Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
How to find the minimum value of this function? How to find the minimum value of
$$\frac{x}{3y^2+3z^2+3yz+1}+\frac{y}{3x^2+3z^2+3xz+1}+\frac{z}{3x^2+3y^2+3xy+1}$$,where $x,y,z\geq 0$ and $x+y+z=1$.
It seems to be hard if we use calculus methods. Are there another method? I have no idea.
Thank you.
| By C-S and AM-GM we obtain:
$$\sum_{cyc}\frac{x}{3y^2+3z^2+3yz+1}=\sum_{cyc}\frac{x}{3y^2+3z^2+3yz+(x+y+z)^2}=$$
$$=\sum_{cyc}\frac{x}{x^2+4y^2+4z^2+2xy+2xz+5yz}=$$
$$=\sum_{cyc}\frac{x^2}{x^3+4xy^2+4xz^2+2x^2y+2x^2z+5xyz}\geq$$
$$\geq\frac{(x+y+z)^2}{\sum\limits_{cyc}(x^3+6x^2y+6x^2z+5xyz)}\geq\frac{(x+y+z)^2}{\sum\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/681153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 2
} |
Rationalizing the denominator 3 It is a very difficult question. How can we Rationalizing the denominator?
$$\frac{2^{1/2}}{5+3*(4^{1/3})-7*(2^{1/3})}$$
| $\frac1{A+B\sqrt[3]{n}+C\sqrt[3]{n^2}}=
\frac{ \begin{vmatrix}
1&\sqrt[3]n&\sqrt[3]{n^2}\\
B&A&Cn\\
C&B&A
\end{vmatrix} }{\begin{vmatrix}
A&Cn&Bn\\
B&A&Cn\\
C&B&A
\end{vmatrix}}=
\frac{A^2-BCn+(C^2n-AB)\sqrt[3]{n}+(B^2-AC)\sqrt[3]{n^2}}
{A^3+B^3n+C^3n^2-3ABCn}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/681873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Why we do division in those steps told, and who invented division? i know how to divide but i dont quit understand why we use those steps told in schools.
like for example
____
3/450 150 quotient
3
-----
15
15
------
000
could someone please tell me why we do these steps is there any other method or c... | Taking your example, write
$$
\frac{450}{3}
= \frac{4 \times 100 + 5 \times 10 + 0 \times 1}{3}
= \frac{4}{3} \times 100 + \frac{5}{3} \times 10 + \frac{0}{3} \times 1.
$$
We will arrange the terms so that the standard method is suggested.
First compute $\frac{4}{3} \times 100 = \left(1 + \frac{1}{3} \right) \times ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Simplfying $\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$ I am trying to simplify the expression:
$\sqrt{31-8\sqrt{15}}+\sqrt{31+8\sqrt{15}}$
I tried to square the expression but I can't do that because it is not an equation so I got stuck. Can someone please give me some pointers on how to proceed?
| Here is another way of doing it.
If we set $a=31+8\sqrt {15}, b=31-8\sqrt {15}$ we have
$a+b=62$ and
$ab=31^2-15\cdot 64=31^2-30\cdot32=1$
We know that $ab$ will work out well, because we know that $(x+y)(x-y)=x^2-y^2$
Then we know that $a$ and $b$ are the roots of $x^2-62x+1=0$
We want the square roots $\alpha$ and ... | {
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"url": "https://math.stackexchange.com/questions/696113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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How to solve : $\lim_{n\rightarrow \infty} \frac{n!}{n\cdot 2^{n}}$ $$\lim_{n\rightarrow \infty} \frac{n!}{n\cdot 2^{n}}$$
I need to solve the limit problem above. I have no idea about what to do. What do you suggest?
Thanks in advance.
| Since
$$n!=1\cdot 2\cdot 3\cdot 4\cdot 5\cdot \cdots \cdot n\ge 2\cdot 3\cdot 4^{n-3}$$
$$\frac{n!}{n\cdot 2^n} \ge 2\cdot 3\cdot \frac{4^{n-4}}{2^n}=\frac{2\cdot3}{4^4} 2^n$$
for $n\ge 4$. Therefore the sequence diverges.
| {
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For all square matrices $A$ and $B$ of the same size, it is true that $(A+B)^2 = A^2 + 2AB + B^2$? The below statement is a true/false exercise.
Statement:
For all square matrices A and B of the same size, it is true that
$(A + B)2 = A^2 + 2AB + B^2$.
My thought process: Since it is not a proof, I figure I can show by ... | A counterexample is all you need, so you're done. You probably could have picked a simpler counterexample, say with most entries of $A$ and $B$ being $0,$ but yours works just fine.
As an alternative, note that if $A$ and $B$ are square matrices of the same size, then $$(A+B)^2=A(A+B)+B(A+B)=A^2+AB+BA+B^2.$$ From this,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Suppose that $2^b-1\mid 2^a+1$. Show that $b = 1$ or $2$. I'm stuck with this one. I would appreciate any idea how to prove this.
| Suppose that $2^b-1 \mid 2^a+1$.
Note that $b$ is the order of $2 \pmod{2^b-1}$. Since $2^{2a} \equiv 1 \pmod{2^b-1}$ we have $b \mid 2a$.
If $b$ is odd then we get $b \mid a$, so $2^a \equiv 1 \pmod{2^b-1}$. However we also have $2^a \equiv -1 \pmod{2^b-1}$, whence $-1 \equiv 1 \pmod{2^b-1}$, so $b=1$.
Otherwise $b$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proving an inequality involving factorials: $(\frac{n}{2})^n \ge n! \geq (\frac{n}{3})^n$ For $n \geq 6$, where $n$ is a natural number, prove that $(\frac{n}{2})^n \ge n! \geq (\frac{n}{3})^n$.
I tried using induction but could not do it.
| Re-trying with induction...assume $(\frac{n}{2})^n \ge n!$ is valid for all natural number $k$ such that $n \ge k \ge 6$, then let's prove it's valid also for $n+1$. For the inductive hypothesis we have
$$ \left(\frac{n}{2}\right)^n \ge n! $$
so
$$ (n+1)\cdot \left(\frac{n}{2}\right)^n \ge (n+1)! $$
but since $\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/705520",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\frac{a+b}{c+b}=\frac{c+d}{a+d}\Rightarrow a=c$ or $a+b+c+d=0$ Please find my mistake or give hint:
If $\frac{c}{a}\geq 1$, then by mediant inequality:
$1\leq\frac{a+b}{c+b}$ and $\frac{c+d}{a+d}\leq 1\Rightarrow c=a$
If $\frac{a}{c}\geq 1$, then:
$\frac{a+b}{c+b}\leq 1$ and $1\leq \frac{c+d}{a+d}\Rightarrow c=a$
thnk... | Note that
$$
\frac{a+b}{c+b} = \frac{c+d}{a+d} \implies
(a+d)(a+b) = (c+d)(c+b)\implies\\
a^2 + ab + ad + db = c^2 + cb + cd + db \implies\\
a^2 - c^2 +ab - cb + ad - cd = 0 \implies\\
(a-c)(a+c) + (a-c)b + (a-c)d = 0 \implies\\
(a-c)(a+b+c+d) = 0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/707456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Center of Mass with two functions
I am having trouble trying to figure out how to go about this problem. I can do problems with single variables but I can not solve this one. I think I would need to subtract the functions from one another but I am not sure which should be subtracted from which. If anyone can help with... | The formula for Centroid
$$f\left(x\right) = 9\cos \left(x\right) ; g\left(x\right) = 9\sin \left(x\right)$$
$$\bar x =\dfrac{ ∫_0^{\frac{{π}}{4}} x\left(f\left(x\right)-g\left(x\right)\right) dx}{∫_0^{\frac{{π}}{4}} \left(f\left(x\right)-g\left(x\right)\right)dx}$$
$$\bar y= \dfrac{ ∫_0^{\frac{{π}}{4}} \frac{\left(f\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/708307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Determining number of solutions with inclusion-exclusion NOTE: I know there are similar questions to this, but the ones on this website are much more complex, and I'd like to get a basic understanding before moving on to them. Please do not mark this as a duplicate.
On a practice quiz:
Use inc-exc to determine the num... | Generating Functions
Look at the coefficient of $x^{15}$ in $\left(1+x+x^2+x^3+\dots+x^6\right)^4$:
$$
\begin{align}
\left(\frac{1-x^7}{1-x}\right)^4
&=\sum_{j=0}^4(-1)^j\binom{4}{j}x^{7j}\sum_{k=0}^\infty(-1)^k\binom{-4}{k}x^k\\
&=\sum_{j=0}^4(-1)^j\binom{4}{j}x^{7j}\sum_{k=0}^\infty\binom{k+3}{k}x^k
\end{align}
$$
wh... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Probability of 5 cards drawn from shuffled deck
Five cards are drawn from a shuffled deck with $52$ cards. Find the probability that
a) four cards are aces
b) four cards are aces and the other is a king
c) three cards are tens and two are jacks
d) at least one card is an ace
My attempt:
a) $\left(13*12*\binom{4}{4}*\... | a) There are $ \binom{52}{5} = 2,598,960 $ ways of choosing 5 cards. There are $\binom{4}{4} = 1 $ way to select the 4 aces. So there are $\binom{48}{1}= 48 $ ways to select the remaining card. Thus there are a total of 48 ways to select 5 cards such that 4 of them are aces, and the probability is:
$\frac{48}{2,598,960... | {
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"timestamp": "2023-03-29T00:00:00",
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Indefinite Integral $\int\frac{1}{1+\tan^{-1}x}\,\text{d}x$ I tried to solve this indefinite integral, $$\int\frac{1}{1+\tan^{-1}x}\,\text{d}x.$$
I tried taking the change of variable $u=\tan^{-1}x$ but failed to reach a solution.
Can anyone help me?
