Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Integrate $\int\frac{x+1}{(x^2-x-5)^3} dx$ can you help with this math problem? I dont know how to start? There is any good method? $$\int\frac{x+1}{(x^2-x-5)^3} dx$$
Thank you
| As in my other answer,
$$\int \frac{x+1}{\left(x^2-x-5\right)^3}\, dx=\frac{1}{2}\left(\int \frac{d\left(x^2-x-5\right)}{\left(x^2-x-5\right)^3}+\int\frac{3}{\left(x^2-x-5\right)^3}\, dx\right)$$
$$=\frac{1}{2}\left(\frac{1}{-2\left(x^2-x-5\right)^2}+3\cdot 2^5\int \frac{d(2x-1)}{\left((2x-1)^2-21\right)^3}\right)$$
Le... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$(a+b)^2+4ab$ and $a^2+b^2$ are both squares I cannot find a complete answer to the following problem (this is the source):
Q. Find all positive integers $(a,b)$ for which $(a+b)^2+4ab$ and $a^2+b^2$ are both squares.
Just something: clearly if $(a,b)$ works then $(a/c,b/c)$ works as well, where $c$ is the greatest... | Let $q^2 = (a+b)^2$, $p^2 = a^2 + b^2$ and $r^2 = (a+b)^2 + 4ab$. Note that $r^2 + p^2 = 2q^2$. Now let $X = r+p$, $Y = r - p$ and $Z = 2q$.
You can easily show that $r^2 + p^2 = 2q^2$ if and only if $X^2 + Y^2 = Z^2$.
So, each solution can be found as follows:
*
*Pick a Pythagorean triple $(X, Y, Z)$ in which $Z$ i... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Equation with different bases (exponential) I seem to be stuck with the following equation right here:
$$2^x + 2^{x+1} = 3^{x+2} + 3^{x+3}$$
| Notice, $$2^x+2^{x+1}=3^{x+2}+3^{x+3}$$
$$2^x+2\cdot 2^x=3^2\cdot 3^x+3^3\cdot 3^x$$
$$2^x+2\cdot 2^x=9\cdot 3^x+27\cdot 3^x$$
$$3\cdot 2^x=36\cdot 3^x$$ $$\frac{3^x}{2^x}=\frac{1}{12}$$
$$\left(\frac{3}{2}\right)^x=\frac{1}{12}$$
$$x\ln\frac{3}{2}=-\ln 12$$
$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{x=-\fr... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Limit of function $x^2-x\cot\left(\frac{1}{x}\right)$ How To compute $\lim_{x \to \infty} x^2 -x \cot(1/x)$? Wolfram says it is $\frac{1}{3}$ and I know it is supposed to converge to a number other than 0 but I keep getting infinity.
| Near $x=0$,
$$
\begin{align}
\tan(x)
&=x+\frac{x^3}3+O\left(x^5\right)\\
&=x\left(1+\frac{x^2}3+O\left(x^4\right)\right)
\end{align}
$$
therefore,
$$
\begin{align}
\cot(x)
&=\frac1x\left(1-\frac{x^2}3+O\left(x^4\right)\right)\\
&=\frac1x-\frac x3+O\left(x^3\right)
\end{align}
$$
So as $x\to\infty$,
$$
\cot\left(\frac1x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1561829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to Evaluate $\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} .$? How to find $$\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} .$$
In fact,
\begin{align*}
&\int_0^1 {\frac{{u\arcsin u}}{{{u^4} + 2{u^2} + 13}}du} \\
=& \int_0^1 {\frac{{u\arcsin u}}{{\left( {{u^2} - \sqrt {2\sqrt {13} - 2} u + \sqrt... | This is probably not an answer
Being too lazy to attack such monsters, I should use Taylor series $$\sin^{-1}(u)=\sum_ 0^\infty\frac{(2n)!}{4^n(n!)^2(2n+1)}u^{2n+1}\qquad (|u|\leq 1)$$ and perform the long division by the denominator. This would lead to $$\frac{{u\sin^{-1} u}}{{{u^4} + 2{u^2} + 13}}=\frac{u^2}{13}+\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1562214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Let $X$ ~ Geometric with $p=\frac{1} {4}$ and $Y$ ~ Uniform on $(1,2,3,4)$. Given X and Y are independent, I need to find $VAR(2X + 2Y)$.
