Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Finding $\iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\,\mathrm dx\mathrm dy$ where $D=\{(x,y)\in\mathbb{R}^2:\ 0\leq y\leq x\,,xy\leq 1\leq x+y-1\}$ How can I evaluate this double integral through a domain transformation?
$$\iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\,\mathrm dx\mathrm dy$$
where $$D=\{(x,y)\in\mathbb{R}^2:\ 0\leq y\leq x\,, xy\leq 1\leq x+y-1\}$$
The region $D$ is the red one in this picture:
I have tried to convert $D$ to polar coordinates but then, it come up the following integral:
$$\frac{1}{3}\int_0^{\frac{\pi}{4}}\cos 2t\left(\frac{1}{\sqrt{(\sin t\cos t)^3}}-\frac{8}{(\sin t+\cos t)^3}\right)dt$$
which I got stuck into.
Does someone any further ideas?
| I have found that is possible to solve this integral by the following transformation:
$$\begin{cases}
u=x+y\\
v=x-y\end{cases}\ \Rightarrow\
\begin{cases}
x=\frac{u+v}{2}\\
y=\frac{u-v}{2}\end{cases}$$
The integral does not converge:
$$\begin{eqnarray}
\iint_D\frac{x^2-y^2}{\sqrt{x^2+y^2}}\ dxdy &=& \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{\frac{(u+v)^2}{4}-\frac{(u-v)^2}{4}}{\sqrt{\frac{(u+v)^2}{4}+\frac{(u-v)^2}{4}}}\cdot \left(-\frac{1}{2}\right)\ dv=\\
&=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}du\int_{\sqrt{u^2-4}}^u\frac{v}{\sqrt{u^2+v^2}}\ dv =\\
&=&-\frac{\sqrt{2}}{4} \int_2^{+\infty}u\ du\int_{\sqrt{u^2-4}}^u 2v(u^2+v^2)^{-1/2}\ dv=\\
&=&-\frac{\sqrt{2}}{2} \int_2^{+\infty}u\cdot [\sqrt{u^2+v^2}]_{\sqrt{u^2-4}}^u\ du =\\
&=&-\int_2^{+\infty}u^2-u\sqrt{u^2-2}\ du=\\
&=&-\int_2^{+\infty}u^2\ du +\frac{1}{2}\int_2^{+\infty}2u(u^2-2)^{1/2}\ du=\\
&=&\lim\limits_{c\to +\infty}\left \{\left[-\frac{u^3}{3}\right]_2^c+\frac{1}{3}\left[\sqrt{(u^2-2)^3}\right]_2^c\right \}=\\
&=&\lim\limits_{c\to +\infty}\left(-\frac{c^3}{3}+\frac{8}{3}+\frac{1}{3}\sqrt{(c^2-2)^3}-\frac{2\sqrt{2}}{3}\right)=\\
&=&\frac{1}{3}\lim\limits_{c\to +\infty}(\sqrt{(c^2-2)^3}-c^3+8-2\sqrt{2})=-\infty\end{eqnarray}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2874801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that, for all $m\geq 2$, there exists $C_m>0$ such that, for all $k\geq 1$, $\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq C_m$ Define $\phi:\mathbb{N}\to \mathbb{N}$ by $\phi(k)=k^m$, where $m\geq 2$ is some fixed number. I want to investigate, if there exists $C_m>0$ such that $$\frac{\phi(k+2)-\phi(k+1)}{\phi(k+1)-\phi(k)}=\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq C_m$$ for all $k\geq 1$. Since
$$
\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}=\frac{\left ( \frac{1+2/k}{1+1/k} \right )^m-1}{1-\left ( \frac{1}{1+1/k} \right )^m},
$$
we define $f:(0,1]\to\mathbb{R}$ by
$$
f(x)=\frac{\left ( \frac{1+2x}{1+x} \right )^m-1}{1-\left ( \frac{1}{1+x} \right )^m}
$$
I suspect that $f(x)$ is increasing, and so $f(1/x)$ must be decreasing (through some calculator graphs), which implies that
$$
\frac{(k+2)^m-(k+1)^m}{(k+1)^m-k^m}\leq \frac{(1+2)^m-(1+1)^m}{(1+1)^m-1^m}=\frac{3^m-2^m}{2^m-1}=:C_m
$$
for all $k\geq 1$. The question is, how do I show the monotonicity of $f(x)$ or $f(1/x)$ in a simplest way? Determining $f'$ and then checking if it's greater than $0$ on $(0,1]$ takes a lot time.
| We can also approach the monotonicity of a bounded approaximation of $f(x)$ as follows.
\begin{eqnarray*}
f(x) &=& \frac{\left(\frac{1+2x}{1+x}\right)^{m} -1}{1-\left(\frac{1}{1+x}\right){m}} \\
&=& \frac{(1+2x)^{m}-(1+x)^{m}}{(1+x)^{m}-1^{m}} \\
&=& \frac{x^{m} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}(1+2x)^{m-1-i}} \right)}{x \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}}\right)} \\
&=& \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}(1+2x)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}}\right)}
\end{eqnarray*}
For $x \ge -\frac{1}{2}$,
\begin{eqnarray*}
f(x) &=& \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}(1+2x)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}}\right)} \\
&\ge& \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i} } \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}}\right)} \\
&=& x^{m-1}.
\end{eqnarray*}
Clearly this bound is monotonically increasing.
Thinking about it, I guess, we can take it further and prove the monotonicity, without relying on the bound.
Upon re-arranging,.
\begin{eqnarray*}
f(x) &=& \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{i}(1+2x)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1-i}{(1+x)^{i}}\right)} \\
&=& \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{(1+x)^{m-1}\left(\frac{1+2x}{1+x}\right)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{\frac{(1+x)^{m-1}}{(1+x)^{m-1}} (1+x)^{i}}\right)} \\
&=& \frac{x^{m-1} \left(\displaystyle\sum_{i=0}^{m-1}{\left(\frac{1+2x}{1+x}\right)^{m-1-i}} \right)}{ \left(\displaystyle\sum_{i=0}^{m-1}{\frac{1}{(1+x)^{m-1-i}} }\right)} \\
&=& x^{m-1} \frac{\displaystyle\sum_{i=0}^{m-1}{\left(1+\frac{x}{1+x}\right)^{m-1-i}} }{ \displaystyle\sum_{i=0}^{m-1}{\left(1-\frac{x}{1+x}\right)^{m-1-i}}} \\
&=& x^{m-1} \frac{\displaystyle\sum_{j=0}^{m-1}{\left(1+\frac{x}{1+x}\right)^{j}} }{ \displaystyle\sum_{j=0}^{m-1}{\left(1-\frac{x}{1+x}\right)^{j}}}
\end{eqnarray*}
Since $\frac{x}{1+x}<1$, the numerator is monotonically increasing and denominator monotonically decreasing. The ratio then is monotonically increasing.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
First Order Separable differential Equation
Problem:
Solve the following differential equation:
\begin{eqnarray*}
6x^2y \, dx - (x^3 + 1) \, dy &=& 0 \\
\end{eqnarray*}
Answer:
This is a separable differential equation.
\begin{eqnarray*}
\frac{6x^2}{x^3+1} \, dx - \frac{dy}{y} &=& 0 \\
\int \frac{6x^2}{x^3+1} \, dx - \int \frac{dy}{y} &=& c_1 \\
2 \ln{|x^3+1|} - \ln{|y|} &=& c_1 \\
\ln{(x^3+1)^2} - \ln{|y|} &=& c_1 \\
\ln{ \Big( \frac{(x^3+1)^2}{|y|} \Big) } &=& c_1 \\
(x^3+1)^2 &=& c|y| \\
\end{eqnarray*}
However, the book gets:
\begin{eqnarray*}
(x^3+1)^2 &=& |cy| \\
\end{eqnarray*}
Is my answer different from the book's answer? I believe it is. What am I missing?Any idea of how to proceed?
Thanks,
Bob
| \begin{align}
-\infty &\lt c_1 &\lt \infty\\
0 &\lt e^{c_1} = c &\lt \infty\\
\text{Therefore, } c&= |c|
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Proof verification of an exercise involving a functional equation Let $f: \mathbb{N} \rightarrow \mathbb{R}$ be a function and $a \in \mathbb{R}$ such that
$$f(m+n) = f(m) + f(n) + a$$
$$f(2) = 10, f(20) = 118$$
Find $a$ and $f$.
I found this exercise at the beginning of a Real Analysis textbook. I've never solved a functional equation before, but here's my solution (attempt):
i) Using induction it's easy to verify that for $m, n \in \mathbb{N}$ we have $f(m \cdot n) = m (f(n) + a)$, since
$$f((m+1)n) = f(mn + n) = f(mn) + f(n) + a$$
ii) Then $118 = f(20) = f(10 \cdot 2) = 10 (f(2) + a) = 10 (10 + a) \Rightarrow a = \frac{9}{5}$
iii) Then $f(0) = f(0 + 0) = f(0) + f(0) + a \Rightarrow f(0) = -a = -\frac{9}{5}$.
We also get $10 = f(2) = f(1+1) = f(1) + f(1) + \frac{9}{5} = 2 f(1) + \frac{9}{5} \Rightarrow f(1) = \frac{41}{10}$
iv) Then finally we can define $f$ recursively by $f(0) = -\frac{9}{5}$ and $$f(n+1) = f(n) + f(1) + \frac{9}{5} = f(n) + \frac{41}{10} + \frac{9}{5} = f(n) + \frac{59}{10}$$
EDIT
Then thanks to the user lulu, the real pattern should be $f(m \cdot n) = m f(n) + (m-1) a$ instead. Using this in ii) then gives $a = 2$. Then we get in iii) that $f(0) = -2$ and $f(1) = 4$, so $f$ is defined by $f(n+1) = f(n) + 6$. And now it's pretty obivous that $f(n) = 6n - 2$ solves the equation.
| $f(m+n)+a=f(m)+f(n)+2a$, Let$g(x)=f(x)+a\rightarrow g(m+n)=g(m)+g(n)$.
So $g$ satisfies Cauchy's functional equation and its solution is $g(n)=kn\rightarrow f(n)=kn-a$
$f(2)=10,f(20)=118 \rightarrow k=6,a=2,f(n)=6n-2$.
It can be sloved as same method if $f:\mathbb R\rightarrow \mathbb R$ and $f$ is lebesgue measurable.
Details about Cauchy's functional equation could refer to https://en.wikipedia.org/wiki/Cauchy%27s_functional_equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How many $4$-digit numbers can be formed using digits $0,1,...6$ such that it contains the digits $3$ and $5$? How many $4$-digit numbers can be formed using digits $0,1,...6$ such that it contains the digits $3$ and $5$?
My try:
All possible $4$-digit numbers $= 7 \cdot 7 \cdot 7 \cdot 6$
$4$-digits numbers NOT containing $3, 5 = 5 \cdot 5 \cdot 5 \cdot 4$
Answer= $7 \cdot 7 \cdot 7 \cdot 6-5 \cdot 5 \cdot 5 \cdot 4 =1558$
Is that OK ? If not, please explain my mistake.
Thank you.
| You also need to subtract out the number of numbers that contain one or more $5$s but no $3$, as well as the number of numbers that contain one or more $3$s but no $5$.
Or, you can also just count directly:
*
*One $3$, one $5$, two something else
*One $3$, two $5$s, one something else
*Two $3$s, one $5$, one something else
*Three $3$s, one $5$
*Two $3$s, two $5$s
*One $3$, three $5$s
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Optimize distance from origin to point on superellipsoid I've been trying to solve this problem with a couple of methods so far (including Lagrange multipliers) but for each method I end up with an unsolvable equation.
My first approach (Lagrange multipliers) starts off with optimizing $x^2+y^2+z^2$ constrained to $\left(\left|\dfrac{x}{a}\right|^r+\left|\dfrac{y}{b}\right|^r\right)^{\frac{t}{r}}+\left|\dfrac{z}{c}\right|^t-1 = 0$ where $a$, $b$, $c$, $r$, and $t$ are variables that tweak the shape of the superellipsoid. From there, I go from this:
$$
\begin{bmatrix}
\dfrac{\partial}{\partial x}\left(x^2 + y^2 + z^2 - \lambda\left(\left(\left|\dfrac{x}{a}\right|^r+\left|\dfrac{y}{b}\right|^r\right)^{\frac{t}{r}}+\left|\dfrac{z}{c}\right|^t-1\right)\right)\\
\dfrac{\partial}{\partial y}\left(x^2 + y^2 + z^2 - \lambda\left(\left(\left|\dfrac{x}{a}\right|^r+\left|\dfrac{y}{b}\right|^r\right)^{\frac{t}{r}}+\left|\dfrac{z}{c}\right|^t-1\right)\right)\\
\dfrac{\partial}{\partial z}\left(x^2 + y^2 + z^2 - \lambda\left(\left(\left|\dfrac{x}{a}\right|^r+\left|\dfrac{y}{b}\right|^r\right)^{\frac{t}{r}}+\left|\dfrac{z}{c}\right|^t-1\right)\right)\\
\dfrac{\partial}{\partial\lambda}\left(x^2 + y^2 + z^2 - \lambda\left(\left(\left|\dfrac{x}{a}\right|^r+\left|\dfrac{y}{b}\right|^r\right)^{\frac{t}{r}}+\left|\dfrac{z}{c}\right|^t-1\right)\right)
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\0\\0
\end{bmatrix}
$$
to
$$
\begin{bmatrix}
2x - \dfrac{t}{r}\lambda\left(\left(\dfrac{x}{a}\right)^r+\left(\dfrac{y}{b}\right)^r\right)^{\dfrac{t-r}{r}}\dfrac{x^{r-1}}{a^r}\\
2y - \dfrac{t}{r}\lambda\left(\left(\dfrac{x}{a}\right)^r+\left(\dfrac{y}{b}\right)^r\right)^{\dfrac{t-r}{r}}\dfrac{y^{r-1}}{b^r}\\
2z-\dfrac{t\lambda}{c^t}z^{t-1}\\
\left(\left(\dfrac{x}{a}\right)^r+\left(\dfrac{y}{b}\right)^r\right)^{\frac{t}{r}}+\left(\dfrac{z}{c}\right)^t-1
\end{bmatrix}
=
\begin{bmatrix}
0\\0\\0\\0
\end{bmatrix}
$$
to
$$
\begin{cases}
2x - \dfrac{t}{r}2z^{2-t}c^tt^{-1}\left(\left(\dfrac{x}{a}\right)^r+\left(\dfrac{y}{b}\right)^r\right)^{\dfrac{t-r}{r}}\dfrac{x^{r-1}}{a^r}=0\\
2y - \dfrac{t}{r}2z^{2-t}c^tt^{-1}\left(\left(\dfrac{x}{a}\right)^r+\left(\dfrac{y}{b}\right)^r\right)^{\dfrac{t-r}{r}}\dfrac{y^{r-1}}{b^r}=0\\
\left(\left(\dfrac{x}{a}\right)^r+\left(\dfrac{y}{b}\right)^r\right)^{\frac{t}{r}}+\left(\dfrac{z}{c}\right)^t=1
\end{cases}
$$
where I'm pretty much stuck, and sympy won't help.
My second attempt is converting the implicit equation of a superellipsoid into its polar form, then using implicit differentiation to get $\dfrac{\partial r}{\partial\theta} = 0$ and $\dfrac{\partial r}{\partial\theta}=0$. I get:
$$
\left(\left(\dfrac{r \sin{\theta} \cos{\phi}}{a}\right)^{n} + \left(\dfrac{r \sin{\phi} \sin{\theta}}{b}\right)^{n}\right)^{\frac{t}{n}} + \left(\dfrac{r \cos{\theta}}{c}\right)^{t} = 1
$$
and sympy gives me:
$$
\begin{cases}
\frac{r \left(\left(\frac{r \cos{\left (\theta \right )}}{c}\right)^{t} \sin^{2}{\left (\theta \right )} - \left(\left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} + \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n}\right)^{\frac{t}{n}} \cos^{2}{\left (\theta \right )}\right)}{\left(\left(\frac{r \cos{\left (\theta \right )}}{c}\right)^{t} + \left(\left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} + \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n}\right)^{\frac{t}{n}}\right) \sin{\left (\theta \right )} \cos{\left (\theta \right )}} = 0 \\
\frac{r \left(\left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} \sin^{2}{\left (\phi \right )} - \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n} \cos^{2}{\left (\phi \right )}\right) \left(\left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} + \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n}\right)^{\frac{t}{n}}}{\left(\left(\frac{r \cos{\left (\theta \right )}}{c}\right)^{t} \left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} + \left(\frac{r \cos{\left (\theta \right )}}{c}\right)^{t} \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n} + \left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} \left(\left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} + \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n}\right)^{\frac{t}{n}} + \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n} \left(\left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} + \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n}\right)^{\frac{t}{n}}\right) \sin{\left (\phi \right )} \cos{\left (\phi \right )}} = 0
\end{cases}
$$
which boils down to:
$$
\begin{cases}
\left(\frac{r \cos{\left (\theta \right )}}{c}\right)^{t} \sin^{2}{\left (\theta \right )} = \left(\left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} + \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n}\right)^{\frac{t}{n}} \cos^{2}{\left (\theta \right )} \\
\begin{cases}
\left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} \sin^{2}{\left (\phi \right )} = \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n} \cos^{2}{\left (\phi \right )} \\
or \\
\left(\frac{r \sin{\left (\theta \right )} \cos{\left (\phi \right )}}{a}\right)^{n} + \left(\frac{r \sin{\left (\phi \right )} \sin{\left (\theta \right )}}{b}\right)^{n} = 0
\end{cases}
\end{cases}
$$
at which point sympy won't help once again, and I've no idea how to solve for $\theta$ and $\phi$.
Anyways, has someone got an alternative method for solving this? Or is it just me who can't solve the equations?
| Use the change of variable
$$\left(\frac xa\right)^r=\rho^{r/t}\cos^2\theta,
\\\left(\frac yb\right)^r=\rho^{r/t}\sin^2\theta.$$
Then the constraint becomes
$$\rho+\left(\frac zc\right)^t-1=0$$ or
$$z=c(1-\rho)^{1/t}.$$
The function to be maximized is now
$$\rho^{2/t}(a^2\cos^{4/r}\theta+b^2\sin^{4/r}\theta)+c^2(1-\rho)^{2/t}.$$
Differentiating on $\theta$ and canceling,
$$-a^2\sin\theta\cos^{4/r-1}\theta+b^2\cos\theta\sin^{4/r-1}\theta=0$$ or
$$\tan^{4/r-2}\theta=\frac{a^2}{b^2}$$ gives you the solutions in $\theta$.
And differentiating on $\rho$,
$$\rho^{2/t-1}(a^2\cos^{4/r}\theta+b^2\sin^{4/r}\theta)-c^2(1-\rho)^{2/t-1}$$
gives
$$\frac{1-\rho}{\rho}=\left(\frac{a^2\cos^{4/r}\theta+b^2\sin^{4/r}\theta}{c^2}\right)^{t/(2-t)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2876537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluate $\int _0^1\frac{dx}{(e^x-1)^{1/3}}$ Study the convergence of the following integral for $\alpha\in\mathbb{R}$ and evaluate it for $\alpha = \frac{1}{3}$ if it converges for that value.
$${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\alpha}} x$$
The integral converges for $\alpha \lt 1$, if I am not wrong, so I can evaluate:
$${\Large\int} _{0}^{1}\frac{dx}{(e^x-1)^{\frac{1}{3}}}$$
whose primitive, using some razionalizations and change of variables, is: $\frac{1}{2}\log|4(e^x-1)^\frac{2}{3}-4(e^x-1)^\frac{1}{3}+4|+\sqrt{3}\arctan\left(\frac{1}{\sqrt{3}}\left(2(e^x-1)^\frac{1}{3}-1\right)\right)-\log|(e^x-1)^\frac{1}{3}+1|+C$
Anyone knows if there is a less intricate way in order of obtain it than almost 3 sheet full of calculations? Thank you
| using $u={{\left( {{e}^{x}}-1 \right)}^{1/3}}$ leads to $\int{\frac{3u}{{{u}^{3}}+1}du}$
$$
\begin{align}
& =\int{-\frac{1}{u+1}+\frac{u+1}{{{u}^{2}}-u+1}du} \\
& =\int{-\frac{1}{u+1}du}+\frac{1}{2}\int{\frac{2u-1}{{{u}^{2}}-u+1}du}+\frac{3}{2}\int{\frac{1}{{{u}^{2}}-u+1}du} \\
& =\ln \left( u+1 \right)+\frac{1}{2}\ln \left( {{u}^{2}}-u+1 \right)+\frac{3}{2}\int{\frac{1}{{{u}^{2}}-u+1}du} \\
\end{align}
$$
For the last integral use $v=2u-1$ and notice that $4\left( {{u}^{2}}-u+1 \right)={{v}^{2}}+3$
$$
\begin{align}
& =\ln \left( u+1 \right)+\frac{1}{2}\ln \left( {{u}^{2}}-u+1 \right)+3\int{\frac{1}{{{v}^{2}}+3}dv} \\
& =\ln \left( u+1 \right)+\frac{1}{2}\ln \left( {{u}^{2}}-u+1 \right)+3\frac{1}{\sqrt{3}}ta{{n}^{-1}}\left( \frac{2u-1}{\sqrt{3}} \right) \\
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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In how many ways can $4$ same oranges and $6$ different apples be distributed to $5$ distinct boxes? If we have $4$ same oranges and $6$ different apples. In how many ways could we distribute them in $5$ different boxes?
I have thought the first part of this problem as saying that the $4$ same oranges have $C(8,4)= 70$ ways of being distributed in $5$ different boxes (using the $C(n+k-1,k)$ formula). But for the $6$ different apples I cannot understand if I should use the $$ n^{k}=5^{6} $$ formula or the $$ \frac{(n+k-1)!}{(n-1)!} $$ formula where $k$ are the apples and $n$ are the boxes. I know the final answer will be the product of $C(8,4)$ and whatever the answer about the apples is, but I am not sure about the formula. If it is the first, why wouldn't the second one work?
There is an extra question, which asks what percentage of the above describes the ways where $2$ exactly fruits are placed in each box. I have thought the answer might be $5! C(9,5)$, because the ways of distributing the first $5$ fruits are $5!$ and the ways of distributing the rest in $5$ different boxes are $C(n+k-1,k)$. Am I correct to assume that? I have also thought the formula $C(n-1,k-1)$ because the $C(n+k-1,k)$ does not give one apple to each $5$ necessarily.
|
In how many ways can $4$ indistinguishable oranges and $6$ different apples be placed in $5$ distinct boxes?
You correctly found that $4$ indistinguishable oranges can be placed in $5$ distinct boxes in
$$\binom{4 + 5 - 1}{5 - 1} = \binom{8}{4}$$
ways. Since there are five ways each of the six apples can be placed in a box, there are $5^6$ ways to distribute the apples. Hence, the number of possible distributions is
$$5^6\binom{8}{4}$$
As for the other formula you considered, if you wanted to arrange four indistinguishable dividers and six different apples in a row, you could choose four of the ten positions for the dividers in $\binom{10}{4}$ ways, then arrange the six apples in the six remaining positions in $6!$ ways, which yields
$$\binom{10}{4}6! = \frac{10!}{4!6!} \cdot 6! = \frac{10!}{4!} = \frac{(6 + 5 - 1)!}{(5 - 1)!}$$
However, we do not care about the order of the apples within each box, so this formula yields too large a number for the number of ways of distributing the apples.
In how many ways can $4$ indistinguishable oranges and $6$ different apples be placed in $5$ distinct so that there are two pieces of fruit in each box?
The restriction that exactly two pieces of fruit are placed in each box means that we have three cases to consider:
*
*Two oranges apiece are placed in each of two boxes.
*Two oranges are placed in one box and one orange apiece is placed in each of two other boxes.
*Each orange is placed in a distinct box.
Two oranges apiece are placed in each of two boxes: Choose which two of the five boxes each receive two oranges. Choose which two of the six apples are placed in the leftmost empty box. Choose which two of the remaining four apples are placed in the leftmost remaining empty box. Place the final two apples in the remaining empty box.
There are $$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2}$$ such distributions.
Two oranges are placed in one box and one orange apiece is placed in each of two other boxes: Choose which of the five boxes receives two oranges. Choose which two of the remaining four boxes each receive one of the other two oranges. Choose which of the six apples is placed in the leftmost box with one apple and which of the remaining five apples is placed in the other box with one apple. Choose which two of the remaining four apples is placed in the leftmost empty box. The other two apples must be placed in the remaining empty box.
There are $$\binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2}$$ such distributions.
Each orange is placed in a distinct box: There are $\binom{5}{4}$ ways to select which four of the boxes each receives one orange. There are $\binom{6}{2}$ ways to select which two apples are placed in the empty box. There are $4!$ ways to distribute the remaining four apples to the four boxes that each contain one orange.
There are $$\binom{5}{4}\binom{6}{2}4!$$ such distributions.
Since these cases are mutually exclusive and exhaustive, the number of ways to distribute four indistinguishable oranges and six distinct apples in such a way that two pieces of fruit are placed in each box is found by adding the results for each case.
$$\binom{5}{2}\binom{6}{2}\binom{4}{2}\binom{2}{2} + \binom{5}{1}\binom{4}{2}\binom{6}{1}\binom{5}{1}\binom{4}{2}\binom{2}{2} + \binom{5}{4}\binom{6}{2}4!$$
| {
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Evaluate $\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{a^2 \cos^2 \theta+b^2 \sin^2 \theta}} d \theta$
$\int_0^{\frac{\pi}{2}} \frac{1}{\sqrt{a^2 \cos^2 \theta+b^2 \sin^2 \theta}} d \theta$
$ = \int_0^{\frac{\pi}{2}} \frac{1}{a}\sec \theta \frac{1}{ \sqrt{1+(b/a)^2 \tan^2 \theta}} d \theta$
But i know $d(\tan \theta) = \sec^2 \theta$. But this is not in that form. Can you tell me how to proceed further?
| $$I(a, b) =\int_0^\frac{\pi}{2} \frac{1}{\sqrt{a^2 \sin^2 t + b^2 \cos^2 t}}dt$$
$$ = \int_0^\frac{\pi}{2} \frac{1}{\sqrt{b^2-(b^2-a^2)\sin^2 t}}dt=\frac1b \int_0^\frac{\pi}{2} \frac{1}{\sqrt{1-\left(1-\frac{a^2}{b^2}\right)\sin^2 t}}dt=\frac1b K\left(1-\frac{a^2}{b^2}\right)$$ Or this can be written as: $I(a, b)=\frac{\pi}{2M(a,b)},$ where $M(a,b)$ is the arithmetic-geometric mean of $a,b$ defined by $M(a,b):=\lim_{n\to\infty}a_n=\lim_{n\to\infty}b_n$ where
$$a_0=a, \ b_0=b, \ a_{n+1}=\frac{a_n+b_n}{2}, \ b_{n+1}=\sqrt{a_nb_n}$$
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Find all pairs of intergers satisfying $x^2+11 = y^4 -xy$ and $y^2 + xy= 30 $
Find all pairs of intergers $(x,y)$ that satisfy the two following equations:
$x^2+11 = y^4 -xy$
$y^2 + xy= 30 $
Here's what I did:
$x^2+11 +(30) = y^4 -xy +(y^2 + xy)$
$x^2+41 = y^4 +y^2 $
$x^2 = y^4 +y^2 - 41$
$x^2 -49 = y^4 +y^2 - 41 - 49$
$(x+7)(x-7)= (y^2 +10) (y^2-9)$
And from here you get that two pairs can be: $(3,7)$ and $(-3,-7)$.