Thanks in advance.
| Let $u=\tan^{-1}x$ ,
Then $x=\tan u$
$\therefore\int\dfrac{1}{1+\tan^{-1}x}~dx$
$=\int\dfrac{d(\tan u)}{u+1}$
$=\dfrac{\tan u}{u+1}-\int\tan u~d\left(\dfrac{1}{u+1}\right)$
$=\dfrac{\tan u}{u+1}+\int\dfrac{\tan u}{(u+1)^2}~du$
$=\dfrac{\tan u}{u+1}+\int\sum\limits_{n=0}^\infty\dfrac{8u}{((2n+1)^2\pi^2-4u^2)(u+1)^2}~du$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/711016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Cholesky decomposition of the inverse of a matrix I have the Cholesky decomposition of a matrix $M$. However, I need the Cholesky decomposition of the inverse of the matrix, $M^{-1}$. Is there a fast way to do this, without first computing $M^{-1}$? In other words, is there a relationship between the Cholesky decomposi... | To add to previous answers, if we view $X$ as covariance matrix of data, the relationship between two decompositions reduces to relationship between coefficients of "right-to-left" autoregressive model and "left-to-right" one (details)
In general, switching order of "inverse" and "Cholesky" gives different results. IE
... | {
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Solve $\tan x =\sqrt{3}$, finding all solutions. My attempt so far:
$$
\tan x = \sqrt3
$$
$$
\frac{\sin x}{\cos x}= \sqrt3
$$
Then I look at the unit circle to find possible solutions. I find two solutions: $$
\frac{\pi}{3} \text{ and }\frac{4\pi}{3}
$$
However, I find the answer to be:
$$
x = \frac{\pi}{3} + \pi k
$... | $$\tan\left(\frac {7\pi}{6}\right) = \dfrac{-\frac 12}{-\frac{\sqrt 3}{2}} = \dfrac 1{\sqrt 3}$$
So $x = \dfrac{7\pi}{6}$ does not solve $\tan x = \sqrt 3$.
The second solution $x$ in the interval $[0, 2\pi]$ is $$x = \dfrac{4\pi}3$$
ADDED after EDITED post:
Note that $$\dfrac {4\pi}{3} = \frac{\pi}{3} + 1\cdot \pi = \... | {
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Factoring $x^5 + x^4 + x^3 + x^2 + x + 1$ without using $\frac{x^n - 1}{x-1}$? I was at a math team meet today and one of the problems was to factor $x^5 + x^4 + x^3 + x^2 + x + 1$. It also gave the hint that it decomposes into two trinomials and a binomial.
The solution they gave was based on the fact that $\frac{x^6 ... | You should be(come) aware of: $\frac{x^n-1}{x-1} = \sum_{k=0}^{n-1} x^k$
It's quite useful.
$$x^5+x^4+x^3+x^2+x+1 \\ = \frac{x^6-1}{x-1} \\ = \frac{(x^3-1)(x^3+1)}{x-1} \\ = (x^2+x+1)(x^3+1) \\ = (x^2+x+1)(x+1)(x^2-x+1)$$
The only other way would be to guess -1 as a root, because you have six terms in ascending polyno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716183",
"timestamp": "2023-03-29T00:00:00",
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What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression What is the sum of $1^4 + 2^4 + 3^4+ \dots + n^4$ and the derivation for that expression using sums $\sum$ and double sums $\sum$$\sum$?
| One way is by using the Euler-Maclaurin summation formula, something you strangely don't find in most standard calculus books.
http://en.wikipedia.org/wiki/Euler%E2%80%93Maclaurin_formula
Let $f(x) = x^4$.
Then
$$f'(x) = 4x^3, f'''(x) = 24 x, f^{(5)}(x) = f^{(7)}(x) = ... = 0 $$
And according to the formula,
$$ \sum_{k... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove if $n^3$ is odd, then $n^2 +1$ is even I'm studying for finals and reviewing this question on my midterm. My question is stated above and I can't quite figure out the proof.
On my midterm I used proof by contraposition by stating:
If $n^2 +1$ is odd then $n^3$ is even.
I let $n^2+1 = (2m+1)^2 + 1$
$= (4m^2 + 4m +... | If $n^3$ is odd, then $n^3-1$ is even. Now, $n^3-1=(n-1)(n^2+n+1)$, since that one of $(n-1)$, $(n^2+n+1)$ (or both) must be even.
'
Suppose that $n-1$ is even, then $n$ is odd, $n^2$ is odd and $n^2+1$ is even.
If $n-1$ is odd, then $n$ is even and $n^2+1$ is also odd. Therefore, $n-1$ must be even.
| {
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Is this matrix decomposition possible? Given a $2\times2$ matrix $S$ with entries in $\mathbb{Z}$ or $\mathbb{Q}$ , when is it possible to write $S=\frac{1}{3}(ABC+CAB+BCA)$ such that $A+B+C=0$, where $A, B, C$ are matrices over the same ground ring as S. Always? How would I find $A, B, C$?
| To challenge this knot, we have no choice but to replace any specific matrices.
As one of the approaches, The rotation matrix is a part of correct answer. Because the matrix has special characteristics. As reasons, variables and equations can be reduced and commutativity is allowed under any two dimension. For example,... | {
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Evaluate the integral: $\int\tan^5 (4x)\,\mathrm dx$. We have to use a trig identity for this. I think it's $\tan^2 x = \sec^2 x - 1$.
But I'm having difficulty setting up the integral.
| The secret is to use $1+\tan^2 w=\sec^2 w$. Indeed:
$$\int \tan^5 4x dx=\int \tan^3 4x\cdot \tan^2 4x dx=\int \tan^3 4x(\sec^2 4x-1)dx= $$
$$=\frac{1}{4}\int (\tan 4x)^3(\sec^2 4x \cdot 4 dx)-\int \tan^3 4x dx= $$
$$=\frac{1}{4}\frac{(\tan 4x)^4}{4} -\int \tan 4x\cdot \tan^2 4xdx=$$
$$=\frac{1}{16}\tan^4 4x-\int \tan 4... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding one sided limits algebraically I was wondering what the best method was for proving this limit algebraically:
$$\lim_{x \to 1}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}$$
I know the answer to this question is ;
$$\lim_{x \to 1^+}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}={-\infty}$$
$$\lim_{x \to 1^-}\frac{3x^4-8x^3+5}{x^3-x^2-x+1}... | Hints:
$$3x^4-8x^3+5=(x-1)(3x^3-5x^2-5x-5)$$
$$x^3-x^2-x+1=(x-1)(x^2-1)=\ldots$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How prove this $\ln{(x+\sqrt{x^2+1})}<\frac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$ let $0<a<1,x<0$,show that
$$\ln{(x+\sqrt{x^2+1})}<\dfrac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$$
My idea:
$$\Longleftrightarrow \ln{(\sqrt{x^2+1}+x)}<\dfrac{x(a^x-1)\ln{a}}{(a^x+1)\ln{(\sqrt{x^2+1}-x)}}$$
Then following I f... | $$\ln{(x+\sqrt{x^2+1})}<\dfrac{x(a^x-1)}{(a^x+1)\log_{a}{(\sqrt{x^2+1}-x)}}$$
Rewrite the inequality in terms of hyperbolic functions:
1.) $\ln{(x+\sqrt{x^2+1})}=\sinh^{-1}x$
2.) $\frac{1}{\log_{a}{(\sqrt{x^2+1}-x)}}=-\frac{\log{a}}{\sinh^{-1}x}$
3.) $\frac{a^x-1}{a^x+1}=\tanh{\left(\frac{x\log{a}}{2}\right)}$
Using th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/724722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Remainder of the polynomial A polynomial function $f(x)$ with real coefficients leaves the remainder $15$ when divided by $x-3$, and the remainder $2x+1$ when divided by $(x-1)^2$. Then the remainder when $f(x)$ is divided by $(x-3)(x-1)^2$ is?
What I have thought-The remainder must be of the form $ax^2+bx+c$. Now app... | Note that
$$
\frac{1}{4}(x-1)^2-\frac{1}{4}(x+1)(x-3)=1.
$$
You can get this by dividing $(x-1)^2$ by $x-3$. So take
$$
f(x)=15\frac{1}{4}(x-1)^2-(2x+1)\frac{1}{4}(x+1)(x-3).
$$
Then the remainder when $f(x)$ is divided by $x-3$ is 15 and by $(x-1)^2$ is $2x+1$ since $(x+1)(x-3)=[(x-1)+2][(x-1)-2]=(x-1)^2-4$.
| {
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Find the Global maximum and minimum values on the closed disk of this function just wondering if I am doing this correctly:
For $f(x,y) = x^3y + xy^3 - xy + 1 $ Find the Global maximum and minimum values on:
$ D = {(x,y) \in R | x^2 + y^2 \le 4} $
I found that the critical points of the function are: $(0,0),(0,1),(0,-... | You may have worked too hard on the circle, and made a little error.
Our function is $xy(x^2+y^2)-xy+1$, which on the circle is $3xy+1$. This is $12\cos\theta\sin\theta+1$, which is $6\sin 2\theta+1$, maximum $7$, minimum $-5$.
Note that parametrization (polar coordinates) could also be useful inside the circle, for o... | {
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"timestamp": "2023-03-29T00:00:00",
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Recurrence relation converting to explicit formula Let $a_n = -2a_{n-1 }+ 15a_{n-2 }$ with initial conditions $a_1 = 10 $ and $a_2 = 70$.
a)Write the first 5 terms of the recurrence relation.
b)Solve this recurrence relation.
c)Using the explicit formula you found in part b, evaluate $a_5$. You must show that you are u... | Use generating functions. Define $A(z) = \sum_{n \ge 0} a_n z^n$, write the recurrence as:
$$
a_{n + 2} = - 2 a_{n + 1} + 15 a_n
$$
Running the recurrence "backwards" gives $a_0 = 6$ (starting at index 0 is nicer all around). Multiply the recurrence by $z^n$, sum over $n \ge 0$, and recognize:
\begin{align}
\sum_{n \ge... | {
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If $a+b+c+d = 2$, then $\frac{a^2}{(a^2+1)^2}+\frac{b^2}{(b^2+1)^2}+\frac{c^2}{(c^2+1)^2}+\frac{d^2}{(d^2+1)^2}\le \frac{16}{25}$ If $a+b+c+d = 2$, prove that
$$\dfrac{a^2}{(a^2+1)^2}+\dfrac{b^2}{(b^2+1)^2}+\dfrac{c^2}{(c^2+1)^2}+\dfrac{d^2}{(d^2+1)^2}\le \dfrac{16}{25}$$
Also $a,b,c,d \ge 0$.
| Let us consider the function
$$
\frac{x^2}{(1+x^2)^2} +
\frac{y^2}{(1+y^2)^2} +
\frac{z^2}{(1+z^2)^2} +
\frac{t^2}{(1+t^2)^2} +
\lambda(x+y+z+t-2)
$$and write the first order conditions:
$$
0 = -\frac{2x(x^2-1)}{(1+x^2)^3} + \lambda,
$$and the same equation for $y,z,t$. In particular,
$$
\frac{x(x^2-1)}{(1+x^2)^3} = \... | {
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"answer_id": 1
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Proof by induction that $n^3 + (n + 1)^3 + (n + 2)^3$ is a multiple of $9$. Please mark/grade. What do you think about my first induction proof? Please mark/grade.
