Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Symmetry of function defined by integral Define a function $f(\alpha, \beta)$, $\alpha \in (-1,1)$, $\beta \in (-1,1)$ as
$$ f(\alpha, \beta) = \int_0^{\infty} dx \: \frac{x^{\alpha}}{1+2 x \cos{(\pi \beta)} + x^2}$$
One can use, for example, the Residue Theorem to show that
$$ f(\alpha, \beta) = \frac{\pi \sin{\left (... | Very interesting question! But, alas, not an answer. Only few representations for the integral obtained. One of them evaluated to the form claimed in the question.
First, transform the integral into a form, symmetric under $\alpha \mapsto -\alpha$:
$$
\int_0^\infty \frac{x^\alpha}{1+2 x \cos(\pi \beta) + x^2} \mathr... | {
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"url": "https://math.stackexchange.com/questions/268789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "177",
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How to calculate the volume obtained by rotating the following around the x axis? The original problem is:
"Find the volume of the solid obtained by rotating about the x axis the region enclosed by the curves $y = \frac{9}{x^2 + 9},y=0,x=0,\,$and $x = 3$"
I set up the following integral $$81\pi\int_0^3\frac{1}{(x^2 + ... | We want to find
$$\int \frac{dx}{(x^2+9)^2}.$$
Let $x=3u$. Apart from a constant factor, we end up with
$$\int\frac{du}{(u^2+1)^2}.$$
Let $u=\tan t$ (we could have gone more directly, by letting $x=3\tan t$). Then $du=\sec^2 t\,dt$, and we end up with
$$\int \frac{\sec^2 t}{\sec^4 t}\,dt.$$
This is the familiar integ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Convergence of the series $\sum_{n=1}^\infty \frac{(2n)!!}{(2n+1)!!} $
Study the convergence of the next series: $$\sum_{n=1}^\infty \frac{(2n)!!}{(2n+1)!!} $$
My solution: since $$\frac{(2n)!!}{(2n+2)!!} \leq \frac{(2n)!!}{(2n+1)!!}$$
forall $n \in \mathbb{N}$ and since $$\sum_{n=1}^\infty \frac{(2n)!!}{(2n+2)!!} = ... | Yes. What you have done is correct. You can attempt to quantify the divergence a bit better. Note that $(2n)!! = 2^n n!$ and $(2n+1)!! = \dfrac{(2n+2)!}{2^{n+1} (n+1)!}$. Hence,
$$\dfrac{(2n)!!}{(2n+1)!!} = 2^{2n+1} \dfrac{(n+1)! n!}{(2n+2)!} = \dfrac{2^{2n+1}}{\dbinom{2n+2}{n+1} \cdot (n+1)}$$
We have that $\dbinom{2n... | {
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Solve this inequality $\prod_{i=1}^{50} \frac {2i-1}{2i} < \frac {1}{10}$ Prove that
$ \frac{1}{2}\cdot \frac{3}{4}\cdot \frac{5}{6}\cdot \frac{7}{8}\cdot \frac{9}{10}\cdot \frac{11}{12}\cdot \frac{13}{14}...\cdot \frac{91}{92}\cdot \frac{93}{94}\cdot \frac{95}{96}\frac{97}{98}\cdot \frac{99}{100} <\frac{1}{10}$
| This is a standard / common problem.
Hint:
$$\prod_{i=1}^{98} \frac {i}{i+1} = \frac {1}{99}$$
$$\sqrt{\frac {1}{99}} \times \frac {99}{100} = \frac {\sqrt{99}}{100} < \frac {1}{10}$$
Hint elaborated slightly:
Let $A = \prod_{i=1}^{49} \frac {2i-1}{2i}$ and $B= \prod_{i=1}^{49} \frac {2i}{2i+1}$. It is clear that $A<... | {
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$ A = x + y + z$, number of solutions in $Z$ if $x, y, z$ are bounded in intervals For the equation $x + y = A$, it's easy, when you notice that when iterating over all possible $x$, the number of solutions for $y$ is $0$ at the beginning, then increases by $1$, then stays constant, then decreases by $1$, and at the en... | The number of non-negative integer solutions to $x+y+z=11$ where $2<x<5, 1<y<5, 3<z<7$ is equal to the number of integer solutions to $a+b+c=2$ where $0\leq a\leq 1$, $0\leq b\leq 2$, $0\leq c\leq 2$ by setting $a=x-3$, $b=y-2$, $c=z-4$.
In this case it is easy to count the number of solutions: $\binom{2+3-1}{3-1}-1=\b... | {
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Optimization of $2x+3y+z$ under the constraint $x^2+ y^2+ z^2= 1$
Let $x, y$ and $z$ be real numbers such that $x^2+ y^2+ z^2= 1$. Find the
maximum and minimum values of $2x + 3y + z$ .
How can I able to solve the problem? Thanks for your help.
| Using the Cauchy-Schwarz inequality, $2x+3y+z\leq \sqrt{2^2+3^2+1^2}\sqrt{x^2+y^2+z^2}=\sqrt{14}$ which is the maximum value.Since $x$,$y$ and $z$ are reals, the minimum value occurs when $x$,$y$ and $z$ are negative which can be used to infer that the minimum value is $-\sqrt{14}$.
| {
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Evaluate $\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$ Evaluate
$$\int_0^1\left(\frac{1}{\ln x} + \frac{1}{1-x}\right)^2 \mathrm dx$$
| @RonGordon last sum becomes:
\begin{align}
&\sum_{k = 1}^{\infty}\left\{
1 + \left(k + 1\right)^{2}\log\left(1 - {1 \over \left[k + 1\right]^2}\right) \right\}=
\sum_{k = 1}^{\infty}\left\{
1 - \left(k + 1\right)^{2}
\int_{0}^{1}{{\rm d}x \over \left(k + 1\right)^{2} - x}\right\}
\\[3mm]&=
-\left[\int_{0}^{1}\sum_{k = ... | {
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"timestamp": "2023-03-29T00:00:00",
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Two different solutions to integral Given the very simple integral
\begin{equation}
\int -\frac{1}{2x} dx
\end{equation}
The obvious solution is
\begin{equation}
\int -\frac{1}{2x} dx = -\frac{1}{2} \int \frac{1}{x} dx = -\frac{1}{2} \ln{|x|} + C
\end{equation}
However, by the following integration rule
\begin{equation... | They are both correct!
$$
-\frac{1}{2}\ln|-2x| + C = -\frac{1}{2}(\ln 2 + \ln |x|) + C = -\frac{1}{2}\ln |x| + (C - \frac{1}{2} \ln 2)
$$
The constant of integration is what "differs" here.
| {
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Prove $\frac{a}{bc}+ \frac{b}{ca}+ \frac{c}{ab} \ge 1$ Let $a,b,c$ be positive real numbers such that $\dfrac{1}{bc}+ \dfrac{1}{ca}+ \dfrac{1}{ab} \ge 1$. Prove that $\dfrac{a}{bc}+ \dfrac{b}{ca}+ \dfrac{c}{ab} \ge 1$.
| I want to give a complete answer.
Assume by the sake of contradiction that $$\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}<1.\tag{1}$$
By Cauchy Schwarz it follows from $(1)$ that $$\frac{3}{abc}>\frac{3}{abc}\left(\frac{a}{bc}+\frac{b}{ca}+\frac{c}{ab}\right)\geq \left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)^2.\tag{2}... | {
"language": "en",
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Integrate $\int_0^\infty \frac{\sqrt{x}}{x^{2}+1}\, \mbox{d} x$ I've been trying to integrate the following
$$\int_{0}^{\infty} \frac{\sqrt{x}}{x^{2}+1} \mbox{d} x$$
on half an annulus in the upper half plane. I keep getting $\frac{\pi}{\sqrt{2}}\ i$, which doesn't give with the numerical approximations I get using Wol... | One way to attack this particular problem is to make the substitution $x=t^2$, as in Ahlfors, Complex Analysis, Third Edition, p. 159; the integral becomes
$$\int_{-\infty}^{\infty} dt \: \frac{t^2}{t^4+1} $$
We may now use a semicircular contour $C$ in the half plane $\Re{z} \ge 0$ rather than the annulus as we have d... | {
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Evaluating integrals such as $\int \frac{1+\cos^{2}x}{1+\cos2x}$ We started integrals not too long ago, I understand it for the most part but I always have a problem figuring out how to solve ones involving trig identities. Like this:
$$\int \frac{1+\cos^{2}x}{1+\cos2x}$$
Indefinite integral of $$\frac{ 1 + \cos^2(x)}{... | Write $$2\cos^2x = 1+ \cos2x$$
$$\cos^2x= \frac{1+ \cos2x}{2}$$
$$=∫1+\frac{1+ \cos2x}{2} * \frac{1}{1+\cos2x} $$
$$=\int\frac{1}{1+\cos2x}+\int\frac{1}{2}$$
Use $$u=2x$$
$$=\frac{1}{2}[\int\frac{1}{1+\cos(u)}]+\frac{1}{2}x$$
$$=\frac{1}{2}[\int\frac{1-cos(u)}{1-\cos^2u}]$$
$$=\frac{1}{2}[\int\frac{1-cos(u)}{\sin^2u}]$... | {
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"timestamp": "2023-03-29T00:00:00",
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Asymptotic expansion of $ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $ I'm trying to compute the asymptotic expansion of
$$ I_n = \int_0^{\pi/4} \tan(x)^n \mathrm dx $$
Here is what I've done:
Change of variable $$ t= \tan x $$
$$ I_n = \int_0^1 \frac{t^n \mathrm dt}{1+t^2} = \int_0^1 \frac{(1-t)^n \mathrm dt}{t^2-2t+2}... | I couldn't follow what you were doing, but I do see that you have the right answer to first order. I can, however, point to a simpler way based on Laplace's Method.
First of all, write the integral as an exponential whose maximum is at $x=0$:
$$I_n = \int_0^{\pi/4} dx \: e^{n \log{\tan{\left ( \frac{\pi}{4} - x \right... | {
"language": "en",
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Floor function inequality I am racking my brain trying to get this problem solved and I can't seem to break it...
Let $m, n$ be positive integers, with $m > 1$. Prove
$$\left\lfloor\frac{n}m\right\rfloor+\left\lfloor\frac{n+1}m\right\rfloor\le\left\lfloor\frac{2n}m\right\rfloor$$
I started trying to use the inequaliti... | Let $a={\sf floor}(\frac{n}{m})$ and $b={\sf floor}(\frac{n+1}{m})$.