I think I've made some progress but don't know if those are the only two pairs that satisfy the equation.
Also I would like to see a different way of solving it since I think subtracting that $49$ was just luck.
| HINT: From the second equation we get that $y \mid 30$. As $30$ has $16$ divisors (both negative and positive) you can take a look at each of them separately. Also this uniquely determines $x$, so you just have to plug it in the first equation to check if the pair is a solution.
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Find the values for $a,b,c,d$
Given $$x^3+4x^2y+axy^2+3xy-bx^cy+7xy^2+dxy+y^2=x^3+y^2$$ for any
real number $x$ and $y$ , find the value of $a,b,c,d$
| Put some values of $x$ and $y$ in it.
Say $x=1$: $$(a+7)y^2+(7-b+d)y =0\;\;\forall y$$ so $a=-7$ and $b-d=7$.
Now we have (for all $x,y$):
$$4x^2y-bx^cy+(d+3)xy=0$$
If $y\ne 0$ we get $$4x^2-bx^c+(d+3)x=0\;\;\;\forall x$$
a) Put $x= 2$: $$\boxed{8+b(2-2^c)=0}$$
b) Put $x=4$: $$\boxed{48+b(4-4^c)=0}$$
Solving this system we get $$-b = {8\over 2-2^c} = {48\over 4-4^c}\implies 2+2^c= 6\implies c = 2$$ so $b=4$ and $d=-3$.
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Find the value of $\alpha,\beta$ for the equation $\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}$ Find the value of $\alpha,\beta$ for the equation $\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}$
$\alpha>0$ & $\beta<\frac{\pi}{2}$
I get the following step after some substitution
$\cos2\alpha + \cos2\beta+\cos2(\alpha+\beta)=-\frac{3}{2}$ from here not able to proceed.
| Suppose $$0<\alpha,\beta<\frac\pi2.\tag{1}$$ From
$$\cos2\alpha + \cos2\beta+\cos2(\alpha+\beta)=-\frac{3}{2}$$
one has
$$ \cos^2\alpha+\cos^2\beta+\cos^2(\alpha+\beta)=\frac34. $$
Note that
$$\cos\alpha \cos\beta \cos(\alpha +\beta)=-\frac{1}{8}\tag{2}$$
implies
$$\cos^2\alpha \cos^2\beta \cos^2(\alpha +\beta)=\frac{1}{64}.$$
By the AM-GM inequality
$$ a+b+c\ge3\sqrt[3]{abc}$$
one has
$$ \frac34=\cos^2\alpha+\cos^2\beta+\cos^2(\alpha+\beta)=3\sqrt[3]{\cos^2\alpha\cos^2\beta\cos^2(\alpha+\beta)}=3\sqrt[3]{\frac{1}{64}}=\frac34 $$
and the equal sign holds if and only if
$$\cos^2\alpha=\cos^2\beta=\cos^2(\alpha+\beta).\tag{3}$$
From (1)(2)(3), it is easy to see
$$ \alpha=\beta=\frac{\pi}{3}. $$
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How to get a implicit equation of this ellipse If I give you the curve $\gamma (t) = ( \cos (t), \cos(t+a))$, how can I obtain an implicit equation? or which change of basis can I do to get the canonical implicit form?
| The canonical form is $A x^2 + B y^2 +2 C x y = 1$
Substitute $x=\cos(t)$ and $y = \cos(a)\cos(t)-\sin(a)\sin(t)$ into the above and try to eliminate all coefficients of $\sin(t)$ and $\cos(t)$.
It helps to use the substitutions $\cos^2(t) = \frac{\cos(2 t)+1}{2}$, $\sin^2(t) = \frac{1-\cos(2 t)}{2}$ and $\sin(t)\cos(t) = \frac{\sin(2 t)}{2}$
You will get in the end the following
$$ \begin{multline} \left( B \cos^2(a) + C \cos(a) + \frac{A-B}{2} \right) \cos(2 t) - \Bigl( B \sin(a)\cos(a)+C \sin(a) \Bigr) \sin(2 t)+ \\ + \left( C \cos(a) + \frac{A+B}{2} \right) = 1 \end{multline} $$
Now we eliminate all coefficients of $\sin(2 t)$ and $\cos(2t)$ to convert the above into a system of equations
$$\begin{aligned}
B \cos^2(a) + C \cos(a) + \frac{A-B}{2} & = 0 \\
B \sin(a)\cos(a)+C \sin(a) & = 0 \\
C \cos(a) + \frac{A+B}{2} & = 1
\end{aligned} $$
Solve the above for $A$, $B$ and $C$.
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Sum of the series $\frac{1}{2.4.6}+\frac{2}{3.5.7}+\frac{3}{4.6.8}+.....+\frac{n}{(n+1).(n+3).(n+5)}$. Sum to n terms and also to infinity of the following series:
$$\frac{1}{2.4.6}+\frac{2}{3.5.7}+\frac{3}{4.6.8}+.....+\frac{n}{(n+1).(n+3).(n+5)}$$
the solution provided by the book is
$$S_n=\frac{17}{96}-[\frac{1}{n+5}+\frac{1}{2(n+4)(n+5)}-\frac{1}{2(n+3)(n+4)}-\frac{3}{4(n+2)(n+3)(n+4)(n+5)}]$$
And $S_\infty=\dfrac{17}{96}$.Can anyone help me to explain how to get $S_n$.
Thanks in Advanced.
| Alternatively, write
$$
\begin{align}\frac{n}{(n+1)(n+3)(n+5)}&=\frac{(n+5)-5}{(n+1)(n+3)(n+5)}\\&=\frac{1}{(n+1)(n+3)}-\frac{5}{(n+1)(n+3)(n+5)}\,.\end{align}$$
Now,
$$\frac{1}{(n+1)(n+3)}=\frac{1}{2}\,\left(\frac{1}{n+1}-\frac{1}{n+3}\right)$$
and
$$\frac{1}{(n+1)(n+3)(n+5)}=\frac{1}{4}\,\left(\frac{1}{(n+1)(n+3)}-\frac{1}{(n+3)(n+5)}\right)\,.$$
That is,
$$\frac{n}{(n+1)(n+3)(n+5)}=\frac{1}{2}\,\left(\frac{1}{n+1}-\frac{1}{n+3}\right)-\frac{5}{4}\,\left(\frac{1}{(n+1)(n+3)}-\frac{1}{(n+3)(n+5)}\right)\,.$$
The rest is straightforward.
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Solving $|z-4i|=2|z+4|$ $$\begin{align}
|z-4i|&=2|z+4|\\[4pt]
|x+yi-4i|&=2|x+yi+4|\\[4pt]
|x+i(y-4)|&=2|(x+4)+iy|\\[4pt]
\sqrt{x^2+(y-4)^2}&=2\sqrt{(x+4)^2+y^2}\\[4pt]
(\sqrt{x^2+(y-4)^2})^2&=(2\sqrt{(x+4)^2+y^2})^2\\[4pt]
x^2+y^2-8y+16&= 4(x^2+8x+16+y^2)\\[4pt]
x^2+y^2-8y+16&= 4x^2+32x+64+4y^2\\[4pt]
0&= 3x^2+3y^2+32x+8y+48
\end{align}$$
Is it okay? Thank you
| Your solution looks fine, but you should realize that it is an equation for the circle with center $\left(-\dfrac{16}{3},-\dfrac{4}{3}\right)$ and with radius $\dfrac{8\sqrt{2}}{3}$. If I were your grader, I would not give you a full credit for simply finding the final equation, yet not realizing it gives a circle. Here is an alternative solution, using Euclidean geometry of the plane.
Let $A$ denote the point $4\text{i}$ of the complex plane $\mathbb{C}\cong\mathbb{R}^2$, whilst $B$ is the point $-4$. Thus, if the point $C$ with complex coordinate $z$ satisfies
$$|z-4\text{i}|=2\,|z+4|\,,$$
then
$$CA=2\,CB\,.$$
On the line $AB$, there are two solutions $D$ and $E$, with complex coordinates
$$\frac{1}{3}\,(4\text{i})+\frac{2}{3}\,(-4)=\frac{-8+4\text{i}}{3}\text{ and }(-1)\,(4\text{i})+2\,(-4)=-8-4\text{i}\,,$$
respectively. Thus, the point $C$ is a point such that $CD$ is the internal angular bisector of $\angle ACB$ and $CE$ is the external angular bisector of $\angle ACB$. We can easily show that the locus of $C$ is a circle $\Gamma$ with diameter $DE$.
Thus, the center $P$ of $\Gamma$ has the complex coordinate
$$\frac{1}{2}\,\left(\frac{-8+4\text{i}}{3}\right)+\frac{1}{2}\,(-8-4\text{i})=\frac{-16-4\text{i}}{3}\,.$$
The radius of $\Gamma$ is $$\frac{1}{2}\,\Biggl|\left(\frac{-8+4\text{i}}{3}\right)-(-8-4\text{i})\Biggr|=\frac{8\sqrt{2}}{3}\,.$$
In other words, the complex coordinate $z$ of $C$ satisfies
$$\Biggl|z-\left(\frac{-16-4\text{i}}{3}\right)\Biggr|=\frac{8\sqrt{2}}{3}\,.$$
We can also write
$$z=\left(\frac{-16-4\text{i}}{3}\right)+\frac{8\sqrt{2}}{3}\,\exp(\text{i}\theta)\,,$$
where $\theta\in\mathbb{R}$.
In general, any solution $z\in\mathbb{C}$ to $|z-a|=r\,|z-b|$, where $r\in\mathbb{R}_{>0}\setminus\{1\}$ and $a,b\in\mathbb{C}$ is given by the circle
$$\left|z-c\right|=\rho\,.$$
Here, $c:=\dfrac{-a+r^2b}{r^2-1}$ and $\rho:=\dfrac{r}{|r^2-1|}\,|a-b|$. In other words,
$$z=c+\rho\,\exp(\text{i}\theta)\,,$$
where $\theta\in\mathbb{R}$. (Note that the solutions in the case where $r=1$ form a degenerate circle---a straight line. This straight line is the perpendicular bisector of the segment joining $a$ and $b$. It is given by the equation $(a-b)\,\bar{z}+(\bar{a}-\bar{b})z=|a|^2-|b|^2$, or equivalently, $\text{Re}\big((\bar{a}-\bar{b})\,z\big)=\dfrac{|a|^2-|b|^2}{2}$. In other words, $z=\dfrac{a+b}{2}+\text{i}(a-b)\,t$, where $t\in\mathbb{R}$.)
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Solving Complex Number Equation with Galois theory Suppose that the complex numbers $\alpha$, $\beta$ and $\gamma$ satisfy
\begin{align*}
\alpha + \beta + \gamma &= 3, \\
\alpha^2 + \beta^2 + \gamma^2 &= 5, \\
\alpha^3 + \beta^3 + \gamma^3 &= 12.
\end{align*}
I want to show that $\alpha^n + \beta^n + \gamma^n \in \mathbb{Z}$ for all $n \in \mathbb{Z}^+$ if possible, using the concept of symmetric polynomials from Galois theory, but I am not sure how.
| Use $(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\iff ab+bc+ca=?p$(say)
$a^3+b^3+c^3-3abc=(a+b+c)\{(a+b+c)^2-3(ab+bc+ca)\}\iff abc=q$(say)
So, $a,b,c$ are the roots of $$t^3-3t^2+pt-q=0$$
$\implies t^{n+3}=3t^{n+2}-pt^{n+1}+qt^{n}$
If $S_m=a^m+b^m+c^m,$
$$S_{m+3}=3S_{m+2}-pS_{m+1}+qS_m$$
Now use Strong induction
| {
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How to find an affine map for given points I am trying to understand affine maps. In the book I am using there is the following example, which I don't understand. Given are the following points:
\begin{array}{lll}
p_0 = \begin{pmatrix} 1 \\ 1 \end{pmatrix},
&
p_1 = \begin{pmatrix} 1 \\ 2 \end{pmatrix},
&
p_2 = \begin{pmatrix} 0 \\ 1 \end{pmatrix},
\\
q_0 = \begin{pmatrix} -2 \\ -1 \end{pmatrix},
&
q_1 = \begin{pmatrix} \phantom{-}0 \\ -1 \end{pmatrix},
&
q_2 = \begin{pmatrix} -2 \\ -2 \end{pmatrix}.
\end{array}
Then the following vectors are computed
\begin{array}{ll}
\overrightarrow{p_0 p_1}
= \begin{pmatrix} 0 \\ 1 \end{pmatrix},
&
\overrightarrow{p_0 p_2}
= \begin{pmatrix} -1 \\ \phantom{-}0 \end{pmatrix},
\\
\overrightarrow{p_0 p_1}
= \begin{pmatrix} 2 \\ 0 \end{pmatrix},
&
\overrightarrow{p_0 p_1}
= \begin{pmatrix} \phantom{-}0 \\ -1 \end{pmatrix}.
\end{array}
I am asked to find an affine map such that $f(p_{i})=q_{i}$.
It now says that the matrix of the map determined by $F(\overrightarrow{p_0 p_i})=\overrightarrow{q_0 q_i}$, $i=1,2$ is given by
$$
A
= \begin{pmatrix}
0 & 2 \\
1 & 0
\end{pmatrix}.
$$
I don't understand how I get this matrix. I would say it is the following one, but I guess I am wrong. I did everything with repect to the bases $\{(0,1),(-1,0)\}$ and $\{(1,0),(0,1)\}$. I am not sure about the second basis, but since there is no other basis mentioned I thought it is the one the author has used too.
$$
\begin{pmatrix}
2 & \phantom{-}0 \\
0 & -1 \\
\end{pmatrix}
$$
Unfortunately I don't know where I making the mistake.
I would really appreciate some help. Thanks a lot in advance!
| There is a neat formula for your case
$$
\vec{f}(x; y) = (-1)
\frac{
\det
\begin{pmatrix}
0 & \vec{q_0} & \vec{q_1} & \vec{q_2} \\
x & p_0^x & p_1^x & p_2^x \\
y & p_0^y & p_1^y & p_2^y \\
1 & 1 & 1 & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
p_0^x & p_1^x & p_2^x \\
p_0^y & p_1^y & p_2^y \\
1 & 1 & 1 \\
\end{pmatrix}
},
$$
where $x$ and $y$ are coordinates of the point you are mapping.
Here's how it works (plugging in initial points)
$$
\vec{f}(x; y) = (-1)
\frac{
\det
\begin{pmatrix}
0 & \vec{q_0} & \vec{q_1} & \vec{q_2} \\
x & 1 & 1 & 0 \\
y & 1 & 2 & 1 \\
1 & 1 & 1 & 1 \\
\end{pmatrix}
}{
\det
\begin{pmatrix}
1 & 1 & 0 \\
1 & 2 & 1 \\
1 & 1 & 1 \\
\end{pmatrix}
} =
(\vec{q}_0 - \vec{q}_2) x +
(\vec{q}_1 - \vec{q}_0) y +
(\vec{q}_0 - \vec{q}_1 + \vec{q}_2) =
$$
now I plug in the final points
$$
= \left[
\begin{pmatrix} -2 \\ -1 \end{pmatrix} -
\begin{pmatrix} -2 \\ -2 \end{pmatrix}
\right] x +
\left[
\begin{pmatrix} 0 \\ -1 \end{pmatrix} -
\begin{pmatrix} -2 \\ -1 \end{pmatrix}
\right] y +
\left[
\begin{pmatrix} -2 \\ -1 \end{pmatrix} -
\begin{pmatrix} 0 \\ -1 \end{pmatrix} +
\begin{pmatrix} -2 \\ -2 \end{pmatrix}
\right]
$$
Simplification yields
$$
\vec{f}(x; y) =
\begin{pmatrix} 0 \\ 1 \end{pmatrix} x +
\begin{pmatrix} 2 \\ 0 \end{pmatrix} y +
\begin{pmatrix} -4 \\ -2 \end{pmatrix}
$$
Or you can write that in canonical form
$$
\vec{f}(x; y) =
\begin{pmatrix}
0 & 2 \\
1 & 0
\end{pmatrix}
\begin{pmatrix} x \\ y \end{pmatrix} +
\begin{pmatrix} -4 \\ -2 \end{pmatrix}.
$$
Now you can easily check that
$$
\vec{f}(1;1) = \begin{pmatrix} -2 \\ -1 \end{pmatrix};~
\vec{f}(1;2) = \begin{pmatrix} 0 \\ -1 \end{pmatrix};~
\vec{f}(0;1) = \begin{pmatrix} -2 \\ -2 \end{pmatrix}.
$$
For more details on how this all works you may check "Beginner's guide to mapping simplexes affinely", where authors of the equation I used elaborate on theory behind it. In the guide there is exactly the same 2D example solved as the one you are interested in.
The same authors recently published "Workbook on mapping simplexes affinely". There are many examples of different problems you can solve with this equation.
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Diophantine equation $x^3+x+a^2=y^2$ Prob: Show that for any positive integer $a$, Diophantine equation $$x^3+x+a^2=y^2$$ has at least one solution $(x, y)$, where $x, y$ are positive integers.
Source: My teacher.
Attempt:
First I tried $a=1$ and found the minimal solution $(x, y)=(72, 611)$, not a friendly one.
Now rewrite the equation as $$x(x^2+1)=(y-a)(y+a).$$ We hope that $x=b_{1}b_{2}, x^2+1=c_{1}c_{2}$ and $b_{2}c_{2}-b_{1}c_{1}=2a$. But how to determine these numbers? I got stuck here. One way promising is to relate to a Pell-type equation.
Another thought is to set $x$ to be some polynomial like $2t^2$ so that the left side can be factorized furthermore. Still little progress.
Please help.
| Your first idea is a great idea! When working this out, I had slightly different variables, but it is equivalent to your starting.
Suppose positive integers $b, c, d, e$ satisfy $y+a=bc$, $y-a=de$, $x=bd$, $x^2+1=ce$, $bc-de=2a$, $ce-(bd)^2 = 1$.
I propose the following lemma:
The sequence $k_n$ is defined as follows: $k \in \mathbb{N}$, $k_0 = 0$, $k_1=1$, $k_{n+2} = kk_{n+1} + k_{n}$ for $n \geq 0$. Then $k_{n-1} k_{n+1} - k_{n}^2 = (-1)^{n} \quad \forall n \geq 1$.
Proof:
For $n=1$, this holds true. Now, assume true for $n$. Then for $n+1$,
$$\begin{aligned}
k_{n} k_{n+2} - k_{n+1}^2 &= k_{n} (kk_{n+1} + k_{n}) - k_{n+1}^2\\
&= -k_{n+1} (k_{n+1} - kk_{n}) + k_{n}^2\\
&= -(k_{n-1} k_{n+1} - k_{n}^2)\\
&= -(-1)^n\\
&= (-1)^{n+1}
\end{aligned}$$
Thus, proven.
Now, if we let $c=k_3 = k^2+1$, $e=k_5=k^4+3k^2+1$, $bd=k_4 = k(k^2+2)$, then from the above lemma, we have $ce-(bd)^2=1$. Letting $d=2a, k=4a^2$, we get $b=2a(k^2+2)$ and $bc-de=2a(k^2+2)(k^2+1)-2a(k^4+3k^2+1) = 2a$. Thus, we have the result
$$x=bd=8a^2(8a^4+1)$$
$$y=4a(8a^4+1)(16a^4+1)-a$$
as a construction for all $a$, and does satisfy the solution you provided for $a=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Find the maximum value of $A$ Let $a;b;c>0$ such that $a+b+c=6$. Find the maximum value of $A=a^2bc+a^2+2b^2+2c^2$
WLOG $b\ge c$. I see maximum value of $A=36$ at $(a;b;c)=(2;1;3)$
So i need to prove $A\le 36$. Or I will prove
$(a+b+c)^4\ge 36a^2bc+(a^2+2b^2+2c^2)(a+b+c)^2$
Or $(2a-b-c)(b^3+c^3+a^2b+a^2c+2ab^2+2ac^2-12abc-b^2c-bc^2)\ge 0$
Then Im stuck here, help me solve it.
| With the help of the Lagrange multipliers and $f(a,b,c) = a^2 b c + a^2 + 2 b^2 + 2 c^2$
$$
L(a,b,c,\lambda) =f(a,b,c)+\lambda(a+b+c-6)
$$
the stationary points are the solutions for
$$
\nabla L = \left\{
\begin{array}{c}
2 b c a+2 a+\lambda =0 \\
c a^2+4 b+\lambda =0 \\
b a^2+4 c+\lambda =0 \\
a+b+c-6=0 \\
\end{array}
\right.
$$
giving
$$
\left[
\begin{array}{ccccc}
a & b & c & \lambda & f(a,b,c)\\
-2 & -1 & 9 & -32 & \color{red}{132} \\
-2 & 9 & -1 & -32 & \color{red}{132} \\
2 & 1 & 3 & -16 & \color{green}{36} \\
2 & 3 & 1 & -16 & \color{green}{36} \\
3 & \frac{3}{2} & \frac{3}{2} & -\frac{39}{2} & \color{green}{\frac{153}{4}} \\
3-\sqrt{5} & \frac{1}{2} \left(3+\sqrt{5}\right) & \frac{1}{2} \left(3+\sqrt{5}\right) & -12 & \color{green}{32} \\
3+\sqrt{5} & \frac{1}{2} \left(3-\sqrt{5}\right) & \frac{1}{2} \left(3-\sqrt{5}\right) & -12 & \color{green}{32} \\
\end{array}
\right]
$$
In red non feasible solutions and in green feasible solutions. The maximum corresponds to $\frac{153}{4}= 38.25$
NOTE
The same solution can be obtained under the tangency condition between $f(a,b,c)-\mu=0$ and $a+b+c-6=0$ by calculating the values for $\mu$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2894315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Summing a series having a geometric component
If $$\alpha=\frac {5}{2!3}+\frac {5\cdot 7}{3!3^2}+\frac {5\cdot 7\cdot 9}{4!3^3}+\dots $$ Find the value of $\alpha^2+4\alpha $.
The possible options are:
*
*21
*23
*25
*27
I think $$\alpha=\sum \frac {(2n+4)!}{2^{n+2}(n+1)!(n+2)!3^{n+1}}$$ but cannot think of what else to do.
| Consider the binomial expansion of $$(1-\frac 23)^{-\frac 32}$$
This has value $3\sqrt{3}$ and when expanded gives the sequence $$1+1+\frac{5}{2!3}+\frac{5\cdot7}{3!3^2}+...$$
So $$\alpha=3\sqrt{3}-2$$ which gives the required expression the value of $23$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is $\int \sqrt{\cos(2Q)} / {\sin(Q)} \,\mathrm{d}Q$?
What is
$$
\int \frac{\sqrt{\cos(2Q)}}{\sin(Q)} \,\mathrm{d}Q?
$$
I have tried all the method which is possible but could not able to find the solution. Can anyone please tell me the solution of this problem.
| First notice that
\begin{equation}
\cos\left(2x\right)=\cos^2\left(x\right)-\sin^2\left(x\right)
\end{equation}
and
\begin{equation}
\sin\left(x\right)=\dfrac{\tan\left(x\right)}{\sec\left(x\right)}
\end{equation}
so
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x
={\displaystyle\int}
\sec^2(x) \cdot{{\dfrac{\sqrt{1-\tan^2\left(x\right)}}{\tan\left(x\right)\left(\tan^2\left(x\right)+1\right)}}}\,\mathrm{d}x
\end{equation}
Change of variable as
\begin{equation}
u = \tan (x)
\end{equation}
we get
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x
={\displaystyle\int}\dfrac{\sqrt{1-u^2}}{u\left(u^2+1\right)}\,\mathrm{d}u
\end{equation}
Another change of variable as
\begin{equation}
v = \sqrt{1 - u^2}
\end{equation}
gives us
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x
=-{\displaystyle\int}\dfrac{v^2}{\left(v^2-2\right)\left(v^2-1\right)}\,\mathrm{d}v
\end{equation}
Now let's factor the denominator as
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x
={\displaystyle\int}\dfrac{v^2}{\left(v-1\right)\left(v+1\right)\left(v^2-2\right)}\,\mathrm{d}v
\end{equation}
Then perform partial fraction decomposition
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x
={\displaystyle\int}\left(\dfrac{2}{v^2-2}+\dfrac{1}{2\left(v+1\right)}-\dfrac{1}{2\left(v-1\right)}\right)\mathrm{d}v
=
2A + \frac{1}{2}B - \frac{1}{2} C
\end{equation}
Let's do $A$,
\begin{equation}
A = {\displaystyle\int}\dfrac{1}{v^2-2}\,\mathrm{d}v
=
={\displaystyle\int}\dfrac{1}{\left(v-\sqrt{2}\right)\left(v+\sqrt{2}\right)}\,\mathrm{d}v
={\displaystyle\int}\left(\dfrac{1}{2^\frac{3}{2}\left(v-\sqrt{2}\right)}-\dfrac{1}{2^\frac{3}{2}\left(v+\sqrt{2}\right)}\right)\mathrm{d}v
\end{equation}
which s
\begin{equation}
A
=
=\dfrac{\ln\left(v-\sqrt{2}\right)}{2^\frac{3}{2}}-\dfrac{\ln\left(v+\sqrt{2}\right)}{2^\frac{3}{2}}
\end{equation}
Now, similarly
\begin{equation}
B =\ln\left(v+1\right)
\end{equation}
and
\begin{equation}
C =\ln\left(v-1\right)
\end{equation}
Plugging all $A,B,C$ back we get
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x
=-\dfrac{\ln\left(v+\sqrt{2}\right)}{\sqrt{2}}+\dfrac{\ln\left(v-\sqrt{2}\right)}{\sqrt{2}}+\dfrac{\ln\left(v+1\right)}{2}-\dfrac{\ln\left(v-1\right)}{2}
\end{equation}
Undoing the change of variable $v = \sqrt{1 - u^2}$, we get
\begin{equation}
{\displaystyle\int}\dfrac{\sqrt{\cos\left(2x\right)}}{\sin\left(x\right)}\,\mathrm{d}x=\dfrac{\ln\left(\sqrt{1-u^2}+\sqrt{2}\right)}{\sqrt{2}}-\dfrac{\ln\left(\sqrt{1-u^2}-\sqrt{2}\right)}{\sqrt{2}}-\dfrac{\ln\left(\sqrt{1-u^2}+1\right)}{2}+\dfrac{\ln\left(\sqrt{1-u^2}-1\right)}{2}
\end{equation}
Undoing the other change of variable $u = \tan (x)$, we get
\begin{equation}
\dfrac{\ln\left(\sqrt{1-\tan^2\left(x\right)}+\sqrt{2}\right)}{\sqrt{2}}-\dfrac{\ln\left(\sqrt{1-\tan^2\left(x\right)}-\sqrt{2}\right)}{\sqrt{2}}-\dfrac{\ln\left(\sqrt{1-\tan^2\left(x\right)}+1\right)}{2}+\dfrac{\ln\left(\sqrt{1-\tan^2\left(x\right)}-1\right)}{2}
\end{equation}
Now, depending where you're integrating, you've got to have absolute values in the arguments of the logarithms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2896148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continued fraction of $ \sqrt{3} $ How can I solve the continued fraction of $\sqrt{3}$?