Does independence mean that $VAR(2X + 2Y) = VAR(2X) + VAR(2Y)$? If so, then $VAR(X)=\dfrac{1-p} {p^2} = \dfrac{\dfrac {3} {4}} {\dfrac{1} {16}}=12$. And $VAR(Y)= \dfrac {1} {4}(1^2 +... | $$Var(2X+2Y) =Var(2(X+Y)) = 4Var(X+Y) = 4[Var(X) + Var(Y)]$$
Or:
If X and Y are independent, then 2X and 2Y are independent.
$$Var(2X+2Y)=Var(2X)+Var(2Y)=4Var(X)+4Var(Y)$$
Assuming your $Var(X)$ and $Var(Y)$ are correct, we have $Var(2X+2Y)=53.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Finding $\lim_{n\to\infty}\sum_{k=1}^n \sin(\frac\pi{\sqrt{n^2+k}})$ Find $$\lim_{n\to\infty}\sum_{k=1}^n \sin(\frac\pi{\sqrt{n^2+k}})$$
I think this has to be evaluated with riemann sums. I tried using $\sin x\approx x$ as n is big, but that got me to $\int_0^1 (1+1/x)^{-1/2}$ but that gives me the wrong answer. Any s... | Use the inequality
$$x - \frac{1}{6}x^3 \leq \sin x \leq x, \quad x \in [0, \pi/2).$$
For sufficiently large $n$, the summand satisfies:
$$\frac{\pi}{\sqrt{n^2 + k}} - \frac{1}{6}\frac{\pi^3}{(n^2 + k)^{3/2}} \leq \sin\left(\frac{\pi}{\sqrt{n^2 + k}}\right) \leq \frac{\pi}{\sqrt{n^2 + k}} \tag{1}.$$
For $k \in \{1, \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1563746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is
$a^2-b^2$ will be a perfect square if the ratio of $a$ and $b$ is
Options are: A) $1:2$, B) $3:4$, C) $5:4$, D) $4:5$, E) None of these
So my first observation was that $|a|$ has to be greater than $|b|$. So only options that suffice this condition a... | In a multiple choice question it can often be quicker to answer it via trial and error as you know one of the choices must be correct. If we ignore the multiple choice part then are looking at: $a^2-b^2=x^2$. This is simplify Pythagoras' Theorem rewritten so if you know solutions to that you have additional solutions. ... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the limit of the recursive sequence Definition of the sequence :
$$a_1=a;\\a_2=b;\\$$ and $$\ \ \ a_{n+2}={{a_n+a_{n+1}}\over2}$$ for $n\geq 1$.
Find the limit of this sequence in terms of $a$ and $b$.
Now in this case , taking $\lim_{n\rightarrow \infty}$ on both sides does not help at all . So what I did w... | If you would like to have a closed form for the individual terms of the recurrence, you can use the method of generating functions. This first step places your sequence as the terms of a power series $G(x) = \sum_{n=1}^\infty a_n x^n$, and then you generate a separate representation using the recursion. It's a long pro... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is it possible to estimate the sign of real part of eigenvalues of a 10 by 10 matrix only by observing all the entries? I have a symbolic 10 by 10 matrix. It is not difficult to get the eigenvalue expressions by using Matlab. But the expressions of some eigenvalues are too long to be analyzed. I was wondering if there ... | Arrange the rows and columns in the order $1,8,4,5,2,3,9,10,6,7$. Your matrix is permutation-similar to the direct sum of
$$
A=\left[\begin{array}{cc|cc}
-k_1-p_1\lambda &-\lambda^2p_1 &0 &0\\
-1 &-\lambda &0 &0\\
\hline
-\frac{\sin{\theta}}{2} &0 &0 &-\frac{\omega}{2}\\
\frac{\cos{\theta}}{2} &0 &\frac{\omega}{2} &0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1566797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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There is no square number that is $3 \mod 4$ Prove that the following cannot be true:
There is no square number that is $3 \mod 4$
$x^2 \equiv 3 \mod 4$, I started with examples:
$1^2 \mod 4=1$
$2^2 \mod 4=0$
$3^2 \mod 4=1$
I am more interested in the proof but Im not seeing how to do it.
