Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $g(x|y=\frac{1}{2})$, the conditional pdf of $X$ given $Y = \frac{1}{2}$ (Need confirmation) Let X and Y be continuous random variables having the joint pdf $$f(x,y) = 8xy , 0\leq{y}\leq{x}\leq{1}$$ I found that the marginal pdf of Y is $f_2(y) = 4y - 4y^3$. Does $g(x|y=\frac{1}{2}) = \frac{f(x,\frac{1}{2})}{f_2(\... | $$f_Y(y) = \int_{y}^{1} 8xy \: \text{dx} = 4y - 4y^3$$
$$ f_Y(y) =
\begin{cases}
4y - 4y^3, & 0 \leq y \leq 1 \\
0, & \text{otherwise}
\end{cases}
$$
$$g(x \mid y=\frac{1}{2}) = \frac{f_{X,Y}(x,y=\frac{1}{2})}{f_Y(y=\frac{1}{2})} = \frac{8x}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1771281",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the Fourier transform of $\sin x^2$. I've tried it by applying integratrion by parts, but I'm not getting the answer correct. Its answer is
$$\frac{1}{\sqrt{2}}\,\sin\left(\frac{k^2}{4} +\frac{\pi}{4}\right).$$
Please help in this.
| Let be
$$
\sin(ax^2)=\frac{\mathrm e^{iax^2}-\mathrm e^{iax^2}}{2i}\qquad\text{and}\qquad\cos(ax^2)=\frac{\mathrm e^{iax^2}+\mathrm e^{iax^2}}{2}
$$
anf use the Fourier transform defined as $$\mathcal F\left\{f(x)\right\}=F(k)=\int_{-\infty}^{\infty} f(x)\mathrm e^{-i2\pi k x}\,\mathrm d x$$
The Fourier transform can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772457",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
finding the number of factors $2^{15}\times3^{10}\times5^6$ The number of factors of $2^{15}\times3^{10}\times5^6$ which are either perfect square or perfect cubes(or both)
I don't know how to start this even!
Plz solve this!
| Case 1: factors that are perfect squares
\begin{align}
A & =\{2^0,2^2,2^4,2^6,2^8,2^{10},2^{12},2^{14}\} \\
B & =\{3^0,3^2,3^4,3^6,3^8,3^{10}\} \\
C & =\{5^0,5^2,6^4,5^6\}
\end{align}
Choose 1 from each set. We have $8\times6\times4=192$ choices.
Case 2: factors that are perfect cubes
\begin{align}
A& =\{2^0,2^3,2^6,2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774503",
"timestamp": "2023-03-29T00:00:00",
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$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove $x^2y+y^2z+z^2x < \frac12$
$x,y,z \geqslant 0$, $x+y^2+z^3=1$, prove
$$x^2y+y^2z+z^2x < \frac12$$
This inequality has been verified by Mathematica. $\frac12$ is not the best bound. I try to do AM-GM for this one but not yet success. The condition $x+y^2+z^3$ is very weird.... | This answer is incomplete.
Let $A=x+y^2$, $B=y^2+z^3$ and $C=z^3+x$.
Claim: $x^2y \leq \dfrac{A}{2}x^{3/2}$,
$y^2z \leq \dfrac{B^{2/3}}{4^{2/3}}y^{4/3}$, $z^2x \leq \dfrac{C^{4/3}}{4^{2/3}}x^{1/3}$.
Assuming the claim, we note that $A+B+C=2$ and we need to prove that the given sum is less than $\dfrac{A+B+C}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
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Olympiad Inequality $\sum\limits_{cyc} \frac{x^4}{8x^3+5y^3} \geqslant \frac{x+y+z}{13}$
$x,y,z >0$, prove
$$\frac{x^4}{8x^3+5y^3}+\frac{y^4}{8y^3+5z^3}+\frac{z^4}{8z^3+5x^3} \geqslant \frac{x+y+z}{13}$$
Note:
Often Stack Exchange asked to show some work before answering the question. This inequality was used as a ... | Not sure, if I missed out anything here. Take a look.
For non negative, $X,Y,Z$,
We can perhaps use Titu's inequality (a mix of Holder and CS), sometimes called Titu's screw lemma (https://en.wikipedia.org/wiki/Nesbitt%27s_inequality).
\begin{equation}
\sum_{k=1}^{n}{\frac{x_{k}^{2}}{a_{k}}} \ge \frac{\left(\sum_{k=1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "124",
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"answer_id": 2
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$x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove $x^{\frac85}+y^{\frac85}+z^{\frac85} \geqslant 3$ $x,y,z \geqslant 0$ and $x^2+y^2+z^2+xyz=4$, prove
$$x^{\frac85}+y^{\frac85}+z^{\frac85} \geqslant 3$$
1) The equality occurs only at $x,y,z=1$. Let's assume $F=x^n+y^n+z^n$, I noticed that $n=1$ then $F \leqslant 3$ an... | Let $x=2\cos\alpha$ and $y=2\cos\beta$, where $\{\alpha,\beta\}\subset\left[0,\frac{\pi}{2}\right]$.
Hence, $z=2\cos\gamma$, where $\alpha+\beta+\gamma=\pi$ and $\gamma\in\left[0,\frac{\pi}{2}\right]$.
Let $f(x)=\left(\cos x\right)^{\frac{8}{5}}$.
Since $f''(x)=\frac{-8(1+4\cos2x)}{25\sqrt[5]{\cos^2x}}\geq0$ for all $x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1777166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2)$
Prove that $$\displaystyle \sum_{k=1}^n \bigg(\dfrac{1}{k}+\dfrac{2}{k+n}\bigg ) \leq \ln(2n) + 2 -\ln(2).$$
I was thinking of using mathematical induction for this. That is,
We prove by induction on $n$. The cas... | First note, that we can rewrite the RHS of your inequality as $\ln(n) +2$.
We proceed by induction. We compute
$$ \sum_{k=1}^{n+1} \left( \frac{1}{k} + \frac{2}{k+n+1}\right)
= \frac{1}{n+1} + \frac{2}{2n+2} + \sum_{k=1}^n \frac{1}{k} + \sum_{k=1}^{n} \frac{2}{k+n+1}$$
shifting the index yields
$$ =\frac{2}{n+1} + \su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1778758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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How to remember sum to product and product to sum trigonometric formulas? They are:
\begin{align}
\cos(a)\cos(b)&=\frac{1}{2}\Big(\cos(a+b)+\cos(a-b)\Big) \\[2ex]
\sin(a)\sin(b)&=\frac{1}{2}\Big(\cos(a-b)-\cos(a+b)\Big) \\[2ex]
\sin(a)\cos(b)&=\frac{1}{2}\Big(\sin(a+b)+\sin(a-b)\Big) \\[2ex]
\cos(a)\sin(b)&=\frac{1}{2}... | How about just restating the LHS. For example, you could restate $\cos a\sin b$ as
$$\frac{\sin a\cos b +\cos a\sin b + \cos a\sin b - \sin a\cos b}{2}$$ and just figure it out from there. For Example, Let's start off with $\cos a\sin b$ and try to derive $\frac{1}{2}[\sin(a+b) - \sin(a-b)]$
$$
\begin{align}
\cos a \si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1779221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Show that $\int_0^\infty\frac{1}{1+x^2+x^{\pi}+x^{\pi-2}}dx = \int_0^\infty\frac{1}{1+x^2+x^{e}+x^{e-2}}dx $ Vladimir Reshetnikov's Identity
(1)
$$\int_0^\infty\frac{1}{1+x^2}\cdot\frac{1}{1+x^{\pi}}dx =\int_0^\infty\frac{1}{1+x^2}\cdot\frac{1}{1+x^{e}}dx $$
Note that $(1+x^2)(1+x^{\pi})=1+x^2+x^{\pi}+x^{2+\pi}$ and
... | Let we do something more general. Let $a>0$ and
$$ I(a) = \int_{0}^{+\infty}\frac{dx}{(1+x^2)(1+x^a)}. $$
Well, $I(a)$ is constant. Don't you believe it? Replace $x$ with $\frac{1}{z}$ to get:
$$ I(a) = \int_{0}^{+\infty}\frac{z^a\,dx}{(1+z^2)(1+z^a)}\,dz $$
from which:
$$ I(a) = \frac{1}{2}\int_{0}^{+\infty}\frac{1+x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1781345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms Find $S=\frac{1}{7}+\frac{1\cdot3}{7\cdot9}+\frac{1\cdot3\cdot5}{7\cdot9\cdot11}+\cdots$ upto 20 terms
I first multiplied and divided $S$ with $1\cdot3\cdot5$
$$\frac{S}{15}=\frac{1}{1\cdot3\cdot5\cdot7}+\frac{1\cdot3}{... | Hint:
$\frac{1}{(2r+5)(2r+3)(2r+1)}=\frac{1}{4}\left(\frac{1}{(2r+3)(2r+1)}-\frac{1}{(2r+5)(2r+3)}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1784481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Let g(x) be a non negative continuous function on R such that $g(x) +g(x+\frac{1}{3})=5$.. Problem :
Let g(x) be a non negative continuous function on R such that $g(x) +g(x+\frac{1}{3})=5$ then calculate the value of integral $\int^{1200}_0 g(x) dx$
My approach :
$g(x) +g(x+\frac{1}{3})=5$.....(1)
put x = x +$\frac... | I don't know whether this solution is the fastest but I am pretty sure it is the most accurate one.
Step 1: multiply the differential infinitesimal (dx) on $g(x) +g(x+\frac{1}{3})=5$
$(g(x) +g(x+\frac{1}{3}))dx= 5dx$
Because of the multiplication's disturbution law:
$g(x)dx +g(x+\frac{1}{3})dx= 5dx$
Step 2: integral t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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solve for $x$: $\frac{\sin(x)}{x}=\frac{5}{6}$ Is it possible to solve for x the following equation without root finding:
$$\frac{\sin(x)}{x}=\frac{5}{6}$$
| Equations which mix polynomial and trigonometric terms do not show analytical solutions (this is already the case for $x=\cos(x)$) and numerical methods should be used.
However, some rather good approximations can be made and, for your curiosity, I give you the links to two questions of mine (here and here).
Using the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1785593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\frac{1}{7\sqrt{2}} \leq \int_{0}^1 \frac{x^6}{\sqrt{1+x^2}}dx \leq \frac{1}{7}$
Prove that $\displaystyle \dfrac{1}{7\sqrt{2}} \leq \int_{0}^1 \dfrac{x^6}{\sqrt{1+x^2}}dx \leq \dfrac{1}{7}$.
