Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Basic question about modular arithmetic applied to the divisor sum function $\sigma(n)$ when $n=5p$ While studying the divisor sum function $\sigma(n)$ (as the sum of the divisors of a number) I observed that the following expression seems to be true always (1):
$\forall\ n=5p, p\in\Bbb P,\ p\gt 5,\ if\ d(5p)=4\ then\... | After reading the answers, I think I have found my own way to the solution. It is based on the property of the prime numbers, for $p\gt 3$ it belongs to Case 1: $1+6x$ or Case 2: $5+6x$. So replacing:
Case 1:
$1+5+p+5p = 1+5+(1+6x)+5(1+6x)=1+5+1+6x+5+30x = 12 + 36x = 12 (1 + 3x)$
And
$12(1+3x)\ mod\ 9 = 3 * (1 + (3\ or... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1249399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$
Find $\int \frac{1+\sin x \cos x}{1-5\sin^2 x}dx$
I used a bit of trig identities to get: $\int \frac {2+\sin (2x)}{-4+\cos(2x)}dx$ and using the substitution: $t= \tan (2x)$ I got to a long partial fractions calculation which doesn't seem right.
Any hints on how t... | We have
$$\int\frac{1+\sin x\cos x}{1-5\sin^2x}\mathrm{d}x
=\int\frac{\mathrm{d}x}{1-5\sin^2x}+\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x$$
and
$$\int\frac{\sin x\cos x}{1-5\sin^2x}\mathrm{d}x
=-\frac{1}{10}\int\frac{\mathrm{d}\left(1-5\sin^2x\right)}{1-5\sin^2x}
=-\frac{1}{10}\ln\left(1-5\sin^2x\right).$$
Then,
$$... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the least degree Polynomial whose one of the roots is $ \cos(12^{\circ})$ Find the least degree Polynomial with Integer Coefficients whose one of the roots is $ \cos(12^{\circ})$
My Try: we know that $$\cos(5x)=\cos^5x-10\cos^3x\sin^2x+5\cos x\sin^4x$$ Putting $x=12^{\circ}$ and Converting $\sin$ to $\cos$ we have... | Without any quartic solving:
Set $y=2x$. The polynomial becomes
$$y^5-5y^3+5y-1=(y-1)(y^4+y^3-4y^2-4y+1)$$
As $2\cos\dfrac \pi{15}\neq 1$, it is a root of $\,f(y)=y^4+y^3-4y^2-4y+1$.
Let's show this polynomial is irreducible over $\mathbf Z$. Indeed, it has no integer root (the only possibilities are $1$ and $-1$, no... | {
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"url": "https://math.stackexchange.com/questions/1252518",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x)$
If $f(x) = \sin^4 x+\cos^2 x\;\forall x\; \in \mathbb{R}\;,$ Then $\bf{Max.}$ and $\bf{Min.}$ value of $f(x).$
My Solution:: Let $$\displaystyle y = \sin^4 x+\cos^2 x \leq \sin^2 x+\cos^2 x=1$$
And for Minimu... | $$\cos^4x-\cos^2x=\cos^2x(\cos^2x-1)=-\sin^2x\cos^2x=-\dfrac{(2\sin x\cos x)^2}4=-\dfrac{\sin^22x}4$$
For real $x,0\le\sin^22x\le1\iff-1\ge-\sin^22x\ge0$
In case you don't know double angle formula, $(\sin x\pm\cos x)^2\ge0\iff-1\le2\sin x\cos x\le1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1252666",
"timestamp": "2023-03-29T00:00:00",
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Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$ Show that $\arctan x+\arctan y = \pi+\arctan\frac{x+y}{1-xy}$, if $xy\gt 1$
I am stuck at understanding why the constraint $xy\gt 1$. Here is my work so far
let $\arctan x =a\implies x=\tan a$
let $\arctan y =b\implies y=\tan b$
therefore $\fr... | I think this might be an idea for you.
Let $$f(x,y)= \arctan(x)+\arctan(y)-\arctan(\frac{x+y}{1-xy})$$ defined for $xy> 1$, then we have
$$ \frac{ \partial f}{\partial x }= \frac{1}{1+x^2} -\frac{(1-xy)+y(x+y)}{(1-xy)^2 +(x+y)^2 }= \frac{1}{1+x^2} -\frac{1+y^2}{(1+y^2)(1+x^2) }=0 $$
Similarly you can show that $ \fr... | {
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"url": "https://math.stackexchange.com/questions/1255968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solutions to $\frac{1}{a} + \frac{1}{b} = \frac{1}{100}$? I encountered this problem yesterday and successfully solved it. I'm interested in seeing other people's approach to solving this problem.
Problem: How many ordered pairs $(a, b)$ are solutions to $\frac{1}{a} + \frac{1}{b} = \frac{1}{100}$ where $a, b \in \mat... | This reminded me of problem 454 on project Euler which I wrote an algorithm for a year ago. (Here is the algorithm if anyone is interested)
So here is how I would solve this question.
Observe that
$$\frac{1}{2} = \frac{1}{3} + \frac{1}{6} \\ \frac{1}{3} = \frac{1}{4} + \frac{1}{12} \\ \frac{1}{4} = \frac{1}{5} + \frac{... | {
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"url": "https://math.stackexchange.com/questions/1256207",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove there do not exist prime numbers $a$, $b$, and $c$ such that $a^3+b^3=c^3$ Prove there do not exist prime numbers $a$, $b$, and $c$ such that $a^3+b^3=c^3$.
From what I understand this proof requires a proof by contradiction or contrapositive...
| $a^3+b^3=(a+b)(a^2-ab+b^2)$ So we have
$(a+b)(a^2-ab+b^2)=c^3$
So $(a+b)=c$ and $(a^2-ab+b^2)=c^2$
Hence $(a+b)^2=a^2+2ab+b^2=a^2-ab+b^2$ which is a contradiction since it implies $3ab=0$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Evaluating trigonometric limit: $\lim_{x \to 0} \frac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2}$
Evaluate $\lim_{x \to 0} \cfrac{ x\tan 2x - 2x \tan x}{(1-\cos 2x)^2} $
This is what I've tried yet:
$$\begin{align} & \cfrac{x(\tan 2x - 2\tan x)}{4\sin^4 x} \\
=&\cfrac{x\left\{\left(\frac{2\tan x}{1-\tan^2 x} \right) - 2\... | Using power series, we only need two:
$$ \tan{x} = x + \frac{1}{3}x^3 + O(x^5), $$
and
$$ \cos{x} = 1- \frac{1}{2}x^2 + O(x^4). $$
Then
$$ x(\tan{2x}-2\tan{x}) = x\left( 2x+\frac{8}{3}x^3 - 2x - \frac{2}{3}x^3 + O(x^5) \right) = 2 x^4 + O(x^5), $$
and
$$ (1-\cos{2x})^2 = \frac{1}{4}(2x)^4 + O(x^6) = 4x^4 + O(x^6), $$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1263968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Write 100 as the sum of two positive integers
Write $100$ as the sum of two positive integers, one of them being a multiple of $7$, while the other is a multiple of $11$.
Since $100$ is not a big number, I followed the straightforward reasoning of writing all multiples up to $100$ of either $11$ or $7$, and then find... | $7x+11y=100$
$7x=100-11y$
$x=\frac{100-11y}7=14-2y+\frac{2+3y}7$
$a=\frac{2+3y}7$
$7a=2+3y$
$3y=-2+7a$
$y=\frac{-2+7a}3=-1+2a+\frac{1+a}3$
$b=\frac{1+a}3$
$3b=1+a$
$a=3b-1$
$y=\frac{-2+7(3b-1)}3=\frac{-9+21b}3=-3+7b$
$x=\frac{100-11(-3+7b)}7=\frac{133-77b}7=19-11b$
$\begin{matrix}
x\gt 0&\to&19-11b\gt 0&\to&11b\lt 19&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1265426",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "33",
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find coefficient of $x^{50}$ Let $f(x)=\frac{1}{(1+x)(1+x^2)(1+x^4)}$then find the coefficient of term $x^{50}$ in $(f(x))^3$.I think that we can set $$(f(x))^3=\frac{a}{(1+x)^3}+\frac{b}{(1+x^2)^3}+\frac{c}{(1+x^4)^3}$$ and find a,b and c then use Taylor seri .
| We have $f(x)=\frac{1}{(1+x)(1+x^2)(1+x^4)}=\frac{1-x}{1-x^8}=(1-x)(1-x^8)^{-1}$ if $|x^8|<1$.
Now note that when $|x|<1$ then \begin{align*}
(1-x)^{-1}=&\sum\limits_{n\geq 0}x^n\\
-(1-x)^{-2}=&\sum\limits_{n\geq 1}nx^{n-1}\\
+2(1-x)^{-3}=&\sum\limits_{n\geq 2}n(n-1)x^{n-2}=\sum\limits_{n\geq 0}(n+2)(n+1)x^{n}\\
(1-x)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1267677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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compute improper integrals using integration by parts Compute
\begin{equation*}
\int_0^\infty \frac{\sin^4(x)}{x^2}~dx\text{ and }\int_0^\infty \frac{\sin (ax) \cos (bx)}{x}~dx.
