Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Minimize $\frac{2}{1-a}+\frac{75}{10-b}$
Let $a,b>0$ and satisfy $a^2+\dfrac{b^2}{45}=1$. Find the minimum
value of $\dfrac{2}{1-a}+\dfrac{75}{10-b}.$
WA gives the result that $\min\left(\dfrac{2}{1-a}+\dfrac{75}{10-b}\right)=21$ with $a=\dfrac{2}{3},b=5$.
Consider making a transformation like this
$$\dfrac{2}{1-a... | The Tangent Line method helps.
Let $a=\frac{2}{3}x$ and $b=5y$.
Thus, $$4x^2+5y^2=9$$ and
$$\frac{2}{1-a}+\frac{75}{10-b}-21=3\left(\frac{2}{3-2x}+\frac{5}{2-y}-7\right)=$$
$$=3\left(\frac{2}{3-2x}-2-2(x^2-1)+\frac{5}{2-y}-5-\frac{5}{2}(y^2-1)\right)=$$
$$=\frac{6(x-1)^2(2x+1)}{3-2x}+\frac{15y(y-1)^2}{2(2-y)}\geq0.$$
T... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the standard matrix of the linear transformation $T$ A linear transformation $T:\mathbb{R}^2 \rightarrow \mathbb{R}^2$ is defined by $$T((2,11))=(8,5)$$ and $$T((1,5))=(5,3)$$ Find the matrix for this linear transformation.
I did this in a very clunky way - since $T$ is linear, I can write that
$$T((2, 11))=T(2e_... | Here is a bit of an elaboration on top of @G Cab's answer.
Consider the linear transformation $T_1$, which maps components of the identity matrix to vectors specified below.
$$T_1 \begin{pmatrix} 1 \\ 0 \end{pmatrix} = \begin{pmatrix} 2 \\ 11 \end{pmatrix}, \, T_1 \begin{pmatrix} 0 \\ 1 \end{pmatrix} = \begin{pmatrix} ... | {
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Prove that $\int_1^a \frac{T_n(x) T_n(x/a)}{\sqrt{a^2 - x^2} \sqrt{x^2 - 1^2}} \frac{a}{x} \mathrm{d}x = \frac{\pi}{2}$ In the paper, Representation of a Function by Its Line Integrals, with Some Radiological Applications, A. M. Cormack, Journal of Applied Physics 34, 2722 (1963), an integral identity is expressed whic... | Using the trigonometric definition of the Chebyshev polynomial of the first kind we can write:
$$\small I_{n}(a)=\int_1^a \frac{T_n(x) T_n\left(\frac{x}{a}\right)}{\sqrt{x^2 - 1}\sqrt{a^2 - x^2}}\frac{a}{x}dx=\int_1^a \frac{\cosh(n\operatorname{arccosh} x) \cos\left(n\arccos \left(\frac{x}{a}\right)\right)}{\sqrt{x^2 ... | {
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Proof by induction:$\frac{3}{5}\cdot\frac{7}{9}\cdot\frac{11}{13}\cdots\frac{4n-1}{4n+1}<\sqrt{\frac{3}{4n+3}}$ In the very beginning I'm going to refer to similar posts with provided answers:
Induction Inequality Proof with Product Operator $\prod_{k=1}^{n} \frac{(2k-1)}{2k} \leq \frac{1}{\sqrt{3k+1}}$ (answered by Öz... | Your calculations are correct.
But I thought it might be helpful also to mention another nice trick to handle such products:
*
*Set $A = \frac 35 \cdot \frac 79 \cdots \frac{4n-1}{4n+1}$
*Let $B = \frac 57 \cdot \frac 9{11} \cdots \frac{4n+1}{4n+3}$
It follows immediately
$$A < B \Rightarrow A^2 < AB = \frac 3{4n+... | {
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Find all values of $m$ such that the equation $ mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$ has nonnegative roots.
Find all values of $m$ such that the equation $$\large mx^4 + x^3 + (8m - 1)x^2 + 4x + 16m = 0$$ has nonnegative roots.
For an equation to have nonnegative roots, it mustn't only have negative roots.
Let $y... | Express $m$ as a function of $x\ge 0$,
$$m(x) = - \frac{x(x^2-x+4)}{(x^2+4)^2}\le0\tag 1$$
It is equivalent to find the lower bound of $m(x)$ for $x>0$. Set $m'(x) = 0$ to get
$$x^4-2x^3+8x-16=(x-2)(x+2)(x^2-2x+4) = 0$$
which shows that the minimum is at $x=2$. Plug it into $(1)$ to get $m(2) = -\dfrac3{16}$. Thus, bel... | {
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Sum of magnitudes of coefficients of polynomial $(x-1)(x-2)(x-3)\cdots(x-(n-1))$ The title says most of it. I have found that the sum of the coefficients of the polynomial$(x-1)(x-2)(x-3)\cdots(x-(n-1))$ yields $n!$.
For example, the coefficients of the polynomial $(x-1)(x-2)$ sum to $1+3+2=6=3!$ and similarly the coe... | Since all roots of the polynomial are positive, Descartes' Rule of signs guarantees that the signs of the coefficients alternate, for example with $n=4$:
$(x-1)(x-2)(x-3)=x^3\color{purple}{-}6x^2\color{purple}{+}11x\color{purple}{-}6$
Because of this sign pattern the absolute values of the coefficients are recovered by... | {
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Show that $a^2$ cannot be congruent to 2 or 3 mod 5 for any integer a) Show that $a^2$ cannot be congruent to $2$ or $3 \bmod 5$ for any integer.
$a^2≡2,3 \pmod5$
$a≡0,±1,±2\pmod5 )⟹a^2≡0,1,4$.
But $2,3$ are not congruent to $0,1,4\pmod5$.
I am not sure if I did it right, please check it for me
(b) Show that if $5\... | For part $(a)$, it suffices to enumerate among all residue classes modulo $5$. So we get:
$$
0^{2} \equiv 0 \pmod{5} \\
1^{2} \equiv 1 \pmod{5} \\
2^{2} \equiv 4 \pmod{5} \\
3^{2} \equiv 4 \pmod{5} \\
4^{2} \equiv 1 \pmod{5} \\
$$
Hence, $a^{2} \not \equiv 2,3 \pmod{5}$, for any integer $a$, since any integer can be wr... | {
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What is $4{(\frac ab )}^2 + \frac ba - a + 1$
What is $4{(\frac ab )}^2 + \frac ba - a + 1$ when $a, b, c$ are natural number and form geometry sequence with $b/a$ is an integer.
the average of a,b,c is b + 1
$a=a, b = ar, c=ar^2 = b^2 /a $ when $ r = b/a$. Also $c/b=b/a \rightarrow ac = b^2$
$a+b+c = 3b + 3 \rightar... | So $r=b/a\in \mathbb{N}$ and thus $b=ar$ and $c=ar^2$. Now we have $$a-2ar+ar^2=3$$
so $$a(1-r)^2 = 3\implies a= 3\;\;\wedge\;\; 1-r = \pm 1\implies r= 2$$
So $b=6$ and now you can finish.
| {
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Find the value of $I=\int_2^3{\frac{dx}{\sqrt{x^3-3x^2+5}}dx}$ Let $I=\int_2^3{\frac{dx}{\sqrt{x^3-3x^2+5}}dx}$ .
Find the value of $[I+\sqrt{3}]$ (where [.] represents the greatest integer function)
I have no idea of calculating such type of problem where integration is complex.
| $x^3-3x^2+3x-1\ge x^3-3x²+5 \ge 1$. (for x>2)
$\frac{1}{\sqrt{(x-1)^3}} \le \frac{1}{\sqrt{x^3-3x^+5}} \le 1$ (for x>2)
$\int_2^3 \frac{1}{\sqrt{(x-1)^3}}dx \le \int_2^3\frac{1}{\sqrt{x^3-3x^+5}}dx \le \int_2^3 1dx$
$2-\sqrt{2} \le I \le 1$
$2-\sqrt{2} + \sqrt{3} \le I+\sqrt{3} \le 1+\sqrt{3}$
$2 \lt I+\sqrt{3} \lt 3$... | {
"language": "en",
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What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
What is $a$ in $\lim_{t \to 0} \left(\frac{a}{t^2} - \frac{\sin 6t}{t^3 \cos^2 3t}\right) = -18$
We have to make this function defined, by change the cosine into sine or tan. So,
$$\lim_{t \to 0} \left(\frac{a}{t^2} - \fr... | There's an error in your computation, and you should find first a Laurent expansion:
$$\frac{a}{t^2} - \frac{2\sin 3t \cos 3t}{t^3 \cos^2 3t}=\frac 1{t^2}\biggl(at-2\frac{\tan 3t}t\biggr)$$
Now the Taylor expansion of $\tan x$ at order $3$ is $\;\tan x=x+\dfrac{x^3}{3}+o\bigl(x^3\bigr)$, so
$$
\frac{a}{t^2} - \frac{2\s... | {
"language": "en",
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Integer solutions to the product of four consecutive integers $ \bullet \textbf{Question} $
The product of four consecutive integers $ x, x + 1, x + 2, x + 3 $ can be written as the product of two consecutive integers, find all integer solutions for $ x $.
$ \bullet \textbf{Rephrasing} $
I decided to name the other int... | $x(x + 1)(x + 2)(x + 3)=(x^2+3x)(x^2+3x+2)$.
Let $N=x^2+3x+1$, then $N^2-1=m(m+1)$.
If $N=0$
There are no solutions since $m(m+1)$ is even.
If $|N|=1$
Then $m=0$ or $-1$.
If $|N|\ge 2$
Then $N^2-1$ is greater than $|N|(|N|-1)$ but less than $|N|(|N|+1)$. There are no solutions.