I know that a continued fraction is [1;1,2,1,2,1,2,....] -> this is the continued fraction of $\sqrt{3}$, but how can I solve? What are the steps?
I don't understand this proof.
Thank you so much.
| Method described by Prof. Lubin at Continued fraction of $\sqrt{67} - 4$
$$ \sqrt { 3} = 1 + \frac{ \sqrt {3} - 1 }{ 1 } $$
$$ \frac{ 1 }{ \sqrt {3} - 1 } = \frac{ \sqrt {3} + 1 }{2 } = 1 + \frac{ \sqrt {3} - 1 }{2 } $$
$$ \frac{ 2 }{ \sqrt {3} - 1 } = \frac{ \sqrt {3} + 1 }{1 } = 2 + \frac{ \sqrt {3} - 1 }{1 } $$
Simple continued fraction tableau:
$$
\begin{array}{cccccccccc}
& & 1 & & 1 & & 2 & \\
\\
\frac{ 0 }{ 1 } & \frac{ 1 }{ 0 } & & \frac{ 1 }{ 1 } & & \frac{ 2 }{ 1 } \\
\\
& 1 & & -2 & & 1
\end{array}
$$
$$
\begin{array}{cccc}
\frac{ 1 }{ 0 } & 1^2 - 3 \cdot 0^2 = 1 & \mbox{digit} & 1 \\
\frac{ 1 }{ 1 } & 1^2 - 3 \cdot 1^2 = -2 & \mbox{digit} & 1 \\
\frac{ 2 }{ 1 } & 2^2 - 3 \cdot 1^2 = 1 & \mbox{digit} & 2 \\
\end{array}
$$
........................
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2898083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding the value of $\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42} $ given a system of three equations Let $a, b, c, x, y, z$ be real numbers that satisfy the three equations
$$ 13x+by+cz=0 $$
$$ ax+23y+cz=0 $$
$$ ax+by+42z=0 $$
Suppose that $ a\neq13 $ and $x\neq0$. What is the value of $$\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42} $$
I tried
$$ (13-a)x+(b-23)y=0 $$
$$ (23-b)y+(c-42)z=0 $$
$$ (13-a)x+(c-42)z=0 $$
$$ (a-13)x=(b-23)y=(c-42)z $$
But I don't know how to continue further
Maybe
$$ \frac{1}{(a-13)x}=\frac{1}{(b-23)y}=\frac{1}{(c-42)z} $$
But any hint will be appreciated
| Your current progress is excellent.
Note that $$\frac{1}{(a-13)x}=\frac{1}{(b-23)y}=\frac{1}{(c-42)z}$$ means that $$\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42}=\frac{13x}{(a-13)x}+\frac{23y}{(b-23)y}+\frac{42z}{(c-42)z}$$ or that $$\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42}=\frac{13x+23y+42z}{(a-13)x}$$ due to the equalities.
Adding the three equations given gives $$13x+23y+42z=-2(ax+by+cz)=-2(ax-13x)=-2x(a-13)$$ from the first equation hence $$\boxed{\frac{13}{a-13}+\frac{23}{b-23}+\frac{42}{c-42}=\frac{-2(a-13)x}{(a-13)x}=-2}$$
P.S. The values of $13,23,42$ are totally arbitrary. This works for any triplet of non-zero integers.
| {
"language": "en",
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"question_score": "7",
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Coefficient of $x^{27}$ in $(1-x^{10})^6/(1-x)^6$. It's been a while since I have been playing with these so excuse me if it is too obvious.
How can I represent
$$ f(x) = \frac{(1-x^{10} ) ^6}{(1-x)^6} $$
as sum of powers of $x$
I am especially interested in the coefficient in front of $x^{27} $ in that sum.
The book I am reading gives this coefficient as obviously being
$${32 \choose 5}- {6 \choose 1} {22 \choose 5}+{6 \choose 2}{12 \choose 5}$$
but I don't know where this comes from.
Many thanks in advance.
| First from the geometric series
$$
\frac{1}{1-x}=\sum_{n=0}^\infty x^n
$$
combined with repeated differentiation yields that
$$
\frac{1}{(1-x)^6}=\sum_{n=0}^\infty\binom{n+5}{5} x^n\tag{1}
$$
(for $|x|<1$, if you don't want to deal with formal series). The binomial theorem implies that
$$
(1-x^{10})^6=1-6x^{10}+\binom{6}{2}x^{20}-\binom{6}{3}x^{30}+\binom{6}{4}x^{40}-\binom{6}{5}x^{50}+x^{60}\tag{2}.
$$
We want the coefficient of $x^{27}$ in the product of (1) and (2). We get an $x^{27}$ term from the products $1\times x^{27}$, $x^{10}\times x^{17}$ and $x^{20}\times x^{7}$. In particular the coefficient is then
$$
\binom{27+5}{5}-6\binom{17+5}{5}+\binom{6}{2}\binom{7+5}{5}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2900106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove that $a+b+c\le 0$ $a,b,c$ are integer numbers different from zero and
$$\frac{a}{b+c^2}=\frac{a+c^2}{b}$$
Prove that $a+b+c\le 0$.
I know how to prove that $a+b<0$ but do not know what about $c$.
| From given equation we get $a+b+c^2 =0$ (since $c\ne 0$), so we have $$a+b+c= c-c^2 = c(1-c) \leq 0$$
If $c>0$ then $1-c\leq 0$ so $c(1-c)\leq 0$ and
if $c<0$ then $1-c>0$ so $c(1-c)<0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
convergence of sequence defined recursively $\frac{2}{a_{n+2}}=\frac{1}{a_{n+1}}+\frac{1}{a_n}$ How to prove that the sequence ${a_n}$ defined by $\frac{2}{a_{n+2}}=\frac{1}{a_{n+1}}+\frac{1}{a_n}$ for $n\geq1$ and $0\lt{a_1}\lt{a_2}$ converges? How to find its limit? I truly do not know how to proceed further. I tried to use A.M>G.M inequality but with no success.
| Note that $2\left(\frac{1}{a_{n+2}}-\frac{1}{a_{n+1}} \right)=-\left(\frac{1}{a_{n+1}}-\frac{1}{a_{n}}\right)$, thus with $b_n = \frac{1}{a_{n+1}}-\frac{1}{a_{n}}$, $b_{n+1}=-\frac 12 b_n$.
Hence $b_{n+1} = \frac{(-1)^n}{2^n}b_1$, so $\sum_n b_n$ converges, thus $\frac1{a_n}$ converges, and so does $a_n$.
Let $\ell=\lim_n a_n$. Then $\sum_{n=1}^\infty b_n = \frac 1\ell -\frac 1{a_1}$, which rewrites as $\frac 23 \left(\frac 1{a_1} - \frac 1{a_2} \right) = \frac 1\ell -\frac 1{a_1}$, hence $$\ell = \frac{1}{\frac 23 \left(\frac 1{a_1} - \frac 1{a_2} \right)+\frac 1{a_1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Writing $(\sqrt{3}+3i)^{18}$ and $(i-1)^{-11}$ in the form $a+bi$ I have two current math problems I just can't solve. I'm to express the following in the form $a+bi$:
$(\sqrt{3}+3i)^{18} \qquad\text{and}\qquad (i-1)^{-11}$
The answer for the first one is $12^9$; and for the second, it's $\frac{1}{64}(1-i)$.
The results I get keep ending up into something that isn't even worth mentioning.
| This is a solution to the problem with as few prerequisites as possible. We have
$$(\sqrt{3}+3\cdot i)^{18}= \big[(\sqrt{3}+3\cdot i)^{3}\big]^{6}.
$$
Note that
\begin{align}
(\sqrt{3}+3\cdot i)^{2}=&(\sqrt{3}+3\cdot i)\cdot (\sqrt{3}+3\cdot i)\\
=& 3 +6\sqrt{3}\cdot i-9\\
=& -6+6\sqrt{3}\cdot i
\end{align}
and
\begin{align}
(\sqrt{3}+3\cdot i)^{3}=&(\sqrt{3}+3\cdot i)\cdot (-6+6\sqrt{3}\cdot i)\\
=& \sqrt{3}\cdot (-6+6\sqrt{3}\cdot i)+3\cdot i\cdot(-6+6\sqrt{3}\cdot i)\\
=& -6\sqrt{3}+18\cdot i-18\cdot i-18\sqrt{3}\\
=& -24\cdot\sqrt{3}\\
\end{align}
Now it's up to you. In the same way, do the other exercise.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$? How can I find the limit of this sequence $(n^2 + 1)^{1/3} - (n^2 + n)^{1/3}$ ? I know that if it is a square root I will multiply by the conjugate. but if it is a cubic root what shall I do?
Thanks
| Use the fact that
$$ a^3-b^3 = (a-b)(a^2+ab+b^2) \quad\Leftrightarrow\quad a-b = \frac{a^3-b^3}{a^2+ab+b^2}. $$
Then
\begin{align}
(n^2+1)^{1/3}-(n^2+n)^{1/3}
&= \frac{(n^2+1)-(n^2+n)}{(n^2+1)^{2/3}+(n^2+1)^{1/3}(n^2+n)^{1/3}+(n^2+n)^{2/3}}\\[0.2cm]
&= \frac{1-n}{(n^2+1)^{2/3}+(n^2+1)^{1/3}(n^2+n)^{1/3}+(n^2+n)^{2/3}}\\[0.2cm]
&= \frac{\frac{1}{n}-1}{\left(n^{1/2}+n^{-3/2}\right)^{2/3} + \left(n+1+\frac{1}{n}+\frac{1}{n^2}\right)^{1/3} + \left(n^{1/2}+n^{-1/2}\right)^{2/3}}\\[0.2cm]
&\to \frac{0-1}{(\infty+0)^{2/3}+(\infty+1+0+0)^{1/3}+(\infty+0)^{2/3}} = 0.
\end{align}
Wolfram confirms.
Now that I think about it, it would have probably been much simpler to use asymptotics.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A problem on transformation of a random variable.
$ f(x) =
\begin{cases}
1 & \text{if $0\le x \le 1$ } \\
0 & \text{otherwise }\end{cases}$
Find the PDF of Random variable $Y=\dfrac{X}{1+X}$
$F(Y\le y)=F\bigg(\dfrac{X}{1+X}\le y\bigg)=F\bigg(X\le \dfrac{y}{1-y}\bigg)$
$U=F_x\bigg(\dfrac{y}{1-y}\bigg)$
$\dfrac{dU}{dy}=f(y)\bigg(\dfrac{1}{(1-y)^2}\bigg)$
$f(y)=\bigg(\dfrac{1}{(1-y)^2}\bigg)$
Please check if I did any mistake. I am not sure how to figure out domain after transformation. Heres my try.
$0\le x \le 1$
$1\le x+1\le2$
$1\ge \dfrac{1}{x+1}\ge\dfrac{1}{2}$
$x\ge \dfrac{x}{x+1}\ge\dfrac{x}{2}$
$x\ge y\ge\dfrac{x}{2}$
Is this a correct way to approach result?
| There are simple and practical reasons why one writes $F_X(x)$ and not $F_x(X)$ or $F_x(x)$ or $F_X(X).$ If one does not know about this, how would one even understand something like $\Pr(X\le x)\text{ ?}$
\begin{align}
\text{For } 0\le y \le \frac 1 2,\text{ we have } & \frac X {1+X} = 1 - \frac 1 {1+X} \le y \\[10pt]
\text{iff } & \frac 1 {1+X} \ge 1 - y \\[10pt]
\text{iff } & 1 + X \le \frac 1 {1-y} \\[10pt]
\text{iff } & X \le \frac 1 {1-y} - 1 = \frac y {1-y}. \\[10pt]
\text{Therefore for } 0 \le y \le \frac 1 2, \quad & \Pr\left( \frac X{1+X} \le y \right) = \Pr\left( X\le \frac y {1-y} \right) \\[10pt]
= {} & \int_0^{y/(1-y)} 1\, dx = \frac y {1-y}. \\[12pt]
\text{Hence } & f_Y(y) = \frac d {dy}\, \frac y {1-y} = \cdots.
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Given $\sin(t) + \cos(t) = a$, derive an expression in '$a$' for $(\cos(t))^4 + (\sin(t))^4$ I received this question from a student's trigonometry review assignment. After spending an embarrassing amount of time on it, I consulted others and learned that nobody has been able to solve this problem for multiple semesters. I wonder if there is a typo in the statement or if I just haven't been rigorous enough? Here is the question:
$$\text{Given } \sin{t} + \cos{t} = a, \text{ find an equivalent expression for } \sin^4{t} + \cos^4{t} \text{ in terms of } a.$$
Has anybody seen this one before? I tried (among other things) this (which is an approximation of an earlier attempt):
$(\sin{t} + \cos{t})^4$ and using whatever identities I could remember;
$(\sin{t} + \cos{t})^4 = \sin^4{t} + \cos^4(t) + 4\sin{t}\cos^3{t} + 6\sin^2{t}\cos^2{t} + 4\sin^3{t}\cos{t}$
so then
$\begin{align}
\sin^4{t} + \cos^4{t} &= (\sin{t} + \cos{t})^4 - 4\sin{t}\cos^3{t} - 6\sin^2{t}\cos^2{t} - 4\sin^3{t}\cos{t}\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(2\cos^2{t} + 3\sin{t}\cos{t} + 2\sin^2{t})\\
&= (\sin{t} + \cos{t})^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)\\
&= a^4 - 2\sin{t}\cos{t}(3\sin{t}\cos{t} + 4)
\end{align}$
Couldn't get further than this, felt like I was overthinking it.
| Alternatively, note that:
$$\begin{align}\sin t+\cos t=a \Rightarrow \sin\left(t+\frac{\pi}{4}\right)=\frac a{\sqrt{2}} \Rightarrow t&=\arcsin \frac a{\sqrt{2}}-\frac{\pi}{4}\\
\sin t=\sin \left(\arcsin \frac a{\sqrt{2}}-\frac{\pi}{4}\right)&=\frac a{\sqrt{2}}\cdot \frac 1{\sqrt{2}}-\sqrt{1-\frac{a^2}{2}}\cdot \frac1{\sqrt{2}};\\
&=\frac12\left(a-\sqrt{2-a^2}\right);\\
\cos t=\cos \left(\arcsin \frac a{\sqrt{2}}-\frac{\pi}{4}\right)&=\sqrt{1-\frac{a^2}{2}}\cdot \frac1{\sqrt{2}}+\frac a{\sqrt{2}}\cdot \frac 1{\sqrt{2}}=\\
&=\frac12\left(\sqrt{2-a^2}+a\right).\end{align}$$
Hence:
$$\begin{align}\sin^4t+\cos^4t&=\frac{2}{16}\cdot \left(a^4+6a^2\left(2-a^2\right)+(2-a^2)^2\right)=\\
&=-\frac{a^4}{2}+a^2+\frac12. \end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the smallest and highest value of the product $xyz$ Find the smallest and highest value of the product $xyz$ assuming that:
$x + y + z = 10$ and
$x^2 + y^2 + z^2 = 36$.
I calculated this:
$x+y+z=10 => (x+y+z)^2=10^2$
$x^2+y^2+z^2+2xy+2yz+2zx=100$
$(x^2+y^2+z^2+2xy+2yz+2zx)-(x^2+y^2+z^2)=100-36$
$2xy+2yz+2zx=64$
$xy+yz+zx=32$
I'm stuck. What is the next step to this exercise?
My idea is to show the equation using one variable and after computing the derivative reach global extremes.
| We have $$x^2+y^2=36-z^2$$ and $$x+y=10-z,$$ which gives
$$(10-z)^2-2xy=36-z^2$$ or
$$xy=32-10z+z^2$$ and $$xyz=32z-10z^2+z^3.$$
Also, $$(x+y)^2\geq4xy,$$ which gives
$$3z^2-20z+28\leq0$$ or
$$2\leq z\leq\frac{14}{3}.$$
Can you end it now?
I got $\max(xyz)=\frac{896}{27}$ and $\min(xyz)=32$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2906604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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CDF of function of two random variables I'm a little confused about how to approach the below question:
Q: Let $\,W=Y-X$ and determine the CDF of $W$.
I've read the solution, here, but this was not very clear and I was wanting an approach from calculus if possible (unless this is too messy).
I understand that $w=-1$ is the point that splits the region but am not sure how to set the integrals up - any help appreciated.
| The question is to find the cdf of $W$ where $W=Y-X$ in order to do that we know that the joint density of $W$ is given by
$$ F_{W}(w) = \iint_{D} f_{X,Y}(x,y) dx dy \tag{1}$$
Where the joint density is split into two uniform densities
$$ f_{X,Y}(x,y) =\begin{align}\begin{cases} \frac{1}{2} & 0 \leq x \leq 1 , 0\leq y\leq 1 \\ \\ \frac{3}{2} & 1 \leq x \leq 2 , 0 \leq y \leq 1 \end{cases} \end{align} \tag{2}$$
It notes the following
$$ P(W \leq w) = P(Y-X \leq w ) =P(Y \leq X +w) \tag{3}$$
Which gives us this picture
While you could use calculus and I actually spent a while preparing an answer with it. There is no point. They are made of triangles it notes that in the answer.
The probabilities of interest can be calculated by taking advantage of
the uniform PDF over the two triangles.
Now instead if you actually look at the picture. I am pasting it
The area of a triangle is $A = \frac{1}{2}b \cdot h $
$$ I_{1} = f_{X,Y}(x,y) \frac{1}{2} \frac{(2+w)}{2} (2+w) \tag{4} $$
$$ I_{1} = \frac{3}{2} \frac{1}{2} \frac{(2+w)}{2} (2+w) \tag{5} $$
$$ I_{1} = \frac{3}{2} \frac{1}{2} \frac{(2+w)^{2}}{2} \tag{6} $$
The other region is $1+w$ which would be only $\frac{1}{2}$ for the joint density. It is the triangular region in the middle.
$$ I_{2} = f_{X,Y}(x,y) \frac{1}{2} (1+w)(1+w) \tag{7} $$
$$ I_{2} = \frac{1}{2} \frac{1}{2} (1+w)^{2} \tag{8} $$
Look at the top now. See the triangle. The base is $-w$ and the height is $\frac{-w}{2}$
$$ I_{3} = \frac{1}{2}\frac{-w}{2} \cdot -w = \frac{w^{2}}{4}\tag{9} $$
We want the region for $w \in (-1,0)$ so now the area of the whole right side is
$$ I_{4} = \frac{1}{2} 1 \cdot 1 = \frac{1}{2} \tag{10}$$
If we subtract off the triangle before we get this area
So we let
$$ I_{5} = f_{X,Y}(x,y)(I_{4} -I_{3}) \tag{11} $$
$$ I_{5} = \frac{3}{2}(\frac{1}{2} -\frac{w^{2}}{4}) \tag{12} $$
Note this a cumulative density. We just added the region for $I_{1}$ now let's add $I_{2}$ call it $I_{6}$ and make our $F_{W}(w)$
$$ I_{6} = I_{2} + I_{5} \tag{13}$$
$$ I_{6} = \frac{1}{2} \frac{1}{2} (1+w)^{2} + \frac{3}{2}(\frac{1}{2} -\frac{w^{2}}{4}) \tag{14}$$
$$ f_{W}(w) =\begin{align}\begin{cases} 0 & \textrm{ if } w < -2 \\ \\ \frac{3}{2} \frac{1}{2} \frac{(2+w)^{2}}{2} & \textrm{ if } -2 \leq w \leq -1 \\ \\ \frac{1}{2} \frac{1}{2} (1+w)^{2} + \frac{3}{2}(\frac{1}{2} -\frac{w^{2}}{4}) & \textrm{ if } -1 < w \leq 0 \\ \\ 1 & \textrm{ if } w > 0 \end{cases} \end{align} \tag{15}$$
fixing this like they did
$$ f_{W}(w) =\begin{align}\begin{cases} 0 & \textrm{ if } w < -2 \\ \\ \frac{3}{8} (2+w)^{2} & \textrm{ if } -2 \leq w \leq -1 \\ \\ \frac{1}{8}\big(-w^{2} +4w +8 \big) & \textrm{ if } -1 < w \leq 0 \\ \\ 1 & \textrm{ if } w > 0 \end{cases} \end{align} \tag{16}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2908704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does the Prime Number Theorem imply for large enough $x$ that $(a+1)\pi(ax) \ge a\pi((a+1)x)$ Give an integer $a \ge 1$, does the Prime Number Theorem imply for large enough $x$ that:
$$(a+1)\pi(ax) \ge a\pi((a+1)x)$$
where $\pi(x)$ is the prime counting function.
Using the formula of Pierre Dusart found here for $x > 598$:
$$\left(\frac{x}{\log x}\right)\left(1 + \frac{0.992}{\log x}\right) \le \pi(x) \le \left(\frac{x}{\log x}\right)\left(1 + \frac{1.2762}{\log x}\right)$$
this is true for $a=1$.
Given a large enough $x$, could it possibly be true for all $a$?
Edit: Changed [] to () to avoid any confusion based on a question received in the comments.
| We know that $ \frac{x}{\ln x}(1+\frac{1}{\ln x} +\frac{2}{\ln^2 x})\leq \pi(x) \leq \frac{x}{\ln x}(1+\frac{1}{\ln x}+ \frac{2.334}{\ln^2 x})$ for all $x\geq 3*10^{11}$
So for all $a\geq 1$ and $a \in \mathbb{R}$ we have that $ (a+1) \pi(a x)-a \pi((a+1)x) > (a+1) \frac{a x}{\ln a x} (1+\frac{1}{\ln ax }+\frac{2}{\ln^2a x})-a \frac{(a+1)x}{\ln(a+1)x}(1+\frac{1}{\ln(a+1)x}+\frac{2.334}{\ln^2 (a+1)x}) >0$ dividing by $a(a+1)$ to get
$ \frac{x}{\ln a x}(1+\frac{1}{\ln ax} +\frac{2}{\ln^2 ax})- \frac{x}{\ln(a+1)x}(1+\frac{1}{\ln(a+1)x}+\frac{2.334}{\ln^2(a+1)x})> (\frac{1}{\ln a x}-\frac{1}{\ln(a+1)x} )x (1+\frac{1}{\ln ax}+\frac{2}{\ln^2 ax })-\frac{0.334x}{\ln^3 (a+1)x} >\frac{\ln(1+1/a)}{\ln^2 (a+1)x} x-\frac{0.334 x}{\ln^3 (a+1)x}$
multiplying by $ \ln^2(a+1)x$ to get $ \ln(1+\frac{1}{a}) x > \frac{0.334x}{\ln a x} => \ln(1+1/a) > \frac{0.334}{\ln a x} => \frac{1}{a+0.5} > \frac{0.334}{\ln ax} $
giving $ \ln a x > 0.334(a+0.5) $ exponentiation both sides gives $ a x > e^{0.334 a+ 0.167} => x > \frac{1.2 e^{0.334a} }{a}$
So for every real number $a \geq 1$ the inequality is valid for all $ x \geq max(3*10^{11},\frac{1.2 e^{0.334a} }{a})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Consecutive zeros in the binary representation of $\sqrt{3}$ $\sqrt3=1.b_1b_2...$is the binary representation of $\sqrt3$.
i.e. $\sqrt3=1+\dfrac{b_1}{2^1}+\dfrac{b_2}{2^2}+...$
Prove that at least one of the digits $b_n,b_{n+1},...,b_{2n}$ is 1.
my attempt:
Square both sides: $$\left(1+\sum_{i=1}^\infty\frac{b_i}{2^i}\right)^2=3$$
Expand and subtract $1$ from both sides
$$2\sum_{i=1}^\infty\frac{b_i}{2^i}+\sum_{i=1}^\infty\frac{b_i^2}{2^{2i}}=2$$
divide both side by 2
$$\sum_{i=1}^\infty\frac{b_i}{2^i}+\sum_{i=1}^\infty\frac{b_i^2}{2^{2i+1}}=1$$
Tidy up
$$\sum_{i=1}^\infty\frac{1}{2^i}\left(b_i+\frac{b_i^2}{2^{i+1}}\right)=1$$
Lemma: $\displaystyle\sum_{i=m}^\infty\frac{1}{2^i}\left(1+\frac{1}{2^{i+1}}\right)<\frac{1}{2^{m-2}}$
Proof: multiply both side by $2^{m-1}$$$\sum_{i=1}^\infty\frac{1}{2^i}+\frac{1}{2^{m+1+i}}<2$$which is trivial for any positive integer m.
Proceed the proof by contradiction: suppose $b_n,b_{n+1},...b_{2n}$ are all equal to $0$. Then the sum $\sum_{i=1}^{n-1}\frac{1}{2^i}\left(b_i+\frac{b_i^2}{2^{i+1}}\right)$ is in the form of $1-\frac{x}{2^{2n-1}}$.
But $$\sum_{i=2n+1}^\infty\frac{1}{2^i}\left(b_i+\frac{b_i^2}{2^{i+1}}\right)<\frac{1}{2^{2n-1}}$$by lemma.
Q.E.D.
Is this proof correct?
I am also happy for an alternative (and hopefully shorter) solution.
| The flaw in your proof was pointed out in the comments, so let me show you my approach
Assume $\sqrt 3$ has all digits $b_i=0$ for $i\in\{n,...,2n\}$ (these are $n+1$ digits). Write
$$\sqrt 3\cdot 2^{n-1}=N + \epsilon= b_1\cdots b_{n-1}.\underbrace{0\cdots0}_{n+1}b_{2n+1}\cdots$$
where $N=b_1\cdots b_{n-1}\in\Bbb N$ is the integer part, and $\epsilon=0.b_{n}b_{n+1}\cdots\in(0,1)$ is the fractional part (which is stricly positive because of the irrationality of $\sqrt{3}$). We have $N< 1.8\cdot 2^{n-1}$ (with $1.8$ being a rough upper bound for $\sqrt 3$). Further, $\epsilon$ must have zeros as the first $n+1$ digits $b_n,...,b_{2n}$. So the largest possible value is
$$\epsilon \le 0.\underbrace{0\dots0}_{n+1}\overline 1=0.\underbrace{0\cdots 0}_n1=\frac1{2^{n+1}}$$
Now observe that
$$\underbrace{3\cdot 2^{2n-2}}_{\text{integer}}=(\sqrt 3\cdot2^{n-1})^2=N^2+2N\epsilon+\epsilon^2.$$
The left side is an integer, and so is $N^2$. Note that $0<2N\epsilon< 0.9$. And finally, since $\epsilon^2<1/2^{2n+2}<0.1$, we cannot close the gap to make the right side a whole integer too.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Evaluate $ \lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right) $ $$\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)$$
Using the L'Hospital Rule I obtained the value $-1/4$, but the answer is given to be $-1/3$. I can't find the mistake. Here's what I did; please point out the mistake.
\begin{align}
\lim_{x\to0}\left({\frac{1}{x^2}}-{\frac{1}{\sin^2x}}\right)&=\lim_{x\to0}\frac{(\sin x+x)(\sin x-x)}{(x\sin x)(x\sin x)} \\[1ex]
&=\lim_{x\to0}\left(\frac{\sin x+x}{x\sin x}\right)\lim_{x\to0}\left(\frac{\sin x-x}{x\sin x}\right) \\[1ex]
&=\lim_{x\to0}\left(\frac{\cos x+1}{\sin x+x\cos x}\right)\lim_{x\to0}\left(\frac{\cos x-1}{\sin x+x\cos x}\right) \\[1ex]
&=\lim_{x\to0}\:(\cos x+1)\,\lim_{x\to0}\left(\frac{\cos x-1}{(\sin x+x\cos x)^2}\right) \\[1ex]
&=\lim_{x\to0}\frac{-\sin x}{(\sin x+x\cos x)(2\cos x-x\sin x)} \\[1ex]
&=-\lim_{x\to0}\left[\frac{1}{1+\cos x\left(\frac{x}{\sin x}\right)}\right]\left(\frac{1}{2\cos x-x\sin x}\right) \\[1ex]
&=-\frac{1}{2}\left[\lim_{x\to0}\,\frac{1}{1+\cos x}\right] \\[1ex]
&=-\frac{1}{4}
\end{align}
| As noticed in the comments, we are allowed to proceed as follows
$$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)= \lim_{x \to 0} \left( \frac{\sin^2 x-x^2}{x^2\sin^2 x} \right)=\lim_{x \to 0} \left( \frac{\sin x+x}{x\sin x} \right)\left( \frac{\sin x-x}{x\sin x} \right)=\ldots$$
but we are not allowed to proceed as follows
$$\ldots=\lim_{x \to 0} \left( \frac{\sin x+x}{x\sin x} \right)\lim_{x \to 0}\left( \frac{\sin x-x}{x\sin x} \right)=\ldots$$
when one or both limits do not exist or the product leads to an undefined expression.
