Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Finding the argument $\theta$ of a complex number I want to find the Argument of $z = -\sqrt{2 - \sqrt{3}} + i\sqrt{2 + \sqrt{3}}$ where $z$ is a complex number of the form $z = a + bi$.
I find that the modulus is $2$, but am having trouble simplifying $\theta = \arctan\left(\frac{\sqrt{2 + \sqrt{3}}}{-\sqrt{2 - \sqrt{... | Note that:
$$
\dfrac{\sqrt{2+\sqrt{3}}}{-\sqrt{2-\sqrt{3}}}=-\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}=-\sqrt{\dfrac{(2+\sqrt{3})(2+\sqrt{3})}{(2-\sqrt{3})(2+\sqrt{3})}}=-(2+\sqrt{3})
$$
since $2+\sqrt{3}=\tan \left(\dfrac{5 \pi}{12}\right)$, and $\theta$ must be in the second quadrant we have $\theta =\left(\dfrac{7 \pi}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1130834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Does $\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} =0$? $$\lim_{x\to0}\left(\frac{\sin(x)}{x}\right)^{\frac{1}{x^2}} \stackrel{?}{=} 0$$
My calculations (usage of L'Hôpital's rule will be denoted by L under the equal sign =):
(Sorry for the small font, but you can zoom in to see better with Firefox)
$$
\b... | Using Taylor series:
$$\frac{\sin(x)}{x}=1- \frac{1}{3!}x^2+o(x^2).$$
So using this formula we get:
$$
\ln \left( \frac{\sin(x)}{x}\right)=\ln \left(1- \frac{1}{3!}x^2+o(x^2) \right).
$$
Now we can use the Taylor expansion for the $\ln$ function so we get:
$$
\ln \left(1- \frac{1}{3!}x^2+o(x^2) \right)= - \frac{1}{3!}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1131069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Evaluating $(1+\cos\frac{\pi}{8})(1+\cos\frac{3\pi}{8}) (1+\cos\frac{5\pi}{8})(1+\cos\frac{7\pi}{8})$
How to find the value of $$\left(1+\cos\frac{\pi}{8}\right)\left(1+\cos\frac{3\pi}{8}\right) \left(1+\cos\frac{5\pi}{8}\right)\left(1+\cos\frac{7\pi}{8}\right)$$
I used: $ 1+ \cos \theta=2\sin^2(\theta/2)$ to get $$... | Here is a solution using complex numbers: Let $\omega = \exp(i\pi/2n)$. Then
\begin{align*}
\prod_{k=1}^{n} \left(1 + \cos \left( \tfrac{2k-1}{2n}\pi \right) \right)
&= \frac{1}{2^{n}} \prod_{k=1}^{n} \left(1 + 2\Re(\omega^{2k-1})+ 1 \right) \\
&= \frac{1}{2^{n}} \prod_{k=1}^{n} (1 + \omega^{2k-1})(1 + \omega^{-(2k-1)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1133590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
limit of $ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $ Hello I am trying to find the limit of
$ \lim\limits_{x \to ∞} \frac {x(x+1)^{x+1}}{(x+2)^{x+2}} $
I've tried applying L'H rule but it ends up getting really messy.
The answer is $ \frac {1}{e} $ so I assume it must simplify into something which I ca... | [\begin{gathered}
\mathop {\lim }\limits_{x \to \infty } \frac{{x{{\left( {x + 1} \right)}^{x + 1}}}}
{{{{\left( {x + 2} \right)}^{x + 2}}}} = \mathop {\lim }\limits_{x \to \infty } \frac{x}
{{x + 2}}.\mathop {\lim }\limits_{x \to \infty } {\left( {\frac{{x + 1}}
{{x + 2}}} \right)^{x + 1}} = 1.\mathop {\lim }\limits... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1134726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Partial differential equation (first order) I don't have ideas to solve the following PDE of the 1st order
$$
(x^2 - y^2 + 1)u_{x} + 2xyu_{y} = 0
$$
Could you give me a hint ?
Thanks,
R.
| $(x^2-y^2+1)u_x+2xyu_y=0$
$\dfrac{x^2-y^2+1}{2xy}u_x+u_y=0$
Follow the method in http://en.wikipedia.org/wiki/Method_of_characteristics#Example:
$\dfrac{dy}{dt}=1$ , letting $y(0)=0$ , we have $y=t$
$\dfrac{dx}{dt}=\dfrac{x^2-y^2+1}{2xy}=\dfrac{x^2-t^2+1}{2xt}$
$2x\dfrac{dx}{dt}=\dfrac{x^2}{t}-\dfrac{t^2-1}{t}$
$\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1135346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
differentiate $(2x^3 + 3x)(x − 2)(x + 4)$ So, I'm really stuck on this problem.
Differentiate $(2x^3 + 3x)(x − 2)(x + 4)$
This is what I come up with $10x^4+16x^3+39x^2+6x-18$.
But, the answer in the book has $16x^4$ as the leading term
Here's my work:
$(2x^3+3x)d/dx(x^2+2x-8)+(x^2+2x-8)d/dx(2x^3+3x)$
$(2x^3+3x)(2x+2)+... | I would do it simply by expanding and then repeatedly using the Power Rule (using the Product Rule would be overkill I think):
$$
(2x^3+3x)(x-2)(x+4) = 2x^5+4x^4-13x^3+6x^2-24x.
$$
Now differentiate with respect to $x$, obtaining
$$
\frac{d}{dx}(2x^5+4x^4-13x^3+6x^2-24x) = 10x^4+16x^3-39x^2+12x-24.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138666",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Trigonometrical limit $$\lim\limits_{x \to 0} \frac{x^4\cos(\frac{1}{x})} {\sin^3x},$$
I have a problem with this limit. I tried to use L'Hôpital's rule but it is not effective. Please help.
| $$\lim\limits_{x → 0} \frac{x^4 \cos \left(x^{-1}\right)}{\sin ^3\left(x\right)} \times \frac{x^{-3}}{x^{-3}}$$
$$\lim\limits_{x → 0} \frac{x\cos \left(x^-1\right)}{x^{-3}\sin ^3\left(x\right)}$$
$$\lim\limits_{x → 0} \frac{ \frac{\cos \left(x^{-1}\right)}{x^{-1}} }{ \frac{\sin ^3\left(x\right)}{x^3} } $$
*
*Using... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1138866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\int \frac{x^3+2}{(x-1)^2}dx$ In order to integrate
$$\int \frac{x^3+2}{(x-1)^2}dx$$
I did:
$$\frac{x^3+2}{(x-1)^2} = x+2+3\frac{x}{(x-1)^2}\implies$$
$$\int \frac{x^3+2}{(x-1)^2} dx= \int x+2+3\frac{x}{(x-1)^2}dx$$
But I'm having trouble integrating the last part:
$$\int \frac{x}{(x-1)^2}dx$$
Wolfram alpra said me ... | First: search for:
$$
\dfrac{x}{(x-1)^2}=\dfrac{1}{(x-1)^2}+\dfrac{A}{x-1}
$$
and find:
$$
x=1+Ax-A \iff x(1-A)=1-A
$$since this must be true $\forall x$ you must have $A-1=0$ and $A=1$.
Now integrates as you have done but be careful that you have a mistake in your work:
$$
\int \dfrac{1}{(x-1)^2} dx = - \dfrac{1}{x-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1139221",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
Integral with un-intuitive U-Substitution On a recent midterm we were given the following integral and were expected to integrate it.
$$\int_{-27}^{-8} \frac{\mathrm dx}{x + x^{2/3}}$$
| $$x+x^{\frac{2}{3}}=\\x^{\frac{3}{3}}+x^{\frac{2}{3}}=\\x^{\frac{2}{3}}(x^{\frac{1}{3}}+1)=\\\sqrt[3]{x^2}(\sqrt[3]{x}+1)$$now apply this$$u=\sqrt[3]{x}\\x=u^3\\dx=3u^2du\\\frac{1}{x+x^{\frac{2}{3}}}dx=\\\frac{1}{u^2(u+1)}*(3u^2du)=\\\frac{3du}{u+1}\\$$so $$\int\frac{1}{x+x^{\frac{2}{3}}}dx=\int \frac{3du}{u+1}=3ln|u+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1139635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
System of 4 tedious nonlinear equations: $ (a+k)(b+k)(c+k)(d+k) = $ constant for $1 \le k \le 4$ It is given that
$$(a+1)(b+1)(c+1)(d+1)=15$$$$(a+2)(b+2)(c+2)(d+2)=45$$$$(a+3)(b+3)(c+3)(d+3)=133$$$$(a+4)(b+4)(c+4)(d+4)=339$$ How do I find the value of $(a+5)(b+5)(c+5)(d+5)$. I could think only of opening each expressio... | Hint:
Use finite differences as was suggested earlier by @user314 and Robert Israel. They behave similar to derivatives.
With $\Delta [p](x) \colon = p(x+1) - p(x)$, we have for $p(x) = a_n x^n + \cdots $ a polynomial of degree $n$,
$$\Delta^n [p(x)] \equiv n ! \cdot a_n$$
Therefore, for our monic polynomial $p(x)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 5,
"answer_id": 0
} |
What is the area of the largest trapezoid that can be inscribed in a semi-circle with radius $r=1$? Steps I took:
I drew out a circle with a radius of 1 and drew a trapezoid inscribed in the top portion of it. I outlined the rectangle within the trapezoid and the two right triangles within it. This allowed me to come t... | I also found the area equation for the trapezoid, but found $h$ in a different way.
The area of a trapezoid is $A=\dfrac{(b_1+b_2)H}{2}$. Base $1$ becomes $2$ units, Base $2$ is $2x$, and using the Pythagorean theorem you can find that $H=\sqrt{1-x^2}$. You then plug these values into the Area formula and find its deri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1140339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
finding the max of $f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$ I need to find the max of $$f(x)=\sqrt{(x^2-4)^2+(x-5)^2}-\sqrt{(x^2-2)^2+(x-1)^2}$$
When $x$ is a real number.
What i did is to simplify: $$f(x)=\sqrt{x^4-7x^2-10x+41}-\sqrt{x^4-3x^2-2x+5}$$.