\end{equation*}
For the first integral I tried letting $u = \sin ^4 x$ and $dv= \frac{1}{x^2}~dx$, which simplified to $\int_0 ^\infty \frac{... | $$\begin{align}\int_0^{\infty} dx \frac{\sin^4{x}}{x^2} &= \int_0^{\infty} dx \frac{\sin^2{x}}{x^2} - \int_0^{\infty} dx \frac{\sin^2{x} \cos^2{x}}{x^2} \\ &= \int_0^{\infty} dx \frac{\sin^2{x}}{x^2} - \frac14 \int_0^{\infty} dx \frac{\sin^2{2 x}}{x^2} \\ &= \int_0^{\infty} dx \frac{\sin^2{x}}{x^2} - \int_0^{\infty} d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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For which values of $a,b,c$ is the matrix $A$ invertible? $A=\begin{pmatrix}1&1&1\\a&b&c\\a^2&b^2&c^2\end{pmatrix}$
$$\Rightarrow\det(A)=\begin{vmatrix}b&c\\b^2&c^2\end{vmatrix}-\begin{vmatrix}a&c\\a^2&c^2\end{vmatrix}+\begin{vmatrix}a&b\\a^2&b^2\end{vmatrix}\\=ab^2-a^2b-ac^2+a^2c+bc^2-b^2c\\=a^2(c-b)+b^2(a-c)+c^2(b-a)... | This is a transpose of the Vandermonde matrix: http://en.wikipedia.org/wiki/Vandermonde_matrix
According to the wikipedia page, the determinant of the matrix will be non-zero precisely when none of the elements $a, b$, or $c$ are the same.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1270699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 3
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Prove $(8k)^{8k}+(8k+1)^{8k+1}$ and $(8k+1)^{8k+1}+(8k+2)^{8k+2}$ are never perfect squares
Prove $$(8k)^{8k}+(8k+1)^{8k+1}\ \ \text{ and } \ \ \ (8k+1)^{8k+1}+(8k+2)^{8k+2}$$ are never perfect squares ($k\ge 1$).
mod $8$ gives $1$ for both, which is a quadratic residue, so doesn't solve it. Found in AoPS.
| We can say something a bit stronger. For $n$ a positive integer, $n^n+(n+1)^{n+1}$ cannot be a perfect square if $n$ is even or $n\equiv 1 \pmod{4}$.
For even $n$ assume $n=2x>0$ and
$$
(2x)^{2x}+(2x+1)^{2x+1}=u^2 \\
(2x+1)^{2x+1} = (u-(2x)^x)(u+(2x)^x)
$$
Let $A=u-(2x)^x,B=u+(2x)^x$ and $g=\gcd(A,B)$. Then
$$
g\mid B-... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Number of ways to throw at most 14 with 4 dice - generating functions Determine the chance to throw at most 14 with 4 normal dice. I will set up the right generating function to determine the number of ways tot thow at most 14 with 4 normal dice and I need some help. I know that:
\begin{align}
x_1 + x_2 + x_3 + x_4 \le... | The probability generating function for rolling a normal die is:
$G(t)=\frac{t}{6} + \frac{t^2}{6} + \frac{t^3}{6} + \frac{t^4}{6} + \frac{t^5}{6} + \frac{t^6}{6}$
To find the probability generating function for the sum of four independent dice then you have
$G_S(t)=\left(\frac{t}{6} + \frac{t^2}{6} + \frac{t^3}{6} + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1278595",
"timestamp": "2023-03-29T00:00:00",
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Quadratic solutions puzzle The equation $x^2+ax+b=0$, where $a\neq b$, has solutions $x=a$ and $x=b$. How many such equations are there?
I'm getting $1$ equation as I can only find $a=b=0$ as an equation, which is not allowed.
$$x=\frac{±\sqrt{a^2-4 b}-a}2$$
$x=a$ or $b$ so these are the equations
$$a=\frac{\sqrt{a^2-4... | You can do this by:
$$(x-a)(x-b)=x^2-(a+b)x+ab=x^2+ax+b$$
Equating constant terms gives $ab=b$ which means $a=1$ or $b=0$
Equating coefficients of $x$ gives $-a-b=a$ or equivalently $2a=-b$
| {
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"url": "https://math.stackexchange.com/questions/1279477",
"timestamp": "2023-03-29T00:00:00",
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Finding Number of Ordered Solutions to Equation $$ A \times B \times C \times D \times E \times F = 7 \times 10^7 $$
How can I find the number of ordered solutions for integers (I mean for integers $A,B,C,D,E,F$) so that they can satisfy the above equation?
| The factorization of $7 \times 10^7 $ is simply $2^7 \times 5^7 \times 7 $.
Suppose for now we restrict the search for positive integer solutions only.
For non-negative integers $x_i, y_i, z_i $, let $A = 2^{x_1} \cdot 5^{y_1} \cdot 7^{z_1}, B = 2^{x_2} \cdot 5^{y_2} \cdot 7^{z_2} , \ldots, F = 2^{x_6} \cdot 5^{y_6} \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1281032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Conic reduction I'm trying to reduce this conic :
$x^2+y^2+2xy+x+y=0$
to a canonical form.
I started with finding the eigenvalues of the matrix associated to the quadratic form $x^2+y^2+2xy$
I found $z_1=2 , z_2=0$
and a basis for the diagonalized matrix $e_1= ({1\over \sqrt2},{1\over \sqrt2})$ $e_2=({1\over \sqrt2},{... | Following the method blindly $x^2+y^2+2xy+x+y={\bf x}^T A{\bf x}+K{\bf x}=\begin{pmatrix}x&y\end{pmatrix}\begin{pmatrix}1&1\\1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}+\begin{pmatrix}1&1\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}=0$,
we use $P=\begin{pmatrix}\frac1{\sqrt{2}}&\frac1{\sqrt{2}}\\\frac1{\sqrt{2}}&-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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$a^{13} \equiv a \bmod N$ - proof of maximum $N$ From Fermat's Little Theorem, we know that $a^{13} \equiv a \bmod 13$. Of course $a^{13} \equiv a \bmod p$ is also true for prime $p$ whenever $\phi(p) \mid 12$ - for example, $a^{13} = a^7\cdot a^6 \equiv a\cdot a^6 = a^7 \equiv a \bmod 7$.
So far I have that the larges... | Putting in $a=2$, we get that $N$ divides $2^{13} - 2 = 2 \cdot 3^2 \cdot 5 \cdot 7 \cdot 13$.
On the other hand, putting in $a=3$, we get that $N$ divides $3^{13} - 3 = 2^4 \cdot 3 \cdot 5 \cdot 7 \cdot 13 \cdot 73$.
Hence $N$ must divide $2 \cdot 3 \cdot 5 \cdot 7 \cdot 13 = 2730$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The box has minimum surface area Show that a rectangular prism (box) of given volume has minimum surface area if the box is a cube.
Could you give me some hints what we are supposed to do??
$$$$
EDIT:
Having found that for $z=\frac{V}{xy}$ the function $A_{\star}(x, y)=A(x, y, \frac{V}{xy})$ has its minimum at $(\s... | Volume $V=xyz$ given. Area $A(x,y,z)=2(xy+yz+zx)$, to minimise, when $x,y,z>0$ and $xyz=V$.
Fact. If $a,b,c>0$, then $a+b+c\ge 3\sqrt[3]{abc}$, and equality holds if and only if $a=b=c$.
Proof. We set $X=\sqrt[3]{a}$, $Y=\sqrt[3]{b}$ and $Z=\sqrt[3]{c}$. Then the identity
$$
X^3+Y^3+Z^3-3XYZ=\frac{1}{2}(X+Y+Z)\big((X-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Arc length of natural log function I am currently trying to find the arc length of $f(x)=ln(x)$, which involves the integral
$$\int \sqrt{1+\frac1{x^2}}dx$$
I managed to solve the integral correctly but I want to know if there is a simpler way as I ended up canceling two terms later on.
For reference this is what I did... | Standard ways to find this integral are to substitute $x=\tan\theta$ or $x=\sinh t$.
You can also let $\frac{1}{x}=\tan\theta$, so $x=\cot\theta$ and
$\displaystyle\int\sqrt{1+\frac{1}{x^2}}\;dx=\int\sec\theta\big(-\csc^{2}\theta\big)d\theta$. $\;\;$Then letting $u=\sec\theta, dv=-\csc^{2}\theta d\theta$ gives
$\disp... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the 2x2 matrix given 2 equations. Find the $2 \times 2$ matrix ${A}$ such that ${A}^2 = {A}$ and
${A} \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}.$
I have tried to express A as a matrix with variables a,b,c, and d, but it gets too messy with the equations. Any help is appreciated... | Notice that $A \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$ and also $A \begin{pmatrix} 6 \\ 2 \end{pmatrix} = A^2 \begin{pmatrix} 7 \\ -1 \end{pmatrix} = A \begin{pmatrix} 7 \\ -1 \end{pmatrix} = \begin{pmatrix} 6 \\ 2 \end{pmatrix}$.
Thus $A \begin{pmatrix} 7 &6 \\ -1 & 2 \end{pmatri... | {
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Find the integer x: $x \equiv 8^{38} \pmod {210}$ Find the integer x: $x \equiv 8^{38} \pmod {210}$
I broke the top into prime mods:
$$x \equiv 8^{38} \pmod 3$$
$$x \equiv 8^{38} \pmod {70}$$
But $x \equiv 8^{38} \pmod {70}$ can be broken up more:
$$x \equiv 8^{38} \pmod 7$$
$$x \equiv 8^{38} \pmod {10}$$
But $x \equiv... | An alternate solution using the Chinese Remainder Theorem.
First off, $2, 3, 5$ and 7 are pairwise relatively prime thus we know the following system of congruences has a unique solution modulo $2\times3\times5\times7=210$.
\begin{cases} x \equiv 4 \pmod {5} \\ x \equiv 0 \pmod {2} \\ x \equiv 1 \pmod {7} \\ x \e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 3
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For which values of $\alpha \in \mathbb{R}$, does the series $\sum_{n=1}^\infty n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1})$ converge? How do I study for which values of $\alpha \in \mathbb{R}$ the following series converges?