Notably in that case by l'Hopital we obtain
$$\ldots=\lim_{x \to 0} \frac {\cos x+1} {\sin x+x\cos x}\cdot \lim_{x \to 0} \frac {\cos x-1} {\sin x+x\cos x}=\ldots$$
and the LHS limit, in the form $\frac 2 0$, doesn't exist while the RHS limit is equal to zero.
Therefore the initial step in that case doesn't work.
Note that in any case also the following step
$$ \ldots=\lim_{x \to 0} (\cos x+1)\,\lim_{x \to 0} \frac {\cos x-1} {(\sin x+x\cos x)^2}=\ldots$$
is not allowed since once we have divided the original limit as the product of two distinct limits we need to operate separetely on each of them when using l'Hopital or Taylor's series. Only when we have calculated the limit for each expression we know whether the initial step was allowed or not.
See also the related Analyzing limits problem Calculus (tell me where I'm wrong).
In that case, following for example the hint given by mrs, a correct way to proceed by l'Hopital is as follows
$$\lim_{x \to 0} \left( {\frac{1}{x^2}} - {\frac{1} {\sin^2 x} }\right)
= \lim_{x \to 0}\left(\frac{\sin^2 x-x^2}{x^4}\cdot\frac{x^2}{\sin^2 x}\right)
\stackrel{?} = \lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}\cdot\lim_{x \to 0}\frac{x^2}{\sin^2 x }=\ldots$$
and since, using l'Hopital for each part, we have
$$\lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}=\lim_{x \to 0}\frac{\sin 2x-2x}{4x^3}=\lim_{x \to 0}\frac{2\cos 2x-2}{12x^2}=\lim_{x \to 0}\frac{-4\sin 2x}{24x}=\lim_{x \to 0}\frac{-8\cos 2x}{24}=-\frac13$$
$$\lim_{x \to 0}\frac{x^2}{\sin^2 x }=\lim_{x \to 0}\frac{2x}{\sin 2x }=\lim_{x \to 0}\frac{2}{2\cos 2x }=1$$
we see that the initial step is allowed and then we can conclude that
$$\ldots= \lim_{x \to 0}\frac{\sin^2 x-x^2}{x^4}\cdot\lim_{x \to 0}\frac{x^2}{\sin^2 x }=-\frac13\cdot 1 =-\frac13$$
Note finally that some intermediate step can be highly simplified using the standard limit $\lim_{x \to 0}\frac{\sin x }x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2910595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 8
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Solve $x,y$ given $1+x^2+2x\sin(\arccos y)=0$ I have to solve $x,y$ given-
$$1+x^2+2x\sin(\arccos y)=0$$
My attempt:
$$1+x^2+2x\sin(\arcsin \sqrt{1-y^2})=0$$ (Is this step valid? Can I convert arccos to arcsin like this?)
$$\implies1+x^2+2x\sqrt{1-y^2}=0$$
Now,I can't procced further. What to do next?
| Solving for x and y, $$1+x^2+2x\sin(\arccos y)=0$$
Firstly, if we take $z=\arccos(y)$ then $y=\cos(z)$ and so $\cos^2(z)=y^2$
we know that $\sin^2(z)+\cos^2(z)=1$ so
$1-\sin^2(z)=y^2\,\therefore \sin(z)=\sqrt{1-y^2}\therefore z=\arcsin(\sqrt{1-y^2})$
If we put this back into our original expression we get:
$$1+x^2+2x\sin\left(\arcsin\left(\sqrt{1-y^2}\right)\right)=0$$
so:
$$1+x^2+2x\sqrt{1-y^2}=0$$
this can be rearranged to give:
$$\sqrt{1-y^2}=-\frac{1+x^2}{2x}$$
This is the same as what you put, so it is correct so far. Now I would rearrange to:
$$y=\sqrt{1-\left(\frac{1+x^2}{2x}\right)^2}$$
For $y=0$,
$$\frac{1+x^2}{2x}=\pm1$$
solving this we get two equations, $x^2+2x+1=0$ and $x^2-2x+1=0$
these can be factorised to $(x+1)^2=0$ and $(x-1)^2=0$ giving the two clear solutions of $x=1$ and $x=-1$
So the answer is $x=\pm1$
| {
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"url": "https://math.stackexchange.com/questions/2910924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of $\sum_{n=1}^{\infty}\left(\left(\sqrt{n}\right)\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right)^{-1}$ Convergence of the following series as $\alpha \in \mathbb{R}$
$$\sum_{n=1}^{\infty}\left(\left(\sqrt{n}\right)\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right)^{-1}$$
As $n \to +\infty$ we have $a_n\sim \left(n^{\frac{1}{2}}\left(n^{\alpha}-n-1\right)\right)^{-1} = \left(n^{\frac{2\alpha+1}{2}}-n^\frac{3}{2}-n^\frac{1}{2}\right)^{-1}$
Hence, as $0<\alpha<1$ we have $a_n =\mathcal{O}\left(\frac{1}{n^\frac{3}{2}}\right)$, as $n \to +\infty$, that converges.
As $\alpha>1$ we have $a_n=\mathcal{O}\left(\frac{1}{n^\frac{2\alpha+1} {2}}\right)$, as $n \to +\infty$, which converges in turn.
Is it right or I got rid of too much informations in the asymptotic expansion?
| We have that
$$\cos\left(\frac{1}{n}\right)=1+O\left(\frac1{n^2}\right)$$
$$n^{\alpha\cos(\frac{1}{n})}=e^{\alpha\cos(\frac{1}{n})\log n}=e^{\alpha\log n+O\left(\frac{\log n}{n^2}\right)}=n^{\alpha}\left(1+O\left(\frac{\log n}{n^2}\right)\right)=n^{\alpha}+O\left(\frac{\log n}{n^{2-\alpha}}\right)$$
$$n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)=n^{\alpha}+O\left(\frac{\log n}{n^{2-\alpha}}\right)-n-1+O\left(\frac1{n^2}\right)$$
and therefore
*
*for $\alpha<1$
$$\left[\sqrt{n}\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right]^{-1}
\sim-\frac1{n\sqrt n}$$
*
*for $\alpha=1$
$$\left[\sqrt{n}\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right]^{-1}
\sim-\frac1{\sqrt n}$$
*
*for $\alpha>1$
$$\left[\sqrt{n}\left(n^{\alpha\cos(\frac{1}{n})}-n-\cos\left(\frac{1}{n}\right)\right)\right]^{-1}
\sim \frac1{n^{\alpha+\frac12}}$$
and the given series converges if and only if $\alpha \neq 1$.
| {
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"url": "https://math.stackexchange.com/questions/2914114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to reduce this to have only one x ... Let $f(x)=\frac{5}{x+4} $
Reduce the difference quotient in the alternate definition of the derivative below so that you only have one x:
$$\frac{f(x) - f(2)}{x-2}$$
I've gotten down to $\frac{5(10-x)}{x-2}$ . but I can't figure out how to reduce it to one $x$. Am I doing something wrong? I have tried long division, which ended up being incorrect, so I have no idea.
| Multiply up and down by $x + 4$
$$\frac{f(x) - f(2)}{x-2}=\frac{5-\frac{5}{6}(x+4)}{(x-2)(x+4)}$$
Multiply up and down by $6$
$$\frac{30-5(x+4)}{6(x-2)(x+4)} = \frac{-5x +10}{6(x-2)(x+4)} = \frac{-5(x-2)}{6(x-2)(x+4)} = -\frac{5}{6}\frac{1}{x+4}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2914788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Inequality, how to know intuition behind it I was solving the following inequality
For $a$, $b$, $c$ and $d$ being positive real numbers
which goes as
$$
\frac{a}{a+b} + \frac{b}{b+c} + \frac{c}{c+d} + \frac{d}{d+a} \leq \frac{a}{b+c} + \frac{b}{c+d} + \frac{c}{d+a} + \frac{d}{a+b}
$$
Which I was successful able to do
But I am not able to understand the intuition behind the inequality, and how someone even came to it.
Can someone help me to intuitively understand it
| By Cauchy-Schwarz inequality,
$$(\frac{a-b}{b+c}+\frac{b-c}{c+d}+\frac{c-d}{a+d}+\frac{d-a}{a+b})(b+c+c+d+a+d+a+b)\ge(a-b+b-c+c-d+d-a)^2=0$$
We need to prove the first bracket is positive, but since $a,b,c,d$ are positive, it's indeed positive.
Equality occurs iff $a-b=b-c=c-d=d-a$, i.e. $a=b=c=d.$
The intuition behind this is that $\sum_{cyc}\frac a{a+b}$ is relatively closer to $1$ than $\sum_{cyc}\frac d{a+b}$. Consider this: the sum of the numerator and the denominator are the same, but RHS is more spread out. Then which side's is bigger? This is not mathematically rigorous, but that's my first instinct. Also, these inequalities can be made by permutating Cauchy-Schwarz, rearrangement inequality, and the Power Mean Inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Calculating the square root of 2 Since $\sqrt{2}$ is irrational, is there a way to compute the first 20 digits of it?
What I have done so far
I started the first digit decimal of the $\sqrt{2}$ by calculating iteratively so that it would not go to 3 so fast. It looks like this:
\begin{align}
\sqrt 2 & = 1.4^{2} \equiv 1.96\\
\sqrt 2 & = 1.41^{2} \equiv 1.9881\\
\sqrt 2 & = 1.414^{2} \equiv 1.999396\\
& \ldots
\end{align}
First I tell whether it passes such that $1.x^{2}$ would be not greater than 3.
If that passes, I will add a new decimal to it. Let's say $y.$ $1.xy^{2}$
If that y fails, I increment $y$ by 1 and square it again.
The process will keep repeating. Unfortunately, the process takes so much time.
| Using the fact that $\sin \frac{\pi}{4} = \frac{\sqrt{2}}{2}$, then we have to find $2 \sin \frac{\pi}{4}$.
We can approximate $\sin x$ using the Taylor series to three terms:
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!} + O(x^6),$$
so we have:
$$\sin \frac{\pi}{4} \approx \frac{\pi}{4} - \frac{(\pi/4)^3}{3!} + \frac{(\pi/4)^5}{5!} .$$
If we approximate $\pi$ as $\frac{22}{7}$, then we have $\frac{\pi}{4} = \frac{11}{14}$, then we have:
$$\sin \frac{\pi}{4} \approx\frac{11}{14} - \frac{(11/14)^3}{3!} + \frac{(11/14)^5}{5!},$$
which when you multiply by $2$ to get $\sqrt{2}$, gives $1.4147$, while the actual value is $1.4142$.
If we expand the Taylor series to more terms, or improve the approximation of $\pi$ (such as $\frac{355}{113}$), then we can get to $20$ correct digits.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2916718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
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"answer_id": 7
} |
There are $4$ numbers. First $3$ make an arithmetic progression. Last $3$ make a geometric progression.
There are four numbers. The first three make an arithmetic progression, and the last three make a geometric progression. The sum of first and last number is $37$. The sum of middle numbers is $36$. Find the numbers.
So I'm trying to solve this problem for about 2 hours now. I can't use any formulas because all I get $0=0$. I don't know all the terms in English so I'll just try to upload my notes here. Eh but it doesn't let me. At first i got that $a_1=-2d$, $d$ was $= -36$ so my numbers were $72$, $36$, $0$, $-35$. The problem is that last 3 numbers doesn't make a geometric progression.
| The first 3 are in arithmetic progression
$a-n, a, a+n, \cdots$
the last 3 are in geometric progression.
$\cdots, a, ra, r^2 a$
There are 4 numbers total.
That means that the 3d element must fit both the geometric and the arithmetic progression
$ra = a+n$
The middle two sum to 36
$a+ra=a(r+1) = 36\\
r = \frac{36}{a} - 1\\
36 -2a = n$
The first and last sum to 37
$a-n + r^2 a = 37$
And now we use the substitutions above to get everything in terms of a.
$a-n + r(ra) = 37\\
a-n + r(a+n) = 37\\
a-(36-2a) + r(a+36-2a) = 37\\
3a-36 + r(36-a) = 37\\
3a-36 + (\frac {36}{a} - 1)(36-a) = 37\\
3a - 36 + \frac {36^2}{a} - 36 - 36 + a = 37\\
4a + \frac{1296}{a} - 145 = 0$
Which multiplies into a quadratic.
$4a^2 -145a + 1269 = 0\\
(a-16)(4a-81) = 0$
$a = 16, a = \frac {81}{4}$
If our sequence is integers....
$12, 16, 20, 25$
If our sequence could include rationals...
$\frac {99}{4}, \frac {81}{4}, \frac {63}{4}, \frac {49}{4}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2920555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 3
} |
Prove an equality with floor function. Let $p\in \Bbb N \ne 0$ and $x\in \Bbb R$.
prove that
$$\left\lfloor \frac {\lfloor px \rfloor}{p} \right\rfloor=\lfloor x\rfloor$$
I tried using the double inequality
$$\lfloor px\rfloor \le px<\lfloor px\rfloor +1$$
and divided by $p$ but a small problem remains.
| $\lfloor px\rfloor \le px<\lfloor px\rfloor +1$
Right. So $\frac {\lfloor px\rfloor}p \le \frac{px}p<\frac {\lfloor px\rfloor +1}p$
$\frac {\lfloor px\rfloor}p \le x<\frac {\lfloor px\rfloor}p + \frac 1p$
...
But perhaps more to the point.
$\lfloor x\rfloor \le x < \lfloor x\rfloor + 1$
$p\lfloor x\rfloor \le px < p\lfloor x\rfloor + p$.
$p\lfloor x\rfloor \le \lfloor px \rfloor \le px < \lfloor px \rfloor + 1 \le p\lfloor x\rfloor + p$
$\lfloor x\rfloor \le \frac {\lfloor px \rfloor}p \le x < \lfloor x\rfloor + 1$
.... or to go back to your original idea:
Not $\lfloor x \rfloor \le x$ so $p\lfloor x \rfloor \le px$ but $\lfloor px \rfloor$ is the largest possible integer equal or less than $px$ so
$p\lfloor x \rfloor \le \lfloor px \rfloor$
So you have $p\lfloor x \rfloor \le \lfloor px \rfloor < px < \lfloor px \rfloor +1 < p\lfloor x \rfloor + p$ ....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2920698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
} |
Solving $x^4 -10x^3 + 26x^2 -10x +1 = 0$. Find the root of the equation
$$x^4 -10x^3 + 26x^2 -10x +1 = 0$$
My attempt:
If $a,b,c$ and $d$ are a roots then $a+b+c+d = 10,$
$(a+b)(c+d) + ab +cd = 26$
$(a+b)cd +ab(c+d) = 10$
$abcd =1$
Now i'm not able procceed further.
| .For palindromic polynomials (whose coefficients read the same from left to right), it is key to realize that for $x \neq 0$, we have $f(x) = 0$ if and only if $f(\frac 1x) = 0$. This is a routine check which I leave to you.
More precisely, note that $x = 0$ is not a root of the polynomial above. Therefore, dividing by $x^2$ and collecting terms, we have :
$$
x^2 + \frac 1{x^2} - 10\left(x + \frac 1x\right) + 26 = 0
$$
Note that $x^2 + \frac 1{x^2} = \left(x + \frac 1x\right)^2 - 2$. Therefore, setting $x + \frac 1x = y$, we get a quadratic polynomial in $y$, which can be solved to get two values of $y$, followed by four values of $x$.
More precisely, $y^2 -10y + 24 = 0$. So we get $y = 4,6$ so $x + \frac 1x = 4$ or $x + \frac 1x = 6$. Solve these to get $2 \pm \sqrt 3$ and $3 \pm 2 \sqrt 2$. Either from here, or from the earlier equations you may notice that they come in pairs whose product is $1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2922660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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} |
Solving $\sin\left(\frac{x}{x^2+1}\right)+\sin\left(\frac{1}{x^2+x+2}\right)=0$
I'm looking for a more elegant way to find the real solution
$$\sin\left(\frac{x}{x^2+1}\right)+\sin\left(\frac{1}{x^2+x+2}\right)=0.$$
Here is my way:
First write the equation as
$$\sin\left(\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+x+2)}\right)\cos\left(\frac{x}{2(x^2+1)}-\frac{1}{2(x^2+x+2)}\right)=0.$$
Then the real solution can be find by
$$\frac{x}{2(x^2+1)}+\frac{1}{2(x^2+x+2)}=0$$
which gives $x=-1$.
| I do not know if this is more elegant.
Let $f(x)=\sin x$, $g(x)=\frac{x}{x^2+1}$, and $h(x)=\frac{1}{x^2+x+2}$.
Then your equation becomes
$$f\left(g(x)\right)+f\left(h(x)\right)=0$$
Since $f(x)=\sin x$ is an odd function,
$$-f\left(h(x)\right)=f\left(-h(x)\right).$$
Then, your equation is that
$$f\left(g(x)\right)=f\left(-h(x)\right).$$
It is known that in the range $\left(-\frac{\pi}{2},\frac{\pi}{2}\right)$ the function $f(x)$ is increasing, and since $-\frac{1}{2}\leq g(x)\leq\frac{1}{2}$ and $0<h(x) \leq \frac{4}{7}$, your equation is equivalent to
$$g(x)=-h(x)\leftrightarrow \frac{x}{x^2+1}=-\frac{1}{x^2+x+2}.$$
Therefore, the real solution comes from
$$x^3+2x^2+2x+1=0,$$
which gives $x=-1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2927206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Extrema of $f(z)=\Big |\bar z(z-2)-2\Re(z) \Big|$ for $z \in \mathbb{C}$ I want to find extrema of $f(z)=\Big |\bar z(z-2)-2\Re(z) \Big|$ for $z \in \mathbb{C}$.
$f(z)=\Big |\bar z(z-2)-2\Re(z) \Big|=\Big| x(x-4)+y^2+i2y\Big|$
Then I defined $g(x,y):=f(z)^2=(x^2-4x+y^2)^2+4y^2$. Because $g$ is strictly monotone the extrema of $f$ and $g$ should be the same.
Now for finding the extrema I looked at the derivatives of $g$.
$g_x(x,y)=2(x^2-4x+y^2)(2x-4)$ and $g_y(x,y)=2(x^2-4x+y^2)2y+8y$. Then I tried to determine where $g_x$ and $g_y$ vanish.
$g_x(x,y)=0 \Leftrightarrow x_1=2$, $x_2=2-\sqrt{4-y^2},x_3=2+\sqrt{4-y^2}$
$g_y(x,y)=0 \Leftrightarrow y_1=0$, $y_2=\sqrt{-x^2+4x-2},y_3=-\sqrt{-x^2+4x-2}$
N0w, I need help with determing the critical points. $(2,0)$ should be one but how can I determine the others?
| Suppose that $(x,y)$ is a critical point of $g$. Since $g_x(x,y)=2(x^2-4x+y^2)(2x-4)$ you have two possibilities:
*
*$x=2$. In that case, $0=g_y(x,y)=2(2^2-4\times2+y^2)2y+8y=4y^3-8y$ and therefore $y=0$ or $y=\pm\sqrt2$.
*$x^2-4x+y^2=0$. Then $0=g_y(x,y)=8y$. So, $y=0$ and therefore $x=0$ or $x=4$.
Therefore, the critical points are $(2,0)$, $\left(2,\pm\sqrt2\right)$, $(0,0)$, and $(4,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2931574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
Find all cosets of $H$ and $K$ in $G$ Let $G = \mathbb{Z}_{10} \times \mathbb{Z}_{4}$ and $H = \langle(3, 2) \rangle$ and $K = \langle(4, 2)\rangle$. Find all cosets to H and K.
We have
\begin{equation*}
\begin{split}
G & = \mathbb{Z}_{10} \times \mathbb{Z}_{4} \\
& = \{0, 1, 2, \ldots, 9 \} \times \{0, 1, 2, 3 \} \\
& = \{(0, 0), (0, 1), (0, 2), (0, 3), (1, 0), \ldots, (1, 3), (2, 0), \ldots, (2, 3), \ldots, (9, 0), \ldots, (9, 3) \} ,
\end{split}
\end{equation*}
and
\begin{equation*}
\begin{split}
H & = \langle(3, 2)\rangle \\
& = \{k \cdot (3, 2) : k \in \mathbb{Z} \} \\
& = \{ (k \cdot 3, k \cdot 2) : k \in \mathbb{Z} \} \\
& = \{(0, 0), (3, 2), (6, 0), (9, 2), (2, 0), (5, 2), (8, 0), (1, 2), (4, 0), (7, 2) \} ,
\end{split}
\end{equation*}
since
\begin{equation*}
\begin{split}
0 \cdot (3, 2) = (0, 0) \quad &\land \quad 1 \cdot (3, 2) = (3, 2) \quad \land \quad 2 \cdot (3, 2) = (6, 0) \\
3 \cdot (3, 2) = (9, 2) \quad & \land \quad 4 \cdot (3, 2) = (2, 0) \quad \land \quad 5 \cdot (3, 2) = (5, 2) \\
6 \cdot (3, 2) = (8, 0) \quad & \land \quad 7 \cdot (3, 2) = (1, 2) \quad \land \quad 8 \cdot (3, 2) = (4, 0) \\
9 \cdot (3, 2) = (7, 2) \quad & \land \quad \ldots
\end{split}
\end{equation*}.
Now, in order to find all cosets of H, do I have to check for every $(a, b) \in G$ what sets I receive when I compute $(a, b) + H = \{(a, b) + h : h \in H \}$? I.e. do I have to compute 40 different sets? Is there a less brute force/tedious method?
| Let me do it for $H$. You have found that $|H|=10$, so that there will be four cosets (by Lagrange).
One of them is $H$. You also know that distinct cosets are disjoint. Thus, pick an element not in $H$, say $(1,0)$; then a new coset is
$$
(1,0)+H=\{(1, 0), (4, 2), (7, 0), (0, 2), (3, 0), (6, 2), (9, 0), (2, 2), (5, 0), (8, 2) \}
$$
Is there an element not yet listed? Yes, $(0,1)$. Then a new coset is
$$
(0,1)+H=\{(0, 1), (3, 3), (6, 1), (9, 3), (2, 1), (5, 3), (8, 1), (1, 3), (4, 1), (7, 3) \}
$$
Now find an element not yet listed and do the same to find the fourth and last coset.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2934265",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
How to know that $2n^3+9n^2+13n+6$ factors into $(n+1)(n+2)(2n+3)$? Apologies in advance, my math is very rusty.
I'm slowly working my way through Schaum's Outline of Discrete Math for some self-study, occasionally filling in large knowledge gaps in my grasp on algebra. In one of the supplementary questions I'm asked to prove (by induction) that:
$$
\sum^n_{i=1}i^2 = \frac{n(n+1)(2n+1)}{6}
$$
In the inductive step I add $(n+1)^2$ to either side and then try to work my way through, multiplying out the factors:
$$
\frac{n(n+1)(2n+1)}{6} + (n+1)^2 \\
= \frac{n(n+1)(2n+1)+6(n+1)^2}{6} \\
= \frac{(n^2+n)(2n+1)+6(n+1)^2}{6} \\
= \frac{2n^3+n^2+2n^2+n+6(n^2+2n+1)}{6} \\
= \frac{2n^3+n^2+2n^2+n+6n^2+12n+6}{6} \\
= \frac{2n^3+9n^2+13n+6}{6}
$$
I think everything up to this point is pretty trivial, but it's around this point that I get a bit lost. I know I need to factor the terms in the 3-degree polynomial in the numerator somehow, but I'm having a hard time with the actual mechanics given the lack of obvious common factors & the fact that 13 is prime.
I'm pretty sure the final result should look something like:
$$
\frac{(n+1)((n+1)+1)(2(n+1)+1)}{6} \\
= \frac{(n+1)(n+2)(2n+3)}{6}
$$
Multiplying the expected result out I can see the two statements are equal and I'm confident that
$$
\frac{2n^3+9n^2+13n+6}{6} = \frac{(n+1)(n+2)(2n+3)}{6}
$$
but I'm just not quite sure how to factor the polynomial myself to arrive at the final result.
Any hints on how to proceed from here, or what I need to be reading up on to get my head around this?
| Find a single root of the polynomial through trial and error and then apply long division. Dividing $(2n^3+9n^2+13n+6)$ by $(n+1)$ yields $(n+2)(2n+3)=2n^2+7n+6$.
A root of $-1$ yields a factor of $(n+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2935977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
} |
By induction showing that $m^3\le2^m$ for $m\ge10$ Show that $m^3\le2^m$ for $m\ge10$
My try:
Base case is true for $m=10$
Inductive Hypothesis: Assume $P(k)$ is true $\implies k^3\le2^k$
Now showing that $P(k+1)$ is true
$(k+1)^3\le2^{k+1}$
$\implies (k+1)^3\le k^3+1+3k^2+3k$
$\le 2^{k+1}+3k^2+3k+1($ from inductive hypothesis$)$
From here I could not proceed.
Can anyone explain how to proceed from here.
| Step: $m+1$:
$m^3= (m+1-1)^3= $
$(m+1)^3 -3(m+1)^2+3(m+1) -1\lt 2^m$
(Hypothesis).
