Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How use Lagrange interpolation prove this inequality How can I use Lagrange interpolation to solve the following problem?
Let $a,b,c,d\in \mathbb{R}$ such that
$$|ax^3+bx^2+cx+d|\le 1 $$
for every $x\in[-1,1]$. Show that
$$|a|+|b|+|c|+|d|\le 7$$
| Let $f(x)=ax^3+bx^2+cx+d$.
From Lagrange Interpolation (or solving a system of linear equations),
we find that $$f(x)=\frac{-2f(-1)+4f(-\frac{1}{2})-4f(\frac{1}{2})+2f(1)}{3}x^3+\frac{2f(-1)-2f(-\frac{1}{2})-2f(\frac{1}{2})+2f(1)}{3}x^2+\frac{\frac{1}{2}f(-1)-4f(-\frac{1}{2})+4f(\frac{1}{2})-\frac{1}{2}f(1)}{3}x+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1547003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How do I find $\liminf$ and $\limsup$ if $a_{2n}=\frac {a_{2n-1}}2$ and $a_{2n+1}=\frac12+\frac {a_{2n}}2$? Its given that $a_1=a>0$ and that for any $n>1$ two things happen:
$$a_{2n}=\frac {a_{2n-1}}2$$
$$a_{2n+1}=\frac12+\frac {a_{2n}}2$$
How do I find $\lim\inf$ and $\lim\sup$
I am trying to look at
$a_{2n+1}$ and ... | It could of course be that I made a mistake in my calculations so proceed with caution!
We'll show by induction that for any $n \geq 1$:
$$
a_{2n} = \frac{a}{2\cdot 4^n} + \sum_{i=1}^{n-1} \frac{1}{4^i}
$$
where a sum $\sum_{i=1}^0\frac{1}{4^n}$ is assumed to be zero.
The first step: $n=1$ is trivial. Now let $n$ be a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1549581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Exponential Diophantine equation $7^y + 2 = 3^x$
Find all positive integer solutions to $$7^y + 2 = 3^x.$$
ATTENTION: MY SOLUTION HAS A TERRIBLE MISTAKE WHICH I HAVE OVERLOOKED!
Obviously, $x > y$. Then, we have $3^x = 7^y + 2 \equiv 0 \pmod {3^y}$. Also, $$7^y = (6 + 1)^y = \sum_{k = 0}^{y} {y \choose k} 6^k \equiv... | Okay so I got this solution after modular bashing for an hour. This better be right.
Since the cases $x,y \le 2$ are already investigated above easily, we look at $x,y \ge 3$.
Rewrite this equation to $$7(7^{y-1}-1)=9(3^{x-2}-1)$$
Now, since $7|3^{x-2}-1$, and the order of $3$ modulo $7$ is $6$, we have $6|x-2$.
This g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1551324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Easy way of Compute this limit Easy way to compute: $$\lim_{x\to \:0\:}\left(\left(\frac{a^x-x\cdot \ln\left(a\right)}{b^x-x\cdot \ln\left(b\right)}\right)^{\frac{1}{x^2}}\right)$$
| Apply $\ln$ to get
$$\tag 1 \frac{\ln (a^x-x\ln a) - \ln (b^x-x\ln b)}{x^2}.$$
Using $e^u = 1 + u + u^2/2 +O(u^3)$ as $u\to 0,$ we get
$$a^x = e^{x\ln a} = 1 + x\ln a + (x\ln a)^2/2 + O(x^3).$$
A similar result holds for $b^x.$ Therefore $(1)$ equals
$$\tag 2 \frac{\ln (1+(x\ln a)^2/2 + O(x^3)) - \ln (1+(x\ln b)^2/2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1552536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\int\arccos(\sqrt{\frac{x-4}{x+6}})dx$ I need integrate:
$$\int\arccos(\sqrt{\frac{x-4}{x+6}})dx$$
How can i solve it? is it good way substitute argument of arccos? $$t=\sqrt{\frac{x-4}{x+6}}$$
| Hint. Here is a route.
By the change of variable
$$
t=\sqrt{\dfrac{x-4}{x+6}}\qquad x= 2\:\dfrac{ 3 t^2+2}{1-t^2} \qquad dx= 2\left(\dfrac{ 3 t^2+2}{1-t^2}\right)'dt
$$ one may obtain
$$
\int\arccos(\sqrt{\frac{x-4}{x+6}})\:dx=2\int\left(\dfrac{ 3 t^2+2}{1-t^2}\right)'\arccos(t)\:dt
$$ then one may integrate by parts ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1553706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Infinite product equality $\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$ Prove the following equation ($|x|<1$)
$$\prod_{n=1}^{\infty} \left(1-x^n+x^{2n}\right) = \prod_{n=1}^{\infty} \frac1{1+x^{2n-1}+x^{4n-2}}$$
I made this question and I have the following answer ... | We have
\begin{align}
\prod_{n = 1}^\infty (1 - x^n + x^{2n}) &= \prod_{n = 1}^\infty \frac{1+x^{3n}}{1+x^n} \\
&= \prod_{n = 1}^\infty \frac{1 - x^{6n}}{1-x^{3n}}\prod_{n=1}^\infty\frac{1-x^{n}}{1-x^{2n}}\\
& = \prod_{n = 1}^\infty \frac{1}{1-x^{3(2n-1)}}\prod_{n = 1}^\infty (1 - x^{2n-1})\\
&= \prod_{n = 1}^\infty \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1555670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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limits of inverse trigonometric functions without L'hospital's rule How do I solve this without using L'Hospital's rule? $$\lim_{h\rightarrow0} \frac{\cos^{-1}(\frac{1}{2}-h) -\cos^{-1}(\frac{1}{2})}{h}$$
I already tried letting $\theta=\cos^{-1}(\frac{1}{2}-h)$ gets $\cos\theta=\frac{1}{2}-h$ then $h=\frac{1}{2}-\cos\... | Your start is good.
$$\lim_{h\to0}\frac{\cos^{-1}\left(\frac{1}{2}-h\right)-\cos^{-1}\left(\frac{1}{2}\right)}{h}$$
Let $\theta=\cos^{-1}\left(\frac{1}{2}-h\right)$ so $h=\frac{1}{2}-\cos h$
$$=\lim_{\theta\to\frac{\pi}{3}}\frac{\theta-\frac{\pi}{3}}{\frac{1}{2}-\cos\theta}$$
Rewrite fraction as a trig value and apply ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1556602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A different approach in distributing $8$ distinct balls into $6$ distinct boxes such that each box has at least $1$ ball Find the number of ways in distributing $8$ distinct balls into $6$ distinct boxes such that there is at least $1$ ball in each box.
We are well acquainted with the traditional Inclusion-Exclusion pr... | On the case 1 you must choose first $2$ boxes over $6$ i.e. $C(6,2)$ and after for the first box you choose $(8\cdot 7)/2!$ and after $(6\cdot 5)/2!$ for the second box, and in third place you count the permutations of the others boxs that is $4!$, so the total is $\binom{6}{2}^2\cdot \binom{8}{2}\cdot4!=15^2\cdot 28\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1559741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor series expansion or any other expansion Can we evaluate this $\lim_{x\to 0}(\frac{3x^2+2}{5x^2+2})^{\frac{3x^2+8}{x^2}}$ using Taylor/Maclaurin series by expanding the function about $x=0?$
I can otherwise solve this limit.This is in the form of $... | Notice, we know $\color{blue}{\lim_{n\to \infty}\left(1-\frac{1}{n}\right)^n=e^{-1}}$
Now we have
$$\lim_{x\to 0}\left(\frac{3x^2+2}{5x^2+2}\right)^{\Large \frac{3x^2+8}{x^2}}$$$$=\lim_{x\to 0}\left(1-\frac{2x^2}{5x^2+2}\right)^{\Large \frac{3x^2+8}{x^2}}$$
let, $\large \frac{2x^2}{5x^2+2}=\frac{1}{t}$$\implies x^2=\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1560950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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The coefficient of $x^3$ in $(1+x)^3 \cdot (2+x^2)^{10}$ Find the coefficient of $x^3$ in the expansion $(1+x)^3 \cdot (2+x^2)^{10}$.
I did the first part, which is expanding the second equation at $x^3$ and I got: $\binom {10} 3 \cdot 2^7 \cdot (x^2)^3 = 15360 (x^2)^3$, but I can't figure out what to do from here.
| Here is a different way. Let $f(x)=(1+x)^3(2+x^2)^{10}$. Then \begin{align}
f'(x)&=3(1+x)^2(2+x^2)^{10}+20x(1+x)^3(2+x^2)^9\\
f''(x)&=...\\
f'''(x)&=6 \left(x^2+2\right)^{10}+360 x (x+1) \left(x^2+2\right)^9+9 (x+1)^2
\left(20 \left(x^2+2\right)^9+360 x^2
\left(x^2+2\right)^8\right)+(x+1)^3 \left(1080 x
\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562811",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
What's $\lim\limits_{n\to\infty} \frac{x^n}{y^n + 1}$ For $x,y \in \mathbb{R}$, x > 1, y >0 what's the limit of $\frac{x^n}{y^n + 1}$?
If $x = y \implies x^n = y^n: $
Let $\varepsilon > 0, y^N > \frac{1}{\varepsilon}, N \in \mathbb{N}$
$|\frac{x^n}{y^n + 1} - 1| = |\frac{x^n}{y^n + 1} - \frac{y^n+1}{y^n+1}| = |\frac{x^... | Assume $y<1$. Then the nominator goes to infinity while the denominator goes to $1$.
Now assume $y>1$. Write the quotient as $\frac{x^n}{y^n} \frac{1}{1+\frac{1}{y^n}}$. Since $y>1$ whe have that $\frac{1}{1+\frac{1}{y^n}}$ goes to $1$. So we only have to worry about $\frac{x^n}{y^n}$. If $x>y$ we see that the limit d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1564990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Solving $\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )$ via partial fractions I'm having issues with the Partial Fractions method:
$$\int \left ( \frac{5x^2+3x-2}{x^3+2x^2} \right )dx$$
I am doing this in a way I was taught, which somehow I feel is a bit different from other methods online.
So, first of all, we must... | $$(3X^2+4X+2X^2-X-2)/(X^3+2X^2 )=(3X^2+4X)/(X^3+2X^2 )+(2X+3)/X^2 -8/(X+2)$$
$ln(X^3+2X^2 )+lnX^2-3/x-8ln(X+2)+c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1565338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
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Show that if $n \equiv 4\pmod 9$, then $n$ cannot be written as the sum of three cubes. This is a repeat of a question asked here.
Show that if $n \equiv 4 (\mod 9)$, then $n$ cannot be written as the
sum of three cubes.