Theorem
The sum of the cubes of three consecutive natural numbers is a multiple of 9.
Proof
First, introducing a predicate $P$ over $\mathbb{N}$, we rephra... | I do not think that this is a real question but if you were my student i would give you an A.
It is all fine to me.
| {
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Prove that if $n(a^2+b^2+c^2)=abc$ then $2\mid n$
Is it true that if $n\in\mathbb N$ and the diophantine equation $$n(a^2+b^2+c^2)=abc,\\(a,b)=(b,c)=(c,a)=1\tag1$$
has positive integer solutions $a,b,c$, then $2\mid n$?
I can prove that $3\mid n:$
1) If $3\not\mid abc$ then $3\mid a^2+b^2+c^2,$ a contradiction.
2)... | It is indeed true. We need the following well known fact:
If $p\equiv3\pmod4$ is prime and $p\mid x^2+y^2$, then $p\mid x$ and
$p\mid y$.
We will prove that
Theorem. If $a,b,c$ is a solution to $(1)$, then exactly one of $a,b,c$ is divisible by $4$.
Proof.
First suppose $a,b,c$ are all odd. Then $a^2+b^2+c^2\equiv3... | {
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Is limits $\lim_{x\to\infty}{(x-2)^2\over2x+1}=\dfrac{1}{2}$? $$\lim_{x\to\infty}{(x-2)^2\over2x+1}=\dfrac{1}{2}$$
I used an online calculator and it said it was actually $=\infty$
Here's how I calculate it:
$$\lim_{x\to\infty}\dfrac{x^2+4-4x}{2x+1}=\lim_{x\to\infty}\dfrac{({x\over x}-{2\over x})({x\over x}-{2\over x})... | You have accidentally divided the numerator by $x^2$ and the denominator only by $x$. For the numerator we get
$$(x-2)^2 = \left[x\left(\tfrac{x}{x}-\tfrac{2}{x}\right)\right]\cdot \left[x\left(\tfrac{x}{x}-\tfrac{2}{x}\right)\right] = x^2\left(1-\tfrac{2}{x}\right)^2,$$
so the fraction is
$$\frac{(x-2)^2}{2x+1} = \fra... | {
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"answer_id": 0
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How to prove series convergence: $\sum \limits_{n=1}^\infty \left(\frac1n+\sqrt{1+n^2}-\sqrt{2+n^2}\right)^2$ I have this series:
$$\sum \limits_{n=1}^\infty \left(\frac1n+\sqrt{1+n^2}-\sqrt{2+n^2}\right)^2$$
I know that it's convergent (from WolframAlpha) but I need to prove it is convergent. How can I do it?
| You have that $$\begin{align*} \dfrac 1 n+\sqrt{1+n^2}-\sqrt{2+n^2}&=\frac1n+\frac{(\sqrt{1+n^2}-\sqrt{2+n^2})\cdot(\sqrt{1+n^2}+\sqrt{2+n^2})}{\sqrt{1+n^2}+\sqrt{2+n^2}}=\\\\&=\frac1n-\frac{1}{\sqrt{1+n^2}+\sqrt{2+n^2}}\\& \le \frac{1}{n} \end{align*}$$ Then the convergence is established by squaring both sides and us... | {
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Prove that the integers $n$ and $n, 2^{2^n} + 1$ are relatively prime. I need help to prove that
$$\gcd(n, 2^{2^n} + 1)=1,\ n = 1,2,\dots$$
I have no idea how start the proof.
| First note the following: For any prime $p$ dividing $2^{2^n}+1$:
$$p|(2^{2^n}+1) \implies 2^{2^n} \equiv_p -1$$
$$\implies \left(2^{2^n}\right)^2 = 2^{2^{n+1}} \equiv_p 1$$
$$2^{2^n} \equiv_p -1; \ 2^{2^{n+1}} \equiv_p 1 \implies 2^{n+1} = Kc,$$
for some odd integer $K$, where $c$ is the smallest positive integer s.t.... | {
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"question_score": "2",
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"answer_id": 3
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Finding the mean with absolute value This question is out of my field and topic that I am teaching myself now, but I was wondering how would you solve this problem if it had the absolute value of it.
My Question:
$$f(x) = \begin{cases}
2/9(x-1), & \text{for $1<x<4$} \\
0, & \text{elsewhere} \\
\end{cases}
$$
I kno... | $\int_1^4|x-\mu|f(x)dx$
$=\int_1^3 (2/9)(x-1)(3-x)dx+\int_3^4 (2/9)(x-1)(x-3)dx$
$2/9\int_1^3(-x^2+4x-3)dx+2/9\int_3^4(x^2-4x+3)dx$
$2/9\int_1^3(-x^3/3+2x^2-3x) + 2/9\int_3^4 (x^3/3-2x^2+3x)$
$(2/9)(-x^3/3+2x^2-3x)|_1^3 + (2/9)(x^3/3-2x^2+3x)|_3^4$
$8/27+8/27$
$16/27$
| {
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Inequality Exercise in Apostol's Calculus I Let p and n denote positive integers. Show that:
$$n^{p} \lt \frac{(n+1)^{p+1} - n^{p+1}}{p+1} < (n+1)^{p}$$
Attempt at Solution
Using the identity $b^{p+1}-a^{p+1} = (b-a)\sum_{k=0}^{p}b^{p-k}a^{k}$, let $b = n+1$ and $a = n$. Then:
$$\frac{(n+1)^{p+1} - n^{p+1}}{p+1} = \fra... | We can use Bernoulli's to strengthen this to $n,p$ positive real numbers. I will show with the LHS.
$(p+1)n^p < (n+1)^{p+1}-n^{p+1}$
$n^p(p+n+p) < (n+1)^{p+1}$
$( p/n+1/n+1) < (1+1/n)^{p+1}$ which just follows from Bernoulli's Inequality. Similar for the RHS. Just write the solution forwards.
| {
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Does the order I multiply the characteristic equation's factors in the homogeneous solution matter? I've been doing a recurrence relation exercise in my book. Doing some steps and comparing them to the ones taken by the book.
$$T(0) = 1$$
$$T(1) = 2$$
$$T(k) - 7T(k-1)+10T(k-2)=6+8k$$
Characteristic equation:
$$a^2-7... | The order doesn't matter.
A general way of solving recurrences like this is as follows. Define the generating function $G(z) = \sum_{n \ge 0} T(n) z^n$, rewrite the recurrence so there aren't subtractions in indices:
$$
T(n + 2) - 7 T(n + 1) + 10 T(n) = 8 n + 22
$$
Multiply the recurrence by $z^n$, sum over $n \ge 0$, ... | {
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Consider the quadratic equation $ax^2-bx+c=0, a,b,c \in N. $ If the given equation has two distinct real root... Problem :
Consider the quadratic equation $~ax^2-bx+c=0, \quad a,b,c \in N. ~$ If the given equation has two distinct real roots belonging to the interval $~(1,2)~ $ then the minimum possible values of $~a~$... | From Vieta's:
$$2<x_1+x_2=\frac ba<4 \Rightarrow 2a<b<4a\\
1<x_1x_2=\frac ca<4 \Rightarrow a<c<4a\\
0<x_2-x_1<1 \Rightarrow 0<\frac{\sqrt{b^2-4ac}}{a}<1 \Rightarrow 0<b^2-4ac<a^2$$
Also:
$$f(1)=a-b+c>0 \Rightarrow c>b-a\\
f(2)=4a-2b+c>0 \Rightarrow c>2b-4a$$
Cases: $a=1$
$$2<b<4 \Rightarrow b=3\\
1<c<4\Rightarrow c=2,3... | {
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Find the value of $x$ for which $ff=gf$. Functions $f$ and $g$ are defined by $f:x \mapsto \frac{1}{2x+1}$, $x \neq \frac{-1}{2}$ and $g:x \mapsto x+1$. Find the value of $x$ for which $ff=gf$.
So I started in this way:
$f[f(x)]=g[f(x)]$
$f(\frac{1}{2x+1}) = g(\frac{1}{2x+1})$
$\frac{1}{2(\frac{1}{2x+1})+1}=\frac{1}{2x... | $\frac{1}{2(\frac{1}{2x+1})+1}=\frac{1}{2x+1}+1 \Leftrightarrow$
$\frac{2x+1}{2x+3}=\frac{2x+2}{2x+1}\Leftrightarrow$
$(2x+1)^2=(2x+2)(2x+3) \Leftrightarrow$
$4x^2+4x+1=4x^2+10x+6 \Leftrightarrow$
$6x+5=0$. So $x=-\frac{5}{6}$. Your book is right.
| {
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Calculation of $\int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$ Calculate
$$ \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2 \cos^2 x}}dx$$
$\bf{My\; Try::}$ Let $\displaystyle I = \int_{0}^{\frac{\pi}{4}}\tan^{-1}\sqrt{\frac{\cos 2x }{2\cos^2 x}}dx = \int_{0}^{\frac{\pi}{4}}\frac{\pi... | Answer: $\displaystyle \int_{0}^{\pi/4}\tan^{-1}\sqrt{\frac{\cos 2x}{2\cos^2 x}}\,dx=\frac{\pi^2}{24}$
Proof:
We are making use of $3$ Lemmas which are (quite ) easy to prove:
1. $\displaystyle \int_{0}^{1}\frac{dx}{\sqrt{x^2+2}\left ( x^2+1 \right )}=\frac{\pi}{6} $
Proof:
$$\begin{align*}
\int_{0}^{1}\frac{dx}{\sqrt... | {
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Polynomials division algebra problem Find sum of coefficients of the quotient obtained in:
$$\frac{2x^n+x^{n-1}+x^{n-2}+...+x^2+x+5}{x-\frac{1}{2}}$$
I got "n" as the answer but according to the book is wrong, I don't know what is wrong exactly, but i want to know why the answer is "2n" ._. , thanks.
| $$\frac{2x^n+x^{n-1}+x^{n-2}+...+x^2+x+5}{x-\frac{1}{2}}=$$
$$\frac{(x-\frac{1}{2})(2x^{n-1}+2x^{n-2}+\dots+2x^2+2x+2)+6}{x-\frac{1}{2}}=$$
$$2x^{n-1}+2x^{n-2}+\dots+2x^2+2x+2+\frac{6}{x-\frac{1}{2}}$$
So the quotient is: $2x^{n-1}+2x^{n-2}+\dots+2x^2+2x+2$
The remainder is: $6$
And the sum of coefficients in the quot... | {
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An Inequality Problem with not nice conditions How to show that $\dfrac{a^3}{a^2+b^2} + \dfrac{b^3}{b^2+c^2} + \dfrac{c^3}{c^2+a^2} \ge \dfrac32$, where $a^2+b^2+c^2=3$, and $a,b,c > 0$ ?
| By weighted AM-HM inequality, $\sum\limits_{cyc}\dfrac{\dfrac{a\cdot a^2}{a^2+b^2}}{3}\ge \dfrac{3}{\sum\limits_{cyc}\dfrac{a^2(a^2+b^2)}{a}}$
Here we consider the weights to be $a^2$,$b^2$ and $c^2$ respectively.
