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How to add compound fractions? How to add two compound fractions with fractions in numerator like this one:
$$\frac{\ \frac{1}{x}\ }{2} + \frac{\ \frac{2}{3x}\ }{x}$$
or fractions with fractions in denominator like this one:
$$\frac{x}{\ \frac{2}{x}\ } + \frac{\ \frac{1}{x}\ }{x}$$
|
The multiplicative inverse of a fraction a/b is b/a. (Wikipedia)
Let us start with the properties:
*
*Division by a number or fraction is the same as multiplication by its inverse or reciprocal.
Division by $r$ is equal to the multiplication by $\dfrac{1}{r}$:
$$\dfrac{\ \dfrac{p}{q}\ }{r}=\dfrac{p}{q}\cdot \dfrac{1}{r}=\dfrac{p\cdot 1}{q\cdot r}=\dfrac{p}{q r}, \quad (1)$$
Division by $\dfrac{t}{u}$ is equal to the multiplication by $\dfrac{u}{t}$:
$$\dfrac{\ s}{\ \dfrac{t}{u}\ }=s\cdot\dfrac{u}{t}=\dfrac{s\cdot u}{t}=\dfrac{su}{t}.\quad (2)$$
*
*Sum of fractions
$$\dfrac{a}{b}+\dfrac{c}{d}=\dfrac{ad+bc}{bd}.\quad (3)$$
Apply $(1)$ to
$$\dfrac{\ \dfrac{1}{x}\ }{2}=\dfrac{1}{x}\cdot\dfrac{1}{2}=\dfrac{1\cdot 1}{x\cdot 2}=\dfrac{1}{2x}$$
and $(2)$ to
$$\frac{\ \dfrac{2}{3x}\ }{x}=\dfrac{2}{3x}\cdot\dfrac{1}{x}=\dfrac{2\cdot 1}{3x\cdot x}=\dfrac{2}{3x^2}.$$
So by $(3)$ we have
$$\dfrac{\ \dfrac{1}{x}\ }{2} + \dfrac{\ \dfrac{2}{3x}\ }{x}=\dfrac{1}{2x}+\dfrac{2}{3x^2}=\dfrac{1\cdot 3x^2+2\times 2x}{2x\cdot 3x^2 }=\dfrac{3x^2+4x}{6x^3}=\dfrac{x(3x+4)}{x(6x^2)}=\dfrac{3x+4}{6x^2}.$$
For
$$\dfrac{x}{\; \dfrac{2}{x}\ } + \dfrac{\; \dfrac{1}{x}\; }{x}$$
we have
$$\dfrac{\; x\; }{\dfrac{2}{x}} + \dfrac{\; \dfrac{1}{x}\; }{x}=\dfrac{x\cdot x}{2} + \dfrac{1}{x\cdot x}=\dfrac{x^2}{2}+\cfrac{1}{x^2}=\dfrac{x^2\cdot x^2+2\cdot 1}{2\cdot x^2}=\dfrac{x^4+2}{2x^2}.$$
We can apply the property Division by a fraction is the same as multiplication by its inverse or reciprocal to the following fraction
$$\dfrac{\;\dfrac{a}{b}\;}{\dfrac{c}{d}}=\dfrac{a}{b}\cdot \dfrac{d}{c}=\dfrac{a\cdot d}{b\cdot c}=\dfrac{ad}{bc}\qquad (4).$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is this Batman equation for real? HardOCP has an image with an equation which apparently draws the Batman logo. Is this for real?
Batman Equation in text form:
\begin{align}
&\left(\left(\frac x7\right)^2\sqrt{\frac{||x|-3|}{|x|-3}}+\left(\frac y3\right)^2\sqrt{\frac{\left|y+\frac{3\sqrt{33}}7\right|}{y+\frac{3\sqrt{33}}7}}-1 \right) \\
&\qquad \qquad \left(\left|\frac x2\right|-\left(\frac{3\sqrt{33}-7}{112}\right)x^2-3+\sqrt{1-(||x|-2|-1)^2}-y \right) \\
&\qquad \qquad \left(3\sqrt{\frac{|(|x|-1)(|x|-.75)|}{(1-|x|)(|x|-.75)}}-8|x|-y\right)\left(3|x|+.75\sqrt{\frac{|(|x|-.75)(|x|-.5)|}{(.75-|x|)(|x|-.5)}}-y \right) \\
&\qquad \qquad \left(2.25\sqrt{\frac{(x-.5)(x+.5)}{(.5-x)(.5+x)}}-y \right) \\
&\qquad \qquad \left(\frac{6\sqrt{10}}7+(1.5-.5|x|)\sqrt{\frac{||x|-1|}{|x|-1}} -\frac{6\sqrt{10}}{14}\sqrt{4-(|x|-1)^2}-y\right)=0
\end{align}
| In fact, the five linear pieces that consist the "head" (corresponding to the third, fourth, and fifth pieces in Shreevatsa's answer) can be expressed in a less complicated manner, like so:
$$y=\frac{\sqrt{\mathrm{sign}(1-|x|)}}{2}\left(3\left(\left|x-\frac12\right|+\left|x+\frac12\right|+6\right)-11\left(\left|x-\frac34\right|+\left|x+\frac34\right|\right)\right)$$
This can be derived by noting that the functions
$$\begin{cases}f(x)&\text{if }x<c\\g(x)&\text{if }c<x\end{cases}$$
and $f(x)+(g(x)-f(x))U(x-c)$ (where $U(x)$ is the unit step function) are equivalent, and using the "relation"
$$U(x)=\frac{x+|x|}{2x}$$
Note that the elliptic sections (both ends of the "wings", corresponding to the first piece in Shreevatsa's answer) were cut along the lines $y=-\frac37\left((2\sqrt{10}+\sqrt{33})|x|-8\sqrt{10}-3\sqrt{33}\right)$, so the elliptic potion can alternatively be expressed as
$$\left(\left(\frac{x}{7}\right)^2+\left(\frac{y}{3}\right)^2-1\right)\sqrt{\mathrm{sign}\left(y+\frac37\left((2\sqrt{10}+\sqrt{33})|x|-8\sqrt{10}-3\sqrt{33}\right)\right)}=0$$
Theoretically, since all you have are arcs of linear and quadratic curves, the chimera can be expressed parametrically using rational B-splines, but I'll leave that for someone else to explore...
| {
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Prove that $\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a$
Let $a, b, c$ and $d$ be positive real numbers such that $a+b+c+d = 4$. Prove that $$\frac4{abcd} \geq \frac a b + \frac bc + \frac cd +\frac d a .$$
How can I approach this using only the AM - GM inequality? Are there any other methods that do not involve the AM-GM inequality?
| This proof uses the rearrangement inequality in addition to AM-GM. After multiplying by $abcd$ our problem is equivalent to solving $$ a^2cd + ab^2d + abc^2 + bcd^2 \leq 4$$ Let $\{a,b,c,d\}=\{w,x,y,z\}$ with $w \ge x \ge y \ge z$. We have $$ a^2cd + ab^2d + abc^2 + bcd^2 = a(acd)+b(abd)+c(abc)+d(bcd)$$ and by the rearrangement equality $$ a(acd)+b(abd)+c(abc)+d(bcd) \le w(wxy)+x(wxz)+y(wyz)+z(xyz)$$ $$ = (wx+yz)(wy+xz)$$ By AM-GM, we get $$ (wx+yz)(wy+xz) \le \left(\frac{wx+yz+wy+xz}{2}\right)^2 = \frac{1}{4} \left((w+z)(x+y)\right)^2$$ Another AM-GM gives us $$ \frac{1}{4} ((w+z)(x+y))^2 \le \frac{1}{4} \left(\left(\frac{w+x+y+z}{2}\right)^2\right)^2 = \frac{1}{4} \left(\left(\frac{4}{2}\right)^2\right)^2 = 4$$ Thus we have $$ a^2cd + ab^2d + abc^2 + bcd^2 \leq 4$$ and furthermore, we have $$ \frac{a}{b} + \frac{b}{c} + \frac{c}{d} + \frac{d}{a} \le \frac{4}{abcd}$$ You might be able to tweak this to avoid the use of rearrangement, but I couldn't find a way.
| {
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$ b_{n + 1} = \frac {b_n^2 + 2b_n}{b_n^2 + 2b_n+2}$ and $ b_1 = 1$, show that $ \left|\frac{2}{n}-\frac{2\ln{n}}{n^2}-b_n\right|\leq\frac{1}{n^2}$ In the recursion $ b_{n + 1} = \frac {b_n^2 + 2b_n}{b_n^2 + 2b_n+2}$, with $ b_1 = 1,$ how can one prove that $ \left|\frac{2}{n}-\frac{2\ln{n}}{n^2}-b_n\right|\leq\frac{1}{n^2}$?
| Note that
$$
b_{n+1}+1=2\frac{(b_n+1)^2}{(b_n+1)^2+1}
$$
Letting $b_n=c_n-1$, we get
$$
c_{n+1}=2\frac{c_n^2}{c_n^2+1}
$$
Letting $c_n=\frac{1}{d_n}$, we get
$$
d_{n+1}=\frac{1}{2}(1+d_n^2)
$$
Note that $d_1=\frac{1}{2}$, and if $0\le d_n\le 1$, then $0\le d_{n+1}\le 1$. Thus, $0\le d_n\le 1$ for all $n$.
Letting $d_n=1-e_n$, we get
$$
e_{n+1}=e_n-\frac{1}{2}e_n^2
$$
Note that $e_n$ is non-increasing and $0\le e_n\le 1$ for all $n$. Therefore, $\lim_{n\to\infty}e_n$ exists. Thus, $\lim_{n\to\infty}\;\frac{1}{2}e_n^2=\lim_{n\to\infty}(e_n-e_{n+1})=\lim_{n\to\infty}\;e_n-\lim_{n\to\infty}\;e_{n+1}=0$. Therefore, $\lim_{n\to\infty}\;e_n=0$.
Letting $e_n=\frac{1}{f_n}$ (whereby $\lim_{n\to\infty}f_n=\infty$), we get
$$
\begin{align}
f_{n+1}-f_n&=\frac{1}{2}\frac{f_{n+1}}{f_n}\\
&=\frac{1}{2}\left(1+\frac{f_{n+1}-f_n}{f_n}\right)
\end{align}
$$
Collecting the $f_{n+1}-f_n$ on the left, we get
$$
(f_{n+1}-f_n)\left(1-\frac{1}{2f_n}\right)=\frac{1}{2}
$$
So that
$$
f_{n+1}-f_n=\frac{1}{2}\left(1+\frac{1}{2f_n}+\frac{1}{4f_n^2}+\frac{1}{8f_n^3}+\dots\right)
$$
We can iteratively apply the Euler-Maclaurin Sum Formula starting with $f_n=\frac{a+n}{2}$. Two passses gives
$$
f_n=\frac{1}{2}((a+n)+\log(a+n)+\frac{\log(a+n)}{a+n})+O\left(\frac{1}{a+n}\right)
$$
Note that $b_n=\frac{1}{f_n-1}$. This yields $b_n=\frac{2}{n}-\frac{2\log(a+n)}{n^2}-\frac{2(a-2)}{n^2}+\dots$.
Had a few minutes, so I added a bit more. Gotta go again; I will finish this later.
| {
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Does the Schur complement preserve the partial order? Let
$$\begin{bmatrix}
A_{1} &B_1 \\ B_1' &C_1
\end{bmatrix} \quad \text{and} \quad \begin{bmatrix}
A_2 &B_2 \\ B_2' &C_2
\end{bmatrix}$$
be symmetric positive definite and conformably partitioned matrices. If
$$\begin{bmatrix}
A_{1} &B_1 \\ B_1' &C_1
\end{bmatrix}-\begin{bmatrix}
A_2 &B_2 \\ B_2' &C_2
\end{bmatrix}$$
is positive semidefinite, is it true
$$(A_1-B_1C^{-1}_1B_1')-(A_2-B_2C^{-1}_2B_2')$$ also positive semidefinite? Here, $X'$ means the transpose of $X$.
| Yes, it does. The assumption $$\begin{bmatrix}
A_{1} &B_1 \\ B_1^T &C_1
\end{bmatrix}-\begin{bmatrix}
A_2 &B_2 \\ B_2^T &C_2
\end{bmatrix} \geq 0$$ implies that for any vector $\begin{pmatrix} x & y \end{pmatrix}$,
$$ \begin{pmatrix} x^T & y^T \end{pmatrix} \begin{bmatrix}
A_{1} &B_1 \\ B_1^T &C_1
\end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} \geq \begin{pmatrix} x^T & y^T \end{pmatrix} \begin{bmatrix}
A_2 &B_2 \\ B_2^T &C_2
\end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix}~~~~~(*)$$ But for any partitioned matrix,
$$\begin{pmatrix} x^T & y^T \end{pmatrix} \begin{bmatrix}
A &B \\ B^T &C
\end{bmatrix} \begin{pmatrix} x \\ y \end{pmatrix} = (x + A^{-1} B y)^T A (x + A^{-1} B y) + y^T(C-B^T A^{-1} B)y.$$ Moreover, if the partitioned matrix on the left-hand side is positive definite, then each of the two terms on the right=hand side is positive. Thus picking arbitrary $y$ and $x = -A_1^{-1}B_1y$ in (*) gives
$$y^T(C_1-B_1^T A_1^{-1} B_1)y \geq \mbox{ something positive} + y^T(C_2-B_2^T A_2^{-1} B_2)y,$$ which implies
$$y^T(C_1-B_1^T A_1^{-1} B_1)y \geq y^T(C_2-B_2^T A_2^{-1} B_2)y,$$
which implies the conclusion you want.
| {
"language": "en",
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Proving $1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$ using induction How can I prove that
$$1^3+ 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
for all $n \in \mathbb{N}$? I am looking for a proof using mathematical induction.
Thanks
| If $S_r= 1^r+2^r+...+n^r=f(n)$ and
$\sigma =n(n+1)$ & $\sigma'=2n+1$
$S_1=\frac{1}{2}\sigma$
$S_2=\frac{1}{6}\sigma\sigma'$
$S_3=\frac{1}{4}\sigma^2$
$S_4=\frac{1}{30}\sigma\sigma'(3\sigma-1)$
$S_5=\frac{1}{12}\sigma^2(2\sigma-1)$
$S_6=\frac{1}{42}\sigma\sigma'(3\sigma^2-3\sigma+1)$
$S_7=\frac{1}{24}\sigma^2(3\sigma^2-4\sigma+2)$
$S_8=\frac{1}{90}\sigma\sigma'(5\sigma^3-10\sigma^2+9\sigma-3)$
$S_9=\frac{1}{20}\sigma^2(2\sigma^3-5\sigma^2+6\sigma-3)$
$S_{10}=\frac{1}{66}\sigma\sigma'(3\sigma^4-10\sigma^3+17\sigma^2-15\sigma+5)$
proof of this is based on the theorem
if r is a positive integer, $s_r$ can be expressed as a polynomial in $n$ of which the highest term in $\frac{n^{r+1}}{r+1}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Some trigonometric formula How to prove that
$1+2(\cos a)(\cos b)(\cos c)-\cos^2 a-\cos^2 b-\cos^2 c=4 (\sin p)(\sin q) (\sin r)(\sin s)$,
where
$p=\frac{1}{2}(-a+b+c)$, $q=\frac{1}{2}(a-b+c)$, $r=\frac{1}{2}(a+b-c)$, $s=\frac{1}{2}(a+b+c)$.
Thanks.
| Use
$$
\begin{eqnarray}
2 (\sin p)(\sin q) &=& \cos(p-q) - \cos(p+q) \\
2 (\sin r)(\sin s) &=& \cos(r-s) - \cos(r+s) \\
\end{eqnarray}
$$
Then use
$$
\begin{eqnarray}
\cos(p-q) \cos(r-s) &=& \frac{1}{2}( \cos(p+s-q-r) + \cos(p+r - s-q)) \\
\cos(p+q) \cos(r-s) &=& \frac{1}{2}( \cos(p+s+q-r) + \cos(p+r - s+q)) \\
\cos(p-q) \cos(r+s) &=& \frac{1}{2}( \cos(p-s-q-r) + \cos(p+r +s-q)) \\
\cos(p+q) \cos(r+s) &=& \frac{1}{2}( \cos(p-s+q-r) + \cos(p+r +s + q))
\end{eqnarray}
$$
Now use expressions for $p$,$q$,$r$,$s$ in terms of $a$,$b$,$c$.
| {
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Complex solutions for Fermat-Catalan conjecture The Fermat-Catalan conjecture is that $a^m + b^n = c^k$ has only a finite number of solutions when $a, b, c$ are positive coprime integers, and $m,n,k$ are positive integers satisfying $\frac{1}{m} + \frac{1}{n} +\frac{1}{k} <1$. There are currently only 10 solutions known (listed at the end of this post). My question concerns the case where $a, b, c$ are positive coprime Gaussian integers. I've found two solutions. Is there a clever method for finding more? I've used brute force techniques.
*
*$(8+5i)^2+(5+3i)^3=(1+2i)^7$
*$(20+9i)^2+(1+8i)^3=(1+i)^{15}$
*$(1+2i)^7+(49+306i)^2=(27+37i)^3$ (Zander)
*$(44+83i)^2+(31+39i)^3=(5+2i)^7$ (Zander)
*$(238+72i)^3+(7+6i)^8=(7347−1240i)^2$ (Oleg567)
Here are the known solutions over integers.
*
*$1^m+2^3=3^2$
*$2^5+7^2=3^4$
*$13^2+7^3=2^9$
*$2^7+17^3=71^2$
*$3^5+11^4=122^2$
*$33^8+1549034^2=15613^3$
*$1414^3+2213459^2=65^7$
*$9262^3+15312283^2=113^7$
*$17^7+76271^3=21063928^2$
*$43^8+96222^3=30042907^2$
| I found the next complex solution! :)
$$(238+72i)^3+(7+6i)^8=(7347−1240i)^2$$
(There is no new method here. Just small contribution to the problem.)
| {
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Solving a Nonlinear System $$\begin{align}
(1/300)a + (-1/200)b &= 5\\
(-1/300)a + ((-1/300) + (1/200))b + (-1/200)c &= -e^b\\
(-1/200)b + (1/200)c &= -e^c
\end{align}
$$
how do I solve for $a, b$ and $c$?
Thanks!
I know if I derive an equation that isolates a variable, like
$kx + e^x = 0$
I can use Newton's method to approximate x.
But still can't figure it out.
(I am just a high school student - so if you can make your answers as easy to understand as possible)
| The first equation is used to substitute all $a \rightarrow \frac{3}{2}b+1500$. The two remaining equations can be collected into a 2x1 vector $f = 0$
$$f = \begin{pmatrix} \hat{e}^b-\frac{b}{300}-\frac{c}{200}-5 \\ \hat{e}^c-\frac{b}{200}+\frac{c}{200} \end{pmatrix} $$
The derivatives with respect to $b$ and $c$ for each part are
$$ f\,' = \begin{pmatrix} \hat{e}^b-\frac{1}{300} & -\frac{1}{200} \\ -\frac{1}{200} & \hat{e}^c+\frac{1}{200} \end{pmatrix} $$
Newton raphson with vectors is $ (b,c) \rightarrow (b,c) - {f\,'}^{-1} f $
$$ \begin{pmatrix}b\\c\end{pmatrix} \rightarrow \begin{pmatrix}b\\c\end{pmatrix} -
\begin{pmatrix} \hat{e}^b-\frac{1}{300} & -\frac{1}{200} \\ -\frac{1}{200} & \hat{e}^c+\frac{1}{200} \end{pmatrix}^{-1} \begin{pmatrix} \hat{e}^b-\frac{b}{300}-\frac{c}{200}-5 \\ \hat{e}^c-\frac{b}{200}+\frac{c}{200} \end{pmatrix} $$
with an initial guess of $\begin{pmatrix}b = 1\\c = 1 \end{pmatrix} $ I get the following iterations
$$ \begin{bmatrix} 1&1\\1.841661&0.003381327\\1.634514&-0.9835724\\1.608907&-1.936174\\1.607725&-2.783955\\1.607126&-3.380392 \end{bmatrix} $$
So in the end we have $ a= \frac{3}{2} 1.607126 + 1500 $, $ b = 1.607126 $, $ c = -3.380392 $.
| {
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Is there a combinatorial way to see the link between the beta and gamma functions? The Wikipedia page on the beta function gives a simple formula for it in terms of the gamma function. Using that and the fact that $\Gamma(n+1)=n!$, I can prove the following formula:
$$
\begin{eqnarray*}
\frac{a!b!}{(a+b+1)!} & = & \frac{\Gamma(a+1)\Gamma(b+1)}{\Gamma(a+1+b+1)}\\
& = & B(a+1,b+1)\\
& = & \int_{0}^{1}t^{a}(1-t)^{b}dt\\
& = & \int_{0}^{1}t^{a}\sum_{i=0}^{b}\binom{b}{i}(-t)^{i}dt\\
& = & \int_{0}^{1}\sum_{i=0}^{b}\binom{b}{i}(-1)^{i}t^{a+i}dt\\
& = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\int_{0}^{1}t^{a+i}dt\\
& = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\left[\frac{t^{a+i+1}}{a+i+1}\right]_{t=0}^{1}\\
& = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\\
b! & = & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(a+b+1)!}{a!(a+i+1)}
\end{eqnarray*}
$$
This last formula involves only natural numbers and operations familiar in combinatorics, and it feels very much as if there should be a combinatoric proof, but I've been trying for a while and can't see it. I can prove it in the case $a=0$:
$$
\begin{eqnarray*}
& & \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{(b+1)!}{0!(i+1)}\\
& = & \sum_{i=0}^{b}(-1)^{i}\frac{b!(b+1)!}{i!(b-i)!(i+1)}\\
& = & b!\sum_{i=0}^{b}(-1)^{i}\frac{(b+1)!}{(i+1)!(b-i)!}\\
& = & b!\sum_{i=0}^{b}(-1)^{i}\binom{b+\text{1}}{i+1}\\
& = & b!\left(1-\sum_{i=0}^{b+1}(-1)^{i}\binom{b+\text{1}}{i}\right)\\
& = & b!
\end{eqnarray*}
$$
Can anyone see how to prove it for arbitrary $a$? Thanks!
| Seven years later I found another way to attack this. Define $f(b, a) = \frac{a!b!}{(a+b+1)!}$ and $h(b, a) = \sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}$. To connect the two, we define $g$ such that $g(0, a) = \frac{1}{a + 1}$ and $g(b + 1, a) = g(b, a) - g(b, a + 1)$ and prove by induction in $b$ that $f = g = h$. In each case the base case is straightforward and we consider only the inductive step.
$$
\begin{eqnarray*}
& & g(b + 1, a) \\
& = & g(b, a) - g(b, a + 1) \\
& = & f(b, a) - f(b, a + 1) \\
& = & \frac{a!b!}{(a+b+1)!} - \frac{(a+1)!b!}{(a+b+2)!} \\
& = & \frac{a!b!(a + b + 2)}{(a+b+2)!} - \frac{a!b!(a+1)}{(a+b+2)!} \\
& = & \frac{a!b!(b+1)}{(a+b+2)!} \\
& = & \frac{a!(b+1)!}{(a+b+2)!} \\
& = & f(b+1, a)\\
\end{eqnarray*}
$$
$$
\begin{eqnarray*}
& & g(b + 1, a) \\
& = & g(b, a) - g(b, a + 1) \\
& = & h(b, a) - h(b, a + 1) \\
& = & \left(\sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\right) - \left(\sum_{i=0}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+2}\right) \\
& = & \frac{1}{a+1} + \left(\sum_{i=1}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\right) - \left(\sum_{i=0}^{b-1}\binom{b}{i}(-1)^{i}\frac{1}{a+i+2}\right) - (-1)^{b}\frac{1}{a+b+2}\\
& = & \frac{1}{a+1} + \left(\sum_{i=1}^{b}\binom{b}{i}(-1)^{i}\frac{1}{a+i+1}\right) + \left(\sum_{i=1}^{b}\binom{b}{i-1}(-1)^{i}\frac{1}{a+i+1}\right) + (-1)^{b+1}\frac{1}{a+b+2}\\
& = & \frac{1}{a+1} + \left(\sum_{i=1}^{b}\left(\binom{b}{i} + \binom{b}{i-1}\right)(-1)^{i}\frac{1}{a+i+1}\right) + (-1)^{b+1}\frac{1}{a+b+2}\\
& = & \frac{1}{a+1} + \left(\sum_{i=1}^{b}\binom{b+1}{i}(-1)^{i}\frac{1}{a+i+1}\right) + (-1)^{b+1}\frac{1}{a+b+2}\\
& = & \sum_{i=0}^{b+1}\binom{b+1}{i}(-1)^{i}\frac{1}{a+i+1} \\
& = & h(b + 1, a) \\
\end{eqnarray*}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/72067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 2
} |
Partial Fractions of form $\frac{1}{(ax+b)(cx+d)^2}$ When asked to convert something like $\frac{1}{(ax+b)(cx+d)}$ to partial fractions, I can say
$$\frac{1}{(ax+b)(cx+d)} = \frac{A}{ax+b} + \frac{B}{cx+d}$$
Then why can't I split $(cx+d)^2$ into $(cx+d)(cx+d)$ then do
$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{cx+d}$$
The correct way is
$$\frac{1}{(ax+b)(cx+d)^2} = \frac{A}{ax+b} + \frac{B}{cx+d} + \frac{C}{(cx+d)^2}$$
| The degree of the denominator is 2, so the numerator has to be of degree 1. So you can either assume it to be $Bx+C$ or (better still), $B(bx+c)+C$, so that
$\frac{B(bx+c)+C}{(bx+c)^2} = \frac{B}{bx+c} + \frac{c}{(bx+c)^2}.$
This generalizes to the case when the denominator has degree $n$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How do I get this matrix in Smith Normal Form? And, is Smith Normal Form unique? As part of a larger problem, I want to compute the Smith Normal Form of $xI-B$ over $\mathbb{Q}[x]$ where
$$
B=\begin{pmatrix} 5 & 2 & -8 & -8 \\ -6 & -3 & 8 & 8 \\ -3 & -1 & 3 & 4 \\ 3 & 1 & -4 & -5\end{pmatrix}.
$$
So I do some elementary row and column operations and get to
$$\begin{pmatrix} 1+x & -2 & 0 & 0 \\ -3(x+1) & x+3 & 0 & 0 \\ 0 & 1 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix}.
$$
Then I work with the upper left 3x3 matrix, and ultimately get:
$$\begin{pmatrix} x-3 & 0 & 0 & 0 \\ 0 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix}.
$$
So now I have a diagonal matrix (and I'm pretty sure I didn't mess anything up in performing row and column operations), except according to http://mathworld.wolfram.com/SmithNormalForm.html, the diagonal entries are supposed to divide each other, but obviously x-3 does not divide x+1. This means that: either I did something wrong, or diagonal matrix is not unique. Any ideas for how to transform my final matrix into a matrix whose diagonal entries divide each other?
| To expand my comment...Add column 2 to column 1. Subtract row 2 from row 1. Now you have a scalar in the (1,1) position -- rescale to 1.
$$\begin{pmatrix} x-3 & 0 & 0 & 0 \\ 0 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim
\begin{pmatrix} x-3 & 0 & 0 & 0 \\ x+1 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim
$$
$$\begin{pmatrix} -4 & -x-1 & 0 & 0 \\ x+1 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim
\begin{pmatrix} 1 & (1/4)(x+1) & 0 & 0 \\ x+1 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim
$$
Now add $(-1/4)(x+1)$ times column 1 to column 2 (to clear everything beside 1).