Solution: Any integer has least residue as either $0,1,2,3,4,5,6,7,8 (\mod 9)$
Now, $$0^3\equi... | HINT.- You have $8\equiv-1\pmod9$. Besides the same property goes for $n\equiv5\pmod9$, I mean
If $n\equiv5\pmod9$ then $n$ cannot be written as the sum of three cubes.
The proof is equal to the same as you have made for $n\equiv4\pmod9$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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How does this determinant calculation work? Given that $a_0, a_1,...,a_{n-1} \in \mathbb{C}$ I am trying to understand how the following calculation for the determinant of the following matrix follows:
$$
\text{det}
\begin{bmatrix}
x & 0 & 0 & ... & 0 & a_0 \\
-1 & x & 0 & ... & 0 & a_1 \\
0 & -1 & x & ... & 0 & ... | The determinant was expanded along the first column, so you get
$$
\begin{bmatrix}
\color{red}{x} & 0 & 0 & ... & 0 & a_0 \\
\color{blue}{-1} & x & 0 & ... & 0 & a_1 \\
0 & -1 & x & ... & 0 & a_2 \\
. \\
. \\
. \\
0 & 0 & 0 & ... & -1 & x + a_{n-1} \\
\end{bmatrix}
\\ =
(\color{red}{x}) \text{ det}
\beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1569936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Definite integral problem involving trigonometric functions
$$\int_0^{\frac{\pi}{4}}\frac{\sec x}{1+2\sin^2x}dx=?$$
Attempt:
$$=\int_0^{\frac{\pi}{4}}\frac{\sec x}{1+(1-\cos 2x)}dx$$
$$=\int_0^{\frac{\pi}{4}}\frac{\sec x}{2-\cos 2x}dx$$
$$=\sqrt{2}\int_0^{\frac{\pi}{4}}\frac{1}{\sqrt{1+\cos 2x}(2-\cos 2x)}dx$$
$$=\fr... | Hint:
$$\begin{align}
I
&=\int_{0}^{\frac{\pi}{4}}\frac{\sec{\left(x\right)}}{1+2\sin^{2}{\left(x\right)}}\,\mathrm{d}x\\
&=\int_{0}^{\frac{\pi}{4}}\frac{\cos{\left(x\right)}}{\cos^{2}{\left(x\right)}\left(1+2\sin^{2}{\left(x\right)}\right)}\,\mathrm{d}x\\
&=\int_{0}^{\frac{\pi}{4}}\frac{\cos{\left(x\right)}}{\left(1-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571727",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Roots of: $2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$ This is maybe a stupid question, but I want to find the roots of:
$$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
What that I did:
$$\underbrace{2(x+2)(x-1)(x-1)(x-1)}_{A}-\underbrace{3(x-1)(x-1)(x+2)(x+2)}_{B}=0$$
So the roots are when $A$ and $B$ are both zeros when $x=1$ and $x=-2$ ... | $$2(x+2)(x-1)^3-3(x-1)^2(x+2)^2=0$$
$$(x-1)^2(x+2)\left[2(x-1)-3(x+2)\right]=0$$
$$(x-1)^2(x+2)(-x-8)=0$$
$$(x-1)^2(x+2)(x+8)=0$$
So this polynomial of degree $4$ has $4$ real roots i.e. $1,1,-2,-8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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A misunderstanding concerning $\pi$ The very well-known expression
$$\frac {\pi} {4} = 1 - \frac {1} {3} + \frac {1} {5} - \frac {1} {7} + \cdots$$
puts me face to face with a contradictory position. Let
$$s_N = \sum_{k = 0}^{N} \frac {1} {4k + 1} - \sum_{k = 0}^{N} \frac {1} {4k + 3}.$$
Then it is obvious that
$$\frac... | As pointed by others, you misused the Euler-Maclaurin formula.
Indeed, the next terms will involve some coefficients times powers of $1,\dfrac13,\dfrac1{4N+1}$ and $\dfrac1{4N+3}$. The constant terms do not vanish as $N\to\infty$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1575290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Sum of n terms of the series $\frac{1}{1 \cdot 3}+\frac{2}{1 \cdot 3 \cdot5}+\frac{3}{1 \cdot 3 \cdot 5 \cdot 7}+\cdots$ I need to find the sum of n terms of the series
$$\frac{1}{1\cdot3}+\frac{2}{1\cdot 3\cdot 5}+\frac{3}{1\cdot 3\cdot 5\cdot 7}+\cdots$$
And I've no idea how to move on. It doesn't look like an arith... | Clearly $$U_{r+1}=\frac{r}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
$$2U_{r+1}=\frac{2r}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
$$2U_{r+1}=\frac{(2r+1)-1}{1 \cdot3\cdot 5\cdot 7 \cdot......\cdot(2r-3)\cdot(2r-1)\cdot(2r+1)}$$
$$2U_{r+1}=\frac{(2r+1)}{1 \cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 2,
"answer_id": 0
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Hessian of a function Given this equation: $f(x,y,z) = \sqrt{1+x^2+y^2+z^2}$
I tried to calculate the Hessian --> for example $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{1+x^2+y^2+z^2}}$ The second derivativ respect to x is hard to calculate for me: I tried the product rule: $x*(1+x^2+y^2+z^2)^{-1/2}$.
Then i get: ... | Indeed $\frac{\partial f}{\partial x} = \frac{x}{\sqrt{1+x^2+y^2+z^2}}$ which you can write as $\frac{\partial f}{\partial x} = x(1+x^2+y^2+z^2)^{-\frac{1}{2}}$ so that $$\frac{\partial^2 f}{\partial x^2} = (1+x^2+y^2+z^2)^{-\frac{1}{2}} -\frac{1}{2} 2x^2 (1+x^2+y^2+z^2)^{-\frac{1}{2}-1} $$ $$ = (1+x^2+y^2+z^2)^{-\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $4x/(x^4-1)$ dx I'm having trouble solving the integral
$$\int_{5/4}^{13/12}\frac{4x}{x^4-1}\,dx$$
I have a feeling it is to do with the log integration identity but can't seem to manipulate it without involving complex numbers, something outside of this module (so I should be able to solve without them).
An... | $$\int \frac{4x}{x^4-1}dx$$
Partial Fraction Decomposition on $\frac{4x}{x^4-1} = \frac{4x}{(x^2+1)(x^2-1)}$ gives $\frac{a}{x^2+1} + \frac{b}{x^2-1} = 4x \implies a(x^2-1)+b(x^2+1) = 4x$. Solving this gives $a=-2x, b=2x$
$$=\int\frac{2x}{x^2-1}-\int\frac{2 x}{x^2+1}dx$$
$$=\log|x^2-1| -\log(x^2+1) + C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1580479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Minimal c satisfying $x+y-(xy)^c \geq 0$ for all $x,y\in [0,1]$ What is the minimal real $c$ satisfying $x+y-(xy)^c \geq 0$ for all $x,y \in [0,1]$?
Experimentally (though my experiments weren't necessarily accurate enough) I reached as low as $c=\tfrac{13}{32}$, where $c=\tfrac{12}{32}$ violates it.
| Your answer is $c = \frac12$.
If $c = 1/2$,
$\begin{array}\\
x+y-(xy)^c
&=x+y-(xy)^{1/2}\\
&=x-(xy)^{1/2}+y/4-y/4+y\\
&=(\sqrt{x}-\frac12\sqrt{y})^2-y/4+y\\
&=(\sqrt{x}-\frac12\sqrt{y})^2+3y/4\\
&\ge 0\\
\end{array}
$
so it is true for
$c \ge \frac12$
(since
$(xy)^{1/2+d}
\le (xy)^{1/2}
$
for
$d \ge 0$).
Suppose $c = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Need help with $\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2x\right)x}\,dx$ Please help me to evaluate this integral
$$\int_0^1\frac{\log(1+x)-\log(1-x)}{\left(1+\log^2x\right)x}\,dx$$
I tried a change of variable $x=\tanh z$, that transforms it into the form
$$\int_0^\infty\frac{4z}{\left(1+\log^2\tanh z\right)\s... | An alternative way to evaluate $$\frac{\pi}{2} \int_{0}^{\infty} \tanh \left(\frac{\pi u}{2} \right) \frac{e^{-u}}{u} \, du ,$$ which is line $3d$ in robjohn's answer, is to add a parameter and then differentiate under the integral sign.
Specifically, let $$I(a) = \frac{\pi}{2}\int_{0}^{\infty} \tanh \left(\frac{\pi u}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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Convergence of $\sum \limits _{n=1}^{\infty} (-1)^{n} \frac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$ $\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}$
How to check this? I've tried using Leibniz test, it's easy to prove that this one is monotonous, but its limit is rather no... | For
$\sum \limits _{n=1}^{\infty} (-1)^{n} \dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}
$,
the ratio of consecutive terms is
$\dfrac{\dfrac{2^{n+1}(n+1)!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)(2n+5)}}{\dfrac{2^nn!}{5 \cdot 7 \cdot \ldots \cdot (2n+3)}}
=\dfrac{2(n+1)}{(2n+5)}
=\dfrac{2n+2}{2n+5}
=1-\dfrac{3}{2n+5}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1583858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How do I calculate $\int_0^ax^2\sqrt{a^2-x^2}dx$ via substitution? I have to calculate the integral $$\int_0^ax^2\sqrt{a^2-x^2}dx$$ using solely substitution (no integration by parts). $a$ is a positive constant.