We have $a \leq \frac{n}{m}$ and $b \leq \frac{n+1}{m}$, so $a+b \leq \frac{2n+1}{m}$. If this were an equality, it would imply that both $a =\frac{n}{m}$ and $b = \frac{n+1}{m}$, so that $m$ would divide both $n$ and $n+1$, and hence divide $n+1-n=1$... | {
"language": "en",
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General expression for exponentiating a matrix? $A= \begin{pmatrix}
\cos(x) & \sin(x) \\
-\sin(x) & \cos(x) \\
\end{pmatrix}$
I found that
$A^2 = \begin{pmatrix}
\cos^2(x)-\sin^2(x) & 2\cos(x)\sin(x) \\
-2\cos(x)\sin(x) & \cos^2(x)-\sin^2(x) \\
\end{pmatrix}$
$A^2 = \beg... | Hint: this is rotation by the angle x
| {
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Problem with simple integral I'm trying to solve this simple integral:
$$\frac12 \int \frac{x^2}{\sqrt{x + 1}} dx$$
Here's what I have done so far:
*
*$\displaystyle t = \sqrt{x + 1} \Leftrightarrow x = t^2 - 1 \Rightarrow dx = 2t dt$
*$\displaystyle \frac12 \int \frac{x^2}{\sqrt{x + 1}} dx = \int \frac{t (t^2 - 1)... | There is a (slightly) more obvious way of solving it: rewrite the numerator as $x^2+1-1$ and then the whole integral as a sum of two integrals:
$$
\int \frac{(x^2-1)dx}{\sqrt{x+1}} + \int \frac{dx}{\sqrt{x+1}}
$$
The second integral is easy, the first one is
$$
\int \frac{(x^2-1)dx}{\sqrt{x+1}} =\int \frac{(x+1)(x-1)... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$? Prove that $3^{2^{k}} - 1$ is divisible by $2^{k+2}$ for $k \ge 1$?
Suppose $p,q,r$ are positive integers satisfying $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}<1$, prove that $\dfrac{1}{p}+\dfrac{1}{q}+\dfrac{1}{r}\le \dfrac{41}{42} $ ?
| For the first one, induction readily gives you the answer.
For $k=1$, we have $2^3 \vert 3^{2^1} - 1$.Assume that it is true for some $k=m$ i.e. $2^{m+2} \vert 3^{2^m} - 1$ i.e. $3^{2^m}-1 = M \cdot 2^{m+2}$We have that $$3^{2^{m+1}}-1 = \left(3^{2^m}+1\right)\left(3^{2^m}-1\right) =\left(3^{2^m}+1\right) \cdot M \cdo... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove the inequality $\frac a{a+b^2+c^2} + \frac b{b+c^2+a^2}+\frac c{c+a^2+b^2} \le 1$ $a, b, c > 0$ such that $abc=1$, prove inequality (Lagrang not allowed here because I am still in grade $9^{th}$):
$$P=\frac a{a+b^2+c^2} + \frac b{b+c^2+a^2}+\frac c{c+a^2+b^2} \le 1$$
I have tried to use AM-GM, that I have:
$(a+b... | Let $a = \frac{x^2}{yz}$ and so forth. Expanding (*):
$$(x^4 yz + y^6 + z^6)(y^4 xz + x^6 + z^6)(z^4 xy + x^6 + y^6) - x^4yz (y^4 xz + x^6 + z^6)(z^4 xy + x^6 + y^6) - y^4xz (x^4yz + y^6 + z^6)(z^4 xy + x^6 + y^6) - z^4xy(x^4yz + y^6 + z^6)(y^4 xz + x^6 + z^6) $$
at quickmath.com gives (**)
$$x^{12}y^6 + x^{12}z^6 + y^... | {
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The Least Natural number $n$ which has $18$ divisors The Least Natural number $n$ which has $18$ divisors
My Try:: $18 = 2\times 3 \times 3 = 3^2 \times 2$
Now How can I solve it
Thanks
| We know the number of divisors of $\prod_{1\le r\le m}p_i^{m_i}$ is $\prod_{1\le r\le m}(m_i+1)$ where $p_i$s are distinct primes, and power $m$s positive integers
As $18=2\cdot3^2$, $m$ can not be $\ge4$
If $m=1$, $m_1+1=18\implies m_1=17\implies n_\text{min}=2^{17}$
If $m=2$, $(m_1+1)(m_2+1)=18=6\cdot3=2\cdot9$
So, e... | {
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Prove the identity $1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$ Here is the problem:
$$1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$$
Can you help me prove that this is an trigonometric identity?
| $$1 + \sin x = 2 \cos^2 \left(45° - \frac{x}{2}\right)$$
$$\cos{2\alpha}=2cos^{2}{\alpha}-1$$
$$\cos{2\cdot \left(45^{\circ}-\frac{x}{2}\right)}=\cos{\left(90^{\circ}-x\right)}=\cos{90^{\circ}\cos{x}}+\sin{90^{\circ}\sin{x}}=0+1\cdot \sin{x}=\sin{x}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Change of Variable (Double Integral) I have been trying to integrate the two following integrands; $$\int \int_{D}(x^{2}+y^{2})dxdy$$ where $D=\{{x^{2}+xy+y^{2}\leq 1}\}$ and $$\int \int_{D}\sqrt{x^{2}+y^{2}}dxdy$$ where $D=\{{x^{2}+y^{2}\leq x}\}$. Now, I have been tempted to use change of variables (polar coordinate... | First, do coordinates transform
$$
x = \frac u{\sqrt 3} - v; \qquad y = \frac u{\sqrt 3} + v
$$
It will reduce your domain $D$ to the inner region of the circle:
$$
x^2+xy+y^2= \frac {u^2}3 - 2\frac{uv}{\sqrt 3} + v^2 + \frac {u^2}3 - v^2 + \frac{u^2}3 + 2 \frac{uv}{\sqrt 3} + v^2 = u^2 + v^2 \leq 1
$$
Now change the s... | {
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Assume $n$ is even. Prove that $323$ divides $20^n+16^n-3^n-1$. I'm unclear what is the best method to teach this with minimum math experience.
| Since $n$ is even:
$20^n + 16^n - 3^n - 1 \equiv 1^n + (-3)^n - 3^n - 1 = 1 + 3^n - 3^n - 1 = 0 \bmod 19$
Also:
$20^n + 16^n - 3^n - 1 \equiv 3^n + (-1)^n - 3^n - 1 = 3^n + 1 - 3^n - 1 = 0 \bmod 17$
So the quantity is divisible by $17\times 19 = 323$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Simplification of $\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$ Simplify the expression $$\sqrt{(1-\cos\alpha \cos\beta)^2-\sin^2\alpha \sin^2\beta}$$
I have done this way : $(1-\cos\alpha \cos\beta)^2 = 1-2\cos\alpha \cos\beta +\cos^2\alpha \cos^2\beta$
Please guide further....
| Let $a=\cos\alpha,b=\cos\beta$
So, the the given expression $$=\sqrt{(1-ab)^2-(1-a^2)(1-b^2)}=\sqrt{1-2ab+a^2b^2-(1-a^2-b^2+a^2b^2)}=\sqrt{a^2+b^2-2ab}=|a-b|=|\cos\alpha-\cos\beta|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/307064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to measure the streakedness of numerical data? Would anyone know how to use C/C++ to calculate the streakedness of data? The definition of streakedness is how many deviations away from the mean(i.e running average a numerical data streak. Thank you for your help.
[EDIT] From our company's chief software architect, ... | $\text{Average} = 13/1 = 13$
B: $\text{Average} = (7+6)/2 = 6.5$
C: $\text{Average} = (7+6)/2 = 6.5$
D: $\text{Average} = (7+3+2)/3 = 4$
E: $\text{Average} = (7+3)/2 = 5$
F: $\text{Average} = 7/1 = 7$
Factoring in non-streaks (counting them as 1’s): | {
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A question on summation notation and Raabe's test So I have a sum
$$\sum_{n=1}^\infty\frac{2\cdot4\cdot...\cdot2n}{5\cdot7\cdot...\cdot(2n+3)}$$
which I thought meant $$\sum_{n=1}^\infty\frac{2n}{2n+3}$$ which is trivially convergent by the limit test, and by Raabe's Test (which i'm currently practicing). But Taylor an... | The series
$$
\sum_{n=1}^\infty\frac{2n}{2n+3}\tag{1}
$$
diverges because the terms don't tend to $0$.
The ratio of the terms of the series
$$
\sum_{n=1}^\infty\frac{2\cdot4\cdot6\cdots2n}{5\cdot7\cdot9\cdots(2n+3)}\tag{2}
$$
is
$$
\frac{2n}{2n+3}\tag{3}
$$
Therefore,
$$
\begin{align}
\lim_{n\to\infty}\frac{2n}{2n+3}
... | {
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Definite integration evaluation of $\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$ OK, so the question says evaluate the integral
$$\int_{0}^{\pi}\frac{x}{(a^2\cos^2x+b^2\sin^2x)^2}dx$$
What I do is use the property that $\int_a^bf(x)dx=\int_a^bf(b+a-x)dx$ and this gives me ($I$ is the value of the integral)
$$\f... | I prefer to the following method:
\begin{align*}
I
:= \int_{0}^{\pi} \frac{x}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \, dx
&= \frac{\pi}{2} \int_{0}^{\pi} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \\
&= \pi \int_{0}^{\frac{\pi}{2}} \frac{dx}{(a^2 \cos^2 x + b^2 \sin^2 x )^2} \\
&= \pi \int_{0}^{\frac{\pi}{2}} \frac{1 + \t... | {
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What are conditions on $x$ that make roots of $y^2 + 2xy - 2 = 0$ greater than 1 in absolute value? Original question: How do we solve $\sqrt{x^2+2}<x-1$? If $x>1$, the solution is $x<-0.5$ which does not make sense. What if $x<1$?
Update: Thank you for all answers. Actually, the original question was: "What are condit... | If you're only interested in real solutions then firstly you must have $x^2 + 2 > 0$, so $x^2 > -2$ but $x^2 > 0$ for all $x \in \mathbb{R}$ so you must have $x>0$.