Then i compute: $$f'(x)=\frac{-5-7x+2x^3}{\sqrt{41-... | here is geo solution which might be make things clear:
let $y=x^2,A(x,y),B(5,4),C(1,2)$,your problem become when $A$ moves on $y=x^2$, find max of $AB-AC$.
it is trivial $AB-AC <CB$, except when $A=D$ then $DB-DC=CB$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1141449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
$\int \sqrt{\frac{x}{x+1}}dx$ In order to integrate
$$\int \sqrt{\frac{x}{x+1}}dx$$
I did:
$$x = \tan^2\theta $$
$$\int \sqrt{\frac{x}{x+1}}dx = \int\sqrt{\frac{\tan^2(\theta)}{\tan^2(\theta)+1}} \ 2\tan(\theta)\sec^2(\theta)d\theta = \int \frac{|\tan(\theta)|}{|\sec^2(\theta)|}2\tan(\theta)\sec^2(\theta)d\theta = \int... | Let $u=\sqrt{\frac{x}{x+1}}$. Then
$$
\begin{align}
\int\sqrt{\frac{x}{x+1}}\,\mathrm{d}x
&=\int u\,\mathrm{d}\frac{u^2}{1-u^2}\\
&=\frac12\int u\,\mathrm{d}\left(\frac1{1-u}+\frac1{1+u}\right)\\
&=\frac12\int u\left(\frac1{(1-u)^2}-\frac1{(1+u)^2}\right)\,\mathrm{d}u\\
&=\frac12\int\left(\frac1{(1-u)^2}-\frac1{1-u}+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1142684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 2
} |
Roll a die $n$ times ($n$ is a natural number). What is the probability that 1 and 6 are observed at least once? Note: This is a homework problem so I cannot accept solutions. I would like suggestions as to how to proceed.
I have that each trial of rolling a die is independent. So I can say:
Let $P(1)$ = Rolling a ... | Let $t_{ab}(n)$ be the number of sequences of length $n$ containing $a$ and $b$ (assuming $a \ne b$).
Let $t_{a}(n)$ be the number of sequences of length $n$ containing $a$.
Let $t(n)$ be the number of sequences of length $n$.
The question is to investigate $t_{16}(n)$. Each sequence starts with a $1$, $6$, or somethi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1143015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
From $\frac{1-\cos x}{\sin x}$ to $\tan\frac{x}{2}$ How can I write $\frac{1-\cos x}{\sin x}$ as $\tan\frac{x}{2}$? I wrote $\sin x$ as $2\sin\frac{x}{2} \cos\frac{x}{2}$ also used the double angle identity for $\cos$ but wasn't able to make much progress
| Also, $\cos x = \cos^2 (x/2) - \sin^2 (x/2)$.
Then letting $y=x/2$,
$$\frac{1-\cos x}{\sin x} = \frac{1 - \cos^2 y + \sin^2 y}{2 \sin y \cos y} = \frac{2 \sin^2 y}{2 \sin y \cos y} = \frac{\sin y}{\cos y} = \tan (x/2).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solve $x^2 y''+(-2x-x^3)y'+5y=0$ Ok so for me I am having trouble solving this equation to get
$y=C_1y_1+C_2y_2$, but I'm having trouble dealing with the (-2x-x^3) part. Usually I would isolate $y''$, then make $y=x^m$, then go from there, but I'm having trouble dealing with the $(-2x-x^3)$ part. Thanks!
When I trie... | Hint:
Let $u=x^2$ ,
Then $\dfrac{dy}{dx}=\dfrac{dy}{du}\dfrac{du}{dx}=2x\dfrac{dy}{du}$
$\dfrac{d^2y}{dx^2}=\dfrac{d}{dx}\left(2x\dfrac{dy}{du}\right)=2x\dfrac{d}{dx}\left(\dfrac{dy}{du}\right)+2\dfrac{dy}{du}=2x\dfrac{d}{du}\left(\dfrac{dy}{du}\right)\dfrac{du}{dx}+2\dfrac{dy}{du}=2x\dfrac{d^2y}{du^2}2x+2\dfrac{dy}{du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144165",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Another method of proving $x^3+y^3=z^3$ has no integral solutions? The equation $$ax^2+by^2=z^3$$ has the following parametrization:
$$y=q(3ap^2-bq^2)$$
$$x=p(ap^2-3bq^2)$$
$$z=ap^2+bq^2$$ can we deduct from that the Diophantine equation $$x^3+y^3=z^3$$ has no nontrivial solutions by choosing $x=a$ and $y=b$?
| That is easy to prove. The proof is rather long but simple.Please disregard any typos.
Case 1: ($p=-1$ and $q=1$)
\begin{eqnarray}
x &=& -x (x (-1)^2 - 3 y (1)^2) \\
2x &=& 3 y \\
\end{eqnarray}
Since $\gcd(x,y)=1$ then $x=y=z=0$
\begin{eqnarray}
y &=& 1 (3x (-1)^2 - y (1)^2) \\
2y &=& 3 x \\
\end{eqnarray}which... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1144745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
prove that : $\cos x \cdot \cos(x-60^{\circ}) \cdot \cos(x+60^{\circ})= \frac14 \cos3x$ I should prove this trigonometric identity.
I think I should get to this point :
$\cos(3x) = 4\cos^3 x - 3\cos x $
But I don't have any idea how to do it (I tried solving $\cos(x+60^{\circ})\cos(x-60^{\circ})$ but I got nothing)
| *
*$ \cos (x-60^\circ ) \times \cos ( x + 60^\circ ) = \frac{1}{2} ( \cos (2x) + \cos 120^\circ) = \frac{1}{2} \cos (2x) - \frac{1}{4}$
*$ (\frac{1}{2} \cos (2x) - \frac{1}{4}) \times \cos x = \frac{1}{4} ( \cos (3x) + \cos (x) ) - \frac{1}{4} \cos (x) = \frac{1}{4} \cos (3x) $
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1145406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
List the elements of $\mathbb Z_2 \times \mathbb Z_3$ and write its operation table (the notation is additive). $\mathbb Z_2 \times \mathbb Z_3 = \{(0, 0), (0, 1), (0, 2), (1, 0), (1, 1), (1, 2)\}.$
$$
\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, ... | $\mathbb Z_3=\{0,1,2\}$
\begin{array}{c|lcr}
+ & 0 &1 & 2\\
\hline
0 & 0 &1 & 2\\
1 & 1 &2 & 0\\
2 & 2 &0 & 1\\
\end{array}
$\mathbb{Z}_3\times\mathbb{Z}_2=\{(0,0),(0,1),(0,2),(1,0),(1,1),(1,2)\}$
\begin{array}{c|lcr}
+ & (0, 0) & (0, 1) & (0, 2) & (1, 0) & (1, 1) & (1, 2)\\
\hline
(0, 0) & (0, 0) & (0, 1) & (0, 2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1149643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculate $\sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$
Calculate the sum $$\displaystyle \sum_{n=1}^{\infty}\frac{(2n-1)!!}{(2n)!!\cdot 2^n}$$
where $(2n-1)!!=1\cdot 3\cdots (2n-1)$, $(2n)!!=2\cdot 4 \cdots 2n$
Using Wolframalpha, the result is $\sqrt{2}-1$. But I don't have any idea to come up with that r... | Since $(2n)!! = 2^n n!$ and $(2n - 1)!! = (2n-1)(2n-3)\cdots 5\cdot 3\cdot 1$,
\begin{align} \sum_{n = 1}^\infty \frac{(2n-1)!!}{(2n)!!2^n} &= \sum_{n = 1}^\infty \frac{(2n-1)(2n-3)\cdots 5\cdot 3\cdot 1}{2^n n!}\frac{1}{2^n}\\
& = \sum_{n = 1}^\infty \frac{\left(n - \frac{1}{2}\right)\left(n - \frac{3}{2}\right)\cdots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150183",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
What is the coefficient of $x^{10}$ in $\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2$? I did (parcially) the following exercise:
There are $10$ identical gift boxes. Each one must be colored with a unique color and there are the colors red, blue, green and yellow. It's po... | Already you have notice that$$(1+x+x^2)(1+x+x^2+x^3)(1+x+x^2+\dots)^2=\left[ \frac{1-x^3}{1-x}\right]\left[ \frac{1-x^4}{1-x}\right]\left[ \frac{1}{1-x}\right]^2.$$ Also $$(1+x+x^2+\dots)^2=(1+2x+3x^2+\cdots+kx^{k-1}+\cdots)$$
$$(1+x+x^2)(1+x+x^2+x^3)=(1+2x+3x^2+3x^3+2x^4+x^5)$$
Therefore coefficient of $x^{10}$ in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1150280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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If $0 \le a \lt b$ prove that $a^2 < b^2$ Below is how I prove it.
Case 1: $a = 0$
*
*$0^2 < b^2$ where $b$ is a positive number.
*$0 < b \times b$
*A positive number times a positive number is always positive.
*It is true.
Case 2: $a > 0$
*
*$a < b \Rightarrow a + x = b$
*$a^2 < b^2 \Rightarrow a^2 < (a+x)... | You don't need cases. Just say:
$a^2 - b^2 = (a - b)(a + b)$.
Since $a \ge 0$ and $ b>0$, $a + b > 0$ and $a - b < 0$ since $a < b$, thus: $(a -b)(a +b) < 0$. So $a^2 - b^2 < 0$. Thus $a^2 < b^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1153191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 0
} |
Euler sequence, limit of an related sequence Study the convergence of the sequence $ \left( a_n\right)_{n\in\mathbb{N^*} }$ defined by
$$
a_{n}+e^{a_{n}}=\left( 1+\frac{1}{n}\right) ^{n}+1,~\forall n\in
\mathbb{N}
^{\ast}
$$
and find its limit. Moreover, deduce the values of limit
$$
\lim_{n\rightarrow\infty} n(1-a_... | Since both the function $f(x) = x + e^x$ and the sequence $\left(1 + \frac{1}{n}\right)^n + 1$ are strictly increasing, it follows that the sequence $a_n$ must be strictly increasing; it is bounded because $\left(1 + \frac{1}{n}\right)^n + 1 \leq e + 1 = f(1)$ is bounded.
If $a = \lim a_n$, then $a$ must satisfy $a + e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1156305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0Describe the image of the set $\{z=re^{it}: 0 \leq t \leq \frac{\pi}{4}, 0<r< \infty\}$ under the mapping $w=\frac{z}{z-1}$.