I'm confused on how to do this?
| Let $x=a\sin\theta\implies dx=a\cos \theta\ d\theta$
$$\int_0^a x^2\sqrt{a^2-x^2}\ dx=\int_{0}^{\pi/2} a^2\sin^2\theta(a\cos\theta)(a\cos \theta\ d\theta)$$
$$=a^4\int_{0}^{\pi/2} \sin^2\theta\cos^2\theta\ d\theta$$
$$=\frac{a^4}{4}\int_{0}^{\pi/2} (2\sin\theta\cos\theta)^2\ d\theta$$
$$=\frac{a^4}{4}\int_{0}^{\pi/2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int \frac{1}{\sin x+\sec x}\,dx $ Evaluate $$\int \frac{1}{\sin x+\sec x}\,dx $$
Expressing $\sin x$ and $\cos x$ in terms of $\tan\frac{x}{2}$ i.e. putting $\sin x=\dfrac{2t}{1+t^2}$, $\cos x=\dfrac{1-t^2}{1+t^2}$ and hence $dx=\dfrac{2\,dt}{1+t^2}$
$$\int \frac{1-t^2}{1+t-t^3}\,dt $$
| Probably a cleaner expression:
$$I=\int\dfrac{\cos x}{\sin x\cos x+1}dx =\int\dfrac{2\cos x}{2+2\sin x\cos x}dx$$
Now write $2\cos x=\cos x+\sin x+(\cos x-\sin x),$
As $\int(\cos x\pm\sin x)dx=\sin x\mp\cos x$
and $(\sin x\mp\cos x)^2=1\mp2\sin x\cos x,$
$$I=\int\dfrac{(\cos x+\sin x)}{3-(\sin x-\cos x)^2}dx+\int\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Uniquely determining complex multiplication I encountered the following problem and it's not entirely clear to me exactly what I am supposed to do:
Show that the following rules uniquely determine complex multiplication on $\mathbb{C}=\mathbb{R^2}$:
*
*(a) $(z_1+z_2)w=z_1w+z_2w$
*(b) $z_1z_2=z_2z_1$
*(c... | Well, using only the rules given, denoting the complex product with $\cdot$, we can say that for any complex numbers $z = a + i \cdot b$ and $w = c + i\cdot d$, we must have (using $(d)$ whenever there are terms involving at least two multiplications and writing $a \cdot b \cdot c$ for $a \cdot (b \cdot c) = (a \cdot b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Sine of argument with large n approximation I have worked an integral and reduced the integral to
$$\frac{n \pi+\sin\left ( \frac{n \pi}{2} \right )-\sin\left ( \frac{3 \pi n}{2} \right )}{2n \pi}$$
I want to show that for $$n\rightarrow \infty$$ the above equation reduces to
$$\frac{1}{2}$$
Evidently, this means the $... | HINT:
$$\lim_{n\to\infty}\space\frac{n\pi+\sin\left(\frac{n\pi}{2}\right)-\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}=$$
$$\lim_{n\to\infty}\space\left(\frac{1}{2}+\frac{\sin\left(\frac{n\pi}{2}\right)}{2\pi n}-\frac{\sin\left(\frac{3\pi n}{2}\right)}{2\pi n}\right)=$$
$$\lim_{n\to\infty}\space\frac{1}{2}+\lim_{n\to\inf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586669",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Write the equation of the parabola that has the vertex at point $(5,0)$ and passes through the point $(7,−2)$. Write the equation of the parabola that has the vertex at point $(5,0)$ and passes through the point $(7,−2)$.
I know how to do it with the $x$ intercepts but I can't solve this.
| The vertex form of the equation of a parabola is
$$f(x) = a(x - h)^2 + k$$
where $(h, k)$ is the vertex of the parabola. In this case, we are given that $(h, k) = (5, 0)$. Hence,
\begin{align*}
f(x) & = a(x - 5)^2 + 0\\
& = a(x - 5)^2
\end{align*}
Since we also know the parabola passes through the point $(7, -2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1587747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to get $a \cos\left( \omega t \right)+b\sin\left( \omega t \right)=A \cos\left( \omega t+\phi \right)$? So this sprung up in a lecture note on quantum harmonic oscillation.
The equation is
$a \cos\left( \omega t \right)+b\sin\left( \omega t \right)=A \cos\left( \omega t+\phi \right)$
I must have forgotten how the ... | Let $A = \sqrt{a^2+b^2}$ and $\phi$ be an angle such that $\cos\phi = \dfrac{a}{\sqrt{a^2+b^2}}$ and $\sin\phi = \dfrac{-b}{\sqrt{a^2+b^2}}$. Then, by using the identity $\cos(x+y) = \cos x\cos y-\sin x\sin y$, we have:
\begin{align}a\cos(\omega t)+b\sin(\omega t) &= \sqrt{a^2+b^2}\left[\dfrac{a}{\sqrt{a^2+b^2}}\cos(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $\frac{AC \times BC}{AD \times BD}$
$AC$ is $2004$. $CD$ bisects angle $C$. If the perimeter of $ABC$ is $6012$, find $\dfrac{AC \times BC}{AD \times BD}$.
Attempt
Let $c = AD+BD$.
We have that $\dfrac{AC}{AD} = \dfrac{BC}{BD}$. Thus, $$\dfrac{BD}{AD}+1 = \dfrac{BC}{AC}+1 \implies \dfrac{BD+AD}{AD} = \dfrac{AC+... | The problem is not well-posed. In fact, $BC$ can be anything between $1002$ and $3006$ (not inclusive), and the quantity in question can be any number greater than $1$ and less than $25$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588383",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the minimum value of $\sqrt { \frac { a }{ b+c } } +\sqrt [ 3 ]{ \frac { b }{ c+a } } +\sqrt [ 4 ]{ \frac { c }{ a+b } }$ If $a, b, c\ge 0$ with $(a+b)(b+c)(c+a) > 0$,
find the minimum of $\sqrt { \frac { a }{ b+c } } +\sqrt [ 3 ]{ \frac { b }{ c+a } } +\sqrt [ 4 ]{ \frac { c }{ a+b } }$.
The minimum is $\fr... | Remark: Here is an ugly solution. Hope to see nice solutions.
Problem: Let $a, b, c \ge 0$ with $(a+b)(b+c)(c+a)\ne 0$. Find the minimum of
$f(a,b,c) = \sqrt{\frac{a}{b+c}} + \sqrt[3]{\frac{b}{c+a}} + \sqrt[4]{\frac{c}{a+b}}$.
Solution:
If $b=0$, by AM-GM inequality, we have $f = \sqrt{\frac{a}{c}} + \sqrt[4]{\frac{c}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1589425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Prove that $\frac{1}{a}+\frac{1}{b}+\frac{1}{c} \geq 2 \left (\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a} \right) \geq \frac{9}{a+b+c}$
Let $a,b,$ and $c$ be positive real numbers, prove that $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c} \geq 2 \left (\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a} \right) \geq \dfrac{9}{a+b+c}$... | The Inequality $$2(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a})\geq \frac{9}{a+b+c}$$ follows by Cauchy Schwarts. Indeed, multiply by $a+b+c$ and write $$[(a+b)+(b+c)+(c+a)]\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}\right)\geq 9$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590720",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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How can I count solutions to $x_1 + \ldots + x_n = N$? I am interested in how many non-negative integer solutions there are to:
$$x_1 + \ldots + x_N = B$$
where at least $K$ of the variables $x_1, \ldots , x_N \geq C$
For example when:
$B = 5, N = 3, K = 2, C = 2$
I want to count the solutions to:
$$x_1 + x_2 + x_3 = 5... | Let $A_i$ be the set of solutions in nonnegative integers to $x_1+\cdots+x_n=B$ with $x_i\ge C$, for $1\le i\le n$,
and let $T_l=\sum\big|A_{i_1}\cap\cdots\cap A_{i_l}\big|$, where the sum is taken over all $l$-subsets of $\{1,\cdots,n\}$, for $1\le l\le n$.
Using Inclusion-Exclusion, the number of elements in at leas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
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Evaluation of limit with confusing moment Evaluate the following limit:
$\lim \limits_{x\to 0}\dfrac{2\cos ^{-3}x\cdot\sin x}{2\sin x+x\cdot \cos x}\color {\blue}{=}\lim \limits_{x\to 0}\dfrac{2\cos ^{-3}x}{2++x\cdot \frac{\cos x}{\sin x}}$ but $$\lim \limits_{x\to 0}x\cdot \frac{\cos x}{\sin x}=1$$ then $\lim \limits_... | Since, $x\ne 0$ hence, one can also cancel $x$ as factor in numerator & denominator $$\lim_{x\to 0}\frac{\cos^{-3}x\cdot \sin x}{2\sin x+x\cdot \cos x}$$
$$=\lim_{x\to 0}\frac{\cos^{-3}x\cdot \frac{\sin x}{x}}{2\frac{\sin x}{x}+\cos x}$$
$$=\lim_{x\to 0}\frac{\frac{\sin x}{x}}{2\frac{\sin x}{x}+\cos x}\left(\frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove that $5^n + 2\cdot3^{n-1} + 1$ is multiple of $8$ Prove that $5^n + 2\cdot3^{n-1}+ 1$ is multiple of $8$.
I've tried using induction (it isn't):
For $n=1$:
$$5^1 + 2\cdot3^{n-1} + 1 = 8$$
If it is true for $n$, then $n+1$?
\begin{align}
5^{n+1} + 2\cdot3^n + 1
=
&(4+1)^n\cdot(4+1)+ 2\cdot(2+1)^n + 1
\\
=& (4^n... | $5^n+2*3^{n-1}+1 = 5+2*1+1=8$ (mod n) for n coprime with 3 and 5. $n=8$ is such a number.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592108",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$ Prove that $\frac{1}{a+1}<\ln \frac{a+1}{a}<\frac{1}{a},a>0$
First inequality using MVT:
$\frac{1}{a+1}<\ln \frac{a+1}{a}:$
$f(a)=\frac{1}{a+1}-\ln \frac{a+1}{a}$
$f(1)=\frac{1-2\ln 2}{2},f^{'}(a)=\frac{1}{a(a+1)^2}>f(1)\Rightarrow f(a)>f(1)$
$\frac{1}{a+1}-\... | METHOD 1: Non-Calculus Based
In This Answer, I showed using basic tools only that the logarithm function satisfies the inequalities
$$\frac{x}{x+1}\le \log(1+x) \le x \tag 1$$
for $x\ge -1$. Note that $\frac{a+1}{a}=1+\frac1a$. Then, setting $x=\frac1a$ in $(1)$ gives the inequalities
$$\frac{1}{a+1}\le \log \left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
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Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$ Find the value of $\sqrt{1+\frac{1}{1^2}+\frac{1}{2^2}}+\sqrt{1+\frac{1}{2^2}+\frac{1}{3^2}}+...+\sqrt{1+\frac{1}{1999^2}+\frac{1}{2000^2}}$
I found the general term of the sequen... | After taking LCM, we get the general term of the series as:
$$\sqrt{\frac{(k^{2}+k+1)^{2}}{k^{2}(k+1)^{2}}} $$
$$=> \frac{k^{2}+k+1}{k^{2}+k}$$
$$=> 1 + \frac{1}{k^{2}+k}$$
So we have
$$\sum_{k=1}^{1999} 1 + \frac{1}{k^{2}+k}$$
$$=> 1999 + \sum_{k=1}^{1999}\frac{1}{k(k+1)}$$
$$=> 1999 + \sum_{k=1}^{1999}\frac{1}{k} -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593591",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve $\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$? I have a problem with this limit, i have no idea how to compute it.
Can you explain the method and the steps used?
$$\lim\limits_{x \to -\infty} \left(x\left(\sqrt{x^2-x}-\sqrt{x^2-1}\right)\right)$$
|
This is the graph of $x(\sqrt{x^2-x}-\sqrt{x^2-1})$.It is tending to $-\infty$ as $x$ is tending to $-\infty$
After rationalising
$L=\lim_{x\to -\infty}\frac{x(1-x)}{\sqrt{x^2}(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}=\lim_{x\to -\infty}\frac{x(1-x)}{|x|(\sqrt{1-\frac{1}{x}}+\sqrt{1-\frac{1}{x^2}})}$
$L=\lim_{x\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593983",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 8,
"answer_id": 1
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Compute $\int\frac{x}{2x^2+x+3}dx$ $$\int\frac{x}{2x^2+x+3}\,dx$$
Well, to approach this kind of exercises I know that I need to check the derivative of the denominator. which is $4x + 1$.