(I have some troubles because of the form [$\infty - \infty$] that arises when taking the l... | First hint: $$\sqrt{n+1}-2\sqrt{n}+\sqrt{n-1} = \sqrt{n+1}-\sqrt{n}+\sqrt{n-1}-\sqrt{n}$$
First term:
$$\sqrt{n+1}-\sqrt{n} = (\sqrt{n+1}-\sqrt{n})\frac{\sqrt{n+1}+\sqrt{n}}{\sqrt{n+1}+\sqrt{n}} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$$
Second term:
$$\sqrt{n-1}-\sqrt{n} = (\sqrt{n-1}-\sqrt{n})\frac{\sqrt{n-1}+\sqrt{n}}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Dividing a Matrix into three parts The matrix $A$ is given by $$\left(\begin{array}{ccc}
1 & 2 & 1 \\
1 & 1 & 2 \\
2 & 3 & 1 \end{array} \right)$$
Given that $A^3$ can be expressed as $A^3$=$aA^2+bA+cI$, find the values of $a,b,c$. Furthermore, express the inverse $A^{-1}$ in the form $A^{-1}$=$dA^2+eA+fI$, where $d$,$... | Calculating $\det(A-\lambda I)=\det\begin{pmatrix}1-\lambda&2&1\\1&1-\lambda&2\\2&3&1-\lambda\end{pmatrix}$ gives
$\hspace{.3 in}(1-\lambda)^3+11-10(1-\lambda)=-(\lambda^3-3\lambda^2-7\lambda-2)$,
so by the Cayley-Hamilton theorem, $\;\;A^3-3A^2-7A-2I=0$
and therefore $\;\;A\big[\frac{1}{2}(A^2-3A-7I)\big]=I$.
Altern... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290440",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Infinite solutions for $(\frac{n+1}{n})^a\cdot (\frac{m+1}{m})^b = 2$ Given $(\frac{n+1}{n})^a\cdot (\frac{m+1}{m})^b = 2$ where a, b, n, and m are all positive integers, are there infinitely many solutions $(a,b,n,m)$?
| Let's try this:
$(n+1)^{a}(m+1)^{b}=2\cdot n^{a}m^{b}$
Obviously either $(n+1)$ or $(m+1)$ must be even but not both. Take
$(n+1)=2^{j}\cdot r;(n+1)^{a}=2^{a\cdot j}r^{a}$
Now $n+1$ is relatively prime to $n$ so if $a\cdot j>1$ then $m=2^{k}s$
and we have
$a\cdot j-b\cdot k=1$
Thus $(a,b),(a,j),(j,k),(j,b)$ are all re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
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Simplifying $\Big[\dfrac{5-\sqrt{a}}{5+\sqrt a}-\dfrac{\sqrt a+5}{\sqrt a-5}+2\Big]^{-2}$ Simplifying $$\Big[\dfrac{5-\sqrt{a}}{5+\sqrt a}-\dfrac{\sqrt a+5}{\sqrt a-5}+2\Big]^{-2}$$
When I try, the numerator cancels out to $0$, yet the answer sheet says $(25-a)^2/10000$. Where am I going wrong & how is getting $10 000... | Let $s = \sqrt{a}$:
$$
\frac{5-s}{5+s}-\frac{s+5}{s-5}
= \frac{-(s-5)^2 - (s+5)^2}
{(5+s)(5-s)}
= \frac{(s-5)^2 + (s+5)^2}
{s^2-5^2}
= \frac{2s^2 + 2 \cdot 5^2}{s^2-5^2}
= \frac{2(a + 25)}{a-25}
$$
Can you finish the problem?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1291367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Math about Geometric series In a geometric series, the sum of $1^{st}$ term $+$ $2^{nd}$ term $+$ $3^{rd}$ term $= 38$,
the sum of $2^{nd}$ term $+ 4^{th}$ term $= 17 \frac{1}{3}$;
how to calculate the common ratio? ( it is for sure the answer is $r= \frac{2}{3}$ and the first term $= 18$)
| $$a(1+r+r^2) = 38\\ar+ar^3 = \frac{52}3 $$ eliminating $a$ leaves us with
$$\frac{1+r+r^2}{r+r^3} = \frac{57}{26}.$$ this can be rewritten as $$ 0=57r^3-26r^2+31r-26 = (3r-2)(19r^2+4r+13).$$ the quadratic equation $19r^2 + 4r+13=0 $ has no real roots, therefore $$r = \frac23, a = 18. $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1296114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Finding the area under $\frac{3}{2x+1}$ and $3x-2$ Find the area between these curves.
$$y=\dfrac{3}{2x+1},\qquad y=3x-2;\qquad x=2\quad \text{et} \quad y=0 $$
Indeed,
I calculate the integral of the blue function between $1$ and $2$. Then, I will calculate the area of the triangle between the yellow line and the x-a... | i think the first integral must be $$\int_{2/3}^{1}(3x-2)dx$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1298499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I simplify $1\times 2 + 2 \times 3 + .. + (n-1) \times n$ progression? I have a progression that goes like this:
$$1\times 2 + 2 \times 3 + .. + (n-1) \times n$$
Is there a way I can simplify it?
| Yes.
\begin{align*}
1\cdot 2 + 2\cdot 3 + \dotsb + (n-1)\cdot n
&= 1\cdot(1+1) + 2\cdot(2+1) + \dotsb + (n-1)\cdot((n-1)+1) \\
&= 1^2 + 1 + 2^2 + 2 + \dotsb + (n-1)^2 + (n-1) \\
&= (1^2+2^2+\dotsb+(n-1)^2) + (1+2+\dotsb+(n-1))
\end{align*}
Both of those sums have well-known closed forms; see here, for example.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Distance between line and a point Consider the points (1,2,-1) and (2,0,3).
(a) Find a vector equation of the line through these points in parametric form.
(b) Find the distance between this line and the point (1,0,1). (Hint: Use the parametric form of the equation and the dot product)
I have solved (a), Forming:
Vecto... | Using this formula and your computations from (a), we get the expression for distance $d$:
$$
d =
\frac{\left\| \
\
\left(\
\begin{bmatrix}
1 \\ 2 \\ -1
\end{bmatrix}
-
\begin{bmatrix}
1 \\ 0 \\ 1
\end{bmatrix}
\
\right)
\times
\,
\begin{bmatrix}
1 \\ -2 \\ 4
\end{bmatrix}
\ \
\right\|}
{
\begin{Vmatrix}
1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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$3^x + 4^y = 5^z$ This is an advanced high-school problem.
Find all natural $x,y$, and $z$ such that
$3^x + 4^y = 5^z$.
The only obvious solution I can see is $x=y=z=2$. Are there any other solutions?
| First mod the whole thing by 3, then $1^y = 2^z$, so $z = 2z_2$ for some $z_2\in\mathbb{N}$. So $3^x = 5^{2z_2} - 2^{2y} = (5^{z_2} + 2^y)(5^{z_2} - 2^y)$.
If $5^{z_2} - 2^y\ne 1$, then by modding each factor by 3 we find that $z_2\ne y \mod 3$ and $z_2 = y \mod 3$, contradiction.
We can invoke Mihăilescu's theorem to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find h in terms of r A sphere and a cylinder have equal volumes. The sphere has a radius 3r.
The cylinder has radius 2r and height h.
Find h in terms of r.
I'm only 15, someone walk me through this as simple as possible :)
| Volume of sphere $= \frac 43 \pi R^3 = \frac 43 \pi (3r)^3$
Volume of cylinder $= \pi R^2 h = \pi(2r)^2h$
These two volumes should be equal so equate them:
$$\frac 43 \pi (3r)^3 = \pi(2r)^2h$$
Now let's solve for $h$ in terms of $r$.
First notice that $(3r)^3 = 3^3r^3 = 27r^3$. Likewise $(2r)^2 = 2^2r^2 = 4r^2$. Ther... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1301348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
} |
If $ x=\frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $y = \frac{\cos^3 t}{\sqrt{\cos 2t}}\;,$ Then $ \frac{dy}{dx}$ in terms of $t$ If $\displaystyle x=\frac{\sin^3 t}{\sqrt{\cos 2t}}$ and $\displaystyle y = \frac{\cos^3 t}{\sqrt{\cos 2t}}\;,$ Then $\displaystyle \frac{dy}{dx}$ in terms of $t$
$\bf{My\; Try::}$ Using The Formul... | You could simplify the intermediate calculations using reduction formulas of the powers of sine and cosine functions $$x=\frac{\sin ^3(t)}{\sqrt{\cos (2 t)}}$$ $$\frac{dx}{dt}=\frac{\sin (2 t) \sin ^3(t)}{\cos ^{\frac{3}{2}}(2 t)}+\frac{3 \sin ^2(t) \cos
(t)}{\sqrt{\cos (2 t)}}=\frac{\sin ^2(t) (2 \cos (t)+\cos (3 t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solve Inequality for $ |x| $ Given $$\big|\frac{(x-2)}{(x+3)}\big| < 4,$$ solve for $x.$
\ My solution
$$|x - 2| < 4|x + 3|$$
Since,
$ |x - 2| \ge |x| - |2| $ and
$ |x + 3| \le |x| + |3| $ according to triangle inequality;
$|x| - |2| < 4|x| + 4|3| $
$-14 < 3|x|$
$|x| > \frac{-14}{3}$
Is this the final answer?
