Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How many solutions has this equation? I've the equation $x - \arctan(2x) = a$ and the question is, how many solution has the equation for different values of $a$, where $a$ is a real number.
I've plotted the graph and found the extreme values at $ x = \pm \frac{1}{2}$.
so naturally I would say:
1 solution for $a < -\frac{1}{2}$ and $a > \frac{1}{2}$
2 solutions for $a = \pm \frac{1}{2}$
and
3 solutions for $ -\frac{1}{2} < a < \frac{1}{2}$
However that's wrong. The solution contains $\pi/4$ too, why? How to solve this?
| Differentiating $f(x) = x - \arctan(2x)-a$, we get ${f}'(x) = 1-\frac{2}{1+4x^{2}} = 0$ when $x = \frac{1}{2}$ or $x= -\frac{1}{2}$. If $|x| > -\frac{1}{2}$, then function is increasing. Otherwise, it is decreasing. Thus, $f(x)$ has at most 3 roots and we just have to check the intervals $x > \frac{1}{2}$, $x < - \frac{1}{2}$, and $ -\frac{1}{2} \leq x \leq \frac{1}{2}$.
In the first interval, $\arctan(2x) > \pi/4$ and $x > \frac{1}{2}$ and $f(x)$ is increasing. Thus, if $a$ is greater than $\frac{1}{2} -\pi/4$ then you get a solution on that interval. Otherwise, you do not.
In second interval, $\arctan(2x) < -\pi/4$ and $x < -\frac{1}{2}$ and $f(x)$ is increasing. So we want $f(\frac{1}{2}) > 0$ for us to have a solution on this interval. Hence, $a < \pi/4 -1/2$ gives you a root on this interval. Otherwise, you do not have a root.
Perhaps you can do the third interval case yourself and put it all together and get a solution?
| {
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"timestamp": "2023-03-29T00:00:00",
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Compute $\zeta(2) = \frac{\pi^2}6$ It is well known that $$\int_0^1\frac{\log{x}}{1 - x}\,\mathrm{d}x = -\frac{\pi^2}{6} $$
This is generally proved by expanding the geometric series.
My question is: can this be done in reverse?
Can we evaluate the integral $\int_0^1\frac{\log{x}}{1 - x}dx$ using other method, for example, differentiation under the integral sign, and thus prove $\zeta(2) = \frac{\pi^2}6$?
| Credits to Daniele Ritelli:
$$\begin{eqnarray*}\zeta(2)&=&\frac{4}{3}\sum_{n=0}^{+\infty}\frac{1}{(2n+1)^2}=\color{red}{\frac{4}{3}\int_{0}^{1}\frac{\log y}{y^2-1}dy}\\&=&\frac{2}{3}\int_{0}^{1}\frac{1}{y^2-1}\left[\log\left(\frac{1+x^2 y^2}{1+x^2}\right)\right]_{x=0}^{+\infty}dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{x}{(1+x^2)(1+x^2 y^2)}dx\,dy\\&=&\frac{4}{3}\int_{0}^{1}\int_{0}^{+\infty}\frac{dx\, dz}{(1+x^2)(1+z^2)}=\frac{4}{3}\cdot\frac{\pi}{4}\cdot\frac{\pi}{2}=\color{red}{\frac{\pi^2}{6}}.\end{eqnarray*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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How find this $\sum_{i=0}^{5}\frac{1}{2+\cos{\left(x+\frac{i\pi}{3}\right)}}\cdot \frac{1}{2+\cos{\left(x+\frac{(i+1)\pi}{3}\right)}}$ Find this follow function $f(x)$ range ,where $x\in R$,
$$f(x)=\sum_{i=0}^{5}\dfrac{1}{2+\cos{\left(x+\dfrac{i\pi}{3}\right)}}\cdot \dfrac{1}{2+\cos{\left(x+\dfrac{(i+1)\pi}{3}\right)}}$$
or
$$f(x)=\dfrac{1}{2+\cos{x}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{2\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{2\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{3\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{3\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{4\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{4\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{5\pi}{3}\right)}}+\dfrac{1}{2+\cos{\left(x+\dfrac{5\pi}{3}\right)}}\cdot\dfrac{1}{2+\cos{\left(x+\dfrac{6\pi}{3}\right)}}$$
I think $f(x)$ have simple form,becasue this is exam problem
I think maybe can use
$$f(2\pi-x)+f(x)=?$$
or maybe have use this
$$\cos{x}\cos{y}=\dfrac{1}{2}[\cos{(x-y)}+\cos{(x+y)}]$$
I use this two idea all solve this problem,
Thank you
| $$\text{Let }y_r=\left(2+\cos\left(x+\frac{r\pi}3\right)\right)\left(2+\cos\left(x+\dfrac{(r+1)\pi}3\right)\right)$$
$$y_r=4+2\left[\cos\left(x+\frac{r\pi}3\right)+\cos\left(x+\dfrac{(r+1)\pi}3\right)\right]+\cos\left(x+\frac{r\pi}3\right)\cos\left(x+\dfrac{(r+1)\pi}3\right)$$
Applying $2\cos A\cos B$ and $\cos C+\cos D$ formula,
$$2y_r=8+8\cos\left(x+\frac{(2r+1)\pi}6\right)\cos\frac\pi6+\left[\cos\frac\pi3+\cos\left(2x+\frac{(2r+1)\pi}3\right)\right]$$
$$=8+4\sqrt3c+\frac12+2c^2-1\text{ where }c=\cos\left(x+\frac{(2r+1)\pi}6\right)$$
$$\implies 4y_r=4c^2+8\sqrt3c+15\ \ \ \ (1)$$
So, we need to find $\displaystyle\sum_{r=0}^5\frac1{y_r}=\frac{\sum_{\text{cyc}} 5 y_r\text{-s at a time}}{\prod _{r=0}^5 y_r}$ which can be easily managed by Vieta's formula
Now, $\displaystyle \cos6\left(x+\frac{(2r+1)\pi}6\right)=\cos\left(6x+\overline{2r+1}\pi\right)=-\cos6x$
So if $\displaystyle \cos6A=-\cos6x=\cos(\pi+6x),$
$\displaystyle\implies6A=2n\pi\pm(\pi+6x)$ where $n$ is any integer
$\displaystyle\implies A=x+\frac{(2n+1)\pi}6$ where $0\le n\le 5\ \ \ \ (2)$
So, as I've explained in the comment,
$(2)$ is the set of roots of $\displaystyle32c^6−48c^4+18c^2+\cos6A-1=0\ \ \ \ (3)$
Now from $\displaystyle (1),c^2=\frac{4y_r-8\sqrt3c-15}{15}\ \ \ \ (4)$
Squaring we get $\displaystyle c^4=\frac{16y_r^2+192c^2+225-(64\sqrt3y_r)c-120y_r+240\sqrt3c}{225} \ \ \ \ (5)$
Put the value of $c^2$ from $(4)$ in $(5)$
Now multiply $(4),(5)$ to get the value of $c^6$
Put the values of $c^6,c^4,c^2$ in $(3)$ to find $c$ in terms of $y_r$
Now put the values of $c,c^2$ in $(4)$ to form a Sextic equation in $y_r$ where we need to apply Vieta's formula mentioned above
| {
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"timestamp": "2023-03-29T00:00:00",
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How to evaluate this integral: $\int \frac{x^4}{(x-1)(x^2-1)}dx\;$? $$\int \frac{x^4}{(x-1)(x^2-1)}dx$$
I tried to decompose the $(x^2-1)$ term into $(x+1)(x-1)$ thus getting $(x-1)^2(x+1)$ as the denominator. I can't use the method of partial fraction because of the $x^4$ term. Should I proceed through normal polynomial(I'm finding it difficult) division or is there any other methods using trigonometric substitution to solve this. Please provide only hints so that I can work it out myself.
| Hint:
$$\frac{x^4}{(x-1)(x^2-1)}=\frac{x^4-1+1}{(x-1)(x^2-1)}=\frac{x^2+1}{x-1}+\frac{1}{(x-1)(x^2-1)}$$
$$=x+1+\frac{2}{x-1}+\frac{1}{2(x-1)}\left(\frac{1}{x-1}-\frac{1}{x+1}\right)$$
$$=x+1+\frac{2}{x-1}+\frac{1}{2(x-1)^2}-\frac{1}{4(x-1)}+\frac{1}{4(x+1)}$$
$$\to \int \frac{x^4}{(x-1)(x^2-1)}dx=\frac{x^2}{2}+x-\frac{1}{2(x-1)}+\frac{7}{4}\ln(x-1)+\frac{1}{4}\ln(x+1)+\text{C}$$
| {
"language": "en",
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Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y$ is a given positive odd integer Let $k$ be a fixed positive odd integer. Find the minimum value of $x^2+y^2$, where $x,y$ are non-negative integers and $x+y=k$
My approaches:
*
*Since $k$ is odd, $x$ and $y$ have different parity. I consider, $k=2m+1$, so that $x+y=2m+1$.
I also consider $x$ to be even and $y$ to be odd.
So, $2p+2q+1=2m+1$.
Also, $4p^2+4q^2+4q+1+8pq+4p=4m^2+4m+1$ which apparently doesn't lead me anywhere.
*$x+y=k$. Then
$x^2+y^2+2xy=k^2$.
Now I am stuck!
Please help!
| Since
$$
x^2+y^2=\frac12\left((x+y)^2+(x-y)^2\right)
$$
the minimum comes when $|x-y|$ is smallest, that is $1$ if $x+y$ is odd. Thus, the minimum is
$$
\frac12\left((x+y)^2+1\right)
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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probability matrix with trace $1$ is square of probability matrix Consider as probability matrix a matrix $M \in [0,1]^{n \times n}$ while every row sums up to $1$.
Statement: Consider a $2\times 2$ probability matrix $M' \in [0,1]^{2 \times 2}$. Show, that the following holds:
$$\exists \text{ probability matrix } M : M^2 = M' \iff \operatorname{Trace}(M') \geq 1$$
I showed $\implies$, i.e. when $M'$ is square of another probability matrix, then one can show, that the trace must be greater or equal to one.
But I got stuck at the $\impliedby$ part. Any tipps?
So far considering user68061's answer: Let $M' = \begin{pmatrix} x & (1-x) \\ (1-y) & y \end{pmatrix}$ and $M = \begin{pmatrix} a & (1-a) \\ (1-b) & b \end{pmatrix}$, then $M^2 = \begin{pmatrix} a^2+(1-a)(1-b) & a(1-a)+b(1-a) \\ a(1-b)+b(1-b) & b^2+(1-a)(1-b)\end{pmatrix}$.
Hence we need to solve
$x = a^2+(1-a)(1-b),\ y=b^2+(1-a)(1-b)$ where $x+y \geq 1$.
But somehow I think I am missing the trickery in calculation to get this solution..can you help?
| I don't know the nice way to do it, but you can brute force it.
Consider probability matrix $ \left( \begin{array}{cc}
a & 1-a \\
b & 1-b \\ \end{array} \right) $ is a probability matrix $ \left( \begin{array}{cc}
a^2+b(1-a) & a(1-a)+(1-a)(1-b) \\
ab+b(1-b) & b(1-a)+(1-b)^2 \\ \end{array} \right) $. So you need to show, that the equation $x=a^2 + b(1-a); y=b(1-a)+(1-b)^2$ has a solution in $a,b$ for $x+y \geq 1$. This can be solved directly (look at $x+y$, find $a-b$ then just substitute to the first equation)
| {
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"timestamp": "2023-03-29T00:00:00",
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Partial sum of a geometric series Consider the geometric series $\sum_{n=0}^\infty ar^n$ where $a=1$
and $r=-\frac{1}{2}$. Since $|r| < 1$, the series converges to
$S = \sum_{n=0}^\infty ar^n = \frac{1}{1+\frac{1}{2}} = \frac{2}{3}$.
I would like to arrive at the same sum by computing
$\lim_{N \to \infty} S_N$ where $S_N$ is the partial sum of $N$ terms of
the goemetric series.
First few terms of the geometric series are: $1, -\frac{1}{2}, \frac{1}{4},
-\frac{1}{8}, \frac{1}{16}, -\frac{1}{32}, \cdots$.
Here are my attempts to find $S_N$:
\begin{align*}
S_1 &= 1 = \frac{2^{(1-1)}}{2^{(1-1)}} \\
S_2 &= 1 - \frac{1}{2} = \frac{1}{2} = \frac{2^{(2-1)} - 1}{2^{(2-1)}} \\
S_3 &= \frac{1}{2} + \frac{1}{4} = \frac{3}{4} =
\frac{2^{(3-1)} - 1}{2^{(3-1)}} \\
S_4 &= \frac{3}{4} - \frac{1}{8} = \frac{5}{8} =
\frac{2^{(4-1)} - 3}{2^{(4-1)}} \\
\cdots \\
S_N &= ??
\end{align*}
\begin{align*}
S_1 &= 1 - \frac{0}{1} \\
S_2 &= 1 - \frac{1}{2} \\
S_3 &= 1 - \frac{1}{2} + \frac{1}{4} = 1 - \frac{1}{4} \\
S_4 &= 1 - \frac{1}{4} - \frac{1}{8} = 1 - \frac{3}{8} \\
\cdots \\
S_N &= ??
\end{align*}
I cannot find a pattern that will help me find $S_N$.
| Note that
$$S_1=1=\color{green}{\bf\frac23}\color{red}{\bf+\frac13},\ S_2=\frac12=\color{green}{\bf\frac23}\color{red}{\bf-\frac1{2\cdot3}},\ S_3=\frac34=\color{green}{\bf\frac23}\color{red}{\bf+\frac1{4\cdot3}},\ S_4=\frac58=\color{green}{\bf\frac23}\color{red}{\bf-\frac1{8\cdot3}},\ldots$$
As computed by @Ravi in a comment below,
$$
S_N=\color{green}{\bf\frac23}+\color{red}{\bf(-1)^{N+1}\frac1{3\cdot2^{N-1}}}=\color{green}{\bf\frac23}-\frac23\cdot\left(-\frac12\right)^N.
$$
| {
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Find recursive forumula for integrals I have to find recursve formulas for solving the following two integrals. The assignment tells one to find an Expression that leads from the calculation of $\dfrac{I_{2n}}{I_{2n+1}}$ to the calculation of $\dfrac{I_{2n-2}}{I_{2n-1}}$
a) $$ I_{2n} = \int_a^b \frac{1}{(1+x^2)^n} dx $$
b) $$ I_{2n+1} = \int_a^b \frac{1}{\sqrt{1+x^2}^{2n+1}} dx $$
This Problem sits in my mind for a few days now and I am really stuck with it. The idea I have come up with for a) is to use Integration by parts:
Let u' be $\frac{1}{1+x^2} $ and v be $\frac{1}{(1+x^2)^{n-1}}$;
u then is $tan^{-1}x$ and I could write:
$$ I_{2n} = \int_a^b \frac{1}{(1+x^2)^n} dx = \int_a^b \frac{1}{1+x^2} \frac{1}{(1+x^2)^{n-1}} dx$$
$$=\tan^{-1}x\cdot\frac{1}{(1+x^2)^{n-1}} - \int_a^b \tan^{-1}x\cdot \frac{d}{dx} \frac{1}{(1+x^2)^{n-1}} dx $$
I could repeat using Integration by parts on the right side but it did not lead me to anything useful...
Do you have any Idea on how to solve this?
Thank you for your help in advance!
FunkyPeanut
| I could show one way to derive the recursive formulas given in one of the comments:
Write
$$I_{2n}=\int\frac{1+x^2-x^2}{(1+x^2)^n}dx= I_{2n-2}-\int\frac{x^2}{(1+x^2)^n}dx
$$
We have to compute the last integral. Using integration by parts (with $u'=x/(1+x^2)^n$) we have
$$
\int\frac{x^2}{(1+x^2)^n}dx= \frac{x}{(2-2n)(1+x^2)^{n-1}}-\frac{1}{2-2n}I_{2n-2}
$$
Hence,
$$
I_{2n}= I_{2n-2}+\frac{x}{(2n-2)(1+x^2)^n}-\frac{1}{2n-2}I_{2n-2}= \frac{2n-1}{2n-2}I_{2n-2}+\frac{x}{(2n-2)(1+x^2)^n}
$$
For the odd indices, we can use the "same trick" to get
$$
I_{2n+1}=\int\frac{1+x^2-x^2}{\sqrt{(1+x^2)^{2n+1}}}dx= I_{2n-1}-\int\frac{x^2}{\sqrt{(1+x^2)^{2n+1}}}dx
$$
Again, integration by parts yields (with $u'=x/\sqrt{(1+x^2)^{2n+1}}$)
$$
\int\frac{x^2}{\sqrt{(1+x^2)^{2n+1}}}dx= \frac{x}{(1-2n)\sqrt{(1+x^2)^{2n-1}}}-\frac{1}{1-2n}I_{2n-1}
$$
Therefore, after some reductions,
$$
I_{2n+1}= \frac{2n}{2n-1}I_{2n-1}+\frac{x}{(2n-1)\sqrt{(1+x^2)^{2n-1}}}.$$
| {
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Integration method for $\int_0^\infty\frac{x}{(e^x-1)(x^2+(2\pi)^2)^2}dx=\frac{1}{96} - \frac{3}{32\pi^2}.$ The following definite integral is obtained directly from Hermite's integral representation of the Hurwitz zeta function. But is it possible to obtain the same result via the residue calculus or another technique?
$$\int_{0}^{\infty}
{x
\over
\left({\rm e}^{x} - 1\right)\left[x^{2} + \left(2\pi\right)^{2}\right]^{2}}
\,{\rm d}x
={1 \over 96} - {3 \over 32\pi^{2}}.
$$
| Binet's second Log Gamma integral formula is
$$
\ln \Gamma(z)=\left(z-\frac{1}{2}\right)\ln z-z+\frac{1}{2}\ln(2\pi)+2\int_0^{\infty} \frac{\tan^{-1}(t/z)}{e^{2\pi t}-1} \ \mathrm{d}t
$$
Differentiating twice w.r.t. $z$ gives
$$
\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + 4\int_0^{\infty} \frac{zt}{(z^2+t^2)^2(e^{2\pi t}-1)} \ \mathrm{d}t.
$$
Let $t \mapsto x/2\pi$, then
$$
\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + \int_0^{\infty} \frac{zx}{\pi^2(z^2+x^2/4\pi^2)^2(e^{x}-1)} \ \mathrm{d}x,
$$
$$
\psi^{(1)}(z) = \frac{1 + 2z}{2 z^2} + 16\pi^2\int_0^{\infty} \frac{zx}{(4\pi^2z^2+x^2)^2(e^{x}-1)} \ \mathrm{d}x.
$$
Evaluate at $z=1$,
$$
{\pi^2 \over 6} = \frac{3}{2}+16\pi^2\int_0^{\infty} \frac{x}{(x^2+4\pi^2)^2(e^{x}-1)} \ \mathrm{d}x,
$$
or, rearranging -
$$
\int_0^{\infty} \frac{x}{(x^2+4\pi^2)^2(e^{x}-1)} \ \mathrm{d}x = \frac{1}{96} - \frac{3}{32\pi^2}.
$$
| {
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Proof Verification: there are infinite values of $a$ such that $n^4 + a$ is composite for all $n \in Z$ Proof Verification: there are infinite values of $a$ such that $n^4 + a$ is composite for all $n \in Z$
Let $n^4 + a = (n^2+xn+p)(x^2+yn+q)$. Knowing that the cubic and quadratic coefficients must be zero, we can show that $x = -y, p+q=x^2,$ and $pq = a$.
One immediate solution is $p = 1, q = 3, (x, y) = \pm2,$ for which $a = 3.$
Assume that there exists a finite, but more than zero, number of $a's$ generated in this manner, and the highest $a$ is generated from values $p, q, x,$ and $y$.
Then $x = -y, p+q=x^2,$ and $pq = a_{highest}$
Let $p_1 = q_1 = 2x^2$. Then $p_1 + q_1 = (2x)^2$. Let $x_1 = 2x$ and $y_1 = -2x$.
Since $p_1 > p$ and $ q_1 > q$, $a_{new} = p_1q_1 > a_{highest}$, which contradicts our assumption. Thus there must be an infinite number of $a$ satisfying the conditions.
Is my proof correct?
| For any integer $k>1$ we have
$$n^4+4k^4=(n^2+2kn+2k^2)(n^2-2kn+2k^2)=((n+k)^2+k^2)((n-k)^2+k^2),$$
which is composite for all $n$, being the product of two integers greater than one.
| {
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"timestamp": "2023-03-29T00:00:00",
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If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
If $\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b}=\frac{1}{a+b}$, then show that $\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
My work:
$(\frac{\sin^4 x}{a}+\frac{\cos^4 x}{b})=\frac{1}{a+b}$
By squaring both sides, we get,
$\frac{\sin^8 x}{a^2}+\frac{\cos^8 x}{b^2}+2\frac{\sin^4 x \cos^4 x}{ab}=\frac{1}{(a+b)^2}$
$\frac{\sin^6 x}{a^2}+\frac{\cos^6 x}{b^2}-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=\frac{1}{(a+b)^2}$
So, now, we have to prove that,
$-2\frac{\sin^4 x \cos^4 x}{ab}-\frac{\sin^6 x \cos^2 x}{a^2}-\frac{\sin^2 x \cos^6 x}{b^2}=0$
I cannot do this. Please help!
| first we square both sides
$\dfrac{1}{(a+b)^2}=\dfrac{\sin^8x}{a^2}+2\dfrac{\sin^4x\cos^4x}{ab}+\dfrac{\cos^8x}{b^2}$
$=\dfrac{\sin^6x(1-\cos^2x)}{a^2}+2\dfrac{\sin^4x\cos^4x}{ab}+\dfrac{\cos^6x(1-\sin^2x)}{b^2}$
$=\dfrac{\sin^6x}{a^2}+\dfrac{\cos^6x}{b^2}-\dfrac{\sin^6x\cos ^2x}{a^2}-\dfrac{\cos^6x\sin^2x}{b^2}+2\dfrac{\sin^4x\cos^4x}{ab}$
it is sufficient to show the last three terms are zero
$=-\dfrac{\sin^6x\cos ^2x}{a^2}-\dfrac{\cos^6x\sin^2x}{b^2}+2\dfrac{\sin^4x\cos^4x}{ab}\\=-\sin ^2x\cos^2x(\dfrac{\sin^4x}{a^2}-\dfrac{2\sin^2 x\cos^2x}{ab}+\dfrac{\cos^4x}{b^2})$
$=-\sin ^2x\cos^2x(\dfrac{\sin^2x}{a}-\dfrac{\cos^2x}{b})^2$
Now consider the initial problem substituting $1=\cos^2x+\sin^2x$
we arrive to
$\dfrac{\sin^4 x}{a}+\dfrac{\cos^4 x}{b}=\dfrac{\sin^2x+\cos^2x}{a+b}$
$\dfrac{\sin^4x}{a}-\dfrac{\sin^2x}{a+b}+\dfrac{\cos^4x}{b}-\dfrac{\cos^2x}
{a+b}=0$
$\dfrac{a\sin^4x+b\sin^4x-a\sin^2x}{a(a+b)}+\dfrac{b\cos^4x+a\cos^4x-b\cos^2x}{b(a+b)}=0$
factorise $a\sin ^2x $ and $b\cos^2x$ to get to
$\dfrac{a\sin^2x(\sin^2x-1)+b\sin^4x}{a}+\dfrac{b\cos^2x(\cos^2x-1)+a\cos^4x}{b}=0$
$\sin^2x(-\cos^2x)+\dfrac{b}{a}\sin^4x+\cos^2x(-\sin^2x)+\dfrac{a}{b}\cos^4x=0\\ \dfrac{b}{a}\sin^4x-2\sin^2x(\cos^2x)+\dfrac{a}{b}\cos^4x=0$
divide by $ab$ here we suppose $a,b\ne 0$
$\dfrac{1}{a^2}\sin^4x-\frac{2}{ab}\sin^2x(\cos^2x)+\dfrac{1}{b^2}\cos^4x=0$
$\dfrac{\sin^4x}{a^2}-\dfrac{2\sin^2 x\cos^2x}{ab}+\dfrac{\cos^4x}{b^2}=(\dfrac{\sin^2x}{a}-\dfrac{\cos^2x}{b})^2=0$
$a\ne -b$
$\fbox{}$
| {
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"answer_count": 5,
"answer_id": 2
} |
Find all $x,y\in\mathbb{Z}$ s.t $2x^3-7y^3=3$ Find all $$x,y\in\mathbb{Z}$$ such that $$2x^3-7y^3=3$$
Solution:
We consider first $$2x^3-7y^3\equiv3 \pmod 2$$ $$5y^3\equiv 1 \pmod 2$$ $$y^3\equiv 1 \pmod2$$ which has solution $y\equiv 1 \pmod 2$
Consider $$2x^3\equiv 3 \pmod 7$$ $$4\cdot 2x^3\equiv 4\cdot3 \pmod 7$$
$$x^3\equiv 5 \pmod 7$$
but none of $x=0,1,2,3,4,5,6\pmod 7$ satisfies the equation
Which would mean that the equation has no integer solutions. Can this be correct?
| Indeed, the equation $2x^3 - 7y^3 = 3$ has no integer solutions. Modulo $7$, we get the congruence
$$2x^3 \equiv 3 \pmod{7}$$
which every solution of the original equation must satisfy. But the group of nonzero remainders modulo $7$ has order $6$, so $x^3 \equiv \pm 1 \pmod{7}$ for all $x\not\equiv 0 \pmod{7}$. Thus, whatever $x$ is, the remainder of $x^3$ modulo $7$ is one of $-1,0,1$, and so $2x^3$ has remainder $0,2$, or $5$, so $$2x^3\not\equiv 3 \pmod{7}$$
for all $x\in\mathbb{Z}$.
| {
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} |
Changing variables for multivariable functions
Let $x=uv$ and $y=\frac{1}{2}(u^2-v^2)$. Substitute $x$ and $y$ with $u$ and $v$ in the expression $z_{x}^2+z_y^2$
My attempt:
$$z_x=u_xz_u+v_xz_v=\frac{1}{v}z_u+\frac{1}{u}z_v$$
Squaring:
$$z_x^2=\frac{1}{v^2}z_u^2+\frac{2}{vu}z_uz_v+\frac{1}{u^2}z_v^2$$
Using the same method for $z_y$
$$z_y=\frac{1}{\sqrt{2y+v^2}}z_u-\frac{1}{\sqrt{u^2-2y}}z_v$$
Squaring again and changing variables and summing:
$$z_x^2+z_y^2=\frac{1}{v^2}z_u^2+\frac{2}{vu}z_uz_v+\frac{1}{u^2}z_v^2+\frac{1}{u^2}z_u^2-\frac{2}{uv}z_uz_v+\frac{1}{v^2}z_v^2=(\frac{1}{v^2}+\frac{1}{u^2})(z_u^2+z_v^2)$$
My book gives the answer $\frac{z_u^2+z_v^2}{u^2+v^2}$. I checked my calculations multiple times and there doesn't seem to be any errors so I assume my reasoning was incorrect somewhere. Any help?
| Tackle it from the other side, express $z_u^2 + z_v^2$ in terms of $z_x,\,z_y$. You have
$$z_u = z_x x_u + z_y y_u = vz_x + uz_y,$$ and $$z_v = z_xx_v + z_y y_v = uz_x - vz_y.$$ That produces $z_u^2 + z_v^2 = (u^2+v^2)(z_x^2+z_y^2)$, agreeing with the book.