The only solutions are when $x(x ... | {
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Solve $xy''-y'-x^3y=0$ I want to solve $xy''-y'-x^3y=0$. My solution:
$y = z\cdot \exp(kx^2)$
$y' = z'\cdot \exp(kx^2) + z\cdot 2kx\exp(kx^2)$
$y'' = z''\cdot \exp(kx^2) + z'\cdot 4kx\exp(kx^2) + z\cdot (2k+4k^2x^2)\exp(kx^2)$
Plug in:
$z''\exp(kx^2)+z'(4kx-\frac{1}{x})\exp(kx^2)+z(2k+4k^2x^2-\frac{1}{x}\cdot2kx-x^2)\e... | Another way to solve it maybe:
$$xy''-y'-x^3y=0$$
Consider $x'$, then the equation becomes:
$$-x''x-x'^2=(x'x)^3y$$
$$-(x'x)'=(x'x)^3y$$
This differential equation is separable:
$$\dfrac {d(x'x)}{(x'x)^3}=-ydy$$
$$\dfrac {1}{(x'x)^2}=y^2+K$$
$$2x'x=\pm \dfrac {2} { \sqrt {y^2+K}}$$
$$(x^2)'= \pm \dfrac {2} { \sqrt {y... | {
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If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$... | If you are dealing with $h(x)=f(x)/g(x)$ then:
$$h'(x_0)=0=\frac{f'(x_0)g(x_0)-f(x_0)g'(x_0)}{[g(x_0)]^2}\to \frac{f'(x_0)}{g'(x_0)}=\frac{f(x_0)}{g(x_0)} \quad \quad (1)$$
but, $f(x)=[g(x)]^2+2$, so
$$f'(x)=2g'(x)g(x)\to \frac{f'(x)}{g'(x)}=2g(x)\quad \quad (2)$$
From $(1)$ and $(2)$,
$$f(x_0)=2[g(x_0)]^2=[g(x_0)]^2... | {
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Limits and infinity minus infinity I understand that we need to rationalize when we have infinity minus infinity like here
$\lim_{x\to \infty}\left(\sqrt{x^2 + 1} - \sqrt{x^2 + 2}\right)$
My question is why can I not just split the limits like this
$\lim_{x\to \infty}\left(\sqrt{x^2 + 1}\right) - \lim_{x\to \infty}\lef... | You can not subtract
$$
\lim_{x\to\infty}(\sqrt{x^2+1}-\sqrt{x^2+2})=\lim_{x\to\infty}\sqrt{x^2+1}-\lim_{x\to\infty}\sqrt{x^2+2},
$$
because the left hand side is $\infty-\infty$ which is not definable.
Instead, you can obtain that
$$
\sqrt{x^2+1}-\sqrt{x^2+2}=\frac{(\sqrt{x^2+1}-\sqrt{x^2+2})(\sqrt{x^2+1}+\sqrt{x^2+2... | {
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Trouble with trig substitution I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here.
Here is the original question: $$\int\frac{-7 x^2}{\sqrt{4x-x^2}} dx$$
And here is the work that I did on it:
Help with trig sub integral
Sorry that the negati... | You do not have a "$\theta$" on your diagram, so it is not possible to verify that $\theta$ and $x$ have a correct relationship. You do not write "$x = \dots$", so I have to guess that you mean
$$ x = 4 \sin^2 \theta \text{.} $$
The original integrand is only defined on $x \in [0,4]$, so happily this choice of subs... | {
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Solve the recurrence relation $a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$ Solve the recurrence
$a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$
with initial conditions $a_1 = 1, a_2 =5$ and $a_3 = 17$ How this was calculated? any hint or idea highly appreciated
| Hint: I'd suggest using characteristic polynomials. This particular recurrence can be made homogeneous linear from:
$$a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$$
$$a_{n-1} - 5a_{n-2} +8a_{n-3} -4a_{n-4} = 3^{n-1}$$
and
$$a_n - 5a_{n-1} +8a_{n-2} -4a_{n-3} = 3^n$$
$$3a_{n-1} - 15a_{n-2} +24a_{n-3} -12a_{n-4} = 3^n$$
and... | {
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Find a closed-form solution for the following recurrence $$T(n) =\begin{cases}5, & \text{if $n=1$} \\2T(n-1)+3n+1, & \text{if $n\geq 2$}\end{cases}$$
Now the answer is as such:
*
*$T(n)=5\cdot 2^{n-1}+\sum_{i=2}^n2^{n-i}(3i+1)$, which I certainly understand.
*$=(3n+6)2^{n-1}-3n-1-3\sum_{i=1}^{n-2}i\cdot 2^i=15\cdo... | A change of variable $j=n-i$ gives:
\begin{align}
T(n)
&=5\cdot 2^{n-1}+\sum_{i=2}^n2^{n-i}(3i+1)\\
&=5\cdot 2^{n-1}+\sum_{j=0}^{n-2}2^j(3(n-j)+1)\\
&=5\cdot 2^{n-1}+(3n+1)\sum_{j=0}^{n-2}2^j-3\sum_{j=0}^{n-2}j2^j\\
&=5\cdot 2^{n-1}+(3n+1)(2^{n-1}-1)-3\sum_{j=0}^{n-2}j2^j\\
\end{align}
Now you need:
\begin{align}
\sum_... | {
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For what values of $a\in \mathbb{R}$ is the integral $\int\int_{\mathbb{R}^2} \frac{1}{(1+x^4+y^4)^a}dxdy$ convergent? Well, first notice that the above integral equals to $2\int_{0}^\infty \int_0^\infty \frac{1}{(1+x^4+y^4)^a}dxdy$ since the integrand is an even function of both $x$ and $y$.
There are two approaches w... | Assume $a > 0$, since divergence is obvious for $a \leqslant 0$. Changing to polar coordinates, $x = r \cos \theta, \,y = r \sin \theta$, we have
$$\int_0^\infty\int_0^\infty \frac{1}{(1+ x^4 + y^4)^a}\, dx\, dy =\int_0^\infty\int_0^{\pi/2} \frac{r\,dr\, d\theta }{(1+ r^4\cos^4 \theta + r^4 \sin^4 \theta)^a}\, $$
Since... | {
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Let $G = \{1,x,y\}$ be a group with identity element $1$. Prove that $x^2 = y, y^2 = x,$ and $xy = 1.$
Let $G = \{1,x,y\}$ be a group with identity element $1$. Prove that $x^2 = y$, $y^2 = x$, and $xy = 1$.
Every time I try this I end up with pages of nonsense. Any help would be appreciated.
| Let's fill in the Cayley table! Recall that Cayley tables are latin squares meaning that every element of the group appears exactly once in every row and every column (not counting the row and column headers). Let's start with a blank canvas:
$$\begin{array}{|c|c|c|}\hline & \color{blue}1 & \color{blue}x & \color{blue}... | {
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A and B commute. Delete the same row and column from each. Do they still commute? Suppose that two matrices $A$ and $B$ commute. If I delete the $i$th row and $i$th column from each, do they necessarily commute? Does the answer change if we guarantee that $A$ and $B$ are Hermitian?
For example, consider the matrices
... | Without loss of generality, you may assume that the first columns and the first rows of the two matrices are deleted. Let
$$
A=\pmatrix{a&u^\ast\\ v&W},\ B=\pmatrix{b&x^\ast\\ y&Z}.
$$
Then
$$
AB=\pmatrix{\ast&\ast\\ \ast&vx^\ast+WZ},
\ BA=\pmatrix{\ast&\ast\\ \ast&yu^\ast+ZW}.
$$
The equality $AB=BA$ therefore implies... | {
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What is the locus of $z^2+\bar{z}^2=2$? I have to prove that it's the equation of an equilateral hyperbola
$$z^2+z^{-2}=2$$
I try this
$z^2+z^{-2} +2 = 2+2$
$(z^2+1/z^2 +2 ) = 4$
$(z+1/z)^2=4$
$ z+1/z= 2 $
$ x+yi + 1/(x+yi) = 2 $
$ ((x+yi)^2+1)/(x+yi)=2$
$ x^2+2xyi-y^2+1=2x+2yi$
$ x^2-2x+1 +2xyi -y^2=2yi $
$(x-1)^2 +2... | Hint One can rewrite this quickly using polar coordinates: For $z = r e^{i \theta}$, we have
$$2 = z^2 + \bar z^2 = (r e^{i\theta})^2 + (r e^{-i\theta})^2 = r^2 (e^{2 i \theta} + e^{-2 i \theta}) = r^2 \cdot 2 \cos 2 \theta .$$
Now, divide both sides by $2$, apply the double angle identity for $\cos$, and rewrite in te... | {
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Evaluate $\lim_{n\to\infty}\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}$ & $\lim_{n\to\infty}n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$. This is a homework question. I have to find two limits:
i. $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}$$
ii. $$\lim_{n\to \infty} n\bigg(\sum_{k=1}^n \frac... | Note first that $\sum_{k=1}^n k^3=\frac14n^2(n+1)^2$. Now note that $\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}=\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+k/n^8}}$ and hence $$\frac1n\sum_{k=1}^n \frac{(k/n)^3}{\sqrt{1+n/n^8}}\leq\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}\leq\frac1n\sum_{k=1}^n (k/n)^3$$ implying that $\lim_{n\t... | {
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Integral $\int \arcsin \left(\sqrt{\frac{x}{1-x}}\right)dx$ $$\int \arcsin\left(\sqrt{\frac{x}{1-x}}\right)dx$$
I'm trying to solve this integral from GN Berman's Problems on a course of mathematical analysis (Question number 1845)
I tried substituting $x$ for $t^2$:
$$2\int t\arcsin\left(\frac{t}{\sqrt{1-t^2}}\right) ... | $$I=\int \sin^{-1} \sqrt{\frac{x}{1-x}} dx$$
Let $$\frac{x}{1-x}=t^2 \implies x=\frac{t^2}{1+t^2} \implies dx=\frac{2t}{(1+t^2)^2}$$
Then $$I=\int\frac{2t \sin^{-1}t}{(1+t^2)^2} ~dt $$
Integrate it by parts taking $\sin^{-1} t$ as the first function.
Then $$I=\sin^{-1} t \int \frac{ 2t dt}{(1+t^2)^2}=\frac{-\sin^{-1}t}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3564971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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Given a right angled triangle ABC, where AP is the median and Q is a point on AB then find AQ in terms of given.
Problem: In a triangle $ABC$, right angled at $B$, $AP$ is the median. $Q$ is a point taken on side $AB$ such that $\angle ACQ =\alpha $ and $\angle APQ =\beta $ and let $CP = BP = x$, Find $AQ$ in terms of... |
Given $\angle QCA=\phi$, $\angle QPA=\psi$ and $x=|BP|=|CP|$, find
$c=|AB|$ and $q=|AQ|$.
Let $|AC|=b,\ |AP|=u,\ |CQ|=v,\ |PQ|=w$.
\begin{align}
b^2&=4\,x^2+c^2
,\quad
v^2=4\,x^2+(c-q)^2
,\\
u^2&=\phantom{4\,}x^2+c^2
,\quad
w^2=\phantom{4\,}x^2+(c-q)^2
,\\
b^2-u^2&=
v^2-w^2=3\,x^2
,\\
b^2-v^2&=u^2-w^2
=2\,c\,q-q... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3565129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the value of $k$ in this question Suppose $x_1$ and $x_2$ are solutions to $x^2+x+k=0$, if $x_1^2+x_1x_2+x_2^2=2k^2$, find the value of $k$.