| from the graph $$f(x) = \frac{x-2}{x+3} = 1-\frac5{x+3}$$ you can see that it has vertical asymptote $x = 3$ and a horizontal one $y = 1.$ the function $f$ is decreasing on $\-infty, -3)$ and increasing on $-3, \infty).$
solving $f(x) = 4$ gives $x = -14/3$ and $f(x) = -4$ gives $x = -2.$
therefore $$ \left|\frac{x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding the point of intersection of two tangents to a circle
19. A circle $C$, of radius $r$, passes through the points $A (a, 0)$, $A_{1} (-a, 0)$ and $B (0, b)$, where $a$ and $b$ are positive and are not equal; a circle $C_{1}$, of radius $r_{1}$, passes through $A, B$ and $B_{1} (0, -b)$. Prove that the centre of... |
Compare the above drawings and determine who's answer is more correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
limits and infinity I'm having trouble wrapping my head around some of the 'rules' of limits. For example,
$$
\lim_{x\to \infty} \sqrt{x^2 -2} - \sqrt{x^2 + 1}
$$
becomes
$$
\sqrt{\lim_{x\to \infty} (x^2) -2} - \sqrt{\lim_{x\to \infty}(x^2) + 1}
$$
which, after graphing, seems to approach zero. My question is how do yo... | Hint
$$\sqrt{x^2 -2} - \sqrt{x^2 + 1}=(\sqrt{x^2 -2} - \sqrt{x^2 + 1})\cdot \frac{\sqrt{x^2 -2} + \sqrt{x^2 + 1}}{\sqrt{x^2 -2} +\sqrt{x^2 + 1}}=\frac{-3}{\sqrt{x^2 -2} + \sqrt{x^2 + 1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1305782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Conditional entropy This is really annoying me: Two players A and B play a best-of-three squash match and the first player to win two games wins the match. A wins a game with probability $p$, B wins with probability $1-p$. Let $X$ be the number of games won by A, let $Y$ be the number of matches played. Show $H(Y|X=x)$... | $\mathsf P(Y=2\mid X=0)$ is the probability that two games in a best out of three match were played, given that player A won no games. It is in face certain that player B will have won two games in a row if Player A won none of the games in a best of three match. Let's examine the possible outcomes and their probab... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding a basis for the nullspace $A$=\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}
I need to find the null space for this matrix. After performing row operations $R_3 + 3R_1$, then $R_{new row 3}$ + $R_2$, I got the reduced row echelon form. (the variable I chose were $... | If matrix
$$
\begin{bmatrix}-2 & 5 & 3 & -1\\ 0 & 1 & -4 & 2\\ 6 & -14 & -13 & 1\\ 0 & 0 &0 &0\end{bmatrix}
$$
Row reduced form of matrix is
$$
\begin{bmatrix}1 & 0 & -\frac{23}{2} & 0\\ 0 & 1 & -4 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 &0 &0\end{bmatrix}
$$
So basis is
$$
\begin{bmatrix}23 \\ 8 \\ 2 \\ 0\end{bmatrix}
$$
If matri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1306305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
$\sin(2x)-\cos(2x)-\sin(x)+\cos(x)=0$ I would like to share a trigonometry question here. Wonder is there another way to solve it or not.
$\sin(2x)-\cos(2x)-\sin(x)+\cos(x)=0$
$2\cos(x)\sin(x)-(1-2\sin^2(x))-\sin(x)+\cos(x)=0$
$2\cos(x)\sin(x)-1+2\sin^2(x)-\sin(x)+\cos(x)=0$
$2\sin^2(x)+2\cos(x)\sin(x)-\sin(x)+\cos(x)-... | If $\cos A-\sin A=\cos B-\sin B$
As $\cos A-\sin A=\cdots=\sqrt2\cos\left(A+\dfrac\pi4\right),$
$\cos\left(A+\dfrac\pi4\right)=\cos\left(B+\dfrac\pi4\right)$
$\implies A+\dfrac\pi4=2m\pi\pm\left(B+\dfrac\pi4\right)$ where $m$ is any integer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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When does this equation have a solutions in integers: $ z ^ x + \bar z ^ y = 1 $? Let $ z $ be a complex number such that $ z = \alpha + i \beta $, where $ \alpha$ and $\beta$ are integers. Let $ \bar z $ be the complex conjugate of $ z $, and let $ x $ and $ y $ be integers.
When does the following equation have a so... | First of all, if you accept that $0^0=1$, then there are some trivial solutions:
Let $z$ be $0$ and let one of $x$ and $y$ be $0$ and the other one be any
positive integer, and you will have $z^x+\bar z^y=0+1=1$. It's easy to see that
if one of $x$, $y$ and $z$ is equal to $0$, then the above cases are the only
possibl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1308206",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\dis... | I don't know if there's still any interest in this question from more than a year ago, but here's an elementary solution:
The numerator in the problem is $H_n=1+\frac12+\frac13+\dots+\frac{1}{n},$ the $n^{\text{th}}$ harmonic number.
The denominator in the problem is
\begin{align}
1+\frac13+\frac15+\dots+\frac1{2n+1}&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 3
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Partial Fraction Using the Long Method I've been trying to express $$\frac{x^2-13}{x^3-7x+6}$$ as a partial fraction, and, so far I have arrived at $$x^2-13=A(x-2)(x+3)+B(x-1)(x+3)+C(x-1)(x-2)$$.
From this onwards, I substituted different values of x to remove unknowns and find the values of the unknowns individually.... | As you have noticed, since $x^3-7x+6= (x-1)(x^2+x-6)=(x-1)(x-2)(x+3) $ we can write $$\frac{x^2-13}{x^3-7x+6}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x+3}$$ to get $$x^2-13=A(x-2)(x+3)+B(x-1)(x+3)+C(x-1)(x-2).$$ Then, to solve this equation for $A, B, C$, it is easier to plug in $x=1,2,-3$, i.e., the roots of denominator,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer... | Define the $2$-adic valuation $\nu_2(r)$ of a nonzero rational number $r = \frac{p}{q}$ to be the number of times $2$ divides $p$ minus the number of times $2$ divides $q$. The $2$-adic valuation of the square of a rational number is even. But the $2$-adic valuation of $2$ is odd. Hence $2$ is not the square of a ratio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 16
} |
Equality with dilogarithms During some calculations with definite integrals I happened to get the equality
\begin{eqnarray}
2\, \textrm{Li}_2(-\frac{1}{2}) - 2 \, \textrm{Li}_2(\frac{1}{4})+ 2\, \textrm{Li}_2(\frac{2}{3})=
3 \log^2 2 - \log^2 3
\end{eqnarray}
Does this follow from some well known equalities for dilog... | I'm going to use the following 3 identities along with the known value $\text{Li}_{2} \left(\frac{1}{2} \right) = \frac{\pi^{2}}{12} - \frac{1}{2} \log^{2}(2)$:
$$\text{Li}_{2}(1-z) = - \text{Li}_{2} \left(1- \frac{1}{z} \right) - \frac{1}{2} \log^{2} (z) , \quad z \notin (-\infty,0] \tag{1}$$
$$\text{Li}_{2}(z) = - \t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1312754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Integral of Trigonometric Identities $$\int(\sin(x))^3(\cos(2x))^2dx$$
I can write $$\sin^3(x)=\sin(x)(1-\cos^2(x)=\sin(x)-\sin(x)\cos^2(x)$$
for $$\cos^2(2x)=(1-\sin^2(x))^2=1-4\sin^2(x)+4\sin^4(x)$$
after simplifying the Trig identities i get:
$$\int(sin^3(x)-4sin^5(x)+4sin^7(x))dx$$
so i need to know how to go furth... | take your integrand $$\sin^3 x-4\sin^5 x+4\sin^7 x $$ and factor the $\sin x$ so you get $$ \sin x\left(\sin^2 x-4\sin^4 x+4\sin^6 x\right) = \sin x\left( (1-\cos^2 x)-4(1-\cos^2 x)^2+4(1-\cos^2 x)^3 \right)$$
now make the substitution $$u = \cos x , \quad du = -\sin x \,dx $$ so you have $$\int\left( \sin^3 x-4\sin^5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $(f^{-1})'(a) = f(x) = 2x^3 + 3x^2+7x+4, a=4 $ Find $(f^{-1})'(a) = f(x) = 2x^3 + 3x^2+7x+4, a=4 $
How do I go about solving a problem like this? What are the steps?
| first you need to find $x$ such that $f(x) = 4.$ you know there is exactly one such $x$ because $f$ has an inverse. by trial and error, you find that $x = 0.$ so $f$ sends $0$ to $4.$ the slope of $f$ at $x = 0$ is $f'(0) = 7.$ therefore the slope of $f^{-1}$ at the point $y = 4$ is $\left(f^{-1}\right)'(4) = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$ Let $x,y,z\ge 0$, show that
$$\dfrac{y}{xy+2y+1}+\dfrac{z}{yz+2z+1}+\dfrac{x}{zx+2x+1}\le\dfrac{3}{4}$$
I had solve
$$\sum_{cyc}\dfrac{y}{xy+y+1}\le 1$$
becasuse After some simple computations, it is equivalent to
$$(1-xyz)^2\ge 0$$
| Because by C-S $$\sum_{cyc}\frac{y}{xy+2y+1}\leq\frac{1}{4}\sum_{cyc}y\left(\frac{1^2}{xy+y}+\frac{1^2}{y+1}\right)=\frac{1}{4}\sum_{cyc}\left(\frac{1}{x+1}+\frac{y}{y+1}\right)=\frac{3}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Convolution of binomial coefficients As part of a (SE) problem I've been working on, I came up with this expression:
$$
\sum_{i=0}^M\binom{M-1+i}{i}\binom{M+i}{i}
$$
I'd like to get a closed form for this, but after a considerable amount of time searching my references and online sources (not to mention the time I've s... | Remark. The Iverson bracket that was used in the first answer is
not quite appropriate as the infinite series it was substituted into
does not converge in a neighborhood of zero.
The appropriate form of the Iverson bracket here is
$$[[0\le q\le M]] =
\frac{1}{2\pi i}
\int_{|z|=\epsilon} \frac{z^{q}}{z^{M+1}}\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Integrate $\int_{0}^{\pi/2}x\sqrt{\tan{x}}\log{(\sin{x})}\,\mathrm dx$ When trying to solve this problem: How to Integrate $ \int^{\pi/2}_{0} x \ln(\cos x) \sqrt{\tan x}\,dx$
I found his sister integral has an interesting closed form provided my calculation is correct. I use an ugly series to find it. Can you use other... | Substitute $t = \sqrt{\tan x}$
\begin{align}
I=&\int_{0}^{\pi/2}x\sqrt{\tan{x}}\ln({\sin x})\ dx\\
= &\int_0^\infty \frac{t^2 \ln\frac{t^4}{1+t^4}\ \tan^{-1}t^2}{1+t^4}dt\\
=&\int_0^\infty \int_0^1 \frac{2y \ t^4 \ln\frac{t^4} {1+t^4}}{(1+t^4)(1+y^4 t^4)}dy \ dt\\
=&\int_0^1 \frac{2y}{1-y^4} \int_0^\infty\bigg(\frac{\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319958",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
trigonometric inequality bound? Suppose that one wants to determine an upper bound of the trigonometric expression
$$
a\sin(x)+b\cos(x),
$$
where $a,b\in\mathbb{R}$. My instinct is to proceed as follows:
$$
a\sin(x)+b\cos(x)\leq |a|+|b|,
$$
which is correct. If one were to drop the modulus signs, would that make the bo... | We can write $a\sin x+b\cos x=C \sin(x+y)$ where $C=\sqrt{a^2+b^2}$ and $y=\arctan (b/a)$ with the arctangent function accounting for the individual signs of $a$ and $b$ (see Note).