What was your mistake? When computing $u_x$ etc., you wrote $u = \frac{x}{v}$ to conclude $u_x = \frac{1}{v}$, as though $v$ was independent of $x$. But it isn't, $v$ is a function of $x$ and $y$, like $u$ is. You would get
$$u_x = \left(\frac{x}{v}\right)_x = \frac{1}{v} - \frac{xv_x}{v^2}$$
correctly from the ansatz $u = \frac{x}{v}$. That doesn't look too nice, the hindsight computation above is much nicer, but of course one can only do that if one knows what one will obtain. Although one knows it is some quadratic combination of $z_u$ and $z_v$, and one only needs to determine the coefficients, which isn't too bad, since there are only $z_u^2,\, z_uz_v,\,z_v^2$.
To compute $u_x,\, u_y,\, v_x,\,v_y$, it is maybe the easiest to invert the matrix
$$\begin{pmatrix}x_u & x_v \\ y_u & y_v \end{pmatrix} = \begin{pmatrix} v & u\\ u & -v \end{pmatrix}$$
and get
$$\begin{pmatrix}u_x & u_y \\ v_x & v_y \end{pmatrix} = \frac{1}{u^2+v^2}\begin{pmatrix} v & u\\ u & -v \end{pmatrix}.$$
| {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Moore-Penrose pseudoinverse of a 3×3 matrix Is there a "simple" formula for computing the Moore-Penrose pseudoinverse of a $3\times 3$ matrix? I mean something like the formula for the inverse (for non-singular matrices), which involves the matrix of minors, etc.
I need that for a computer program, and I feel that using LAPACK's SVD is a bit of an overkill.
| You are probably thinking of the formula
$$
\begin{align}
\mathbf{A}^{-1}
&= \frac{\text{adj } \mathbf{A}} {\det \mathbf{A} } \\
&= \frac{\left( \text{cof } \mathbf{A}\right)^{\mathrm{T}}} {\det \mathbf{A} } \\
\end{align}
$$
The matrix $\text{adj } \mathbf{A}$ is the adjugate of $\mathbf{A}$ and is the transpose of $\mathbf{C}$, the matrix of cofactors of $\mathbf{A}$.
For a nonsingular $\mathbf{A}\in\mathbb{R}^{3 x 3}$,
$$
\mathbf{A} =
\left[
\begin{array}{ccc}
a_{11} & a_{12} & a_{13} \\
a_{21} & a_{22} & a_{23} \\
a_{31} & a_{32} & a_{33}
\end{array}
\right],
$$
the matrix of cofactors is composed of the determinants
$$
\mathbf{C} =
\left[
\begin{array}{ccc}
%
+ \left| \begin{array}{cc} a_{22} & a_{23} \\ a_{32} & a_{33} \end{array} \right| &
- \left| \begin{array}{cc} a_{21} & a_{23} \\ a_{31} & a_{33} \end{array} \right| &
+ \left| \begin{array}{cc} a_{21} & a_{22} \\ a_{31} & a_{32} \end{array} \right| \\
%
- \left| \begin{array}{cc} a_{12} & a_{13} \\ a_{32} & a_{33} \end{array} \right| &
+ \left| \begin{array}{cc} a_{11} & a_{13} \\ a_{31} & a_{33} \end{array} \right| &
- \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{31} & a_{32} \end{array} \right| \\
%
+ \left| \begin{array}{cc} a_{12} & a_{13} \\ a_{22} & a_{23} \end{array} \right| &
- \left| \begin{array}{cc} a_{11} & a_{13} \\ a_{21} & a_{23} \end{array} \right| &
+ \left| \begin{array}{cc} a_{11} & a_{12} \\ a_{21} & a_{22} \end{array} \right| \\
\end{array}
\right].
$$
Because you specified that $\mathbf{A}$ is nonsingular, the matrix inverse exists and is the same as the Moore-Penrose pseudoinverse:
$$
\mathbf{A}^{-1} = \mathbf{A}^{\dagger}.
$$
Example
Pencil and paper exercise of confirmation:
$$
\mathbf{A} =
\left[
\begin{array}{ccc}
1 & 0 & 1 \\
0 & 1 & 1 \\
1 & 0 & 0
\end{array}
\right],
$$
The determinant is $\det \mathbf{A} = 1$, and the matrix of cofactors is
$$
\mathbf{C} =
\left[
\begin{array}{rrr}
0 & 1 & -1 \\
0 & -1 & 0 \\
-1 & -1 & 1
\end{array}
\right]
$$
The inverse matrix is
$$
\mathbf{A}^{-1} = \frac{\left( \text{cof } \mathbf{A}\right)^{\mathrm{T}}} {\det \mathbf{A} } = \frac{\mathbf{C}^\mathrm{T}} {-1} =
\left[
\begin{array}{rrr}
0 & 0 & 1 \\
-1 & \phantom{-}1 & 1 \\
1 & 0 & -1
\end{array}
\right].
$$
| {
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} |
Sign problem in proof that $\Gamma'(1) = -\gamma$ Convergence of $(u_n)_{n \in \mathbb{N}} := \sum\nolimits_{k=1}^n \frac{1}{k} - \log(n)$ was proven before
The egality $\Gamma'(1) = \lim_{n \to \infty}\int_0^n \log(t) (1-\frac{t}{n})^n dt$ was also proven before
Let $\gamma$ denote the limit of $(u_n)_{n \in \mathbb{Z}_{>0}}$. Show
\begin{align*}
\int_0^1 (1-u)^n \log(u)du = \frac{1}{n + 1} \int_0^1 \frac{(1-u)^{n+1} - 1}{u} du
\end{align*}
and conclude $\Gamma'(1) = -\gamma$.
I managed to prove the Integral identity but then I get lost !
Thanks in advance !
Calculation
\begin{align*}
\Gamma'(1) &= \lim_{n \to \infty}\int_0^n \log(t)\left(1-\frac{t}{n}\right)^ndt
\end{align*}
Substitution: $u = \frac{t}{n}$, $du = \frac{1}{n} dt$
\begin{align*}
\Gamma'(1) &= \lim_{n \to \infty}\int_0^1 [\log(u) + \log(n)](1-u)^n n du \\
&= \lim_{n \to \infty}\Big( n\int_0^1 \log(u)(1-u)^n du + n \log(n) \int_0^1 (1-u)^{n} du \Big) \\
&= \lim_{n \to \infty}\Big( \frac{n}{n+1} \int_0^1 \frac{(1-u)^{n+1} - 1}{u} du - \frac{n}{n+1} \log(n)
\left.(1-u)^{n+1}\right|_{u=0}^1 \Big)
\end{align*}
Second substitution $x = 1-u$, $du = -dx$:
\begin{align*}
\Gamma'(1) &= \lim_{n \to \infty}\Big( \frac{-n}{n+1} \int_0^1 \frac{x^{n+1} - 1}{1-x} dx + \frac{n}{n+1} \log(n) \Big) \\
&= \lim_{n \to \infty} \frac{n}{n+1} \Big( \int_0^1 \frac{1-x^{n+1}}{1-x} dx + \log(n)\Big)
\end{align*}
We use that $H_n = \sum_{k=1}^n \frac{1}{k} = \int_0^1 \frac{1-x^n}{1-x}dx$:
\begin{align*}
\Gamma'(1) &= \lim_{n \to \infty}\Big( \frac{n}{n+1} \Big) \cdot \lim_{n \to \infty}
\Big( \sum_{k=1}^{n+1}\frac{1}{k} + \log(n) \Big) \\
&= \lim_{n \to \infty} \Big( \sum_{k=1}^{n+1}\frac{1}{k} + \log(n) \Big) \\
\end{align*}
But I should get !!!
\begin{align*}
-\lim_{n \to \infty} \Big( \sum_{k=1}^{n+1}\frac{1}{k} - \log(n) \Big) \\
\end{align*}
| When you substitute $u = 1-x$ in
$$\int_0^1 \frac{(1-u)^{n+1}-1}{u}\,du,$$
you apparently forgot to adapt the integration limits (or maybe the sign from $du = -dx$),
$$\begin{align}
\int_0^1 \frac{(1-u)^{n+1}-1}{u}\,du &= \int_{1-0}^{1-1} \frac{(1-(1-x))^{n+1}-1}{1-x}\,d(1-x)\\
&= \int_1^0\frac{1-x^{n+1}}{1-x}\,dx\\
&= -\int_0^1\frac{1-x^{n+1}}{1-x}\,dx.
\end{align}$$
It turns out right then.
| {
"language": "en",
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"answer_count": 1,
"answer_id": 0
} |
Determine if $\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$ converges or diverges. Another series I found I'm struggling with.
Determine if the following series converges or diverges.$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1}$$
Ratio test and n-th root test are both inconclusive, Leibniz - criterion cannot be applied since the sequence given is not in the form of $(-1)^na_n$. I am sure the problem can be solved with the limit comparison test, though, the $n^{(...)}$ look pretty inviting after all. Let $a_n:=\frac{(-1)^nn^2+n}{n^3+1}$ then $|a_n|= \frac{n^2+n}{n^3+1}$. A try showing that the series diverges using the divergence of $\sum\frac{1}{n}$. For $n≥1$ it is clear that
$$ \frac{n^3+n^2}{n^3+1} ≥ 1 $$ dividing by $n$ yields:
$$ \frac{n^2+n}{n^3+1} = |a_n| ≥ \frac{1}{n}$$
Thus the series diverges. (?)
EDIT: Hints you gave me yield:
$$\sum_{n=1}^{\infty}\frac{(-1)^nn^2+n}{n^3+1} = \sum_{n=1}^{\infty}(-1)^n\frac{n^2}{n^3+1} +
\sum_{n=1}^{\infty}\frac{n}{n^3+1}$$ With the first series converging by Leibniz-theorem and the second by limit comparison test with $\frac{1}{n^2}$.
| Let denote
$$\frac{(-1)^nn^2+n}{n^3+1}=\underbrace{\frac{(-1)^nn^2}{n^3+1}}_{=u_n}+\underbrace{\frac{n}{n^3+1}}_{=v_n}$$
The series $\displaystyle \sum_n v_n$ is convergent since $v_n\sim_\infty \frac 1 {n^2}$.
We have
$$u_n=\frac{(-1)^nn^2}{n^3+1}=\frac{(-1)^n}{n}\frac{1}{1+\frac{1}{n^3}}=\frac{(-1)^n}{n}\left(1-\frac{1}{n^3}+o\left(\frac{1}{n^3}\right)\right)$$
so we can see that $\displaystyle\sum_n u_n$ is a sum of convergent series hence it's convergent. Conclude.
| {
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"source": "stackexchange",
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"answer_count": 4,
"answer_id": 1
} |
Double integral -- tricky?
If $f(x,y) = x^2+y^2$ and $D=\{(x,y)\in\mathbb{R}^2:x^2+y^2\geq1, x^2+y^2-2x\leq0 \text{ and } y\geq0\}$, find $\displaystyle\int\displaystyle\int_D f$.
$D$ looks like the intersection between two circumferences, as I draw below:
Polar coordinates seems a obvious choice, but using them I'd have unwanted bits, this is, if I split the region in two sections (by $x=\frac{1}{2}$) after computing the red and blue region I would have twice $R1$ and $R2$, a problem that could be solved computing each one of them and then substract them of the final answer -this is, after adding the integral of the red and blue sections-.
I believe the answer would be given by
$$\displaystyle\int \displaystyle\int_D f = \displaystyle\int\displaystyle\int _{\text{RED}} f + \displaystyle\int\displaystyle\int _{\text{BLUE}} f - \displaystyle\int\displaystyle\int _{\text{R1}} f - \displaystyle\int\displaystyle\int _{\text{R2}} f$$
$R1$ and $R2$ could be described as follows:
$$R1 = \{(x,y):0\leq y\leq \sqrt{3}x,\;0\leq x\leq 1/2\}$$
$$R2 = \{(x,y):0\leq y\leq \sqrt{3}-\sqrt{3}x,\;1/2 \leq x\leq 1\}$$
Which gives
$$\displaystyle\int\displaystyle\int _{\text{R1}} f =\displaystyle\int_0^{1/2}\displaystyle\int_0^{\sqrt{3}x} (x^2+y^2)\; dy dx = 2\sqrt{3}\displaystyle\int_0^{1/2}x^3 = \displaystyle\frac{1}{52\sqrt{3}}.$$
$$\displaystyle\int\displaystyle\int_{R2} f = \displaystyle\int_{1/2}^1\displaystyle\int_0^{\sqrt{3}-\sqrt{3}x}(x^2+y^2)\; dydx \\ = \displaystyle\int_{1/2}^1(x^2(\sqrt{3}-\sqrt{3}x)dx + \frac{1}{3}\int_{\frac{1}{2}}^1 (\sqrt{3}-\sqrt{3}x)^3 dx \\ = \left[\sqrt{3}\displaystyle\int_{1/2}^1 x^2 -\sqrt{3}\displaystyle\int_{1/2}^1 x^3\right] + \frac{1}{3}\int_{\frac{1}{2}}^1 (\sqrt{3}-\sqrt{3}x)^3 dx \\ = \sqrt{3}\left[\left(\displaystyle\frac{1}{3}-\displaystyle\frac{1}{24}\right)+\left(\displaystyle\frac{1}{4}-\displaystyle\frac{1}{64}\right) \right] + \displaystyle\frac{1}{3}\left(\displaystyle\frac{1}{2}\right) \\= \frac{\sqrt{3}}{16}+\frac{1}{6} = \frac{1}{48}(3\sqrt{3}+8).$$
Now changing to polar coordinates $x=r\cos \theta$, $y= r\sin \theta$ to compute the red and blue:
$$\displaystyle\int\displaystyle\int _{\text{RED}} f = \displaystyle\int_0^{\pi/3}\displaystyle\int_0^1 r^3\;drd\theta = \frac{1}{4}\displaystyle\int_0^{\pi/3}d\theta = \frac{1}{12}\pi$$
Setting auxiliary axis $u = x-1, v = y$ I have $\displaystyle\frac{\partial (x,y)}{\partial (u,v)}=1$ and after with $u=r\cos \theta$, $v=r\sin \theta$
$$\displaystyle\int\displaystyle\int _{\text{BLUE}} f = \displaystyle\int_{2\pi/3}^{\pi}\displaystyle\int_0^1 r^3\;drd\theta = \frac{1}{12}\pi.$$
Finally
$$\displaystyle\int \displaystyle\int_D f = \frac{1}{6}\pi -\displaystyle\frac{1}{52\sqrt{3}} -\frac{1}{48}(3\sqrt{3}+8) .$$
Now I'd like to know if the solution is right, but fundamentally, can this integral be computed without splitting the $D$?
P.S: Legitimate question, but couldn't resist the pun.
| The integral's basically:
$$\int\limits_{1/2}^1\int\limits_{\sqrt{1-x^2}}^{\sqrt{1-(x-1)^2}}\,f(x,y)dydx+\int\limits_1^2\int\limits_0^{\sqrt{1-(x-1)^2}}f(x,y)\,dydx$$
| {
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"answer_id": 0
} |
Finding a least common multiple (LCM) My Algebra 2 book explains how to find a least common multiple:
Find the least common multiple of $4x^2 - 16$ and $6x^2 - 24x + 24$.
Solution
Step 1 Factor each polynomial. Write numerical factors as products of primes.
$4x^2 - 16 = 4(x^2 - 4) = (2^2)(x + 2)(x - 2)$
$6x^2 - 24x + 24 = 6(x^2 - 4x + 4) = (2)(3)(x-2)^2$
Step 2 Form the LCM by writing each factor to the highest power it occurs in either polynomial.
LCM = $(2^2)(3)(x + 2)(x - 2)^2 = 12(x + 2)(x - 2)^2$
I don't understand their wording, and I don't want to go onto the rest of my assignment that includes finding the least common denominators until I know how to do it correctly, instead of going back and doing it over when I find out I'm doing it wrong.
| It works just like with integers. If you want to find the LCM of $84=2^2\cdot 3 \cdot 7$ and $90=2\cdot 3^2 \cdot 5$ you collect the highest power of each prime, getting $2^2 \cdot 3^2 \cdot 5 \cdot 7 = 1260$ The LCM has to incorporate all the polynomial factors to the highest power in any of the things you are taking the LCM of.
| {
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} |
What is the probability no slots contain more than two balls given I am trying to sort 5 balls into 6 slots? I am having a difficult time understanding how to approach this problem. Suppose I have $6$ total slots and $5$ balls. Now, I assign the balls at random to the slots. What is the probability that no slot will contain more than two balls?
My approaches so far:
I recognized that $\binom{5+(6-1)} 5$ ($10$ choose $5$) represents the total combinations per the combinations with repetitions formula.
Next, I realized that maybe I can find the probability a slot has exactly $5$, $4$, $3$ balls in one spot, then take $1$ - the sum of those probabilities.
The correct answer is around $80\%$ so I am way off with this approach. Do you guys have any idea? Thanks!
| Five balls can be distributed in 6 slots following the GF
$(1+b+b^2+b^3+b^4+b^5)(1+b+b^2+b^3+b^4+b^5)...(1+b+b^2+b^3+b^4+b^5)$ (six factors)
The coefficient of $b^5$ is $252$ and represents the ways of placing 5 balls without restrictions in six slots. With the given restriction, the GF is
$(1+b+b^2)(1+b+b^2)...(1+b+b^2)$
The coefficient of $b^5$ is now $126$, that gives a probability of $50 \% $, as pointed above.
| {
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How prove this the equation $\{x^3\}+\{y^3\}=\{z^3\}$has infinitely many rational non-integers solutions, Show that the equation
$$\{x^3\}+\{y^3\}=\{z^3\}$$
has infinitely many rational non-integers solutions,Here,$\{a\}$ denotes the fractional part of $a$
I have solve this follow two problem
the equation 1:
$$\{x\}+\{y\}=\{z\}$$
has infinitely many rational non-integers solutions.
I take $$x=n+0.2,y=n+0.3,z=n+0.5,n\in N^{+}$$
the equation 2:
$$\{x^2\}+\{y^2\}=\{z^2\}$$
has infinitely many rational non-integers solutions.
then I take
$$x=10n+0.3,y=10n+0.4,z=10n+0.5$$
because
$$x^2=100n^2+6n+0.09,y^2=100n^2+8n+0.16,z^2=100n^2+10n+0.25$$
$$\Longrightarrow \{x^2\}=0.09,\{y^2\}=0.16,\{z^2\}=0.25$$
$$\Longrightarrow \{x^2\}+\{y^2\}=\{z^2\}$$
But for
the equation
$$\{x^3\}+\{y^3\}=\{z^3\}$$
has infinitely many rational non-integers solutions
I can't.Thank you,
and I gues this follow problem maybe is true.and maybe can prove it?
the equation
$$\{x^4\}+\{y^4\}=\{z^4\}$$
has infinitely many rational non-integers solutions?
the equation
$$\{x^5\}+\{y^5\}=\{z^5\}$$
has infinitely many rational non-integers solutions?
and so on
| For any $n > 2$, pick any $m > 1$ such that $\gcd(m,n) = 1$. Notice
$$\gcd(m,n) = 1 \implies \gcd(m^n,n) = 1$$
We can define a number $\lambda$ by
$$\lambda = \text{mod}( n^{\varphi(m^n)-1}, m^n )$$
where $\varphi(x)$ is the Euler's totient function and $\lambda$ will satisfy $$\lambda n = 1 \pmod{m^n}$$
Let $k \in \mathbb{Z}_{+} \text{ s.t. } \lambda n = k m^n + 1$, we have:
$$\begin{align}
\left(\frac{1}{m^2} + \lambda m^{n-2}\right)^n
= & \frac{1}{m^{2n}}
+ n \lambda \frac{m^{n-2}}{m^{2(n-1)}}
+ \underbrace{\binom{n}{2}\lambda^2 \frac{m^{2(n-2)}}{m^{2(n-2)}} + \cdots}_{\in \mathbb{Z}}\\
= & \frac{1}{m^{2n}} + \frac{1}{m^n} + \underbrace{k + \binom{n}{2}\lambda^2 \frac{m^{2(n-2)}}{m^{2(n-2)}} + \cdots}_{\in \mathbb{Z}}
\end{align}$$
This implies
$$\left\{\frac{1}{m^{2n}}\right\} +
\left\{\frac{1}{m^{n}}\right\}
= \left\{\left(\frac{1}{m^2} + \lambda m^{n-2}\right)^n\right\}
$$
For example, when $n = 3$, we can take $m = 2$,
$$\lambda = 3\quad\longrightarrow\quad
\left\{\frac{1}{4^3}\right\} + \left\{\frac{1}{2^3}\right\} = \left\{\left(\frac{25}{4}\right)^3\right\}$$
When $n = 4$, we can take $m = 2$,
$$\lambda = 61\quad\longrightarrow\quad
\left\{\frac{1}{9^4}\right\} + \left\{\frac{1}{3^4}\right\} = \left\{\left(\frac{4942}{9}\right)^4\right\}$$
When $n = 5$, we can take $m = 2$,
$$\lambda = 13\quad\longrightarrow\quad
\left\{\frac{1}{4^5}\right\} + \left\{\frac{1}{2^5}\right\} = \left\{\left(\frac{417}{4}\right)^5\right\}$$
Since for any $n > 2$, there are infinitely many $m$ relative prime to it. This implies there are infinitely many non-integral rational solutions for $\{ x^n \} + \{ y^n \} = \{ z^n \}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/649738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Determine the largest power of 10 that is a factor of $50!\,$? How would one find the largest power of 10 which is a factor of $50!\,{}$?
| This problem is equivalent to find the number of trailing zeros in $n!=50!$. I adapt this old answer of mine to the question Derive a formula to find the number of trailing zeroes in $n!$.
For every integer $n$ the exponent of the prime $p$ in the prime
factorization of $n!$ is given by de Polignac's formula $$\displaystyle\sum_{i\ge 1}\left\lfloor \frac{n}{p^{i}}
\right\rfloor =\left\lfloor \frac{n}{p}\right\rfloor+\left\lfloor \frac{n}{p^2}\right\rfloor+\cdots+\left\lfloor \frac{n}{p^k}\right\rfloor ,\qquad p^k\le n,\ p^{k+1}\gt n.$$
because this exponent is obtained by adding to the numbers between $1$ and $n$ which are
divisible by $p$ the number of those divisible by $p^{2}$, then the number
of those divisible by $p^{3}$, and so on. The process terminates at the
greatest power $p^{i}\leq n$.
For $n=50$ the exponent of $5$ in in the prime factorization of $50!$ is $$e_5(50!)=\sum_{i\geq 1}\left\lfloor \dfrac{50}{5^{i}}\right\rfloor=\left\lfloor \dfrac{50}{5}\right\rfloor +\left\lfloor \dfrac{50}{5^{2}}\right\rfloor =10+2=12<47.$$
Similarly the exponent of $2$ in the prime factorization of $50!$ is
$$\begin{align}
e_2(50!)=\sum_{i\geq 1}\left\lfloor \dfrac{50}{2^{i}}\right\rfloor &= \left\lfloor \dfrac{50}{2}\right\rfloor+\left\lfloor \dfrac{50}{2^{2}} \right\rfloor + \left\lfloor \dfrac{50}{2^{3}} \right\rfloor + \left\lfloor\dfrac{50}{2^{4}}\right\rfloor +\left\lfloor \dfrac{50}{2^{5}}\right\rfloor
\\
&=25+12+6+3+1
\\
&=47>12.
\end{align}$$
So $50!=2^{47}5^{12}m=(2^{35}m)10^{12}$, with $\gcd(m,10)=1$, as pointed out by robjohn. Consequently, the number of trailing zeroes in $50!$ is $12$ and the power is $10^{12}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/650436",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Computing $\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right) \, dx$
For $a\ge 0$ let's define $$I(a)=\int_{0}^{\pi}\ln\left(1-2a\cos x+a^2\right)dx.$$ Find explicit formula for $I(a)$.
My attempt: Let
$$\begin{align*}
f_n(x)
&= \frac{\ln\left(1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2\right)-\ln\left(1-2a\cos x+a^2\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(\displaystyle\frac{1-2
\left(a+\frac{1}{n}\right)\cos x+\left(a+\frac{1}{n}\right)^2}{1-2a\cos x+a^2}\right)}{\frac{1}{n}}\\
&=\frac{\ln\left(1+\dfrac{1}{n}\left(\displaystyle\frac{2a-2\cos x+\frac{1}{n}}{1-2a\cos x+a^2}\right)\right)}{\frac{1}{n}}.