From the inequality above, we can have $\sum\limits_{cyc}\dfrac{a\cdot a^2}{a^2+b^2}\ge \dfrac{9}{\sum... | {
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For $1-r<|\theta|<1/2$, $|\frac{2r\sin{2\pi\theta}}{1-2r\cos{2\pi\theta}+r^2}-\frac{\cos\pi\theta}{\sin\pi\theta}|
For $1-r<|\theta|<1/2$ show that
$$|\frac{2r\sin{2\pi\theta}}{1-2r\cos{2\pi\theta}+r^2}-\frac{\cos\pi\theta}{\sin\pi\theta}|<C\frac{(1-r)^2}{|\theta|^3}$$
This inequality shows that the integral of the l... | Just a small trick: write $\frac{\cos \pi \theta}{\sin \pi \theta} = \frac{2\cos^2 \pi \theta}{\sin 2 \pi \theta}$ and use $1+ \cos 2 \pi \theta = 2\cos^2 \pi \theta.$ Now proceed by taking l.c.m, and all that.
| {
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How do I solve this definite integral: $\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$? $$\int_0^{2\pi} \frac{dx}{\sin^{4}x + \cos^{4}x}$$
I have already solved the indefinite integral by transforming $\sin^{4}x + \cos^{4}x$ as follows:
$\sin^{4}x + \cos^{4}x = (\sin^{2}x + \cos^{2}x)^{2} - 2\cdot\sin^{2}x\cdot\cos^{2... | We have $\displaystyle\cos^4x+\sin^4x=(\cos^2x+\sin^2x)^2-2\cos^2x\sin^2x=1-2\cos^2x\sin^2x$
$\displaystyle=\frac{2-\sin^22x}2=\frac{2(1+\tan^22x)-\tan^22x}{2\sec^22x}=\frac{\tan^22x+2}{2\sec^22x}$
$$\int\frac{dx}{\cos^4x+\sin^4x}=\int\frac{2\sec^22x}{\tan^22x+2}dx$$
Setting $\tan2x=u,$ $$\int\frac{2\sec^22x}{\tan^22x... | {
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Equality of integrals: $ \int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \int_{0}^{1} \frac {1}{1+x^2} \, \mathrm{d}x $ In Street-Fighting Mathematics (page 16), Prof. Sanjoy Mahajan states that $$ \displaystyle\int_{0}^{\infty} \frac {1}{1+x^2} \, \mathrm{d}x = 2 \cdot \displaystyle\int_{0}^{1} \frac {1}{1... | I assume you mean
$$\int_0^{\infty} \dfrac{dx}{1+x^2} = 2 \int_0^1 \dfrac{dx}{1+x^2}$$
This is true because
$$\underbrace{\int_1^{\infty} \dfrac{dx}{1+x^2} = \int_1^0 \dfrac{-du/u^2}{1+1/u^2}}_{x = 1/u} = \int_0^1 \dfrac{du}{u^2+1}$$
Hence,
$$\int_0^{\infty} \dfrac{dx}{1+x^2} = \int_0^{1} \dfrac{dx}{1+x^2} + \int_1^{\i... | {
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Prove this identity: $\sin^4x = \frac{1}{8}(3 - 4\cos2x + \cos4x)$ The problem reads as follows.
Prove this identity: $$\sin^4x = \frac{1}{8}(3 - 4\cos2x + \cos4x)$$
I started with the right side and used double angles identities for $\cos2x$ and a sum and then then double angle identity for the $\cos4x...$ It all got ... | $\sin^4x = (1 - \cos^2x)^2 = \left(1 - \dfrac{1 + \cos2x}{2}\right)^2 = \dfrac{(1 - \cos2x)^2}{4} = \dfrac{1 - 2\cos2x + \cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\cos^22x}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{\dfrac{1 + \cos4x}{2}}{4} = \dfrac{1}{4} - \dfrac{\cos2x}{2} + \dfrac{1 + \cos4x}{8} ... | {
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integer solutions to $x^2+y^2+z^2+t^2 = w^2$ Is there a way to find all integer primitive solutions to the equation $x^2+y^2+z^2+t^2 = w^2$? i.e., is there a parametrization which covers all the possible solutions?
| Bradley’s excellent paper on equal sums of squares gives the complete parameterization of the equation
$$
x_1^2 + x_2^2 + x_3^2 + x_4^2 = y_1^2
$$
as
\begin{align}
x_1 &= (uz+vy+wz)^2 -m^2(m^2+x^2+y^2+z^2-u^2-v^2-w^2) \\
x_2 &= 2m(um^2+uz^2+xvm-ywm+xwz+yvz) \\
x_3 &= 2m(vm^2+vy^2+zwm-xum+zuy+xwy) \\
x_4... | {
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Prove that $7^{100}+3^{10}=8^{100}$ or $7^{100}+3^{10}<8^{100}$
Prove that $7^{100}+3^{10}=8^{100}$ or $7^{100}+3^{10}<8^{100}$
I tried using some theorems of divisibility, to show that one divides the other, and the other also divides the first, but could not.
| If you list $7^x$ and $8^x$ you will notice that when $x$ is divisible by $4$, $7^x$ ends in $1$ and $8^x$ ends in $6$. $$7^0=8^0=1$$ $$7^{4+j}=7^j*7^4$$ $$7^4 = 2401$$ $2401$ ends in $1$ as does $7^0$ and any number that ends in $1$ times any number than ends in 1 results in a number that also ends in $1$. This works... | {
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Evaluate the indefinite integral $\int \frac{\sqrt{9-4x^{2}}}{x}dx$ $$\int \frac{\sqrt{9-4x^{2}}}{x}dx$$
How Can I attack this kind of problem?
| $$(9-4x^2)^{1/2} = \sum_{n=0}^{\infty}\binom{1/2}{n}9^{1/2-n}(-4x^2)^{n}$$
$$\dfrac{(9-4x^2)^{1/2}}{x} = \sum_{n=0}^{\infty}\binom{1/2}{n}(-4)^{n}\left(\dfrac{9}{x^2}\right)^{1/2-n}$$
$$\int \left(\dfrac{9}{x^2}\right)^{1/2-n}(-4)^{n} \;\mathrm{d}x =(-4)^{n}9^{1/2-n}\int {x}^{1-2n}\; \mathrm{d}x = (-4)^{n}9^{1/2-n}\lef... | {
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Maclaurin series: $\frac{x^3}{3!}+\frac{x^4}{4!}+\frac{x^7}{7!}+\frac{x^8}{8!}+\frac{x^{11}}{11!}+\frac{x^{12}}{12!}+...$ The Taylor series of a real or complex-valued function ƒ(x) that is infinitely differentiable at a real or complex number a is the power series
$$f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a... | While I've never seen the series you define here, it can be shown to be an elementary function. The process for doing this goes under the name of multisectioning: the key concept is that if $\zeta$ is an $n$th root of unity, then the Maclaurin coefficients of the functions $e^{\zeta^kx}$ for $0\leq k\lt n$ all have $n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/767995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$2 \cos^2 x − 2 \cos x− 1 = 0$ Find the solutions if $0^\circ \le x < 360^\circ$
Find the solutions of $$2 \cos^2 x − 2 \cos x− 1 = 0$$ for all $0^\circ ≤ x < 360^\circ$.
For $0^\circ \le x < 360^\circ$, I'm getting $x=111.5^\circ$ and $x=248.5^\circ$.
Is this correct? Thanks!
| Observe that $2 \cos^2 x - 2 \cos x - 1$ is irreducible and cannot be factored.
Apply the quadratic formula:
$$\cos x = \frac{-(-2)\pm\sqrt{(-2)^2-4(2)(-1)}}{2(2)}=\frac{2\pm\sqrt{12}}{4}=\frac{2\pm2\sqrt{3}}{4}=\frac{1\pm\sqrt{3}}{2}$$
But $\frac{1 + \sqrt{3}}{2}$ is not in between $-1$ and $1$, so we can discard it a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768401",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How prove this inequality $\frac{1}{n}\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}\le\frac{1}{a+b+1}\binom{n}{a}\binom{n}{b}$
let $a,b,n$ be positive integer numbers,and such $a,b\le n$, show that
$$\dfrac{1}{n}\sum_{k=0}^{n-1}\binom{k}{a}\binom{k}{b}\le\dfrac{1}{a+b+1}\binom{n}{a}\binom{n}{b}$$
this inequality maybe c... | Since the inequality is symmetric in $a$ and $b$ we may (and will), suppose $a\leq b\leq n$.
Moreover, the inequality is trivially satisfied if $n=b$. So let us suppose that the inequality is proved for some $n\geq b$. It follows that
$$\eqalign{
\sum_{k=0}^n\binom{k}{a}\binom{k}{b}&=\sum_{k=0}^{n-1}\binom{k}{a}\binom{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/768645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
set problem of integers Consider the following set $F=\{F^0, F^1, F^2, \ldots\}$. This set consists of positive integers which satisfy the following properties:
*
*$F^0= F^1=1$
AND
*$F^n= F^{n-1} + F^{n-2}$ for all positive integers $n\geq2$.