Assuming $x\geq1$ and squaring both sides gives $x^2 +2 < (x-1)^2$ expanding gives $x^2 + 2 < x^2 - 2x + 1$ which can be written as $1 < -2x$ and clearly t... | {
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What is the remainder when $x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41$ is divided by: $x^2-8x+12$. What is the remainder when: $$x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41$$ is divided by: $$x^2-8x+12$$
$x^2-8x+12$ $\leftrightarrow (x-2)(x-6)$
This gives me the polynomial:... | It is worth noting that
$$
\begin{multline}
x^{100} -8x^{99}+12x^{98}-3x^{10}+24x^{9}-36x^{8}+3x^{2}-29x + 41\\
=x^{98}(x^2-8x+12)-3x^8(x^2-8x+12)+3(x^2-8x+12)-5x+5
\end{multline}
$$
so that the remainder is $-5x+5$.
| {
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"answer_count": 3,
"answer_id": 2
} |
Find the standard matrix for a linear transformation If $T: \Bbb R^3→ \Bbb R^3$ is a linear transformation such that:
$$
T \Bigg (\begin{bmatrix}-2 \\ 3 \\ -4 \\ \end{bmatrix} \Bigg) = \begin{bmatrix} 5\\ 3 \\ 14 \\ \end{bmatrix}$$ $$T \Bigg (\begin{bmatrix} 3 \\ -2 \\ 3 \\ \end{bmatrix} \Bigg) = \begin{bmatrix}-4 \\ 6... | You can put into a matrix given vectors and their images. If you then do elementary row operations, this property is not changed. (After each step you have in each row a vector and its image. This is because of linearity.)
If you manage to obtain the identity matrix on the left, then you know the images of the vectors ... | {
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Help with particular solution to solving $z^4 - 2z^3 + 9z^2 - 14z + 14 = 0$. I asked a question about how to solve: $$z^4−2z^3+9z^2−14z+14 = 0$$
When all you know is that there is a root with the real part of 1. I was given great answers and you can find the question here.
(If you feel that this question is redun... | Right as far as it goes. Just that you made a mistake: The remainder is $(2 - 2 y^2) z + (y^4 - 7 y^2 + 6)$.
Now, for the proposed factor really to be a factor, this remainder must be the zero polynomial, that is:
$$
\begin{align*}
2 - 2 y^2 &= 0 \\
y^4 - 7 y^2 + 6 &= 0
\end{align*}
$$
From the first one you get $y = 1... | {
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Prove that $\lim\limits_{n\rightarrow \infty}\int_1^3\frac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and evaluate it. I am trying to show that $\lim_{n\rightarrow \infty}\int_1^3\dfrac{nx^{99}+5}{x^3+nx^{66}} d x$ exists and what its value is. I know that to do this I must show that $\dfrac{nx^{99}+5}{x^3+nx^{66}}\rightarrow... | The integrand simplifies to $\dfrac{nx^{96}}{1+nx^{63}}+\dfrac{5}{x^3+nx^{66}}$.
The second term is less than $\dfrac{5}{n}$ on our interval, so it is harmless.
For the first term, divide. We get $x^{33}-\dfrac{x^{33}}{1+nx^{63}}$. The function $\dfrac{x^{33}}{1+nx^{63}}$ is less than $\dfrac{1}{n}$ on our interval.
| {
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Help understanding the solution to #83 Section 9.1 Calculus 9e, by Larson.
Find $a_n$ of $1, \; \frac{-1}{1\times 3}, \; \frac{1}{1\times3\times5}, \frac{-1}{1\times 3\times 5 \times 7}$
I solved it as $\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!}$
The solution guide says: $\dfrac{\left(-1\right)^{n-1}2^n n!}{\... | It is $(2n-1)!!$, not $(2n-1)!$. Use that $(2n)!!=2^n n!$
If you don't know: $(2n-1)!!=(2n-1)(2n-3)\cdot\ldots \cdot3 \cdot 1$ and $(2n)!!=(2n)(2n-2)\cdot \ldots 4 \cdot 2$
Hint: multiple and divide by $(2n)!!$, note that $(2n-1)!! \cdot (2n)!!= (2n)!$
Full solution:
$\dfrac{\left(-1\right)^{n-1}}{\left(2n-1\right)!!}... | {
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Range of $z,$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$ If $x+y+z = 3$ and $xy+yz+zx = -9$ and $x,y,z\in \mathbb{R}$. Then value of $z$ lie in the interval.
My Try:: Let $x,y,z$ be the roots of the quadratic equation
$t^3-(x+y+z)t^2+(xy+yz+zx)t-xyz = 0$
Let $xyz = p, $ Then let $f(t) = t^3-3t^2-9t-... | Equating the values of $x$
$$3-y-z=-\frac{9+yz}{y+z}\implies y^2+y(z-3)+z^2-3z-9=0$$
As $y$ is real, the discriminant $(z-3)^2-4\cdot1\cdot(z^2-3z-9)$ must be $\ge0$
$$\implies z^2-2z-15\le0\implies (z-5)(z+3)\le0$$
As $(x-a)(x-b)\le 0$ for $a\le b\implies a\le x\le b$
So,$-3\le z\le 5$
Observe that the given condition... | {
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 1
} |
How to get the sum of the values in a $N \times N$ table? How to get the sum of the values in a $N \times N$ table (without adding repeating products such as $6 \times 7$ and $7 \times 6$ twice and without counting perfect squares)?
Figured out that
$1 \cdot 0+2 \cdot 1+3 \cdot (1+2)+4 \cdot (1+2+3)+\dots+n\cdot (1+2+3... | If I understand correctly, you want
$$\begin{align*}
\sum_{i=1}^{N-1}\sum_{k=i+1}^Nik&=\sum_{k=2}^N\sum_{i=1}^{k-1}ik\\
&=\sum_{k=2}^Nk\sum_{i=1}^{k-1}i\\
&=\sum_{k=2}^Nk\left(\frac{k(k-1)}2\right)\\
&=\frac12\sum_{k=2}^N\left(k^3-k^2\right)\\
&=\frac12\sum_{k=2}^Nk^3-\frac12\sum_{k=2}^Nk^2\\
&=\frac12\left(\frac14N^2... | {
"language": "en",
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"source": "stackexchange",
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"answer_count": 2,
"answer_id": 0
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Show that the curve $x^2+y^2-3=0$ has no rational points
Show that the curve $x^2+y^2-3=0$ has no rational points, that is, no points $(x,y)$ with $x,y\in \mathbb{Q}$.
Update: Thanks for all the input! I've done my best to incorporate your suggestions and write up the proof. My explanation of why $\gcd(a,b,q)=1$ is a... | Here is a more geometric flavored proof:
$x^2+y^2-3=0 \iff x^2+y^2 = \sqrt{3}^2$ is a circle with radius $\sqrt{3}$ centered at the origin. Think of the points along the circle as polar coordinates $(r, \theta)$, i.e.
\begin{equation}(\sqrt{3}, \theta) \text{ where } 0 \leq \theta \leq 2\pi \end{equation}
The formulas... | {
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"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
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Integrating $\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$ Could someone help with the following integration:
$$\int^0_\pi \frac{x \sin x}{1+\cos^2 x}$$
So far I have done the following, but I am stuck:
I denoted $ y=-\cos x $ then:
$$\begin{align*}&\int^{1}_{-1} \frac{\arccos(-y) \sin x}{1+y^2}\frac{\mathrm dy}{\sin x}\\&... | $$I=\int_0^{\pi} \frac{-x\sin x}{1+\cos^2 x}\,dx=\int_0^{\pi} \frac{(x-\pi)\sin x}{1+\cos^2 x}dx\quad(x\to \pi-x)$$
$$\Rightarrow I=\frac{\pi}{2}\int_0^{\pi}\frac{-\sin x}{1+\cos^2 x}\,dx$$
Let $t=\cos x:$
$$I=\frac{\pi}{2}\int_{-1}^{1}-\frac{1}{1+t^2}\,dt=-\frac{\pi^2}{4}$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "12",
"answer_count": 4,
"answer_id": 2
} |
$\mathcal{D}$-classes Let $$\alpha = \left(\begin{array}{@{\,}c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\;\; }c@{\,}}
1&2&3\\
1&3&3
\end{array}\right) \in \mathcal{T}_3\text{.}$$
(a) Show that the $\mathcal{D}$-class of $\alpha$ contains all those elements of $\mathcal{T}_3$ which have the same rank (cardinality of their ... | I've had to change accounts as I can't access my email account so I can't reply to any posts because I'm not using the account I originally asked the question from!
So:
$\beta \in R_{\alpha} \iff \text{Ker}(\beta)$ has classes $\{1\},\{2,3\}$ and $\beta \in L_{\alpha} \iff \text{Im}(\beta) = \{1,3\}$
$$R_{\alpha} = \le... | {
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"answer_count": 2,
"answer_id": 1
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Fourier series of function $f(x)=0$ if $-\pi$$f(x) = \begin{cases}0 & \text{if }-\pi<x<0, \\
\sin(x) & \text{if }0<x<\pi.
\end{cases}$$
My attempt:
I went the route of expanding this function with a complex Fourier series.
$$f(x) = \sum_{n=-\infty}^{+\infty} C_{n}e^{inx}$$
$$C_{n} = \frac {1}{2\pi} \int_{0}^{\pi} \frac... | Since the function $\phi(t)$ is defined on $-L=-\pi<t<\pi=L$ , the Fourier series can be expressed as shown below :
The numerical tests of the formula are well consistent with a good accuracy.