\end{align*}$$
Now it is easy to see that $f_n(x) \to \frac{2a-2\cos x}{1-2a\cos x+a^2}$ as $n \to \infty$. $|f_n(x)|\le \frac{2a+2}{(1-a)^2}$ RHS is integrable so $\lim_{n\to\infty}\int_0^\pi f_n(x)dx = \int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2} dx=I'(a)$. But
$$\int_0^\pi \frac{2a-2\cos x}{1-2a\cos x+a^2}=\int_0^\pi\left(1-\frac{(1-a)^2}{1-2a\cos x+a^2}\right)dx.$$ Consider
$$\int_0^\pi\frac{dx}{1-2a\cos x+a^2}=\int_0^\infty\frac{\frac{dy}{1+t^2}}{1-2a\frac{1-t^2}{1+t^2}+a^2}=\int_0^\infty\frac{dt}{1+t^2-2a(1-t^2)+a^2(1+t^2)}=\int_0^\infty\frac{dt}{(1-a)^2+\left((1+a)t\right)^2}\stackrel{(*)}{=}\frac{1}{(1-a)^2}\int_0^\infty\frac{dt}{1+\left(\frac{1+a}{1-a}t\right)^2}=\frac{1}{(1-a)(1+a)}\int_0^\infty\frac{du}{1+u^2}=\frac{1}{(1-a)(1+a)}\frac{\pi}{2}.$$
So $$I'(a)=\frac{\pi}{2}\left(2-\frac{1-a}{1+a}\right)\Rightarrow I(a)=\frac{\pi}{2}\left(3a-2\ln\left(a+1\right)\right).$$
It looks too easy, is there any crucial lack?
$(*)$ — we have to check $a=1$ here by hand and actually consider $[0,1), (1,\infty)$ but result on these two intervals may differ only by constant - it may be important but in my opinion not crucial for this proof.
| Considering the following diagram:
$\hspace{2cm}$
we get that
$$
\begin{align}
\int_0^\pi\log\left(1-2r\cos(\theta)+r^2\right)\,\mathrm{d}\theta
&=\int_0^{2\pi}\log\sqrt{1-2r\cos(\theta)+r^2}\,\mathrm{d}\theta\\
&=\mathrm{Re}\left(\int_\gamma\log(z-1)\frac{\mathrm{d}z}{iz}\right)
\end{align}
$$
along the path $\gamma=r\,e^{i[0,2\pi]}$.
If $r\le1$, the singularity at $z=0$ has residue $0$. Thus, the integral is $0$.
If $r\gt1$, then we need to modify the path to avoid the branch cut for $\log(z-1)$ along $\{t\in\mathbb{R}:t\ge1\}$. That is, in addition to the circular contour $\gamma=r\,e^{i[0,2\pi]}$, we need to follow the two contours $[r,1]$ below the real axis and $[1,r]$ above the real axis. The sum of the integrals along these two contours is
$$
\int_r^12\pi i\frac{\mathrm{d}z}{iz}+\int_1^r0\frac{\mathrm{d}z}{iz}=-2\pi\log(r)
$$
Since the integral along all three contours is $0$, the integral along the circular part must be $2\pi\log(r)$.
Putting these two cases together, we get
$$
\int_0^\pi\log\left(1-2r\cos(\theta)+r^2\right)\,\mathrm{d}\theta
=\left\{\begin{array}{l}
0&\text{if }r\le1\\
2\pi\log(r)&\text{if }r\gt1
\end{array}\right.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/650513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 4
} |
Find bases for the subspaces $U_1, U_2, U_1 \cap U_2, U_1 + U_2$ Let $$U_1 = \left\{ \left(
\begin{array}{cc}
x_1\\
x_2\\
x_3\\
x_4\\
\end{array}
\right) \in \mathbb{R} :-x_1-x_2+x_3=0 \right\}$$
$$U_2 = span \left( \left( \begin{array}{cc}
-1 \\
1 \\
1 \\
2 \\ \end{array} \right), \left( \begin{array}{cc}
2 \\
-1 \\
-1 \\
2 \\ \end{array} \right) \right) $$
, find the bases for $U_1, U_2, U_1 \cap U_2, U_1 + U_2$ and give the dimension of each subspace.
Now $U_1$ can be written as $$\left\{ \left(
\begin{array}{cc}
x_1\\
x_2\\
x_1 + x_2\\
x_4\\
\end{array}
\right) \right\}$$ and so a basis would be $$B= \left( \left( \begin{array}{cc}
1 \\
0 \\
1 \\
0 \\ \end{array} \right), \left( \begin{array}{cc}
0 \\
1 \\
1 \\
0 \\ \end{array} \right), \left( \begin{array}{cc}
0 \\
0 \\
0 \\
1 \\ \end{array} \right) \right)$$
the vectors in the span of $U_2$ form a basis since they are linearly independent.
Now to find the basis of $U_1 + U_2$ I could take the span of all the basis vectors in $U_1$ and $U_2$, but since there are only four linearly independent vectors in that span, only four of those vectors form the basis of $U_1 + U_2$.
I cannot find a basis for the intersection of $U_1$ and $U_2$. I can only determine the dimension of that subspace by using the dimension formula for subspaces. So $U_1 \cap U_2$ must have dimension 1, but I cannot find a spanning vector. Can anybody help please?
| I tend to use a bare-hands approach like this:
Suppose $x \in U_1 \cap U_2$. Since $x \in U_2$, this implies that
$$x= s \left(
\begin{array}{cc}
-1\\
1\\
1\\
2\\
\end{array}
\right)
+t \left(
\begin{array}{cc}
2\\
-1\\
-1\\
2\\
\end{array}
\right) =
\left(
\begin{array}{cc}
-s + 2t\\
s-t\\
s-t\\
2s+2t\\
\end{array}
\right)
$$
Since we require this to belong to $U_1$, we need $x_3 = x_1 + x_2$, so that $s=2t$.
So we can just take (for example) $t=1$ and $s=2$. This gives a vector in the intersection:
$$x = \left(
\begin{array}{cc}
0\\
1\\
1\\
6\\
\end{array}
\right)$$
Since this is non-zero, it is a spanning set (in fact a basis) for the intersection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/654856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
is there a nicer way to $\int e^{2x} \sin x\, dx$? I have to solve this:
$\int e^{2x} \sin x\, dx$
I managed to do it like this:
let $\space u_1 = \sin x \space$ and let $\space \dfrac{dv}{dx}_1 = e^{2x}$
$\therefore \dfrac{du}{dx}_1 = \cos x \space $ and $\space v_1 = \frac{1}{2}e^{2x}$
If I substitute these values into the general equation:
$\int u\dfrac{dv}{dx}dx = uv - \int v \dfrac{du}{dx}dx$
I get:
$\int e^{2x} \sin x dx = \frac{1}{2}e^{2x}\sin x - \frac{1}{2}\int e^{2x}\cos x\, dx$
Now I once again do integration by parts and say:
let $u_2 = \cos x$ and let $\dfrac{dv}{dx}_2 = e^{2x}$
$\therefore \dfrac{du}{dx}_2 = -\sin x$ and $v_2 = \frac{1}{2}e^{2x}$
If I once again substitute these values into the general equation I get:
$\int e^{2x}\sin x dx =\frac{1}{2}e^{2x}\sin x - \frac{1}{4}e^{2x}\cos x - \frac{1}{4}\int e^{2x}\sin x dx$
$\therefore \int e^{2x}\sin x dx = \frac{4}{5}(\frac{1}{2}e^{2x}\sin x -
\frac{1}{4}e^{2x}\cos x) + C$
$\therefore \int e^{2x}\sin x\, dx = \frac{2}{5}e^{2x}\sin x - \frac{1}{5}e^{2x}\cos x + C$
I was just wondering whether there was a nicer and more efficient way to solve this?
Thank you :)
| A different method (though not really easier) is to use the fact that $\sin x$ can be expressed using exponents:
$$\sin x = \frac{e^{ix} - e^{-ix}}{2i}$$
So that your integral is:
$$\int e^{2x} \sin x\ dx = \int \frac{e^{2x+ix} - e^{2x-ix}}{2i} dx= \frac{e^{2x+ix}}{4ix-2x} - \frac{e^{2x-ix}}{4ix+2x}+C$$
$$=\frac{-1}{10}(2i+1)e^{2x+ix} +\frac{1}{10} (2i-1)e^{2x-ix}+C$$
$$=\frac{2}{5}e^{2x}\sin x-\frac{1}{5}e^{2x}\cos x +C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/657389",
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"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
} |
Simplify the expression where $x>y>0$ Simplify $$\frac{\sqrt {x^2+y^2}+x}{y+\sqrt {x^2-y^2}}\cdot\frac{x-\sqrt {x^2+y^2}}{\sqrt {x^2-y^2}-y}$$
Help please, I tried but the answer doesn't match. I did it by multiplying of course and then normally simplifying.
I tried and what came is $\dfrac{-2xy -y^2}{2xy + x^2}$ and the correct answer is $\dfrac{y^2}{2y^2-x^2}$
| $(a+b)(a-b)=a^2-b^2$
So,
$[(\sqrt {x^2+y^2})+x]\cdot [x-(\sqrt {x^2+y^2})]=[x+(\sqrt {x^2+y^2})]\cdot [x-(\sqrt {x^2+y^2})]=x^2-(x^2+y^2)=-y^2$
Can you proceed?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/661114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Closed-Form Solution to Infinite Sum Does the convergent infinite sum
$$
\sum_{n=0}^{\infty} \frac{1}{2^n + 1}
$$
have a closed form solution? Quickly coding this up, the decimal approximation appears to be $1.26449978\ldots$
| Your series can be re-written in terms of the q-polygamma function $\psi_q(z)$ which is simply the logarithmic derivative of the q-gamma function $\Gamma_q(z)$. Both of which are special functions related to the theory of q-series: $$\sum_{n=0}^\infty\frac{1}{2^n+1}=\frac{\psi_{1/4}(1)-\psi_{1/2}(1)-\ln(3)}{\ln(2)}-\frac{3}{2}$$
Also as a consequence of several papers written by Erdos your sum is irrational. Erdos investigated similar series' when studying and also proving the irrationality of an analogous convergent series known as the "Erdős–Borwein constant" - the sum of the reciprocals of all the Mersenne numbers.
It also has several other series representations:
$$\sum_{n=0}^\infty\frac{1}{2^n+1}=\frac{1}{2}+\sum_{n=1}^\infty\frac{1}{2^n-1}-\sum_{n=1}^\infty\frac{2}{2^{2n}-1}=\frac{1}{2}-\sum_{n=1}^\infty\frac{(-1)^{n}}{2^n-1}=\frac{1}{2}-\sum_{n=1}^\infty\frac{\sum_{d\mid n}(-1)^d}{2^n}$$
$$=\frac{1}{2}+\sum_{n=1}^\infty\frac{1}{2^{n^2}}\frac{(2^n+1)}{(2^n-1)}-2\sum_{n=1}^\infty\frac{1}{4^{n^2}}\frac{(4^{n}+1)}{(4^{n}-1)}$$
$$=\frac{1}{2}+\sum_{n=1}^\infty\frac{1}{2^{n^2}}+2\sum_{n=1}^\infty\frac{1}{2^{n^2}(2^n-1)}-2\sum_{n=1}^\infty\frac{1}{4^{n^2}}-4\sum_{n=1}^\infty\frac{1}{4^{n^2}(4^{n}-1)}$$
$$=1+\frac{1}{2}\sum_{n=-\infty}^\infty\frac{1}{2^{n^2}}-\sum_{n=-\infty}^\infty\frac{1}{4^{n^2}}+2\sum_{n=1}^\infty\frac{1}{2^{n^2}(2^n-1)}-4\sum_{n=1}^\infty\frac{1}{4^{n^2}(4^{n}-1)}$$
$$=1+\frac{1}{2}\prod_{n=1}^\infty\frac{(2^{2n-1}+1)^2}{2^{6n-2}(2^{2n}-1)^{-1}}-\prod_{n=1}^\infty\frac{(2^{4n-2}+1)^2}{2^{12n-4}(2^{4n}-1)^{-1}}+\sum_{n=1}^\infty\frac{2}{2^{n^2}(2^n-1)}-\sum_{n=1}^\infty\frac{4}{4^{n^2}(4^{n}-1)}$$
Which comes from the Jacobi triple product identity.
Though in terms of computation, I would say the third series I gave which came as a result of a Lambert series identity would converge the quickest relative to the other expressions I listed:
$$\sum_{n=0}^\infty\frac{1}{2^n+1}=\frac{1}{2}+\sum_{n=1}^\infty\frac{1}{2^{n^2}}\frac{(2^n+1)}{(2^n-1)}-2\sum_{n=1}^\infty\frac{1}{4^{n^2}}\frac{(4^{n}+1)}{(4^{n}-1)}$$
Now in terms of a different "closed form" then the one at the top that uses the q-polygamma function. I would say that it's unlikely you're going to find another such representation in terms of anything other then a similar type of q-series based special function. Though similar lambert series' and q series expressions can take some very nice closed form values when their input is evaluated at exponentials of scaled values of $\pi$, like the series given in the link provided by Mhenni Benghorbal.
Also on a somewhat unrelated note, the inner partial sum appearing in the third series representation I gave for your sum: $$\sum_{d|n}(-1)^d=\frac12(-1)^n\sum_{a^2-b^2=n\\(a,b)\in \Bbb{Z}^2}1$$
Is equivalent to one half of negative one to the $n^{th}$ power multiplied by the number of representations of $n$ as a difference of the squares of two integers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/662795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Evaluate integral as a logarithm plus an arctangent. Evaluate the integral as a logarithm plus an arctangent.
$$ \int \frac{x}{3x^2-18x+45} \ dx $$
I just completed the square and couldn't continue.
$$ \int \frac{x}{3(x-3)^2+18} \ dx $$
Fixed a typo $18$ changed to $18 x$
| Write the numerator as $x-3+3$ to get
$$
\int \frac{x-3}{3 (x-3)^2 + 18}dx + \int \frac{3}{3 (x-3)^2 + 18} dx$$
This gives
$${{\log \left(3\,\left(x-3\right)^2+18\right)}\over{6}}+{{\arctan \left({{x-3}\over{\sqrt{6}}}\right)}\over{\sqrt{6}}}$$
Here are the details.
In the first integral let $(x-3)^2 = u$. Then $2 (x-3) dx = du$ and hence
$$
\int \frac{x-3}{3 (x-3)^2 + 18}dx = \frac{1}{2} \int \frac{du}{3 u + 1}$$
If you do not see the integral right-away, make the second substitution $v=3 u +1$ to get
$dv = 3 du$ and
$$\int \frac 12\frac{du}{3 u + 1} =\frac 12 \frac{1}{3} \int \frac{dv}{v} = \frac{1}{6}\log(v)$$
To get the second integral
$$
\int \frac{3}{3 (x-3)^2 + 18} dx =\int \frac{1}{ (x-3)^2 + 6} dx =
\frac{1}{6} \frac{1}{ (\frac{x-3}{\sqrt{6}})^2 + 1} dx $$
Now make the substitution
$$
\frac{x-3}{\sqrt{6}} = v$$
and use $dx = \sqrt{6} dv$ to ge the second integral as
$$
\frac{1}{\sqrt{6}} \int \frac{dv}{v^2+1} = \frac{1}{\sqrt{6}} \arctan(v)
$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
For which $x$ does the series $∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n})\, x^n$ converge? Determine for what value of $x$ the series converges
$$\sum_{n=1}^\infty \left(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}\right) x^n $$
Observe that
$∑_{n=1}^∞(1+\frac{1}{2}+ \frac{1}{3}+⋯+\frac{1}{n}) x^n =∑_{n=1}^∞ (∑_{n=1}^∞\frac{1}{n}) x^n$
Can I consider $(\sum_{n=1}^\infty\frac{1}{n})$ as $a_n$ and use the ratio test here or I must use the Hadamard?
| Let $H_n=1+\frac{1}{2}+\cdots+\frac{1}{n}$,
Then $\frac{H_{n+1}}{H_n}=1+\frac{1}{(n+1)H_n}$ tends to 1, Thus the ratio of convergence is $R=1$ (D'Alembert criterion).
Now note that the Cauchy product of
$$
\sum_{n=0}^\infty x^n, \quad \sum_{n=1}^\infty \frac{x^n}{n}
$$
is exactly
$$
\sum_{n=1}^\infty H_n\, x^n
$$
Therefore,
$$
\forall x\in (-1,1), \sum_{n=0}^\infty H_n\, x^n=-\frac{ln(1-x)}{1-x}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/668423",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Integrating Factor and the Exact Equation $(y-\frac{1}{x})dx+\frac{1}{y}dy=0$.
$$(y - \frac{1}{x})dx + \frac{1}{y}dy = 0 \,\,\, , \,\,\, y(\sqrt[]{2}) = \sqrt[]{2}$$
So I need to find the integrating factor. $\frac{M_y - N_x}{N} = \frac{1 - 0}{0}$ which we cannot have so that leaves us with only $\frac{N_x - M_y}{M} = -1$. This last condition says if the result is dependent on $y$, then the integrating factor is $u(y) = exp(\int \frac{N_x - M_y}{M})$, but our result was a constant. It relies neither on $x$ or $y$. What now?
| We can get an exact form
$$\left(y - \frac{1}{x} \right)dx + \frac{1}{y}dy = 0 \implies ydx-xdy=xy^2 dx$$
$$\frac{ydx-xdy}{y^2}=xdx \implies d\left(\frac{x}{y} \right)=xdx$$
On integration we get
$$\frac{x}{y} =\frac{x^2}{2} +k$$
$k$ is some constant
Now using $y(\sqrt{2})=\sqrt{2}$ we get $k=0$
Therefore the specific solution is $y=\frac{2}{x}$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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} |
A closed form for the antiderivative of $\frac 1{\sin^5 x +\cos^5 x} $ Does there exist a closed form expression for $$\int \dfrac {dx}{\sin^5 x +\cos^5 x}? $$
| Given $\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx$
First we will simplify $\sin^5 x+\cos^5 x = \left(\sin^2 x+\cos^2 x\right)\cdot \left(\sin^3 x+\cos^3 x\right) - \sin ^2x\cdot \cos ^2x\left(\sin x+\cos x\right)$
$\displaystyle \sin^5 x+\cos^5 x= (\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)$
So Integral is $\displaystyle \int\frac{1}{\sin^5 x+\cos^5 x}dx $
$\displaystyle = \int\frac{1}{(\sin x+\cos x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$
$\displaystyle = \int \frac{(\sin x+\cos x)}{(\sin x+\cos x)^2\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$
$\displaystyle = \int \frac{(\sin x+\cos x)}{(1+\sin 2x)\cdot (1-\sin x\cdot \cos x-\cos^2 x\cdot \sin^2x)}dx$
Let $(\sin x-\cos x) = t\;,$ Then $(\cos +\sin x)dx = dt$ and $(1-\sin 2x) = t^2\Rightarrow (1+\sin 2x) = (2-t^2)$
So Integral Convert into $\displaystyle = 4\int\frac{1}{(2-t^2)\cdot(5-t^4)}dt = 4\int\frac{1}{(t^2-2)\cdot (t^2-\sqrt{5})\cdot (t^2+\sqrt{5})}dt$
Now Using partial fraction, we get
$\displaystyle = 4\int \left[\frac{1}{2-t^2}+\frac{1}{(2-\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}-t^2)}+\frac{1}{(2+\sqrt{5})\cdot 2\sqrt{5}\cdot (\sqrt{5}+t^2)}\right]dt$
$ = \displaystyle \sqrt{2}\ln \left|\frac{\sqrt{2}+t}{\sqrt{2}-t}\right|+\frac{1}{(2-\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \ln \left|\frac{5^{\frac{1}{4}}+t}{5^{\frac{1}{4}}-t}\right|+\frac{2}{(2+\sqrt{5})\cdot 5^{\frac{3}{4}}}\cdot \tan^{-1}\left(\frac{t}{5^{\frac{1}{4}}}\right)+\mathbb{C}$
where $t=(\sin x-\cos x)$
| {
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"url": "https://math.stackexchange.com/questions/670916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Partial fractions: why does $\int dt \implies t + C$ I am working on a partial fraction problem here, I understand everything in the problem except $t+C$, so I'd like to know where did the $t+C$ come from ?
I want to solve this integral
$$
\int \frac{dy}{(y+2)(1-y)} = \int dt
$$
$$\begin{align}
1 &= \frac{A}{y+2} + \frac{B}{1-y} \\
1 &= A(1-y) + B(y+2)
\end{align}$$
Let $y=1$, then $1=B(3)$ and $B=1/3$.
Let $y=-2$, then $1=A(3)$ and $A=1/3$
$$\begin{align}
\int \frac{1/3}{y+2} + \frac{1/3}{1-y} \;dy &= \int dt\\
\frac{1}{3}(\ln|y+2| - \ln|1-y|) &= \color{red}{t+C} & \text{?}\\
\ln\left|\frac{y+2}{1-y}\right| &=3(t+C)\\
\frac{y+2}{1-y} = Ce^{3t}
\end{align}$$
| Think of $\int dt$ as $\int1dt$. The integral of $1$ with respect to any variable is that variable, so in this case, $\int1dt$ is $t+C$. Where $C$ is a constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/671252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I check the convergence of the sequence? Does it diverge? How can I check the convergence of the sequence $\frac{1}{\sqrt{n^2+1}}+\frac{2}{\sqrt{n^2+2}}+\cdots+\frac{n}{\sqrt{n^2+n}}$? I think that it diverges,because it is bounded below from $\frac{n(n+1)}{2\sqrt{n^2+n}} $ and above from $\frac{n(n+1)}{2\sqrt{n^2+1}}$..Is this correct?
| Since $n^2\le n^2+k\le\left(n+\frac12\right)^2$ for $0\le k\le n$, we have
$$
\frac1{n+\frac12}\sum_{k=1}^nk\le\sum_{k=1}^n\frac{k}{\sqrt{n^2+k}}\le\frac1n\sum_{k=1}^nk
$$
Thus,
$$
\frac n2\le\frac{n(n+1)}{2n+1}\le\sum_{k=1}^n\frac{k}{\sqrt{n^2+k}}\le\frac{n+1}2{}
$$
That is, the sequence diverges.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Given positive real numbers $a, b, c$ with $aI am trying to prove the following:
$$\frac{a}{1+a} < \frac{b}{1+b} + \frac{c}{1+c}$$
given that $a, b, c > 0$ and $a < b+c$. I tried various rearrangements but can't seem to get anywhere with it.
| Let us prove the equivalent inequality
$$\frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a} > 0.$$
Since $\frac{n}{1+n} = 1 - \frac{1}{1+n}$, observe that
$$\begin{align*}
\frac{b}{1+b} + \frac{c}{1+c} - \frac{a}{1+a}
&= 1 - \frac{1}{1+b} - \frac{1}{1+c} + \frac{1}{1+a}\\
&> 1 - \frac{1}{1+b} - \frac{1}{1+c} + \frac{1}{1+b + c + bc}\\
&= \left ( 1 - \frac{1}{1+b} \right ) \left ( 1 - \frac{1}{1+c} \right ) \\
&> 0.
\end{align*}$$
The first inequality used is
$$\frac{1}{1+a} > \frac{1}{1 + b + c + bc},$$
which is true since $a < b + c$ and $0 < bc$.
The final inequality holds since $b, c > 0$, so both terms in parentheses are positive.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Expected value for $X, Y \sim \mathcal{U}[0,10]$ vs. expected value of $E(2x)$ $X, Y$ are independent events both $\sim \mathcal{U}[0,10]$.
I know that $$E[X+Y] = E[X] + E[Y] = 5 + 5 = 10$$
and $$E[2X] = 2 E[X] = 10 \Rightarrow E[X+Y] = E[2X].$$
However, is $E\left[(X+Y-10)_+\right] = E\left[(2X-10)_+\right]$ ?
I believe the answer is no, but why?
| The right one is
$$
\frac{1}{10} \int_0^{10} (2x-10)^+dx
= \frac{1}{10} \int_5^{10} (2x-10)dx = 2.5
$$
and the left one is
$$
\begin{split}
\frac{1}{100} \int_0^{10} \int_0^{10} (x+y-10)^+ dydx
&= \frac{1}{100} \int_0^{10} \int_{10-x}^{10} (x+y-10) dydx \\
&= \frac{1}{100} \int_0^{10} \left[ x(x-10) + 50 - (10-x)^2/2 \right] dx \\
&= \frac{1}{100} \frac{500}{3} = \frac{5}{3}.
\end{split}
$$
the intuition is that in the $X+X$ case, knowing the left also fixes the right, but for $X+Y$ the left and the right are independent so there is more possibility for successful outcomes.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/673909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Permutations 1-line notation, and inverse Write (15)(286)(479) in 1-line notation. Find the inverse of (15)(286)(479).
Can anyone please help?
Thank you.
| Let me consider a similar permutation: $$\sigma:=(1\:8\:3)(2\:9)(4\:6\:7).$$ Now, since $9$ is the greatest number that appears in the disjoint cycles, then we can assume that $\sigma$ is a permutation on $\{1,2,3,4,5,6,7,8,9\}.$ In particular, $\sigma$ takes $1\mapsto 8\mapsto 3\mapsto 1,$ $2\mapsto 9\mapsto 2,$ $4\mapsto6\mapsto7\mapsto4,$ and (by process of elimination) $5\mapsto 5.$ The inverse of $\sigma$ will undo all of these mappings--meaning it needs to send $1\mapsto 3\mapsto 8\mapsto1,$ $2\mapsto 9\mapsto 2,$ $4\mapsto 7\mapsto 6\mapsto 4,$ and $5\mapsto 5$. Put back into disjoint cycle notation, $$\sigma^{-1}=(1\:3\:8)(2\:9)(4\:7\:6),$$ or equivalently, $$\sigma^{-1}=(1\:8\:3)^{-1}(2\:9)^{-1}(4\:6\:7)^{-1}.$$ There are other ways we could show this, of course, since $(1\:3\:8)=(3\:8\:1)=(8\:1\:3),$ for example, but this does the trick.
As for rewriting $\sigma$ in one-line notation, it ultimately amounts to figuring out where $\sigma$ takes $1,2,3,4,5,6,7,8,$ and $9,$ respectively. Since $\sigma(1)=8,$ $\sigma(2)=9,$ $\sigma(3)=1,$ $\sigma(4)=6,$ $\sigma(5)=5,$ $\sigma(6)=7,$ $\sigma(7)=4,$ $\sigma(8)=3,$ and $\sigma(9)=2,$ then the one-line notation is: $$8,9,1,6,5,7,4,3,2$$ This is simply a condensed form of the two-line notation: $$\begin{pmatrix}1 & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9\\8 & 9 & 1 & 6 & 5 & 7 & 4 & 3 & 2\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/674046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Ratio and proportion problem
Question:
If $$(a+b):(b+c):(c+a)=6:7:8$$ and $a+b+c=14$, then find the value of $c$.