My book says let $f(x) = \dfrac{1}{\sqrt{1+x^2}}$ and $g(x) = x^6$. Then $\displaystyle \int_0^1 \dfrac{x^6}{\sqrt... | We have $$\frac{1}{\sqrt 2} \le \frac{1}{\sqrt{1+x^2}} \le 1$$ for all $x \in [0,1]$ since $0 \le x^2 \le 1$. Indeed, then $1 \le 1+x^2 \le 2$ and so $$\frac{1}{2} \le \frac{1}{1+x^2} \le 1.$$ Thus $$\frac{1}{\sqrt 2} \le \frac{1}{\sqrt{1+x^2}} \le 1.$$
Then just multiply by $x^6$ and integrate.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$? Evaluate this limit
$$\lim_{x \to 0} \frac{x^{6000} - (\sin x)^{6000}}{x^2(\sin x)^{6000}}$$
What's the method?
The answer is $1000$.
| This answer is pretty useless for the results per se, much shorter proofs have been given. Honestly, it took me some time to write it, and I could non just abandon it in the forest.
[EDIT] Its intent is to play with standard inequalities, that are useful to keep in mind, and keep them as long as possible, without using... | {
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"url": "https://math.stackexchange.com/questions/1786112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 8,
"answer_id": 3
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Finding the limit of $\lim_{n\rightarrow\infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+...[\text{upto n times}])$ The question is as follows-
Evaluate the limit.
$$\lim_{n\rightarrow\infty}(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+...[\text{upto n times}])$$
I have no idea on how to solve this... | Use formula:
$$\frac{1}{n^3-n}=\frac1{(n-1)n(n+1)}=\frac12 \left(\frac1{n(n-1)}-\frac1{n(n+1)} \right)$$
Then
$$\lim_{n\rightarrow\infty}\left(\frac{1}{6}+\frac{1}{24}+\frac{1}{60}+\frac{1}{120}+...+\frac{1}{n^3-n}\right)=$$
$$=\lim_{n\rightarrow\infty}\frac12\left(\frac{1}{2\cdot1}-\frac{1}{2\cdot3}+\frac{1}{3\cdot2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1787375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $k_{max}$ if such $\frac{2(a^2+kab+b^2)}{(k+2)(a+b)}\ge \sqrt{ab}$ Let $a,b>0$ then we have
$$\color{crimson}{\dfrac{2(a^2+kab+b^2)}{(k+2)(a+b)}\ge \sqrt{ab}}$$ Find $k_{\max}$
Everything I tried has failed so far.
Here is one thing I tried, but obviously didn't work.
Consider the Special case $a=b$
then
$$\color... | The inequality can be rewritten as $$2(a-b)^2\ge (k+2)(\sqrt{a}-\sqrt{b})^2$$ which is equivalent when $a\neq b$ to $$2(\sqrt{a}+\sqrt{b})^2\ge (k+2)\sqrt{ab}$$ or $$2(\sqrt{a}-\sqrt{b})^2\ge (k-6)\sqrt{ab}$$ This shows that the inequality always holds for $k=6$. For any $k>6$, we may take $a=1, b=1+\epsilon$ for suffi... | {
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"url": "https://math.stackexchange.com/questions/1788931",
"timestamp": "2023-03-29T00:00:00",
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show that $\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $ Let $a,b,c$ are positive numbers,if $$a+b+c=1$$
show that $$\frac { 1 }{ 1-a } +\frac { 1 }{ 1-b } +\frac { 1 }{ 1-c } \ge \frac { 2 }{ 1+a } +\frac { 2 }{ 1+b } +\frac { 2 }{ 1+c } $... | We shall use the sigma sign ($\sum$) for cyclic sum.
By changing every number 1 into $a+b+c$, what we are trying to prove is equivalent to :
$$\sum \frac {1}{a+b} \ge \sum \frac{2}{(a+b)+(c+a)}$$
Let
$$a+b=x$$
$$b+c=y$$
$$c+a=z$$
We have to prove :
$$\sum \frac {1}{x} \ge \sum \frac{2}{x+y}$$
We have $$ \frac 1x + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1790541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Inequality problem $\frac 1x + \frac 1y + \frac 1z > 5$ prove Prove that:
$$\frac 1x + \frac 1y + \frac 1z > 5$$
where $x+y+z=1$, $x, y, z$ are real numbers not equal $0$
and $x\neq y \neq z $
| Note: This somewhat longish answer contains several proofs of the inequality. I'm leaving things in the order in which they accreted, but I recommend skipping to the boldface "added yet later" section for the simplest of the proofs here. I also highly recommend Macavity's answer as the simplest of all.
Using AGM twi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Tricky Multivariable Inequality Given 100 positive real numbers $x_1, x_2, \cdots, x_n$ that satisfy
$x_1^2+x_2^2+\cdots+x_n^2>10000$ and
$x_1+x_2+\cdots x_n\le 300$,
prove that there exist three numbers from this set such that the sum of these three numbers is larger than 100.
I have tried using Lagrange Multipliers a... | I shall prove the following statement, from which the problem is solved by setting $B=300$ and $n=100$.
Let $S=\{x_1,x_2,\ldots,x_n\}$ be a set of $n \geq 3$ non-negative numbers and let $t=\sum_{i=1}^{n}x_i^2$ and
$s=\sum_{i=1}^{n} x_i$.
Suppose that $s \geq \dfrac{B^2}{9}$ and $t \leq B$. Then there exist three numb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1792511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Parametrize the "alpha curve" I wanted to check if my solution would be correct for the question:
Q. Parametrize the "alpha curve"
$y^2$=$x^3$+$x^2$
A. y=tx
$(tx)^2$=$x^3$+$x^2$
$t^2$$x^2$=$x^3$+$x^2$
$t^2$$x^2$-$x^2$=$x^3$
$x^2$($t^2$-1)=$x^3$
x=$t^2$-1
Substitute x=$t^2$-1 into $y^2$=$x^3$+$x^2$
$y^2$=$x^3$+$x^2$
= $... | In polar coordinates,
$$r^2\sin^2(\theta)=r^3\cos^3(\theta)+r^2\cos^2(\theta),$$
then
$$r=\frac{\sin^2(\theta)-\cos^2(\theta)}{\cos^3(\theta)}=\frac{1-2\cos^2(\theta)}{\cos^3(\theta)}.$$
Then numerator cancels for the four angles $\theta=\pm\frac\pi4,\pm\frac{3\pi}4$, corresponding to a double point at the origin, the ... | {
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"url": "https://math.stackexchange.com/questions/1792903",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Error in solving $\int \sqrt{1 + e^x} dx$ . I want to solve this integral for $1 + e^x \ge 0$
$$\int \sqrt{1 + e^x} dx$$
I start by parts $$\int \sqrt{1 + e^x} dx = x\sqrt{1 + e^x} - \int x \frac{e^x}{2\sqrt{1 + e^x}} dx $$
Substitute $\sqrt{1 + e^x} = t \implies dt = \frac{e^x}{2\sqrt{1 + e^x}} dx$
So I remain with $... |
Where is the mistake hidden?
In the following part :
So I remain with $$x\sqrt{1 + e^x} - \int \ln((t-1)(t+1)) dt = x\sqrt{1 + e^x} + 2t +(-t+1)\ln(t-1)+ (-t+1)\ln(t+1) $$
It should be the following (a sign mistake) :
$$x\sqrt{1 + e^x} - \int \ln((t-1)(t+1)) dt = x\sqrt{1 + e^x} + 2t +(-t+1)\ln(t-1)+ (-t\col... | {
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"timestamp": "2023-03-29T00:00:00",
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Show that if $n\equiv 3, 6 \pmod9 $ then $n$ is not a sum of two squares Show that if $n\equiv 3, 6 \pmod9 $ then $n$ is not a sum of two squares.
I started by: Assume $n=a^2+b^2$ a sum of two squares. Then $a^2,b^2\equiv 0,1,4,7 \pmod9$, and no combination these numbers can yield $3$ or $6$ so that $a^2+b^2\equiv 3,6 ... | It suffices to show that if $3\mid a^2+b^2$, then $3\mid a$ or $3\mid b$. There is a separate question about this: Prove that if $3\mid a^2+b^2$ then $3\mid a$ and $3\mid b$. (And you might find something also in the questions which are linked there.) Unfortunately, that question is not in a very good state. It was edi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluation of $\int_{0}^{1}\frac{\ln x}{x^2-x-1}dx$
Evaluation of $\displaystyle \int_{0}^{1}\frac{\ln x}{x^2-x-1}dx$
$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\infty}\frac{\ln x}{x^2-x-1}dx=\int_{0}^{\infty}\frac{\ln(x)}{\left(x-\frac{1}{2}\right)^2-\left(\frac{\sqrt{5}}{2}\right)^2}dx$
Now Put $\displaysty... | So, by the comment of the OP, the integral we have to study is $$\int_{0}^{1}\frac{\log\left(x\right)}{x^{2}-x-1}dx.$$ We note that $$\begin{align} I= & \int_{0}^{1}\frac{\log\left(x\right)}{x^{2}-x-1}dx \\ = & \int_{0}^{1}\frac{\log\left(x\right)}{\left(x-\frac{1-\sqrt{5}}{2}\right)\left(x-\frac{1+\sqrt{5}}{2}\right)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1794496",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
} |
To show that the variables in the system are same in magnitude I am stuck with this interesting problem,
If for non-negative integers $a, b, \text{and} c$, $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$ are both integers then show that $|a|=|b|=|c|$.
I could only make a few observations... | Let $m=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}$ and $n=\frac{b}{a}+\frac{c}{b}+\frac{a}{c}$. Then $\frac{a}{b},\frac{b}{c},\frac{c}{a}$ are the roots of $x^3-mx^2+nx-1=0$.
Hence we have $a^3-ma^2b+nab^2-b^3=0$, where $a,b,m,n$ are all integers, with $a,b$ positive. Let $p^r$ be the highest power of $p$ dividing $a$, and $p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1795246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $\ln(2^5)-\pi=8\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{e^{n\pi}+1}-\frac{1}{e^{2n\pi}+1}\right)$ We took this idea from Simon Plouffe see here
$$\ln(2^5)-\pi=8\sum_{n=1}^{\infty}\frac{1}{n}\left(\frac{1}{e^{n\pi}+1}-\frac{1}{e^{2n\pi}+1}\right)$$
Can anyone prove this identiy?