$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ x+1 & x+1-(1/4)(x+1)^2 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim
$$
Add $-(x+1)$ times row 1 to row 2 (to clear everything below 1) & simplify the (2,2)-entry. Then rescale row 2 (so the polynomial is monic).
$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & -(1/4)(x+1)(x-3) & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim
\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & (x+1)(x-3) & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & x+1\end{pmatrix} \sim
$$
Finally swap columns 2 and 4 and then rows 2 and 4 to switch the positions of $(x+1)(x-3)$ and $x+1$. We are left with the Smith normal form.
$$\begin{pmatrix} 1 & 0 & 0 & 0 \\ 0 & x+1 & 0 & 0 \\ 0 & 0 & x+1 & 0 \\ 0 & 0 & 0 & (x+1)(x-3)\end{pmatrix}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove the identity $ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$ $$ \sum\limits_{s=0}^{\infty}{p+s \choose s}{2p+m \choose 2p+2s} = 2^{m-1} \frac{2p+m}{m}{m+p-1 \choose p}$$
Class themes are: Generating functions and formal power series.
| Let $d_s = \binom{p+s}{s} \binom{2p+m}{2p+2s}$. Using the recurrence relations for binomial, the ratio of successive terms is:
$$
\frac{d_{s+1}}{d_s} = \frac{\left(s - m/2\right)\left(s -(m-1)/2\right)}{ (s+1)(s+p+1/2) } = \frac{(s+a)(s+b)}{(s+1)(s+c)}
$$
The hypergeometric certificate above means that
$$
\sum_{s=0}^\infty d_s = d_0 \sum_{s=0}^\infty \frac{(a)_s (b)_s}{s! (c)_s} = \binom{2p+m}{2p} {}_2 F_1\left( -\frac{m}{2}, -\frac{m-1}{2} ; p+\frac{1}{2} ; 1\right)
$$
where $a = -\frac{m}{2}$, $b=-\frac{m-1}{2}$ and $c=p+\frac{1}{2}$.
Using Gauss's theorem, valid for $c>a+b$:
$$
{}_2 F_1\left( a, b; c; 1\right) = \frac{\Gamma(c) \Gamma(c-a-b)}{\Gamma(c-a) \Gamma(c-b)}
$$
we obtain the required identity:
$$
\sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} = \binom{2p+m}{2p} \frac{\Gamma\left(p+\frac{1}{2}\right) \Gamma\left( p+m \right)}{ \Gamma\left( p+\frac{m+1}{2} \right) \Gamma\left( p+\frac{m}{2} \right) }
$$
Applying the duplication formula for $\Gamma(2p+m+1)$ and $\Gamma(2p+1)$ arising from $\binom{2p+m}{2p}$ we arrive at the result:
$$
\sum_{s=0}^\infty \binom{p+s}{s} \binom{2p+m}{2p+2s} = 2^{m-1} (m+2p) \frac{\Gamma(m+p)}{\Gamma(m+1) \Gamma(p+1)} = 2^{m-1} \frac{m+2p}{m+p} \binom{m+p}{p}
$$
| {
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"url": "https://math.stackexchange.com/questions/77949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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How do I prove equality of x and y? If $0\leq x,y\leq\frac{\pi}{2}$ and $\cos x +\cos y -\cos(x+y)=\frac{3}{2}$, then how can I prove that $x=y=\frac{\pi}{3}$?
Your help is appreciated.I tried various formulas but nothing is working.
| $cos x +cos y -cos(x+y)=2\cos(\frac{x+y}{2})\cos(\frac{x-y}{2})-(2\cos^2(\frac{x+y}{2})-1)=2\cos(\frac{x+y}{2})\cdot\left(\cos(\frac{x-y}{2})-cos(\frac{x+y}{2})\right)+1\le $
$2\cos(\frac{x+y}{2})\cdot\left(1-cos(\frac{x+y}{2})\right)+1\le 1/2+1=\frac{3}{2}$, where the inequality is by $ab\le (\frac{a+b}{2})^2$, so $\cos(\frac{x+y}{2})\cdot\left(1-cos(\frac{x+y}{2})\right)\le 1/4$ . The equality holds iff $x=y=\pi/3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/78629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
If $2^n+n^2$ is prime number then $n \equiv 0 \pmod 3 $? Is it true that :
$((2^n+n^2) \in \mathbf{P} \land n \geq 3)\Rightarrow n\equiv 0 \pmod 3 $
I have checked this statement for the following consecutive values of $n$ : $3,9,15,21,33,2007,2127,3759$
Note that $2^n+n^2$ is special case of the form $2^n+k\cdot n$. One can easily show using Dirichlet's theorem that for any odd $n$ sequence $a_k=2^n+k\cdot n$ contains infinitely many primes.
Any idea how to prove or disprove statement above ?
| If $n =1 \mod 6$ then $2^n+n^2 = 0 \mod 3$.
If $n =2 \mod 6$ then $2^n+n^2 = 0 \mod 2$.
If $n =4 \mod 6$ then $2^n+n^2 = 0 \mod 2$.
If $n =5 \mod 6$ then $2^n+n^2 = 0 \mod 3$.
P.S. Looking mod 6 is natural, since $2^n \mod 3$ repeats after 2 steps. So to decide what is $2^n \mod 3$, we need to know if $n$ is odd or even. Thus we need to look $\mod 2$ and $\mod 3$ at the same time....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/79564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Is the integral $\int_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$ equal for all $a \neq 0$? Let $a$ be a non-zero real number. Is it true that the value of $$\int\limits_0^\infty \frac{\mathrm{d} x}{(1+x^2)(1+x^a)}$$ is independent on $a$?
| $$
\begin{align}
I & = \int_0^{\infty} \frac{dx}{(1+x^2)(1+x^a)}\\
\frac{dI}{da} & = -\int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}
\end{align}
$$
Let $\displaystyle J = \int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}$
$$
\begin{align}
J & = \int_0^{\infty} \frac{x^a \log(x) dx}{(1+x^2)(1+x^a)^2}\\
& \stackrel{x=1/y}{=} \int_{0}^{\infty} \frac{1/y^a \log(1/y) d(1/y)}{(1+(1/y)^2)(1+(1/y)^a)^2}\\
& = \int_{\infty}^{0} \frac{y^a \log(y) dy}{(1+y^2)(1+y^a)^2}\\
& = -J
\end{align}
$$
Hence, $\frac{dI}{da}=0$. Hence, $I$ is independent of $a$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/87735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "42",
"answer_count": 4,
"answer_id": 3
} |
What are some "natural" interpolations of the sequence $\small 0,1,1+2a,1+2a+3a^2,1+2a+3a^2+4a^3,\ldots $? (This is a spin-off of a recent question here)
In fiddling with the answer to that question I came to the set of sequences
$\qquad \small \begin{array} {llll}
A(1)=1,A(2)=1+2a,A(3)=1+2a+3a^2,A(4)=1+2a+3a^2+4a^3, \ldots \\
B(1)=1,B(2)=1+3a,B(3)=1+3a+6a^2,B(4)=1+3a+6a^2+10a^3, \ldots \\
C(1)=1,C(2)=1+4a,C(3)=1+4a+10a^2,C(4)=1+4a+10a^2+20a^3, \ldots \\
\ldots \\
\end{array} $
with some indeterminate a .
We had the discussion often here in MSE, that interpolation to fractional indexes, say A(1.5)=?? is arbitrary, considering, that an initial solution composed with any 1 -periodic function satisfies the condition. But here the embedding in a set of sequences, which are constructed from binomial-coefficients might suggest some "natural" interpolation, such as for
$\qquad \small K(1)=1, K(2)=1+a, K(3)=1+a+a^2, \ldots $
the interpolation $\small K(r) = {a^r-1 \over a-1}$ seems the most "natural" which even can smoothly be defined for a=1. This observation made me to refer to "q-analogues" $\small [r]_a $ in my answer in the initiating MSE-question, but it's not obvious how to interpolate the shown sequences of higher orders A , B , C (I think they're not related to the "q-binomial-analogues" , for instance ).
Q: So what would be some "natural" interpolation to fractional indexes for the sequences A, B, C, and possibly in general for sequences generated in the obvious generalized manner?
Agreeing mostly with Henning's ansatz I got now the general form as
$$ A_m(n) = {1 \over (1-a)^m} - \sum_{k=0}^{m-1} \binom{n+m}{k}{a^{n+m-k} \over (1-a)^{m-k} } $$
I do not yet see, whether some examples of fractional indexes agree with the solutions of all three given answers so far, for instance: given a=2.0 what is A(1.5), B(4/3), C(7/5)? With my programmed version I get now
$\qquad \small A(1.5)\sim 9.48528137424 $
$\qquad \small B(4/3) \sim 11.8791929545 $
$\qquad \small C(7/5) \sim 18.4386165488 $
(No interpolation for fractional m yet)
[update 2] the derivative-versions of Sivaram and Michael arrive at the same values so I think, all versions can be translated into each other and mutually support each other to express a "natural" interpolation.
[update 3] I had an index-error in my computation call. Corrected the numerical results.
| Here is an attack leading to a closed forms for the $A$s and $B$s that can be evaluated for fractional $n$:
First, let $ A_n = \sum_{k=0}^n (k+1)a^k $.
For $n>0$ we can write
$$ A_n = a A_{n-1} + 1 + a + \cdots + a^n = a A_{n-1} + \frac{a^{n+1}-1}{a-1} $$
We can then use a base case of $A_0=1=\frac{a-1}{a-1}$ to unfold the entire recurrence into
\begin{align}
A_n &= \frac{1}{a-1}\Bigl( (a^{n+1}-1) + a(a^n-1) + a^2(a^{n-1}-1) + \cdots + a^n(a-1) \Bigr)
\\&= \frac{1}{a-1}\Bigl( (n+1)a^{n+1} - (1+a+\cdots+a^n) \Bigr)
\\&= \frac{1}{a-1}\Bigl( (n+1)a^{n+1} - \frac{a^{n+1}-1}{a-1} \Bigr)
\end{align}
We can push this into the $B$s by noting that
$$ B_n=A_n + a A_{n-1} + a^2 A_{n-2} + \cdots + a^n A_0 $$
so
\begin{align}
B_n &= \frac{1}{a-1}\left(\sum_{k=0}^n k+1\right)a^{n+1}
- \frac{1}{(a-1)^2} \Bigl( (a^{n+1}-1)+a(a^{n}-1)+\cdots+a^n(a-1) \Bigr)
\\&= \frac{1}{a-1} \frac{(n+1)(n+2)}{2} a^{n+1}
- \frac{1}{(a-1)^2} \Bigl( (n+1)a^{n+1} - (1+a+\cdots+a^n) \Bigr)
\\&= \frac{1}{a-1} \frac{(n+1)(n+2)}{2} a^{n+1}
- \frac{1}{(a-1)^2} (n+1)a^{n+1}
+ \frac{1}{(a-1)^3} (a^{n+1}-1)
\end{align}
A pattern is emerging ... I will leave the derivation for $C_n$ to the reader, but the only difficulty should be in keeping the algebra straight.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/88107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Completing the square How might I find linear combinations $$\begin{align*}
A&=a_1x+a_2y+a_3z\\
B&=b_1x+b_2y+b_3z\\
C&=c_1x+c_2y+c_3z
\end{align*}$$
Such that I can transform the two polynomials
$$2x^2+3y^2-2yz+3z^2\text{ and }x^2+6xy+3y^2+2yz-6zx+3z^2$$
into
$A^2+B^2+C^2$ and $\alpha A^2+\beta B^2+\gamma C^2$ respectively for some $\alpha, \beta,\gamma\in \mathbb R$?
I think I should be completing the square, but I can't see how to.
| Your question is equivalent to finding a matrix $S$ such that $S^\top XS=I$ and $S^\top YS$ is diagonal, where
$$
X=\begin{pmatrix}2&0&0\\ 0&3&-1\\ 0&-1&3\end{pmatrix},
Y=\begin{pmatrix}1&3&-3\\ 3&3&1\\ -3&1&3\end{pmatrix}.
$$
In a related question (BTW, the $X,Y$ here are exactly the same as the matrices $A,B$ in that question; is this a homework question or something else?), I have explained how to find a matrix $M$ such that both $M^\top XM=\Lambda_1$ and $M^\top YM=\Lambda_2$ for some diagonal matrices $\Lambda_1$ and $\Lambda_2$. Concretely,
$$
M=\frac{1}{\sqrt{\lambda+4}}
\begin{pmatrix}
2&1&0\\
-1&1&1\\
1&0&1
\end{pmatrix}
\begin{pmatrix}1&0&0\\0&\lambda-4&-2\\0&2&\lambda-4\end{pmatrix},
\quad \lambda=\frac{9+\sqrt{17}}{2}.
$$
Now you may take $S=M\Lambda_1^{-1/2}$. The coefficients of $A$ can be read off from the first row of $S^\top$, and vice versa for $B$ and $C$. The coefficients $\alpha,\beta,\gamma$ are the diagonal entries of $S^\top YS$.
Another way of solving the problem is to find the square root of $X$, then orthogonally diagonalize $(\sqrt{X}^\top)^{-1}Y\sqrt{X}^{-1}$. In other words, if $\sqrt{X}$ is a matrix such that $\sqrt{X}^\top\sqrt{X}=X$, and $Q$ is an orthogonal matrix such that $Q^\top(\sqrt{X}^\top)^{-1}Y\sqrt{X}^{-1}Q$ is diagonal, then you may set $S=\sqrt{X}^{-1}Q$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/88279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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$\int \cos^{-1} x \; dx$; trying to salvage an unsuccessful attempt $$
\begin{align}
\int \cos^{-1} x \; dx &= \int \cos^{-1} x \times 1 \; dx
\end{align}
$$
Then, setting
$$\begin{array}{l l}
u=\cos^{-1} x & v=x \\
u' = -\frac{1}{\sqrt{1-x^2}} & v'=1\\
\end{array}$$
Then by the IBP technique, we have:
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &=\cos^{-1} (x) \cdot x - \int x \cdot -\frac{1}{\sqrt{1-x^2}} \; dx\\
&= x \cos^{-1} (x) - \int -\frac{x}{\sqrt{1-x^2}} \; dx\\
\end{array}$$
Now at this point suppose I have overlooked the possibility of using integration by substitution (setting $u=1-x^2$) to simplify the second integral. Instead, I attempt to reapply IBP to the second integral $\int -\frac{x}{\sqrt{1-x^2}} \; dx$.
I let
$$\begin{array}{l l}
u= -\frac{1}{\sqrt{1-x^2}} = -(1-x^2)^{-\frac{1}{2}} \qquad & v= \frac{x^2}{2} \\
u' = - \left( -\frac{1}{2} \right) (1-x^2)^{-\frac{3}{2}} \times -2x = -x(1-x^2)^{-\frac{3}{2}} \qquad & v'=x\\
\end{array}$$
Then by IBP again,
$$\begin{array}{l l}
\int \cos^{-1} x \times 1 \; dx &= x \cos^{-1} (x) - \left( -\frac{1}{\sqrt{1-x^2}} \cdot \frac{x^2}{2} - \int \frac{x^2}{2} \cdot -x(1-x^2)^{-\frac{3}{2}} \; dx \right) \\
\end{array}$$
At this stage, I can see no way to proceed. Can anyone see a reasonable way to salvage this solution, continuing along this line of reasoning? Or was approaching the second integral by IBP doomed to fail?
| So, it looks like you are just having problems with $- \int \frac{x^2}{2} \cdot -x(1-x^2)^{-\frac{3}{2}} \; dx =\frac{1}{2}\int \frac{x^3}{(1-x^2)^{\frac{3}{2}}}dx $.
Lets look at $\int \frac{x^3}{(1-x^2)^\frac{3}{2}}dx$
$$
u=1-x^2 , \text{ then }
$$
$$
du=-2xdx \text{ and }
$$
$$
u-1=x^2
$$
Rewriting,
$$
\begin{array}{l l}
\int \frac{x^3}{(1-x^2)^\frac{3}{2}}dx &=-\frac{1}{2}\int \frac{x^2(-2xdx)}{(1-x^2)^\frac{3}{2}}\\
\\
\end{array}
$$
So,
$$
\begin{array}{l l}
-\frac{1}{2}\int \frac{1-u}{u^\frac{3}{2}}du &=-\frac{1}{2}\int (u^{-\frac{3}{2}}-u^{-\frac{1}{2}})du\\
&= \frac{1}{\sqrt{u}}+\sqrt{u}+C\\
\end{array}
$$
or, going back to x
$$
\frac{1}{\sqrt{1-x^2}}+\sqrt{1-x^2}+C
$$
getting a common denominator gives the equivalent
$$
\frac{2-x^2}{\sqrt{1-x^2}}+C
$$
I hope that helps. When you first start learning it is tough to know which method to use! Practice is really one of the best ways to get better at integration problems. I didn't try it, but I bet you could solve this integral by trig-sub too.
| {
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"timestamp": "2023-03-29T00:00:00",
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Is this a proper use of induction? ($(n^2+5)n$ is divisible by 6) Just want to get input on my use of induction in this problem:
Question. Use mathematical induction to prove that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$.
Proof by mathematical induction.
(1) show base case ($n=1$) is true:
$$
((1)^2 + 5) (1) = 6
$$
$6$ is indeed divisible by $6$, so base case ($n=1$) is true
(2a) Assume case $n$ is true: $(n^2+5)n$ is divisible by $6$.
(2b) Show that case $n$ $\implies$ case $(n+1)$
$$
\begin{align*}
((n+1)^2+5)(n+1)
&\rightarrow
((n^2+2n+1)+5)(n+1)
\\ &\rightarrow
[(n^2+5)+(2n+1)](n+1)
\\ &\rightarrow
(n^2+5)n + (n^2+5)+(2n+1)n+ (2n+1)
\\ &\rightarrow
(n^2+5)n + [(n^2+5)+(2n^2+n)+ (2n+1)]
\\ &\rightarrow
(n^2+5)n + [(3n^2+3n)+6]
\end{align*}
$$
Now we can see case $(n+1)$ $= (n^2+5)n + (3n^2+3n)+6$.
We know $6$ is divisible by $6$ and we are assuming $(n^2+5)n$ is divisible by $6$ already, so all we need to do is show $(3n^2+3n)$ is divisible by $6$:
Letting $n=1$ for $(3n^2+3n)$ gives: $(3(1)^2+3(1)) = 6$
Thus, it has been demonstrated that $(n^2+5)n$ is divisible by $6$ for all integers $n \geqslant 1$.
I'm not sure if letting $n=1$ for that last expression is enough to prove it is divisible by $6$
| $(n^{2} + 5)n \equiv (n^{2}-1)n \equiv (n-1)n(n+1) \pmod 6$. Since these are three consecutive integers, one of them must be congruent to $0\pmod 3$ and one must also be even, or congruent to $0 \pmod 2$. Then as $\gcd (2,3) = 1$, the product $(n-1)n(n+1) \equiv 0 \mod 6$. Altogether meaning that $6 | (n^{2}+5)n$ for all $n$.
| {
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Finding any digit (base 10) of a binary number $2^n$ I was doing some Math / CS work, and noticed a pattern in the last few digits of $2^n$.
I was working in Python, in case anyone is wondering.
The last digit is always one of 2, 4, 8, 6; and has a period of 4:
n = 1, str(2 ^ 1)[-1] = 2
n = 2, str(2 ^ 2)[-1] = 4
n = 3, str(2 ^ 3)[-1] = 8
n = 4, str(2 ^ 4][-1] = 6
I was interested, so I looked further. The second to last digit also has a repeating pattern, this time with a period of 20:
n = 2, str(2 ^ 2)[-2] = 0
n = 3, str(2 ^ 3)[-2] = 0
n = 4, str(2 ^ 4)[-2] = 1
n = 5, str(2 ^ 5)[-2] = 3
n = 6, str(2 ^ 6)[-2] = 6
n = 7, str(2 ^ 7)[-2] = 2
n = 8, str(2 ^ 8)[-2] = 5
n = 9, str(2 ^ 9)[-2] = 1
n = 10, str(2 ^ 10)[-2] = 2
n = 11, str(2 ^ 11)[-2] = 4
n = 12, str(2 ^ 12)[-2] = 9
n = 13, str(2 ^ 13)[-2] = 9
n = 14, str(2 ^ 14)[-2] = 8
n = 15, str(2 ^ 15)[-2] = 6
n = 16, str(2 ^ 16)[-2] = 3
n = 17, str(2 ^ 17)[-2] = 7
n = 18, str(2 ^ 18)[-2] = 4
n = 19, str(2 ^ 19)[-2] = 8
n = 20, str(2 ^ 20)[-2] = 7
n = 21, str(2 ^ 21)[-2] = 5
Is there a way I can generalize this so I can find any digit of any $2^n$ with out having to actually calculate the value of $2^n$?
If not, is there a way to find the period of the pattern, given an index (from the back of the integer) to look at?
| The reason for the observer behavior is periodicity of $n \mapsto 2^n \bmod 10$, or $n \mapsto 2^n \bmod 10^k$ for any fixed $k$. This wikipedia article, as well as this, among others, math.SE answer are relevant.
Specifically, for a fixed positive integer $k$, $m \mapsto 2^m \bmod 10^k$ is quasi-periodic with period $p = \phi(5^k) = 4 \cdot 5^{k-1}$, i.e. for all $m \geq k$, $2^m \equiv 2^{m+p} \mod 10^k$.
Thus the last digit has period $4$, the one before last, has period $20$, the third one from the end has period $100$ and so on.
| {
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Planetary Motion: Integral I am solving central force problem to deduce equations of orbit of planets. During the calculation, I am stuck over an integral which I am unable to solve. Can anyone help or guide me in this?
The integral is:
$$\int \dfrac{dr}{r \left(2\mu Er^2 + 2 \mu Cr - l^2 \right)^{1/2}} .$$
Here, $\mu$, $E$, $C$ and $l$ are constants.
Thanks in advance!
| Make the substitution $\frac1r = x$. Then,
$$
\begin{align*}
\int \frac{dr}{r \left(2\mu Er^2 + 2 \mu Cr - l^2 \right)^{1/2}}
&= \int \frac{dr}{r^2 \left( 2\mu E + \frac{2 \mu C}{r} - \frac{l^2}{r^2} \right)^{1/2}}
\\ &= - \int \frac{dx}{\left( 2\mu E + 2 \mu C x - l^2x^2 \right)^{1/2}}
\\ &= - \int \frac{dx}{\left( 2\mu E+ \frac{\mu^2 C^2}{l^2} - \left( lx - \frac{\mu C}{l} \right)^2 \right)^{1/2}}
\end{align*}
$$
Substitute $lx - \frac{\mu C}{l} = \sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \sin \theta$. Then $l dx = \sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \cos \theta d \theta$. Therefore, the integral simplifies to
$$
\begin{align*}
- \frac{1}{l} \int \frac{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \cos \theta d \theta}{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}} \cos \theta}
&= - \frac{1}{l} \int d \theta = - \frac{\theta}{l}
\\ &= - \frac{1}{l} \arcsin \left( \frac{lx - \frac{\mu C}{l}}{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}}} \right)
\\ &= - \frac{1}{l} \arcsin \left( \frac{\frac{l}{r} - \frac{\mu C}{l}}{\sqrt{2 \mu E + \frac{\mu^2 C^2}{l^2}}} \right)
\end{align*}
$$
Intuition for some of the steps. The above answer might appear a bit too slick, but it really is natural. In the step where we got
$$
- \int \frac{dx}{\left( 2\mu E+ \frac{\mu^2 C^2}{l^2} - \left( lx - \frac{\mu C}{l} \right)^2 \right)^{1/2}} ,
$$
the natural thing to do is to substitute $lx - \frac{\mu C}{l} = y$. Moreover, we will denote the constant $2\mu E+ \frac{\mu^2 C^2}{l^2}$ by $A^2$. Then the integral becomes
$$
- \frac{1}{l} \int \frac{dx}{\sqrt{A^2 - y^2}} .
$$
The reason for calling that quantity $A^2$ instead of $A$ is to make the expression inside the square root look homogeneous (in physical terms, now $A$ has the same dimension as $y$). The last integral is a standard integral. It is fairly common to try substitutions involving trigonometric or hyperbolic functions. After some trial and error, we find that $y = A \sin \theta$ and $y = A \cos \theta$ will both work. [Of course, in this case, substituting $y$ as $A \tan \theta$, $A \sec \theta$, $A \cosh \theta$, $A \sinh \theta$ etc. do not seem to give anything.] My solution uses $y = A \sin \theta$.
| {
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"source": "stackexchange",
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Deduce that $\mathbb E(X^3)=1^3+2^3+3^3+4^3+5^3+6^3$ A fair die is tossed and let the random variable $X$ be the number that appears.
Deduce that
$$
\mathbb E(X^3)=\frac{1^3+2^3+3^3+4^3+5^3+6^3}6.
$$
First of all, I would like to know the probability distribution of this random variable $X$.
| $X$ takes the values $1$, $2$, $3$ ,$4$, $5$, and $6$ (assuming a six-sided die). Since the die is fair, outcomes are equally likely; so $X$ takes the value $i$ with probability $1/6$ for each $i=1, 2,3,4,5,6$.