On the figure below, small values of $m$ are taken in order to make clear the deviations in case of series limited to not enou... | {
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"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
} |
$T(n) = 2T(n/2) +\lg(n!) $ asymptotic bounds $T(n) = 2T(n/2) + \lg(n!)$
What are the upper and lower bounds of this equation?
| $$\log(n!) \sim n\log(n)$$
Hence, we have
$$T(n) \sim 2T(n/2) + n \log(n)$$
Setting $n=2^k$, and calling $T(2^k) = g(k)$, we get that
\begin{align}
g(k) & \sim 2g(k-1) + k2^k \log(2) \sim k 2^k \log(2) + 2((k-1)2^{k-1} \log(2) + 2g(k-2))\\
& \sim k 2^k \log(2) + (k-1)2^k \log(2) + 4 g(k-2)\\
& \sim k 2^k \log(2) + (k-1... | {
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On prime numbers let $q$ be a prime
let $p = 2^q -1 $
is p must be prime always for any prime q ?
is this is true always ?
or it is false for some prime q ?
if it is false , give an example to show that there is a prime q such that
$2^q -1$ is not a prime
thanx
| Here's some example computations to suggest a way you might try to prove that $q$ must be prime:
$$x^2 - 1 = (x - 1)(x + 1)$$
So:
$$x^6 - 1 = (x^3)^2 - 1 = (x^3 - 1)(x^3 + 1)$$
And so:
$$2^6 - 1 = 63 = (2^3 - 1)(2^3 + 1) = 7 \cdot 9$$
Or:
$$x^3 - 1 = (x - 1)(x^2 + x + 1)$$
So:
$$x^{15} - 1 = (x^5)^3 - 1 = (x^5 - 1)(x^{... | {
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"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Prove $\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x=e$ I need to prove that $$\lim_{x\to\infty}\left(\sin\frac1x+\cos\frac1x\right)^x=e.$$ I tried to activate the identity of $\sin^2x+\cos^2x=1$ but I'm still stuck with $$\left(\cfrac{1}{\sin\frac1x-\cos\frac1x}\right)^x.$$
Can I get a hint?
| My attempted solution without using asymptoptic arguments:
$$
\sin{\frac{1}{x} } = \frac{1}{x} - \frac{1}{3!x^3}+\frac{1}{5!x^5}-\ldots \\
\cos{\frac{1}{x} } = 1 -\frac{1}{2!x^2} + \frac{1}{4!x^4} -\ldots\\
e^{x^{-1}}- \left( \sin{\frac{1}{x} } + \cos{\frac{1}{x} }\right) = 2\left( \frac{1}{2!x^2} + \frac{1}{3!x^3} +... | {
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"answer_count": 5,
"answer_id": 3
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$(x-a)(x-b)(x-c)(x-d)=ex$ We can verify that $x=125,162,343$ are the roots of equation $(x-105)(x-210)(x-315)=2584x$.
My question is,Could you find five positive integers $a,b,c,d,e$, which $(x-a)(x-b)(x-c)(x-d)=ex$ has four positive integer roots?
Thanks in advance.
| Thank you for your attention, some one has offered me some examples:
$(x-5)(x-6)(x-16)(x-18)=84x,x=4,9,12,20;$
$(x-3)(x-4)(x-16)(x-22)=280x,x=2,8,11,24;$
$(x-3)(x-6)(x-32)(x-44)=2520x,x=2,11,24,48.$
| {
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"question_score": "16",
"answer_count": 2,
"answer_id": 1
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How to calculate volume of $z = 0, z = 1, x+y+z=2, x = 0, y = 0$ area by tripple integral? I am calculating volume of body that is defined by $z = 0, z = 1, x+y+z=2, x = 0, y = 0$ to do this I have two possible ways:
*
*$$\int\limits_0^2\int\limits_0^{2-x}\int\limits_0^{2-x-y}1\mathrm{d}z\,\mathrm{d}y\,\mathrm{d}dx-... | If you look at a cross-section of the volume in a plane $z=\text{constant}$, then the region is a right triangle of area
$$\frac{1}{2} (2-z)^2$$
The volume is then
$$\int_0^1 dz \: \frac{1}{2} (2-z)^2 = \left [ \frac{1}{6} (2-z)^3\right ]_1^0 = \frac{7}{6}$$
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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$\textrm{Res}\left(\frac{\log z}{z^3+8}; z_k\right) = \frac{-z_k \log z_k}{24}$ when $z_k$ solves $z^3+8=0$ The problem in the book is
Compute $\int_0^\infty \frac{dx}{x^3+8}$.
I set up the keyhole contour, apply the residue theorem, and go through the tedious algebra. I get stuck in doing so, but looking at the solu... | OK, I am convinced that your mistake lies in calling one of the poles $z=e^{-i \pi/3}$. For your keyhole contour, it should be $z=e^{i 5 \pi/3}$.
To see this, I am just going to write down the equation you get after applying the contour and the residues:
$$-\int_0^{\infty} \frac{dx}{x^3+8} = \frac{\log{2}+i \pi/3}{12 ... | {
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"answer_count": 2,
"answer_id": 1
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Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$ Establish convergence of the series: $1-\frac{1}{2}-\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+\frac{1}{6}-...$
The number of signs increases by one in each "block".
I have an idea. Group the series like this: $1-(\frac{1}... | $$
\underbrace{\vphantom{\frac11}+1}_{\text{length }1}
\underbrace{-\frac12-\frac13}_{\text{length }2}
\underbrace{+\frac14+\frac15+\frac16}_{\text{length }3}
\underbrace{-\frac17-\frac18-\frac19-\frac1{10}}_{\text{length }4}
\underbrace{+\frac1{11}+\frac1{12}+\frac1{13}+\frac1{14}+\frac1{15}}_{\text{length }5}-\ldots
... | {
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"url": "https://math.stackexchange.com/questions/336035",
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"source": "stackexchange",
"question_score": "18",
"answer_count": 6,
"answer_id": 2
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Why $\sum_{k=1}^{\infty} \frac{k}{2^k} = 2$? Can you please explain why
$$
\sum_{k=1}^{\infty} \dfrac{k}{2^k} =
\dfrac{1}{2} +\dfrac{ 2}{4} + \dfrac{3}{8}+ \dfrac{4}{16} +\dfrac{5}{32} + \dots =
2
$$
I know $1 + 2 + 3 + ... + n = \dfrac{n(n+1)}{2}$
| It's fairly simple. Look at it this way:
$$
\sum_{i=1}^\infty \frac{i}{2^i} = \sum_{i=1}^\infty \frac{\sum_{k=1}^i 1}{2^i} = \sum_{i=1}^\infty \sum_{k=1}^i \frac{1}{2^i}
$$
From here, we just change the order of addition. Rather than adding along $k$, and then $i$, we add along $j=i-k$, and then along $k$. This turns o... | {
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"source": "stackexchange",
"question_score": "35",
"answer_count": 12,
"answer_id": 9
} |
4-th derivative of $(1+x+x^2) / (1-x+x^2) $ using Taylor polynomial for $1/(1-x)$ Using $n$-th Taylor polynomial for $f_1(x)=\frac{1}{1-x}$ with center in $0$, find $4$-th derivative of $f_2(x)=\frac{1+x+x^2}{1-x+x^2}$ in the point $0$ without calculating it's $1$,$2$ or $3$ derivative.
I'm looking for hints, it's a ... | $$f_2(x)=\frac{1+x+x^2}{1-x+x^2}=\frac{(1+x+x^2)(1+x)}{(1-x+x^2)(1+x)}=\\
=\frac{(1+x+x^2)(1+x)}{1+x^3}=
=\frac{1+2x+2x^2+x^3}{1+x^3}=\\
=1+2\frac{1+x}{1+x^3}=1+2 \frac{1}{1+x^3}+2x\frac{1}{1+x^3}.$$
Added:
From Taylor's polynomial for
$$\frac{1}{1-x}=\sum\limits_{k=1}^{n}{\;x^k}+o(x^n)$$
we have
$$\frac{1}{1+x^3}=\s... | {
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"answer_count": 2,
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Can't find solutions for $\tan{2x} = \tan{x}$ Solving $\tan{2x}=\tan{x}$
Reducing the left side: $$\frac{\sin{2x}}{\cos{2x}} = \frac{2\sin{x}\cos{x}}{2\cos^{2}(x)-1}.$$
Reducing the right side: $$\tan{x} = \frac{\sin{x}}{\cos{x}}.$$
therefore:
$$\frac{2\sin x\cos x}{2\cos^2x-1} = \frac{\sin x}{\cos x}$$
$$\frac{2\cos x... | Hint:
$$\tan(2x) = \tan(x) \quad \Rightarrow \\
\frac{\sin(2x)}{\cos(2x)} = \frac{\sin(x)}{\cos(x)} \quad \Rightarrow \\
\frac{2\sin(x)\cos(x)}{2\cos^2(x) -1} = \frac{\sin(x)}{\cos(x)} \quad \Rightarrow \\
\frac{2\cos^2(x)}{2\cos^2(x) - 1} = 1 \quad \text{ or }\quad \sin(x) = 0.
$$
It looks like you got the "other solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/344015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solve inequality with $x$ in the denominator Solve for $x$ when it is in the denominator of an inequality
$$\frac{4}{x+4}\leq2$$
I believe the first step is the multiply both side by $(x+4)^2$
$$4(x+4)\leq 2(x+4)^2$$
$$4x+16\leq 2(x^2+8x+16)$$
$$4x+16\leq 2x^2+16x+32$$
$$0 \leq 2x^2+12x+16$$
$$0 \leq (2x+8)(x+2)$$
Stu... | Why not just do it this way:
For $x + 4 > 0$ the following is legal:
$$
\begin{align*}
\frac{4}{x + 4} &\le 2 \\
\frac{x + 4}{4} &\ge \frac{1}{2} \\
x + 4 &\ge 2 \\
x &\ge -2
\end{align*}
$$
If $x + 4 < 0$, i.e. $x < -4$, the inequality is satisfied.
The solution is $(-\infty, -4) \cup [-2, \infty)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/344268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 6
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If $x^9-x^5+x-2=0$ , how to prove $\sqrt[11]{3}If $x^9-x^5+x-2=0$ is known,
how to prove $\sqrt[11]{3}<x<\sqrt[10]{3}$ ，$\sqrt[7]{2}<x<\sqrt[6]{2}$。
| Hint: $x^9-x^5+x=2 \implies \dfrac{((x^4)^3+1)}{x^4+1}=\dfrac{2}{x}$
Spoiler:
Wolframalpha gives only real solution to this, i.e $x \approx 1.112$, which is in between $3^{\frac{1}{11}}$ and $3^{\frac{1}{10}}$ OR(as Thomas Andrews says) $2^{\frac{1}{6}}$ and $3^{\frac{1}{7}}$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/344350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$ How can I show:
$$\lim_{x\to\infty}x^{2/3}((x+1)^{1/3}-x^{1/3})=1/3$$
I've tried multiplying with its "conjugate" but that doesn't seem to help that much.
Thanks!
| $\displaystyle(x+1)^{1/3}-x^{1/3}=\frac{1}{3c^{2/3}},c\in(x,x+1)$(By mean value theorem)
$\displaystyle \frac{1}{3(x+1)^{2/3}}\le\frac{1}{3c^{2/3}}\le\frac{1}{3x^{2/3}},\forall c\in(x,x+1)$
So we have ,
$\displaystyle\frac{1}{3(x+1)^{2/3}}\le(x+1)^{1/3}-x^{1/3}\le \frac{1}{3x^{2/3}}$
$\Rightarrow \displaystyle\frac{x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/344585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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$f(x - 1) + f(x − 2) $ and the sum of coeficients If $f(x-1)+f(x-2) = 5x^2 - 2x + 9$
and
$f(x)= ax^2 + bx + c$
what would be the value of $a+b+c$?
