it is approached.
It is even a good approximation of the function. If you use $x=\frac \pi 6$ the exact value would be $$A=\left(\frac{3}{\pi }\right)^{2 \left(2+\sqrt{3}\right)}\approx 0.708768$$ while the truncated series would give $$\frac{6480-7 \pi ^2}{6480 \sqrt[3]{e}}\approx 0.708892$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2342498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve in integers the equation, $(x^2-y^2)^2=1+16y$? How to solve in integers the equation,
$$(x^2-y^2)^2=1+16y$$
By observation we can take $y=3,5$
Is there another method to solve the equation?
| The integers solution I found are: $(\pm1,0),(\pm4,3),(\pm4,5)$.
Here's how I did it:
\begin{eqnarray}
(x^2-y^2)^2&=&1+16y\\ &=& 1\textrm{ mod }16 \qquad \textrm{(means when you divide a number by 16, remainder is 1)}\\x^2-y^2 &=& \pm\sqrt{1\textrm{ mod }16},
\end{eqnarray}
The only values of $y$ that give an integer square root for the above are 0,3 and 5.
For example, $y=3$ works because, $16\times3 + 1 = 49, \pm\sqrt{49} = \pm7$.
Now, $x^2-3^2 = \pm 7 \implies x^2 = \pm7+9 \implies x = \pm4$ or $x = \pm \sqrt{2}$. But since we are ol,y interested in integer solutions we only take the integer value of $x$. Hence, one possible integer solution is $x = \pm 4, y = 3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2345086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 2
} |
How do I solve this fraction question? If $a = -1/5$, how do I calculate:
$$3 a + 2 a^2$$
I did $3\times(-1/5) + (-1/5) \times (-1/5) \times 2$, but can't figure out what the right way to solve this is.
| A helpful tip:
You can only add fractions with the same denominator (the numbers at the bottom). If the denominators are not equal, you have to multiply them by a constant number $k$ to add them.
Starting from $$\frac{3 \times -1}{5} + \frac{-1 \times -1 \times 2}{5\times5},$$
since $+a \times -b = -ab$, and $-a \times -b = ab$ for arbitrary numbers $a,b$, we have:
$$-\frac{3}{5} + \frac{2}{25}$$
Since $\frac{1}{5}$ is $5$ times that of $\frac{1}{25}$, $k=5$, and we have to multiply both numerator and denominator by $5$ to simplify:
$$-\frac{15}{25} + \frac{2}{25}$$
Now you can simplify the expression by cancelling $\frac{10}{25}$ and adding the $2$ fractions together:
$$-\frac{15}{25} + \frac{2}{25} = \frac{-15+2}{5}$$
I'll leave the last step for you to figure out.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $\frac{a^3+b^3+c^3}{abc}$ If $ax^2+bx+c = 0$ and $bx^2+cx + a = 0$ have a common root and $a\neq 0$, then find $$\frac{a^3+b^3+c^3}{abc}$$
I tried that for both equations to have a common root, the expression on left hand sides must be equal, ie $$ax^2+bx+c = bx^2+cx + a$$
for this we must have $x=1$ (i cannot prove this, but it appears to be true). Also both of these must be equal to $0$, so we have:
$$a+b+c=0$$
So using this we say
$$\frac{a^3+b^3+c^3}{abc} = \frac{a^3+b^3+c^3-3abc+3abc}{abc}=\frac{(a+b+c)(...)}{abc}+3$$
So we get the answer as $3$.
How do we say that $x =1$ is the commmon root? Thanks!
| We have $$f(x)=ax^2+bx+c\\g(x)=bx^2+cx+a$$
If we set $t$ as the common root, then we know that $f(t)=g(t)=0$:
\begin{align}at^2+bt+c&=bt^2+ct+a\end{align}
We can equate coefficents to conclude that $a=b=c$
Therefore, we can say that \begin{align}\frac{a^3+b^3+c^3}{abc}&=\frac{a^3+a^3+a^3}{aaa}\\
&=\frac{3a^3}{a^3}\\
&=3\end{align}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Evaluating the expression $\cos(\frac{1}{2}\tan^{-1}(-\frac{4}{3}))$ I can prove that $\cos(\tan^{-1}(x)) = +\dfrac{1}{\sqrt{1+x^2}}$, set $y=\tan^{-1}(x)$ and therefore $y=\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
Dividing both sides by $2$:
$\dfrac{y}{2}=\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}})$
$\cos(\dfrac{1}{2}\tan^{-1}(x))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{1}{\sqrt{1+x^2}}))$
$\cos(\dfrac{1}{2}\tan^{-1}(-\frac{4}{3}))=\cos(\dfrac{1}{2}\cos^{-1}(\frac{3}{5}))$
This is my approach to the problem, are there any better ways to simplify the initial expression and perhaps find an approximation even without using a calculator?
|
Let $\theta = \arctan\left(-\dfrac 43 \right)$
Then, from the picture, $\cos \theta = \dfrac 35$ and $\theta$ is in the fourth quadrant.
So $\cos(\frac 12 \theta)
= \sqrt{\dfrac{1 + \cos \theta}{2}}
= \sqrt{\dfrac{1 + \frac 35}{2}}
= \dfrac{2}{\sqrt 5}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2351812",
"timestamp": "2023-03-29T00:00:00",
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What is the value of the expression $[1 + \cos(\frac{\pi}{8})][1 + \cos(\frac{3\pi}{8})][1 + \cos(\frac{5\pi}{8})][1 + \cos(\frac{7\pi}{8})]$? This is rather a simple problem that I'm posting ; looking forward not for the solution of it but the different ways it could be solved.
What is the value of the following expression?
$$\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 5\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 7\pi }{ 8 } \right) } \right) $$
Edit: Do I manually put the values of $\cos\frac{\pi}{8}$ and all other cosine terms? Is there a better and shorter way?
| $$\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 5\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 7\pi }{ 8 } \right) } \right) =\\ =\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \pi -\frac { 3\pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \pi -\frac { \pi }{ 8 } \right) } \right) =\\ =\left( 1+\cos { \left( \frac { \pi }{ 8 } \right) } \right) \left( 1+\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1-\cos { \left( \frac { 3\pi }{ 8 } \right) } \right) \left( 1-\cos { \left( \frac { \pi }{ 8 } \right) } \right) =\\ =\left( 1-\cos ^{ 2 }{ \frac { \pi }{ 8 } } \right) \left( 1-\cos ^{ 2 }{ \frac { 3\pi }{ 8 } } \right) =\sin ^{ 2 }{ \frac { \pi }{ 8 } } \sin ^{ 2 }{ \frac { 3\pi }{ 8 } } =\frac { \left( 1-\cos { \frac { \pi }{ 4 } } \right) }{ 2 } \cdot \frac { \left( 1-\cos { \frac { 3\pi }{ 4 } } \right) }{ 2 } =\\ =\frac { \left( 1-\cos { \frac { \pi }{ 4 } } \right) \left( 1-\cos { \left( \pi -\frac { \pi }{ 4 } \right) } \right) }{ 4 } =\frac { \left( 1-\cos { \frac { \pi }{ 4 } } \right) \left( 1+\cos { \frac { \pi }{ 4 } } \right) }{ 4 } =\frac { 1-\cos ^{ 2 }{ \frac { \pi }{ 4 } } }{ 4 } =\frac { 1-\frac { 1 }{ 2 } }{ 4 } =\frac { 1 }{ 8 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2352508",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.