The probability distribution of $X$ is therefore
$$
p_X(i)=\textstyle{1\over 6},\quad i=1, 2, \ldots, 6.
$$
As a warm up to your problem, let's find $\Bbb E(X)$. The expected value of $X$ is
$$
\Bbb E(X)=\sum_{i=1}^6 \,i\, p_X(i)= 1\cdot {1\over6}+2\cdot {1\over6}+3\cdot {1\over6}+4\cdot {1\over6}
+5\cdot {1\over6}+6 \cdot {1\over6}=3.5.
$$
To find the expected value of a function of $X$, you could use the following fact:
Fact:
For a discrete variable $X$ that takes the values $x_1, x_2,\ldots,x_n$,
the expectation of a function $h(X)$ of $X$ is
$$\Bbb E\bigl(h(X)\bigr) =\sum_{i=1}^n h(x_i) p_X(x_i).$$
To find $\Bbb E(X^3)$, we apply the above fact with $h(x)=x^3$:
$$
\Bbb E( X^3)=\sum_{i=1}^6\, i^3 p_X(i)= ( 1)^3\cdot {1\over6}+( 2)^3\cdot {1\over6}+( 3)^3\cdot {1\over6}+( 4)^3\cdot {1\over6}
+( 5)^3\cdot {1\over6}+( 6)^3 \cdot {1\over6} .
$$
| {
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Need help deriving recurrence relation for even-valued Fibonacci numbers. That would be every third Fibonacci number, e.g. $0, 2, 8, 34, 144, 610, 2584, 10946,...$
Empirically one can check that:
$a(n) = 4a(n-1) + a(n-2)$ where $a(-1) = 2, a(0) = 0$.
If $f(n)$ is $\operatorname{Fibonacci}(n)$ (to make it short), then it must be true that $f(3n) = 4f(3n - 3) + f(3n - 6)$.
I have tried the obvious expansion:
$f(3n) = f(3n - 1) + f(3n - 2) = f(3n - 3) + 2f(3n - 2) = 3f(3n - 3) + 2f(3n - 4)$
$ = 3f(3n - 3) + 2f(3n - 5) + 2f(3n - 6) = 3f(3n - 3) + 4f(3n - 6) + 2f(3n - 7)$
... and now I am stuck with the term I did not want. If I do add and subtract another $f(n - 3)$, and expand the $-f(n-3)$ part, then everything would magically work out ... but how should I know to do that? I can prove the formula by induction, but how would one systematically derive it in the first place?
I suppose one could write a program that tries to find the coefficients x and y such that $a(n) = xa(n-1) + ya(n-2)$ is true for a bunch of consecutive values of the sequence (then prove the formula by induction), and this is not hard to do, but is there a way that does not involve some sort of "Reverse Engineering" or "Magic Trick"?
| By inspection $f(3n+3)=4f(3n)+f(3n-3)$, as you’ve already noticed. This is easily verified:
$$\begin{align*}
f(3n+3)&=f(3n+2)+f(3n+1)\\
&=2f(3n+1)+f(3n)\\
&=3f(3n)+2f(3n-1)\\
&=3f(3n)+\big(f(3n)-f(3n-2)\big)+f(3n-1)\\
&=4f(3n)+f(3n-1)-f(3n-2)\\
&=4f(3n)+f(3n-3)\;.
\end{align*}$$
However, I didn’t arrive at this systematically; it just ‘popped out’ as I worked at eliminating terms with unwanted indices.
Added: Here’s a systematic approach, but I worked it out after the fact.
The generating function for the Fibonacci numbers is $$g(x)=\frac{x}{1-x-x^2}=\frac1{\sqrt5}\left(\frac1{1-\varphi x}-\frac1{1-\hat\varphi x}\right)\;,$$ where $\varphi = \frac12(1+\sqrt5)$ and $\hat\varphi=\frac12(1-\sqrt5)$, so that $f(n)=\frac1{\sqrt5}(\varphi^n-\hat\varphi^n)$. Thus, $f(3n)=\frac1{\sqrt5}(\varphi^{3n}-\hat\varphi^{3n})$. Thus, we want
$$\begin{align*}
h(x)&=\frac1{\sqrt5}\sum_{n\ge 0}(\varphi^{3n}-\hat\varphi^{3n})x^n\\
&=\frac1{\sqrt5}\left(\sum_{n\ge 0}\varphi^{3n}x^n-\sum_{n\ge 0}\hat\varphi^{3n}x^n\right)\\
&=\frac1{\sqrt5}\left(\frac1{1-\varphi^3 x}-\frac1{1-\hat\varphi^3 x}\right)\\
&=\frac1{\sqrt5}\cdot\frac{(\varphi^3-\hat\varphi^3)x}{1-(\varphi^3+\hat\varphi^3)x+(\varphi\hat\varphi)^3x^2}\;.
\end{align*}$$
Now $\varphi+\hat\varphi=1$, $\varphi-\hat\varphi=\sqrt5$, $\varphi\hat\varphi=-1$, $\varphi^2=\varphi+1$, and $\hat\varphi^2=\hat\varphi+1$, so
$$\begin{align*}
h(x)&=\frac1{\sqrt5}\cdot\frac{(\varphi^3-\hat\varphi^3)x}{1-(\varphi^3+\hat\varphi^3)x+(\varphi\hat\varphi)^3x^2}\\
&=\frac{(\varphi^2+\varphi\hat\varphi+\hat\varphi^2)x}{1-(\varphi^2-\varphi\hat\varphi)x-x^2}\\
&=\frac{(\varphi^2-1+\hat\varphi^2)x}{1-(\varphi^2+1+\hat\varphi^2)x-x^2}\\
&=\frac{(\varphi+\hat\varphi+1)x}{1-(\varphi+3+\hat\varphi)x-x^2}\\
&=\frac{2x}{1-4x-x^2}\;.
\end{align*}$$
It follows that $(1-4x-x^2)h(x)=2x$ and hence that $h(x)=4xh(x)+x^2h(x)+2x$. Since the coefficient of $x^n$ in $h(x)$ is $f(3n)$, this tells me that
$$\begin{align*}
\sum_{n\ge 0}f(3n)x^n&=h(x)=4xh(x)+x^2h(x)+2x\\
&=\sum_{n\ge 0}4f(3n)x^{n+1}+\sum_{n\ge 0}f(3n)x^{n+2}+2x\\
&=\sum_{n\ge 1}4f(3n-3)x^n+\sum_{n\ge 2}f(3n-6)x^n+2x\;,
\end{align*}$$
which by equating coefficients immediately implies that $f(3n)=4f(3n-3)+f(3n-6)$ for $n\ge 2$. It also gets the initial conditions right: the constant term on the righthand side is $0$, and indeed $f(3\cdot 0)=0$, and the coefficient of $x$ is $4f(0)+2=2=f(3)$, as it should be.
| {
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Finding the positive integer solutions $(x,y)$ of the equation $x^2+3=y(x+2)$
Finding the positive integer solutions $(x,y)$ of the equation
$x^2+3=y(x+2)$
Source: Art of Problem Solving Vol. 2
Any help would be appreciated.
| Divide the polynomial $x^2+3$ by the polynomial $x+2$. We get
$$\frac{x^2+3}{x+2}=x-2+\frac{7}{x+2}.$$
For an integer $x$, the value of $x-2+\frac{7}{x+2}$ is an integer if and only if $x+2$ divides $7$. The only positive integer $x$ for which this is true is given by $x=5$.
Comment: If we are interested in integer solutions, not necessarily positive, note that $x-2+\frac{7}{x+2}$ is an integer also at $x=-1$, $x=-3$, and $x=-9$.
| {
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Compute: $\int_{0}^{1}\frac{x^4+1}{x^6+1} dx$ I'm trying to compute: $$\int_{0}^{1}\frac{x^4+1}{x^6+1}dx.$$
I tried to change $x^4$ into $t^2$ or $t$, but it didn't work for me.
Any suggestions?
Thanks!
| The denominator of the integrand $f(x):=\dfrac{x^{4}+1}{x^{6}+1}$ may be factored as
\begin{eqnarray*}
x^{6}+1 &=&\left( x^{2}+1\right) \left( x^{4}-x^{2}+1\right) \
&=&\left( x^{2}+1\right) \left( x^{2}-\sqrt{3}x+1\right) \left( x^{2}+\sqrt{3
}x+1\right)
\end{eqnarray*}
If you expand $f(x)$ you get
$$\begin{eqnarray*}
f(x) &=&\frac{2}{3}\frac{1}{x^{2}+1}+\frac{1}{6}\frac{1}{x^{2}-\sqrt{3}x+1}+
\frac{1}{6}\frac{1}{x^{2}+\sqrt{3}x+1} \\
&=&\frac{2}{3}\frac{1}{x^{2}+1}+\frac{2}{3}\frac{1}{\left( 2x-\sqrt{3}
\right) ^{2}+1}+\frac{2}{3}\frac{1}{\left( 2x+\sqrt{3}\right) ^{2}+1}.
\end{eqnarray*}$$
Since
$$
\int \frac{1}{x^{2}+1}dx=\arctan x
$$
and
$$
\begin{eqnarray*}
\int \frac{1}{\left( ax+b\right) ^{2}+1}dx &=&\int \frac{1}{a\left(
u^{2}+1\right) }\,du=\frac{1}{a}\arctan u \\
&=&\frac{1}{a}\arctan \left( ax+b\right),
\end{eqnarray*}
$$
we have
$$
\begin{eqnarray*}
\int_{0}^{1}\frac{x^{4}+1}{x^{6}+1}dx &=&\frac{2}{3}\int_{0}^{1}\frac{1}{%
x^{2}+1}dx+\frac{2}{3}\int_{0}^{1}\frac{1}{\left( 2x-\sqrt{3}\right) ^{2}+1}%
dx \\
&&+\frac{2}{3}\int_{0}^{1}\frac{1}{\left( 2x+\sqrt{3}\right) ^{2}+1}dx \\
&=&\frac{2}{3}\arctan 1+\frac{2}{3}\left( \frac{1}{2}\arctan \left( 2-\sqrt{3%
}\right) -\frac{1}{2}\arctan \left( -\sqrt{3}\right) \right) \\
&&+\frac{2}{3}\left( \frac{1}{2}\arctan \left( 2+\sqrt{3}\right) -\frac{1}{2}%
\arctan \left( \sqrt{3}\right) \right) \\
&=&\frac{1}{6}\pi +\frac{1}{3}\left( \arctan \left( 2-\sqrt{3}\right)
+\arctan \left( \sqrt{3}\right) \right) \\
&&+\frac{1}{3}\left( \arctan \left( 2+\sqrt{3}\right) -\arctan \left( \sqrt{3%
}\right) \right) \\
&=&\frac{1}{6}\pi +\frac{1}{3}\left( \arctan \left( 2-\sqrt{3}\right)
+\arctan \left( 2+\sqrt{3}\right) \right) \\
&=&\frac{1}{6}\pi +\frac{1}{6}\pi \\
&=&\frac{1}{3}\pi,
\end{eqnarray*}
$$
because$^1$
$$
\arctan \left( 2-\sqrt{3}\right) +\arctan \left( 2+\sqrt{3}\right) =\frac{1}{
2}\pi.
$$
$^1$We apply the arctangent additional formula to $u=2-\sqrt{3}$ and $v=2+\sqrt{3}$
$$
\arctan u+\arctan v=\arctan \frac{u+v}{1-uv}.
$$
Since the product $uv=1$ and $\arctan \left( 2-\sqrt{3}\right) >0,\arctan
\left( 2+\sqrt{3}\right) >0$, we get on the right $\arctan \dfrac{4}{1-1}=
\dfrac{\pi }{2}.$
| {
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Simplifying/Converting limits I have this in my lecture:
How did
$$\lim_{x\rightarrow \infty} x^3 \left(\tan{\frac{1}{x}}\right)\left(\sin{\frac{3}{x^2}}\right)$$
become
$$\lim_{x\rightarrow \infty}3\left(\frac{\tan{\frac{1}{x}}}{\frac{1}{x}}\right)\left(\frac{\sin{\frac{3}{x^2}}}{\frac{3}{x^2}}\right)$$
Note the $x^3$ and the $3$ and denominator of the 2nd equation
UPDATE
The question is how come the $x^3$ became $3\left(\frac{1}{\frac{1}{x}}\right)\left(\frac{1}{\frac{3}{x^2}}\right)$
| It is likely a typo. It should probably read:
$$
3 \cdot \frac{\tan(\frac{1}{x})}{\frac{1}{x}} \cdot \frac{\sin(\frac{3}{x^2})}{\frac{3}{x^2}}
$$
which by multiplying the denominator is
$$
\frac{3}{\frac{1}{x} \cdot \frac{3}{x^2}} \cdot \tan\left(\frac{1}{x} \right) \sin\left(\frac{3}{x^2}\right) = x^3 \tan\left(\frac{1}{x} \right) \sin\left(\frac{3}{x^2}\right).
$$
More importantly, do you see where these fractions come from? That is, do you wee why we would want to divide $\tan(1/x)$ by $1/x$ and $\sin(3/x^2)$ by $3/x^2$?
| {
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Prove that $x_1^2+x_2^2+x_3^2=1$ yields $ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $ Prove this inequality, if $x_1^2+x_2^2+x_3^2=1$:
$$ \sum_{i=1}^{3}\frac{x_i}{1+x_i^2} \le \frac{3\sqrt{3}}{4} $$
So far I got to $x_1^4+x_2^4+x_3^4\ge\frac{1}3$ by using QM-AM for $(2x_1^2+x_2^2, 2x_2^2+x_3^2, 2x_3^2+x_1^2)$, but to be honest I'm not sure if that's helpful at all.
(You might want to "rename" those to $x,y,z$ to make writing easier):
$$ \frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt3}4$$
| Here's a solution with tangent line method.
Note that $$\frac{x}{1+x^2} \le \frac{3\sqrt{3}}{16}(x^2+1) \iff \frac{1}{48}(3x-\sqrt{3})^2(\sqrt{3}x^2+2x+3\sqrt{3}) \ge 0 \text{ (true by discriminant) }$$
Now we have $$\frac{x}{1+x^2}+\frac{y}{1+y^2}+\frac{z}{1+z^2} \le \frac{3\sqrt{3}}{16}(x^2+y^2+z^2+3) = \frac{3\sqrt{3}}{4}$$ as desired. $\blacksquare$
| {
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Showing $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$ Given that n is a positive integer show that $\gcd(n^3 + 1, n^2 + 2) = 1$, $3$, or $9$.
I'm thinking that I should be using the property of gcd that says if a and b are integers then gcd(a,b) = gcd(a+cb,b). So I can do things like decide that $\gcd(n^3 + 1, n^2 + 2) = \gcd((n^3+1) - n(n^2+2),n^2+2) = \gcd(1-2n,n^2+2)$ and then using Bezout's theorem I can get $\gcd(1-2n,n^2+2)= r(1-2n) + s(n^2 +2)$ and I can expand this to $r(1-2n) + s(n^2 +2) = r - 2rn + sn^2 + 2s$ However after some time of chasing this path using various substitutions and factorings I've gotten nowhere.
Can anybody provide a hint as to how I should be looking at this problem?
| Playing around along the lines you were exploring should work. Let's do it somewhat casually, aiming always to reduce the maximum power of $n$.
If $m$ divides both $n^3+1$ and $n^2+2$, then $m$ divides $n(n^2+2)-(n^3+1)$, so it divides $2n-1$.
(You got there.)
But if $m$ divides $n^2+2$ and $2n-1$, then $m$ divides $2(n^2+2)-n(2n-1)$, so it divides $n+4$. (This move is slightly risky. In some cases this kind of move could introduce spurious common divisors. But (i) In this case it doesn't; and (ii) We will be checking at the end anyway whether our list of candidates is correct.)
But if $m$ divides $n+4$ and $2n-1$, then $m$ divides $2(n+4)-(2n-1)$, so it divides $9$.
We did not work explicitly with greatest common divisors, so we are not finished. We must show that $1$, $3$ and $9$ are all achievable. For $\gcd$ $1$, let $n=0$. For $\gcd$ $3$, let $n=2$. For $\gcd$ $9$, let $n=4$.
| {
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"url": "https://math.stackexchange.com/questions/109876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Direct proof that for a prime $p$ if $p\equiv 1 \bmod 4$ then $l(\sqrt{p})$ is odd.
Definition: Assume $p$ is a prime. $l(\sqrt{p})=$ length of period in simple continued fraction expansion of $\sqrt{p}$.
The standard proof of this uses the following:
*
*$p$ is a prime implies $p \equiv 1 \bmod 4$ iff $x^2-py^2=-1$ has integer solutions.
*$p$ is a prime implies $x^2-py^2=-1$ has integer solutions iff $l(\sqrt{p})$ is odd.
I have proofs of (1) and (2) so that I have a proof of the stated question.
What I would like is a proof that does not use any equivalences. That is,
Assume $p \equiv 1 \bmod 4$. Show that $l(\sqrt{p})$ is odd.
I will accept contradiction or contrapositive as well. I know this may seem strange but I think there is a lot to learn from this proof.
Also, I asked this on mathoverflow link but got no definative answer. I think a proof may exist using Farey graphs and/or Ford Circles.
| We can deduce both of (1) and (2), and the fact that $p$ is a sum of two squares, by playing with quadratic forms. I have the impression this is all "well known to those who know it well". But I've never seen this elementary argument written down in elementary language, so here it is.
We begin with some general discussion of quadratic equations. Let the roots of $a x^2+bx+c$ be $\lambda_1$ and $\lambda_2$. Given a triple of integers $(a,b,c)$, we will think of them as encoding the quadratic equation $a x^2+bx +c$ and we will write the roots of that equation as $\lambda_1$ and $\lambda_2$. We will be performing operations on the coefficients of the equation and seeing how the roots change. For example, the roots of $a (x-1)^2 + b (x-1) + c=a x^2 + (-2a+b) x + (a-b+c)$ will be $\lambda_1+1$ and $\lambda_2+1$. We'll write $E(a,b,c) = (a, -2a+b, a-b+c)$ and $E(\lambda)= \lambda+1$, using the number of inputs of $E$ to make it clear which we are talking about. Here are the operators we will need, as they act on triples $(a,b,c)$ and on the roots $\lambda_i$:
$$\begin{array}{|c|l@{}c@{}r|l|}
\hline
E & (a,&-2a+b,& a-b+c) & 1+\lambda \\
\hline
E^{-1} & (a,& 2a+b,& a+b+c) & -1+\lambda \\
\hline
F & (a-b+c,& b-2c,& c) & \frac{\lambda}{1+\lambda} \\
\hline
F^{-1} & (a+b+c,& b+2c,& c) & \frac{\lambda}{1-\lambda} \\
\hline
\end{array}$$
Observe that all of these operators preserve:
*
*The discriminant $\Delta = b^2 - 4 ac$
*The parity of $b$ (actually, this follows from the previous line, since $\Delta \equiv b^2 \bmod 4$).
*$GCD(a,b,c)$.
We'll be interested in quadratics which can be obtained from $x^2-p$, so we will take $\Delta = 4p$, $b \equiv 0 \bmod 2$ and $GCD(a,b,c) = 1$.
We define $(a,b,c)$ to be reduced if $a>0$ and $c<0$. In a reduced quadratic, we have $b^2 < b^2+4a(-c)=4p$, so there are only finitely many possible values for $b$ and, for each of these values, only finitely many possible values for $a$ and $c$. Write $Q$ for the finite set of reduced triples $(a,b,c)$ with $b^2+4a(-c) = 4p$, $b \equiv 0 \bmod 2$ and $GCD(a,b,c) = 1$.
Let $(a,b,c)$ in $Q$. The roots of $ax^2+bx+c$ are real and of opposite sign; call them $\lambda_- < 0 < \lambda_+$. The following lemmas are left as exercises:
Lemma We have $E (a,b,c) \in Q$ if and only if $\lambda_- < -1$, if and only if $a-b+c<0$. We have $F (a,b,c) \in Q$ if and only if $\lambda_- > -1$, if and only if $a-b+c>0$. Exactly one of these two cases occurs.
Lemma We have $E^{-1} (a,b,c) \in Q$ if and only if $\lambda_+ > 1$, if and only if $a+b+c<0$. We have $F^{-1} (a,b,c) \in Q$ if and only if $\lambda_+ < 1$, if and only if $a+b+c>0$. Exactly one of these two cases occurs.
Draw a directed graph $G$ with vertex set $Q$ and an edge $(a,b,c) \to (a',b',c')$ if $(a',b',c')$ is $E(a,b,c)$ or $F(a,b,c)$. By the above lemmas, every vertex of $G$ has out-degree $1$ and in-degree $1$, so $G$ is a disjoint union of cycles. We will "color" the edges of $G$ with the symbols $E$ and $F$ accordingly. I encourage you to draw this graph for a few small values of $p$ (non-primes, and primes that are $3 \bmod 4$, are good too).
Consider the bijection $\sigma: (a,b,c) \mapsto (a,-b,c)$ from $Q$ to itself. This induces a symmetry of $G$, preserving the coloring of edges but reversing their directions. Now, $(1,0,-p)$, corresponding to $x^2-p$, is fixed by $\sigma$, so $\sigma$ must take the cycle through $(1,0,-p)$ to itself. Call this cycle $C$. We consider two cases:
Case 1 $C$ has odd length, so the point antipodal to $(1,0,-p)$ is the midpoint of an edge. Let us suppose that it is an $E$-edge $(a,b,c) \to (a,-2a+b,a-b+c)$; the case of an $F$-edge is similar. Then $\sigma$ swaps $(a,b,c)$ and $(a,-2a+b,a-b+c)$, so we deduce $-b=-2a+b$ and $c=a-b+c$, implying $a=b$. Then $b^2-4ac = b(b-4c)=4p$. Since $b$ is even, $b-4c$ is as well, and the two factors are $2$ and $2p$; since $c<0$ we have $b=2$ and $b-4c=2p$, giving $c = (1-p)/2$. But then $GCD(a,b,c) = GCD(2,2,(1-p)/2)=2$ (using that $p \equiv 1 \bmod 4$) and contradicting our choice that $GCD=1$.
We have a contradiction and deduce that we are instead in:
Case 2 $C$ has even length. Then the point $(a,b,c)$ antipodal to $(1,0,-p)$ must be fixed by $\sigma$, so $b=0$. Then $4a(-c) = 4p$ so $a(-c) = p$. We are looking for the solution that is not $(1,0,-p)$, so it must be $(p,0,-1)$. In short, we conclude that $G$ contains a path from $(1,0,-p)$ to $(p,0,-1)$. Moreover, the edge coming out of $(1,0,-p)$ is an $E$-edge and the edge into $(p,0,-1)$ is an $F$-edge, so we have
$$(p,0,-1) = F^{a_{1}} E^{a_{2}} \cdots F^{a_{2r-1}} E^{a_{2r}} (1,0,-p)$$
for some positive integers $(a_1, a_2, \ldots, a_{2r})$.
Converting to roots of quadratics,
$$\frac{1}{\sqrt{p}} = F^{a_{1}} E^{a_{2}} \cdots F^{a_{2r-1}} E^{a_{2r}} \left( \sqrt{p} \right).$$
We have $E^a(\lambda) = a+\lambda$ and $F^a(\lambda) = \frac{1}{a+\frac{1}{\lambda}}$.
So
$$\frac{1}{\sqrt{p}} = \frac{1}{a_1 +}\frac{1}{a_2 + } \cdots \frac{1}{a_{2r-1}+} \frac{1}{a_{2r}+\sqrt{p}}$$
or
$$\sqrt{p} = a_1 + \frac{1}{a_2 + } \cdots \frac{1}{a_{2r-1}+} \frac{1}{a_{2r}+\sqrt{p}}.$$
We deduce that
$$\sqrt{p} = [a_1, \overline{a_2, a_3, \ldots, a_{2r-1}, a_{2r}+a_1}].$$
The period is odd, as desired.
We can also use this result to solve the negative Pell equation. Let the Mobius transformation $F^{a_1} E^{a_2} \cdots F^{a_{2r-1}} E^{a_{2r}}(x)$ be equal to $\frac{ax+b}{cx+d}$. We have
$$\frac{a \sqrt{p} + b }{c \sqrt{p} + d} = \frac{1}{\sqrt{p}}.$$
Using that $(a,b,c,d)$ are integers and $p$ is not square, we have $b=c$ and $d=ap$.
Since $\det \left( \begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \right) = 1$, we deduce that $a^2 p - b^2 =1$ or, in other words, the negative Pell equation $b^2 - p a^2 = -1$ is solvable.
While we're at it, we can deduce more by also using the symmetry $\tau: (a,b,c) \mapsto (-c,b,-a)$. This swaps $(1,0,-p)$ and $(p,0,-1)$, and it reverses edge coloring and direction. So it reverses the path $(1,0,-p) \to \cdots \to (p,0,-1)$. Since $\tau$ switches edge coloring, it can not fix any edge, and we deduce that the mid point of the path must be a $\tau$ fixed vertex. We have $\tau(a,b,c) = (a,b,c)$ if and only if $a=-c$; in this case, $b^2 - 4 ac = b^2 + (2a)^2 = 4p$ and we deduce that $p = (b/2)^2 + a^2$ is a sum of squares.
Finally, since $\tau$ reverses the path, we deduce that $(a_1, a_2, \ldots, a_{2r})$ is a palindrome.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A tricky double integral What is
$$\int_0^1 \int_0^1 \frac{ dx \; dy}{1+xy+x^2y^2} ? $$
Can you do one of the integrals and turn it into a single integral?
I get lost in a sea of inverse tangents.
| The first step would be one $\arctan$ only.
The denominator is
$$1+xy+x^2y^2 =\left(\frac{1}{2}+xy\right)^2 + \frac{3}{4}$$
Hence, with $a=\frac{2}{\sqrt{3}}$ we have
$$\int\frac{dy}{1+xy+x^2y^2} =\int\frac{dy}{\left(\frac{1}{2}+xy\right)^2 + 1/a^2}=\qquad\qquad\qquad\text{ }\\
a^2\int\frac{dy}{a^2\left(\frac{1}{2}+xy\right)^2 + 1}=
a^2\frac{\arctan(\frac{a}{2}+axy)}{ax} + C=\\a\frac{\arctan\left(\frac{1}{\sqrt{3}}+\frac{2xy}{\sqrt{3}}\right)}{x} + C$$
so that $$\int_0^1\frac{dy}{1+xy+x^2y^2} = a\frac{\arctan(\frac{1}{\sqrt{3}}+\frac{2x}{\sqrt{3}})}{x} -\frac{\pi}{3\sqrt{3}}\frac{1}{x}$$
The integral of $\arctan(a+bx)/x$ is not that easy, it ends up in dilogarithms...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Show that $\tan 3x =\frac{ \sin x + \sin 3x+ \sin 5x }{\cos x + \cos 3x + \cos 5x}$ I was able to prove this but it is too messy and very long. Is there a better way of proving the identity? Thanks.
| The identities for the sum of sines and the sum of cosines yield
$$
\frac{\sin(x)+\sin(y)}{\cos(x)+\cos(y)}=\frac{2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}{2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)}=\tan\left(\frac{x+y}{2}\right)\tag{1}
$$
Equation $(1)$ implies that
$$
\frac{\sin(x)+\sin(5x)}{\cos(x)+\cos(5x)}=\tan(3x)=\frac{\sin(3x)}{\cos(3x)}\tag{2}
$$
We also have that if $b+d\not=0$, then
$$
\frac{a}{b}=\frac{c}{d}\Rightarrow\frac{a}{b}=\frac{c}{d}=\frac{a+c}{b+d}\tag{3}
$$
Combining $(2)$ and $(3)$ yields
$$
\tan(3x)=\frac{\sin(x)+\sin(3x)+\sin(5x)}{\cos(x)+\cos(3x)+\cos(5x)}\tag{4}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/113451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Integers and fractions How would I write this as an integer or a fraction in lowest terms?