My solution:
*
*$$\frac{(a+b)(c+a)}{(b+c)}=\frac{(6)(8)}{7}$$$$\Rightarrow \frac{ac + a^2 + bc + ba}{b+c} = \frac{48}{7}$$$$\Rightarrow \frac{a(b+c)+a^2+bc}{b+c}=\frac{48}{7}$$$$\Rightarrow ????$$
*$$\Rightarrow a+b=6x \space\space \text{and} \space \space b+c=7x$$$$\Rightarrow b=6x-a\space\space\space\text{and}\space\space\space b=7x-c$$$$\Rightarrow \text{solving we get}\space\space x = c-a$$$$\Rightarrow ????$$
My query:
I am totally stuck on this problem. Please help.
Thanks a lot!
| $$\frac{a+b}6=\frac{b+c}7=\frac{c+a}8=\frac{-(a+b)+(b+c)+(c+a)}{-6+7+8}=\frac{2c}9$$
Similarly, each ratio is equal to $$\frac{a+b+(b+c)+(c+a)}{6+7+8}=\frac{2(a+b+c)}{21}$$
$$\implies \frac{2c}9=\frac{2(a+b+c)}{21}$$
Now, we have $a+b+c=14$
| {
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"timestamp": "2023-03-29T00:00:00",
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If $x+\frac1x=5$ find $x^5+\frac1{x^5}$. If $x>0$ and $\,x+\dfrac{1}{x}=5,\,$ find $\,x^5+\dfrac{1}{x^5}$.
Is there any other way find it?
$$
\left(x^2+\frac{1}{x^2}\right)\left(x^3+\frac{1}{x^3}\right)=23\cdot 110.
$$
Thanks
| Using the factorization identity $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ we obtain
\begin{align}
x^5+\frac{1}{x^5}&=\left(x+\frac{1}{x}\right)\left(x^4-x^2+1-\frac{1}{x^2}+\frac{1}{x^4}\right) \\ &=\left(x+\frac{1}{x}\right)\left(\left(x+\frac{1}{x}\right)^4-5x^2-5-5\frac{1}{x^2}\right) \\&=\left(x+\frac{1}{x}\right)\left(\left(x+\frac{1}{x}\right)^4
-5\left(x+\frac{1}{x}\right)^2+5\right)=5(5^4-5\cdot5^2+5)=2525,
\end{align}
since
$$
\left(x+\frac{1}{x}\right)^4=x^4+4x^2+6+\frac{4}{x^2}+\frac{1}{x^4}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/678650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 10,
"answer_id": 4
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Proof of Divisibility of $n(n^2+20)$ by 48. This is a question from Bangladesh National Math Olympiad 2013 - Junior Category that still haunts me a lot. I want to find an answer to this question. Please prove this.
If $n$ is an even integer, prove that $48$ divides $n(n^2+20)$.
| Let $n = 2k$ for some integer $k$. Then,
$$\begin{align}2k((2k^2) + 20) &= 2k(4k^2 + 20)\\
&=8k(k^2 + 5)\end{align}$$
But
$$\begin{align}k(k^2 + 5) &\equiv k(k^2-1)\mod 2\\
&=(k+1)(k)(k-1) \mod 2\end{align}$$
Since $2$ must divide at least one of these three consecutive integers, we have $2 \mid k(k^2 + 5) \implies 16 \mid 8k(k^2 + 5)$.
Similarly,
$$\begin{align}k(k^2 + 5) &\equiv k(k^2-1)\mod 3\\
&=(k+1)(k)(k-1) \mod 3\end{align}$$
Using the same argument, $3$ must divide at least one of these three consecutive integers. Hence $3 \mid k(k^2 + 5) \implies 3 \mid 8k(k^2 + 5)$.
Since both $3$ and $16$ divides $n(n^2 + 20)$, we get that $48$ must also divide $n(n^2 + 20)$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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On average, how many times does one need to roll three fair dice to get a sum of 10? Three fair dice are rolled simultaneously. On average, how many times does one need to roll three fair dice to get a sum of $10$?(All rolls are independent of each other, and a roll of the three dice counts as one time.)
So I found the probability that the three dice sum to $10$ to be $.125$. I don't know what to do to find the number of rolls. I feel like it has something to do with expected value but I'm not sure how to set it up.
thanks for the help!
| The generating function for the 3 die is
$$ \frac{x^3}{216}+\frac{x^4}{72}+\frac{x^5}{36}+\frac{5 x^6}{108}+\frac{5 x^7}{72}+\frac{7 x^8}{72}+\frac{25 x^9}{216}+\frac{x^{10}}{8}+\frac{x^{11}}{8}+\frac{25 x^{12}}{216}+\frac{7 x^{13}}{72}+\frac{5 x^{14}}{72}+\frac{5 x^{15}}{108}+\frac{x^{16}}{36}+\frac{x^{17}}{72}+\frac{x^{18}}{216}$$
this looks tough but can be done by pencil and paper. From it we see that the probability is 1 / 8 for one roll of 3 die. From the formula for expectation we get:
$$\frac{1}{8}\cdot 1+\frac{1}{8}\cdot\frac{7}{8}\cdot 2+\frac{1}{8} \left(\frac{7}{8}\right)^2 \cdot 3+\frac{1}{8} \left(\frac{7}{8}\right)^3 \cdot 4 \ +...+ = \frac{1}{8} \sum _{k=1}^{\infty } k \left(\frac{7}{8}\right)^{k-1}=8$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Determining which sets are spanning sets for P3? I have a list of sets:
(a): {$1, x^2, x^2 - 2$}
(b): {$2, x^2, x, 2x + 3$}
(c): {$x + 2, x + 1, x^2 - 1$}
(d): {$x + 2, x^2 -1$}
I am supposed to determine which of the following are spanning sets in $P_3$.
My attempted solution:
(A): Ax^2 + bx + c = A(1) + B(x^2) +C(x^2-2)
Ax^2 + bx + c = (A-2C) + Bx^2 + C(x^2)
This is pretty much as far as I've got and I am not exactly sure what I'm doing or solving for. Any clarification would be wonderful.
| I found b) and c) are the spanning sets.
For b) You only need to show that you can find reals m, n, p such that:
ax^2 + bx + c = 2m + nx^2 + px. You don't need 2x + 3 since 2x + 3 = 2x + (3/2)*2 is in the span{2,x}. So choose n = a, p = b, and m = c/2.
For c) again ax^2 + bx + c = m(x+2) + n(x+1) + p(x^2 - 1). Choose p = a, and :
m = b - a - c, and n = a + c.
For a) x^2 + 2x + 1 is not in the span{1, x^2, x^2 - 2}.
For d) ax^2 + bx + c = m(x+2) + n(x^2 -1) ==> n = a , m = b, c = 2m - n = 2b - a. So you only need to find a quadratic polynomial like x^2 + x - 5, then it is not in the span{x+2, x^2 - 1}.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve equation: $5^x = -2x + 7$ How to solve that equation: $$5^x = -2x + 7$$
I already have the answer $x=1$. Can anyone please explain to me?
| The equation can be solved in terms of the Lambert W function.
Writing $5^x$ as ${\rm e}^{x \ln 5}$ the equation becomes
\begin{eqnarray*}
{\rm e}^{x \ln 5} &=& -2x + 7 \\
\Rightarrow (-2x + 7) {\rm e}^{-x \ln 5} &=& 1 \\
\frac{\ln 5}{2} (-2x + 7) {\rm e}^{-x \ln 5} &=& \frac{\ln 5}{2} \\
(-x \ln 5 + \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5} &=& \frac{\ln 5}{2} \\
(-x \ln 5 + \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5 + \frac{7}{2} \ln 5 - \frac{7}{2} \ln 5} &=& \frac{\ln 5}{2} \\
(-x \ln 5+ \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5 + \frac{7}{2}\ln 5} {\rm e}^{-\frac{7}{2} \ln 5} &=& \frac{\ln 5}{2} \\
\Rightarrow (-x \ln 5 + \frac{7}{2} \ln 5) {\rm e}^{-x \ln 5 + \frac{7}{2} \ln 5} &=& \frac{\ln 5}{2} {\rm e}^{\frac{7}{2} \ln 5}
\end{eqnarray*}
The last equation is now in the form for the defining equation for the Lambert W function, namely
$${\rm W}(x) {\rm e}^{{\rm W}(x)} = x$$
Solving for W one has
$$-x \ln 5 + \frac{7}{2} \ln 5 = {\rm W}_0 \left (\frac{\ln 5}{2} {\rm e}^{\frac{7}{2} \ln 5} \right )$$
Note the principal branch ${\rm W}_0(x)$ for the Lambert W function is chosen since its argument is positive.
Finally, solving for $x$ yields
\begin{equation}
x = \frac{7}{2} - \frac{1}{\ln 5} {\rm W}_0 \left (\frac{\ln 5}{2} {\rm e}^{\frac{7}{2} \ln 5} \right )
\end{equation}
which is the solution we sought.
Getting the above solution into a form more easily recognised we note that since
$${\rm e}^{\frac{7}{2} \ln 5} = {\rm e}^{(\frac{5}{2} + 1) \ln 5} = {\rm e}^{\frac{5}{2} \ln 5} {\rm e}^{\ln 5} = 5 {\rm e}^{\frac{5}{2} \ln 5}$$
the equation for $x$ can be re-written as
$$x = \frac{7}{2} - \frac{1}{\ln 5} {\rm W}_0 \left (\frac{5 \ln 5}{2} {\rm e}^{\frac{5 \ln 5}{2}} \right )$$
Now as the Lambert W function is the inverse of the function $y = x {\rm e}^x$, one has the following simplification
$${\rm W}_0 \left (\frac{5 \ln 5}{2} {\rm e}^{\frac{5 \ln 5}{2}} \right ) = \frac{5 \ln 5}{2}$$
from which we get
$$x = \frac{7}{2} - \frac{1}{\ln 5} \frac{5 \ln 5}{2} = \frac{7}{2} - \frac{5}{2} = 1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/684864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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geometric series used to work out big O notation for resizing an array in a stack It's a geometric series
$$
1 + 2 + 4 + \cdots + 2^k = \frac{1 - 2^{k+1}}{1 - 2}
$$
Here, $2^k$ = N. You get $1 + 2 + 4 + \cdots + N = \frac{1 - 2N}{-1}$. Therefore, $2 + 4 + \cdots + N = 2N−2$. When $N$ is big, you can just drop the $−2$ to get big $O$ notation.
Above is the working out I was given. The array is doubled once it is full. So when an array of size 1 has 1 item, the array doubles to size $2$. What I am trying to find out is why
$$
1 + 2 + 4 + \cdots + 2^k = \frac{1 - 2^{k+1}}{1 - 2}
$$
and $2^k = N$. I was wondering if step by step workings can be shown and an explanation.
| $$S=1 + 2 + 4 + \cdots + 2^k
\\2S= 2 + 4 + \cdots + 2^k+2^{k+1}
\\S-2S=1-2^{k+1}
\\S(1-2)= 1-2^{k+1}
\\S= \frac{1 - 2^{k+1}}{1 - 2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/686999",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\mathbb{E}[B^4(t)]$ with $B$= brownian motion Can anyone help me to find:
$\mathbb{E}[B^4(t)]$ where $B$ is a brownian motion?
I thought using this density function:
$f_{B_t}(x) = \frac{1}{\sqrt{2 \pi t}} e^{-\frac{x^2}{2t}}$,
but I don't know how to apply it.
| Note that
$$\mathbb{E}(B_t^4) = \frac{1}{\sqrt{2\pi t}} \int_{\mathbb{R}} x^4 \exp \left(- \frac{x^2}{2t} \right) \, dx = \frac{2}{\sqrt{2\pi t}} \int_{0}^{\infty} x^4 \exp \left(- \frac{x^2}{2t} \right) \, dx.$$
If we set $y := x^2/2t$, then
$$\begin{align*} \mathbb{E}(B_t^4) &= \frac{2}{\sqrt{2\pi t}} \int_0^{\infty} (2t y)^2 \cdot e^{-y} \frac{\sqrt{t}}{\sqrt{2y}} \, dy = \frac{4t^2}{\sqrt{\pi}} \int_0^{\infty} y^{3/2} e^{-y} \, dy \\ &= \frac{4t^2}{\sqrt{\pi}} \Gamma \left( \frac{5}{2} \right) \end{align*}$$
where $\Gamma$ denotes the Gamma function. Using
$$\Gamma \left( \frac{5}{2} \right) = \frac{3}{2} \Gamma \left( \frac{3}{2} \right) = \frac{3}{2} \frac{1}{2} \Gamma \left( \frac{1}{2} \right) = \frac{3}{2} \frac{1}{2} \sqrt{\pi}$$
we get
$$\mathbb{E}(B_t^4) = 3t^2.$$
Remark This calculation holds for any Gaussian random variable with mean $0$ and variance $t$. Since the moments of Gaussian random variables are known you do not need to perform this calculations - unless you are interested in the proof of this well-known result.
| {
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Checking multivariable function's differentiability at $(0,0)$ I have a function: $$f(x,y) = \begin{cases} (x^2+y^2)\sin (x^2+y^2)^{-1} \ \ \ &\text{, if } x^2+y^2 \neq 0 \\ 0 &\text{, if } x^2+y^2=0 \end{cases}$$
I calculated the partial derivatives:
$$f_x(x,y)=2x\left( \sin (x^2+y^2)^{-1}-\frac{(x^2+y^2)\cos (x^2+y^2)^{-1}}{(x^2+y^2)^2} \right)$$
$$f_y(x,y)=2y\left( \sin (x^2+y^2)^{-1}-\frac{(x^2+y^2)\cos (x^2+y^2)^{-1}}{(x^2+y^2)^2} \right)$$
I'm stuck at checking the function's differentiability at point $(0,0)$, I use following definition :
$f$ is differentiable at point $(a,b)$ and $M,N\in \mathbb{R}$ $\leftrightarrow$
$\lim_{h_1,h_2\to 0} \frac{f(a+h_1,b+h_2)-f(a,b)-Mh_2-Nh_2}{\sqrt{h_1^2+h_2^2}}=0$
Here's what I've done so far:
$$\lim_{h_1,h_2\to 0} \frac{(h_1^2+h_2^2)\sin(h_1^2+h_2^2)^{-1}-Mh^1-Nh_2}{\sqrt{h_1^2+h_2^2}}=$$
$$= \lim_{h_1,h_2\to 0} \frac{\sqrt{h_1^2+h_2^2}\sqrt{h_1^2+h_2^2}\sin(h_1^2+h_2^2)^{-1}-Mh^1-Nh_2}{\sqrt{h_1^2+h_2^2}}=$$
$$=\lim_{h_1,h_2\to 0} \frac{\sqrt{h_1^2+h_2^2}\left(\sqrt{h_1^2+h_2^2}\sin(h_1^2+h_2^2)^{-1}-\frac{Mh_1}{\sqrt{h_1^2+h_2^2}}-\frac{Nh_2}{\sqrt{h_1^2+h_2^2}}\right)}{\sqrt{h_1^2+h_2^2}}=$$
$$=\lim_{h_1,h_2\to 0} \sqrt{h_1^2+h_2^2}\sin(h_1^2+h_2^2)^{-1}-\frac{Mh_1-Nh_2}{\sqrt{h_1^2+h_2^2}}=$$
| If a radial function $f(x,y) = g(x^2+y^2)$ is differentiable, its differential should be zero, by simmetry.
So you should check whether
$$
\lim_{(x,y)\to(0,0)}\frac{\sqrt{x^2+y^2}}{\sin(x^2+y^2)} = 0.
$$
But this is false even for $y=0$ where you get
$$
\lim_{x\to 0}\frac{|x|}{\sin(x^2)} = +\infty.
$$
So your function is not differentiable.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"answer_count": 2,
"answer_id": 0
} |
Proof by mathematical induction - Fibonacci numbers and matrices Using mathematical induction I am to prove:
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^n $ =
$ \left( \begin{array}{ccc}
F_{n+1} & F_n \\
F_n & F_{n-1} \end{array} \right) $
where $F_k$ represents the $k^{th}$ Fibonacci number.
my base case is $n =2$
LHS: $ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right) \times$
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right) $$=$
$ \left( \begin{array}{ccc}
2 & 1 \\
1 & 1 \end{array} \right) $
RHS: $ \left( \begin{array}{ccc}
F_3 & F_2 \\
F_2 & F_1 \end{array} \right) $$=$
$ \left( \begin{array}{ccc}
2 & 1 \\
1 & 1 \end{array} \right) $
So $n = k + 1$
$ \left( \begin{array}{ccc}
F_{k+2} & F_{k+1} \\
F_{k+1} & F_k \end{array} \right) $
So for my inductive step I did:
$ \left( \begin{array}{ccc}
F_{k+1} & F_k \\
F_k & F_{k-1} \end{array} \right) $ $+$
$ \left( \begin{array}{ccc}
1 & 1 \\
1 & 0 \end{array} \right)^{k+1} $
And now I'm not sure where to proceed from here. Can anyone point me in the right direction? Assuming my previous work is correct.
| Inductive proof:
For $n=1$ is true, as the OP correctly observed.
Assume that it is true for $n=k$. Then
$$
\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)^k=\left(\begin{matrix} F_{k+1}& F_k \\
F_k& F_{k-1}\end{matrix}\right)
$$
Then for $n=k+1$ we have
\begin{align}
\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)^{k+1}&=\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)^{k}\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)
=\left(\begin{matrix} F_{k+1}& F_k \\
F_k& F_{k-1}\end{matrix}\right)\left(\begin{matrix} 1& 1 \\ 1& 0\end{matrix}\right)=
\left(\begin{matrix} F_k+F_{k+1}& F_{k+1} \\ F_{k-1}+F_{k}& F_k\end{matrix}\right)\\ &=
\left(\begin{matrix} F_{k+2}& F_{k+1} \\ F_{k+1}& F_k\end{matrix}\right),
\end{align}
and hence it is true for $n=k+1$. Note that in the last equality above we used the recursive definition of Fibonacci sequence.
| {
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How to show whether a $3\times 4$ matrix has no solution, a unique solution or infinitely many solutions? The system is :
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
-2 & 5 & -4 & -1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
After Gaussian elimination, I found that
$$
\begin{array}{cccc|cc}
1 & -4 & 6 & a & & 0 \\
0 & 1 & -\tfrac{8}{3} & - \left( 2a- \tfrac{1}{3} \right) & & - \tfrac{1}{3}b \\
0 & 0 & 0 & 10-5a & & c-2b
\end{array}
$$
Is it correct and I can continue to determine whether there is no solution, a unique solution or infinitely many solutions?
Here are the operations:
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
-2 & 5 & -4 & -1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
$R_2+2R_1\rightarrow R_2$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & -3 & 8 & 2a-1 & | & b \\
1 & -10 & 22 & 8 & | & c
\end{matrix}
$$
$R_3-R_1\rightarrow R_3$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & -3 & 8 & 2a-1 & | & b \\
0 & -6 & 16 & 8-a & | & c
\end{matrix}
$$
$R_3-2R_2\rightarrow R_3$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & -3 & 8 & 2a-1 & | & b \\
0 & 0 & 0 & 10-5a & | & c-2b
\end{matrix}
$$
$-\frac 13(R_2)\rightarrow R_2$
$$
\begin{matrix}
1 & -4 & 6 & a & | & 0 \\
0 & 1 & -\frac 83 & -\tfrac{2a-1}{3} & | & -\frac 13b \\
0 & 0 & 0 & 10-5a & | & c-2b
\end{matrix}
$$
| There's no solution if there's a row of the form $0 0 0 0 | Q$ where $Q$ is not zero. Which row could possibly look like that? What would have to happen (to $a$, $b$, and $c$) for the row to look like that?
If there is a solution, then the variable represented by the 3rd column is arbitrary, so there are infinitely many solutions.
| {
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} |
Infinite series for partial sums of square roots. Can you prove these infinite series for partial sums of square roots?
$$\sqrt{1}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\right)$$
$$\sqrt{1}+\sqrt{2}=\sum\limits_{n=1}^{\infty}\left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}-\frac{2}{\sqrt{n+2}}\right)$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}=\sum\limits_{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}-\frac{3}{\sqrt{n+3}}\right)$$
$$\sqrt{1}+\sqrt{2}+\sqrt{3}+\sqrt{4}=\sum _{n=1}^{\infty } \left(\frac{1}{\sqrt{n}}+\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}-\frac{4}{\sqrt{n+4}}\right)$$
$$\cdots$$
And is there some easy cancellation that I have missed on the right hand side?
Mathematica:
Clear[s, i, n, j]
s = 1/2;
i = 1;
j = 0;
Sum[1/(n + 0)^s - 1/(n + 1)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s - 2/(n + 2)^s + 1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s - 3/(n + 3)^s + 1/(n + 0)^s,
{n, 1, Infinity}]
N[%, 20]
Sum[1/(n + 1)^s + 1/(n + 2)^s + 1/(n + 3)^s - 4/(n + 4)^s +
1/(n + 0)^s, {n, 1, Infinity}]
N[%, 20]
N[Accumulate[Sqrt[Range[4]]], 20]
| $$
\begin{align}
\sum_{n=1}^\infty\left(\left(\sum_{k=0}^{m-1}\frac1{\sqrt{n+k}}\right)-\frac{m}{\sqrt{n+m}}\right)
&=\lim_{N\to\infty}\sum_{n=1}^N\sum_{k=0}^{m-1}\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\
&=\lim_{N\to\infty}\sum_{k=0}^{m-1}\sum_{n=1}^N\left(\frac1{\sqrt{n+k}}-\frac1{\sqrt{n+m}}\right)\\
&=\lim_{N\to\infty}\sum_{k=0}^{m-1}\left(\sum_{n=k+1}^{k+N}\frac1{\sqrt{n}}-\sum_{n=m+1}^{m+N}\frac1{\sqrt{n}}\right)\\
&=\lim_{N\to\infty}\sum_{k=0}^{m-1}\left(\sum_{n=k+1}^m\frac1{\sqrt{n}}-\sum_{n=k+N+1}^{m+N}\frac1{\sqrt{n}}\right)\\
&=\sum_{k=0}^{m-1}\sum_{n=k+1}^m\frac1{\sqrt{n}}\\
&=\sum_{n=1}^m\sum_{k=0}^{n-1}\frac1{\sqrt{n}}\\
&=\sum_{n=1}^m\sqrt{n}
\end{align}
$$
| {
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Do not understand formula... How does;
$x^{n} -y^{n}=(x-y)(x^{n-1} + x^{n-2}y+...+x y^{n-2}+ y^{n-1} )$
work on $x^{2} - y^{2}$
When I attempt to apply the formula on $x^{2} - y^{2}$
I get the following
$x^{2} - y^{2} =(x-y)( x^{1} + x^{0}y+...+x y^{0} + y^{1} )$
$x^{2} - y^{2} =(x-y)(2x+2y)$
which is obviously false. What is the correct way to use the formula?
| If you use proper notation, instead of dots, you get the correct result:
$x^n-y^n=(x-y)\sum\limits_{k=0}^{n-1}x^{n-1-k}y^k$
Then $x^2-y^2=(x-y)(x^{1-0}y^0+x^{1-1}y^1)=(x-y)(x+y).$
| {
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} |
Express a vector as a combination of a linearly dependent set So I have to express the vector
\begin{pmatrix}
-2 \\
-4 \\
1 \\
0 \\
\end{pmatrix}
as a combination of these vectors:
\begin{pmatrix}
1 \\
0 \\
2 \\
1 \\
\end{pmatrix}
\begin{pmatrix}
1 \\
1 \\
0 \\
2 \\
\end{pmatrix}
\begin{pmatrix}
-1 \\
-3 \\
4 \\
-4 \\
\end{pmatrix}
\begin{pmatrix}
1 \\
2 \\
-1 \\
1 \\
\end{pmatrix}
But when I solved the augmented matrix, I got
\begin{pmatrix}
1 & 0 & 2 & 0 & -1 \\
0 & 1 & -3 & 0 & 2 \\
0 & 0 & 0 & 1 & -3 \\
0 & 0 & 0 & 0 & 0 \\
\end{pmatrix}
Meaning the system is linearly dependent. How should I go about expressing a linear combination without a linearly independent set?
| I got the following set of equations:
$$\begin{align}
a_1 + 2a_3 &= -1 \\
a_2 - 3a_3 &= 2 \\
a4 &= -3
\end{align}$$
There were infinitely many solutions. I set $a_3 = 0$ and got the following as a valid linear combination:
$$-v_1 + 2v_2 + -3v_4$$
I tried this and got a failing answer before, but it was because I used $v_3$ in place of $v_4$.
| {
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} |
Square root of x plus square root of x plus ... I want to confirm if my answer in this problem is correct:
$$\sqrt{(x + \sqrt{(x + ...))}} = (1 + \sqrt{53}) / 2 $$
Solution:
$$x + \sqrt{(x + \sqrt{(x + ...))}} = (1 + \sqrt{53})^2 / 4 $$
$$x + (1 + \sqrt{53}) / 2 = (1 + 2\sqrt{53} + 53) / 4 $$
$$ x + (1 + \sqrt{53}) / 2 = (54 + 2\sqrt{53}) / 4 $$
$$x + (1 + \sqrt{53}) / 2 = (27 + \sqrt{53}) / 2 $$
$$2x + 1 + \sqrt{53} = 27 + \sqrt{53}$$
$2x = 26$
$x = 13$
I also tried solving something like
$$\sqrt{x + \sqrt{x + \sqrt{x + \sqrt(x)}}} = (1 + \sqrt{53}) / 2 $$
and the answer is $x = 13.0006$
Please confirm if my answer is correct or not. Thanks!
| Yes your answer is correct. Let's write it more readable:
Let
$$y=\sqrt{x+\sqrt{x+\cdots}}$$
hence the equation is
$$y=\frac{1+\sqrt{53}}{2}$$
so squaring the terms of the last equality gives
$$y^2=x+y=\frac{(1+\sqrt{53})^2}{4} $$
hence we have
$$x=y^2-y=\frac{(1+\sqrt{53})^2}{4}-\frac{1+\sqrt{53}}{2}=\frac{1+2\sqrt{53}+53-2-2\sqrt{53}}{4}=\frac{52}{4}=13$$
| {
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"answer_id": 0
} |
How to prove $\sin{b}\sin{c}\sin{(b-c)}(\sin^2{b}+\sin^2{c}+\sin^2{(b-c)})+\dots=0$
If $a,b,c\in (0,\pi)$ and $a+b+c=\pi$, show that:
$$\sin b \sin c \sin(b-c) \left(\sin^2 b + \sin^2 c + \sin^2(b-c)\right) \\
+ \sin c \sin a \sin(c-a) \left(\sin^2 c + \sin^2 a + \sin^2(c-a)\right) \\
+ \sin a \sin b \sin(a-b) \left(\sin^2 a + \sin^2 b + \sin^2(a-b)\right) \\
+ \sin(b-c)\sin(c-a)\sin(a-b) \left(\sin^2(b-c) + \sin^2(c-a) + \sin^2(a-b)\right) = 0$$
This problem is from my friend. He asked me and I can't solve it. Thank you for your help.