I was doing
$f(x-1)+f(x-2)= f(x-3)$
then
$f(x)$
a = 5
b = -2
c = 9
$(5-3)+(-2-3)+(9-3)$
But do not think is is correct
What would be correct approach?
| We know $$f(x-1)=a(x-1)^2+b(x-1)+c\quad\text{and}\quad f(x-2)=a(x-2)^2+b(x-2)+c.$$ Expand these terms, add them, and combine like terms via powers of $x$. Now you can get three equations in three variables by equating the coefficients of the left with the right since you know $$f(x-1)+f(x-2)=5x^2-2x+9.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 3
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Multivariable limit $\lim_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1}$ $$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 +y^2 + 1} - 1} $$
According to my textbook the limit equals $2$.
What I have tried:
Using the squeeze theorem:
$$ \lim \limits_{(x,y)\to (0,0)} \frac {x^2 + y^2}{\sqrt{x^2 ... | $$ \displaystyle \lim_{(x,y)\to(0,0)} \frac{(x^2+y^2)}{(x^2-y^2)}+ix $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/345262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
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2 test for convergence problems: $\sum^{\infty}_{k=1}5^k/(3^k+4^k)$and $\sum^{\infty}_{n=1}\tan\left(1/n\right)$ For the first problem: $$\sum^{\infty}_{k=1}\frac{5^k}{3^k+4^k}$$
I tried to take the ratio test but was having trouble simplifying $$\frac{3^k + 4^k}{3^{k+1} + 4^{k+1}}$$
For the second problem: $$\sum^{\in... | For the first problem
$$\frac{\frac{5^{k+1}}{3^{k+1}+4^{k+1}}}{\frac{5^k}{3^k+4^k}}$$
$$=\frac{5^{k+1}(3^k+4^k)}{5^k(3^{k+1}+4^{k+1})}$$
$$=5\cdot\frac{\left(\frac34\right)^k+1}{3\cdot\left(\frac34\right)^k+4}$$
So, using Ratio Test: $$\lim_{k\to\infty}\frac{\frac{5^{k+1}}{3^{k+1}+4^{k+1}}}{\frac{5^k}{3^k+4^k}}=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/346353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Finding the limiting distribution of a $3 \times 3$ Markov chain This is a question from a book.
Find $\lim_{n\rightarrow \theta}P^n$ where $$P=\begin{pmatrix}0 & 1 & 0\\
\frac{1}{6} & \frac{1}{2} & \frac{1}{3}\\
0 & \frac{2}{3} & \frac{1}{3}
\end{pmatrix}$$
I assumed that there was a typo and in fact what is being a... | There are two issues:
*
*Your third equation has a wrong coefficient. It should be $0\pi_{0}+\color{red}{\frac{1}{3}}\pi_{1}+\frac{1}{3}\pi_{2}=\pi_{2}$.
*The textbook answer doesn't seem to be correct. The solution should be $(\pi_0,\,\pi_1,\,\pi_2)=(0.1,\, 0.6,\, 0.3)$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate: $\sum_{n=1}^{\infty}\frac{1}{n k^n}$ How to evaluate this series for $k > 1$?
$$\sum_{n=1}^{\infty}\frac{1}{n k^n}$$
For $k = 2$, I tried to evaluate $\displaystyle \sum_{n = 0}^\infty \int_{1}^{2} x^{-(n+1)}dx = \int_{1}^{2} \sum_{n = 0}^\infty x^{-(n+1)}dx = \int_1^{2}\frac{1}{x(x-1)}dx$ $\displaystyle =... | $$\frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{nk^{n-1}} =\frac{1}{k} \sum_{n=1}^{\infty} \int^1_0 (x/k)^{n-1} dx=\frac{1}{k} \int^1_0 \sum_{n=1}^{\infty} (x/k)^{n-1} dx = \int^1_0 \frac{1}{k-x} dx= \log \left( \frac{k}{k-1} \right).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
} |
Circles of radius $2$ passing through origin with centers on $x=1$
There are two circles of radius $2$ that have centers on the line $x=1$
and pass through the origin. Find their equations.
Please explain to me what the problem is really saying.
| A circle with centre $(a,b)$ and radius $r>0$ has equation $(x-a)^2+(y-b)^2=r^2$. Since these circles have centre on the line $x=1$ and radius $2$, they have equation $(x-1)^2+(y-b)^2=4$. Since the circles pass through the point $(0,0)$, we have $(0-1)^2+(0-b)^2=4$, so $b^2+1=4$. Solving for $b$ gives $b=\pm\sqrt 3$, s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/347344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Formula for the $1\cdot 2 + 2\cdot 3 + 3\cdot 4+\ldots + n\cdot (n+1)$ sum Is there a formula for the following sum?
$S_n = 1\cdot2 + 2\cdot 3 + 3\cdot 4 + 4\cdot 5 +\ldots + n\cdot (n+1)$
| $k(k+1) = \frac{1}{3}((k+1)^3-k^3-1)$. So all the $k$'s cancel, except the first and last. We get:
$\sum_1^n k(k+1) = \frac{1}{3}\sum_1^n ((k+1)^3-k^3-1) = \frac{1}{3}((n+1)^3-n-1) = \frac{1}{3}n(n+1)(n+2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/347585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then $3$ divides both $a$ and $b$. If $a$ and $b$ are integers such that $9$ divides $a^2 + ab + b^2$ then show that $3$
divides both $a$ and $b$.
can anyone tell me please how to solve these types of problem oe which formula is required
| Starter: Since $a^3-b^3=(a-b)(a^2+ab+b^2)$, and by Fermat'little theorem, we conclude that, under our condition, $a\equiv b\pmod 3$.
Suppose the contrary: $a, b$ are relatively prime to $3$. So $a=3k\pm1, b=3l\pm1$. Then $a^2=9k^2\pm6k+1$,
$b^2=9l^2\pm6l+1$,
$ab=9kl\pm3(k+l)+1$.
Summing together, we find: $a^2+ab+b^2=9... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Showing that a $3^n$ digit number whose digits are all equal is divisible by $3^n$
Let $c$ be a $3^n$ digit number whose digits are all equal. Show that $3^n$ divides
$c$.
I have no idea how to solve these types of problems. Can anybody help me please?
| Because the digits sum to a multiple of $3$, a block of $3$ identical digits is divisible by $3$.
$3$ of those blocks of $3$ digits, having been divided by $3$, is also divisible by $3$.
$3$ of those blocks of $3^2$ digits, having been divided by $3^2$, is also divisible by $3$.
$3$ of those blocks of $3^3$ digits, hav... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/349952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Identity involving partial sums of Fourier series Suppose $f$ is a continuous periodic function and $S_Nf(x) = \sum^N_{n=−N} \hat f(n) e^{inx}$, where $$\hat f(n)= \frac{1}{2\pi}\int_0^{2\pi}f(x)e^{-inx} dx.$$
How can I show that $$\sum_{j=0}^{N-1}S_jf(x)= \int_{-\pi}^{\pi} \frac{\sin^2(\frac{1}{2}Ny)}{\sin^2(\frac{1}{... | This relies on switching the order of summation and integration. For one particular value of $S_k$:
$$\begin{align}S_k &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \sum_{n=-k}^k e^{i k (x-x')} \\ &= \frac{1}{2 \pi} \int_0^{2 \pi} dx' \: f(x') \frac{e^{i (k+1)(x-x')} - e^{-i k (x-x')}}{e^{i (x-x')} -1}\\ &= \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/351942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simple Modulo Question $6x = 9 \pmod{11}$ I am trying to solve $6x = 9\pmod{11}$
The solution suggests notice that $2*6 = 12 = 1 \pmod {11}$
$$6x = 9 \pmod {11}\\12x = 18 \pmod {11}\\x = 7 \pmod{11}$$
I don't get the final step from $12x = 18\pmod{11}$ to the solution.
| The fact that $2\cdot 6 \equiv 12 \equiv 1 \pmod {11}$ tells you that $2$ is the multiplicative inverse of $6 \pmod{11}$, i.e., $\;2\cdot 6 = 12 \equiv 1 \pmod{11}$
$$2\cdot 6 x\equiv 2\cdot 9 \pmod{11} \implies 1\cdot x \equiv 18 \pmod{11}$$ $$\iff \underbrace{x\equiv 18 - 11 \pmod{11} \iff x\equiv 7 \pmod{11}}_{\larg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/353600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Evaluate $\int \frac{x^2 + x+3}{x^2+2x+5}\ dx$ How can we evaluate $$\displaystyle\int \frac{x^2 + x+3}{x^2+2x+5} dx$$
To be honest, I'm embarrassed. I decomposed it and know what the answer should be but
I can't get the right answer.
| You can decompose your integrand as follows:
$$ \frac{ x^2 + 2x + 5 - x - 1 - 1}{x^2 + 2x + 5} = 1 - \frac{x + 1}{x^2 + 2x + 5} - \frac{1}{(x+1)^2 + 4}$$
You can integrate the first term directly, the second term after the substitution $u = x^2 + 2x + 5$, and the third term by recalling that $(\arctan{x})' = 1/(x^2 + 1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Basic calculation with root roots and power Sorry for the boring question but I just need someone to remind me the way to calculate this:
$\displaystyle \left(\frac{a}{2}\right)^2 + x^2 = a^2$ (i used carrot sign cause i dont know how to do factorial in my mac keyboard)
The answer is $\displaystyle \frac{\sqrt {3}}{2}a... | $$\left(\frac a2\right)^2=\frac{a^2}{2^2}=\frac{a^2}4.$$ Thus, using the difference of squares identity $$y^2-z^2=(y-z)(y+z),$$ the following are equivalent: $$\left(\frac a2\right)^2+x^2=a^2\\x^2+\frac{a^2}4-a^2=0\\x^2-\frac{3a^2}4=0\\x^2-\left(\frac{a\sqrt{3}}2\right)^2=0\\\left(x-\frac{a\sqrt{3}}2\right)\left(x+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/355926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Riemann-Stieltjes integration with floor function Evaluate $$\int_{\frac{2}{3}}^8 f(x)d\alpha(x)$$ where $\alpha$ is continuous and $f$ is the floor function, that is $f(x)$ is the greatest integer less than or equal $x$.