Then, I can re-write the integral: $\int\frac{0.25(4x+1) - 0.25}{2x^2+x+3}\,dx$.
Then, I get:
$$0.25\int\frac{(4x+1)}{2x^2+x+3}\,dx... | $$\int \frac{x}{2x^2+x+3}dx$$
$$=\frac{1}{4} \int \frac{4x+1}{2x^2+x+3} dx- \frac{1}{4} \int \frac{1}{2x^2+x+3}$$
I presume you know how to solve the first integral [Do so using the sub: $u=2x^2+x+3$].
Complete the square of the second and substitute $m=\sqrt{2}x+\frac{1}{2\sqrt{2}}$to get:
$$\frac{-1}{4}\int\frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Given matrix A. To find $A^{2010}$
Let $\theta = 2\pi/67$. Now consider the matrix
$$
A =
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \\
-\sin \theta & \cos \theta
\end{pmatrix}.
$$
Then the matrix $A^{2010}$ is
\begin{align*}
&\text{(A)}\;
\begin{pmatrix}
\phantom{-}\cos \theta & \sin \theta \... | Note that $A$ is a rotation matrix that takes a vector and rotates it by $1/67$th of a revolution. Now notice that $2010 = 67 \times 30$, so the answer is indeed (B).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1594802",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the area of the region determined by the system: \begin{align} y & \ge |x| \\ y & \le -|x+1| +4 \\ \end{align}
Find the area of the region determined by the system: \begin{align} y
& \ge |x| \\ y & \le -|x+1| +4 \\ \end{align}
My attempt
*
*Assuming $x>0$
I have the system
$$\begin{cases}\begin{align}
y &... | For $x > 0$, your area is the area enclosed by the graphs of $y = x$ and $y = -x + 3$ between $0$ and their intersection point. Thus, determine $x = -x +3$ which gets you $2x = 3 \Leftrightarrow x = {3\over 2}$ . So the first area is
$$\int_{0}^{3 \over 2} 2x - 3 dx$$
Process for $x < 0$ analogously.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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How do I evaluate the sum $\sum_{k=1}^\infty\left(\ln\big(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\big)\right)$ How do I evaluate the sum
$$\sum_{k=1}^\infty\left(\ln\Big(1+\frac{1}{k+a}
\Big)-\ln\Big(1+\frac{1}{k+b}\Big)\right)$$
where $0 <a<b<1$?
Hints will be appreciated
Thanks
| We have
$$
\sum_{k=1}^n \ln\left(1+\frac{1}{k+a}\right)=\ln\left(\prod_{k=1}^n\Big(1+\frac{1}{k+a}\Big)\right)=\ln\left(\prod_{k=1}^n\frac{k+1+a}{k+a}\right)=\ln\left(\frac{n+1+a}{1+a}\right)
$$
and thus
$$
\sum_{k=1}^n \ln\left(1+\frac{1}{k+a}\right)-\ln\left(1+\frac{1}{k+b}\right)=
\ln\left(\frac{n+1+a}{1+a}\right)-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$ Find the maximum possible value of $8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$,where $x>0$
Let $P(x)=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3$
By using $AM-GM$ inequality on the first two terms give me
$8(27)^{\log_{6}x}+27(8)^{\log_... | Hint:
Let $E=8(27)^{\log_{6}x}+27(8)^{\log_{6}x}-x^3=8(3^3)^{\log_{6}x}+27(2^3)^{\log_{6}x}-x^3$
$$=8(3)^{\log_{6}x^3}+27(2)^{\log_{6}x^3}-x^3$$
Now put $\log_{6}x^3=t$
$ \Rightarrow 6^t=x^3$
So we have that $E=8(3)^{t}+27(2)^{t}-6^t$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1595906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Question about cancellations in a series resulting in $\frac {m+1}{m-k+1}$ I have the following series:
$$\displaystyle\sum_{i=1}^{k+1}iP_i = 1\left(1-\frac k m\right)+2\frac k m \left(1-\frac{k-1}{m-1}\right)+3\frac k m \frac {k-1}{m-1}\left(1-\frac{k-2}{m-2}\right)\\+...+(k+1)\frac k m \frac {k-1}{m-1}...\left(\frac{... | This looks harder than it is.
Notice that your series can be written as:
\begin{align}
1 - & \frac{k}{m} + 2\frac{k}{m} - 2\frac{k}{m} \cdot \frac{k-1}{m-1}+3\frac{k}{m}\cdot \frac{k-1}{m-1}-3\frac{k}{m}\cdot \frac{k-1}{m-1}\cdot \frac{k-2}{m-2}+\dots+ \\ & (k-1)\cdot \frac{k}{m} \cdot \frac{k-1}{m-1} \cdot \dots \cdot... | {
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"url": "https://math.stackexchange.com/questions/1597189",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$?
Find the Maclaurin series for $(\cos x)^6$ using the Maclaurin series for $\cos x$ for the terms up till $x^4$.
Here is what I've worked out:
Let $f(x) = \cos x,\ g(x) = (\cos x)^6$.
$$g(x) = (f(x))^6$$
$$\cos x = 1 - \frac{1}{... | $$\cos x=\frac {e^{ix}+e^{-ix}}2$$
so
$$32\cos^6x=\cos 6x+6\cos 4x+15\cos 2x+10=$$$$=(1-18x^2+54x^4+\dots)+(6-48x^2+64x^4-\dots)+(15-30x^2+10x^4-\dots)+10$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 3
} |
Evaluating the improper integral $\int_0^{\infty}\frac{dx}{1+x^3}$
Evaluate $$\int_{0}^{\infty}\frac{dx}{1+x^3}.$$
I tried integration by partial fraction. My work is below:
$$\int_{0}^{\infty}\frac{dx}{1+x^3}=\frac{1}{3} \int_{0}^{\infty}\frac{1}{x+1}+\frac{1}{3} \int_{0}^{\infty}\frac{2-x}{x^2-x+1}.$$
It seems that... | \begin{align}
\int_{0}^{\infty}\frac{1}{1+x^3}dx&\overset{x\to\frac1x}=\int_{0}^{\infty}\frac{x}{1+x^3}dx
= \frac12\int_{0}^{\infty}\frac{1+x}{1+x^3}dx\\
& = \frac12\int_{0}^{\infty}\frac{1}{x^2-x+1}dx
= \frac12\int_{0}^{\infty}\frac{d(x-\frac12)}{(x-\frac12)^2+\frac34}\\
&=\frac1{\sqrt3}\tan^{-1}\frac{2x-1}{\sqrt3}\bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598581",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Prove that that $\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c} \geq \frac{(x+y+z)^2}{a+b+c}.$
Prove that that $\dfrac{x^2}{a}+\dfrac{y^2}{b}+\dfrac{z^2}{c} \geq \dfrac{(x+y+z)^2}{a+b+c}.$ with $a,b,c$ positive real numbers.
Attempt
I tried using Cauchy-Schwarz, but I can't find the correct $a_i$ and $b_i$. How would you ... | More generally, the following is called Titu's Lemma or Engel's form of Cauchy-Schwarz inequality:
For all $a_i\in\mathbb R$, $b_i\in\mathbb R^+$:
$$\frac{a_1^2}{b_1}+\frac{a_2^2}{b_2}+\cdots+\frac{a_n^2}{b_n}\ge \frac{(a_1+a_2+\cdots+a_n)^2}{b_1+b_2+\cdots+b_n}$$
Proof: by Cauchy-Schwarz: $$(b_1+b_2+\cdots+b_n)\left(\... | {
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"url": "https://math.stackexchange.com/questions/1599990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
$\lim \frac{x^2-\sin^2{x}}{\tan(3x^4)}$ as $x$ goes to $0$
Calculate $\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)}$
How does one calculate this limit?
Is it valid to say, since $\sin^2{x}$ is approximated by $x^2$ as $x \to 0$, we have:
$\displaystyle \lim_{x \to 0} \frac{x^2-\sin^2{x}}{\tan(3x^4)} =... | $$L=\lim_{x \to 0}\frac{x^2-sin^2x}{3x^4} \times \lim_{x \to 0} \frac{3x^4}{tan(3x^4)}=\frac{1}{3}\lim_{x \to 0}\frac{x^2-sin^2x}{x^4}$$
$$L=\frac{1}{3}\lim_{x \to 0}\frac{x-sinx}{x^3} \times \lim_{x \to 0}\frac{x+sinx}{x}=\frac{2}{3}\lim_{x \to 0}\frac{x-sinx}{x^3}$$
Now you can use Lhopital's rule
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$. Use the Cauchy-Schwarz Inequality to prove that $a^2+b^2+c^2 \ge ab+ac+bc $ for all positive $a,b,c$.
That's what I've tried:
Let a Cauchy-Schwarz Inequality be :
\begin{array}
(((\sqrt{a} )^2+(\sqrt{b})^2+(\sqrt{c})^2... | By C-S $\sum\limits_{cyc}(1+1)(a^2+b^2)\geq\sum\limits_{cyc}(a+b)^2$
and we are done!
| {
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"url": "https://math.stackexchange.com/questions/1601949",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Prove that $(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$
For the non-negative real numbers $a, b, c$ prove that $$(a^2+2)(b^2+2)(c^2+2)\geq 3(a+b+c)^2$$
What I did is applying Holder's inequality in LHS:$$(a^2+(\sqrt{2})^2)(b^2+(\sqrt{2})^2)(c^2+(\sqrt{2})^2) \geq (abc + 2\sqrt{2})^2$$
Then it suffices to prove that $$(abc... | By C-S $ (a^{2}+2)\left(1+\frac{(b+c)^{2}}{2}\right)\geq(a+b+c)^{2}$.
Hence, it remains to prove that $ (b^{2}+2)(c^{2}+2)\geq3\left(1+\frac{(b+c)^{2}}{2}\right)$,
which is equivalent to $ (b-c)^{2}+2(bc-1)^{2}\geq0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Why this system have one solution Let $b\in (1,2),x\in (0,\frac{\pi}{2})$,if such
$$\begin{cases}
2b^2+b-4=2\sqrt{4-b^2}\cos{x}\\
2b^2-4=2b\cos{(x+\frac{\pi}{18})}-2\sqrt{4-b^2}\cos{\frac{5\pi}{18}}
\end{cases}$$
show that:$$x=\dfrac{\pi}{6}$$
Here is what I already got.