We found this identity via a s... | We can calculate the sum $$\sum_{n = 1}^{\infty}\frac{1}{n(e^{nx} + 1)}$$ as follows.
Let $q = e^{-x}$ and then we have
\begin{align}
F(q) &= \sum_{n = 1}^{\infty}\frac{1}{n(e^{nx} + 1)}\notag\\
&= \sum_{n = 1}^{\infty}\frac{q^{n}}{n(1 + q^{n})}\notag\\
&= \sum_{n = 1}^{\infty}\frac{1}{n}\sum_{k = 1}^{\infty}(-1)^{k - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1796001",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solution of integral $\int \frac{\sin (x)}{\sin (5x) \sin (3x)}\,\mathrm dx.$ Find the following integral:
$$\int \frac{\sin (x)}{\sin (5x) \sin (3x)}\,\mathrm dx.$$
I don't know how to deal with the $\sin (x)$ in the numerator. If it had been $\sin (2x)$ then we could have used $\sin (2x)= \sin (5x-3x)$. How to deal... | I think it's a little easier if you leave in in terms of Chebyshev polynomials of the second kind, $U_n(\cos x)=\frac{\sin(n+1)x}{\sin x}$. Then the integral looks like
$$\begin{align}\int\frac{\sin x}{\sin5x\sin3x}dx&=\int\frac{\sin x}{\sin^2xU_4(\cos x)U_2(\cos x)}dx\\
&=\int\frac{-dv}{(1-v^2)U_4(v)U_2(v)}\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1798423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Solving a trig equation that is quadratic? I have to solve for $x$ given
$$\tan^2 x = 2 + \tan x\;\;\;\;\;\;0≤x≤2\pi$$
I brought it all to one side and set it all equal to zero like:
$$\tan^2 x - \tan x - 2 = 0$$
What am i supposed to do from there to solve for x?
| You began with:
$ \tan^2 x = 2 + \tan x $.
As you mentioned, bring all terms to the L.H.S.:
$ \tan^2 x - \tan x - 2 = 0 $
Now let:
$ y = \tan z $.
As a result:
$ y^2 - y - 2 = 0 $
Use the quadratic formula to solve for $y$.
$ y = \frac{-(-1) \pm \sqrt{(-1)^2 - 4(1)(-2)}}{2(1)} $
$ = \frac{1 \pm \sqrt{9}}{2} $
$ = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1801366",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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If $a, b, c >0$ prove that $ [(1+a)(1+b)(1+c)]^7 > 7^7a^4b^4c^4 $. I solved it using AM, GM inequalities and reached to $[(1+a)(1+b)(1+c)]^7 > 2^{21}(abc)^\frac72 $ please help how to get $7^7(abc)^4$ in the inequality.
| $\dfrac{1 + a + b + c + ab + bc + ac + abc}{7} > \dfrac{a + b + c + ab + bc + ac + abc}{7} \geq \sqrt[7]{a^4b^4c^4}$
where the second inequality follows from the AM-GM inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1802976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Integrating trig function I'm stuck at this problem:
$$ \int{\sqrt{(\sin^2 x)^2 + (2\sin x \cos x)^2}dx} = \int{\sqrt{\sin^2 x \sin^2 x + 4\sin^2 x \cos^2 x} dx}$$
I tried a few trig identities: $\sin^2 x = \frac{1-\cos 2x}{2} $ and $ \cos^2 x = \frac{1+\cos 2x}{2} $ and $\sin^2 x + \cos^2 x = 1$. Keep hitting dead en... | $$\begin{align} &\int{\sqrt{\sin^2x \sin^2x + 4\sin^2 x \cos^2 x} dx} \\ &
=\int{\sqrt{\sin^2x( \sin^2x + 4\cos^2 x)} dx} \\ &
=\begin{cases}\int{\sin x\sqrt{( \sin^2x + 4\cos^2 x)} dx} & & & &, \sin x \geq 0 \\ -\int{\sin x\sqrt{( \sin^2x + 4\cos^2 x)} dx} & & & &, \sin x < 0\end{cases} \\ &
=\begin{cases}\int{\sin x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1804708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
Solve for integers $x, y, z$ such that $x + y = 1 - z$ and $x^3 + y^3 = 1 - z^2$.
I think we'll have to use number theory to do it. Simply solving the equations won't do.
If we divide the second equation by the first, we get:
$$x^2 - xy... | Substituting the first in the second gives
$$x^3+y^3+(x+y)^2-2(x+y)=0$$
so $x+y=0$ (giving $z=1$) or
$$x^2-xy+y^2+x+y-2=0,$$
that is,
$$\left(x+\frac12-\frac y2\right)^2+\frac34(y+1)^2-3=0$$
so $(y+1)^2\leq4$, which leaves to check $y\in\{-3,-2,-1,0,1\}$.
All solutions are given by
$$\begin{align*}(x,y,z)\in\{&(a,-a),\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1805528",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
} |
Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$.
Find the volume of the region enclosed by $x^2+y^2+z^2=2$ and $x^2+y^2=z$.
I tried to solve the problem above by doing a change of variables to the spherical coordinate system, that is,
$x= \rho \cos \theta \sin \varphi$
$y = \rho \sin \theta ... | I can't understand why this solid is getting chopped up into two pieces. We know the $\theta$ goes all the way around, from $0$ to $2\pi$, then the boundary conditions can be solved as $z=r^2$, $r^2+z^2=r^2+r^4=2$, so $r^4+r^2-2=(r^2+2)(r^2-1)=(r^2+2)(r+1)(r-1)=0$ leads to $r=1$ as the only positive solution. So we kno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Range of function $f(x) = \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
Range of function $f(x) = \sqrt{x+27}+\sqrt{13-x}+\sqrt{x}$
$\bf{My\; Try::}$ For $\min$ of $f(x)$
$$\left(\sqrt{13-x}+\sqrt{x}\right)^2=13-x+x+2\sqrt{x}\sqrt{13-x}= 13+2\sqrt{x}\sqrt{13-x}\geq 13$$
Now $$\sqrt{x+27} + \sqrt{13-x}+\sqrt{x} \geq \sqrt{27} + \... | Hint:by Cauchy-Schwarz inequality $$121=[(x+27)+3(13-x)+2x][1+\dfrac{1}{3}+\dfrac{1}{2}]\ge f^2(x)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1806945",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\int_{0}^{1}{\ln(x) \over 2-x}\mathrm dx={\ln^2(2)-\zeta(2)\over 2}$ $$I=\int_{0}^{1}{\ln(x) \over 2-x}\,\mathrm{d}x={\ln^2(2)-\zeta(2)\over 2}$$
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}$$
Using binomial series here
$$(2-x)^{-1}={1\over2}\left(1-{x\over 2}\right)^{-1}={1\over2}+{x\over 4}+{x^2\o... | On the path of user1952009,
Perform the change of variable $y=1-x$,
$\displaystyle I=\int_0^1 \dfrac{\ln(1-x)}{1+x}dx$
Perform the change of variable $y=\dfrac{1-x}{1+x}$,
$\begin{align}
I&=\int_0^1 \dfrac{\ln\left(\tfrac{2x}{1+x}\right)}{1+x}dx\\
&=\int_0^1 \dfrac{\ln 2}{1+x}dx+\int_0^1 \dfrac{\ln x}{1+x}dx-\int_0^1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Extreme values of $\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}$ Let $a,b,c$ be side lengths of a triangle. What are the minimum and maximum of $$\frac{(a+b+c)(2ab+2bc+2ca-a^2-b^2-c^2)}{abc}?$$
When $a=b=c$, the value is $9$. In addition, we can write $a=x+y,b=y+z,c=z+x$ since they are side lengths of a triangle. The e... | We will prove that
$$8 \leqslant \frac{8(x+y+z)(xy+yz+zx)}{(x+y)(y+z)(z+x)} \leqslant 9$$
For the Left Hand Side:
$$(x+y+z)(xy+yz+zx) - (x+y)(y+z)(z+x)=xyz \geqslant 0$$
For the Right Hand Side
\begin{align*}
\ & 9(x+y)(y+z)(z+x)-8(x+y+z)(xy+yz+zx)
\\ &=x^2y+x^2z+xy^2+xz^2+y^2z+yz^2-6xyz
\\ &\geqslant 0
\end{align*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1807729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find all pairs (x, y) of real numbers such that $16^{x^2+y} + 16^{x+y^2}=1$ Find all pairs (x, y) of real numbers such that
$$16^{x^2+y} + 16^{x+y^2}=1$$
| A very nice problem!
By application of inequality $a+b\ge2\sqrt{ab}$ we have
$$16^{x^2+y}+16^{x+y^2}\ge 2\sqrt{16^{x^2+y}\cdot 16^{x+y^2}}=2\cdot 4^{x^2+x+y^2+y}$$
Since $t^2+t\ge-\frac{1}{4}$ holds for all $t\in\mathbb{R}$, for all $x,y\in\mathbb{R}$ we have
$$16^{x^2+y}+16^{x+y^2}\ge2\cdot 4^{x^2+x+y^2+y}\ge 2\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $|2x-1| -|x+5| = 3$
Problem : Solve the equation $|2x-1| - |x+5| = 3$
In my attempt to solve the problem, I only manage to get one of the solutions.
Attempted Solution
$$\begin{equation}
\begin{split}
|x|-|y| & \leq |x-y| \\
\implies |2x-1| - |x+5| & \leq |(2x-1)-(x+5)| \\
\implies 3 &\leq |x-6| \\... | Remember that $|x|=\begin{cases}x&\text{if}~x\geq 0\\ -x&\text{if}~x<0\end{cases}$
Now, we attempt to rewrite the original expression without absolute value signs. To do so, we ask ourselves when do we need to change the sign for each? On what regions will we need to change the sign of $|2x-1|$? The sign of $|x+5|$?... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1808972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Proof of the second symmetric derivative Prove that if $f''(a)$ exists, then $$f''(a)=\lim_{h\to0}\frac{f(a+h)+f(a-h)-2f(a)}{h^{2}}.$$
I really have no idea on this one. Am I supposed to apply the mean value theorem?
| Let us make the problem more general trying to approximate the second derivative based on three points $x+ah$, $x+bh$, $x+ch$.