| $$I = \int_{2/3}^8 \lfloor x\rfloor d \alpha(x) = \int_{2/3}^1 \lfloor x\rfloor d \alpha(x) + \sum_{k=1}^7 \int_k^{k+1} \lfloor x\rfloor d \alpha(x) = 0 + \sum_{k=1}^7 \int_k^{k+1} \lfloor x\rfloor d \alpha(x)$$
Now $\lfloor x \rfloor = k$ for $x \in [k,k+1)$. Hence, we get that
\begin{align}
I & = \sum_{k=1}^7 k (\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/356628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How can I prove that $xy\leq x^2+y^2$? How can I prove that $xy\leq x^2+y^2$ for all $x,y\in\mathbb{R}$ ?
| First,
$$
a^2+3b^2\geq 0
$$
Now,
$$
a^2+2ab+b^2+a^2-2ab+b^2\geq a^2-b^2
$$
Thus,
$$
(a+b)^2+(a-b)^2\geq (a+b)(a-b)
$$
Let $a+b=x, a-b=y$, i.e., $a=\frac{x+y}{2}, b=\frac{x-y}{2}$ which is one-to-one correspondence between $(x,y)$ and $(a,b)$, then,
$$
x^2+y^2\geq xy
$$
Q.E.D
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/357272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 25,
"answer_id": 11
} |
How can I determine the Jordan Form of a matrix? How can I go about proving that the characteristic polynomial, minimal polynomial, and the dim(eigenspace) is enough to determine the Jordan Form of a matrix for n<7?
| Let's give it a try:
$$A=\begin{pmatrix}\color{red}0&\color{red}1&\color{red}0&0&0&0&0\\
\color{red}0&\color{red}0&\color{red}1&0&0&0&0\\
\color{red}0&\color{red}0&\color{red}0&0&0&0&0\\
0&0&0&\color{red}0&\color{red}1&0&0\\
0&0&0&\color{red}0&\color{red}0&0&0\\
0&0&0&0&0&1&0\\
0&0&0&0&0&0&1\end{pmatrix}\;,\;\;\;
B=\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/358777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Find this $a,b,c$ such that $\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$ It is known that$$\sqrt{9-8\sin 50^{\circ}}=a+b\sin c^{\circ}$$
for exactly one set of positive integers $(a,b,c)$ where $0<c<90$
find the value
$$\dfrac{b+c}{a}$$
my idea,$ \sin 50^\circ >\sin 45^\circ >\frac{_5}{^8} $
so$\sqrt{9-8\sin 50^{\cir... | Well the most forward way I think of is using the fact that $cos(2c)=1-2sin^2(c)$. This gives you some equation like:
$2X^2+X-(1+sin(50°))=0$ where $X=sin(c)$
Once you have $sin(c)$ you can get $c$ quite easily given your condition over $c$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
} |
An equation to solve in natural numbers Some clues needed for the equation
$$(3+\sqrt{2})^x-(5-3\sqrt{2})^y=6+9\sqrt{2}, \quad x,y \in \mathbb{N}$$
| Since you want to solve in natural numbers, notice that $0 < 5 - 3\sqrt{2} < 1$. This implies that $ 0 < ( 5 - 3 \sqrt{2} ) ^ y < 1$, and hence
$$6 + 9 \sqrt{2} < (3 + \sqrt{2} )^x < 7 + 9 \sqrt{2} $$
This forces $x = 2$, from which $y = 1$ (from Brian).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/359803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$, what is $3x^3 - 9x$?
Suppose $x = \sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}$. Then what is $3x^3 - 9x$?
I tried factorizing $3x^3 - 9x = 3x(x^2 - 3)$ then substituting the values which gives me something very lengthy. I eventually got to the answer — $10$ — after working a bit. A... | $x^3=\left(\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}\right)^3$
$=\left(\sqrt[3]{3}\right)^3+\left(\frac{1}{\sqrt[3]{3}}\right)^3+3\cdot\sqrt[3]{3}\cdot\frac{1}{\sqrt[3]{3}}\left(\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}\right)$
$=3+\frac13+3\cdot\sqrt[3]{3}\cdot \frac{1}{\sqrt[3]{3}}\cdot x$ as $\sqrt[3]{3} + \frac{1}{\sqrt[3]{3}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360087",
"timestamp": "2023-03-29T00:00:00",
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Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$ Calculate $\sum_{n=2}^\infty ({n^4+2n^3-3n^2-8n-3\over(n+2)!})$
I thought about maybe breaking the polynomial in two different fractions in order to make the sum more manageable and reduce it to something similar to $\lim_{n\to\infty}(1+{1\over1!}+{1\over2... | Express $$n^4+2n^3-3n^2-8n-3=(n+2)(n+1)n(n-1)+B(n+2)(n+1)n+C(n+2)(n+1)+D(n+2)+E--->(1)$$
So that $$T_n=\frac{n^4+2n^3-3n^2-8n-3}{(n+2)!}=\frac1{(n-2)!}+\frac B{(n-1)!}+\frac C{(n)!}+\frac D{(n+1)!}+\frac E{(n+2)!}$$
Putting $n=-2$ in $(1), E=2^4+2\cdot2^3-3\cdot2^2-8\cdot2-3=1$
Similarly, putting $n=-1,0,1$ we can fin... | {
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Complex analysis graphing an ellipse. Give a geometric argument that $|z-4i|+|z+4i|=10$.
My attempt: $|z-4i|$ represents the distance from an arbitrary point to the coordinate $(0,4i)$ while $|z+4i|$ is the distance from an arbitrary point to the coordinate $(0,-4i)$.
Then stuck; I don't know how this relates to an ell... | An ellipse is a set of points such that the sum of the distances from each point in the set to a pair of fixed points is a constant.
Here is a derivation of the equation of an ellipse from this definition:
Imagine you have tacks at the points $(-c,0$ and $(c,0)$, which hold each end of a string of length $2 a$. We dra... | {
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"answer_id": 0
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Proving that a critical point is a minimum I am solving a problem, which is
"Find the point of the paraboloid $P:=\{(x,y,z)\in\mathbb{R}^3 | x^2+y^2=z\}$ which is the nearest to the point $(1,1,\frac12)$."
I have already determined (using the Lagrange multipliers method) that $p=(2^{-\frac23},2^{-\frac23},2^{-\frac13})... | Prof. Cook is using the multivariate index $D$ applied to the distance-squared function for points on the paraboloid measured from $(1,1,\frac{1}{2})$. It's not the easiest thing to read, but it does establish that the point you found is at the minimal distance from the point external to the paraboloid.
Alternatively,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/360832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
To find the logarithm of $1728$ to the base $2 \sqrt{3}$
Find the logarithm of: $1728$ to base $2\sqrt{3}$.
Let, $\log_{2\sqrt{3}} 1728 = y$, then
$$\begin{align} (2\sqrt{3})^y &= 1728\\
2^y(\sqrt3)^y &= 1728\\2^y(3^\frac12)^y &= 1728\\2^y(3^\frac y2)
&= 1728\\2^y × 3^\frac y2 &= 2^6 × 3^3 \end{align}$$
What should... | Or: $1728 = 12^{3}$ and $12 = (2\sqrt{3})^{2},$ so $1728 = (2\sqrt{3})^{6}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/361461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
series involving $\log \left(\tanh\frac{\pi k}{2} \right)$ I found an interesting series
$$\sum_{k=1}^\infty \log \left(\tanh \frac{\pi k}{2} \right)=\log(\vartheta_4(e^{-\pi}))=\log \left(\frac{\pi^{\frac{1}{4}}}{2^{\frac{1}{4}}\Gamma \left( \frac{3}{4}\right)} \right)$$
*
*Does anybody know how to approach this se... | It may be of interest to note that the sum
$$ g(x) = \sum_{k=1}^\infty \log\tanh (kx)$$
is harmonic and may be evaluated using Mellin transforms, yielding an asymptotic expansion about zero.
The Mellin transform $f^*(s)$ of the base function
$$ f(x) = \log\tanh x$$
may be computed as follows
\begin{align}
f^*(s) & = \m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/363004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 0
} |
Value of summation $ij$ What is the value of $\sum \limits_{1 \le i < j \le 10} ij$?
| $$\sum_{1\le r\le n}r\left(\sum_{r+1\le s\le n}s\right)$$
$$=\sum_{1\le r\le n}r\{\sum_{1\le s\le n}s-\sum_{1\le s\le r}s\}$$
$$=\sum_{1\le r\le n}r\{\frac{n(n+1)}2-\frac{r(r+1)}2\}$$
$$=\frac{n(n+1)}2\sum_{1\le r\le n}r- \frac12\{\sum_{1\le r\le n}r^3+\sum_{1\le r\le n}r^2\}$$
Now, you know
$\sum_{1\le r\le n}r=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/363276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$ , and find the point on the curve closest to $P$.
Find the shortest distance from the point $P(0,1)$ to a point on the curve $x² - y² = 1$, and find the point on the curve closest to $P$.
What I did so far is :
plot the y var fro... | Your approach is fine. It is a little easier to minimize $d^2$ than $d$ as that gets rid of one square root. You lost a sign when you went from $d=\sqrt{x^2+(1-y)^2}$ to $d=\sqrt{2-2\sqrt{1-x^2}}$ as the $y^2$ term is positive. Plotting the curve shows that the distance to either branch from $(0,1)$ is the same, so ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $ \int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x}\,dx $ using substitution $t=\frac{\pi}{2}-x$ This problem is given on a sample test for my calculus two class. $$ \int^{\frac{\pi}{2}}_0 \frac{\sin x}{\sin x+\cos x}\,dx $$ I can find the value of this integral using other substitutions which lead to partia... | There is an easy way to solve. Let
$$ A=\int_0^{\frac{\pi}{2}}\frac{\sin x}{\sin x+\cos x}dx, B=\int_0^{\frac{\pi}{2}}\frac{\cos x}{\sin x+\cos x}dx. $$
Then
$$ A+B=\frac{\pi}{2} $$
and
$$ A-B=\int_0^{\frac{\pi}{2}}\frac{\sin x-\cos x}{\sin x+\cos x}dx=\int_0^{\frac{\pi}{2}}\frac{-d(\sin x+\cos x)}{\sin x+\cos x}=-\ln(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/364923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
With the binomial expansion of $ (3+x)^4$ express $(3 - \sqrt 2)^4$ in the form of $p+q\sqrt 2$ I know how to expand the binomial expansion but I have no idea how to do this second part.