Thus, the maximum is $C=\sqrt{a^2+b^2}$.
NOTE:
We can see that this is correct using only the addition angle formula for the sine fun... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Evaluate $\int_{\pi/6}^{\pi/2} (\frac{1}{2} \tan\frac{x}{2}+\frac{1}{4} \tan\frac{x}{4}+ \cdots+\frac{1}{2^n} \tan\frac{x}{2^n}+\cdots) dx$ $$f(x)=\frac{1}{2} \tan\frac{x}{2}+\frac{1}{4} \tan\frac{x}{4}+....+\frac{1}{2^n} \tan\frac{x}{2^n}+...$$
Check the function $f(x)$ is continuous on $[\frac{\pi}{6},\frac{\pi}{2}]$... | I didn't manage to finish this problem, and I would post it as a comment, but it's too long and I figured that it might help someone to help the OP find an answer, so I have posted my solution thus far.
So we begin by noting that $\tan(x)$ is continuous in the region $[-\frac{\pi}{2},\frac{\pi}{2}]$. We further note t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Showing that $(a^2-a+1)(b^2-b+1)(c^2-c+1) \leq 7$ How can I show that $(a^2-a+1)(b^2-b+1)(c^2-c+1) \leq 7$ given that $a+b+c = 3?$
Attempt: Setting $x=2a-1, y=2b-1, z=2c-1,$ we obtain that
$[(2a-1)^2+3][(2b-1)^2+3][(2c-1)^2+3] \leq 448,$
$s=x+y+z = 3, q=xy+yz+zx,$ and $p=xyz.$
So $(x^2+y^2+z^2) = 9-2q,$ and $x^2y^2+y... | The LHS is convex in each of the variables $a,b,c$ and $a, b,c \in [0,3]$ with $a+b+c=3$. So the LHS gets maximised when $a,b,c \in \{0,3\}$.
It isn't hard to verify that the maximum is when any two variables are zero and the other $3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit of a function with exponentiation In my assignment I have to calculate the following limit:
$$\lim_{x \to 0} \left(1+\frac {1-\cos x} {x} \right)^\frac{1}{x}$$
According to wolfram alpha, the result is $\sqrt{e} $.
However in my calculations I got a different one. Can you please let me know where did I get it wr... | The best approach is to take logs. If $L$ is the desired limit then we have
\begin{align}
\log L &= \log\left(\lim_{x \to 0}\left(1 + \frac{1 - \cos x}{x}\right)^{1/x}\right)\notag\\
&= \lim_{x \to 0}\log\left(1 + \frac{1 - \cos x}{x}\right)^{1/x}\text{ (by continuity of log)}\notag\\
&= \lim_{x \to 0}\frac{1}{x}\log\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
} |
Summation of series in powers of x with certain combinations as coefficients How can I find the sum: $$\sum_{k=0}^{n} \binom{n-k}{k}x^{k}$$
Edit: The answer to this question is: $$\frac{{(1+\sqrt{1+4x})}^{n+1}-{(1-\sqrt{1+4x})}^{n+1}}{2^{n+1}\sqrt{1+4x}}$$ I don't know how to arrive at this answer.
| Introduce the generating function
$$f(z) = \sum_{n\ge 0} z^n \sum_{k=0}^n {n-k\choose k} x^k.$$
This becomes
$$f(z) = \sum_{k\ge 0} x^k
\sum_{n\ge k} z^n {n-k\choose k}
\\ = \sum_{k\ge 0} x^k
\sum_{n\ge 0} z^{n+k} {n\choose k}
= \sum_{k\ge 0} x^k
\sum_{n\ge k} z^{n+k} {n\choose k}
\\ = \sum_{k\ge 0} x^k
\sum_{n\ge ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Using the triangle inequality to bound $\frac{x^3 + 3x + 1}{10-x^3}$ for $|x+1|<2$ How do I use the triangle inequality to bound the function $$f(x) = \frac{x^3 + 3x + 1}{10 - x^3}$$
on the interval $|x+1|<2$? I understand how the triangle inequality works, but using fractions with triangle inequality is confusing me.
| Hint.
Use $\vert \vert x \vert - \vert y \vert \vert \le \vert x-y \vert$ with $y=-1$. You'll get $\vert x \vert < 3$ with your hypothesis. Hence you can bound
$$\vert x^3 + 3x + 1 \vert \le \vert x \vert^3 + 3\vert x \vert + 1 < 37$$
Also for $x \le 0$ you have $-x^3 \ge 0$. Therefore for $x \le 0$: $0 \le \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Integral ${\large\int}_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}$ How to prove the following conjectured identity?
$$\int_0^\infty\frac{dx}{\sqrt[4]{7+\cosh x}}\stackrel{\color{#a0a0a0}?}=\frac{\sqrt[4]6}{3\sqrt\pi}\Gamma^2\big(\tfrac14\big)\tag1$$
It holds numerically with precision of at least $1000$ decimal digits.
Are... | This is a partial answer to the second question. Mathematica could evaluate
$$\int_0^\infty\frac{dx}{\sqrt[4]{a+\cosh x}},$$
in term of the following Appell function:
$$
F_1\left(\frac{1}{4},\frac{1}{4},\frac{1}{4},\frac{5}{4},\sqrt{a^2-1}-a,\frac{1}{\sqrt{a^2-1}-a}\right).
$$
For $a=0$ and $a=1$ there is a closed-form... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 4,
"answer_id": 0
} |
High School Trigonometry ( Law of cosine and sine) I am preparing for faculty entrance exam and this was the question for which I couldn't find the way to solve (answer is 0). I guess they ask me to solve this by using the rule of sine and cosine:
Let $\alpha$, $\beta$ and $\gamma$ be the angles of arbitrary triangle... | Step 1:
Write cosines relations for a triangle:
$$a^2=b^2+c^2-2bc \cos(A)$$
$$b^2=a^2+c^2-2ac \cos(B)$$
$$c^2=a^2+b^2-2ab \cos(C)$$
Step 2:
Write sinus Area relation for the triangle:
$$Area(ABC)=\frac{ab \sin(C)}{2}=\frac{bc \sin(A)}{2}=\frac{ac \sin(B)}{2}$$
Step 3:
Write cosines relations as you have in your questio... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 7
} |
If $f(x)=8x^3+3x$ , $x\in\mathbb{R}$, how do I find $\lim_{x \to \infty}\frac {f^{-1}(8x)-f^{-1}(x)}{x^{1/3}}$? Let $f: \mathbb R \to \mathbb R$ be defined as $f(x)=8x^3+3x$. Then $f$ is continuous , strictly increasing, and
$\lim _{x\to \infty}f(x)=\infty , \lim_{x \to -\infty}f(x)=-\infty$ , so $f$ is injective and ... | I got something different than the one in the prevous answer:
$f^{-1}(8x)=z$ and $f^{-1}(x)=y$ so
$8x=8z^{3}+3z$ and $x=8y^{3}+3y$
Therefore,
$8z^{3}\left ( 1+\frac{3}{z^{2}} \right )=8x\Rightarrow z\left ( 1+\frac{3}{z^{2}} \right )^{1/3}=x^{1/3}$ and similarly
$2y\left ( 1+\frac{3}{8y^{2}} \right )^{1/3}=x^{1/3}$.
N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 0
} |
Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equa... | Argument:
Every $k^3$ is the sum of $k$ consecutive odd numbers.
E.g.: $8 = 3+5$, $\quad 27 = 7+9+11$, $\quad 64 = 13+15+17+19$.
The sum of $n$ consecutive $k^3$ numbers, starting from $k = 1$, is the sum of
$n(n+1)/2$ consecutive odd numbers.
But the sum of a number of consecutive odd numbers is the square of
that n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 1
} |
How to compute the derivative of $\sqrt{x}^{\sqrt{x}}$? I know have the final answer and know I need to use the natural log but I'm confused about why that is.
Could someone walk through it step by step?
| Let $y= \sqrt{x}^{\sqrt x}$. Then $\ln(y) = \sqrt{x} \ln(\sqrt x) = \frac{1}{2} \sqrt x \ln(x)$. So,
$$
\frac{d}{dx} \ln(y) = \frac{d}{dx} \left(\frac{1}{2}\sqrt x\ln(x)\right)
$$
$$
\frac{1}{y} \frac{dy}{dx} = \frac{1}{2} \left(\frac{\sqrt x}{x} + \frac{1}{2\sqrt x}\ln(x) \right)
$$
So,
$$
\frac{dy}{dx} = y \cdo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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List the elements of the field $K = \mathbb{Z}_2[x]/f(x)$ where $f(x)=x^5+x^4+1$ and is irreducible Since $\dim_{\mathbb{Z}_2} K = \deg f(x)=5$, $K$ has $2^5=32$ elements. So constructing the field $K$, I get:
\begin{array}{|c|c|c|}
\hline \text{polynomial} & \text{power of $x$} & \text{logarithm} \\\hline
0 & 0 & -\... | The reason you have a problem is that $x^5+x^4+1$ is not irreducible in $\mathbb{Z}_2[x]$.
Thus, if $R=\mathbb{Z}_2[x]/(x^5+x^4+1)$, then $R^\times$ is not (as you would have expected if the polynomial were irreducible) a cyclic group of order $31$, so the fact that $x^{21}\equiv 1\bmod (x^5+x^4+1)$ is not surprising a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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problem in solving this problem from olympiad(use of invariant) Start with the set $\{3, 4, 12\}$. In each step you may choose two of the numbers $a$, $b$
and replace them by $0.6a − 0.8b$ and $0.8a + 0.6b$. Can you reach $\{4, 6, 12\}$
in finitely many steps:
Invariant here is that $a^2+b^2$ remains constant. Till he... | Well, $c$ does not change, so since $a^2+b^2$ does not change and $c$ does not change either we conclude $a^2+b^2+c^2$ does not change. In $(3,4,12)$ We have $a^2+b^2+c^2=169=13^2$.