My idea: let $\sin{a}=x,\sin{b}=y,\sin{c}=z$. Then $LHS=\cdots$.
I fell very hard, so I can't do any work. Thank you.
| Consider the following configuration:
where $ABC$ is an acute triangle and $X'$ is the symmetric of $X$ with respect to the perpendicular bisector of $YZ$. The six depicted points are clearly concyclic, and by assuming that the diameter of the circumcircle is one we have:
$$ z = XY = \sin\widehat{Z},\quad XX'=\left|\sin(\widehat{Y}-\widehat{Z})\right|.$$
Without loss of generality, we can further assume that $\widehat{A}\geq\widehat{B}\geq\widehat{C}$, as depicted above. By applying the Ptolemy's theorem to the isosceles trapezoid $AA'CB$ we get:
$$ AA'\cdot a = AA'\cdot BC = AC^2 - AB^2 = b^2-c^2,$$
in the same way:
$$ BB'\cdot b = BB'\cdot AC = BC^2 - AB^2 = a^2-c^2,$$
$$ CC'\cdot c = CC'\cdot AB = BC^2 - AC^2 = a^2-b^2.$$
Now we have to prove:
$$ bc AA'(b^2+c^2+AA'^2) - ac BB'(a^2+c^2+BB'^2) + ab CC' (a^2+b^2+CC'^2) - AA'BB'CC'(AA'^2+BB'^2+CC'^2)=0.$$
We multiply everything by $a^3 b^3 c^3$ in order to remove $AA',BB'$ and $CC'$ through the previous Ptolemy's identities. We end with:
$$\begin{eqnarray*}
&b^4 c^4 a AA'(a^2b^2+a^2c^2+a^2 AA'^2)\\
-&a^4 c^4 b BB'(a^2b^2+b^2c^2+b^2 BB'^2)\\
+&a^4 b^4 c CC'(a^2c^2+b^2c^2+c^2 CC'^2)\\
-&a AA' b BB' c CC'(b^2c^2 a^2 AA'^2+a^2 c^2 b^2 BB'^2+a^2 b^2 + c^2 CC'^2)=0,
\end{eqnarray*}$$
or:
$$\sum_{cyc}b^4 c^4(b^2-c^2)(a^2b^2+a^2c^2+(b^2-c^2)^2)+\left(\prod_{cyc}(b^2-c^2)\right)\cdot\sum_{cyc}b^2c^2(b^2-c^2)^2 = 0.$$
$$\sum_{cyc}a^2 b^4 c^4(b^4-c^4)+ \sum_{cyc}b^4 c^4(b^2-c^2)^3+\left(\prod_{cyc}(b^2-c^2)\right)\cdot\sum_{cyc}b^2c^2(b^2-c^2)^2 = 0.$$
Since $\prod_{cyc}(x-y)=-\sum_{cyc}xy(x-y)$, the last identity is tedious to check but straightforward.
| {
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"source": "stackexchange",
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Values of the limit $\lim\limits_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$ Hi I have a question regarding finding the values of limit for the following question.
Let $a, b \in \mathbb R$. Find the limit
$$\lim_{x\to+\infty}\left(\sqrt{(x+a)(x+b)}-x\right)$$
| Hint: Multiply the numerator and denominator by the conjugate of the expression.
That is, you can try working with $$\dfrac{\sqrt{(x+a)(x+b)} - x}1\cdot\frac{\sqrt{(x + a)(x + b)} + x}{\sqrt{(x + a)(x + b)} + x} = \dfrac{(x+a)(x + b) - x^2}{\sqrt{(x + a)(x+b)} + x}$$
Expanding $(x + a)(x + b)$ in the numerator and simplifying gives us
$$ \dfrac{(a+b)x + ab}{\sqrt{x^2 + (a+b)x + ab} + x}$$
Now divide numerator and denominator by $x$ to find your limit.
$$ \lim_{x\to \infty} \dfrac{(a+b) + \frac{ab}x}{\sqrt{1 + \frac{(a+b)}x + \frac{ab}{x^2}} + 1} = \dfrac{a+b}{2}$$
| {
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"answer_id": 3
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Curve parametrization When plotting following implicit function $x^4 + y^4 = 2xy$, what is the best way to parametrize it? I tried $y=xt$, but then I get $x=\sqrt{2t\over {1+t^4}}$ and $y=\sqrt{2t^3\over {1+t^4}}$, which means I am restricted by $x>0, y>0, t>0$. It is easy to prove that the curve is symmetric about the origin, but still I can't say anything about the second a fourth quadrant...
Thanks!
| That relation requires $xy \ge 0$, which translates to
$$
\left [
\begin{array}{c}
\left \{
\begin{array}{c}
x \ge 0 \\ y \ge 0
\end{array}
\right . \\
\left \{
\begin{array}{c}
x < 0 \\ y < 0
\end{array}
\right .
\end{array}
\right .
$$
So, the curve defined by the relation doesn't go through second and fourth quadrants. Simple $x \to -x$ and $y \to -y$ substitutions show that curve is symmetric w.r.t. origin. So, indeed you may consider only first quadrant by the parametrization you suggested and simply reflect it from the origin due to the symmetry. Otherwise, if you want rigorous derivation, you need to keep in mind that $x^2 = a$ means that $x = \pm \sqrt a$, for $a \ge 0$, not just $\sqrt a$.
$$
\left [
\begin{array}{l}
\left \{
\begin{array}{c}
x = \sqrt{\frac {2t}{1+t^4}} \\
y = \sqrt{\frac {2t^3}{1+t^4}}
\end{array}
\right . \\
\left \{
\begin{array}{c}
x = -\sqrt{\frac {2t}{1+t^4}} \\
y = -\sqrt{\frac {2t^3}{1+t^4}}
\end{array}
\right .
\end{array}
\right .
$$
and $t \ge 0$.
Update
You can also use angle-based parametrization per @user86418' post
$$
x = \pm \sqrt{r \cos \theta} \\
y = \pm \sqrt{r \sin \theta}
$$
so curve equation becomes
$$
r = 2\sqrt{\sin \theta \cos \theta} = \sqrt{2 \sin 2\theta}
$$
By substituting to original equation one can find that
$$
\begin{array}{c}
x = \pm \sqrt[4]{2\sin 2\theta \cos^2 \theta} \\
y = \pm \sqrt[4]{2\sin 2\theta \sin^2 \theta}
\end{array}, \qquad \theta \in \left [ 0, \frac \pi 2\right ]
$$
| {
"language": "en",
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"source": "stackexchange",
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Is there a formula for $1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$?
Given
$$f(n) = 1 + (1 + 2) + (1 + 2 + 3)+ \cdots + (1 + 2 + 3 +\cdots + n)$$
I am wondering if there is a straightforward formula to compute $f(n)$ and how it may be derived.
The only reduction I thought about so far would be:
$$n\cdot 1 + (n - 1)\cdot 2 + (n - 3)\cdot 3 +\cdots $$
which seems symmetrical; for example, an odd and even n
5 4 3 2 1 4 3 2 1
* *
1 2 3 4 5 1 2 3 4
but I'm not sure if and how it may help derive a formula.
| I think the initial steps towards a formula look a bit simpler than seen so far here.
Beginning with
$$ H^{(2)}(n) = 1 + (1+2) + (1+2+3) + ... + (1+2+3+4+...+n) $$
we count the number of occurences of the $1$, of the $2$
$$ H^{(2)}(n) = 1\cdot n + 2 \cdot (n-1) + 3\cdot (n-2) + ... + n \cdot 1 $$
and rewrite a bit
$$ \begin{align} H^{(2)}(n) &= 1\cdot ((n+1)-1) + 2 \cdot ((n+1)-2) + 3\cdot ((n+1)-3) + ... + n \cdot ((n+1)-n) \\ \qquad \\
&= (1+2+3+...+n)(n+1) - (1^2+2^2+3^2+...+n^2) \\ \qquad \\
&= {n(n+1) \over 2}(n+1) - {2n^3 + 3n^2 + 1n\over 6} \end{align}$$
The last steps stems from the faulhaber formulae for sums-of-like-powers.
Final step, expanding & collecting this gives the same result as derived by other answerers:
$$ H^{(2)}(n) = {3n(n+1)(n+1)-2n^3 - 3n^2 - 1n \over 6} \\
= {n^3+3n^2+2n \over 6} \\
= {n(n+1)(n+2) \over 3!} $$
For $n=5$ we get the sum $H^{(2)}(5)=35$
| {
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How prove this inequality $\frac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\le\sqrt[4]{\frac{a(a+b)(a+b+c)(a+b+c+d)}{24}}$ let $a,b,c,d>0$, show that
$$\dfrac{a+\sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}}{4}\le\sqrt[4]{\dfrac{a(a+b)(a+b+c)(a+b+c+d)}{24}}$$
This post three -varible $a,b,c>0,a+b+c=21$ prove that $a+\sqrt{ab} +\sqrt[3]{abc} \leq 28$
How prove it Four-varible ? I think use AM-GM inequality to solve it,But I can't.Thank you
I guess this follow is also true:
let $a_{1},a_{2},\cdots,a_{n}>0$,show that
$$\dfrac{a_{1}+\sqrt{a_{1}a_{2}}+\sqrt[3]{a_{1}a_{2}a_{3}}+\sqrt[4]{a_{1}a_{2}a_{3}a_{4}}+\cdots+\sqrt[n]{a_{1}a_{2}\cdots a_{n}}}{n}\le\sqrt[n]{\dfrac{a_{1}(a_{1}+a_{2})(a_{1}+a_{2}+a_{3})\cdots(a_{1}+a_{2}+\cdots+a_{n})}{n!}}$$
Thank you
| \begin{align}
a& \frac{a+b}2 \frac{a+b+c}3 \frac{a+b+c+d}4 \\
&= \frac1{4^4} \left(a+a+a+a \right) \left(a+a+b+b \right) \left(a+b+\tfrac{a+b+c}3+c \right) \left(a+b+c+d \right) \\
&\ge \frac1{4^4} \left(a+a+a+a \right) \left(a+a+b+b \right) \left(a+b+\sqrt[3]{abc}+c \right) \left(a+b+c+d\right) \\
&\ge \frac1{4^4} \left(a + \sqrt{ab}+\sqrt[3]{abc}+\sqrt[4]{abcd}\right)^4 \quad \text{by Holder}
\end{align}
Not sure if one can extend the pattern to general $n$ though. For that, you may want to see http://www.jstor.org/stable/2975630
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/714101",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
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if $f'''(x)$ is continuous everywhere and $\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$ Compute $f''(0)$
if $f'''(x)$ is continuous everywhere and $$\lim_{x \to 0}(1+x+ \frac{f(x)}{x})^{1/x}=e^3$$ Compute $f''(0)$
The limit equals to $$\begin{align} \lim_{x \to 0} \frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=0. \end{align}$$
From $$\frac{\log(1+x+ \frac{f(x)}{x})}{x}-3=o(1)$$ as $x \to 0$, I get $$1+x+\frac{f(x)}{x} = e^{3x+o(x)},$$ and $$f(x)=x(e^{3x+o(x)}-x-1),\frac{f(x)}{x}=e^{3x+o(x)}-x-1$$ as $x \to 0$. So both $f(0)$ and $f'(0)$ are $0$. Approximating $e^{3x+o(x)}=1+3x+o(x)$ I get $$\begin{align} f(x) &= x(1+3x+o(x)-x-1) =2x^2+o(x^2). \end{align}$$
Now I try to use the definition of derivative to calculate the $f''(x)$
$$f''(x)=\lim_{x \to 0}\frac{f'(x)-f'(0)}{x}=\lim_{x \to 0} \frac{f'(x)}{x}$$
I'm not sure whether I can get $f'(x)$ by differentiating the approximation $2x^2+o(x^2)$ and how to differentiate $o(x^2)$.
| Taking logs (as done in OP's post) we can see that $$\lim_{x \to 0}\dfrac{\log\left(1 + x + \dfrac{f(x)}{x}\right)}{x} = 3$$ or $$\lim_{x \to 0}\log\left(1 + x + \frac{f(x)}{x}\right) = 0$$ or $$\lim_{x \to 0}1 + x + \frac{f(x)}{x} = 1$$ or $$\lim_{x \to 0}\frac{f(x)}{x} = 0 \Rightarrow \lim_{x \to 0}f(x) = 0$$ and hence by continuity $f(0) = 0$. Now we can see that $$f'(0) = \lim_{x \to 0}\frac{f(x) - f(0)}{x} = \lim_{x \to 0}\frac{f(x)}{x} = 0$$ Now let $f''(0) = a$ and we have via Taylor's theorem $$f(x) = f(0) + xf'(0) + \frac{x^{2}}{2}f''(0) + o(x^{2})$$ This assumes only the existence of $f''(0)$ and nothing more. We have thus finally $$f(x) = \frac{ax^{2}}{2} + o(x^{2})$$ and therefore
\begin{align}
3 &= \lim_{x \to 0}\dfrac{\log\left(1 + x + \dfrac{f(x)}{x}\right)}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\log\left(1 + x + \dfrac{ax}{2} + o(x)\right)}{x}\notag\\
&= \lim_{x \to 0}\dfrac{\log\left(1 + x + \dfrac{ax}{2} + o(x)\right)}{x + \dfrac{ax}{2} + o(x)}\cdot\dfrac{x + \dfrac{ax}{2} + o(x)}{x}\notag\\
&= \lim_{x \to 0}\left(1 + \frac{a}{2} + o(1)\right)\notag\\
&= 1 + \frac{a}{2}\notag
\end{align}
and hence $a = f''(0) = 4$.
| {
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"url": "https://math.stackexchange.com/questions/714564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How find this $\frac{yf_{y}-z}{f_{x}}+\frac{xf_{x}-z}{f_{y}}-xf_{x}-yf_{y}+x+y+z=C$ solution In plane $R^3$,Find $z=f(x,y)$, such the length of the portion of any tangent line to the astroid
$$z=f(x,y)$$
cut off by the coordinate axes is constant $C$,
This problem is from this post (when I answer it) How find a solution to this PDE $\frac{xf'_{x}}{f'_{y}}+\frac{yf'_{y}}{f'_{x}}+x+y=C$
my idea: I think this is anser is $$\sqrt{x}+\sqrt{y}+\sqrt{z}=C$$,because we easy to find this function is such it,
follow is my partial answer:
let $z=f(x,y)$,then the tangent plane is
$$f_{x}[X-x]+f_{y}[Y-y]=Z-z$$
so
$$X=\dfrac{yf_{y}-z}{f_{x}}+x,Y=\dfrac{xf_{x}-z}{f_{y}}+y,Z=z-xf_{x}-yf_{y}$$
so
$$X+Y+Z=C\Longrightarrow \dfrac{yf_{y}-z}{f_{x}}+\dfrac{xf_{x}-z}{f_{y}}-xf_{x}-yf_{y}+x+y+z=C$$
then How find $z=f(x,y)?$
I know this PDE have one solution
$$\sqrt{x}+\sqrt{y}+\sqrt{z}=C $$ is such it,But Now How prove it?
Thank you,I fell this problem is interesting. I hope someone can help me,Thank you
| my answer is :
firstly,we assume that:
$$X=\dfrac{yf_{y}-z}{f_{x}}+x=P+x,Y=\dfrac{xf_{x}-z}{f_{y}}+y=Q+y,Z=z-xf_{x}-yf_{y}=z-R$$
then,
$C^2=x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}=C+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}-P-Q+R$
$C(C-1)=(X+Y+Z)(X+Y+Z-1)=(X+Y+Z)^2-(X+Y+Z)=(x+y+z+P+Q-R)^2-(x+y+z+P+Q-R)=2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}-P-Q+R$$\Longrightarrow$$x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{xz}=C^2$
$\Longrightarrow\sqrt{x}+\sqrt{y}+\sqrt{z}=x+y+z+P+Q-R$$\Longrightarrow$$\frac{f_{x}}{f_{y}}-{f_{x}}=\frac{1}{\sqrt{x}}-1$$,\frac{f_{y}}{f_{x}}-{f_{y}}=\frac{1}{\sqrt{y}}-1$$,\frac{{1}}{f_{x}}+\frac{1}{f_{y}}=1-\frac{1}{\sqrt{z}}$$\Longrightarrow$$f_{x}=-\frac{\sqrt{z}}{\sqrt{x}},f_{y}=-\frac{\sqrt{z}}{\sqrt{y}}$
hence, $C^2=C^2$ holds
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/715357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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what is the proof for $\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} = \frac{1}{4} $ Can someone provide a proof for the solution of this series
$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)} = \frac{1}{4} $
| $$\dfrac{1}{n(n+1)(n+2)}=\dfrac{1}{2}\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)$$
so
\begin{align*}\sum_{n=1}^{\infty}\dfrac{1}{n(n+1)(n+2)}&=\dfrac{1}{2}\sum_{n=1}^{\infty}\left(\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}\right)\\
&=\dfrac{1}{2}\lim_{n\to\infty}\left(\dfrac{1}{1\times 2}-\dfrac{1}{(n+1)(n+2)}\right)\\
&=\dfrac{1}{4}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/721749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Two circles inside a right angled triangle! The other day I was playing with Ms Paint drawing circles here and there - I coincidentally drew a circle inside a right angled triangle which I already drew. Strangely A problem struck to my mind and I tried solving it , but I was unable to do so. I put forward the statement of the problem which I managed to frame myself:
Problem: The legs of a right angled triangle are of length $a$ and $b$. Two circles with equal radii are drawn such such that they touch each other and sides of the triangle as shown in the figure. Find the radius of the circle in terms of $a$ and $b$.
Figure (of course my MS Paint one)
Further Scope - Is there any way to generalize this for other shapes or for any other triangle?
-------EDIT---------------------
Now to make things interesting : Say we have a right angled triangle which is given . Then is there a method by which we can construct those two circles with a straightedge and a compass?
|
Writing $a := |BC|$, $b := |CA|$, $c := |AB| = \sqrt{a^2+b^2}$, and $r = |PE| = |PF|$ (so that $|PD| = 3r$), we have
$$\begin{align}
|\triangle ABC| &= |\triangle ABP| + |\triangle BCP| + |\triangle CAP| \\[4pt]
\implies \qquad \frac{1}{2} |BC||CA| &= \frac{1}{2} \left(\; |AB| |PF| + |BC||PD| + |CA||PE| \;\right) \\[4pt]
\implies \qquad a b &= c r + 3 a r + b r = r ( 3 a + b + c )\\[6pt]
\implies \qquad r &= \frac{ab}{3 a + b + c} = \frac{ab}{3 a + b + \sqrt{a^2+b^2}}
\end{align}$$
To address @DanielV's suggestion of generalizing to higher dimensions, consider a right-corner tetrahedron $OABC$, with right corner at $O$ and edge lengths $a := |OA|$, $b := |OB|$, $c := |OC|$. (Note that I'm changing notation slightly from the above.) Let a sphere with center $P$ and radius $r$ be tangent to the faces around vertex $A$, and let a congruent sphere (tangent to the first) be tangent to the faces around vertex $O$. Then $P$ has distance $r$ from faces $\triangle OAB$, $\triangle OCA$, $\triangle ABC$ (the ones touching $A$), and distance $3r$ from face $\triangle OBC$ (the one opposite $A$).
Here's a poor attempt at a diagram:
(In this case, the altitudes from $P$ are color-coded to match their parallel counterparts through $O$. The black altitude is to face $\triangle ABC$.)
Thus,
$$\begin{align}
|OABC| &= |OABP| + |OBCP| + |OCAP| + |ABCP| \\[4pt]
\implies \qquad \frac{1}{6}a b c &= \frac{1}{3}\left(\; r\;|\triangle OAB| + r \;|\triangle OCA| + r\;|\triangle ABC| + 3r\;|\triangle OBC| \;\right) \\[4pt]
&= \frac{1}{3}r \cdot \frac{1}{2} \left(\; a b + c a + 3 b c + 2\;|\triangle ABC| \;\right) \\[6pt]
\implies \qquad r &= \frac{abc}{3bc + ab + ca + 2\;|\triangle ABC|} \qquad (\star)
\end{align}$$
Fun fact: The Pythagorean Theorem for Right-Corner Tetrahedra says that
$$|\triangle ABC|^2 = |\triangle OBC|^2 + |\triangle OCA|^2 + |\triangle OAB|^2$$
so that we have
$$|\triangle ABC| = \frac{1}{2} \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}$$
and $(\star)$ becomes
$$r = \frac{abc}{3bc + ab + ca + \sqrt{\; b^2 c^2 + c^2 a^2 + a^2 b^2 \;}}$$
In $4$-dimensional space (where there's an analogous Pythagorean Theorem, as there is in any-dimensional space), we have
$$r = \frac{abcd}{3bcd + acd + abd + abc + \sqrt{\;b^2 c^2 d^2 + a^2 c^2 d^2 + a^2 b^2 d^2 + a^2 b^2 c^2\;}}$$
and so forth.
Incidentally, the matching-notation version of the initial answer is
$$r = \frac{ab}{3b + a + \sqrt{\;b^2 + a^2\;}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/721885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "30",
"answer_count": 4,
"answer_id": 1
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solve partial differential equation
$$y^2u_x + xu_y = \sin(u^2) \\
u(x,0)=x$$
I get the projected characteristic curve on xy plane easily.
However, cannot get the other one.
actually the problem is getting the value of $U_{xx}, U_{xy}, U_{yy}, U_x, U_y$ on x-axis. How can I do this?
| $y^2u_x+xu_y=\sin u^2$
$\dfrac{u_x}{x}+\dfrac{u_y}{y^2}=\dfrac{\sin u^2}{xy^2}$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=\dfrac{1}{y^2}$ , letting $y(0)=0$ , we have $y^3=3t$
$\dfrac{dx}{dt}=\dfrac{1}{x}$ , letting $x(0)=x_0$ , we have $x^2=2t+x_0^2$
$\dfrac{du}{dt}=\dfrac{\sin u^2}{xy^2}=\dfrac{\sin u^2}{3^\frac{2}{3}~t^\frac{2}{3}\sqrt{2t+x_0^2}}$ , we have $\int^u\csc u^2~du=\int^t\dfrac{dt}{3^\frac{2}{3}~t^\frac{2}{3}\sqrt{2t+x_0^2}}+f(x_0^2)$ , i.e. $\int^u\csc u^2~du=\int^\frac{y^3}{3}\dfrac{dt}{3^\frac{2}{3}~t^\frac{2}{3}\sqrt{2t+x_0^2}}+f\left(x^2-\dfrac{2y^3}{3}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/721967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Find the critical numbers of the function. $$\sin^2(x) + \cos(x)$$
$$\{0 < x < 2\pi\}$$
I thought the answer would be $\pi$, but it is not. Can anyone explain why the answer is not $\pi$?
| Taking the derivative:
$$\dfrac{d}{dx}\left(\sin^2 x + \cos x\right)=2\sin x \cos x -\sin x$$
We want $2\sin x\cos x-\sin x$ to equal $0$ (provided that $0 < x < 2\pi$). We immediately see that when $\sin x = 0$, $2\sin x \cos x-\sin x =0$. The only value of $x$ that will make $\sin x = 0$ and is in the interval $0<x<2\pi$ is $\pi$.
Now, when $\cos x=\frac{1}{2}$, $2\sin x \cos x-\sin x = \sin x-\sin x =0$. There are two values of $x$ that will make $\cos x =0$ and are in the interval. You should know that they are: $x=\frac{\pi}{3}$ and $x=\frac{5\pi}{3}$. Therefore your critical points are:
$$\displaystyle \color{green}{\boxed{x=\pi, \ \dfrac{\pi}{3}, \ \dfrac{5\pi}{3}}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/723778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What's the easiest way to factor $5^{10} - 1$? What's the easiest way to factor $5^{10} - 1$?
I believe $5 - 1$ is a factor based off the binomial theorem.
From there I do not know.
We are using congruence's in this class.
| Consider $x^{10} - 1.$
The first trivial factorization is $(x^5-1)(x^5+1)$.
Note that $1+x+x^2+x^3+x^4 = \frac{x^5-1}{x-1}$ by the geometric series formula.
Secondly, replace $x$ with $-x$ to see that $1-x+x^2-x^3+x^4 = \frac{x^5+1}{x+1}.$
$$x^{10} - 1 = (1+x+x^2+x^3+x^4)(1-x+x^2-x^3+x^4)(x-1)(x+1).$$
It should be easy to factor $5^{10} - 1$ into $4\times 6\times 781 \times 521.$
Another trick is to see that $11$ clearly divides $781$ because $7+1 = 8$ (cf. multiplying by $11$ shortcuts).
Simple division gives us $$\boxed{5^{10} - 1 = 2^3 \times 3 \times 11 \times 71 \times 521}.$$
The only thing left is to show that $521$ is prime.
All we have to check is primes less than $\sqrt{521}$ because if there are factors other than $1$ and $521$ it must be less than (or equal to) $\sqrt{521}$. (Note that $23^2 = 529$).
Clearly $2,3,5$ and $ 11$ do not work so now we must check $7, 13, 17$ and $19$.
$7$ does not work because if $7$ divided $521$ it would divide $521-21 = 500$.
$13$ does not work because $521-26 = 495 = 99\times5$.
$17$ does not work because $521-17*3 = 470 = 47\times10$.
$19$ does not work because $521-19*9 = 350 = 35\times10$.
