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An alternative solution for $\displaystyle \int_0^{\pi/2}\sin (2nx)\cot x dx$
Evaluate $\displaystyle \int_0^{\pi/2}\sin (2nx)\cot x dx$
$$\sin (2nx)\cot x=\frac{\sin (2nx)}{\sin x}\cos x$$
I know that
$$\frac{\sin (\frac{k\beta}{2})}{\sin (\frac{\beta}{2})}\cos\left(\alpha+\frac{k-1}{2}\beta\right)=\cos\alpha+\cos(\alpha+\beta)+\cos(\alpha+2\beta)+\cdots+\cos(\alpha+(k-1)\beta)$$
So comparing between $\frac{\sin (2nx)}{\sin x}\cos x$ and $\frac{\sin (\frac{k\beta}{2})}{\sin (\frac{\beta}{2})}\cos\left(\alpha+\frac{k-1}{2}\beta\right)$ I got $\beta=2x$, $k=2n$ and $\alpha=-(2n-2)x$.
Therefore,
$$\begin{align*}\frac{\sin (2nx)}{\sin x}\cos x&=\cos(-[2n-2]x)+\cos(-[2n-2]x+2x)+\cos(-[2n-2]x+4x)+\cdots+\cos(2nx)\\&=\cos0+\cos(2nx)+2[\cos(2x)+\cos(4x)+\cdots+\cos((2n-2)x)]\\&=1+\cos(2nx)+2[\cos(2x)+\cos(4x)+\cdots+\cos((2n-2)x)]\end{align*}$$
$$\int_0^{\pi/2}\cos 2kx dx=0$$
$$\implies\int_0^{\pi/2}\sin (2nx)\cot x dx=\int_0^{\pi/2}dx=\frac{\pi}{2}$$
Can someone provide an alternative solution without using summation of cosines. Thanks in advance.
| Let
$$f_n=\int_{0}^{\frac \pi 2} \sin (2nx)\cot (x)\;\text{d}x$$
We will first show that $f_{n+1}-f_n=0$ for any $n\in \mathbb N$.
\begin{align*}
f_{n+1}-f_n&=\int_{0}^{\frac \pi 2}(\sin(2(n+1)x)-\sin(2nx))\cot(x)\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\sin\left(\frac{2(n+1)x-2nx}2\right)\cot(x)\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\sin(x)\frac{\cos (x)}{\sin (x)}\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\cos (x)\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} 2\cos\left(\frac{2(n+1)x+2nx}2\right)\cos\left(\frac{2(n+1)x-2nx}2\right)\;\text{d}x\\
&=\int_{0}^{\frac \pi 2} \cos(2(n+1)x)+\cos(2nx))\;\text{d}x\\
&=\frac 1{2(n+1)}\sin(2(n+1)x)+\frac 1{2n}\sin(2nx)\bigg|_{0}^{\frac \pi 2}\\
&=\frac 1{2(n+1)}\sin((n+1)\pi)+\frac 1{2n}\sin(n\pi)-\frac 1{2(n+1)}\sin(0)-\frac 1{2n}\sin(0)\\
&=0
\end{align*}
Now, just calculate $f_1=\frac \pi 2$ and complete the proof
| {
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"url": "https://math.stackexchange.com/questions/4208063",
"timestamp": "2023-03-29T00:00:00",
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Find the intervals in which the function decreasing. $f(x)= \log{(1+x)}-{2x\over 1+x^2}$ $f(x)= \log{(1+x)}-{2x\over 1+x^2}$ find the intervals in which the function decreasing.
My solution: $$f'(x)={1\over 1+x}- {2(1-x^2) \over (1+x^2)^2} \\ ={x^4+2x^3+4x^2-2x-1 \over (1+x)(1+x^2)^2}$$
Since we need to find the interval for decreasing hence $f'(x)<0$.
Therefore, $${x^4+2x^3+4x^2-2x-1<0}$$
Now how to proceed to find the interval. Thanks for help.
| The first derivative must be less than zero for a function to be decreasing.
$f'(x)={x^4+2x^3+4x^2-2x-1 \over (1+x)(1+x^2)^2}$,
The system of inequalities must be resolved by:
$x^4+2x^3+4x^2-2x-1 <0$,
$x+1>0$,
for the existence of the logarithm in the real field.
In addition, it must be $x≠-1$.
The solution to the problem is as follows:
$x_{1}<x<x_{2}$,
$x_{1}=-\frac{\sqrt{6}\sqrt{3\sqrt{R}-S}}{6}+\frac{\sqrt{3T}}{6}-\frac{1}{2}$,
$x_{2}=+\frac{\sqrt{6}\sqrt{3\sqrt{R}-S}}{6}+\frac{\sqrt{3T}}{6}-\frac{1}{2}$;
the values of $R$, $S$, and $T$ are:
$R=4A^{1/3}+4B^{1/3}+4a^{1/3}+4b^{1/3}+1$,
$S=6C^{1/3}+6c^{1/3}+5$,
$T=12D^{1/3}+12d^{1/3}-5$;
and again
$A=\frac{4375}{5832}-\frac{125\sqrt{129}}{1944}$,
$a=\frac{4375}{5832}+\frac{125\sqrt{129}}{1944}$,
$B=\frac{2386}{729}-\frac{70\sqrt{129}}{243}$,
$b=\frac{2386}{729}+\frac{70\sqrt{129}}{243}$,
$C=\frac{35}{216}-\frac{\sqrt{129}}{72}$,
$c=\frac{35}{216}+\frac{\sqrt{129}}{72}$,
$D=C$,
$d=c$.
| {
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"url": "https://math.stackexchange.com/questions/4212135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate angle $x$ in the figure For reference:
My progress:
\begin{align*}
& AH \perp FD \\
& \triangle AFD \text{ is isosceles}
\quad
\therefore \measuredangle BFA = \measuredangle FDA = x \\
& AF = FD \\
& \measuredangle HBA = 180-135^\circ = 45^\circ
\quad
\therefore \triangle HBA \text{ is isosceles}
\end{align*}
I drew some auxiliary lines but it wasn't enough to reach the solution.
| Once again, Pythagoras theorem does the job.
Say, $BC = a, AC = b$. Drop a perp from $B$ to $AD$. Notice that right triangles $\triangle ABC \cong \triangle ABH$, given $\angle ABC = \angle BAH = 45^\circ - x$ and $AB~$ is hypotenuse to both triangles.
So, $AH = BC = a, BH = AC = b$
Similarly notice that $\triangle DBE \cong \triangle BDH$. So, $DE = BH = b, DH = BE$ but $BE = CF = DF = a + b$
That leads to $DF = a + b, AF = a + 2b, AD = 2a + b$
Applying Pythagoras,
$(a+b)^2 + (a+2b)^2 = (2a+b)^2 \implies (a + b) (a-2b) = 0$
So we get $a = 2b$ and that shows $\triangle ADF$ is $3:4:5$ triangle.
Hence, $2x \approx 37^\circ$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Centroid and midpoints cyclic then prove that $AC^2 + BC^2 = 2AB^2.$
Let $ABC$ be a non-equilateral triangle with integer sides. Let $D$ and $E$ be respectively the mid-points of $BC$ and $CA$ ; let $G$ be the centroid of $\Delta{ABC}$. Suppose, $D$, $C$, $E$, $G$ are concyclic.Prove that $AC^2 + BC^2 = 2AB^2.$
My progress:
*
*By angle chase we get, $$\angle GBA=\angle GED=\angle GCD\implies (CGB)~~\text{is tangent to}~~AB.$$
*Similarly, we get $$\angle GAB=\angle GDE=\angle GCE\implies (CGA)~~\text{is tangent to}~~AB.$$
*By power of point, we have $$\frac{AB^2}{4}=FA^2=FB^2=FG\cdot FC=\frac{1}{3}FG^2. $$
| You were nearly done:
$$
\frac{AB^2}{4}=\frac{1}{3}FC^2 \tag{1}
$$
From Appolonius' Theorem:
$$b^2 + a^2=2 FC^2 + \frac{c^2}{2}$$
$$\implies FC^2=\frac{2a^2+2b^2-c^2}{4}$$
Substituting for $FC^2$ in $Eq(1)$:
$$\frac{c^2}{4}=\frac{1}{3}\frac{2a^2+2b^2-c^2}{4}$$
A little algebra yields the answer.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Partial fraction expansion question I have to integrate following expression (but integration is not the problem):
$$\frac{x^2+3x-2}{(x-1)(x^2+x+1)^2}$$
It is pretty obvious that:
$$\frac{x^2+3x-2}{(x-1)(x^2+x+1)^2}=\frac{A}{x-1} + \frac{Mx+N}{x^2+x+1} + \frac{Px+Q}{(x^2+x+1)^2}$$
The first and the easiest step is to find an $A$:
$$A=\frac{x^2+3x-2}{(x^2+x+1)^2}, x=1$$
$$A=\frac{2}{9}$$
And then there comes a problem - I don't know how to do the rest. I tried to multiply the whole thing by $(x^2+x+1)^2$ and differentiate, but it didn't seem to be useful at all, especially because $(x^2+x+1)^2$ doesn't have real roots.
As popping900 suggested. I can take just four different x values and solve system of for equations, but i would like to see a more elegant or shorter solution, if one exists
| Another easy coefficient is $M$. Multiply both sides by $x$ and take the limit as $x\to +\infty$:
$$0=A+M\implies M=-A=-2/9$$
For $x=0$ we find
$$2=-A+N+Q\implies N+Q=A+2=20/9.$$
For $x=i$ we find
$$-3=\frac{A}{i-1} + \frac{Mi+N}{i} + \frac{Pi+Q}{-1}\implies Q=8/3, N+P=-1/9$$
Therefore $N=20/9-Q=-4/9$ and $P=-N-1/9=1/3$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do I evaluate this limit that seems to be indeterminate? I've been trying to evaluate the below limit, which Mathematica claims is equals to $1$.
$$\underset{n\to \infty }{\text{lim}}\frac{\frac{1}{2} (n+1) \sin \left(\frac{2 \pi }{n+1}\right)-\pi }{\frac{1}{2} n \sin \left(\frac{2 \pi }{n}\right)-\pi }$$
However, my attempts, of which I attempted using L'Hopital's rule, always end with the indeterminate form of $\frac{0}{0}$, because taking the limit to infinity of both the non-$\pi$ sections of the numerator and denominators gives $\pi$.
Is Mathematica correct, and if so, what is the solution?
| HINT
We have that
$$\frac{\frac{1}{2} (n+1) \sin \left(\frac{2 \pi }{n+1}\right)-\pi }{\frac{1}{2} n \sin \left(\frac{2 \pi }{n}\right)-\pi }=
\frac{{\sin \left(\frac{2 \pi }{n+1}\right)}-\frac{2 \pi }{n+1} }{\left(\frac{2 \pi }{n+1}\right)^3}
\frac{\left(\frac{2 \pi }{n} \right)^3}{{\sin \left(\frac{2 \pi }{n}\right)}-\frac{2 \pi }{n} }
\frac {n^2} {(n+1)^2}$$
then it suffices to show that as $x \to 0$
$$\frac{\sin x-x}{x^3}\to-\frac 16 $$
| {
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"timestamp": "2023-03-29T00:00:00",
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Simplifying a fraction by only multiplying one side by the denominator? Given the problem: $x\sqrt{3} = 6$
We can solve it this way:
$ x = \frac{ 6 } {\sqrt{3} } \times \frac{ \sqrt{3} } { \sqrt{3} } $
$ x = \frac{ 6 \sqrt{3} } {3} $
$ x = 2 \sqrt{3} $
In the first step, why can we multiply only the right side by $\sqrt{3}$? Shouldn't both sides be modified in the same way to keep them balanced?
| To get from here:
$x\sqrt{3} = 6$
to here:
$ x = \frac{ 6 } {\sqrt{3} } \times \frac{ \sqrt{3} } { \sqrt{3} }, $
both sides of the equation were divided by $\sqrt{3},$ not multiplied by $\sqrt{3}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Need help to evaluate $\int\limits_{\frac{1}{\sqrt{2}}}^{1} \int\limits_{\sqrt{1-x^{2}}}^{x}\frac{1}{\sqrt{x^2+y^2}}dydx$ I have this double integral $\int\limits_{\frac{1}{\sqrt{2}}}^{1} \int\limits_{\sqrt{1-x^{2}}}^{x} \frac{1}{\sqrt{x^2+y^2}}dydx$
I tried to transform into polar coordinates using $x = r\cos \theta$ , $y=r\sin\theta$ with $\left | J \right |= r$. Getting something like
$\int \int \frac{1}{r}rdrd\theta$ , but unable to define the upper and lower bounds of the integral.
Any help with that?
| Note that\begin{align}\sqrt{1-x^2}\leqslant y\leqslant x&\iff1-x^2\leqslant y^2\leqslant x^2\\&\iff1\leqslant x^2+y^2\leqslant2x^2.\end{align}In polar coordinates, the final pair of inequalities becomes$$1\leqslant r^2\leqslant2r^2\cos^2\theta.$$So, take $\theta\in\left[0,\frac\pi4\right]$, so that $x,y\geqslant 0$ and that $\cos^2\theta\geqslant\frac12$. You also know that $r\geqslant1$. But you also know that $x\leqslant1$; in other words, $r\leqslant\frac1{\cos\theta}$. So, compute$$\int_0^{\pi/4}\int_1^{1/\cos\theta}1\,\mathrm dr\,\mathrm d\theta.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Stationary points of a multivariable function This question might just be a quick one but I'm slightly confused by the answer I've been provided for this question.
I have the function: $f(x,y) = (x^2+2y^2)e^{-y^2 - x^2}$
I found the partial derivatives:
$f_x = (2x-2x(x^2+2y^2))e^{-y^2-x^2}$
$f_y = (4y-2y(x^2+2y^2))e^{-y^2-x^2}$
I know that the stationary points can be found where $f_x=0$ and/or where $f_y=0$, so that's what I did, I set both $f_x$ and $f_y$ to equal $0$:
\begin{align}
\tag{1}
(2x-2x(x^2+2y^2))e^{-y^2-x^2}&=0\\
\tag{2}
(4y-2y(x^2+2y^2))e^{-y^2-x^2}&=0
\end{align}
I then divided both $(1)$ and $(2)$ by $e^{-y^2-x^2}$ and simplified both equations, this is what I was left with:
\begin{align}
\tag{1'}
x(1-x^2-2y^2)=0\\
\tag{2'}
y(2-x^2-2y^2)=0
\end{align}
This is where I was slightly confused. On the answer sheet I'm provided, the only stationary points they find are $(0,0)$, $(1,0)$ and $(-1,0)$
I understand how $(0,0)$ was found but I'm not sure how $(\pm1,0)$ was found. Did they make the two equations equal to each other then solve?
| If you have $$AB=0\\CD=0$$
then you have the following solution $A=C=0$, $A=D=0$, $B=C=0$, $B=D=0$. So in your case, $A=x$, $B=1-x^2-2y^2$, $C=y$, $D=2-x^2-2y^2$. Note that you can't have $B=D=0$ since they differ by $1$. So you need to check $$x=y=0\\x=2-x^2-2y^2=0\\y=1-x^2-2y^2=0$$
The solutions are $(0,0)$, $(0,\pm1)$, $(\pm1,0)$
| {
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"timestamp": "2023-03-29T00:00:00",
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Prove integral inequality including n-order derivative Let $f$ be $n$ times continuously differentiable on $[0,1]$, with $f(\frac{1}{2})=0$ and $f^{(i)}(\frac{1}{2})=0$ when $i$ is even and less than $n$. Prove
$$
\left( \int_0^1{f\left( x \right) \mathrm{d}x} \right) ^2\leqslant \frac{1}{\left( 2n+1 \right) 4^n\left( n! \right) ^2}\int_0^1{\left( f^{\left( n \right)}\left( x \right) \right) ^2\mathrm{d}x}.
$$
$f\left( x \right) =f\left( \frac{1}{2} \right) +f'\left( \frac{1}{2} \right) \left( x-\frac{1}{2} \right) +\frac{1}{2!}f''\left( \frac{1}{2} \right) \left( x-\frac{1}{2} \right) ^2+\cdots +\frac{1}{n!}f^{\left( n-1 \right)}\left( \frac{1}{2} \right) \left( x-\frac{1}{2} \right) ^{n-1}+\frac{1}{n!}f^{\left( n \right)}\left( \frac{1}{2}+\theta \left( x-\frac{1}{2} \right) \right) \left( x-\frac{1}{2} \right) ^n=\frac{1}{n!}f^{\left( n \right)}\left( \frac{1}{2}+\theta \left( x-\frac{1}{2} \right) \right) \left( x-\frac{1}{2} \right) ^n$，$\theta\in(0,1)$.
so $\int_0^1{f\left( x \right) \mathrm{d}x}=\frac{1}{n!}\int_0^1{f^{\left( n \right)}\left( \frac{1}{2}+\theta \left( x-\frac{1}{2} \right) \right) \left( x-\frac{1}{2} \right) ^n\mathrm{d}x}$
How to deal with $\int_0^1{\left( f^{\left( n \right)}\left( x \right) \right) ^2\mathrm{d}x}$ to prove this inequality?
| This almost achieves the desired bound, maybe I have some calculations error in here and it actually works.
For $n=0$ this is just Cauchy-Schwarz, so assume $n>0$.
Let $a=\frac{1}{2}$ for clarity and $x\in[0,1]$. Define $$R_{n-1}(x)=\int_a^x\frac{f^{(n)}(t)}{(n-1)!}(x-t)^{n-1}dt$$ This is the error term in Taylor's theorem, i.e. $$f(x) = f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\dots+\frac{f^{(n-1)}(a)}{(n-1)!}(x-a)^{n-1}+R_{n-1}(x)$$
Therefore:$$\int_0^1f(x)=\sum_{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}\int_0^1(x-a)^{k}dx+\int_0^1R_{n-1}(x)dx$$
Note that every term in the sum vanishes: For even $k$ because of $f^{(k)}(a)=0$ and for odd $k$ because the integral is zero due to symmetry.
Thus $$\int_0^1f(x)dx=\int_0^1R_{n-1}(x)dx=\int_0^1\int_a^x\frac{f^{(n)}(t)}{(n-1)!}(x-t)^{n-1}dtdx$$
Now by Cauchy-Schwarz we have \begin{align*}
\left\vert\int_a^x{f^{(n)}(t)}(x-t)^{n-1}dt\right\vert^2\leq\left\vert\int_a^x{f^{(n)}(t)}^2dt\right\vert\cdot\left\vert\int_a^x(x-t)^{2(n-1)}dt\right\vert
\end{align*}
The absolute values are necessary on the right side to include the case $x<a$. Now we have$$\left\vert\int_a^x{f^{(n)}(t)}^2dt\right\vert\leq\int_0^1f^{(n)}(t)^2dt$$ and \begin{align*}
\left\vert\int_a^x(x-t)^{2(n-1)}dt\right\vert=\left\vert\frac{(x-a)^{2n-1}}{2n-1}\right\vert
\end{align*}
Hence
\begin{align*}
(n-1)!\int_0^1f(x)dx&\leq\int_0^1\left\vert\int_a^xf^{(n)}(x-t)^{n-1}dt\right\vert dx\\
&\leq\int_0^1\sqrt{\int_0^1f^{(n)}(t)^2dt}\sqrt{\left\vert\frac{(x-a)^{2n-1}}{2n-1}\right\vert}dx
\end{align*}
And:
\begin{align*}
\int_0^1\sqrt{\left\vert\frac{(x-a)^{2n-1}}{2n-1}\right\vert}dx&=\frac{2}{\sqrt{2n-1}}\int_0^{a}x^{(2n-1)/2}dx\\
&=\frac{2}{\sqrt{2n-1}}\frac{\left(\frac{1}{2}\right)^{(2n+1)/2}}{\frac{2n+1}{2}}\\
&=\frac{4}{\sqrt{2n-1}(2n+1)}\frac{1}{2^{(2n+1)/2}}
\end{align*}
Putting all together then gives us:
\begin{align*}
\left(\int_0^1f(x)dx\right)^2&\leq \frac{1}{(n-1)!^2}\frac{4^2}{(2n-1)(2n+1)^2}\frac{1}{2^{2n+1}}\int_0^1f^{(n)}(t)^2dt\\
&=\frac{1}{(n-1)!^2}\frac{8}{(4n^2-1)(2n+1)}\frac{1}{4^n}\int_0^1f^{(n)}(t)^2dt\\
&\leq\frac{1}{(n-1)!^2}\frac{8}{4n^2(2n+1)}\frac{1}{4^n}\int_0^1f^{(n)}(t)^2dt\\
&=\frac{2}{n!^2(2n+1)4^n}\int_0^1f^{(n)}(t)^2dt\\
\end{align*}
Unfortunately this is worse than what we wanted to show by a factor of $2$, perhaps one can improve one of the above estimates or I just made a mistake somewhere?
| {
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"timestamp": "2023-03-29T00:00:00",
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Proving that the sum of $n^{th}$ powers of the roots of a quadratic with integer coefficients is also an integer I am trying to prove that if a quadratic equation $x^2+px+q$ has roots $\alpha$ & $\beta$, then $$\alpha^n+\beta^n\in Z\ \text{for all }\ \ n\in N $$
My attempt:
$$\alpha^n+β^n=(α+β)^n-\sum_{k=1}^{n-1}\ {n\choose k} \ a^{n-k}\ b^k$$
For odd $n$:
$$\begin{aligned}
\sum_{k=1}^{n-1}\ {n\choose k} \ a^{n-k}\ b^k=&\ \ \ \ {n\choose 1}a^{n-1}b\quad +\quad{n\choose 2}a^{n-2}b^2 \quad\ \ \ ...\quad {n\choose \frac{n-1}{2}}a^{\frac{n+1}{2}}b^{\frac{n-1}{2}}\\
&\quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad +\\
\\
&{n\choose n-1}a^{1}b^{n-1}+{n\choose n-2}a^{2}b^{n-2\quad}...{n\choose \ n- \frac{n-1}{2}}a^{\frac{n-1}{2}}b^{\frac{n+1}{2}}\\
\\
=& {n\choose 1}αβ\ (α^{n-2}+β^{n-2})+{n\choose 1}(αβ)^2\ (α^{n-4}+β^{n-4})...\\
\\
=& \sum_{k=1}^{\frac{n-1}{2}}{n\choose k}(αβ)^k\ (α^{n-2k}\ +\ β^{n-2k})
\end{aligned}$$
Defining $I_n:=α^n\ +\ β^n$, therefore, $$I_n=(-p)^n-\sum_{k=1}^{\frac{n-1}{2}} {n\choose k}\cdot q^k\cdot I_{n-2k}$$
Since $(-p)^n$, $q$ and $n\choose k$ are integers, $I_N$ is an integer if $I_n$ is an integer for all odd $n<N$
Base step:
$$\alpha+β=-p\ \Rightarrow integer$$
$$\alpha^3+β^3=(α+β)^3-3αβ(α+β)=\ -p^3+3pq\ \Rightarrow integer$$
Strong induction:
Assuming $I_n$ is an integer for all odd $n$ unto some odd natural number $k$,
$$I_{k+2}=(-p)^{K+2}-[{k+2\choose 1}\ q\ I_k\ +\ {k+2\choose 2}\ q^2\ I_{k-2}\ {k+2\choose 3}\ q^3\ I_{k-4}\ ...\ {k+2\choose \frac{k+1}{2}}\ q^{\frac{k+1}{2}}\ I_1]$$
$ I_{k+2} $ is also an integer. Since it has already been shown that $I_1$ and $I_3$ are integers, $I_n$ must be an integer for all odd natural numbers.
For even $n$:
Using a similar process, it can be shown that for even $n$,
$$\sum_{k=1}^{n-1}\ {n\choose k}\ α^{n-k}\ β^k=\ {n\choose \frac{n}{2}}\ (αβ)^{\frac{n}{2}}\ +\ \sum_{k=1}^{\frac{n}{2}-1}{n\choose k}(αβ)^k\ (α^{n-2k}\ +\ β^{n-2k})
$$
and thus$$I_n=(-p)^n-{n\choose \frac{n}{2}}\ (q)^{\frac{n}{2}}\ -\sum_{k=1}^{\frac{n-1}{2}} {n\choose k}\cdot q^k\cdot I_{n-2k}$$
Base step:
$$\alpha^2+β^2=(α+β)^2-2αβ=\ p^2-2q\ \Rightarrow integer$$
$$\alpha^4+β^4=(α+β)^4-4αβ(α^2+β^2)-6(αβ)^2 \Rightarrow integer$$
Strong induction:
Assuming $I_n$ is an integer for all even $n$ upto some even natural number $k$,
$$I_{k+2}=(-p)^{K+2}-{k+2\choose \frac{k+2}{2}}(p)^{\frac{K+2}{2}}\ -\ [{k+2\choose 1}\ q\ I_k\ +\ {k+2\choose 2}\ q^2\ I_{k-2}\ {k+2\choose 3}\ q^3\ I_{k-4}\ ...\ {k+2\choose \frac{k+1}{2}}\ q^{\frac{k+1}{2}}\ I_1]$$
$ I_{k+2} $ is also an integer. Since it has already been shown that $I_2$ and $I_4$ are integers, $I_n$ must be an integer for all even natural numbers.
Is this proof correct? Please point out anything that I have missed or should mention explicitly. Moreover, is there any way to prove the statement for both even and odd $n$ directly or is it necessary to deal with them separately?
| Here is a "cheat" method. The Fundamental theorem of symmetric polynomials says that any symmetric polynomial $f(x,y)$ with integer coefficients (meaning $f(x,y)=f(y,x)$) can be written as an integer-coefficient polynomial of $xy$ and $x+y$ (the elementary symmetric polynomials.)
In your case, since $f(x,y)=x^n+y^n$ is a symmetric polynomial, it can be written as a polynomial $g(x+y,xy)$ with integer coeffcients. Now, we have $\alpha^n+\beta^n=g(\alpha+\beta,\alpha\beta)=g(-p,q)$, which is clearly an integer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Geometric proof of $\cos A + \cos B + \cos C + \cos (A+B+C) = 4 \cos \frac{A+B}{2} \cos \frac{B+C}{2} \cos \frac{A+C}{2}$
Prove:
$$\cos A + \cos B + \cos C + \cos (A+B+C) = 4 \cos \frac{A+B}{2} \cos \frac{B+C}{2} \cos \frac{A+C}{2}$$
Geometrically.
I found the algebraic solutions here but I want to figure out how to calculate the above using geometrical arguments of complex numbers/ vectors.
Some hints of geometry I figured out already:
*
*$\frac{A+B}{2}$ is equidistant in angle mangitude from $A$ and $B$, that is $|\frac{A+B}{2} -A| = | \frac{A+B}{2} - B|$, similar results hold for the other angle averages.
*We may think of the above as summing the projections of three unit vectors onto the x-axis: $ \left( \tau(A) + \tau(B) + \tau(C) + \tau( A+B+C)\right) \cdot \hat{i}$, where $\tau(\phi)$ is the unit vector making angle $\phi$ with the x-axis
Related , related
| Let
$$\,\bar e=\dbinom01,\;\bar a=\dbinom{\sin A}{\cos A},\;
\bar b=\dbinom{\sin(A+B)}{\cos(A+B)},\;
\bar c=\dbinom{\sin(A+B+C)}{\cos(A+B+C)},\tag1$$
where $\dbinom yx$ means the radius-vector with the cartesian coordinates $\,x,\,y.$
Then the scalar productions are
$$\dbinom{\sin\alpha}{\cos\alpha}\cdot\dbinom{\sin\beta}{\cos\beta}
=\cos\alpha\cos\beta+\sin\alpha\sin\beta=\cos(\alpha-\beta),\tag2$$
$$\bar e\cdot\bar a=\cos A,\quad
\bar a\cdot\bar b=\cos B,\quad
\bar b\cdot\bar c=\cos C,\quad
\bar e\cdot\bar c=\cos(A+B+C).\tag3$$
Also, are known the identities
$$\sin x +\sin y = 2\sin\dfrac{x+y}2\,\cos\dfrac{x-y}2,\tag4$$
$$\cos x +\cos y = 2\cos\dfrac{x+y}2\,\cos\dfrac{x-y}2.\tag5$$
From $(1)-(5)$ should
$$\cos A+\cos B+\cos C+\cos(A+B+C)=\bar e\cdot\bar a+\bar a\cdot\bar b+
\bar b\cdot\bar c+\bar e\cdot\bar c=(\bar e+\bar b)\cdot(\bar a+\bar c)$$
$$=\dbinom{\sin(A+B)}{1+\cos(A+B)}\cdot\dbinom{\sin A+\sin(A+B+C)}{\cos A+\cos(A+B+C)}$$
$$=2\cos\dfrac{A+B}2\,
\begin{pmatrix}\sin\dfrac{A+B}2\\ \cos\dfrac{A+B}2\end{pmatrix}
\cdot2\cos\dfrac{B+C}2
\begin{pmatrix}\sin\dfrac{2A+B+C}2\\ \cos\dfrac{2A+B+C}2\end{pmatrix}$$ $$\color{green}{\mathbf{=4\cos\dfrac{A+B}2\,\cos\dfrac{B+C}2\,\cos\dfrac{C+A}2.}}$$
At the same time,
$$\cos x \cos y = \dfrac12(\cos(x-y)+\cos(x+y)),$$
and the approach
$$4\cos\dfrac{A+B}2\cos\dfrac{B+C}2\cos\dfrac{C+A}2
=2\left(\cos\dfrac{A-C}2+\cos\dfrac{A+2B+C}2\right)\cos\dfrac{C+A}2$$
$$=\cos C+\cos A+\cos B+\cos(A+B+C)$$
looks the best.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4227398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 3,
"answer_id": 1
} |
Integration by Substitution with $~ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) ~$ The angle was halved I've been encountering the problem of the below integration.
$$ \int_{0 }^{2\pi } \frac{ d\theta }{ R+ r \cdot \cos^{}\left(\theta_{} \right) } \tag{1} $$
The official description states that the above integration formula can be calculated using substitution of integration.
$$ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \therefore ~~ \int_{0 }^{2\pi } \frac{ d\theta }{ R+ r \cdot \cos^{}\left(\theta_{} \right) } = \frac{ 2\pi }{ \sqrt{ R ^{2} -r ^{2} } } $$
Currently I can't derive the above RHS.
I think firstly find out the form of result of calculations of indefinite integral of eqn1 is wiser way.
What I tried so far are as below.
$$ t= \tan^{}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \frac{ \theta_{} }{ 2 }= \tan^{-1} \left( t \right) ~~ \leftarrow~~ \text{Thought that this approach won't work} $$
$$ \frac{ dt }{ d\theta } = \sec^{2}\left( \frac{ \theta_{} }{ 2 } \right) \cdot \frac{1}{2} $$
$$ \frac{ 2dt }{ d\theta } = \sec^{2}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \frac{ d\theta }{ 2 dt } =\sec^{-2}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ \frac{ d\theta }{ 2 dt } = \left( \sec^{}\left( \frac{ \theta_{} }{ 2 } \right) \right)^{-2} $$
$$ \frac{ d\theta }{ 2 dt } = \left( \cos^{-1}\left( \frac{ \theta_{} }{ 2 } \right) \right)^{-2} $$
$$ \frac{ d\theta }{ 2 dt } = \cos^{2}\left( \frac{ \theta_{} }{ 2 } \right) $$
$$ d\theta = 2 dt \cdot \cos^{2}\left(\frac{ \theta_{} }{ 2 } \right) $$
First things to first, the equation1 has $~ \cos^{}\left(\theta_{} \right) ~$ however how can I handle $~ t=\tan^{}\left( \frac{ \theta_{} }{ 2 } \right) ~$ ??
| Try to solve the problem using the following two equations:
\begin{equation}
\cos(2x)= 2\cos^2(x)-1, \\
\cos^2(x)= \frac{1}{1+ \tan^2(x)}.
\end{equation}
It would be better if you would work on the problem by yourself now, however, I'll write a detailed answer below this line. If you don't want to read it then don't scroll further.
We want to calculate
$$
\int _0^{2\pi}\frac{d\theta}{R+ r\cos\theta}.
$$
Set
$$
t= \tan\left(\frac\theta2 \right),
$$
then $\theta= 2\arctan(t)$, and hence
$$
d \theta= \frac{2}{1+t^2}d t.
$$
Moreover,
$$
\cos\theta= \cos(2\arctan t)= 2\cos^2(\arctan t)-1 = \frac{2}{1+ \tan^2(\arctan t)}-1= \frac{2}{1+t^2}-1= \frac{1-t^2}{1+t^2}.
$$
Putting these together we have
\begin{equation}
\int _0^{2\pi}\frac{d\theta}{R+ r\cos\theta}= 2\int _0^{\pi}\frac{d\theta}{R+ r\cos\theta}= 2\int_0^{+\infty} \frac{\frac{2}{1+t^2}}{R+ r\frac{1-t^2}{1+t^2}} dt= 2\int_0^{\infty} \frac{2}{R(1+t^2)+r (1-t^2)}dt= 4\int_0^\infty\frac{dt}{(R+r)+ t^2(R-r)}= \frac{4}{R+r}\int_0^\infty\frac{dt}{1+ \frac{R-r}{R+r}t^2}= \frac{4}{\sqrt{R+r}\sqrt{R-r}}\int_0^\infty \frac{du}{1+u^2}= \frac{4}{\sqrt{R^2-r^2}}[\arctan u]\Big|_0^{+\infty}= \frac{2\pi}{\sqrt{R^2-r^2}},
\end{equation}
where we used the substitution $u= \frac{\sqrt{R-r}}{\sqrt{R+r}}t$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Calculate the volume between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$. Calculate the volume between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$.