First of all, one should notice equation $x=\d... | Hint $cos50\approx 0.6$ so create $cos(x+π/18)$ in terms of b from second equation and create $cosx$ in terms of b from first equation creating them and simplifying you get a huge equation! Which is $$8b^4+4b^3-18b^2+4b^2\sqrt{4-b^2}+6.6b\sqrt{4-b^2}-26.4\sqrt{4-b^2}+4=0$$ solving it you get roots but only $1.15023$ sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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What's wrong with this use of Taylor's expansions? I'm trying to find the value of the following limit:
$$
\lim_{x \to 0} \frac{x^2\cos x - \sin(x\sin x)}{x^4}
$$
Which I know equals to $-\dfrac13$.
I tried to do the following:
$$
\lim_{x \to 0} \frac{x^2(1 - \frac{x^2}{2} + o(x^4)) - \sin(x(x + o(x^3)))}{x^4}\\
= \lim... | Since you have $x^4$ in the denominator, you need to expand the numerator up to $x^4$. You did that for $x^2\cos x$ but not for $\sin (x \sin x)$.
So, you need to include terms up to $x^4$ in the expansion of $\sin (x \sin x)$:
$$\sin (x \sin x) = x^2-x^4/6+o(x^6)$$
This comes for
$$\sin (x) = x-x^3/6+o(x^5)$$
$$x\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
} |
Solve $\log_{1/3}(x^2-3x+3)≥0$ I want to solve $$\log_{1/3}(x^2-3x+3)≥0$$
Now I know the result is: $x ∈ <1;2>$, but i am not sure how to get it.
My thoughts: $\frac{1}{3}$ to the power of positive number $= (x^2-3x+3)$, now I would solve $x^2-3x+3$ with the help of discriminant to get the points where $x$ is zero, and... | Raise $\frac{1}{3}$ to the power of both sides, and flip the inequality because $\frac{1}{3}^x$ is a decreasing function:
$$
\begin{aligned}
\log_{1/3}(x^2 - 3x + 3 ) &≥ 0\\
x^2 - 3x + 3 &\le \left(\frac{1}{3}\right)^0\\
x^2 - 3x + 3 &\le 1\\
x^2 - 3x + 2 &\le 0\\
\end{aligned}
$$
Now factor to get $(x-1)(x-2) \le 0$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605538",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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How to calculate the expected return in this coin toss game? In a coin tossing game that is made of two rounds, there are two cases.
In first round, if heads comes up, case 1 will be played in second round, if tails comes up, case 2 will be played in second round.
In case 1, two coins are tossed. If at least one of the... | If we land on case 1 in round two, our game is guaranteed to end with either a win or a loss. The probability of a win is simply $\frac{1}{2} \cdot \frac{3}{4} = \frac{3}{8}.$
However, if we land on case 2 in round two, our game could last for an infinite amount of time. To solve for the probability of a win, we sum th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1607271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How many integer solutions does the following system have? $$y_1 + y_2 + y_3 = 20$$
$$1 \le y_1 \le 5$$
$$y_2 \ge 5$$
$$y_3 \ge 5$$
I know how to solve it if it were:
$$y_1 + y_2 + y_3 = 20$$
$$y_1 \ge 1$$
$$y_2 \ge 5$$
$$y_3 \ge 5$$
then I would do:
$$\begin{align}
x_1 + x_2 + x_3 &= (y_1 - 1) + (y_2 - 5) + (y_3 - 5) ... | You wish to solve the equation
$$y_1 + y_2 + y_3 = 20 \tag{1}$$
in the integers subject to the restrictions $1 \leq y_1 \leq 5$, $y_2 \geq 5$, $y_3 \geq 5$. By making the substitutions of $x_1 + 1$ for $y_1$, $x_2 + 5$ for $y_2$, and $x_3 + 5$ for $y_3$, you obtained
$$x_1 + x_2 + x_3 = 9 \tag{2}$$
and correctly note... | {
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"url": "https://math.stackexchange.com/questions/1607360",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Extreme of $\cos A\cos B\cos C$ in a triangle without calculus.
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.
I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing helps.
On this topic we learned also about Cauchy inequality, but I have... | Due to the inequality of means and the law of cosines, we know that:
\begin{equation*}
4a^4=((a^2+b^2-c^2)+(a^2+c^2-b^2))^2\geq 4(a^2+b^2-c^2)(a^2+c^2-b^2)=\\ 16a^2bc\cos\beta \cos\gamma \Rightarrow a^2\geq 4bc\cos\beta \cos\gamma
\end{equation*}
\begin{equation*}
4b^4=((a^2+b^2-c^2)+(b^2+c^2-a^2))^2\g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 7
} |
Is mathematical Induction possible in this situation? Is mathematical Induction possible with this sigma sign?
$\sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) = \frac{b^{n}+1}{b+1}$
with $n = 2s+1 ; s \epsilon \mathbb{N}$
Statement: $\sum_{k=1}^{n} ((-1)^{n-k} * b^{n-k}) = \frac{b^{n}+1}{b+1}$
Assumption: $\sum_{k=1}^{n+2} ((... | Let's pose
$$A_s = \sum_{k=1}^{2s+1} (-b)^{2s+1-k} = \frac{b^{2s+1}+1}{b+1}.$$
We start from $A_1$:
$$A_1 = b^2-b+1 = \frac{b^3+1}{b+1} = \frac{b^{2s+1}+1}{b+1}.$$
Now we need to find the inductive rule.
$$A_{s+1} = \sum_{k=1}^{2(s+1)+1} (-b)^{2(s+1)+1-k} = \sum_{k=1}^{2s+3} (-b)^{2s+3-k} = \\
= \sum_{k=1}^{2s+1} (-b)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1609814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Convexity of the natural exponential fuction - directly from the definition Without using the Second Derivative Test, can the convexity of the natural exponential function be shown directly from the definition of convexity? The expression
\begin{equation*}
e^{t} = \sum_{n=0}^{\infty} \frac{t^{n}}{n!}
\end{equation*}
c... | Demonstration in the case that $x > 0$, $y > 0$, and $s, \, t \in \mathbb{Q}^{+}$
Since $s$ is a positive, proper, rational number, there are positive integers $j < k$ such that $s = j/k$, and since $s$ and $t$ are a pair of conjugate numbers, $t = (k - j)/k$. For any positive, real numbers $z_{1} , \, z_{2} , \, \ldo... | {
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"url": "https://math.stackexchange.com/questions/1609904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
If the $81$ digit number $111\cdots 1$ is divided by $729$, the remainder is?
$729=9^3$
For any number to be divisible by $9$, the sum of the digits have to be divisible by $9$. The given number is divisible by $9$.
Then I tried dividing th... | We are looking for
$ x = 10^{80} + 10^{79} + \dots + 10^2 + 10 + 1 \mod{9^3}$
$ 10^k = (9 + 1)^k = \sum\limits_{i=0}^{k}{k \choose i}9^i $ (Binomial theorem)
Considering the second equality $\mod{9^3} $, we only need first three summands of the sum. So $ 10^k \equiv 1 + k\cdot 9 + \frac{k\cdot(k-1)}{2}\cdot 9^2\mod{9^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
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Evaluate $ \lim_{x \rightarrow - \infty} \frac{\sqrt{9x^6-x}}{x^3+1} $ I have started learning limits in calculus and I came across this question:
Evaluate $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9x^6-x}}{x^3+1} $ .
I rewrite the above as $ \displaystyle \lim_{x \rightarrow \infty} \dfrac{\sqrt{9-\df... | Dividing by $|x^3|$ we get
$$\frac{\sqrt{9x^6-x}}{x^3+1}=\frac{\frac{\sqrt{9x^6-x}}{|x^3|}}{\frac{x^3+1}{|x^3|}}$$
As $x\to-\infty$ we have, for $x<0$,
\begin{align}
\frac{\frac{\sqrt{9x^6-x}}{|x^3|}}{\frac{x^3+1}{|x^3|}}&=\frac{\sqrt{\frac{9x^6-x}{x^6}}}{\frac{x^3+1}{-x^3}}=\frac{\sqrt{9-\frac{1}{x^5}}}{-1-\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Solving the absolute value inequality $\big| \frac{x}{x + 4} \big| < 4$ I was given this question and asked to find $x$:
$$\left| \frac{x}{x+4} \right|<4$$
I broke this into three pieces:
$$
\left| \frac{x}{x+4} \right| = \left\{
\begin{array}{ll}
\frac{x}{x+4} & \quad x > 0 \\
-\frac{x}... | Divide the interval into 3 pieces.
X is between (-4,0)
X is greater than 0
X is less than -4
On each interval try to figure out whether the sign is positive or negative.
$$\left (\frac{|x|}{|x+4|}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1614060",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
What is the inverse of $f(x) = \dfrac{x}{x^2 - 1}$ I need to find a continuous inverse of $$f(x) = \dfrac{x}{x^2 - 1}$$
Let $y = f(x) = \dfrac{x}{x^2 - 1} \Rightarrow y(x^2-1) = x \Rightarrow yx^2-x = y$
How should I proceed from here?
| A plot of the function looks like this
To be invertible, a function of x and y must pass the horizontal line test. That is, any horizontal line cannot pass through more than one point of the graph.
This function fails the horizontal line test. But if you break the graph up into the blue and red branches shown (where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $7^{101}+8^{101}$ divisible by 25? If not, what is $ 7^{101} + 8^{101} \bmod 25$ What I derived is: $$\begin{align}7^{101}+8^{101} &\equiv (5+2)^{101}+ (5+3)^{101} \\
&\equiv 2^{101}+101\cdot5\cdot2^{100}+3^{101}+101\cdot 5\cdot 3^{100} \\
&\equiv 2^{101}+(100+1)\cdot5\cdot 2^{100}+3^{101}+(100+1)\cdot5\cdot3\cdot2^... | Use the lifting the exponent lemma:
$v_5(7+8)^{101}=v_5(7+8)+v_5(101)=1+0=1$, so $5^2\not\mid (7+8)^{101}$
The link is for the pdf-file called Lifting The Exponent Lemma by Amir Hossein Parvardi.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615171",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Linear combinations of vectors that depend on a parameter
Let $s \in \mathbb{R}$ and
$$\mathbf{u} = \begin{bmatrix} 1 \\ 4 \\ 0 \end{bmatrix}, \quad
\mathbf{v}=\begin{bmatrix} 1 \\ 1 \\ 1 \end{bmatrix}, \quad
\mathbf{w}=\begin{bmatrix} s^2 \\ 9 \\ s^2 \end{bmatrix}$$
for which values $s \in \mathbb{R}$ is
$$\mat... | Note that $$\mathrm{det}\left(\begin{array}{rrr} 1 & 1 & s^2 \\ 4 & 1 & 9 \\ 0 & 1 & s^2\end{array}\right)=s^2-9.$$ Thus, if $s\ne \pm 3,$ the determinant is different from zero. Thus, the columns form a basis and so it is possible to write the vector $\bf{b}$ as a linear combination of the columns.