$x_1+x_2 = -1$
$x_1x_2 = k$
$(x_1+x_2)^2 = x_1^2+2(x_1x_2)+x_2^2=2k^2$
$=\frac {x_1^2}{2}+x_1x_2+\frac {x_2^2}{2}=2k^2=\frac{1}{2}$
$2k^2=\frac{1}{2}$
$k^2=\frac{1}{4}$
$k... | You are wrong in this line $~(x_1+x_2)^2 = x_1^2+2(x_1x_2)+x_2^2\color{red}{=2k^2}~$. The value in the left hand side is not equal to $~{2k^2}~$ due to the present of $~\bf 2~$ in the middle term. Here is the complete solution (as per the process you follow).
$$(x_1+x_2)^2 = x_1^2+2x_1x_2 +x_2^2$$
$$\implies (-1)^2=(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3566177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Rabin Cryptosystem Proof The system is explained here: https://cryptography.fandom.com/wiki/Rabin_cryptosystem
I am trying to prove that the roots
$r \equiv y_p \cdot p \cdot m_q + y_q \cdot q \cdot m_p \pmod{N}\\
-r \equiv N - r \pmod{N}\\
s \equiv y_p \cdot p \cdot m_q - y_q \cdot q \cdot m_p \pmod{N}\\
-s \equi... | Note that $m_p^2 \equiv c \pmod{p}$ by construction and also $m_q^2 \equiv c \pmod{q}$.
Also by EEA, we chose $x_p,x_q$ so that $x_p p + y_q q=1 $ so it follows that
$$x_p \cdot p \equiv 1 \pmod{q} \text{ and } x_q \cdot q \equiv 1 \pmod{p}$$
So using the CRT, look at what $r^2 \pmod{p}$, $r^2 \pmod{q}$ equal:
$$r^2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3567321",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Does $\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $ converge? Does the following integral converge? I will post my solution, but I am unsure if it is true.
$$\int_{0}^{\infty} \frac{2x +3}{\sqrt {x^3 + 2x + 5}} \,dx $$
My solution
Let
$$ g(x) = \frac{2x}{\sqrt {x^3}}$$
$$ f(x) = \frac{2x +3}{\sqrt {x^3... | Notice that:
$$\frac{2x+3}{\sqrt{x^3+2x+5}}\approx x^{-0.5}$$ for large $x$ and we know that:
$$\int_0^\infty\frac{1}{x^n+c}dx$$
only converges for $c>0,n>1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Find an equation for the tangent line of $y = e^{4x}$ that is parallel to the linear equation $y = 5 x + 17$. I don't know how to solve this problem:
Find an equation for a tangent line of $y = e^{4x}$ that is parallel to the line given by $y = 5x + 17$.
I have tried with multiple approaches and have gotten
$$y-e^4(... | The tangent line will have gradient $5$, so you have to find an $x$ such that
$\frac{d (e^{4x})}{dx} = 5$. Since $\frac{d (e^{4x})}{dx} = 4e^{4x}$, setting
$x = \frac{\log\left(\frac{5}{4}\right)}{4}$ works. Since $\frac{dy}{dx} = 4y$ when $y = e^{4x}$, then $y$ will be $\frac{5}{4}$ when the gradient is $5$.
The line... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$ solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$. what is the general solution for $x$.
I wrote $\cos2x$ as $2\cos^2x-1$ and $\cos3x$ as $4\cos^3x-3\cos x$. Then the expression gets reduced to $\cos x(2\cos^2x-1)(4\cos^3x-3\cos x)=\frac{1}{4}$. substituted $\cos x=t$ and... | Expanding the equation in $t$ gives
$$32 t^6-40 t^4+12 t^2-1=0$$ Let $y=t^2$ to make
$$32 y^3-40 y^2+12y-1=0$$ By inspection $y=\frac 14$ is a solution; so,
$$\left(y-\frac 14 \right)(8y^2-8y+1)=0$$
I am sure that you will easily finish.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Closed formula for $\sum_{k=0}^{n}\binom{n+k}{k}x^k$ What would be the following sum:
$$\sum_{k=0}^{n}\binom{n+k}{k}x^k$$
I want this closed formula to prove
:
$$\sum_{k=0}^{n}\binom{n+k}{k}\frac{1}{2^k}=2^n$$
My try:
I could not find any closed form of this expression, but a recurrence relation would be like that:... | $$\sum_{k=0}^{n}\binom{n+k}{k}x^k = \sum_{k=0}^{n}x^k\cdot[t^n](1+t)^{n+k}=[t^n](1+t)^n\sum_{k=0}^{n}(x(1+t))^k=[t^n](1+t)^n\frac{1-x^n(1+t)^n}{1-x(1+t)}$$
if $x=\frac{1}{2}$ equals
$$ [t^n](1+t)^n \frac{1-\left(\frac{1+t}{2}\right)^n}{\frac{1-t}{2}}=\frac{2}{2^n}[t^n]\frac{2^n(1+t)^n-(1+t)^{2n}}{1-t}=\frac{2}{2^n}\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Integral $\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx$ I would like to learn more about this integral:
$$\int_{0}^{\frac{\pi}{4}}x\ln(\cos x)dx=\frac{1}{128}\Big(16\pi \operatorname{C}-21\zeta(3)-4\pi^2\ln(2)\Big)$$
where $\operatorname{C}$ is the Catalan's constant.
I have tried integration by parts , substitution $y=\cos ... | Not an answer but some transformations which might be useful. Didn't go beyond double series.
$$\int_{0}^{\frac{\pi}{4}}x\ln(\cos(x))dx= \color{blue}{(\cos x=t)} $$
$$ =\int_{1/\sqrt{2}}^1 \arccos t \ln t \frac{dt}{\sqrt{1-t^2}}=\color{blue}{(1-t^2=u^2)} $$
$$ = \frac12 \int_0^{1/\sqrt{2}} \frac{du}{\sqrt{1-u^2}} \arcs... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3572727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Given unit roots $ω$, find $\frac ω{1+ω^2} + \frac{ω^2}{1+ω^4} + \frac{ω^3}{1+ω} + \frac{ω^4}{1+ω^3}$ Let ω be a complex number such that $ω^5=1$ and $ω≠1$. Find
$$\frac ω{1+ω^2} + \frac{ω^2}{1+ω^4} + \frac{ω^3}{1+ω} + \frac{ω^4}{1+ω^3}$$
I've been having trouble with this unit, need help with solving this problem. (So... | Note that,
$$\frac ω{1+ω^2}\cdot\frac{ω^3}{ω^3} = \frac{ω^4}{1+ω^3}, \>\>\>\>\>\>\>
\frac{ω^3}{1+ω}\cdot\frac{ω^4}{ω^4} = \frac{ω^2}{1+ω^4}$$
Thus,
$$\begin{aligned}
& \frac ω{1+ω^2} + \frac{ω^2}{1+ω^4} + \frac{ω^3}{1+ω} + \frac{ω^4}{1+ω^3} \\
= &2\left(\frac ω{1+ω^2} + \frac{ω^3}{1+ω} \right)
=2\cdot \frac{ω(1+ω)+ω^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3574376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that $a^{2^n} - 1$ is divisible by $4\cdot2^n$ by induction Given that $a$ is any odd number and $n$ is any integer
I got to $(2k + 3)^{2^k}\cdot(2k + 3)^{2^k} - 1$ at the $(k+1)$-th step.
| For $n=1$ this is obvious, because $a^2-1=(a-1)(a+1)$, and both factors are even, so $2^2 = 4\cdot 2^0$ divides $a^2-1$.
Assume this is true for $n$, so $4\cdot 2^n$ divides $a^{2^n}-1$. Then
\begin{align*}
a^{2^{n+1}}-1 & = a^{2^n\cdot2}-1 \\
& = (a^{2^n}-1)(a^{2^n}+1) \\
& = 4\cdot 2^n \cdot k\cdot(a^{2^n}+1),
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3576201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to solve this quartic? I have the following equation, which I've put into the form of a normal quartic:
$$x^4+4ax^3+(4a^2+1)x^2-1=0$$
I'm trying to find the solutions for $x$ in terms of $a$. Is there an easier way then using the quartic formula?
| Solving by Resultant.
$Res_x(x^4 + 4 a x^3 + (4 a^2 + 1) x^2 - 1, x^2 + (y + a) x + A)=\\
1 - a^2 - a^4 + 2 A - A^2 + 5 a^2 A^2 + 4 a^4 A^2 - 2 A^3 + 4 a^2 A^3 + A^4\\
-2 a (1 + 2 A + A^2 + 4 a^2 A^2 + 2 A^3)y + (-1 + 2 a^2 + 4 A + A^2 + 4 a^2 A^2) y^2 - y^4$
Let linear term equal zero, then:
1) find $A$ from cubic $1 ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integer Solution to $x^3+y^2=z^2$ What is the non-zero integer general solution to $x^3+y^2=z^2$ ? I guess it is already solved in some book or paper, in that case plz help me to find that.
Edit:
*
*One solution is -
$n^3 = [(n)(n+1)/2]^2 - [(n)(n-1)/2]^2$
Found in here.
*$(x,y,z)=\left(abuv,\frac{ab(bu^3-av^3)}{... | There are a lot of solutions. For example, parameterize a set of such solutions.
We have modulo $7$ the only cubes are $0$ and $\pm1$ and besides $t^6=1$ for all $t$ non divisible by $7$. Take for example the cube $1$ so we have
$$1\equiv(z-y)(z+y)\pmod7$$ We can choose for example $z+y=t^5$ and $z-y=t$ so we get
$$z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3578993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Manual square root method explained I was investigating some methods of computing square roots by hand and I found a certain script which presents the following method of computing:
NOTE: the following notation represents positional notation of a number, e.g. $$\overline{ab} = 10a + b$$
The algorithm is presented with ... |
the author just says that a=3 with no further logic behind it.
Why couldn't it be a=2 for example?
Pairing from the implied decimal point, the number is "14 17 52 25".
The value of "a" is chosen as the largest digit whose square isn't greater than 14.
4×4 is 16, which is too big, but 3×3 is 9, which fits, so a = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582965",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{... | Clearly $A,B,2B-A>0$.
Note that $B=2+\dfrac{1}{B}$ and $A=1+\dfrac{1}{A}\iff 2A=2+\dfrac{2}{A}$.
It follows, $2A-B=\dfrac{2}{A}-\dfrac{1}{B}=\dfrac{2B-A}{AB}>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 4
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Calculate limit of this sequence $I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$ I need to find the limit of the sequence :
$$I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$$
The only thing I have done is : take substitution $\tan x = z$
this gives
$$dx = \dfrac{1}{1 + z^2}\,dz$$
So, integral becomes :
$$I_n = \int_{0}^{\inft... | While the estimation done above is sufficient, it might be interesting to see how to obtain the analytical expression by elementary principles from which the result is immediately seen. Rewrite the integral using the symmetry
\begin{align}
I_n &= \int_0^1 \frac{x^{1/n} + x^{-1/n}}{x^2+1} \, {\rm d}x \\
&=\sum_{k=0}^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
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The maximum for $xy \sin \alpha + yz \sin \beta +zx \sin \gamma$. Question:
Deduce the maximum of $xy \sin \alpha + yz \sin \beta +zx \sin \gamma$ if $x,y,z$ are real numbers that satisfy $x^2+3y^2+4z^2=6$ with $0<\alpha,\beta,\gamma<\pi$ such that $\alpha+\beta+\gamma=2\pi$.