Here is what I got so far. First I got the reverse function
$$z= \frac{w}{w-1}=\frac{2u+2iv}{u-1+iv}$$
I did some algebr... | It's easy to see that the $x$-axis is mapped to itself, but note that the positive part is mapped to $(-\infty,0) \cup (1,\infty)$
To see what $re^{i\frac{\pi}{4}}$ is mapped to try to work with $f(re^{i\frac{\pi}{4}})$ and get a general formula of a curve in $r$. You should get a circle around $\frac{1}{2}-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1159067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Limit of $x^2\sin\left(\ln\sqrt{\cos\frac{\pi}{x}}\right)$ Find $$\lim_{x\to\infty}x^2\sin\left(\ln\sqrt{\cos\frac{\pi}{x}}\right).$$
I tried substituting $x=1/t$ with $t$ approaching $0$ but the term inside the bracket is not giving me ideas on how to compute the limit.
| Recall that, as $u \to 0$, we have
$$
\begin{align}
\cos u& =1-\frac {u^2}{2}+\mathcal{O}(u^3)\\
\sin u& =u+\mathcal{O}(u^3)\\
\ln (1+u)&=u-\frac {u^2}{2}+\mathcal{O}(u^3)
\end{align}
$$ giving, as $x \to \infty$,
$$
\cos\frac{\pi}{x}=1-\frac{\pi^2}{2x^2}+\mathcal{O}\left(\frac{1}{x^3}\right)
$$
$$
\begin{align}
\log ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1167629",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
} |
Summation of sine by considering the imaginary part of exp(ik*seta)
prove that
$$
\sum\limits_{k=o}^{n} {\sin k\theta} = \frac{\cos{\frac {1} {2}}\theta - \cos(n + {\frac 1 2}\theta)} {2\sin \frac{1} {2}\theta}
$$
I would like to solve this by considering the imaginary part of $\sum {\exp(ik\theta)}$.
This is an a... | Since
$$\sum_{k = 0}^n e^{ik\theta} = \frac{1 - e^{i(n+1)\theta}}{1 - e^{i\theta}} = \frac{e^{i(n+1)\theta/2}}{e^{i\theta/2}}\frac{e^{-i(n+1)\theta/2} - e^{i(n+1)\theta/2}}{e^{-i\theta/2} - e^{i\theta/2}} = e^{in\theta/2} \frac{\sin [(n+1)\theta/2]}{\sin \theta/2},$$
Taking imaginary parts yields
\begin{align}\sum_{k =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Please help finding positive integers less than $1000$ which satisfy the constraint: $x=7k , x=4l+2 , x=3m+1$ Please help finding positive integers less than $1000$ which satisfy the constraint: $x=7k , x=4l+2 , x=3m+1$
| We want to find integers $x$ that satisfy$$\begin{align}x&=7k\\x&=4l+2\\x&=3m+1\end{align}$$
Substitute $x=7k$ into the second equation and obtain
$$7k = 4l+2\\\implies 8k-k=4l+2\\\implies -k=4\heartsuit+2\\\implies k = 4n+2 $$
So we have $k=4n+2$, where $n$ is an integer. Then
$$x=7(4n+2) = 28n+14$$
For $x$ to satisfy... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1168841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve this sum problem?
For the first radical section.
$$\sqrt{1\times 2\times 3\times 4 + 1} - 1 = 1 + 3 + 1 - 1= 5 - 1 = 4$$
The second radical section.
$$(\sqrt{2\times 3\times 4\times 5 + 1}) = 4 = 4 + 6 + 1 - 4 = 7$$
The third radical (not shown)
$$\sqrt{3\times 4\times 5\times 6 + 1} - 9 = 9 + 9 + 1 - 9 =... | Split your sum up:
$$3\sum\limits_{n=1}^{97}n + \sum\limits_{n=1}^{97}1$$
$$3 {97 \choose 2} + 97$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1169429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding the degree and basis of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$ I'm having a difficult time finding the degree and basis of $\mathbb{Q}(\sqrt{5}, \sqrt{7})$ over $\mathbb{Q}$.
I know that $\sqrt{7}$ satisfies $f(x) = x^2 - 7$ and is irreducible over $\mathbb{Q}$. So $[\mathbb{Q}(\sqrt{7}) : \mathbb{Q}] = 2$. I also kn... | Some of the "hard part" done for you: It should be clear that
$\Bbb Q(\sqrt{5} + \sqrt{7}) \subseteq \Bbb Q(\sqrt{5},\sqrt{7})$ since $\sqrt{5} + \sqrt{7} \in \Bbb Q(\sqrt{5},\sqrt{7})$.
The other direction is a bit "trickier", note that:
$(\sqrt{5} + \sqrt{7})^3 = 5\sqrt{5} + 15\sqrt{7} + 21\sqrt{5} + 7\sqrt{7} = 26\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1171268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Is $53$ expressible in this form? It seems as if prime numbers may always be expressed in the form $a\cdot 2^b+c \cdot 3^d$ for some nonnegative integers $b,d$ and $a,c\in \{-1,0,1\}$.
Examples:
$2=1\cdot 2^1+0\cdot 3^d$
$3=0\cdot 2^b+1\cdot 3^1$
$5=1\cdot 2^1+1\cdot 3^1$
$7=1\cdot 2^2+1\cdot 3^1$
$11=1\cdot 2^3+1\cdot... | You can check that for any $k,l$ we have that
$$2^k \not\equiv 3^l+53 \pmod{117}$$
and that
$$2^k+53 \not\equiv 3^l\pmod{117}$$
Just enumerating the different powers.
Edit: If you consider the sets $Q_t = \{2^{k_1}3^{k_2}\dots p_t^{k_t}, k_1,\dots,k_t \ge 0\}$ and ask for integers not in $Q_t$ or sum or difference o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1173460",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
} |
Shortcuts for integrating $\int \sqrt{x^2-36} \,dx$ Is there any simple method to solve this integral
$$\int \sqrt{x^2-36} \,dx$$
Because my method is a bit complicated and long:
| I think integration by parts is the shortest way here. Let $u = \sqrt{x^2 - 36}$ and $\mathrm dv = \mathrm dx$. Then,
$$\begin{align}
\int\sqrt{x^2 - 36}\,\mathrm dx &= x\sqrt{x^2 - 36} - \int\frac{x^2}{\sqrt{x^2 - 36}}\,\mathrm dx =\\[0.3em]
&= x\sqrt{x^2 - 36} - \int\frac{x^2 - 36 + 36}{\sqrt{x^2 - 36}}\,\mathrm dx =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1174951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Does every positive rational number appear once and exactly once in the sequence $\{f^n(0)\}$ , where $f(x):=\frac1{2 \lfloor x \rfloor -x+1} $ Consider the map $f:\mathbb Q^+ \to \mathbb Q^+$ defined as $f(x):=\dfrac1{2 \lfloor x \rfloor -x+1} , \forall x \in \mathbb Q^+$ ; then is the function
$g:\mathbb Z^+ \to \ma... | As Henning noted in the comments,
$$\begin{align*}
f\left(\frac{p}q\right)&=\frac1{2\left\lfloor\frac{p}q\right\rfloor-\frac{p}q+1}\\
&=\frac{q}{2q\left(\frac{p-(p\bmod q)}q\right)-p+q}\\
&=\frac{q}{p+q-2(p\bmod q)}\;.
\end{align*}$$
If we let $p_0=q_0=1$ and define $p_n$ and $q_n$ by the recurrences
$$\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1179591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Proving that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational I've been struggling to show that the number $\sqrt[3]{7 + \sqrt{50}} + \sqrt[3]{7 - 5\sqrt{2}}$ is rational. I would like to restructure it to prove it, but I can't find anything besides $\sqrt{50} =5 \sqrt{2}$. Could anybody give ... | Let $(a+b\sqrt{2})^3=7+5\sqrt{2}$, then $(a-b\sqrt{2})^3=7-5\sqrt{2}$. This gives us
\begin{align}
a^3+6ab^2 & = 7\\
2b^3+3a^2b & = 5
\end{align}
From the two equations listed above we get.
$$\frac{a(a^2+6b^2)}{b(2b^2+3a^2)}=\frac{7}{5}.$$
Now let $x=\frac{a}{b}$. Then
$$5x(x^2+6)=7(3x^2+2).$$
Finally we get
$$5x^3-21... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1180599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 0
} |
Lagrange Multipliers, maximize $f=xy$ restricted to $g=x^2+y^2=r^2$ So I have to solve the system of equations
$$\cases{\nabla f = \lambda \nabla g\\x^2+y^2 = r^2}.$$
Then $y=2\lambda x, x=2\lambda y$. Sorry if this is obvious, but how can I get $x$ and $y$ only as a function of $\lambda$? Otherwise I'm not able to fin... | Since $y = 2\lambda x$ and $x = 2\lambda y$, $x^2 + y^2 = 4\lambda^2(x^2 + y^2)$, i.e., $r^2 = 4\lambda^2 r^2$. Thus $4\lambda^2 = 1$, yielding $\lambda = \pm 1/2$. If $\lambda = 1/2$, then $y = x$, so from the condition $x^2 + y^2 = r^2$, we obtain $x = y = \pm r/\sqrt{2}$. If $\lambda = -1/2$, then $y = -x$, so the c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1181307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integral fraction of polynomials I have this problem:
$$\int \frac{-2x^2+6x+8}{x^4-4x+3}dx$$
I have tried using partial fractions, but I can't get solution. Thank you for any advice.
| As one can obviously see, $x=1$ is a root of the denominator, and thus, it can be divided by $(x-1)$. Using long division, we get $(x-1)(x^3+x^2+x-3)$
Again, we can see that $x=1$ is a solution, of the second factor, so, using long division again, we get $(x-1)^2(x^2+2x+3)$
$$\frac{-2x^2+6x+8}{(x-1)^2(x^2+2x+3)}=\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1182910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Show that there exists a polynomial $p(X) ∈ F[X]$ satisfying $c^n + c^{−n} = p(c + c^{−1})$. Let $c$ be a nonzero element of a field $F$ and let $n > 1$ be
an integer. Show that there exists a polynomial $p(X) ∈ F[X]$ satisfying
$c^n + c^{−n} = p(c + c^{−1})$.
I made some particular cases, but not seems to have a form ... | Complete solution:
We can prove this by induction, starting with a base case of $n=0$. If $n=0$, then taking $p(X)=2$ is a solution to the problem. Suppose $n>0$ and that we have polynomials $p_k(X)\in F[X]$ such that $p_k(c+c^{-1})=c^k+c^{-k}$ for all $k<n$. Then notice that $(c+c^{-1})^n=c^n+nc^{n-1}c^{-1}+\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1183161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Solve $\arg(-1/z)=-2\pi/3$ and $|1-\frac2z|=1$ $\arg(\frac{-1}z)=\frac{-2\pi}3$
what does this mean? how to i get the $\arg(z)$ from this.