$(m+1)^3 \lt $
$2^m +1-3(m+1) +3(m+1)^2<$
$2^m +1 + 3(m+1)^2 <2^m +2^m =2^{m+1};$
Used :
$1+3(m+1)^2 =$
$3m^2+6m +4 \lt $
$6m^2+4 < 2^m$ , for $m \ge 10.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Double summation with improper integral So my friend sent me this really interesting problem. It goes:
Evaluate the following expression:
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \int_{0}^\infty \frac{x^{b}}{e^{ax} \ b!} \ dx .$$
Here is my approach:
First evaluate the integral:
$$ \frac{1}{b!} \int_0^\infty \frac{x^b}{e^{ax}}\ dx.$$
This can be done using integration by parts and we get:
$$ \frac{1}{b!} \frac{b}{a} \int_0^\infty \frac{x^{b-1}}{e^{ax}}\ dx.$$
We can do this $ b $ times until we get:
$$ \frac{1}{b!} \frac{(b)(b-a).....(b-b+1)}{a^b} \int_0^\infty \frac{x^{b-b}}{e^{ax}}\ dx.$$
and hence we end up with:
$$ \frac{1}{b!} \frac{b!}{a^b}\qquad\left(\frac{-1 \ e^{-ax}}{a}\Big|_0^\infty\right) = \frac{1}{a^{b+1}}.$$
Now we can apply the sum of GP to infinity formula and we get:
$$ \sum_{a=2}^\infty \sum_{b=1}^\infty \frac{1}{a^{b+1}} = \sum_{a=2}^\infty \frac{\frac{1}{a^{2}}}{1-\frac{1}{a}}.$$
This is a telescoping series and we end up with $$ \frac{1}{a-1} = \frac{1}{2-1} = 1.$$
Do you guys have any other ways of solving this problem? Please do share it here.
| For the first part I often use the Laplace transform:
$$\frac{1}{b!} \int_0^\infty \frac{x^b}{e^{ax}}\ dx = \frac{1}{b!} \int_0^\infty x^be^{-ax}\ dx = \frac{1}{b!} {\cal L}(x^b)\Big|_{s=a} = \frac{1}{b!} \frac{b!}{a^{b+1}} = \frac{1}{a^{b+1}}$$
this make it easier.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2937751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 4,
"answer_id": 0
} |
Find the limit of $\begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}$
Find the following limit: \begin{equation*} \lim_{x \rightarrow 4} \frac{\sqrt{1 + 2x} -3}{\sqrt{x} - 2} \end{equation*}
I have tried to divide the numerator and denominator by $\sqrt{x}$, but it did not work.
I have tried to multiply by the conjugates of the numerator and denominator simultaneously but it did not work.
I have tried to multiply by the conjugates of the numerator only but it did not work.
So what shall I do?
| Hint:$$\frac{\sqrt{1+2x}-3}{\sqrt{x}-2} = \frac{1+2x-9}{(\sqrt{x}-2)(\sqrt{1+2x}+3)} = \frac{2(\sqrt{x}-2)(\sqrt{x}+2)}{(\sqrt{x}-2)(\sqrt{1+2x}+3)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2938884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Prove that: $\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$ Prove that for nonegative $x,y,z$ we have: $$\sqrt{x+y} + \sqrt{y+z} + \sqrt{z+x} \leq \sqrt{6(x+y+z)}$$
I prove that using the tangent line method. We may assume that $x+y+z=1$, so you we have to prove $$\sqrt{1-x}+\sqrt{1-y}+\sqrt{1-z}\leq \sqrt{6}$$
A tangent on $f(x)=\sqrt{1-x}$ at $x={1\over 3}$ is $$y=-{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$
So we have, for all $x\in[0,1]$:
$$\sqrt{1-x} \leq -{\sqrt{6}\over 4}x+{5\sqrt{6}\over 12}$$ and we are done...
I wonder if there is elegant method avoiding calculus?
| Hint: This is just
$$
(a+b+c)^2\leq 3(a^2+b^2+c^2)\iff ab+bc+ca\leq a^2+b^2+c^2
$$
which is proved easily enough. You can start by $a\equiv\sqrt{x+y}$, $b\equiv\ldots$, etc.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2941228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Determinants of products of binary matrices and binomial coefficients Consider two binary semi-infinite matrices with obvious patterns:
$$
C=
\begin{bmatrix}
1 &0 &0 &0 &0 &0 &0 &\cdots\\
1 &0 &0 &0 &0 &0 &0 &\cdots\\
0 &1 &0 &0 &0 &0 &0 &\cdots\\
0 &1 &0 &0 &0 &0 &0 &\cdots\\
0 &0 &1 &0 &0 &0 &0 &\cdots\\
0 &0 &1 &0 &0 &0 &0 &\cdots\\
0 &0 &0 &1 &0 &0 &0 &\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots
\end{bmatrix}
$$
and
$$
T=
\begin{bmatrix}
1 &1 &0 &0 &0 &0 &0 &\cdots\\
0 &1 &1 &0 &0 &0 &0 &\cdots\\
0 &0 &1 &1 &0 &0 &0 &\cdots\\
0 &0 &0 &1 &1 &0 &0 &\cdots\\
0 &0 &0 &0 &1 &1 &0 &\cdots\\
0 &0 &0 &0 &0 &1 &1 &\cdots\\
0 &0 &0 &0 &0 &0 &1 &\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots
\end{bmatrix}
$$
Let $A_n=T^n C$, then the non-zero entries of $A_n$ are the ${n+1}\choose{m}$, $0\le m \le n+1$ binomial coefficients. For example,
$$
A_3=
\begin{bmatrix}
4 &4 &0 &0 &0 &0 &\cdots\\
1 &6 &1 &0 &0 &0 &\cdots\\
0 &4 &4 &0 &0 &0 &\cdots\\
0 &1 &6 &1 &0 &0 &\cdots\\
0 &0 &4 &4 &0 &0 &\cdots\\
0 &0 &1 &6 &1 &0 &\cdots\\
\vdots &\vdots &\vdots &\vdots &\vdots &\vdots &\ddots
\end{bmatrix}
$$
Is there a simple formula for the determinants of the $k\times k$ upper left blocks of the matrices $A_n$? That is: what is $\det(B_n^k)$, where $$B_n^k(ij)=A_n(ij),$$ $1\le i,j \le k\le n+1$?
NOTES:
*
*The computer factorisation of $\det(B_n^k)$ shows that
$\det(B_n^{(n+1)}=2^{n(n+1)/2}$ and the determinants factorisations
have only factors less than $2(n+1)$, which suggests that the
determinants are products of binomial coefficients of the form
${l}\choose{k}$, $0\le k,l\le 2(n+1)$.
*This question is motivated by the continued fraction approximation of
the square root function, the matrices $A_n$ being the Hurwitz
matrices of the continued fractions.
| It follows from the answer to Coefficients of binomial continued fractions, since the matrices in the question are Hurwitz matrices of the continued fractions that:
$$ \det B_n^1=n,$$
$$ \det B_n^2=\frac{n(n^2-1)}{3},$$
$$ \det B_n^3=\frac{n^2(n^2-1)(n^2-2^2)}{3^2\cdot5},$$
$$ \det B_n^4=\frac{n^2(n^2-1)^2(n^2-2^2)(n^2-3^2)}{3^3\cdot5^2\cdot7},$$
$$ \det B_n^5=\frac{n^3(n^2-1)^2(n^2-2^2)^2(n^2-3^2)(n^2-4^2)}{3^4\cdot5^3\cdot7^2\cdot9},$$
$$ \dots$$
$$ \det B_n^l=n^{\lceil l/2\rceil}\prod_{k=1}^{l-1}\frac{(n^2-k^2)^{\lfloor (n-k+1)/2\rfloor}}{(2k+1)^{n-k}}.$$
The fact that $$\det B_n^n=2^{n(n-1)/2}$$ follows from Rational fraction expression for triangular powers of 2
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
"answer_count": 1,
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} |
Prove $\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \le \frac{3}{2}$ If $a.b,c \in \mathbb{R^+}$ and $ab+bc+ca=1$
Then Prove $$S=\frac{a}{\sqrt{a^2+1}}+\frac{b}{\sqrt{b^2+1}}+\frac{c}{\sqrt{c^2+1}} \le \frac{3}{2}$$
My try we have $$S=\sum \frac{a}{\sqrt{a^2+ab+bc+ca}}=\sum \frac{a}{\sqrt{a+b}\sqrt{a+c})}$$
any hint here?
| By AM-GM
$$\sum_{cyc}\frac{a}{\sqrt{a^2+1}}=\sum_{cyc}\frac{a}{\sqrt{(a+b)(a+c)}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{a}{a+c}+\frac{a}{a+b}\right)=$$
$$=\frac{1}{2}\sum_{cyc}\left(\frac{b}{b+a}+\frac{a}{a+b}\right)=\frac{3}{2}.$$
| {
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} |
Then value of $\alpha^2 +4\alpha$ in Infinite series
If $\displaystyle \alpha = \frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots \infty.$
Then value of $\alpha^2 +4\alpha$ is
Try: Let $$S = \frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots $$
$$S+1 = 1+\frac{5}{2!\cdot 3}+\frac{5\cdot 7}{3!\cdot 3^2}+\frac{5\cdot 7 \cdot 9}{4!\cdot 3^3}+\cdots \cdots $$
Now camparing with $$(1+x)^n = 1+nx+\frac{n(n-1)x^2}{2!}+\frac{n(n-1)(n-2)x^3}{6\cdot 3!}+\cdots \cdots$$
So $\displaystyle nx=\frac{5}{6}$ and $\displaystyle \frac{n(n-1)x^2}{2}=\frac{35}{27}$
So $$\frac{nx(nx-x)}{2}=\frac{5}{12}\cdot \frac{5-6x}{6}=\frac{35}{27}$$
So $\displaystyle x=-\frac{41}{18}$ and $\displaystyle n=-\frac{15}{41}$
I am getting $\displaystyle S+1=\bigg(1-\frac{41}{18}\bigg)^{-\frac{15}{41}}$
but answer of $\alpha^2+4\alpha = 23$
which is not possible from my answer. could some help me how can i solve it, thanks
| $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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\newcommand{\ic}{\mathrm{i}}
\newcommand{\mc}[1]{\mathcal{#1}}
\newcommand{\mrm}[1]{\mathrm{#1}}
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Indeed, $\ds{\alpha}$ is given by
\begin{align}
\alpha & \equiv
\sum_{n = 1}^{\infty}{\prod_{k = 1}^{n}\pars{2k + 3} \over
\pars{n + 1}!\,3^{n}} =
\sum_{n = 1}^{\infty}{2^{n}\prod_{k = 1}^{n}\pars{k + 3/2} \over
\pars{n + 1}!\,3^{n}}
\\[5mm] & =
\sum_{n = 1}^{\infty}{\pars{5/2}^{\large \overline{n}} \over
\pars{n + 1}!}\pars{2 \over 3}^{n} =
\sum_{n = 1}^{\infty}{\Gamma\pars{5/2 + n}/\Gamma\pars{5/2} \over
\pars{n + 1}!}\pars{2 \over 3}^{n}
\\[5mm] & =
\sum_{n = 1}^{\infty}{\pars{n + 3/2}! \over
\pars{n + 1}!\pars{3/2}!}\pars{2 \over 3}^{n} =
\sum_{n = 1}^{\infty}{1 \over n + 1}{n + 3/2 \choose n}
\pars{2 \over 3}^{n}
\\[5mm] & =
\sum_{n = 1}^{\infty}\pars{\int_{0}^{1}t^{n}\,\dd t}
\bracks{{-5/2 \choose n}\pars{-1}^{n}}
\pars{2 \over 3}^{n}
\\[5mm] & =
\int_{0}^{1}\sum_{n = 1}^{\infty}{-5/2 \choose n}
\pars{-\,{2 \over 3}\,t}^{n}\,\dd t =
\int_{0}^{1}\bracks{\pars{1 - {2 \over 3}\,t}^{-5/2} - 1}\,\dd t
\\[5mm] &
\implies \bbx{\alpha = 3\root{3} - 2} \implies
\bbx{\alpha^{2} + 4\alpha = \color{red}{\large 23}}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2944522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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} |
Is there an elementary method of finding this missing angle? Let a point $P$ lie in a triangle $\triangle ABC$ such that $\angle BCP = \angle PCA = 13^\circ$, $\angle CAP = 30^\circ$, and $\angle BAP = 73^\circ$. Compute $\angle BPC$.
I have an ugly trig solution that looks something like this:
Let $\angle PBC = \theta$. It follows that $\angle PBA = 51-\theta$. From trig Ceva, we see that:
$$\frac{\sin(30)}{\sin(73)}*\frac{\sin(51-\theta)}{\sin(\theta)}*\frac{\sin(13)}{\sin(13)} = 1$$
Observe that $90-73=17$, and conveniently $17*3=51$. This inspires the following manipulations:
$$\frac{1}{2\sin(73)} * \frac{\sin(51-\theta)}{\sin(\theta)} = 1$$
$$\frac{1}{2\cos(17)} * \frac{\sin(51)\cos(\theta)-\cos(51)\sin(\theta)}{\sin(\theta)} = 1$$
$$\sin(51)\cos(\theta)-\cos(51)\sin(\theta)= 2\cos(17)\sin(\theta) $$
$$\sin(51)\cos(\theta) = 2\cos(17)\sin(\theta) + \cos(51)\sin(\theta)$$
$$\sin(51)\cos(\theta) = \sin(\theta)(2\cos(17) + \cos(51))$$
$$\tan(\theta) = \frac{\sin(51)}{2\cos(17) + \cos(51)}$$
Proceeding with triple-angle formulae:
$$\tan(\theta) = \frac{3\sin(17)-4\sin^3(17)}{2\cos(17) + 4\cos^3(17)-3\cos(17)}$$
$$\tan(\theta) = \frac{\sin(17)}{\cos(17)} * \frac{3-4\sin^2(17)}{4\cos^2(17)-1}$$
$$\tan(\theta) = \tan(17) * \frac{3-4(1-\cos^2(17))}{4\cos^2(17)-1}$$
$$\tan(\theta) = \tan(17) * \frac{4\cos^2(17)-1}{4\cos^2(17)-1}$$
$$\tan(\theta) = \tan(17)$$
We conclude that $\theta = 17$ and $\boxed{\angle BPC = 150}$.
This is simply horrific. Is there a more elegant method? I notice that $73 = 13 + 60$, but I don't see where I would put an equilateral triangle.
| Hint. Reflect $AC$ along $PC$ onto $BC$ to get the line $A'C$. To show that $\angle BPC=150^\circ$, it suffices to show that $\triangle BA'P$ is similar to $\triangle BPC$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Smooth parametrization of curve $\cosh x+\cosh y = \operatorname{constant}$ I have a curve (image below)
$$
\cosh x+\cosh y = C,\qquad C>2.
$$
I would like to get its smooth parametrization of form
$$
x = f(t),\qquad y=g(t),\qquad t\in[a,b],
$$
so for every point on the curve there is a corresponding parameter $t$. (In the same manner, as for curve $x^2+y^2=1$, there is a smooth parametrization $x=\cos t$, $y = \sin t$).
I would appreciate any help. Thanks!
| Here's something more symmetric. I'll write the steps if anyone wants to modify the idea to get something else.
First, we can remove the pesky hyperbolic functions simply by letting $x = \ln t,\ y = \ln s$ so $$\cosh x + \cosh y = c \iff t + \frac 1t + s + \frac 1s = 2c.$$
Now, we'd like to "invert" $z\mapsto z + 1/z$. More precisely, we want $t + 1/t = 2u^2+2$ and $s + 1/s = 2v^2 + 2$, so the whole thing becomes $ u^2 + v^2 = c - 2$ which we know how to parametrize.
Solving the appropriate quadratics, we get $$t = 1 + u^2\pm \sqrt{2u^2+u^4},\ s = 1+v^2 \pm \sqrt{2v^2 + v^4}.$$ Noting that we want $ u^2 + v^2 = c - 2$, we further choose $u = \sqrt {c-2} \cos\varphi$ and $v = \sqrt {c-2}\sin\varphi$.
All in all, $$x = \ln \left( 1 + (c-2)\cos^2\varphi \pm \sqrt{2(c-2)\cos^2\varphi + (c-2)^2\cos^4\varphi} \right),\\
y = \ln \left( 1 + (c-2)\sin^2\varphi \pm \sqrt{2(c-2)\sin^2\varphi + (c-2)^2\sin^4\varphi} \right).
$$
Now you must be wondering what to do with these $\pm$ signs. Well, if you choose one of the four possibilities, you will get the part of the curve in the corresponding quadrant. So, to deal with that:
$$x = \ln \left( 1 + (c-2)\cos^2\varphi + \cos\varphi \sqrt{2(c-2) + (c-2)^2\cos^2\varphi} \right),\\
y = \ln \left( 1 + (c-2)\sin^2\varphi + \sin\varphi \sqrt{2(c-2) + (c-2)^2\sin^2\varphi} \right),
$$ and we are finally done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2947044",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that Ratio Test gives no information for $3^{-1} + 5^{-1} + 3^{-2} + 5^{-2} + 3^{-3} + 5^{-3} + \cdots$
Q: Examine the series:
\begin{align*}
\frac{1}{3} + \frac{1}{5} + \frac{1}{3^2} + \frac{1}{5^2} + \frac{1}{3^3} + \frac{1}{5^3} + \frac{1}{3^4} + \frac{1}{5^4} + \cdots \\
\end{align*}
Prove that the Root Test shows that the series converges while the Ratio test gives no information.
I can use the Root Test to show convergence as the question requests. I will omit that. My issue is that when I apply the Ratio Test, it also seems to show convergence, when the problem asks me to show that the Ratio Test gives no information.
Applying the Ratio Test:
\begin{align*}
\lim\limits_{j \to \infty} \left| \frac{a_{j+1}}{a_j} \right| &= \lim\limits_{j \to \infty} \frac{\frac{1}{3^{j+1}} + \frac{1}{5^{j+1}}}{\frac{1}{3^j} + \frac{1}{5^j}} \\
&= \lim\limits_{j \to \infty} \frac{\frac{1}{3} + \frac{1}{5} \cdot \left( \frac{3}{5} \right)^j}{1 + \left( \frac{3}{5} \right)^j} \\
&= \frac{1}{3} < 1 \\
\end{align*}
And therefore the ratio test says that this series will converge which is not what the problem was asking for.
| Any sum of the form
$\sum_{n=1}^{\infty} (a_n+b_n)
$
where
$\dfrac{b_n}{a_n} < 1$
and
$\dfrac{a_{n+1}}{b_n} > 1$
will behave this way.
If
$a_n = u^{2n-1}$
and
$b_n = v^{2n}$
then
$\dfrac{b_n}{a_n}
=\dfrac{v^{2n}}{u^{2n-1}}
=\frac1{u}(v/u)^{2n}
$
and
$\dfrac{a_{n+1}}{b_n}
=\dfrac{u^{2n+1}}{v^{2n}}
=u(u/v)^{2n}
$
so,
if $u > v$,
the ratio test will not work.
However,
if $0 < u, v < 1$,
the root test shows that
the sum converges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2947946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Formalization/Verification of a beginner combinatorics problem I have the following task:
To build a computer chip, 4 non-distinct components are needed. Of 12 existing components, 2 are defective. Given that 4 components are chosen, what are the probabilities of the following combinations?
*
*No defective components
*Exactly one defective component
*Exactly two defective components
Progress so far:
The general approach I use is to consider the number of non-defective components over the total number of components for repeated decrements of each set.
*
*$\frac{10}{12} \cdot \frac{9}{11} \cdot \frac{8}{10} \cdot \frac{7}{9} \approx 0.4242 $
For the next question, I sum the probabilities of each case, depending on if the defective component is chosen last, second last, etc.
*$\frac{10}{12} \cdot \frac{9}{11} \cdot \frac{8}{10} \cdot \frac{2}{9} \ + \frac{10}{12} \cdot \frac{9}{11} \cdot \frac{2}{10} \cdot \frac{8}{9} \ + \frac{10}{12} \cdot \frac{2}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} \ + \frac{2}{12} \cdot \frac{10}{11} \cdot \frac{9}{10} \cdot \frac{8}{9} $
I could employ a similar intuitive approach for 3. , however I feel like this is messy and not particularly rigorous. How do I answer these questions formally?
| Method 1: The number of ways of selecting a subset of size $k$ from a set with $n$ elements (the number of unordered selections of $k$ elements from a set of $n$ elements) is
$$\binom{n}{k} = \frac{n!}{k!(n - k)!}$$
There are $$\binom{12}{4}$$ ways to select four of the twelve components.
The number of ways of selecting exactly $k$ of the $2$ defective components and $4 - k$ of the $12 - 2 = 10$ good components is
$$\binom{2}{k}\binom{10}{4 - k}$$
Therefore, the probability of selecting exactly $k$ defective components is
$$\Pr(\text{exactly}~k~\text{defective}) = \frac{\dbinom{2}{k}\dbinom{10}{4 - k}}{\dbinom{12}{4}}$$
Hence,
\begin{align*}
\Pr(\text{no defective components}) & = \frac{\dbinom{2}{0}\dbinom{10}{4}}{\dbinom{12}{4}}\\
\Pr(\text{exactly one defective component}) & = \frac{\dbinom{2}{1}\dbinom{10}{3}}{\dbinom{12}{4}}\\
\Pr(\text{exactly two defective components}) & = \frac{\dbinom{2}{2}\dbinom{10}{2}}{\dbinom{12}{4}}
\end{align*}
Method 2: You have correctly calculated the first two answers. There are only three possibilities. The number of defective components selected is either zero, one, or two. Therefore,
$$\Pr(\text{exactly two defective components}) = 1 - \Pr(\text{no defective components}) - \Pr(\text{exactly one defective component})$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Conditional Expectation of E(2X+Y|X-Y=1) I met these two problems. The first question reads: what is $E(X|X+Y=1)$
given that $X$ and $Y$ are both independent standard normal random variables.
The second reads that $X$ and $Y$ are correlated standard normal random variables at a correlation of $\rho$, then what is $E(2X+Y|X-Y=1)?$
I tried to first find the sample space of given condition that $X+Y=1$ and $X-Y=1$. I realized that they are both sample space with $0$ probability since the two space are reduced to $\rm I\!R^1$ from $\rm I\!R^2$, therefore the conditions should have $0$ probability. In that case
$E(X|X+Y=1) = E(1-Y) = 1-0\quad$ since $\quad Y\subset \rm I\!R^1$
$E(2X+Y|X-Y=1) = E(2X+X-1) = E(3X-1) = -1$ since $\quad X\subset \rm I\!R^1$
are these answers correct?
| We assume that $X$ and $Y$ are jointly normal. Then $X-Y$ and $X+Y$ are independent. Moreover,
\begin{align*}
E\big(X \mid X+Y=1 \big) &= \frac{1}{2}E\big( (X+Y)+(X-Y) \mid X+Y=1\big)=\frac{1}{2},
\end{align*}
and
\begin{align*}
E\big(2X+Y \mid X-Y=1 \big) &=E\big(X \mid X-Y=1 \big) +E\big(X+Y \mid X-Y=1 \big)\\
&=\frac{1}{2} + E(X+Y) = \frac{1}{2}.
\end{align*}
| {
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Prove $\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $ using Cauchy-Schwarz How do I prove this inequality
$$\ 1<\frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1}<2 $$
I've tried to prove that $$\ \frac{1}{n+1}+\frac{1}{n+2}+ ... + \frac{1}{3n+1} \ $$ is less than $$\ \frac{1}{1}+\frac{1}{2}+ ... + \frac{1}{n} \ $$ and that this sum is either less than $2$ or equal to $2$ using C-S.
| Here's an alternative answer using $AM-HM$ inequality: $$\frac{a_1+a_2+\cdots+a_n}{n} \geq \frac{n}{\frac{1}{a_1}+\frac{1}{a_2}+\cdots+\frac{1}{a_n}}$$
take $a_1=n+1, a_2=n+2,\cdots,a_{2n+1}=3n+1$, then by above inequality $$\frac{(n+1)+(n+2)+\cdots+(3n+1)}{2n+1} \geq \frac{2n+1}{\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n+1}}$$ Consequently, $$\frac{4n^2+4n+1}{2n+1} \geq \frac{2n+1}{\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n+1}}$$ which implies $$\frac{1}{n+1}+\frac{1}{n+2}+\cdots+\frac{1}{3n+1} \geq \frac{(2n+1)^2}{4n^2+4n+1}=1$$
| {
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"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Find the smallest positive integer $X$ such that $478^{870} \equiv X \ (\text{mod} \ 273)$ Appreciate if one could advise if my solution is correct. Here is my attempt of the problem:
Since $(273, 478) =1,$ by Euler's theorem, $478^{\phi(273)}=478^{144} \equiv 1 \ \ (\text{mod} \ 273) \implies 478^{864} \equiv 1 \ \ (\text{mod} \ 273).$
Next, $478^{2} \equiv 22 \ \ (\text{mod} \ 39) \implies 478^{6} \equiv 22^{3} \equiv 1 \ \ (\text{mod} \ 39)$ and $478^{\phi(7)} = 478^{6} \equiv 1 \ \ (\text{mod} \ 7) $
Hence, $478^{6} \equiv 1 \ \ (\text{mod} \ 273)$ and $478^{864+6}= 478^{870} \equiv 1 \ \ (\text{mod} \ 273)$
| Hint $\ \ \begin{align}478\ \ &\equiv 1,2,-3\,\bmod 3,7,13\\ \Rightarrow\, 478^{\large\color{#0b0}3}&\equiv 1,1,\color{#c00}{{-}1}\, \bmod 3,7,13\end{align}\ $ and $\ \ \color{#c00}2\cdot \color{#0b0}{3\ }\mid 870$
Remark $ $ The idea is that $\,6 = \color{#c00}2\cdot \color{#0b0}{3\ }$ is a common multiple of the orders of $a=478$ mod $\,3,7,13$ therefore $\,a^6\equiv 1$ for each modulus, hence also modulo their lcm = product $= 273$, i.e. $\,3,7,13\mid a^6-1\,\Rightarrow\, 273\mid a^6-1,\,$ This is the idea behind the Carmichael approach in lhf's answer, but it uses Euler's $\,\phi(m)\,$ as its multiple of the order $\!\bmod m$
| {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$ is rational or irrational?
The number $x$ defined below is rational or irrational?
$$x=\sqrt[3]{\sqrt{5}+2}+\sqrt[3]{\sqrt{5}-2}$$
From: IMO 1973 - Longlist
My attempt (my real question is at the end):
the identity $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^3-ab-ac-bc)$ when $a+b+c=0$, leads to $$a^3+b^3+c^3=3abc \tag{1}$$
Now considering
$$a=\sqrt[3]{\sqrt{5}+2},b=\sqrt[3]{\sqrt{5}-2},c=-x$$
from (1) it is true that
$$x^3-3x-2\sqrt{5}=0 \tag{2}$$
That is the number $x$ is a root from (2).
Note: By trial and error I've found that answer is $x=\sqrt{5}$ (the other 2 roots are complex), that is irrational. But my question is more subtle.
Question: Can I conclude just inspecting (2), judging by the coefficient $2\sqrt{5}$, that $x$ is irrational, without actually solving the equation? In a math contest that might be helpful, if possible, as it would avoid extra steps.
| Let $\alpha=\sqrt[3]{\sqrt{5}+2}$ and $\beta=\sqrt[3]{\sqrt{5}-2}$. Then $\alpha\beta=1$ and
$$ 2\sqrt{5} = \alpha^3+\beta^3 = (\alpha+\beta)((\alpha+\beta)^2-3\alpha\beta)=(\alpha+\beta)^3-3(\alpha+\beta),$$
so the rationality of $\alpha+\beta$ would imply the rationality of $2\sqrt{5}$. It follows that $\alpha+\beta\not\in\mathbb{Q}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "14",
"answer_count": 4,
"answer_id": 1
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Operator norm of a family of matrices Let $c$ be a complex number. Consider the family of $n\times n$ matrices $M_n$ which have $c$'s on one off-diagonal, $\bar{c}$'s on the other off-diagonal, and zero everywhere else. So $M_4$ looks like this:$$\left(\begin{array}{cccc}0&c&0&0\\\bar{c}&0&c&0\\0&\bar{c}&0&c\\0&0&\bar{c}&0\end{array}\right)$$How does one find a general formula for the operator norm of $M_n$?
| Write $c=\alpha|c|$, with $|\alpha|=1$. Then
$$
M_4=|c|\,\begin{bmatrix}
1&0&0&0\\
0&\alpha&0&0\\
0&0&\alpha^2&0\\
0&0&0&\alpha^3\end{bmatrix}
\begin{bmatrix}
0&1&0&0\\
1&0&1&0\\
0&1&0&1\\
0&0&1&0
\end{bmatrix}
\begin{bmatrix}
1&0&0&0\\
0&\alpha&0&0\\
0&0&\alpha^2&0\\
0&0&0&\alpha^3\end{bmatrix}^*,
$$
and so $M_4(c)$ is unitarily equivalent to $|c|\,M_4(1)$. Thus we only need to determine the spectrum of $M_4(1)$. The same argument applies for $M_n$ for any $n$.