Attempt
We project on the $xy$ plane the intersection between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$, which is the circle $x^2+y^2=1, z=1$.
We can conclude that the region between $z=\sqrt{x^2+y^2}$ and $z=x^2+y^2$ can be described by
$$-1\leq x\leq 1, -\sqrt{1-x^2}\leq y \leq \sqrt{1-x^2}, x^2+y^2\leq z \leq \sqrt{x^2+y^2}$$
The volume is given by
$$V=\int \int \int_W dxdydz=\int_{-1}^{1} \int_{-\sqrt{1-x^2}}^{\sqrt{1-x^2}} \int_{x^2+y^2}^{\sqrt{x^2+y^2}} dxdydz$$
When I try to solve this, I get a difficult expression and cannot calculate it. So I think, everything I have done is wrong.
| The surfaces intersect at the origin and at the point such that $z=1$, so the wanted volume is the difference between the volume of a paraboloid and the volume of a cone, which equals $\frac{\pi}{3}$. The volume of the paraboloid equals (by integrating the areas of the $z$-sections)
$$ \int_{0}^{1}\pi z\,dz = \frac{\pi}{2} $$
so the outcome is $\color{red}{\frac{\pi}{6}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4231184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
$(1+x+x^2)^n=P_0+P_{1}x + P_{2}x^2+ \cdots +P_{2n}x^{2n}$ Prove that,$ P_0+P_{3}+P_{6}+ \cdots =3^{(n-1)}$ Let's say $$ S_n = (1+x+x^2)^n $$
n=1 $$S_1=1+x+x^2$$
n=2 $$S_2=1+2x+3x^2+2x^3+x^4$$
n=3 $$S_3=1+3x+6x^2+7x^3+6x^4+3x^5+x^6$$
n=4 $$S_4=1+4x+10x^2+16x^3+19x^4+16x^5+10x^6+4x^7+x^8$$
By taking coefficients of the $S_n$ we can form this type of triangle similar to Pascal's Trinagle
$$\begin{matrix}
&&&&&&&&&1\\
&&&&&&&1&&1&&1\\
&&&&&1&&2&&3&&2&&1\\
&&&&1&&3&&6&&7&&6&&3&&1\\
&&1&&4&&10&&16&&19&&16&10&&4&&1
\end{matrix}$$
| We define $\xi:=e^{\frac{2\pi i}{3}}$ and $P(x)=(1+x+x^2)^n$. Observe, that $$P(1)+P(\xi)+P(\xi^2)=3(P_0+P_3+\ldots).$$
This holds for all real polynomials. Just prove it for monomials. Further we know $P(1)=3^n$ and $P(\xi)=P(\xi^2)=0$, because $\xi$ and $\xi^2$ are roots of $1+x+x^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Are there positive integers $x$, $y$ and prime numbers $p$ so that $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p}$ I have a solution for this, but I'm not really sure about that:
We have: $\sqrt[3]{x}+\sqrt[3]{y}=\sqrt[3]{p}$, multiplying both sides by $\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2}$ we get $x+y=\sqrt[3]{p}(\sqrt[3]{x^2}-\sqrt[3]{xy}+\sqrt[3]{y^2})$ so $\sqrt[3]{px^2}-\sqrt[3]{pxy}+\sqrt[3]{py^2}$ is an integer.
Suppose $\sqrt[3]{px^2}$, $\sqrt[3]{pxy}$, and $\sqrt[3]{py^2}$ are all integers, we can easily see that $p|x$ and $p|y$.
Let $x^2=p^2a^3$, $y^2=p^2b^3$ where $a$ and $b$ are positive integers. Then we can see that $a$ and $b$ also have to be perfect squares since $x=p\sqrt{a^3}$ and $y=p\sqrt{b^3}$ are integers. Since $\gcd(2,3)=1$, we can let $x^2=p^2a'^6$, $y^2=p^2b'^6$, or $x=pa'^3$, $y=pb'^3$ where $a'$ and $b'$ are integers. Subbing that to the original equation we get $\sqrt[3]{pa'^3}+\sqrt[3]{pb'^3}=\sqrt[3]{p}$, so $\sqrt[3]{p}(a'+b')=\sqrt[3]{p}$. But $a'+b'>1$, so $\sqrt[3]{p}(a'+b')>\sqrt[3]{p}$, hence a contradiction.
I can see I made a lot of assumptions here, like "Let $x^2=p^2a^3$, $y^2=p^2b^3$", or assuming all $\sqrt[3]{px^2}$, $\sqrt[3]{pxy}$, and $\sqrt[3]{py^2}$ are integers. Are there any problems with my work or is it good to go? And moreover, do you have a better solution than this? I appreciate your time and effort for this and thanks a lot in advance.
| The fundamental problem is on your “Suppose that…” line. You need to prove that, and it can be a bit of work. The rest of the proof is fine.
A similar more direct argument just cubes both sides to get:
$$x+3\sqrt[3]{x^2y}+ 3\sqrt[3]{xy^2}+y=p$$
Letting $u= 3\sqrt[3]{x^2y},v= 3\sqrt[3]{xy^2}$ then $u+v=p-x-y$ and $uv=9xy.$
This means $u$ is a root of the two integer polynomials:
$$f(w)=w^3-27x^2y\\g(w)=w^2-(p-x-y)w+9xy$$
If $u$ is not an integer, then $g$ must be the minimal polynomial for $u,$ so $g$ must divide $f.$ The only possible factorization is:
$$f(w)=(w-3x)g(w).$$
Then $$0=f(3x)=27x^3-27x^2y=27x^2(x-y),$$ and we get $x=0$ or $x=y.$ But $x>0$ and, if $x=y,$ we’d have $$ \sqrt[3]x+\sqrt[3]y=\sqrt[3]{8x}=\sqrt[3]p$$ or $p=8x,$ contradicting that $p$ is prime.
So $u$ must be an integer, and the same for $v.$
So $x^2y$ and $xy^2$ must be perfect cubes. Then $$\frac xy=\frac{x^2y}{xy^2}=\left(\frac ab\right)^3$$ for positive integers $a,b.$ This means $a^3x=b^3y.$
Then $$\begin{align} a\sqrt[3]p&=a(\sqrt[3]x+\sqrt[3]{y})\\
&=(a+b)\sqrt[3]y\end{align}$$
Cubing both sides:
$$a^3p=(a+b)^3y.\tag1$$
So $p$ must divide $y.$ But then $y\geq p.$
Similarly $x\geq p.$
But then $$\sqrt[3]x+\sqrt[3]y>\sqrt[3]p,$$ a contradiction.
The general question of integer roots to:
$$\sqrt[3]x+\sqrt[3]y=\sqrt[3]{z}$$ works the same, but you have to allow $x=0,y=z$ and $x=y,$ and $z=8x.$
When you get to (1), you have to write $z=z_0z_1^3,$ where $z_0$ has non nontrivial cube divisors. Then for (1) you get:
$$(az_1)^3z_0=(a+b)^3y.$$ This means $y=z_0y_1^3$ and then, similarly, $x=z_0x_1^3,$ and you get $x_1+y_1=z_1,$ so all solutions can be written:
$$(x,y,z)=\left(z_0x_1^3,z_0y_1^3,z_0(x_1+y_1)^3\right)$$
For all positive cube-free $z_0$ and integers $x_1,y_1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4232735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Expressing the $n$–th derivative of $y=\frac{1}{(1+x^2)^2}$ In an attempt to find a way to express the n-th derivative of $y=\frac{1}{(1+x^2)^2}$ using the binomial theorem I got stuck in the last computation.
I'll explain what I did:
$$f(x)=\frac{1}{(1+x^2)^2}=\frac{1}{(x+i)^2(x-i)^2}=\\
\frac{i}{4}\left( \frac{1}{x+i} \right)-\frac{1}{4}\left( \frac{1}{(x+i)^2} \right)-\frac{i}{4}\left( \frac{1}{x-i} \right)-\frac{1}{4}\left( \frac{1}{(x-i)^2} \right)$$
Then I used $y=\frac{1}{ax+b} \Rightarrow y_{n}=\frac{(-1)^nn!a^n}{(ax+b)^{n+1}}$ and $y=\frac{1}{(ax+b)^2} \Rightarrow y_{n}=\frac{(-1)^n(n+1)!a^n}{(ax+b)^{n+2}}$ to get:
$$\frac{(-1)^n}{4}\left( in!\left( \frac{(x-i)^{n+1}-(x+i)^{n+1}}{(x^2+1)^{n+1}} \right)-(n+1)!\left( \frac{(x-i)^{n+2}+(x+i)^{n+2}}{(x^2+1)^{n+2}} \right) \right)$$
Then I stopped for a second and I analized $(x-i)^{n+1}-(x+i)^{n+1}$ and $(x-i)^{n+2}+(x+i)^{n+2}$ using the binomial theorem:
-
$(x-i)^{n+1}-(x+i)^{n+1}$ = $$\sum_{k=0}^{n+1}\binom{n+1}{k}x^{n+1-k}\cdot i^k((-1)^k-1)$$
$i^k((-1)^k-1) =0$ for $k=2n$ ,whereas for $k=2n+1$ $\longrightarrow$$-2i(-1)^{\frac{k-1}{2}}$
-
$(x-i)^{n+2}+(x+i)^{n+2}$ = $$\sum_{k=0}^{n+2}\binom{n+2}{k}x^{n+2-k}\cdot i^k((-1)^k+1)$$
$i^k((-1)^k+1) =0$ for $k=2n+1$ , whereas for $k=2n$ $\longrightarrow$$2(-1)^{\frac{k}{2}}$.
And now I got stuck sorting out the indeces $n,k$ of summations; what I obtain is:
$$\frac{d^n}{dx^n}(f(x))=\\
\frac{(-1)^nn!}{2(x^2+1)^{n+1}}\cdot \sum_{k=1}^{n+1}\binom{n+1}{k}x^{n+1-k}(-1)^{\frac{k-1}{2}} - \frac{(-1)^n(n+1)!}{2(x^2+1)^{n+2}}\cdot \sum_{k=2}^{n+2}\binom{n+2}{k}x^{n+2-k}(-1)^{\frac{k}{2}}$$
But testing it I can see it’s wrong.
Can anyone help me please sorting the $n,k$ indices and perhaps have a look at my computations also?
Thank you
|
Update (28/08/2021): I eventually managed to solve my problem.
This is the (tested) result obtained:
$$\frac{d^n}{dx^n}\left(\frac{1}{(1+x^2)^2} \right)=$$
$$ \fbox{$\begin{array}{rcl}\displaystyle \frac{(-1)^{n+1}n!}{2(1+x^2)^{n+2}}\left\{ (n+1)\sum_{k=0/even}^{n+2}\binom{n+2}{k}x^{n+2-k}(-1)^{\frac{k}{2}}-(1+x^2)\sum_{k=0/odd}^{n+1}\binom{n+1}{k}x^{n+1-k}(-1)^{\frac{k-1}{2}} \right\}\end{array}$} $$
Thanks a lot to everyone who invested time and effort into answering me.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4233047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to find the limit of $\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$ How to find the limit of the following function?
$$\lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}$$
What I tried is as follows.
$$\tan^{-1} x = y \implies \tan y = x $$
$$\frac{x \tan^{-1} x}{1-\cos x}=\frac{\tan y \cdot y}{1-\cos(\tan y) }$$
But it didn't work.
Please Help me.
| Without using L'Hopital's rule or Taylor series:
$\begin{align} \lim_{x \to 0} \frac{x \tan^{-1} x}{1-\cos x}&=\lim_{x \to 0}\frac{ \frac{\tan^{-1} x}{x}}{\frac{1-\cos x}{x^2}} \; \text{(Divided Numerator and Denominator by $x^2$)}\\
&=\frac{\lim\limits_{x \to 0} \frac{\tan^{-1} x}{x}}{\lim\limits_{x \to 0}\frac{1-\cos x}{x^2}}\\
&=\frac{1}{1 \over 2}=2 \end{align}$
$\begin{align}\because \lim_{x \to 0} \frac{\tan^{-1} x}{x}&=\lim_{y \to 0} \frac{y}{\tan y} \;\;(\text{substituted}\; y=\tan^{-1} x, \;y\to 0\text{ as}\; x \to 0)\\
&=\frac1{\lim\limits_{y \to 0} \frac{\tan y}{y}}\\
&= \frac{1}{1}=1 \;\quad \left(\because \lim\limits_{y \to 0} \frac{\tan y}{y} =1\right)\end{align} $
And,
$\begin{align}\lim\limits_{x \to 0}\frac{1-\cos x}{x^2}&=\lim\limits_{x \to 0}\frac {2\sin^2\left(\frac{x}{2}\right)}{4\left(\frac{x}{2}\right)^2}\\&={1\over 2}\left(\lim\limits_{\frac{x}{2}\to 0}\frac{\sin \frac{x}{2}}{\frac{x}{2}}\right)^2\\&={1\over 2}(1)^2 \quad \; \left(\because \lim\limits_{y\to 0}\frac{\sin y}{y}=1\right)\\&={1\over 2}\end{align}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4234781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 0
} |
Simplify $\sum_{m=1}^{6} \frac{1}{(\sin k)(\sin k - \cos k)}$ where $k = \theta + \frac{m \pi}{4}$ Good Day, I was trying to solve the below problem:
Simplify $$\sum_{m=1}^{6} \frac{1}{(\sin k)(\sin k - \cos k)}$$ where $k = \theta + \frac{m \pi}{4} \text{and } 0 < \theta < \frac{\pi}{2}.
$
I was thinking of decomposing the fraction somehow and getting a telescoping sum or something, but was unable to do so. I am absolutely clueless and there is nothing that I know that simplifies or progresses on the problem. The only thing that I think works is $$\frac{1}{(\sin k)(\sin k - \cos k)} = \frac{1}{\sin^2k-\sin k \cos k} = \frac{2}{1 - \cos 2k-\sin2k}$$ but again I've no idea how to proceed.
Any help would be appreciated. Thanks.
| $\displaystyle\sum_{m=1}^{6} \frac{1}{(\sin k)(\sin k - \cos k)}$
$\displaystyle\implies\frac{1}{\sqrt2}\sum_{m=1}^{6} \frac{1}{(\sin k)(\sin (k -\frac{\pi}{4})}$
$\displaystyle\implies\sum_{m=1}^{6} \frac{\sin\frac{\pi}{4}}{\sin k\sin (k -\frac{\pi}{4})}$
$\displaystyle\implies\sum_{m=1}^{6} \frac{\sin\bigg(k-(k-\frac{\pi}{4})\bigg)}{(\sin k)(\sin (k -\frac{\pi}{4})}$
$\displaystyle\implies\sum_{m=1}^{6} \frac{\sin\bigg(A-B\bigg)}{(\sin A)(\sin B)}$
$\displaystyle\implies\displaystyle\sum_{m=1}^{6}(\cot B-\cot A)$
$\displaystyle\implies\sum_{m=1}^{6}(\cot(k -\frac{\pi}{4}) -\cot k)$
$\displaystyle\implies\sum_{m=1}^{6}\bigg(\cot(\theta+(m-1)\frac{\pi}{4}) -\cot (\theta +m\frac{\pi}{4})\bigg)$
$\displaystyle\implies\sum_{m=1}^{6}g(m-1)-g(m)$ where $g(m)=\cot (\theta+\frac{m\pi}{4})$
Now it can be telescoped to $g(0)-g(6)$
| {
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Quartic polynomials having no real roots
Let $$ f(n,x) = 1 + \dfrac{x}{2^n} + \dfrac{x^2}{3^n} + \dfrac{x^3}{4^n} + \dfrac{x^4}{5^n}$$ The value(s) of the positive integer $n$ such that $f(n, x) = 0$ has no real roots is/are
*
*A) $2021$
*B) $2022$
*C) $4$
*D) $7$
My attempt
I tried to compute the derivatives of $f(n, x)$ with respect to $x$ and found that the second derivative was strictly increasing for such values of $n$ given in the options. That gives me $f'(n,x)$ has only one real root. I am unable to observe anything about $f(n, x)$. Any constructive hint is appreciated.
| It is clear that the quartic polynomial $ \ 1 + \dfrac{x}{2^n} + \dfrac{x^2}{3^n} + \dfrac{x^3}{4^n} + \dfrac{x^4}{5^n} \ \ $ has no positive real zeroes by the "Rule of Signs" (or even just "by inspection") and that any possible negative real zeroes would be due to the terms with odd powers of $ \ x \ $ .
However, we can look at the behavior of pairs of terms to show that the function $ \ f(n,x) \ $ is positive for all real $ \ x \ \ . $ For the second and third terms in ascending order,
$$ \frac{x^2}{3^n} \ + \ \frac{x}{2^n} \ \ = \ \ \frac{1}{3^n} · \left( \ x^2 \ + \ \frac{3^n}{2^n}·x \ \right) \ \ = \ \ \frac{1}{3^n} · \left( \ x \ + \ \frac12·\frac{3^n}{2^n} \ \right)^2 \ - \ \frac{3^n}{2^{2n+2}} \ \ . $$ The minimal value for this pair of terms as a function of $ \ n \ $ forms a geometric sequence with term ratio $ \ r \ = \ \frac{3}{2^2} \ \ , \ $ so the minimum has an absolute-value less than $ \ \dfrac{3^4}{2^{10}} \ = \ \dfrac{81}{1024} \ \ \approx \ \ 0.0791 \ \ $ for all natural numbers $ \ n \ \ge \ 4 \ \ . $
For the fourth and fifth terms in ascending order, we wish to establish the inequality
$$ \frac{x^4}{5^n} \ + \ \frac{x^3}{4^n} \ \ \ge \ \ \frac{x^4}{5^n} \ - \ \frac{x^2}{3^n} \ \ ; $$ the function on the right side of the inequality is an even-symmetry function and we will consider its behavior for $ \ x \ < \ 0 \ \ . $ From this, we have
$$ \frac{x^3}{4^n} \ + \ \frac{x^2}{3^n} \ \ = \ \ x^2 · \left(\frac{x }{4^n} \ + \ \frac{1}{3^n} \right) \ \ \ge \ \ 0 \ \ , $$
which holds in the interval $ \ -\left(\frac43 \right)^n \ \le \ x \ \le \ 0 \ \ . $ For the even-symmetry "comparison function", we obtain
$$ \frac{x^4}{5^n} \ - \ \frac{x^2}{3^n} \ \ = \ \ \frac{1}{5^n} · \left( \ x^4 \ - \ \frac{5^n}{3^n}·x^2 \ \right) \ \ = \ \ \frac{1}{5^n} · \left( \ x^2 \ - \ \frac12·\frac{5^n}{3^n} \ \right)^2 \ - \ \frac{5^n}{4·3^{2n}} \ \ . $$
(The squared-$x \ $ in the parenthetical factor is the result of the symmetry of the function.) Since $ \ x \ = \ -\frac{1}{\sqrt2}·\left(\sqrt{\frac53} \right)^n \ $ is within the interval $ \ -\left(\frac43 \right)^n \ \le \ x \ \le \ 0 \ \ , $ we are able to say that
$$ \frac{x^4}{5^n} \ + \ \frac{x^3}{4^n} \ \ \ge \ \ - \ \frac{5^n}{4·3^{2n}} \ \ . $$
This minimum as a function of $ \ n \ $ forms a geometric sequence with term ratio $ \ r \ = \ \frac{5}{3^2} \ \ , \ $ so it has an absolute-value less than $ \ \dfrac{5^4}{4·3^8} \ = \ \dfrac{625}{26,244} \ \ \approx \ \ 0.0238 \ \ $ for all natural numbers $ \ n \ \ge \ 4 \ \ . $
Hence, $ \ f(n,x) \ = \ 1 + \dfrac{x}{2^n} + \dfrac{x^2}{3^n} + \dfrac{x^3}{4^n} + \dfrac{x^4}{5^n} \ \ge \ 1 - 0.103 \ \ $ for all real $ \ x \ $ at all natural numbers $ \ n \ \ge \ 4 \ \ ; $ its four zeroes thus occur in two complex-conjugate pairs.
| {
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The area of the blue circle centered at the origin, which is inscribed in the parabola $y = x ^2 − 100$, can be expressed as $\frac{a}{b}\pi$ The area of the blue circle centred at the origin, which is inscribed in the parabola $$y = x^{2}-100$$ can be expressed as $\frac{a}{b}\pi$, where $\frac{a}{b}$
is a
fraction in its lowest terms (i.e. a, b are coprime positive integers). What is
the value of $a + b$?
I did this far. Can anyone please check and let me know if its correct or not?
Here the equation of Parabola is,$$y=x^2−100$$
As, the blue circle is centered at the origin so the circle equation would be,$$x^{2}+y^{2}=r^{2}$$
As, the circle is inscribed in the parabola the radius of the circle is equal to the shortest distancefrom the origin to the any point on the parabola
∴The radius of the circle,$$r=\sqrt{x^{2}+y^{2}}=\sqrt{x^{2}+(x^{2}-100)^2}$$
By differentiating r with respect to x we get,
$$\frac{dr}{dx}=\frac{d}{dx}\sqrt{x^2+(x^2-100)^2}\\\Rightarrow\hspace{.1cm}\frac{dr}{dx}=\frac{2x^3-199x}{\sqrt{x^4-199x^2+10000}}$$
Let,$$\frac{dr}{dx}= 0$$
So,$\hspace{.1cm}2x^3−199x= 0$ or $\sqrt{x^4-199x+10000}= 0$
$⇒x= 0,±\sqrt{\frac{199}{2}}$
$\sqrt{x^4-199x+10000}$ has no real solutions for x.
∴$x= 0,±\sqrt{\frac{199}{2}}$
∴If x=±\sqrt{\frac{199}{2}},
∴$y^2=x^2−100= (±\sqrt{\frac{199}{2}})^2−100\hspace{.5cm}\text{[Implementing x value]}\\=−12$
∴At $x=\pm\sqrt{\frac{199}{2}}$,
radius, $r=\sqrt{x^2+y^2}\\⇒r=\frac{\sqrt{399}}{2}$
∴The area of the circle is= $\displaystyle πr^2=π(\frac{\sqrt{399}}{2})^2=\frac{399\pi}{4}$
The circle area was expressed as $\frac{a}{b}\pi$.
∴$a= 399 b= 4$
∴$(a+b)=399+4=403$
| You can combine your first two equations and get
$ y^2 + y + 100 - r^2 = 0$. Then
$$y = \frac{-1 \pm \sqrt{1 - 400+4r^2} }{2}$$If the circle intersects the parabola at two points - single value of y - then the square root term must be $0$, or $4r^2 = 399$.
| {
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Solving the recurrence $a_{k+2} = \frac {k - 4}{(k+1)(k+2)}a_k$? I am solving the ODE $y'' - xy' + 4y = 0$ via power series, which leads me to the following recurrence:
$$a_{k+2} = \frac {k - 4}{(k+1)(k+2)}a_k \tag{1}$$
where $k \ge 1$ and $a_0, a_1$ are given. I would like to ask a question about one of the steps in my solution ($\color{red}{\text{in red below}}$), and whether there is a better solution.
My Solution:
With $k = 2$ we see that $a_4 = 0$, and as a result $a_n = 0$ for all even $n \ge 4$. So we concentrate on $a_n$ with $n$ odd.
Define $b_k = a_{2k + 1}$ for $k \ge 0$. Letting $k = 2k-1$ in $(1)$,
$$b_k = \frac {2k - 5}{2k(2k+1)}b_{k-1}.$$
Now
$$
\begin{align*}
b_k &= \frac {2k - 5}{2k(2k+1)}b_{k-1} \\
&= \frac {2k - 5}{2k(2k+1)} \cdot \frac {2k - 7}{2(k-1)(2k-1)}b_{k-2}\\
& \ \ \vdots\\
&= \frac {2k - 5}{2k(2k+1)} \cdot \frac {2k - 7}{2(k-1)(2k-1)} \cdots \frac {-3}{2(1)(1)}b_0
\end{align*}.
$$
$\color{red}{\text{(trouble above)}}$
After cancelling, we have
$$b_k = \frac {3}{2^k \cdot k! \cdot (2k+1)(2k-1)(2k-3)}b_0$$
and we are essentially done.
Trouble: I am worried about the part where I used the vertical dots. I always have difficulty with figuring out what the final coefficient is once we get to $b_0$. This is how I currently think about it:
Consider $b_k = \frac {2k - 5}{2k(2k+1)}b_{k-1}$. To get to $b_0$ on the RHS, we must subtract $1$ a total number of $k-1$ times. Every time we subtract a $1$, we subtract a $2$ from the $(2k-5)$ term, so the (numerator of the) final term should be $2k-5 - 2(k-1) = -3$. Similarly for the $(2k)$ and $(2k+1)$ terms in the denominator.
In which way do you figure out the last term? (I think my way is not the best)
| Addressing the red part: when you expand
$$
b_k = \frac {2k - 5}{2k(2k+1)}b_{k-1} \tag 1
$$
the penultimate step is
$$b_k = \frac {2k - 5}{2k(2k+1)} \cdots b_1$$
and then at the last step, we substitute $b_1$ with what we find from $(1)$. So the last factor should be $\frac {-3}{2 \cdot (3)} b_0$ and thus
$$b_k = \frac {2k - 5}{2k(2k+1)} \cdots \frac {-3}{2 \cdot 3}$$
| {
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Confusion in Cauchy Integral Formula I am trying to calculate a simple contour integral in three different ways and am getting three different results.
$$\int_{\vert z \vert = 2} \frac{1}{z^2+1}dz$$
Method $1$:
Write $\frac{1}{z^2 + 1} = \frac{1}{(z+i)(z-i)} = \frac{1/(z+i)}{z-i}$. Since $i \in B_2(0)$, we can apply the Cauchy Integral Formula to get
$$\int_{\vert z \vert = 2} \frac{1}{z^2+1}dz = \int_{\vert z \vert = 2} \frac{1/(z+i)}{(z-i)}dz = 2\pi i f(i)$$
where $f(z) = \frac{1}{z+i}$. Hence, the integral evaluates to $\pi$ since $f(i) = \frac{1}{2i}$.
Method $2$:
This is almost identical to the above. Write $\frac{1}{z^2 + 1} = \frac{1}{(z+i)(z-i)} = \frac{1/(z-i)}{z-(-i)}$. Since $-i \in B_2(0)$, we can apply the Cauchy Integral Formula to get
$$\int_{\vert z \vert = 2} \frac{1}{z^2+1}dz = \int_{\vert z \vert = 2} \frac{1/(z-i)}{(z-(-i))}dz = 2\pi i g(-i)$$
where $g(z) = \frac{1}{z-i}$. Hence, the integral evaluates to $-\pi$ since $g(-i) = -\frac{1}{2i}$.
Method $3$:
Using partial fractions, $\frac{1}{z^2+1} = \frac{1}{(z+i)(z-i)} = -\frac{1}{2i}\left(\frac{1}{z+i} - \frac{1}{z-i}\right)$. So,
$$\int_{\vert z \vert = 2} \frac{1}{z^2+1}dz = -\frac{1}{2i} \left(\int_{\vert z \vert = 2} \frac{1}{z+i}dz - \int_{\vert z \vert = 2}\frac{1}{z-i}dz\right) = 0$$
since both integrals are $2\pi i$.
Which of these three methods is correct and why are the other two wrong? (By symmetry I feel like the third one is correct, though I also think it may be that the first two are correct but somehow represent integrating in opposite directions along the contour.)
| There are two poles inside $|z|=2$, so Cauchy's integral formula leads to
$$ \oint_{|z|=2}\frac{dz}{z^2+1}=2\pi i \sum_{a\in\{-i,i\}}\operatorname*{Res}_{z=a}\left(\frac{1}{z^2+1}\right)=0. $$
A simpler way to provide a quick reply is to notice that $\frac{1}{z^2+1}$ is holomorphic in the annulus $2\leq |z|\leq R$ for any $R>2$, so
$$ \oint_{|z|=2}\frac{dz}{z^2+1} = \oint_{|z|=R}\frac{dz}{z^2+1}. $$
On the other hand for any $z$ such that $|z|=R$ we have $|z^2+1|\geq R^2-1$, so
$$\left|\oint_{|z|=R}\frac{dz}{z^2+1}\right| \leq \frac{2\pi R}{R^2-1} $$
where the RHS converges to zero as $R\to +\infty$. In particular the original integral equals zero.
| {
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What is the formal way find $a$ such that $\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4} = \frac{3}{4}$? So I got this problem
Determine the value of $a$ that satisfies the following limit
$$\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4} = \frac{3}{4}$$
If we substitute $x = 4$, the value of the limit will be $\frac{4a + 4}{0}$ which is undefined. So, the numerator must have the same root as the denominator. With $x - 4 = (\sqrt{x} - 2)(\sqrt{x} + 2)$, and the one which make the limit undefined is $(\sqrt{x} - 2)$. From here, I conclude that the numerator must have ($\sqrt{x} - 2$) as a root. And from here, I can do polynomial division and got that $a = -1$.
The other way is by argue that "if the limit have a value, then make the result indeterminate form: $\frac{0}{0}$. Then, we'll get $4a + 4 = 0$ and $a = -1$.
The problem is, I find my method is not good enough for essay problems.
| By $y=x-4 \to 0$ we have
$$\lim_{x\to4} \frac{ax-\sqrt{x}+6}{x-4}=\lim_{y\to 0} \frac{a(y+4)-\sqrt{y+4}+6}{y}=\lim_{y\to 0} \frac{ay+4a-\sqrt{y+4}+6}{y}$$
and to guarantee that limit exists we need
$$4a-\sqrt{y+4}+6 \to 0 \iff a=-1$$
and for $a=-1$
$$ \frac{-y-\sqrt{y+4}+2}{y}=-1-\frac{\sqrt{y+4}-2}{y} \to -1-\frac14=-\frac 5 4$$
indeed by $f(y)=\sqrt{y+4} \implies f'(y)=\frac1{2\sqrt{y+4}}$ we have that by the definition of derivative
$$\lim_{y\to 0}\frac{\sqrt{y+4}-2}{y}=f'(0)=\frac14$$
| {
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Need help on remainder factor theorem question. Source--(https://brilliant.org/wiki/remainder-factor-theorem/)
Question--In an attempt to discover a formula for the Fibonacci numbers, Alex finds a cubic polynomial $h(x)$ such that $h(1)=1$, $h(2)=1$, $h(3)=2$ and $h(4)=3$. What is the value of $h(5)$?
answer by brilliant.org--
Consider the cubic polynomial $j(x)=h(x)−x+1$. Then $j(1) = 1$, $j(2) = 0, j(3) = 0$ and $j(4)=0$. By the remainder factor theorem, we have
$j(x)=A(x) (x-2)(x-3)(x-4),$
where A(x) is a polynomial. Since j(x) is a cubic, it follows that A(x) has degree 0 and thus is a constant which we denote by A. Substituting x=1, we obtain
$1 = j(1) = A(1-2)(1-3)(1-4)$ implies $A = -1/6$
Thus, $h(x) = j(x)+x-1 = -1/6 (x-2)(x-3)(x-4)+x-1$. Hence,
$h(5)=−1/6(5−2)(5−3)(5−4)+5−1=3$
Note: The closed form of the Fibonacci sequence is an exponential function. This cannot be approximated using a polynomial function for large values of n.
my doubt- how $j(x)=h(x)-x+1$ and what is $-x+1$ here? Please explain this in details . I am new in this topic . Thats why I need help.
| Forget $h(x)$.... just forget it. (for now).
Suppose we ask an entirely different question:
How do we find a cubic polynomial $j(x)$ so $j(1) = 1$ but $2,3,4$ are the three roots of $j(x)$?
Well as $x = 2, 3,4$ are the three roots of $j(x)$ we can use the remainder theorem to find that $j(x) = A(x) (x-2)(x-3)(x-4)$ but as we want $j(x)$ to be a cubic we have $A(x)$ is just a constant we can call $A$. And as $j(1) = A(1-2)(1-3)(1-4) = 1$ we find $A = -\frac 16$ and $j(x) = -\frac 16(x-2)(x-3)(x-4)$.
That gives use a cubic polynomial $j(x)$ where $j(1) =1; j(2)=0;j(3)=0; j(4) = 0$.
But we didn't want that!
We wanted a cubic polynomial $h(x)$ where $h(x)=1; h(2)=1;h(3)=2; h(4)=3$.
How do we find that?
Well notice that $D(x) = h(x) - j(x)$ is a polynomial so that
$D(1) = h(1) - j(1) =1-1 = 0; D(2)=h(2)-j(2)=1-0=1;D(3)=h(3) - j(3)= 2-0=2$ and $D(4) = h(4)-j(4) = 3-0 =3$.