I'm thinking of reciprocals.
$\left|1-\frac2z\right|=1$ how do i solve for this as well.. i'm confused when i negative sign appears
| Note that:
$$\frac{1}{z} = \frac{x -i y}{x^2+y^2}$$
Now you have:
$$\arg\left(-\frac{1}{z}\right) = -\frac{2\pi}{3}\implies \arg \left(\frac{-x +i y}{x^2+y^2}\right) = \tan^{-1}{\left(-\frac{y}{x}\right)} = -\frac{2\pi}{3}$$
$$\implies \frac{y}{x} = \tan(\frac{2\pi}{3})$$
The other condition you can write as:
$$ \vert ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1187763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Coordinate transformation (or conversion) into yards Following is a soccer field with its dimensions.
There is a similar field, but I am capturing coordinates via mouse-movement. So, what (115,75) shows here, is 567.5,369 when I capture the coordinates when the mouse moves. How could I convert a general coordinate (x... | So you want an affine transformation $L\colon\Bbb R^2\to\Bbb R^2$ that maps
*
*$\begin{pmatrix} 0\\ 0\end{pmatrix}$ to $\begin{pmatrix} a_0\\ b_0\end{pmatrix}$
*$\begin{pmatrix} x_1\\ 0\end{pmatrix}$ to $\begin{pmatrix} a_1\\ b_1\end{pmatrix}$ where $x_1\neq 0$ (for you example, take $x_1=115$)
*$\begin{pmatrix} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1188003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Geometric Interpretation of the Basel Problem? Does the Pi in the solution to the Basel problem have any geometric significance? Every time I see Pi, I have to think of a circle. I would love to see a nice intuitive picture connecting the Basel problem with geometrical figures. Anyone here been lucky enough to stumb... | The Montreal interpretation of the Basel problem
Suppose you randomly pick a street in the left window, one which is parallel to that window (red streets). Similarly, you pick a blue street in the right window:
Then, from this perspective the cross section between the two streets has 50% chance of being visible in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1189023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 5,
"answer_id": 3
} |
1985 Putnam A1 Solution
I dont see what they mean by bijection of triples of subsets of $\{1, \ldots, 10\}$ and the $10\times3$ matrix with $0, 1$ entries?
How is that created?
| For example, suppose the three sets are
\begin{align}
A_1 & = \{1,5,6,8\} \\
A_2 & = \{1,2,3,4,10\} \\
A_3 & = \{2,4,5,7,9,10\}
\end{align}
(Note that $A_1\cup A_2\cup A_3=\{1,2,3,4,5,6,7,8,9,10\}$ and $A_1\cap A_2\cap A_3=\varnothing$.)
Then the $10\times3$ matrix is
$$
\begin{bmatrix}
1 & 1 & 0 \\
0 & 1 & 1 \\
0 & 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 2,
"answer_id": 1
} |
Find the derivative of the function. y = $\sqrt{7x+\sqrt{7x+\sqrt{7x}}} $ This question is really tricky. I am wondering if I am right?
| Build from inside:
$$
\begin{align}
\frac{d}{dx}\sqrt{7x}&=\frac{7}{2\sqrt{7x}}\\
\frac{d}{dx}\sqrt{7x+\sqrt{7x}}&=\frac{1}{2\sqrt{7x+\sqrt{7x}}}\cdot\left(7+\frac{d}{dx}\sqrt{7x}\right)\\
&=\frac{1}{2\sqrt{7x+\sqrt{7x}}}\cdot\left(7+\frac{7}{2\sqrt{7x}}\right)\\
\frac{d}{dx}\sqrt{7x+\sqrt{7x+\sqrt{7x}}}&=\frac{1}{2\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1190891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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Determine the coefficient for fourier sine series of f(x) = cos ${\pi x\over L}$ Determine the coefficient for Fourier sine series of f(x) = cos ${\pi x\over L}$
I know how to solve it but I get stuck at the end.
$$B_n = \int_{0}^{L} \cos\frac{\pi x}{L} \, \sin\frac{n\pi x}{L}$$
We use the identity $$\sin a \, \cos ... | Consider
\begin{align}
B_{n} = \frac{2}{L} \int_{0}^{L} \cos\left( \frac{\pi x}{L} \right) \sin\left( \frac{n \pi x}{L} \right) \, dx.
\end{align}
For the case $n=1$ it is seen that
\begin{align}
B_{1} &= \frac{1}{L} \int_{0}^{L} \sin\left( \frac{2 \pi x}{L} \right) \, dx \\
&= \frac{1}{L} \left[ - \frac{L}{2 \pi} \co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Show that $-\frac{2yx^3}{(x^2+y^2)^2}$ is bounded.
Show that $-\frac{2yx^3}{(x^2+y^2)^2}$ is bounded.
I'm approaching this starting with
$$ \left| \frac{2yx^3}{(x^2+y^2)^2} \right| = \left| \frac{2yx^3}{x^4+2x^2y^2+y^4} \right| \leq \left| \frac{2yx^3}{2x^2y^2} \right| = \left| \frac{x}{y}\right|.$$
However, this doe... | By AM-GM,
$$ |yx^3| = 27\left|y\cdot\frac{x}{3}\cdot\frac{x}{3}\cdot\frac{x}{3}\right| \leq 27\left(\frac{|x|+|y|}{4}\right)^4 = \frac{27}{64}(|x|+|y|)^4$$
while by AM-QM:
$$ \frac{x^2+y^2}{2} \geq \left(\frac{|x|+|y|}{2}\right)^2 $$
hence:
$$\left|\frac{2yx^3}{(x^2+y^2)^2}\right|\leq \frac{\frac{27}{32}}{\frac{1}{4}}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1191477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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The limit : $ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $ The limit:
$ \lim _{x \to \infty } \sqrt{x^2 +x} - \sqrt{x^2 +1} $
(A) is 0 (B) is $\frac 1 2 $ (C) is 2 (D) does not exist
Is doing it with Binomial expansion and cancelling the terms only way?
| \begin{array} \\
\displaystyle \lim_{x \rightarrow \infty} \sqrt{x^2 + x} - \sqrt{x^2 + 1} &= \displaystyle \lim_{x \rightarrow \infty} \left(\sqrt{x^2 + x} - \sqrt{x^2 + 1}\right) \frac{\sqrt{x^2 + x} + \sqrt{x^2 + 1}}{\sqrt{x^2 + x} + \sqrt{x^2 + 1}} \\
&= \displaystyle \lim_{x \rightarrow \infty} \frac{(x^2 + x) - (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1193965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find a formula for $\sin(3a)$ and use to calculate $\sin(π/3)$ and $\cos(π/3)$? Problem:
Find a formula for $\sin(3a)$ in terms of $\sin(a)$ and $\cos(a)$. Use this to calculate $\sin(π/3)$ and $\cos(π/3)$.
My attempt:
\begin{align}
\sin(3a) &= \sin(2a + a) = \sin(2a)\cos(a) + \cos(2a)\sin(a) \\
&= \sin(a + a)\cos(... | hint: Put $x = \sin \left(\frac{\pi}{3}\right)$, then the equation $0=\sin\left(3\cdot\dfrac{\pi}{3}\right)= 3\sin\left(\frac{\pi}{3}\right) - 4\sin^3(\frac{\pi}{3})= 3x - 4x^3$ gives: $4x^3-3x = 0 \to x(4x^2-3) = 0$. Can you solve for $x$. Note that $x \neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1196178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding an isomorphism between rings Find an isomorphism between $T: \mathbb{F_3}[x]/(x^2+x+2) \to \mathbb{F_3}[x]/ (x^2 + 1)$. We have that both of these are fields. The answer states the map $T$ is given by $T(a+bx) = (a+b) + bx$ so $T = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix}$. Could someone explain how they go... | There is a mechanical way to find isomorphisms like this. We know that such an isomorphism exists by the uniqueness of finite fields of a given order. Furthermore, such an isomorphism is completely determined by the image of $x$. Finally, we know that the image of $x$ can be represented by $a + bx$ such that $(a + bx)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1197653",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Count the number of one pair hands in a standard deck In an attempt to answer this question, I tried the following solution:
*
*Let $A$ denote the number of one-pair hands
*Let $B$ denote the number of two-pair hands
*Let $C$ denote the number of three-of-a-kind hands
*Let $D$ denote the number of full-house hand... | To choose a pair, you choose one of the thirteen kinds, choose two of the four suits from that kind, choose three of the remaining kinds, and choose one of the four suits from each of those kinds, which yields
$$\binom{13}{1}\binom{4}{2}\binom{12}{3}\binom{4}{1}^3 = 1,098,240$$
hands with one pair.
Say your hand is $\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1199370",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving a goal with an existential quantifier and making sure it covers all cases I'm trying to prove the following theorem:
Suppose x is a real number. Prove that if x $\neq$ 1 then there is a
real number y such that $\frac{y + 1}{y - 2}$ = x.
The logical structure of the sentence is:
x $\neq$ 1 $\implies$ $\exist... | You should solve for $y$ in terms of $x$. Once you find it, you can then go back and choose $y$ to be this number and show that it works. Messing around, we see that:
\begin{align*}
\frac{y + 1}{y - 2} &= x \\
y + 1 &= xy - 2x \\
2x + 1 &= xy - y = y(x - 1) \\
y &= \frac{2x + 1}{x - 1}
\end{align*}
We're now ready to p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1200408",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Limit of a two variable function $\frac{x^2\sin^2{y}}{2x^2+y^2}$ I am trying to find the limit as $(x,y) \rightarrow (0,0)$ of $\frac{x^2\sin^2{y}}{2x^2+y^2}$
I tried this:
$\frac{x^2}{2x^2+y^2}<1/2$ so using the squeeze theorem $\frac{x^2}{2x^2+y^2}<1/2 \sin^2{y}$
and the limit of the right handside is 0, so the limit... | Since $$2x^2+y^2 \geq 2\sqrt{2}\,|xy|$$
it follows that:
$$\left|\frac{x^2\sin^2 y}{2x^2+y^2}\right|\leq\left|xy\cdot\frac{xy}{2x^2+y^2}\right|\leq\frac{|xy|}{2\sqrt{2}}$$
so the limit is zero.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1202880",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Solving $3\sin x - \cos x = 2$ for $x \in [0, 2\pi)$ Problem:
Solve $$3\sin x - \cos x = 2, \ \ \ x \in [0, 2\pi)$$
My attempt:
I am able to solve it using Weierstrass substitutions and a good bit of patience, but the problem was given at an exam at a level where such substitutions are not part of the curriculum.