Hence Express $$(3 - \sqrt 2)^4$$ in the form of $$P+Q\sqrt 2$$
Where P and Q are integers, and then how to state the values of P and... | Here is the binomial formula with two complex numbers $a$ and $b$ :
$$(a+b)^n = \sum_{k=0}^n {n \choose k}\ a^{n-k} b^k$$
For $n=4$ you get :
$$\begin{array}{rcl} (a+b)^4 & = & \sum_{k=0}^4 {4 \choose k}\ a^{4-k} b^k \\ & =& {4 \choose 0}\ a^4b^0+{4 \choose 1}\ a^3 b^1+{4\choose 2}\ a^2b^2+{4 \choose 3}\ a^1b^3+{4\cho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Series of Functions - Pointwise and Uniform Convergence. I was hoping for some help for the following questions.
Prove that the series $\sum_{n=1}^\infty x^n(1-x)$ converges pointwise but not uniformly on $[0,1]$.
Prove that the series $\sum_{n=1}^\infty (-1)^nx^n(1-x)$ converges uniformly on $[0,1]$.
| $f(x)=\sum_{n=1}^\infty x^n(1-x)=x-x^2+x^2-x^3+x^3-x^4+$..
Note that $f(x)=x $ except $f(1)=0$ and $|S_N-x|=x^{N+1}$ which has supremum $1$ on $[0,1)$
for all $N$.So we don't have uniform convergence .[Pointwise convergence follows from
$x^{N+1} \to 0$ as $ N\to \infty$]
Or you may use the fact that uniform converge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/366707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Deriving Closed Form for a Recursion via Generating Functions Consider (1) $a_{n+2} = 2a_{n+1} - a_n + 4n3^n$ with $a_0 = a_1 = 1$.
Using generating functions and setting $A(x) = \sum a_nx^n$ we obtain
$$\begin{align*}&\quad\sum a_{n+2}x^{n+2} = \sum2a_{n+1}x^{n+2} - \sum a_nx^{n+2} + \sum 4n3^nx^{n+2}\\ &\implies [A(x... | Use Wilf's techniques (see "generatingfunctionology"). Your recurrence is:
$$
a_{n + 2} = 2 a{n + 1} - a_n + 4 n 3^n \qquad a_0 = a_1 = 1
$$
Define $A(z) = \sum_{n \ge 0} a_n z^n$, multiply by $z^n$ and sum over $n \ge 0$, recognizing the resulting sums gives:
$$
\frac{A(z) - a_0 - a_1 z}{z^2}
= 2 \frac{A(z) - a_0}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/368111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
$\sum_{m=1}^\infty \frac {2^{\widehat m}+2^{-\widehat m}}{2^{m}} =3$ For any positive integer $n$ , let $\widehat n$ denote the integer nearest to $\sqrt n$. Then how to prove that $$\sum_{m=1}^\infty \frac {2^{\widehat m}+2^{-\widehat m}}{2^{m}} =3$$?
| Let $f(m)=m-\widehat m, g(m)=m+\widehat m$, then
\begin{equation}
\sum_{m=1}^{\infty}{\frac{2^{\widehat m}+2^{-\widehat m}}{2^{m}}}=\sum_{m=1}^{\infty}{\left(\left(\frac{1}{2}\right)^{m-\widehat m}+\left(\frac{1}{2}\right)^{m+\widehat m}\right)}=\sum_{m=1}^{\infty}{\left(\frac{1}{2}\right)^{f(m)}}+\sum_{m=1}^{\infty}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What's the value of $ y^{(n)}$when $ y=\frac{x^n}{(x+1)^2(x+2)^2}$ What's the value of $\displaystyle y^{(n)}$when $\displaystyle y=\frac{x^n}{(x+1)^2(x+2)^2}$?
My Try:Let $\displaystyle y_n=\frac{x^n}{(x+1)^2(x+2)^2}$,so $\displaystyle y_n=xy_{n-1}$.According to Leibniz's formula,$$y_n^{(n)}=ny_{n-1}^{(n-1)}+xy_{n-1... | You can prove inductively that $$y_n = p_n(x) + (-1)^n\left(\frac{-(n+2)}{x+1}+\frac{1}{(x+1)^2} + \frac{2^{n+1}-n2^{n-1}}{x+2} + \frac{2^n}{(x+2)^2}\right)$$
Where $p_n$ is a polynomial of degree less than $n$.
That would let you compute $y_n^{(n)}$. It doesn't appear to give a pretty result, however.
I suspect Andre... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/372426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is $\sqrt[3]{p+q\sqrt{3}}+\sqrt[3]{p-q\sqrt{3}}=n$, $(p,q,n)\in\mathbb{N} ^3$ solvable? In this recent answer to this question by Eesu, Vladimir
Reshetnikov proved that
$$
\begin{equation}
\left( 26+15\sqrt{3}\right) ^{1/3}+\left( 26-15\sqrt{3}\right) ^{1/3}=4.\tag{1}
\end{equation}
$$
I would like to know if this resu... | Here's a way of finding, at the very least, a large class of rational solutions. It seems plausible to me that these are all the rational solutions, but I don't actually have a proof yet...
Say we want to solve $(p+q\sqrt{3})^{1/3}+(p-q\sqrt{3})^{1/3}=n$ for some fixed $n$. The left-hand side looks an awful lot like th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/374619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "49",
"answer_count": 6,
"answer_id": 5
} |
How do you determine the local extrema points for $y=\sqrt{3}\cos(3x)+\sin(3x)$ $$y=\sqrt{3}\cos(3x)+\sin(3x); 0\le{x}\le{\frac{2\pi}{3}}$$
I know that the local extrema can be determined by using the first derivative test. I took the derivative of $y$ and got $$y'= -3\sqrt{3}\sin(3x)+3\cos(3x)$$ I then solved the der... | $ y=\sqrt{3}\cos(3x)+\sin(3x)=2(\dfrac{\sqrt{3}}{2}\cos(3x)+\dfrac{1}{2}\sin(3x)=2(\sin\dfrac{\pi}{3}\cos(3x)+cos\dfrac{\pi}{3}\sin(3x))=2\sin(3x+\dfrac{\pi}{3}) $
$\dfrac{\pi}{3} \leq 3x+\dfrac{\pi}{3} \leq 2\pi+\dfrac{\pi}{3}$, so there is 2 peaks when $3x+\dfrac{\pi}{3}=\dfrac{\pi}{2}$ or $\dfrac{3 \pi}{2} $, anoth... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/375870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Analytic geometry straight line problem Prove that two straight lines represented by the equation $x^3+y^3+bx^2y+cxy^2=0$ will be at right angles if $b+c=-2$
I didn't know that even straight lines like planes can be represented by a combined equation. can someone please explain how this happens and how to find the indi... | Suppose you have two lines through the origin, one of slope $\alpha$ and the other of slope $\beta$:
\begin{align*}
l_1:\quad y &= \alpha x & \alpha x - y &= 0 \\
l_2:\quad y &= \beta x & \beta x - y &= 0
\end{align*}
Then you can multiply (powers of) these euqations to obtain a combined equation, which will be zero if... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/377134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the limit of $ x_n = \prod_{j=2}^{n} \left(1 - \frac{2}{j(j+1)}\right)^2$ I am stuck on the following problem:
Let $x_n=(1-\frac{1}{3})^2(1-\frac{1}{6})^2(1-\frac{1}{10})^2 \ldots...(1-\frac{1}{n(n+1)/2})^2, \text{where} \space n \geq 2$. Then $\lim_{n \to \infty}x_n=?$
I see that $x_n^{\frac{1}{2}}=\frac{2}{3... | $$
\begin{align}
\prod_{j=2}^n\left(1-\frac2{j(j+1)}\right)
&=\prod_{j=2}^n\left(\frac{(j-1)(j+2)}{j(j+1)}\right)\\
&=\color{#00A000}{\prod_{j=2}^n\frac{j-1}{j}}
\color{#C00000}{\prod_{j=2}^n\frac{j+2}{j+1}}\\
&=\color{#00A000}{\frac1n}\color{#C00000}{\frac{n+2}{3}}\\
&=\frac13\frac{n+2}{n}
\end{align}
$$
Taking it to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/379511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Prove $\left(\frac{2}{5}\right)^{\frac{2}{5}}<\ln{2}$ Inadvertently, I find this interesting inequality. But this problem have nice solution?
prove that
$$\ln{2}>(\dfrac{2}{5})^{\frac{2}{5}}$$
This problem have nice solution? Thank you.
ago,I find this
$$\ln{2}<\left(\dfrac{1}{2}\right)^{\frac{1}{2}}=\dfrac{\sqrt{2}}{2... | The Newton-Raphson method for $x=\left(\frac{2}{5}\right)^{2/5}$ is $x_{n+1}=\frac45x_n+\frac{4}{125x_n^4}$. Take $x_0=1$ and assume that each $x_n\geq\left(\frac{2}{5}\right)^{2/5}$. Compute the iterations up to $x_5=0.693145$, using a sufficient level of precision, around $6$ decimals. Finally, compute the upper boun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/380302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "95",
"answer_count": 8,
"answer_id": 7
} |
What did I do wrong in u-substitution The problem: find the indefinite integral of $x(x-1)^2$
I used u-substitution, $u = x-1, x = u+1, du = dx$.
which gave me $(u+1)u^2$. I distributed and got $u^3 + u^2$, and took the integral to get $[(u^4)/4] + [(u^3)/3]$ replacing $u$ gave me an answer of $[((x-1)^4)/4] + [((x-1)^... | You did nothing "wrong" with your u-substitution. Your evaluation of the indefinite integral is correct. To see this,
Suggestion:
Expand the binomials in your answer: Expand $(x-1)^4$ and $(x - 1)^3$ in the numerators, respectively, simplify, and take into account the constant value (absorbed by the constant of integr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/383971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove the following: $$\sum_{k=1}^{\infty }\frac{1}{(2k-1)^{2}}=\frac{\pi ^{2}}{8}$$
I don't really know how to prove this, will assuming that $$cos(x)=\sum_{k=0}^{\infty }(-1)^{k}\frac{x^{2k}}{(2k)!}$$ help?