Now, if $s=-3$ it i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that $ \sum\limits_{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2$
Let $ \triangle ABC$ be an acute-angled triangle. Prove that
$ \sum\limits_\text{cyc} \sqrt{\cot{A}+\cot{B}} \ge 2\sqrt2 $
Attempt
Since $\triangle{ABC}$ is acute, we may say that $A,B,C \in (0, \frac{\pi}{2})$. Now, we have that by a result for ... | hint:
$\sqrt{\cot{A}+\cot{B}}=\sqrt{\dfrac{sin(A+B)}{sinAsinB}}=\sqrt{\dfrac{sinC}{sinAsinB}}=\dfrac{sinC}{\sqrt{sinAsinBsinC}}$
$3\sqrt[3]{sinAsinBsinC} \le sinA+sinB+sinC \le \dfrac{3\sqrt{3}}{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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The tangent at a point $P$ on the curve $y=\ln(\frac{2+\sqrt{4-x^2}}{2-\sqrt{4-x^2}})-\sqrt{4-x^2}$ meets the $y-$axis at $T,$then find $PT^2.$ The tangent at a point $P$ on the curve $y=\ln(\frac{2+\sqrt{4-x^2}}{2-\sqrt{4-x^2}})-\sqrt{4-x^2}$ meets the $y-$axis at $T,$then find $PT^2.$
Let the point of tangency be $P... |
I found $\frac{dy}{dx}=\frac{2\sqrt{4-x^2}+x^2}{x\sqrt{4-x^2}}$
This is not correct. $$\small\begin{align}\frac{dy}{dx}&=\frac{2-\sqrt{4-x^2}}{2+\sqrt{4-x^2}}\cdot\frac{(2+\sqrt{4-x^2})'(2-\sqrt{4-x^2})-(2+\sqrt{4-x^2})(2-\sqrt{4-x^2})'}{(2-\sqrt{4-x^2})^2}-\frac{-2x}{2\sqrt{4-x^2}}\\&=\frac{\frac{-4x}{\sqrt{4-x^2}}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616705",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to simplify this? What did I do wrong? $$(3-\sqrt3) (2-\sqrt3)-\sqrt3\cdot\sqrt{27}.$$
So I simplified the $\sqrt3\cdot\sqrt{27}$ part into
$$\sqrt 3\cdot\sqrt{9\cdot3} = \sqrt3\cdot3\cdot\sqrt3=4\cdot\sqrt3$$
Then I multiplied the brackets :
$$6+\sqrt3-2\cdot\sqrt3-3\cdot\sqrt3 =6-4\cdot\sqrt3.$$
Then minus the $... | Second line, second equal sign is wrong.
$\sqrt{3}\cdot 3 \cdot \sqrt{3} = 9$
Multiplying the brackets yelds:
$6 - 2\sqrt{3} - 3\sqrt{3} +3 = 9 - 5\sqrt{3}$
Putting it all together:
$9 - 5\sqrt{3} - 9 = -5\cdot \sqrt{3}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617971",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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I need to solve $\phi (x,y) = \frac{2V}{\pi} \int_{0}^{\infty} \frac{\sin(kx)\cosh(ky) dk}{k\cosh(ka)}$ I start with a integral in complex plane $$\oint_c \frac{e^{izx} e^{zy} dz}{z\cosh(za)}$$ where $c$ is a countour starting in $z = -R$ along the real axis and jumping the pole at origin and continuing to $z = R$ and ... | Assuming your summation is correct we have
$$-\sum_{n=1}^\infty (-1)^ne^{-\frac{\pi n x}{2a}}\frac{e^{\frac{\pi n y}{2a}i}+e^{-\frac{\pi n y}{2a}i}}{2n}=-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{e^{-\frac{\pi n x}{2a}+\frac{\pi n y}{2a}i}}{n}-\frac{1}{2}\sum_{n=1}^\infty (-1)^n\frac{e^{-\frac{\pi n x}{2a}-\frac{\pi n y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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How to prove that all odd powers of two add one are multiples of three
For example
\begin{align}
2^5 + 1 &= 33\\
2^{11} + 1 &= 2049\ \text{(dividing by $3$ gives $683$)}
\end{align}
I know that $2^{61}- 1$ is a prime number, but how do I prove that $2^{61}+1$ is a multiple of three?
| $2 = 3-1$
$2^k = (3 - 1)^k = 3^k - k*3^{k - 1} ..... = \sum_{n = 0}^k {k \choose n}3^{k - n}(-1)^n = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n + {k \choose 3}3^{k - n}(-1)^k = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n \pm 1$
Since $k$ is odd:
$2^k = \sum_{n = 0}^{k - 1} {k \choose n}3^{k - n}(-1)^n - 1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "41",
"answer_count": 11,
"answer_id": 9
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is this correct $\lim_{ n \to \infty} \sum_{k=2^n}^{2^{n+1}} \frac{1}{k}= \ln 2$? I think i might have seen this result somewhere $\lim_{ n \to \infty} \sum_{k=2^n}^{2^{n+1}} \frac{1}{k}= \ln 2$ but cant remember for sure.
Is there a name for sums having limits in both lower and upper bound?
is there anything similar f... | Let $\alpha=\lim\limits_{n\to\infty}\frac1{n+1}+\frac1{n+2}+\dots+\frac1{2n}$
The following is taken from this answer:
In this answer, It is shown that $\left(1+\frac1n\right)^n$ increases and $\left(1+\frac1n\right)^{n+1}$ decreases to $e$. Thus,
$$
\left(1+\frac1n\right)^n\le\left(1+\frac1{n+1}\right)^{n+1}\le\dots\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove that if $a,b,c$ are positive real numbers and $a+b+c = 1$ then $(3abc)^2 \ge (1-2a)(1-2b)(1-2c)(a^2+b^2+c^2).$
Prove that if $a,b,c$ are positive real numbers and $a+b+c = 1$ then $(3abc)^2 \ge (1-2a)(1-2b)(1-2c)(a^2+b^2+c^2).$
Cauchy-Schwarz doesn't really help here and neither does rearrangement so I see not... | We need to prove that
$$9a^2b^2c^2\geq(a^2+b^2+c^2)(a+b-c)(a+c-b)(b+c-a)(a+b+c)$$
Let $a+b-c=z$, $a+c-b=y$ and $b+c-a=x$.
Since $\prod\limits_{cyc}(a+b-c)>0$, we can assume that $x$, $y$ and $z$ are non-negatives.
Indeed, let $x<0$ and $y<0$. Hence, $x+y<0$, which gives $c<0$. Contradiction.
Thus, we need to prove tha... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving a differential equation with Bernoulli's Method What approach do you take to solve the differential equation $ y' + (6y/x) = (y^3)/ x^5\ $ through the use of Bernoulli's method?
I've assumed u = y^(-2) for substitution, but I don't know where to go from there. The answer is y $ ((C\x^2\) + (1/8x^4))^(-.5) \ $
| $\frac{dy}{dx}+6\frac{y}{x}=\frac{y^3}{x^5}$
Multiply by $\frac{1}{y^3}$
$\frac{1}{y^3}\frac{dy}{dx}+6\frac{1}{y^2x}=\frac{1}{x^5}$
Let $\frac{1}{y^2}=z$
$\frac{-2}{y^3}\frac{dy}{dx}=\frac{dz}{dx}$
$-\frac{dz}{dx}-12\frac{z}{x}=-2\frac{1}{x^5}$
$P(x)=-\frac{12}{x}$
$\int e^{-\frac{12}{x}}dx=x^{-12}$
${z}{x^{-12}}=\int ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solution of $ (dy/dx)^2 = (x^2+y^2) $ How do I find the solution of:
$$ (dy/dx)^2 = (x^2+y^2). $$
I tried $ y=tx$ and $x=r\cos(t) ,y=r\sin(t)$ substitutions but they did not help. I am not able to change it into any variable separable form and the only meaning I can make out is that the slope at any point is equal to... | $\left(\dfrac{dy}{dx}\right)^2=x^2+y^2$
$\dfrac{dy}{dx}=\pm\sqrt{x^2+y^2}$
Apply the Euler substitution:
Let $u=y\pm\sqrt{x^2+y^2}$ ,
Then $y=\dfrac{u}{2}-\dfrac{x^2}{2u}$
$\dfrac{dy}{dx}=\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac{du}{dx}-\dfrac{x}{u}$
$\therefore\left(\dfrac{1}{2}+\dfrac{x^2}{2u^2}\right)\dfrac... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $3 \le a+b+c \le 2\sqrt{3}$ in a triangle
Let $a,b,$ and $c$ be the lengths of the sides of a triangle satisfying $ab+bc+ca = 3.$ Prove that $3 \le a+b+c \le 2\sqrt{3}$.
The idea I had was $(a+b+c)^2 = a^2+b^2+c^2+2(ab+bc+ca) = a^2+b^2+c^2+6 \geq 9$ by rearrangement. That takes care of the first inequalit... | While working on a non-variational approach, I present a variational apporach.
By the given condition, $ab+bc+ca=3$, we get
$$
(b+c)\,\delta a+(c+a)\,\delta b+(a+b)\,\delta c=0\tag{1}
$$
For interior critical points maximizing $a+b+c$, we want
$$
\delta a+\delta b+\delta c=0\tag{2}
$$
To get $(2)$ for all variations th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 1
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Finding $(a+\sqrt b)^n+(a-\sqrt b)^n$ where $n$ is natural For the expression $\left(a+\sqrt{b}\right)^n+\left(a-\sqrt{b}\right)^n$ where $n \in \mathbb{N}$, and $a,b, \in \mathbb{Q}$, the radical is always ends up cancelled, and the result is always in $\mathbb{Q}$. Is there any way that this could be reexpressed wit... | Note: $(x + y)^n = x^n + {n \choose 1} x^{n-1}y + \dotsc + y^n$
While $(x - y)^n = x^n - {n \choose 1} x^{n-1}y + \dotsc \pm y^n$
The even terms are opposite parity while the odd terms are the same so
$(x + y)^n + (x - y)^n = 2x^n + 2{n \choose 2}x^{n-2}y^2 + \dotsc$
So $(a + \sqrt{b})^n + (a - \sqrt{b})^n = 2(a^{n} +... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Compute all roots of $(-8)^{\frac{1}{3}}$ $$(-8)^{\frac{1}{3}}$$
The problem states to compute all roots of the complex number above. Below is my attempt, but my inquiries are if I did it right and why it doesn't match Wolfram. Wolfram only returns $1$ result - was I only supposed to have $1$ as well??
$$\text{modulus}... | We can use elementary factorization to find all roots. Let $x = \left(-8\right)^{\frac{1}{3}}\Rightarrow x^3 = -8\Rightarrow x^3 + 8 = 0\Rightarrow (x+2)(x^2-2x+4) = 0\Rightarrow (x+2)((x-1)^2 + 3) = 0\Rightarrow x = -2, (x-1)^2 = -3\Rightarrow x = -2, x-1 = \pm i\sqrt{3}\Rightarrow x = -2, 1 \pm i\sqrt{3}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1624180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Path needed for solving these linear equations in Zn (my example Z105) So these are two equations :
$$49x \equiv 21 \pmod {105}$$
$$64x \equiv 21 \pmod {105}$$
I should find the multiplicative inverse of $64$ and then $49$ that gets the result of $1$
so....