Prove that for all positive integers $n$, the elements of the set $F$ sa... | Hint:
$$\begin{pmatrix} F_{n+1}&F_n\\F_n&F_{n-1}\end{pmatrix}\begin{pmatrix} 1&1\\1&0
\end{pmatrix}=\begin{pmatrix} F_{n+1}+F_n&F_{n+1}\\F_{n}+F_{n-1}&F_n\end{pmatrix}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/769404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate the product ST, and infer from it the inverse of T. S=\begin{pmatrix}
1/2 & 1/2 & 0\\
1 & 0 & 0\\
-3/2 & 0 & 1/2
\end{pmatrix}
T=
\begin{pmatrix}
0 & 1 & 0\\
2 & -1 & 4\\
0 & 3 & 2
\end{pmatrix}
I have calculated ST to be =
\begin{pmatrix}
1 & 0 & 2\\
0 & 1 & 0\\
0 & 0 & 1
\end{pmatrix}
But i'm now unsu... | Think in terms of Gaussian elimination. It should be clear that
$$
\pmatrix{
1&0&-2\\
0&1&0\\
0&0&1
}
\pmatrix{
1&0&2\\
0&1&0\\
0&0&1
} = I
$$
That is, we have
$$
\pmatrix{
1&0&-2\\
0&1&0\\
0&0&1
}
\;(ST) =
\left(\pmatrix{
1&0&-2\\
0&1&0\\
0&0&1
}
\;S \right)\cdot T = I
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/771988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How prove this $100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$
let $a,b,c\ge 0$, and such $$a+b+c=6$$
show that
$$100+5(a^2+b^2+c^2)-2(a^2b^2+b^2c^2+a^2c^2)-a^2b^2c^2\ge 0$$
My idea: since
$$a^2+b^2+c^2=(a+b+c)^2-2(ab+bc+ac)=36-2(ab+bc+ac)$$
$$a^2b^2+b^2c^2+a^2c^2=(ab+ca+bc)^2-2abc(a+b+c)=(ab+bc+ac)^2-12a... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, the condition does not depend on $w^3$ and we need to prove that $f(w^3)\geq0$,
where $f$ is a concave function.
But the concave function gets a minimal value for an extremal value of $w^3$,
which happens for an equality case of two variables or when maybe $w^3=0$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/772429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
First order Cauchy problem I'm trying to solve the PDE
\begin{align*}
au_x + (bx+cu-1)u_y = d \\
u(x,0) = 0.
\end{align*}
So far, I got the solution for the case $cd-ab \neq 0$ as
\begin{align*}
u(x,y) = \frac{d}{cd-ab}\left( 1-bx-\sqrt{(1-bx)^2+2(cd+ab)y} \right)
\end{align*}
Now, I'm stuck on the case $ab = cd$. Can... | Solve the equation by the method of characteristics.
The characteristic ODE are as follows.
$\dfrac{d x}{ d t} =a$, $\dfrac{d u}{ d t} = d$, $\dfrac{d y}{ d t} =b x +c u -1$,
with initial conditions $x(0) = x_0$, $u(0) = 0 $, $y(0) = 0$ at $t=0$.
By solving the first two characteristic ODE, there are
$x = x_0 + a t $ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/773909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Calculation of $f_{\omega^3}(2)$ in the fast growing hierarchy How is the number
$$\large f_{\omega^3}(2)$$
in the fast growing hierarchy calculated ?
My only idea is to convert to
$$\large f_{\omega^2 2}(2)$$
but now I have no idea how to continue.
| Too large to calculate!
I assume you are referring to the definition on Wikipedia: http://en.wikipedia.org/wiki/Fast-growing_hierarchy
If so, then, as the article points out, $f_{\omega + 1}(64)$ is already larger than Graham's number. $f_{\omega^3}$ is much larger than this.
But perhaps you're asking what an algorithm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to calculate $\sum_{k=1}^n \left(k \sum_{i=0}^{k-1} {n \choose i}\right)$ How do I calculate the following summation?
$$\sum_{k=1}^n \left[k \sum_{i=0}^{k-1} {n \choose i}\right]$$
| (Note: I have attempted to incorporate
OP's edit to the problem.)
$\begin{array}\\
\sum_{k=1}^n (k \sum_{i=0}^{k-1} {n \choose i})
&=\sum_{k=1}^n \sum_{i=0}^{k-1} (k {n \choose i})\\
&=\sum_{i=0}^{n-1} \sum_{k=i+1}^n (k {n \choose i})\\
&=\sum_{i=0}^{n-1} {n \choose i}\sum_{k=i+1}^n k \\
&=\sum_{i=0}^{n-1} {n \choose i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/781645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Picard Approximation Consider the initial value problem
$$\frac{dy}{dx} = y^2 + 3x^2 - 1, \\
y(1) = 1$$
on D = {|x-1| <= 1, |y-1| <= 1}
Find the second approximation to the solution and estimate the error term
If someone could maybe walk me through the steps that would be most appreciated =)
| We are given:
$$\tag 1 \frac{dy}{dx} = y^2 + 3x^2 - 1, y(1) = 1, ~\mbox{on}~~ D = {|x-1| \le 1, |y-1| \le 1}$$
The Picard-Lindelöf iteration is given by:
$$\tag 2 \displaystyle y_0(x) = y_0, ~~y_{n+1}(x) = y_0 + \int^x_{x_0} f(s, y_n(s))ds$$
For $(1)$, we have: $f(s, y_n(s)) = y_n^2+3s^2-1$ and using $(2)$, yields:
*... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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calculus radical integration question Does anyone know how to calculate integral of $\sqrt{ 1-\cos (x)}$ ? I tried several methods resulting in $-2\sqrt2 \cos (x/2) + c$, but this is wrong in accordance with the text book, so i dont know how to proceed...
| \begin{align*}
\int \sqrt{ 1-\cos (x)} dx &= \int\sqrt 2 \left| \sin \left( \frac x 2 \right ) \right | dx\\
&= -2\sqrt 2 \cos \left( \frac x 2 \right ) \mathrm{sgn} \left( \sin \left( \frac x 2 \right )\right)\\
&= -2\sqrt 2 \cot \left( \frac x 2 \right ) \left| \sin \left( \frac x 2\right)\right|\\
&= -2 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/783547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Finding the inverse Laplace transform of $\frac{s^2-4s-4}{s^4+8s^2+16}$ $$F(s) = \frac{s^2-4s-4}{s^4+8s^2+16}$$
My work is as follows,
$$\frac{s^2-4s-4}{(s^2+4)^2}=\frac{s^2+4}{(s^2+4)^2}-\frac{8}{(s^2+4)^2}-\frac{4s}{(s^2+4)^2}$$
The inverse laplace of the first term is, $\frac{1}{2} \sin(2t)$
The second one has no di... | Note that
$$
\mathcal{L}^{-1}\left[\frac{a}{s^2+a^2}\right]=\sin at
$$
and
$$
\mathcal{L}^{-1}\left[\frac{s^2-a^2}{(s^2+a^2)^2}\right]=t\cos at,
$$
then
$$
\frac{a}{s^2+a^2}-a\cdot\frac{s^2-a^2}{(s^2+a^2)^2}=\frac{2a^3}{(s^2+a^2)^2}.
$$
Hence
$$
\mathcal{L}^{-1}\left[\frac{2a^3}{(s^2+a^2)^2}\right]=\sin at-at\cos at
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/784688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How to integrate $\sqrt{1-\sin 2x}$? I want to solve the following integral without substitution:
$$\int{\sqrt{1-\sin2x}} \space dx$$
I have:
$$\int{\sqrt{1-\sin2x}} \space dx = \int{\sqrt{1-2\sin x\cos x}} \space dx = \int{\sqrt{\sin^2x + \cos^2x -2\sin x\cos x}} \space dx$$
but this can be written in two ways: $... | The result in wolfram is given by :
$$\frac{\sqrt{1-\sin(2x)}(\cos x+\sin x)}{\cos x-\sin x}\tag{1}$$
Note your results from simplifying the integration. You want to check that $(1)$ is equal to either$$\int\sqrt{(\sin x-\cos x)^2}\mathrm dx=\int\sin x\mathrm dx-\int\cos x\mathrm dx=-\cos x-\sin x.\tag{2}$$
or
$$\int\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/785271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Infinite Sum of products What is the infinite sum
$$S = {1 + \frac{1}{3} + \frac{1\cdot 3}{3\cdot 6} + \frac{1\cdot 3\cdot 5}{3\cdot 6\cdot 9}+ ....}?$$
I attempted messing around with the $n$ th term in the series but didnt see any solution.
How should I proceed?
| HINT:
$n^{th}$ term is $$\frac{1\cdot3\cdot5\cdots(2n-1)}{3^n\cdot n!}$$
$$=\frac{2^n\cdot n!\cdot1\cdot3\cdot5\cdots(2n-1)}{6^n\cdot n!\cdot n!}$$
$$=\frac{2n!}{6^n\cdot n!\cdot n!}$$
$$=\frac{^{2n}C_n}{6^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/786162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$4$ variable system of equations Find all the solutions to this system of equations:
$$\begin{cases}
-2d^3+3a^2d+d+2c^3-3a^2c-c=0\\
2d^3-3db^2-d-2c^3+3b^2c+c=0\\
-6c^2b+b^3+b+6c^2a-a^3-a=0\\
6d^2b-b^3-b-6d^2a+a^3+a=0
\end{cases}$$
After factoring:
$$\begin{cases}
d(-2d^2+3a^2+1)+c(2c^2-3a^2-1)=0\\
d(2d^2-3b^2-1)-c(2c... | Adding $1$st and $2$nd equations, one can get
$$
3d(a^2-b^2)-3c(a^2-b^2)=0,
$$
$$
(d-c)(a-b)(a+b)=0\tag{1};
$$
adding $3$rd and $4$th equation, one can get
$$
b(6d^2-6c^2)-a(6d^2-6c^2)=0,
$$
$$
(b-a)(d-c)(d+c)=0\tag{2}.
$$
Equations $(1), (2)$ can help here?
Other (similar) way:
write system in form:
$$\begin{cases}
... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a=\dfrac{3+\sqrt{5}}{2}.$ Show that $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$.
Let $a=\dfrac{3+\sqrt{5}}{2}.$
Show that for all $n\in\mathbb N$, $\lfloor a \lfloor an \rfloor \rfloor+n$ is divisible by $3$.
My teacher solve this problem with induction, I am just curious if we can do this exerc... | Let $\alpha=\dfrac{1+\sqrt{5}}{2}, \beta=\dfrac{3+\sqrt{5}}{2}$, then
$$\beta= \alpha^2= \alpha+1$$
Since
$$\alpha\lfloor\beta n\rfloor-\lfloor\beta n\rfloor=\alpha n-(\alpha-1)\{\alpha n\}=\lfloor\alpha n\rfloor+(2-\alpha)\{\alpha n\}$$
so $\lfloor\alpha n\rfloor\lt \alpha\lfloor\beta n\rfloor-\lfloor\beta n\rfloor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/789343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Positive integer solutions of $a^3 + b^3 = c$ Is there any fast way to solve for positive integer solutions of $$a^3 + b^3 = c$$ knowing $c$?