By Taylor, we have $$f(x+kh)=f(x)+h k f'(x)+\frac{1}{2} h^2 k^2 f''(x)+\frac{1}{6} h^3 k^3
f^{(3)}(x)+\frac{1}{24} h^4 k^4
f^{(4)}(x)+O\left(h^5\right)$$ Now, consider $$F=Af(x+ah)+Bf(x+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Decomposition of a rational function in partial fractions I am trying to decompose the following rational function: $\dfrac{1}{(x^2-1)^2}$ in partial fractions (in order to untegrate it later).
I have notices that $(x^2-1)^2 = (x+1)^2(x-1)^2$
Therefore $\exists A, B, C, D$ s.t: $\dfrac{1}{(x^2-1)^2} = \dfrac{1}{(x-1)^... | Hint: substitute $x=i $ to get two additional equations one for the real part and one for the imaginary part
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1809467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Intersection of plane How would I do this question.
Find a plane that contains the point A(3,1,−1) and touches the cylinder with radius 3 whose axis is the line p : x = 0, y = z.
It was on a test I did a few days ago and has been bugging me since as I don't even know where to approach this from. Any help would be much ... | Assume the plane is $\pi:ax+by+cz+d=0$. The vector $(0,1,1)$ is on the cylinder axis so it is perpendicular to the normal of the plane:
$$(0,1,1)\cdot(a,b,c)=0\implies c=-b$$
Also, $A(3,1,-1)$ is on the plane so: $$3a+b-c+d=0 \implies b=-\frac{1}{2}d-\frac{3}{2}a$$
We know $(0,0,0)$ is on the axis of the cylinder so th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1810979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrating the following $\int \sqrt{\tan x+1}\,dx$
Question: Integrate the following, $$\int\sqrt{\tan x+1}\;dx.$$
Wolfram Alpha returns a non-elementary answer. Can someone please spot the mistake I have made here:
First consider this integral:
$$\int \frac{1}{(x+1)\sqrt{x+3}} \, dx = -\sqrt{2}\tanh^{-1}\frac{\sq... | This integral is not non-elementary. Following from @Harry Peter's hint, we have
$$ \begin{align} \frac{2u^2}{u^4 - 2u^2 + 2} &= \frac{2}{u^2 + \dfrac{2}{u^2} - 2} \\ &= \frac{1 - \dfrac{\sqrt{2}}{u^2}}{\left(u+\dfrac{\sqrt{2}}{u}\right)^2-2\sqrt{2}-2} + \frac{1 + \dfrac{\sqrt{2}}{u^2}}{\left(u - \dfrac{\sqrt{2}}{u}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1812219",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
vector of eigenvalues is an eigenvector When is it the case that the vector $\begin{bmatrix} \lambda_1 \\ \lambda_2 \\ ... \end{bmatrix}$ of eigenvalues of a matrix is in fact an eigenvector of that matrix?
| To make our lives easier, let's try $2\times 2$:
$$\left \{ \begin{align*}
\begin{pmatrix} a & b \\ c & d \end{pmatrix}
\begin{pmatrix} u \\ v \end{pmatrix} = u
\begin{pmatrix} u \\ v \end{pmatrix} \\[5pt]
\begin{pmatrix} a & b \\ c & d \end{pmatrix}
\begin{pmatrix} v \\ u \end{pmatrix} = v
\begin{pmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to prove $|a_{2n}-a_{n}|<\frac{10}{27}$ Let sequence $$a_{1}=1,a_{n+1}=\dfrac{2}{2a_{n}+1}$$
show that
$$\left|a_{2n}-a_{n}\right|<\dfrac{10}{27}$$
and the constant $\frac{10}{27}$ A smaller number instead
$$a_{n+1}+\dfrac{1+\sqrt{17}}{4}=2(1+\sqrt{17})\cdot\dfrac{a_{n}+\frac{1+\sqrt{17}}{4}}{4(2a_{n}+1)}\tag{1}$$... | The function $f(x)=\frac{1}{x+1/2}$ maps $[2/3,1]$ to $[2/3,6/7] \subset [2/3,1]$, since $1/3 < 10/27$ the result follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1814625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the limit $\lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$
Evaluate the limit $\displaystyle \lim_{n\rightarrow \infty}\frac{1}{\sqrt{n}}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\ldots+\frac{1}{\sqrt{n}}\right)$, without using a ... | By Cesaro-Stolz
$$\ldots = \lim_{n \to \infty} \frac{\sum_{k=1}^{n+1} \frac{1}{\sqrt{k}}- \sum_{k=1}^{n} \frac{1}{\sqrt{k}} }{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}}{\sqrt{n+1}-\sqrt{n}} = \lim_{n \to \infty} \frac{\frac{1}{\sqrt{n+1}}(\sqrt{n+1} + \sqrt{n})}{n+1 - n} = 2 $$
Also, there i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 3,
"answer_id": 2
} |
Suppose $-a\sin(s) - b\cos(s) = 0$, then $a^2 + b^2 = 1$?
Suppose $-a\sin(s) - b\cos(s) = 0$ and given that $a^2 + b^2 \le 1$, then $a^2 + b^2 = 1$?
I am having trouble getting the above identity. I vaguely recall that
$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$
where $$\tan(\theta) = \frac{b}{a}$$
But I st... | Let us see the identity you vaguely recall:
$$a\sin(s) + b\cos(s)=\sqrt{a^2+b^2}\sin(s+\theta)$$
with $\tan\theta=\frac ba$. To get it, we factor out the square root, then set $\sin\theta=\frac{b}{\sqrt{a^2+b^2}}$ and $\cos\theta=\frac{a}{\sqrt{a^2+b^2}}$, which is licit since the squares of those numbers sum to 1. Now... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1815474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluate $\int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx$
\begin{equation}
\int_0^\infty\frac{1-e^{-x}(1+x )}{x(e^{x}-1)(e^{x}+e^{-x})}dx
\end{equation}
My colleague got this problem from his friend but he didn't know the answer so he asked my help. Unfortunately, after hours of tired effort I was una... | I answer my own OP instead of improving it as a proof to user @mickep that I did give a try to this problem but I didn't want to post some useless efforts like "I tried substitution $x=\tan y$ then I failed... miserably". I consider putting this kind of effort is a a complete joke as shown in some posts with integratio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1816901",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 5,
"answer_id": 4
} |
Integrate $\int \frac {x^{2}} {\sqrt {x^{2}-16}}dx$ $\displaystyle\int \dfrac {x^{2}} {\sqrt {x^{2}-16}}dx$
Effort 1:
Let be $x=4\sec u$
$dx=4.\sin u.\sec^2u.du$
Then integral;
$\displaystyle\int \dfrac {\sec^2u \; .4.\sin u.\sec^2u.du} {\sqrt {16\sec^2u-16}}=\displaystyle\int \sec^3.du$
After I didn't nothing.
Effort ... | Note that
\begin{align}
\int \frac{x^2}{\sqrt{x^2-16}} dx &= \int \frac{(x^2-16)+16}{\sqrt{x^2-16}}dx \\
&=\int \sqrt{x^2-16}dx + 16 \int \frac{dx}{\sqrt{x^2-16}} \\
&= \sqrt{x^2-16} x - \int \frac{x^2}{\sqrt{x^2-16}}dx+ 16 \int \frac{dx}{\sqrt{x^2-16}}dx
\end{align}
Hence
$$\int \frac{x^2}{\sqrt{x^2-16}} = \frac{1}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1817677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
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Sum of combinatorics sequence $\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1}$ I need to find sum like $$\binom{n}{1} + \binom{n}{3} +\cdots+ \binom{n}{n-1},\qquad \text{ for even } n$$
Example:
Find the sum of $$\binom{20}{1} + \binom{20}{3} +\cdots+ \binom{20}{19}=\ ?$$
| Hint:
Consider the difference between:
$$ \sum_{k=0}^{20}\binom{20}{k}1^k = (1+1)^{20}\quad\text{and}\quad \sum_{k=0}^{20}\binom{20}{k}(-1)^k = (1-1)^{20}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1818946",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
What is the meaning of $X^{2}$? Can you let me know the meaning of $X^{2}$ where X is a random variable? (Please, give me an example in real life and explain the meaning of $X^{2}$?)
| Here's a simple example. Suppose I give you an unusual six-sided die. It's not loaded, so any of the six sides is equally likely to be rolled, but the faces are numbered $$1, 1, 1, 2, 2, 3.$$ So, for any given roll of the die, you assign the random variable $X$ to the outcome you observe. Then you would say $$\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find $3^{333} + 7^{777}\pmod{ 50}$ As title say, I need to find remainder of these to numbers. I know that here is plenty of similar questions, but non of these gives me right explanation. I always get stuck at some point (mostly right at the beginning) and don't have idea how to start.
Thanks in advance.
| First, let us compute $\varphi(50)=\varphi(2\cdot 25)=\varphi(2)\varphi(25)=20$. By Euler's theorem, if $a$ is relatively prime to 50, $a^{20}\equiv 1 \pmod{50}$. Therefore
$$3^{333}+7^{777}\equiv 3^{16*20+13}+7^{38*20+17}=1^{16}*3^{13}+1^{38}*7^{17}\pmod{50}.$$
Thus, we have reduced the problem to computing $3^{13}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1819423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Solutions to $\frac1{\lfloor x\rfloor}+\frac1{\lfloor 2x\rfloor}=\{x\}+\frac13$
Find all solutions to $$\dfrac{1}{\lfloor x\rfloor}+\dfrac{1}{\lfloor 2x\rfloor}=\{x\}+\dfrac{1}{3}$$
$$$$
Unfortunately I have no idea as to how to go about this. On rearranging, I got $$3\lfloor 2x\rfloor = 3\lfloor x\rfloor\{x\}-2\lflo... | I decided to do @mathlove's case $2$.
CASE $2$
We see that $n \gt 0$.
$$\dfrac{1}{n} + \dfrac{1}{2n + 1}= \delta + \frac 13$$
$$\dfrac{3n+1}{2n^2+n} - \dfrac 13 = \delta$$
$$ \delta = \frac{-2n^2 + 8n + 3}{6n^2 + 3n}$$
\begin{align}
0 \le \frac{-2n^2 + 8n + 3}{6n^2 + 3n} \lt 1 \\
0 \le -2n^2 + 8n + 3 \lt 6n^2 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 4
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$25$ men are employed to do a work $25$ men are employed to do a work, which they could finish it in $20$ days but the drop off by $5$ men at the end of every $10$ days. In what time will the work be completed?