So what polynomial has
$\begin{cases}D(1)= 0\\D(2)=1\\D(3)=2\\D(4)=3\end{cases}$
?
It's easy to see that is just a linear progression and is the polynomial $D(x) = x-1$.
So $h(x) - j(x) = D(x) = x-1$ so
$j(x) = h(x) -x+1$. And $h(x) = j(x)+x -1$.
And we had $j(x) = -\frac 16(x-2)(x-3)(x-4)$ and we have $h(x) = j(x) + x -1$ so
$h(x) = -\frac 16(x-2)(x-3)(x-4) + x-1$.
And that's that. That is the cubic polynomial where $h(1)=1; h(2)=1; h(3)=2; h(4)=4$.
And $h(5)=-\frac 16(5-2)(5-3)(5-4) + 5-1 = -\frac 16\cdot 3\cdot 2\cdot 1 + 4=-1 + 4 =3$
=====
In general if we want to find a polynomial where $h(w_1)=c_1; h(w_2)=c_2; ..... h(w_k)=c_k$ a strategy is to use the remainder theorem to find a polynomial $j_1(x)$ where $j_1(w_1) = c_1$ but $j_1(w_i) = 0$ for all other $w_i$.
The we can reduce the question to finding a polynomial $D_1(x) = h(x) - j_1(x)$ where $D_1(w_1) = 0$ and $D_1(w_i) = c_i$ for all the other $w_i$. That might not seem any easier, but notice we did reduce the number of non-zero values by $1$ and if we reiterate we will eventually find a $D_k(x)$ where $D_k(w_{1...,k-1}) = 0$ and $D_k(w_k) = c_k$ and we can find that with the remainder theorem and the putting it all together we will have found $h(x)$.
| {
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"question_score": "2",
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Combinatorics using averages
How many solutions exist for $$x+2y+4z=100$$ in non-negative integers?
The author, Martin Erickson, in his book, Aha! Solutions, published by MAA, gives the following brief solution :
There are $26$ choices for $z$, namely, all integers from $0$ to $25$. Among these choices, the average value of $4z$ is $50$. So, on average, $x+2y=50$. In this equation, there are $26$ choices for $y$, namely, all integers from $0$ to $25$. The value of $x$ is determined by the value of $y$. Hence, altogether there are $26^2=676$ solutions to the original equation.
Can somebody explain to me why this mindblowing solution works using averages?
| There are $26$ choices for $z$ in $\{0,1,\ldots,25\}$.
Given $z$, we have $x+2y=100-4z$, which leaves $\frac{1}{2}(100-4z)+1=51-2z$ choices for $y$ in $\{0,1,\ldots,\frac{1}{2}(100-4z)\}$. Having fixed $y$, we must have $x=100-4z-2y$.
The total number of solutions is therefore
$$\sum_{z=0}^{25} (51-2z) = 1 + 3 + \cdots + 49 + 51.$$
Note that the average value of the addends here is $26$. Because the addends form an arithmetic sequence, you can replace the addends with $26$ via
$$1 + 3 + \cdots + 49 + 51 = 26 + 26 + \cdots + 26 + 26 = 26 \cdot 26.$$
This can be seen by noting that $1+51=26+26$ and $3 + 49 = 26 + 26$, and so on.
Although the solution in the book is correct, it omits justification of several steps.
| {
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Checking for orthogonality of $\cos(\pi x)$ on the set of functions ${\sin(n\pi x)}$ in the $[0,1]$ interval I'm trying to verify the orthogonality of the function $\cos(\pi x)$ to the set of functions $\{\sin(n\pi x)\}, n=1,2,3...$,on the $[0,1]$ interval, for which I have taken the integral obtaining the following result:
$$\int_0^1 \cos(\pi x)\sin(n\pi x) dx= \begin{cases}
0 & \text{if $n \gt 1$, and if n odd} \\ \frac{n(1+(-1)^n)}{\pi(n^2-1)} & \text{if $n \gt 1$, and if n even} \end{cases}$$
But the source I'm reading states (with no proving) that the result is:
$$\int_0^1 \cos(\pi x)\sin(n\pi x) dx=
0 \text{ if $n \neq 1$} $$
Is this the case? I've verified multiple times my calculations by hand and software and still my result differs from the one in the book.
More info:
The original part of the problem was the following claim about the fourier coefficients of the function $\cos(\pi x)$ respect the set of eigenfunctions $\varphi_n(x)=\sin(n\pi x)+n\pi\cos(n\pi x)$ with $n=1,2,3...$ in the interval $[0,1]$
$$a_n=\int_0^1 \cos(\pi x)[\sin(n\pi x)+n\pi\cos(n\pi x)]= \begin{cases}
0 & \text{if $n \neq 1$} \\ \frac{\pi}{2} & \text{if $n = 1$} \end{cases}$$
| $$\cos (\pi x) \sin (n \pi x) = \frac{1}{2} (\sin (\pi x + n \pi x) - \sin (\pi x - n \pi x)) = \frac{1}{2}(\sin (\pi (n + 1) x) + \sin (\pi (n - 1)x)$$
We consider only the case where $n \neq 1$.
Taking the integral from $0$ to $1$ gives $\frac{1}{2\pi} (\frac{1}{n + 1} (1 - \cos((n + 1) \pi)) + \frac{1}{n - 1} (1 - \cos((n - 1) \pi)))$.
When $n$ is odd, we have $\cos((n + 1) \pi) = \cos((n - 1) \pi) = 1$ and thus the integral is $0$.
When $n$ is even, we have $\cos((n + 1) \pi) = \cos((n - 1) \pi) = -1$, and thus the integral is $\frac{1}{\pi} (\frac{1}{n + 1} + \frac{1}{n - 1}) = \frac{2n}{\pi(n^2 - 1)}$.
Finally, in the case that $n = 1$, we see that we're integrating only $\frac{1}{2} \sin(2 \pi x)$ from $0$ to $1$. This integral works out to be 0.
So the integral is always equal to $0$ when $n$ is odd, and $\frac{2n}{\pi (n^2 - 1)}$ when $n$ is even.
Thus, both you and the book are wrong in the case $n = 1$. You are correct on all other cases.
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"language": "en",
"url": "https://math.stackexchange.com/questions/4251412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Question about exact ODE. Why in $h'(y)$ contain $x$? Find the solution ODE
$$\left(x^3e^xy+4x^2e^xy+2xe^xy\right)dx+(x^3e^x+x^2e^x)dy=0.$$
Let $M(x,y)=x^3e^xy+4x^2e^xy+2y$ and $N(x,y)=x^3e^x+x^2e^x$.
\begin{align*}
\dfrac{\partial M}{\partial y}&=\dfrac{\partial}{\partial y}\left(x^3e^xy+4x^2e^xy+2xe^xy\right) \\
&=x^3e^x+4x^2e^x+2xe^x.\\
\dfrac{\partial N}{\partial x}&=\dfrac{\partial}{\partial x}\left(x^3e^x+x^2e^x\right) \\
&=3x^2e^x+x^3e^x+2xe^x+x^2e^x\\
&=x^3e^x+4x^2e^x+2xe^x.
\end{align*}
This is exact ODE since $\dfrac{\partial M}{\partial y}= \dfrac{\partial N}{\partial x}$.
Now,
\begin{alignat}{2}
&&\dfrac{\partial F(x,y)}{\partial x}&=M(x,y)\nonumber\\
\Longleftrightarrow\quad
&&\dfrac{\partial F(x,y)}{\partial x}&=x^3e^xy+4x^2e^xy+2y\nonumber\\
\Longleftrightarrow\quad
&&\int\partial F(x,y)&=\int\left(x^3e^xy+4x^2e^xy+2y\right) \partial x\nonumber\\
\Longleftrightarrow\quad
&&\int\partial F(x,y)&=\int x^3e^xy\partial x+\int4x^2e^xy \partial x+\int2y \partial x.\label{ijoet}
\end{alignat}
Consider that
\begin{align}
\int x^3e^xy\partial x&=x^3e^xy-\int e^xy 3x^2 \partial x\nonumber\\
&=x^3e^xy-\int 3x^2 e^x y \partial x\nonumber\\
&=x^3e^xy-\left(3x^2e^x y-\int e^x y 6x \partial x\right)\nonumber\\
&=x^3e^xy-3x^2e^x y+\int 6x e^x y\partial x\nonumber\\
&=x^3e^xy-3x^2e^x y+6x e^x y-\int e^x y 6\partial x\nonumber\\
&=x^3e^xy-3x^2e^x y+6x e^x y- 6 e^x y+h_1(y).\label{meong}
\end{align}
\begin{align}
\int 4x^2e^xy\partial x&=4x^2e^xy-\int e^xy 8x \partial x\nonumber\\
&=4x^2e^xy-\int 8x e^xy \partial x\nonumber\\
&=4x^2e^xy-\left(8x e^xy-\int e^xy 8 \partial x\right)\nonumber\\
&=4x^2e^xy-8x e^xy+\int 8e^xy \partial x\nonumber\\
&=4x^2e^xy-8x e^xy+ 8e^xy +h_2(y).\label{meong1}
\end{align}
\begin{align}
\int2y \partial x&= 2xy+h_3(y).\label{meong2}
\end{align}
So, we have
\begin{alignat}{2}
&&\int\partial F(x,y)&=x^3e^xy-3x^2e^x y+6x e^x y- 6 e^x y+h_1(y)\nonumber\\
&&&\quad +4x^2e^xy-8x e^xy+ 8e^xy +h_2(y)+2xy+h_3(y)\nonumber\\
\Longleftrightarrow\quad
&&F(x,y)&=x^3e^xy+x^2e^x y-2 x e^x y+2e^x y+2xy +h(y)\nonumber
\end{alignat}
which $h(y)=h_1(y)+h_2(y)+h_3(y)$.
Next, consider that
\begin{alignat*}{2}
&&\dfrac{\partial F(x,y)}{\partial y}&=N(x,y)\\
\Longleftrightarrow\quad
&&\dfrac{\partial}{\partial y}\left(x^3e^xy+x^2e^x y-2 x e^x y+2e^x y+2xy +h(y)\right)&=x^3e^x+x^2e^x\\
\Longleftrightarrow\quad
&&x^3e^x+x^2e^x -2 x e^x +2e^x +2x +h'(y)&=x^3e^x+x^2e^x\\
\Longleftrightarrow\quad
&&h'(y)&=2 x e^x -2e^x -2x \\
\end{alignat*}
When I want to find $h(y)$, I have found $h'(y)=2 x e^x -2e^x -2x$.
Why in $h'(y)$ contain $x$? What my mistake?
| You could group the terms of $M$ as you have seen in the $x$-derivative of $N$ as
$$
x^3e^xy+4x^2e^xy+2xe^xy=[x^3e^xy+3x^2e^xy]+[x^2e^xy+2xe^xy]
\\
=(x^3e^xy)_x+(x^2e^xy)_x
$$
This gives the integral of $M$ in $x$-direction directly as
$$
\int M(x,y)dx =x^3e^xy+x^2e^xy+h(y).
$$
For some reason your partial integration resulted in extra terms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4256062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find $\int{\sqrt[5]{\frac{x}{x+1}}\frac{1}{x^3}}\>dx$ How to integrate
$$\int{\sqrt[5]{\frac{x}{x+1}}\>\frac{1}{x^3}}\>dx$$
My work:
$$ \sqrt[5]{\frac{x}{x+1}}=t$$
$$t^5=\frac{x}{x+1}$$
$$5t^4dt=\frac{1}{(x+1)^2}dx$$
But here, I don't know how to get $\frac{1}{x^3}$.
| $t^5=\frac{x}{x+1} \Rightarrow xt^5+t^5=x \Rightarrow x=\frac{t^5}{1-t^5}$
$5t^4 dt = \frac{1}{(x+1)^2} dx \Rightarrow dx=5t^4 (x+1)^2 dt$
$x+1=\frac{t^5}{1-t^5}+1=\frac{1}{1-t^5}$
$\frac{(x+1)^2}{x^3}=\frac{(1-t^5)^{-2}}{t^{15}(1-t^5)^{-3}}=\frac{1-t^5}{t^{15}}$
$\frac{t dx}{x^3}=\frac{5t^4 (x+1)^2}{x^3} dt=\frac{5 (1-t^5)}{t^{11}} dt=(5 t^{-11}-5t^{-6})dt$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to invert these matrix via Gauss method? Im trying to find the inverse matrix based on
$\left[\begin{array}{ccc}
2 & 2 & -1\\
0 & 4 & -1\\
-1 & -2 & 1
\end{array}\right]$
$R_{1}=$ First row
$R_{2}=$ Second row
$R_{3}=$Third row
but at a certain step I can't reduce it further. Which must be the next operation?
$\left[\begin{array}{ccc}
2 & 2 & -1\\
0 & 4 & -1\\
-1 & -2 & 1
\end{array}\mid\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]$ $\rightarrow\frac{R_{1}}{2}$
$\left[\begin{array}{ccc}
1 & 1 & \frac{-1}{2}\\
0 & 4 & -1\\
-1 & -2 & 1
\end{array}\mid\begin{array}{ccc}
\frac{1}{2} & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\right]$ $\rightarrow R_{3}+R_{1}$
$\left[\begin{array}{ccc}
1 & 1 & \frac{-1}{2}\\
0 & 4 & -1\\
0 & -1 & \frac{1}{2}
\end{array}\mid\begin{array}{ccc}
\frac{1}{2} & 0 & 0\\
0 & 1 & 0\\
\frac{1}{2} & 0 & 1
\end{array}\right]$$\rightarrow\frac{R_{2}}{4}$
$\left[\begin{array}{ccc}
1 & 1 & \frac{-1}{2}\\
0 & 1 & \frac{-1}{4}\\
0 & -1 & \frac{1}{2}
\end{array}\mid\begin{array}{ccc}
\frac{1}{2} & 0 & 0\\
0 & \frac{1}{4} & 0\\
\frac{1}{2} & 0 & 1
\end{array}\right]$$\rightarrow R_{1}-R_{2}$
$\left[\begin{array}{ccc}
1 & 0 & \frac{-1}{4}\\
0 & 1 & \frac{-1}{4}\\
0 & -1 & \frac{1}{2}
\end{array}\mid\begin{array}{ccc}
\frac{1}{2} & -\frac{1}{4} & 0\\
0 & \frac{1}{4} & 0\\
\frac{1}{2} & 0 & 1
\end{array}\right]$$\rightarrow R_{3}+R_{2}$
$\left[\begin{array}{ccc}
1 & 0 & \frac{-1}{4}\\
0 & 1 & \frac{-1}{4}\\
0 & 0 & \frac{1}{4}
\end{array}\mid\begin{array}{ccc}
\frac{1}{2} & -\frac{1}{4} & 0\\
0 & \frac{1}{4} & 0\\
\frac{1}{2} & \frac{1}{4} & 1
\end{array}\right]$$\rightarrow4R_{3}$
$\left[\begin{array}{ccc}
1 & 0 & \frac{-1}{4}\\
0 & 1 & \frac{-1}{4}\\
0 & 0 & 1
\end{array}\mid\begin{array}{ccc}
\frac{1}{2} & -\frac{1}{4} & 0\\
0 & \frac{1}{4} & 0\\
2 & 1 & 4
\end{array}\right]$$\rightarrow?$
| Continuing with $R_{1}\rightarrow R_{1}+\frac{1}{4}R_3$ and $R_{2}\rightarrow R_{2}+\frac{1}{4}R_{3}$ we have
$$\left[\begin{array}{ccc}
1 & 0 & 0\\
0 & 1 & 0\\
0 & 0 & 1
\end{array}\mid\begin{array}{ccc}
1 & 0 & 1\\
\frac{1}{2} & \frac{1}{2} & 1\\
2 & 1 & 4
\end{array}\right]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Deriving the derivative of $(1+x^m)^n$ from first principles and the binomial expansion? From the chain rule:
$$\frac{d}{dx} (1+x^m)^n = nmx^{m-1}(1+x^m)^{n-1}$$
And I am trying to prove an this result using only the first two terms of the binomial expansion and first principles.
Here is my attempt:
$$(1+x^m)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \dots$$
Keeping just the first two terms and using differentiation by first principles:
$$
\begin{align}
\frac{d}{dx} (1 + x^m)^n &= \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{f(x + \Delta x) - f(x)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + (x + \Delta x)^m)^n - (1 + x^m)^n}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + x^m(1 + \frac{\Delta x}{x})^m)^n - (1 + nx^m)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + x^m(1 + \frac{\Delta x}{x})^m)^n - (1 + nx^m)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + x^m(1 + m \frac{\Delta x}{x}))^n - (1 + nx^m)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{(1 + nx^m(1 + m \frac{\Delta x}{x})) - (1 + nx^m)}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{nx^m(m \frac{\Delta x}{x}))}{\Delta x} \\
&= \lim_{\Delta x \to 0} \frac{nmx^{m-1}\Delta x}{\Delta x} \\
&= nmx^{m-1} \\
\end{align}
$$
| For $h\to 0$, $$\frac{(1+(x+h)^m)^n -(1+x^m)^n}{h} \\ = \frac{\left(1+x^m(1+\frac hx)^m \right)^n-(1+x^m)^n}{h} \\ \to \frac{\left(1+x^m(1+\frac{mh}{x}) \right)-(1+x^m)^n}{h} \\ = \frac{(1+x^m +mhx^{m-1})^n -(1+x^m)^n}{h} \\ = \frac 1h \sum_{r=1}^{\infty} \binom nr (mhx^{m-1})^r(1+x^m)^{n-r} \\ \to nmx^{m-1} (1+x^m)^{n-1}$$
as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260771",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Extra solutions when solving $\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$ The problem:
$$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$$
My solution:
$$\sin\theta+\cos\theta=\sqrt{2\sin2\theta}$$
$$\sin\theta+\cos\theta=\sqrt{2\times2\sin\theta\cos\theta}$$
$$\sin\theta+\cos\theta=2\sqrt{\sin\theta\cos\theta}$$
$$\sin\theta-2\sqrt{\sin\theta\cos\theta}+\cos\theta=0$$
$$(\sqrt{\sin\theta}-\sqrt{\cos\theta})^2=0$$
$$\sqrt{\sin\theta}=\sqrt{\cos\theta}$$
Either $\cos\theta$ or $\sin\theta$ can be equal to zero. Both can't be equal to zero at the same time. As we can see that $\sqrt{\sin\theta}=\sqrt{\cos\theta}$, neither of $\cos\theta$ and $\sin\theta$ is equal to zero. So, dividing both sides by $\cos\theta$ is valid.
$$\sqrt{\tan\theta}=1$$
$$\tan\theta=1$$
$$\theta=n\pi+\frac{\pi}{4}\tag{1}$$
My question:
*
*We've stumbled upon an interesting solution in (1). Here, $\theta$ satisfies our original equation only when $n$ is even or $0$. Why is that? Isn't $n$ supposed to belong to the set of integers?
PS: This might help you in answering the question.
| We need that $\sin2\theta\ge 0$ that is
$$0+2n\pi\le2\theta \le \pi+2n\pi \quad \iff \quad n\pi\le \theta \le \frac \pi 2+n\pi$$
therefore only solutions in the first or third quadrant are allowed but since in the third quadrant $\sin \theta + \cos \theta <0$, only solutions in the $\color{red}{\text{first quadrant}}$ are allowed that is by $n=2k$
$$\color{red}{2k\pi\le \theta \le \frac \pi 2+2k\pi}$$
Then, since all terms involved in the expression are positive, and we have that for $A,B\ge 0$
$$A=\sqrt B \iff A^2=B$$
we can then proceed by squaring both side to obtain an $\color{magenta}{\text{equivalent equation}}$
$$\sin\theta+\cos\theta=\sqrt{2\sin2\theta} \iff 1+\sin 2\theta =2 \sin 2\theta \iff \color{magenta}{\sin 2\theta =1} $$
$$\iff2\theta=\frac \pi 2+2n\pi \iff \theta=\frac \pi 4+n\pi\iff \color{magenta}{\theta=\frac \pi 4+2k\pi}$$
Edit
As an alternative approach, since all terms involved in the expression are positive, by AM-GM we obtain
$$\sin\theta+\cos\theta \ge 2\sqrt{\sin \theta \cos \theta}=\sqrt{2\sin 2\theta}$$
with equality for $\sin\theta=\cos\theta \implies \theta=\frac \pi 4+2k\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4261477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "22",
"answer_count": 5,
"answer_id": 1
} |
Answer to an integral is wrong I have an integral question, to which I found the same answer at Mathway and Integral but it is labeled wrong by someone:
Question: find $$\int\frac{1}{\sqrt{9-\frac{9x^2}{4}}}dx.$$
Answer: $$\frac{2}{3}\arcsin\left(\frac{x}{2}\right) + C.$$
Is the answer really wrong?
| Your answer is definitely right!
Let $x=2 \sin \theta$, then $d x=2 \cos \theta d \theta$
$$
\begin{aligned}
I: &=\int \frac{1}{\sqrt{9-\frac{9x^{2}}{4}}} d x \\
&=\int \frac{1}{\sqrt{9-9 \sin ^{2} \theta}} \cdot 2 \cos \theta d \theta \\
&=\frac{2}{3} \int 1 d \theta \\
&=\frac{2}{3} \theta+C \\
&=\frac{2}{3} \sin ^{-1}\left(\frac{x}{2}\right)+C
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4268122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Five Porismatic Equations. Here is a really tough problem.
If
$$\boldsymbol{a\cos\alpha\cos\beta+b\sin\alpha\sin\beta+c=0}$$
$$\boldsymbol{a\cos\gamma\cos\delta+b\sin\gamma\sin\delta+c=0}$$
$$\boldsymbol{a\cos\beta\cos\gamma+b\sin\beta\sin\gamma+c=0}$$
$$\boldsymbol{a\cos\delta\cos\epsilon+b\sin\delta\sin\epsilon+c=0}$$
$$\boldsymbol{a\cos\epsilon \cos\alpha+b\sin\epsilon\sin\alpha+c=0}$$
prove that
$$\boldsymbol{\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}=
\left(\frac{1}{b}+\frac{1}{c}\right)
\left(\frac{1}{c}+\frac{1}{a}\right)
\left(\frac{1}{a}+\frac{1}{b}\right)}$$
where all angles are unequal and between $0$ and $2\pi$.
I cannot work out the algebra on this problem.
This is a system of porsimatic equations. Meaning that it only has distinct solutions if some condition on the variables holds.
The method is the following, in the case of a chain of three equations,
$$a\cos\alpha\cos\beta+b\sin\alpha\sin\beta+c=0$$
$$a\cos\beta\cos\gamma+b\sin\beta\sin\gamma+c=0$$
$$a\cos\gamma\cos\alpha+b\sin\gamma\sin\alpha+c=0$$
We can show
$$\tan\frac{1}{2}(\alpha+\beta)=\frac{b}{a}\tan \gamma$$
either by setting an equation in variable $t$ with solutions, $\tan\frac{1}{2}\alpha$ and $\tan\frac{1}{2}\beta$ and using Vietas formulas or more straighforwardly solving for $\sin\gamma$ and $\cos\gamma$ to get $\tan\gamma$.
Similarly
$$\tan\frac{1}{2}(\alpha+\gamma)=\frac{b}{a}\tan \beta$$
$$\tan\frac{1}{2}(\gamma+\beta)=\frac{b}{a}\tan \alpha$$
and using these we get
$$\tan\frac{1}{2}(\alpha-\beta)=\frac{ab\sin(\alpha-\beta)}{a^2\cos\alpha\cos\beta+b^2\sin\alpha\sin\beta}$$ then using the definition of $\tan=\frac{\sin}{\cos}$ and dividing by $\sin\frac{1}{2}(\alpha-\beta)$ we see that
$$a\cos\alpha\cos\beta+b\sin\alpha\sin\beta+c=c-\frac{ab}{a+b}$$
In the case of a chain of five equations we get the fomulas,
$$\tan\frac{1}{2}(\alpha+\gamma)=\frac{b}{a}\tan\beta$$
$$\tan\frac{1}{2}(\beta+\delta)=\frac{b}{a}\tan\gamma$$
$$\tan\frac{1}{2}(\gamma+\epsilon)=\frac{b}{a}\tan\delta$$
$$\tan\frac{1}{2}(\alpha+\delta)=\frac{b}{a}\tan\epsilon$$
$$\tan\frac{1}{2}(\epsilon+\beta)=\frac{b}{a}\tan\alpha$$
But how to proceed from there ?
I guess you want $\tan\frac{1}{2}(\alpha-\epsilon)
$ as a function of $\alpha$ and $\epsilon$.
| This is only a small correction to the question statement, but is still too long for the usual comment format. You probably forgot to include the condition that no two of the angles are opposites. Otherwise, there are several simple enough counterexamples, such as the following one.
For simplicity, I write $\theta_1,\ldots,\theta_5$ instead of $\alpha,\beta,\gamma,\delta,\epsilon$, and I also put $c_k=\cos(\theta_k),s_k=\sin(\theta_k)$. The counterexample has $a=-1,b=c=1$ (so $a^{-3}+b^{-3}+c^{-3}=1 \neq 0 = (a^{-1}+b^{-1})(a^{-1}+c^{-1})(b^{-1}+c^{-1})$)
$$
\begin{array}{|c|c|c|c|c|}
\hline
k & c_k & s_k & c_kc_{k+1} & s_ks_{k+1} \\
\hline
1 & -\frac{35}{37} & -\frac{12}{37} & \frac{1925}{2701} & -\frac{576}{2701} \\
\hline
2 & -\frac{55}{73} & \frac{48}{73} & -\frac{275}{949} & \frac{576}{949} \\
\hline
3 & \frac{5}{13} & \frac{12}{13} & -\frac{3}{13} & -\frac{48}{65} \\
\hline
4 & -\frac{3}{5} & -\frac{4}{5} & \frac{21}{37} & -\frac{48}{185} \\
\hline
5 & -\frac{35}{37} & \frac{12}{37} & \frac{1225}{1369} & -\frac{144}{1369} \\
\hline
\end{array}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4268371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
show this inequality $\sqrt{\frac{a^b}{b}}+\sqrt{\frac{b^a}{a}}\ge 2$ let $a,b>0.$ Show that
$$\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2\tag{1}$$
I known How to prove $a^b+b^a>1$,where $a,b>0.$ See $x^y+y^x>1$ for all $(x, y)\in \mathbb{R_+^2}$
to prove $（1）$, I want use AM-GM inequality
$$\sqrt{\dfrac{a^b}{b}}+\sqrt{\dfrac{b^a}{a}}\ge 2\left(\dfrac{a^b}{b}\cdot\dfrac{b^a}{a}\right)^{1/4}=2\left(a^{b-1}b^{a-1}\right)^{1/4}$$
But $a^{b-1}b^{a-1}$ is not always $>1$
| Using again Generalized Young inequality we have :
$$\left(\frac{1}{a^{b}}+\frac{1}{x^{b}}\right)\left(a^{\sqrt{2^{-1}}x^{\left(1-b\right)}}\cdot x^{\left(\sqrt{2^{-1}}\left(a\right)^{\left(1-b\right)}\right)}\cdot a^{\frac{\left(b-0.5\right)\sqrt{2}}{a^{b}}}\cdot x^{\frac{\left(b-0.5\right)\sqrt{2}}{x^{b}}}\right)^{\frac{1}{\sqrt{2}a^{-b}+\frac{\sqrt{2}}{x^{b}}}}\leq \left(\sqrt{\frac{x^{a}}{a}}+\sqrt{\frac{a^{x}}{x}}\right)$$
where $b\to1$ and $a,x>0$
Final conjecture :
Let $a,x>0$ then it seems we have :
$$\left(\frac{1}{a^{b}}+\frac{1}{x^{b}}\right)\left(a^{\sqrt{2^{-1}}x^{\left(1-b\right)}}\cdot x^{\left(\sqrt{2^{-1}}\left(a\right)^{\left(1-b\right)}\right)}\cdot a^{\frac{\left(b-0.5\right)\sqrt{2}}{a^{b}}}\cdot x^{\frac{\left(b-0.5\right)\sqrt{2}}{x^{b}}}\right)^{\frac{1}{\sqrt{2}a^{-b}+\frac{\sqrt{2}}{x^{b}}}}\geq 2$$
Where $b\to 1$
I think we can settle $b=1$ it works also .
Using a bit of algebra (introducing log and invert the variables)
We need to show :
$x,a > 0$, we need to prove that
$$f\left(x\right)=(x+a)\ln\frac{a+x}{2}-\frac{\left(a+1\right)\ln a+\left(x+1\right)\ln x}{2} \ge 0.$$
A proof of this fact can be found here :
Prove $2(x + y)\ln \frac{x + y}{2} - (x + 1)\ln x - (y + 1)\ln y \ge 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4268913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "23",
"answer_count": 4,
"answer_id": 2
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Simplify the following expression!
Simplify the following expression
$$\sqrt[3]{a+\frac{a+8}{3}\sqrt{\frac{a-1}{3}}}+\sqrt[3]{a-\frac{a+8}{3}\sqrt{\frac{a-1}{3}}}$$
I tried using the form $\displaystyle\frac{a^3+b^3}{a^2-ab+b^2}$, and i also tried to assume the requested expression with $c$ and use the form $a^3+b^3-c^3=-3abc$, if $a+b-c= 0$, but haven't obtained its simple form yet,
i get this problem from a book but i doesn't get the solution
if anyone can complete it it's really amazing
| A first (direct) approach:
Formula :
$$E:=\sqrt[3]{a+\frac{a+8}{3}\sqrt{\frac{a-1}{3}}}+\sqrt[3]{a-\frac{a+8}{3}\sqrt{\frac{a-1}{3}}},\tag{1}$$
contains expression
$$u:=\sqrt{\frac{a-1}{3}}\tag{2}$$
which is only meaningful under condition $a \ge 1$.
We set apart the case $a=1$ where $E=2$.
Let us invert (2) under the form
$$a=3u^2+1\tag{3}$$
Then:
$$\frac{a+8}{3}=\frac{3u^2+9}{3}=u^2+3$$
In this way, (1) becomes:
$$E=\sqrt[3]{(3u^2+1)+(u^2+3)u}+\sqrt[3]{(3u^2+1)-(u^2+3)u}$$
$$E=\sqrt[3]{(u+1)^3}+\sqrt[3]{-(u-1)^3}$$
$$E=u+1-(u-1) = \color{red}{2}$$
which is a constant independent from $a$.
A second approach:
Formula (1) has the same "structure" as Cardano formula, expressing a root of the general reduced cubic equation:
$$x^3+px+q=0\tag{4}$$
under the form
$$\sqrt[3]{-\frac{q}{2}+\sqrt{\frac{D}{4 \times 27}}}+\sqrt[3]{-\frac{q}{2}-\sqrt{\frac{D}{4 \times 27}}} \ \text{where} \ D:=4p^3+27q^2,\tag{5}$$
($D$ is the so-called discriminant of equation (4)).
Identifying (1) and (5), one gets:
$$a=-\frac{q}{2} \tag{6}$$
and:
$$\frac{(a+8)^2(a-1)}{27}=\frac{4p^3+27q^2}{4 \times 27} \tag{7}$$
Plugging (6) into (7), we get:
$$(a+8)^2(a-1)=p^3+27a^2\tag{8}$$
Expanding (8), one gets:
$$p^3=a^3-12a^2+48a-64=(a-4)^3 \ \ \implies \ \ p=a-4\tag{9}$$
Plugging (6) and (9) into (4):
$$x^3+(a-4)x-2a=0 \ \ \iff \ \ (x-2)(x^2+2x+a)=0$$
with roots:
$$\begin{cases}x_1&=&\color{red}{2}\\
x_2&=&-1-\sqrt{1-a^2}\\
x_3&=&-1+\sqrt{1-a^2}\end{cases}$$
We find back the solution $\color{red}{x=2}$.
The two other solutions $x_2, x_3$ are in fact complex in general because $1-a^2<0$ which is not possible because expression $E$ in (1) is assumed to be real.
Remark: Here is a formula with 2 variables with a complicated LHS, similar to (1) and a very simple RHS, free from one of the two variables:
$$\sqrt[3]{n\left(3m-n^2\right) +m\sqrt{8m-3n^2}} + \sqrt[3]{n\left(3m-n^2\right) -m\sqrt{8m-3n^2}} =n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Finding all pairs of integer solutions to ${x^y = (x+y)^2}$ How do I find all solutions for ${x^y = (x+y)^2}$, where $x$ and $y$ are positive integers and at least one of them is prime.
For example, ${64 = 2^6 = (2+6)^2}$
A friend of mine pointed out that $x|y^2$ then $x^2|y^2$, so, when $x$ is prime and $y$ is not prime, we can write $y=xk$. Thus $x^{xk} = x^2 (1+k^2)$ and $x^{xk-2} = (1+k^2)$.
| I'll ignore the "prime" requirement. It just spoils the fun. And I assume positive and negative integers.
x=0 is not possible. $0^y$ is only defined if $y>0$, but then $(0 + y)^2 > 0 = 0^y$.