My current method is checking if $c - a^3$ is a perfect cube for a range of numbers for $a$, but this takes a lot of time for larger numbers. I know $c = (a+b)(a^2-ab+b^2)$, but ... | $c=(a+b)(a^2−ab+b^2)$ should actually be quite helpful. After factoring $c$, you find all possibly ways of writing $c = xy$. Let $x = a + b$ or $b = x - a$, so
$$
a^2-ab+b^2 = a^2 - a(x-a) + (x-a)^2 = 3a^2 - 3ax + x^2 = y
$$
Solve this quadratic equation for a and check that a is an integer.
Obviously many factors nee... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/792085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
Inverse of Symmetric Matrix Plus Diagonal Matrix if Square Matrix's Inverse Is Known Let $A$ be an $n \times n$ symmetric matrix of rank $n$ with known inverse $A^{-1}$. Let $D$ be a diagonal matrix with the same dimensions and rank. What is the fastest way to compute $(A+D)^{-1}$? Assume all diagonal elements of $A... | Since $(A+D)^{-1} = (I+ A^{-1} D)^{-1} A^{-1}$, and $A^{-1}$ is known, we just need to find out how to calculate $B ^{-1}$ where $B$ is the (known) matrix $1+A^{-1} D$.
Following Greg Martin's observation, consider first the case where the only nonzero entry of $D$ is in the upper left corner. In this case, all columns... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/792477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 1,
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Proving $\prod_{i=1}^np_i+1$ is not a perfect square
Let $m=\displaystyle{\prod_{i=1}^np_n}$ be the product of the first $n$ primes $(n>1)$. prove that $m+1$ cannot be a perfect square.
I think that the opposite it correct: $m+1$ is not a complete square iff $(m+1)^{\frac {p-1}{2}}\equiv -1\pmod p$ but for each prime... | Every prime $p$ greater than $2$ is congruent to $1$ or $3 \pmod{4}$. Next, note that $3^2 \equiv 1 \pmod{4}$. Hence, $\prod_{i=2}^n p_i \equiv 1$ or $3 \pmod{4}$. Once we include the first prime, $2$, we get $\prod_{i=i}^n p_i \equiv 2 \pmod{4}$. Thus:
$$\prod_{i=i}^n p_i + 1 \equiv 3 \pmod{4}$$
This cannot be a p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/796567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Integral $\int \sqrt{x^2-3x+2}\ dx$ How to evaluate $$\int_3^{17} \sqrt{x^2-3x+2}\ dx \ ?$$ I tried Euler's substitution, that is $$\sqrt{x^2-3x+2}=x+t \Longleftrightarrow \frac{t^2-2}{-3-2t}+t=\frac{t^2+3t+2}{2t+3}\ ,$$ which I obtained from $$x^2+2tx+t^2=x^2-3x+2\Longleftrightarrow x=\frac{t^2-2}{-3-2t}$$ $$dx=\frac{... | An Euler Substitution will work, remember that there are several such substitutions and you have the root of a quadratic which factors. Another approach in addition to the excellent idea above is to write the integral as
$$\int (x-1)\sqrt{\frac{x-2}{x-1}}$$
and let
$$y^2=\frac{x-2}{x-1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/799164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Arc Length polar curve $$r=a\sin^3\left(\frac{\theta}{3}\right) $$
I tried solving it using the equation for arc length with $dr/d\theta$ and $r^2$. Comes out messy and complicated.
| The portion of the arclength calculation in which we determine the infinitesimal arclength element $ \ ds \ $ just requires applying the Chain Rule carefully and using the Pythagorean Identity:
$$ \frac{dr}{d\theta} \ = \ \frac{d}{d\theta} \left[ \ a \ \sin^3 \left(\frac{\theta}{3}\right) \right] \ = \ a \ \cdot \ 3 \ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is this a theorem (is it correct?)? My instruction notes have specified a theorem of matrix transpose that be there two compatible matrices $A$ and $B$ in respect of their sums and products, then:
$(AB)^T =A^TB^T$
So I set on to verify if indeed this is correct:
Let $A = \begin{pmatrix}2 & 2 \\ 6 & 4\end{pmatrix} $
Le... | The assertion is false; what is true is that
$$
(AB)^T=B^TA^T
$$
which can be easily proved. Actually, the product $A^TB^T$ need not be defined when the product $AB$ is defined, so the equality doesn't make sense unless both matrices are square. However, the equality doesn't generally hold in this case.
Let's see what ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Lagrange's Trigonometric Identity Lagrange's Trig identity is
$$
1+\cos\theta+\cos 2\theta +\cdots + \cos n \theta=\frac{1}{2}+\frac{\sin\frac{(2n+1)\theta}{2}}{2\sin \frac{\theta}{2}},\quad (0<\theta <2\pi).
$$
How can we prove this identity using series method and complex variables? I tried to use
$$
\sum_{n=0}^\in... | If $S = 1+z+z^2+\cdots+z^n$, then $$S-zS=(1+z+z^2+\cdots+z^n)-(z+z^2+z^3+\cdots+z^{n+1})=1-z^{n+1}.$$
Therefore, $S=\frac{1-z^{n+1}}{1-z}$, with $z \not=1$. Equating both expressions of $S$, we have $$1+z+z^2+\cdots+z^n=\frac{1-z^{n+1}}{1-z}$$
Substitute $z=e^{i\theta}$, with $0 < \theta < 2\pi$, into the expression, a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/800489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Solving $df/dx = x\log x$ I have the following differential equation
$$\frac{df}{dx} = Kx\log x,$$
$K$ a constant. I'm wandering how one might solve for $f$.
| Integrating by parts we know that:
$$fg = \int f'g+\int fg'$$
Now take $f' = x,\; g=\log x$ (I use $e$ as the base for the logarithm). This means that $f= \frac{1}{2}x^2$ and $g'= \frac{1}{x}$ , so we get:
$$\int f'g = \int x \log x \;dx = \frac{1}{2}x^2\log x - \int \frac{1}{2}x^2\frac{1}{x}\;dx$$
$$= \frac{1}{2}x^2\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Show how to compute $2^{343}$ using the least multiplication. Show how to compute $2^{343}$ using the least multiplication.
| If you have $x$, you can calculate $x^7$ in four integer multiplications:
*
*$x \cdot x = x^2$
*$x^2 \cdot x^2 = x^4$
*$x^4 \cdot x^2 = x^6$
*$x^6 \cdot x = x^7$.
Since $343=7^3$, this means you can calculate $2^{343}$ in twelve multiplications: four to go from $2$ to $2^7$, four to go from $2^7$ to $(2^7)^7=2^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Double integral for $\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$ I'm trying to evaluate this
$$\int_{0}^{1} \int_{-1}^{0} \frac {xy}{x^2 + y^2 + 1}\ dy\ dx$$
tried substition
$$ u = {(x^2+y^2+1)}^{-1} \ \ du = \ln {(x^2+y^2+1)}$$
but du is not found in the given equation. I have the feeling that I sh... | $$\int_0^1\int_{-1}^0\frac{xy}{x^2+y^2+1}dydx = \int_0^1x\int_{-1}^0y(x^2+y^2+1)^{-1}dydx\\= \int_0^1\frac{x}{2}\int_{-1}^0(x^2+y^2+1)^{-1}(2y)dydx \\= \int_0^1\frac{x}{2}\log(x^2+y^2+1)_{y=-1}^{0} dx \\= \int_0^1\frac{x}{2}(\log(x^2+1)-\log(x^2+2))dx \\= \frac{1}{4}\int_0^1\log(x^2+1)(2x)dx-\frac{1}{4}\int_0^1\log(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
First order recurrence relation I have to solve this relation:
$$a_1 = k \\ a_n = \frac{10}{9} a_{n-1} + k + 1 - n$$
(k is constant)
How can I do it??
| Use the generating function $A(x) = \sum_{n \ge 1}a_nx^n$ to capture the sequence $\{a_n\}$.
We are given (after adjusting indices $n \rightarrow n + 1$)
$$a_{n+1} = \frac{10}{9}a_n + k - n \text{ for } n \ge 1$$
Multiply by $x^n$ throughout and sum for all values of $n$ for which the recurrence holds.
$\begin{eqnarray... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/802635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Cant solve this differential equation using laplace transforms $$\left\{\begin{array}{ccc}
y''(t) &=& x(t) - 2y(t)\\
x''(t) &=& - 2x(t) + y(t)
\end{array}\right.$$
$y(0)=1,\;x(0)=1,\; y'(0) = \sqrt3,\; x'(0) = -\sqrt3$.
I am trying to solve this differential system using laplace transforms but the terms make it very ... | Taking the LT of both equations and initial values yields
$$(s^2+2) Y(s)-X(s) = s+\sqrt{3}$$
$$(s^2+2) X(s)-Y(s) = s-\sqrt{3}$$
Solution yields
$$X(s) = \frac{s}{s^2+1} - \frac{\sqrt{3}}{s^2+3} $$
$$Y(s) = \frac{s}{s^2+1} + \frac{\sqrt{3}}{s^2+3} $$
Inversion of these, individually, yields
$$x(t) = \cos{t} - \sin{\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/804010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Using the substitution $u=x^3$, find the general solution of $xy''+y'+9x^5=0$. Using the substitution $u=x^3$, find the general solution of $xy''+y'+9x^5=0$.
I have no idea about above question? Could somebody suggest me a solution or resource for such problems?
| HINT:
$$u=x^3\implies\frac{du}{dx}=3x^2$$
Using Chain Rule, $$\frac{dy}{dx}=\frac{dy}{du}\cdot\frac{du}{dx}=3x^2\cdot\frac{dy}{du}$$
$$\implies x\frac{dy}{dx}=3u\cdot\frac{dy}{du}$$
Differentiating wrt $x$ $$x\frac{d^2y}{dx^2}+\frac{dy}{dx}=3\left[\frac{du}{dx}\cdot\frac{dy}{du}+u\cdot\frac{d\left(\dfrac{dy}{du}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
A logarithmic integral $\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx$ How to prove the following
$$\int^1_0 \frac{\log\left(\frac{1+x}{1-x}\right)}{x\sqrt{1-x^2}}\,dx=\frac{\pi^2}{2}$$
I thought of separating the two integrals and use the beta or hypergeometric functions but I thought these are ... | Denote the integral as $I$. Define
$$ I(a)=\int_0^1 \frac{\log\left(\frac{1+ax}{1-x}\right)}{x\sqrt{1-x^2}} dx. $$
Then $I(-1)=0$, $I(1)=I$ and
\begin{eqnarray}
\frac{\partial I(a)}{\partial a}&=&\int_0^1 \frac{1}{(1+ax)\sqrt{1-x^2}}dx\\
&=&\int_0^{\frac{\pi}{2}} \frac{1}{1+a\sin t}dt \quad(x=\sin t).