My Attempt
In $20$ days, $25$ men can do $1$ work.
In $1$ day, $25$ men can do $\frac {1}{20}$ work.
In $1$ ... | As you've said in one day a man can complete $\frac{1}{500}$ of the job.
The amount of work done every 10 days is then given by $\frac{1}{500}\times10\times(30-5t)$ where $t$ is how many lots of 10 days have passed (starting from $t=1$).
Adding this from $t=1$ to $t=n$ gives:
$$\sum_{t=1}^n\frac{1}{500}\times10(30-5t)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1821645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $9 ≥ 4x + 1$, which inequality represents the possible range of values of $12x + 3?$ If $9 ≥ 4x + 1$, which inequality represents the possible range of values of $12x + 3$?
I've been trying to do SAT prep, and I came across this question. It allowed me to show an explanation and it still didn't make any sense to me.... | $\begin{array}{llr}12x+3 &= 1\cdot (12x + 3)&\text{due to properties of 1}\\
&=3\cdot\frac{1}{3}\cdot(12x+3)&\text{by replacing 1 with}~3\cdot\frac{1}{3}\\
&=3\cdot(\frac{1}{3}\cdot 12x + \frac{1}{3}\cdot 3)&\text{by distributivity of multiplication over addition}\\
&=3\cdot(4x+1)&\text{by evaluating}~\frac{1}{3}\cdot ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1824989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Why does an argument similiar to $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$ show that $2+4+8+...=-2$ See how to prove $\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...=1$
$x=\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
$2x=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...$
Then:
$x=1$
Now I use the same argument to prove $2+4+8+...=-2$
$x... | You can't subtract divergent series.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1826986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Minimizing $\cot^2 A +\cot^2 B + \cot^2 C$ for $A+B+C=\pi$
If $A + B + C = \pi$, then find the minimum value of $\cot^2 A +\cot^2 B + \cot^2 C$.
I don't know how to solve it. And can you please mention the used formulas first.
What I can see is that if one of the angles $A$, $B$, $C$ is small, then the value $\cot^2A... | Put $x=\cot A,y=\cot B$. Using the standard formulae we have $z=\cot C=-\cot(A+B)=\frac{1-xy}{x+y}$. So we want to minimize $$f(x,y)=\frac{(x^2+y^2)(x+y)^2+(xy-1)^2}{(x+y)^2}$$ where $x,y$ can take any real values.
If we want to minimize $(x^2+y^2)(x+y)^2+x^2y^2-2xy+1$ subject to $x+y=k$, then using Lagrange multiplier... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1827361",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Area of the triangle formed by circumcenter, incenter and orthocenter Lets say we have $\triangle$$ABC$ having $O,I,H$ as its circumcenter, incenter and orthocenter. How can I go on finding the area of the $\triangle$$HOI$.
I thought of doing the question using the distance (length) between $HO$,$HI$ and $OI$ and then ... | Given any triangle $\triangle ABC$, we will
abuse notation and use the same letter to represent
both a vertex and the angle at that vertex. Let
*
*$a, b, c$ be the side lengths $|BC|$, $|CA|$ and $|AB|$.
*$L = a + b + c$ be the perimeter.
*$c_X, s_X, t_X$ be $\cos X, \sin X, \tan X$ for any angle $X \in \{ A, B, C... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1828297",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Proof from Matrix I have tried upto where I could. please anyone help me to complete my proof..
Help much appreciated
| I therefore ends your calculations:
$\left(
\begin{array}{cc}
1 & \frac{\sin \frac \alpha 2}{\cos \frac \alpha 2} \\
-\frac{\sin \frac \alpha 2}{\cos \frac \alpha 2} & 1
\end{array}
\right) \left(
\begin{array}{cc}
\cos \alpha & -\sin \alpha \\
\sin \alpha & \cos \alpha
\end{array}
\right) =\left(
\begin{array}{cc}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1829422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to arrive at Ramanujan's nested radicals? Ramanujan found that $\sqrt[3]{\cos\left(\frac {2\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {4\pi}{7}\right)}+\sqrt[3]{\cos\left(\frac {8\pi}{7}\right)}=\sqrt[3]{\frac {1}{2}\left(5-3\sqrt[3]{7}\right)}$ on the last page of Ramanujan's second notebook.
He also found that $\sq... | Those identities are not difficult to prove once established. For instance, $\alpha=\cos\left(\frac{2\pi}{7}\right),\beta=\cos\left(\frac{4\pi}{7}\right),\gamma=\cos\left(\frac{6\pi}{7}\right)=\cos\left(\frac{8\pi}{7}\right)$ are algebraic conjugates, roots of the polynomial:
$$ p(x) = 8x^3+4x^2-4x-1.$$
If follows that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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how to solve $x^{113}\equiv 2 \pmod{143}$ I need to solve $x^{113} \equiv 2 \pmod{143}$
$$143 = 13 \times 11$$
I know that it equals to $x^{113}\equiv 2 \pmod{13}$ and $x^{113}\equiv 2 \pmod{11}$
By Fermat I got
1) $x^{5} \equiv 2 \pmod{13}$
2) $x^{3} \equiv 2 \pmod{11}$
Now I'm stuck..
| Brute-force approach:
We want to find an $x$ such that $$x^{113} \equiv 2 \pmod{143}$$
We split $\mathbb{Z}_{143}$ into $\mathbb{Z}_{13} \times \mathbb{Z}_{11}$ since $13\cdot 11 = 143$ and $11,13$ are prime. Then, we compute the result of the congruences in $\pmod{13}$ and $\pmod{11}$, that is we solve
$$x^{113} \eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1830303",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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When is $\sum_{i=1}^n a_i^{-2}=1$?
For which natural numbers $n$ do there exist $n$ natural numbers $a_i\ (1\le i\le n)$ such that $\displaystyle\sum_{i=1}^n a_i^{-2}=1$?
I didn't see an easy way of solving this. There is a solution for $n=1$ since we just take $a_i = 1$ and a solution for $n=4$ if we take $a_i = 2$.... | If $n$ works, then $n+3$ and $n+8$ works.
Proof. Take a term, $a_i$, and replace that with $4$ copies of $2a_i$. Take a term, $a_i$, and replace that with $9$ copies of $3a_i$. Both work.
So, since $1$ works, all numbers $\equiv1\pmod 3$ work.
Since $9$ works ($a_i=3\forall i$), all numbers $\equiv0\pmod 3$ greater tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1832793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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Determine matrix of linear map Linear map is given through:
$\phi\begin{pmatrix} 3 \\ -2 \end{pmatrix} =\begin{pmatrix} -3 \\ -14 \end{pmatrix} $
$\phi\begin{pmatrix} 3 \\ 0 \end{pmatrix} =\begin{pmatrix} -9 \\ -6 \end{pmatrix}$
Determine matrix $A$ linear map.
Here I have solution but I dont understand how to get it.
... | One possibility is "brute force" solution: you know $A\pmatrix{3 \\ -2} = \pmatrix{-3 \\ -14}, A\pmatrix{3 \\ 0} = \pmatrix{-9 \\ -6}$, that's four equations for four unknown elements, solvable in a number of ways. This approach is inelegant and quickly becomes hard for bigger dimensionality, but is perfectly doable.
T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1833369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Evaluation of Irrational Integral
Evaluation of $$\int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx$$
$\bf{My\; Try::}$ Let $$I = \int\frac{x^4}{(1-x^4)^{\frac{3}{2}}}dx = -\frac{1}{4}\int x\cdot \frac{-4x^3}{(1-x^{4})^{\frac{3}{2}}}dx$$
Using Integration by parts, We get
$$I =\frac{x}{2(1-x^4)^{\frac{1}{2}}}-\int\frac{1}{(1-x... | $$
\int\frac{1}{\sqrt{1-x^4}}\,dx=\int\frac{1}{\sqrt{1+x^2}}\frac{1}{\sqrt{1-x^2}}\,dx
$$
Let $u=\arcsin x$. Then $du=1/\sqrt{1-x^2}\,dx$, and we get
$$
\int\frac{1}{\sqrt{1+\sin^2 u}}\,du=F(-1\,|\, u)+C=F(-1\,|\,\arcsin x)+C
$$
where $F$ denotes an elliptic integral of first kind. Also, see wiki for notational variati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Deriving formula from Fourier series: $\frac{\pi^2}{12} = \sum\limits_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$ The equation/formula $$ \frac{\pi^2}{12} = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}}{n^2}$$ is to be derived.
I know that the Fourier expansion of $f(x)=x$ for $x \in (-\pi,\pi)$ is $$f(x)=x=\sum_{n=1}^{\infty} \fra... |
This does not directly answer the question posed in the OP regarding the use of Fourier Series to prove that
$$\frac{\pi^2}{12}=\sum_{n=1}^\infty \frac{(-1)^{n+1}}{n^2} \tag 1$$
But, I thought it might be instructive to present an approach that relies only on the Basel Problem
$$\frac{\pi^2}{6}=\sum_{n=1}^\infty \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834674",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
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If $x= m-m^2-2$ then find $x^4+3x^3+2x^2-11x+6$ where m is a cube root of unity If $$x= m-m^2-2$$ then find $$x^4+3x^3+2x^2-11x+6$$ where $m$ is a cube root of unity.
My try:
Since $ m+ m^2+1=0$ the value of $x$ is $-1$.
Let $f(x)=x^4+3x^3+2x^2-11x+6$
then $ f(-1)=5$ So the answer is $5$.
Am I correct? In my book it sa... | If $m$ is a primitive cube root of unity, $m^2+m+1=0$ is correct, but that just gives
$$ x = m-m^2-2 = -2\pm i\sqrt{3}. $$
Given $p(z)=z^4+3z^3+2z^2-11z+6$, we have:
$$ q(z) = p(z-2) = z^4-5z^3+8z^2-15z+28 $$
and
$$ p(x) = q(\pm i\sqrt{3}) =\color{red}{13}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find a matrix $B$ such that $B^3 = A$
$$A=\begin{pmatrix} 1 & -1 \\ -2 & 1 \end{pmatrix}$$
Find a matrix $B$ such that $B^3$ = A
My attempt:
I found $\lambda_1= 1+{\sqrt 2}$ and $\lambda_2= 1-{\sqrt 2}$
I also found their corresponding eigenvectors $\vec v_1 =\begin{pmatrix} \frac{-\sqrt 2}{2} \\ 1 \end{pmatrix}$ an... | you find a matrix $P=\left(
\begin{array}{cc}
\frac{-\sqrt{2}}2 & \frac{\sqrt{2}}2 \\
1 & 1
\end{array}
\right) $ diagonalizes $ A $ that is
$P^{-1}AP=\left(\begin{array}{cc}
1+\sqrt{2} & 0 \\
0 & 1-\sqrt{2}
\end{array}\right) $,
then we look
for a simple matrix $ M =PBP^{-1} $ such that it is diagonal and $
M^3 = D$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1834826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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How to sketch the region on the complex plane? I am going through a basic course on complex analysis. I have a problem in understanding the following.