I've... | When you arrive at
$$
9\sin^2x - 6\sin x\cos x + \cos^2x = 4
$$
just note that $4=4\sin^2x+4\cos^2x$ so you can rewrite your equation as
$$
5\sin^2x-6\sin x\cos x-3\cos^2x=0
$$
Since $\cos x=0$ is not a solution, divide by $\cos^2x$ and get
$$
5\tan^2x-6\tan x-3=0
$$
so
$$
\tan x=\frac{3\pm\sqrt{24}}{5}
$$
You need to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1203112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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What is the probability of getting at least one $3$ when rolling two standard dice? What is the probability of getting at least one $3$ when rolling two standard ($6$-sided) dice?
I was thinking the answer would be $11$ out of $36$, but the textbook says it is $10$ out of $36$.
Please help.
| The correct answer is: $\frac {1}{6} \times \frac{5}{6} + \frac {5}{6} \times \frac{1}{6} + \frac {1}{6} \times \frac{1}{6} = \frac {11}{36}$
The test book is in error.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1203774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$. Prove by induction that $n^3 + 11n$ is divisible by $6$ for every positive integer $n$.
I've started by letting $P(n) = n^3+11n$
$P(1)=12$ (divisible by 6, so $P(1)$ is true.)
Assume $P(k)=k^3+11k$ is divisible by 6.
$P(k+1)=(k+1)^... | Since
$n^{3} + 11n = 6m$
for some integer $m,$
we have
$$(n+1)^{3} + 11(n+1) = 6m + 3n^{2} + 3n + 12 = 6m + 12 + 3(n^{2} + n).$$
It suffices to prove that $3(n^{2} + n)$ is a multiple of $6$. But, since if $n$ is odd then $n^{2} + n = 2m'$ for some integer $m'$ and if $n$ is even then of course $n^{2} + n = 2m''$ for s... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating over the two form Let $A=(0,1)^2$. Let $\alpha:A\to\Bbb R^3$ be given by the equation
$$\alpha(u,v)=(u,v,u^2+v^2+1)$$
Let $Y$ be the image set of $\alpha$. Evaluate the integral over $Y_\alpha$ of the 2-form
$x_2dx_2\land dx_3+x_1 x_3 dx_1\land dx_3$.
Can someone please give me a useful hint for solving th... | Given parametrization:
$$\phi (u,v) = (u,v,{u^2} + {v^2} + 1)$$
Setting
$$x = u,y = v,z = {u^2} + {v^2} + 1$$
Then calculate differential resp. to parametrization:
$$\begin{gathered}
dx = du \hfill \\
dy = dv \hfill \\
dz = 2udu + 2vdv \hfill \\
\end{gathered} $$
Form in old coordinates:
$$\omega (x,y,z) = ydy \... | {
"language": "en",
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"source": "stackexchange",
"question_score": "4",
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If $x^2 +px +1$ is a factor of $ ax^3 +bx+c$ then relate $a,b,c$ Suppose If $x^2 +px +1$ is a factor of $ax^3 +bx+c$ then relate $a,b,c$ such that $a,b,c \in R$
I can write $$ax^3 +bx+c=(x^2 +px +1)(\lambda x +D)$$
$$\implies ax^3 +bx+c =\lambda x^3 + x^2.p\lambda + x(\lambda+pD)+D $$
and then compare coefficient to fi... | Following the path you have taken, and correcting the algebra, one Gets
$$(\lambda X+D)(X^2+pX+1)=\lambda X^3+(\lambda p+D)X^2+(Dp+\lambda)X+D$$
And this is valid for all $X$. So by identifying the coefficients we get
$$\begin{cases} \lambda=a\\ \lambda p+D=0\\ Dp+\lambda=b\\ D=c \end{cases}$$
which reads
$$\begin{case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1205540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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Dynamical system $x_{n+1} = \frac{1}{2}(x_n - \frac{1}{x_n}) \ \ , \ \ n = 0, 1 , 2,...$
Consider the dynamical system
$$
x_{n+1} = \frac{1}{2}(x_n - \frac{1}{x_n}) \ \ , \ \ n = 0, 1 , 2,...
$$
So by using the substitution $x_n = \cot(y_n)$, I have found:
$$
x_n = \cot(\cot^{-1} (x_0) \cdot 2^n )
$$
but the next p... | For $x_0=\cot(\theta)$ we have that
$x_1=\frac{1}{2}\Big[\frac{\cos(\theta)}{\sin(\theta)}-\frac{\sin(\theta)}{\cos(\theta)}\Big]= \cot(2\theta)$
and inductively
$x_{n}=\cot(2^n\theta)$
Thus when $\theta=\frac{\pi}{2^p-1}$ we obtain
$x_{n+p}=\cot\big(\frac{2^{n+p}\pi}{2^p-1}\big)=\cot\Big(\frac{2^n(2^{p}-1)\pi}{2^p-1}... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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the sum of the squares I think it is interesting, if we have the formula
$$\frac{n (n + 1) (2 n + 1)}{6} = 1^2 + 2^2 + \cdots + n^2 .$$
If the difference between the closest numbers is smaller (let's call is a) we obtain, for example, if a=0.1
$$\frac{n (n + 0.1) (2 n + 0.1) }{6 \cdot 0.1} = 0.1^2 + 0.2^2 + \cdots ... | I think this is correct, as you are essentially integrating $x^2$, to get the result of $\frac{x^3}3$. this, if you don't know, is adding together rectangles of increasingly smaller width (a) and heights which are the result of a function ($x^2$).
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Set A = {1,2,3,4,5} Pick randomly one digit and remove it. What is the prob. that we pick an odd digit the 2nd time. The probability that we pick any number for the first time is $\dfrac{1}{5}$
the sample space of sample spaces after the first event is then
{2,3,4,5}
{1,3,4,5}
{1,2,4,5}
{1,2,3,5}
{1,2,3,4}
prob. to pic... | There are simpler ways. There is absolutely no preference between the numbers, so each number is equally likely to be the second one drawn. Since 3 of 5 numbers are odd, the probability is 3/5 that an odd number is drawn.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1206670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Integral of rational function with trigonometric functions $$
\int \frac{dx}{(\sqrt{\cos x}+ \sqrt{\sin x})^4}
$$
I saw this problem online and it looked like an interesting/difficult problem to try and tackle. My attempt so far is to use tangent half-angle substitution.
Let $t= \tan^2 (\frac{x}{2})$, then $dt= \frac{... | To make the integral easier to evaluate. Write the integral as
$$ I = \int \frac{dx}{\cos^2(x)( 1+ \sqrt{\tan(x)} )^4} $$
and the use the substitution $\tan(x)=u^2$ which makes the integral falls a part
$$ I=2 \int \frac{u}{(1+u)^4} du .$$
You can use another substitution $1+u=t$ to finish the problem.
| {
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"url": "https://math.stackexchange.com/questions/1207121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Roots to the quartic equation, $(x+1)^2+(x+2)^3+(x+3)^4=2$ Solving with Mathematica gives me the four roots, $$x=-4,-2,\dfrac{-7\pm\sqrt5}{2}$$ Is there some trick to solving this that doesn't involve expanding and/or factoring by grouping?
| Set $y=x+2$. The equation rewrites as:
$$(y-1)^2+y^3+(y+1)^3==2\iff y(y^3+5y^2+7y^+2)=0$$
The second factor has an integer root: $-2$, hence the factorisation:
$$y^3+5y^2+7y +2=(y+2)(y^2+3y+1).$$
Finally, the roots (in $y$) are: $0, -2, \dfrac{-3\pm\sqrt 5}{2} $, whence the roots of the initial equation:
$$ \Bigl\{ -2,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving $\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$ for all $n\geq 1$ by induction How prove the following equality:
$a_n$:=$\sum\limits_{k=1}^{2n} {(-1)^k \cdot k^2}=(2n+1)\cdot n$
$1$.presumption: $(-1)^1 \cdot 1^2+(-1)^2\cdot2^2=(2 \cdot 1+1) \cdot 1=3$ that seems legit
$2$.precondition:
$a_{n-1}$= $(2... | $$a_n=\sum_{k=1}^{2(n-1)}(-1)^kk^2+\sum_{k=2n-1}^{2n}(-1)^kk^2$$
$$=a_{n-1}+\sum_{k=2n-1}^{2n}(-1)^kk^2$$
$=2n^2-3n+1+(2n)^2-(2n-1)^2=\cdots$
W/O using induction,
$$a_n=\sum_{k=1}^{2n}(-1)^kk^2=\sum_{k=1}^n[(2k)^2-(2k-1)^2]=4\sum_{k=1}^nk-\sum_{k=1}^n1=4\cdot\dfrac{n(n+1)}2-n$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Bonferroni Inequalities
Let $k$ and $m$ be positive integers with $k>m$.
Then the partial sums of
$$
1-\binom{k}{1} + \binom{k}{2} - \cdots (-1)^m\binom{k}{m}
$$
has alternating signs.
(The partial sums of the given sum are $P_1=1$, $P_2=1-\binom{k}{1}$, $P_3=1-\binom{k}{1}+\binom{k}{2}$, etc)
I arrived at the ... | $$\begin{align}
\require{cancel}
&1-\binom{k}{1} + \binom{k}{2} - \cdots (-1)^m\binom{k}{m}\\
&=1+\sum_{r=1}^{m}(-1)^k\binom kr\\
&=1+\sum_{r=1}^{m}(-1)^r\left[\binom {k-1}{r-1}+\binom {k-1}r\right]\\
&=\color{lightgrey}{\cancel{1}-\left[\cancel{{\binom {k-1}0}}+\bcancel{\binom {k-1}1}\right]
+\left[\bcancel{\binom {k-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1217459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Scale invariant Image Moments - not scale variant? I came across a problem working with image moments [1].
It is stated that
$\eta_{ij} = \frac{\mu_{ji}}{\mu_{00}^{k}}$
where $k = 1 + \frac{i+j}{2}$
is scale invariant.
However, if I try to reproduce this, it does not appear scale invariant at all.
Consider a simple ex... | The invariance is exact in the continuous domain. Your problem here is how you scale, which is limited by the discrete grid.