$(1-\frac12)(1+\frac 12)(1-\frac13)(1+\frac13)(1-\frac14)(1+\frac14).....(1-\frac1{99})(1+\frac1{99})$
I really need to understand where to start and the process if anyone can help me.
| Hint: telescopy, note how the adjacent like-colored terms all cancel out of the products below
$$\rm \left(1-\frac{1}2\right)\left(1-\frac{1}3\right)\cdots \left(1-\frac{1}n\right)\ =\ \frac{1}{\color{red}2} \frac{\color{red}2}{\color{green}3} \frac{\color{green}3}{\color{blue}4} \frac{\color{blue} 4}{}\: \cdots\: \frac{}{\color{brown}{n-1}}\frac{\color{brown}{n-1}}n\ =\ \frac{1}n $$
$$\rm \left(1+\frac{1}2\right)\left(1+\frac{1}3\right)\cdots \left(1+\frac{1}n\right)\ =\ \frac{\color{red}3}2 \frac{\color{green}4}{\color{red}3} \frac{\color{blue}5}{\color{green}4}\frac{}{\color{blue}5}\: \cdots\: \frac{\color{brown} n}{}\frac{n+1}{\color{brown}n}\ =\ \frac{n+1}2$$
Hence $$\rm \prod_{k\:=\:2}^n\: \left( 1-\frac{1}{k^2}\right)\ =\ \frac{n+1}{2\:n} $$
Follow the above link for many more examples of telescopic proofs (additive and multiplicative).
| {
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"timestamp": "2023-03-29T00:00:00",
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Bennett's Inequality to Bernstein's Inequality Bennett's Inequality is stated with a rather unintuitive function,
$$
h(u) = (1+u) \log(1+u) - u
$$
See here. I have seen in multiple places that Bernstein's Inequality, while slightly weaker, can be obtained by bounding $h(u)$ from below,
$$
h(u) \ge \frac{ u^2 }{ 2 + \frac{2}{3} u}
$$
and plugging it back into Bennett's Inequality. However, I can't see where this expression comes from. Could someone point me in the right direction?
| While the answer by @Arash outlines the general strategy of deriving the inequality,
the explicit computation is missing.
I am providing it here in the hope that it will be helpful for future readers.
The actual computation of the derivative is highly simplified by first simplifying
the last term in the definition of
$$
f(x)
= (1 + x) \ln(1 + x) - x - \frac{x^2}{2 + \frac{2}{3} x}
$$
as follows:
$$
\frac{x^2}{2 + \frac{2}{3} x}
= \frac{3}{2} \frac{x^2}{3 + x}
= \frac{3}{2} \frac{x \cdot (3 + x) - 3 x}{3 + x}
= \frac{3}{2} x - \frac{9}{2} \frac{x}{x + 3}.
$$
Therefore, we see
\begin{align*}
\require{cancel}
f(x)
& = (1 + x) \ln (1 + x) - \frac{5}{2} x + \frac{9}{2} \frac{x}{x + 3} , \\
f'(x)
& = \ln(1 + x) + 1 - \frac{5}{2} + \frac{9}{2} \frac{x+3 - x}{(x+3)^2} \\
& = \ln(1 + x) - \frac{3}{2} + \frac{3^3}{2} (x+3)^{-2}, \\
f''(x)
& = \frac{1}{1 + x} - 3^3 \cdot (x+3)^{-3} \\
& = \frac{(x+3)^3 - 27 \cdot (1 + x)}{(1 + x) (3+x)^3} \\
& = \frac{x^{3} + 9 x^{2} + \bcancel{27 x} + \cancel{27} - \cancel{27} - \bcancel{27 x}}
{(1 + x) (3+x)^3} \\
& = \frac{x^{3} + 9 x^{2}}
{(1 + x) (3+x)^3}
\geq 0
\end{align*}
for $x \geq 0$, which is the only case we are interested in.
Since $f(0) = 0$ and $f'(0) = 0$, this implies $f' \geq 0$ and then $f \geq 0$
on $[0,\infty)$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Ratio test for $\frac{\sqrt{n^n}}{2^n}$ Before I tried root test, I did ratio test for $\frac{\sqrt{n^n}}{2^n}$ and got:
$$\lim_{n\to\infty}\frac{\sqrt{(n+1)^{(n+1)}}}{2^{n+1}}\cdot \frac{2^n}{\sqrt{n^n}} = \lim_{n\to\infty} \frac{1}{2}\cdot \sqrt{\frac{(n+1)^{(n+1)}}{n^n}} = \frac{1}{2}$$
But correct answer with ratio test is $\frac{\sqrt{n}}{2}$. So I must have did something wrong
UPDATE: Ratio Test
$$\sqrt[n]{\frac{\sqrt{n}^n}{2^n}}=\frac{\sqrt{n}}{2}$$
| The ratio of consecutive terms is indeed $$\frac{1}{2}\cdot \sqrt{\frac{(n+1)^{(n+1)}}{n^n}}$$ but that second factor does not tend to $1$ like you seem to have assumed. We actually have $$ \sqrt{\frac{(n+1)^{(n+1)}}{n^n}} = \sqrt{n+1} \cdot \sqrt{ \left( 1+ \frac{1}{n} \right)^n } .$$
Since $\left( 1+ \frac{1}{n} \right)^n \to e , $ we have that $$ \frac{a_{n+1} }{a_n} \sim \frac{\sqrt{ne} }{2} $$
which tells you the sequence diverges to infinity. In fact it also tells you $$ a_n\approx \frac{\sqrt{n! e^n} }{2^n} $$
which is also seen more directly if we use $ n! \approx (n/e)^n .$
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\lim_{x \to 0}\frac{1-\cos (1- \cos x)}{x^4}$ $$\lim_{x \to 0}\frac{1-\cos (1- \cos x)}{x^4}$$
I don't think L'Hôpital's rule is a good idea here.
I will not finish this until the evening and it's easy to make mistake. Maybe I can expand $\cos$ in a series?
But I don't know how to use this trick...
| You can expand cos in a series, like you said:
$$1 - \cos\left(1 - \cos x\right) = 1 - \cos\left(1 - \left(1 - \frac{x^2}{2!} + \frac{x^4}{4!} - \cdots\right)\right) $$
$$= 1 - \cos\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right) $$
$$= 1 - \left[1 - \frac{1}{2!}\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right)^2 + \frac{1}{4!}\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right)^4 \cdots \right]$$
$$= \frac{1}{2!}\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right)^2 - \frac{1}{4!}\left(\frac{x^2}{2!} - \frac{x^4}{4!} + \cdots\right)^4 $$
Now since we are taking the limit as $x \to 0$ of that over $x^4$, all terms of fifth degree or higher go to $0$. So the limit is just $\frac{1}{x^4}\frac{1}{2!}\left(\frac{x^2}{2!}\right)^2 = \frac{1}{8}$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Trigonometric identity, possible error I need to prove the following trigonometric identity:
$$ \frac{\sin^2(\frac{5\pi}{6} - \alpha )}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2(-\alpha - \frac{13\pi}{2}) =\sin^2(\alpha)$$
I cannot express $\sin(\frac{5\pi}{6}-\alpha)$ as a function of $\alpha$. Could it be a textbook error?
| Some important translations:
$$\tag 1\sin(x\pm 2 \pi) = \sin x $$
$$\tag {1'}\cos(x\pm 2 \pi) = \cos x $$
$$\tag 2\cot(x\pm \pi)= \cot x$$
$$\tag {2'}\tan(x\pm \pi)= \tan x$$
$$\tag 3 \sin \left(\frac \pi 2 -x \right)=\cos x$$
$$\tag 4 \cos \left(\frac \pi 2 -x \right)=\sin x$$
$$\tag 5 \sin(\pi-x)=\sin x$$ and $$\tag 6\cos(\pi-x)=-\cos x$$
and $$\tag 7 \sin(-x)=-\sin x$$
$$\tag 8 \cos (-x) = \cos x$$
Although taking $\alpha =0$ reveals that the equality doesn't hold, assume that there is no typo, then, we could move on as follows.
You have that
$$ \frac{\sin^2 \left(\frac{5\pi}{6} - \alpha \right)}{\cos^2(\alpha - 4\pi)} - \cot^2(\alpha - 11\pi)\sin^2 \left(-\alpha - \frac{13\pi}{2}\right) =\sin^2(\alpha)$$
Using the above, we can write
$$\eqalign{
& {\sin ^2}\left( {\frac{{5\pi }}{6} - \alpha } \right) = {\left[ { - \sin \left( {\alpha - \frac{{5\pi }}{6}} \right)} \right]^2} = {\sin ^2}\left( {\alpha - \frac{{5\pi }}{6}} \right) \cr
& {\cot ^2}(\alpha - 11\pi ) = {\cot ^2}\left( {\alpha - 10\pi } \right) = \cdots = {\cot ^2}\alpha \cr
& {\sin ^2}\left( { - \alpha - \frac{{13\pi }}{2}} \right) = {\left[ { - \sin \left( {\alpha + \frac{{13\pi }}{2}} \right)} \right]^2} = \sin {\left( {\alpha + \frac{{13\pi }}{2}} \right)^2} \cr
& {\cos ^2}(\alpha - 4\pi ) = {\cos ^2}(\alpha - 2\pi ) = {\cos ^2}\alpha \cr} $$
so that we have
$$\frac{{{{\sin }^2}\left( {\alpha - \frac{{5\pi }}{6}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\sin ^2}\left( {\alpha + \frac{{13\pi }}{2}} \right) = {\sin ^2}\alpha $$
Now
$${\sin ^2}\left( {\alpha - \frac{{5\pi }}{6}} \right) = {\sin ^2}\left( {\alpha - \frac{{3\pi }}{6} - \frac{{2\pi }}{6}} \right) = {\sin ^2}\left( {\alpha - \frac{\pi }{3} - \frac{\pi }{2}} \right) = {\left( { - 1} \right)^2}{\sin ^2}\left( {\frac{\pi }{2} - \left( {\alpha - \frac{\pi }{3}} \right)} \right) = {\cos ^2}\left( {\alpha - \frac{\pi }{3}} \right)$$
and
$$\eqalign{
& {\sin ^2}\left( {\alpha + \frac{{13\pi }}{2}} \right) = {\sin ^2}\left( {\alpha + \frac{{12\pi }}{2} + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha + 6\pi + \frac{\pi }{2}} \right) = {\sin ^2}\left( {\alpha + \frac{\pi }{2}} \right) \cr
& = {\sin ^2}\left( {\frac{\pi }{2} - \left( { - \alpha } \right)} \right) = {\cos ^2}\left( { - \alpha } \right) = {\cos ^2}\alpha \cr} $$
so that you have
$$\frac{{{{\cos }^2}\left( {\alpha - \frac{\pi }{3}} \right)}}{{{{\cos }^2}\alpha }} - {\cot ^2}\alpha {\cos ^2}\alpha = {\sin ^2}\alpha $$
Now, solving for $${{{\cos }^2}\left( {\alpha - \frac{\pi }{3}} \right)}$$
gives
$${\cos ^2}\left( {\alpha - \frac{\pi }{3}} \right) = {\cos ^2}\alpha {\sin ^2}\alpha + {\cos ^6}\alpha \frac{1}{{{{\sin }^2}\alpha }}$$
There is some typo in your excercise, since letting $\alpha =0$ gives $1/4$ on the LHS and is not defined for the RHS. When you discover what the typo is, move on with the listed translations.
| {
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} |
How to integrate $\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$? I'm trying to evaluate the integral $\displaystyle\int_0^\infty\frac{x^{1/3}dx}{1+x^2}$.
My book explains that to evaluate integrals of the form $\displaystyle\int_0^\infty x^\alpha R(x)dx$, with real $\alpha\in(0,1)$ and $R(x)$ a rational function, one first starts with a substitution $x=t^2$, to transform the integral to
$$
2\int_0^\infty t^{2\alpha+1} R(t^2) dt.
$$
It then observes that
$$
\int_{-\infty}^\infty z^{2\alpha+1}R(z^2)dz=\int_0^\infty(z^{2\alpha+1}+(-z)^{2\alpha+1})R(z^2)dz.
$$
Since $(-z)^{2\alpha}=e^{2\pi i\alpha}z^{2\alpha}$, the integral then equals
$$
(1-e^{2\pi i\alpha})\int_0^\infty z^{2\alpha+1}R(z^2)dz.
$$
How are you suppose to apply the residue theorem to this integral if the integrand is not a rational function?
In my case, I have
$$
\int_0^\infty x^{1/3}R(x)dx=2\int_0^\infty t^{5/3}R(t^2)dt.
$$
Also
$$
\int_0^\infty z^{5/3} R(z^2)dz=\frac{1}{1-e^{(2\pi i)/3}}\int_{-\infty}^{\infty}\frac{z^{5/3}}{1+z^4}dz.
$$
I don't know what to do after that, since this last integrand still has a fractional power in the numerator. How does this work? Thanks.
(This is part (g) of #3 on page 161 of Ahlfors' Complex Analysis, part of some self-study.)
| Suppose $\alpha \in (-1, 1)$. Consider a keyhole contour $\Gamma$ of outer radius $R$ and inner radius $\varepsilon$ about the positive real axis. Let $\gamma_R$ denote the outer arc, $\gamma_\varepsilon$ denote the inner arc, and $\gamma_\pm$ denote the segment going away from and towards the origin. Note that this contour contains two singularities of $f(z) = \frac{z^\alpha}{z^2 + 1}$, namely $z = \pm i = \exp\left(\frac{\pi i}{2}\right ), \exp\left(\frac{3\pi i}{2}\right )$. We have
\begin{align*}
\int_\Gamma \frac{z^\alpha}{z^2 + 1} \ \mathrm{d}z &= 2\pi i \left (\text{Res}_f\left(\exp\left(\frac{\pi i}{2}\right ) \right ) + \text{Res}_f\left(\exp\left(\frac{3\pi i}{2}\right ) \right )\right )\\
&= \pi \left (\exp\left(\frac{\alpha \pi i}{2}\right ) - \exp\left(\frac{3\alpha \pi i}{2}\right )\right )
\end{align*}
On the other hand,
\begin{align*}
\int_{\gamma_R} \frac{z^\alpha}{z^2 + 1} \ \mathrm{d}z &= \int_0^1 \frac{R^\alpha \exp(2\alpha\pi i t)}{R^2 \exp(2\pi i t) + 1} 2\pi i R \exp(2\pi i t) \ \mathrm{d}t\\
&= 2\pi i R \int_0^1 \frac{R^\alpha \exp(2\alpha\pi i t)\exp(2\pi i t) }{R^2 \exp(2\pi i t) + 1} \ \mathrm{d}t\\
\end{align*}
using $z = R\exp(2\pi it)$. Bounding, we get
\begin{align*}
\left | 2\pi i R \int_0^1 \frac{R^\alpha \exp(2\alpha\pi i t)\exp(2\pi i t) }{R^2 \exp(2\pi i t) + 1} \ \mathrm{d}t\right | &\leq 2\pi R \int_0^1 \left | \frac{R^\alpha \exp(2\alpha\pi i t)\exp(2\pi i t) }{R^2 \exp(2\pi i t) + 1} \ \right |\mathrm{d}t\\
&\leq 2\pi R \int_0^1 \frac{R^\alpha}{R^2 - 1} \ \mathrm{d}t = \frac{2\pi R^{\alpha + 1}}{R^2 - 1}
\end{align*}
Since $\alpha \in (-1, 1)$, as $R \to \infty$, this tends to $0$. Further note that for $\gamma_\varepsilon$, this should be identical except $R$ becomes $\frac{1}{R}$, so we have bound $\frac{2\pi R^{-1-\alpha}}{1-R^{-2}} = \frac{2\pi R^{1-\alpha}}{R^2-1}$. But again, $\alpha \in (-1, 1)$ implies that this expression tends to $0$ as $R \to \infty$ (or $\varepsilon \to 0$), so this also disappears.
For $\gamma_+$, we have
\begin{align*}
\int_{\gamma_+} \frac{z^\alpha}{z^2 + 1} \ \mathrm{d}z &= \int_\varepsilon^R \frac{x^\alpha}{x^2 + 1} \ \mathrm{d}x
\end{align*}
For $\gamma_-$, we substitute $z = x \exp(2\pi i)$, so we have
\begin{align*}
\int_{\gamma_-} \frac{z^\alpha}{z^2 + 1} \ \mathrm{d}z &= \int_R^\varepsilon \frac{x^\alpha \exp(2\alpha \pi i)}{x^2 + 1} \ \mathrm{d}x
\end{align*}
and so by adding all the integrals together (and sending $R \to \infty$, $\varepsilon \to 0$), we get
\begin{align*}
\pi \left (\exp\left(\frac{\alpha \pi i}{2}\right ) - \exp\left(\frac{3\alpha \pi i}{2}\right )\right ) &= \left (1 - \exp(2\alpha\pi i) \right ) \int_0^\infty \frac{x^\alpha}{x^2 + 1} \ \mathrm{d}x
\end{align*}
Hence, if $\omega = \exp\left(\frac{\alpha \pi i}{2}\right )$, we get
\begin{align*}
\int_0^\infty \frac{x^\alpha}{x^2 + 1} \ \mathrm{d}x &= \frac{\pi \left(\omega - \omega^3 \right )}{1 - \omega^4}\\
&= \frac{\pi}{\omega + \omega^{-1}}\\
&= \frac{\pi}{2}\sec\left (\frac{\alpha \pi}{2} \right )
\end{align*}
Therefore,
\begin{align*}
\boxed{\int_0^\infty \frac{x^\alpha}{x^2 + 1} = \frac{\pi}{2}\sec\left (\frac{\alpha \pi}{2} \right )}
\end{align*}
Substituting $\alpha = \frac{1}{3}$ gives $\frac{\pi}{\sqrt{3}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/121976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Find the largest positive value of $x$ at which the curve $y = (2x + 7)^6 (x - 2)^5$ has a horizontal tangent line. I need help with the following question.
Find the largest positive value of x at which the curve:
$$y = (2x + 7)^6 (x - 2)^5$$
has a horizontal tangent line.
| Hint: $\displaystyle{\frac{dy}{dx} = \left(12(2x+7)^5(x-2)^5+5(2x+7)^6(x-2)^4 \right) = 0}$ , in other words
$\displaystyle{\frac{dy}{dx} = (2x+7)^5(x-2)^4(22x+11) = 0}$ at what points?
$\displaystyle{\frac{dy}{dx} = 0}$ at $x=-\frac{7}{2}, x=2, x=-\frac{1}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/122226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Laurent Series of rational function about $z=0$ Find the Laurent expansion for $$\frac{\exp(\frac 1{z^2})}{z-1}$$ about $z=0$. I know that $\exp( \frac 1{z^2}) = \sum_{n=0}^\infty \frac{z^{-2n}}{n!}$
and $ \frac 1{z-1}=-\sum_{n=0}^\infty z^n$
| Write $$\exp(1/z^2) = \sum_{n=-\infty}^\infty a_n z^n \qquad\text{and}\qquad \frac{1}{z-1} = \sum_{n=-\infty}^\infty b_n z^n,$$
i.e. $a_{-2n} = 1/n!$ and $a_k = 0$ otherwise, and $b_n = -1$ for $n \ge 0$ and $b_n = 0$ for $n < 0$. Multiply the two series:
$$ \left(\sum_{n=-\infty}^\infty a_n z^n\right)\left(\sum_{n=-\infty}^\infty b_n z^n\right) = \left(\sum_{n=-\infty}^\infty c_n z^n\right)$$
and start identifying the $c_n$:s. We get
$$ c_0 = a_0 b_0 + a_{-2}b_2 + a_{-4}b_4 + \cdots = -\left( \frac1{0!} + \frac{1}{1!} + \frac1{2!} + \cdots \right) = -e,$$
$$ c_1 = a_0 b_1 + a_{-2}b_3 + a_{-4}b_5 + \cdots = -\left( \frac1{0!} + \frac{1}{1!} + \frac1{2!} + \cdots \right) = -e,$$
etc. In other words, $c_n = -e$ for all $n \ge 0$. For the negative indices you get
$$ c_{-1} = a_{-2} b_1 + a_{-4}b_3 + a_{-6}b_5 + \cdots = -\left( \frac{1}{1!} + \frac1{2!} + \frac1{3!}+ \cdots \right) = -(e-1),$$
$$ c_{-2} = a_{-2} b_0 + a_{-4}b_2 + a_{-6}b_4 + \cdots = -\left( \frac{1}{1!} + \frac1{2!} + \frac1{3!}+ \cdots \right) = -(e-1),$$
$$ c_{-3} = a_{-4} b_1 + a_{-6}b_3 + a_{-8}b_5 + \cdots = -\left( \frac{1}{2!} + \frac1{3!} + \frac1{4!}+ \cdots \right) = -(e-1-1),$$
$$ c_{-4} = a_{-4} b_0 + a_{-6}b_2 + a_{-8}b_4 + \cdots = -\left( \frac{1}{2!} + \frac1{3!} + \frac1{4!}+ \cdots \right) = -(e-1-1).$$
By now, the pattern should be obvious, if $n = -2k$,
$$ c_{-2k} = c_{-2k+1} = - \left(e - 1 - \frac1{1!} - \frac1{2!} - \cdots \frac1{k!}\right) = -e + \sum_{j=0}^k \frac{1}{j!}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/122368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Exponential Equation with mistaken result I'm on my math book studying exponential equations, and I got stuck on this Problem:
What is sum of the roots of the equation:
$$\frac{16^x + 64}{5} = 4^x + 4$$
I decided to changed: $4^x$ by $m$, so I got: $$\frac{m^2 + 64}{5} = m + 4$$
working on it I've got: $m^2 - 5m + 44 = 0$
but solving this equation the roots were: $x = \frac{5 \pm \sqrt{151i}}{2}$
which isn't even close from the possible answers: 1, 3, 8, 16 or 20.
What's the mistake ? thanks in advance;
| Perhaps you copied the equation wrong. For the equation
$$ \frac{16^x + 64}{5} = 4^x + b$$
(where presumably $x$ is supposed to be real), substituting $m = 4^x$ we get
$$m = \frac{5 \pm \sqrt{20 b - 231}}{2}$$
where we want both solutions to be real, so $11.55 \le b < 12.8$. Now
$x_1 + x_2 = k$ where $m_1 m_2 = 4^{x_1 + x_2} = 4^k$. In this case
$$m_1 m_2 = \frac{5^2 - (20 b - 231)}{4} = 64 - 5 b$$
Thus if $b=12$ you get $4^k = 4$ so $k = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/124456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Binomial Coefficients in the Binomial Theorem - Why Does It Work Question to keep it simple: Given
$(a+b)^3=\binom{3}{0}a^3+\binom{3}{1}a^2b+\binom{3}{2}ab^2+\binom{3}{3}b^3$
Could you please give me an intuitive combinatoric reason to why the binomial coefficients are here?
for instance, what does $\binom{3}{2}ab^2$ mean combinatorially?
Thank you very much.
| This isn't as rigorous as the accepted answer, but explores the question a bit from another perspective.
Take your example,
$(a+b)^3=\binom{3}{0}a^3+\binom{3}{1}a^2b+\binom{3}{2}ab^2+\binom{3}{3}b^3$
Let us consider this small rewrite of the same example:
$(a+b)^3=\binom{3}{0}{a}\cdot{a}\cdot{a}+
\binom{3}{1}{a}\cdot{a}\cdot{b}+
\binom{3}{2}{a}\cdot{b}\cdot{b}+
\binom{3}{3}{b}\cdot{b}\cdot{b}$
Consider that the binomial if expanded would take on the following product:
$(a+b)^3 = (a+b)(a+b)(a+b)$
To compute this product we have to pick one summand in each of the $(a+b)(a+b)(a+b)$ and multiply them together. We then do this in all possible ways adding each term together. Since there's $2$ possibilities for each of $3$ choices there's $2^3=8$ total possibilities. In essence, we are forming all length 3 strings over the alphabet $\{a,b\}$.
That may be hard to follow, but it's easy to see it written out:
$(a+b)^3 = (a+b)(a+b)(a+b) = $
${a}\cdot{a}\cdot{a} +
{a}\cdot{a}\cdot{b} + {a}\cdot{b}\cdot{a} + {b}\cdot{a}\cdot{a} +
{b}\cdot{b}\cdot{a} + {b}\cdot{a}\cdot{b} + {a}\cdot{b}\cdot{b} + {b}\cdot{b}\cdot{b}$
So there's $8$ possibilities, so let's think about it intuitively as you asked:
*
*How many length 3 strings can we make over $\{a,b\}$ where we are allowed to choose $0 b$'s? Well, there's $\binom{3}{0}$ of those: $\{aaa\}$. Hence the first term, $\binom{3}{0}a^3$.
*How many length 3 strings can we make over $\{a,b\}$ where we are allowed to chose $1 b$? Well, there's $\binom{3}{1}$ of those: $\{aab,aba,baa\}$. Hence the second term, $\binom{3}{1}a^2b$.
*Same as above, but we choose $2 b$'s? $\{bba,bab,abb\}$. To get third term, $\binom{3}{2}ab^2$. This is what Marc van Leeuwen's terse answer is conveying, and is in essence the "combinatorial meaning" you asked for in your example.
*Now we choose $3 b$'s? $\{bbb\}$. To get fourth term, $\binom{3}{0}b^3$.
Dave L. Renfro makes a good point:
none of the current answers explain why (a+b)(a+b)(a+b) can be
expanded by considering the sum of all possible ordered products of
three elements, where the first element comes from the first factor of
(a+b), the second element comes from the second factor of (a+b), and
the third element comes from the third factor of (a+b).