E $\subset\mathbb{C}$ is defined as $$E := \{z\in\mathbb{C}:\vert z+i \vert = 2\vert z\vert \}$$ I want to know if this set is connected, closed, bounded but I do not kn... | I would write $z=x+\mathrm i y$ and then see what equation I got.
\begin{eqnarray*}
|z+\mathrm i| &=& 2|z| \\ \\
|(x+\mathrm iy)+\mathrm i| &=& 2|x+\mathrm iy| \\ \\
|x+(1+y)\mathrm i|&=& 2|x+\mathrm iy| \\ \\
\sqrt{x^2+(1+y)^2} &=& 2\sqrt{x^2+y^2} \\ \\
x^2+(1+y)^2 &=& 4(x^2+y^2) \\ \\
x^2+1+2y+y^2 &=& 4x^2+4y^2 \\ \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1835912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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When is $2^m3^n +1$ the square of some integer?
Find all pairs of natural numbers $(m, n)$ for which $2^m3^n +1$ is the square of some integer.
The powers of $2$ modulo $10$ cycle as $2,4,8,6,\ldots$ and the powers of $3$ as $3,9,7,1,\ldots$. The square of am integer needs to end in $0,1,4,5,6,$ or $9$. Thus, $2^m3^n... | $2^m3^n+1=k^2$ for some $k\in\mathbb Z^+$ is true if and only if $2^m3^n=(k+1)(k-1)$.
Natural numbers may or may not include $0$ (see Wikipedia or MathWorld). I'll assume $0\in\mathbb N$.
By Euclidean algorithm: $$\gcd(k+1,k-1)$$
$$=\gcd((k+1)-(k-1),k-1)$$
$$=\gcd(2,k-1)$$
$$=\gcd(2,k+1)$$
If $m=0$, then $\gcd(k+1,k-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Roots of $y=x^3+x^2-6x-7$ I'm wondering if there is a mathematical way of finding the roots of $y=x^3+x^2-6x-7$?
Supposedly, the roots are $2\cos\left(\frac {4\pi}{19}\right)+2\cos\left(\frac {6\pi}{19}\right)+2\cos\left(\frac {10\pi}{19}\right)$, $2\cos\left(\frac {2\pi}{19}\right)+2\cos\left(\frac {14\pi}{19}\right)+... | By the translation $x\to x-\frac13$, you cancel the square term.
$$(x-\frac13)^3+(x-\frac13)^2-6(x-\frac13)-7=x^3-\frac{19x}3-\frac{133}{27}.$$
Then multiplying by $27$ and replacing $x\to3x$, you get
$$x^3-57x-133=0.$$
Next, setting $x=2\sqrt{19}\cos(\theta)$ (with the aim to let $4\cos^3(\theta)-3\cos(\theta)=\cos(3\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1837628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Prove inequality $\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}>1$
For any $n\ge2, n \in \mathbb N$ prove that
$$\sqrt{\frac{1}n}-\sqrt{\frac{2}n}+\sqrt{\frac{3}n}-\cdots+\sqrt{\frac{4n-3}n}-\sqrt{\frac{4n-2}n}+\sqrt{\frac{4n-1}n}>1$$
My work s... | Another proof is this:
Note that
$$
2 = \sqrt{\frac{4n}{n}} = \frac{1}{\sqrt{n}}\sum_{j=0}^{4n-1}\sqrt{j+1}-\sqrt{j}
$$
where the RHS can be expressed as
$$
\frac{1}{\sqrt{n}}\left(\sum_{j=1}^{2n}(\sqrt{2j}-\sqrt{2j-1})+\sum_{j=1}^{2n-1}(\sqrt{2j-1}-\sqrt{2j-2})\right)
$$
Using that the function $f:(0,+\infty)\to \m... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Let $a, b, c>0$, such that $a+b+c=1$, prove that $\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\ge\frac{9}{4}$ Let $a, b, c>0$, such that $a+b+c=1$, prove that:
$$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}\ge\frac{9}{4}$$
| Hint: $$\frac{a}{(b+c)^2}+\frac{b}{(a+c)^2}+\frac{c}{(a+b)^2}=$$
$$=\frac{a}{(1-a)^2}+\frac{b}{(1-b)^2}+\frac{c}{(1-c)^2}$$
Let $f(t)=\frac t{1-t^2}$
Use Jensen's Inequality
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1841134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Express the function $ f $ without using absolute value signs $\left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|}$? Good evening to everyone:
This is the equation $$ f(x) = \left|\frac{x-2}{x+3}\right|e^{\left|x-2\right|} $$ What I've tried is: $$ \frac{x-2}{x+3}\ge 0 => x-2 \ge 0 => x \ge 2$$ Then $$ \frac{-x+2}{-x-3} ... | $$\frac{x-2}{x+3}\ge 0\stackrel{\text{Mult. by}\;(x+3)^2}\iff (x-2)(x+3)\ge0\;,\;\;x\neq-3\iff$$
$$x<-3\;\;\text{or}\;\;x\ge2$$
and then
$$\left|\frac{x-2}{x+3}\right|e^{|x-2|}=\begin{cases}\frac{x-2}{x+3}e^{x-2},&x\ge2\\{}\\\frac{x-2}{x+3}e^{-(x-2)},&x<-3\end{cases}$$
And thus
$$-3<x<2\implies \left|\frac{x-2}{x+3}\r... | {
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If $x^2+y^2 \equiv 0\pmod{p}$, then $p \equiv 1 \pmod{4}$
Prove that if $x^2+y^2 \equiv 0\pmod{p}$ where $p$ is a prime and $x,y$ are not both divisible by $p$, then $p \equiv 1 \pmod{4}$.
I tried using that $x^2 \equiv -y^2 \pmod{p}$ and conjectured that $-1$ must a quadratic residue modulo $p$, but I am not sure ho... | Since $p \not \mid y$, there exists integer $y'$ such that $yy' \equiv 1 \pmod p$. Then multiply both side of the congruence with $y'$ to get that:
$$(xy')^2 \equiv -(yy')^2 \equiv - 1 \pmod p$$
So hence $-1$ is a quadratic residue modulo $p$. To prove that $p \equiv 1 \pmod 4$ from this note that:
$$1 \equiv x^{p-1} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1843792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Evaluation of $\int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$
Evaluation of $$\int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$$
$\bf{My\; Try:::}$ Let $$I(a) = \int_{0}^{\sqrt{2}-1}\frac{\ln(1+ax^2)}{1+x}dx$$
Now $$I'(a) = \int_{0}^{\sqrt{2}-1}\frac{x^2}{(1+ax^2)(1+x)}dx = \frac{1}{a}\int_{0}^{\sqrt{2}-1}\frac{(1+ax^2)... | Since the number $\sqrt{2}-1$ is invariant under the transformation $x \mapsto \frac{1-x}{1+x}$, it is natural to make that substitution. Doing so, we find:
$$\begin{align} I &= \int_0^{\sqrt{2}-1} \frac{\ln(1+x^2)}{1+x} dx
\\&= \int_{\sqrt{2}-1}^1 \cfrac{\ln(1+x^2)+\ln\left(\frac2{(1+x)^2}\right)}{1+x}dx
\\&=\frac12 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What are the differences between: $\sqrt{(-3)^2}$, $\sqrt{-3^2}$ and $(\sqrt{-3})^2$. First, is $\sqrt{-3}$ is equal to $-3$ or is it imaginary?
What is the difference between:
*
*$\sqrt{(-3)^2}$
*$\sqrt{-3^2}$
*$(\sqrt{-3})^2$
Can I write $(\sqrt{-3})^2 = -3$?
And, given the rule that $\sqrt{a^n}$ is equal to ... | $\sqrt{(-3)^2}=\sqrt{9}=3$,
$\sqrt{-3^2}=\sqrt{-9}=3i$,
$(\sqrt{-3})^2=(\sqrt{3}i)^2=-3$
The rule is true only in the case $a>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1845348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Compute $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta$ I''m stuck in a exercise in complex analysis concerning integration of rational trigonometric functions. I have the solution but I don't understand a specific part. Here it goes:
We want to find $\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\t... | $$I=\int_{0}^{2\pi}\frac{\sin^2\theta}{5+3\cos\theta}\,d\theta=\int_{0}^{\pi}\sin^2(\theta)\left(\frac{1}{5+3\cos\theta}+\frac{1}{5-3\cos\theta}\right)\,d\theta$$
hence:
$$ I = 10 \int_{0}^{\pi}\frac{\sin^2\theta}{25-9\cos^2\theta}\,d\theta = 20\int_{0}^{\pi/2}\frac{\cos^2\theta}{25-9\sin^2\theta}d\theta $$
and by sett... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1846308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$
Find all the numbers $n$ such that $3\cdot 2^n+2\cdot 3^n\equiv 1 \pmod 7$
Attempt:
$\star$ denotes $3\cdot 2^n+2\cdot 3^n$
$$\text{for }n=1:\quad\star\equiv 5\not\equiv 1\\
\text{for }n=2:\quad\star\equiv 5\not\equiv 1\\
\text{for }n=3:\qu... | As $2^3\equiv1,3^3\equiv-1\pmod7$
let us start with $n=3a,3a+1,3a+2$
Case$\#1:$ $n=3a$
$$1\equiv3\cdot2^{3a}+2\cdot3^{3a}\equiv3+2(-1)^a$$
$\iff2(-1)^a\equiv-2\iff(-1)^a\equiv-1$ as $(2,7)=1$
$\implies a$ is odd $=2b+1$(say) $\implies n\equiv3\pmod6$
Case$\#2:$ $n=3a+1$
$$1=3\cdot2^{3a+1}+2\cdot3^{3a+1}\equiv6(1+(-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1849904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Factorization of $x^n + y^n$, what sort of coefficients show up? We know that$$a^2 + b^2 = (a + bi)(a - bi).$$ What are the complete factorizations of $a^3 + b^3$, $a^4 + b^4$, $\ldots$ , $a^k + b^k$, etc.? What sort of coefficients show up?