If y < 0 then $x^y$ is not an integer unless $x = ±1$ and $x^y = ±1$. Since $x^y = (x+y)^2 ≥ 0$ we must have $x^y = +1$ and $x+y = ± 1$. $x = -1$ is not possible because it makes $x + y ≤ -2$ and $(x+y)^2 ≥ 4$. $x=1$ implies $x+y = ±1$, therefore $y=0$ or $y=-2$, so $1^{-2} = (1-2)^2 = 1$ is the only solution with $y<0$.
If $y=0$ then $x^y = 1$, so $(x+0)^2 = 1$ so $x = ±1$. ${±1}^0 = (±1 + 0)^2 = 1$.
If $y=1$ then $x^y = x = (x+1)^2$. We must have $x>0$, and any $x > 0$ is less than $(x+1)^2$, so no solution.
If $y=2$ then $x^y = x^2 = (x+2)^2$, so $x + 2 = ±x$. This is the case only for x = -1, y = 2: ${-1}^2 = (-1 + 2)^2 = 1$.
If $x = ±1$ then $x^y$ must be 1, so $x+y = ±1$. Not possible with $y ≥ 3$.
If $y ≥ 3$ and $|x| ≥ 2$ then $x^y$ grows faster than $(x + y)^2$; the only values where $(x+y)^2 ≥ x^y$ are:
$2^3 = 8 < 25 = (2+3)^2$
$2^4 = 16 < 36 = (2+4)^2$
$2^5 = 32 < 49 = (2+5)^2$
$2^6 = 64 = 64 = (2+6)^2$
$3^3 = 27 < 36 = (3+3)^2$
and the only solution is $2^6 = (2+6)^2 = 64$.
Summary: The only solutions with positive or negative integers are $1^{-2} = (1-2)^2 = 1$, $1^0 = (1+0)^2 = 1$, ${-1}^0 = (-1 + 0)^2 = 1$, ${-1}^2 = (-1 + 2)^2 = 1$, and $2^6 = (2+6)^2 = 64$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$ The Equation
How can I analytically show that there are no real solutions for $\sqrt[3]{x-3}+\sqrt[3]{1-x}=1$?
My attempt
With $u = -x+2$
$\sqrt[3]{u-1}-\sqrt[3]{u+1}=1$
Raising to the power of $3$
$$(u+1)^{2/3}(u-1)^{1/3} - (u+1)^{1/3}(u-1)^{2/3}=1\\(u+1)^{1/3}(u^2-1)^{1/3} - (u-1)^{1/3}(u^2-1)^{1/3}=1\\(u^2-1)^{1/3}\cdot\boxed{\left[(u+1)^{1/3}-(u-1)^{1/3}\right]}=1$$
Raising to the power of $3$:
$$(u^2-1)\cdot\left[3(u+1)^{1/3}(u-1)^{2/3}-3(u+1)^{2/3}(u-1)^{1/3}+2\right]=1\\(u^2-1)\cdot\left[3(u^2-1)^{1/3}(u-1)^{1/3}-3(u+1)^{1/3}(u^2-1)^{1/3}+2\right]=1$$
Thus: $(u^2-1)\cdot\left[3(u^2-1)^{1/3}\boxed{\left[(u-1)^{1/3}-(u+1)^{1/3}\right]}+2\right]=1$
And with $y = (u-1)^{1/3}-(u+1)^{1/3}$, we can say that:
$y^3=-3y(u^2-1)^{1/3}+2$
I am stuck... Any tips for this radical equation?
| If you're looking for real solutions, I think you can look more closely at $\sqrt[3]{u-1} - \sqrt[3]{u+1} = 1$. It seems difficult for the left side to be positive, and indeed we would be done if it is the case that it is non-positive for all real $u$. Indeed, we have that $1 > -1$ and $u+1 > u-1$, hence $\sqrt[3]{u+1} \geq \sqrt[3]{u-1}$ because $\sqrt[3]{x}$ is an increasing function. Now if $u$ were a real solution then $1 = \sqrt[3]{u-1} - \sqrt[3]{u+1} \leq 0$, contradiction.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4272214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Question on finding derivative for a piecewise function with absolute value Let $f(x)=\begin{cases}\frac{1}{2}x^2 ~~~~~&\text{if } |x|\leq c \\ c|x| - \frac{1}{2}c^2 ~~~~~&\text{if } |x|> c \end{cases}$, where $c>0$ is just a constant value in $\mathbb{R}$
Then I find out $f'(x) = \begin{cases}x ~~~~~&\text{if } |x| < c \\ \frac{cx}{|x|} ~~~~~&\text{if } |x| > c \end{cases}$.
Since, $c>0$ is a constant in $\mathbb{R}$.
Then, $\lim_{h\rightarrow c^{-}}\frac{f(h)-f(c)}{h}=\lim_{h\rightarrow c^{-}}\frac{\frac{1}{2}h^2-\frac{1}{2}c^2}{h} = \frac{0}{c} = 0$
Also, $\lim_{h\rightarrow c^{+}}\frac{f(h)-f(c)}{h}=\lim_{h\rightarrow c^{+}}\frac{c|h|-\frac{1}{2}c^2-\frac{1}{2}c^2}{h} = \frac{0}{c} = 0$
So, in this case is $f'(x)=\begin{cases}x ~~~~~&\text{if } |x| < c \\ \frac{cx}{|x|} ~~~~~&\text{if } |x| > c \\ 0 ~~~~~&\text{if } |x| = c \end{cases}$ the derivative of $f(x)$ above ?
| It does not seem quite right to me. The denominator should be $h-c$ in each calculation.
For $x = c > 0$, the left derivative is,
\begin{align}
\lim_{h \to c^-} \frac{f(h)-f(c)}{h-c} = c
\end{align}
and the right derivative,
\begin{align}
\lim_{h \to c^+} \frac{f(h)-f(c)}{h-c} = c
\end{align}
so that the derivative at $c$ exists but is $c$, not $0$. Similar arguments apply when $x = -c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273358",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
unable to solve Lagrange PDE $U_x +xyU_y +(2x^2z\ln|y|)U_z = 0$ Hi I need to solve $$u_x +xyu_y +(2x^2z\ln|y|)u_z = 0$$
Μy try:
I wrote the equations:
$$ dx = \frac{dy}{xy} = \frac{dz}{2x^2z\ln|y|} $$
so the first surface is $\phi_1 = u$
Second surface:
$$ dx\cdot x = \frac{dy}{y} \Rightarrow 0.5x^2 + C_1= \ln|y| \Rightarrow C = y\cdot e^{-0.5x^2}$$
$$ \phi_2 = ye^{\frac{x^2}{2}}$$
Third surface (I think here lies the problem):
$$ 2x^2\ln|y| = \frac{dz}{z} \Rightarrow \ln|y|(\frac{2x^3}{3} + C) = \ln|z| \Rightarrow \\
\Rightarrow y^{\frac{2x^3}{3}}\cdot y^C = z $$
Now I substitue y:
$$ z \cdot y^{\frac{-2x^3}{3}} = Ce^{0.5x^2} \Rightarrow z \cdot y^{\frac{-2x^3}{3}}e^{-0.5x^2} = C $$
So the third surface is:
$$ \phi_3 = zy^{\frac{-2x^3}{3}}e^{-0.5x^2} $$
Which is a mistake...
according to the solution: $\phi_3 = zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}$
because the solution is $$u(x, y) = f(zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}, ye^{\frac{x^2}{2}})$$
I can't figure out where I went wrong, any help/hints would be appreciated!
| $$ dx = \frac{dy}{xy} = \frac{dz}{2x^2z\ln|y|} \quad\text{is OK.}$$
I agree with $ \quad 0.5x^2 + C_1= \ln|y| \Rightarrow C = y\cdot e^{-0.5x^2}$
$$ \phi_2 = ye^{-\frac{x^2}{2}}$$
$\phi_2 = ye^{\frac{x^2}{2}}$ is not correct.
For the third surface, you forgot $dx$ in $ 2x^2\ln|y| = \frac{dz}{z} $ which should be
$$ 2x^2\ln|y|dx = \frac{dz}{z}$$
with $\ln|y|=0.5x^2 + C_1$
$$ 2x^2(0.5x^2 + C_1)dx = \frac{dz}{z}$$
$$ (x^4 + 2C_1x^2)dx = \frac{dz}{z}$$
Introducing another constant $C\neq C_1$ is a cause of confusion.
I suppose that you can take it from here.
Note : There is a typo in the expected solution $u(x, y) = f(zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}, ye^{\frac{x^2}{2}})$ which should be :
$$u(x, y) = f(zy^{\frac{-2x^3}{3}}e^{\frac{2x^5}{15}}, ye^{-\frac{x^2}{2}})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4274655",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $0
If $0<a<b$ and $0<c<d$ then $\frac{c+a}{d+a} <\frac{c+b}{d+b}.$
I get to $$d+a<d+b \Longrightarrow \frac{1}{d+b} < \frac{1}{d+a}$$ but that inequality seems opposite of what I am trying to prove. Any advice is appreciated.
| $$ \dfrac{c+a}{d+a} =\dfrac{d+a+c-d}{d+a} = 1 + \dfrac{c-d}{d+a}$$
$$ \dfrac{c+b}{d+b} =\dfrac{d+b+c-d}{d+b} = 1 + \dfrac{c-d}{d+b}$$
You have already proved that
$$ \frac{1}{d+b} < \frac{1}{d+a}$$
As $c-d<0$,
$$ \frac{c-d}{d+b} > \frac{c-d}{d+a}$$
$$ \implies1+ \frac{c-d}{d+b} > 1+\frac{c-d}{d+a}$$
$$ \implies \dfrac{c+b}{d+b} > \dfrac{c+a}{d+a}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4275780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
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Change of variables for summation fails Suppose I have the sum $ \displaystyle S_1 = \sum_{i=1}^5 \cfrac{1}{(i+1)^2} $
I can make a change of variables by letting $ j=i+1 $. When $ i=1 \implies j=2 $, and when $ i=5 \implies j=6 $
$ \displaystyle \therefore S_1 = \sum_{j=2}^6 \cfrac{1}{j^2} $.
But if I were to make a change of variables with $ j=(i+1)^2 $ and use the same logic as above, the resulting sum would be: $ \displaystyle \sum_{j=4}^{36}\cfrac{1}{j} $. Which is not equal to the original sum $ S_1 $.
In theory, why does this change of variable fail?
| $\displaystyle \sum_{j=4}^{36}\cfrac{1}{j}=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}+....+\frac{1}{36}$ whereas $\displaystyle \sum_{j=2}^{6}\cfrac{1}{j^2}=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+\frac{1}{5^2}+\frac{1}{6^2}=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$. So their sums are not matching. Instead you can write $$\displaystyle \sum_{j=2}^{6}\cfrac{1}{j^2}=\displaystyle\sum_{j\in\{4,9,16,25,36\}}\cfrac{1}{j}$$. $\displaystyle\sum_{j\in\{4,9,16,25,36\}}\cfrac{1}{j}$ means that it is the sum of all values $\frac{1}{j}$ where $j$ belongs to the set {$4,9,16,25,36$}. This implies that $\displaystyle\sum_{j\in\{4,9,16,25,36\}}\cfrac{1}{j}$ becomes $\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+\frac{1}{25}+\frac{1}{36}$ which is the required value.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4276075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to simplify $\sqrt{\tan^2 x + \cot^2x }$? How to simplify :
$$\sqrt{\tan ^2 x + \cot ^2x }$$
the option are :
(i) $ \tan x \cdot \sin x$
(ii) $\sin x \cdot \cos x $
(iii) $ \sec x \cdot \csc x $
(iv) $ \frac{1}{\tan x - \cot x}$
(v) $ \csc^2 x - \sec ^2 x$
My approach :
Since $\tan x = \frac{\sin x}{\cos x}$ and $\cot x = \frac{\cos x}{\sin x}$,
$$
\begin{align}
\sqrt{\tan ^2 x + \cot ^2x } &= \sqrt{\frac{\sin ^2 x}{\cos ^2 x} + \frac{\cos ^2 x}{\sin ^2 x}} \\
&= \sqrt{\frac{\sin ^4 x + \cos ^4 x}{\sin ^2 x \cdot \cos ^2 x} }\\
&= \sqrt{\frac{(\sin ^2 x + \cos ^2 x)^2 - 2 \sin x \cdot \cos x}{\sin ^2 x \cdot \cos ^2 x} }\\
&= \sqrt{\frac{1 - 2\sin x \cdot \cos x}{\sin ^2 x \cdot \cos ^2 x} }\\
&= \sqrt{\sec ^2 x \cdot \csc^2 x - 2 \sec x \cdot \csc x}\\
&= \sqrt{\sec x \cdot \csc x ( \sec x \cdot \csc x - 2)}\\
\end{align}
$$
from this point, I don't have any idea how should I approach this problem to get another form of this equation available on the option.
Another approach I have in mind is from changing $\cot x = \frac{1}{\tan x}$
$$
\begin{align}
\sqrt{\tan ^2 x + \cot ^2x } &= \sqrt{\tan ^2 x + \frac{1}{\tan ^2 x}} \\
&= \sqrt{\frac{\tan ^4 x + 1}{\tan ^2 x} }\\
\end{align}
$$
From this point, I don't have any idea.
What am I missing or what approach should you suggest to change the form to the option available on the option?
| You made a mistake in one of your steps. You should have
$$\sin^4x + \cos^4x = (\sin^2x+\cos^2x)^2-2\sin^2x\cos^2x$$
This then helps you achieve
$$\sqrt{\sec^2x\csc^2x-2}$$
Other than that, none of the choices are correct (as far as I can tell, the above expression can't be simplified any further).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\lim_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}=\quad?\;\;$ (without using Taylor's Theorem) It is clear that this problem could use Taylor's Theorem to expand the numerator as series, which leads to the solution easily.
But I don't think this method is beautiful enough.
Is there any better method to solve this problem? Any answer would be highly appreciated.
| This may not be as beautiful as you were hoping, but here is a solution:
Add and subtract $e^e$ and break the limit into two:
\begin{align*}
\lim_{x\to 0}\frac{e^{(1+x)^{1/x}}-e^e}{x^2}-\frac{(1+x)^{e/x}-e^e}{x^2}&=\lim_{x\to 0}e^e\left[\frac{e^{(1+x)^{1/x}-e}-1}{x^2}-\frac{e^{e/x\ln(1+x)-e}-1}{x^2}\right]
\end{align*}
For the first term, set $\alpha=(1+x)^{1/x}-e$. Note that $\alpha\to 0$ as $x\to 0$. Then we have
\begin{align*}
\lim_{x\to0}\frac{e^{\alpha}-1}{\alpha}\cdot\frac{\alpha}{x^2}=\lim_{x\to0}\frac{\alpha}{x^2}
\end{align*}
For the second term, set $\beta=\frac ex\ln(1+x)-e$. Note that $\beta\to 0$ as $x\to 0$. Then we have
\begin{align*}
\lim_{x\to0}\frac{e^{\beta}-1}{\beta}\cdot\frac{\beta}{x^2}=\lim_{x\to0}\frac{\beta}{x^2}
\end{align*}
Note that I am able to evaluate the first factors of the above limits to $1$, since both terms of the original limit have the "same" factor. Therefore, our original limit becomes
\begin{align*}
\lim_{x\to 0}e^e\left[\frac{\alpha-\beta}{x^2}\right]&=\lim_{x\to 0}e^e\left[\frac{(1+x)^{1/x}-e}{x^2}-\frac{\frac ex\ln(1+x)-e}{x^2}\right]\\
&=\lim_{x\to0}e^e\left[\frac{e^{1/x\ln(1+x)}-e}{x^2}-\frac{\frac ex\ln(1+x)-e}{x^2}\right]\\
&=\lim_{x\to0}e^{e+1}\left[\frac{e^{1/x\ln(1+x)-1}-1}{x^2}-\frac{\frac1x\ln(1+x)-1}{x^2}\right]
\end{align*}
We will only look at the first term. Set $\gamma=\frac1x\ln(1+x)-1$. Then we have
\begin{align*}
\lim_{x\to0}\frac{e^\gamma-1}{\gamma}\cdot\frac{\gamma}{x^2}
\end{align*}
So, our original limit becomes
\begin{align*}
\lim_{x\to0}e^{e+1}\left[\frac{e^\gamma-1}{\gamma}\cdot\frac{\gamma}{x^2}-\frac{\gamma}{x^2}\right]&=\lim_{x\to0}e^{e+1}\cdot\frac{\gamma}{x}\cdot\left(\frac{e^\gamma-1-\gamma}{\gamma x}\right)\\
&=\lim_{x\to0}e^{e+1}\cdot\frac{\ln(1+x)-x}{x^2}\cdot\left(\frac{e^\gamma-1-\gamma}{\gamma x}\right)
\end{align*}
The first factor give us $e^{e+1}$, the middle one can be found using l'Hopitals rule twice, which gives us $-1/2$. The final term evaluates to $-1/4$, but I don't have time to try and find out how at the moment.
Edit: The limit is finished below.
To get the final term, apply l'Hopital's rule twice to get
\begin{align*}
\lim_{x\to 0}\frac{e^\gamma(\gamma'^2+\gamma'')-\gamma''}{\gamma''x+2\gamma'}.
\end{align*}
Now,
\begin{align*}
\gamma'&=\frac{1}{1+x}\cdot\left(\frac{x-\ln(1+x)}{x^2}-\frac{\ln(1+x)}{x}\right)\\
\gamma''&=\frac{1}{(1+x)^2}\cdot\left(\frac{2\ln(1+x)}{x}+\frac{2\ln(1+x)+4x\ln(1+x)-3x^2-2x}{x^3}\right)
\end{align*}
Therefore,
\begin{align*}
\lim_{x\to0}\gamma'&=1\cdot\left(\frac12-1\right)=-\frac12\\
\lim_{x\to0}\gamma''&=1\cdot\left(2-\frac43\right)=\frac23
\end{align*}
Hence,
$$
\lim_{x\to 0}\frac{e^\gamma(\gamma'^2+\gamma'')-\gamma''}{\gamma''x+2\gamma'}=\frac{e^0((-1/2)^2+2/3)-2/3}{2/3(0)+2(-1/2)}=-\frac14.
$$
Putting this all together gives us our final answer:
\begin{align*}
\lim_{x\to0}e^{e+1}\cdot\frac{\ln(1+x)-x}{x^2}\cdot\left(\frac{e^\gamma-1-\gamma}{\gamma x}\right)=\frac{e^{e+1}}8
\end{align*}
| {
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"timestamp": "2023-03-29T00:00:00",
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Proper notation for equation solving. What is the proper notation when solving one equation by inserting another known relationship?
$$y=x^2, y+2x^2=1$$$$y+2x^2=1\Leftrightarrow x^2+2x^2=1 \Leftrightarrow x=\pm\sqrt{\frac{1}{3}}$$
I assume the first $\Leftrightarrow$ is wrong since we are inserting a relationship, and the $y+2x^2=1$ equation is not logically equal to $x^2+2x^2=1$ by it self.
Edit: I guess what I'm asking is: Is there a way of notating $f(x),g(x)\Rightarrow h(x)$ cleanly?
| \begin{align}&&y=x^2 \;&\text{ and }\; y+2x^2=1\\&\iff &y=x^2 \;&\text{ and }\; x^2+2x^2=1 \\&\iff &y=x^2 \;&\text{ and }\; x=\pm\sqrt{\frac{1}{3}}\\&\iff &y=\frac13 \;&\text{ and }\; x=\pm\sqrt{\frac{1}{3}}.\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Extraneous solution when solving $x^2+x+1=0$ by getting $x^2=1/x$ Let's assume that we have$$x^2+x+1=0.\tag1$$
Substituting $x=0$, we get $1=0$, so $0$ is not a root for the quadratic equation and thus, $x\neq0$. Therefore, there exists $\frac{1}{x}$, which we'll multiply by both sides of $(1)$, giving us $$x+1+\frac{1}{x}=0.$$
We will, then, move $\frac{1}{x}$ to the other side and get $$x+1=-\frac{1}{x}.$$ If we add $x^2$ to both sides and note that $x^2+x+1=0$, we will have $x^2-\frac{1}{x}=0$. The real root of this equation is $x=1$, which is not a root of $(1)$.
I was wondering at which step did I do something that was incorrect and resulted in this supposed root.
| The first step when we got an extraneous root is the equation $x^2 - \frac{1}{x} = 0$. It is a corollary of equation(1). But it's only corollary: it's not equivalent to it.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4293874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Calculating $B^{10}$
Calculate $B^{10}$ when $$B = \begin{pmatrix} 1 & -1\\ 1 & 1 \end{pmatrix}$$
The way I did it was
$$ B = I + A $$
where
$$A = \begin{pmatrix} 0 & -1 \\ 1 & 0 \end{pmatrix}$$
and $A^2=-I$. Since $A$ and $I$ are commutative,
$$\begin{aligned} B^2 &= (I+A)^2 = 2A \\ B^3 &= (I+A)2A = 2A-2I\\ B^4 &= (I+A)(2A-2I) = -4I\\ \vdots \\ B^{10} &= 32A \end{aligned}$$
Is there a simpler method or a smarter approach if you want to do this for, e.g., $B^{100}$?
| I would use the fact that$$B=\sqrt2\begin{bmatrix}\cos\left(\frac\pi4\right)&-\sin\left(\frac\pi4\right)\\\sin\left(\frac\pi4\right)&\cos\left(\frac\pi4\right)\end{bmatrix},$$and that therefore$$(\forall n\in\Bbb N):B^n=2^{n/2}\begin{bmatrix}\cos\left(\frac{n\pi}4\right)&-\sin\left(\frac{n\pi}4\right)\\\sin\left(\frac{n\pi}4\right)&\cos\left(\frac{n\pi}4\right)\end{bmatrix}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4294619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Solving $X + \frac{X}{1+2} + \frac{X}{1+2+3} + \frac{X}{1+2+3+4} +\dots+ \frac{X}{1+2+3+\dots+2017} = 2017$ $$X + \frac{X}{1+2} + \frac{X}{1+2+3} + \frac{X}{1+2+3+4} +\dots+ \frac{X}{1+2+3+\dots+2017} = 2017$$
I’ve noticed that when you substitute around a bit, you get that
$yX$, where $y$ is the amount of $X$, is equal to $2017 \cdot (1+2+\dots+2017)$.
Does this actually help?
| We can prove the following via induction. $$ \dfrac1{1} + \dfrac1{1+2} + \dfrac1{1+2+3} + \cdots + \dfrac1{1+2+\cdots+n} = \dfrac {2n}{n+1} $$
So the solution to the equation
$$X + \frac{X}{1+2} + \frac{X}{1+2+3} + \frac{X}{1+2+3+4} +\dots+ \frac{X}{1+2+3+\cdots+n} = n$$ is $$ X = \dfrac{n+1}2 .$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit $\lim_{x\to\infty} x(\sqrt{x^2 +4} - \sqrt{x^2 + 2})$ So i have this limit:
$$\lim_{x\to\infty} x(\sqrt{x^2 +4} - \sqrt{x^2 + 2})$$
Why can't I just divide by $\sqrt{x^2}$? Outside of the square roots you'd divide by $x$ (or $-x$). If I do this I'd get 1(1-1) which could be an answer right? The answer is $-\frac{3}{2}$ and I saw the solutions which make sense but I don't get why my method isn't right.
It has to be done without l'Hopital by the way!
EDIT sorry i wrote down the wrong answer from a different exercise
This is what I did:
$$\lim_{x\to\infty} x(\sqrt{x^2 +4} - \sqrt{x^2 + 2}) = \lim_{x\to\infty} 1(\sqrt{1 + \frac{4}{x^2}} - \sqrt{1 + \frac{2}{x^2}})$$
And since $\frac{4}{x^2}$, $\frac{2}{x^2}$ are $0$, I thought the answer would be $1(1-1) = 0 $
| We have
$$\frac {\sqrt{x^2+4}+\sqrt{x^2+2}} {\sqrt{x^2+4}+\sqrt{x^2+2}}=1$$
so
$$x(\sqrt{x^2+4}-\sqrt{x^2+2})\\=x(\sqrt{x^2+4}-\sqrt{x^2+2})\cdot \frac {\sqrt{x^2+4}+\sqrt{x^2+2}} {\sqrt{x^2+4}+\sqrt{x^2+2}} \\
=\frac {x((\sqrt{x^2+4})^2-(\sqrt{x^2+2})^2)}{\sqrt{x^2+4}+\sqrt{x^2+2}}\\
=\frac {2x} {\sqrt{x^2+4}+\sqrt{x^2+2}} $$
and therefore
$$\lim x(\sqrt{x^2+4}-\sqrt{x^2+2})= \lim \frac {2x} {\sqrt{x^2+4}+\sqrt{x^2+2}}$$
if $x$ tends to an arbitrary value.
We have
$$x(\sqrt{x^2+4}-\sqrt{x^2+2})\\
=x^3 (\sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}})$$
and therefore
$$\lim x(\sqrt{x^2+4}-\sqrt{x^2+2})\\
=\lim x^3 (\sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}})$$
but we have
$$x(\sqrt{x^2+4}-\sqrt{x^2+2})
\ne 1 (\sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}})$$
except for $x=1$, so there is no reason to assume
that the limits of the RHS and LHS of this inequality are equal except if $x\to 1$
A different example is
$$\lim_{x\to \infty}\frac {\sqrt{x^2+4}-\sqrt{x^2+2}} x $$
here you have
$$=\lim_{x\to\infty}\frac{x \sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}}} x\\
=\lim_{x\to \infty}(\sqrt{1+\frac 4 {x^2} }-\sqrt{1+\frac 2 {x^2}})\\
=\sqrt{1+\lim_{x\to \infty} \frac 4 {x^2}} - \sqrt{1+\lim_{x\to \infty}
\frac 2 {x^2}} =0$$
But you also can use the conjugate trick from the first example:
$$\frac {\sqrt{x^2+4}-\sqrt{x^2+2}} x \\
=\frac {\sqrt{x^2+4}-\sqrt{x^2+2}} x \frac {\sqrt{x^2+4}+\sqrt{x^2+2}} {\sqrt{x^2+4}+\sqrt{x^2+2}} \\
= \frac {(\sqrt{x^2+4})^2-(\sqrt{x^2+2})^2} {x (\sqrt{x^2+4}+\sqrt{x^2+2})}\\
=\frac {2} {x (\sqrt{x^2+4}+\sqrt{x^2+2})} \in [\frac 2 {x \cdot 2(x+1)},\frac 2 {x \cdot 2x}] {\to} [0,0] \quad ({x\to \infty})
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the area of a trapezium with perpendicular diagonals and non-parallel sides meeting at an angle of $45^{\circ}$.
Consider the trapezium $ABCD$ such that the sides $AB$ and $DC$ are parallel, $AB=4$, $DC=10$, and the diagonals $AC$ and $BD$ are perpendicular to each other. Sides $DA$ and $CB$ are extended to meet each other at $E$. Angle $DEC$ is equal to $\frac{\pi}{4}$. Find the area of the trapezium.
Source: Pathfinder for Mathematics
I could solve using coordinate geometry but I wanted to know if there was an alternate solution using similarity of triangles since the formulas I used have not been taught till this exercise in the book. My solution:
Consider trapezium ABCD. Let $D(0,0), C(10,0), B(x+4,y)$ and $A(x,y)$. Let $m_1, m_2, m_3$ and $m_4$ be slopes of $BD, AC, AD$ and $BC$ respectively. Therefore,
$$m_1m_2 = -1$$ $$\tan\left(\frac{\pi}{4}\right)=\left|\frac{m_3-m_4}{1+m_3m_4}\right|$$
Solving will give
$y=\frac{20}{3}$ and we do not have to calculate $x$ since it will automatically cancel out when calculating area. Applying basic formula $$\Delta = [ADC]+[ABC]=\frac12\cdot (AB+DC) \cdot h= \frac12\cdot ((x+4-x)+(10)) \cdot y=\frac{140}{3}$$
| $\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;$
Define point $X$ as the intersection of diagonals $AC$ and $BD$.
Let $AX=x$ and $BX=y$.
$$\triangle AXB\sim DXC \implies CX=\frac{5y}{2} \;\;\text{and}\;\; DX=\frac{5x}
{2}.$$
Using the Pythagorean Theorem ($AC\perp BD$),
$$x^2+y^2=16\implies x^4+y^4=256-2x^2y^2\tag{1}$$
$$AD=\frac{\sqrt{4x^2+25y^2}}2 \;\;\text{and}\;\;BC=\frac{\sqrt{4y^2+25x^2}}2$$
Since $AB\parallel CD$,
$$\triangle AEB\sim \triangle DEC \implies AE=\frac{\sqrt{4x^2+25y^2}}3 \;\;\text{and}\;\;BE=\frac{\sqrt{4y^2+25x^2}}3$$
Using the cosine rule in $\triangle AEB$,
$$\begin{align*}
16&=\frac{4x^2+25y^2}{9}+\frac{4y^2+25x^2}{9}-2\cdot\frac{\sqrt{4x^2+25y^2}}3 \cdot\frac{\sqrt{4y^2+25x^2}}3\cdot \cos 45^{\circ} \\
16&=\frac{29}{9}(x^2+y^2)-\frac{\sqrt{100(x^4+y^4)+641x^2y^2}}{9}\cdot \sqrt{2}
\end{align*}$$
Simplifying the above equation using $(1)$, we have,
$$x^2y^2=\frac{160}{21}$$
Area of trapezium is ,
$$[ABCD]=\frac{1}{2}\cdot \frac{7x}{2}\cdot \frac{7y}{2}=\frac{49}{8}x^2y^2=\frac{140}{3}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4297305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Growth/decay rate of $a_1 = \frac{5}{2}, a_{n + 1} = \frac{1}{5}(a_n^2 + 6)$ this might be a vague question but I am interested in the recurrence relation defined by $a_1 = \frac{5}{2}$ and $a_{n + 1} = \frac{1}{5}(a_n^2 + 6)$. Some properties I have proven include that
*
*$2 < a_n \leq \frac{5}{2}$
*is strictly decreasing
*converges to $2$.
The next thing I did was looking at the difference $r_i = a_i - 2$, for which I found to 0.5, 0.45, 0.4005, 0.35248, 0.3068325, 0.2642952, 0.2254066, 0.1904869, 0.1596466, 0.1328146. Clearly there is some "near-"exponential decay going on here. One more result I have proven is that $\frac{8}{10}r_i < r_{i + 1}\leq \frac{9}{10}r_i$. However I am unable to show any further results about the error term $r_i$ - obviously it tends to $0$ exponentially with $0.8 < r \leq 0.9$ but can we do better?
So my question is, is there more precise bounds for $a_{i + 1} - 2$, whether by relating to $a_i - 2$ or in general?
(Also note that the original quadratic recurrence doesn't seem to be solvable by existing techniques.)
| Let $r_n = a_n-2$. One has
\begin{align*}
r_{n+1} &= a_{n+1}-2 \\
&= \frac{1}{5}(a_n^2+6)-2 \\
&= \frac{1}{5}((r_n+2)^2+6)-2\\
&=\frac{1}{5}(r_n^2+4r_n)
\end{align*}
so $(r_n)$ satisfies the relation
$$\boxed{r_{n+1} =\frac{1}{5}(r_n^2+4r_n)}$$
You deduce that
\begin{align*} \log(r_{n+1})-\log(r_n) &= \log\left(\frac{4}{5}\right)+ \log\left(1+\frac{r_n}{4}\right)
\end{align*}
and because $(r_n)$ tends to $0$, one has the following limit
$$\lim_{n \rightarrow +\infty} \log(r_{n+1})-\log(r_n) = \log\left(\frac{4}{5}\right)$$
By Cesaro's theorem, you get that
$$\log(r_n) \sim n\log\left(\frac{4}{5}\right)$$
which gives you
$$\boxed{\lim_{n \rightarrow +\infty} r_n^{1/n} = \frac{4}{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4297479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Two quite messy sequences Consider the following $50-$term sums$:$
$$ S=\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....\frac{1}{99\cdot100}$$ and
$$T=\frac{1}{51\cdot100}+\frac{1}{52\cdot99}+\frac{1}{53\cdot98}+....+\frac{1}{100\cdot51}$$
Express $\frac{S}{T}$ as an irreducible fraction.