\end{eqnarray}
Fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/805893",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 10,
"answer_id": 2
} |
Proving $\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1$ $A,B,C$ are positive reals with product 1. Prove that $$\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1$$
How can I prove this inequality. I just need a hint to get me started. Thanks
| Given that (w.l.o.g.) $0\lt A\le B\le C, A\cdot B\cdot C=1,$ we wish to show
$$\left(A-1+\frac1B\right)\left(B-1+\frac1C\right)\left(C-1+\frac1A\right)\leq1\tag 1$$
First, noting that $A-1+\dfrac 1B=AC+A-1$ and also $AC+A-1=(A-1)(C+1)+C,$ we transform $(1)$ as follows:
$$(AC+A-1)(AB+B-1)(BC+C-1)\\
=(A-1)(C+1)(B-1)(A+1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/806673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Power Factoring Contest Question The question was as follows:
Compute the smallest positive integer $n$ such that $n^n$ has at least $1,000,000$ positive divisors.
I did some work, finding that if $n=2^a*3^b*5^c*7^d$ then the $n^n= 2^{na}*3^{nb}*5^{nc}*7^{nd}$ and that the number of positive divisors is $(an+1)(bn+1)(c... | Already $n=2\cdot 3\cdot 17= {102}$ gives us $(n+1)^3>10^6$ divisors.
Any number with at least four distinct prime divisors is $\ge 2\cdot 3\cdot 5\cdot 7>102$ and can thus be ignored: We need to check only up to three prime divisors.
If $n=p^kq^m$, we get $(kn+1)(nm+1)$ divisors (the case $m=0$ is not excluded). To m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/806769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to prove this inequality $\frac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\frac{\pi}{n+1}}$
Let $a_{1},a_{2},\cdots,a_{n},n\ge 2$ be real numbers,show that
$$\dfrac{a_{1}a_{2}+a_{2}a_{3}+\cdots+a_{n-1}a_{n}}{a^2_{1}+a^2_{2}+\cdots+a^2_{n}}\le\cos{\dfrac{\pi}{n+1}}$$
I thi... | Here is a derivation using the calculus of variations.
Given that
$$
\sum_{k=1}^na_k^2=1\tag{1}
$$
we want to find the maximum of
$$
\sum_{k=1}^na_ka_{k-1}\tag{2}
$$
where $a_0=a_{n+1}=0$.
We want to find $a_k$ so that $(2)$ is stationary, that is,
$$
\begin{align}
0
&=\sum_{k=1}^na_k\,\delta a_{k-1}+a_{k-1}\,\delta a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/807326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 4,
"answer_id": 3
} |
Prove that the gradient of a unit vector equals 2/magnitude of the vector Let $\vec r=(x,y,z)$
Firstly find $\vec \nabla (\frac 1 r)$ where r is the magnitude of $\vec r$.
I think I've done this correctly to get $-x(x^2+y^2+z^2)^{-\frac32} \hat i-y(x^2+y^2+z^2)^{-\frac32} \hat j-z(x^2+y^2+z^2)^{-\frac32} \hat k$
Second... | I'm unsure of the context to this problem for you, but it first came up in my E+M physics class, but I'll try to create a more rigorous proof than I was initially shown.
First we have
$$
\vec{r} = (x,y,z) \implies r = \|r\| = \sqrt{x^2+y^2+z^2} \\
\implies \frac{\vec{r}}{r} =
\left(\frac{x}{\sqrt{x^2+y^2+z^2}}, \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/807853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Log trig integral with radical Show that:
$$\int_{0}^{\Large\frac{\pi}{2}}\frac{\log(\sin\theta)}{\sqrt{1+\sin^{2}\theta}}\ d\theta=-\frac{\Gamma^{2}\left(\dfrac14\right)\sqrt{\dfrac\pi2}}{16},$$ or some other equivalent form.
| Consider the integral
\begin{align}
I = \int_{0}^{\pi/2} \frac{\ln(\sin\theta)}{\sqrt{1+\sin^{2}\theta}} \ d\theta.
\end{align}
By making the substitution $t = \sin\theta$ the integral is reduced to
\begin{align}
I = \int_{0}^{1} \ln(t) \ (1-t^{4})^{-1/2} \ dt.
\end{align}
Making the substitution $u = t^{4}$ yields
\be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What is the difference to compute double integral? I see a double integral on web with a strange way to calculate. Please help me to make it clear.
Here is the integral:
$$\int_1^2\int_1^2(x+y)\ dx\ dy$$
As my way, I calculate it:
$$\int_1^2\int_1^2(x+y)\ dx\ dy= \int_1^2\left. {\left( {\frac{{{x^2}}}{2} + xy} \right)}... | The second one approach is using substitution $u=x+y\;\Rightarrow\;du=dx$, then the inner integral turns out to be
$$
\int_{1}^2 (x+y)\ dx = \int_{x=1}^2 u\ du = \left.\frac12u^2\right|_{x=1}^2 = \left.\frac12(x+y)^2\right|_{x=1}^2=\frac12\left[(y+2)^2-(y+1)^2\right].
$$
Similarly, the next integral is solved by using ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808536",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find values of the parameter a so that equation has equal roots. $x^2+2a\sqrt{a^2-3}x+4=0$
My final result was 2 and -0.5. Was it correct?
| To get a pair of equal roots, we need to have $2a\sqrt{a^2 - 3} = 4$.
$$\begin{align} 2a\sqrt{a^2-3}=4 &\iff \sqrt{a^4 - 3a^2}= 2 \\ \\ &\iff a^4 - 3a^2 = 4\\ \\ & \iff a^4 - 3a^2 - 4 = 0 \\ \\ & \iff (a^2 - 4)(a^2 +1) = 0\end{align} $$
$a^2 + 1 \geq 1$, so can never be zero. So we solve $$a^2 - 4 = (a+2)(a-2)=0$$ $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/808989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Asymptotic expansions for the roots of $\epsilon^2x^4-\epsilon x^3-2x^2+2=0$ I'm trying to compute the asymptotic expansion for each of the four roots to the following equation, as $\epsilon \rightarrow 0$:
$\epsilon^2x^4-\epsilon x^3-2x^2+2=0$
I'd like my expansions to go up through terms of size $O(\epsilon^2)$.
I´ve... | Two of the roots will be near the roots $1$ and $-1$ of $-2x^2 + 2$, the other two will be near $\infty$. Solving numerically for $\epsilon = .01$,
those roots are approximately $-100$ and $200$.
So it looks like you want $x \sim c \epsilon^{-1}$. Indeed with $x = y/\epsilon$ the equation becomes $y^4 - y^3 - 2 y^2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/810031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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Solving the Integral using $\ln|u|$ Can Someone help me solve this
$$
\int\frac{19\tan^{-1}x}{x^{2}}\,dx
$$
We have been told to use $\ln|u|$ and $C$.
Thanks!
| Using integration by parts we see:
$$ \int \frac{ \arctan(x) }{x^2} dx = -\frac{\arctan(x) }{x} + \int \frac{dx}{ (x^2+1) x} $$
Now if we use a partial fraction decomposition we obtain:
$$\int \frac{dx}{ (x^2+1) x} = \int \left ( \frac{1}{x} - \frac{x}{x^2 +1} \right ) dx $$
From here it should be clear that
$$\int \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/813209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is there always an alternative way to compute integral other than complex integral? Sometimes we have to compute integrals that are not easy to calculate so that we need to depend on the method of complex integrals like the residue method. But I became curious about possibility of alternative method of evaluation of in... | For $1 <a <2$,
$$ \begin{align} \int_{0}^{\infty} x^{1-a} \cos (wx)\ dx &= w^{a-2} \int_{0}^{\infty} u ^{1-a} \cos (u) \ du \\ &= \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty} \int_{0}^{\infty} \cos (u) \ t^{a-2} e^{-ut} \ dt \ du
\\ &= \frac{w^{a-2}}{\Gamma(a-1)} \int_{0}^{\infty}t^{a-2} \int_{0}^{\infty} \cos (u)e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/814854",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Compute and find 2009th decimal(2009th digit after the point), without automation, the following sum Compute and find 2009th decimal of (2009th digit after the point), without automation, the following sum
$$\frac{10}{11}+\frac{10^2}{1221}+\frac{10^3}{123321}+ \cdots +\frac{10^9}{123456789987654321}$$
| Let
$$a_n=\sum\limits_{b=1}^n\frac{10^b}{12\ldots bb\ldots 21}$$
where $b$ is a digit less than or equal to $9$.
We prove inductively that $a_n=1-\dfrac{1}{\underbrace{11\ldots11}_{n+1}}$.