I found these terms to subtract by trying to eliminate the $1$ at the end of $571$ by turning it into a $5$ or a $0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/724109",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Evaluate the summation involving binomials. $\sum _{ i=0 }^{ 100 }{\binom{k}{i}}*{\binom{M-k}{100-i}*\frac{k-i}{M-100}}/{\binom{M}{100}}$
I wrote the first few terms but couldn't find any pattern and how to club the terms. Help.
| Let's generalize a bit by introducing indices:
$$
\sum_i \binom{k}{i}
\cdot \binom{M - k}{r - i}
\cdot \frac{k - i}{M - r}
/ \binom{M}{r}
= \frac{1}{(M - r) \binom{M}{r}}
\cdot \sum_i \binom{k}{i} \binom{M - k}{r - i} \cdot (k - i)
$$
We can split the $k - i$ to simplify.
So we are interested in sums of the forms:
$$
\sum_i \binom{a}{i} \binom{b}{r - i}
$$
and
$$
\sum_i i \binom{a}{i} \binom{b}{r - i}
$$
Now remember:
$$
(1 + z)^a = \sum_i \binom{a}{i} z^i \\
z \frac{\mathrm{d}}{\mathrm{d} z} (1 + z)^a = \sum_i i \binom{a}{i} z^i
$$
and also:
$$
\left(\sum_i u_i \right) \cdot \left( \sum_i v_i \right)
= \sum_i \sum_{0 \le j \le i} u_j v_{i - j}
$$
In particular:
\begin{align}
\sum_{r \ge 0} \sum_i \binom{a}{i} \binom{b}{r - i} z^r
&= \left(\sum_i \binom{a}{i} z^i \right)
\cdot \left(\sum_i \binom{b}{i} z^i \right) \\
&= (1 + z)^a \cdot (1 + z)^b \\
&= (1 + z)^{a + b}
\end{align}
Comparing coefficients you get Vandermonde's convolution:
$$
\sum_i \binom{a}{i} \binom{b}{r - i} = \binom{a + b}{r}
$$
For the other half:
\begin{align}
\sum_r \sum_i i \binom{a}{i} \binom{b}{r - i} z^r
&= \left(\sum_i i \binom{a}{i} z^i \right)
\cdot \left(\sum_i i \binom{b}{i} z^i \right) \\
&= a z (1 + z)^{a - 1} \cdot (1 + z)^b \\
&= a z (1 + z)^{a + b - 1}
\end{align}
The coefficient of $z^r$ here is a bit trickier:
$$
[z^r] a z (1 + z)^{a + b - 1}
= a [z^{r - 1}] (1 + z)^{a + b - 1}
= a \binom{a + b - 1}{r - 1}
$$
I'm sure you can take it from here.
| {
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Minimum area of a triangle In triangle inscribed circle with radius $r = 1$ and one of it sides $a=3$. Find the minimum area of triangle? Ans = 5.4
My reasonings:
$BC = a$, $AC = b$, $AB = c$
$AD=AF=x$
$FC=CE=y$
$BD=BE=z$
$a=z+y$, $b=x+y$, $c=x+z$
The radius of the incircle is $$r =\frac{A_{ABC}}{s}$$ where $s = \frac{a+b+c}{2} = x + y+ z$. By condition $z+y=3$ so $s=x+3$
By Heron's formula, the area of the triangle is $A=\sqrt{s(s-a)(s-b)(s-c)}$
In other side $A=sr = x+3$.
What is next? I think that I should get a fucntion for which I will can find a minimum, but I don't know how.
| You were on the right track.
Using your notation, $A = \sqrt{s(s - a)(s - b)(s - c)} = \sqrt{(x + 3)xyz}$, but also $A = x + 3$. So
$$\begin{align}
\sqrt{(x + 3)xyz} &= x + 3, \\
xyz &= x + 3, \\
x &= \frac{3}{yz - 1}.
\end{align}$$
Substituting that into $A$ we get
$$A = \sqrt{\left(\frac{3}{yz - 1} + 3\right)\frac{3}{yz - 1}yz} = \sqrt{\frac{3yz}{yz - 1}\cdot\frac{3}{yz - 1}yz} = \frac{3yz}{yz - 1} = \frac{3}{yz - 1} + 3.$$
Finally, by AM-GM we get lower bound:
$$A = \frac{3}{yz - 1} + 3 \geqslant \frac{3}{\frac{(y + z)^2}{4} - 1} + 3 = \frac{3}{\frac{9}{4} - 1} + 3 = \frac{12}{5} + 3 = \frac{27}{5} = 5.4$$
Because it is AM-GM, equality is reached when $y = z = 1.5$
P.S. You may wonder, why for some values of $y$ and $z$ expression $\frac{3}{yz - 1}$ may become infinite or negative. Isn't it strange? Besides, I myself silently assumed that it is positive. And there's a reason for that.
Positive values of $\frac{3}{yz - 1}$ correspond to the situation you describe: a circle inscribe in a triangle.
When it becomes infinite two sides $AB$ and $AC$ become parallel.
And finally, when it's negative, your circle is no longer an incircle, it becomes excircle.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/725056",
"timestamp": "2023-03-29T00:00:00",
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Solve $\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$ In solving the definite integral $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx$$I tried to do $$\int{\frac{x-1}{x\sqrt{3x^2-4}}}\;dx=\int{\frac{1}{\sqrt{3x^2-4}}}\;dx-\int{\frac{1}{x\sqrt{3x^2-4}}}\;dx\Longrightarrow$$ If $x=\frac{2}{\sqrt{3}}\sec(u)\Longrightarrow dx=\frac2{\sqrt3}\tan(u)\sec(u)\;du$, then
$$\int{\frac{1}{\sqrt{3x^2-4}}}\;dx-\int{\frac{1}{x\sqrt{3x^2-4}}}\;dx=$$ $$\frac{2}{\sqrt3}\int\frac{\tan(u)\sec(u)}{2\sqrt{\left(\frac{\sqrt3}{2}\sec(u)\right)^2-1}}\;du-\frac1{2}\int\frac{\frac{\sqrt3}{2}}{\frac{\sqrt3}{2}x\sqrt{\left(\frac{\sqrt3}{2}x\right)^2-1}}\;dx=$$ $$\frac{1}{\sqrt3}\int\sec{u}\;du-\frac12\sec^{-1}\left(\frac{\sqrt3}{2}x\right)$$
In my feedback says that the result is $$\frac{1}{\sqrt3}\log\left|\sqrt3x+\sqrt{3x^2-4}\right|-\frac12\sec^{-1}{\left(\frac{\sqrt3}{2}x\right)}+C$$
It also says that I have to use this method of trigonometric substitution.
The path is this?
I know $\int\sec(t)dt=\log|\sec t+\tan t|$
| You are essentially finished. We have $\sec u=\frac{\sqrt{3}x}{2}$ and $\tan u=\frac{\sqrt{3x^2-4}}{2}$. Thus your procedure gives as first term
$$\frac{1}{\sqrt{3}}\log\left|\frac{\sqrt{3}x+\sqrt{3x^2-4}}{2}\right|.$$
This is equal to
$$\frac{1}{\sqrt{3}}\log\left|\sqrt{3}x+\sqrt{3x^2-4}\right|-\frac{1}{\sqrt{3}}\log 2.$$
So the two versions differ by a constant, which can be absorbed in the constant of integration.
| {
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"timestamp": "2023-03-29T00:00:00",
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Help finishing proof via induction for a summation So I have to prove the following equation using induction for n >= 2:
$$
\sum\limits_{i=1}^n 4/5^i < 1
$$
However the question asks me to prove something stronger such as this:
$$
\sum\limits_{i=1}^n 4/5^i <= 1 - \frac{1}{5^n}
$$
first to imply the first equation is true.
So far I have the following:
Base Case:
Let n = 2
$$
\sum\limits_{i=1}^2 4/5^i = \frac{4}{5} + \frac{4}{25} = \frac {24}{25}
$$
then I also applied it to
$$ 1 - \frac{1}{5^n} \rightarrow 1 - \frac{1}{5^2} = \frac{24}{25}$$
Therefore I can make the following assumptions yes?
Inductive Hypothesis
for all 2 <= n <= k it is
$$
\sum\limits_{i=1}^n 4/5^i = 4\frac{\frac{1}{5^n} - 1}{\frac{1}{5} - 1} = 1 - \frac{1}{5^n} < 1
$$
Inductive Step
Hopefully I'm ok up to here, I'll show what I have so far for this step.
$$
\sum\limits_{i=1}^{k+1} 4/5^i = \frac{\frac{1}{5^{k+1}} - 1}{\frac{1}{5} - 1} = 4\frac{(\frac{1}{5^k}-1) * \frac{1}{5} - \frac{4}{5}}{\frac{1}{5} -1} $$
$$
= \frac{1}{5} * 4\frac{(\frac{1}{5^k}) - 1}{\frac{1}{5} -1} - 4\frac{\frac{4}{5}}{\frac{1}{5} - 1}
$$
so here I have:
$$
4\frac{(\frac{1}{5^k}) - 1}{\frac{1}{5} -1}
$$
which I know is:
$$
= \sum\limits_{i=1}^k 4/5^i
$$
which is my inductive hypothesis, I am unsure of how to finish my proof from here... any help correcting or finishing the proof is very much appreciated
| Hint:
$$\sum_{i=1}^{k+1}4/5^i=\sum_{i=1}^k 4/5^i+4/5^{k+1}$$
Use the induction hypothesis on the sum from $1$ to $k$ and simplify.
| {
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A definite integral $$\int_{-\frac\pi2}^{\frac\pi2}\frac{\ln(1+b\sin x)}{\sin x} dx \\|b|<1$$
I tried putting $-x$ using properties of definite integral, but that doesn't really help.
$$I=\int_{-\frac\pi2}^{\frac\pi2}\frac{-\ln(1-b\sin x)}{\sin x}dx$$
I don't think adding these 2 equations would yield anything. And I just cannot think of a good substitution.
Edit : Plugging b=1, it gives 4.9xxx . as $\pi ^2 \approx 10$, the value is $\pi \arcsin b$. Other values for b confirm this.
| From the Taylor series expansion of $\ln(1+x)$ we know that
$$\ln(1+b\sin x) = b\sin x - \frac{b^2\sin^2x}{2} +\frac{b^3\sin^3x}{3} - \cdots$$
Therefore
$$\frac{\ln(1+b\sin x)}{\sin x} = b - \frac{b^2\sin x }{2} +\frac{b^3\sin^2 x }{3}-\cdots$$
Integrating both sides of the equation
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \int^{\pi/2}_{-\pi/2} b - \frac{b^2\sin x }{2} +\frac{b^3\sin^2 x }{3}-\cdots \mathrm{d}x$$
Observe that $\int^{\pi/2}_{-\pi/2}\sin^nx\mathrm{d}x = 0$ if n is an odd number. The above equation is then equivalent to
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \int^{\pi/2}_{-\pi/2} b +\frac{b^3\sin^2 x }{3}+\frac{b^5\sin^4 x }{5}\cdots \mathrm{d}x$$
As $\sin^{2n}x$ is an even function therefore
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x =2 \int^{\pi/2}_0 b +\frac{b^3\sin^2 x }{3}+\frac{b^5\sin^4 x }{5}\cdots \mathrm{d}x$$
Using the fact that $\int_0^{\pi/2}\sin^nx\mathrm{d}x = \frac{(n-1)\cdot(n-3)\cdots3\cdot1}{n\cdot(n-2)\cdots4\cdot2}$ if n is even (See this page for the proof)
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x =2\cdot\frac{\pi}{2}\Bigg[ b+\frac{b^3}{3}\frac{1}{2}+\frac{b^5}{5}\frac{3\cdot1}{4\cdot2}\cdots\Bigg]$$
$$=\pi\Bigg[ b+\frac{b^3}{3}\frac{1}{2}+\frac{b^5}{5}\frac{3\cdot1}{4\cdot2}\cdots\Bigg]$$
The term in the parenthesis is precisely the Taylor series expansion (Thanks to your edit. I was completely clueless at this point) for $\sin^{-1} b $. Therefore the integral is
$$\int^{\pi/2}_{-\pi/2}\frac{\ln(1+b\sin x )}{\sin x }\mathrm{d}x = \pi\sin^{-1} b $$
| {
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Fibonacci Proof Using Induction
$$f(n) = \left\{\begin{matrix}
0 & n=1\\
1 & n=2\\
f_{n-1} + f_{n-2} &
n\geqslant 2\end{matrix}\right.$$
How can I prove by induction that $$f_{n} \leq \left ( \frac{1+\sqrt{5}}{2} \right )^{n-1}$$ for all$$ n\geq l_{a}$$, I have to find the smallest value for $$l_{a}$$
| $f_1 = 0 \leq \left(\frac{1+\sqrt{5}}{2}\right)^0 = 1$
$f_2 = 1 \leq \frac{1+ \sqrt{5}}{2} \approx 1.618$
Suppose $f_k \leq \left(\frac{1+\sqrt{5}}{2}\right)^{k-1}, f_{k+1} \leq \left( \frac{1 + \sqrt{5}}{2} \right)^{k}$
Then $f_{k+2} = f_k + f_{k+1} \leq \left(\frac{1+\sqrt{5}}{2}\right)^{k-1} + \left( \frac{1 + \sqrt{5}}{2} \right)^{k} = \left(\frac{1 + \sqrt{5}}{2}\right)^{k+1}$
Proof of last statement: $a^{k+1} = a^k + a^{k-1} \rightarrow a^2 - a - 1 \rightarrow a = \frac{1 + \sqrt{5}}{2}$
| {
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Let $G$ be a group, where $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$. Prove that $G $ is an abelian group. Let G be a group, where $(ab)^3=(a^3)(b^3)$ and $(ab)^5=(a^5)(b^5)$. Prove that $G$ is abelian group.
Thank you in advance. Any help is appreciated.
| From $(ab)^3=a^3b^3$ and $(ab)^5=a^5b^5$ it follows that $(ba)^2=a^2b^2$, and $(ba)^4=a^4b^4$. This implies $(ba)^4=(ba)^2(ba)^2=a^2b^2a^2b^2=a^4b^4$, so $b^2a^2=a^2b^2$, hence all squares commute. Thus $(ba)^2=baba=a^2b^2=b^2a^2=bbaa$ and it follows that $ab=ba$.
| {
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What is this trigonometric expression equal to? If $\frac{cos{x}}{cos{y}}=\frac{a}{b}$, then $(a\times tan{x}+b\times tan{y})$ equals
(A)$(a+b)cot{\frac{x+y}{2}}$
(B)$(a+b)tan\frac{x+y}{2}$
(C)$(a+b)(tan\frac{x}{2}+tan\frac{y}{2})$
(D)$(a+b)(cot\frac{x}{2}+cot\frac{y}{2})$
I believe know the trigonometric formulas but I am just not able to reduce the given expression to any of the options. Help.
| If $x=y$ then the expression is $(a+b)\tan x$, so the only alternative that could possibly be true is (B): that is,
$$\hbox{if}\quad\frac{\cos x}{\cos y}=\frac{a}{b}\quad\hbox{then}\quad
a\tan x+b\tan y=(a+b)\tan\frac{x+y}{2}\ .$$
It remains to prove it.
We have
$$\frac{a}{a+b}=\frac{\cos x}{\cos x+\cos y}\ ,\quad \frac{b}{a+b}=\frac{\cos y}{\cos x+\cos y}\ ,$$
so
$${\rm LHS}
=(a+b)\frac{\sin x+\sin y}{\cos x+\cos y}
=(a+b)\frac{\sin\frac{x+y}{2}\cos\frac{x-y}{2}}{\cos\frac{x+y}{2}\cos\frac{x-y}{2}}
={\rm RHS}\ .$$
| {
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How to compute $\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$? How to compute this integral? :
$$\int_0^{\infty}\sqrt x \exp\left(-x-\frac{1}{x}\right) \, dx$$
Wolframalpha gives the answer $\dfrac{3\sqrt{\pi}}{2e^2}$, but how to compute this?
| A little roundabout, but here goes. Write
$$\begin{align}I &= \underbrace{\int_0^{\infty} dx \, \sqrt{x} \, e^{-\left (x+\frac1{x} \right )}}_{x=u^2} \\ &= 2 e^2 \underbrace{\int_0^{\infty} du \, u^2 \, e^{-\left (u+\frac1{u} \right )^2}}_{v=u+\frac1{u}}\\ &= e^2 \int_{\infty}^2 dv \, \left (1-\frac{v}{\sqrt{v^2-4}} \right )\left (\frac{v^2}{2}-\frac{v}{2} \sqrt{v^2-4}-1 \right )\, e^{-v^2}\\&+e^2 \int_2^{\infty} dv \, \left (1+\frac{v}{\sqrt{v^2-4}} \right )\left (\frac{v^2}{2}+\frac{v}{2} \sqrt{v^2-4}-1 \right )\, e^{-v^2}\\ &= 2 e^2 \underbrace{\int_2^{\infty} dv \,v \left ( \sqrt{v^2-4}+\frac1{\sqrt{v^2-4}}\right )\, e^{-v^2}}_{y=v^2-4}\\ &= e^{-2} \int_0^{\infty} dy \, \left (\sqrt{y}+\frac1{\sqrt{y}} \right ) e^{-y}\\ &= e^{-2} \left (\frac{\sqrt{\pi}}{2} + \sqrt{\pi} \right ) \\ &= \frac{3 \sqrt{\pi}}{2 e^2}\end{align}$$
For a little more background on how the integral over $v$ gets split, see this answer. Also note that, in the third step, I made use of the fact that $u^2=v u-1$.
| {
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Invalid subtraction when solving system of equations? I'm trying to solve these two equations:
$$\begin{cases}
1-4x(x^2+y^2)=0 \\
1-4y(x^2+y^2)=0
\end{cases}$$
and I tried to do it by subtracting the first equation from the second, yielding $(4x-4y)(x^2+y^2)=0$. Clearly this is satisfied when $x=y$, which gives $(x,y)=(\frac{1}{2},\frac{1}{2})$, or when $x^2+y^2=0$, which gives $(x,y)=(0,0)$.
But when I plug them back in, it's obvious that $(x,y)=(0,0)$ is not a solution. Where did I go wrong?
| First $x=y$ yields $8x^3=1$
$x=\dfrac{1}{2},\dfrac{-1+\sqrt3i}{4},\dfrac{-1-\sqrt3i}{4}$
$\therefore(x,y)=\left(\dfrac{1}{2},\dfrac{1}{2}\right),\left(\dfrac{-1+\sqrt3i}{4},\dfrac{-1+\sqrt3i}{4}\right),\left(\dfrac{-1-\sqrt3i}{4},\dfrac{-1-\sqrt3i}{4}\right)$
But $x^2+y^2=0$ yields $1=0$ which is not useful.
Note than $x^2+y^2=0$ should not immediately implies $x=y=0$ , which should have no concept abut the false solution $(x,y)=(0,0)$ .
| {
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Limit of: $\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$ I calculated this limit with L'Hospital's rule, but at the end it got rather complicated.
$\lim_{x\to0}\frac{1-\sqrt[3]{\cos{x}}}{1-\cos{\sqrt[3]{x}}}$
Is there some other more effective way for this limit?
| Well the idea is very simple. We have to use the fundamental limit $$\lim_{y \to 0}\frac{1 - \cos y}{y^{2}} = \frac{1}{2}$$ This is a pretty standard result which can easily be proved by simplifying $(1 - \cos y)$ as $2\sin^{2}(y/2)$ and then applying $\lim\limits_{y \to 0}\dfrac{\sin y}{y} = 1$.
For the current question we first put $\sqrt[3]{\cos x} = t$ so that $\cos x = t^{3}$ and then $$1 - \sqrt[3]{\cos x} = 1 - t = \dfrac{1 - t^{3}}{1 + t + t^{2}} = \frac{1 - \cos x}{1 + \sqrt[3]{\cos x} + \sqrt[3]{\cos^{2}x}}$$ We can now proceed as follows: $$\begin{aligned}L &= \lim_{x \to 0}\frac{1 - \sqrt[3]{\cos x}}{1 - \cos\sqrt[3]{x}}\\
&= \lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt[3]{x}}\cdot\frac{1}{1 + \sqrt[3]{\cos x} + \sqrt[3]{\cos^{2}x}}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{1 - \cos\sqrt[3]{x}}\\
&= \frac{1}{3}\lim_{x \to 0}\frac{1 - \cos x}{x^{2}}\cdot\frac{x^{2}}{1 - \cos\sqrt[3]{x}}\\
&= \frac{1}{3}\cdot\frac{1}{2}\lim_{x \to 0}\frac{x^{2}}{1 - \cos\sqrt[3]{x}}\\
&= \frac{1}{6}\lim_{x \to 0}\dfrac{x^{2}}{\dfrac{1 - \cos\sqrt[3]{x}}{\sqrt[3]{x^{2}}}\cdot\sqrt[3]{x^{2}}}\\
&= \frac{1}{6}\lim_{x \to 0}x^{4/3}\cdot\frac{x^{2/3}}{1 - \cos (x^{1/3})}\\
&= \frac{1}{6}\cdot 0\cdot 2 = 0\end{aligned}$$
| {
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if $abc=1$, then $a^2+b^2+c^2\ge a+b+c$ This is supposed to be an application of AM-GM inequality.
if $abc=1$, then the following holds true: $a^2+b^2+c^2\ge a+b+c$
First of all,
$a^2+b^2+c^2\ge 3$
by a direct application of AM-GM.Also,we have
$a^2+b^2+c^2\ge ab+bc+ca$
Next,we consider the expression
$(a+1)(b+1)(c+1)=a+b+c+ab+bc+ca+abc\le a+b+c+a^2+b^2+c^2+1$
but that hardly helps.I know that
$3(a^2+b^2+c^2)\ge (a+b+c)^2$
From the first derived inequality,we know that the left hand side of the above inequality is greater than or equal to 9.But I can't see how that can be used here.I know that we can get $a+b+c\ge 3$ using AM-GM but that does not take me a step closer to finding the solution(from what I can understand).Also,
$a^3+b^3+c^3\ge a^2b+b^2c+c^2a$
$(a+b+c)[(a-b)^2+(b-c)^2+(c-a)^2]\ge a^2b+b^2c+c^2a$
A hint will be appreciated at this point.
| Since $(2,0,0)\succ\left(\frac{4}{3},\frac{1}{3},\frac{1}{3}\right)$, our inequality it's just Muirhead.
| {
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How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$ How can I prove that $2(\cos^6(x)-\sin^6(x))-3(\cos^4(x)+\sin^4(x))=-4\sin^6(x)-1$
I tried to factor and I got $2\cos^4(x)+(-2\sin^2(x)-3)(\cos^4(x)+\sin^4(x))$ but that doesn't lead me to my goal.
I also tried to write all the cosines in terms of sines to have: $-3\sin^6(x)-9\sin^2(x)-2-3\sin^4(x)$
But I don't see how to continue
Any hint is welcome! thnxx
| According to the equation:
$$2(cos^6(x) - sin^6(x)) - 3(cos^4(x) + sin^4(x)) = -4sin^6(x) - 1$$
this means that:
$$\frac{2(cos^6(x) - sin^6(x)) - 3(cos^4(x) + sin^4(x)) + 1}{-4} = sin^6(x)$$
And since we are dividing by a negative where $x\in \mathbb{Q}$ I assume, this means that:
$$3(cos^4(x) + sin^4(x)) - 1 > 2(cos^6(x) - sin^6(x))$$
Meaning that:
$$\frac{3(cos^4(x) + sin^4(x)) - 1}{2} - cos^6(x) + 2sin^6(x) > sin^6(x) = \frac{2(cos^6(x) - sin^6(x)) - 3(cos^4(x) + sin^4(x)) + 1}{-4}$$
Therefore when we do some of the simplifications, we find that:
$$-4sin^6(x) - 1 < 2 - 6(cos^4(x) + sin^4(x)) + 4cos^6(x) + 8sin^6(x)$$
$$\implies sin^6(x) < \frac{3 - 6(cos^4(x) + sin^4(x)) + 4cos^6(x) + 8sin^6(x)}{-4}$$
$$\therefore \frac{3(cos^4(x) + sin^4(x)) - 1}{2} - cos^6(x) + 2sin^6(x) = \frac{3 - 6(cos^4(x) + sin^4(x)) + 4cos^6(x) + 8sin^6(x)}{-4}$$
So there's your proof. Since there is no contradiction and we basically come to a fact something like $-4n = \frac{-4n}{-4}$ then we cannot go any further with this, meaning that every single equation in this answer is true! So there ya go. Hope I helped! :)
| {
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How does Laplace expansion work?
\begin{bmatrix} 1 & 2 & 0 & 0 & a\\ 0 & 1 & 0 & 0 & 0\\ 0 & 0 & 1 & 1 & 0\\ 0 & 0 & 0 & 1 & 0\\ 1 & 0 & 1 & 1 & 1 \end{bmatrix}
Caculate its determinant.
Although it has small numbers, it's still fairly long to calculate the determinant of a large matrix. Peeking at the solution to this exercise, it says
Applying Laplace expansion on the second row and then on the third, we
obtain a matrix of size $3$ where we can apply Sarrus. The determinant
is $1 - a$.
I don't really know either of those terms, Laplace and Sarrus. Looking up at Wikipedia, Sarrus is pretty clear to me now, but Laplace is still fairly confusing.
How does Laplace expansion work?
| Hint: Expand along the rows that have the most zeros.
Expanding along row 2, we have:
$$\begin{vmatrix} 1 & 0 & 0 & a\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1 \end{vmatrix}$$
Note, what I mean by expanding along row $2$ using Laplace Expansion (see link below), is:
$$0 \begin{vmatrix} 2 & 0 & 0 & a\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 0 & 1 & 1 & 1 \end{vmatrix} + 1 \begin{vmatrix} 1 & 0 & 0 & a\\ 0 & 1 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 1 & 1 & 1 & 1 \end{vmatrix} + 0 \begin{vmatrix} 1 & 2 & 0 & a\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 1 & 0\\ 1 & 0 & 1 & 1 \end{vmatrix} + 0 \begin{vmatrix} 1 & 2 & 0 & a\\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 0\\ 1 & 0 & 1 & 1 \end{vmatrix} + 0 \begin{vmatrix} 1 & 2 & 0 & 0\\ 0 & 0 & 1 & 1 \\ 0 & 0 & 0 & 1\\ 1 & 0 & 1 & 1 \end{vmatrix}$$
Notice all of the zero terms that are leading? Only one item with a $1$ in front survives. Repeating this process two more times yields the result as follows.