My attempt$:$
The first equation can be written as $$S=\frac11-\frac12+\frac13-\frac14+\frac15-\frac16+....+\frac1{99}-\frac1{100}$$ $\implies$ $$S=(1+\frac13+\frac15+....+\frac1{99})-(\frac12+\frac14+\frac16+....+\frac1{100})$$ or $$S=1+\frac12\operatorname{ln}50-\frac12\operatorname{ln}\frac{103}{2}$$ and $$T=2(\frac{1}{51\cdot100}+\frac{1}{52\cdot99}+\frac{1}{53\cdot98}+....+\frac{1}{75\cdot76})$$ After this I am not able to do anything. And also the value of S which I get doesn't seems to be correct(though I have calculated it using formula)as it is quite messy and I don't think S can be further reduced. Any help will be greatly appreciated.
| $$H'_n = \sum_{k = 1}^n \frac {(-1)^n}{k}$$
*
*Harmonic Function : ($4$)
$$\color{green}{S} =\frac{1}{1\cdot2}+\frac{1}{3\cdot4}+\frac{1}{5\cdot6}+....\frac{1}{99\cdot100}= \color{green}{H'_{100}} $$
$$\begin{align*}T
&= \frac{1}{51\cdot100}+\frac{1}{52\cdot99}+\frac{1}{53\cdot98}+....+\frac{1}{100\cdot51}\\
& = \frac 1{151} \left[\frac 1{51} + \frac 1{100} +\frac 1{52} +\frac 1{99}....... \frac {1}{99}+\frac 1{52} + \frac1{100} + \frac 1{51}\right]\\
& = \frac 2{151}\left(\frac 1{51} + \frac 1{52} +.....\frac 1{100}\right)\\
& = \frac 2{151}(H_{100}-H_{50})\\
& = \frac 2{151}H'_{100} \\
\end{align*}$$
$$\color{blue}{T = \frac 2{151}S \implies \frac S{T} = \frac {151}{2}}$$
$H'_{2n} = H_{2n} -H_{n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Limit of multivariare function at the origin Consider the function $f:\mathbb{R}^2\to \mathbb{R}$ defined as:
$$
f(x,y)=
\begin{cases}
\frac{|x|^{\frac{5}{2}}y}{(x^2+y^4)\sqrt{x^2+y^2}}.\quad&\text{ if } (x,y)\neq 0\\
0\quad& \text{ if }(x,y)= 0\,.
\end{cases}
$$
Is this function continuous at the origin?
If the limit exists it has to be $0$ since, for example, if we take the restriction $x=y$ we obtain:
$$
\lim_{x\to0}\frac{|x|^{\frac{5}{2}}x}{\sqrt{2}(x^2+x^4)|x|}=0
$$
Restrictions to any kind of powers seem to give the same result suggesting that the function has to be continuous at the origin (as the graph also seems to confirm) but I'm not able to find an estimate for $f$ to use the squeeze theorem and actually prove continuity.
Any help will be greatly appreciated.
| We have
$$
\frac{|x|^{\frac{5}{2}}y}{(x^2+y^4)\sqrt{x^2+y^2}}=
|x|^{\frac{1}{2}}\cdot\frac{|x|^{2}}{x^2+y^4}\cdot
\frac{y}{\sqrt{x^2+y^2}}.
$$
Since
$$
\left|\frac{|x|^{2}}{x^2+y^4}\right|=\frac{x^{2}}{x^2+y^4}\leq 1
$$
and
$$
\left|\frac{y}{\sqrt{x^2+y^2}}\right|=\frac{|y|}{\sqrt{x^2+y^2}}\le1
$$
we obtain
$$
\left| \frac{|x|^{\frac{5}{2}}y}{(x^2+y^4)\sqrt{x^2+y^2}} \right| \leq |x|^{\frac{1}{2}}
$$
or
$$
-|x|^{\frac{1}{2}}\leq \frac{|x|^{\frac{5}{2}}y}{(x^2+y^4)\sqrt{x^2+y^2}} \leq |x|^{\frac{1}{2}}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Number of solutions in non-negative integers to $x_1 + x_2 + x_3 + x_4 + x_5 = 9$ where $x_1, x_2,x_3,x_4,x_5 \neq 1$ using generating functions Using generating functions, the answer is the coefficient of $x^9$ in the expression $(1+x^2 +x^3 +x^4 \dots)^5$
Since $1+x+x^2+x^3 + \dots = \frac{1}{1-x}$ we get:
$(1+x^2 +x^3 +x^4 \dots)^5 = (\frac{1}{1-x} - x)^5 = (\frac{1 - x + x^2}{1-x})^5 = (\frac{1}{1-x})^5 \cdot (1+x^2-x)^5$
Since we generally have $(\frac{1}{1-x})^n = \sum_{k=0}^n {n+k-1 \choose k}x^k$
We get $(\frac{1}{1-x})^5 = \sum_{k=0}^5 {5+k-1 \choose k}x^k = {4 \choose 0} + {5 \choose 1}x+{6 \choose 2}x^2+{7 \choose 3}x^3+{8 \choose 4}x^4+{9 \choose 5}x^5$
And by the binomial theorem we get $(1+x^2-x)^5 = \sum_{k=0}^5 {5 \choose k}(x^2-x)^k = {5 \choose 0} + {5 \choose 1}(x^2-x) + {5 \choose 2}(x^2-x)^2 + {5 \choose 3}(x^2-x)^3 + {5 \choose 4}(x^2-x)^4 + {5 \choose 5}(x^2-x)^5$
So we are left with calculating the coefficient of $x^9$ in
$({4 \choose 0} + {5 \choose 1}x+{6 \choose 2}x^2+{7 \choose 3}x^3+{8 \choose 4}x^4+{9 \choose 5}x^5) \cdot ({5 \choose 0} + {5 \choose 1}(x^2-x) + {5 \choose 2}(x^2-x)^2 + {5 \choose 3}(x^2-x)^3 + {5 \choose 4}(x^2-x)^4 + {5 \choose 5}(x^2-x)^5)$
The problem is that the $(x^2-x)^k$ parts are really long to compute and it seems like there should be an easier way. Any advice on how to solve this without having to open up these expressions?
| Hint: we have to find the coefficient of $x^9$ in $${(1-x)}^{-5}{(x^2-x+1)}^5$$ which can be transformed to by multiplying denominator and numerator by ${(1+x)}^5$ $${(x^3+1)}^5{(1-x^2)}^{-5}$$ or $${(1+5x^3+10x^6+10x^9..)}{(1+5x^2+15x^4+35x^6...)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4300088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$x_1=\sqrt{2}, x_{n+1}=x_n+\frac 1 {x_n}$ then $\sum_{i=1}^{2021} \frac {x_i^2}{2x_ix_{i+1}-1}>\frac{2021^2}{x_{2021}^2+\frac 1 {x_{2021}^2}}.$
$x_1=\sqrt{2}, x_{n+1}=x_n+\frac 1 {x_n}$ then show that $\displaystyle \sum_{i=1}^{2021} \dfrac {x_i^2}{2x_ix_{i+1}-1}>\dfrac{2021^2}{x_{2021}^2+\dfrac 1 {x_{2021}^2}}.$
My Attempt:
\begin{align}
&2x_ix_{i+1}-1=2x_i\bigg(x_i+\frac 1 {x_i} \bigg)-1=2x_i^2+1. \\
&\therefore \sum_{i=1}^{2021} \frac {x_i^2}{2x_1x_{i+1}-1} = \sum_{i=1}^{2021} \frac{x_i^2}{2x_i^2+1} = \frac 1 2 \sum_{i=1}^{2021} 1-\frac 1 {2x_i^2+1} = \frac {2021}{2}-\sum_{i=1}^{2021}\frac 1 {4x_i^2+2}. \\
&x_{2021}^2+\dfrac {1} {x_{2021}^2} = \bigg(x_{2021}+\dfrac 1 {x_{2021}} \bigg)^2-2 = x_{2022}^2-2. \\
\ \\
&\therefore ETS) \ \dfrac {2021} {2} - \sum_{i=1}^{2021} \dfrac 1 {4x_i^2+2} > \dfrac {2021^2}{x_{2022}^2-2}.
\end{align}
| The OP wrote that $\ \displaystyle \sum_{i=1}^{2021} \dfrac{x_i^2}{x_ix_{i+1}-1} = \sum_{i=1}^{2021} \dfrac{x_i^2}{2x_i^2+1} \ $
If $f(x)=\dfrac{1}{x}$, we can write that
$\ \displaystyle \sum_{i=1}^{2021} \dfrac{x_i^2}{x_ix_{i+1}-1} = \sum_{i=1}^{2021} f\left( 2+\dfrac{1}{x_i^2}\right) = \sum_{i=1}^{2021} f\left( x_{i+1}^2-x_i^2\right)$
$f$ is strictly convex on $(0,+\infty )$ because $f''$ is positive, then:
$\ \displaystyle \sum_{i=1}^{2021} \dfrac{x_i^2}{x_ix_{i+1}-1} > 2021 f\left( \dfrac{\sum_{i=1}^{2021} (x_{i+1}^2-x_i^2)}{2021}\right) = 2021 f\left( \dfrac{x_{2022}^2-2}{2021}\right)$
So we get
$\ \displaystyle \sum_{i=1}^{2021} \dfrac{x_i^2}{x_ix_{i+1}-1} > \dfrac{2021^2}{x_{2022}^2- 2} = \dfrac{2021^2}{x_{2021}^2+\dfrac{1}{x_{2021}^2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$abc+a+b+c=ab+bc+ca+5,$ $a, b, c \in \mathbb{R}$. Find the minimum of $a^2+b^2+c^2$.
$abc+a+b+c=ab+bc+ca+5,$ $a, b, c \in \mathbb{R}$. Find the minimum of $a^2+b^2+c^2$.
I will let $\displaystyle \sum_{c} = \sum_{cyc}.$
$\displaystyle \sum_c a^2 - \sum_c ab = \frac 1 2 \bigg( \sum_c(a-b)^2 \bigg) \geq 0.$
$\displaystyle \sum_c a^2 \geq \frac {\displaystyle \bigg(\sum_c a \bigg)^2}{3}$.
$\displaystyle \bigg( \sum_c a^2 \bigg) \bigg( \sum_c b^2 \bigg) \bigg( \sum_c c^2 \bigg) \bigg( \sum_c a^2b^2c^2 \bigg) \geq \bigg( \sum_c abc \bigg)^4.$
$\therefore \displaystyle \bigg( \sum_c a^2 \bigg)^3 \geq 27a^2b^2c^2.$
Let $a+b+c=X$, $ab+bc+ca=Y$, $abc=Z$, $a^2+b^2+c^2=W$.
\begin{align}
& W \geq \dfrac {X^2} {3} & (1) \\
& W \geq Y & (2) \\
& W^3 \geq 27Z^2 & (3)
\end{align}
According to $(1)$ and $(3)$,
\begin{align} W^2 \geq 3XZ.& & (4) \end{align}
According to $(2)$ and $(4)$,
\begin{align} W \geq \sqrt [3] {3XYZ}. \end{align}
| You have $(a-1)(b-1)(c-1)=4$ and you seek to minimise
$a^2+b^2+c^2$. WLOG let $a$ be the largest among $a,b,c$, then $x=a-1>0$.
$\begin{align}
a^2+b^2+c^2&=(a^2-2)+(b^2+c^2+1+1)\\
&\geqslant (a^2-2)+\tfrac14(b+c-1-1)^2\tag{CS inequality}\\
&\geqslant (x^2+2x-1)+(b-1)(c-1)\tag{$(p+q)^2\geqslant 4pq$}\\
&=\color{red}{x^2+2x+4/x}-1
\end{align}$
Now the univariate expression in red can be easily minimised using calculus or $7$-term AM-GM to get $a^2+b^2+c^2\geqslant 6$.
Equality is when $x=1,b=c=-1$ or in general when $(a,b,c)\sim (2,-1,-1)$ where $\sim$ denotes any permutation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4301229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove that 1 is a triple root of a polynomial I'm studying for an exam and trying to prove whether 1 is a triple root for the polynomial:
$$x^{2n+1}-(2n+1)x^{n+1}+(2n+1)x^n-1$$
for every $n\geq1$.
In our math class we never solved such a problem. So far we only used horner's scheme to prove that someone is a root, double root or triple root.
Can you please help me solve this problem?
| $$\begin{align*}
x^{2n+1}&-(2n+1)x^{n+1}+(2n+1)x^n-1\\
&=x^{2n+1}-1-(2n+1)[ x^{n+1}- x^n]\\
&=(x-1)(x^{2n}+x^{2n-1}+\cdots+1)-(2n+1)(x-1)x^n\\
&=(x-1)(x^{2n}+x^{2n-1}+\cdots+x^{n+1}+x^{n-1}+\cdots+1-2nx^n)\\
&=(x-1)[x^n(x^n-1)+x^{n-1}(x^n-x)+\cdots+x^n(x-1)\\
&\qquad\qquad+x^{n-1}(1-x)+x^{n-2}(1-x^2)+\cdots+x(1-x^{n-1})+1-x^n]
\end{align*}$$
Combining corresponding terms from opposite ends of the expression inside the square bracket we have:
$$
=(x-1)[(x^n-1)^2+x(x^{n-1}-1)^2+...+x^{n-1}(x-1)^2]
$$
It is clear that each term inside the square bracket is divisible by $(x-1)^2$ so the whole expression is divisible by $(x-1)^3$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_count": 3,
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} |
Give a complete non-redundant list of elements $x\in\mathbb{C}$ such that $x^{10}+x^5+1=0$ Since, $x^{10}+x^5+1 = (x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)=0$.
I know that $x^2+x+1=0$ has root $x=-\frac{1}{2}+i\frac{\sqrt{3}}{2}$ and $\:x=-\frac{1}{2}-i\frac{\sqrt{3}}{2}$.
How am I supposed to find the roots in $\mathbb{C}$ for $x^8-x^7+x^5-x^4+x^3-x+1$ ?
| We might also pretend we don't know about cyclotomic polynomials and try constructing the roots of the equation from properties of complex numbers. From the equation $ \ z^{10} + z^5 + 1 \ = \ 0 $ $ \ \Rightarrow \ z^{10} + z^5 \ = \ -1 \ \ , $ we observe that $ \ w \ = \ z^5 \ $ and $ \ w^2 \ $ sum to a real number, so we have $ \ w^2 \ = \ \overline{w} \ \ . $ If we write $ \ w \ = \ \alpha + \beta·i \ \ , $ then this equation tells us that
$ (\alpha^2 \ - \ \beta^2) \ + \ 2·\alpha·\beta·i \ \ = \ \ \alpha \ - \ \beta·i $
$\Rightarrow \ \ \alpha \ = \ -\frac12 \ , \ \beta \ \neq \ 0 \ \ $ by equating imaginary parts (we know $ \ w \ $ is complex since $ \ w^2 + w + 1 \ = \ 0 \ $ has a negative discriminant)
$ \Rightarrow \ \ \left(-\frac12 \right)^2 \ - \ \beta^2 \ = \ -\frac12 \ \Rightarrow \ \beta^2 \ = \ \frac34 \ \ . $
We do this to satisfy ourselves that $ \ |w| \ = \ 1 \ \ . $ Thus, $ \ w \ = \ -\frac12 + \frac{\sqrt3}{2}·i \ = \ e^{ \ i·2 \pi/3} \ \ , $ $ w^2 \ = \ \overline{w} \ = \ -\frac12 - \frac{\sqrt3}{2}·i \ = \ e^{ \ i·4 \pi/3} \ \ . $
The ten roots of the original equation are then the five complex fifth-roots each of $ \ w \ $ and $ \ \overline{w} \ \ , $
$$ z \ \ = \ \ \large{e^{ \ i·\left(\frac{2 \pi}{15} + \frac{2·k·\pi}{3} \right)} } \ \ = \ \ e^{ \ i·\left(\frac{(2 + 6k) \pi}{15} \right)} \ \ , \ \ \large{e^{ \ i·\left(\frac{4 \pi}{15} + \frac{2·k·\pi}{3} \right)} } \ \ = \ \ e^{ \ i·\left(\frac{(4 + 6k) \pi}{15} \right)} \ \ , $$
or $$ z \ \ = \ \ \large{e^{ \ i·\frac{2 \pi}{15} } \ , \ e^{ \ i·\frac{4 \pi}{15} } \ , \ e^{ \ i·\frac{8 \pi}{15} } \ , \ e^{ \ i·\frac{10 \pi}{15} } \ = \ e^{ \ i·\frac{2 \pi}{3} } \ , \ e^{ \ i·\frac{14 \pi}{15} } \ , \ e^{ \ i·\frac{16 \pi}{15} } \ ,} $$ $$ \large{e^{ \ i·\frac{20 \pi}{15} } \ = \ e^{ \ i·\frac{4 \pi}{3} } \ , \ e^{ \ i·\frac{22 \pi}{15} } \ , \ e^{ \ i·\frac{26 \pi}{15} } \ , \ e^{ \ i·\frac{28 \pi}{15} } \ \ . } $$
On the Argand diagram, these roots marks the vertices of two regular pentagons rotated 24º relative to one another. Note that the fifth-roots of $ \ 1 \ $ are excluded from the set of implied fifteenth-roots of $ \ 1 \ \ . $
[I acknowledge the much nicer and more sophisticated arguments by MH.Lee and dxiv ; I was interested here in what the least means employed to answer the question might be.]
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $(a+b)^n \leq 2^{n-1}(a^n+b^n)$ by induction. I want to prove that that $(a+b)^n \leq 2^{n-1}(a^n+b^n)$ is true with the help of induction.
base case: for $n=0$ we get $(a+b)^0 \leq 2^{-1}(a^0 +b^0) \Longleftrightarrow 1 \leq 1$. Thus the inequality for $n=0$ is correct.
IH: For a any $n \in \mathbb{N}$ and $a,b \geq 0$, $(a+b)^n \leq 2^{n-1}(a^n+b^n)$ holds.
IS: For $n \rightarrow n + 1$ we get.
\begin{align*}
&\hspace{0.55cm} (a+b)^{n+1} \leq 2^n(a^{n+1} + b^{n+1}) \\
&= (a+b)^n (a+b) \leq 2^n(a^na + b^nb) \\
&= (a+b)^n (a+b) \leq 2^n(a^n+b^n) (a+b) \\
&= (a+b)^n \leq 2^n(a^n+b^n) \\
&= \frac{(a+b)^n}{2} \leq 2^{n-1}(a^n+b^n) \\
&\rightarrow \frac{(a+b)^n}{2} \leq (a+b)^n \leq 2^{n-1}(a^n+b^n)
\end{align*}
Because of IH $(a+b)^n \leq 2^{n-1}(a^n+b^n)$, $0.5(a+b)^n \leq 2^{n-1}(a^n+b^n)$ must also hold for $a,b \geq 0$ and $n \in \mathbb{N}$. Thus the formula holds for all $n$
Is this a valid proof? I am not sure if the last part is correct and if I can simply divide by $(a+b)^n$, because it can be zero
2nd try:
\begin{align*}
&\hspace{0.95cm} (a+b)^{n+1} = (a+b)^n(a+b)\\
&\stackrel{\text{IV}}{\longrightarrow} (a+b)^n(a+b) \leq 2^{n-1}(a^n + b^n)(a+b) \\
&\Longleftrightarrow 2{(a+b)^{n+1}} \leq 2^n(a^n+b^n) (a+b) \\
&\Longleftrightarrow 2{(a+b)^{n+1}} \leq 2^n(a^na+b^n*b + a^nb + b^na) \\
&\Longleftrightarrow 2{(a+b)^{n+1}} \leq 2^n(a^{n+1}+b^{n+1} + a^nb + b^na) \\
&\longrightarrow (a+b)^{n+1} < 2{(a+b)^{n+1}} \leq 2^n(a^{n+1}+b^{n+1} + a^nb + b^na) \\
&\longrightarrow (a+b)^{n+1} \leq 2^n(a^{n+1}+b^{n+1} + a^nb + b^na)
\end{align*}
It remains to show that $a^{n+1} + b^{n+1} \geq a^nb + b^na$ holds, because then the statement also holds for $n+1$ $\rightarrow$ $(a+b)^{n+1} \leq 2^n(a^{n+1}+b^{n+1})$ and thus all $n \in \mathbb{N}$. It holds from above
\begin{align*}
\hspace{1cm} a^{n+1} + b^{n+1} &\geq a^nb + b^na \\
\Longleftrightarrow a^{n+1} + b^{n+1} - a^nb - b^na &\geq 0 \\
\Longleftrightarrow a^n(a-b) - b^n(a - b) &\geq 0 \\
\Longleftrightarrow (a^n-b^n)(a-b) &\geq 0 \\
\end{align*}
$(a^n-b^n)(a-b) \geq 0$ is satisfied for all $n \in \mathbb{N}$ and $a,b \geq 0$ and thus the inequality from the problem also holds for all $n$
| Another solution which avoids induction is using Jensen's inequality. I assume from your attempt that $a, b \geq 0$.
Let $f(x) = x^n$ for some $n \in \mathbb{N}$. Note that $f$ is a convex function over the interval $(0, \infty)$. Then by Jensen we have
$$
\frac{(a+b)^n}{2^n} = f\bigg(\frac{a+b}{2}\bigg) \leq \frac{f(a)+f(b)}{2} = \frac{a^n+b^n}{2} \\
\iff (a+b)^n \leq 2^{n-1}\big(a^n+b^n\big).
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Help with finding the equation of line tangent to the semi circle with the equation: $y=\sqrt{1-x^2}$
Ok so I need to show that the equation of the tangent above to the semi-circle with the equation: $y=\sqrt{1-x^2}$, is $y=-\frac{1}{\sqrt{3}}x$ + $\frac{2}{\sqrt{3}}$
What we from the question:
The tangent intersects the $x$-axis at $(2,0)$, (assume that you do not know any other points of intersection)
What I tried to do so far:
Since its a tangent, I decided to differentiate the semi-circle function using the chain rule, getting:
$\frac{dy}{dx}$ = $\frac{-x}{\sqrt{1-x^2}}$
Then using the equation of a line formula:
$y=mx+c$
--> $y=\frac{-x}{\sqrt{1-x^2}}x +c$
Substituting the point in:
-->$0=\frac{-2}{\sqrt{1-2^2}}(2) +c$
Now clearly something is wrong because I will get the square root of a negative, which does not make any sense. Need help from here.
*Note: I know that there are other ways to solve this, but I would prefer if calculus was used to solve this question.
| If we construct the radius from the center of the circle to the point of intersection, this radius, the line and the x axis forms a right triangle with angle $\theta$ at the origin.
$\sec \theta = 2$
The point of intersection is $(\cos \theta, \sin\theta) = (\frac 12, \frac {\sqrt 3}{2})$
The line intersects the y-axis at $(0,\csc \theta) = \frac {2}{\sqrt 3}$
In intercept-intercept form: $\frac {x}{2} + \frac {y\sqrt 3}{2} = 1$
Or in standard form $x + y\sqrt 3 = 2$
If you really want to use calculus, we have:
$\frac {dy}{dx} = -\frac {x}{\sqrt {1-x^2}} = -\frac {x}{y}$
And this slope must equal the slope of the line from $(2,0)$ to $(x,y)$
$\frac {y-0}{x-2} = -\frac {x}{y}\\
y^2 = -x^2 + 2x\\
1-x^2 = -x^2 + 2x\\
x = \frac {1}{2}$
$y = -\frac {1}{\sqrt 3}(x-2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $f^{(n)}(0)$ where $f(x) = e^{x^2}$ The problem is stated as:
Calculate $f^{(n)}(0)$ where $f(x) = e^{x^2}$ and $n \in \mathbb{N}$
Attempt:
We know that $e^{x} = \sum_{k=0}^{n} \frac{1}{k!}x^k +\frac{x^{n+1}e^{\theta x}}{(n+1)!}$ for some $\theta = \theta (x) \in (0,1)$
Hence, $e^{x^2}=\sum_{k=0}^{n} \frac{1}{k!}x^{2k} +\frac{x^{2n+1}e^{\theta x^2}}{(n+1)!}$
Comparing terms with the general Maclaurin polynomial, we get that:
$\frac{1}{k!}x^{2k} = \frac{f^{(n)}(0)}{n!}x^{n}$
First, we have to equate the exponent of the $x$ - terms, so $2k = n \iff k = n/2$
Therefore, we have that:
$\Rightarrow f^{n}(0) = \frac{n!}{(n/2)!}$
However, since we had natural numbers in our expression, I wonder what would happen if $n$ was odd in this case. How could I generalize this even further?
Thanks!
| We have
\begin{align*}
e^{x^2}&=\sum_{k=0}^\infty \frac{(x^2)^k}{k!}\\
&=\frac{x^0}{0!}+\frac{x^2}{1!}+\frac{x^4}{2!}+\frac{x^6}{3!}+\dots\\
&=\frac{x^0}{0!}+0\frac{x^1}{1!}+2!\frac{x^2}{2!}+0\frac{x^3}{3!}+\frac{4!}{2!}\frac{x^4}{4!}+0\frac{x^5}{5!}+\frac{6!}{3!}\frac{x^6}{6!}+\dots\\
\end{align*}
So,
$$e^{x^2}=\sum_{k=0}^\infty a_k\frac{x^k}{k!},$$
where
\begin{align*}
a_k=\begin{cases}
\frac{k!}{(k/2)!}&\text{if $k$ is even}\\
0&\text{if $k$ is odd}
\end{cases}
\end{align*}
| {
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"url": "https://math.stackexchange.com/questions/4305189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding a closed formula for calculating $\frac{d^n}{{dx}^n}f\left(x\right)=\frac{d^n}{{dx}^n}e^{x^2}$ For reasons I can't even remember, the other day I wanted to find out if there was a closed formula for calculating the $n$-th derivative $\frac{d^n}{{dx}^n}f\left(x\right)=\frac{d^n}{{dx}^n}e^{x^2}$ for the function $f\left(x\right)=e^{x^2}$. Where I ended up after some trial and error is the formula $$\frac{d^n}{{dx}^n}e^{x^2}=c_n\left(\sum_{0 \leq i \leq \lfloor\frac{n}{2}\rfloor} {p_i x^{n-2i}}\right)e^{x^2}=c_n\left(c_{n-1}x^n + \sum_{1 \leq i \leq \lfloor\frac{n}{2}\rfloor} {p_i x^{n-2i}}\right)e^{x^2},$$ with $c_n=2^{n-\lfloor\frac{n}{2}\rfloor}$.
The $p_{i \geq1}$ turn out to be as follows:
$n$
$p_{i=1}$
$p_{i=2}$
$p_{i=3}$
$p_{i=4}$
$...$
0
–
–
–
–
$...$
1
–
–
–
–
$...$
2
1
–
–
–
$...$
3
3
–
–
–
$...$
4
12
3
–
–
$...$
5
20
15
–
–
$...$
6
60
90
15
–
$...$
7
84
210
105
–
$...$
8
224
840
840
105
$...$
9
288
1512
2520
945
$...$
$...$
$...$
$...$
$...$
$...$
$...$
I have not yet understood the rule behind the $p_i$ sequences $$p_{i=1}:\left(1,3,12,20,60,84,224,288,...\right),$$
$$p_{i=2}:\left(3,15,90,210,840,1512,...\right),$$
$$p_{i=3}:\left(15,105,840,2520,...\right),$$
$$p_{i=4}:\left(105,945,...\right),$$
$$...$$ I suspect it has something to do with binomial coefficients, since the coefficients $p_i$ arise from multiplying binomials during the derivation. One regularity I've noticed so far is that starting at $p_{i=2}$, the first values always correspond to the second ones of the previous $p$ sequence.
Do any of you have an idea how I can formalize the coefficients $p_i$ and the timing of their occurrence and integrate them into the above closed formula? Or do you know if there even already exists a known solution to the problem, namely finding a closed formula to calculate the $n$-th derivative $\frac{d^n}{{dx}^n}f\left(x\right)=\frac{d^n}{{dx}^n}e^{x^2}$?
Thank you and best regards!
| Hint: Using Wolfram Alpha we find for small values of $n$
\begin{align*}
\frac{d}{dx}e^{x^2}&=2x^2e^{x^{2}}\\
\frac{d^2}{dx^2}e^{x^2}&=2\left(2x^2+1\right)e^{x^{2}}\\
\frac{d^3}{dx^3}e^{x^2}&=4\left(2x^3+3x\right)e^{x^{2}}\tag{1}\\
\frac{d^4}{dx^4}e^{x^2}&=4\left(4x^4+12x^2+3\right)e^{x^{2}}\\
\end{align*}
Another expression besides Faa di Brunos formula is stated as identity (3.56) in H.W. Gould's Tables of Combinatorial Identities, Vol. I and called:
Hoppe Form of Generalized Chain Rule
Let $D_x$ represent differentiation with respect to $x$ and $y=y(x)$. Hence $D^n_x g(y)$ is the $n$-th derivative of $g$ with respect to $x$. The following holds true
\begin{align*}
D_x^n g(y)=\sum_{k=0}^nD_y^kg(y)\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}y^{k-j}D_x^ny^j
\end{align*}
In the special case
\begin{align*}
g(y(x))=e^{y(x)}=e^{x^{2}}
\end{align*}
we have $$D_y^kg(y)=D_y^k e^y=e^y$$ and obtain
\begin{align*}
D_x^ne^y=e^y\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}y^{k-j}D_x^ny^j\tag{2}
\end{align*}
With $y=y(x)=x^2$ we obtain from (2)
\begin{align*}
\color{blue}{D_x^ne^{x^2}}&=e^{x^2}\sum_{k=0}^n\frac{(-1)^k}{k!}\sum_{j=0}^k(-1)^j\binom{k}{j}x^{2(k-j)}D_x^nx^{2j}\tag{3.1}\\
&=e^{x^2}\sum_{k=\left\lfloor\frac{n+1}{2}\right\rfloor}^n\frac{(-1)^k}{k!}\sum_{j=\left\lfloor\frac{n+1}{2}\right\rfloor}^k(-1)^j\binom{k}{j}x^{2(k-j)}D_x^nx^{2j}\tag{3.2}\\
&\,\,\color{blue}{=e^{x^2}\sum_{k=\left\lfloor\frac{n+1}{2}\right\rfloor}^n\frac{(-1)^k}{k!}\sum_{j=\left\lfloor\frac{n+1}{2}\right\rfloor}^k(-1)^j\binom{k}{j}(2j)^{\underline{n}}x^{2k-n}}\tag{3.3}\\
\end{align*}
Comment:
*
*In (3.1) we apply Hoppe's formula with $y=y(x)=x^2$.
*In (3.2) we observe that we differentiate $x^{2j}$ $n$ times. This implies that terms with indices $k,j< \left\lfloor\frac{n+1}{2}\right\rfloor$ vanish.
*In (3.3) we calculate $D_x^nx^{2j}$ using the falling factorial notation $(2j)^{\underline{n}}=(2j)(2j-1)\cdots(2j-n+1)$.
Let's look at a small example in order to see formula (3.3) in action.
Example: $n=3$ We obtain
\begin{align*}
e^{x^2}&\sum_{k=2}^{3}\frac{(-1)^k}{k!}\sum_{j=2}^k(-1)^j\binom{k}{j}(2j)(2j-1)(2j-2)x^{2k-3}\\
&=4e^{x^2}\sum_{k=2}^{3}\frac{(-1)^k}{k!}\sum_{j=2}^k(-1)^j\binom{k}{j}j(j-1)(2j-1)x^{2k-3}\\
&=4e^{x^2}\left(\frac{1}{2}\left(\sum_{j=2}^2(-1)^j\binom{2}{j}j(j-1)(2j-1)\right)\right)x\\
&\qquad+4e^{x^2}\left(\frac{(-1)}{6}\left(\sum_{j=2}^3(-1)^j\binom{3}{j}j(j-1)(2j-1)\right)\right)x^3\\
&=4e^{x^2}\left(\frac{1}{2}\binom{2}{2}\cdot 6\right)x
+4e^{x^2}\left(-\frac{1}{6}\binom{3}{2}\cdot 6+\frac{1}{6}\binom{3}{3}\cdot 30\right)x^3\\
&\,\,\color{blue}{=4e^{x^2}\left(2x^3+3x\right)}
\end{align*}
in accordance with (1).
| {
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Rank of a pair of coprime integers Let's say two pairs of coprime integers $(a, b)$ and $(c, d)$ are connected if $ac + bd = 1$.
A connected chain, or just a chain, is a sequence of coprime pairs in which every two consecutive pairs are connected.
Clearly, if $(a, b)$ is connected to $(c, d)$, then $(a, b)$ is connected to $(c + nb, d - na)$ for an integer $n$.
Using the property, we may find a connected pair $(x, y)$ such that $max(|x|,|y|) < max(|a|,|b|)$.
Continuing the process we may build a chain that ends on $(0, 1)$ or $(1, 0)$ in absolute values.
Examples (in absolute values):
*
*$(41, 53)$, $(22, 17)$, $(7, 9)$, $(4, 3)$, $(1, 1)$, $(0, 1)$.
*$(41, 53)$, $(31, 24)$, $(7, 9)$, $(4, 3)$, $(1, 1)$, $(0, 1)$.
The described algorithm may give chains of different length:
*
*$(41, 61)$, $(58, 39)$, $(2, 3)$, $(1, 1)$, $(0, 1)$.
*$(41, 61)$, $(3, 2)$, $(1, 1)$, $(0, 1)$.
Let's define the rank of a pair of coprime integers $(a, b)$ as the minimal length within all possible chains connecting $(a, b)$ with $(0, 1)$ or $(1, 0)$ in absolute values.
Let's call a chain with the minimal length a minimal chain for $(a, b)$.
Questions:
*
*Does the rank exist for any pair of coprime or prime integers?
*Is there a maximal rank within all pairs of coprime or prime integers with rank?