In the first equation i know that the $\gcd(49, 21)$ is $7$
So... | The inverse of $64$ modulo $105$ can be found with the Euclidean algorithm:
\begin{align}
105&=64\cdot 1+41\\
64&=41\cdot 1+23\\
41&=23\cdot 1+18\\
23&=18\cdot 1+5\\
18&=5\cdot 3+3\\
5&=3\cdot 1+2\\
3&=2\cdot 1+1
\end{align}
Thus
\begin{align}\def\c#1{\color{red}{#1}}
1&=\c{3}-\c{2}\\
&=\c{3}-(\c{5}-\c{3})=(-1)\cdot\c{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1628027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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A sequence to achieve $\frac{1}{a_{2016}}$ It is given that $a_ka_{k-1} + a_{k-1}a_{k-2} = 2a_k a_{k-2}$ , $k\geq3$ and $a_1=1$.
We have $S_q= \sum_{k=1}^{q} \frac{1}{a_k} $ and given that $\frac{S_{2q}}{S_{q}}$ is independent of q then $\frac{1}{a_{2016}}$ is = ?
I think no information is given to find $a_2$. How ... | You can find $a_2$ by solving three simultaneous equations ($a_1$ fixed at $1$):
$$
a_3 a_2 + a_2 = 2_a3 \implies a_2 = \frac{2a_3}{a_3+1} \\
a_4a_3 + a_3a_2 = 2a_4a_2\implies a_4a_3 + a_3\frac{2a_3}{a_3+1} = 2a_4\frac{2a_3}{a_3+1}\implies a_4 = \frac{2a_3}{3-a_3} \\
\frac{S_4}{S_2} = \frac{S_2}{S_1} \implies \frac{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Integration by parts: $\int{\frac{dx}{(x^2 + a^2)^n}}$. I need to show that the following holds using integration by parts:
\begin{equation}
\int{\frac{dx}{(x^2 + a^2)^n}} = \frac{x}{2a^2(n-1)(x^2 + a^2)^{n-1}} + \frac{2n - 3}{2a^2(n-1)} \int{\frac{dx}{(x^2 + a^2)^{n-1}}}
\end{equation}
I really just don’t know where ... | Let $I_{n}=\int \frac{dx}{(x^{2}+a^{2})^{n}}$, then
\begin{align*}
I_{n} &=
\frac{x}{(x^{2}+a^{2})^{n}}-
\int x \, d\left[\frac{1}{(x^{2}+a^{2})^{n}} \right] \\
&=\frac{x}{(x^{2}+a^{2})^{n}}+2n\int \frac{x^{2}dx}{(x^{2}+a^{2})^{n+1}} \\
&=\frac{x}{(x^{2}+a^{2})^{n}}+
2n\int \left[ \frac{1}{(x^{2}+a^{2})^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1631747",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Integral of a trig function divided by the square root of a polynomial: $\int_a^b\frac{\sin x}{\sqrt{(x-a)(b-x)}}dx$? I was trying to help some physics students with an integral on their homework and they've presented me with something that has me stumped.
The integral they are working on is:
$$\int_a^b\frac{\sin x}{\... | Rewrite as
$$\begin{align}\int_0^{b-a} dx \frac{\sin{(x+a)}}{\sqrt{x (b-a-x)}} &= \int_0^1 du \frac{\sin{[(b-a) u+a]}}{\sqrt{u (1-u)}}\\ &= \cos{a}\int_0^1 du \frac{\sin{[(b-a) u]}}{\sqrt{u (1-u)}} + \sin{a} \int_0^1 du \frac{\cos{[(b-a) u]}}{\sqrt{u (1-u)}}\\ &= 2 \cos{a} \int_0^1 du \frac{\sin{[ (b-a) u^2 ]}}{\sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1635411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find the real values of the parameter a, so that the inequality doesn't have positive solutions? The inequality is the following:
$$
\frac{x^2 - 6ax + 2x - 5a - 1}{x+a+1} < 0
$$
If we write this inequality like $\frac{a}{b} < 0$, I may say that for $a=0$, I found that
$X_1=3a - 1 - \sqrt{9a^2-a+2}$
$X_2=3a - 1 ... | For a fraction to be negative we need both the numerator and the denominator to have the opposite sign. For example, $(-2)/(+4) = -1/2$ and $(+3)/(-4) = -3/4$.
You need to solve the simultaneous inequalities:
Either $x^2+2(1-3a)x-(1+5a) > 0$ and $x+a+1<0$,
or $x^2+2(1-3a)x-(1+5a) < 0$ and $x+a+1>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1636162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that if $|x| < 1$ then $\frac{\ln(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n-1}s_nx^n,\,\,\,\,\,s_n\sum_{k=1}^n\frac{1}{k}$. By multiplying power series, show that if $|x| < 1$ then
$$\frac{\ln(1+x)}{1+x}=\sum_{n=1}^\infty (-1)^{n-1}s_nx^n,\,\,\,\,\,s_n\sum_{k=1}^n\frac{1}{k}.$$
So I know that:
$$\ln(1+x)=-\sum\limits_... | The following is not an answer but cannot be a comment either.
Let me take from your attempt:
\begin{align*}
\frac{\ln(1+x)}{1+x}={}&\left(-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}\right)\left(1+\sum\limits_{n=1}^\infty (-1)^nx^n \right)={} \\
{}={}&-\sum\limits_{n=1}^\infty\frac{(-1)^nx^n}{n}+\left(-\sum\limits_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1637949",
"timestamp": "2023-03-29T00:00:00",
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Find a Mobius Transformation that carries the points $ -1, i, 1+i$ to the following: My goal is to find a Mobius transformation that transforms $-1, i, 1+i$ onto the points
a) $0, 2i, 1-i$
b) $i, \infty, 1$
For part a, I know that the Mobius transformation $M$ will be such that $M(-1) = 0$. So, this means that it shoul... | A Mobius transformation is an automorphism of the Riemann sphere $S = \Bbb C \cup \{ \infty \}$. It comes from a rational fraction of degree $1$ on $\Bbb C$, which is a map of the form $z \mapsto \frac {az+b}{cz+d}$ where $a,b,c,d$ are complex numbers.
In order for the map to be a bijection you need a condition on $a,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the Jordan normal form of a nilpotent matrix $N$ given the dimensions of the kernels of $N, N^2, N^3$
Let $N\in \text{Mat}(10 \times 10,\mathbb{C})$ be nilpotent. Furthermore let $\text{dim} \ker N =3 $, $\text{dim} \ker N^2=6$ and $\text{dim} \ker N^3=7$. What is the Jordan Normal Form?
The only thing I know is... | Since $N$ is nilpotent, it has $0$ as unique eigenvalue. Since $\dim\ker N=3$, by the rank nullity theorem, the rank of $N$ is $7$. Hence the JNF of $N$ has seven 1's. Assume that there are $k$ blocks, and denote by $n_i$ the size of the $i$-th block. Of course, $n_1+\cdots+n_k=10$. Now, the rank of $N$ is $(n_1-1)+(n_... | {
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"url": "https://math.stackexchange.com/questions/1645809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
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integrate $\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx$
$$\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx$$
$$\int \frac{(16-9x^2)^{\frac{3}{2}}}{x^6}dx=\int \frac{3\left(\frac{16}{9}-x^2\right)^{\frac{3}{2}}}{x^6}dx$$
$x=\frac{4}{3}\sin\theta$
$dx=\frac{4}{3}\cos\theta d\theta$
$$\int \frac{3\left(\frac{16}{9}-\frac{16}{9... | Notice, you have made mistake while substituting
let $3x=4\sin\theta\implies dx=\frac{4}{3}\cos\theta\ d\theta$
$$\int \frac{(16-9x^2)^{3/2}}{x^6}\ dx$$$$=\int\frac{(16-16\sin^2\theta)^{3/2}}{\left(\frac 43\sin\theta\right)^6} \left(\frac{4}{3}\cos\theta\ d\theta\right)$$
$$=\frac{243}{16}\int \frac{\cos^4\theta}{\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1646079",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate $I=\int_0^{1}\frac{1+x}{x^2+x+1}\log\left({\frac{x}{1-x}}\right)\,\mathrm dx$ without using complex analysis
Calculate $$I=\int_0^{1}\frac{1+x}{x^2+x+1}\log\left({\frac{x}{1-x}}\right)\,\mathrm dx$$
without using complex analysis.
How to calculate without using the residue theorem?
The correct answer is
$... | \begin{align}
I=&\int_0^{1}\frac{(1+x)\ln{\frac{x}{1-x}}}{x^2+x+1}dx\\
=&\ \frac12\int_0^1 \frac{\ln{\frac{x}{1-x}}}{x^2+x+1}
+ \frac{(2x+1)\ln x}{x^2+x+1}-\frac{(2x+1)\ln (1-x)}{x^2+x+1} \ dx\\
=&\ \frac12(I_1+I_2-I_3)
\end{align}
where
\begin{align}
I_1&=\int_0^1 \frac{\ln{\frac{x}{1-x}}}{x^2+x+1}\overset{\frac {\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
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Number of ways of selecting 3 numbers from $\{1,2,3,\cdots,3n\}$ such that the sum is divisible by 3
Find the Number of ways of selecting 3 numbers from $\{1,2,3,\cdots,3n\}$ such that the sum is divisible by 3. (Numbers are selected without replacement).