Currently, I am not very sure how to appro... | Okay I finally arrived at an answer. First, lets reduce the cases we need to look at. Clearly $\sin \alpha, \sin \beta, \sin \gamma >0$. So, if one/two of the variables $x,y$ or $z$ is negative, the expression $xy \sin \alpha +yz \sin \beta + zx \sin \gamma$ will not be maximum. Clearly, if all three $x,y,z$ are negati... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
My attempt is as follows:
$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\righ... | hint
Observe that
$$x_2=\frac{k+\sqrt{k^2-5}}{5}=\frac{1}{k-\sqrt{k^2-5}}$$
$ x_2 $ is integral if $ (k -\sqrt{k^2-5})=\pm 1$ ,
By the same, $ x_1 $ is integral if
$ (k+\sqrt{k^2-5}) =\pm 1$.
This gives the possible values for $ k$ :
$-4, -3, 3, 4$.
the sum of positives values is $7$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that: $\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}$ Pythagoras's constant in Fibonacci number!
How do I show that?
$$\sqrt{2}=\frac{F_n^2+F_{n+1}^2+F_{n+2}^2}{\sqrt{F_n^4+F_{n+1}^4+F_{n+2}^4}}\tag1$$
Where $F_n$ is Fibonacci sequence.
$$2(F_n^4+F_{n+1}^4+F_{n+2}^4)^2=F_n^2+F_{n+1}... | Let $x=F_{n+2}$,$y=F_{n+1}$ and $z=F_{n}$. Squaring both sides, we have:
$$2(x^4+y^4+z^4)-(x^2+y^2+z^2)^2=0$$
Factoring out, we have:
$$(x-y-z)(x+y-z)(x-y+z)(x+y+z)=0$$
Now, by the product cancellation law, we can say that all the product is equal to zero when: $$x-y-z=0\leftrightarrow x=y+z$$
Remebring the substituito... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Isosceles triangle $ABC$ with an inside point $M$, find $\angle BMC$ We have an isosceles $\triangle ABC, AC=BC, \measuredangle ACB=40^\circ$ and a point $M$ such that $\measuredangle MAB=30^\circ$, $\measuredangle MBA=50^\circ$.
Find $\measuredangle BMC$.
Starting with $\angle ABC=\angle BAC=70^\circ \Rightarrow \ang... |
Construct the equilateral triangle $AHB$. Given that $AC = BC, AH = BH$ and the shared $CH$, the triangles $AHC$ and $BHC$ are congruent. Then, $\angle BCH = \dfrac12\angle ACB = 20^\circ$.
Since $AH = BH$ and $\angle BAM = \angle HAM = 30^\circ$, the triangles $BAM$ and $HAM$ are congruent, which yields $\angle HBM =... | {
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"timestamp": "2023-03-29T00:00:00",
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How to show that : $4(-1)^nL_n^2+L_{4n}-L_n^4=2$ How can we prove that:
$$4(-1)^nL_n^2+L_{4n}-L_n^4=2$$
Where $L_n$ is Lucas number
We got $L_n=\phi^n+(-\phi)^{-n}$
$4(-1)^nL_n^2=8(-1)^n\phi^{2n}+8$
$L_{4n}=\phi^{4n}+(-\phi)^{-4n}$
$L_n^4=4\phi^{4n}+4(-1)^n\phi^{2n}+4$
$$8(-1)^n\phi^{2n}+8+\phi^{4n}+(\phi)^{-4n}-4\phi^... |
$4(-1)^nL_n^2=8(-1)^n\phi^{2n}+8$
$L_n^4=4\phi^{4n}+4(-1)^n\phi^{2n}+4$
These are incorrect.
We have
$$\begin{align}4(-1)^nL_n^2&=4(-1)^n(\phi^n+(-\phi)^{-n})^2
\\\\&=4(-1)^n(\phi^{2n}+2\phi^n(-\phi)^{-n}+(-\phi)^{-2n})
\\\\&=4(-1)^n(\phi^{2n}+2(-1)^n+\phi^{-2n})
\\\\&=4(-1)^n\phi^{2n}+8+4(-1)^n\phi^{-2n}
\\\\&=4(-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588885",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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On the integral $\int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$ I'm having a difficult time evaluating the integral
$$\mathcal{J} = \int_0^{\sqrt{2}/2} \frac{\arctan \sqrt{1-2t^2}}{1+t^2} \, \mathrm{d}t$$
This is integral arose after simplifying the integral $\displaystyle \int_{0}^{\pi/4 } \a... | \begin{align}J&=\int_0^{\frac{1}{\sqrt{2}}} \frac{\arctan\left(\sqrt{1-2x^2}\right)}{1+x^2}\,dx\\
&\overset{x=\frac{1}{\sqrt{2}}\sin u}=\frac{1}{\sqrt{2}}\int_0^{\frac{\pi}{2}}\frac{\cos u\arctan(\cos u)}{1+\frac{1}{2}\sin^2 u}\,du\\
&=\sqrt{2}\int_0^{\frac{\pi}{2}}\frac{\cos u\arctan(\cos u)}{2+\sin^2 u}\,du\\
&=\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3592972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Find the locus of the complex number from the given data If $|\sqrt 2 z-3+2i|=|z||\sin (\pi/4+\arg z_1) + \cos (3\pi/4-\arg z_1)|$, where $z_1=1+\frac{1}{\sqrt 3} i$, then the locus of $z$ is ...
$$\arg~z_1 =\frac{\pi}{6}$$
So RHS of the equation is $\;|z|\dfrac{1}{\sqrt 2}$.
Then
$\;|\sqrt 2 z-3+2i|=|z|\dfrac{1}{\sqr... | $Arg[1+i/\sqrt{3}]= \pi/6$
$$|z\sqrt{2}-3-2i|=|z||\sin(\pi/4+\pi/6)+cos(3\pi/4-\pi/6)|$$
$$\implies |z\sqrt{2}-3-2i|=|z|\sqrt{\frac{1}{2}} \implies
\left|\frac{2z-(3+2i)\sqrt{2}}{z}\right|=1$$
The locus of $z$ is a circle. Let $z=x+iy$,
we have $$(2x-3\sqrt{2})^2+(2y-2\sqrt{2})^2=(x^2+y^2)$$
tHe simplified eq. of circl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$ for all positive integers
I am trying to prove
$$(n+1)^{2/3} -n^{2/3} <\frac{2}{3} n^{-1/3}$$
for all positive integers.
My attempts so far have been to Taylor expand the left hand side:
$$(n+1)^{2/3} -n^{2/3}\\
=n^{2/3}\big((1+1/n)^{2/3} -1\big)\\
=n^{2/3}... | It seems nobody can resist this one :)
$$\begin{align}(n+1)^{2/3}-n^{2/3}&=\left((n+1)^{1/3}-n^{1/3}\right)\left((n+1)^{1/3}+n^{1/3}\right)\\
&=\frac{\left(n+1-n\right)\left((n+1)^{1/3}+n^{1/3}\right)}{(n+1)^{2/3}+n^{1/3}(n+1)^{1/3}+n^{2/3}}\\
&=\frac1{n^{1/3}\left(\frac{(n+1)^{2/3}}{n^{1/3}(n+1)^{1/3}+n^{2/3}}+1\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3593439",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
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How to show that $\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}=\frac{1}{2N!^2}$?
How to show that $\displaystyle\sum_{n=0}^{N}\frac{(-1)^n}{{(N+n)!}(N-n)!}=\frac{1}{2N!^2}$?
How show that $(1)$ is:
$$\sum_{n=0}^{N}\frac{(-1)^n{2n \choose n}{N+n \choose N-n}}{[(n+1)(n+1)(n+2)(n+3)\cdots(n+N)]^2}=\frac{1}{2N!^2}\tag1$$... | Multiplying both sides by $2(2N)!$, we see the equality is equivalent to
\begin{align}
&2\sum_{n=0}^{N}(-1)^n\binom{2N}{{N+n}}=\binom{2N}{N}.
\end{align}
Now write the left side
\begin{align}
2\sum_{n=0}^{N}(-1)^n\binom{2N}{{N+n}} &=\left(\binom{2N}{N}-\binom{2N}{N+1}+\dots\right) +\left(\binom{2N}{N}-\binom{2N}{N+1}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3596234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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The maximum an complex expression If Z is a complex number such that :
$|Z+4|\leq3$ , find the maximum value of $|Z+1|$
My turn:
$|(x+4) + yi|\leq 3$
represents the surface of acircle whose center is (-4 ,0) and its radius is 3 , then
$-7\leq x \leq -1$ , then
$-6 \leq (x+1)\leq 0$
Then
$0\leq(x+1)^2 \leq 36 $
And
$-3... |
As seen from the graph, the maximum $|z+1|$ is at $z=-7$, i.e.
$$|z+1| \le |-7+1| = 6$$
which can also be shown analytically as follows,
$$|z+1|=|z+4+(-3)| \le |z+4| + |-3| \le 3 +3 = 6$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrating $\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx$ When i came across this one i thought of using the substitution $x=\frac{1-t}{1+t}$ that lead to an easy expression.
$$\int _0^1\frac{\ln \left(1-x\right)}{x^2+1}\:dx\overset{x=\frac{1-t}{1+t}}=\ln \left(2\right)\int _0^1\frac{1}{t^2+1}\:dt+\int _0^1\frac{\l... | You are not allowed to switch the integral and derivative. The conditions of the theorem about switching those are not met. Denote $\frac{\ln(1-ax)}{x^2+1}$ by $f(x,a)$, one of the conditions is that there must exist a function $g:[0,1] \rightarrow \mathbb{C} $ such that $\int_0^1 |g(x)|dx < \infty$ and $| \frac{\part... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3599967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Given that $a+b+c=0$, show that $2(a^4+b^4+c^4)$ is a perfect square Given that $a+b+c=0$. Show that: $2(a^4+b^4+c^4)$ is a perfect square
MY ATTEMPTS:
I found that when $a+b+c=0$, $a^3+b^3+c^3=3abc$
So I did:
$(a^3+b^3+c^3)(a+b+c)$ -- $a^4+b^4+c^4=-(a^3c+a^3b+ab^3+b^3c+c^3a+c^3b)$
And then I tried to substitute $a^4+b... | From $a+b+c=0$ it follows $c=-a-b$, and then
\begin{align*}
c^4&=a^4+4a^3b+6a^2b^2+4ab^3+b^4\\
\implies \qquad 2(a^4+b^4+c^4)&=4a^4+8a^3b+12a^2b^2+8ab^3+4b^4\\
&=(2a^2+2ab+2b^2)^2
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3602704",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Prove by mathematical induction that $3^n>2n^3$ I'm having trouble with this question:
"Prove by mathematical induction that for all integers $n\ge 6$, $3^n>2n^3$".