| You could always write $a^n+b^n = a^n - c^n$ where $c = \xi_n b$, where $\xi_n = e^{\pi i/n}$ (so that $\xi_n^n = -1$). Then fixing $a$, the roots of $a^n - c^n$ are solutions to $a^n = c^n$ and so there are $n$ solutions for $c$; namely, the numbers in the form of $\zeta_n^k a$ where $0\leq k\leq n-1$ and $\zeta = e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1850294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If A and B are positive real numbers and each of the equations: $x^2+ax+2b=0$ and $x^2+2bx+a=0$ has real roots, what is the smallest value of A+B Problem
In the equation:$x^2+ax+2b=0$ and $x^2+2bx+a=0$, we have to figure out the sum of a and b by using the following identity:
$P(x)=Q(x)*D(x)+R(x)$
where P(x) is the equ... | HINT:
We need discriminant $\ge0,$
$\implies a^2\ge8b, 4b^2\ge4a\iff b^2\ge a$
$a^4\ge64a\iff a(a-4)(a^2+4a+4^2)\ge0\iff a(a-4)\ge0$ as $a^2+4a+4^2=(a+2)^2+2^2\ge2^2>0$
$\implies$ either $a\le0$ or $a-4\ge0\iff a\ge4$
But $a$ is positive.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Find all the numbers $n$ such that $\frac{4n-5}{60-12n}$ can't be reduced.
Attempt:
$$\gcd(4n-5,60-12n)=(4n-5,-8n+55)=(4n-5,-4n+50)=(4n-5,45)$$
$$n=1: (4-5,45)=1\quad \checkmark\\
n=2: (3,45)=3\quad \times\\
n=3: (7,45)=1\quad \checkmark\\
n=4... | We have that $\gcd(pq + r,p) = \gcd(p,r)$ and since $12n - 60 = 3(4n-5)+45$, we have that $\frac{4n-5}{60-12n}$ can't be reduced if and only if $\gcd(4n-5,45) = 1$ and since $45=3^2\cdot 5$, this happens if and only if $3\nmid 4n-5$ and $5\nmid 4n-5$.
Now,
$4n-5\equiv 0 \pmod 3\iff 4n \equiv 8 \pmod 3 \iff n\equiv 2\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1852984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find this function, and what method to use? The function is $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}$ and they are asking us to find out what $f(-x)$ is?
| Observe that $f(x-\frac{1}{x})= x^3-\frac{1}{x^3}=(x-\frac{1}{x})^3+3\cdot x\cdot \frac{1}{x}\cdot(x-\frac{1}{x})=(x-\frac{1}{x})^3+3(x-\frac{1}{x})$
Hence, we can say that $$f(z)=z^3+3z$$ where $z$ is any real variable.
So we have that $$\color{red}{f(-x)}=\color{blue}{-x^3-3x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1853364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximize $k=x^2+y^2$ Subject to $x^2-4x+y^2+3=0$ Question
Let $x$ and $y$ be real numbers satisfying the equation $x^2-4x+y^2+3=0$. Find the maximum and minimum values of $x^2+y^2$.
My work
Let $k=x^2+y^2$
Therefore, $x^2-4x+y^2+3=0$ ---> $k-4x+3=0$ .
What do I do next? How do I find an expression in terms of $k$ tha... | $F(x,y,\lambda)=x^2+y^2+\lambda (x^2-4 x+y^2+3)$ is lagrange multiplier. We need to solve system of equations $$F'_x=0$$ $$F'_y=0,$$ $$x^2-4 x+y^2+3=0$$ or $$2x+2\lambda x-4\lambda=0,$$ $$2y+2\lambda y=0,$$ $$x^2-4 x+y^2+3=0.$$ We have $(x,y,\lambda)\in\{(1,0,1),(3,0,-3)\}$. For $\lambda=1$ is $d^2 F>0$ so $(x,y)=(1,0)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1854240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Factorial sum estimate $\sum_{n=m+1}^\infty \frac{1}{n!} \le \frac{1}{m\cdot m!}$ Prove that:
$$\displaystyle \sum_{n=m+1}^\infty \dfrac{1}{n!} \le \dfrac{1}{m\cdot m!}$$
I have tried induction on $m$ but it does not work very well. Any suggestion?
| $$\sum_{n=m+1}^{\infty} \frac{1}{n!}=\frac{1}{m!} \left(\frac{1}{m+1}+\frac{1}{(m+1)(m+2)}+\frac{1}{(m+1)(m+2)(m+3)}+\ldots \right)$$ $$ \leq \frac{1}{m!} \left(\frac{1}{m+1}+\frac{1}{(m+1)^2}+\frac{1}{(m+1)^3}+\ldots \right)=\frac{1}{m!}\frac{1}{m+1}\frac{1}{1-\frac{1}{m+1}}=$$ $$\frac{1}{m!}\frac{1}{m+1-1}=\frac{1}{m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1856779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$ This series represents the sum of reciprocal pentagonal numbers (multiplied by $3$). I got its value from Wolfram Alpha:
$$\sum_{n=1}^\infty \frac{6}{n(3n-1)}=9 \ln 3-\sqrt{3} \pi$$
My attempt: Turn the series into an integral:
$$\sum_{n=1}^\infty \frac{x... | We can do the indefinite integral. Start by the obvious guess
$$\frac{d}{dx} \frac{-\log(1-x^3)}{x} = \frac{\log(1-x^3)}{x^2}+\frac{3x}{1-x^3}$$
Now we have to wipe out that second term.
$$
\frac{3x}{1-x^3}=\frac{1}{1-x}+\frac{x-1}{1+x+x^2} \\
\int\frac{1}{1-x}= -\log(1-x)
$$
leaving the term $\frac{x-1}{1+x+x^2}$ to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Find $x$ in $1!+2!+\ldots+100!\equiv x \pmod{19}$ Here I come from one more (probably again failed) exam. We never did congruence with factorials; there were 3 of 6 problems we never worked on in class and they don't appear anywhere in scripts or advised literature.
First is this one:
Find $x$ in $1!+2!+\ldots+100! \... | $18! \equiv -1 \pmod{19}$ by Wilson's Theorem. Then you can find the modular inverse of $18$ modulo $19$, multiply both sides by it and you will get $17! \equiv 1 \pmod{19}$. You can continue simularly for the big numbers that are left.
Also you can go step by step in the multiplication. For example for $13!$ you can s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1857449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the highest point of intersection
Find the highest point of intersection of the sphere $x^2+y^2+z^2=30$ and the cone $x^2+2y^2-z^2=0$.
Am I supposed to use the Lagrange multiplier for this?
EDIT:
So this is what I've tried...
$z^2=-x^2-y^2+30$ and $z^2=x^2+2y^2$. Then $-x^2-y^2+30 = x^2+2y^2$.
This gives us $3y... | As suggested by John Molokach: solving for $z^2$ yields
$$
2x^2+3y^2=30,
$$
so the intersection of both curves projected in the $x,y$ plane is the ellipse
$$
\begin{cases}
x=\sqrt{15}\cos t\\
y=\sqrt{10}\sin t
\end{cases}
$$
Replacing these equations in the cone yields
$$
z=\sqrt{15+5\sin^2t}
$$
which has maximum $z_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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How do we prove that $4(3\sqrt2-4)=\prod_{n=1}^{\infty}\left({e^{2\pi(2n-1)}-1\over e^{2\pi(2n-1)}+1}\right)^8?$ How do we prove that
$$4(3\sqrt2-4)=\prod_{n=1}^{\infty}\left({e^{2\pi(2n-1)}-1\over e^{2\pi(2n-1)}+1}\right)^8\tag1$$
Rewrite as, to keep it simple
Let $a=e^{2\pi(2n-1)}$
$$4(3\sqrt2-4)=\prod_{n=1}^{\inf... | This one is similar to this answer of your previous question and you should have guessed it. Based on the definition of Ramanujan's class invariants
\begin{align}G_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 + e^{-\pi\sqrt{n}})(1 + e^{-3\pi\sqrt{n}})(1 + e^{-5\pi\sqrt{n}})\cdots\tag{1}\\
g_{n} &= 2^{-1/4}e^{\pi\sqrt{n}/24}(1 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Is it possible to have a power series for arctan(x) centered at 1? My Calculus 2 professor referenced that such a series is impossible, but why? I understand how to properly find the power series of arctan(x) centered at 0.
| Since
$$
\begin{align}
\tan(\arctan(1+x)-\arctan(1))
&=\frac{(1+x)-1}{1+(1+x)\cdot1}\\
&=\frac x{2+x}\tag{1}
\end{align}
$$
we have
$$
\begin{align}
\arctan(1+x)
&=\frac\pi4+\arctan\left(\frac{x}{2+x}\right)\\
&=\frac\pi4+\sum_{k=0}^\infty\frac{(-1)^k}{2k+1}\left(\frac x{2+x}\right)^{2k+1}\tag{2}
\end{align}
$$
Expandi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1858869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Conjectured value of $\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln x}\right)\frac{\mathrm{d}x}{x^2+1}$ I was curious whether this integral has a closed form expression :
$$\int_{0}^{\infty}\left(\frac{x-1}{\ln^2 x}-\frac{1}{\ln
x}\right)\frac{\mathrm{d}x}{x^2+1}$$
The integrand has a singularity at $x=1$,... | Though using the residue method is somewhat straightforward, but not everyone can understand it. So, here is a residue-free method:
Split the integral into two terms where each term is in the interval $0<x<1$ and $1<x<\infty$, then use the substitution $x\mapsto\frac{1}{x}$ to the second term. We will get
$$
\left[\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "34",
"answer_count": 5,
"answer_id": 4
} |
Minimal polynomial and possible Jordan forms
Let $A$ be an $8\times 8$ complex matrix with characteristic polynomial $$p_A(x)=(x-1)^4(x+2)^2(x^2+1)$$ and minimal polynomial $$m_A(x)=(x-1)^2(x+2)^2(x^2+1).$$ Determine all possible Jordan canonical forms of $A$. Also, what is the dimension of the eigenspace for $\lambda... | You identified the blocks and eigenspace dimensions correctly. The blocks for a given form, however, can be permuted. Thus, your first matrix represents 5!/2! = 60 forms, and your second matrix, 6!/2! = 360 forms for a total of 420 forms.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1863819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Proof about Pythagorean triples Show that if $(x,y,z)$ is a Pythagorean triple, then $10\mid xyz$
Proof
First, if $x$, $y$, $z$ are all odd, then so are $x^2$, $y^2$, $z^2$, so $x^2+y^2$ is even, which means that $x^2+y^2 \neq z^2 $. Hence, at least one of $x$, $y$, $z$ is even, so $2\mid xyz$ (clear).