Scaling by a factor 2 of your original 4 squares would result in 4 squares, centered at (-1,-1), (-1,1), (1,-1) and (1,1), each with a weight of 4:
$$
\mu_{20} = 4 \cdot 1^2 + 4 \cdot 1^2 + 4 \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1218329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do you find the inflection point of this graph? The graph is this:
$$
\frac{(x+1)^3 - 4(x+1)^2 + 4(x+1)}{(x+1)^2 - 2(x+1) + 1}
$$
I know you can find second derivative and then solve for values that make it undefined or 0, but I was told apparently there is another faster way to get the inflection point. What is t... | Let $$y=\frac{(x+1)^3 - 4(x+1)^2 + 4(x+1)}{(x+1)^2 - 2(x+1) + 1}$$
Then using the substitution, $u=x+1$, we obtain
$$y=\frac{u^3 - 4u^2 + 4u}{u^2 - 2u + 1}$$
This simplifies to $$y=\frac{u(u-2)^2}{(u-1)^2}$$
Differentiating with respect to $u$, we obtain
$$\frac{\partial y}{\partial u}=\frac{u^3-3u^2+4u-4}{(u-1)^3}$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int_0^\infty \left( \frac{x^{10}}{1+x^{14}} \right)^{2} \, dx$ This is a integration question from a previous calculus exam:
Evaluate $$\int_0^\infty \left( \frac{x^{10}}{1+x^{14}} \right)^{2} \, dx$$
I rewrote it as $$\lim \limits_{b \to \infty} \int_0^b \left(\frac{x^{10}}{1+x^{14}} \right)^{2} \, dx,$$ ... | Hint:
$$\begin{align}\frac{1}{14}\int \frac{x^7 14x^{13}}{(1 + x^{14})^2}dx& \underbrace{=}_{\color{red}{\text{by parts}}}\frac{1}{14}\bigg( -x^7\frac{1}{1 + x^{14}} + 7\color{#05f}{\int \frac{x^6}{1+x^{14}}dx}\bigg) \\&= \frac{1}{14}\bigg(\arctan(x^7) -\frac{x^7}{1 + x^{14}}\bigg) + C\end{align}$$
For $$\color{#05f}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1220359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find a volume using a triple integral.. This is the problem:
Find the volumen of de solid bounded by $x^2+y^2=2$, $z=0$ and $x+2y+2z=2$.
I have set the parameters to:
$0 \leq z \leq \dfrac{2-x-y}{2}$
$ 0\leq y\leq 2-x$
$0 \leq x \leq 2$
and evaluated:
$
\displaystyle{ \int\limits_{0}^{2} \int\limits_{0}^{2-x} \int\l... | There is no way in which a 2d-circle and a plane may bound a solid, so I assume you are trying to find the volume of the intersection between the cylinder $\{x^2+y^2\leq 2,z\geq 0\}$ and the half-space $x+2y+2z\leq 2$. It is worth to apply a rotation around the $z$-axis, by setting:
$$ u=\frac{x+2y}{\sqrt{5}},v=\frac{2... | {
"language": "en",
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Proving gcd($a,b$)lcm($a,b$) = $|ab|$
Let $a$ and $b$ be two integers. Prove that $$ dm = \left|ab\right| ,$$ where $d = \gcd\left(a,b\right)$ and $m = \operatorname{lcm}\left(a,b\right)$.
So I went about by saying that $a = p_1p_2...p_n$ where each $p_n$ is a prime. Same applies to $b = q_1q_2 ... q_c$. So then $m ... |
If $a=b=0$, then any integer is a common divisor of $a$ and $b$.
So, there is no greatest common divisor of $a$ and $b$.
So we don't consider the case in which $a=b=0$.
If $a=0$ and $b\neq 0$, then the set of the common divisors of $a$ and $b$ is equal to the set of divisors of $b$.
Obviously, the greatest divisor o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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What does a positive definite matrix have to do with Cauchy-Schwarz Inequality? In my text book, Cauchy-Schwarz Inequality is extended to a positive definite matrix.
But I neither understand what the relationship between Cauchy-Schwarz Inequality and a positive definite matrix nor the sentence underlined in red, I am n... | The sentence you underlined in red abbreviates these steps:
$$(\mathbf{B}^{1/2} \mathbf{b} \cdot \mathbf{B}^{-1/2} \mathbf{d})^2 \leq (\mathbf{B}^{1/2} \mathbf{b} \cdot \mathbf{B}^{1/2} \mathbf{b})(\mathbf{B}^{-1/2} \mathbf{d} \cdot \mathbf{B}^{-1/2} \mathbf{d}) $$
This is the Cauchy-Schwarz inequality applied as requ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1226455",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "10",
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When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble - trouble understanding proof Theorem: When $p=2$ or $p$ prime, with $p=1\pmod{4}$, $x^2\equiv -1\pmod{p}$ is soluble
Proof: When $p=2$, the statement is clear.
Assume $p\equiv 1\pmod{4}$, let $r=\frac{p-1}{2}$ and $x=r!$
Then since $r$ is e... | For any $i$, $p-i\equiv -i\pmod{p}$, so the right-hand side of your final congruence is
\begin{equation*}
(1\cdot 2\cdot...\cdot r)((p-1)\cdot(p-2)\cdot...\cdot(p-r))
\equiv (1\cdot 2\cdot...\cdot r)((-1)\cdot(-2)\cdot...\cdot(-r))
\equiv r!(-1)^rr! \equiv (-1)^r(r!)^2\pmod{p}.
\end{equation*}
But since $p\eq... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the following integral: $ \int \frac{x^2}{x^2+x-2} dx $ Solve the integral: $ \int \frac{x^2}{x^2+x-2} dx $
I was hoping that writing it in the form $ \int 1 - \frac{x-2}{x^2+x-2} dx $ would help but I'm still not getting anywhere.
In the example it was re-written as $ \int 1 - \frac{4}{3x+6} - \frac{1}{3x-3} dx ... | when you get the $\int 1 - \frac{x-2}{x^2+x-2} dx$, you can use $x^2+x-2 = (x-1)(x+2)$.
then it become
$$
\int 1 - \frac{x+2-4}{(x+2)(x-1)}dx
$$
which equal to
$$
\int 1 - \frac{1}{x-1}-\frac{4}{(x+2)(x-1)}dx
$$
$$
\int 1 - \frac{1}{x-1}-4/3(\frac{1}{(x-1)}+\frac{1}{(x+2)})dx
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1231509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Help with double simultaneous equations and roots I am in the course of a project, in which I need to solve these two simultaneous equations:
\begin{equation}
\sqrt{(1000-y)^2 + x^2} - \sqrt{y^2 + x^2} = 342.371
\end{equation}
\begin{equation}
\sqrt{(2000-x)^2 + y^2} - \sqrt{y^2 + x^2} = 961.674
\end{equation}
I know t... | $\sqrt{(1000-y)^2+x^2}-\sqrt{y^2+x^2}=Q$.
$\sqrt{(1000-y)^2+x^2}=\sqrt{y^2+x^2}+Q$.
$(1000-y)^2+x^2=y^2+x^2+2Q\sqrt{y^2+x^2}+Q^2$.
$(1000-y)^2+x^2-y^2-x^2-Q^2=2Q\sqrt{y^2+x^2}$.
$1000000-2000y-Q^2=2Q\sqrt{y^2+x^2}$.
Now square both sides to get an equation quadratic in $x$ and $y$. Then go through the sam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1233578",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Dice roll probability, at least 9 total? If I have two dice with $6$ sides each, what is the probability of me rolling atleast $9$ total? I think I'm correct when thinking that the probability of rolling a $9$ is $\frac{4}{36}$, that is $11.1...\%$, but how do I go from here to calculate the "at least" part?
| $\begin{array}{|c|c|c|c|}
\hline
&\overrightarrow{ D2} & \color{blue}{1} & \color{blue}{2} & \color{blue}{3} & \color{blue}{4} & \color{blue}{5} & \color{blue}{6} \\ \hline D1\downarrow\\\hline
\color{blue}{1} &&2 &3 &4 & 5 & 6 & 7\\ \hline
\color{blue}{2}&& 3 & 4&5 & 6 & 7 & 8\\ \hline
\color{blue}{3} &&4 &5 &6 & 7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1235970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$? What is the value of $1^2 + 2^2 + 3^2 + \cdots + (p-1)^2\pmod{p}$?
Let's try a several primes greater than 3...
If $p=5$, then we have $1^2 + 2^2 + 3^2 + 4^2 = 30$, so that $30\pmod{5} = 0$
If $p=7$, then we have $1^2 + 2^2 + 3^2 + 4^2 + 5^2 + 6^2= 91$... | Assuming the formula for sum of squares (which has to be an integer) we see that the 6 in the denominator has to divide the numerator: Now 6 is coprime to $p$ (for $p >3$). Now you can figure it out yourself.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1236894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Question about matrix multiplication notation I have the following matrices:
$A=\begin{pmatrix}
-\frac{2}{3} & \frac{1}{3} & 0 \\
\frac{1}{6} & -\frac{1}{3} & \frac{1}{2} \\
\frac{1}{6} & \frac{1}{3} & \frac{1}{2} \\
\end{pmatrix}$
; $B=\begin{pmatrix}
2 & 3\\
2 & 0 \\
... | Yes. Note that you can perform the usual rows-columns product of two matrices $AB$ only if the number of columns of $A$ is the same as the number of rows of $B$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Find multivariable limit $\frac{x^2y}{x^2+y^3}$ Find multivariable limit of: $$\lim_{ \left( x,y\right) \rightarrow \left(0,0 \right)}\frac{x^2y}{x^2+y^3}$$
How to find that limit? I was trying to do the following, but i am not able to find a proper inequality:
$$| \frac{x^2y}{x^2+y^3} | = |y-\frac{y^4}{x^2+y^3}| \le$$... | Late answer since it was marked as duplicate somewhere else but the way I suggest was not among the answers.
Looking at the denominator of $\frac{x^2y}{x^2+y^3}$ it makes sense to consider a path of approaching $(0,0)$ close to the curve $x^2+y^3=0$.
So, consider for $x> 0$ and $\alpha > 0$ the path
$$(x,-\sqrt[3]{x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1237309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Combinatorics Recurrence relation
Let $h_n$ be a number sequence where $h_n = 3h_{n-1} - 2h_{n-2}$ with $h_0 = 0$ and $h_1 = 1$. Compute the ordinary generating function of $h_n$ and then using the generating function compute a formula for $h_n$.