On the other hand in $(4,6,12)$ we have $a^2+b^2+c^2=196=14^2$. Since $196$ is different from $169$ we cannot reach the desired target.
Yo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How to evaluate $\int { \frac { dx }{ { 2x }^{ 4 }+{ 2x }^{ 2 }+1 } }$? How to integrate something like
$\int { \frac { dx }{ { 2x }^{ 4 }+{ 2x }^{ 2 }+1 } }$
I know the method of partial fractions.
But what if the denominator is not factorizeable like the one in the question?
| First pull out any constant coefficient of $x^4$ to get a monic polynomial.
In $$ x^4 + B x^2 + C $$ with $$ B^2 - 4 C < 0, $$ so that $C > 0,$ we get
$$ x^4 + B x^2 + C = (x^2 + \lambda x + \sqrt C) (x^2 - \lambda x + \sqrt C), $$
where
$$ \lambda = \sqrt {2 \sqrt C - B} $$
is also real.
The original problem was ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Finding the area remaining after flipping a rectangle inside a rectangle Let $r$ be the inside rectangle of base $b$ and height $h$.
Let $R$ be the outside rectangle of base $B$ and height $H$
The dimensions of $r$ and $R$ are related in the following way:
I want to find the area left when you flip $r$ against the wal... | The untouched area for $b=h$ (ie, when $B H$ is a square), is $$\text{ area}=b^2 \left(-2 \sqrt{3}-\sqrt{7}+\frac{2 \pi }{3}+7-4 \csc ^{-1}\left(\frac{4}{\sqrt{7}}\right)\right):$$
Here, $B:H=1:1.$ It is likely this is either optimal, or close to optimal.
The area $=0$ for $h\geq\left(\sqrt{7}-2\right) b$ - or, of co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$ How can we find the value of $x$ in $\sin^{-1}(x)+\sin^{-1}(1-x)=\cos^{-1}(x)$?
Note that $\sin^{-1}$ is the inverse sine function.
I'm asking for the solution $x$ for this equation.
Please work out the solution.
| My answer is similar to @juantheron.
First we know that the equation is valid for $0\le\theta\le\dfrac{\pi}2,$ for some $\theta$ in a right triangle. Now taking the cosine of both sides we have:
$$x=\cos(\sin^{-1}(x)+\sin^{-1}(1-x).$$
Using the angle sum formula for cosine and the Pythagorean Theorem, we get
\begin{al... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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solve system equation: $ a \cdot b = 3 \cdot a-b+1, b \cdot c = 3 \cdot b - c + 1, c \cdot a = 3 \cdot c - a + 1$ I want to solve this system of equations but i'm stuck.
Here is it:
$$ a \cdot b = 3 \cdot a - b + 1 $$
$$ b \cdot c = 3 \cdot b - c + 1 $$
$$ c \cdot a = 3 \cdot c - a + 1 $$
| Hint: If $a\neq-1\wedge b\neq-1\wedge c\neq -1$
$$\begin{cases}b=\frac{3a+1}{1+a}\\c=\frac{3b+1}{1+b}\\a=\frac{3c+1}{1+c}\end{cases}$$
What happens if you substitute (1) in (2) and, after that, (2) in (3)?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Fourier sine transform of $\frac{1}{2}+\frac{1-x^2}{4x}\ln\vert\frac{1+x}{1-x}\vert$ Show that
$$
\int_0^{\infty} kF(k)\sin(ka)\,dk = \frac{\pi}{2}aG(a)
$$
where
$$
F(x) = \frac{1}{2}+\frac{1-x^2}{4x}\ln\vert\frac{1+x}{1-x}\vert
$$
and
$$
G(x) = \frac{\sin x-x\cos x}{x^4}
$$
EDIT: The source can be found here. One shou... | The integral can be written as
\begin{align}
I&=\frac{1}{2}\int_0^{\infty} k\sin(ka)\left[1+\frac{1-k^2}{2k}\ln\left|\frac{1+k}{1-k}\right|\right]\,dk\\
&=\frac{1}{2}\int_0^{\infty} \left( 1-k^2 \right)\sin(ka)\left[\frac k{1-k^2}+\frac{1}{2}\ln\left|\frac{1+k}{1-k}\right|\right]\,dk\\
&=\frac{1}{2}\left( 1+\frac{d^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1340986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Finding an isomorphism between polyomial quotient rings Let $F_1 = \mathbb{Z}_5[x]/(x^2+x+1)$ and $F_2 = \mathbb{Z}_5[x]/(x^2+3)$. Note neither $x^2+x+1$ nor $x^2+3$ has a root in $\mathbb{Z}_5$, so that each of the above are fields of order 25, and hence they are isomorphic from elementary vector space theory, howeve... | HINT:
You want $(\mathbb{Z}/5)[x]/(x^2+x+1) \to (\mathbb{Z}/5)[x]/(x^2+3)$, $x \mapsto a x + b$, so $x^2+x+1 \mapsto (ax+b)^2+ (ax + b) +1$, and you want the latter a multiple of $x^2+3$. Divide $(ax+b)^2+ (ax + b) +1$ by $x^2+3$ and you get the remainder
$$a\,(2 b+1)\, x + (b^2 + b+1 - 3 a^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1341458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Find the last two digits of $33^{100}$
Find the last two digits of $33^{100}$
By Euler's theorem, since $\gcd(33, 100)=1$, then $33^{\phi(100)}\equiv 1 \pmod{100}$. But $\phi(100)=\phi(5^2\times2^2)=40.$
So $33^{40}\equiv 1 \pmod{100}$
Then how to proceed?
With the suggestion of @Lucian:
$33^2\equiv-11 \pmod{100}$ t... | You can use fast exponentiation: modulo $100$
$$33^2\equiv -11,\quad 33^4\equiv 21,\quad 33^8\equiv 441\equiv 41,\quad 33^{16}\equiv1681\equiv -19$$
whence $\,33^{20}\equiv -19\cdot 21 =-(20-1)(20+1)\equiv 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 0
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Find the least positive angle satisfying the trigonometric equation $\sin^3 x+\sin^3 2x+\sin^3 3x=(\sin x+\sin 2x+\sin 3x)^3$. I did solve the question, but my method is highly tedious. I combined the sin and then opened the cubic.... Is there some trick? Something I am missing? Thanks.
| For $\sin ax+\sin bx=0$
Method $\#1:$
$\sin ax+\sin bx=0\iff\sin ax=-\sin bx=\sin(-bx)$ as $\sin(-A)=-\sin A$
$\implies ax=n\pi+(-1)^n(-bx)$ where $n$ is any integer
If $n$ is even $=2m$(say), $ax+2m\pi-bx\iff x=\dfrac{2m\pi}{a+b}$
Similarly for odd $n=2m+1$(say)
Method $\#2:$
Using Prosthaphaeresis Formula, $$\sin ax+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $x=(9+4\sqrt{5})^{48}=[x]+f$ . Find $x(1-f)$.
If $x=(9+4\sqrt{5})^{48}=[x]+f$, where $[x]$ is defined as integral part of $x$
and $f$ is a fraction, then $x(1-f)$ equals .
$\color{green}{a.)\ 1} \\
b.)\ \text{less than}\ 1 \\
c.)\ \text{more than}\ 1 \\
d.)\ \text{between}\ 1 \text{and }\ 2 \\
e.)\ \text{none of th... | Consider
$$P=(9+4\sqrt 5)^{48}+(9-4\sqrt 5)^{48}.$$
Note that $P$ is an integer.
Now we have $0\lt 9-4\sqrt 5\lt 1$. Hence we have
$$0\lt (9-4\sqrt 5)^{48}\lt 1.$$
Hence, we have
$$x=(9+4\sqrt 5)^{48}=P-1+1-(9-4\sqrt 5)^{48}.$$
This implies that $f=1-(9-4\sqrt 5)^{48}$.
Thus, we have
$$x(1-f)=(9+4\sqrt 5)^{48}(9-4\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1342467",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$ \lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\frac{1}{\sqrt{n}}$
Knowing that :
$$I_n=\int_0^{\frac{\pi}{2}}\cos^n(t) \, dt$$
$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1$$
$... | We are given
$$I_{2n}=\frac{\pi}{2}\frac{(2n-1)!!}{(2n)!!} \tag 1$$
and
$$I_n\sim \sqrt{\frac{\pi}{2n}} \tag 2$$
From $(2)$ is trivial to see that
$$I_{2n}\sim \frac12 \sqrt{\frac{\pi}{n}} \tag 3$$
Then, using $(1)$ and $(3)$, we find that
$$\begin{align}
\frac{(2n+1)!!}{\sqrt{n}(2n)!!}&=\frac{(2n+1)(2n-1)!!}{\sqrt{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Differentiate the Function $f(x)= \sqrt{x} \ln x$ Differentiate the Function $f(x)= \sqrt{x} \ln x$
| $f(x) = \sqrt{x} \ln x$. Using the product rule we have that $$f'(x) = \ln x \cdot \frac{\mathrm{d}}{\mathrm{d}x}\sqrt{x} + \sqrt{x}\cdot\frac{\mathrm{d}}{\mathrm{d}x} \ln x$$
Hence $$f'(x) = \frac{\ln x}{2\sqrt{x}} + \frac{\sqrt{x}}{x}.$$
Further simplification results in $$f'(x) = \frac{\ln x}{2\sqrt{x}} + \frac{2}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1344727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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A proof by induction and trigonometry Do you know how to prove that $\displaystyle\cos\left(\frac{x}{2}\right) + \cos\left(\frac{3x}{2}\right)+\cdots + \cos\left(\frac{(2n-1)x}{2}\right) = \frac{\sin nx}{2\sin\left(\frac x 2\right)}$ using induction?