I got to $P(k)=2k^3<3^k$ and $P(k+1)=2(k+1)^3<3^{k+1}=2k^3+6k^2+6k+2<3^k*3$,
but I dont know how I can get $P(k+1)$ from $P(K)$...
Thanks
| Note that $f(k) = \frac{k}{k+1}$ is a positive, strictly increasing function for positive $k$ (since $\frac{k}{k+1} = 1 - \frac{1}{k+1}$ and $\frac{1}{k+1}$ is strictly decreasing). Thus, for $k \ge 6$, you have
$$\begin{equation}\begin{aligned}
f^3(k) & = \left(\frac{k}{k+1}\right)^3 \\
& \ge \left(\frac{6}{7}\right)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3604289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding all Lines Passing Through a Point Given the Product of Intercepts How does one find the equation of all lines passing through a point (Ex. $(6, -1)$), satisfying the condition that the product of their $x$ and $y$ intercepts must equal some number $c$ (Ex. $3$)?
As far as I understand this can be conceptualized... | So I came across a similar issue and I couldn't find any answers which were satisfactory. So I dug up as much information as I could and I found a solution. Here goes...
We have a known point, $(6,-1)$. For any line that passes through this point, it must have an x-intercept of $(a,0)$ and a y-intercept of $(0,b)$. We ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3607423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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What are the integer solutions to $a^{b^2} = b^a$ with $a, b \ge 2$ I saw this in quora.
What are all the
integer solutions to
$a^{b^2} = b^a$
with $a, b \ge 2$?
Solutions I have found so far:
$a = 2^4 = 16, b = 2,
a^{b^2}
= 2^{4\cdot 4}
=2^{16},
b^a = 2^{16}
$.
$a = 3^3, b = 3,
a^{b^2} = 3^{3\cdot 9}
=3^{27},
b^a = 3... | IMO 1997, Problem B2
Find all pairs $(a, b)$ of positive integers that satisfy $a^{b^2} = b^a$.
Answer
$(1,1)$, $(16,2)$, $(27,3)$.
Solution
Notice first that if we have $a^m = b^n$, then we must have $a = c^e$, $b = c^f$, for some $c$, where $m=fd$, $n=ed$ and $d$ is the greatest common divisor of $m$ and $n$.
[Proof... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How can I solve this geometry problem without trigonometry?
Let $ABC$ be a triangle with $D$ on the side $AC$ such that $\angle DBC=42^\circ$ and $\angle DCB=84^\circ$. If $AD = BC$, find $x = \angle DAB$.
It's supposed to be solved with constructions, but I couldn't figure out. See the trigonometric version here.... | Let $E$ be such that $BCDE$ is an isosceles trapezoid with bases $BC$, $DE$. Then
$$\angle EBD = \angle EBC - \angle DBC = 84^\circ - 42^\circ = 42^\circ = \angle DBC = \angle BDE$$
thus triangle $EBD$ is isosceles with $BE=DE$.
Note that $ED=BE=DC$, $AD=BC$, and $\angle EDA = \angle BCD$. Hence triangle $EDA$ is cong... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3611833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$ in terms of elementary constants How can we evaluate
$$\sum _{n=1}^{\infty } \frac{1}{n^5 2^n \binom{3 n}{n}}$$
Related: $\sum _{n=1}^{\infty } \frac{1}{n^4 2^n \binom{3 n}{n}}$ is solved recently (see here for the solution) by elementary method, so I wo... | This is by no means a solution but an extended comment which shows that this sum - generalized to a generating function - belongs to the well-known class of special functions, the hypergeometric functions. Maybe this can be of some use.
Defining the generating function
$$g(q,z) = \sum _{n=1}^{\infty } \frac{z^n}{n^q\bi... | {
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"url": "https://math.stackexchange.com/questions/3611964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. If $\log_8 3 = P$ and $\log_3 5 = Q$, express $\log_{10} 5$ in terms of $P$ and $Q$. Your answer should no longer include any logarithms.
I noted that $\log_5 10=\frac{1}{\log_{10} 5}.$
I also noted that $\log_{5} 10=\log_5 2+\log_5 5=... | An attempt. But I'm not sure whether the assumption I make at the beginning is correct, or whether it is arbitrary.
We have :
$P = \log_8 3 \iff 8^P = 3$
and
$Q = log_3 5 \iff 3^Q = 5$.
Suppose that : $log_{10} 5 = kPQ$
$log_{10} 5 = kPQ$
$\rightarrow 10^{log_{10} 5} = 10^{kPQ}$
$\rightarrow 5 = 10^{kPQ}$
$\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3614454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If a,b,c be roots of $2x^3+x^2+x-1=0$ show that some expression is equal to 16. If $a,b,c$ are roots of $2x^3+x^2+x-1=0,$ show that
$$\bigg(\frac{1}{b^3}+\frac{1}{c^3}-\frac{1}{a^3}\bigg)\bigg(\frac{1}{c^3}+\frac{1}{a^3}-\frac{1}{b^3}\bigg)\bigg(\frac{1}{a^3}+\frac{1}{b^3}-\frac{1}{c^3}\bigg)=16$$
My attempt:
Let $\fra... | Hint:
Set $x^3=y$
$$(2x^3-1)^3=-(x^2+x)^3$$
$$(2y-1)^3=-(y)^2-y-2y\iff?\ \ \ \ (2)$$ whose roots are $a^3,b^3,c^3$
Now $$z=\dfrac1{a^3}+\dfrac1{b^3}-\dfrac1{c^3}=\dfrac1{a^3}+\dfrac1{b^3}+\dfrac1{c^3}-\dfrac2{c^3}=\dfrac{a^3b^3+b^3c^3+c^3a^3}{(abc)^3}-\dfrac2{c^3}$$
$$\implies\dfrac2{c^3}=?\iff c^3=?$$
Now as $c^3$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3615742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Integration of Exponential with Polynomial in its Exponent I seek to solve the following integral:
$$
\int_0^T dt \frac{1}{\sqrt{4 \pi s^2 t}}\cdot \exp\left({-\frac{(x-vt)^2}{4 s^2t}}\right)
$$
My first idea was to substitute $1/\sqrt{t}$ using $u=\frac{1}{2} \sqrt{t}$ and $du = dt \frac{1}{\sqrt{t}}$, yielding
$$
\in... | For $a,b>0$ we have
\begin{align}
\int \limits_0^a \mathrm{e}^{-b \left(x-\frac{1}{x}\right)^2} \, \mathrm{d} x &= \frac{1}{2} \int \limits_0^a \mathrm{e}^{-b \left(x-\frac{1}{x}\right)^2} \left(1 + \frac{1}{x^2} + 1 - \frac{1}{x^2}\right) \mathrm{d} x \\
&= \frac{1}{2} \int \limits_0^a \mathrm{e}^{-b \left(x-\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3617144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proving $\sum\limits_{\rm cyc} 1/(a^2 -b+4) \geq3/4$ Suppose $a,b,c\in\mathbb R^+$ with $a+b+c=3.$ Prove that $$\frac{1}{a^2- b+4}+\frac{1}{b^2-c+4}+\frac{1}{c^2- a+4}\geqslant\frac{3}{4}.$$
I tried various approaches, but nothing seems to work. Whenever I give a lower bound using $C-S$ or $AM-GM$ it becomes very weak.... | By C-S
$$\sum_{cyc}\frac{1}{a^2-b+4}=\sum_{cyc}\frac{(a+5b+3c)^2}{(a+5b+3c)^2(a^2-b+4)}\geq\frac{81(a+b+c)^2}{\sum\limits_{cyc}(a+5b+3c)^2(a^2-b+4)}.$$
Thus, it's enough to prove that:
$$\frac{81(a+b+c)^2}{\sum\limits_{cyc}(a+5b+3c)^2(a^2-b+4)}\geq\frac{3}{4},$$ which is true and nice!
After homogenization we need to p... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Proving the limit of $\frac{n+2}{n^3 -4n-2}$ using first principles. I'm trying to prove the limit of the sequence of $x_n=\frac{n+2}{n^3 -4n-2}$ using first principles. Here's what I've managed.
First I find the proposed limit using the algebra of limits.
$$x_n=\frac{n+2}{n^3 -4n-2} = \frac{\frac{1}{n^2}+\frac{2}{n^3}... | Your proof is fine, except you have not provided details of how $\frac{n+2}{n^3-4n-2} \leq \frac 2n$ for $n \geq 3$.
How would I prove this? Cross multiply first, to see that it is enough to prove $n^2+2n \leq 2n^3-8n -4$ for $n \geq 3$. At $n=3$, this inequality is true by checking, and if it is true for $n=k$, then w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3618533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of $a^2+b^2+c^2+2abc$ when $a+b+c=3$ and $a,b,c\geq0$.
Given $a,b,c\geq0$ such that $a+b+c=3$, find the minimum value of $$P=a^2+b^2+c^2+2abc.$$
It seems like the minimum value of $P$ is $5$ when $a=b=c=1$, but I can find at least one example where $P<5$.
My attempt:
Without loss of generality... | It occurs for $c=0$ and $a=b=\frac{3}{2}$ and we obtain a value $\frac{9}{2}$.
Also, you proved that it's a minimal value.
I like the following way.
We need to prove that
$$(a+b+c)(a^2+b^2+c^2)+6abc\geq\frac{1}{2}(a+b+c)^3$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$ which is true by Schur.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3620461",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Fastest way to solve $x^3\equiv x \pmod{105}$ $$x^3\equiv x \pmod{105}$$
I'm trying to solve this equation. Here's what I tried so far:
$$x^3\equiv x \pmod{105} \iff x^2\equiv 1 \pmod{105}$$
Then, applying the Chinese remainder theorem, I got the system:
$$\cases{x^2 \equiv 1 \pmod{5}\\x^2 \equiv 1 \pmod{7}\\x^2 \equiv... | Hint $ $ It's always true mod $3,\,$ so by CRT we need only combine all roots $\{0,\pm1\}$ mod $5$ and $7,\,$ and $\,x\equiv a\pmod{\!5},\,x\equiv b\pmod{\!7}\!\iff\! x\equiv b+14(b-a)\pmod{\!35}.\,$ For $\,a,b\in \{0,\pm1\}$ this yields $\,x\equiv \pm \{0,1,6,14,15\}\pmod{\!35}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3623260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find all real solutions for $x$ in $ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $ Find all real solutions for $x$ in
$$ 2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2 . $$
I noticed that $$2(2^x- 1) x^2 + (2^{x^2}-2)x = 2^{x+1} -2=2(2^x−1)(x^2-1)+(2^{x^2}−2)x=0.$$
I know that $2^{n+1}-1$ will always be greater than $2^n-1$... | $1$, $-1$ and $0$ are roots.
Let $x\notin\{0,1,-1\}$.