I'll do my best to explain this one. You see, we're actually repeatedly applying the distributive law of multiplication over addition.
For this example, consider this:
$(a+b)(a+b)(a+b)$
$= ((a+b)a + (a+b)b)(a+b)$ Here you distribute the first factor into the second.
$= (aa + ba + ab + bb)(a+b)$ Distribute again.
$= (aa(a+b) + ba(a+b) + ab(a+b) + bb(a+b))$ Distribute again.
$= aaa + aab + baa + bab + aba + abb + bba + bbb$
Intuitively as we distribute an $(a+b)$ we are "choosing" a or b, but recording both results... in a way.
In effect if we say distribute $c$ over $(a+b)$ we are effectively recording what happens to c when we choose a, and also when we choose b, e.g. $c(a+b) = ca + cb$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/127926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Square Root Of A Square Root Of A Square Root Is there some way to determine how many times one must root a number and its subsequent roots until it is equal to the square root of two or of the root of a number less than two?
sqrt(16)=4
sqrt(4)=2
sqrt(2) ... 3
--
sqrt(27)=5.19615...
sqrt(5.19615...)=2.27950...
sqrt(2.27950...)=1.509803...
sqrt(1.509803...) ... 4
--
Also, using the floor function...
sqrt(27)=floor(5.19615...)
sqrt(5)=floor(2.23606...)
sqrt(2) ... 3
| Taking the square root $n$ times is taking the $2^n$-th root. If the $2^n$-th root of $x$ is $\le\sqrt2$, but the $2^{n-1}$-st root of $x$ is $>\sqrt2$, then
$$x^{1/2^n}\le \sqrt2<x^{1/2^{n-1}}\;,$$
which, after raising everything to the $2^n$ power, is equivalent to
$$x\le (2^{1/2})^{2^n}<x^2\;,$$
or $$x\le 2^{2^{n-1}}<x^2\;.$$ Now take logs base $2$ (which I write $\lg$):
$$\lg x\le 2^{n-1}<2\lg x\;,$$ which after a little massage can be written $$2^{n-2}<\lg x\le 2^{n-1}\;.$$
Take your examples: $\lg 16=4$, so $2^1<\lg 16\le 2^2$, and we must have $n=3$, while $\lg 27$ is clearly between $4$ and $5$, so $2^2<\lg 27\le 2^3$, and we must have $n=4$.
Taking the floor of the square root each time complicates matters slightly, but not much. $\lfloor\sqrt x\rfloor\le\sqrt2$ iff $x<2^2$ iff $\lg x<2$, $\lfloor\sqrt x\rfloor<2^2$ iff $\sqrt x<2^2$ iff $x<2^{2^2}$ iff $\lg x<2^2$, $\lfloor \sqrt x\rfloor\le 2^{2^2}$ iff $\sqrt x<2^{2^2}$ iff $x<2^{2^3}$ iff $\lg x<2^3$, and so on. If $\frac12<\lg x<2$, one step is required. If $2\le\lg x<2^2$, two steps are required. In general, $n$ steps are required if $2^{n-1}\le\lg x<2^n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/129446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Number of positive integral solutions of equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$ $$\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$$
This is one of the popular equation to find out the number of solutions. From Google, here I found that for equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$, number of solutions are
$\psi(n)=\text{number of divisors of } n$.
$$\frac{\psi(n^2)+1}{2}$$
but when i turned here, it says that for equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$, number of solutions are
$$\frac{\psi(n!^2)-1}{2}$$
However in the first link they explained the equation for $n=4$ and their formula correctly suits on that. But I want to confirm the correct answer.
If there exist any better way to get the number of positive integral solutions for the equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$, then please suggest me.
Would prime factorization help here? I have an algorithm to find out prime factors of any number. But not have any exact idea about implementing it over here.
My objective is to find out total number of positive integral solutions of the equation $\dfrac 1 x+ \dfrac 1 y= \dfrac 1 {n!}$.
Thank you.
| If we let $N! = M$,
$$ \frac{1}{x}+\frac{1}{y} = \frac{1}{N!} = \frac{1}{M}$$
Let $x=M+a$ and $y=M+b$ where $a$ and $b$ are integers (positive or negative)
$$ \Rightarrow \frac{2M+a+b}{(M+a)(M+b)} = \frac{1}{M}$$
$$ 2M^2+M(a+b) = M^2+M(a+b)+ab$$
$$ \Rightarrow M^2 = ab$$
Now look at all the divisors of $M^2 = (N!)^2$ and find the values of $a$ and $b$ and hence find the values of $x$ and $y$.
It will be easier if you pick an example such as
$$ \frac{1}{x}+\frac{1}{y} = \frac{1}{3!} = \frac{1}{6} \tag{1}$$
and show that
$$\fbox{(7, 42), (8, 24), (9, 18), (10, 15), (12, 12), (15, 10), (18, 9), (24, 8),(42, 7),(5,-30),}\\
\fbox{(4,-12),(3,-6), (2, -3), (-3, 2), (-6, 3), (-12, 4), (-30, 5)} $$
are $(x,y)$ pairs that satisfy $(1)$, which when counted are $17$ pairs.
Hint: The number of divisors of $36=6^2 = d(6^2) = 9$, and they are $1,2,3,4,6,9,12,18, 36$.
What is the relationship between $d(6^2)$ and the number of solutions?
Note: I am just giving you an approach. It was not explicitly mentioned that $(x,y)$ pairs have to be positive. If that is the case, then you have to only consider positive pairs. (Henry rightly pointed out that it should be clear whether you are looking for only positive integer solutions).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/131050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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"answer_id": 1
} |
Fastest increase of a function For the function $f(x,y,z) = \frac{1 }{ x^2+y^2+z^2}$ what is the direction of the fastest increase at $(1,1,1)$?
| In the direction of the gradient of $f$ evaluated at $(1,1,1)$.
This is a general fact concerning functions $f=f(x,y,z)$: The direction in which a differentiable function $f$ increases most rapidly at the point $(a,b,c)$ is in the direction of $\nabla f(a,b,c)$.
You'll need to find the general formula for $\nabla f(x,y,z)$ first. Then simply evaluate $\nabla f(1,1,1)$. The solution is below, but it would be to your benefit to try the computations before looking at it.
Solution:
The first thing you have to do is compute the gradient of $f$ in general; that is compute $\nabla f(x,y,z)$:
$$\eqalign{
\nabla f(x,y,z) &=f_x(x,y,z)\,{\bf i} +f_y(x,y,z)\,{\bf j} + f_z(x,y,z)\,{\bf k} \cr
&={\partial \over\partial x} {1\over x^2+y^2+z^2}\,{\bf i}
+{\partial \over\partial y} {1\over x^2+y^2+z^2} \,{\bf j}
+{\partial \over\partial z} {1\over x^2+y^2+z^2}\,{\bf k} \cr
&= -(x^2+y^2+z^2)^{-2}\cdot 2x\,{\bf i}
-(x^2+y^2+z^2)^{-2}\cdot 2y\,{\bf j}
-(x^2+y^2+z^2)^{-2}\cdot 2z\,{\bf k}\cr
&={ -2x\over (x^2+y^2+z^2)^{2}}\,{\bf i}
+ { -2y\over (x^2+y^2+z^2)^{2}}\,{\bf j}
+ { -2z\over (x^2+y^2+z^2)^{2}}\,{\bf k}.
}
$$
Now evaluate the gradient at the point $(1,1,1)$:
$$\eqalign{
\nabla f(1,1,1)
&={ -2\cdot 1\over (1^2+1^2+1^2)^{2}}\,{\bf i}
+{ -2\cdot 1\over (1^2+1^2+1^2)^{2}}\,{\bf j}
+ { -2\cdot 1\over (1^2+1^2+1^2)^{2}}\,{\bf k}\cr
&={-2\over9}\,{\bf i}+{-2\over9}\,{\bf j}+{-2\over9}\,{\bf k} .
}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/132146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
What is $\lim\limits_{k \to 0}{f(k) = 2 + k^{\frac{3}{2}}\cos {\frac{1}{k^2}}}$ Just want to check this one:
I got:
$$\displaystyle \lim_{k \to 0}{f(k) = 2} \;+\; \lim_{k \to 0}{k^{\frac{3}{2}}\cos {\frac{1}{k^2}}}$$
Since $\lim\limits_{k \to 0}\cos{\frac{1}{k^2}} = 0$, using the squeeze theorem, I have $\lim\limits_{k \to 0} k^{\frac{3}{2}}\cos{\frac{1}{k^2}} = 0$.
So
$$\begin{align*}
\lim_{k \to 0}f(k) &= 2 + \lim_{k \to 0}k^{\frac{3}{2}}\cos\left(\frac{1}{k^2}\right)\\
&= 2 + 0\\
&= 2
\end{align*}$$
Is this correct?
Thanks!
| Almost. Since $\lim\limits_{k\rightarrow 0^+} k^{3/2}=0$ (note the one-sided limit) and since $-1\le \cos(x)\le1$ for all $x$, it follows from the Squeeze Theorem that $\lim\limits_{k\rightarrow 0^+} \bigl[\,k^{3/2}\cos(1/k^2)\,\bigr]=0$.
Thus, $\lim\limits_{k\rightarrow 0^+} \bigl[2+k^{3/2}\cos(1/k^2)\,\bigr]=2+0=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/135614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to calculate $\int{\frac{dx}{3x^2+2}}$? I've started doing $$\displaystyle\int{\dfrac{dx}{3x^2+2}}$$
but I only get
$$\displaystyle\int{(3x^2+2)^{-1}dx}\\
\frac{1}{6}\displaystyle\int{\frac{6x(3x^2+2)^{-1}}{x}dx}\\
$$
And I don't know how to do solve this.
| If we put $\sqrt{\frac{3}{2}}x=t$ then it means that $\mathrm dx=\sqrt{\frac{2}{3}}\mathrm dt$
$$\begin{eqnarray*}
\int{\dfrac{\mathrm dx}{3x^2+2}} &=& \displaystyle \frac{1}{2}\int{\dfrac{\mathrm dx}{(\sqrt{\frac{3}{2}}x)^2+1}}\\
&=& \displaystyle \frac{1}{2}\sqrt{\frac{2}{3}}\int{\dfrac{\mathrm dt}{t^2+1}}\\
&=& \displaystyle \frac{1}{2}\sqrt{\frac{2}{3}} \arctan(t)+\text C\\
&=&\frac{\sqrt{2}}{2\sqrt{3}}\arctan\left(\sqrt{\frac{3}{2}}x\right)+\text C\\
&=&\frac{1}{\sqrt{6}}\arctan{\left(\sqrt{\frac{3}{2}}x\right)}+\text C
\end{eqnarray*}$$
| {
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How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$ How can I show using mathematical induction that $\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \frac{2^n - 1}{2^n}$
Edit: I'm specifically stuck on showing that $\frac{2^n - 1}{2^n} + \frac{1}{2^{n+1}} = \frac{2^{n+1}-1}{2^{n+1}}$.
What should the approach be here?
| To prove that
$$\frac{1}{2} + \frac{1}{4} + \cdots + \frac{1}{2^n} = \sum_{k=1}^n \frac{1}{2^k} = \frac{2^n - 1}{2^n}$$
We can use two steps with telescopic induction.
This form of induction works as follows: To prove that
$$\sum_{k=1}^n f(k)=g(n)$$
Induction can be used to prove this equality by confirming the following two steps:
*
*See if $f(1)=g(1)$ (base case)
*See if $g(n+1)-g(n)=f(n+1)$
We shall let $f(x)=\frac{1}{2^x}$ and $g(x)=\frac{2^x - 1}{2^x}$. We confirm that
$$f(1)=\frac{1}{2^1}=g(1)=\frac{2^1-1}{2^1}=\frac{1}{2}$$
We then confirm that step 2 is true:
$$
g(n+1)-g(n)=
\frac{2^{n+1}-1}{2^{n+1}}-\frac{2^{n}-1}{2^{n}}=
\frac{2^{n+1}-1}{2^{n+1}}-\frac{2(2^{n}-1)}{2^{n+1}}=
\frac{2^{n+1}-1-2^{n+1}+2}{2^{n+1}}=
\frac{1}{2^{n+1}}=
f(n+1)
$$
Thus confirming the equality.
| {
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If $x$ and $y$ are rational numbers and $x^5+y^5=2x^2y^2,$ then $1-xy$ is a perfect square. Prove that if $x, y$ are rational numbers and
$$ x^5 +y^5 = 2x^2y^2$$
then $1-xy$ is a perfect square.
| I would be so tempted to divide by $x^2 y^2$, so I would consider the following cases:
Case a:
if $x=0$ then that implies $y=0$.
Case b:
if $y=0$ then that would imply $x=0$
Case c:
$x \neq 0, y \neq 0$
Thus dividing by $x^2y^2$ will be legal
$$
\begin{align*}
\frac{x^3}{y^2} - 2 + \frac{y^3}{x^2} = 0 \\
x\left(\frac{x}{y}\right)^2 - 2 +y\left(\frac{y}{x}\right)^2 = 0
\end{align*}
$$
Substitute $u=\left(\frac{x}{y}\right)^2$
$$
\begin{align*}
xu - 2 + \frac{y}{u} = 0\\
xu^2-2u+y=0
\end{align*}
$$
This is a quadratic in $u$ and since $x$ and $y$ are rationals, $u$ is rational. The discriminant is $4(1-xy)$, which has to be a perfect square for $u$ to be rational.
Thus $1-xy$ is a perfect square.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Show that $\forall n \in \mathbb{N} \left ( \left [(2+i)^n + (2-i)^n \right ]\in \mathbb{R} \right )$ Show that $\forall n \in \mathbb{N} \left ( \left [(2+i)^n + (2-i)^n \right ]\in \mathbb{R} \right )$
My Trig is really rusty and weak so I don't understand the given answer:
$(2+i)^n + (2-i)^n $
$= \left ( \sqrt{5} \right )^n \left (\cos n\theta + i \sin n\theta \right ) + \left ( \sqrt{5} \right )^n \left (\cos (-n\theta) + i \sin (-n\theta) \right ) $
$= \left ( \sqrt{5} \right )^n \left ( \cos n\theta + \cos (-n\theta) + i \sin n\theta + i \sin (-n\theta) \right ) $
$= \left ( \sqrt{5} \right )^n 2\cos n\theta$
Could someone please explain this?
| There are two ways to write a complex number: rectangular form, e.g., $x+iy$, and polar form, e.g., $re^{i\theta}$. The conversion between them uses trig functions: $$re^{i\theta}=r\cos\theta+ir\sin\theta\;.\tag{1}$$ Going in the other direction, $$x+iy=\sqrt{x^2+y^2}\,e^{i\theta}\;,$$ where $\theta$ is any angle such that $$\cos\theta=\frac{x}{\sqrt{x^2+y^2}}\;\text{ and }\sin\theta=\frac{y}{\sqrt{x^2+y^2}}\;.$$ The important thing for your argument is that $r=\sqrt{x^2+y^2}$.
The $r$ corresponding to $2+i$ is therefore $\sqrt{2^2+1^2}=\sqrt5$, and that corresponding to $2-i$ is $\sqrt{2^2+(-1)^2}=\sqrt5$ as well. The angles for $2+i$ is an angle $\theta$ whose cosine is $\frac2{\sqrt5}$ and whose sine is $\frac1{\sqrt5}$, while the angle for $2-i$ is an angle whose cosine is $\frac2{\sqrt5}$ and whose sine is $-\frac1{\sqrt5}$. It doesn’t matter exactly what they are; the important thing is that if we let the first be $\theta$, the second is $-\theta$, since $$\cos(-\theta)=\cos\theta\;\text{ and }\sin(-\theta)=-\sin\theta\;.$$
Substituting into $(1)$ gives you $$2+i=\sqrt5\cos\theta+i\sqrt5\sin\theta=\sqrt5(\cos\theta+i\sin\theta)=\sqrt5 e^{i\theta}$$ and $$2-i=\sqrt5\cos(-\theta)+i\sqrt5\sin(-\theta)=\sqrt5(\cos\theta-i\sin\theta)=\sqrt5 e^{-i\theta}\;.$$
Now use the fact that it’s easy to raise an exponential to a power:
$$\begin{align*}
(2+i)^n+(2-i)^n&=(\sqrt5)^n\left(e^{i\theta}\right)^n+(\sqrt5)^n\left(e^{-i\theta}\right)^n\\
&=(\sqrt5)^n\left(e^{in\theta}+e^{-in\theta}\right)\\
&=(\sqrt5)^n\Big(\big(\cos n\theta+i\sin n\theta\big)+\big(\cos(-n\theta)+i\sin(-n\theta)\big)\Big)\\
&=(\sqrt5)^n\Big(\cos n\theta+i\sin n\theta+\cos n\theta-i\sin n\theta\Big)\\
&=(\sqrt5)^n 2\cos n\theta\;.
\end{align*}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
Calculating $\tan15\cdot \tan 30 \cdot \tan 45 \cdot \tan 60 \cdot \tan 75$ What is $\tan15\cdot \tan 30 \cdot \tan 45 \cdot \tan 60 \cdot \tan 75$ equal to (in degrees)?
Here is how I tried to solve it:
I assumed $x = \tan 15 (x= 0.27)$, so I rewrote it as:
$x\cdot 2x \cdot 3x \cdot 4x \cdot 5x = 120x^5$
$0.27^5 = 0.001323$
$120 \cdot 0.001323 = 0.16$
But Google Calculator gives me $-1.19576279$
What is the right way to calculate this trig expression?
| EDIT : Arturo's solution is really cool. Use that instead, I'll let my solution stay just as a reference.
$$\tan(2A) \neq 2\times \tan(A)$$
You might want to read/revise double angle formulas.
Assuming you need to do this without a calculator,
$$\begin{align*}
\tan 30 &= \frac{1}{\sqrt{3}}\\
\tan 45 &= 1\\
\tan 60 &= \sqrt{3}\\
\tan 60 &= \sqrt{3}\\
\tan(A+B) &= \frac{\tan(A)+\tan(B)}{1-\tan(A)\tan(B)},\\
\tan(15+15) &= \tan(30) \\
&= \frac{1}{\sqrt{3}} \\
&= \frac{\tan(15)+\tan(15)}{1-\tan(15)\tan(15)}
\end{align*}$$
Solving $\frac{2x}{1-x^2} = \frac{1}{\sqrt{3}}$ will give you x = 0.2679
Similarly, you can write $\tan(75) = \tan(90-15)$ and you can work out the product.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating $ \int_0^{2\pi} \frac{5}{2}|\sin(2t)| dt$ I'm stuck on integration of the following function:
$$ \int_0^{2\pi} \frac{5}{2}|\sin(2t)| dt$$
I understand the few first steps with substitution, etc., but I can't get the end result which is "10".
Could someone give me a step by step solution for it?
| We have
$$\begin{equation}
\int^{t=2 \pi}_{t=0} \frac{5}{2} |\text{sin} 2t | \, dt = \frac{5}{2} \int^{t=2 \pi}_{t=0} |\text{sin} 2t | \, dt
\end{equation}$$
Now use the substitution $y = 2t$, $dy = 2dt$. The boundaries shift as follows: if $t = 2\pi$ then $y = 4\pi$ and if $t = 0$ then $y = 0$
We plug this into the integral to get
$$\begin{equation}
\frac{5}{2} \int^{t=2 \pi}_{t=0} |\text{sin} 2t | \, dt = \frac{5}{2} \int^{y=4 \pi}_{y=0} |\text{sin } y | \, \frac{dy}{2} = \frac{5}{4} \int^{y=4 \pi}_{y=0} |\text{sin } y | \, dy
\end{equation}$$
It remains to remember that sin$x \geq 0$ for $x \in [0,\pi]$, and sin$x \leq 0$ when $ x \in [\pi, 2\pi]$. Finally, we can also use that sin$x$ is periodic, which means that sin$(x + 2\pi) = $ sin$x$.
So we get
$$\begin{equation}
\frac{5}{4} \int^{y=4 \pi}_{y=0} |\text{sin } y | \, dy = 2 \left(\frac{5}{4} \int^{y=2 \pi}_{y=0} |\text{sin } y | \, dy \right) = 2 \left(\frac{5}{4} \int^{y= \pi}_{y=0} \text{sin } y \, dy - \frac{5}{4} \int^{y= 2 \pi}_{y=\pi} \text{sin } y \, dy \right)
\end{equation}$$
You can now compute the integral to obtain the desired result.
| {
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If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$. In this (btw, nice) answer to Twin primes of form $2^n+3$ and $2^n+5$, it was said that:
If $n\equiv 2\pmod 3$, then $7\mid 2^n+3$?
I'm not familiar with these kind of calculations, so I'd like to see, if my answer is correct:
*
*Let $n=3k+2$ so then
$2^{3k+2}+3\equiv 2^{3k+2}-4 \equiv 4\left(2^{3k}-1\right)\equiv \phantom{4\cdot } 0 \pmod 7$.
*Reading a bit about Modular arithmetics, I found that
$
a_1 \equiv b_1 \pmod N $ and $ a_2 \equiv b_2 \pmod N
$
then
$
a_1 a_2 \equiv b_1 b_2 \pmod n$, if $a_i$ and $b_i$ are integers.
*Since we have $4\equiv 4 \pmod 7$, I conclude that $2^{3k}-1 \equiv 0 \pmod 7$.
*Finally we use that
$
2^{3n}-1=(2^3-1)\cdot \left(1+2^3+2^{2\cdot 3}+2^{3\cdot 3}+\cdots+2^{(k-1)3}\right)
$
and are done.
Are there alternative ways to prove it?
| In fact, we can prove a stronger result and the proof is easier. The result we will prove is that $$x^2+x+1 \text{ divides }x^{3k+2} + x+1$$ for all $k \in \mathbb{N}$. Setting $x=2$ gives the result, you are looking for.
The proof follows immediately from the remainder theorem since $(x^2+x+1) = (x-\omega)(x-\omega^2)$, where $\omega$ is the complex cube-root of unity.
(Remember that $(x-a)$ divides $f(x)$ if and only if $f(x) = 0$)
Plugging in $\omega$ in $x^{3k+2}+x+1$ gives us $\omega^{3k+2} + \omega + 1 = \omega^2 + \omega + 1 = 0$. This gives us that $(x- \omega)$ divides $x^{3k+2}+x+1$.
Similarly, plugging in $\omega^2$ in $x^{3k+2}+x+1$ gives us $\omega^{6k+4} + \omega^2 + 1 = \omega + \omega^2 + 1 = 0$. This gives us that $(x- \omega^2)$ divides $x^{3k+2}+x+1$.
Hence, $(x-\omega)(x-\omega^2) = x^2 + x + 1$ divides $x^{3k+2}+x+1$.
Setting $x=2$ gives us the result you want i.e. $2^2 + 2 + 1 = 7$ divides $2^{3k+2}+2+1 = 2^{3k+2}+3$.
EDIT (Deleted the other answer and merged with this)
Another way to prove $x^2+x+1 \text{ divides }x^{3k+2} + x+1$ for all $k \in \mathbb{N}$ is by induction. All that is needed for induction is that $$x^{3k+5} + x + 1 = x^3 \left(x^{3k+2} + x + 1\right) - x^4 - x^3 +x + 1\\
= x^3 \left(x^{3k+2} + x + 1\right) - x^3 (x+1) +x + 1\\
= x^3 \left(x^{3k+2} + x + 1\right) - (x^3-1) (x+1)$$
So if $x^2 + x +1$ divides $\left(x^{3k+2} + x + 1\right)$, then it also divides $x^{3k+5} + x + 1$ since $x^2 + x +1$ divides $x^3-1$. Also, the base case $k=0$ is trivially true since $x^2 + x + 1$ divides $x^{3 \times 0 + 2} + x + 1 = x^{2} + x + 1$.
Hence, $x^2 + x + 1$ divides $x^{3k+5} + x + 1$ forall $k \in \mathbb{N}$. Setting $x=2$ as before gives you what you want.
| {
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"timestamp": "2023-03-29T00:00:00",
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Folding a rectangular paper sheet You are given a rectangular paper sheet. The diagonal vertices of the sheet are brought together and folded so that a line (mark) is formed on the sheet. If this mark length is same as the length of the sheet, what is the ratio of length to breadth of the sheet?
This is my first question on this site, so if this is not a good question please help.
| $\hskip 2.2in$
The above figure was done using grapher on mac osx.
Let $l$ be the length (i.e. the sides $AD$ and $BC$) and $b$ be the breadth (i.e. the sides $AB$ and $CD$). Once you get the diagonal vertices together, i.e. when $D$ coincides with $B$, the length $EB$ is the same as the length $ED = l-a$.
Hence, for the right triangle, we have that $$a^2 + b^2 = (l-a)^2\\ b^2 = l^2 - 2al\\ a = \frac{l^2 - b^2}{2l}$$
The length of $BF$ is $l-a$ and is given by $$l-a = l - \frac{l^2 - b^2}{2l} = \frac{2l^2 - l^2 + b^2}{2l} = \frac{l^2 + b^2}{2l}$$
The distance between the two points is $$EF^2 = r^2 = b^2 + (l-2a)^2 = b^2 + \left(l - \frac{l^2 - b^2}{l} \right)^2 = b^2 + \frac{b^4}{l^2}$$
This is so since the vertical distance between $E$ and $F$ is $l-2a$.
You are given that $r = l$ and hence $$l^2 = b^2 + \frac{b^4}{l^2}$$If we let $$\frac{l}{b} = x,$$ then we get that $$x^2 = 1 + \frac1{x^2}\\ x^4 = x^2 + 1$$ which gives us that $$x = \sqrt{\frac{1}{2} (1+\sqrt{5})} = \sqrt{\phi}$$ where $\phi$ is the golden ratio.
| {
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Is $k=3$ the only solution to $\sum_{r=1}^n r^k =\left( \sum_{r=1}^nr^1 \right)^2 $? $f(k) = \sum_{r=1}^{n} r^k$. Find an integer $x$ that solves the equation $f(x) = \bigl(f(1)\bigr)^2$.
Problem credit: http://cotpi.com/p/2/
I understand why $x = 3$ is a solution. $1^3 + 2^3 + \dots + n^3 = \left(\frac{n(n + 1)}{2}\right)^2 = (1 + 2 + \dots + n)^2$. But how can we prove that there is no other solution?