I have tried with $n = 1$ which gives $\cos \frac{x}{2} = \frac{\sin... | I see this is quite an old question but here is a solution that assumes only basic high school compound angle formulas:
If we assume that $\displaystyle\sum_{k=1}^n \cos\left(\frac{(2k-1)x}{2}\right)=\frac{\sin(nx)}{2\sin\left(\frac x 2\right)}$ then it follows that:
$$\begin{align}
\sum_{k=1}^n \cos\left(\frac{(2k-1)x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to solve $(z^1+z^2+z^3+z^4)^3$ using Pascals Triangle? In an exercise it seems I must use Pascal's triangle to solve this $(z^1+z^2+z^3+z^4)^3$. The result would be $z^3 + 3z^4 + 6z^5 + 10z^ 6 + 12z^ 7 + 12z^ 8 + 10z^ 9 + 6z^ {10} + 3z^ {11} + z^{12}$. But how do I use the triangle to get to that result? Personally... | $$(z+z^2+z^3+z^4)^3 = z^3\cdot\left(\frac{1-z^4}{1-z}\right)^3=z^3(1-3z^4+3z^8-z^{12})\sum_{n\geq 0}\binom{n+2}{2}z^n$$
hence:
$$\begin{eqnarray*}(z+z^2+z^3+z^4)^3&=&(z^3-3z^7+3z^{11}-z^{15})\sum_{n\geq 0}\binom{n+2}{2}z^n\\&=&\sum_{n\geq 3}\binom{n-1}{2}z^n-3\sum_{n\geq 7}\binom{n-5}{2}+3\sum_{n\geq 11}\binom{n-9}{2}z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $n(n+1)(n+2)=6m^3$ in positive integers
How to find all positive integers $m,n$ such that $n(n+1)(n+2)=6m^3$ ?
I can see that $m=n=1$ is a solution, but is it the only solution ?
| The paper Rational approximation to algebraic numbers of small height: the Diophantine equation $\left|ax^n-by^n\right|=1$ by Michael A. Bennett (J. reine angew. Math. 535 (2001),1–49) shows that the equation in the title has at most one solution in positive integers $x,y$ for given positive integers $a,b,n$ with $n\ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345732",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
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Quadratic polynomials describe the diagonal lines in the Ulam-Spiral I'm trying to understand why is it possible to describe every diagonal line in the Ulam-Spiral with an quadratic polynomial $$2n\cdot(2n+b)+a = 4n^2 + 2nb +a$$ for $a, b \in \mathbb{N}$ and $n \in 0,1,\ldots$.
It seems to be true but why?
Wikipedia s... | Note that the first "ring" of numbers has just 1 number in it, the next ring has $9-1=8$ numbers, then $25-9=16$, $49-25=24$, $81-49=32$, and so on. These numbers (aside from the first) are increasing by 8. When you start somewhere and go out along a diagonal, with each step you increase by 8 more than you did with the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1347560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find the measurement of line BD So I was trying to find the measurement of $BD$ I drew green lines to make myself some angles, the measurement $3$ is from the point A to C, If only I can line $AE$ or $CE$ then I will just use the cosine law to get $BE$ and $DE$, however I can't figure it out.
| Hint: Let's call $H_B$ the foot of the line from $B$ to $AC$ and $H_D$ the foot of the line from $D$ to $AC$
Then you have $AH_B=AB \cos 30$, $BH_B=AB \sin 30$
and $CH_D=CD \cos 60$, $DH_D=CD \sin 60$
Thales gives us: $\dfrac{EB}{ED}=\dfrac{EH_B}{EH_D}=\dfrac{BH_B}{DH_D}$
But $H_BH_D=AC-AH_B-CH_D=H_BE+EH_D$
You can p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1348201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$
I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help w... | I like the answer by @Steven Gregory because of the way the dominos fall and it seems the only one presented that a precalculus student could hope to find. Using the reflection and multiplication formulas for the gamma function a fairly compact proof is possible.
$$\begin{align}\prod_{r=1}^7\cos\frac{r\pi}{15} & = \pro... | {
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"url": "https://math.stackexchange.com/questions/1351337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
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Determine whether $\sum \frac{2^n + n^2 3^n}{6^n}$ converges For the series $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n},$$ I was thinking of using the root test? so then I would get $(2+n^2/n+3)/6$ but how do I find the limit of this?
| $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n}=\sum_{n=1}^{\infty}\left(\frac{1}{3^n}+\frac{n^2}{2^n}\right),$$
where
$$\sum_{n=1}^{\infty}\frac{1}{3^n}=\frac{1}{1-\frac{1}{3}}-1=\frac{1}{2}$$
and, by ratio test,
$$\frac{\frac{(n+1)^2}{2^{n+1}}}{\frac{n^2}{2^n}}=\frac{1}{2}\left(1+\frac{1}{n}\right)^2 \to \frac{1}{2}<1 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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How do you take the integral of $\int \frac{\mathrm{d}y}{y(3-y)}$? How do you evaluate the following? $$\int \frac{\mathrm{d}y}{y(3-y)}$$
I have looked at the solution and I don't understand how they are taking the integral of this? They go from:
$$\int \frac{\mathrm{d}y}{y(3-y)}$$
to
$$\int \frac13 \left(\frac1y + \... | Partial fraction decomposition gives:
\begin{align*}
\dfrac{1}{y (3-y)} & = \dfrac{A}{y} + \dfrac{B}{3-y}\\
1&=A(3-y) + By
\end{align*}
Looking at the $y$ coefficients and the constant coefficients, we obtain the following two equations:
$$0 = A - B $$
$$1 = 3A$$
Which solve to give $A = B = \dfrac{1}{3}$.
Hence
$$\df... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to rewrite $\pi - \arccos(x)$ as $2\arctan(y)$? I get the following results after solving the equation $\sqrt[4]{1 - \frac{4}{3}\cos(2x) - \sin^4(x)} = -\,\cos(x)$, :
$$
x_{1} = \pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}\\
x_{2} = \pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}
$$
Wo... | You can see in the figure the angle $\alpha$ corresponding to $\arccos(\frac{\sqrt{6}}{3})$ and the tangent of the same $\alpha$ from which you can easily deduce your own conclusion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is this formula true for $n\geq 1$:$4^n+2 \equiv 0 \mod 6 $? Is this formula true for $n\geq 1$:$$4^n+2 \equiv 0 \mod 6 $$.
Note :I have tried for some values of $n\geq 1$ i think it's true such that
:I used the sum digits of this number:$N=114$,$$1+1+4\equiv 0 \mod 6,1²+1²+4²\equiv 0 \mod 6,1^3+1^3+4^3\equiv 0 \mod 6,... | For a less number theoretic approach, let $S_n$ be the statement that $4^n + 2$ is divisible by 6. Clearly $S_1$ is true since $4^1 + 2 = 6$. Now assuming that $4^k + 2$ is divisible by 6,
\begin{align*}
4^{k+1} + 2 &= 4\cdot4^n + 2 \\
&= 4\cdot4^n + 4\cdot2 - 6 \\
&= 4(4^n + 2) - 6,
\end{align*}
we immediately have t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1356911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
Let $0\le a\le b\le c,abc=1$, then show that
$$a+b^2+c^3\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}$$
Things I have tried so far:
$$\dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}=\dfrac{b^2c^3+ac^3+ab^2}{bc^2}$$
Since $abc=1$, it suffices to ... | Write
$$a = \frac{p^3}{p q r},\, b= \frac{q^3}{p q r}, \,
c= \frac{r^3}{p q r}
$$
with $0 < p\le q \le r$ and substitute to get the equivalent inequality:
$$
p q^8 r^4 + p^5 q^3 r^5 + q r^{12}-p^6 q^7 - p q^5 r^7 - p^5 r^8 \ge 0$$
Now write $p=u,\, q=u+v, \,r=u+v+w$, with $u>0$, $\ v,w\ge 0$ and substitute to get an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Possible values of z? Let $f:[-2,2]\to \mathbb {R} $ where $$f(x)=x^3+2(\sin x)^5+3(\tan x)^7+\left\lfloor\frac{x^2+1}{z}\right\rfloor $$ is an odd function then what are possible values of $z$?
$\lfloor\cdot\rfloor $ is the floor function.
| We know that the functions $x^n$ for $n$ odd, $\sin x$ and $\tan x$ are all odd functions. Furthermore, if $h$ and $g$ are odd, then $h(g(-x)) = h(-g(-x))=-h(g(-x))$, so the composition of two odd functions is again odd.
And we can therefore conclude that $x^3+ 2 (\sin x)^5 + 3 ( \tan x)^7$ is an odd function. So for $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $a^2+b^2$ is a multiple of three, then a and b are multiples of three I have attempted to prove the above. I am uncertain about the correctness of my proof:
Both numbers have to be multiples of three, i.e. $3a+3b=3n$, $\ 3(a+b)=3n$
It is not possible to arrive at an integer that is a multiple of three wit... | If $a\ne 0\ (\ mod\ 3)$, then $a^2\equiv 1\ (\ mod\ 3) $
If $a\equiv 0\ (\ mod\ 3)$, then $a^2\equiv 0\ (\ mod\ 3)$
If $b\ne 0\ (\ mod\ 3)$, then $b^2\equiv 1\ (\ mod\ 3) $
If $b\equiv 0\ (\ mod\ 3)$, then $b^2\equiv 0\ (\ mod\ 3)$
So, the only way to get $a^2+b^2\equiv 0\ (\ mod\ 3)$ is $a\equiv b\equiv 0\ (\ mod\ 3)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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$A$ and $B$ similar if $A^2=B^2=0$ and dimension of range $A$ and $B$ are equal Suppose $A$ and $B$ are linear transformations on finite dimensional vector space $V$,s.t. $A,B\neq 0$ and $A^2=B^2=0$. Suppose the dimension of range $A$ and $B$ are equal, can $A$ and $B$ be similar?
| that is surely possible. Take for example $A= \begin{pmatrix}0 & 0 \\ 1& 0 \end{pmatrix}$ and $ B=\begin{pmatrix}0 & 1 \\ 0& 0 \end{pmatrix}$ and then it holds that
$$
A^2=B^2=0
$$
and
$$
A=SBS^{-1} \text{ with } S=\begin{pmatrix}0 & 1 \\ 1& 0 \end{pmatrix}
$$
so $A$ and $B$ are indeed similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$
My try:I have just normally squared the expression and got
$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2... | Assume WLOG $a > b > 0$, $A = \sqrt{a^2\cos^2x+b^2\sin^2x}, B = \sqrt{a^2\sin^2x+b^2\cos^2x}\Rightarrow A^2+B^2 = a^2+b^2\Rightarrow u^2 = (1\cdot A+1\cdot B)^2\leq (1^2+1^2)(A^2+B^2)=2(a^2+b^2)\Rightarrow u^2_{max} = 2(a^2+b^2)$. To find $u^2_{min}$, you need to find the min of $(a^2\cos^2x+b^2\sin^2x)(a^2\sin^2x+b^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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How can I prove the following equality I have the following equality :
$$I_1=-\frac{ab}{2\pi}\int_0^\pi \frac{\cos(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=\frac{a}{a+b}-\frac{1}{2}$$
$$I_2=-\frac{ab}{2\pi}\int_0^\pi \frac{\sin(2t)}{a^2\sin^2(t)+b^2\cos^2(t)}dt=0$$
where $0 <b \leq a$.