Thus, $$2(2^x-1)x^2+2x(2^{x^2-1}-1)=2(2^x-1)$$ or
$$(2^x-1)(x^2-1)+(2^{x^2-1}-1)x=0$$ or
$$\frac{2^x-1}{x}+\frac{2^{x^2-1}-1}{x^2-1}=0,$$ which has no real roots because for any $a\neq0$ we have $$\frac{2^a-1}{a}>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Question about $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$ $\lim\limits_{x \to2} \frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}$
Let $x=t^6$, then it becomes $\lim\frac{t^2 - (\sqrt[6]{2})^2}{t^3 - (\sqrt[6]{2})^3}$=$\lim\frac{(t+\sqrt[6]{2})(t-\sqrt[6]{2})}{(t-\sqrt[6]{2})(t^2+... | Let $x=y+2$
$$\frac{\sqrt[3]{x} - \sqrt[3]{2}}{\sqrt{x} - \sqrt{2}}=\frac{\sqrt[3]{y+2} - \sqrt[3]{2}}{\sqrt{y+2} - \sqrt{2}}$$ and use the binomial theorem or Taylor series around $y=0$. Using Taylor, you would have
$$\frac{\frac{y}{3\ 2^{2/3}}-\frac{y^2}{18\ 2^{2/3}}+O\left(y^3\right) } {\frac{y}{2 \sqrt{2}}-\frac{y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3624692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\int \frac{dx}{(a+b\cos x)^n}=\frac{A\sin x}{(a+b\cos x)^{n-1}}+B\int {dx \over (a+b\cos x)^{n-1}}+C\int {dx \over (a+b\cos x)^{n-2}}$ I've been trying to develop this integral in parts twice, but I can't come up with anything meaningful,
Could you give me some hints on how to develop this exercise?
Prove t... | Note,
$$\frac b{\sin x} \left(\frac{\sin^2x}{(a+b\cos x)^{n-1}}\right)'
=\frac{(n-1)(b^2-a^2)}{(a+b\cos x)^n}+\frac{2a(n-2)}{(a+b\cos x)^{n-1}}
-\frac{n-3}{(a+b\cos x)^{n-2}}$$
Integrate both sides and denote $I_n= \int \frac{dx}{(a+b\cos x)^n}$,
$$(n-1)(b^2-a^2)I_n + 2a(n-2)I_{n-1} -(n-3)I_{n-2} \\
= \int \frac b{\sin... | {
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"url": "https://math.stackexchange.com/questions/3628255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is there a simple proof for the behaviour of this solution? Let $0 <s \le 1$, and suppose that $0 <b \le a$ satisfy
$$ ab=s,a+b=1+\sqrt{s}.$$
Then $a \ge 1$.
I have a proof for this claim (see below), but I wonder if there are easier or alternative proofs.
In particular, my proof is based on explicit computation of... | Hint: $(a-1)(b-1) = ab - a - b + 1 = s - \sqrt{s} \leq 0 $.
Hence, conclude that $ a \geq 1 \geq b $, with equality when $ s = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3629503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluating $\cos\frac{17\pi}{18} \cos\frac{7\pi}{9}+ \sin\frac{17\pi}{18} \sin\frac{7\pi}{9}$ with a sum or difference formula
Use a sum or difference formula to find the exact value of the following:
$$\cos\frac{17\pi}{18} \cos\frac{7\pi}{9}+ \sin\frac{17\pi}{18} \sin\frac{7\pi}{9}$$
The answer I got for this was $0... | Yes; using the difference formula, it's $\cos\left(\dfrac{17\pi}{18}-\dfrac{14\pi}{18}\right)=\cos
\left(\dfrac\pi6\right)=\dfrac{\sqrt3}2.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3630182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Finding $\lim_{n\to\infty}\prod_{n=1}^{\infty}\left(1-\frac{1}{n(n+1)}\right)$ I have a trouble with this limit of the infinite product:
$$\lim _{n \to\infty}\left(1-\frac{1}{1 \cdot 2}\right)\left(1-\frac{1}{2 \cdot 3}\right) \cdots\left(1-\frac{1}{n(n+1)}\right)$$
My attempt:
We have
$$\prod_{n=1}^{\infty}\left(1-\f... | By means of Weierstrass product of the gamma function，observe that
$$\frac{1}{\Gamma(z)}=e^{\gamma z} z \prod_{n=1}^{\infty}\left(1+\frac{z}{n}\right) e^{-z / n}$$
where $\gamma $ is the Euler–Mascheroni constant.
Then
$$\prod_{n=1}^{\infty} \frac{\left(n-a_{1}\right)\left(n-a_{2}\right)}{n(n+1)}=\frac{\Gamma\left(1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3636067",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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How to prove the number of three-order $3\times 3$ matrices with row and column sums both equal to $r$ is $H_3(r) = \binom{r+5}{5} - \binom{r+2}{5}$? The combinatorial problem is as follows:
Let $H_3(r)$ denote the number of $3\times 3$ matrices with nonnegative integer
entries such that each row and each column sum to... | Let
$$A=\begin{bmatrix}1&0&0\\0&1&0\\0&0&1\end{bmatrix}\qquad
B=\begin{bmatrix}0&1&0\\0&0&1\\1&0&0\end{bmatrix}\qquad
C=\begin{bmatrix}0&0&1\\1&0&0\\0&1&0\end{bmatrix}$$
$$D=\begin{bmatrix}0&0&1\\0&1&0\\1&0&0\end{bmatrix}\qquad
E=\begin{bmatrix}0&1&0\\1&0&0\\0&0&1\end{bmatrix}\qquad
F=\begin{bmatrix}1&0&0\\0&0&1\\0&1&0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3638113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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What is $\frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$ for $x \rightarrow 0$? I am trying to find the limit
$\lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}$
for $a,b \in \rm{I\!R}_{+}$. Applying ... | Note
$$\begin{align}
L = & \lim_{x \rightarrow0} \frac{a b \sin x}{2\sqrt{(a^2 + b^2 + 2 ab \cos x ) \cdot (a+b - \sqrt{a^2+b^2+ 2 ab \cos x})}}\\
= & \lim_{x \rightarrow0} \frac{1}{2\sqrt{a^2 + b^2 + 2 ab \cos x }}
\cdot \lim_{x \rightarrow0} \frac{a b \sin x}{\sqrt{ a+b - \sqrt{a^2+b^2+ 2 ab \cos x}}}\\
=&\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3640785",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Does this limit converge on e? Playing around with some math in python today I came across what appears to be an interesting pattern:
Starting at n=1 as n approaches positive infinity, take (n+1)^(n+2)/n^(n+1) and get a list of ratios of exponential expressions. At first glance the ratios between the numbers appeared t... | We have
\begin{align*}
&\left( {1 + \frac{1}{{n + 1}}} \right)^{n + 2} (n + 2) - \left( {1 + \frac{1}{n}} \right)^{n + 1} (n + 1) \\ & = \left( {1 + \frac{1}{n}} \right)^{n + 1} \left[ {\left( {\frac{{(n + 2)n}}{{(n + 1)^2 }}} \right)^{n + 1} \frac{{(n + 2)^2 }}{{(n + 1)^2 }} - 1} \right](n + 1)
\\ &=
\left( {1 + \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3641036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find the first four non zero terms of the Taylor series for $f(x) = \int_0^x \frac{1-\cos(t)}{t}dt$
Find the first four non zero terms of the Taylor series for $f(x) =
\int_0^x \frac{1-\cos(t)}{t}dt$
$$f(x) = \int_0^x \frac{1-\cos(t)}{t}dt = \int_0^x \frac{1- (1 + \sum_{n=1}^{\infty}\frac{(-1)^n t^{2n}}{(2n)!})}{t} d... | It is correct, and a quite fast way of solving it. Another (not as optimal) way of doing it is by differentiating the function and using the fundamental theorem of calculus.
$$f(x) = \int_0^x \frac{1-\cos(t)}{t}dt \Rightarrow f'(x) = \frac{1-\cos(x)}{x}.$$
Evaluating the term above at $x=0$ yields zero, and therefore t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3642587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluate $\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$ without L'Hôpital This limit is one of the "Problems Plus" from Stewart Calculus:
$$\lim_{x \to 0} \frac{\sin\left(a+2x\right)-2\sin\left(a+x\right)+\sin\left(a\right)}{x^{2}}$$
Note that the limit is of the indetermi... | Let's first expand the $\sin$ functions using the addition formulae,
$$\sin(a+2x) = \sin(a) \cos(2x) + \cos(a) \sin(2x)$$
and similarly
$$\sin(a+x) = \sin a \cos x + \cos a \sin x$$
Expand $x$-dependent $\sin$ and $\cos$ terms to second order,
$$\sin x = x + O(x^3), \quad \cos x = 1 - \frac{x^2}{2} + O(x^4)$$
so that w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3643583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for all positives $a, b$ and $c$, $(\sum_{cyc}\frac{c + a}{b})^2 \ge 4(\sum_{cyc}ca)(\sum_{cyc}\frac{1}{b^2})$.
Prove that for all positives $a, b$ and $c$, $$\left(\frac{b + c}{a} + \frac{c + a}{b} + \frac{a + b}{c}\right)^2 \ge 4(bc + ca + ab) \cdot \left(\frac{1}{a^2} + \frac{1}{b^2} + \frac{1}{c^2}\righ... | The inequality follows from the famous inequality Ukraine MO 2001.
If $ a,\,b,\,c$ and $x,\,y,\,z$ non-negative real numbers, then
$$ [x(b+c) + y(c+a) + z(a+b)]^2 \geqslant 4(ab+bc+ca)(xy+yz+zx). \quad (1)$$
Now, using $(1)$ with $x=a^2,\,y=b^2,\,z=c^2$ we get
$$ [a^2(b+c) + b^2(c+a) + c^2(a+b)]^2 \geqslant 4(ab+bc+ca)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3649363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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How to find the expected value of infinite discrete random variable? Question: A fair six-sided die is rolled repeatedly. Let $X$ be the number of times it is rolled until two different numbers are seen (so, for example, if the first two rolls were each $4$ and the third roll was $5$ then $X=3$).