Expanding along row 3, we have:
$$\begin{vmatrix} 1 & 0 & a\\ 0 & 1 & 0 \\ 1 & 1 & 1 \end{vmatrix}$$
Expanding along row 2, we have:
$$\begin{vmatrix} 1 & a\\ 1 & 1 \end{vmatrix}$$
Result:
$$\mbox{det} ~ = ~ (1-a)$$
I am basically using Laplace Expansion, because all the leading numbers for the expansion are zero. The site includes an example of where this is not the case.
| {
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Suppose $A=\langle a,b\mid ab^2a^{-1}b^{-3},ba^2b^{-1}a^{-3}\rangle.$ Show that $A \cong \{1\}$. Suppose $A=\langle a,b\mid ab^2a^{-1}b^{-3},ba^2b^{-1}a^{-3}\rangle$, where $\langle a_1,\ldots, a_n \mid R\rangle$ is the group generated by $a_1,\ldots, a_n$ with relations in $R$. Show that $A \cong \{1\}$.
After carrying out some algebras, I obtain $ab^2=b^3a$ and $ba^2=a^3b$. From here I don't know how to proceed.
Can anyone give some hints?
| $ab^2a^{-1}=b^3 \Rightarrow ab^4a^{-1}=b^6 \Rightarrow a^2b^4a^{-2}=ab^6a^{-1} = b^9$.
Now $a^2 = b^{-1}a^3b$, and substituting for $a^2$ in the equation $a^2b^4a^{-2}=b^9$ gives $a^3b^4a^{-3}=b^9$.
But $a^3b^4a^{-3}=b^9 = a(a^2b^4a^{-2})a^{-1} = ab^9a^{-1}$, so $ab^9a^{-1}=b^9$.
But $a^{-1}b^9a=b^6$, so $b^9=b^6$ and $b^3=1$, from which it is easy to prove that $a=b=1$.
| {
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Calculating $1+\frac13+\frac{1\cdot3}{3\cdot6}+\frac{1\cdot3\cdot5}{3\cdot6\cdot9}+\frac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $
How to find infinite sum How to find infinite sum $$1+\dfrac13+\dfrac{1\cdot3}{3\cdot6}+\dfrac{1\cdot3\cdot5}{3\cdot6\cdot9}+\dfrac{1\cdot3\cdot5\cdot7}{3\cdot6\cdot9\cdot12}+\dots? $$
I can see that 3 cancels out after 1/3, but what next? I can't go further.
| From the OGF of Catalan numbers we have that:
$$ \sum_{n\geq 0}\binom{2n}{n}x^n = \frac{1}{\sqrt{1-4x}} $$
where the radius of convergence of the LHS is $\frac{1}{4}$ since $\frac{\binom{2n+2}{n+1}}{\binom{2n}{n}}=\frac{4n+2}{n+1}$.
By evaluating the previous identity at $x=\frac{1}{6}$ it follows that:
$$ \color{red}{\sqrt{3}} = 1+\sum_{n\geq 1}\frac{(2n)!}{n! n! 6^n} = 1+\sum_{n\geq 1}\frac{(2n-1)!!}{3^n n!} $$
where the RHS is exactly our sum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/746388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
"answer_count": 7,
"answer_id": 2
} |
Lagrange multipliers from hell I was asked to solve this question, decided to try and solve it with lagrange multipliers as I see no other way:
"Find the closest and furthest points on the circle made from the intersection of the ball $(x-1)^2+(y-2)^2+(z-3)^2=9$ and the plane $x-2z=0$ from the point $(0,0)$".
What I did:
the distance for any point $(x,y,z)$ from the origin is $d(x,y,z)=\sqrt{x^2+y^2+z^2}$. so using lagrange multipliers we have:
$d(x,y,z)=\sqrt{x^2+y^2+z^2}$
$C_1(x,y,z)=(x-1)^2+(y-2)^2+(z-3)^2-9$
$C_2(x,y,z)=x-2z$
$L(x,y,z) = d-\lambda_1C_1-\lambda_2C_2 $ meaning:
$L(x,y,z)=\sqrt{x^2+y^2+z^2}-\lambda_1[(x-1)^2+(y-2)^2+(z-3)^2-9]-\lambda_2(x-2z)$
Let's derive and solve when derivatives are zero:
$\frac{\partial L}{\partial x}= \frac{x}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(x-1)-\lambda_2=0$
$\frac{\partial L}{\partial y} = \frac{y}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(y-2)=0$
$\frac{\partial L}{\partial z} = \frac{z}{\sqrt{x^2+y^2+z^2}}-2\lambda_1(z-3)+2\lambda_2=0$
$\frac{\partial L}{\partial \lambda_1} = -(x-1)^2-(y-2)^2-(z-3)^2+9=0$
$\frac{\partial L}{\partial \lambda_2} = 2z-x=0$
Solving this monstrous system seems very unlikely, and very difficult, and not how the question is meant to be solved. am I missing something?
| If you eliminate, say, $x$ using the second constraint, you are left with a two-variate function and a single constraint: Minimize/Maximize $d^2 = 5z^2 + y^2$ subject to $(y-2)^2 + 5(z-1)^2 = 4$. Now, a Lagrange solution is not bad. From here, you can play also with basic algebra and trig to avoid calculus altogether.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$ I am trying to calculate:
$$\int_0^1 \frac{\ln(1-x+x^2)}{x-x^2}dx$$
I am not looking for an answer but simply a nudge in the right direction. A strategy, just something that would get me started.
So, after doing the Taylor Expansion on the $\ln(1-x+x^2)$ ig to the following: Let $x=x-x^2$ then $\ln(1-x)$ then,
\begin{align*}
=&-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-...\\
=&-(x-x^2)-\frac{(x-x^2)^2}{2}-...\\
=&-x(1-x)+\frac{x^2}{2}(1-x)^2-\frac{x^3}{3}(1-x)^3\\
\text{thus the pattern is:}\\
=&\frac{x^n(1-x)^n}{n}
\end{align*}
Am I right?
Then our Integral would be: $$\sum_{n=0}^{\infty} \frac{1}{n+1} \int_0^1 x^n(1-x)^n$$
Am I on the right track? Suggestions, tips, comments?
$\underline{NEW EDIT:}$
SO after integrating the function I got the following after a couple of iterations:
\begin{align*}
\frac{n(n-1)...1}{(n+1)(n+2)...(2n)}\int_0^1 x^{2n} dx
\end{align*}
This shows a pattern:
\begin{align*}
=&\frac{(n!)^2}{(2n)!} (\frac{1}{2n+1})\\
=& \frac{(n!)^2}{(2n+1)!}
\end{align*}
So my question is, what to do from here. I have done all this but still have no clue how to actually solve the integral. Can somebody shed some light on this!
Thanks
| Here is a relatively simple way that only relies on knowing the following Maclaurin series expansion for the square of the inverse sine function (for several different proofs of this, see here)
$$(\sin^{-1} x)^2 = \frac{1}{2} \sum_{n = 1}^\infty \frac{(2x)^{2n}}{n^2 \binom{2n}{n}}, \qquad |x| \leqslant 1.$$
Note that if we set $x = 1/2$ one obtains:
$$\sum_{n = 1}^\infty \frac{1}{n \binom{2n}{n}} = \frac{\pi^2}{18}. \qquad (*)$$
Now
\begin{align}
\int_0^1 \frac{\ln (1 - x + x^2)}{x - x^2} \, dx &= \int_0^1 \frac{\ln [1 - (x - x^2)]}{x - x^2} \, dx\\
&= -\int_0^1 \sum_{n = 1}^\infty \frac{(x - x^2)^n}{n} \frac{dx}{x - x^2} \tag1\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \int_0^1 (x - x^2)^{n - 1} \, dx \tag2\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \int_0^1 x^{n - 1} (1 - x)^{n - 1} \, dx\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \operatorname{B} (n,n) \tag3\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \frac{\Gamma (n) \Gamma (n)}{\Gamma (2n)} \tag4\\
&= -\sum_{n = 1}^\infty \frac{1}{n} \frac{(n - 1)! (n - 1)!}{(2n - 1)!}\\
&= -2\sum_{n = 1}^\infty \frac{1}{n} \frac{(n!)^2}{(2n)!}\\
&= -2 \sum_{n = 1}^\infty \frac{1}{n \binom{2n}{n}}\\
&= -\frac{\pi^2}{9} \tag5
\end{align}
Explanation
(1): Maclaurin series expansion for $\ln (1 - z)$.
(2): The dominated convergence theorem allows the summation and integration signs to be interchanged.
(3): Integral representation for the beta function.
(4): Using the property $\operatorname{B} (x,y) = \frac{\Gamma (x) \Gamma (y)}{\Gamma (x + y)}$.
(5): Using the result given above in ($*$).
| {
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"question_score": "13",
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"answer_id": 2
} |
Evaluate $\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x$ My try, using $x = \sec(u)$ substitution:
$$
\begin{eqnarray}
\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x &=& \int \frac{\sqrt{\sec^2(u) - 1}}{\sec(u)}\tan(u)\sec(u) \mathrm{d}u \\
&=& \int \tan^2(u) \mathrm{d}u \\
&=& \tan(u) - u + C \\
&=& \tan(arcsec(x)) - arcsec(x) + C
\end{eqnarray}
$$
However, according to Wolfram Alpha, the answer should be:
$$
\int \frac{\sqrt{x^2-1}}{x} \mathrm{d}x = \sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C
$$
When I derive this last answer I don't get back the integrand, but rather:
$$
\frac{\mathrm d}{\mathrm d x}\left(\sqrt{x^2-1}+\arctan \left( \frac{1}{\sqrt{x^2-1}} \right)+C\right) = \frac{x}{\sqrt{x^2-1}}- \frac{x}{(x^2-1)^{3/2}\left(1+\frac{1}{x^2-1}\right)}
$$
I don't know how to simplify this expression more. Also, I am unable to check whether my answer is correct because I don't know how to find the derivative of $arcsec(x)$.
Can someone check my calculations and tell me where I've done something wrong and how one can simplify the last expression to get back the integrand?
| $$\frac{x}{\sqrt{x^2-1}} - \frac{x}{(x^2-1)^{3/2} \left(1+\frac{1}{x^2-1}\right)} = \frac{x}{\sqrt{x^2-1}} - \frac{x}{\sqrt{x^2-1} (x^2-1+1)}$$
$$ = \frac{x}{\sqrt{x^2-1}} - \frac{x}{x^2 \sqrt{x^2-1}} = \frac{x^3 - x}{x^2 \sqrt{x^2-1}} = \frac{x (x^2-1)}{x^2 \sqrt{x^2-1}} = \frac{\sqrt{x^2-1}}{x}$$
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 5,
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} |
Solving integral $ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x $ there is integral $$ \int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\:\mathrm{d}x$$
i am trying to separate this :
$$=\int \mathrm{d}x -\int \frac{\mathrm{d}x}{1+x+\sqrt{1+x+x^2}} $$ but have no idea about second
| We have the algebraic form:
$$
\frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}.
$$
Multiplying by $\dfrac{(1+x)-\sqrt{1+x+x^2}}{(1+x)-\sqrt{1+x+x^2}}$ yields
$$
\frac{\sqrt{1+x+x^2}-1}{x}=\frac{\sqrt{1+x+x^2}}{x}-\frac1x.
$$
The integral becomes
$$
\int \frac{x+\sqrt{1+x+x^2}}{1+x+\sqrt{1+x+x^2}}\ dx=\int \frac{\sqrt{1+x+x^2}}{x}\ dx-\int \frac1x\ dx.
$$
The left part integral in RHS can be solved using Euler substitution by letting $t-x=\sqrt{1+x+x^2}$, you will get
$x=\dfrac{t^2-1}{2t+1}$, $dx=\dfrac{2(t^2+t+1)}{(2t+1)^2}\ dt$, and $\sqrt{x^2+x+1}=\dfrac{t^2+t+1}{2t+1}$, then it becomes
$$
\int \frac{\sqrt{1+x+x^2}}{x}\ dx=\int \dfrac{2(t^2+t+1)^2}{(t^2-1)(2t+1)^2}\ dt.
$$
The last part can be solved by using partial fraction decomposition
$$
\dfrac{2(t^2+t+1)^2}{(t^2-1)(1+2t)^2}=-\frac1{t+1}+\frac1{2t+1}-\frac3{2(2t+1)^2}+\frac1{t-1}+\frac12.
$$
I hope I don't mess up and this helps you.
| {
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"source": "stackexchange",
"question_score": "10",
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} |
$\iiint_Dz \;dxdydz$ where D is $z \ge 0 , z^2 \ge 2x^2+3y^2-1,x^2+y^2+z^2\le3.$ I asked a questions about regions and then tried to compute a tripple integral:
$$\iiint_Dz \;dxdydz$$ D is $z \ge 0 , z^2 \ge 2x^2+3y^2-1,x^2+y^2+z^2\le3.$
I tried, but now I am stuck: how do I calculate the volume of $D_{1b}$? Since I don't have many of these problems left, I decided not to look at the solution photo1, photo2.
Attempt: Here is a photo of my calculations
| $$ z>0$$ What implies:$$z\geq\sqrt{2x^2+3y^2-1}$$
Volume I need to calculate is the interior of sphere: $$ x^2+y^2+z^2=3$$ and above (where we have orthogonal projection to the xy plane) the hyperboloid.
Excuse me for lousy picture!
So I have:
$$ U_{tot}= \int\int\int_{D_1} zdV + \int\int\int_{D_2}z dV=U_1+U_2$$
Define orthogonal projection with $z=0$:
$$z=0=\sqrt{2x^2+3y^2-1}\rightarrow 2x^2+3y^2=1$$
Sphere projection is: $x^2+y^2=(\sqrt{3})^2$
Orthogonal projection $D_1$ is now described with the area between the eclipse and a circle, we need to define some parametrization in order to simplify. For example I can try something like this:
$$x=r\cos\theta,y=r\sin\theta$$
Equation of circle is now: $r=\sqrt{3}$ and equation of eclipse is: $2r^2\sin^2\theta+3r^2\cos^2\theta=1 \rightarrow r=\sqrt{\frac{1}{2cos^2\theta+3\sin^2\theta}}=\sqrt{\frac{1}{2+\sin^2\theta}}$
On the orthogonal projection $D_1$ we have a volume from hyperboloid to the sphere.
$$ U_1=\int^{2\pi}_0d\theta \int^{\sqrt{3}}_{\sqrt{\frac{1}{2+\sin^2\theta}}}rdr\int^{\sqrt{3-r^2}}_{\sqrt{2r^2\cos^2\theta+3r^2\sin^2\theta-1}=\sqrt{2r^2+r^2\sin^2\theta-1}}zdz =$$
$$=\frac{1}{2}\int^{2\pi}_0d\theta \int^{\sqrt{3}}_{\sqrt{\frac{1}{2+\sin^2\theta}}} \left( 3-r^2-2r^2-r^2\sin^2\theta+1\right) rdr=\\ =\frac{1}{2}\int^{2\pi}_0d\theta \int^{\sqrt{3}}_{\sqrt{\frac{1}{2+\sin^2\theta}}} \left(4r-3r^3-r^3\sin^2\theta\right)dr=\\
=\frac{1}{2}\int^{2\pi}_0 \left( 2r^2-\frac{r^4}{4}(3+\sin^2\theta) \right)|^{\sqrt{3}}_{\sqrt{\frac{1}{2+\sin^2\theta}}} d\theta=\\
=\frac{1}{2}\int^{2\pi}_0 \left( 6-\frac{9}{4}(3+\sin^2\theta)\right)-\left( \frac{2}{2+\sin^2\theta}-\frac{3+\sin^2\theta}{4(2+\sin^2\theta)^2}\right)d\theta=...
$$
On the orthogonal projection $D_2$ we have a volume from plane $z=0$ to the sphere.
$$U_2=\int^{2\pi}_0 d\theta \int^{\sqrt{3}}_0 rdr \int^{\sqrt{3-r^2}}_0 zdz =\\
= \frac{1}{2}\int^{2\pi}_0 d\theta \int^{\sqrt{3}}_0 r \left( 3-r^2\right)dr=\\
= \frac{1}{2}\int^{2\pi}_0 \left( 3r-r^3/3\right)|^{\sqrt{3}}_0d\theta=\\
= \frac{1}{2}\int^{2\pi}_0 3\sqrt{3}-\sqrt{3} d\theta=\\
= \sqrt{3}\int^{2\pi}_0 d\theta=2\sqrt{3}\pi.
$$
Sorry if a have mistaken somewhere in the calculus I will check it once again...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $f(x) = \frac{1}{x^2}$ is uniformly continuous on $[1, \infty]$ I am trying to prove this function is uniformily continuous on $[1, \infty]$, so far i have;
$$|f(x) - f(x)| = |\frac{1}{x^2} - \frac{1}{y^2}| = |\frac{(x-y)(x+y)}{x^2y^2}|$$
and then,
$$|x-y||\frac{x+y}{x^2y^2}|$$
I am not really sure where to go from this point?
| Since $x, y\ge 1$ then
$$|x-y|\frac{x+y}{x^2y^2}\le |x-y|\frac{x^2+y^2}{x^2y^2}= |x-y|\left(\frac{1}{x^2}+\frac{1}{y^2}\right)$$
$$ =|x-y|\left(\frac{1}{x^2}+\frac{1}{y^2}\right)\le 2|x-y|$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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solving $X^2 - 3X - A = 0$ where $A,X \in \mathbb{M_2(\mathbb{R})}$ Given $A = \begin{pmatrix} 7 & 3 \\ 3 & 7 \end{pmatrix}$ find a $2\times 2$ matrix $X$ s.t. $X^2 - 3X - A = 0$, in the previous parts I have diagonalised $A$ and got $P^{-1}AP = \begin
{pmatrix} 4 & 0 \\ 0 & 10 \end{pmatrix}$, but I'm not sure how this relates to finding $X$.
An idea was that because $A$ has $7,3$ repeating then $X = \begin{pmatrix} a & b \\ b & a \end{pmatrix}$ but this was only an idea and I'm not sure why $X$ would be in that form.
| $$X^2-3X-A=0$$
$$P(X^2-3X-A)P^{-1}=0$$
Which using the distruibitive property and the fact that $PX^2P^{-1}=(PXP^{-1})^2$, yields:
$$(PXP^{-1})^2-3(PXP^{-1})-PAP^{-1}=0$$
Now complete the square:
$$(PXP^{-1})^2-3(PXP^{-1})+\frac{9}{4}I_2=\frac{9}{4}I_2+PAP^{-1}$$
$$(PXP^{-1}-\frac{3}{2}I_2)^2=\frac{9}{4}I_2+PAP^{-1}= \begin{pmatrix} \frac{9}{4} & 0 \\ 0 & \frac{9}{4} \end{pmatrix}+\begin{pmatrix} 4 & 0 \\ 0 & 10 \end{pmatrix}$$
$$(PXP^{-1}-\frac{3}{2}I_2)^2= \begin{pmatrix} 6.25 & 0 \\ 0 & 12.25 \end{pmatrix}=\begin{pmatrix} 2.5 & 0 \\ 0 & 3.5 \end{pmatrix}^2$$
Thus, one of the solutions is obtained by solving
$$PXP^{-1}-\frac{3}{2}I_2= \begin{pmatrix} 2.5 & 0 \\ 0 & 3.5 \end{pmatrix}$$
Which is easy to solve.
To determine all solutions you just need to find all square roots of the diagonal matrix:$$\begin{pmatrix} 6.25 & 0 \\ 0 & 12.25 \end{pmatrix}$$
I put one of four square roots I know, I am not sure if there are others. I think you can easily determine this
| {
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} |
Number of integer solutions by generating functions method I'm stuck in the middle of a problem and not sure where to go next. The original problem is:
Find the number of integer solutions to the equation
$$2x + 3y + 4z + w + s + t = n$$ with $$0 \le w \le 2$$ $$2 \le s \le 5$$ $$0 \le t \le 3$$
Now I was able to create my generating functions equations to get my overall equation to this:
$$ G(x) = (\frac{1}{1-x^2})(\frac{1}{1-x^3}) (\frac{1}{1-x^4})(1+x+x^2)(x^2+x^3+x^4+x^5)(1+x+x^2+x^3) $$
Which with the help of the finite geometric series and some cancelling I was able to simplify down to:
$$ G(x) = (\frac{1}{1-x^2})(\frac{x^2}{1}) (\frac{1}{(1-x)^3})(\frac{1-x^4}{1}) $$
But now I'm stuck. Any help would be appreciated. Thanks.
Edit: With the extra step of simplification
$$ G(x) = (\frac{x^2}{1}) (\frac{1}{(1-x)^3})(\frac{1+x^2}{1}) $$
| You end up with:
\begin{align}
[x^n] G(x)
&= [x^n] \frac{x^2 + x^4}{(1 - x)^3} \\
&= [x^{n - 2}] (1 - x)^{-3} + [x^{n - 4}] (1 - x)^{-3} \\
&= \binom{-3}{n - 2} (-1)^{n - 2} + \binom{-3}{n - 4} (-1)^{n - 4} \\
&= \binom{n - 2 + 3 - 1}{3 - 1} + \binom{n - 4 + 3 - 1}{3 - 1} \\
&= \frac{n (n - 1)}{2} + \frac{(n - 2) (n - 3)}{2} \\
&= n^2 - 3 n + 3
\end{align}
| {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
} |
Analytic Geometry How does one solve:
Find the equation of the circle which has it's center on the line $y= 3-x$ , and which has as tangents the lines $ 2y-x = 22, $ $ 2x+y=11 $ ?
| Let $(a,b)$ be the center of the circle. Since the center lies on the line $y=3-x$, then we have $b=3-a$. The equation of the circle if the center on $(a,b)$ and radius $r$ is
$$
(x-a)^2+(y-b)^2=r^2\tag1
$$
and the equation of its tangent line is
$$
(x_c-a)(x-a)+(y_c-b)(y-b)=r^2,\tag2
$$
where $(x_c,y_c)$ is point of contact. Equation $(2)$ can be written as
$$
(x_c-a)x+(y_c-b)y=r^2+a(x_c-a)+b(y_c-b).\tag3
$$
The tangent lines are
$$
-x+2y=22\tag4
$$
and
$$
2x+y=11.\tag5
$$
Using $b=3-a$, comparing $(3)$ and $(4)$ yields $x_1-a=-1$, $y_1-b=2$, and
$$
\begin{align}
r^2+a(x_1-a)+b(y_1-b)&=22\\
r^2+a(-1)+b(2)&=22\\
r^2-a+2(3-a)&=22\\
r^2-3a&=16.\tag6
\end{align}
$$
Similarly, comparing $(3)$ and $(5)$ yields $x_2-a=2$, $y_2-b=1$, and
$$
\begin{align}
r^2+a(x_2-a)+b(y_2-b)&=11\\
r^2+a&=8.\tag7
\end{align}
$$
Solving $(6)$ and $(7)$ yields $a=-2$, $b=5$, and $r^2=10$. Thus, using $(1)$, the equation of the circle is
$$
\Large\color{blue}{(x+2)^2+(y-5)^2=10}.
$$
$$\\$$
$$\Large\color{blue}{\text{# }\mathbb{Q.E.D.}\text{ #}}$$
| {
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Proving uniform convergence
Prove uniform convergence of function series: $$ \sum_{n=0}^\infty \frac{1}{n^2 + x} \sin \frac{1}{n^2 + x}$$ on $ \Bbb R $
I'm stuck with a problem, because I've proven that is uniformly convergent(proof below), while a friend of mine, gave a counterexample to this problem, $$ x = \frac{1 - \frac{\pi n^2}{2}}{\pi/2} $$ and then it is only $$\sum_{n=0}^\infty \frac{\pi}{2}\sin \frac{\pi}{2} = \sum_{n=0}^\infty \frac{\pi}{2} $$ , so it not uniformly convergent.
Here is my proof of uniform convergence:
$$ \frac{1}{n^2 + x} \sin \frac{1}{n^2 + x} < \frac{1}{n^2 + x} $$ , since $$ \sin {x} < 1 $$
and
$$ \frac{1}{n^2 + x} < \frac{1}{n^2} $$ , which is convergent.
Hence, I claim that by Weierstrass M-Test : $$ \sum_{n=0}^\infty \frac{1}{n^2 + x} \sin \frac{1}{n^2 + x} $$ is uniformly convergent on $$ \Bbb R $$ .
So, where is the mistake? I'd appreciate any help.
P.S Sorry, for any mistakes in latex.
EDIT: I'm awfully sorry it should be : $$ \sum_{n=0}^\infty \frac{1}{n^2 + x} \sin \frac{1}{n^2 + x}$$
| As $x$ approaches $-n^2$, $\frac1{n^2+x}\sin\left(\frac1{n^2+x}\right)$ oscillates wildly between very large numbers. Since uniform convergence only worries about the tail, we can fix this by assuming that $x$ is not the negative of a perfect square and that $x\gt-m^2$.