| If $x$ is even then $x=2k$ so the square of $x$ is equal $4k^2$ and leaves no remainder when divided by $4$.
If $x$ is odd then $x=2k-1$ so the square of $x$ is equal $4k^2-4k+1$ and leaves a remainder $1$ when divided by $4$.
So we have either $x^2 \equiv 0 \mod 4$ or $x^2 \equiv 1 \mod 4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1571402",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find $x,y,z>0$ such that $x+y+z=1$ and $x^2+y^2+z^2$ is minimal How can I find $3$ positive numbers that have a sum of $1$ and the sum of their squares is minimum?
So far I have:
$$x+y+z=1 \qquad \implies \qquad z=1-(x+y)$$
So,
$$f(x,y)=xyz=xy(1-x-y)$$
But I'm stuck from here. Hints?
| By C-S $$3(x^2+y^2+z^2)=(1+1+1)(x^2+y^2+z^2)\geq(x+y+z)^2=1.$$
Thus, $x^2+y^2+z^2\geq\frac{1}{3}$.
The equality occurs for $x=y=z=\frac{1}{3}$, which is also the answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1572128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 10,
"answer_id": 9
} |
8-puzzle which has the numbers in order but has gap in between If the numbers of the 8-puzzle are all in order, but the blank tile is somewhere in between, is this puzzle solvable? (for example, $\begin{bmatrix} 1 & 0 & 2 \\ 3 & 4 & 5 \\ 6 & 7 & 8 \end{bmatrix}$, where 0 represents the blank tile)
| Count how far the gap tile is displaced. In your case it has been moved up 2 and over 1. So that is a displacement of 3. Call that "odd" or 1.
Now do that for every square.
1 is in place. So that's displaced 0.
2 is over 1 so that's displaced 1.
3 is down one and over 3 so that's 3. odd. 1.
4 is over 1. 1.
5 is ove... | {
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"url": "https://math.stackexchange.com/questions/1576254",
"timestamp": "2023-03-29T00:00:00",
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On $p^2 + nq^2 = z^2,\;p^2 - nq^2 = t^2$ and the "congruent number problem" (Much revised for brevity.) An integer $n$ is a congruent number if there are rationals $a,b,c$ such that,
$$a^2+b^2 = c^2\\
\tfrac{1}{2}ab = n$$
or, alternatively, the elliptic curve,
$$y = x^3-n^2x = x(x-n)(x+n)\tag1$$
is solvable in the rati... | (Edited 2018): After 2 years finally found the answer.
I. For Question $1$, the answer is yes.
If one has a solution to $x^4-n^2y^4 = z^2$, then,
$$p^2-nq^2 = (z^2-2nx^2y^2)^2\\
p^2+nq^2 = (z^2+2nx^2y^2)^2$$
where $p=x^4+n^2y^2$ and $q=2xyz$.
II. For Question $2$, the answer is no.
For example, in this Mathworld li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1576861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find all primes $p$ such that $ p^3-4p+9 $ is a perfect square.
Find all primes $p$ such that $ p^3-4p+9 $ is a perfect square.
I tried a few different values for $p$, namely $2,3,5,7,$ and $11$. The prime $p =2,7,11$ all worked but $p =13$ didn't so it makes me wonder. How can I find all primes such that it is a pe... | If $p^3-4p+9=n^2$ then $n^2\equiv 9 \pmod{p}$ and $\pm n \equiv 3 \pmod{p}$ since $n^2-9\equiv 0$ can only have two roots modulo a prime.