I made a list like this:
The sum of all elements along such ... | Let $S = \{1, 2, 3, \ldots, 3n - 2, 3n - 1, 3n\}$. Let
\begin{align*}
A & = \{k \in S \mid k \equiv 0 \pmod{3}\}\\
B & = \{k \in S \mid k \equiv 1 \pmod{3}\}\\
C & = \{k \in S \mid k \equiv 2 \pmod{3}\}
\end{align*}
Observe that $|A| = |B| = |C| = n$. We can choose three numbers from $S$ in the following ways:
*
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is $\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} = \infty$? It was asked in our test, and below is what I did:
$$\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{\sqrt{x^2+16}-5}\times\frac{\sqrt{x^2+16}+5}{\sqrt{x^2+16}+5} $$
$$=\lim_{x\to -3}\frac{x^2+9}{x^2-9}\times \left(\sqrt{x^2+16}+5\r... | The numerator is positive for both LHL and RHL, but the denominator is $\rm +ve$ for one and $\rm -ve$ for other:
$$\begin{align}\text{Since }\\
&\text{as }x\to-3^-,\ \ \sqrt{x^2+16}-5 >0 \\
&\lim_{x\to-3^-}f(x)=\infty
\end{align}$$
$$\begin{align}\text{also,}\\
&\text{as }x\to-3^+,\ \ \sqrt{x^2+16}-5 <0 \\
&\lim_{x\to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651727",
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Finding coefficient of polynomial? The coefficient of $x^{12}$ in $(x^3 + x^4 + x^5 + x^6 + …)^3$ is_______?
My Try:
Somewhere it explain as:
The expression can be re-written as: $(x^3 (1+ x + x^2 + x^3 + …))^3=x^9(1+(x+x^2+x^3))^3$
Expanding $(1+(x+x^2+x^3))^3$ using binomial expansion:
$(1+(x+x^2+x^3))^3 $
$= 1+3(x+... | It's basically the number of ways you can write $12$ as a sum of three integers all greater than or equal to $3$, where order matters. So $12=3+3+6$, $12=3+4+5$, $12=4+4+4$. The first can be rearranged three ways, the second can be rearranged six ways and the last can only be arranged one way. So yes, I believe the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 4,
"answer_id": 3
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Integration of $\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx$ How can we integrate:
$$
\int\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}dx
$$
Using simple algebraic identities I deduced it to
$$
\int\frac{1-2\sin^2x\cdot\cos^2x}{(\sin x+\cos x)(1-\sin x\cdot\cos x)}dx
$$ but can't proceed further. Please provide some dire... | $\displaystyle\frac{\sin^4x+\cos^4x}{\sin^3x+\cos^3x}=\sin x+\cos x-\frac{\sin x\cos x}{\sin^3 x+\cos^3x}=\sin x+\cos x-\frac{\sin x\cos x}{(\sin x+\cos x)(1-\sin x\cos x)}$,
and $\;\;\displaystyle\frac{\sin x\cos x}{(\sin x+\cos x)(1-\sin x\cos x)}=A\left(\frac{\sin x+\cos x}{1-\sin x\cos x}\right)+B\left(\frac{1}{\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Distribution of the number of times a player wins a game with random numbers Five distinct numbers are randomly distributed to players numbered
$1$ through $5$. Whenever two players compare their numbers, the one
with higher one is declared the winner. Initially, players $1$ and $2$
compare their numbers; the winner th... | In the following I denote the numbers that are distributed with $1,2,3,4$ and $5$ since the only thing that matters is their order and not their exact values. In total, there are $5!=120$ ways for these numbers to be distributed among the $5$ Players, each of these ways being equally probable. So, $$P(X=i)=\frac{\text{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Convergence of the series $\sum_n (-1)^{\lfloor \sqrt{n-1}\rfloor} \frac 1 n$ Let's consider the series $\sum a_n$, with $$a_n = (-1)^{\lfloor \sqrt{n-1}\rfloor} \frac 1 n$$
It looks absolutely like an example for the Leibniz test but here the signs don't interchange one for one. How can we prove the convergence of thi... | You can reduce the question to a question about Liebniz series by inserting parenthesis around the terms that have the same sign. The original series $\sum_{n=1}^{\infty} a_n$ is:
$$ \frac{1}{1} - \frac{1}{2} - \frac{1}{3} - \frac{1}{4} + \frac{1}{5} + \dots + \frac{1}{9} - \frac{1}{10} - \dots. $$
Denote the partial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661525",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Find the factors of $(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$ Find the factors of
$$(a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$$
the answer is $24abc$
Let $E = (a+b+c)^3-(b+c-a)^3-(c+a-b)^3-(a+b-c)^3$
Since $b+c = a $ makes $E = 0, \therefore (b+c-a)$ is one factor, similarly $(c+a-b)$ and $(a+b-c)$ are factors, but ca... | Let $\displaystyle b+c-a=x,c+a-b=y,a+b-c=z\implies x+y+z=a+b+c$
We need $\displaystyle(x+y+z)^3-x^3-y^3-z^3=\{x+(y+z)\}^3-x^3-y^3-z^3$
$\displaystyle=x^3+(y+z)^3+3x(y+z)\{x+(y+z)\}-x^3-y^3-z^3$
$\displaystyle=3yz(y+z)+3x(y+z)\{x+(y+z)\}$
$\displaystyle=3(y+z)\{yz+x(x+y+z)\}$
$$\displaystyle\implies(x+y+z)^3-x^3-y^3-z^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663930",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Calculation of $\min$ distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$
Calculation of $\min$ distance between $x^2+y^2=9$ and $2x^2+10y^2+6xy=1$
$\bf{My\; Try::}$ Using the fact that the distance between two curve is independent of shifting.
So put $x=u+v$ and $y=u-v$ in thsese two curves, We get $\displaystyle u^... | The original ellipse is congruent with the ellipse whose equation is $x^2+11y^2=1$.
And all the ellipse which is similar to it has the equation form:$x^2+11y^2=\lambda$ Make it and $x^2+y^2=9$ as a equation group and solve it: $ x = \sqrt \frac{99-\lambda}{10} or -\sqrt \frac{99-\lambda}{10}, y = \sqrt \frac{\lambda -9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1664049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 0
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Remainder when $x^{2016}+x^{2013}+\cdots+x^6+x^3$ is divided by $x^2+x+1$. I am currently preparing for a certain quiz show when I encountered this question:
What is the remainder when $x^{2016}+x^{2013}+\cdots+x^6+x^3$ is divided by $x^2+x+1$?
I know for a fact that $x^3-1=(x-1)(x^2+x+1)$.
And I got
$$x^{2016}+x^{2013... | As $x^3=1\implies x^{3n}=(x^3)^n=1\implies x^{3n}\equiv1\pmod{x^2+x+1}$
and $2016=3\cdot672$
$$\sum_{r=1}^{672}x^{3r}\equiv\sum_{r=1}^{672}1\pmod{x^2+x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
limit of sum defined sequence Let $x_n=\displaystyle \sum_{k=1}^n \sqrt{1+\frac{k}{n^2}}, n\ge1$. Prove that $\displaystyle \lim_{n \rightarrow \infty} n (x_n-n-\frac{1}{4})=\frac{5}{24}$.
What I've done:it's easy to show that $\displaystyle \lim_{n \rightarrow \infty} \frac{x_n}{n}=1$ and $\displaystyle \lim_{n \right... | My answer is wrong, but I would like to know why.
$$n(x_n-n-\frac{1}{4})=n \sum (\sqrt{(1+\frac{k}{n^2})}-(1+\frac{1}{4n}))$$
$$=n \sum \frac{(1+\frac{k}{n^2})-(1+\frac{1}{4n} )^2}{
\sqrt{(1+\frac{k}{n^2})}+1+\frac{1}{4n}
}$$
$$=n \sum \frac{
\frac{k}{n^2} - \frac{1}{16n^2} -\frac{1}{2n}}{
\sqrt{(1+\frac{k}{n^2})}+1+\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669217",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$ How do you go about factoring $({16+4t^2+\frac{t^4}{4}})^{1/4}$
Which factoring method would be the best?
| $$\left({16+4t^2+\frac{t^4}{4}}\right)^{1/4}=\left({4^2+2\cdot4\cdot\frac{t^2}{2}+\left(\frac{t^2}{2}\right)^2}\right)^{1/4}=$$
$$=\left(\left(4+\frac{t^2}{2}\right)^2\right)^{1/4}=\left(4+\frac{t^2}{2}\right)^{2\cdot{1/4}}=\left(4+\frac{t^2}{2}\right)^{{1/2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1670726",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that for positive integer $ n$ we have $169| 3^{3n+3}-26n-27$.
Prove that for positive integer $n$ we have $169| 3^{3n+3}-26n-27$.
It is easy to show that if we take the expression modulo $13$ we have $3^{n+3}-26n-27 \equiv 1^{n+1}-1 \equiv 1-1 \mod 13 = 0 \mod 13$. How do I prove it is divisible by $13^2$?
| Use congruences.
$3$ has order $3$ modulo $13$, hence
$$3^{3n+3}-26n-27\equiv 1^{n+1}-0-1=0.$$
This proves it is divisible by $13$. More precisely, we have
\begin{align*}
3^{3n+3}-26n-27&=27(27^n-1)-26n=27\cdot26(27^{n-1}+27^{n-2}+\dots+27+1)-26n\\
&=2\cdot 13\bigl(27(27^{n-1}+27^{n-2}+\dots+27+1)-n\bigr).
\end{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1672413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Find the number of distinct integers with non-decreasing digits formed from one or more of the digits $2, 2, 3, 3, 4, 5, 5, 5, 6, 7$ Suppose integers are formed by taking one or more digits from the
following: $2, 2, 3, 3, 4, 5, 5, 5, 6, 7$. For example, $355$ is a possible choice while $44$ is not. Find the number of... | An integer with non-decreasing digits selected from the digits $2, 2, 3, 3, 4, 5, 5, 5, 6, 7$ is a selection of one of more digits of the string $2233455567$. For instance, the selection $22\color{blue}{33}4\color{blue}{5}55\color{blue}{67}$ corresponds to the integer $33567$.
Since the digits of the integer are non... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proof for $\forall x\in \mathbb{Z}^+, \exists t \in \mathbb{Z}, 5 \nmid x \to ((x^2= 5t + 1) \vee (x^2 = 5t – 1))$ I am trying to write a proof for the following: If $x$ is a positive integer that is not divisible by $5$, then $x^2$ can be written either as $x^2 = 5t+1$ or $x^2 = 5t−1$ for some integer $t$. So far, I h... | Once you get that far, just check every case:
$x = 5n+1$:
$$x^2 = 25n^2 - 10n + 1 = 5(5n^2 + 2n) + 1 = 5k + 1$$
$x = 5n + 2$:
$$x^2 = 25n^2 + 20n + 4 = 25n^2 + 20n + 5 - 1 = 5(5n^2 + 4n + 1) - 1 = 5t - 1$$
$x = 5n + 3$:
$$x^2 = 25n^2 + 30n + 9 = 25n^2 + 30n + 10 - 1 = 5(5n^2 + 6n + 2) -1 = 5l - 1$$
$x = 5n + 4$:
$$x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675055",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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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.