How to I find the expe... | There are faster ways, but one relatively simple approach is
$\mathsf{E}[X]=2\left(\frac 56\right) + 3\left(\frac 56\right)\left(\frac 16\right) + 4\left(\frac 56\right)\left(\frac 16\right)^2 + 5\left(\frac 56\right)\left(\frac 16\right)^3 + \cdots$
so $\frac16 \mathsf{E}[X]=2\left(\frac 56\right)\left(\frac 16\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3654422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Repeated roots: $f^{1992}(x) = x$ for $f(x) = 4x(1-x)$ Given a function $f : [ 0 , 1 ] \to \ [ 0 , 1 ]$ defined by $f(x) = 4x(1-x)$ for all $x \in [ 0 , 1 ]$. How many distinct roots does the equation $f^{1992}(x) = x$ has. ($f^n(x) = f\left(f^{n-1}(x)\right)$)
I've already know that $\deg\left(f^{1992}(x)\right) = 2^{... | It is possible to solve $f_n(x) = x$, $ \ \ f_n(x)=f(f_{n-1}(x))$ in a general way, for any integer $n$:
$ $
$$f_{n+1}(x)=4 f_{n}(x)(1-f_{n}(x))$$
Solving recursively for $f_n(x)$:
$$f_{n}(x)=\frac{1}{2}-\frac{1}{2}\cos(2^n c_x)$$
$$f_1(x)=\frac{1}{2}-\frac{1}{2}\cos(2 c_x)=4x(1-x).$$
solving for $c_x$ and $x\in [0,1]$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655004",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $x, y \in \mathbb R$ such that $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2$
Let $x, y \in \mathbb R$ s. t . $x^2+y^2=2x-2y+2$. Find the largest possible value of $x^2+y^2-\sqrt{32}$
I know this is a duplicate of another question, but that question has solutions involving calculus and geometry,... | We have: $$6+ \underbrace{\left( 1+\frac{\,\sqrt {2}}{2} \right) \left( {x}^{2}+{y}^{2}-2\,x+2\,y-2
\right)}_{=\, 0} -({x}^{2}+{y}^{2}-\sqrt {32})$$
$$=\frac{\,\sqrt {2}}{4} \left( \sqrt {2}\,x-\sqrt {2}-2 \right) ^{2}+\frac{\sqrt{2}}{4}\,
\left( \sqrt {2}y+\sqrt {2}+2 \right) ^{2}\geqq 0$$
Therefore: $$x^2 +y^2 -\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3655727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Decomposing an inverse Laplace transform The given inverse Laplace transform is:
$$\mathscr{L}^{-1}\left[\frac{5s^2+12s-4}{s^3-2s^2+4s-8} \right]$$
First split it up into three separate fractions and factorize the denominator
$$\mathscr{L}^{-1} \left[\frac{5s^2}{(s-2)(s^2+4)} \right]+\mathscr{L}^{-1} \left[\frac{12s}{... | Note that the denominator is:
$$(s-2)(s^2+4)$$
The numerator can be rewritten as :
$$f(s)=5s^2+12s-4$$
$$f(s)=5(s^2+4)-24+12s$$
$$f(s)=5(s^2+4)+12(s-2)$$
Back to our function :
$$h(t)=\mathscr{L}^{-1}\left[\frac{5s^2+12s-4}{s^3-2s^2+4s-8} \right]
$$
$$h(t)=\mathscr{L}^{-1}\left[\dfrac 5 {s-2}+\dfrac{12}{s^2+4} \right]$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3656751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Using triangle Inequality to solve inequality. Question from a textbook.
Show that for non negative $x,y,z$ that
$$ (x+y+z) \sqrt{2} \leq \sqrt{x^2 + y^2} + \sqrt{ y^2 + z^2} + \sqrt{x^2 +z^2} $$
and that for $ 0< x \leq y \leq z$,
$$ \sqrt{y^2 +z^2} \leq x\sqrt{2} + \sqrt{(y-x)^2 +(z-x)^2} $$
,which the hint was t... | *
*$(\sqrt{x^2 + y^2} + \sqrt{ y^2 + z^2} + \sqrt{x^2 +z^2} )^2= 2(x^2 + y^2+ z^2) + \;\;non-negative\;\; terms\geq 2(x^2 + y^2+ z^2)$
which yields the first inequality.
*$(y-x)^2 + (z-x)^2=y^2+z^2+2x^2-2x(y+z)\geq y^2+z^2+2x^2-2\sqrt{2}x(\sqrt{y^2+z^2})=(\sqrt{y^2+z^2}-\sqrt{2}x)^2$
yields the second inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3657747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Lower bound $\frac{\det(A+B+C)}{\det(A+C)}$ in terms of $\frac{\det(A+B)}{\det(A)}$ Assume matrix $A$ is symmetric and positive definite, and matrices B and C are symmetric and positive semi-definite. Originally I have ratio between determinants:
$$\frac{\det(A+B)}{\det(A)}$$
By adding another matrix C inside the deter... | Turns out it is very easy to show the inequality holds.
\begin{aligned}
\frac{\det(A+B)}{\det(A)} &= \frac{det(A+C)}{det(A)}\frac{det(A+B)}{det(A+C)}\\
& \leq \frac{det(A+C)}{det(A)}\frac{det(A+B+C)}{det(A+C)}\\
\end{aligned}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3661007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$
If $a, b$ and $c$ (all distinct) are the sides of a triangle ABC opposite to the angles $A, B$ and $C$, respectively, then $\frac{c\sin(A-B)}{a^2-b^2}-\frac{b\sin(C-A)}{c^2-a^2}$ is equal to $?$
By opening, $\sin(A-B)$ as $\sin A\cos B-\cos A\sin B$ and ... | One can easily show that
$$
\frac{c\sin(A-B)}{a^2-b^2}=\frac1{2R},\tag1
$$
where $R$ is the radius of the circumscribed circle.
Indeed substituting in LHS of (1)
$$a=2R\sin A,\quad b=2R\sin B,\quad c=2R\sin C $$
one obtains:
$$\begin{align}
\frac{c\sin(A-B)}{a^2-b^2}&=\frac1{2R}\frac{\sin C\sin(A-B)}{\sin^2A-\sin^2B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3663019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Find the minimum value of $\frac ab+\frac {b}{a+b+1}+\frac {b+1}{a}$ when $a,b>0$ I was trying to solve this question and I observed that if I use partial derivative then it would be calculating. Moreover, I observed that $\frac ab\times\frac {b}{a+b+1}\times\frac {b+1}{a}=\frac {b+1}{a+b+1}$. So I thought to use AM-GM... | For $a>0$ and $b>0$ let $1=t$ and $f(t)=\frac{a}{b}+\frac{b}{a+b+t}+\frac{b+t}{a},$ where $t\geq0.$
Thus, $f'(t)=\frac{1}{a}-\frac{b}{(a+b+t)^2}>0,$ which says $$f(t)\geq f(0)=\frac{a}{b}+\frac{b}{a+b}+\frac{b}{a}.$$
Now, let $a=bx$, where $x>0$.
Thus, we need to find $\min\limits_{x>0}g,$ where $$g(x)=x+\frac{1}{x+1}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3670740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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On Radical Mersenne Prime Containing Series, and why the number 5 doesn't appear here? As part of a project on operators I was studying the function:
$$ L =\left( \sqrt{x^3} - x \right) + \left( \sqrt{\sqrt{x^3}} - \sqrt{x} \right) + \left( \sqrt{\sqrt{\sqrt{x^3}}} - \sqrt{\sqrt{x}}\right) + ... $$
It can be written ... | $$\prod_{k=0}^{n-1} \left(\frac{3}{2^i}-k\right) = \sum_{j=1}^n \frac{3^j}{2^{ij}} \left[{n \atop j}\right]$$
where $\left[{n \atop j}\right]$ is a Stirling number of the first kind.
Then
$$\sum_{i=1}^\infty \prod_{k=0}^{n-1} \left(\frac{3}{2^i}-k\right) = \sum_{j=1}^n \frac{3^j}{2^j-1} \left[{n \atop j}\right] $$
Simi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3672228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Probability problem on umbrellas
Two absent-minded roommates, mathematicians, forget their umbrellas in some way or another. $A$ always takes his umbrella when he goes out, while $B$ forgets to take his umbrella with probability $1/2$. Each of them forgets his umbrella at a shop with probability $1/4$. After visiting ... | A forgets his umbrella along the way with probability $p=1-\left(\frac{3}{4}\right)^3=\frac{37}{64}$. B forgets his umbrella along the way with probability $q=\frac{1}{2}\left(1-\left(\frac{3}{4}\right)^3\right)=\frac{37}{128}$. Therefore, the probability that exactly one umbrella remains when both A and B return home ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3673481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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$a+b+c+d=1, a,b,c,d ≠ 0$, then prove that $(a + \frac{1}{a})^2 + (b + \frac{1}{b})^2 + (c + \frac{1}{c})^2 + (d + \frac{1}{d})^2 \ge \frac{289}{4} $ If $a+b+c+d=1$, $a,b,c,d ≠ 0$, prove that $$\left(a + \frac{1}{a}\right)^2 + \left(b + \frac{1}{b}\right)^2 + \left(c + \frac{1}{c}\right)^2 + \left(d + \frac{1}{d}\right)... | If you consider $a,b,c,d$ are all positive
According to cauchy-schwarz inequality
$(1+1+1+1)*((a+1/a)^2+(b+1/b)^2+(c+1/c)^2+(d+1/d)^2)>=$
$(a+b+c+d+1/a+1/b+1/c+1/d)^2=$
$(a+b+c+d+4+b/a+a/b+c/a+a/c+d/a+a/d+c/d+d/c+c/b+b/c+d/b+b/d)^2>=(5+12)^2$[because $a/b+b/a>=2 and a+b+c+d=1$]
So we get.... $(a+1/a)^2+(b+1/b)^2+(c+1/c... | {
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"url": "https://math.stackexchange.com/questions/3674014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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If $z^2+z+2=0$, then $z^2 + \frac{4}{z^2} = -3$
If $z^2+z+2=0$, for $z\in\mathbb C$, demonstrate:
$$z^2 + \frac{4}{z^2} = -3$$
$z^2 = - 2 - z$, but it didn't help me.
Is there any other elegant solution?
| $z^2+z+2=0$ means $z^2=-z-2$ and $1+z+\dfrac 2z=0$ so $\dfrac2z=-1-z$ so $\dfrac4{z^2}=z^2+2z+1$.
Therefore, $z^2+\dfrac4{z^2}=z^2+z-1=z^2+z+2-3=-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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The polynomial $x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$ has no real roots.
Prove that the polynomial $$x^{2n} - 2x^{2n-1} + 3x^{2n-2}- \cdots -2nx + 2n + 1$$ has no real roots.
Here is the solution
Transcribed from this image
*For $x \leq 0$ we have obviously $p(x)>0$. Let $x>0$. We transform the p... | Solution shows that $p(x)>0, \text{ for all } x$, which means that there is no any value of $x$ such that $p(x)=0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability: 3 Urns, pulling out 3 balls Question: There are 3 urns: A, B and C.
Urn A has 2 white balls and 4 red balls
Urn B has 8 white balls and 4 red balls
Urn C has 1 white ball and 3 red balls
We pull one ball from each urn, and we get that 2 of them are white.
What is the probability that the ball we pulled ... | Let $P(E)$ be the probability that the ball we pulled from urn A is white. There are $6 \cdot 12 \cdot 4 = 288$ possible combinations of balls drawn in our sample space.