Does this start looks right?
We write the recurrence relation in form:... | Once you have:
$$ g(x)=\sum_{n\geq 0}h_n x^n = \frac{\color{red}{x}}{(1-2x)(1-x)} = \frac{1}{1-2x}-\frac{1}{1-x}\tag{1}$$
it is straightfoward to check that $\color{red}{h_n=2^n-1}.$
Just check your computations since the value in zero of $\frac{1+x}{(1-2x)(1-x)}$ is one, while you have $h_0=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1238840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Compute $\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$ Given
$$\int_{0}^{\infty}\frac{x \log(x)}{(1+x^2)^2}dx$$
I couldn't evaluate this integral. My only idea here was evaluating this as integration by parts.
\begin{align}
\int\frac{x \log(x)}{(1+x^2)^2}dx & = \frac{1}{2} \int\frac{ \log(x)}{(x^2+1)^2}d(x^2+1)\\
& =... | You can evaluate using the residue theorem. In this case, by considering the contour integral
$$\oint_C dz \frac{z \log^2{z}}{(1+z^2)^2} $$
where $C$ is a keyhole contour about the positive real axis of outer radius $R$ and inner radius $\epsilon$, and letting $R \to \infty$ and $\epsilon \to 0$, we have
$$-i 4 \pi I_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1239241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Help in finding the sum of the series $$\sum_{n=1}^\infty \frac{1}{n^4+n^2+1}$$
I tried breaking into factors but it is not telescoping.
$$\frac {1}{(n^2+n+1)(n^2-n+1)} = \frac {1}{2n} \left(\frac {1}{n^2-n+1} - \frac {1}{n^2+n+1}\right)$$
| Let $\omega=\exp\frac{2\pi i}{3}$. Then:
$$\sum_{n\geq 1}\frac{1}{n^4+n^2+1}=\frac{1}{i\sqrt{3}}\left(\sum_{n\geq 1}\frac{1}{n^2-\omega}-\sum_{n\geq 1}\frac{1}{n^2-\omega^2}\right)=\color{red}{-\frac{1}{2}+\frac{\pi\sqrt{3}}{6}\tanh\frac{\pi\sqrt{3}}{2}}$$
since we can deal with series like the ones appearing in the mi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Every tetrahedron has a vertex whose adjacent edges can form a triangle
Prove that in every tetrahedron there exists a vertex $v$ such that the three edges incident at $v$ have lengths that can form a triangle.
It can be proved using a tedious casework: assume by contradiction that all vertex have edges that do not s... | Lemma: Let $x,y,z>0$. Then there exists a triangle with sides $x,y,z$ if and only if $2x^2y^2+2y^2z^2+2z^2x^2>x^4+y^4+z^4$.
Proof of lemma: Observe that $2x^2y^2+2y^2z^2+2z^2x^2-x^4+y^4+z^4=(x+y+z)(-x+y+z)(x-y+z)(x+y-z)$. If there exists a triangle with sides $x,y,z$, then each factor is positive, so their product is p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1240756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Simplifying$\left|\frac{z-3}{z+3} \right|=2$ I want to graph the following, but simplifying is the question here:
$$\left|\frac{z-3}{z+3} \right|=2$$
Now I can do this : $$\frac{|z-3|}{|z+3|}=2 $$
$$|z-3|=2|z+3|$$
$$|x+iy-3|=2|x+iy+3|$$
What manipulation do I use here? I have the answer is a circle of radius $4$ center... | $$|x+iy-3|^2=4|x+iy+3|^2$$
$$|x+iy-3|^2=(x-3)^2+y^2$$
$$4|x+iy+3|^2=4(x+3)^2+4y^2$$
So:
$$(x-3)^2+y^2=4(x+3)^2+4y^2$$
$$x^2-6x+9+y^2=4x^2+24x+36+4y^2$$
$$0=3x^2+30x+27+3y^2$$
$$0=x^2+10x+9+y^2$$
$$0=(x+5)^2-16+y^2$$
$$16=(x+5)^2+y^2$$
It's an equation of circle of radius $4$ centered at $(-5,0)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1241626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
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Non-standard question about random variables I am not sure which subbranch of mathematics this is, so I cannot give a precise tag. I am doing research, and this suddenly popped out of no where. So, please hear me out.
$x$ is a variable that takes on random values. $x$ is said to 'stable' if $|x| \leq 1$.
Here is an ... | You want to find $c$ such that $$-1\le \dfrac{\sqrt{(Y-2/c)^2-4}}{2} \le 1$$ (and in particular is real) for $0 \le Y \le 1$. Equivalently,
$$ 4 \le (Y - 2/c)^2 \le 8$$
for $0 \le Y \le 1$. So either $2 \le Y - 2/c \le 2 \sqrt{2}$ in that interval, or $-2\sqrt{2} \le Y - 2/c \le -2$. Now unfortunately,
$2\sqrt{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is the tip for this exact differential equation? $$ xdx + ydy = \frac{xdy - ydx}{x^2 + y^2} $$
I have multiplied the left part $x^2+y^2$ for $x dx + y dy$ getting
$$(x^3+xy^2+y)dx+(x^2y+y^3-x)dy=0$$
And the derivative test give me:
$\frac{dM}{dy}= 0+2xy+1$ and $\frac{dN}{dx} = 2xy+0-1$.
Where´s my mistake?
| In terms of polar coordinate $(x,y) = (r\cos\theta,r\sin\theta)$, we have
$$\begin{align}
xdx + ydy
&=
r\cos\theta(\cos\theta dr - r\sin\theta d\theta) +
r\sin\theta(\sin\theta dr + r\cos\theta d\theta) = r dr\\
\frac{xdy - ydx}{x^2+y^2}
&=
\frac{r\cos\theta(\sin\theta dr + r\cos\theta d\theta) - r\sin\theta(\cos\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1242665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to solve $\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}dx$? I need to compute $$\int \frac{(x-1)\sqrt{x^4+2x^3-x^2+2x+1}}{x^2(x+1)}\ dx.$$ I tried it on wolfram but it timed out, maybe because I am on a mobile device. Any hint is appreciated.
| Hint:
$$\begin{align}
I
&=\int\frac{x-1}{x^2\left(x+1\right)}\sqrt{\left(x^2+3x+1\right)\left(x^2-x+1\right)}\,\mathrm{d}x\\
&=\int\frac{y}{\left(\frac{1+y}{1-y}\right)^2}\sqrt{\left(\frac{5-y^2}{\left(1-y\right)^2}\right)\left(\frac{1+3y^2}{\left(1-y\right)^2}\right)}\,\frac{2\,\mathrm{d}y}{\left(1-y\right)^2};~~~\sma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243264",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
How is $\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$? As the title states, how is: $$\frac{(10^{4})^{6}-1}{10^4-1} = 1 + 10^{4} + 10^{8} + 10^{12} + 10^{16} + 10^{20}$$
I can't see the pattern. Can someone please help? Thanks.
| It is true in general that $x^6-1=(x-1)(1+x+x^2+\cdots+x^5)$. Just substitute $x=10^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1243585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Given $r>0$, find $k>0$ such that $\sqrt{(x-2)^2+(y-1)^2}Using the axioms, theorem, definitions of high school algebra concerning the real numbers, then prove the following:
Given $r>0$, find a $k>0$ such that:
$$\text{for all }x, y: \sqrt{(x-2)^2+(y-1)^2}<k\implies|xy-2|<r $$
I tried with several values given to $k$ a... | The center of the circle is at $(2,1)$, which is also on the hyperbola $xy = 2$. So it makes sense to write $x = 2 + a$ and $y = 1 + b$. Then your problem can be restated as: Given $r > 0$, find a $k > 0$ such that if $\sqrt{a^2 + b^2} < k$, then $|a + 2b + ab| < r$.
If $\sqrt{a^2 + b^2} < k$, then both $|a|$ and $|b|$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1245474",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Upper and Lower Darboux integral of a piecewise function $f(x)=x$ and $f(x)=0$. Let $0<a<b$. Find the upper and lower Darboux integrals for the function $$f(x)=x$$ if $x\in[a,b]\cap\mathbb{Q}$ and $$f(x)=0$$ if $x\in[a,b]-\mathbb{Q}$.
I am so lost on this problem. Any hints or solutions are greatly appreciated.
| Any interval in any subdivision of $[a,b]$ must contain a rational and irrational point, since the intervals are not singletons nor empty. This is because the irrationals and rationals are both dense in $\mathbb{R}$. We then have that the supremum of $f$ on any non-empty, non-singleton closed interval $[x,y]$ is $\sup[... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1250251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof of $\sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n}$ Prove that for $n\geq 2, \: \sum_{k = 1}^{n} \frac{1}{k^{2}} < 2 - \frac{1}{n} $
I used induction and I compared the LHS and the RHS but I'm getting an incorrect inequality
| By induction assume it is true for $n$.
Then, by induction,
$\sum_{k = 1}^{n+1} \frac{1}{k^{2}} =\sum_{k = 1}^{n} \frac{1}{k^{2}} +\frac{1}{(n+1)^2} < 2-\frac{1}{n} +\frac{1}{(n+1)^2}$.