The eigenvalues of $M_n(1)$ are $\lambda_k=2\cos\frac{k\pi}{n+1}$, where an eigenvector for $\lambda_k$ is $\sum_{s=1}^n\sin\frac{ks\pi}{n+1}e_s$, and $e_1,\ldots,e_n$ is the canonical basis.
As $M_n(1)$ is selfadjoint, its norm is the greatest eigenvalue in absolute value, so
$$
\|M_n(c)\|=|c|\,\|M_n(1)\|=2|c|\,\cos\frac{\pi}{n+1}.
$$
$\newcommand\abajo{\\ \ \\}$
Edit: The Eigenvalues of $M_n(1)$.
Let $e_1,\ldots,e_n$ be the canonical basis of $\mathbb C^n$. Let $\eta_k=\sum_{\ell=1}^n\sin\frac{k\ell\pi}{n+1}\,e_\ell$, $k=1,\ldots,n$. Then
\begin{align*}
M_n(1)\eta_k
&=\sum_{s=1}^{n-1}\sin\frac{sk\pi}{n+1}\,e_{s+1}+\sum_{s=2}^{n}\sin\frac{sk\pi}{n+1}\,e_{s-1} \abajo
&=\sin\frac{2k\pi}{n+1}\,e_1+\sum_{s=2}^{n-1}\left(\sin\frac{(s-1)k\pi}{n+1}+\sin\frac{(s+1)k\pi}{n+1}\right)\,e_{s}
+\sin\frac{(n-1)k\pi}{n+1}\,e_{n} \abajo
&=2\cos\frac{k\pi}{n+1}\sin\frac{k\pi}{n+1}\,e_1+\sum_{s=2}^{n-1}2\cos\frac{k\pi}{n+1}\sin\frac{sk\pi}{n+1}\,e_s+2\cos\frac{k\pi}{n+1}\sin\frac{nk\pi}{n+1}\abajo
&=2\cos\frac{k\pi}{n+1}\,\eta_k.
\end{align*}
All the above requires is the sine of a sum formula.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Matrix equation using determinant This question got me thinking and confused on how to solve it.
If $$\begin{align}k(x-a)+2x-z&=0\\k(y-a)+2y-z&=0 \\ k(z-a)-x-y+2z&=0\end{align}$$
Show that $$x = \frac{ak(k+3)}{k^2+4k+2}.$$
So far so good, I could only solve it simultaneously.
| Straightforward, using Cramer’s rule, see https://en.wikipedia.org/wiki/Cramer%27s_rule#Explicit_formulas_for_small_systems:
$$\begin{align}
{}&\det\begin{pmatrix}
2+k&0&-1\\
0&2+k &-1\\
-1&-1&2+k
\end{pmatrix}\\
&\qquad=(2+k)^2\det\begin{pmatrix}
1&0&-1/(2+k)\\
0&1 &-1/(2+k)\\
-1&-1&2+k
\end{pmatrix}\\
&\qquad=(2+k)^2\det\begin{pmatrix}
1&0&-1/(2-k)\\
0&1 &-1/(2+k)\\
0&-1&2+k-1/(2+k)
\end{pmatrix}\\
&\qquad=(2+k)^2\det\begin{pmatrix}
1&0&-1/(2-k)\\
0&1 &-1/(2+k)\\
0&0&2+k-2/(2+k)
\end{pmatrix}\\
&\qquad=(2+k)^2(2+k-2/(2+k))\\
&\qquad=(k+2)(k^2+4k +2)=:D.
\end{align}$$
Now
$$\begin{align}
\det\begin{pmatrix}
ka&0&-1\\ ka &2+k&-1\\ ka &-1&2+k
\end{pmatrix}
&=ka\det\begin{pmatrix}
1&0&-1\\ 1&2+k&-1\\ 1&-1&2+k
\end{pmatrix}\\
&=ka\det\begin{pmatrix}
1&0&0\\ 1&2+k&0\\ 1&-1&3+k
\end{pmatrix}\\
&=ka(2+k)(3+k)=:D_x.
\end{align}
$$
Hence $x=D_x/D$ in case $k\notin\{-2,-2\pm\sqrt2\}$.
| {
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Does the series $\sum_{n=1}^\infty\frac{(2n)!}{2^{2n}(n!)^2}$ converge or diverge.
$$\sum_{n=1}^\infty\frac{(2n)!}{2^{2n}(n!)^2}$$
Can I have a hint for whether this series converges or diverges using the comparison tests (direct and limit) or the integral test or the ratio test?
I tried using the ratio test but it failed because I got 1 as the ratio. The integral test seems impossible to use here.
| For a more direct approach, you might directly expand the terms as follows:
$$\begin{align}
\frac{(2n)!}{4^n (n!)^2} &= \frac{1}{4^n}\frac{2n(2n-1)}{n^2}\frac{(2n-2)(2n-3)}{(n-1)^2}\cdots\frac{(4)(3)}{2^2} \frac{(2)(1)}{1^2} \\
&= \frac{2^n}{4^n}\frac{2n-1}{n} \frac{2n-3}{n-1}\cdots\frac{3}{2} \frac{1}{1} \\
&= \frac{4^n}{4^n} \frac{n - 1/2}{n} \frac{n - 3/2}{n-1} \cdots \frac{3/2}{2} \frac{1/2}{1}.
\end{align}$$
This is almost a telescoping product. By subtracting $1/2$ from each numerator (except the last), we get a smaller term that does telescope. Thus
$$
\frac{(2n)!}{4^n (n!)^2}
\geq
\frac{n-1}{n}\frac{n-2}{n-1} \cdots \frac{1}{2} \cdot (1/2) = \frac{1}{2n}.
$$
Thus the $n$th term of your series is bigger than $1/2n$, and diverges by comparison with the harmonic series
$$ \sum_{n \geq 1} \frac{1}{n}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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what is the probability that drawn balls have the same number? In the urn, there is one ball with number $1$, $2$ with number $2$, and so on until $n$ balls with the number $n$.
From the urn, we draw two balls.
Calculate the probability that the two drawn balls have the same number.
| Total number of balls is $$\frac{(n)\cdot (n+1)}{2}$$
So, sample space is $$^{(\frac{(n)\cdot (n+1)}{2})}C_2$$
You want to select 2 balls of similar number. This starts from ball numbered 2.
Number of ways to select 2 balls from each group of same numbered balls is
$$^2C_2+^3C_2+.....+^nC_2$$
So, final solution is:
$$\frac{(^2C_2+^3C_2+.....+^nC_2)}{^{(\frac{(n)\cdot (n+1)}{2})}C_2}$$
Further simplification: $$ (^2C_2+^3C_2+.....+^nC_2)$$
can be written as $$\frac{1}{2} \cdot (2\cdot1+3\cdot2+4\cdot3+....+n\cdot(n-1))$$
Let $$2S=(2\cdot1+3\cdot2+4\cdot3+....+n\cdot(n-1))$$
$$2S=\Sigma_1^n n\cdot(n-1)$$
$$2S=\frac{n\cdot(n^2-1)}{3}$$
So, $$S=\frac{n\cdot(n^2-1)}{6}$$
$$ ^\frac {n\cdot(n+1)}{2}C_2=\frac{1}{2}\cdot(\frac{n\cdot (n+1)}{2})\cdot(\frac{n^2+n-1}{2})$$
Giving you $$\frac{4}{3({n+2})} $$
| {
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Solving recurrence by generating function Notice: This is an ungraded practice problem. My solution is correct.
Question: How can I solve the problem below using generating functions which make it clear it uses the sum of squares forumla?
Problem: Solve $a_{n+1} - a_n = n^2$, $n \ge 0$, $a_0 = 1$ by the method of generating functions.
My solution:
$$
\begin{array}{r c l}
a_{n+1} - a_n & = & n^2 \\
a_{n+1} x^{n+1} - x a_n x^n & = & n^2x^{n+1} \\
\sum_{i = 0}^{\infty}a_{i+1} x^{i+1} - \sum_{i = 0}^{\infty}a_{i+1} x a_i x^i & = & \sum_{i = 0}^{\infty} i^2 x^{i+1} \\
\end{array}
$$
$$\text{Let } f(x) = \sum_{i = 0}^{\infty}a_i x^i \text{. Then:}$$
$$
\begin{array}{r c l l}
(f(x) - a_0) - xf(x) & = & \sum_{i = 0}^\infty i^2 x^{i+1} \\
f(x) & = & \frac{1}{1-x} (a_0 + \sum_{i = 0}^\infty i^2 x^{i+1}) \\
& = & (1 + x + x^2 + ...) (1 + x^2 + 4x^3 + 9x^4 + ...) \\
& = & 1 + x + 2x^2 +6x^3 + ... & \text{expansion by hand}
\end{array}
$$
What routes can I take that make it clear that the last line is equivalent to $\sum_{i = 0}^\infty 1 + \frac{i(i-1)(2i-1)}{6}$? It must be done using generating functions.
|
We obtain
\begin{align*}
\color{blue}{f(x)}&=\frac{1}{1-x}\left(a_0+\sum_{i=0}^\infty i^2x^{i+1}\right)\\
&=\frac{1}{1-x}\left(1+x\sum_{i=0}^\infty i^2 x^i\right)\\
&=\frac{1}{1-x}+x\left(\sum_{j=0}^\infty x^j\right)\left(\sum_{i=0}^\infty i^2x^i\right)\\
&=\frac{1}{1-x}+x\sum_{n=0}^\infty\left(\sum_{{i+j=n}\atop{i,j\geq 0}}i^2\right)x^n\tag{1}\\
&=\frac{1}{1-x}+x\sum_{n=0}^\infty\left(\sum_{i=0}^n i^2\right)x^n\\
&=\frac{1}{1-x}+\sum_{n=0}^\infty\frac{n(n+1)(2n+1)}{6}x^{n+1}\\
&=\sum_{n=0}^\infty x^n+\frac{1}{6}\sum_{n=1}^\infty(n-1)n(2n-1)x^n\\
&\,\,\color{blue}{=\sum_{n=0}^\infty\left(1+\frac{1}{6}(n-1)n(2n-1)\right)x^n}
\end{align*}
The essence is applying the Cauchy product formula in (1): Multiplication of $\sum_{i=0}^\infty c_i x^i$ with $\frac{1}{1-x}$ transforms the coefficient $c_i \to \sum_{i=0}^n c_i$, here with $c_i=i^2$.
| {
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Your evil probability professor has an urn with 9 balls. Your evil probability professor has an urn with 9 balls: 2 red, 3 white and 4 blue. He draws two balls from the urn without replacement. Let X be the number of red balls drawn and Y the number of white balls.
a) Determine the joint probability mass function of X and Y.
b) Are X and Y independent random variables?
c) Compute the covariance between X and Y.
For point A:
$P(0,0)=\frac{4}{9} \cdot \frac{3}{8} = \frac{1}{6}$ That is correct for the solution.
$P(0,1)=\frac{4}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{4}{8} = \frac{1}{3}$ That is correct for the solution.
$P(1,0)=\frac{2}{9} \cdot \frac{4}{8} + \frac{4}{9} \cdot \frac{2}{8}= \frac{2}{9}$ That is correct for the solution.
$P(1,1)=\frac{2}{9} \cdot \frac{3}{8} + \frac{3}{9} \cdot \frac{2}{8}= \frac{1}{6}$ That is correct for the solution.
$P(2,0)=\frac{2}{9} \cdot \frac{1}{8} = \frac{1}{36}$ That is correct for the solution.
$P(0,2)=\frac{3}{9} \cdot \frac{2}{8} = \frac{1}{12}$ That is correct for the solution.
For point B to check the independancy I have just to check if for example $P(X=0,Y=0) = P(X=0) \cdot P(Y=0)$.
$\frac{1}{6} \neq (\frac{7}{9} \cdot \frac{6}{8}) \cdot (\frac{6}{9} \cdot \frac{5}{8})$.
So X and Y are not independent.
For point C I know that the covariance $Cov(X,Y)=E[X \cdot Y]-E[X]\cdot E[Y]$, but how can compute the expectations, do I have to figure put with distribution is? How can I do it?
Can someone help me?
Thanks in advance, Fabio!
| For a) keep in mind that the order you draw the balls in matters. For example, for $P(0,1)$ we can draw a blue ball then a white ball or a white ball then a blue ball.
For b) independence says that for every $x$ and $y$, $P(x,y) = P_X(x)P_Y(y)$. It's not enough to check a single pair $(x,y)$ to prove independence. On the other hand, if there is some $x$ and $y$ for which $P(x,y) \ne P_X(x)P_Y(y)$ then $X$ and $Y$ are not independent.
For c) recall that $\displaystyle \mathbf E[f(X,Y)] = \sum_{(x,y)} f(x,y)P(x,y)$.
| {
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Solve this question Question : $$\begin{align}a+b+c &= 0\\
a^3 +b^3 +c^3 &= 12\\
a^5 +b^5 +c^5 &= 40\end{align}$$
Then , $a^4 + b^4 + c^4 = ?$
My try : As a common perspective I just went to find any trick related to it and my first step is as usual as a common man will think
$3abc = 12$ and got $abc = 4$
After that I am unable to link it with any other equation
| Denote $S_n = a^n + b^n + c^n$. So we have $S_0 = 3, S_1 = 0, S_3 = 12$ and $S_5 = 40$.
Consider $f(x) = (x-a)(x-b)(x-c)$. Write $f(x) = x^3 - ux^2 - vx - w$. Then it is easy to see $u = 0$ and $v = -(ab + bc + ca) = \frac{S_2}2$. This means $a^3 = va + w$, so $a^n = va^{n-2} + wa^{n-3}$, and do the same on $b, c$ and summing together gives you $S_n = vS_{n-2} + wS_{n-3}$.
This should give you enough information set up a system of equations to see $S_4 = 8$. For instance, the first thing you can do is
$S_3 = vS_1 + wS_0$, so $12 = 3w$ and $w = 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2967980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$a^n-a + 1 $ divisible by $n$
Problem. Given $a$ is a positive integer greater than 3, are there infinitely many positive integers $n$ satisfying $a^n-a + 1 $
divisible by $n$?
| $N=a^n-a+1$
$a^{p_1} ≡ a \mod p_1$
$a^{p_2 } ≡a \mod p_2$
$(a^{p_1})^{p_2} ≡ a^{p_2} \mod p_1 ≡(a\ mod p_2) \mod p_1= k_1 p_1 + k_2 p_2 + a$
$a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$
⇒ $a^{p_1p_2}-a+1=k_1p_1 +k_2p_2 +1$
If $n=p_1p_2 | a^n-a+1$ then we must have:
$p_1p_2 | k_1p_1 + k_2p_2+1$
So we have following linear equation:
$k_1p_1 + k_2 p_2 = m p_1p_2-1$
For certain value of $p_1$ and $p_2$ and m,there can be infinitely many solutions for $k_1$ and $k_2$. For example:
with $p_1=5$, $p_2=7$ and $m=3$ the equation has one solution like $k_1=11$ and $k_2=7$ and all other solutions can be found by:
$k_1= 7 t + 11$ and $k_2= -5 t+7$.
Now in first step problem reduces to:
Find n so that there exist a common divisor between $n$ and $N=a^n-a+1$.
For example for $a=3$, $p_1=5$ and $p_2=7$ we have:
$3^{35}-3+1=105$ and $(35, 105)=5$
In second step we must find m, $k_1$ and $k_2$ for certain amount of $p_1$ and $p_2$ so that $(n, N)=n$.
Relation $a^{p_1p_2}=k_1 p_1 + k_2 p_2 +a$ shows that $a|k_1 p_1 + k_2 p_2 $; if $k_1=u+1$ and $k_2=v-1$, $p_1= a .b+1$ and $p_2=a.c +1$, i.e. $p_1≡1\mod a$ and $p_2≡1\mod a$, then:
$k_1p_1+k_2p_2= M(a)$
Or: $a | k_1p_1+k_2p_2$
We can see this in solution $n=409\times 9831853$ for $a=6$; $409=48\times 6 +1$ and $9831853=1638642 \times 6 +1$
This can help us in choosing a and primes $p_1$ and $p_2$
I see no reason for the lack of more solutions.
| {
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Find the value of $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$
Suppose $(a, b, c)\in\Bbb R^3$, $a,b,c$ are all nonzero, and we have $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$. Find the value of $$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}.$$
Here is my attempt:
$$\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{bc+ac+ab}{abc}.$$
I am having trouble in figuring out the best approach to simplify $
\sqrt{a+b}+\sqrt{b+c}=\sqrt{c+a}$
so that I can find the value of $
\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$ . Hope somebody has an idea.
| Take the square of the condition :
$$a+b+b+c+2\sqrt{a+b}\sqrt{b+c}=a+c$$
This implies $b=-\sqrt{a+b}\sqrt{b+c}$, so $b^2=(a+b)(b+c)$, hence
$$ab+bc+ac=0$$
So your answer should be $0$ (what else could it be ?) :-)
EDIT :
I thought about this a little (well, OK, a big !) more, and especially the feasibility of the initial condition.
We have $(1)\,:\,ab+ac+bc=0$, so $a$, $b$ and $c$ can't be all positive. Which asks the question : can $a+b$, $a+c$ and $b+c$ be positive ?
Now we can't have two negative values, because the sum would be negative. So only one of the three variables can be negative. And it can't be $a$ or $c$, because, for example, if $a<0$, $a+c<c<b+c$, and the left member of the condition is strictly greater than the right member.
So $b<0$, and solving for $b$ in $(1)$, we find $b=-\frac{ac}{a+c}=-\frac{1}{1/a+1/c}$ (which, BTW, is half the opposite of the harmonic mean of $a$ and $c$).
Remains to prove that $a+b$ and $b+c$ are both positive. For example
$$a-\frac{ac}{a+c}=\frac{a^2}{a+c}>0$$
Now I can sleep without thinking anymore at this funny problem :-)
END EDIT.
And BTW : welcome to MSE !
| {
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Prove that $\int^{4\pi}_0(3x^2\sin\frac{1}{x}-x\cos\frac{1}{x})\,dx=\frac{32\sqrt2}{\pi^3}$ Prove that $$\int^{4\pi}_0(3x^2\sin\frac{1}{x}-x\cos\frac{1}{x})\,dx=\frac{32\sqrt2}{\pi^3}.$$
This question is regarding improper integrals of the second kind. I'm stuck at it. Letting $1/x=u$, I have,
$$\int^\frac{1}{4\pi}_\infty(-3\sin u+u\cos u)\, du = [4\cos u+u\sin u]\Bigm|^\frac{1}{4\pi}_\infty$$ and after this I'm stuck.
Comments and hints appreciated!
| Your integral is $[x^3\sin\frac{1}{x}]_0^{4\pi}=64\pi^3\sin\frac{1}{4\pi}-\lim_{x\to 0}x^3\sin\frac{1}{x}$. Since $|\sin\frac{1}{x}|\le 1$, $\lim_{x\to 0}x^3\sin\frac{1}{x}=0$. The final result is $64\pi^3\sin\frac{1}{4\pi}$.
| {
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Prove that $1/n+1/(n+1)+\dots+(1/2n)>2/3$
Prove that $\displaystyle\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}$
I tried to use mathematical induction, but I'm not able to prove that:
$$
\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3}.
$$
My method was:
Assumption:
$$\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}$$
$$\frac{1}{n} + \frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}$$
$$\frac{1}{n+1} + \frac{1}{n+2} + \dots + \frac{1}{2n} + \frac{1}{2n+1} + \frac{1}{2n+2}>\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}-\frac{1}{n}$$
Now in order to prove the thesis I have to prove that
$$\frac{2}{3} + \frac{1}{2n+1} + \frac{1}{2n+2}-\frac{1}{n} > \frac{2}{3}$$
But it's a contradiction. Did I make a mistake somewhere? How can I solve this problem? I'd appreciate your help.
| HINT
As suggested by Wojowu in the comment, sometimes induction works for a stronger hypothesis, in that case let try with
$$\displaystyle\frac{1}{n} + \frac{1}{n+1} + \dots + \frac{1}{2n}>\frac{2}{3}+\frac1{4n}>\frac23$$
Refer also to the related
*
*Proof by induction, $1/2 + ... + n/2^n < 2$
*If $x>0$ real number and $n>1$ integer, then $(1+x)^n>\frac{1}{2}n(n-1)x^2$
*Show that $\left(1+\frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\left(1+\frac{1}{3^3}\right)...\left(1+\frac{1}{n^3}\right) < 3$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2975250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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} |
Help Factoring A Difficult Expression Given A = $\begin{bmatrix}a & b & c \\c & a & b \\b & c & a \\\end{bmatrix}$, B =$\begin{bmatrix}a & b & c \\a^2 & b^2 & c^2 \\a^4 & b^4 & c^4 \\\end{bmatrix}$
Suppose $C =AB=\begin{bmatrix}
a^2 + a^2b + a^4c & ab + b^3 + b^4c & ac +bc^2 + c^5 \\
\\
ca + a^3 +a^4b & cb + ab^2 + b^5 & c^2 + ac^2 +bc^4 \\
\\
ba + a^2c + a^5 & b^2 + b^2c + ab^4 & bc + c^3 + ac^4
\end{bmatrix}$
I have been asked to show that the following holds:
$det(C)= -abc(b-c)(a-c)(a-b)(a^2-ab+b^2 -bc +c^2 -ac)(a+b+c)^2$
The way i proceeded to go about this is using the fact that $det(AB) = det(A)det(B)$.Now finding the determinant of A and B i have the following.
$det(A)= a^3 -3abc +b^3 +c^3$
$det(B) = abc(c^3b - cb^3 - c^3a + ca^3 + b^3a - ba^3)$
The problem i have is factoring out the necessary factors from the product of the determinants. I have tried loads of methods now and i have got nowhere and am a bit depressed by it. Any ideas would be really appreciated.
Best Regards
| Hint: $$a^3+b^3+c^3=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
| {
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Showing that $2^{2^{2^x}}<100^{100^x}$ for large $x$ I wish to show that $2^{2^{2^x}}<100^{100^x}$ for $x$ sufficiently large. I have taken logs (base 10) of both sides to get $2^{(2^x-1)}\log_{10} 2$ and $100^x$. It is not immediately clear how I can prove that the second term here is larger than the first.
Any help appreciated.
| $100^{100^x}= 2^{[\log_2{100}]*100^x} $. Let $k = \log_2{100}$. ($6< k < 7$)
$100^{100^x} = 2^{k*2^{kx}}= 2^{2^{\log_2k}*2^{kx}}$. Let $m = \log_2 k$. ($2< m < 3$)
$100^{100^x} = 2^{2^m*2^{kx}}=2^{2^{m + kx}}$
Now $2^{2^{m+kx}}< 2^{2^{2^x}} \iff m + kx < 2^x$ for sufficiently large $x$.
Which should be enough to be convincing.
But if not:
If $x > m$ then $m + kx < x +kx = (k+1)x< 8x=2^3*x$.
And if we can show $x < 2^{x-3}$ for sufficiently large $x$ we are done.
which... of course it is.
$(x)' = 1$ and $(2^{x-3})'= \ln 2* 2^{x-3}> 1$ for all $x > ...$ well $x > 3 + \log_2 (\frac 1{\ln 2})= 3 + \frac {\ln \frac 1{\ln 2}}{\ln 2}\approx 3.52$. then $2^{x-3}$ is increasing faster than $x$. So at $x = 6>3 + \frac {\ln \frac 1{\ln 2}}{\ln 2}$ we have $x = 6 < 8 = 2^{x-3}$ and $2^{x-3}$ is increasing faster than $x$ so for $x \ge 6$ we will have our result.
| {
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Geometrically showing $\frac{\alpha}{\beta} > \frac{\sin\alpha}{\sin\beta}$, for $0 < \beta < \alpha < 90^\circ$
I attempting to prove the show trig identity, i.e. $\frac{\alpha}{\beta} > \frac{\sin\alpha}{\sin\beta}$. I have found that the radius is $1$ so there are three line segments equal to $1$, but I am unsure which segment would be $\frac{\sin\alpha}{\sin\beta}$?
| $$\sin x < x < \tan x \qquad \forall~x \in [0, \frac{\pi}2]$$
If one agrees to start with the above known fact (which is purely an intuitive geometric fact as detail in e.g. this answer and NOT about algebraic analysis of functions), then:
On one hand, we have
$$ \frac{\alpha - \beta}2 > \sin \bigl( \frac{ \alpha -\beta}2\bigr) \quad \implies\quad \alpha - \beta > 2 \sin \bigl( \frac{ \alpha -\beta}2\bigr) \tag*{Eq.(1)}$$
On the other hand, $ \alpha > \beta$ and cosine being a decreasing function yield
$$ \beta < \tan\beta = \frac{ \sin\beta }{ \cos\beta } < \frac{ \sin\beta }{ \cos\bigl( \frac{ \alpha + \beta}2 \bigr) } \quad \implies \quad \sin\beta > \beta \cdot \cos\bigl( \frac{ \alpha + \beta}2 \bigr) \tag*{Eq.(2)}
$$
The product of the two inequalities (all terms are positive) gives us
\begin{align}
&& (\alpha - \beta) \cdot \sin\beta &> \beta \cdot 2 \sin\bigl( \frac{ \alpha - \beta}2 \bigr)\cos\bigl( \frac{ \alpha + \beta}2 \bigr) \\
&\implies & (\alpha - \beta) \cdot \sin\beta &> \beta \cdot (\sin\alpha - \sin\beta) \\
&\implies & \frac{\alpha - \beta}{ \beta } &> \frac{\sin\alpha - \sin\beta}{ \sin\beta}\qquad \text{, then $+1$ on both sides} \\
&\implies & \frac{ \alpha }{ \beta } &> \frac{ \sin\alpha }{ \sin\beta}
\end{align}
Note that the whole derivation can be viewed as a purely geometric proof expressed via algebra. One can make a sketch and find ALL the relevant lengths (including e.g. $\frac{ \sin\beta }{ \cos( \frac{ \alpha + \beta}2 ) }$).
| {
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Equivalence relation with the floor function Let us consider a function $f:\ \Bbb{R}\ \longrightarrow\ \Bbb{R}$ and we define the equivalence relation $\sim$ on $\Bbb{R}$ such that:
$$x\ \sim\ y\qquad\Leftrightarrow\qquad f(x)=f(y).$$
Note: The following $\lfloor x\rfloor$ is the floor function of $x$. It gives you the first integer smaller or equal to $x$. For example $\lfloor2.2\rfloor=2$, $\lfloor12\rfloor=12$, and $\lfloor−2.2\rfloor=−3$.