| If you must use $\cos (x)$, you can focus on zeros of it. $(...,-\frac{(2n-1)\pi}{2},....,-\frac{3\pi}{2},-\frac{\pi}{2},\frac{\pi}{2},\frac{3\pi}{2},...,\frac{(2n-1)\pi}{2},..)$ where $n$ is a positive integer
Hint:
$$\cos x = \cdots\left(1+\frac{x}{\frac{3\pi}{2}}\right)\left(1+\frac{x}{\frac{\pi}{2}}\right)\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/384319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to prove this trigonometric expression? How would you go about proving the following?
$${1- \cos A \over \sin A } + { \sin A \over 1- \cos A} = 2 \operatorname{cosec} A $$
This is what I've done so far:
$$LHS = {1+\cos^2 A -2\cos A + 1 - \cos^2A \over \sin A(1-\cos A)}$$
....no idea how to proceed .... X_X
| You did everything thus far correctly, I just pick up with where you left off in the second line:
$$\begin{align}(1 - \cos A)^2 + \sin^2 A \over \sin A(1 - \cos A)
& = \dfrac{1 - 2 \cos A + \cos^2 A + \sin^2 A}{\sin A(1 - \cos A)} \\ \\
& = {1 \color{blue}{\bf + \cos^2 A} -2\cos A + 1 \color{blue}{\bf - \cos^2A} \over... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/385537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
How to derive $\cos\frac{n\pi}{3}=\frac{1+3(-1)^{[\frac{n+1}{3}]}}{4}$ Consider the following formula to calculate a trigonometric function:
$$\cos\frac{n\pi}{3}=\frac{1+3(-1)^{[\frac{n+1}{3}]}}{4}$$ $[x]$ denotes the integer part of $x$. The formula is valid for $n=0,2,4,6,...$
I'm curious how this formula is de... | First, observe that
$$\cos\frac{(n+6)\pi}{3}=\cos\left(\frac{n\pi}{3}+2\pi\right)=\cos\frac{n\pi}{3}$$
and
$$(-1)^{\lfloor\frac{(n+6)+1}{3}\rfloor}=(-1)^{\lfloor\frac{n+1}{3}\rfloor+2}=(-1)^{\lfloor\frac{n+1}{3}\rfloor}$$
Hence we only need to consider $n=0$, $n=2$, and $n=4$. These cases are trivial to check to be tru... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/386726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving $\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$ without induction I was looking at: $$\sum_{k=1}^n{k^2}=\frac{n(n+1)(2n+1)}{6}$$
It's pretty easy proving the above using induction, but I was wondering what is the actual way of getting this equation?
| HINT:
To find $\sum_{1\le r\le n}r^m$ we can utilize the identity
$$(r+1)^{m+1}-r^{m+1}$$
$$=\sum_{1\le t\le m+1}\binom {m+1}t r^{m+1-t}=(m+1)r^m+\sum_{2\le t\le m+1}\binom {m+1}t r^{m+1-t}$$ and need to know $\sum_{1\le r\le n}r^s$ for $0\le s\le m-1$
For $m=2,$ $$(r+1)^3-r^3=3r^2+\sum_{2\le t\le 3}\binom 3t r^{3-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/387664",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
On the Hurwitz Zeta Function In my mathematics course in Uni. (I'm a physics student) my prof. gave us the following exercise: to express the Hurwitz Zeta function $\zeta(2k+1,\frac{1}{4})$ with $k=1,2,3,\dots$ in terms of the Riemann zeta function. He says there is a closed form for this, something like
$\zeta(2k+1,\f... | \begin{equation}
\begin{array}{c}
\left.
\begin{array}{c}
\zeta (2n+1,\frac{1}{4}) \\
\zeta (2n+1,\frac{3}{4})%
\end{array}%
\right\} =2^{2n}(2^{2n+1}-1){\zeta }(2n+1) \\
\pm \frac{1}{2\pi }\left( 2n+2+4^{2n+2}\right) {\zeta }(2n+2)-2\sum
\limits_{l=0}^{n-1}4^{2n-2l}{{{\zeta }(2n-2l)\zeta }}(2l+2)%
\end{array}
\tag*{(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Proof of the identity $2^n = \sum\limits_{k=0}^n 2^{-k} \binom{n+k}{k}$ I just found this identity but without any proof, could you just give me an hint how I could prove it?
$$2^n = \sum\limits_{k=0}^n 2^{-k} \cdot \binom{n+k}{k}$$
I know that $$2^n = \sum\limits_{k=0}^n \binom{n}{k}$$ but that didn't help me
| I tried to generalize, but equation $(2)$ shows why $x=\frac12$ is needed to make a simple equation.
Note that
$$
\begin{align}
\sum_{k=0}^{n+1}x^k\binom{n+k+1}{k}
&=\sum_{k=0}^{n+1}x^k\left[\binom{n+k}{k}+\binom{n+k}{k-1}\right]\\
&=\sum_{k=0}^nx^k\binom{n+k}{k}+x^{n+1}\binom{2n+1}{n+1}\\
&+\sum_{k=0}^{n+1}x^{k+1}\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 8,
"answer_id": 1
} |
Solving recurrence equation using generating functions $$a_{0} = 0, a_{1} = 1, a_{2} = 2, a_{3} = 6$$
$$a_{n} = a_{n+3} - a_{n+2}$$
$\sum = a_{n}x^{n}$
$A(x) = \frac{A(X) - x - 2x^{2}}{x^{3}} - \frac{A(x) - x}{x^{2}}$
$A(x) = \frac{x^{3} + x - 1}{-x^{2} - x}$
Is this the correct solution? The answer is different as fa... | If we make a blanket assumption that $a_n=0$ for all $n<0$, we can write the recurrence as
$$a_n=a_{n-1}+a_{n-3}+[n=1]+[n=2]+4[n=3]\;,\tag{1}$$
where the last three terms are Iverson brackets, and it will be correct for all $n\ge 0$.
If you’ve not seen this idea before, try evaluating $(1)$ for $n=0,1,2,3$ on the ass... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/389308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving ODE using frobenius method. 3 coefficients I'm trying to learn frobenius method by solving some problems (ODEs).
For example:
$$xy''+(2x+1)y'+(x+1)y=0$$
Let $y=\sum\limits_{n=0}^\infty a_nx^{n+r}$. Then, I took derivatives and put into the equation:
$$\sum\limits_{n=0}^\infty a_n(n+r)^2x^{n+r-1}+2\sum\limits_{n... | You state:
After I shifted to make their orders same:
$$
\sum_{k = -2}^\infty a_{k + 2}(k + r + 2)^2 x^{k + r + 1} + 2 \sum_{k = -1}^\infty a_{k + 1}(k + r + 2)x^{k + r + 1}+\sum_{k = 0}^\infty a_k x^{k + r + 1}
$$
What you do is the following. Given that
$$
\sum_{k = 0}^\infty a_k (k + r)^2 x^{k + r - 1} + \sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/393157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Integrate by parts: $\int \ln (2x + 1) \, dx$ $$\eqalign{
& \int \ln (2x + 1) \, dx \cr
& u = \ln (2x + 1) \cr
& v = x \cr
& {du \over dx} = {2 \over 2x + 1} \cr
& {dv \over dx} = 1 \cr
& \int \ln (2x + 1) \, dx = x\ln (2x + 1) - \int {2x \over 2x + 1} \cr
& = x\ln (2x + 1) - \int 1 - {1 \over... | Why don't you put $u = 2x + 1$ so that $du = 2 \,dx$? Then we'd have $$\int \log (2x + 1) \, dx = \frac {1} {2} \int \log u \, du = \frac {1} {2} (u \log u - u) = \frac {1} {2} (2x + 1) \log (2x + 1) - x - \frac {1} {2} + C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/393929",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Polynomials - The sum of two roots If the sum of two roots of
$$x^4 + 2x^3 - 8x^2 - 18x - 9 = 0$$ is
$0$, find
the roots of the equation
| Let those two roots be $a,-a$ and the other two roots are $b,c$
$$\text{Now, }(x-a)(x+a)(x-b)(x-c)=(x^2-a^2)\{x^2-(b+c)x+bc\}$$
$$=x^4-x^3(b+c)+(bc-a^2)x^2+a^2(b+c)x-a^2bc$$
Comparing the coefficients of $x^3$ of this with that of $x^4 + 2x^3 - 8x^2 - 18x - 9 = 0,$
$b+c=-2$
Comparing the coefficients of $x,a^2(b+c)=-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/396015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
How prove this $\sum_{n=1}^{\infty}\frac{\zeta_{2}}{n^4}=\zeta^2(3)-\frac{1}{3}\zeta(6)$ show that
$$\sum_{n=1}^{\infty}\dfrac{\zeta_{2}}{n^4}=\zeta^2(3)-\dfrac{1}{3}\zeta(6)$$
where
$$\zeta_{m}=\sum_{k=1}^{n}\dfrac{1}{k^m},\zeta(m)=\sum_{k=1}^{\infty}\dfrac{1}{k^m}$$
is true？
because This result is my frend tell me.
T... | Consider $$f(z) = \frac{\Big(\psi_{1}(-z) \Big)^{2}}{z^{\color{red}{3}}}$$
(where $\psi_{1}(z)$ is the trigamma function) and integrate around a circle centered at the origin of complex plane of radius $N + \frac{1}{2}$ where $N$ is a positive integer.
Then $$\lim_{N \to \infty } \int_{|z| = N+\frac{1}{2}} f(z) \ dz = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/397055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Compute limit of the sequence $x_n$ given by $x_{n+2}=-\frac{1}{2}(x_{n+1}-x_n^2)^2+x_n^4$ Let $(x_n)$ be a real sequence such that $x_0=a\in\mathbb{R},x_1=b\in\mathbb{R},x_{n+2}=-\dfrac{1}{2}\left(x_{n+1}-x_{n}^2\right)^2+x_{n}^4\;\forall n\in\mathbb{N} $ and $|x_n|\leq \dfrac{3}{4},\forall n\in\mathbb{N}$. The sequen... | As Greg Martin stated in comment, the limit satisfies the equation $\displaystyle L=-\frac{1}{2}(L-L^2)^2+L^4$, not hard to find(by rearranging and factoring the equation) the solutions of the equation are $0,1,-1,-2$.
Because $\displaystyle |x_n|\leq \frac{3}{4}$ the sequence cannot converge to $1, -1, -2$. The sequen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/397850",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
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