If $P(F)$ is the probability that exactly two balls are chosen from the sample space then: $$P(F) = \frac{\text{Number of combinations of exactly 2 wh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this inequality with $xyz=1$ let $x,y,z>0$ and such $xyz=1$,show that
$$f(x)+f(y)+f(z)\le\dfrac{1}{8}$$
where $f(x)=\dfrac{x}{2x^{x+1}+11x^2+10x+1}$
I try use this $2x^x\ge x^2+1$,so we have
$$2x^{x+1}+11x^2+10x+1\ge x^3+11x^2+11x+1=(x+1)(x^2+10+1)$$
It need to prove
$$\sum_{cyc}\dfrac{x}{(x+1)(x^2+10x+1)}\le\dfr... | Now, let $x=\frac{a}{b},$ $y=\frac{b}{c}$, where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{c}{a}$ and since $$x^x\geq x,$$ it's enough to prove that:
$$\sum_{cyc}\frac{ab}{13a^2+10ab+b^2}\leq\frac{1}{8},$$
which is true by BW.
Indeed, let $a=\min\{a,b,c\}$, $b=a+u$ and $c=a+v$.
Thus, we need to prove that:
$$384(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3674817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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} |
Find solutions of recursive equations Stuck... Find solutions of recursive equations using generating functions.
$$
x_{n+2} = 14x_{n+1} - 49x_n + n7^n, n\ge 0\\
x_0 = 1\\
x_1=14
$$
What I tried:
$$
x_{n} = 14x_{n-1} - 49x_{n-2} + (n-2)7^{n-2}, n\ge 2\\
a_{n} = 14a_{n-1} - 49a_{n-2} + (n-2)7^{n-2}\\
F(x) = \sum_{n=0}a_... | This looks correct so far. To proceed, note that $$\sum_{n \ge 0} n z^n = \sum_{n \ge 1} n z^n = z \sum_{n \ge 1} n z^{n-1} = z \frac{d}{dz}\sum_{n \ge 1} z^n = z \frac{d}{dz}\frac{z}{1-z} = \frac{z}{(1-z)^2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3676876",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Suppose $f \circ g(x) = \frac{x^2-6x+2}{x+1}$ and $g(x) = 1 - x$. Then $f(-1)$ is equal to... I tried to substitute the value of $g(x)$ to every $x$ in $f \circ g(x) = \frac{x^2-6x+2}{x+1}$ and ended up with $\frac{x^2+4x-3}{-x+2}$.
Although honestly I do not know if that helps, or what I could do next.
I thought mayb... | Given that $f(1-x)=\frac{x^2-6x+2}{x+1}$
Put $x=1-u$, we get:
$f(u)=\frac{(1-u)^2-6(1-u)+2}{(1-u)+1}$
Set $u=-1$, we get:
$f(-1)=\frac{2^2-2\cdot 6+2}{2+1}=\frac{-6}{3}=-2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3681089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How to integrate $\int\frac{dx}{\sqrt{x+1}+\sqrt{x-1}}$? Firstly, I tried to multiply the denominator and numerator by $\sqrt{x+1}$ but to no avail. Then I tried to take $\sqrt{x+1} = u$; now $\sqrt{x-1}=u-2$ but I couldn't find any success here either.
How can I integrate this expression?
| Setting $\small\left\lbrace\begin{aligned}x&=\cosh{\left(2y\right)}\\ \mathrm{d}x&=2\sinh{\left(2y\right)}\,\mathrm{d}y\end{aligned}\right. $, we get : \begin{aligned}\int{\frac{\mathrm{d}x}{\sqrt{1+x}+\sqrt{x-1}}}&=\frac{1}{\sqrt{2}}\int{\frac{2\sinh{\left(2y\right)}}{\sinh{y}+\cosh{y}}\,\mathrm{d}y}\\ &=\frac{1}{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3688344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Technique for simplifying, e.g. $\sqrt{ 8 - 4\sqrt{3}}$ to $\sqrt{6} - \sqrt{2}$ How to find the square root of an irrational expression, to simplify that root. e.g.:
$$
\sqrt{ 8 - 4\sqrt{3} } = \sqrt{6} - \sqrt{2}
$$
Easy to verify:
\begin{align}
(\sqrt{6} - \sqrt{2})^2 = 6 - 2\sqrt{12} +2
= 8 - 4 \sqrt{3}
\end{alig... | I don't think there's a name for this procedure but let's apply it to $\sqrt{8-4\sqrt3}$.
If you suspect this equals $\sqrt a\pm\sqrt b$ with rationals $a$ and $b$, then
$$8-4\sqrt3=(\sqrt a\pm\sqrt b)^2=(a+b)\pm2\sqrt{ab}$$
so you want to solve simultaneously $a+b=8$ and $-4\sqrt{3}=\pm2\sqrt{ab}$.
So you need the min... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3689064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
The binomial coefficient $\left(\begin{array}{l}99 \\ 19\end{array}\right)$ is $ 107,196,674,080,761,936, x y z $ , Find $x y z$
The binomial coefficient $\left(\begin{array}{l}99 \\ 19\end{array}\right)$ is a 21 -digit number:
$
107,196,674,080,761,936, x y z
$
Find the three-digit number $x y z$
I showed that $... | Since $99 \equiv -1 \pmod {25}$, we have $99 \cdot 98 \cdots 81 \equiv (-1)^{19}19! \pmod {25}$. What we would like to do is to simply divide by $19!$ and be done, but you'll notice that $19! \equiv 0 \pmod{25}$ because of the multiples of $5$. So instead, we treat the multiples of $5$ separately and this gives
$$ \bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3690411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How would you find the number of solutions to $m_1+2m_2+3m_3=n$? Assuming $m_i$ are nonnegative integers
I understand that we need to use $C(n+k-1,n-1)$ here but I am not sure how the coefficients of $m_i$ affect the equation?
For example to find the number of solutions to $m_1+m_2+m_2=n$, we could use the equation abo... | We can write a statement that is not easy to calculate and try to simplify:
We know $0\le m_3 \le \left\lfloor \dfrac{n}{3}\right\rfloor$ and once we have chosen $m_3$, $0\le m_2 \le \left\lfloor\dfrac{n-3m_3}{2}\right\rfloor$. And for every choice of $m_3$ and $m_2$, there is exactly one choice of $m_1$, which yields ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3695874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integrability of $\frac{1}{(x^2+y^2+z^2)^a}$ on $E=\{(x,y,z)\in \mathbb{R}^3: z>1, \ z^2(x^2+y^2)<1 \}$ Let $E=\{(x,y,z)\in \mathbb{R}^3: z>1, \ z^2(x^2+y^2)<1 \}$ and $$f_{a}(x)=\frac{1}{(x^2+y^2+z^2)^a}$$
I need to find all $a\in \mathbb{R}$ such that $f_a\in L^1(E).$
I already know a solution to this problem, and ... | Estimating $\int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a}$ correctly is how you would get the correct range for convergence. For $a\ge 0$, the integrand is decreasing, so we can estimate
$$ \int_{z^2}^{z^2+z^{-2}} \frac{dt}{t^a} \le \frac1{(z^{2})^a}(z^2 + z^{-2} - z^2) = \frac1{z^{2a+2}}\in L^1_{\text{loc}}(1,\infty),$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3697911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
I don’t understand how to reduce this fraction to the stated solution: The fraction is as follows:
$$
\frac{9 \cdot 11 + 18 \cdot 22 + 27 \cdot 33 + 36 \cdot 44 }{
22 \cdot 27 + 44 \cdot 54 + 66 \cdot 81 + 88 \cdot 108}
$$
That’s all fine. What I don’t get is that my textbook says this reduces to the following:
$$\frac... | There seems to be a mistake:
\begin{align*}
\frac{9\cdot 11 + 18 \cdot 22 + 27 \cdot 33 + 36 \cdot 44}{22\cdot 27 + 44\cdot 54 + 66 \cdot 81 + 88 \cdot 108} & = \frac{9\cdot 11(1\cdot 1+2\cdot 2+3\cdot 3 + 4\cdot 4)}{22\cdot 27(1\cdot 1+2\cdot 2+3\cdot 3+4\cdot 4)} \\
& = \frac{9\cdot 11(1^2+2^2+3^2 + 4^2)}{22\cdot 27... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3702238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Using the Maclaurin series for $\frac{1}{1-x}$ to find $\frac{x}{1+x^2}$ Suppose I know the Maclaurin series for $$\frac{1}{1-x}=1+x+x^2+x^3+...= \sum_{n=0}^{\infty}x^n \tag{1}$$
then I can find the Maclaurin series for $\frac{1}{(1-x)^2}$ by the substitution $x\to x(2-x)$, which is obtained by solving the following eq... | Your series $(1)$ is valid for $x\in(-1,1)$.
So series $(2)$ is valid for $x(2-x)\in(-1,1)$, i.e. for $x\in(1-\sqrt 2,1)\cup(1,1+\sqrt 2)$.
But series $(3)$ is only valid if $1-\frac{1}{x}-x\in(-1,1)$. And it turns out that this is not true for any $x\in\Bbb R$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3703913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove $(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$ For $a,b,c>0; abc=1.$ Prove$:$ $$(a^2+b^2+c^2)^3 \geqq 9(a^3+b^3+c^3)$$
My proof by SOS is ugly and hard if without computer$:$
$$\left( {a}^{2}+{b}^{2}+{c}^{2} \right) ^{3}-9\,abc \left( {a}^{3}+{b} ^{3}+{c}^{3} \right)$$
$$=\frac{1}{8}\, \left( b-c \right) ^{6}+{\frac {117... | Suppose $a = \max\{a,b,c\}.$ By the AM-GM inequality we have
$$9abc(a^3+b^3+c^3) \leqslant \left(ab+ca+\frac{a^3+b^3+c^3}{3a}\right)^3.$$
Therefore, we need to prove
$$a^2+b^2+c^2 \geqslant ab+ca+\frac{a^3+b^3+c^3}{3a},$$
equivalent to
$$\frac{(2a-b-c)(a^2+b^2+c^2-ab-bc-ca)}{3a} \geqslant 0.$$
which is true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3704808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Find three vectors in ${R^3}$ such that the angle between all of them is pi/3? Is there a simple way to do this? I have found that ${(a.b)/|a||b|}$ must be equal to ${1/2}$ but from there I am stuck how to proceed. Any help?
P.s. This is from the MIT 2016 Linear Algebra course and is not homework.
| Take an orthonormal basis $(i,j,k)$ of $\mathbb R^3$, $v_1 = \cos \frac{\pi}{6} i + \sin \frac{\pi}{6} j$ and $v_2 = \cos \frac{\pi}{6} i - \sin \frac{\pi}{6} j$. We have by construction $\angle(v_1, v_2) = \frac{\pi}{3}$.
Now let's find $\alpha$ such that $v_3 = \cos \alpha i + \sin \alpha k$ solve the problem.
That w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3705930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
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.