So to finish we need $ 2-\frac{1}{n} +\frac{1}{(n+1)^2} \leq 2 - \frac{1}{n+1}$,
or $\frac{1}{n+1} \leq \frac{1}{n}-\frac{1}{(n+1)^2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1251544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Integral with residues $\int_0^\infty \tfrac{\sin^2(x)}{1+x^4}dx$ I am trying to calculate
$\displaystyle\int_0^\infty \dfrac{\sin^2(x)}{1+x^4}dx$
using method of residues. I have already seen this post, "Integrating $\int_{-\infty}^\infty \frac{1}{1 + x^4}dx$ with the residue theorem" And am integrating a quarter circ... | First note that the integrand is even, so
$$
\int_{0}^\infty \frac{\sin^2 x}{1+x^4}\,dx =
\frac12 \int_{-\infty}^\infty \frac{\sin^2 x}{1+x^4}\,dx.
$$
Furthermore $\sin^2 x = \dfrac{1-\cos 2x}{2}$ so
$$
\int_{0}^\infty \frac{\sin^2 x}{1+x^4}\,dx =
\frac12 \int_{-\infty}^\infty \frac{\sin^2 x}{1+x^4}\,dx =
\frac14 \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Find the integral: $\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x$ I have to find the following integral:
$$\int ( 4x -1 +3 \sqrt{x})\mathrm{d}x$$
My answer is $2x^2 -\ 1x + \frac{2\sqrt{27}}{3}$. Am I right?
| $\bf{My \; Solution::}$ $\displaystyle \int (4x-1+3\sqrt{x})dx = 4\int xdx - \int 1 dx+3\int x^{\frac{1}{2}}dx = 4\cdot \frac{x^2}{2}-x+3\cdot \frac{2}{3}x^{\frac{3}{2}}+\mathcal{C}$
Above we have used the formula $\displaystyle \int x^ndx = \frac{x^{n+1}}{n+1}+\mathcal{C}\;,$ Where $n\neq -1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\lim _{n\to \infty }\int_1^2\:\frac{x^n}{x^n+1}dx$ We have $$I_n=\int _1^2\:\frac{x^n}{x^n+1}dx$$ and we need to find $\lim _{n\to \infty }I_n$. Have any ideea how we can evaluate this limit?
| Note that $\frac{x^n}{x^n+1}=1-\frac{1}{x^n+1}$. Thus,
$$\int_1^2 \frac{x^n}{x^n+1}dx=1-\int_1^2 \frac{1}{1+x^n}dx$$
For $n>1$, the integral on the right-hand side satisfies the inequality
$$\left|\int_1^2 \frac{1}{1+x^n}dx\right|\le \int_1^2 x^{-n}dx=\frac{1-2^{1-n}}{n-1}$$
which clearly goes to zero as $n \to \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1253353",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find with proof all solutions to $2^n = a! + b!$, where $a$, $b$, $n$ are positive integers and $a \leq b$. So far I have looked at $n=1$ with $a=1$ and $b=1$ which is $$2^1 = 1! + 1! = 2 $$
and $n=2$ with $a=2$ and $b=2$, $$2^2 = 2!+ 2! = 4$$
and finally $n=3$ with $a=3$ and $b=2$, $$2^3 = 3! + 2! = 8$$
I have also tr... | *
*If $a,b \geq 3$, we have $3$ divides $a!$ and $3$ divides $b!$. Hence, $3$ divides $a!+b!$. However, $3$ doesn't divide $2^n$.
*$a \geq 3$, $b=1 \text{ or }2$. If $b=1$, then $a!+1$ is odd and cannot be $2^n$. If $b=2$, then $a!+2 = 2(a!/2+1)$, where $a!/2+1$ is odd for $a\geq 4$, and hence cannot equal $2^n$. For... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$ Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$
I am stuck at understanding why the constraint $xy\gt 1$. Here is my work so far
let $\arctan x =a\implies x=\tan a$
let $\arctan y =b\implies y=\tan b$
therefore $\fr... | Here is an insight given by a purely geometrical proof. Consider the Figure below.
Supposing $x>0$, the right-angled triangle $\triangle ABC$ has sides $\overline{AB} = 1$ and $\overline{BC} = x$, so that
$$\alpha = \arctan x.$$
Extend side $AB$ with a segment $\overline{BD}=\frac{x}{y}$, giving
$$\gamma = \arctan y.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1255968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
How can I integrate $\int {dx \over \sqrt{3^2+x^2}} $ using Trigonometric Substitution? $$\int {dx \over \sqrt{9+x^2}} = \int {dx \over \sqrt{3^2+x^2}} $$
$$ x =3\tan\theta$$
$$dx = 3\sec^2\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2 + 3^2\tan^2\theta}} d\theta$$
$$\int {3\sec^2\theta \over \sqrt{3^2(1+\tan^2\theta)}... | The two results are equivalent, up to choice of constant.
\begin{align}
\ln\left({\sqrt{9+x^2}+x \over 3}\right)+C &= \ln\left(\sqrt{9+x^2} + x\right) - \ln(3) + C \\
&= \ln\left(\sqrt{9+x^2} + x\right) + C'
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1259731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Power series properties. The sum of two series. Let $\sum a_nz^n$ and $\sum b_nz^n$ power series with radiuses of convergences $R_1,R_2$ respectively. Suppose the radius of convergence of $(\sum a_n+b_n)z^n$ is $R$. Find an example in which $\infty>R>\min \{R_1,R_2\}$, given $R_1=R_2$.
What am I to do here? Previously ... | $$1+x+\frac{x^2}{2^2}+x^3 + \frac{x^4}{2^4} + x^5 + \frac{x^6}{2^6}+\cdots $$
and
$$1-x+\frac{x^2}{2^2}-x^3 + \frac{x^4}{2^4} - x^5 + \frac{x^6}{2^6}- \cdots$$
both have radius of convergence $1.$ But the sum of these series is
$$2+ \frac{x^2}{2} + \frac{x^4}{2^3}+ \frac{x^6}{2^5}+\cdots ,$$
which has radius of conver... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1262267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
} |
How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge \sqrt{x^2+y^2}-2$ How to calculate the volume of the solid described $\frac{x^2}{4}+ \frac{y^2}{4}+z^2 \le 1$ and $z \ge \sqrt{x^2+y^2}-2$?
I try
$x=2r \cos \phi$,
$y=2r \sin \phi$,
$z=z$, but but probably not the wa... | Substitute $r^2 = x^2 + y^2$ and make $r^2$ the subject.
The inequalities are then
$4(1 - z^2) \geq r^2$ and
$(z+2)^2 \geq r^2$.
Now we compare the LHS of the two inequalities, to determine which gives a stricter upper bound on $r^2$.
$$(z+2)^2 - 4(1-z^2) = 5z^2 +4z$$
which has roots $z=-\frac45$ and $z=0$.
Thus the se... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1264885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Computing $\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx$ I wish to compute
$$\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3}dx, \quad a>0$$
but have no contour to work with. Does anyone have ideas on how to compute this integral?
| \begin{align*}
\int_{0}^{\infty} \frac{x^2 + a}{x^6 + a^3} \, dx
&= \frac{1}{a^{3/2}} \int_{0}^{\infty} \frac{x^2 + 1}{x^6 + 1} \, dx \\
&= \frac{1}{a^{3/2}} \int_{0}^{1} \frac{x^2 + 1}{x^6 + 1} \, dx+\frac{1}{a^{3/2}} \int_{1}^{\infty} \frac{x^2 + 1}{x^6 + 1} \, dx\\
&= \frac{1}{a^{3/2}} \int_{0}^{1} \frac{x^2 + 1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Finding The Number Of Inversions In A Permutation
Let the be the following permutation:
$(1 5 4)(3 6)\in S_6$
How do I count the number of inversions to calculate the sign of the permutation?
$(1 5 4)(3 6)=(1 5)(1 4),(3 6)=3$ so it has an odd sign
\begin{pmatrix}
1 & 2 & 3 & 4 & 5 & 6 \\
5 & 2 & 6 & 1 & ... | You have not done anything wrong except in thinking the permutation is even. There are indeed 9 inversions as you have calculated.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Power Series of $\frac{3}{(1-3x)^2}$ The problem is to find the power series of this function $$\frac{3}{(1-3x)^2}$$ centered at $x = 0$.
Normally you convert it into $\frac{1}{1-x}$ form. Since the denominator is squared do you multiply that out first then convert it into standard form?
| Using the standard Taylor approach, compute the successive derivatives:
$$f(x)=\frac3{(1-3x)^2},f(0)=3$$
$$f'(x)=\frac{3\cdot3\cdot2(1-3x)}{(1-3x)^4}=\frac{3\cdot3\cdot2}{(1-3x)^3},f'(0)=3\cdot3\cdot2$$
$$f''(x)=\frac{3\cdot3^2\cdot3\cdot2(1-3x)^2}{(1-3x)^6}=\frac{3\cdot3^2\cdot3\cdot2}{(1-3x)^4},f''(0)=3\cdot3^2\cdot3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1266789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Represent the transformation with respect to the standard basis
Consider a linear transformation from $R^2$ to $R^2$ defined via:
$$\left(\begin{matrix} 1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matrix} 3 \\ 1\end{matrix} \right)$$ and $$\left(\begin{matrix} -1 \\ 3\end{matrix} \right) \mapsto \left(\begin{matri... | A vector can be decomposed regarding a basis, esp. the standard basis:
$$
\left(
\begin{matrix}
x \\
y
\end{matrix}
\right)
=
x
\left(
\begin{matrix}
1 \\
0
\end{matrix}
\right)
+
y
\left(
\begin{matrix}
0 \\
1
\end{matrix}
\right)
= x e_1 + y e_2
$$
so a linear transformation $A$ acting on a vector can be calculated b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1268156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Use Lagrange Interpolation polynomial to find this $\sum_{cyc}\frac{x^3}{(x^2-y^2)(x^2-z^2)}$
Let $x,y,z$ be the solutions of the equation $t^3-t^2+2t-3=0$. Find the sum
$$\dfrac{x^3}{(x^2-y^2)(x^2-z^2)}+\dfrac{y^3}{(y^2-x^2)(y^2-z^2)}+\dfrac{z^3}{(z^2-x^2)(z^2-y^2)}$$
How can I use the Lagrange
interpolation polyn... | Yes, with some modification, we can use the Lagrange interpolation method. First, note that $(y+z)(z+x)(x+y)=(x+y+z)(yz+zx+xy)-xyz$. As $x+y+z=1$, $yz+zx+xy=2$, and $xyz=3$, $(y+z)(z+x)(x+y)=1\cdot 2-3=-1$. The required expression is then
$$\begin{align}
\sum_{\text{cyc}}\frac{x^3}{\left(x^2-y^2\right)\left(x^2-z^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1269254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How can I find the indefinite integral of $\int\sin^3x \cos^3x dx $? I am looking to evaluate the indefinite integral
\begin{equation*}
\int\sin^3x \cos^3x dx.
\end{equation*}
I'm not sure if I started this right but I broke the terms up like this:
\begin{equation*}
\int\sin^2x \sin x \cos^2x \cos x
\end{equation*}
E... | $$\begin{align}
\int \sin^3x \cos^3xdx
& = \int\sin^3x(1-\sin^2x)\cos x dx \\
& (\sin x = t , \cos x dx = dt)\\
& = \frac{\sin^4x}{4} - \frac{\sin^6x}{6} + C
\end{align}$$
$$\begin{align}
\int \sin^3x \cos^3xdx
& = \int\cos^3x(1-\cos^2x)\sin x dx \\
& (\cos x = t , \sin x dx = -dt)\\
& = -\frac{\cos^4x}{4} + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.