*Is there an algorithm of constructing a minimal chain or calculating the rank of a given coprime pair?
| Your question (3), begin with your coprime pair $(a,b)$ with $a > b > 0,$ the minimal chain comes from using the Extended Euclidean Algorithm and writing the (simple) continued fraction for $\frac{a}{b},$ then let each convergent $\frac{p}{q}$ be part of the chain, as pair $(p,q)$
$$ \gcd( 53, 41 ) = ??? $$
$$ \frac{ 53 }{ 41 } = 1 + \frac{ 12 }{ 41 } $$
$$ \frac{ 41 }{ 12 } = 3 + \frac{ 5 }{ 12 } $$
$$ \frac{ 12 }{ 5 } = 2 + \frac{ 2 }{ 5 } $$
$$ \frac{ 5 }{ 2 } = 2 + \frac{ 1 }{ 2 } $$
$$ \frac{ 2 }{ 1 } = 2 + \frac{ 0 }{ 1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccccc}
& & 1 & & 3 & & 2 & & 2 & & 2 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 4 }{ 3 } & & \frac{ 9 }{ 7 } & & \frac{ 22 }{ 17 } & & \frac{ 53 }{ 41 }
\end{array}
$$
$$ $$
$$ 53 \cdot 17 - 41 \cdot 22 = -1 $$
$$ \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc \bigcirc $$
$$ \gcd( 61, 41 ) = ??? $$
$$ \frac{ 61 }{ 41 } = 1 + \frac{ 20 }{ 41 } $$
$$ \frac{ 41 }{ 20 } = 2 + \frac{ 1 }{ 20 } $$
$$ \frac{ 20 }{ 1 } = 20 + \frac{ 0 }{ 1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccc}
& & 1 & & 2 & & 20 & \\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 3 }{ 2 } & & \frac{ 61 }{ 41 }
\end{array}
$$
$$ $$
$$ 61 \cdot 2 - 41 \cdot 3 = -1 $$
| {
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"url": "https://math.stackexchange.com/questions/4306147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do we solve $x^2 + \{x\}^2 = 33$ without computer? This is a problem taken from a group on Facebook. I wonder how to solve this without numerical process.
$x^2 + \{x\}^2 = 33\tag{1}$
My unfinished attempt:
$$\begin{align}
x^2 + \{x\}^2 &= 33\\
x^2 + \left(x - \lfloor x \rfloor\right)^2 &= 33\\
x^2 + x^2 - 2x \lfloor x \rfloor + \lfloor x \rfloor^2 &= 33\\
2x\left(x - \lfloor x \rfloor\right) + \lfloor x \rfloor^2 &= 33\\
2x\{x\} + \lfloor x \rfloor^2 = 33
\end{align}$$
I'm stuck at there. Also we know that the fractional part of $x$ is bounded i.e.:
$$0\leq\{x\}<1$$
From that, I can predict where the two solutions are placed at:
$$\begin{align}
0&\leq \{x\} < 1\\
0 &\leq \{x\}^2 < 1\\
x^2 &\leq x^2 + \{x\}^2 < x^2 + 1\\
x^2 &\leq 33 < x^2 + 1 \tag{$x^2 + \{x\}^2 = 33$}\\
S &= -\sqrt{33}\leq x \leq \sqrt{33} \quad \bigcap \quad x< -\sqrt{32} \lor x > \sqrt{32}\\
\therefore S &= -\sqrt{33}\leq x < -4\sqrt2 \quad \bigcup \quad 4\sqrt{2} < x \leq \sqrt{33}\\
S &\approx -5.74456 \leq x < -5.65685 \quad \bigcup \quad 5.65685 < x \leq 5.74456
\end{align}$$
I don't know if that's even going to help. Anyway, here are the two solutions (to the original problem) that's given by Wolfram Alpha:
$$\begin{align}
x_1 &= \frac12 \left(3\sqrt{17} - 1\right) \approx -5.815\\
x_2 &= \frac12 \left(-1 - 3\sqrt{113}\right) \approx 5.685
\end{align}$$
You see $x_1$ isn't at the interval. I'm confused.
| ${x}^{\mathrm{2}} +\left\{{x}\right\}^{\mathrm{2}} =\mathrm{33} \\ $
$-\left\{{x}\right\}\pm\sqrt{\mathrm{33}−\left\{{x}\right\}^{\mathrm{2}} }=\left[{x}\right]\in\left\{-\mathrm{6},\mathrm{5}\right\} \\ $
$\mathrm{33}−{x}^{\mathrm{2}} \in\left\{\left({x}+\mathrm{6}\right)^{\mathrm{2}} ,\left({x}−\mathrm{5}\right)^{\mathrm{2}} \right\} \\ $
$\left(\mathrm{2}{x}^{\mathrm{2}} +\mathrm{12}{x}+\mathrm{3}\right)\left({x}^{\mathrm{2}} −\mathrm{5}{x}−\mathrm{4}\right)=\mathrm{0} \\ $
${x}\in\left\{-\mathrm{3}−\frac{\sqrt{\mathrm{30}}}{\mathrm{2}},\frac{\mathrm{5}+\sqrt{\mathrm{41}}}{\mathrm{2}}\right\} \\ $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4308317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find $\textstyle{\frac{1\cdot 2}{3!} +\frac{2\cdot2^2}{4!}+\frac{3\cdot2^3}{5!}+\frac{4\cdot2^4}{6!}+\cdots}$ up to n terms? $$
\frac{(1)2}{3!} + \frac{(2)2^2}{4!} + \frac{(3)2^3}{5!} + \frac{(4)2^4}{6!} + \cdots =\sum\limits_{k=1}^{n}\frac{k\cdot 2^k}{(k+2)!}
$$
My attempt:
$$
\begin{align}
e^x&=\sum_{n=1}^{\infty}\frac{x^n}{n!} \\
(e^x)'&=\sum_{n=1}^{\infty}\frac{nx^{n-1}}{n!} \\
x\cdot(e^x)'&=\sum_{n=1}^{\infty}\frac{nx^{n}}{n!} \\
x\cdot(e^x)'&=\sum_{n=1}^{\infty}\frac{n(n+1)(n+2)x^{n}}{(n+2)!}
\end{align}
$$
After this attempt I realized exponential series is for infinite terms whereas the question concerns finite terms so approach may not work.
Can you please give any hints on the right approach to be tried?
| Using my hint in the comments,
\begin{align*}
\sum\limits_{k = 1}^n {\frac{{k2^k }}{{(k + 2)!}}} & = \sum\limits_{k = 1}^n {\frac{{(k + 2)2^k - 2^{k + 1} }}{{(k + 2)!}}} = \sum\limits_{k = 1}^n {\frac{{2^k }}{{(k + 1)!}}} - \sum\limits_{k = 1}^n {\frac{{2^{k + 1} }}{{(k + 2)!}}} \\ & = \sum\limits_{k = 1}^n {\frac{{2^k }}{{(k + 1)!}}} - \sum\limits_{k = 2}^{n + 1} {\frac{{2^k }}{{(k + 1)!}}} = 1 - \frac{{2^{n + 1} }}{{(n + 2)!}}.
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Calculating square roots to 1 decimal place Square roots to 1 d.p.
Hi I'm a trainee teacher with a background in engineering.
As part of the UK GCSE mathematics syllabus, students are expected to calculate, without a calculator the square root of a number to 1 d.p. The method expected is shown below, with the example of $$ Find\;\sqrt{32}\;to\;1\;d.p. $$
$$
\sqrt{25}<\sqrt{32}<\sqrt{36} \\
5<\sqrt{32}<6 \\
\text {Try}\;5.6^2 = 31.36 \\
\therefore\;5.6<\sqrt{32}<6 \\
\text {Try}\;5.7^2 = 32.49\\
\therefore 5.6<\sqrt{32}<5.7\\
\text {Consider midpoint};5.65^2=31.9225\\
\therefore\;5.65<\sqrt{32}<5.7\\
\therefore\sqrt{32}=5.7\;\text {to 1 d.p.}
$$
The last step is the sections that is puzzling me, if we can see that 32.49 is closer to 32 than 31.36, can we not say our answer is 5.7 without having to create an inequality. My mentor has said that its because x^2 is not linear so although 5 is half way between 1 and 9, the square root is 5 is not halfway between the square roots of 1 and 9, which I do understand. But this is not the same as the problems, because you would say
$$
\sqrt{4}<\sqrt{5}<\sqrt{9}\\
2<\sqrt{5}<3\\
\text {as 4 is closer to 5 then } \sqrt{5} \text { is closer to }\sqrt{4}=2\\
\sqrt{5}=2.2360...\\
\text {so the above is true.}
$$
I'm struggling to see this in context as to why it's not true but cannot figure it out.
| Try this problem: Find $\sqrt{31.924}$ to one decimal place.
$$
\sqrt{25} < \sqrt{31.924} < \sqrt{36} \\
5 < \sqrt{31.924} < 6 \\
\text{Try } 5.6^2 = 31.36 \\
5.6 < \sqrt{31.924} < 6 \\
\text{Try } 5.7^2 = 32.49\\
5.6 < \sqrt{31.924} < 5.7
$$
And now we observe that $31.924 - 31.36 = 0.564$ while $32.49 - 31.924 = 0.566$.
Therefore $31.924$ is closer to $5.6^2$ than to $5.7^2$.
But $\sqrt{31.924} \approx 5.65013$ rounded to one decimal place is $5.7.$
Here's an interesting twist, however: working with decimal numbers, the input number $x$ (that you are taking a square root of) must have at least three decimal places in order to set up an example in which $x$ is closer to $a^2$ than to $b^2$ (where $a$ and $b$ are two consecutive one-decimal-digit numbers) and yet $\sqrt{x}$ is closer to $b$ than to $a.$
Let $a = n/10,$ where $n$ is an integer, and let $b = a + 0.1 = (n+1)/10.$
Then $a^2 = n^2/100$ and
$$b^2 = \frac{(n+1)^2}{100} = a^2 + \frac{2n+1}{100}.$$
Consider all the two-place decimal numbers between $a^2$ and $b^2$;
all such numbers up to $a^2 + \frac{n}{100}$ are closer to $a^2,$
and all such numbers from $a^2 + \frac{n+1}{100}$ upwards are closer to $b^2.$
But also
$$ (a + 0.05)^2 = \frac{\left(n+ \frac12\right)^2}{100}
= \frac{n^2 + n + \frac14}{100} = a^2 + \frac{n + \frac14}{100}.$$
So the square roots of all numbers from $a^2$ to $a^2 + \frac{n}{100}$
are closer to $a$ than to $b$ and
the square roots of all numbers from $a^2 + \frac{n+1}{100}$ to $b^2$
are closer to $b$ than to $a.$
The only case in which it is incorrect to use the "closest square" method to decide which way to round $\sqrt{x}$ is when
$$ a^2 + \frac{n + \frac14}{100} \leq x < a^2 + \frac{n + \frac12}{100}.$$
And it takes at least three decimal places to write such a number $x$ in decimal notation.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find the value of ${{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right)}$ I am trying to solve:
${\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + {{\cos }^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) + \csc{^{ - 1}}\left( {\sqrt 2 } \right)} \right)$
My solution is as follow:
$T = {\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + {{\cos }^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) + \csc{^{ - 1}}\left( {\sqrt 2 } \right)} \right) $
Since:
$\csc{^{ - 1}}\left( {\sqrt 2 } \right) = {\sin ^{ - 1}}\left( {\frac{1}{{\sqrt 2 }}} \right) = \frac{\pi }{4};{\cos ^{ - 1}}\left( {\frac{{\sqrt {12} }}{4}} \right) = {\cos ^{ - 1}}\left( {\frac{{\sqrt 3 }}{2}} \right) = \frac{\pi }{6}$
Then:
$T = {\sin ^{ - 1}}\cot \left( {{{\cos }^{ - 1}}\left( {\sqrt {\frac{{2 + \sqrt 3 }}{4}} } \right) + \frac{\pi }{4} + \frac{\pi }{6}} \right)$
I am not able to proceed further.
| Hint:
$$y=\dfrac{2+\sqrt3}4=\dfrac{(\sqrt3+1)^2}8$$
$$\sqrt y=\dfrac{\sqrt3+1}{2\sqrt2}=\cos\dfrac\pi6\cos\dfrac\pi4+\sin\dfrac\pi6\sin\dfrac\pi4=\cos\left(\dfrac\pi4-\dfrac\pi6\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4312851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Calculating the Integral for the Brachistochrone Problem I am trying to prove that the shortest possible time of descent in the Brachistochrone problem is:
$$ T = \sqrt\frac{c_1}{2g} \theta_1 $$
I have the following 2 parametric equations:
$$ x = \frac{c_1}{2} (\theta - \sin\theta) \qquad y = \frac{c_1}{2} (1 - \cos\theta) $$
Solving $$ \frac{dx}{d\theta} = \frac{c_1}{2} (1-\cos\theta) \qquad \frac{dy}{d\theta} = \frac{c_1}{2} (\sin\theta) $$
I know that $$ T = \int_0^{x_1} \frac{\sqrt{1 + (y')^2}}{\sqrt{2gy}} dx \tag{1} $$
Now solve $$y' = \frac{dy}{dx} = \frac{dy / d\theta}{ dx / d\theta } = \frac{\sin\theta}{1-\cos\theta} $$
Therefore $$ (y')^2 = \frac{\sin^2\theta}{(1-\cos\theta)^2} = \frac{1+\cos\theta}{1-\cos\theta} $$
Now plugging everything back into \ref{1} and putting the integral in terms of $\theta$:
$$T = \int_0^{\theta_1} \frac{\sqrt{1 + \frac{1+\cos\theta}{1-\cos\theta}}}{\sqrt{2g\frac{c_1}{2} (1 - \cos\theta)}} \frac{c_1}{2} (1-\cos\theta) d\theta$$
I tried solving this integral but I am having a hard time with it. I would appreciate some guidance on how to tackle this integral.
| I did not check your calculations, but there are a few things which make the problem simpler
$$\sqrt{1+\frac{1+\cos (\theta)}{1-\cos (\theta)}}=\csc \left(\frac{\theta}{2}\right)$$ Simplifying,we have
$$I=\frac 12 \sqrt{\frac{c_1}{g}}\int \csc \left(\frac{\theta}{2}\right)\sqrt{1-\cos (\theta)} \,d\theta=\frac 12 \sqrt{\frac{c_1}{g}}\int \csc \left(\frac{\theta}{2}\right) \sqrt{2 \sin ^2\left(\frac{\theta}{2}\right) }\,d\theta$$
$$I=\sqrt{\frac{c}{2 g}} \int d\theta=\sqrt{\frac{c}{2 g}}\, \theta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Finding $\int \frac{d x}{x+\sqrt{1-x^{2}}}$.
I have to calculate the following integral:
$$
\int \frac{d x}{x+\sqrt{1-x^{2}}}
$$
An attempt:$$
\begin{aligned}
\int \frac{d x}{x+\sqrt{1-x^{2}}} & \stackrel{x=\sin t}{=} \int \frac{\cos t}{\sin t+\cos t} d t \\
&=\int \frac{\cos t(\cos t-\sin t)}{\cos 2 t} d t
\end{aligned}
$$
I find the solution is
$$\frac{\ln{\left(x + \sqrt{1 - x^{2}} \right)}}{2} + \frac{\sin^{-1}{\left(x \right)}}{2}+C$$
How can I get this without trigonometric substitution?
| Well, i was try with this
$\int \frac{dx}{x +\sqrt{1-x^2}}=\int \frac{x-\sqrt{1-x^2}}{(x +\sqrt{1-x^2})(x -\sqrt{1-x^2})}= \int \frac{x-\sqrt{1-x^2}}{x^2-(1-x^2)}= \int \frac{x-\sqrt{1-x^2}}{2x^2-1}= \int \frac{x}{2x^2-1} -\int \frac{\sqrt{1-x^2}}{2x^2-1}$.
For the first integral we use $u=2x^2-1$, $du=4xdx$ and so on...
I believe that for the second integral we can use $t=\sqrt{1-x^2}$ then $x^2=1-t^2$, $dt=-\frac{\sqrt{1-t^2}}{t}dx$, so
$-\int \frac{\sqrt{1-x^2}}{2x^2-1}=-\int \frac{t}{2(1-t^2)-1}(\frac{-t}{\sqrt{1-t^2}})dt=\int \frac{t^2}{(1-2t^2)\sqrt{1-t^2}}= \frac{-1}{2}\int \frac{-2t^2}{(1-2t^2)\sqrt{1-t^2}}=\frac{-1}{2}\int \frac{1-2t^2 -1}{(1-2t^2)\sqrt{1-t^2}}$
$=\frac{-1}{2}\int \frac{dt}{\sqrt{1-t^2}}+\frac{1}{2}\int \frac{dt}{(1-2t^2) \sqrt{1-t^2}}=\frac{-1}{2}\int \frac{dt}{\sqrt{1-t^2}}+\frac{1}{2}\int \frac{dt}{(1-t^2-t^2) \sqrt{1-t^2}}=\frac{-1}{2}\int \frac{dt}{\sqrt{1-t^2}}+\frac{-1}{2}\int \frac{dt}{(1-t^2)^{\frac{3}{2}}-t^2\sqrt{1-t^2}}$
and, again for $\int \frac{dt}{(1-t^2)^{\frac{3}{2}}-t^2\sqrt{1-t^2}}$
If we put $u=\sqrt{1-t^2}$ then $u^3=(1-t^2)^{\frac{3}{2}} $, $t^2=1-u^2$ and $\frac{-udu}{\sqrt{1-u^2}}=dt$ implies
$\int \frac{dt}{(1-t^2)^{\frac{3}{2}}-t^2\sqrt{1-t^2}}= \int \frac{-udu}{\sqrt{1-u^2}(u^3- (1-u^2)u)}=\int \frac{du}{(1-u^2)^{\frac{3}{2}}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4316023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$PQ ∥ BC$ for isosceles $\triangle ABC$ and inscribed equilateral $\triangle PQR$ with $R$ being midpoint of $BC$
Triangle $ABC$ is isosceles. An equilateral triangle $PQR$ is inscribed in it with $R$ being the midpoint of $BC$. How can you prove $PQ \parallel BC$?
| Please see the below diagram which gives a counter-example for obtuse angled isosceles triangle ($120$-$30$-$30$) as mentioned by John Omielan.
For $\angle BRP = 60^\circ + x, \angle CRQ = 60^\circ - x$ or vice versa with $0 \lt x \lt 30^\circ$ will give us points $P$ and $Q$ on sides $AB$ and $AC$ such that $\triangle PQR$ is equilateral but $PQ$ is not parallel to $BC$.
By law of sines, we can show that $120$-$30$-$30$ is the only isosceles triangle for which $PQ$ is not necessarily parallel to $BC$.
Say $\angle B = \angle C = y$ and $\angle BRP = 60^\circ + x, \angle CRQ = 60^\circ - x$
By law of sines in $\triangle BPR$,
$ \displaystyle \frac{\sin (180^\circ - (60^\circ + x+y))}{BR} = \frac{\sin y}{PR} \tag1$
By law of sines in $\triangle CQR$,
$ \displaystyle \frac{\sin (180^\circ - (60^\circ - x + y))}{CR} = \frac{\sin y}{QR} \tag2$
As $BR = CR$ and $PR = QR$, from $(1)$ and $(2)$ we obtain
$\sin (60^\circ - x + y) = \sin (60^\circ + x + y)$
So we either have $60^\circ - x + y = 60^\circ + x + y ~$ i.e. $ ~x = 0$. That leads to $\angle BRP = \angle CRQ = 60^\circ ~$ and $ ~PQ \parallel BC$.
Or we have,
$(60^\circ - x + y) + (60^\circ + x + y) = 180^\circ \implies y = 30^\circ$ and $\triangle ABC$ is $120$-$30$-$30$ triangle. In this case, it is not necessary that $ \angle BRP = \angle CRQ$. I have demonstrated this case in the first part of my answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4317359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving function I tried to prove $f(n) = 4^n + 5n^2 \log n$ is not $O(2^n ).$ by using contradiction.
$4^n + 5n^2 \log n \le C \cdot 2^n$ for $n\ge k$
Then, divide by $2^n$
$2^n \le C - \frac{(5n^2 \cdot \log n)}{2^n}$ for $n\ge k$
but I stuck here. How can I move?
| You can prove that $f(n) \notin O(g(n))$ by saying that $f(n) \in \omega(g(n))$.
To do this we will have to find $\omega$ by using its definition which is stated as: $\omega(g(n))$ is found when $C >0, \exists k, k\ge 1$ such that $0 \le f(n) > C \cdot g(n), \forall n \ge k$. In English we need to show all of $C$ and $k$ are not in $O(g(n))$.
We can do this by the following:
$f(n) < C \cdot g(n) \space \forall n \ge k \\
4^n + 5 n^2\cdot \log n > C \cdot 2^n \space \forall n \ge k\\
\text{We see that there is an inequality so we can say,} \\
4^n + 5n^2\log n > 4^n + 5n^2 > 4^n > C\cdot 2^n \\
\text{Then, } 4^n > C \cdot 2^n \to 2^n > C \to \log_2(2^n) > \log_2(C) \to n > \log_2(C) \\
\text{We have shown that we have a $n$ and now we will choose a k that will be greater than our $C$.} \\
\text{Say, } k = \log_2(C)+1, \text{then } n > \log_2(C) \space \forall n \ge \log_2(C) +1 \\
\therefore 4^n + 5n^2 \cdot \log n \in \omega(2^n) \text{ meaning } 4^n + 5^n \cdot \log n \notin O(2^n).$
| {
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"url": "https://math.stackexchange.com/questions/4319530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding the range of $y =\frac{x^2+2x+4}{2x^2+4x+9}$ (and $y=\frac{\text{quadratic}}{\text{quadratic}}$ in general) I had this problem in an exam I recently appeared for:
Find the range of
$$y =\frac{x^2+2x+4}{2x^2+4x+9}$$
By randomly assuming the value of $x$, I got the lower range of this expression as $3/7$. But for upper limit, I ran short of time to compute the value of it and hence couldn't solve this question.
Now, I do know that one way to solve this expression to get its range is to assume the whole expression as equals to K, get a quadratic in K, and find the maximum/minimum value of K which will in turn be the range of that expression. I was short on time so avoided this long winded method.
Another guy I met outside the exam center, told me he used an approach of $x$ tending to infinity in both cases and got the maximum value of this expression as $1/2$. But before I could ask him to explain more on this method, he had to leave for his work.
So, will someone please throw some light on this method of $x$ tending to infinity to get range, and how it works. And if there exists any other efficient, and quicker method to find range of a function defined in the form of a ( quadratic / quadratic ).
| In general, if $\deg f = 0$ where $$f(x) = \frac{a_nx^n + a_{n - 1}x^{n - 1} + \cdots + a_1x + a_0}{b_nx^n + b_{n - 1}x^{n - 1} + \cdots + b_1x + b_0},$$ the limit of $f$ as $x$ increases/decreases without bound is $a_n/b_n$.
In your case, $a_2 = 1$ and $b_2 = 2$. Hence, $a_2/b_2 = 1/2$.
We'll factor $f$ as $$\frac{x^2+2x+4}{2x^2+4x+9} = \frac{(x + 1)^2 + 3}{2(x + 1)^2 + 7}.$$
Notice that for all $x \in \mathbb{R}$, $f > 0$. Also, we can see that $(x+1)^2 + 2 < 2(x + 1)^2 + 7$. This means that the range should be a part of $(0,1/2)$. Since both numerator and denominator have $(x + 1)^2$ without any remaining $x$'s, we can see that this will be at its minimum when $x = -1$. Then, $$f(-1) = \frac{(-1 + 1)^2 + 3}{2(-1 + 1)^2 + 7} \\ = \frac{(0)^2 + 3}{2(0)^2 + 7} \\ \frac{3}{7}$$
Therefore, the range is $[3/7, 1/2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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How to convert a quadratic solution to an unusual format I'm looking at old past papers and found this question:
"Solve the quadratic equation $3x^2 + 4x - 5$ giving your answer in the form $\frac{a}{b\pm\sqrt{19}}$, where $a$ and $b$ are integers."
I've never seen a quadratic solution in this form, with the surd root on the bottom. Does anybody have any hints on how to rearrange into this format?
Edit: For clarity, I've got $x = \frac{-2 \pm \sqrt{19}}{3}$, I just can't figure out how to convert that answer into the form they want.
| Another approach would be to apply Viete's relations: the sum of the roots of $ \ 3x^2 + 4x - 5 \ = \ 0 \ $ is
$$ \frac43 \ \ = \ \ - \left(\frac{a}{b + \sqrt{19}} \ + \ \frac{a}{b - \sqrt{19}} \right) \ \ = \ \ -\frac{2ab}{b^2 - 19} $$
and the product of the roots is
$$ \frac{-5}{3} \ \ = \ \ \frac{a}{b + \sqrt{19}} \ · \ \frac{a}{b - \sqrt{19}} \ \ = \ \ \frac{a^2}{b^2 - 19} \ \ . $$
The ratio of these equations produces
$$ \frac{4/3}{-5/3} \ \ = \ \ -\frac45 \ \ = \ \ \frac{-2ab}{a^2} \ \ = \ \ -\frac{2b}{a} \ \ \Rightarrow \ \ a \ = \ \frac52 · b \ \ . $$
Inserting this into, say, the first equation will lead to $ \ 19b^2 \ = \ 76 \ \ ; $ using the second equation yields a similar result. It will suffice to use the positive value for $ \ b \ \ , $ as using the negative value just gives an equivalent pair of roots in $ \ \frac{a}{b \ \pm \ \sqrt{19}} \ \ . $
| {
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"timestamp": "2023-03-29T00:00:00",
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How to solve this system of equations by hand: $(x-6)^2 + y^2 = 50, x^2 + (y+2)^2 = 50$ I'm having trouble solving the following system of equations by hand:
$$(x-6)^2 + y^2 = 50 \\x^2 + (y+2)^2 = 50$$
I've tried expanding and removing the square terms, but then I'm left with 2 unknown linear terms and only 1 equation. I've also tried substituting $x^2 = 50 - (y+2)^2$ into the first equation, but the result is
EDIT: I got: $y^2 + 2y -35 = 0$ which yields the correct answer. I just made an arithmetic error. Thanks!
I also looked up how to solve for the intersection of circles because that's what these equations remind me of, but this reference (https://mathworld.wolfram.com/Circle-CircleIntersection.html) assumes that one of them is centred at $(0, 0)$. I had an idea to substitute $u = y+2$, but that just moved the linear terms to another spot in the problem.
Any hints are appreciated to nudge me in the right direction. Thanks in advance!
| Since the circles have the same radius, their intersection points will lie on the perpendicular bisector of the segment connecting the centers of the circles. That midpoint is $ \ \left(\frac{6 \ + \ 0}{2} \ , \ \frac{0 \ + \ [-2]}{2} \right) \ = \ (3 \ , \ -1) \ \ . $ The slope of the line segment is $ \ \frac{0 \ - \ [-2]}{6 \ - \ 0} \ = \ \frac13 \ \ , $ so the slope of the perpendicular bisector is $ \ -3 \ \ , $ making its equation
$$ y \ - \ [-1] \ = \ -3·(x \ - \ 3) \ \ \Rightarrow \ \ y \ = \ -3x \ + \ 8 \ \ . $$
Inserting this result into either circle equation produces
$$ ( x - 6)^2 \ + \ (-3x + 8 )^2 \ = \ 50 \ \ or \ \ x^2 \ + \ (-3x + 10 )^2 \ = \ 50 $$ $$\Rightarrow \ \ 10x^2 \ - \ 60x \ + \ 50 \ = \ 0 \ \ \rightarrow \ \ x^2 \ - \ 6x \ + \ 5 \ = \ 0 \ \ . $$
This factors pretty easily to yield the $ \ x-$coordinates for the intersections; using those in the bisector line equation then gives us the $ \ y-$ coordinates. (Both intersection points have integer coordinates.)
$ \ \ $
ADDENDUM (12/5) --
What if the two radii are unequal? The intersection points of the two circles still lie on a line perpendicular to the segment connecting their centers, but that line does not bisect the segment. If we call the larger circle's radius $ \ R \ \ , $ the smaller radius $ \ r \ \ , $ and the distance between the centers $ \ D \ \ , $ then the triangle formed by those two centers and one of the intersection points resembles the following:
In this diagram, $ \ h \ $ is the distance of an intersection point from the center-linking segment; we wish to determine the point where the intersection-linking line meets this segment, with $ \ x \ $ being the distance of that point from the center of the smaller circle. Because the triangle is divided into two smaller right triangles, we can write
$$ h^2 \ \ = \ \ r^2 \ - \ x^2 \ \ = \ \ R^2 \ - \ (D - x)^2 \ \ \Rightarrow \ \ r^2 \ - \ x^2 \ \ = \ \ R^2 \ - \ D^2 \ + \ 2Dx \ - \ x^2 $$
$$ \Rightarrow \ \ x \ \ = \ \ \frac{D^2 \ - \ (R^2 \ - \ r^2)}{2D} \ \ \ \text{or} \ \ \ \frac{x}{D} \ \ = \ \ \frac{D^2 \ - \ (R^2 \ - \ r^2)}{2D^2} \ \ . $$
[We note that this confirms the earlier remark that the line of intersections bisects the center-linking segment when the two circles are of equal radius.]
As an illustration for this method, we find the intersections of the two circles $ \ (x-4)^2 + y^2 \ = \ 36 \ \ $ and $ \ x^2 + (y+3)^2 \ = \ 25 \ \ . $ As the centers of the circles are $ \ (4 \ , \ 0 ) \ $ and $ \ (0 \ , \ -3) \ \ , $ the distance separating them is $ \ D = 5 \ \ . $ The relation we derived gives us
$$ \frac{x}{D} \ \ = \ \ \frac{25 \ - \ (36 \ - \ 25)}{2 \ · \ 25} \ \ = \ \ \frac{25 \ - \ 11}{50} \ \ = \ \ \frac{7}{25} \ \ . $$
So the intersection of the two lines of interest is $ \ \frac{7}{25} \ $ of the way from $ \ (0 \ , \ -3) \ $ to $ \ (4 \ , \ 0) \ \ $ or $ \ \left(\frac{18·0 \ + \ 7·4}{25} \ , \ \frac{18·[-3] \ + \ 7·0}{25} \right) \ = \ \left(\frac{28}{25} \ = \ 1.12 \ , \ -\frac{54}{25} \ = \ -2.16 \right) \ \ . $ The slope of the segment connecting the circle centers is $ \ \frac{0 \ - \ [-3]}{4 \ - \ 0} \ = \ \frac34 \ \ . $ Hence, the slope of the line of intersections is $ \ -\frac43 \ \ $ and its equation is
$$ y \ - \ \left[-\frac{54}{25} \right] \ = \ -\frac43·\left(x \ - \ \frac{28}{25} \right) \ \ \Rightarrow \ \ y \ = \ -\frac43x \ - \ \frac23 \ \ . $$
Upon inserting this relation into, say, the first line equation yields
$$ (x \ - \ 4)^2 \ + \ \left(-\frac43x \ - \ \frac23 \right)^2 \ = \ 36 \ \ \Rightarrow \ \ 25x^2 \ - \ 56x \ - \ 176 \ \ = \ \ 0 $$ $$ \Rightarrow \ \ x \ = \ \frac{28 \ \pm \ 72}{25} \ \ . $$
Using the equation of the line of intersections, we find the coordinates of the intersection points to be
$$ (4 \ , \ -6 ) \ \ \ \text{and} \ \ \ \left(-\frac{44}{25} \ = \ -1.76 \ , \ \frac{42}{25} \ = \ 1.68 \right) \ \ . $$ The graph above presents the geometrical situation.
[This is not presented as a more efficient method of solving for the circle intersections, but as an alternative way of viewing the problem.]
| {
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Using AM-GM Inequalities: For a,b,c all positive and $a+b+c=1$, prove that $ab+bc+ca\le 1/3$ Using AM-GM Inequalities.
For a,b,c all positive and $a+b+c=1$, prove that $ab+bc+ca\le 1/3$.
My attempt:
I took AM-GM of $a+b$ and $c$, and then $a$ and $b+c$ and so on.
Using that i got $\le 3/8.$
| $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)=1$
but you know from AM-GM $\frac{a^2+b^2}{2}\ge ab$
Note that,
$a^2+b^2+c^2=\frac{a^2+b^2}{2}+\frac{b^2+c^2}{2}+\frac{c^2+a^2}{2}\ge ab+bc+ca$
$\Rightarrow a^2+b^2+c^2+2(ab+bc+ca)\ge 3(ab+bc+ca)$
$\implies ab+bc+ca\le \frac13$
| {
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Number of Quadratic equation with different condition If the number of quadratic polynomials $ax^2+2bx+c$ which satisfy the following conditions:
(i) a, b, c are distinct
(ii) a, b, c $\in$ ${1, 2, 3,. 2001, 2002}$
(iii) x + 1 divides $ax^2+2bx+c$
is equal to $1000 \lambda$, then find the value of $\lambda$.