Base case: $\dfrac{10}{11}=1-\dfrac{1}{11}$
Now assume
$$a_k=1-\dfrac{1}{\underbrace{11\ldots11}_{k+1}}$$
then
$$\begin{align}
a_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/816589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Why does $\frac{(x^2 + x-6)}{x-2} = \frac{(x+3)(x-2)}{x-2}$? I'm not the best at algebra and would be grateful if someone could explain how you can get from,
$$\frac{x^2 + x-6}{x-2}$$
to,
$$\frac{(x+3)(x-2)}{x-2}$$
| Try working backwards. Notice that the denominator does not change, so let's focus on the numerator:
$$(x+3)(x-2) = x\cdot x-2x+3x-2\cdot 3 \,\,\,\,\,\,\text{FOIL}.$$
If we look at like terms (meaning equal powers of $x$), we see that we have two pieces that have $x$ that we can simplify: $-2x+3x = (-2+3)x = x.$ As for... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/817424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
How to find the following sum? $\sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} - \frac{1}{{4n + 7}}} \right)} $ I want to calculate the sum with complex analysis (residue)
$$
1 - \frac{1}{7} + \frac{1}{9} - \frac{1}{{15}} + \frac{1}{{17}} - ...
$$ $$
1 + \sum\limits_{n = 0}^\infty {\left( {\frac{1}{{4n + 9}} -... | Here is a method without complex analysis. I use the following two:
$$\int_0^1 x^{4n+8}\,dx=\frac{1}{4n+9}$$
$$\int_0^1 x^{4n+6}\,dx=\frac{1}{4n+7}$$
to get:
$$\sum_{n=0}^{\infty} \left(\frac{1}{4n+9}-\frac{1}{4n+7}\right)=\int_0^1 \sum_{n=0}^{\infty} \left(x^{4n+8}-x^{4n+6}\right)\,dx=\int_0^1 \frac{x^8-x^6}{1-x^4}\,d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/817911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
Limits of trig functions How can I find the following problems using elementary trigonometry?
$$\lim_{x\to 0}\frac{1−\cos x}{x^2}.$$
$$\lim_{x\to0}\frac{\tan x−\sin x}{x^3}.
$$
Have attempted trig identities, didn't help.
| HINT :
\begin{align}
\lim_{x\to 0}\frac{1−\cos x}{x^2}&=\lim_{x\to 0}\frac{1−\cos x}{x^2}\cdot\frac{1+\cos x}{1+\cos x}\\
&=\lim_{x\to 0}\frac{1−\cos^2 x}{x^2(1+\cos x)}\\
&=\lim_{x\to 0}\frac{\sin^2 x}{x^2(1+\cos x)}\\
&=\lim_{x\to 0}\frac{\sin x\cdot\sin x}{x\cdot x\cdot(1+\cos x)}\\
&=\lim_{x\to 0}\frac{\sin x}{x}\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$ Find the value of $\sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7}$
My approach :
I used $\sin A +\sin B = 2\sin(A+B)/2\times\cos(A-B)/2 $
$\Rightarrow \sin\frac{2\pi}{7}+\sin\frac{4\pi}{7}+\sin\frac{8\pi}{7} = 2\sin(3\pi)/7\times\co... | You can evaluate this by using complex methods.
Let $\alpha=e^{2\pi i/7}$. Then
$$\sin\frac{2\pi}{7}=\frac{\alpha-\alpha^{-1}}{2i}\ ,\quad
\sin\frac{4\pi}{7}=\frac{\alpha^2-\alpha^{-2}}{2i}\ ,\quad
\sin\frac{8\pi}{7}=\frac{\alpha^4-\alpha^{-4}}{2i}\ .$$
Now let $S$ be the sum of these three numbers. Write $S$ in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/818749",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 4
} |
Finding the equation of the tangent I was working on the following question:
The curve $C$ has equation $y=(x+3)^2$ and the point $A$, with $x$ coordinate $-5$, and lies on $C$.
* $a)$ Find the equation of the tangent line to $C$ at $A$, giving your answer in the form $y= mx+c$
* $b)$ Another point $B$ also lies on ... | The slope of the tangent line, at $A$ is equal to $y\prime(-5)$. Then you can just plug that back into the equation $y-y_1=m(x-x_1)$ and solve for $y$.
$$\frac{d}{dx}(x+3)^2 = 2(x+3) = 2x + 6 \implies y\prime(-5) = -10+6 = -4$$
$$y-y(-5) =y\prime(-5)(x+5)$$
$$y-4 = -4(x+5) \implies y-4=-4x-20$$
$$\boxed{y=-4x-16}$$
For... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Answer to simple algebraic formula manipulation I have to show that
$2y+(x + 1) = 3 \cdot 2^{x+1}− (x + 1) − 2$
is equal to
$y=3⋅2^x−x−2$
I can get this far:
$2y+(x+1)=3⋅2^{x+1}−(x+1)−2$
$2y+(x+1)=3⋅2^{x+1}−x−1−2$
$2y+(x+1)=3⋅2^{x+1}−x−3$
$2y+x+1=3⋅2^{x+1}−x−3$
$2y=3⋅2^{x+1}−x−3−1−x$
$2y=3⋅2^{x+1}−2x−4$
Now I sh... | I think you missed one $x$ to be an exponent. Go ahead and divide by $2$ to get:
$$
y= 3\cdot \frac{2^{x+1}}2+x-2=3\cdot {2^{x}}+x-2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/819299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Factoring in the derivative of a rational function Given that
$$
f(x) = \frac{x}{1+x^2}
$$
I have to find
$$\frac{f(x) - f(a)}{x-a}$$
So some progressing shows that:
$$
\frac{\left(\frac{x}{1+x^2}\right) - \left(\frac{a}{1+a^2}\right)}{x-a} =
\frac{(x)(1+a^2)-(a)(1+x^2)}{(1+x^2)(1+a^2)}\cdot\frac{1}{x-a} =
\frac{x+xa^2... | By the quotient rule for the difference $\ f'(x)\, :=\, \dfrac{f(x)-f(a)}{x-a}$
$$\quad \begin{eqnarray} (g/h)'(x) &=\,\ & \dfrac{\color{#c00}{g'(x)} h(a) - g(a)\color{#0a0}{h'(x)}}{h(a)h(x)}\\
\begin{array}{l}\ g(x)=x\qquad\Rightarrow\,\color{#c00}{g'(x) = 1}\\ h(x) = 1+x^2\,\Rightarrow\,\color{#0a0}{h'(x) = x+a}\\\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/819527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Integrate : $\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}dx$
$$\int \frac{x^2}{(x\cos x -\sin x)(x\sin x +\cos x)}\ dx$$
My approach :
Dividing the denominator by $\cos^2x$ we get $\dfrac{x^2\sec^2x }{(x -\tan x)(x\tan x +1)}$ then
$$\int \frac{x^2\sec^2x}{x^2\tan x -x\tan^2x+x-\tan x}\ dx=\int \frac{x^2(1+\t... | $\displaystyle u=x\sin x+\cos x,v=x\cos x-\sin x \Rightarrow \displaystyle du=x\cos xdx,dv=-x\sin xdx$
$\begin{align}\displaystyle\int\frac{x^{2}dx}{(x\sin x+\cos x)(x\cos x-\sin x)}&=\int\left(\frac{x\cos x}{x\sin x+\cos x}-\frac{-x\sin x}{x\cos x-\sin x}\right)dx\\&=\displaystyle\int\frac{du}{u}-\int\frac{dv}{v}\\&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/821862",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 5,
"answer_id": 4
} |
Quick induction proof I am trying to prove $n^3<n!$ for all integers $n\geq 6.$ It would be trivial to do this by induction if $(n+1)^3<(n+1)n^3$ holds. I looked this up, and I found this is true for integers $n\geq 3.$ However, does anybody know of any more insightful method of proof for either statement, ie. not indu... | You can avoid induction if you allow division in the following way:
Let $n\geq 6$. Notice that
\begin{align*}
n!&=n(n-1)(n-2)(n-3)(n-4)(n-5)!\\
&\geq n(n-1)(n-2)3\cdot 2\cdot 1\tag{1}\\&=6n(n-1)(n-2).
\end{align*}
Now, we can make the following evaluation:
\begin{align*}
\dfrac{n^3}{n(n-1)(n-2)}=\dfrac{n}{n-1}\dfrac{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/825506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Random Variable Problems? Can someone show me how to work this out? I can't get the answers in the boxes.
| Let's $Y$ be the number of spots on the dice.
Then:
$$Y = 1 \Rightarrow X = (1 - 4)^2 = 9$$
$$Y = 2 \Rightarrow X = (2 - 4)^2 = 4$$
$$Y = 3 \Rightarrow X = (3 - 4)^2 = 1$$
$$Y = 4 \Rightarrow X = (4 - 4)^2 = 0$$
$$Y = 5 \Rightarrow X = (5 - 4)^2 = 1$$
$$Y = 6 \Rightarrow X = (6 - 4)^2 = 4$$
You can write that:
$$P(X = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/827065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove that $a+b$ is a perfect square $a, b, c$ are natural numbers such that $1/a + 1/b = 1/c$ and $gcd(a,b,c)=1$. Prove $a+b$ is a perfect square.
| $\dfrac{a+b}{ab} = \dfrac{1}{c} \to a+b = \dfrac{ab}{c}$. Thus:
$c|ab$. Write: $a = dp$, and $b = dq$ with $(p,q) = 1$. Thus: $\dfrac{ab}{c} = d^2\cdot \dfrac{pq}{c}$.
Claim: $c = pq$.
Proof: We have: $b-c = \dfrac{bc}{a}$. Thus: $a|bc \to dp|dqc \to p|qc$. Similarly:
$a - c = \dfrac{ac}{b}$. Thus: $b|ac \to dq|dpc \to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/828918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Prove infinite series $$
\frac{1}{x}+\frac{2}{x^2} + \frac{3}{x^3} + \frac{4}{x^4} + \cdots =\frac{x}{(x-1)^2}
$$
I can feel it. I can't prove it. I have tested it, and it seems to work. Domain-wise, I think it might be $x>1$, the question doesn't specify. Putting the LHS into Wolfram Alpha doesn't generate the RHS (it... | Let's put this together cleanly:
For the sum to converge, $|x| > 1$ which means
$$\sum_{n=0}^\infty x^{-n} = \frac{x}{x-1}$$
Squaring this gives
$$\frac{x^2}{(x-1)^2} = \sum_{n=0}^\infty \sum_{m=0}^\infty x^{-(n+m)} $$
$$ = \sum_{r=0}^\infty (r+1)x^{-r}$$
(because each value of $(n+m)$ occurs $(n+m+1)$ times in the i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/829168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
} |
Prove by induction that $(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}$ Prove by induction that $$(n+1)^2 + (n+2)^2 + ... + (2n)^2 = \frac{n(2n+1)(7n+1)}{6}.$$
I got up to:
$n=1$ is true, and assuming $n=k$ prove for $n=k+1$.
Prove...
$$\frac{k(2k+1)(7k+1)+6(2(k+1))^2}{6} = \frac{(k+1)(2k+3)(7k+8)}{6}$$
I... | Inductive step spelled out in detail:
\begin{align*}
&\hphantom{=}\frac{n(2n+1)(7n+1)}6+(2n+1)^2+(2n+2)^2-(n+1)^2\\
&=\frac{n(2n+1)(7n+1)}6+(2n+1)^2+3(n+1)^2\\
&=(2n+1)\left(\frac{n(7n+1)}6+2n+1\right)+3(n+1)^2\\
&=(2n+1)\frac{n(7n+1)+12n+6}6+3(n+1)^2\\
&=(2n+1)\frac{7n^2+13n+6 }6+3(n+1)^2\\
&=\frac{(2n+1)(n+1)(7n+6)}6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/831521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
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.