Will we have other solutions if $x$ is real? What if $x$ is complex?
| I assume that you want to find values of $x$ for which $f(x) = (f(1))^2$ as functions in $n$. In this case, we can simply substitute in $n=1$, $n=2$, and $n=3$. Namely, we need to satisfy $$1^x=1 \qquad \text{and} \qquad 1^x + 2^x = 9 \qquad \text{and} \qquad 1^x + 2^x + 3^x = 36$$ Subtracting successive equations yields $$2^x = 8 \qquad \text{and} \qquad 3^x = 27$$ Obviously, $x = 3$ is the only real value which works. You might wonder whether there are any complex values. Suppose $x=a+bi$. Then $$8=2^x = 2^a2^{bi} = 2^ae^{ib\ln{2}}$$ Matching magnitudes, we need $2^a=3$, so that $a = 3$. Meanwhile, notice that $b$ must take the form $\frac{2\pi}{\ln{2}}m$ for some integer $m$. If we continue this analysis for the equation $3^x=27$, we also need that $b = \frac{2\pi}{\ln{3}}n$ for some integer $n$. Setting these expressions for $b$ equal and solving yields $$m \ln{3} = n\ln{2},$$ or equivalently, $$3^m = 2^n,$$ which is only true when $m=n=0$. Therefore, $b=0$, and so $x=3$ is the only value for which $f(x) = (f(1))^2$ for every $n$.
| {
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How to show that $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $ using induction? I am attempting a question, where I have to show $n(n^2 + 8)$ is a multiple of 3 where $n\geq 1 $.
I have managed to solve the base case, which gives 9, which is a multiple of 3.
From here on,
I have $(n+1)((n+1)^2 + 8)$
$n^3 + 3n^2 + 11n + 9$
$n(n^2 + 8) + 3n^2 + 3n + 9$
How can I show that $3n^2 + 3n + 9$ is a multiple of 3?
| Since you have proven that this formula is available for $n=1$ we suppose that formula is available for all natural numbers n=k
$$k(k^2+8)$$ then according to axiomme of mathematical induction we need to prove that formula is valid for $n=k+1$ or
$$(k +1)((k+1)^2+8)$$ is multiple of 3
now we have
$$(k+1)((k+1)^2+8)=(k+1)(k^2+2k+1+8)=k^3+3k^2+11k+9=(k^3+8k)+(3k^2+3k+9)=k(k^2+8)+3(k^2+k+3)$$
first part of expression is factor of 3 by assumption an second part evidently is factor of 3
| {
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Show that $11^{n+1}+12^{2n-1}$ is divisible by $133$. Problem taken from a paper on mathematical induction by Gerardo Con Diaz. Although it doesn't look like anything special, I have spent a considerable amount of time trying to crack this, with no luck. Most likely, due to the late hour, I am missing something very trivial here.
Prove that for any integer $n$, the number $11^{n+1}+12^{2n-1}$ is divisible by $133$.
I have tried multiplying through by $12$ and rearranging, but like I said with meager results. I arrived at $11^{n+2}+12^{2n+1}$ which satisfies the induction hypothesis for LHS, but for the RHS I got stuck at $12 \times 133m-11^{n+1}-11^{n+3}$ or $12^2 \times 133m-12 \times11^{n+1}-11^{n+3}$ and several other combinations none of which would let me factor out $133$.
| In fact, we can prove a stronger result and the proof is easier. The result we will prove is that $$x^2+x+1 \text{ divides }x^{n+1} + (x+1)^{2n-1}$$ for all $n \in \mathbb{N}$. Setting $x=11$ gives the result, you are looking for.
The proof follows immediately from the remainder theorem since $(x^2+x+1) = (x-\omega)(x-\omega^2)$, where $\omega$ is the complex cube-root of unity.
(Remember that $(x-a)$ divides $f(x)$ if and only if $f(x) = 0$)
Plugging in $\omega$ in $x^{n+1} + (x+1)^{2n-1}$ gives us $$\omega^{n+1} + (\omega+1)^{2n-1} = \omega^{n+1} + (-\omega^2)^{2n-1} = \omega^{n+1} - \omega^{4n-2} = \omega^{n+1} - \omega^{3n} \omega^{n-2}\\
=\omega^{n+1} - 1 \times \omega^{n-2} = \omega^{n-2} \left( \omega^3 - 1\right) = 0$$
This gives us that $(x- \omega)$ divides $x^{n+1} + (x+1)^{2n-1}$.
Similarly, plugging in $\omega^2$ in $x^{n+1} + (x+1)^{2n-1}$ gives us $$\omega^{2n+2} + \left( \omega^2 + 1 \right)^{2n-1} = \omega^{2n+2} + (-\omega)^{2n-1} = \omega^{2n+2} - \omega^{2n-1}\\ = \omega^{2n-1} \left( \omega^3 - 1\right) = 0$$ This gives us that $(x- \omega^2)$ divides $x^{n+1} + (x+1)^{2n-1}$.
Hence, $(x-\omega)(x-\omega^2) = x^2 + x + 1$ divides $x^{n+1} + (x+1)^{2n-1}$.
Setting $x=11$ gives us the result you want i.e. $11^2 + 11 + 1 = 133$ divides $11^{n+1} + 12^{2n-1}$.
Note that the above result can also be proved by inducting on $n$.
| {
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Real roots of $3^{x} + 4^{x} = 5^{x}$ How do I show that $3^{x}+4^{x} = 5^{x}$ has exactly one real root.
| Rewrite our equation as
$$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x=1.$$
We have the familiar solution $x=2$.
If $x>2$, then $\left(\frac{3}{5}\right)^x \lt \left(\frac{3}{5}\right)^2$ and
$\left(\frac{4}{5}\right)^x \lt \left(\frac{4}{5}\right)^2$, and therefore
$$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x \lt 1.$$
Similarly, if $x<2$ then $\left(\frac{3}{5}\right)^x \gt \left(\frac{3}{5}\right)^2$
and $\left(\frac{4}{5}\right)^x \gt \left(\frac{4}{5}\right)^2$, so
$$\left(\frac{3}{5}\right)^x+\left(\frac{4}{5}\right)^x \gt 1.$$
| {
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Long division in integration by partial fractions I am trying to figure out what my book did, I can't make sense of the example.
"Since the degree of the numberator is greater than the degree of the denominator, we first perform the long division. This enables us to write
$$\int \frac{x^3 + x}{x -1} dx = \int \left(x^2 + x + 2 + \frac{2}{x-1}\right)dx = \frac{x^3}{3} + \frac{x^2}{2} + 2x + 2\ln|x-1| + C$$
I am mostly concerned with the transformation of the problem by long division I think.
I attempt to do this on my own.
$(x+1)$ and $(x^3 + x)$ inside the long division bracket
I am left with $x^2 - 1$ on top and a leftover -1
This is not in their answer, I do not know how they did that.
| You want to express $x^3+x$ as $(x-1)(\text{something})+r$.
I want to show you a way of solving this problem with a homemade technique. We see that the "something" must be a polynomial of degree $2$, or either we'll be getting $x^4$ which we don't want.
$$x^3+x=(x-1)(ax^2+bx+c)+r$$
If we multiply out we get
$$x^3+x=ax^3+bx^2+cx-ax^2-bx-c+r$$
Now, we equate the coefficients in each side:
$$1x^3=ax^3$$
$$0x^2=bx^2-ax^2$$
$$1x=cx-bx$$
$$0=r-c$$
What the above means is that two polynomials are equal iff their coefficients are equal.
From the above we get $a=1$, so
$$0x^2=bx^2-x^2$$
$$1x=cx-bx$$
$$0=r-c$$
Thus $b=1$.
$$1x=cx-1x$$
Then $c=2$, and finally
$$0=r-3$$
So$r=2$. So we get what we wanted
$$x^3+x=(x-1)(x^2+x+2)+2$$
Dividing by $x-1$ gives what your book has
$$\frac{x^3+x}{x-1}=x^2+x+2+\frac{2}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154008",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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} |
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle
This question came up in a miscellaneous problem set I have been working on to refresh my memory on several topics I did earlier this year. I have tried changing $4\sin(A)\sin(B)\sin(C)$ to $$4\sin(B+C)\sin(A+C)\sin(A+B)$$ by making substitutions by reorganizing $A+B+C=\pi$. I then did the same thing to the other side to get $$-2(\sin(B+C)\cos(B+C)+\sin(A+C)\cos(A+C)+\sin(A+B)\cos(A+B))$$ and then tried using the compound angle formula to see if i got an equality. However the whole thing became one huge mess and I didn't seem to get any closer to the solution. I am pretty sure there is some simpler way of proving the equality, but I can't seem to figure it out. Maybe there is a geometric interpretation or maybe it can be done using just algebra and trig. Any hint's would be appreciated (I would prefer an algebraic approach, but it would be nice to see some geometric proofs as well)
| Since $A,B,C$ are angles of a triangle, we have that $A+B+C = \pi$. Recall the following trigonometric identities
\begin{align}
\sin(\pi-\theta) & = \sin(\theta)\\
\cos(\pi-\theta) & = -\cos(\theta)\\
\sin(2\theta) + \sin(2\phi) & = 2 \sin(\theta + \phi) \cos(\theta-\phi)\\
\sin(2\theta) & = 2\sin(\theta) \cos(\theta)\\
\cos(\theta - \phi) - \cos(\theta + \phi) & = 2 \sin(\theta) \sin(\phi)
\end{align}
We have that
\begin{align}
\sin(2A) + \sin(2B) & = 2 \sin(A+B) \cos(A-B)\\
& = 2 \sin(\pi-C) \cos(A-B)\\
& = 2 \sin(C) \cos(A-B)
\end{align}
Hence,
\begin{align}
\sin(2A) + \sin(2B) + \sin(2C) & = 2 \sin(C) \cos(A-B) + 2 \sin(C) \cos(C)\\
& = 2 \sin(C) \left(\cos(A-B) + \cos(C) \right)\\
& = 2 \sin(C) \left(\cos(A-B) + \cos(\pi-(A+B)) \right)\\
& = 2 \sin(C) \left(\cos(A-B) - \cos(A+B) \right)\\
& = 2 \sin(C) \times 2 \sin(A) \sin(B)\\
& = 4 \sin(A) \sin(B) \sin(C)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 3,
"answer_id": 1
} |
Polar equation of a circle A very long time ago in algebra/trig class we did polar equation of a circle where
$r = 2a\cos\theta + 2b\sin\theta$
Now I forgot how to derive this. So I tried using the standard form of a circle.
$$(x-a)^2 + (y - b)^2 = a^2 + b^2$$
$$(a\cos\theta - a)^2 + (b\sin\theta - b)^2 = a^2 + b^2$$
$$(a^2\cos^2 \theta + a^2 - 2a^2\cos\theta) + (b^2\sin^2 \theta + b^2 - 2b^2\sin\theta) = a^2 + b^2$$
$$a^2\cos^2 \theta + b^2\sin^2 \theta - 2a^2 \cos\theta - 2b^2 \sin\theta = 0$$
Now I am stuck, I think I was supposed to complete the square or something. Could someone finish my thought?
| Although it's already been answered, it seemed like a fun way to procrastinate on my homework.
$(x-a)(x-a) + (y-b)(y-b) = aa + bb$
Substitute $x = rcos(θ)$ and $y = rsin(θ)$.
$(rcos(θ) - a)(rcos(θ) - a) + (rsin(θ) - b)(rsin(θ) - b) = aa + bb$
Multiply using FOIL.
$rrcos(θ)^2 - 2arcos(θ) + aa + rrsin(θ)^2 - 2brsin(θ) + bb = aa + bb$
Rearrange the terms.
$rrcos(θ)^2 + rrsin(θ)^2 -2arcos(θ) - 2brsin(θ) + aa + bb = aa + bb$
Factor out $r^2$. Subtract $a^2$ from both sides. Subtract $b^2$ from both sides:
$rr(cos(θ)^2 + sin(θ)^2) -2arcos(θ) - 2brsin(θ) = 0$
Simplify left side by using the trig identity $sin(θ)^2 + cos(θ)^2 = 1$.
$rr(1) - 2arcos(θ) - 2brsin(θ) = 0$
Add $2arcos(θ)$ and $2brsin(θ)$ to both sides.
$rr(1) = 2arcos(θ) + 2brsin(θ)$
Simplify the left side. Factor $r$ out of the right side.
$rr = r(2acos(θ) + 2bsin(θ))$
Divide both sides by $r$.
$(rr)/r = 2(acos(θ) + bsin(θ))$
Simplify the left side.
$r = 2acos(θ) + 2bsin(θ)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/154550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 4
} |
How to solve this quartic equation? For the quartic equation:
$$x^4 - x^3 + 4x^2 + 3x + 5 = 0$$
I tried Ferrari so far and a few others but I just can't get its complex solutions. I know it has no real solutions.
| Let $f(x) = x^4 - x^3 +4x^2 +3x+5$. Once you know that $f(x)$ doesn't have real solutions, try some easy complex numbers like $i$, $\omega$, other $n^{th}$ roots of unity etc. Note that if we plug in $x = \omega$, where $\omega$ is the complex cube-root of unit, we get that $$f(\omega) = \omega^4 - \omega^3 +4 \omega^2 + 3 \omega + 5 = \omega - 1 + 4 \omega^2 + 3 \omega +5 = 4(\omega^2 + \omega + 1) = 0.$$
Hence, $x^2 + x + 1$ divides $f(x)$. Hence, $f(x) = (x^2 + x + 1) (x^2 + ax + b)$. This gives us that $$f(x) = x^4 + (a+1)x^3 + (1 + a + b)x^2 +(a+b)x + b$$
Hence, $a=-2$ and $b=5$. Hence, $$f(x) = (x^2 + x + 1) (x^2 - 2x + 5)$$
Hence, the roots are $$x = \omega, \omega^2, 1 \pm 2i$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Simplify these expressions with radical sign 2 My question is
1) Rationalize the denominator:
$$\frac{1}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$
My answer is:
$$\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{18}$$
My question is
2) $$\frac{1}{\sqrt{2}+\sqrt{3}-\sqrt{5}}+\frac{1}{\sqrt{2}-\sqrt{3}-\sqrt{5}}$$
My answer is: $$\frac{1}{\sqrt{2}}$$
I would also like to know whether my solutions are right.
Thank you,
| *
*Your answer is almost correct.
Multiplying by$$\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{\sqrt{2}+\sqrt{3}-\sqrt{5}}$$ and simplifying will give your that answer:
$$\frac{\sqrt{2}+\sqrt{3}-\sqrt{5}}{2 \sqrt 6}=\frac{\sqrt{12}+\sqrt{18}-\sqrt{30}}{12}$$
2. Your answer is correct.
Multiplying the first fraction by $$\frac{\sqrt{2}+\sqrt{3}+\sqrt{5}}{\sqrt{2}+\sqrt{3}+\sqrt{5}}$$ And the second by $$\frac{\sqrt{2}-\sqrt{3}+\sqrt{5}}{\sqrt{2}-\sqrt{3}+\sqrt{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the square root of the polynomial My question is:
Find the square root of the polynomial-
$$\frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left(\frac{x}y + \frac{y}x\right) + 3$$
| $$\frac{x^2}{y^2} + \frac{y^2}{x^2} - 2\left(\frac{x}y + \frac{y}x\right) + 3$$
$$=\frac{{x^4} + {y^4} - 2\left({x^2} +{y^2}\right)xy + 3{x^2}{y^2}}{{x^2}{y^2}}$$ $$=\frac{{x^4} + {y^4} + 2{x^2}{y^2}+ {x^2}{y^2} - 2\left({x^2} +{y^2}\right)xy }{{x^2}{y^2}}$$ $$=\frac{({x^2}+{y^2})^2+ {(xy)^2} - 2\left({x^2} +{y^2}\right)xy }{{x^2}{y^2}}$$ $$=\frac{({x^2}+{y^2} - xy)^2}{{(xy)^2}}$$
the square root of the polynomial = $$\pm \frac{{x^2}+{y^2} - xy}{xy}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/155981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Leslie Matrix characteristic polynomial I´m having problems to prove the Leslie matrix characteristic polynomial.
I have to prove that the characteristic polynomial is:
$$
\ λ^{n}-a_{1}λ^{n-1}-a_{2}b_{1}λ^{n-2}-a_{3}b_{1} b_{2}λ^{n-3} - ... -
a_{n}b_{1} b_{2}...b_{n-1}\
$$
I would apreciate some light!
| I'm assuming (from Wikipedia) that the matrix is $$\begin{pmatrix} a_1 & a_2 & a_3 & a_4 & ... & a_n \\ b_1 & 0 & 0 & 0 & ... & 0 \\ 0 & b_2 & 0 & 0 & ... & 0 \\0 & 0 & b_3 & 0 & ... & 0 \\ 0 & 0 & 0 & .. & .. & 0 \\ 0 & 0 & 0 & ... & b_{n-1} & 0\end{pmatrix}$$
Use induction on $n$. For $n = 1$ the proof is easy.
For $n > 1$ use Laplace expansion on the farthest right column to get
$$\det \begin{pmatrix} t - a_1 & -a_2 & -a_3 & -a_4 & ... & -a_n \\ -b_1 & t & 0 & 0 & ... & 0 \\ 0 & -b_2 & t & 0 & ... & 0 \\0 & 0 & -b_3 & t & ... & 0 \\ 0 & 0 & 0 & .. & .. & 0 \\ 0 & 0 & 0 & ... & -b_{n-1} & t\end{pmatrix}$$
= $$t *\det \begin{pmatrix} t - a_1 & -a_2 & -a_3 & -a_4 & ... & -a_{n-1} \\ -b_1 & t & 0 & 0 & ... & 0 \\ 0 & -b_2 & t & 0 & ... & 0 \\0 & 0 & -b_3 & t & ... & 0 \\ 0 & 0 & 0 & .. & .. & 0 \\ 0 & 0 & 0 & ... & -b_{n-2} & t\end{pmatrix}$$
$$ + (-1)^n * \det \begin{pmatrix} -b_1 & t & 0 & ... & 0 \\ 0 & -b_2 & t & ... & 0 \\ 0 & 0 & -b_3 & ... & 0 \\ 0 & 0 & 0 & ... & t \\ 0 & 0 & 0 & ... & -b_{n-1} \end{pmatrix}$$ $$= t(t^{n-1} -a_1 t^{n-2} - a_2 b_1 t^{n-2} - ... - a_{n-1}b_1b_2...b_{n-2}) - b_1 b_2...b_{n-1}$$ $$= t^n - a_1 t^{n-1} - a_2 b_1 t^{n-2} - ... - a_n b_1 b_2 ... b_{n-1}$$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/158877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Euler's product formula for $\sin(\pi z)$ and the gamma function I want to derive Euler's infinite product formula
$$\displaystyle \sin(\pi z) = \pi z \prod_{k=1}^\infty \left( 1 - \frac{z^2}{k^2} \right)$$
by using Euler's reflection equation $\Gamma(z)\Gamma(1-z) \sin(\pi z) = \pi$ and the definition of $\Gamma(z)$ as an infinite product, namely
$$\displaystyle \Gamma(z) := \frac{1}{z} \prod_{k=1}^\infty \frac{(1+\frac{1}{k})^z}{1+\frac{z}{k}}.$$
To be precise, I obtain that
$$\sin(\pi z) = \pi z(1-z) \left( \prod_{k=1}^\infty \frac{1+\frac{z}{k}}{(1+\frac{1}{k})^z} \right) \left( \prod_{k=1}^\infty \frac{1+\frac{1-z}{k}}{(1+\frac{1}{k})^{1-z}} \right)$$
hence I wish to prove
$$(1-z) \left( \prod_{k=1}^\infty \frac{1+\frac{z}{k}}{(1+\frac{1}{k})^z} \right) \left( \prod_{k=1}^\infty \frac{1+\frac{1-z}{k}}{(1+\frac{1}{k})^{1-z}} \right) = \prod_{k=1}^\infty \left( 1 - \frac{z^2}{k^2} \right).$$
I multiplied things out and got it to the form
$$(1-z) \prod_{k=1}^\infty \frac{1 + \frac{1}{k} + \frac{z(1-z)}{k^2}}{1 + \frac{1}{k}} = (1-z) \prod_{k=1}^\infty \left( 1 + \frac{\frac{z(1-z)}{k}}{1+\frac{1}{k}}\right)$$ however the $(1-z)$ factor out front is giving me some trouble; I'm not sure how to proceed.
| Use the fact that $\Gamma (1-z) = -z\, \Gamma(-z)$ and then:
$$\Gamma(1-z)\Gamma(z) = -z \, \Gamma(-z)\Gamma(z) = -z \cdot \frac{1}{-z}\cdot \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{\left(1 + \frac{z}{k} \right)\left(1 - \frac{z}{k} \right) } = \frac{1}{z} \prod_{k=1}^{+\infty} \frac{1}{1 - \frac{z^2}{k^2}} = \frac{\pi}{\sin \pi z}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
A simple quadratic inequality For positive integers $n\ge c\ge 5$, why does
$$c+2(n-c)+\frac{(n-c)^2}{4}\le\frac{(n-1)^2}{4}+1\text{ ?}$$
| To avoid fractions, we multiply the left-hand side by $4$, obtaining
$$(n-c)^2+8(n-c)+4c.$$
Complete the square. We get
$$(n-c+4)^2 +4c -16.$$
Now calculate $[(n-1)^2 +4]-[(n-c+4)^2 +4c -16]$.
The difference of squares factors as $(c-5)(2n-c+3)$, so
$$\begin{align}[(n-1)^2 +4]-[(n-c+4)^2 +4c -16]&=(c-5)(2n-c+3)-4(c-5)\\
&=(c-5)(2n-c-1).\end{align}$$
The condition $c \ge 5$ ensures that $c-5\ge 0$. And since $n \ge c$, the term $2n-c-1$ is positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/159997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Linear algebra: power of diagonal matrix? Let A = $\begin{pmatrix} 3 & -5 \\ 1 & -3 \end{pmatrix}$. Compute $A^{9}$. (Hint: Find a matrix P such that $P^{-1}AP$ is a diagonal matrix D and show that $A^{9}$= $PD^{9}P^{-1}$
Answer: $\begin{pmatrix} 768 & -1280 \\ 256 & -768 \end{pmatrix}$
I keep getting $\begin{pmatrix} -768 & 1280 \\ -256 & 768 \end{pmatrix}$ but could it be still right? I have D=$\begin{pmatrix} -2 & 0\\ 0& 2\end{pmatrix}$ and P =$\begin{pmatrix} 5&1 \\1 & 1 \end{pmatrix}$ with $P^{-1}$= $\begin{pmatrix} \frac{1}{4} & \frac{1}{-4} \\ \frac{1}{-4} & \frac{5}{4} \end{pmatrix}$
| Your answer is not correct. Please note that the eigenvectors should be corresponding to the eigenvalues. So, if you choose $$D=\left(
\begin{array}{cc}
-2 & 0 \\
0& 2
\end{array}
\right),$$ then your $P$ should be $$P=\left(
\begin{array}{cc}
1 & 5 \\
1& 1
\end{array}
\right),$$
because $(1,1)$ is the eigenvector corresponding to $-2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/162516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How to find the sum of this infinite series. How to find the sum of the following series ?
Kindly guide me about the general term, then I can give a shot at summing it up.
$$1 - \frac{1}{4} + \frac{1}{6} -\frac{1}{9} +\frac{1}{11} - \frac{1}{14} + \cdots$$
| Using the principal value for the doubly infinite harmonic series yields
$$
\begin{align}
\sum_{k=0}^{\infty} \left(\dfrac1{5k+1} - \dfrac1{5k+4}\right)
&=\frac15\sum_{k=-\infty}^\infty\frac{1}{k+1/5}\\
&=\frac{\pi}{5}\cot\left(\frac{\pi}{5}\right)\\
&=\frac{\pi}{5}\sqrt{\frac{5+2\sqrt{5}}{5}}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/163165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Given number of trailing zeros in n!, find out the possible values of n. It's quite straightforward to find out number of trailing zeros in n!.
But what if the reverse question is asked?
n! has 13 trailing zeros, what are the possible values of n ?
How should we approach the above question ?
| Write $n$ in base $5$ as $n = a_0 + 5a_1 + 25a_2 + 125a_3 + \cdots$ where $0 \leq a_k \leq 4$. $$\left\lfloor \dfrac{n}5\right\rfloor = a_1 + 5a_2 + 25a_3 + \cdots$$
$$\left\lfloor \dfrac{n}{25}\right\rfloor = a_2 + 5a_3 + \cdots$$
$$\left\lfloor \dfrac{n}{125}\right\rfloor = a_3 + \cdots$$
Hence, $$\left\lfloor \dfrac{n}5\right\rfloor + \left\lfloor \dfrac{n}{25}\right\rfloor + \left\lfloor \dfrac{n}{125}\right\rfloor + \cdots = a_1 + 6 a_2 + 31 a_3 + \cdots$$
Note that the coefficients of the terms from $a_3$ are greater than $13$. Since the sum must give us $13$, we get that $a_k = 0$, for all $k \geq 3$. Hence, we need to find the number of integer solutions to $a_1 + 6a_2 =13$ with $0 \leq a_1,a_2 \leq 4$. The constraint $0 \leq a_1,a_2 \leq 4$ further implies that $a_2$ cannot be $0$ and $1$. Hence, $a_2 = 2$. This gives us $a_1 = 1$.
Hence, $n = a_0 + 5a_1 +25a_2 = a_0 + 5 \times 1 + 25 \times 2 = 55+a_0$ where $a_0 \in \{0,1,2,3,4\}$. Hence, the desired values of $n$ are $$\{55,56,57,58,59\}$$
The same idea in principle will work when the trailing number of zeros is any number not just $13$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/163388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Solving a complex integral I need help solving an integral from John Conway book.
Lets $\alpha$ complex number different from 1 find integral $$\int\frac{dx}{1-2\alpha\cos{x}+{\alpha}^2}$$ from 0 to $2\pi$ in unit circle $$(z-\alpha)^{-1}(z-\frac{1}{\alpha})^{-1}$$
| Let $t=\tan(x/2) $ so that $\cos x = \frac{1-t^2}{1+t^2}$ and $dx=\frac{2 dt}{1+t^2}.$ The integral becomes $$ \int \frac{1}{1-2\alpha \frac{1-t^2}{1+t^2} + \alpha^2} \frac{2 dt}{1+t^2}=2 \int \frac{dx}{(\alpha-1)^2+ (\alpha+1)^2 t^2}= \frac{1}{\alpha^2-1} \tan^{-1} \left( \frac{(\alpha+1)t}{\alpha-1} \right)+ C= \frac{1}{\alpha^2-1} \tan^{-1} \left( \frac{(\alpha+1)\tan(x/2)}{\alpha-1} \right)+ C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/164192",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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A problem dealing with even perfect numbers. Question: Show that all even perfect numbers end in 6 or 8.