My solution is as follow
x + 1 divides hence $f(x)=ax^2+2bx+c$, hence $f(-1)=0$, therefore $2b=a+c$
As a.b.c are distinct therefore $a\ne c$
Maximum value of b is 1000 and minimum value of b is 1 and all the value increases by 1
Minimum value of a+c=2 and Maximum value of a+c=2000
It is like distribution of 2b identical balls into 2 distinct boxes so that each box has at least 1 balls ${}^{2b - 1}{C_{2 - 1}} \Rightarrow {}^{2b - 1}{C_1}$
b=1; ${}^{2 - 1}{C_{2 - 1}} = 1$
b=2; ${}^{4 - 1}{C_{2 - 1}} = 3$
b=1000, ${}^{2000 - 1}{C_{2 - 1}} = 1999$
The total number of ways=$\frac{n}{2}\left( {2a' + \left( {n - 1} \right)d'} \right) = \frac{{1000}}{2}\left( {2 + \left( {1000 - 1} \right)2} \right) = {1000^2}$
Now as a,b and c are distinct removing the following cases {1,1},{2,2},....{1000,1000}
Hence total number of cases are ${\left( {1000} \right)^2} - 1000 = 1000 \times 999, \lambda = 999$ but answer is $2002$, i cannot figure out my mistake
| (This problem actually has very little to do with quadratic polynomials...)
I guess I've been looking at a lot of arithmetic progression problems lately, so I'm currently primed to think of them when I see $ \ a + c \ = \ 2b \ \ . $ Since $ \ a \ , \ b \ , \ c \ $ are to be distinct integers, the progressions must have non-zero integer differences between their terms. Also, as MyMolecules observes, the parity of $ \ a \ $ and $ \ c \ $ must be the same.
If we start with ascending progressions, those with initial term $ \ 1 \ $ are $ \ 1 , 2 , 3 \ \ , \ \ 1 , 3 , 5 \ \ , \ \ldots \ , $ $ 1 , 1001 , 2001 \ \ , \ \ $ which number $ \ 3 + (n - 1)·2 \ = \ 2001 \ \Rightarrow \ n \ = \ 1000 \ \ . $ Those with initial term $ \ 3 \ \ -- \ \ 3 , 4 , 5 \ \ , \ \ 3 , 5 , 7 \ \ , \ \ldots \ , \ \ 3 , 1002 , 2001 \ \ -- \ $ number $ \ 999 \ \ , $ and so forth up to $ \ 1999 , 2000 , 2001 \ \ . $ So there are $ \ 1000 + 999 + 998 + \ \ldots \ + 1 \ = \ \frac{1000·1001}{2} \ \ $ of these. The ascending progressions with even initial terms are $$ 2 , 3 , 4 \ \ , \ \ 2 , 4 , 6 \ \ , \ \ldots \ , \ \ 2 , 1002 , 2002 \ \ , \ \ 4 , 5 , 6 , \ \ldots \ , \ \ 4 , 1003 , 2002 , \ \ldots \ , \ \ 2000 , 2001 , 2002 \ \ . $$
There are also $ \ \frac{1000·1001}{2} \ $ of these.
We form descending progressions by reversing the order of $ \ a \ $ and $ \ c \ \ , $ so the total number of these is the same as the number of ascending progressions. Hence, the number of sets of quadratic coefficients satisfying the given conditions is $ \ 2 · 2 · \frac{1000·1001}{2} \ = \ 2002·1000 \ \ , $ or $ \ \mathbf{\lambda \ = \ 2002 } \ \ . $
| {
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Find the number of solutions to $n_1 +n_2 + n_3 + n_4 = 12$ when values of $n_k$ have restrictions Find the number of solutions in integers to
$$n_1 + n_2 + n_3 + n_4 = 12$$
satisfying $0 \le n_1 \le 4, 0 \le n_2 \le 5, 0 \le n_3 \le 8, \text{and } 0\le n_4 \le 9$
I'm having trouble with this. I can see that $C(12 + 4 - 1, 12)$ is the solution when the only restriction is $n_k \ge 0$. However, I'm unsure how to get to (the textbook solution):
$C(12+4-1,12) - [C(7+4-1,7)+C(6+4 -1, 6)+C(3+4-1,3)+C(2+4-1,2)-C(1+4-1,1)]$
| Let $A$ denote nonnegative solutions to $n_1+n_2+n_3+n_4=12$ and $A_i$ denote those solutions where the upper bound on $n_i$ is violated, then inclusion/exclusion gives
$$|\overline{A_1\cup A_2\cup A_3\cup A_4}|=|A|-(|A_1|+|A_2|+|A_3|+|A_4|)+(|A_1\cap A_2|+\cdots)-\cdots\\
=\binom{12+4-1}{12}-\left(\binom{7+4-1}7+\binom{6+4-1}6+\binom{3+4-1}3+\binom{2+4-1}2\right)+\binom{1+4-1}1$$
where e.g. $|A_1|=\binom{7+4-1}7$ because the solutions there can be recast as solutions to $(m_1+5)+n_2+n_3+n_4=12$ with $m_1,n_i\ge0$ or $m_1+n_2+n_3+n_4=7$, and omitted terms are because they are empty (i.e. they reduce to a sum of nonnegative integers to a negative integer).
| {
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Finding the sum of $1+4k+9k^2+...+n^2k^{n-1}$ I'm having trouble using Abel's summation formula as $a_1b_1+a_2b_2+...+a_nb_n=(a_1-a_2)(b_1)+(a_2-a_3)(b_1+b_2)+...+(a_{n-1}-a_n)(b_1+b_2+...+b_{n-1})+a_n(b_1+b_2+...+b_n)$
to find the sum of $1+4k+9k^2+...+n^2k^{n-1}$.
I know $1+2k+3k^2+...+nk^{n-1}=\frac{nk^n}{k-1}-\frac{k^n-1}{(k-1)^2}$ from a previous part.
Applying Abel's formula once I get $(-3)(1)+(-5)(1+k)+...-(2n-1)(1+k+k^2+...+k^{n-2})+n^2(1+k+k^2+...+k^{n-1})$
Then I think the next step is to apply it again to get $-2(1)-2(1+1+k)-...-2(1+(1+k)+(1+k+k^2)+...+(1+k+k^2+...+k^{n-3}))+(2n-1)(1+(1+k)+...+(1+k+...+k^{n-2}))+n^2\left(\frac{k^n-1}{k-1}\right)$
$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+(2n-1)((n-1)+(n-2)k+...+(n-(n-1))k^{n-2})+n^2\left(\frac{k^n-1}{k-1}\right)$
$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+(2n-1)\left(n\left(\frac{k^{n-1}-1}{k-1}\right)-\frac{(n-1)k^{n-1}}{k-1}+\frac{k^{n-1}-1}{(k-1)^2}\right)+n^2\left(\frac{k^n-1}{k-1}\right)$
$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+(2n-1)\left(\frac{k^{n-1}-n}{k-1}+\frac{k^{n-1}-1}{(k-1)^2}\right)+n^2\left(\frac{k^n-1}{k-1}\right)$
$=-2[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]+\frac{2n-1}{(k-1)^2}\left(k^n-nk+n-1\right)+n^2\left(\frac{k^n-1}{k-1}\right)$
My problem arises when trying to find $[1+(1+1+k)+(1+1+k+1+k+k^2)+...+(1+(1+k)+...+(1+k+..+k^{n-3}))]$, it feels way too contrived and the sums within a sum thing is confusing me, I feel like I'm not taking the right approach, if anyone has a faster way using this formula that would be greatly appreciated.
By the way the solution is $\frac{n^2k^n}{k-1}-\frac{(2n-1)k^n+1}{(k-1)^2}+\frac{2k^n-2}{(k-1)^3}$
| Hint
In one hand,
$$S_n(x)=\sum_{k=1}^nkx^k=x\cdot \frac{d}{dx}\sum_{k=0}^nx^k.$$
Finally,
$$\text{Your sum}=\frac{d}{dx}S_n(x).$$
| {
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Find the centre and radius of $\arg{\left(\frac{z-(5+7i)}{z-(7+9i)}\right)}=\frac{\pi}{4}$? I'm having huge trouble trying to find the centre and radius of the circle that contains the segment $\arg{\left(\frac{z-(5+7i)}{z-(7+9i)}\right)}=\frac{\pi}{4}$
At first I tried to solve for the intersection of the two lines running from the centre to $(5,7)$ and $(7,9)$. That is, I tried to solve for the intersection of $(y-9)=m_1(x-7)$ and $(y-7)=m_2(x-5)$, but there are too many unknown variables so that didn't work.
Next I tried to find the cartesian equation to $\arg{\left(\frac{z-(5+7i)}{z-(7+9i)}\right)}=\frac{\pi}{4}$ by letting $z=x+iy$.
After substituting that in and realising the denominator I got $\arg{\left(\frac{(x-5)(x-7)-(y-7)(y-9)}{(x-7)^2+(y-9)^2}+i\frac{(x-5)(y-9)+(y-7)(x-7)}{(x-7)^2+(y-9)^2}\right)}=\frac{\pi}{4}$
Which I think then means $\tan^{-1}{\left(\frac{(x-5)(y-9)+(y-7)(x-7)}{(x-5)(x-7)-(y-7)(y-9)}\right)}=\frac{\pi}{4}$
Which gave the curve $(x-5)(y-9)+(y-7)(x-7)=(x-5)(x-7)-(y-7)(y-9)$, however when I plugged that into desmos I got a hyperbola with the minimum of the upper branch being $(7,9)$ and the max of the lower branch being $(5,7)$. So I've clearly done something wrong when finding the cartesian equation.
Nevertheless does anyone know the most efficient way to find the centre and radius? I'm not well versed in circle geometry theorems so maybe there is something there I have missed?
| Your mistake is in simplifying the contents of the arg bracket.
The real part is $$\frac{(x-5)(x-7)+(y-7)(y-9)}{(x-7)^2+(y-9)^2}$$ and the imaginary part is $$\frac{(y-7)(x-7)\color{red}{-}(x-5)(y-9)}{(x-7)^2+(y-9)^2}$$
Since the arg is $\frac{\pi}{4}$ we can set these equal to each other and this leads to the equation of the circle $$(x-7)^2+(y-7)^2=4,$$
with the obvious conclusion.
The result can be arrived at very easily just by drawing a picture and no algebra (as has been indicated by others)
| {
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prove: $15(a+b)\ge17+14\sqrt{2ab}$
Let $a,b\ge0: a^4+b^4=17$. Prove that: $$15(a+b)\ge17+14\sqrt{2ab}$$
I am looking for a nice approach. My approach is ugly by replace: $b=\sqrt[4]{17-a^4} $ and the rest is working with fuction.
I am quite sure $(1,2)$ is the only case equality so I guess we can use AM-GM in someway. Is there any better idea for this problem?
Thanks for your help!
| Marginally less ugly, let $s=a+b, p=ab$, then:
$$
a^4+b^4 = 17 \;\;\iff\;\; (s^2-2p)^2-2p^2=17 \;\;\iff s^4 - 4ps^2 + 2p^2 - 17 = 0
$$
Solving the quadratic in $\,p\,$, and retaining the root which satisfies $p \le s^2$:
$$
p = s^2 - \sqrt{\frac{s^4+17}{2}}
$$
The inequality to prove is equivalent to:
$$
(15s-17)^2 \ge 2\cdot 14^2 \,p = 14^2\left(2s^2-\sqrt{2\left(s^4+17\right)}\right)
$$
Rearranging with positive quantities on both sides and squaring:
$$
2 \cdot 14^4 \left(s^4+17\right) \ge \left(2 \cdot 14^2 s^2 - (15s-17)^2\right)^2
$$
After expanding, collecting and "luckily" finding the rational root $s=3$:
$$
17 (2879 s^4 - 10020 s^3 - 9622 s^2 + 17340 s + 71919) \ge 0
\\ \iff\;\;\;\; (s - 3)^2 (2879 s^2 + 7254 s + 7991) \ge 0
$$
The quadratic factor has no real roots, so the inequality holds true, with equality iff $s=3\,$, which then gives $p=2$ i.e. $\{a,b\}=\{1,2\}$.
| {
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Suppose $f(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n$. Then $\Delta^nf(x)=n!h^n$ and $\Delta^{n+r}f(x)=0,$ for $r=1,2,3,4,5... \infty$. Suppose $f(x)=x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n$. Then $\Delta^nf(x)=n!h^n$ and $\Delta^{n+r}f(x)=0,$ for $r=1,2,3,4,5... \infty$.(Where we have equally spaced points $x,x+h,...,x+nh$ with corresponding values $f(x),f(x+h),...f(x+nh)$. Forward difference operator is defined as $\Delta f(x)=f(x+h)-f(x)$)
My attempt:-
When $f(x)=x,f(x)=x^2+a_1x+a_2$ , I could prove, How do I prove generally?
Is the argument true?
$\Delta f(x)=f(x+h)-f(x)=(x+h)^n+a_1(x+h)^{n-1}+a_2(x+h)^{n-2}+...+a_{n-1}(x+h)+a_n-(x^n+a_1x^{n-1}+a_2x^{n-2}+...+a_{n-1}x+a_n).$ By binomial expansion, we get a polynomial of degree $n-1$. Following the step n times, we landed upon $\Delta^nf(x)=$constant. But I am not able to prove it is $\Delta^nf(x)=n!h^n$.
| Let $e_n(x) = x^n$. We can write a polynomial $p(x) = \sum_{k=0}^n a_k x^k$ as $p = \sum_{k=0}^n a_k e_k$
Note that $\Delta^r$ is a linear operator, so
$\Delta^r p = \sum_{k=0}^n a_k \Delta^r e_k$, in particular, we
need only focus on computing $\Delta^n e_m$ for $m =n,n+1,...$.
Note that $(\Delta^1 e_1 )(x) = h$ and $\Delta e_0 = 0$ (that is $\Delta $ applied to constants results in zero).
So, suppose that for $k=1,...,n$ that $(\Delta^n e_n)(x) = n! h^n$
and $\Delta^n e_r = 0$ for $r=1,...,n-1$.
Suppose $n \ge 1$, then
\begin{eqnarray}
(\Delta e_{n+1}) (x) &=& \sum_{k=1}^{n+1} \binom{n+1}{k} x^{n+1-k} h^k \\
&=& \sum_{k=1}^{n+1} \binom{n+1}{k} e_{n+1-k} (x) h^k \\
&=& (n+1)h e_{n}(x)+\sum_{k=2}^{n+1} \binom{n+1}{k} e_{n+1-k} (x) h^k
\end{eqnarray}
Then $\Delta^{n+1} e_{n+1} = \Delta^{n} (\Delta e_{n+1}) = (n+1)\Delta^{n} e_n = (n+1)! h^{n+1}$.
Note that since $\Delta^n e_n$ is a constant, it follows that $\Delta^{n+r} e_n = \Delta^r \Delta^{n} e_n =0$ for $r=1,2,..$.
| {
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Solve for $x$: $\frac{x^2-10x+15}{x^2-6x+15}=\frac{3x}{x^2-8x+15}$ Solve the equation: $\dfrac{x^2-10x+15}{x^2-6x+15}=\dfrac{3x}{x^2-8x+15}$.
When $x\ne3$ and $x\ne5$ we get $$(x^2-10x+15)(x^2-8x+15)=3x(x^2-6x+15)\\(x^2-9x+15-x)(x^2-9x+15+x)-3x(x^2-6x+15)=0\\(x^2-9x+15)^2-x^2-3x(x^2-6x+15)=0.$$ I am stuck here. The Rational Root Theorem won't be useful as the equation does not have such roots.
I got $-9x$ by averaging $-10x$ and $-8x$.
I don't know if it makes sense.
| Let $y:=x^2-6x+15$ so$$\frac{y-4x}{y}=\frac{3x}{y-2x}\implies y^2-6xy+8x^2=3xy\implies (y-x)(y-8x)=0.$$That gives you two quadratics to solve.
| {
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Using the Chinese Remainder Theorem, $17x \equiv 9 \pmod{276}$ I want to uses the Chinese Remainder Theorem to solve $17x \equiv 9 \pmod{276}$ by breaking it up into a system of three linear congruences,
$$17x \equiv 9 \pmod{3}$$
$$17x \equiv 9 \pmod{4}$$
$$17x \equiv 9 \pmod{23}$$
For that I reduced it to
$$x \equiv 0 \pmod{3}$$
$$x \equiv 1 \pmod{4}$$
$$17x \equiv 9 \pmod{23}$$
So for converting this In terms of chinese reminder Theorem , I calculate The solution Of last linear Congurence as
$$x \equiv 13 \pmod{23}$$
So Our System Of Linear Congurence is now :
$$x \equiv 0 \pmod{3}$$
$$x \equiv 1 \pmod{4}$$
$$x \equiv 13 \pmod{23}$$
And now I apply the Chinese Remainder Theorem on it such that
$$92b_1 \equiv 1 \pmod{3}$$
$$69b_2 \equiv 1 \pmod{4}$$
$$12b_3 \equiv 1 \pmod{23}$$
So $b_1$ = 2 , $b_2$ = 1 , $b_3$ = 2
So simultaneous solution be
$$92\cdot2\cdot0 + 69\cdot1\cdot1 + 13\cdot2\cdot5 = 199$$
But it's wrong (@_@)༎ຶ‿༎ຶ . Can please Please Someone can Correct me.
| $17x\equiv 9\bmod 276$
$276=16\times 17 +4\Rightarrow 17x=9+(16(17)+4)k$
$\Rightarrow 17(x-16k)=9+4k$
For $k=2$ we have:
$17(x-16\times2)=9+4\times 2=17$
$\Rightarrow x-32=1\rightarrow x=33$
$k=53\rightarrow x-53\times 16=13\times 17\rightarrow x=861$
K makes an arithmetic progression with common difference $d=51$:
$k= 2, 53, 104, 155\cdot\cdot\cdot$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$\sum_{k=0}^{n-1} {n-1-k\choose k}\left(\frac{1}{2}\right)^{n-1-k}+\sum_{k=0}^{n-2} {n-2-k\choose k}\left(\frac{1}{2}\right)^{n-2-k} $ How can you find
$$
\sum_{k=0}^{n-1} {n-1-k\choose k}\left(\frac{1}{2}\right)^{n-1-k}+\sum_{k=0}^{n-2} {n-2-k\choose k}\left(\frac{1}{2}\right)^{n-2-k}
$$
?
I found the value via interpritting the above formula combinatorially. (If I am correct, it is $\frac{2}{3}+\frac{1}{3}\left(-\frac{1}{2}\right)^n$)
But I want to know how to solve it by way of complex integrals or formal power series or any algebraic manipulations.
| In trying to evaluate
$$\sum_{k=0}^{n-1} {n-1-k\choose k} \frac{1}{2^{n-1-k}}
+ \sum_{k=0}^{n-2} {n-2-k\choose k} \frac{1}{2^{n-2-k}}$$
we introduce
$$S_n = \sum_{k=0}^n {n-k\choose k} \frac{1}{2^{n-k}}$$
which is
$$\frac{1}{2^n} \sum_{k=0}^n {n-k\choose n-2k} 2^k
= \frac{1}{2^n} [z^n] (1+z)^n
\sum_{k\ge 0} \frac{z^{2k}}{(1+z)^k} 2^k.$$
Here we have extended the sum to infinity because the coefficient
extractor enforces the upper limit. Even more, it enforces $n\ge 2k.$
Continuing,
$$\frac{1}{2^n} [z^n] (1+z)^n
\frac{1}{1-2z^2/(1+z)}
\\ = \frac{1}{2^n}
\; \underset{z}{\mathrm{res}} \;
\frac{1}{z^{n+1}} (1+z)^{n+1}
\frac{1}{1+z-2z^2}.$$
Now we put $z/(1+z) = w$ so that $z = w/(1-w)$ and $dz = 1/(1-w)^2 \;
dw$ to get
$$\frac{1}{2^n}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}}
\frac{1}{1+w/(1-w)-2w^2/(1-w)^2}
\frac{1}{(1-w)^2}
\\ = \frac{1}{2^n}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}}
\frac{1}{(1-w)^2+w(1-w)-2w^2}
\\ = \frac{1}{2^n}
\; \underset{w}{\mathrm{res}} \;
\frac{1}{w^{n+1}}
\frac{1}{1-w-2w^2}.$$
We have
$$\frac{1}{1-w-2w^2}
= \frac{1}{3} \left[\frac{1}{1+w} + \frac{2}{1-2w}\right]$$
so that extracting the coefficient we find
$$ S_n = \frac{1}{2^n \times 3} [ (-1)^n + 2^{n+1} ]$$
or alternatively
$$\bbox[5px,border:2px solid #00A000]{
S_n = \frac{1}{3} \left(-\frac{1}{2}\right)^n
+ \frac{2}{3}.}$$
We also have
$$S_{n-1} + S_{n-2}
= \frac{1}{3} \left(-\frac{1}{2}\right)^{n-2}
\left(1-\frac{1}{2}\right)
+ \frac{4}{3}
= - \frac{1}{3} \left(-\frac{1}{2}\right)^{n-1}
+ \frac{4}{3} = 2 S_n.$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 3,
"answer_id": 0
} |
Maclaurin series of $(1+x^3)/(1+x^2)$ I can't seem to figure out the Maclaurin series of $(1+x^3)/(1+x^2)$
I started with $ 1/(1-x)= \sum_{n=1}^{\infty} x^n $
$ 1/(1-(-x^2)) = \sum_{n=1}^{\infty}(-1)^n x^{2n} $
$ (1+x^3) \sum_{n=1}^{\infty}(-1)^n x^{2n}= [\sum_{n=1}^{\infty}(-1)^n x^{2n}]+x^3 [\sum_{n=1}^{\infty}(-1)^n x^{2n}] $
$ [\sum_{n=1}^{\infty}(-1)^n x^{2n}]+[\sum_{n=1}^{\infty}(-1)^n x^{2n+3}] $ ....
Here I'm stuck. I don't know how to merge the two sums. Maybe I have made a mistake. When I write out the sum I get:
$ 1 -x^2 +x^3 + x^4 - x^5 - x^6 + x^7 + x^8 -x^9 - x^{10} +... $
So it seems that you need to write out a 'simple' $\sum_{n=1}^{\infty} x^n $ but I don't know how to do the start and how to distribute the minuses. Or maybe I made a mistake.
| The answer you are conjecturing can be written as
$$\frac{1+x^3}{1+x^2}=1-x^2+\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4})$$
Multiplying both sides by $1+x^2$ gives us
\begin{align*}
1+x^3&=(1+x^2)(1-x^2)+(1+x^2)\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4})\\
&=1-x^4+\sum_{n=0}^\infty (-1)^n(1+x^2)(x^{2n+3}+x^{2n+4})\\
&=1-x^4+\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4}+x^{2n+5}+x^{2n+6})
\end{align*}
The sum on the RHS is telescoping; that is, $\sum_{n=0}^\infty (-1)^n(x^{2n+3}+x^{2n+4}+x^{2n+5}+x^{2n+6})$ is equal to
\begin{align*}
(x^3+x^4+x^5+x^6)-(x^5+x^6+x^7+x^8)+(x^7+x^8+x^9+x^{10})-\dots=x^3+x^4.
\end{align*}
Therefore,
$$1+x^3=1-x^4+x^3+x^4=1+x^3.$$
You should be able to work backwards to fomulate a proof.
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
For $a, b, c$ with $a+b+c=0$ prove $\frac15\sum a^5=\frac13\sum a^3\cdot\frac12\sum a^2$ and $\frac17\sum a^7=\frac15\sum a^5\cdot\frac12\sum a^2$ Consider the following problem:
Problem. Suppose that real numbers $a$, $b$ and $c$ satisfy the condition $a+b+c=0$. Prove the following identities:
$$
\frac{a^5+b^5+c^5}{5}=\frac{a^3+b^3+c^3}{3}\cdot\frac{a^2+b^2+c^2}{2},
\\
\frac{a^7+b^7+c^7}{7}=\frac{a^5+b^5+c^5}{5}\cdot\frac{a^2+b^2+c^2}{2}.
$$
Perhaps, the shortest solution I can think of is as follows: plug $c=-a-b$ into the equation but instead of expanding everything use the following identities
$$
(a+b)^3-a^3-b^3=3ab(a+b),
\\
(a+b)^5-a^5-b^5=5ab(a+b)(a^2+ab+b^2),
\\
(a+b)^7-a^7-b^7=7ab(a+b)(a^2+ab+b^2)^2.
$$
However, these identies are coming out of nowhere and moreover, in order to prove them one still needs to do some computations.
Question. Is it possible to solve this problem in a "smart" way (i.e. avoiding computations and preferrably elementary since it is almost a high school problem)? Any other solutions are also welcome.
Comment. It should be also noted that it is unclear (at least for me) how those identies were invented. It seems there is no nice similar identities for $\frac{a^p+b^p+c^p}{p}$ for $p$ other than 2, 3, 5, 7.
| This is a problem about elementary symmetric functions. Let
$$s_1=a=b+c$$
$$s_2=ab+bc+ac$$
$$s_3=abc$$
Now assuming that $s_1=a+b+c=0$ we show that
\begin{equation*}
\begin{aligned}
-\frac{a^2+b^2+c^2}{2}&=s_2=ab+bc+ca\\
\frac{a^3+b^3+c^3}{3}&=s_3=abc\\
-\frac{a^4+b^4+c^4}{4}&=-\frac{1}{2}s_2^2=-\frac{(a^2+b^2+c^2)^2}{2}\\
\frac{a^5+b^5+c^5}{5}&=-s_2s_3\\
-\frac{a^6+b^6+c^6}{6}&=-\frac{1}{2}s_3^2+\frac{1}{3}s_2^3\\
-\frac{a^7+b^7+c^7}{7}&=s_2^2s_3\\
\end{aligned}
\end{equation*}
And the results follow.
The calculational scheme for deriving the above equalities is
to note that
$$\ln (1+ax)=ax-\frac{a^2}{2}x^2+\frac{a^3}{3}x^3+\cdots $$
and adding,
$$\ln (1+ax)(1+bx)(1+cx)=
(a+b+c)x-\frac{a^2+b^2+c^2}{2}x^2+
\frac{a^3+b^3+c^3}{3}x^3+\cdots $$
on the other hand,
$$(1+ax)(1+bx)(1+cx)= 1+s_1x+s_2x^2+s_3x^3$$
so we have under the assumption that $s_1=0$,
$$\ln (1+s_2x^2+s_3x^3)=(s_2x^2+s_3x^3)-\frac{1}{2}
(s_2x^2+s_3x^3)^2+\frac{1}{3}(s_2x^2+s_3x^3)^3\cdots $$
from which the above equalities follow easily by equating the coefficients of the two power series.
In particular this shows that every $\frac{a^n+b^n+c^n}{n}$ is a polynomial in $n=2,3$ assuming $s_1=0$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Find minimum $|z|$ satisfying $|z + 1/z |= 2$. When I tried this using normal complex inequalities like $|z_{1} - z_{2}| \ge ||z_{1}| - |z_{2}||$. $\sqrt 2 - 1$ came up but the real answer seems to be $(3 - 2\sqrt 2)^{1/2}$. Some online answers on other sites support my answer as well, but I am confused which ones correct. If the latter is correct please explain how.
| Render $\sqrt{3-2\sqrt2}=\sqrt x-\sqrt y$ for $x$ and $y$ presumed rational. Then
$3-2\sqrt2=(\sqrt x-\sqrt y)^2=(x+y)-2\sqrt{xy}$
where the quadratic surds, for rational $x$ and $y$, are equal only if the rational components and square-root components are separately equal. This leads to
$x+y=3$
$xy=2$
$x(3-x)=2, x^2-3x+2=0, x\in{1,2}$
Since $x>y$ for a positive square root $\sqrt x-\sqrt y$, we must have $x>3/2$ so we take $x=2,y=3-x=1$, and...
$\sqrt{3-2\sqrt2}=\sqrt 2-1.$
The claimed disagreement between your answers does not exist.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Showing $\frac{3\sqrt{3}}{2\pi}\sum_{z\in\Lambda}\frac1{1-\left(\frac{z}{\sqrt3}-1\right)^3}=1$, with $\Lambda$ a lattice Consider the sum
$$\frac{3 \sqrt{3}}{2\pi}\sum_{z\in \Lambda} \frac{1}{1-\left(\dfrac{z}{\sqrt{3}}-1\right)^3} \overset{?}=1$$
with $\Lambda=3\mathbb Ze^{\pi i/6}+3\mathbb Ze^{-\pi i/6}$, then it is numerically not too difficult to see that this sum is equal to $1.$ I am however looking for an analytic argument of this nice fact. I assume it must rely on some subtle symmetries?
Please let me know if you have any questions.
| Let $\Lambda' = (e^{i\pi/6}\mathbb{Z} \oplus e^{-i\pi/6}\mathbb{Z}) - \frac{1}{\sqrt{3}}$. Then by substituting $z = 3(\omega + \frac{1}{\sqrt{3}})$,
$$ S := \sum_{z \in \Lambda} \frac{1}{1-((z/\sqrt{3})-1)^3}
= \sum_{\omega \in \Lambda'} \frac{1}{1-(\sqrt{3} \omega)^3}. $$
Our goal is to show that
Claim. $\displaystyle S = \frac{2\pi}{3\sqrt{3}}$.
Now by noting that $\Lambda'$ can be decomposed into concentric "discrete regular triangles" centered at the origin, let $\Lambda'_N$ be the union of the first $N$ smallest triangles. For example, $\Lambda'_5$ is
In particular, each $\Lambda'_N$ is symmetric about a $\frac{2\pi}{3}$-rotation about the origin. Then by using the partial fraction decomposition
$$ \frac{1}{1 - w^3} = \frac{1}{3} \sum_{\xi \, : \, \xi^3 = 1} \frac{1}{1 - \xi w}, $$
we find that
\begin{align*}
S
= \lim_{N\to\infty} \sum_{\omega \in \Lambda'_N} \frac{1}{1-(\sqrt{3} \omega)^3}
= \lim_{N\to\infty} \sum_{\omega \in \Lambda'_N} \frac{1}{1-\sqrt{3}\omega}
= \frac{1}{\sqrt{3}} \lim_{N\to\infty} \sum_{\tau \in \Lambda''_N} \frac{1}{\tau},
\end{align*}
where $\tau = \frac{1}{\sqrt{3}} - \omega$ and $\Lambda''_N = \frac{1}{\sqrt{3}} - \Lambda'_N$. For example, $\Lambda''_5$ is:
Now by regrouping the dots in $\Lambda''_N$ into concentric regular triangles centered at the origin, we are left with $N-1$ triangles plus two extra "discrete lines", which we denote by $\gamma_N$. For example, the next picture demonstrates the decomposition of $\Lambda''_5$ into the triangles and $\gamma_5$:
Since the sum along each concentric triangle vanishes by symmetry, we are left with
\begin{align*}
S
= \frac{1}{\sqrt{3}} \lim_{N\to\infty} \sum_{\tau \in \gamma_N} \frac{1}{\tau}
= \frac{1}{\sqrt{3}} \lim_{N\to\infty} \sum_{\tau \in \gamma_N} \frac{1}{(\tau/N)} \cdot \frac{1}{N}
\end{align*}
Now the last one can be recognized as a Riemann sum for a contour integral. Indeed, if we set
$$ z_1 = -\frac{\sqrt{3}}{2} - \frac{3i}{2}, \qquad z_2 = \sqrt{3}, \qquad z_3 = -\frac{\sqrt{3}}{2} + \frac{3i}{2} $$
so that the polygonal line $\overline{z_1 z_2} \cup \overline{z_2 z_3}$ is the "limit of the rescaled discrete line $N^{-1}\gamma_N$" as $N\to\infty$, then
\begin{align*}
S
&= \frac{1}{\sqrt{3}} \int_{\overline{z_1 z_2} \cup \overline{z_1 z_2}} \frac{|\mathrm{d}z|}{z} \\
&= \frac{1}{\sqrt{3}} \biggl( \frac{1}{e^{i\pi/6}} \int_{[z_1, z_2]} \frac{\mathrm{d}z}{z} + \frac{1}{e^{i5\pi/6}} \int_{[z_2, z_3]} \frac{\mathrm{d}z}{z} \biggr) \\
&= \frac{1}{\sqrt{3}} \biggl( \frac{1}{e^{i\pi/6}} \cdot \frac{2\pi i}{3} + \frac{1}{e^{i5\pi/6}} \cdot \frac{2\pi i}{3} \biggr) \\
&= \frac{2\pi}{3\sqrt{3}}.
\end{align*}
Therefore the desired conclusion follows.
| {
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"source": "stackexchange",
"question_score": "15",
"answer_count": 1,
"answer_id": 0
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Find the area of the shaded region in the figure below For reference:
In the figure $O$ is the center of the circle and its radius measures $a$ and $AQ = QB$. Calculate the area of the shaded region.(Answer: $\frac{a^2}{4}(\pi-2)$)
correct figure
My progress:
If $AQ = BQ \implies \angle AQB=90^\circ$
Complete the square $AQBD$.
incorrect figure. incorrect figure, please do not consider it for any effect
$OC = r$ and $QC =R = AC.$
$O$ is centre of square.