For the function $f(x)=\lfloor x/10\rfloor$, only one of the following statement is true.
For $x=a_0+10a_1+10^2a_2+...+10^na_n$, with all $a_j\in\{0,1,2,\ldots,9\}$ and for all $c\in\{0,1,2,\ldots,9\}$:
1) $x\ \sim\ x-c$ if $a_0-c<5$,
2) $x\ \sim\ x+c$ if $a_0+c<10$.
My working:
Take $a_0=9$ and $c=2$ then the first statement is false as $9-2>5$ but if I apply to the second statement $9+2>10$ which is wrong too, so I think my method is wrong.
I am guessing the second one is true from the function $f(x)=\lfloor x/10\rfloor$, but I'm really not sure about this. Any help will be appreciated!
Thanks.
| It's just decimal representation. If you subtracta a $c$ from a number do you have to carry/borrow a one or not.
to wit....
So $x = 10^na_n + 10^{n_1}a_{n-1} + ..... + 10a_1 + a_0$
$\frac x{10} = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ..... + a_1 + \frac {a_0}{10}$
$f(x) = \lfloor \frac x{10} \rfloor = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ..... + a_1$
And $x - c = 10^na_n + 10^{n_1}a_{n-1} + ..... + 10a_1 + (a_0-c)$
And $\frac {x-c}{10} = 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 + \frac {(a_0-c)}{10}$
To find the least integer depends upon the value of $\frac {(a_0-c)}{10}$.
$-9 \le a_0 -c \le 9$ and $-.9\le \frac {(a_0-c)}{10} \le .9$
If $\frac {(a_0-c)}{10} \ge 0$ then
$\rfloor \frac {x-c}{10}\lfloor = 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 + \rfloor\frac {(a_0-c)}{10}\lfloor= 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 = f(x)$
If $\frac {(a_0-c)}{10} \le 0$ then
$\rfloor \frac {x-c}{10}\lfloor = 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 + \rfloor\frac {(a_0-c)}{10}\lfloor= 10^{n-1}a_n + 10^{n_2}a_{n-1} + ..... + a_1 - 1= f(x)-1$
So $x$~$x-c$ if $a_0 -c \ge 0$. But $x$ !~ $x-c$ if $a_0 - c < 0$. So if $a_0 - c < 0 < 5$ this is not true.
....
You can do 2) the same way.
$f(x+c) = 10^{n-1}a_n + 10^{n-2}a_{n-1} + ..... + a_1 + \lfloor \frac {a_0+c}{10}\rfloor = f(x) + \lfloor \frac {a_0+c}{10}\rfloor$.
And $\lfloor \frac {a_0+c}{10}\rfloor = 0$ if $a_0+c< 10$ but $\lfloor \frac {a_0+c}{10}\rfloor = 1$ if $a_0 + c \ge 10$.
So 2) is true.
| {
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Show that $x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n$ is a bounded sequence.
Let $n\in \mathbb N$ and:
$$
x_n = \sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n
$$
Show that $\{x_n\}$ is a bounded sequence.
This sequence appears a bit tricky because it involves harmonic series. Below are steps I take.
Lower bound:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \ge \sqrt{\sum_{k=1}^n\left(\frac{k}{k}\right)^2} = \sqrt{\sum_{k=1}^n1}=\sqrt{n}\implies\\
\implies x_n \ge \sqrt n - \sqrt n \ge 0
$$
Lower bound is simple.
Upper bound:
To get rid of radical lets use Cauchy-Schwarz (note the below is incorrect as shown in Zvi's answer):
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} \le \sqrt{\left(\sum_{k=1}^n\left(\frac{k+1}{k}\right)\right)^2} =\sum_{k=1}^n\left(\frac{k+1}{k}\right) = n+\sum_{k=1}^n{1\over k} = n + H_n
$$
So this doesn't show $x_n$ is bounded above. I've tried another approach:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} - \sqrt n = \frac{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n}
$$
Consider nominator:
$$
\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2 - n=n+\sum_{k=1}^n{2\over k}+\sum_{k=1}^n{1\over k^2}-n = \sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k}
$$
For denominator:
$$
\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2} + \sqrt n = \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n
$$
So $x_n$ is:
$$
x_n = \frac{\sum_{k=1}^n{1\over k^2} + \sum_{k=1}^n{2\over k}}{ \sqrt{n + \sum_{k=1}^n{1\over k^2}+\sum_{k=1}^n{2\over k}} + \sqrt n }
$$
But i don't see how to proceed from this point. What else could i try? How to show $x_n$ is bounded above?
Please note the precalculus tag.
| Jakobian gave a very good hint. I am giving somewhat sharper bounds for $x_n$. Note that the OP incorrectly used the Cauchy–Bunyakovsky–Schwarz inequality. Indeed, when applied correctly, we get
$$\sqrt{n}\left(x_n+\sqrt{n}\right)=\sqrt{\sum_{k=1}^n1^2}\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2}\geq \sum_{k=1}^n1\cdot\left(\frac{k+1}{k}\right)=n+H_n,$$
where $H_n=\sum_{k=1}^n\frac1k$ is the $n$th harmonic number. That is,
$$x_n\geq \frac{H_n}{\sqrt{n}}.$$
For the upper bound, the OP got
$$x_n=\frac{\sum_{k=1}^n \frac{1}{k^2}+2H_n}{\sqrt{\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2}+\sqrt{n}}.$$
We can show that $\sum_{k=1}^n\frac{1}{k^2}$ is bounded above by $2$ via
$$\sum_{k=1}^n\frac{1}{k^2}<1+\sum_{k=2}^n\frac{1}{k(k-1)}=1+\left(1-\frac{1}{n}\right)<2.$$
However, a sharper upper bound is $\sum_{k=1}^n\frac{1}{k^2}<\zeta(2)=\frac{\pi^2}{6}$. Also, $$\sum_{k=1}^n\left(\frac{k+1}{k}\right)^2\geq \sum_{k=1}^n1^2=n.$$ This proves that
$$x_n<\frac{\frac{\pi^2}{6}+2H_n}{\sqrt{n}+\sqrt{n}}=\frac{H_n}{\sqrt{n}}+\frac{\pi^2}{6}\left(\frac{1}{2\sqrt{n}}\right).$$
(You can replace $\frac{\pi^2}{6}$ by $2$ if you want a precalculus solution.)
That is, we have
$$\frac{H_n}{\sqrt{n}}\leq x_n<\frac{H_n}{\sqrt{n}}+\frac{\pi^2}{6}\left(\frac{1}{2\sqrt{n}}\right).$$
So, $x_n$ has the asymptotic behavior of $\frac{H_n}{\sqrt{n}}\approx \frac{\ln n}{\sqrt{n}}$. But to show that $x_n$ is bounded, we don't need to know that $H_n\approx \ln n$ (well, to be even more precise, $H_n\approx \gamma+\ln n$, where $\gamma$ is the Euler–Mascheroni constant). Clearly,
$$\frac{\pi^2}{6}\left(\frac{1}{2\sqrt{n}}\right)\leq \frac{\pi^2}{6}\left(\frac{1}{2}\right).$$
Furthermore,
$$H_n=\sum_{k=1}^n\frac{1}{k}\leq \sum_{k=1}^n\frac{1}{\sqrt{k}}<\sum_{k=1}^n\frac{2}{\sqrt{k}+\sqrt{k-1}}=2\sum_{k=1}^n(\sqrt{k}-\sqrt{k-1})=2\sqrt{n}.$$
Hence,
$$x_n<\frac{2\sqrt{n}}{\sqrt{n}}+\frac{\pi^2}{6}\left(\frac{1}{2}\right)=2+\frac{\pi^2}{6}\left(\frac{1}{2}\right)<\infty.$$
Again, replace $\frac{\pi^2}{6}$ by $2$ if you don't want any non-precalculus knowledge in the proof. So, you would get $x_n<3$ for all $n$.
| {
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Can we proof that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?; given $0 < a < b < 1$ I have found in some test problem:
Given $0 < a < b < 1$, can we conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$?
I divide the combination of a and b into a few cases. Then, see what happens.
case 1 : $a \rightarrow 0$ and $b \rightarrow 0$
When $a$ approaches to zero and $b$ approaches to $a$ (which is zero), the expression will be evaluated as $\sqrt{0 + 0}$ and $\sqrt{0} + \sqrt{0}$. Since $0 = 0$, this case cannot conclude that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 2 : $a \rightarrow 1$ and $b \rightarrow 1$
When $b$ approaches to one and $a$ approches to $b$ (which is one), the expression will be evaluated as $\sqrt{1 + 1}$ and $\sqrt{1} + \sqrt{1}$. Since
$\sqrt{2} < 2$, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 3 : $a \rightarrow 0$ and $b \rightarrow 1$
When $a$ approaches to zero and $b$ approches to one, the expression will be evaluated as $\sqrt{0 + 1}$ and $\sqrt{0} + \sqrt{1}$. Since
$1 = 1$, then this case cannot concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
case 4 : $a \rightarrow b$
When $a$ approaches to $b$ the expression will be evaluated as $\sqrt{b + b}$ and $\sqrt{b} + \sqrt{b}$. Since $\sqrt{2b} < 2\sqrt{b} $, then this case concludes that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$
From above cases, am I still missing some point? If not, How should I write the conclusion from that pieces of thinking mathematically? Because I instinctively believe that $\sqrt{a + b} < \sqrt{a} + \sqrt{b}$ should be true (and not $\sqrt{a + b} \leq \sqrt{a} + \sqrt{b}$).
| We know that $$0 < 2\sqrt{ab}$$
add $a+b$ to both sides, we have
$$a+b < a+2\sqrt{ab}+b=(\sqrt{a}+\sqrt{b})^2$$
then we can take square root on both sides.
| {
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Evaluating $\int \frac{t}{1+t^2+t^4+t^8}dt$ I am stuck to evaluate the following integral in terms of finite terms (with out using power series technique).
$$\int \frac{t}{1+t^2+t^4+t^8}dt$$
Could anyone help me?
| First, change variables to $u = t^2$, because we can and 8th order polynomials are awful.
$$
\int \frac{t\,dt}{1+t^2+t^4+t^8} = \frac{1}{2}\int\frac{du}{1+u+u^2+u^4}
$$
Next, since $1+u+u^2+u^4$ has no real zeroes and no $u^3$ term, there exist $a,b,c\in\mathbb R^+$ such that $1+u+u^2+u^4 = [(u+a)^2 + b^2][(u-a)^2+c^2]$. These numbers have a very complicated closed form, but their numeric values are $a \approx 0.547$, $b \approx 0.586$, $c \approx 1.121$.
Next comes partial fraction time. There are four real numbers $A,B,C,D$ such that
\begin{multline}
\frac{1}{[(u-a)^2 + b^2][(u+a)^2+c^2]} = \frac{Ab}{[(u+a)^2 + b^2]} + \frac{Bc}{[(u-a)^2 + c^2]} \\ + \frac{2C(u+a)}{[(u+a)^2 + b^2]} + \frac{2D(u-a)}{[(u-a)^2 + c^2]}
\end{multline}
This turns out to have the solution
$$
A = \frac{4a^2-b^2 + c^2}{b\Delta}\;\;\;;\;\;\;B = \frac{4a^2 + b^2 - c^2}{c\Delta}\;\;\;;\;\;\; C = -D = \frac{2a}{\Delta}
$$
where $\Delta = (b^2-c^2)^2 + 8a^2(b^2+c^2) + 16a^4$. These numbers also have a very complicated closed form, and their numeric values are $A \approx 0.0417$, $B \approx 0.591$, $C = -D\approx0.179$.
Lastly, each term in the partial fraction can be integrated using the formulae
$$
\int \frac{b\, du}{(u+a)^2 +b^2} =\tan^{-1}\left(\frac{u+a}{b}\right)\;\;\;\;;\;\;\;\;\int\frac{2(u+a)du}{(u+a)^2+b^2} = \ln\left[(u+a)^2+b^2\right]
$$
Doing this and using $u = t^2$ again gives us the final antiderivative:
$$
\int\frac{tdt}{1+t^2+t^4+t^8} = \frac{1}{2}\left(A\tan^{-1}\left[\frac{t^2-a}{b}\right] + B\tan^{-1}\left[\frac{t^2+a}{c}\right] + C\ln\left[\frac{(t^2+a)^2+b^2}{(t^2-a)^2+c^2}\right]\right)
$$
You're not really going to do any better than this. The roots of the polynomial in the denominator have no simple form.
| {
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Integral of $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ I have this simple integral: $\int \frac{-x^2+2x-3}{x^3-x^2+x-1}dx$ and I can't come up with the correct answer.
Here's what I did:
I found the roots for $x^3-x^2+x-1$ so I could do partial fractions:
$\frac{-x^2+2x-3}{x^3-x^2+x-1}=\frac{A}{x-1}+\frac{Bx+C}{x^2+1}=\frac{A(x^2+1)+(Bx+C)(x-1)}{(x-1)(x^2+1)}$
I then found the values for $A$, $B$ and $C$:
$x^2(A+B)+x(-B+C)+A-C=-x^2+2x-3$
$
\left\{
\begin{array}{c}
A+B=-1 \\
-B+C=2 \\
A-C=-3
\end{array}
\right.
$
$A=-1$, $B=0$, $C=2$
So now back again with the integral:
$\int \frac{-1}{x-1}dx+\int \frac{2}{x^2+1}dx=-\int \frac{1}{x-1}dx + 2 \int \frac{1}{x^2+1}dx=-\ln|x-1|+2|x^2+1|=2\ln|\frac{x^2+1}{x-1}|+C$
But why the correct answer is $2 \arctan(x)-\ln|x-1|+C$? Where does the $arctan$ come from?
| You have calculated the problem correctly but you made a small mistake while taking the integral of $\int\dfrac{2}{x^2+1}$
To solve such integrals,
take the constant out and then simplify:
$$2\cdot \int \:\frac{1}{x^2+1}dx$$
We know, $$\int \:\frac{1}{x^2+1}dx = \arctan(x) $$
So,
$$\int\dfrac{2}{x^2+1} = 2\cdot \arctan \left(x\right)$$
As you progress in Calculus, you have to have these kinds of integrals memorized. See derivatives of inverse trig functions.
| {
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Proving $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$
Prove for all $n \in \mathbb{Z}^+$ that $5 \mid (n^5-n)$
My proof
Basis step: Since $5 \mid (1^5-1) \iff 5 \mid 0$ and $5 \mid 0$ is true, the statement is true for $n=1$.
Inductive step: Assume the statement is true for $n = k$; that is, assume that $5 \mid (k^5-k)$ is true. Then there is $m \in \mathbb{Z}^+$ such that $$k^5 - k = 5m.$$ We must show that this statement is true for $n = k+1$, i.e. show that there is $\ell \in \mathbb{Z}^+$ such that $$(k+1)^5 - (k+1) = 5\ell.$$ Note that $(k+1)^5 - (k+1)$ expands as $k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$.
We can try to find a polynomial $P(x)$ such that $$(k^5-5) + P(x) =(k+1)^5 - (k+1) = k^5 + 5k^4 + 10k^3 + 10k^2 + 4k$$ so as to try to add $P(x)$ to both sides of the assumption. We find that
$$\begin{align}P(x) &= k^5 + 5k^4 + 10k^3 + 10k^2 + 4k - (k^5-5) \\
&=5k^4 + 10k^3 + 10k^2 + 5k \end{align}$$ and we can also observe that since $k\in\mathbb{Z}^+$, we have that $\frac{1}{5}P(x) = k^4 + 2k^3 + 2k^2 + k$ is a positive integer. Thus we add this to both sides of our assumption $$ \begin{align}
k^5 - k &= 5m \\
(k^2 - k) + P(x) &= 5m + P(x) \\
(k+1)^5 - (k+1) &= 5\left(m + \tfrac{1}{5}P(x)\right)
\end{align}.$$ Since $\frac{1}{5}P(x),m \in \mathbb{Z}^+$, it follows that $m + \tfrac{1}{5}P(x) \in \mathbb{Z}^+$. Thus $5 \left| \big[ (k+1)^5 - (k+1) \big] \right.$
By PMI, $5 \mid (n^5-n)$ for all $n \in \mathbb{Z}^+$.
My questions
*
*Is this proof valid?
*What other ways can this be proved by induction? The polynomial expansions took a while to deal with, so I was wondering if there are any alternate methods. (Just FYI, I only have college first-year-level knowledge)
| $n^5 - n = n(n^4 - 1) = n(n^2 - 1)(n^2 + 1) = n(n - 1)(n+1)(n^2 + 1)$.
There is a unique $q, r$ so that $n = 5q + r; 0\le r < 5$. This is basically dividing $n$ by $5$ and finding the remainder.
If $r = 0$ then $n = 5q$ and $n^5 - n = 5q(n-1)(n+1)(n^2 + 1)$ and we are done.
If $r = 1$ then $n = 5q +1$ and $n-1 = 5q$ and $n^5 -n = n(5q)(n+1)(n^2 + 1)$ and we are done.
If $r = 2$ then $n = 5q + 2$ and $n^2 + 1 = (5q+2)^2 + 1 = 25q^2 + 20q + 4 + 1 = 5(5q^2 + rq + 1)$ and $n^5 -n = n(n-1)(n+1)5(q^2 + 4q + 1)$ and we are done.
If $r = 3$ then $n = 5q + 3$ and $n^2 + 1 = (25q^2 + 30q + 9) + 1 = 5(q^2 + 6q + 2)$ and $n^5 -n = n(n-1)(n+1)5(q^2 + 6q + 2)$ and we are done.
If $r = 4$ then $n = 5q + 5$ and $n + 1 = 5q +5=5(q+1)$ and $n^5 - n = n(n-1)5(q+1)(n^2 + 1)$ and we are done.
Those are the only $5$ options.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2985270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Area bounded by $y=\sqrt{\frac{1+\sin{x}}{\cos{x}}}$ and $y=\sqrt{\frac{1-\sin{x}}{\cos{x}}}$
The area of the region between the curves
$y=\sqrt{\frac{1+\sin{x}}{\cos{x}}}$ and
$y=\sqrt{\frac{1-\sin{x}}{\cos{x}}}$ bounded by the lines $x=0$ and
$x=\frac{\pi}{4}$ is:
a) $\int_{0}^{\sqrt{2}-1}\frac{t}{(1+t^2)\sqrt{1-t^2}}dt$
b) $\int_{0}^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
c) $\int_{0}^{\sqrt{2}+1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$
d) $\int_{0}^{\sqrt{2}+1}\frac{t}{(1+t^2)\sqrt{1-t^2}}dt$
My Attempt:
So I started with,
$$\int_{0}^{\frac{\pi}{4}}\sqrt{\frac{1+\sin{x}}{\cos{x}}}-\sqrt{\frac{1-\sin{x}}{\cos{x}}}dx$$
$$\int_{0}^{\frac{\pi}{4}}\frac{2\sin{x}}{\sqrt{\cos{x}}(\sqrt{1+\sin{x}}+\sqrt{1-\sin{x}})}dx$$
Putting $\cos{x}=u$
$$\int_{1}^{\frac{1}{\sqrt{2}}}\frac{-2}{\sqrt{u}(\sqrt{1+\sqrt{1-u^2}}+\sqrt{1-\sqrt{1-u^2}})}du$$
Putting $u=\frac{1}{w}$ and further $w-1=t$,
I end up with,
$$\int_{0}^{\sqrt{2}-1}\frac{2}{(1+t)(\sqrt{(1+t)+\sqrt{(1+t)^2-1}})+(\sqrt{(1+t)-\sqrt{(1+t)^2-1}})}dt$$
But I am not able to reduce this to any of the given options? Am I miscalculating something or this is reducible?
Any hints would be helpful.
Thank you.
| $$I=\int_0^{\pi/4}\sqrt{\frac{1+\sin(x)}{\cos(x)}}-\sqrt{\frac{1-\sin(x)}{\cos(x)}}dx$$
using Weierstrass substitution:
https://en.wikipedia.org/wiki/Tangent_half-angle_substitution
we know that:
$$\sin(x)=\frac{2t}{1+t^2},\,\cos(x)=\frac{1-t^2}{1+t^2},\,dx=\frac{2dt}{1+t^2}$$
so we can obtain:
$$I=\int_0^{\sqrt{2}-1}\left(\sqrt{\frac{1+t^2+2t}{1-t^2}}-\sqrt{\frac{1+t^2-2t}{1-t^2}}\right).\frac{2dt}{1+t^2}$$$$
=\int_0^{\sqrt{2}-1}\left(\sqrt{\frac{(1+t)^2}{(1-t)(1+t)}}-\sqrt{\frac{(1-t)^2}{(1-t)(1+t)}}\right).\frac{2dt}{1+t^2}$$$$
=\int_0^{\sqrt{2}-1}\left(\sqrt{\frac{1+t}{1-t}}-\sqrt{\frac{1-t}{1+t}}\right).\frac{2dt}{1+t^2}$$$$
=\int_0^{\sqrt{2}-1}\frac{(1+t)-(1-t)}{\sqrt{1-t^2}}.\frac{2dt}{1+t^2}$$$$
=\int_0^{\sqrt{2}-1}\frac{4t}{(1+t^2)\sqrt{1-t^2}}dt$$
so it is $(b)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2987706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Proving $\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$
Prove the identity
$$8\cos^4 \theta -4\cos^3 \theta-8\cos^2 \theta+3\cos \theta +1=\cos4\theta-\cos3\theta$$
If $7\theta $ is a multiple of $2\pi,$ Show that $\cos4\theta=\cos3\theta$ and deduce,
$$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}=-\frac{1}{2}$$
My Work
I was able to prove identity using half angle formula and $\cos3\theta $ expansion.
Since
$$7\theta=2n\pi$$ $$4\theta=2n\pi-3\theta$$
$$\therefore \cos4\theta=\cos3\theta$$
I cannot prove the final part.
Please help me. Thanks in advance.
| Here is one way to do it. It doesn't use your previous work, though.
$\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7}\\
\frac {\sin\frac \pi7(\cos\frac{2\pi}{7}+\cos\frac{4\pi}{7}+\cos\frac{6\pi}{7})}{\sin
\frac \pi7}\\
\frac {\sin\frac \pi7\cos\frac{2\pi}{7}+\sin\frac \pi7\cos\frac{4\pi}{7}+\sin\frac \pi7\cos\frac{6\pi}{7})}{\sin
\frac \pi7}\\
\sin A\cos B = \frac12 \sin(A+B) - \frac12\sin(B-A)\\
\frac {\sin\frac {3\pi}7 - \sin \frac {\pi}{7}+\sin\frac {5\pi}7 - \sin\frac{3\pi}{7} +\sin\frac {7\pi}7 - \sin\frac {5\pi}{7}}{2\sin
\frac \pi7}\\
\frac {\sin \pi - \sin \frac {\pi}{7}}{2\sin
\frac \pi7} = -\frac 12\\
$
Using the info from part 1.
Let $x = \cos \theta$
if $\theta = \frac{2n\pi}{7}$
$8x^4-4x^3 -8x^2+3x+1 = 0$
and $1, \cos \frac{2\pi}{7}, \cos \frac{4\pi}{7}, \cos \frac{6\pi}{7}$ are roots of the polynomial.
$8(x - 1)(x - \cos{2\pi}{7})(x - \cos{4\pi}{7})(x - \cos{6\pi}{7}) = 8x^4-4x^3 -8x^2+3x+1$
By Viteta's rules
$8(1+\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7}) = 4\\
1+\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7} = \frac 12\\
\cos{2\pi}{7}+\cos{4\pi}{7}+\cos{6\pi}{7} = -\frac 12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2991542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Show that the following inequality holds when $x>0$ $\require{cancel}$
Show that the following inequality holds for $x>0$
$$1+\frac{x}{2}-\frac{x^2}{8}<\sqrt{x+1}<1+\frac{x}{2}.$$
I proceeded as follows
$$\sqrt{x+1}=1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}},\quad \xi\in[0,x]$$
and substituing in the inequality yields
$$1+\frac{x}{2}-\frac{x^2}{8}<1+\frac{x}{2}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<1+\frac{x}{2}$$
which is equivalent to
$$\begin{cases}
\cancel{1+\frac{x}{2}-\frac{x^2}{8}}\cancel{<1+\frac{x}{2}-\frac{x^2}{8}}+\frac{x^3}{16 (\xi+1)^{5/2}}\quad(1)\\
\cancel{1+\frac{x}{2}}-\frac{x^2}{8}+\frac{x^3}{16 (\xi+1)^{5/2}}<\cancel{1+\frac{x}{2}}\qquad\,\,\,\,\,\, (2)
\end{cases}$$
Inequality $(1)$ holds for the values given by the problem statement. Instead for $(2)$
$$\frac{x^3}{16 (\xi+1)^{5/2}}<\frac{x^2}{8}$$
But $$\displaystyle{\max_{0\leq\xi\leq x}\Bigg\{\frac{1}{16 (\xi+1)^{5/2}}\Bigg\}}=\displaystyle{\min_{0\leq\xi\leq x}{\Big\{16 (\xi+1)^{5/2}}\Big\}}$$
which occurs at $\xi=0$, and thus, for $(2)$, it is left to prove that
$$\frac{x^3}{16}<\frac{x^2}{8}$$
but this happens for $x<0\,\vee\,0<x<2$.
Instead, if I use $\xi=x$ then inequality $(2)$ becomes
$$\frac{x^3}{16 (x+1)^{5/2}}<\frac{x^2}{8}$$
which holds for $-1<x<0\,\vee\,x>0$, and thus coincides with the restriction given by the problem.
Would it be correct to take $\xi=x$ rather than $0$? Is this approach correct at all?
| For the left inequality:
$$
1 + \frac{x}{2} - \sqrt{1 + x} = \frac{\left(1 + \frac{x}{2}\right)^2 - (1 + x)}{1 + \frac{x}{2} + \sqrt{1 + x}} < \frac{\frac{x^2}{4}}{2} = \frac{x^2}{8}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2997403",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Calculate $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$ Calculate $\sum_{k=1}^n (-1)^{k+1} \binom{n}{k}\frac{1}{k}$, I do not know hot get rid of that $k$, for me it is similar like $\binom{n}{k}=\frac{k}{n} \binom{n-1}{k-1}$, do you have some idea?
| This problem and its type appear at MSE regularly. Suppose we seek to
compute
$$S_n = \sum_{k=1}^n {n\choose k} \frac{(-1)^{k+1}}{k}.$$
With this in mind we introduce the function
$$f(z) = n! (-1)^{n+1} \frac{1}{z^2} \prod_{q=1}^n \frac{1}{z-q}.$$
We then obtain for $1\le k\le n$
$$\mathrm{Res}_{z=k} f(z) =
(-1)^{n+1} \frac{n!}{k^2} \prod_{q=1}^{k-1} \frac{1}{k-q}
\prod_{q=k+1}^n \frac{1}{k-q}
\\ = (-1)^{n+1} \frac{n!}{k}
\frac{1}{k!} \frac{(-1)^{n-k}}{(n-k)!}
= {n\choose k} \frac{(-1)^{k+1}}{k}.$$
This means that
$$S_n = \sum_{k=1}^n \mathrm{Res}_{z=k} f(z)$$
and since residues sum to zero we have
$$S_n + \mathrm{Res}_{z=0} f(z) + \mathrm{Res}_{z=\infty} f(z) = 0.$$
We can compute the residue at infinity by inspection (it is zero) or
more formally through
$$\mathrm{Res}_{z=\infty}
n! (-1)^{n+1} \frac{1}{z^2} \prod_{q=1}^n \frac{1}{z-q}
\\ = - n! (-1)^{n+1} \mathrm{Res}_{z=0} \frac{1}{z^2}
z^2 \prod_{q=1}^n \frac{1}{1/z-q}
\\ = - n! (-1)^{n+1} \mathrm{Res}_{z=0}
\prod_{q=1}^n \frac{z}{1-qz}
\\ = - n! (-1)^{n+1} \mathrm{Res}_{z=0} z^n
\prod_{q=1}^n \frac{1}{1-qz} = 0.$$
We get for the residue at $z=0$ that
$$\mathrm{Res}_{z=0} f(z) =
n! (-1)^{n+1}
\left. \left(\prod_{q=1}^n \frac{1}{z-q}\right)'\right|_{z=0}
\\ = - n! (-1)^{n+1} \left.