This is what I have. All even perfect numbers are of the form $n=2^{p-1}(2^p -1)$ where $p$ is prime and so is $(2^p -1)$.
What I did was set $2^{p-1}(2^p -1)\equiv x\pmod {10}$ and proceeded to show that $x=6$ or $8$ were the only solutions.
Now, $2^{p-1}(2^p -1)\equiv x\pmod {10} \implies 2^{p-2}(2^p -1)\equiv \frac{x}{2}\pmod {5}$, furthermore there are only two solutions such that $0 \le \frac{x}{2} < 5$. So I plugged in the first two primes and only primes that satisfy. That is if $p=2$ then $\frac{x}{2}=3$ when $p=3$ then $\frac{x}{2}=4$. These yield $x=6$ and $x=8$ respectively. Furthermore All solutions are $x=6+10r$ or $x=8+10s$.
I would appreciate any comments and or alternate approaches to arrive at a good proof.
| Note that the powers of $2$ are congruent to $2$, $4$, $8$, or $6$, according to whether the exponent is congruent to $1$, $2$, $3$, or $0$ modulo $4$.
Assume $p\equiv 1\pmod{4}$. Then $2^{p-1}\equiv 6\pmod{10}$, and $2^p-1\equiv 1\pmod{10}$, so the product is congruent to $6$ modulo $10$.
If $p\equiv 3\pmod{4}$, then $2^{p-1}\equiv 4\pmod{10}$, and $2^p-1\equiv 7\pmod{10}$, so the product is congruent to $8$ modulo $10$.
Finally, if $p=2$, then $2(3)=6$.
| {
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How to calculate $\int_{-a}^{a} \sqrt{a^2-x^2}\ln(\sqrt{a^2-x^2})\mathrm{dx}$ Well,this is a homework problem.
I need to calculate the differential entropy of random variable
$X\sim f(x)=\sqrt{a^2-x^2},\quad -a<x<a$ and $0$ otherwise. Just how to calculate
$$
\int_{-a}^a \sqrt{a^2-x^2}\ln(\sqrt{a^2-x^2})\,\mathrm{d}x
$$
I can get the result with Mathematica,but failed to calculate it by hand.Please give me some idea.
| This integral can be performed via differentiation under the integral sign. First note that for $|x|\leq1$ we have $\ln \sqrt{1-x^2} = \frac12\ln (1-x^2)$. Moreover, simple application of the chain rule yields
$$ \frac{d}{d\alpha} (1-x^2)^\alpha = (1-x^2)^\alpha \ln(1-x^2) .$$
The remaining integral is a special case of the beta function with $x=1/2$ and $y=\alpha+1$. Thus, we have
$$\int_0^1\!dx\,(1-x^2)^\alpha =
\frac12\int_0^1\!dy\,y^{1/2} (1-y)^\alpha= \frac{\sqrt{\pi} \Gamma(1+\alpha)}{2 \Gamma(\frac{3}{2} + \alpha)}.$$
The original integral, we obtain by taking the derivative with respect to $\alpha$ and afterwards setting $\alpha=1/2$;
$$
\begin{align}\int_{-a}^a\!dx\,\sqrt{a^2-x^2} \ln\sqrt{a^2-x^2}
&=a^{2} \int_{0}^1\!dx\,\sqrt{1-x^2} [\ln a^2 + \ln(1-x^2)]\\
&= \frac{a^2 \pi \log a}{2}+ a^{2} \frac{d}{d\alpha} \int_0^1\!dx\,(1-x^2)^\alpha \Big|_{\alpha=1/2}\\
&= \frac{a^2 \pi \log a}{2}+ a^{2} \frac{d}{d\alpha} \frac{\sqrt{\pi} \Gamma(1+\alpha)}{2 \Gamma(\frac{3}{2} + \alpha)} \Big|_{\alpha=1/2} \\
&= \frac{a^2 \pi \log a}{2}+ a^{2} \frac{\Gamma(1+\alpha)}{\Gamma(3/2+\alpha)} [\psi(1+\alpha)-\psi(\tfrac{3}{2}+\alpha)]\Big|_{\alpha=1/2} \\
&=\frac{a^2 \pi \log a}{2}+ \frac{a^{2}\pi}4(1-2 \ln 2).
\end{align}$$
Where we used the special values of the $\Gamma$ and the $\psi = (\log \Gamma)'$ at integer and half-integer values.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Finding two numbers given their sum and their product
Which two numbers when added together yield $16$, and when multiplied together yield $55$.
I know the $x$ and $y$ are $5$ and $11$ but I wanted to see if I could algebraically solve it, and found I couldn't.
In $x+y=16$, I know $x=16/y$ but when I plug it back in I get something like $16/y + y = 16$, then I multiply the left side by $16$ to get $2y=256$ and then ultimately $y=128$. Am I doing something wrong?
| We are trying to solve the system of equations $x+y=16$, $xy=55$. Here are a couple of systematic approaches that work in general.
Approach $1$: We will use the identity $(x+y)^2-4xy=(x-y)^2$. In our case, we have $(x+y)^2=256$, $4xy=220$, so $(x-y)^2=36$, giving $x-y=\pm 6$.
Using $x+y=16$, $x-y=6$, we get by adding that $2x=22$, and therefore $x=11$. It follows that $y=5$.
The possibility $x+y=16$, $x-y=-6$ gives nothing new. Adding, we get $2x=10$, so $x=5$, and therefore $y=11$.
Approach $2$: From $x+y=16$, we get $y=16-x$. Substitute for $y$ in $xy=55$. We get $x(16-x)=55$. Simplification gives $x^2-16x+55=0$. The quadratic factors as $(x-5)(x-11)$, so our equation becomes $(x-5)(x-11)=0$, which has the solutions $x=5$ and $x=11$.
But we cannot necessarily rely on there being such a straightforward factorization. So in general after we get to the stage $x^2-16x+55=0$, we would use the Quadratic Formula. We get
$$x=\frac{16\pm\sqrt{(-16)^2-4(55)}}{2}.$$
Compute. We get the solutions $x=5$ and $x=11$. The corresponding $y$ are now easy to find from $x+y=16$.
Remarks: $1,$ Recall that the Quadratic Formula says that if $a\ne 0$, then the solutions of the quadratic equation $ax^2+bx+c=0$ are given by
$$x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}.$$
Your approach was along reasonable lines, but things went wrong in the details. From $xy=55$ we get $x=\frac{55}{y}$. Substituting in the formula $x+y=16$, we get
$$\frac{55}{y}+y=16.$$
A reasonable strategy is to multiply through by $y$, getting $55+y^2=16y$, or equivalently $y^2-16y+55=0$. Now we have reached a quadratic equation which is basically the same as the one we reached above.
$2.$ The first approach that we used (presented as an algorithm, and stripped of algebraic notation) goes back to Neo-Babylonian times. The "standard" problem was to find the dimensions of a door, given its perimeter and area.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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} |
Find all real solutions to $8x^3+27=0$
Find all real solutions to $8x^3+27=0$
$(a-b)^3=a^3-b^3=(a-b)(a^2+ab+b^2)$
$$(2x)^3-(-3)^3$$ $$(2x-(-3))\cdot ((2x)^2+(2x(-3))+(-3)^2)$$ $$(2x+3)(4x^2-6x+9)$$
Now, to find solutions you must set each part $=0$. The first set of parenthesis is easy $$(2x+3)=0 ; x=-\left(\frac{3}{2}\right)$$
But, what I do not know is how to factor a trinominal (reverse of the FOIL method)
I know that $(a+b)(c+d)=(ac+ad+bc+bd)$. But coming up with the reverse does not make sense to me. If someone can only tell me how to factor a trinomial that would be great.
| You are working too hard. Note that
$$8x^3+27=0\iff x^3=\frac{-27}{8}\iff x=-\sqrt[3]{27/8}\iff x=-\frac{3}{2}$$
and so the only real solution is $x=-3/2$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
Solve for $x$; $\tan x+\sec x=2\cos x;-\infty\lt x\lt\infty$
$$\tan x+\sec x=2\cos x$$
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x$$
$$\left(\dfrac{\sin x+1}{\cos x}\right)=2\cos x$$
$$\sin x+1=2\cos^2x$$
$$2\cos^2x-\sin x+1=0$$
Edit:
$$2\cos^2x=\sin x+1$$
$$2(1-\sin^2x)=\sin x+1$$
$$2\sin^2x+\sin x-1=0$$
$\sin x=a$
$$2a^2+a-1=0$$
$$(a+1)(2a-1)=0$$
$$a=-1,\dfrac{1}{2}$$
$$\arcsin(-1)=-90^\circ=-\dfrac{\pi}{2}$$
$$\arcsin\left(\dfrac{1}{2}\right)=30^\circ=\dfrac{\pi}{6}$$
$$180^\circ-30^\circ=150^\circ=\dfrac{5 \pi}{6}$$
$$x=\dfrac{\pi}{6},-\dfrac{\pi}{2},\dfrac{5 \pi}{6}$$
I actually do not know if those are the only answers considering my range is infinite:$-\infty\lt x\lt\infty$
| You correctly applied the definitions of $\tan$ and $\sec$ to go from the equation in the problem statement to
$$\left(\dfrac{\sin x}{\cos x}\right)+\left(\dfrac{1}{\cos x}\right)=2\cos x.$$
However, you combined the two fractions incorrectly; in general
$$\frac{a}{c}+\frac{b}{c}=\frac{a+b}{c}$$
For example, you know that
$$\frac{1}{4}+\frac{1}{4}=\frac{2}{4}=\frac{1}{2}$$
and that $$\frac{1}{4}+\frac{1}{4}\neq\frac{1}{4}.$$
The correct form of the equation should be
$$2\cos^2x-\sin x-1=0.$$
You will want to use the fact that for any $x$,
$$\sin^2(x)+\cos^2(x)=1.$$
Thus, any occurrence of $\cos^2(x)$ can be replaced with $(1-\sin^2(x))$. $$2(1-\sin^2(x))-\sin(x)-1=0$$ $$[2\cdot1-2\cdot\sin^2(x)]-\sin(x)-1=0$$ $$-2\cdot\sin^2(x)+(-1)\sin(x)-1+2=0$$ $$(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$$
Then, you will need to use the quadratic formula to solve for $\sin(x)$. It may help to write $y=\sin(x)$ temporarily, to prevent confusion. Keep in mind that there may be two different possible values of $\sin(x)$ that result.
$$(-2)\sin^2(x)+(-1)\sin(x)+(1)=0$$ $$(-2)y^2+(-1)y+(1)=0$$ $$y=\frac{-(-1)\pm\sqrt{(-1)^2-4(-2)(1)}}{2(-2)}=\frac{1\pm\sqrt{1+8}}{-4}=\frac{1\pm 3}{-4}=\begin{cases}\frac{1+3}{-4}=\frac{4}{-4}=\fbox{$-1$} &\text{ or }\\\\ & \\ \frac{1-3}{-4}=\frac{-2}{-4}=\fbox{$\frac{1}{2}$}.\end{cases}$$
Lastly, you need to find those values of $x$ that produce that value (or those values) of $\sin(x)$. Keep in mind that the function $\sin(x)$ is periodic; here is part of its graph to illustrate:
(and it continues similarly out to infinity in each direction) so that there are always infinitely many values of $x$ that will produce a given value of $\sin(x)$.
For any real number $x$, the values of
$$\sin(x+2\pi k)$$
are all identical for any integer $k$,. There are usually even more values you can plug into $\sin$ that will give the same number (look at the graph for an idea).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/171792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving:$\tan(20^{\circ})\cdot \tan(30^{\circ}) \cdot \tan(40^{\circ})=\tan(10^{\circ})$ how to prove that : $\tan20^{\circ}.\tan30^{\circ}.\tan40^{\circ}=\tan10^{\circ}$?
I know how to prove
$ \frac{\tan 20^{0}\cdot\tan 30^{0}}{\tan 10^{0}}=\tan 50^{0}, $
in this way:
$ \tan{20^0} = \sqrt{3}.\tan{50^0}.\tan{10^0}$
$\Longleftrightarrow \sin{20^0}.\cos{50^0}.\cos{10^0} = \sqrt{3}.\sin{50^0}.\sin{10^0}.\cos{20^0}$
$\Longleftrightarrow \frac{1}{2}\sin{20^0}(\cos{60^0}+\cos{40^0}) = \frac{\sqrt{3}}{2}(\cos{40^0}-\cos{60^0}).\cos{20^0}$
$\Longleftrightarrow \frac{1}{4}\sin{20^0}+\frac{1}{2}\sin{20^0}.\cos{40^0} = \frac{\sqrt{3}}{2}\cos{40^0}.\cos{20^0}-\frac{\sqrt{3}}{4}.\cos{20^0}$
$\Longleftrightarrow \frac{1}{4}\sin{20^0}-\frac{1}{4}\sin{20^0}+\frac{1}{4}\sin{60^0} = \frac{\sqrt{3}}{4}\cos{60^0}+\frac{\sqrt{3}}{4}\cos{20^0}-\frac{\sqrt{3}}{4}\cos{20^0}$
$\Longleftrightarrow \frac{\sqrt{3}}{8} = \frac{\sqrt{3}}{8}$
Could this help to prove the first one and how ?Do i just need to know that $ \frac{1}{\tan\theta}=\tan(90^{\circ}-\theta) $ ?
| In a word, yes. You already know that (in degrees) $\tan 20\cdot\tan30=\tan10\cdot\tan50$ so
$$\tan20\cdot\tan30\cdot\tan40 = \tan10\cdot\tan50\cdot\tan40$$
and your observation that
$$\frac{1}{\tan40}=\tan50$$
is all you need.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/172182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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} |
show that $x^2+y^2=z^5+z$ Has infinitely many relatively prime integral solutions How to show that this equation:
$$x^2+y^2=z^5+z$$
Has infinitely many relatively prime integral solutions
| The number $z^4+1$ is a sum of two relatively prime squares. Let $z$ be the sum of two relatively prime squares. Then the product $(z^4+1)z$ is a sum of two squares, by the Brahmagupta identity
$$(s^2+t^2)(u^2+v^2)=(su\pm tv)^2+(sv\mp tu)^2.$$
Now we take care of the relatively prime part. Suppose that $m$ has a representation as a sum of two squares, but no primitive representation. Then $m$ is divisible by $4$ or by some prime of the form $4k+3$. In our case, primes of the form $4k+3$ are irrelevant. And if $z$ is a sum of two relatively prime squares, then $(z^4+1)z$ cannot be divisible by $4$.
So pick for example $z$ a power of $5$, or a prime of the form $4k+1$.
| {
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Prove that $\left \{ \frac{n}{n+1}\sin\frac{n\pi}{2} \right \}$ is divergent. I want to prove whether the sequence $\{a_n\} = \left \{ \dfrac{n}{n+1}\sin\dfrac{n\pi}{2} \right \}$ (defined for all positive integers $n$) is divergent or convergent.
I suspect that it diverges, because the $\sin\frac{n\pi}{2}$ factor oscillates between -1, 0 and 1.
Here is my attempt to prove that it is divergent:
Suppose that the sequence is convergent. This should lead to a contradiction. If it is convergent, then, for every $\epsilon>0$, there is an $N>0$ such that if $n>N$, then $\left|\dfrac{n}{n+1}\sin\dfrac{n\pi}{2} -L\right|<\epsilon$.
In particular, for $\epsilon = \frac{1}{2}$:
$$-\frac{1}{2}<\frac{n}{n+1}\sin\frac{n\pi}{2} -L<\frac{1}{2} \text{ for every integer }n>N$$
Notice that the sequence of values of $\sin\dfrac{n\pi}{2}$ is $1, 0, -1, 0, 1, 0, -1, \cdots$ for $n=1, 2, 3, 4, 5, 6, 7, \cdots$. If $n = 1, 5, 9 \cdots$, $\sin\frac{n\pi}{2}$ is $1$, so, if we choose a particular $n$ from this list of values, the value of $a_n$ is $\dfrac{n}{n+1}$. In this case, $\sin\dfrac{(n+2)\pi}{2}$ will be $-1$, therefore $a_{n+2} = -\dfrac{n+2}{n+3}$. So, we have that:
$$-\dfrac{1}{2}<\dfrac{n}{n+1}-L<\dfrac{1}{2} \text{ and } -\dfrac{1}{2}<-\dfrac{n+2}{n+3}-L<\dfrac{1}{2}$$
Rearranging the terms:
$$\dfrac{1}{2}>-\dfrac{n}{n+1}+L>-\dfrac{1}{2} \text{ and } \dfrac{1}{2}>\dfrac{n+2}{n+3}+L>-\dfrac{1}{2}$$
$$\dfrac{n}{n+1}+\dfrac{1}{2}>L>\dfrac{n}{n+1}-\dfrac{1}{2} \text{ and } \dfrac{1}{2}-\dfrac{n+2}{n+3}>L>-\dfrac{n+2}{n+3}-\dfrac{1}{2}$$
The first inequality says that $L$ is positive, but the second inequality says that $L$ is negative; if this is correct so far, then I found a contradiction, proving that the sequence is divergent.
My question is: is this correct or is there some inconsistency?
| Here is another approach:
show that there exists infinitely many terms within $\delta$ neighbourhood of 0,1 and -1.
in other words 0,1 and -1 are all limits so there is no single limit for this sequence.
| {
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"timestamp": "2023-03-29T00:00:00",
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Evaluating $\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$ Consider $z \in \mathbb{C}$ and
$$\int_{|z|=1} \frac{\sin(z^2)}{ \left( \sin(z) \right)^2} dz.$$
How would we integrate this?
| $$\text{If}\,\,f(z)=\frac{\sin z^2}{\sin^2z}\,\,,\,\text{then}\,\,Res_{z=0}(f)=\lim_{z\to0}\frac{d}{dz}\left(z^2\frac{\sin z^2}{\sin^2z}\right)=0$$
So the singularity at $\,z=0\,$ is in fact a removable one and thus the integral equals zero.
We can also use power series in a tiny neighbourhood of zero:
$$\frac{\sin z^2}{\sin^2z}=\frac{z^2-\frac{z^6}{3!}+\cdots}{\left(z-\frac{z^3}{3!}+\cdots\right)^2}=\frac{z^2\left(1-\frac{z^4}{3!}+\cdots\right)}{z^2\left(1-\frac{z^2}{3}+\cdots\right)}=\left(1-\frac{z^4}{3!}+\cdots\right)\left(1+\frac{z^4}{3}+\cdots\right)$$
the rightmost parentheses being the power series for $\,\displaystyle{\frac{1}{1-z}\,\,,\,|z|<1}\,$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Minimum value of given expression What is the minimum value of the $$ \frac {x^2 + x + 1 } {x^2 - x + 1 } \ ?$$
I have solved by equating it to m and then discriminant greater than or equal to zero and got the answer, but can algebraic manipulation is possible
| For $x\geq0$ we have $$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}\geq1.$$
For $x<0$ by AM-GM we obtain:
$$\frac{x^2+x+1}{x^2-x+1}=1+\frac{2x}{x^2-x+1}=1+\frac{2}{x+\frac{1}{x}-1}=$$
$$=1-\frac{2}{-x+\frac{1}{-x}+1}\geq1-\frac{2}{2\sqrt{-x\cdot\frac{1}{-x}}+1}=\frac{1}{3}.$$
The equality occurs for $-x=\frac{1}{-x}$ or for $x=-1,$ which says that we got a minimal value.
By the same way we can get a maximal value if you want.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/174905",
"timestamp": "2023-03-29T00:00:00",
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Gradient And Hessian Of General 2-Norm Given $f(\mathbf{x}) = \|\mathbf{Ax}\|_2 = (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}$,
$\nabla f(\mathbf{x}) = \frac {\mathbf{A}^\mathrm{T} \mathbf{Ax}} {\|\mathbf{Ax}\|_2} = \frac {\mathbf{A}^\mathrm{T} \mathbf{Ax}} {(\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}}$
$\nabla^2 f(\mathbf{x}) = \frac { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2} \cdot \mathbf{A}^\mathrm{T} \mathbf{A} - (\mathbf{A}^\mathrm{T} \mathbf{Ax})^\mathrm{T} (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{-1/2} \mathbf{A}^\mathrm{T} \mathbf{Ax} } {(\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} ) } = \frac { \mathbf{A}^\mathrm{T} \mathbf{A} } { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{1/2}} - \frac {\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{A} \mathbf{A}^\mathrm{T} \mathbf{Ax} } { (\mathbf{x}^\mathrm{T} \mathbf{A}^\mathrm{T} \mathbf{Ax} )^{3/2} }$
I guess I am looking for confirmation that I have done the above correctly. The dimensions match up except for the second term of the Hessian is a scalar, which makes me think that something is missing.
Edit: Also, the last equality reduces to
$\nabla^2 f(\mathbf{x}) = \frac {\mathbf{A}^\mathrm{T} \mathbf{A} - \nabla f(\mathbf{x})^\mathrm{T} \nabla f(\mathbf{x})} {\|\mathbf{Ax}\|_2}$
| It is easier to work with $\phi(x) = \frac{1}{2} f^2(x)$. Just expand $\phi$ around $x$.
$\phi(x+\delta) = \frac{1}{2} (x + \delta)^T A^T A (x + \delta) = \phi(x) + x^TA^TA \delta + \frac{1}{2} \delta^T A^T A \delta$. It follows from this that the gradient $\nabla \phi(x) = A^T A x$, and the Hessian is $H = A^TA$.
To finish, let $g(x) = \sqrt{2x}$, and note that $f = g \circ \phi$. To get the first derivative, use the composition rule to get
$D f(x) = Dg(\phi(x)) D \phi(x)$, which gives $Df(x) = \frac{1}{\sqrt{2 \phi(x)}} x^T A^T A = \frac{1}{\|Ax\|} x^T A^T A$.
Let $\eta(x) = \frac{1}{\|Ax\|}$, and $\gamma(x) = x^T A^T A$, and note that $D f(x) = \eta(x) \cdot \gamma(x)$, so we can use the product rule. Let $h(x) = Df(x)$ then the product rule gives $D h(x) (\delta) = (D \eta(x) (\delta)) \gamma(x) + \eta(x) D \gamma(x) (\delta)$.
Expanding this yields: $Dh(x)(\delta) = (- \frac{1}{\|Ax\|^2} \frac{1}{\|Ax\|} x^T A^T A \delta) x^T A^T A + \frac{1}{\|Ax\|} \delta^T A^T A $. Noting that $x^T A^T A \delta = \delta^T A^T A x$, we can write this as:
$$Dh(x)(\delta) = \delta^T(\frac{1}{\|Ax\|} A^T A - \frac{1}{\|Ax\|^3 } A^T A x x^T A^T A),$$
or alternatively:
$$D^2 f(x) = \frac{1}{\|Ax\|} A^T A - \frac{1}{\|Ax\|^3 } A^T A x x^T A^T A .$$
The only difference with the formula given in the question is that the latter dyad was written incorrectly (instead of the dyad $g g^T$, you have the scalar $g^T g$).
| {
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Proving Quadratic Formula purplemath.com explains the quadratic formula. I don't understand the third row in the "Derive the Quadratic Formula by solving $ax^2 + bx + c = 0$." section. How does $\dfrac{b}{2a}$ become $\dfrac{b^2}{4a^2}$?
| I see no visual proof, so I will add one.
The rectangle below is broken up into to parts: a square and another rectangle. Note that the area of the original rectangle is given by the sum of the two smaller areas: $$x^2 + bx$$
The next step is to divide the smaller vertical rectangle by two, rotate the strip and add it to the bottom:
The new figure is a square with sides $x + \frac{b}{2}$ and a bite out of the bottom right. The bite is also a square, and has sides $\frac{b}{2}$. The area of this figure, then, is the area of the square, less the area of the bite:
$$(x + \frac{b}{2})^2 - (\frac{b}{2})^2$$
But this "square less a bite" figure must have the same area as the original figure, since it was obtained by cutting the original figure and rearranging the pieces. So we can equate the expressions for both these areas:
$$x^2 + bx = (x + \frac{b}{2})^2 - (\frac{b}{2})^2$$
At this point, you can satisfy yourself of the above equality by multiplying out the business on the right, simplifying, and showing that it becomes the expression on the left.
Now you have a way of rewriting quadratics of the form $x^2 + bx$ which will prove to be very useful very soon.