$QO$ is angle bisector, therefore $\angle AQO$ is $45^\circ.$
$QD = R\sqrt2$
Considering $\triangle OCQ$,
$\displaystyle r^2+\left(\frac{R}{2}\right)^2=OQ^2\implies r^2+\frac{R^2}{4}=(R\sqrt2)^2$
$\therefore R = \dfrac{2r\sqrt7}{7}$
I don't see a solution...is it missing some information?
The book has another similar question but in this question $a = 2$ and answers match if we replace $a$ with $2$. Diagram below -
| Here is a solution that I have so far. It does turn out after all the lengthy work that $\angle HOF \approx 90^\circ$. So there should be a way to show $\angle HOF = 90^\circ$ when $O$ lies on $AE$. But I have not been able to see a geometric solution yet.
If $R$ is the radius of the quarter circle and radius of the smaller circle is $a$,
$AS^2 = AG \cdot AE~$ i.e. $~(R-a)^2 = (AO + a) \cdot (AO-a)$ $\implies AO^2 = R^2 - 2aR + 2a^2$
Also, $~QF \cdot QE = QT^2~$ or $~(R-FE) \cdot R = a^2 $
$\implies FE = \dfrac{R^2-a^2}{R}, ~QF = \dfrac{a^2}{R}$
If $M$ is the midpoint of $FE$, $OM \perp FE$. Notice that $\triangle OEM \sim \triangle OAS~$. So,
$\displaystyle \frac{ME^2}{OE^2} = \frac{AS^2}{AO^2} \implies \frac{(R^2-a^2)^2}{4 a^2 R^2} = \frac{(R-a)^2}{R^2-2aR+2a^2}$
Simplifying, $2a^4 + 2a^3R + R^4 - 5a^2R^2 = 0$
$(R-a) (2a^3 + 4a^2R - aR^2 - R^3) = 0$
As $R = a$ is not a solution that we are interested in, we solve $2a^3 + 4a^2R - aR^2 - R^3 = 0$. WolframAlpha gives an approximate form solution of $R \approx 1.81361 a$. We then find $AO \approx 1.28917a, QF \approx 0.551386 a$
Next, $ \displaystyle AK = AS \cdot \frac{AE}{AO} = (1.81361 a - a) \cdot \frac {AO + a}{AO} \approx 1.44472a$
So, $QK = R - AK \approx 0.36889a$
$ \displaystyle \cos \angle AQE = \frac{QK}{QE} \approx 0.2034$
Applying law of cosine, $AF^2 = AQ^2 + QF^2 - 2 AQ \cdot QF \cdot \cos \angle AQE \approx 3.18641 a^2$
$AS^2 = AH \cdot AF = (AF - HF) \cdot AF$
We obtain $~HF \approx 1.41422 a, ~$ which is approximately $a \sqrt2$. With $HF \approx a \sqrt2$ and $OF = OH = a$, we have $\angle HOF \approx 90^\circ$.
That leads to the shaded area $ \displaystyle A \approx \frac{\pi}{4} \cdot a^2 - \frac 12 \cdot a^2 = \frac {a^2}{4} (\pi - 2)$
| {
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"answer_count": 3,
"answer_id": 1
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Solve the equation $\sin x+\cos x=k \sin x \cos x$ for real $x$, where $k$ is a real constant. As I had solved the equation when $k=1$ in Quora and MSE by two methods, I started to investigate the equation for any real constant $k$:
$$
\sin x+\cos x=k \sin x \cos x,
$$
I first rewrite the equation as
$$
\sqrt{2} \cos \left(x-\frac{\pi}{4}\right)=\frac{k}{2} \sin (2 x)
$$
Letting $ \displaystyle y=x-\frac{\pi}{4}$ yields
$$
\begin{array}{l}
\sqrt{2} \cos y=\frac{k}{2}\left(2 \cos ^{2} y-1\right) \\
2 k \cos ^{2} y-2 \sqrt{2} \cos y-k=0
\end{array}
$$
When $k\neq 0$, using quadratic formula gives
$$
\cos y=\frac{1 \pm \sqrt{1+k^{2}}}{\sqrt{2} k}
$$
For real $y$, we have to restrict $\displaystyle \frac{1 \pm \sqrt{1+k^{2}}}{\sqrt{2} k}$ in $[-1,1]$. Then I found that
$$
-1 \leqslant \frac{1+\sqrt{1+k^{2}}}{\sqrt{2} k} \leqslant 1 \Leftrightarrow \quad|k| \geqslant 2 \sqrt{2}
$$ and $$
-1 \leqslant \frac{1-\sqrt{1+k^{2}}}{\sqrt{2} k} \leqslant 1 \Leftrightarrow \quad k<0 \text { or } k>0
$$
Now we can conclude that
A. When $k\neq0$
$$x=n \pi-\frac{\pi}{4}$$
B. When $0\neq|k| \geqslant 2 \sqrt{2}, $
$$x=\frac{(8 n+1) \pi}{4} \pm \arccos \left(\frac{1\pm \sqrt{1+k^{2}}}{\sqrt{2} k}\right)$$
C. When $ 0 \neq|k|<2 \sqrt{2},$
$$ x= \frac{(8 n+1) \pi}{4} \pm \arccos \left(\frac{1-\sqrt{1+k^{2}}}{\sqrt{2} k}\right)
$$
where $n\in Z.$
I am looking forward to seeing other methods to solve the equation.
Furthermore, how about $$a\sin x+b\cos x+c\sin x\cos x=0?$$
| The case of $k=0$ is elementary. Then squaring both members,
$$2\cos x\sin x+1=k^2(\cos x\sin x)^2$$ immediately gives
$$\cos x\sin x=\frac{1\pm\sqrt{k^2+1}}{k^2}$$ and
$$\cos x+\sin x=\frac{1\pm\sqrt{k^2+1}}k.$$
This is a classical sum/product problem, solved with
$$\cos x-\sin x=\pm\sqrt{(\cos x+\sin x)^2-4\cos x\sin x}\\
=\pm\frac{\sqrt{(1\pm\sqrt{k^2+1})^2-4(1\pm\sqrt{k^2+1})}}k.$$
| {
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"answer_count": 4,
"answer_id": 2
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Given $a_n = {(1+\frac{1}{n})}^{1/n}-1$, does $\sum_{n=1}^∞ a_n$ converge? I am a calculus student, And I'm trying to find out If the following series converges.
Given
$a_n = {(1+\frac{1}{n})}^{1/n}-1$
Does the following series converges?
$\sum_{n=1}^∞ a_n$
My thoughts:
It looks like $a_n$ converges to $0$ very quickly as $n$ goes to infinity, So my intuition says yes.
I've been trying to show convergence for some time now, and didn't had much progress. No candidates for comparison test, Other theorems did not work either.
Any hints/Tips for beginners?
| According to the Cauchy convergence criterion of the series:
For $ \forall \epsilon>0,\exists N>0, \forall n>N$, for $\forall p \in \mathbb N_{+}$，
\begin{align}\left|a_{n+1}+a_{n+2}+\cdots+a_{n+p}\right| &=\left|\left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}-1+\left(1+\frac{1}{n+2}\right)^{\frac{1}{n+2}}-1\right. \\ & \qquad\left. +\cdots+\left(1+\frac{1}{n+p}\right)^{\frac{1}{n+p}}-1 \right|\\ &<\varepsilon \tag{1}
\end{align}
If this condition is satisfied, then $\sum_{n=1}^{\infty} a_{n}$ converges.
*
*$\lim_{n \to \infty}a_n=0$
$$\begin{aligned} & \because \lim _{n \rightarrow \infty} a_{n}=\lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{\frac{1}{n}}-1 \\ \because & \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{\frac{1}{n}}=\lim _{n \rightarrow \infty} e^{\ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}} \\ & \lim _{n \rightarrow \infty} \ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}=\lim _{n \rightarrow \infty} \frac{1}{n} \cdot \ln \left(1+\frac{1}{n}\right) \\ &\qquad\qquad\qquad\qquad\; =\lim _{t \rightarrow 0} t \ln (1+t)=0 \\ \therefore & \lim _{n \rightarrow \infty}\left(1+\frac{1}{n}\right)^{\frac{1}{n}}=1, \lim _{n \rightarrow \infty} a_{n}=1-1=0 \end{aligned}$$
*$a_n=\left(1+\frac{1}{n}\right)^{\frac{1}{n}}$ is monotonically decreasing
Using the inequality $(2)$:
$$\quad \frac{1}{n+1}<\ln \left(1+\frac{1}{n}\right)=\ln (n+1)-\ln n<\frac{1}{n} \quad(n=1,2, \cdots) \tag{2}$$
$$\because \ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}=\frac{1}{n} \ln \left(1+\frac{1}{n}\right),$$
$$\therefore \quad \frac{1}{n(n+1)}<\ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}<\frac{1}{n^{2}}$$
$$\because \quad \ln \left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}=\frac{1}{n+1} \ln \left(1+\frac{1}{n+1}\right)=\frac{1}{n+1}[\ln (n+2)-\ln (n+1)]$$
$$\frac{1}{n+2}<\ln (n+2)-\ln (n+1)<\frac{1}{n+1}$$
$$ \therefore \quad \ln \left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}<\frac{1}{(n+1)^2}<\frac{1}{n(n+1)}<\ln \left(1+\frac{1}{n}\right)^{\frac{1}{n}}$$
So, $$\left(1+\frac{1}{n}\right)^{\frac{1}{n}}>\left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}.\tag{3}$$
*So the right end of equation $(1)$ can be deflated into：
\begin{align}\left|a_{n+1}+a_{n+2}+\cdots+a_{n+p}\right| &=\left|\left(1+\frac{1}{n+1}\right)^{\frac{1}{n+1}}-1+\left(1+\frac{1}{n+2}\right)^{\frac{1}{n+2}}-1 \right. \\ & \qquad \left. +\cdots+\left(1+\frac{1}{n+p}\right)^{\frac{1}{n+p}}-1 \right|\\ &< p \left|\left(1+\frac{1}{n}\right)^{\frac{1}{n}}-1 \right|\\&<\varepsilon \tag{4}
\end{align}
($n$ is sufficiently large to satisfy $\left(1+\frac{1}{n}\right)^{\frac{1}{n}}-1<\frac{\varepsilon}{p}$.)
Since the limit $a_n$ is convergent, equation $(4)$ holds, so equation $(1)$ holds, so the series satisfies the Cauchy convergence criterion, so the series converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to show positivity of Fresnel C integral? The Fresnel $C$-integral is defined as follows.
$$C(x) = \int_0^x \cos(t^2) \, dt $$
From the plot found on Wikipedia it seems to be non-negative for all $x \geq 0$ however it is not obvious to me why this is. For example, if you make the change of variables $t = u^\frac{1}{2}$ then you find
$$C(x) = \frac{1}{2} \int_0^{x^2} u^{-\frac{1}{2}} \cos(u) \, du$$
My first thought was to split it into positive and negative regions and make a crude bound. For example, $\cos(u)$ will be positive on $[0,\pi/2]$ and negative on $[\pi/2,3\pi/2]$. A crude lower bound on the first region would be $\left(\frac{\pi}{2}\right)^{-\frac{1}{2}} \int_0^\frac{\pi}{2} \cos(u) \, du$ and on the second region would be $\left(\frac{\pi}{2}\right)^{-\frac{1}{2}} \int_\frac{\pi}{2}^\frac{3\pi}{2} \cos(u) \, du$. However adding these two bounds together yields something negative and so it doesn't work (however, this approach does work to show the non-negativity of the Fresnel $S$-integral).
| As DionelJamie states, the claim is true so long as you can show $\int_0^\frac{3\pi}{2} u^{-\frac{1}{2}} \cos(u) \,du \geq 0$. I will take this as obvious.
To show this let us split up the integral $J = \int_0^\frac{3\pi}{2} u^{-\frac{1}{2}} \cos(u) \,du$ into four regions. $J=J_1 + J_2 + J_3 + J_4$.
\begin{align}
J_1&= \int_0^\frac{\pi}{4} u^{-\frac{1}{2}} \cos(u) \,du \geq \cos\left(\frac{\pi}{4}\right)\int_0^\frac{\pi}{4} u^{-\frac{1}{2}} \,du = \sqrt{\frac{\pi}{2}} \\
J_2&= \int_\frac{\pi}{4}^\frac{\pi}{2} u^{-\frac{1}{2}} \cos(u) \,du \geq \left( \frac{\pi}{2} \right)^{-\frac{1}{2}}\int_\frac{\pi}{4}^\frac{\pi}{2} \cos(u) \,du = \left( \frac{\pi}{2} \right)^{-\frac{1}{2}} \left( 1 - \frac{1}{\sqrt{2}}\right)\\
J_3&= \int_\frac{\pi}{2}^\pi u^{-\frac{1}{2}} \cos(u) \,du \geq \left(\frac{\pi}{2}\right)^{-\frac{1}{2}}\int_\frac{\pi}{2}^\pi \cos(u) \,du = -\left(\frac{\pi}{2}\right)^{-\frac{1}{2}} \\
J_4&= \int_\pi^\frac{3\pi}{2} u^{-\frac{1}{2}} \cos(u) \,du \geq \pi^{-\frac{1}{2}}\int_\pi^\frac{3\pi}{2} \cos(u) \,du = - \pi^{-\frac{1}{2}}
\end{align}
Adding up these bounds we find that $J \geq \sqrt{\frac{\pi}{2}}- \frac{2}{\sqrt{\pi}} = \frac{1}{\sqrt{2\pi}}\left( \pi - 2 \sqrt{2} \right)$. Numerically one finds that $\pi - 2 \sqrt{2} > 0$ though I haven't tried to think of a proof of this.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$( 1 - \sin x ) ( 1 + \csc x ) = \cos x \; , \; 0 \leq x \leq 2 \pi$ I found some solutions to this problem however I appear to be missing some solutions. Could anyone help me find the whole solution set?
$$( 1 - \sin x ) ( 1 + \csc x ) = \cos x \; , \; 0 \leq x \leq 2 \pi$$
$$1 + \csc x - \sin x - \sin x \csc x = \cos x $$
$$ 1 + \frac {1}{\sin x} - \sin x - 1 = \cos x $$
$$ \frac {1}{\sin x} - \sin x = \cos x $$
$$ \frac {1}{\sin x} = \cos x + \sin x $$
$$ 1 = \sin x (\cos x + \sin x) $$
$$ \sin^2 x + \cos^2 x = \sin x \cos x + \sin^2 x $$
$$ \cos^2 x = \sin x \cos x $$
$$ \cos x = \sin x $$
$$ \frac {\sin x}{\cos x} = 1 $$
$$\tan x = 1$$
$$ x = \frac \pi 4, \, \frac {5 \pi}{4}$$
| *
*
$$( 1 - \sin x ) ( 1 + \csc x ) = \cos x \; , \; 0 \leq x \leq 2 \pi\tag{*}\\\ldots$$
$$ \cos^2 x = \sin x \cos x \tag A$$
$$ \cos x = \sin x $$
$$ \frac {\sin x}{\cos x} = 1 \tag B$$
$$\tan x = 1$$
$$ x = \frac \pi 4, \, \frac {5 \pi}{4}$$
You divided equation $(A)$ by $\cos^2x$ to obtain equation $(B).$ Notice that this move implicitly assumes that the former has no solution satisfying $\cos^2x=0.$ But this assumption is unjustifiable.
You ought to have factorised it out instead:
$$ \cos^2 x = \sin x \cos x \tag A$$
$$ \cos^2 x \big(\ 1-\tan x \big)=0\\
\cos x=0 \;\text{or}\;\tan x =1$$
$$x=\frac \pi 4\:\text{or}\:\frac \pi 2\:\text{or}\:\frac {5 \pi}4\:\text{or}\:\frac {3 \pi}2\tag#$$
*But we are not quite done. Notice that our work so far has the following structure
if Line $1$ is true, then so is Line $2$
if Line $2$ is true, then so is Line $3$
...
if Line $n{-}1$ is true, then so is Line $n$
but nothing indicates that Line $n$ (the solution statement) actually implies Line $1$ (the given equation)?
This means that we have potentially introduced extraneous solutions. Thus, to finish our work, we must check each solution in $(\#)$ to see whether it actually satisfies the given equation $(*).$ In this example, all the four solutions happen to be genuine.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
$\sqrt{x^2}=|x|$ or $\sqrt{x^2}=x$ in an indefinite integral Question:
Find the following integral: $\int{\sqrt{1+\cos(x)}}dx$
My attempt:
$$\int{\sqrt{1+\cos(x)}}dx$$
$$=\int{\sqrt{2\cos^{2}\frac{x}{2}}}dx$$
$$=\int{\sqrt{2}\ \left|\cos\frac{x}{2}\right|}dx$$
$$=\sqrt{2}\int{\left|\cos\frac{x}{2}\right|}dx\tag4$$
$$=\sqrt{2}\int{\pm\cos\frac{x}{2}}dx$$
$$=\pm\sqrt{2}\int{\cos\frac{x}{2}}dx$$
$$=\pm2\sqrt{2}\sin\frac{x}{2}+C$$
My book's attempt:
$$\int{\sqrt{1+\cos(x)}}dx$$
$$=\int{\sqrt{2\cos^{2}\frac{x}{2}}}dx$$
$$=\int{\sqrt{2}\cos\frac{x}{2}}dx$$
$$=\sqrt{2}\int{\cos\frac{x}{2}}dx$$
$$=2\sqrt{2}\sin\frac{x}{2}+C$$
Basically, my book didn't put $\pm$ sign, while I did. My book did this essentially: $\sqrt{x^2}=x$, while I did this: $\sqrt{x^2}=|x|$. Is my process more correct?
| *
*$\displaystyle\pm\cos\frac{x}{2}$ is actually less descriptive
than
$\displaystyle\left|\cos\frac{x}{2}\right|.$
*$$\int_0^{2\pi}{\left|\cos\frac{x}{2}\right|}\,\mathrm
dx = 4,$$ but $$\pm\int_0^{2\pi}{\cos\frac{x}{2}}\,\mathrm dx=0= \int_0^{2\pi}{\cos\frac{x}{2}}\,\mathrm dx.$$
The correct way to continue from line $(4)$ is to split up the integral such that each of the integrand's $x$-intercepts is a limit of integration.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4364920",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 0
} |
Solving $\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$
I've been asked to solve the limit.
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Here's my approach:
$$\lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}$$
Using the identity, $\cos(x) =\sin(90^{\circ} - x)$
\begin{aligned}\implies \lim_{x\to0}\frac{\cos\left(\frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
& = \lim_{x\to0}\frac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}
\\& = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} - \dfrac{\pi}{2\cos(x)}\right)\cdot\dfrac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}}{\sin(\sin(x^2))}
\\ & = \lim_{x\to0}\dfrac{\left(\dfrac{\pi}{2} - \dfrac{\pi}{2\cos(x)}\right)}{\sin(\sin(x^2))}\cdot \underbrace{\lim_{x\to0}\dfrac{\sin\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}{\left(\frac{\pi}{2} - \frac{\pi}{2\cos(x)}\right)}}_{1}
\\ & = \dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(\sin(x^2))}{\sin(x^2)}}_1\cdot\sin(x^2)}
\\ & =\dfrac{\lim\limits_{x\to0}\dfrac{\pi}{2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}{\underbrace{\lim\limits_{x\to0}\dfrac{\sin(x^2)}{x^2}}_1\cdot x^2}
\\ & = \color{blue}{\boxed{\lim\limits_{x\to0}\dfrac{\pi}{2x^2}\left(\dfrac{\cos(x) - 1}{\cos(x)}\right)}}
\end{aligned}
Now, I'm unable to think of anything to do with this boxed part. Can anyone check my above method and tell me what to do further with this question? Any other shorter method is also most welcomed!
| $$
\frac{\cos x - 1}{x^2}
= \frac{\cos x - 1}{x^2} \times \frac{\cos x + 1}{\cos x + 1}
= \frac{\cos^2 x - 1}{x^2(\cos x + 1)}
= \frac{-\sin^2 x}{x^2} \times \frac{1}{\cos x + 1}
$$
and the left fraction can be computed since $\sin(x)/x \to 1$...
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4366903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 0
} |
Number of integer solutions of linear equation. I have the following problem.
Assume we have an unlimited number of blocks of 1cm, 2cm and 3cm height. Ignoring the position of the blocks, how many towers of 15cm height can we build?
I know I must find the coefficient of $x^{15}$ of the function
$$
f(x)=\frac{1}{1-x}\frac{1}{1-x^2}\frac{1}{1-x^3}=(1+x+x^2)\frac{1}{1-x^2}\frac{1}{(1-x^3)^2}
$$
How do I find it?
| Denote the coefficient of $x^n$ in $(\frac{1}{1-x} \frac{1}{1-x^2})$ by f(n). Then $f(n) = \lceil \frac{n+1}{2} \rceil$. Obviously, the answer is
$f(0) + f(3) +...+f(12) + f(15) = 1 + 2 + 4 + 5 + 7 + 8 = 27 $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4368201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
The diagonals of a rhombus, given area and tangent The area of the rhombus $ABCD$ is $24$ $cm^2$, if $\tan\measuredangle ABC=\dfrac{24}{7}$, find the diagonals $AC$ and $BD$.
I think we can say that $\measuredangle ABC$ is an acute angle. Is that true? Then $$\begin{cases}\tan\beta=\dfrac{24}{7}\\\sin^2\beta+\cos^2\beta=1\end{cases}$$ gives $\cos\beta=\dfrac{7}{25},\sin\beta=\dfrac{24}{25}.$ The area of $ABCD$ is $$S_{ABCD}=a^2\sin\beta=24\\a^2\cdot\dfrac{24}{25}=24\\a=5>0.$$ Now the Cosine Rule in triangle $ABC$ gives $$AC^2=2\cdot5^2-2\cdot5^2\dfrac{7}{25}=36,AC=6$$ The relationship $d_1d_2=48$ (from the area with the formula $S_{ABCD}=\frac{d_1d_2}{2}$) is very "clear". Can we come up with something else with the diagonals to make the solution better?
| If $O$ is the intersection of $AC,BD$
$2AO=AC,2BO=BD$
$24=\dfrac{AC\cdot BD}2$
$2\angle ABO=\angle ABC=2x$(say)
Now $\tan x=\dfrac{AO}{BO}$
But $\tan2x=\dfrac{24}7>0\implies0<2x<\dfrac\pi2$
Use $\tan2x=\dfrac{2\tan x}{1-\tan^2x}$ to find out $\tan x$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4369214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Find a closed form for $f_n=f_{n-1}+2f_{n-2}, n\ge 3$ i) find a suitable matrix $A \in M_{2,2}( \mathbb{Q})$
$
\left( \begin{array}{cc}
f_{n} \\
f_{n-1}
\end{array} \right)
%
= A \left( \begin{array}{cc}
f_{n-1} \\
f_{n-2}
\end{array} \right)
$
at this point I used the recursion equation and I got the result that:
$ A =\left( \begin{array}{cc}
1 & 2 \\
1 & 0
\end{array} \right)
$
ii) diagonalize the matrix $A$ which means that find a matrix $s \in GL_{2}(\mathbb{Q})$ so that $S^{-1}AS$ is a diagonal matrix.
I have calculated the eigenvalues and eigenvectors of matrix A as usual and I got that
$ S =\left( \begin{array}{cc}
2 & -1 \\
1 & 1
\end{array} \right)
$
iii) combining i) and ii) find a closed formel to calculate the value of $f_{n}$
at this point I have no clue how to show the last step iii)
this is my findig to iii)
$
\left( \begin{array}{cc}
f_{n} \\
f_{n-1}
\end{array} \right)
%
= A^n \left( \begin{array}{cc}
f_{n-1} \\
f_{n-2}
\end{array} \right)
$
but how can I show that I have to put $A^n$ to the equation above?
thus
$
\left( \begin{array}{cc}
f_{3} \\
f_{2}
\end{array} \right)
%
= A^n \left( \begin{array}{cc}
f_{2} \\
f_{1}
\end{array} \right)
$
is that correct?
| You found that $S^{-1}AS=D$ where $D$ is diagonal, so that $A=SDS^{-1}$. That is,
you've shown that
$$\left( \begin{array}{cc}
f_{n} \\
f_{n-1}
\end{array} \right)
%
= SDS^{-1} \left( \begin{array}{cc}
f_{n-1} \\
f_{n-2}
\end{array} \right)$$
where $S$, and $D$ are known fixed matrices, that is, independent of $n$. Therefore, substituting again we get
$$\left( \begin{array}{cc}
f_{n} \\
f_{n-1}
\end{array} \right)
%
= SDS^{-1} \left( \begin{array}{cc}
f_{n-1} \\
f_{n-2}
\end{array} \right) = SDS^{-1} SDS^{-1} \left( \begin{array}{cc}
f_{n-2} \\
f_{n-3}
\end{array} \right) = SD^2S^{-1} \left( \begin{array}{cc}
f_{n-2} \\
f_{n-3}
\end{array} \right)$$
and repeating the process $k$ times we get
$$\left( \begin{array}{cc}
f_{n} \\
f_{n-1}
\end{array} \right)
%
= SD^kS^{-1} \left( \begin{array}{cc}
f_{n-k} \\
f_{n-(k+1)}
\end{array} \right)$$
so for $k=n-1$ we get
$$\left( \begin{array}{cc}
f_{n} \\
f_{n-1}
\end{array} \right)
%
= SD^{n-1}S^{-1} \left( \begin{array}{cc}
f_{1} \\
f_{0}
\end{array} \right)$$
Now since $D$ is diagonal then you can easily find $D^{n-1}$, and so if you are given $f_1$ (are you?) then you get an explicit formula for $f_n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4370529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can we evaluate the integral $\int_{0}^{\infty} \frac{d x}{1+x+x^{2}+\ldots+x^{n-1}}$, where $n\geq 3$? We are going to investigate the integral
$$
I_{n}=\int_{0}^{\infty} \frac{d x}{1+x+x^{2}+\ldots+x^{n-1}} \text {, where } n \geqslant 3.
$$
Let’s start with the simpler cases.
$$
\begin{aligned}
I_{3} &=\int_{0}^{\infty} \frac{d x}{1+x+x^{2}} \\
&=\int_{0}^{\infty} \frac{d x}{\left(x+\frac{1}{2}\right)^{2}+\left(\frac{\sqrt{3}}{2}\right)^{2}} \\
&=\frac{2}{\sqrt{3}}\left[\tan ^{-1}\left(\frac{2 x+1}{\sqrt{3}}\right)\right]_{0}^{\infty} \\
&=\frac{2}{\sqrt{3}}\left(\frac{\pi}{2}-\frac{\pi}{6}\right) \\
&=\frac{2 \pi}{3 \sqrt{3}}
\end{aligned}
$$
and
$$
\begin{aligned}
I_{4} &=\int_{0}^{\infty} \frac{1}{(1+x)\left(1+x^{2}\right)} d x \\
&=\frac{1}{2} \int_{0}^{\infty}\left(\frac{1}{x+1}+\frac{1-x}{x^{2}+1}\right) d x \\
&=\frac{1}{2}\left[\ln (x+1)+\tan ^{-1} x-\frac{1}{2} \ln \left(x^{2}+1\right)\right]_{0}^{\infty} \\
&=\frac{1}{4}\left[\ln \frac{(x+1)^{2}}{x^{2}+1}\right]_{0}^{\infty}+ \left[\frac{1}{2} \tan ^{-1} x\right]_{0}^{\infty} \\
&=\frac{\pi}{4}
\end{aligned}
$$
But the integrals are difficult when $n\geq 5$.
My question: Is there any elementary method to evaluate it?
Your suggestions and solutions are warmly welcome.
| You must have an error somewhere in the indices because
$$I_n=\int_0^\infty \frac{x-1}{x^{n+1}-1}\,dx = \frac{\pi }{n+1}\csc \left(\frac{2 \pi }{n+1}\right)$$
There is another (but not elementary) method to find another form of the result using hypergeometric functions.
$$J_n=\int\frac{x-1}{x^{n+1}-1}\,dx =x \, _2F_1\left(1,\frac{1}{n+1};\frac{n+2}{n+1};x^{n+1}\right)-$$ $$\frac{1}{2} x^2 \,
_2F_1\left(1,\frac{2}{n+1};\frac{n+3}{n+1};x^{n+1}\right)$$ which leads to
$$I_n=\cos \left(\frac{\pi }{n+1}\right) \Gamma \left(\frac{n}{n+1}\right) \Gamma
\left(\frac{n+2}{n+1}\right)-$$ $$\frac{1}{2} \cos \left(\frac{2 \pi }{n+1}\right)
\Gamma \left(\frac{n-1}{n+1}\right) \Gamma \left(\frac{n+3}{n+1}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4375240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Let C1 and C2 be circles with center (−8, 0), (8, 0) respectively and radius 6, 2 respectively. Find the locus of a point P outside both C1 and C2 : Q) Fill in the blanks. Let $C_1$ be the circle with center $(−8, 0)$ and radius $6$. Let $C_2$ be the circle with center $(8,0)$ and radius $2$. Given a point $P$ outside both circles, let $L_i(P)$ be the length of a tangent segment from $P$ to circle $C_i$. The locus of all points $P$ such that $L_1(P) = 3 L_2(P)$ is a circle with radius _____ and center at _____,_____o--->
A) I found this question intriguing while solving the CMI past papers. Here is the work that I've done to attempt to fill in the blanks however the Complete equation of the Locus of the Circle is missing and I was unable to derive the complete equation.
Firstly,
I Decided to take the Locus as a moving point $(x,y)$ and then I expressed, using Pythagoras, the Length of the tangents $L_1$ and $L_2$ when $y=0$.
This Results in,
$$(x+8)^2 - 6^2 = 9((x-8)^2 -2^2)$$
Further Implying,
$$8x^2 - 160x +512 = 0$$
Giving $x=4$ and $16$, when $y=0$.
Hence the midpoint of $4$ and $16$ is $10$, which would be its center, Thus, giving Center $(10,0)$ and Radius $6$.
Although it may not be simple to use this elementary method to find the radius and the center of a circle all the time. Due to symmetry this question wasn't a challenge However Finding the general equation of the circle eludes me when the case isn't so simple. Any Help in Finding the General equation of a Locus would be Appreciated.
Thank you! :)
| Let us consider the general case with the following notations : $(C_1),(C_2)$ for the circles, $C_1=(a_1,0), \ C_2=(a_2,0)$ for their centers and $r_1,r_2$ for their radii, resp. Let $P=(x,y)$. Let $T_1$ and $T_2$ be two points of tangency (of one of the tangents issued from $P$) onto circles $(C_1),(C_2)$.
Fig. 1: Case $a_1=-4, r_1=2, a_2=1, r_2=1, k=2$. The locus is the red circle.
$$L_1(P)=kL_2(P) \ \ \text{that can be written by squaring:} \ \ (PT_1)^2=k^2 (PT_2)^2\tag{1}$$
Applying Pythagoras theorem in triangles $PT_1C_1$ and $PT_2C_2$ resp.:
$$PC_1^2=PT_1^2+r_1^2 \ \ \text{and} \ \ PC_2^2=PT_2^2+r_2^2$$
As a consequence:
$$PT_1^2=PC_1^2-r_1^2 \ \ \text{and} \ \ PT_2^2=PC_2^2-r_2^2 $$
$$PT_1^2=((x-a_1)^2+y^2)-r_1^2 \ \ \text{and} \ \ PT_2^2=((x-a_2)^2+y^2)-r_2^2 $$
Plugging these expressions into (1) gives:
$$((x-a_1)^2+y^2)-r_1^2=k^2[((x-a_2)^2+y^2)-r_2^2]$$
that can be simplified into:
$$(k^2-1)(x^2+y^2)-2(k^2a_2-a_1)x+(k^2a_2^2-a_1^2)-(k^2r_2^2-r_1^2)=0$$
which, by division by $(k^2-1)$ gives indeed the equation of a circle,
$$(x^2+y^2)-2\underbrace{\frac{k^2a_2-a_1}{k^2-1}}_a x+\underbrace{\frac{(k^2a_2^2-a_1^2)-(k^2r_2^2-r_1^2)}{k^2-1}}_c=0\tag{2}$$
The center of this circle is therefore :
$$\left(\frac{k^2a_2-a_1}{k^2-1},0 \right)$$
which is a weighted combination of the coordinates of the initial centers.
In order to determine its radius, we have to refer to the classical equation of a circle:
$$(x-a)^2+(y-b)^2=R^2 \ \ \iff \ \ x^2+y^2-2ax-2by+\underbrace{a^2+b^2-R^2}_c=0\tag{3}$$
Therefore $R$ is given by
$$R^2=a^2+b^2-c=\left(\frac{k^2a_2-a_1}{k^2-1}\right)^2- \frac{(k^2a_2^2-a_1^2)-(k^2r_2^2-r_1^2)}{k^2-1}$$
The RHS must fulfill a positivity constraint, otherwise the locus, instead of being a circle will be empty... This the case for example when $a_1=-1,a_2=1, r_1=4, r_2=1/2, k=3$.
Remark: This exercice generalizes the "circles of Apollonius"
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
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