I used the residues but I could not prov... | One trick to facilitate analysis is to write
$$\sin^2x=\frac{1-\cos 2x}{2}$$
and
$$\cos^2x=\frac{1+\cos 2x}{2}$$
Thus,
$$a^2\sin^2(t)+b^2\cos^2(t)=\frac{b^2+a^2}{2}+\frac{b^2-a^2}{2}\cos 2t$$
For the integral of interest, we can write
$$\begin{align}
I_1&=-\frac{ab}{2\pi}\int_0^{\pi} \frac{\cos(2t)}{a^2\sin^2(t)+b^2\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How to integrate $5^{5^{5^x}}5^{5^x}5^x$? How to integrate $5^{5^{5^x}}5^{5^x}5^x$ ?
I guess it's simple but can't figure it out !
| Let $u=5^{5^x}$, then $x = \frac{\frac{\ln(u)}{\ln(5)}}{\ln{5}}$ and $du = 5^{x+5^x}\ln^2(5)\hspace{.1cm} dx \implies dx = \frac{5^{-x-5^x}}{\ln^2(5)} \hspace{.1cm} du$. Thus we transform
$ \begin{align}
\int 5^{5^{5^x}} 5^{5^x} 5^x \hspace{.1cm} dx &= \int 5^uu5^x \frac{5^{-x-5^x}}{\ln^2(5)} \hspace{.1cm} du \\
&= \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| Start by noticing that $(2 + \sqrt{5})^{3} = 38 + 17 \sqrt{5}$ and $(2 - \sqrt{5})^{3} = 38 - 17 \sqrt{5}$. Now,
\begin{align}
\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38} &= \sqrt[3]{(2 + \sqrt{5})^{3}} - \sqrt[3]{(\sqrt{5} - 2)^{3}} \\
&= (2 + \sqrt{5}) - (-2 + \sqrt{5}) \\
&= 4.
\end{align}
What seems to ha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 5
} |
Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1) \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I c... | I think you are overthinking this.
To show that $x$ satisfies the quadratic equation $3x^2−2px+(p+3)=0$, you have two cases :
(1) $b^2−4ac > 0$ :
Therefore $4(p^2−3(p+3)) > 0$ i.e. $p^2−3(p+3) > 0$
(2) $b^2−4ac = 0$ :
Then $4(p^2−3(p+3)) = 0$ i.e. $p^2−3(p+3) = 0$
The third case $b^2−4ac < 0$ here is irrevelant since $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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$\sum\limits_{i=1}^n \frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$, for all $x_i>0.$ Can you prove the following new inequality? I found it experimentally.
Prove that, for all $x_1,x_2,\ldots,x_n>0$, it holds that
$$\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod\limits _{j=1}^nx_j}} \ge
1\,.... | The inequality $\displaystyle\sum_{i=1}^n\frac{x_i}{\sqrt[n]{x_i^n+(n^n-1)\prod \limits_{j=1}^nx_j}} \ge 1$ is trivial given the claim below. Of course, the equality occurs if and only if $x_1=x_2=\ldots=x_n$.
Claim: For
every $i=1,2,\ldots,n$, we have $\displaystyle\frac{x_i}{\sqrt[n]{x_i^n+\left(n^n-1\right)\,\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
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Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find th... | As I've suggested in the comment yesterday,
let $x=2m\cos y\implies(2m\cos y)^3-(2m\cos y)+1=0\ \ \ \ (1)$
As $\cos3y=4\cos^3y-3\cos y,$
$2m^3(\cos3y+3\cos y)-(2m\cos y)+1=0$
$\iff2m^3\cos3y+2m\cos y(m^2-1)+1=0\ \ \ \ (2)$
WLOG choose $m^2-1=0\iff m=\pm1$
Let $m=1$
$(1)$ reduces to $8\cos^3y-6\cos y+1=0 \ \ \ \ (3)$
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 4
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Is it possible to evaluate this binomial sum? Would it be possible to evaluate this sum?
$$\sum_{k=0}^{N/2}k\binom{N+1}{k},$$
where $N$ is even? I know that the sum
$$\sum_{k=0}^{N+1}k\binom{N+1}{k}=2^N(N+1)$$ (by http://mathworld.wolfram.com/BinomialSums.html, equation $(21)$), but I can't figure out a way to evaluate... | Since for $k$ such that $1 \leq k \leq N + 1$, we have
$$k\binom{N + 1}{k} = k \frac{(N + 1)!}{k! (N + 1 - k)!} = \frac{(N + 1)!}{(k - 1)!(N + 1 - k)!} = (N + 1)\binom{N}{k - 1}.$$
If $N$ is even, $\binom{N}{0} = \binom{N}{N}, \binom{N}{1} = \binom{N}{N - 1}, \binom{N}{2} = \binom{N}{N - 2}, \ldots, \binom{N}{N/2 - 1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1372836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Given $\tan A + \tan B = 3x$ and $\tan A \tan B = 2x^2$, find $\tan A - \tan B$ Given $$\tan A + \tan B = 3x$$ and $$\tan A \tan B = 2x^{2},$$ how can one find $\tan A - \tan B$? I have tried substitution, but failed to find the answer.
Edit: Can this problem be solved using the formulas for sums and differences of tan... | We know that $$(a-b)^2=a^2+b^2-2ab=(a+b)^2-4ab$$ $$\implies a-b=\pm \sqrt{(a+b)^2-4ab}$$ Now, we have
$$\tan A-\tan B=\sqrt{(\tan A+\tan B)^2-4\tan A\tan B}$$ $$=\sqrt{(3x)^2-4(2x^2)}$$ $$=\sqrt{x^2}$$$$=\pm x=|x|$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1374369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
| Add the first two
$$\frac 1{1+\sqrt3+\sqrt2} + \frac 1{1+\sqrt3-\sqrt2}=2\frac{1+\sqrt3}{(1+\sqrt3)^2-2}=2\frac{1+\sqrt3}{2+2\sqrt3}=1.$$
Similarly, add the last two
$$2\frac{1-\sqrt3}{2-2\sqrt3}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
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remainder of $a^2+3a+4$ divided by 7
If the remainder of $a$ is divided by $7$ is $6$, find the remainder when $a^2+3a+4$ is divided by 7
(A)$2$ (B)$3$ (C)$4$ (D)$5$ (E)$6$
if $a = 6$, then $6^2 + 3(6) + 4 = 58$, and $a^2+3a+4 \equiv 2 \pmod 7$
if $a = 13$, then $13^2 + 3(13) + 4 = 212$, and $a^2+3a+4 \equiv 2 \pmod ... | $a^2 + 3a + 4 \equiv a^2 - 4a + 4 \equiv (a-2)^2 \pmod 7$
If $a\equiv b \pmod 7$, then $a^2 + 3a + 4 \equiv (b-2)^2 \pmod 7$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Modular Quadratic Equation I'm trying to solve that equation:
$x^2-3x-5\equiv0\pmod{343}$
I've completed the square as follows:
$x^2-3x-5 \equiv x^2+340x-5\equiv(x+170)^2-170^2-5\pmod{343}\\
(x+170)^2 \equiv 93\pmod{343}\\
y^2 \equiv 93 \pmod{343}$
But I have no idea how to move on. How can I use the fact that $343=7^3... | HINT:
Start with $x^2-3x-5\equiv0\pmod7\iff0\equiv x^2-3x-5+7=(x-1)(x-2)$
If $x-1\equiv0\pmod7, x=7a+1$ where $a$ is any integer
Now $x^2-3x-5=(7a+1)^2-3(7a+1)-5=49a^2-7a-7$
which $\equiv0\pmod{7^2}\iff7|(a+1)\implies a\equiv-1\pmod7, a=7b-1$ where $b$ any integer
$\implies x=7a+1=7(7b-1)+1=49b-6$
Now $x^2-3x-5=(49b-6)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Application of Jensen's inequality to $x^x+y^y+z^z$ Claim: If $x, y, z >0$ and $x+y+z = 3\pi, $ then $x^x + y^y + z^z > 81.$
My attempt: Let $f(w) = w^w$, so $f$ is convex on $(0, \infty).$ By Jensen's inequality, $f(x\frac{x}{3\pi}+ y\frac{y}{3\pi} + z\frac{z}{3\pi}) \leq \frac{x}{3\pi}f(x) + \frac{y}{3\pi}f(y) +... | You rather just want to use that $$\frac{1}{3}f(x)+\frac{1}{3}f(y)+\frac{1}{3}f(z) \geq f\left(\frac{1}{3}x+\frac{1}{3}y+\frac{1}{3}z\right)=f(\pi)=\pi^\pi > 3^3 =27$$
Multiply by 3 to get the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1377930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 0
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.