Let $M,N\gt2m$, then
$$
\begin{align}
\sum_{n=M}^N\frac1{n^2+x}
&\le\sum_{n=M}^N\frac1{n^2-m^2}\\
&\le\sum_{n=M}^N\frac1{n^2-(n/2)^2}\\
&\le\frac43\sum_{n=M}^N\frac1{n^2}\\
&\le\frac43\sum_{n=M}^N\frac1{n(n-1)}\\
&=\frac43\left(\frac1{M-1}-\frac1N\right)
\end{align}
$$
This is independent of $x$ as long as $x\gt-m^2$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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inner product of trivector and bivector in geometric algebra Hestenes's "New Foundations for Classical Mechanics" book (page 47, 1.1c) sets a problem to show:
$\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \cdot B=\mathbf{a} \left( \left( \mathbf{b} \wedge \mathbf{c} \right) \cdot B \right)+\mathbf{b} \left( \left( \mathbf{c} \wedge \mathbf{a} \right) \cdot B \right)+\mathbf{c} \left( \left( \mathbf{a} \wedge \mathbf{b} \right) \cdot B \right).\end{aligned}$
I'm having trouble proving this. Using an antisymmetric expansion of the wedge within a grade one selection, we have
$\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \cdot B&=\frac{1}{{6}}{\left\langle \left( \mathbf{a} \mathbf{b} \mathbf{c}+\mathbf{b} \mathbf{c} \mathbf{a}+\mathbf{c} \mathbf{a} \mathbf{b}-\mathbf{a} \mathbf{c} \mathbf{a}-\mathbf{b} \mathbf{a} \mathbf{c}-\mathbf{c} \mathbf{b} \mathbf{a}\right)B \right\rangle}_1 \\ &=\frac{1}{{3}}{\left\langle \mathbf{a} \left( \mathbf{b} \wedge \mathbf{c} \right) B+\mathbf{b} \left( \mathbf{c} \wedge \mathbf{a}\right) B+\mathbf{c} \left( \mathbf{a} \wedge \mathbf{b}\right) B\right\rangle}_1 \\ &=\frac{1}{{3}}\left(\mathbf{a} \left( \left( \mathbf{b} \wedge \mathbf{c} \right) \cdot B \right)+\mathbf{b} \left( \left( \mathbf{c} \wedge \mathbf{a}\right) \cdot B \right)+\mathbf{c} \left( \left( \mathbf{a} \wedge \mathbf{b}\right) \cdot B \right)\right) \\ &\quad +\frac{1}{{3}}\left(\mathbf{a} \cdot {\left\langle \left( \mathbf{b} \wedge \mathbf{c} \right) B \right\rangle}_2+\mathbf{b} \cdot {\left\langle \left( \mathbf{c} \wedge \mathbf{a}\right) B \right\rangle}_2+\mathbf{c} \cdot {\left\langle \left( \mathbf{a} \wedge \mathbf{b}\right) B \right\rangle}_2\right).\end{aligned}$
The first three terms have the desired form, but are off by a factor of 3. My attempts to eliminate the latter three terms have all gone in circles. Any tips, or suggestions for a different approach?
| I figured it out. The key is trying not to be so clever, instead just expanding in successive dot products and then regrouping. With $B = \mathbf{u} \wedge \mathbf{v}$
$\begin{aligned}\left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right)\cdot B&={\left\langle{{ \left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \left( \mathbf{u} \wedge \mathbf{v} \right) }}\right\rangle}_{1} \\ &={\left\langle{{ \left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \left( \mathbf{u} \mathbf{v} - \mathbf{u} \cdot \mathbf{v} \right) }}\right\rangle}_{1} \\ &=\left( \left( \mathbf{a} \wedge \mathbf{b} \wedge \mathbf{c} \right) \cdot \mathbf{u} \right) \cdot \mathbf{v} \\ &= \left( \mathbf{a} \wedge \mathbf{b} \right) \cdot \mathbf{v} \left( \mathbf{c} \cdot \mathbf{u} \right)+\left( \mathbf{c} \wedge \mathbf{a} \right) \cdot \mathbf{v} \left( \mathbf{b} \cdot \mathbf{u} \right)+\left( \mathbf{b} \wedge \mathbf{c} \right) \cdot \mathbf{v} \left( \mathbf{a} \cdot \mathbf{u} \right) \\ &=\mathbf{a} \left( \mathbf{b} \cdot \mathbf{v} \right)\left( \mathbf{c} \cdot \mathbf{u} \right)-\mathbf{b} \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) \\ &\quad +\mathbf{c} \left( \mathbf{a} \cdot \mathbf{v} \right)\left( \mathbf{b} \cdot \mathbf{u} \right)-\mathbf{a} \left( \mathbf{c} \cdot \mathbf{v} \right)\left( \mathbf{b} \cdot \mathbf{u} \right) \\ &\quad +\mathbf{b} \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right)-\mathbf{c} \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) \\ &=\mathbf{a}\left( \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) - \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{b} \cdot \mathbf{u} \right) \right)\\ &\quad +\mathbf{b}\left( \left( \mathbf{c} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) - \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{c} \cdot \mathbf{u} \right) \right) \\ &\quad +\mathbf{c}\left( \left( \mathbf{a} \cdot \mathbf{v} \right) \left( \mathbf{b} \cdot \mathbf{u} \right) - \left( \mathbf{b} \cdot \mathbf{v} \right) \left( \mathbf{a} \cdot \mathbf{u} \right) \right) \\ &=\mathbf{a} \left( \mathbf{b} \wedge \mathbf{c} \right)\cdot \left( \mathbf{u} \wedge \mathbf{v} \right) \\ &\quad +\mathbf{b} \left( \mathbf{c} \wedge \mathbf{a} \right)\cdot \left( \mathbf{u} \wedge \mathbf{v} \right) \\ &\quad +\mathbf{c} \left( \mathbf{a} \wedge \mathbf{b} \right) \cdot \left( \mathbf{u} \wedge \mathbf{v} \right) \\ &=\mathbf{a} \left( \mathbf{b} \wedge \mathbf{c} \right) \cdot B+\mathbf{b} \left( \mathbf{c} \wedge \mathbf{a} \right) \cdot B+\mathbf{c} \left( \mathbf{a} \wedge \mathbf{b} \right) \cdot B.\end{aligned}$
| {
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$99$th derivative of $\sin x$ Can someone help me calculate the $99$th derivative of $\sin(x)$?
Calculate $f^{(99)}(x) $ for the function $f(x) = \sin(x) $
| Notice if $f(x) = \sin x$
$f'(x) = \cos x = \sin( x + \frac{\pi}{2}) $
$f''(x) = - \sin x = \sin( x + \pi) = \sin( x + 2 (\frac{\pi}{2})) $
$f'''(x) = - \cos x = \sin(x + 3( \frac{\pi}{2} ))$
$f''''(x) = \sin x $
Hence, we can say that
$$ f^{(n)} (x) = \sin \left( x + n \cdot\frac{\pi}{2} \right) $$
| {
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Can an odd perfect number be divisible by $5313$? I know that an odd perfect number cannot be divisible by $105$ or $825$. I wonder if that's also the case for $5313$.
| It is quite easy to prove that an odd perfect prime has a factorisation into powers of odd primes, where exactly one power is a prime of the form 4k+1, raised to a power 4n+1 (n can be 0), and all the other powers are even. This implies an odd perfect number which is a multiple of $5313 = 3 * 7 * 11 * 23$ must be a multiple of $3^27^211^223^2$. $3^2 7^2 11^2 23^2$ has an abundancy of 1.9037 < 2, so multiples of this number could a priori have an abundance of 2.
However, since $1 + 3 + 9 = 13$, if the number is not a multiple of $3^4$ then it must be a multiple of 13 (it doesn't have to be a multiple of $13^2$ because 13 = 4k+1). Since 1 + 7 + 49 = 57, if the number is not a multiple of $7^4$ then it must be a multiple of 3 (which it is) and of 19, and therefore a multiple of 19^2.
Each of the four possibilities $3^4 7^4 11^2 23^2$, $3^4 7^2 11^2 23^2 19^2$, $3^2 7^4 11^2 23^2 13$ and $3^2 7^2 11^2 23^2 19^2 13$ has an abundancy just above 2, so these numbers and any multiples of them cannot be perfect numbers.
| {
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"timestamp": "2023-03-29T00:00:00",
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Use $\sum_{n=0}^{\infty} \frac {2n +1} {2^n} = 6 $ to show $\sum_{n=0}^{\infty} \frac {2n +1} {2^n} i^n = \frac 4 {25} + i \frac {22} {25}$. I've shown that the series $$\sum_{n=0}^{\infty} \frac {2n +1} {2^n}$$ converges to $6$ using elementary series operations.
However how can I use this to show that $$\sum_{n=0}^{\infty} \frac {2n +1} {2^n} i^n = \frac 4 {25} + i \frac {22} {25}$$
and $$\sum_{p=0}^{\infty} (-1)^p\frac {4p +1} {4^p}$$ is convergent and has a sum which should be found ?
I don't see the connection, except that the second series is a product of a series I know is convervent (first series) and a series that is divergent.
| Hints:
$$i^n=\begin{cases}\;\;\,i&,\;\;n=1\pmod4\\-1&,\;\;n=2\pmod4\\-i&,\;\;n=3\pmod4\\\;\;\,1&,\;\;n=0\pmod4\end{cases}$$
Added on request: Using the above and the sum of a geometric series with ratio $\;r\;,\;\;|r|<1\;$
$$\sum_{k=0}^\infty\frac{2n+1}{2^n}i^n=2\sum_{n=0}^\infty \frac {i^nn}{2^n}+\sum_{n=0}^\infty\left(\frac i2\right)^n=$$
$$=2\sum_{n=0}^\infty\frac{4n+1}{2^{4n+1}}-2\sum_{n=0}^\infty\frac{4n+2}{2^{4n+2}}-2i\sum_{n=0}^\infty\frac{4n+3}{2^{4n+3}}+2\sum_{n=0}^\infty\frac{4n}{2^{4n}}+\frac1{1-\frac i2}=$$
Now use that (yes, again geometric, but this time power, series):
$$\frac1{1-x}=\sum_{n=0}^\infty x^n\implies\frac1{(1-x)^2}=\sum_{n=1}^\infty nx^{n-1}\;,\;\;|x|<1\;\;\;(**)$$
so for example
$$\sum_{n=0}^\infty\frac{4n+1}{2^n(=2\cdot2^{n-1})}=2\sum_{n=1}^\infty n\left(\frac12\right)^{n-1}+\frac1{1-\frac12}=2\frac1{\left(1-\frac12\right)^2}+2=10$$
$\color{red}{\text{But now I'm realizing there's a much easier way to do this...!}}$
Using (**) above:
$$\sum_{k=0}^\infty\frac{2n+1}{2^n}i^n=i\sum_{n=0}^\infty n\left(\frac i2\right)^{n-1}+\sum_{n=0}^\infty\left(\frac i2\right)^n=i\frac1{\left(1-\frac i2\right)^2}+\frac1{1-\frac i2}=$$
$$=\frac{4i}{3-4i}+\frac2{2-i}=\frac{-16+12i}{25}+\frac{20+10i}{25}=\frac4{25}+\frac{22}{25}i$$
| {
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How to solve this equation $2\cos(\frac {x^2+x}{6})=2^x+2^{-x}$ How do I solve for $x$ from this equation?
$$2\cos\frac {x^2+x}{6}=2^x+2^{-x}$$
| Let $y=2^x$, then
$$
\begin{align}
\ln y&=\ln2^x\\
\ln y&=x\ln 2\\
y&=e^{x\ln 2}.
\end{align}
$$
Consequently, $2^{-x}=e^{-x\ln 2}$ and
$$
\begin{align}
2\cos\left(\frac{x^2+x}{6}\right)&=e^{x\ln 2}+e^{-x\ln 2}\\
\cos\left(\frac{x^2+x}{6}\right)&=\frac{e^{x\ln 2}+e^{-x\ln 2}}{2}
\end{align}
$$
Now, let $x=i\theta$, then
$$
\begin{align}
\cos\left(\frac{(i\theta)^2+i\theta}{6}\right)&=\frac{e^{i\theta\ln 2}+e^{-i\theta\ln 2}}{2}\\
\cos\left(\frac{-\theta^2+i\theta}{6}\right)&=\cos(\theta\ln 2)\\
\frac{-\theta^2+i\theta}{6}&=\theta\ln 2\\
\theta^2+(6\ln2-i)\theta&=0\\
\theta(\theta+6\ln2-i)&=0\\
\theta_1=0&\text{ or }\ \theta_2=i-6\ln2.
\end{align}
$$
Thus, $\large x_1=0$ and $\large x_2=-(1+6i\ln2)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Integration by parts order question + integration question In integration by parts, does it matter which of your terms is v (or, rather $f(x)$) and which term is du (or, rather $ f'(x) $?
Also, I'm having trouble finding $\int e^{2x} \cos(3x)dx $. I keep getting $$\frac {2e^{2x}(6\sin(3x) - \cos(3x))}{37}$$ when $u=e^{2x}$ and $dv = \cos(3x) dx $.
Thanks!
| Yes, it does matter. You can check Wikipedia's page on integration by parts, and in particular the LIATE rule. In general, you want $u$ to be easier to differentiate, whereas $dv$ to be easier to integrate.
In your example, however, if you follow the rule above, both functions should be $u$. You can overcome this by applying the same choice multiple times. For example, let's set $u = e^{2x}$ and $dv = \cos (3x)\ dx$. Then $du = 2e^{2x}\ dx$ and $v = \frac{\sin 3x}{3}$. So:
$$\int e^{2x} \cos (3x)\ dx = \frac{1}{3}e^{2x}\sin 3x - \frac{2}{3} \int \sin(3x) e^{2x}\ dx \tag 1$$ Integrating again, with $u = e^{2x}$ and $dv = \sin(3x)\ dx$ we get that $du = 2 e^{2x}\ dx$ and $v = - \frac{1}{3} \cos 3x$. The latter integral becomes:
$$\int \sin(3x) e^{2x}\ dx = -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3}\int e^{2x}\cos(3x)\ dx$$
Substituting back into $(1)$ we get:
$$\int e^{2x} \cos (3x)\ dx = \frac{1}{3}e^{2x}\sin 3x - \frac{2}{3} \left ( -\frac{1}{3}e^{2x}\cos 3x + \frac{2}{3}\int e^{2x}\cos(3x)\ dx\right )$$
and then
$$\int e^{2x} \cos (3x)\ dx = \frac{1}{3}e^{2x}\sin 3x + \frac{2}{9} e^{2x}\cos 3x - \frac{4}{9} \int e^{2x}\cos(3x)\ dx$$
Adding $\displaystyle \frac{4}{9} \int e^{2x}\cos(3x)\ dx$ to both sides of the equation:
$$\frac{13}{9} \int e^{2x} \cos (3x)\ dx = \frac{1}{3}e^{2x}\sin 3x + \frac{2}{9} e^{2x}\cos 3x$$
Finally:
$$\begin{align*}
\int e^{2x} \cos (3x)\ dx &= \frac{3}{13}e^{2x}\sin 3x + \frac{2}{13} e^{2x}\cos 3x + C = \\
&= \frac{1}{13} e^{2x} (3 \sin 3x + 2 \cos 3x) + C
\end{align*}$$.
| {
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Solve $f(x) = ax^2 + bx + c$ to find the value of $K$ $f(x)=ax^2+bx+c$, where $a=-9$, $b=12$ and $c=16$. If $$-1<f'(x)<1$$ then $h<x<k$. To $2$ decimal places, what is the value of $k$?
Hi, this is working for solving $f(x) = ax^2 + bx + c$ to find the value of $K$
I used Quadratic Equation to solve for x and got the answer for
$x = -0.824045318$ or $2.157378652$
Hence, the value of K is -0.82
Did I used the correct formula to solve this question?
| well, suppose $f(x) = ax^2 + bx + c $, then $f'(x) = 2ax + b $. If $-1 < f'(x) < 1$, then
$$ -1 < 2ax + b < 1 \iff -b-1 < 2ax < 1 - b \iff - \frac{b+1}{2a} < x < \frac{1-b}{2a}$$
| {
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using substitution wrongly Solving integral, first way:
$$\int \frac{du}{u^2-9}=-\int \frac{du}{9-u^2}$$
$$u={3\sin v}$$
$$du=3\cos vdv$$
$$-\int \frac{3\cos vdv }{9-9\sin^{2}v}=-\frac 13\int\frac{dv}{\cos v}=-\frac 13\ln\left(\sec v+\tan v\right)=-\frac13 \ln \frac {\frac u3}{\sqrt{1-\frac{u^2}{9}}}$$
$$-\frac13 \ln \frac{u}{\sqrt{9-u^2}}=-\frac16 \ln \frac {u^2}{9-u^2}=\frac16 \ln \frac{9-u^2}{u^2}$$
second way:
$$\int \frac{du}{u^2-9}=\frac 16\int \frac{du}{u-3}-\frac16 \int \frac{du}{u+3}=\frac16 \ln \frac {u-3}{u+3}$$
what's wrong with first way?
Thanks all. $1$ from $\sec$ is missed, must be $1+\frac u3$
| $-\dfrac 13 \ln|\sec v + \tan v|$ $ = -\dfrac 13 \ln \left| \dfrac{1+\sin v}{\cos v} \right| \\ = -\dfrac 13 \ln \left|\dfrac{3+u}{\sqrt{9-u^2}}\right| \\ = - \dfrac 16 \ln \left|\dfrac{(3+u)^2}{(3-u)(3+u)}\right| \\ = \dfrac 16 \ln\left|\dfrac{3-u}{3+u}\right|$
| {
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How many different systems are there? A Football team has 10 players besides the Keeper, who are allocated into 3 positions, the defense, the center and the attack. A system is the allocation of the players into these positions.
For example the system 4 4 2 means that there are 4 players at the defense, 4 players at the center and 2 players at the attack.
There are 5 players available for the defense, 6 players for the center and 4 players for the attack.
How many different systems can be formed, when we know that at each position must be at least one player??
I found the following sentence:
$$ \text{ The number of integer solutions of the equation} x_1+x_2+ \dots + x_n=r , x_1>a_1 , x_2>a_2 , \dots, x_n>a_n \text{ can be found with the formula } \displaystyle{\frac{r-a_1-a_2- \dots -a_n-1}{n-1}} $$
Can I do it as followed?
$$x_1+x_2+x_3=10 , x_1 \leq 5, x_2 \leq 6, x_3 \leq 4, x_1,x_2,x_3 \geq 1 $$
$$y_1=x_1-1, 0 \leq y_1 \leq 4 $$
$$y_2=x_2-1, 0 \leq y_2 \leq 5 $$
$$y_3=x_3-1, 0 \leq y_3 \leq 3 $$
$$ y_1+1+y_2+1+y_3+1=10 \Rightarrow y_1+y_2+y_3=7 , y_1,y_2,y_3 \geq 0$$
There are $\binom{3+7-1}{7}=36$ integer solutions of the equation $ y_1+y_2+y_3=7 , y_1,y_2,y_3 \geq 0$
We have to substract the cases:
*
*$x_1>5: \binom{10-5-1}{2}=6 $
*$x_2>6: \binom{10-6-1}{2}=3$
*$x_3>4: \binom{10-4-1}{2}=10 $
So,there are $36-6-3-10=17$ different systems that can be formed.
Or couldn't we do it this way?
| i have another idea. I used a generating function. Here it is $(x+x^2+x^3+x^4+x^5+x^6)*(x+x^2+x^3+x^4+x^5)*(x+x^2+x^3+x^4)$. If you expand this expression you can read off the number of systems can be formed. The Expansion is $x^3+3 x^4+6 x^5+10 x^6+14 x^7+17 x^8+18 x^9+17 x^{10}+14 x^{11}+10 x^{12}+6 x^{13}+3 x^{14}+x^{15}$
The exponents are the number of team members. In your case 10. The related coefficient is the number of systems can be formed.
greetings,
calculus
| {
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Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$. Let $f(x)=ax^2+bx+c$ where $a,b,c$ are real numbers. Suppose $f(-1),f(0),f(1) \in [-1,1]$. Prove that $|f(x)|\le \frac{3}{2}$ for all $x \in [-1,1]$.
I made quite a few attempts but could not explicitly derive the required condition. I can understand that it happens, but cannot derive that. Please help.
| The proof for 3/2 follows, though intuitively, I think that you can limit it to 5/4: it's a quadratic and if you draw an extreme graph then it should go through (-1, 1), (0, -1), (1, -1) with a minimum at (1/2, -5/4), or some reflection of this. Back to 3/2 ...
If $f$ is monotonic in [-1, 1] then trivially, $|f| \le 1$, so assume not.
Then $f$ is differentiable and (being quadratic) can only have one extreme point at $f'(x_0) = 0$, i.e $2ax_0 + b$ = 0, so that $ax_0 = -b/2$. The extreme value of the function at this point is therefore $f(x_0) = x_0(-b/2) + bx_0 + c = bx_0/2 + c$.
Taking the possible values at -1, 0, and 1 gives
$|f(-1)| = |a - b + c| \le 1$; $|f(0)| = | c| \le 1$; $|f(1)| = |a + b + c| \le 1$
So $|c| \le 1 $ and it follows from these inequalites that $|b| \le 1$
Substituting in the value of $f(x_0)$,
$|f(x_0)| = |bx_0/2 + c| \le |b|.|x_0|/2 + |c| \le 1.1/2 + 1 = 3/2$
Continuation of proof to show that $|f(x)| \le 5/4$
(1) If $|a| \ge 1$
At the extremum, $x_0 = -b/2a$ so by substitution $f(x_0) = ab^2/4a^2 -b^2/2a +c$
$|f(x_0)| = |c - b^2/4a| \le |c| + |b|^2/4|a|$. We already proved $|c|, |b| \le 1$ and by assumption $|a| \ge 1$ so $|f(x_0)| \le 1 + 1/4 = 5/4$
(2) If $|a| \le 1$
Take the Taylor expansion of $f(x)$ about $x_0$: $f(x) = f(x_0) + (x-x_0) f'(x_0) + (x-x_0)^2 f''(z)/2$ for $z$ between $x$ and $x_0$. Note that $f'(x_0) = 0$ at the extremum, and $f''(z) = 2a $ for all $z$, so
$f(x) = f(x_0) + a(x-x_0)^2$, and $|f(x_0)| = |f(x) - a(x - x_0)^2| \le |f(x)| + |a|(x-x_0)^2$. By assumption, $|a| \le 1$, so
$|f(x_0)| \le |f(x) | + (x-x_0)^2$
For $x = -1, 0, or +1$ and $x_0$ in [-1, 1], one of these values makes $|x - x_0| \le 1/2$ (take 0 if $-1/2 \le x_0 \le +1/2 $ ; +1 if $x_0 > 1/2$; -1 if $x_0 \le -1/2$), and in all cases the corresponding function value satisfies $|f(x)| \le 1$, so at this x,
$|f(x_0)| \le 1 + (1/2)^2 = 5/4.$
Footnote in response to comment 24 April 2022....
A) $|f(-1)| = |a - b + c| \le 1$
B) $|f(0)| = | c| \le 1$
C) $|f(1)| = |a + b + c| \le 1$
From (A), $ -1 \le a - b + c \le 1$
So, $-1 \le - a + b - c \le 1$
And, (C) $-1 \le a + b + c \le 1$
Then $-2 \le 2b \le 2 $
Giving $|b| \le 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/774069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\lim_{n \to \infty} \frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5}$ Is my solution correct?
By Faulhaber's formula, $1^4 + 2^4 + \ldots + n^4 = \frac{6n^5+15n^4+10n^3-n}{30} $.
$$\frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5} = \frac{1}{n^5} \left[\sum_{k=1}^{2n} k^4 - \sum_{k=1}^{n} (2k)^4 \right] = \frac{1}{n^5} \left[\frac{6\cdot(2n)^5 + o(n^5) - 2^4 \cdot 6 \cdot n^5 - o(n^5)}{30} \right] = \frac{6\cdot2^5 - 2^4 \cdot 6}{30} + o(1) = \frac{16}{5} + o(1)$$
So, $\lim_{n \to \infty} \frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5} = 3.2$
| Using Stolz:
$$
\lim_{n\to\infty}\frac{1^4 + 3^4 + \ldots + (2n-1)^4}{n^5}=
\lim_{n\to\infty}\frac{(2n+1)^4}{(n+1)^5-n^5}=
\lim_{n\to\infty}\frac{16n^4+\cdots}{5n^4+\cdots}=\cdots
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/775490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral$\int_0^{\pi/4} \log \tan \left(\frac{\pi}{4}\pm x\right)\frac{dx}{\tan 2x}=\pm\frac{\pi^2}{16}$ Hi I am trying to prove $$
\int_0^{\pi/4} \log \tan \left(\frac{\pi}{4}\pm x\right)\frac{dx}{\tan 2x}=\pm\frac{\pi^2}{16}.
$$
What an amazing result and a clever one this is. I tried writing
$$
\int_0^{\pi/4} \log \sin \left(\frac{\pi}{4}\pm x\right)\frac{dx}{\tan 2x}-\int_0^{\pi/4} \log \cos \left(\frac{\pi}{4}\pm x\right)\frac{dx}{\tan 2x}.
$$
Changing variables $y=2x$ I obtained
$$
\frac{1}{2}\int_0^{\pi/2} \log \sin \left(\frac{\pi}{4}\pm \frac{y}{2}\right)\frac{dy}{\tan y}-\frac{1}{2}\int_0^{\pi/2} \log \cos \left(\frac{\pi}{4}\pm \frac{y}{2}\right)\frac{dy}{\tan y}.
$$
I would rather work with the log sine/cosines for $y\in [0,\pi/2]$ since we can use $\int_0^{\pi/2} \log \sin x dx=-\frac{\pi}{2} \ln 2.$ But I am stuck here. Thanks
| We can write the integral as:
\begin{align*}
\int_0^{\pi/4} \frac{\log{\tan{(\frac{\pi}{4}-x)}}}{\tan{(2x)}}\, dx &= \int_0^{\pi/4} \log{\tan{(x)}}\tan{(2x)} \, dx \\
\end{align*}
Let
\begin{align*}
I(a) &= \int_0^{\pi/4} \tan{(x)}^a\, \tan{(2x)} dx \\
&= \int_0^{\pi/4} \frac{2\, \tan{(x)}^{a+1}}{1-\tan{(x)}^2} dx\\
&=\int_0^1 \frac{2\, t^{a+1}}{1-t^4}\, dt\\
&=\int_0^1 2\, t^{a+1}\, \sum_{n\ge 0} t^{4n}\, dt\\
&=\sum_{n\ge 0} \int_0^1 2\, t^{a+1+4n}\, dt\\
&=\sum_{n\ge 0} \frac{2}{a+2+4n}
\end{align*}
and the required integral is:
\begin{align*}
I'(0) &= \sum_{n\ge 0} -\frac{2}{(4n+2)^2}\\
&= -\frac{2}{4}\cdot \frac{3}{4} \zeta{(2)} = -\frac{\pi^2}{16}
\end{align*}
and in general,
\begin{align*}
I^{(n)}(0) = \boxed{\displaystyle \int_0^{\pi/4} \frac{\left(\log{\tan{\left(\frac{\pi}{4}- x\right)}}\right)^n}{\tan{(2x)}}\, dx = \frac{\left(-1\right)^{n} n!}{2^n} \left(1-\frac{1}{2^{n + 1}}\right) \zeta(n + 1)}
\end{align*}
and proceeding similarly for the other case:
\begin{align*}
\boxed{\displaystyle \int_0^{\pi/4} \frac{\left(\log{\tan{\left(\frac{\pi}{4}+ x\right)}}\right)^n}{\tan{(2x)}}\, dx = \frac{n!}{2^n} \left(1-\frac{1}{2^{n + 1}}\right) \zeta(n + 1)}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/779262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
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