\left(\prod_{q=1}^n \frac{1}{z-q}\right)
\sum_{q=1}^n \frac{1}{z-q} \right|_{z=0}
\\ = n! (-1)^n \frac{(-1)^{n}}{n!} \left(-H_{n}\right)
= -H_n.$$
We thus have $S_n - H_n = 0$ or
$$\bbox[5px,border:2px solid #00A000]{
S_n = H_n = \sum_{k=1}^n \frac{1}{k}.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Maximum value of $x$ when equality is given $$ x + y = \sqrt{x} + \sqrt{y} $$
Find maximum value of $x$. $x$ and $y$ are reals.
| Since $x+ y = \sqrt{x} + \sqrt{y}$ is symmetric, WLOG, we can consider $x\ge y\ge 0$ or $y=ax, 0\le a\le 1$. Then:
$$x+ ax = \sqrt{x} + \sqrt{ax} \Rightarrow \color{red}{\sqrt{x}}=\frac{1+\sqrt{a}}{1+a} \to \text{max s.t. $0\le a\le 1$}\\
f'(a)=\left(\frac{1+\sqrt{a}}{1+a}\right)'=0 \Rightarrow \frac{\frac1{2\sqrt{a}}(1+a)-(1+\sqrt{a})}{(1+a)^2}=0 \Rightarrow \\
a+2\sqrt{a}-1=0 \Rightarrow a=3-2\sqrt{2}\approx 0.17.$$
Note that the function $f(a)$ is continuous at $0\le a\le 1$ and $f(0)=f(1)=1$ and $f(3-2\sqrt{2})=\frac12(1+\sqrt{2})\approx 1.207$. Hence it must be the maximum point (see Desmos graph).
So, the maximum value of $x$ is:
$$\color{red}x=\left(\frac{1+\sqrt{3-2\sqrt{2}}}{1+(3-2\sqrt{2})}\right)^2=\frac14\left(1+\sqrt{2}\right)^2\approx 1.457.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 3
} |
Fast way to solve $(s-2i)^2 (s+2i)^2$ $(s-2i)^2 (s+2i)^2=$
$(s^2-4si-4) (s^2+4si-4)=$
$s^4+4s^3i-4s^2-4s^3i+16s^2+16si-4s^2-16si+16 =$
$(s^4-8s^2+16) =$
$(s^2+4)^2$
Is there a quicker way to see that $(s-2i)^2 (s+2i)^2= (s^2+4)^2$
| notice:
$$(a+b)(a-b)=a^2-b^2$$
so:
$$(a+bi)(a-bi)=a^2-(bi)^2=a^2+b^2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3002841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What am I doing wrong finding $\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{1/x^2}$? It has an answer here, but I'd like to know where my solution went wrong.
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x}{1+x\cdot3^x} \right)^{\frac{1}{x^2}} $$
$$\lim_{x\to 0} \left( \frac{1+x\cdot2^x +x\cdot 3^x-x\cdot 3^x}{1+x\cdot3^x} \right)^\frac{1}{x^2} $$
$$\lim_{x\to 0} \left( 1 + \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x} \right)^\frac{1}{x^2} $$
$$\lim_{x\to 0} \left( 1 + \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x} \right)^{\frac{1}{x^2}\cdot \frac{1+x\cdot 3^x}{x\cdot2^x-x\cdot 3^x}\cdot\frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x^2}\cdot \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x}\cdot \frac{2^x-3^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x}\cdot \frac{(1+1)^x-(2+1)^x}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{1}{x}\cdot \frac{1+x+o(x)-1-x2-o(x)}{1+x\cdot3^x}} $$
$$\lim_{x\to 0} e^{\frac{-1}{1+x\cdot3^x}} $$
$$e^{-1}$$
The answer in the book is $\frac{2}{3}$.
| You went wrong in this step
$$\ldots=\lim_{x\to 0} \left( 1 + \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x} \right)^{\frac{1}{x^2}\cdot \frac{1+x\cdot 3^x}{x\cdot2^x-x\cdot 3^x}\cdot\frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x}} =\lim_{x\to 0} \color{red}{e^{\frac{1}{x^2}\cdot \frac{x\cdot2^x-x\cdot 3^x}{1+x\cdot3^x}}}= \ldots$$
since you take the limit for a part of the expression.
Refer also to the related
*
*Problem with limit solving
*Analyzing limits problem Calculus (tell me where I'm wrong).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3003184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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bound on $\biggr\rvert \frac{x+2y-2}{x^2+y^2+1} \biggr\lvert $? Question: How to get a good bound on $\biggr\rvert \frac{x+2y-2}{x^2+y^2+1} \biggr\lvert $?
Context: I want to show that $\frac{x+2y-2}{x^2+y^2+1}$ attains maximum and minimum value on $\Bbb R^2$. So I need to find some compact set $K$ which contains the minimum and maximum, and show that on $\Bbb R^2 - K$ , the absolute value of the function is small.
Attempt:
$\biggr\rvert \frac{x+2y-2}{x^2+y^2+1} \biggr\lvert
\leq \frac{|x+2y-2|}{x^2+y^2+1} \leq \frac{\sqrt{x^2+y^2}+2(\sqrt{x^2+y^2})-2}{x^2+y^2+1} \leq \frac{3\sqrt{x^2+y^2}-2}{x^2+y^2}
$
This is good enough to bound the function when $x^2+y^2$ is large, but is there a better way?
But I am not sure how to get rid of the $-2$ in the numerator.
| For the upper bound observe
\begin{align}
c(x^2+y^2+1)-x-2y+2 =&\ c\left(x^2-\frac{1}{c}x\right)+c\left(y^2-\frac{2}{c}y\right)+(2+c)\\
=&\ c\left(x-\frac{1}{2c} \right)^2+c\left(y-\frac{1}{c} \right)^2+2+\frac{4c^2-5}{4c} \geq 0
\end{align}
for all $x, y$ iff
\begin{align}
2+\frac{4c^2-5}{4c}=0 \ \ \Leftrightarrow \ \ c=\frac{1}{2}.
\end{align}
Hence it follows
\begin{align}
x+2y-2 \leq \frac{1}{2}(x^2+y^2+1) \ \ \Longleftrightarrow \ \ \frac{x+2y-2}{x^2+y^2+1}\leq \frac{1}{2}
\end{align}
for all $x, y$. Maximum is attained when $(x, y) = (1, 2)$.
For the lower bound, we consider
\begin{align}
c(x^2+y^2+1)-x-2y+2 =&\ c\left(x^2-\frac{1}{c}x\right)+c\left(y^2-\frac{2}{c}y\right)+(2+c)\\
=&\ c\left(x-\frac{1}{2c} \right)^2+c\left(y-\frac{1}{c} \right)^2+2+\frac{4c^2-5}{4c} \leq 0
\end{align}
which needs to hold for all $x, y$. In particular, the inequality has to hold when $x=\frac{1}{2c}$ and $y=\frac{1}{c}$. Hence
\begin{align}
2+\frac{4c^2-5}{4c}\leq 0
\end{align}
if $c \in (-\infty, -5/2]$. Hence
\begin{align}
-\frac{5}{2} \leq \frac{x+2y-2}{x^2+y^2+1}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3005868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
An AMM-like integral $\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx$
How can we evaluate $$I=\int_0^1\frac{\arctan x}x\ln\frac{(1+x^2)^3}{(1+x)^2}dx=0?$$
I tried substitution $x=\frac{1-t}{1+t}$ and got
$$I=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} \arctan \frac{t-1}{t+1}}{t^2-1}dt\\
=\int_0^1\frac{2 \ln \frac{2 (t^2+1)^3}{(t+1)^4} (\arctan t-\frac\pi4)}{t^2-1}dt$$
I'm able to evaluate $$\int_0^1\frac{\ln \frac{2 (t^2+1)^3}{(t+1)^4}}{t^2-1}dt$$
But I have no idea where to start with the rest one.
| A solution by Cornel Ioan Valean. The problem is similar to the problem AMM $12054$. Using the well-known result in $4.535.1$ from Table of Integrals, Series and Products by I.S. Gradshteyn and I.M. Ryzhik:
$$\int_0^1 \frac{\arctan(y x)}{1+y^2x}\textrm{d}x=\frac{1}{2y^2}\arctan(y)\log(1+y^2)$$
We have:
$$\frac{1}{2}\int_0^1\frac{\arctan(y)\log(1+y^2)}{y}dy=\int_0^1\left(\int_0^1 \frac{y\arctan(y x)}{1+y^2x}\textrm{d}x\right)\textrm{d}y$$
$$\overset{yx=t}{=}\int_0^1\left(\int_0^y \frac{\arctan(t)}{1+y t}\textrm{d}t\right)\textrm{d}y=\int_0^1\left(\int_t^1 \frac{\arctan(t)}{1+y t}\textrm{d}y\right)\textrm{d}t$$
$$=\int_0^1\frac{\arctan(t)\log\left(\frac{1+t}{1+t^2}\right)}{t} \textrm{d}t\overset{t=y}=\int_0^1\frac{\arctan(y)\log\left(\frac{1+y}{1+y^2}\right)}{y} \textrm{d}y$$
And the result is proved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3006106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
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Calculate $\sum_{k=1}^{2007} \frac{5^{2008}}{25^k+5^{2008}}$ $\sum_{k=1}^{2007} \frac{5^{2008}}{25^k+5^{2008}}$. I know that solution is $\frac{2007}{2}$ and that I can use a Gaussian method.
If I use Gaussian method I do not get some sum that help me. if I put $n=2008$, then $Sn=\sum_{k=1}^n \frac{5^{n+1}}{25^k+5^{n+1}}$. and $Sn=\sum_{k=1}^n \frac{5^{n}}{25^{k-n}+5^{n+1}}$. So then $2Sn=\sum_{k=1}^n \frac{5^{n+1}}{25^k+5^{n+1}}+\frac{5^{n+1}}{25^{n-k}+5^{n+1}}$. do you have something better?
| Using $\frac{1}{1+q}+\frac{1}{1+1/q}=1$, $$\sum_{k=1}^n\frac{1}{1+5^{2k-n-1}}=\frac{1}{2}\sum_{k=1}^n\bigg(\frac{1}{1+5^{2k-n-1}}+\frac{1}{1+5^{2(n+1-k)-n-1}}\bigg)\\=\frac{1}{2}\sum_{k=1}^n\bigg(\frac{1}{1+5^{2k-n-1}}+\frac{1}{1+5^{n+1-2k}}\bigg)=\frac{1}{2}\sum_{k=1}^n 1=\frac{n}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3008088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find all real numbers $x,y,z\in [0,1]^3$ such that $(x^2+y^2)\sqrt{1-z^2}\ge z$.... Such that:
$$(x^2+y^2)\sqrt{1-z^2}\ge z$$
and
$$(z^2+y^2)\sqrt{1-x^2}\ge x$$
and
$$(x^2+z^2)\sqrt{1-y^2}\ge y$$
Since $x,y,z$ $\in ]0,1[^3$
then , there are some real numbers $a,b,c$ such that
$\cos a=x, \cos b=y , \cos c=z$
After some manipulations , we find that :
$$\frac{1}{1+\tan^2 a}+\frac{1}{1+\tan^2 b}\ge \frac{1}{\tan c}$$
.... same for other inequalities
I don't know what i must do now
| You were very close, but you made a mistake. You should have
$$\frac{1}{1+\tan^2 a}+\frac{1}{1+\tan^2b}\geq \frac{1}{\tan c}$$
instead. Here is a similar approach.
First, we assume that $x,y,z>0$. So, we can define $p,q,r\geq 0$ to be $\frac{\sqrt{1-x^2}}{x}$, $\frac{\sqrt{1-y^2}}{y}$, and $\frac{\sqrt{1-z^2}}{z}$, respectively. The three inequalities become
$$\frac{1}{1+p^2}+\frac{1}{1+q^2}\geq \frac1r\wedge \frac{1}{1+q^2}+\frac{1}{1+r^2}\geq \frac1p \wedge \frac{1}{1+r^2}+\frac{1}{1+p^2}\geq \frac1q.$$
Adding all of these and then dividing the result by $2$ yield
$$\frac{1}{1+p^2}+\frac{1}{1+q^2}+\frac{1}{1+r^2}\geq \frac{1}{2p}+\frac{1}{2q}+\frac{1}{2r}.\tag{1}$$
However, by AM-GM, $1+p^2\ge 2p$, $1+q^2\ge 2q$, and $1+r^2\ge 2r$. That is,
$$\frac{1}{1+p^2}+\frac{1}{1+q^2}+\frac{1}{1+r^2}\leq \frac{1}{2p}+\frac{1}{2q}+\frac{1}{2r}.\tag{2}$$
From (1) and (2), we must have
$$\frac{1}{1+p^2}+\frac{1}{1+q^2}+\frac{1}{1+r^2}= \frac{1}{2p}+\frac{1}{2q}+\frac{1}{2r},$$
which implies $1+p^2=2p$, $1+q^2=2q$, $1+r^2=2r$, so $p=q=r=1$ and $x=y=z=\frac1{\sqrt{2}}$.
Now, WLOG, if $x=0$, then $y^2\sqrt{1-z^2}\geq z$ and $z^2\sqrt{1-y^2}\geq y$. Multiplying the two inequalities gives
$$y^2z^2\sqrt{1-y^2}\sqrt{1-z^2}\geq yz.$$
If $yz\neq 0$, then dividing by $yz$, we have
$$yz\sqrt{1-y^2}\sqrt{1-z^2}\geq 1.$$
But by AM-GM, $y\sqrt{1-y^2}=\sqrt{y^2(1-y^2)}\leq \frac{y^2+(1-y^2)}{2}=\frac12$ and similarly, $z\sqrt{1-z^2}\leq \frac12$. So,
$$yz\sqrt{1-y^2}\sqrt{1-z^2}\leq \frac12\cdot\frac12=\frac14<1.$$
This is a contradiction, so $yz=0$, so $y=0$ or $z=0$, so $x=y=z=0$. Therefore, if any of the variables is $0$, all of them are $0$. There are then two solutions $$x=y=z=0 \wedge x=y=z=1/\sqrt{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3010735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Limit of $\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$ (No L'Hôpital)
$\lim_{x \to 0} (\cot (2x)\cot (\frac{\pi }{2}-x))$
I can't get to the end of this limit. Here is what I worked out:
\begin{align*}
& \lim_{x \to 0} \frac{\cos 2x}{\sin 2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\lim_{x \to 0}\frac{\frac{\cos2x }{2x}}{\frac{\sin 2x}{2x}}\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \frac{\cos 2x}{2x} \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \frac{{\cos^2 (x)}-{sin^2 (x)}}{2x}\cdot\frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{\cos^2(x)}{2x}-\frac{sin^2 x}{2x}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1-\sin^2 x}{2x}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin^2 x}{2x}-\frac{sin x}{2}\right) \cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\frac{\sin x}{2}-\frac{sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
= \lim_{x \to 0} \left(\frac{1}{2x}-2\frac{\sin x}{2}\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )} \\
= & \lim_{x \to 0} \left(\frac{1}{2x}-\sin x\right)\cdot \frac{\cos(\frac{\pi }{2}-x )}{\sin(\frac{\pi }{2}-x )}
\end{align*}
Here is where I can't seem to complete the limit, the 2x in the denominator is giving me a hard time and I don't know how to get rid of it. Any help would be appreciated. (In previous questions I got a really hard time because of my lack of context, I hope this one follows the rules of the site. I tried.)
| $$F(x)=\cot2x\cot\left(\dfrac\pi2-x\right)=\dfrac{\tan x}{\tan2x}=\dfrac{\tan x(1-\tan^2x)}{2\tan x}$$
For $\tan x\ne0,F(x)=\dfrac{1-\tan^2x}2$
As $x\to0,x\ne0\implies\lim_{x\to0}F(x)=\lim_{x\to0}\dfrac{1-\tan^2x}2=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3011543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to find limit of a sequence $\lim_{n\to \infty} \frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}$? I have no idea how to solve this: $$\lim_{n\to \infty} \frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}$$ for: $|a|<1$, $|b|<1 $
I would be happy for any advice.
Thank you
| The limit is $$\lim_{n\to \infty}\left(\frac{1+a+a^2+...+a^n}{1+b+b^2+...+b^n}\cdot\frac{1-a}{1-b}\cdot\frac{1-b}{1-a}\right)=\lim_{n\to \infty} \frac{1-a^{n+1}}{1-b^{n+1}}\cdot\frac{1-b}{1-a}=\frac{1-b}{1-a}$$
EDIT
My thinking (pedagogic insert):
$1+a+a^2+...+a^n\;$ is the sum of first $n+1$ terms of geometric series.
The calculation that simplifies the sum and helps when looking for the limit, is multiplying by $\frac{1-a}{1-a}.$
Similarly for $\;1+b+b^2+...+b^n.$
Both numerator and denominator converge (since $|a|,|b|<1$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Number of Non negative integer solutions of $x+2y+5z=100$ Find Number of Non negative integer solutions of $x+2y+5z=100$
My attempt:
we have $x+2y=100-5z$
Considering the polynomial $$f(u)=(1-u)^{-1}\times (1-u^2)^{-1}$$
$\implies$
$$f(u)=\frac{1}{(1-u)(1+u)}\times \frac{1}{1-u}=\frac{1}{2} \left(\frac{1}{1-u}+\frac{1}{1+u}\right)\frac{1}{1-u}=\frac{1}{2}\left((1-u)^{-2}+(1-u^2)^{-1}\right)$$
we need to collect coefficient of $100-5z$ in the above given by
$$C(z)=\frac{1}{2} \left((101-5z)+odd(z)\right)$$
Total number of solutions is
$$S(z)=\frac{1}{2} \sum_{z=0}^{20} 101-5z+\frac{1}{2} \sum_{z \in odd}1$$
$$S(z)=540.5$$
what went wrong in my analysis?
| I will find number of solutions of equation $5x+2y+z=10 n$ in general:
clearly the positive solutions $x_0, y_0, z_0$ of this equation are corespondent to the solution $x_0+2,y_0, z_0$ of equation $5x+2y+z=10(n+1)$.Clearly for $x=>2$, finding the solutions of $5x+2y+z=10(n+1)$ will lead to finding the solution of first equation,provided we consider $x-2$ in first equation.
If the number of solutions of equation $5x+2y+z=10(n+1)$ is $\phi(n+1)$ and that of equation $5x+2y+z=10n$ is $\phi(n)$ the difference of $\phi(n+1)$ and $\phi(n)$ is equal to the number of solutions of equation $5x+2y+z=10(n+1)$ for $x=0$ and $x=1$. But this equation has $5n+6$ solutions for $x=0$, (i.e. $0=<y=<5n+5)$ and it has $5n+3$ solutions for $x=1$, (i.e $0=<y=<5n+2)$. Therefore we have:
$\phi(n+1)-\phi(n)=10n+9$
We can also search and find that $\phi(1)=10$, so we can write:
$\phi(1)=10$
$\phi(2)-\phi(1)=10\times 1+9$
$\phi(3)-\phi(2)=10\times 2+9$
.
.
.
$\phi(n)-\phi(n-1)=10(n-1)+9$
Summing theses relations gives:
$\phi(n)=5n^2 +4n +1$
In your question $n=10$, therefore number of solutions is $\phi(10)=5.10^2+4.10+1=541$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3014438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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} |
An Inconsistency in Numerical Approximation Consider the expression
$$
10^5 - \frac{10^{10}}{1+10^5}.
$$
Using the elementary properties of fractions we can evaluate the expression as
$$
10^5 - \frac{10^{10}}{1+10^5} = \frac{10^5 + 10^{10} - 10^{10}}{1+10^5} = \frac{10^5}{1+10^5}\approx 1.
$$
Note that the approximation $10^5+1 \approx 10^5$ is used in the last step. Now suppose we use the same approximation, but apply it before we perform the subtraction. We get
$$
10^5 - \frac{10^{10}}{1+10^5} \approx 10^5 - \frac{10^{10}}{10^5} = 0.
$$
The same logic works for
$$
10^p - \frac{10^{2p}}{1+10^p}
$$
for arbitrary large $p$, so it cannot be simply an issue with the accuracy of the approximation.
Is there an easy explanation of what's going on here?
| There is no paradox.
When you approximate $$\frac{10^5}{1+10^5}=1-0.000099999000\cdots$$ with $1$, the error is on the order of $10^{-5}$.
But in the second case, the same error is multiplied by $10^5$, so that it is no more negligible.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3017585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $ how can I find sum of series $ \sum_{n=1}^{\infty} (n\cdot \ln \frac{2n+1}{2n-1} - 1) $?
It is so weird for me because I put this to Mathematica and it tells me that sum does not converge...
Let consider sum no to infinity, but to n
$$ \sum_{k=1}^{n} (k\cdot \ln \frac{2k+1}{2k-1} - 1) =$$
$$ ln \frac{3}{1}\cdot \left(\frac{5}{3}\right)^2 \cdot...\cdot \left(\frac{2n+1}{2n-1}\right)^n - n = ln \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1} - n $$
but $$ n = ln e^n $$
so
it will be $$ln\frac{1}{e^n} \cdot \frac{1}{1}\cdot \frac{1}{3}\cdot \frac{1}{5}\cdot ... \frac{1}{2n-1}$$
So the limit of it is $-\infty$
Have I done this well or I missed sth?
| Let $f(x)=\displaystyle\sum_{n=1}^{\infty}\left(n\ln\frac{n+x}{n-x}-2x\right)$ for $x\in(-1,1)$. Then
$$f'(x)=\displaystyle\sum_{n=1}^{\infty}\frac{2x^2}{n^2-x^2}=1-\pi x\cot\pi x$$
(termwise differentiation is admissible because of uniform convergence of the latter series in $[-a,a]$ for any $0<a<1$; the second equality is known). Thus,
$$f(x)=x-\frac{1}{\pi}\int_{0}^{\pi x}t\cot t\,dt=x(1-\ln\sin\pi x)+\frac{1}{\pi}\int_{0}^{\pi x}\ln\sin t\,dt.$$
Your sum is $f(1/2)=(1-\ln2)/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3019050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Maximum of $ab+2bc+3ca$ with $a^4+b^4+c^4=1$ Let $a,b,c\in \mathbb R^+$ with $a^4+b^4+c^4=1$. What is the maximal value $ab+2bc+3ca$ can take?
I tried using Cauchy-Schwarz several different ways and the best upper bound I got was $\sqrt{14}$, but it was never sharp.
Numerical search suggests that the maximum occurs at about $a=0.763316$, $b=0.697312$, $c=0.80698$ with $ab+2bc+3ca=3.505647$, though I couldn't find any valuable relation between these numbers and the rationals.
| It's enough to look for non-negative variables.
Let $f(a,b,c)=ab+2bc+3ac+\lambda(a^4+b^4+c^4-1)$ and $a=xb$.
Thus, in the critical point we have
$$b+2c+4\lambda a^3=a+2c+4\lambda b^3=2b+3a+4\lambda c^3=0,$$ which gives
$$\frac{b+3c}{a^3}=\frac{a+2c}{b^3}=\frac{2b+3a}{c^3}.$$
From the first equation we obtain:
$$c=\frac{a^4-b^4}{3b^3-2a^3},$$ which after substitution in the second gives
$$\frac{(x^4-1)^3}{(3-2x^3)^3}\left(x+\frac{2(x^4-1)}{3-2x^3}\right)=2+3x$$ or
$$45x^{13}+34x^{12}-288x^{10}-183x^9-6x^8+648x^7+432x^6-9x^5-642x^4-432x^3+246x+160=0,$$
which gives $x=1.09465...$ or $x=1.26369...$ and we can show that $x=1.09465...$ gives a maximal value.
| {
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"url": "https://math.stackexchange.com/questions/3020440",
"timestamp": "2023-03-29T00:00:00",
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What will be the pdf of $X+Y$ if $X$ and $Y$ are iid from Cauchy? Suppose $X$ and $Y$ follow Cauchy distribution independent of each other. What will be the pdf of $X+Y$?
What I got by using convolution theorem is that the density $g$ of $X+Y$ is $:$
$$g(x) = \int_{-\infty}^{\infty} f(y) f(x-y)\ \mathrm {dy}$$ where $f$ is the density of the Cauchy distribution given by $f(x)=\frac {1} {\pi ({1+ x^2})},x \in \Bbb R$. Then the whole integration becomes $$\frac {1} {\pi^2} \int_{-\infty}^{\infty} \frac {\mathrm {dy}} {(1+y^2)(1+(x-y)^2)}.$$ Now how do I solve this integral? Please help me in this regard.
Thank you very much.
| You can try partial fraction decomposition:
\begin{align}\frac1{(1+y^2)(1+(x-y)^2)} &= \frac{x+2y}{x(x^2+4)(1+y^2)} + \frac{3x-2y}{x(x^2+4)(1+(x-y)^2)} \\
&= \frac{1}{(x^2+4)(1+y^2)}+
\frac{2y}{x(x^2+4)(1+y^2)} + \frac{2(x-y)}{x(x^2+4)(1+(x-y)^2)} + \frac{1}{(x^2+4)(1+(x-y)^2)}\\
&= \frac1{x^2+4}\left(\frac1{1+y^2} + \frac1{1+(x-y)^2}\right) + \frac1{x(x^2+4)}\left(\frac{y}{1+y^2} + \frac{x-y}{1+(x-y)^2}\right)
\end{align}
so we have
\begin{align}
\frac1{\pi^2}\int_{-\infty}^\infty \frac{dy}{(1+y^2)(1+(x-y)^2)} &=
\frac1{\pi^2(x^2+4)}\left[\int_{-\infty}^\infty\left(\frac1{1+y^2} + \frac1{1+(x-y)^2}\right)dy + \frac1x\int_{-\infty}^\infty\left(\frac{y}{1+y^2} + \frac{2(x-y)}{1+(x-y)^2}\right)dy
\right]\\
&= \frac1{\pi^2(x^2+4)}\left[\Big(\arctan(1+y^2) + \arctan(1+(x-y)^2)\Big)\Big|_{-\infty}^\infty + \frac1x\Big(\ln(1+y^2) - \ln(1+(x-y)^2)\Big)\Big|_{-\infty}^\infty
\right]\\
&= \frac1{\pi^2(x^2+4)}\left[2\pi + \frac2x \lim_{y \to \infty} \ln \left(\frac{1+y^2}{1+(x-y)^2}\right)\right]\\
&= \frac{2}{\pi(x^2+4)}
\end{align}
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.