Next, for any... | No problem to give a "theoretical" proof. Just for fun let me work with a table in $\mathbb F_5^2$.
By the quite known parametrization of the Pythagorean triple, $xyz$ is even so we must prove in
$\mathbb F_5$ that
$$xyz=mn(m^2-n^2)(m^2+n^2)=0\qquad (1)$$
If $m=n$ and if $mn=0$ it is clear so we prove $(1)$ with $m\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to find $ \tan \left(\frac{x}{2}\right) $ knowing that $\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $ Good evening to everyone. I don't know how to find $ \tan \left(\frac{x}{2}\right) $ knowing that $$\cos \left(x\right)+\sin \left(x\right)=\frac{7}{5} $$ and x$\in (0,\frac{\pi}{3})$ Here's what I've tried... | \begin{align}
\cos x +\sin x &= \frac 75 \\
\cos^2 x + 2 \cos x \sin x + \sin^2 x &= \frac{49}{25} \\
2 \cos x \sin x &= \frac{24}{25} \\
\hline
\cos^2 x - 2 \cos x \sin x + \sin^2 x &= 1 - 2 \cos x \sin x \\
(\cos x - \sin x)^2 &= \dfrac{1}{25} \\
\cos x - \sin x &= \pm \frac 15 \\
\hline
(\cos x,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1865712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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How to prove by induction that $3^{3n}+1$ is divisible by $3^n+1$ for $(n=1,2,...)$ So this is what I've tried:
Checked the statement for $n=1$ - it's valid.
Assume that $3^{3n}+1=k(3^n+1)$ where $k$ is a whole number (for some n). Proving for $n+1$: $$3^{3n+3}+1=3^33^{3n}+1=3^3(3^{3n}+1)-26=3^3k(3^n+1)-26=3^3k(3^n+1)-... | It's true for $n=1$. Assume it holds for $n$, i.e: $3^{3n} + 1 = k(3^n +1)$ then consider $$3^{3(n+1)} +1 = 27 \cdot 3^{3n} + 1 = 27 (3^{3n} +1) - 26 = 27k(3^n +1) - 26$$
Let's get it into a more amenable form $$\begin{equation}3^{3(n+1)} +1 = 9k3^{n+1} + 27k - 26 = 9k(3^{n+1}+1) + 18k - 26 \end{equation} \tag{1}$$
So ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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If $\sin x + \sin y = 1$ and $\cos x + \cos y = 0$, solve for $x$ and $y$
*
*$\sin x + \sin y = 1$
*$\cos x + \cos y = 0$
Any valid pair of $(x, y)$ is fine, as the restrictions on the board in the image below are obscured.
I got the question from chapter 26 of a comic called Yamada-kun.
How can I solve this equa... | There are identities for sum of sin and cos:
$$\sin(x)+\sin(y) = 2\sin\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right)_.$$
$$\cos(x) + \cos(y) = 2\cos\left(\frac{x+y}{2}\right)\cos\left(\frac{x-y}{2}\right).$$
Using the first equation tells us that $\cos\left(\frac{x-y}{2}\right)\neq 0.$ Therefore by the second... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1866796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Compute $1 \cdot \frac {1}{2} + 2 \cdot \frac {1}{4} + 3 \cdot \frac {1}{8} + \cdots + n \cdot \frac {1}{2^n} + \cdots $ I have tried to compute the first few terms to try to find a pattern but I got
$$\frac{1}{2}+\frac{1}{2}+\frac{3}{8}+\frac{4}{16}+\frac{5}{32}+\frac{6}{64}$$
but I still don't see any obvious pattern... | First consider the partial sum by showing
$$ \sum_{n=1}^k \frac{n}{2^n} = 2^{-k}(-k+2^{k+1}-2). $$
Now $k\to \infty$ gives the result $2$.
Edit: Proof of the partial sum by induction
$k=1$:
$$\sum_{n=1}^1 \frac{1}{2} = 1/2 = 2^{-1}(-1+2^{2}-2) \quad \checkmark$$
Let $\sum_{n=1}^k \frac{n}{2^n} = 2^{-k}(-k+2^{k+1}-2)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1867561",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 4
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If $ax^2 +bxy+cy^2+5x-2y+3$ divided by $x-y+1$ has remainder $0$, determine $a$, $b$, and $c$.
If $ax^2 +bxy+cy^2+5x-2y+3$ divided by $x-y+1$ has remainder $0$, determine $a$, $b$, and $c$.
I do not know how to approach this problem and would appreciate advice how to proceed.
| Since there is no remainder the original equation will be the product of $x-y+1$ with another unknown equation. In order to create the $x^2$, $y^2$, and $xy$ terms we expect the form to be $dx+ey+f$.
$$ax^2+bxy+cy^2+5x−2y+3 = (x−y+1)(dx+ey+f)$$
$$ax^2+bxy+cy^2+5x−2y+3 = dx^2+(e-d)xy-ey^2+(d+f)x+(e-f)y+f$$
working rig... | {
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"url": "https://math.stackexchange.com/questions/1868114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate the expression $\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$ $$\sqrt{6-2\sqrt5} + \sqrt{6+2\sqrt5}$$
Can anyone tell me the formula to this expression.
I tried to solve in by adding the two expression together and get $\sqrt{12}$ but as I insert each expression separately in calculator the answer is above $\sqrt{12}$.... | Hint. Observe that
$$
(\sqrt{5}-1)^2=6-2\sqrt{5},\quad (\sqrt{5}+1)^2=6+2\sqrt{5}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871377",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Simplify the expression $\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k}{k}$ Simplify the expression $\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots +\binom{n+k}{k}$
My attempt: Using the formula $\binom{n+1}{k}=\binom{n}{k}+\binom{n}{k-1}$
$\binom{n}{0}+\binom{n+1}{1}+\binom{n+2}{2}+\cdots \binom{n+k... | First show that $\large{n\choose r}={{n-1}\choose {r-1}}+{{n-1}\choose r}$,
from which we get $\large{{n+r+1}\choose r}={{n+r}\choose r}+{{n+r}\choose {r-1}}$
By the same rule, $\large{{n+r}\choose {r-1}}={{n+r-1}\choose {r-1}}+{{n+r-1}\choose {r-2}}$
$\large{{n+r-1}\choose {r-2}}={{n+r-2}\choose {r-2}}+{{n+r-3}\choose... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1871953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to formally use Taylor expansions for $n$th derivatives and generating functions? When deriving Catalan numbers, the generating function takes on this form:
$$C(x) = \frac{1}{2} (1 - \sqrt{1-4x}) = \frac{1}{2} (1 - f(x))$$
where $f(x) = \sqrt{1-4x}$
How does one formally show what it evaluates to? I can do it somew... | May be, you could use the generalized binomial expansion $$(1+x)^a=\sum_{n=0}^\infty \binom{a}{n} x^n$$ which makes $$f(x)=\sqrt{1-4x}=\sum_{n=0}^\infty (-1)^n\binom{\frac 12}{n} (4x)^n=1+\sum_{n=1}^\infty (-1)^n\binom{\frac 12}{n} (4x)^n$$ and then $$C(x) = \frac{1}{2} (1 - \sqrt{1-4x})=\frac{1}{2}\sum_{n=1}^\infty (-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1872322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Which answer is correct? Finding the limit of a radical as $x$ approaches infinity. When I take $$\lim_{x \to -∞} \sqrt{x^2+7x}+x,$$ I multiply by the conjugate over the conjugate to get
$$\lim_{x \to -∞}\frac{7x}{\sqrt{x^2+7x}-x},$$ and multiply by either $\frac{\frac{1}{x}}{\frac{1}{x}}$ or $\frac{\frac{1}{-x}}{\frac... | Let's see what happens to $$f(x) = \sqrt{x^2 + 7x} + x$$ when $x$ takes on actual negative values. When $x = -100$, we get $$f(-100) = 10 \sqrt{93} - 100 \approx -3.56349.$$ When $x = -10000$, we get $$f(-10000) = 100 \sqrt{9993} - 10000 \approx -3.50061.$$ So, from a numerical standpoint, this seems to suggest that... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Minimum value of $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$ If $f(x) = \sqrt{x^2 + (1-x)^2} + \sqrt{(1-x)^2 +(1+x)^2}$
Find the minimum value of the function
I tried using the AMGM inequality and differentiation but didn't know how to solve it any ideas?
This is from a math competition. ( I would like to ... |
Let $A=(0,1)$, $B=(1,1)$, and $C=(-x,x)$ as in the picture. Then
$$AC=\sqrt{x^2+(1-x)^2}, BC=\sqrt{(1+x)^2+(1-x)^2}.$$
Let $D$ be symmetric to $A$ about $x+y=0$ (trajectory of $C$). Apparently
$$AC+BC\ge BD=AE+BE.$$
Minimal is attained at $C=E$. I leave you figure the coordinates of $E$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is it possible to $\int \sqrt{\cot x}$ by hand $$\int \sqrt{\cot x}{dx}$$
$$\int \sqrt{\frac{\cos x}{\sin x}}{dx} $$
Using half angle formula
$$\int \sqrt{\frac{1-\tan^2 \frac{x}{2}}{2\tan \frac{x}{2}}}{dx}$$
But I am not getting any lead from here .I think it is not possible to integrate $\sqrt{\cot x}$ by hand .
I c... | Let $$I = \int \sqrt{\cot x}dx = \frac{1}{2}\int 2\sqrt{\cot x}dx = \frac{1}{2}\int \left[\left(\sqrt{\cot x}+\sqrt{\tan x}\right)+\left(\sqrt{\cot x}-\sqrt{\tan x}\right)\right]dx$$
Now Let $$J = \int \left(\sqrt{\cot x}+\sqrt{\tan x}\right)dx = \sqrt{2}\int\frac{\cos x+\sin x}{\sqrt{\sin 2x}}dx = \sqrt{2}\int\frac{(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1873601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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