When we're looking for the roots of a quadratic equation, we are looking for the values of $x$ that make it zero. The equation will take on this form: $$Ax^2 + Bx + C=0$$ so let's make it look like $x^2 + bx$. First, divide everything by $a$:
$$x^2 + \frac{B}{A}x + \frac{C}{A} = 0$$
In the previous equations, the coefficent of $x$ was $b$. In this new equation, it's $\frac{B}{A}$. Let's rewrite the above equations, substituting
$\frac{B}{A}$ for $b$, and adding in $\frac{C}{A}$ to each side, so it all equals zero:
$$x^2 + \frac{B}{A}x +\frac{C}{A}= (x + \frac{B}{2A})^2 - (\frac{B}{2A})^2 +\frac{C}{A}=0$$
Rearrange a bit and begin to solve for $x$:
$$(x + \frac{B}{2A})^2 = (\frac{B}{2A})^2 -\frac{C}{A}= (\frac{B}{2A})^2 -\frac{4AC}{4A^2}=\frac{B^2-4AC}{4A^2}$$
Take the square root of both sides:
$$x + \frac{B}{2A} =\pm\frac{\sqrt{B^2-4AC}}{2A}$$
And, finally, bring $\frac{B}{2A}$ to the right:
$$x =\frac{-B\pm\sqrt{B^2-4AC}}{2A}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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What's wrong with this conversion? I need to calculate the following limes:
$$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
$$
My first intuition was that the answer is $x$, but after a bit of fiddling with the root I got thoroughly confused. I know that below conversion goes wrong somwhere, but where?
$$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
= \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2*n^2}{n^2}}
= \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2*n^2}}{n}
= \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n}
= 0
$$
| $$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
= \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}}
= \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n}
\neq \lim_{n\rightarrow\infty} \frac{\sqrt{\frac{1}{n^2}+x^2}}{n}
$$
$$
\lim_{n\rightarrow\infty} \sqrt{\frac{1}{n^2}+x^2}
= \lim_{n\rightarrow\infty} \sqrt{\frac{1+x^2 \cdot n^2}{n^2}}
= \lim_{n\rightarrow\infty} \frac{\sqrt{1+x^2 \cdot n^2}}{n}
= \lim_{n\rightarrow\infty} \frac{n\sqrt{\frac{1}{n^2}+x^2}}{n}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/177182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Prove $\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$ How would I prove the following?
$$\cos^2 x \,\sin^3 x=\frac{1}{16}(2 \sin x + \sin 3x - \sin 5x)$$
I do not know how to do do the problem I do know $\sin(3x)$ can be $\sin(2x+x)$ and such yet I am not sure how to commence.
| We know $e^{iy}=\cos y+i\sin y$
So, $\cos 5x+i\sin 5x=e^{i5x}=(e^{ix})^5=(\cos x+i \sin x)^5$
Using binomial theorem & equating imaginary parts,
$sin 5x = sin^5x − 10\cdot sin^3x cos^2x + 5\cdot sin x cos^4x$
$=sin^5x − 10\cdot sin^3x(1 - sin^2x) + 5\cdot sin x (1 - sin^2x)^2$
So, $sin 5x= 16\cdot\sin^5x − 20\cdot\sin^3x + 5\cdot\sin x$
We know, $\sin3x=3\sin x -4sin^3x $
$\sin^3x\cos^2x=\sin^3x(1-\sin^2x)=\sin^3x-\sin^5x$
Let $A\sin5x+B\sin3x+C\sin x=\sin^3x-\sin^5x$
Or, $A(16\cdot\sin^5x − 20\cdot\sin^3x + 5\cdot\sin x) + B(3\sin x -4sin^3x)+C\sin x$
$=\sin^3x-\sin^5x$
Comparing the coefficients of different powers of sinx,
5th power=>16A=-1=>A=$-\frac{1}{16}$
3rd power=>-20A-4B=1=>B=$\frac{1}{16}$
1st power=>5A+3B+C=0=>C=-(5A+3B)=$\frac{1}{8}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/179294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
How many positive values of $a$ are possible in $2^{3}\le a\lfloor a\rfloor \le 4^{2} + 1$ How many positive values of $a$ are possible in the following case?
$$2^{3}\le a\lfloor a\rfloor \le 4^{2} + 1$$
where $a\lfloor a\rfloor$ such that $a[a]$ is an integer.
| It is clear that $\lfloor a\rfloor$ must be $\ge 3$ and $\le 4$.
In the case $\lfloor a\rfloor=3$, the product can be $9$, $10$, or $11$. For $8$ is too small, since $a\ge \lfloor a\rfloor$. And $12$ is too big, since then $a=4$, giving the wrong value for the floor function. So $a$ can have values $3$, $10/3$, and $11/3$.
In the case $\lfloor a\rfloor=4$ there are two possible values of the product, $16$ and $17$.
So the number of possible values of $a$ is $5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/181092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
} |
$\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$. Find the positive integer solutions (x,y). $\frac{1}{x}+\frac{4}{y} = \frac{1}{12}$, where $y$ is and odd integer less than $61$.
Find the positive integer solutions (x,y).
| $y=\frac{48x}{x-12}$
Now $x-12$ must be multiple of 16, else y will be even.
Let x-12=16k=>$y=\frac{48(16k+12)}{16k}=>\frac{3(16k+12)}{k}=48+\frac{36}{k}$
So, k must divide 36 and must be of the form 4r, where r is an odd integer.
$1≤y≤61=>1≤48+\frac{36}{k}≤61$
$\frac{36}{k}≤13=>k≥3$.
But $16k=x-12=>x=12+16k>0=>k>-1$
$=>k≥3$
So, the possible values of k are 4,12,36.
$k=4=>y=48+9=57, x=16\cdot 4+12=76.$
$k=12=>y=48+3=51, x=16\cdot 12+12=204$
$k=36=>y=48+1=49, x=16\cdot 36+12=588$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Spivak Calculus 3rd ed. $|a + b| \leq |a| + |b|$ I'm working through the first chapter of Michael Spivak's Calculus 3rd ed.
Towards the end of the chapter he proves $ |a + b| ≤ |a| + |b| $ using the observation that $|a|= \sqrt{ a^2 }$ when $a$ is $ ≥ 0 $ .
$ |a + b| ≤ |a| + |b| $
$$ (|a + b|)^2 = (a + b)^2 $$ $$= a^2 + 2ab + b^2 $$
$$ ≤ a^2 + 2|a| |b| + b^2 $$
$$ = |a|^2 + 2|a| |b| + |b|^2 $$
$$ = (|a| + |b|)^2 $$
I am unsure about what's going on with the equality sign. How does it go from $=$ to $≤$ on line 3 when a and b are changed to their absolute value and back to $=$ again on line 4 when $a^2$ and $b^2$ are changed to their absolute values?
| Let $a=r_1(\cos A+i\sin A)$ and $b=r_2(\cos B+i\sin B)$
So, $|a|=r_1$ and $|b|=r_2$
Now, $|a+b|=\sqrt{(r_1\cos A + r_2\cos B)^2+(r_1\sin A + r_2\sin B)^2}$
$=\sqrt{r_1^2+r_2^2+2r_1r_2\cos(A-B)}$
$≤\sqrt{r_1^2+r_2^2+2r_1r_2}\ $ as $\cos(A-B)≤1$ for real A,B
$=r_1+r_2=|a|+|b|$
Also observe, $|a+b|=\sqrt{r_1^2+r_2^2+2r_1r_2\cos(A-B)}≥\sqrt{r_1^2+r_2^2-2r_1r_2}=||r_2|-|r_1||$ as $\cos(A-B)≥-1$ for real A,B
$|a+b|≥||b|-|a||$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to prove Lagrange trigonometric identity I would to prove that
$$1+\cos \theta+\cos 2\theta+\ldots+\cos n\theta =\displaystyle\frac{1}{2}+
\frac{\sin\left[(2n+1)\frac{\theta}{2}\right]}{2\sin\left(\frac{\theta}{2}\right)}$$
given that
$$1+z+z^2+z^3+\ldots+z^n=\displaystyle\frac {1-z^{n+1}}{1-z}$$
where $z\neq 1$.
I put $z=e^{i\theta}$. I already got in left hand side cos exp in real part, but there is a problem in the right hand side, I can't split imaginary part and real part. Please help me. Thanks in advance.
| Here's a variation of @lab's variation of @DonAntonio's solution:
$$\begin{align}
\frac{e^{(n+1)i\theta}-1}{e^{i\theta-1}-1}
&= \frac{e^{(n+1)i\theta}-1}{e^{i\theta/2} \left(e^{i\theta/2}-e^{-i\theta/2}\right)} \\
&= \frac{e^{(n+1/2)i\theta}-e^{-i\theta/2}}{2i\sin\frac{\theta}{2}} \\
&= \frac{\mathrm{cis}\frac{(2n+1)i\theta}{2}-\mathrm{cis}\frac{-\theta}{2}}{2i\sin\frac{\theta}{2}} \\
&= \frac{-i\;\left(\mathrm{cis}\frac{(2n+1)i\theta}{2}-\mathrm{cis}\frac{-\theta}{2}\right)}{2\sin\frac{\theta}{2}} \\
\end{align}
$$
where, of course, "$\mathrm{cis}\,x := \cos x + i \sin x$". The real part is then
$$\begin{align}
\frac{1}{2\sin\frac{\theta}{2}}\left(-i^2\sin\frac{(2n+1)\theta}{2}+i^2\sin\frac{-\theta}{2}\right)
&= \frac{1}{2\sin\frac{\theta}{2}}\left(\sin\frac{(2n+1)\theta}{2}+\sin\frac{\theta}{2}\right) \\
&= \frac{1}{2}\left(\frac{\sin\frac{(2n+1)\theta}{2}}{\sin\frac{\theta}{2}}+1\right)
\end{align}$$
In this variation, we avoid appealing to the lesser-known sine-times-cosine prosthaphaeresis identity (although that identity is helpful to know!).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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} |
Prove the inequality $a^2bc+b^2cd+c^2da+d^2ab \leq 4$ with $a+b+c+d=4$ Let $a,b,c$ and $d$ be positive real numbers such that $a+b+c+d=4.$
Prove the inequality
$$a^2bc+b^2cd+c^2da+d^2ab \leq 4 .$$
Thanks :)
| Let $\{a,b,c,d\}=\{x,y,z,t\}$, where $x\geq y\geq z\geq t$.
Hence, since $(x,y,z,t)$ and $(xyz,xyt,xzt,yzt)$ are the same ordered,
by Rearrangement and AM-GM we obtain:
$$a^2bc+b^2cd+c^2da+d^2ab=a\cdot abc+b\cdot bcd+c\cdot cda+d\cdot dab\leq$$
$$\leq x\cdot xyz+y\cdot xyt+z\cdot xzt+t\cdot yzt=xy(xz+yt)+zt(xz+yt)=$$
$$=(xy+zt)(xz+yt)\leq\left(\frac{xy+xz+zt+yt}{2}\right)^2=$$
$$=\left(\frac{(x+t)(y+z)}{2}\right)^2\leq\left(\frac{\left(\frac{x+y+z+t}{2}\right)^2}{2}\right)^2=4.$$
Done!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Finding all integers such that $a^2+4b^2 , 4a^2+b^2$ are both perfect squares Are there any integers $a,b$, such that:
$$a^2+4b^2 , 4a^2+b^2$$
are both perfect squares?
| Integers $m$ such that,
$$ma^2+b^2=d^2\\a^2+mb^2 = c^2\tag1$$
have solutions unfortunately do not have an official name. However, the $n$ of a similar system,
$$a^2+b^2 = c^2\\a^2+nb^2=d^2\tag2$$
are called concordant numbers and it is well-established it has positive solutions $a,b$ for the sequence,
$$n=1, 7, 10, 11, 17, 20, 22, 23, 24, 27, 30, 31, 34, 41, 42, 45, 47, 49,\dots$$
Let $m=r^2,$ and $a=p/r,\; b= q$, then,
$$p^2+q^2=s_1^2\\p^2+r^4q^2 = s_2^2\tag3$$
But since in your case we have $m=r^2=4,$ and $n=r^4=16$ is not a concordant number, then $(1)$ has no positive integer solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/184659",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
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} |
Prove inequality $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq 6$ Let $a,b,c>0$.
What is the proof that:
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{9\sqrt[3]{abc}}{a+b+c}\geq 6$$
| Take
$$ \frac{a}{b}+\frac{a}{b}+\frac{b}{c}\geq 3\frac{a}{ (abc)^\frac{1}{3}}$$
$$ \frac{b}{c}+\frac{b}{c}+\frac{c}{a}\geq 3\frac{b}{ (abc)^\frac{1}{3}}$$
$$ \frac{c}{a}+\frac{c}{a}+\frac{a}{b}\geq 3\frac{c}{ (abc)^\frac{1}{3}}$$
from AM-GM
and then add them and you get
$$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b+c}{(abc)^\frac{1}{3}}$$
So it suffices to prove that
$$\left ( \frac{(a+b+c)}{(abc)^\frac{1}{3}}\, \, \, \, \, \, +\frac{9 (abc)^\frac{1}{3}}{a+b+c}\, \, \, \, \,\right )\geq 6$$
which holds
from the basic inequality $x^2+y^2 \geq 2xy$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 1,
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Finding the sum of this alternating series with factorial denominator. What is the sum of this series?
$$ 1 - \frac{2}{1!} + \frac{3}{2!} - \frac{4}{3!} + \frac{5}{4!} - \frac{6}{5!} + \dots $$
| Alternatively, write it as:
$$1-\frac{1}{1!} +\frac{1}{2!} - \frac{1}{3!}... +\\
\left(-\frac{1}{1!}+ \frac{2}{2!} - \frac{3}{3!}...\right)$$
The first line is $e^{-1}$ and the second line, after cancelling terms, you see is $-e^{-1}$
More generally, if $$(z)_i = z(z-1)...(z-(i-1))$$ is the falling factorial, and $p(z) = a_0(z)_0 + a_1(a)_1 + ... a_k(z)_k$, then:
$$\sum_{n=0}^\infty \frac{p(n)}{n!} x^n =
e^x (a_0 + a_1x + a_2x^2 + ... a_k x^k)$$
In this case, $p(z) = 1 + z = (z)_0 + (z)_1$ so
$$\sum_{n=0}^\infty \frac{n+1}{n!} x^n =
e^x (1+x)$$
And, in particular, for $x=-1$, $$\sum_{n=0}^\infty \frac{(-1)^n(n+1)}{n!} = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/185915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Inequality:$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$ For $ a,b,c,d\geq 0 $ with $ a+b = c+d = 2 $, how to prove that $$ (a^{2}+c^{2})(a^{2}+d^{2})(b^{2}+c^{2})(b^{2}+d^{2})\leq 25$$
| We will make repeated use of the identity $(p^2+q^2)(r^2+s^2)=(pr-qs)^2+(ps+qr)^2$.
Applying it to the first two terms yields
$$
(a^2+c^2)(a^2+d^2)=(a^2-cd)^2+(a(c+d))^2=(a^2-cd)^2+(2a)^2 \, .
$$
Similarly, the third and fourth terms multiply to $(b^2-cd)^2+(2b)^2$.
Applying the identity again to these two quantities, we have
$$
\left[(a^2-cd)^2+(2a)^2\right]\left[(b^2-cd)^2+(2b)^2\right] \\
= \left[(a^2-cd)(b^2-cd)-4ab\right]^2+\left[(a^2-cd)(2b)+(b^2-cd)(2a)\right]^2 \\
=: Q^2+R^2 \, ,
$$
and we'll now work on simplifying $Q$ and $R$ separately.
First,
\begin{eqnarray}
Q &=& (a^2-cd)(b^2-cd)-4ab \\
&=& a^2b^2-(a^2+b^2)cd+c^2d^2-4ab \\
&=& a^2b^2-(a^2+2ab+b^2)cd+2abcd + c^2d^2-4ab \\
&=& a^2b^2-4cd+2abcd+c^2d^2-4ab \\
&=& ab(ab+cd-4)+cd(ab+cd-4)\\
&=& (ab+cd)(ab+cd-4) \, .
\end{eqnarray}
Next,
\begin{eqnarray}
R &=& (a^2-cd)(2b)+(b^2-cd)(2a) \\
&=& 2ba^2-2bcd+2ab^2-2acd \\
&=& 2ab(a+b)-2cd(a+b)\\
&=&4(ab-cd) \, .
\end{eqnarray}
So the quantity we're trying to bound is equal to
$$
\left[(ab+cd)(ab+cd-4)\right]^2+[4(ab-cd)]^2 \, .
$$
Let $x=ab$, $y=cd$. Then the constraints on $a,b,c,d$ imply that $x,y$ each vary freely over $[0,1]$. So, in order to prove the inequality, it suffices to show that
$$
f(x,y)=(x+y)^2(x+y-4)^2+16(x-y)^2
$$
is bounded above by 25 on the unit square.
To show this, we first make the temporary change of variables $s=x+y$, $t=x-y$; let $F(s,t)=s^2(s-4)^2+16t^2$ be the pullback of $f$ under this coordinate change. Then $F_s=2s(s-4)(2s-4)$ is nonvanishing for $0<s<2$. It follows that $F$ has no critical points on this strip, and thus that $f$ has no critical points on the interior of the unit square. So $f$ will attain its maximum somewhere on the boundary.
Since $f$ is symmetric in its arguments, it suffices to examine the functions $f_0(y)=f(0,y)$ and $f_1(x)=f(x,1)$. We have
$$f_0(y)=y^2(y-4)^2+16y^2=y^4-8y^3+32y^2 \\
f_0'(y)=4y^3-24y^2+64y=4y(y^2-6y+16)=4y((y-3)^2+7) \, ;
$$
thus $f_0'$ is nonzero on $(0,1)$, so $f_0$ has no critical points on $(0,1)$. Also,
$$
f_1(x)=(x+1)^2(x-3)^2+16(x-1)^2=((x-1)^2+1)^2\\
f_1'(x)=4(x-1)((x-2)^2+1)^2 \, ;
$$
thus $f_1'$ is also nonzero on $(0,1)$, and $f_1$ has no relevant critical points. So $f$ is maximized at one of the corners of the square.
Finally,
$$
f(0,0)=16 \\
f(0,1)=25 \\
f(1,1)=16
$$
and so the maximum value of $f$ is 25, as desired.
| {
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"timestamp": "2023-03-29T00:00:00",
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Second order non-linear differential equation $ y_n'' -nx\frac{1}{\sqrt {y_n}}=0$
Is there any known method to solve such second order non-linear differential equation?
What I tried to solve:
$ 2y_n'y_n''= 2nx\frac{y_n'}{\sqrt {y_n}}$
$ y_n'^2= 4nx\sqrt {y_n}-4n\int\sqrt {y_n} dx$
After that I could not see any way how to proceed.
Please advice what to do to solve the differential equation.
Many thanks
| In fact it belongs to an Emden-Fowler equation.
First, according to http://eqworld.ipmnet.ru/en/solutions/ode/ode0302.pdf or http://www.ae.illinois.edu/lndvl/Publications/2002_IJND.pdf#page=6 , all Emden-Fowler equations can be transformed into Abel equation of the second kind.
Let $\begin{cases}u=\dfrac{x^3}{y_n^\frac{3}{2}}\\v=\dfrac{x}{y_n}\dfrac{dy_n}{dx}\end{cases}$ ,
Then $\dfrac{dv}{du}=\dfrac{\dfrac{dv}{dx}}{\dfrac{du}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{1}{y_n}\dfrac{dy_n}{dx}-\dfrac{x}{y_n^2}\left(\dfrac{dy_n}{dx}\right)^2}{\dfrac{3x^2}{y_n^\frac{3}{2}}-\dfrac{3x^3}{2y_n^\frac{5}{2}}\dfrac{dy_n}{dx}}=\dfrac{\dfrac{x}{y_n}\dfrac{d^2y_n}{dx^2}+\dfrac{v}{x}-\dfrac{v^2}{x}}{\dfrac{3u}{x}-\dfrac{3uv}{2x}}=\dfrac{\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2}{3u\left(1-\dfrac{v}{2}\right)}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}=\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}+v-v^2$
$\dfrac{x^2}{y_n}\dfrac{d^2y_n}{dx^2}=3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v$
$\dfrac{d^2y_n}{dx^2}=\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)$
$\therefore\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)-\dfrac{nx}{\sqrt{y_n}}=0$
$\dfrac{y_n}{x^2}\left(3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v\right)=\dfrac{nx}{\sqrt{y_n}}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=\dfrac{nx^3}{y_n^\frac{3}{2}}$
$3u\left(1-\dfrac{v}{2}\right)\dfrac{dv}{du}+v^2-v=nu$
$3u\left(\dfrac{v}{2}-1\right)\dfrac{dv}{du}=v^2-v-nu$
Let $w=\dfrac{v}{2}-1$ ,
Then $v=2w+2$
$\dfrac{dv}{du}=2\dfrac{dw}{du}$
$\therefore6uw\dfrac{dw}{du}=(2w+2)^2-(2w+2)-nu$
$6uw\dfrac{dw}{du}=4w^2+6w+2-nu$
$w\dfrac{dw}{du}=\dfrac{2w^2}{3u}+\dfrac{w}{u}+\dfrac{2-nu}{6u}$
In fact, all Abel equation of the second kind can be transformed into Abel equation of the first kind.
Let $w=\dfrac{1}{z}$ ,
Then $\dfrac{dw}{du}=-\dfrac{1}{z^2}\dfrac{dz}{du}$
$\therefore-\dfrac{1}{z^3}\dfrac{dz}{du}=\dfrac{2}{3uz^2}+\dfrac{1}{uz}+\dfrac{2-nu}{6u}$
$\dfrac{dz}{du}=\dfrac{(nu-2)z^3}{6u}-\dfrac{z^2}{u}-\dfrac{2z}{3u}$
Please follow the method in http://www.hindawi.com/journals/ijmms/2011/387429/#sec2.
| {
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"timestamp": "2023-03-29T00:00:00",
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how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Possible Duplicate:
Proof that $\sum\limits_{k=1}^nk^2 = \frac{n(n+1)(2n+1)}{6}$?
Summation of natural number set with power of $m$
How to get to the formula for the sum of squares of first n numbers?
how can one find the value of the expression, $(1^2+2^2+3^2+\cdots+n^2)$
Let,
$T_{2}(n)=1^2+2^2+3^2+\cdots+n^2$
$T_{2}(n)=(1^2+n^2)+(2^2+(n-1)^2)+\cdots$
$T_{2}(n)=((n+1)^2-2(1)(n))+((n+1)^2-2(2)(n-1))+\cdots$
| In general, $$\sum_{i=1}^{n}i^{2} = \frac{n(n+1)(2n+1)}{6}.$$
A collection of proofs of this fact can be found here.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/188602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Proving inequality $\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}$ In the pdf which you can download here I found the following inequality which I can't solve it.
Exercise 2.1.11 Let $a,b,c \gt 0$. Prove that
$$\sqrt{\frac{2a}{b+c}}+\sqrt{\frac{2b}{c+a}}+\sqrt{\frac{2c}{a+b}} \leq \sqrt{3 \left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)}.$$
Thanks :)
| By C-S
$$\left(\sum_{cyc}\sqrt{\frac{a}{b+c}}\right)^2\leq\sum_{cyc}\frac{a}{a+c}\sum_{cyc}\frac{a+c}{b+c}.$$
Thus, it remains to prove that
$$\frac{3}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)\geq\sum_{cyc}\frac{a}{a+c}\sum_{cyc}\frac{a+c}{b+c}$$ or
$$\sum_{cyc}(3a^6c^3+3a^5b^4+6a^5c^4+a^6b^2c+2a^6c^2b+4a^5b^3c+4a^5c^3b+7a^4b^4c+$$
$$+a^5b^2c^2-11a^4b^3c^2-12a^4c^3b^2-8a^3b^3c^3)\geq0,$$
which is obviously true.
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/189140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Sum of product of Fibonacci numbers I want to calculate the sum of product of Fibonacci number for a given $n$. That is, for given $n$, say
$$F_1 F_n + F_2 F_{n-1} + F_3 F_{n-2} + F_4 F_{n-3} + F_5 F_{n-4} + \cdots.$$
what should be my approach.
| Let's use
$$
F_n = \frac{1}{\sqrt{5}} \left( \phi^n - (-1)^n \phi^{-n} \right)
$$
where $\phi$ is a Golden ratio constant $\phi = \frac{\sqrt{5}+1}{2}$. Now the sum reduces to the combination of geometric sums:
$$
\begin{eqnarray}
\sum_{k=1}^{n-1} F_k F_{n-k} &=& \frac{1}{5}\sum_{k=1}^{n-1} \left(\phi^k - (-1)^k \phi^{-k}\right)\left(\phi^{n-k} - (-1)^{n-k} \phi^{k-n}\right) \\ &=&
\frac{1}{5}\sum_{k=1}^{n-1} \left(\phi^n - (-1)^k \phi^{n-2k} - (-1)^{n-k} \phi^{2k-n} + (-1)^n \phi^{-n} \right) \\
&=& \frac{n-1}{5} \left(\phi^{n} + (-1)^n \phi^{-n}\right) - \frac{\phi^{n}}{5} \sum_{k=1}^{n-1} \left( -\phi^{-2}\right)^{k} - \frac{(-1)^{n}}{5} \phi^{-n} \sum_{k=1}^{n-1} \left(-\phi^2\right)^k
\end{eqnarray}
$$
Using the geometric sum $\sum_{k=1}^{n-1} x^k = \frac{x-x^n}{1-x}$ we get:
$$\begin{eqnarray}
\sum_{k=1}^{n-1} F_k F_{n-k} &=& \frac{n-1}{5} \left(\phi^{n} + (-1)^n \phi^{-n}\right) + \frac{2}{5} \frac{\phi^{n-1} + (-1)^n \phi^{1-n}}{\phi + \phi^{-1}} \\
&=& \frac{n-1}{5} L_n + \frac{2}{5} F_{n-1}
\end{eqnarray}
$$
where $L_n$ denotes $n$-th Lucas number.
Quick sanity check in Mathematica:
In[84]:= Table[
Sum[Fibonacci[k] Fibonacci[n - k], {k, 1, n - 1}] == (n - 1)/
5 LucasL[n] + 2/5 Fibonacci[n - 1], {n, 1, 6}]
Out[84]= {True, True, True, True, True, True}
The sum stated in the edited question is obtained by replacing $n$ with $n+1$ in the sum evaluated above:
$$
\sum_{k=1}^{n} F_k F_{n+1-k} = \frac{n}{5} L_{n+1} + \frac{2}{5} F_n
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/190443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
Solve $\sqrt{x-4} + 10 = \sqrt{x+4}$ Solve: $$\sqrt{x-4} + 10 = \sqrt{x+4}$$
Little help here? >.<
| We will assume that $x$ ranges over the reals $\ge 4$, to make sure that the square roots are real. Note that
$$\sqrt{x+4}-\sqrt{x-4}=\frac{(\sqrt{x+4}-\sqrt{x-4})(\sqrt{x+4}+\sqrt{x-4})}{\sqrt{x+4}+\sqrt{x-4}} =\frac{8}{\sqrt{x+4}+\sqrt{x-4}} .$$
For $x\ge 4$, $\sqrt{x+4}+\sqrt{x-4}\ge 2\sqrt{2}$. It follows that $\sqrt{x+4}-\sqrt{x-4}\le \dfrac{8}{2\sqrt{2}}=2\sqrt{2}$ for all $x\ge 4$. In particular, $\sqrt{x+4}-\sqrt{x-4}$ cannot be equal to $10$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
How to solve $x^3=-1$? How to solve $x^3=-1$? I got following:
$x^3=-1$
$x=(-1)^{\frac{1}{3}}$
$x=\frac{(-1)^{\frac{1}{2}}}{(-1)^{\frac{1}{6}}}=\frac{i}{(-1)^{\frac{1}{6}}}$...
| Set $\displaystyle x=re^{i \theta}$. So $\displaystyle r^3e^{i3\theta}= x^3= -1= e^{i \pi}$, hence $r^3=1$ and $3 \theta= \pi [2\pi]$. Finally, $r=1$ and $\displaystyle \theta = \frac{\pi}{3} \left[\frac{2\pi}{3} \right]$ so you get three solutions: $$x_1= e^{i \pi/3}= \frac{1}{2}+ i \frac{\sqrt{3}}{2}, \ x_2=e^{-i 2 \pi /3}= \frac{1}{2} - i \frac{\sqrt{3}}{2}, \ x_3= e^{i \pi}=-1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/192742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "27",
"answer_count": 8,
"answer_id": 0
} |
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