By choosing the appropriate sign we can write
$$
\begin{align}
n^2=(-n)^2 &= (kp+3)^2 \\
p^3-4p+9 &= k^2p^2+6kp+9 \\
0 &= p^2-k^2p-(6k+4) \\
p & = \frac{k^2\pm \sqrt{k^4+24k+16}}{2}
... | {
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"timestamp": "2023-03-29T00:00:00",
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What are all the concordant forms $n$ such that $a^2+b^2 = c^2,\,a^2+nb^2=d^2$ for $n<1000$? Part I. The list of congruent numbers $n<10^4$ such that the system,
$$a^2-nb^2 = c^2$$
$$a^2+nb^2 = d^2$$
has a solution in the positive integers is known (A003273)
$$n = 5, 6, 7, 13, 14, 15, 20, 21, 22, 23, 24, 28, 29, 30, 31... | For the system of equations:
$$\left\{\begin{aligned}&a^2+b^2=c^2\\&a^2+qb^2=w^2\end{aligned}\right.$$
Solution 1:
$$a=p-s,\quad b=2t,\quad c=p+s$$
$$w=\mp2q+p+s\pm2$$
$$q=(p\pm1)(s\pm1)$$
$$ps=t^2$$
Solution 2:
$$a=t^2-1,\quad b=2t,\quad c=t^2+1$$
$$w=3t^2-1$$
$$q=2t^2-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1584485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Integrate: $\int \frac{2x^2-3x+8}{x^3+4x} \, dx$ $$\int \frac{2x^2-3x+8}{x^3+4x}\,dx$$
My main problem is calculating the $B$ and $C$. This is the algebra part. Thus, what is a technique I can use that is in line with what I did to calculate $A$?
| Multiplying both sides of $$\frac{2x^2-3x+8}{x^3+4x}=\frac Ax+\frac{Bx+C}{x^2+4}$$
with $x(x^2+4)$, you can remove denominators, and get
$$2x^2-3x+8=A(x^2+4)+(Bx+C)x. $$
Then set $x=\;$ the poles of the fraction:
*
*$x=0$ yields $8=4A+0$, whence $A=2$.
*$x=2\mathrm i$ yields $-8-6\mathrm i+8=-6\mathrm i=0+(2B\mat... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to find the exponent of a prime in $n!$ Let the positive integer $n$ be written as powers of prime $p$ so that we have $n=a_kp^k+....+a_2p^2+a_1p+a_0,$ where $0\leq a_i<p.$ Show that the exponent of the highest power of $p$ appearing in the prime factorization of $n!$ is
$\frac{n-(a_k+....+a_1+a_0)}{p-1}$.
I know t... | $$
\begin{align}
\sum_{i=1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor
&= \left\lfloor \frac{n}{p} \right\rfloor + \left\lfloor \frac{n}{p^2} \right\rfloor + \dotsb + \left\lfloor \frac{n}{p^k} \right\rfloor + \sum_{i=k+1}^{\infty} \left\lfloor \frac{n}{p^i} \right\rfloor \\
&= \left\lfloor \frac{n}{p} \right\rf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1586557",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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If $a,b,c>0$ and $a+b+c=1$, prove $\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{\sqrt{abc}}{a+ba}\le 1+\frac{3\sqrt{3}}{4}$ If $a,b,c>0$ and $a+b+c=1$, then prove
$$\frac{a}{a+bc}+\frac{b}{b+ca}+\frac{\sqrt{abc}}{a+ba}\le 1+\frac{3\sqrt{3}}{4}$$
| HINT: set $a=xy,b=yz,c=zx$ and after this use the $\tan(\alpha/2)$ etc substitution4}
after simplification we get
$$\frac{1}{1+z^2}+ \frac {1}{1+x^2}+\frac {y}{1+y^2}\le 1+\frac{3\sqrt{3}}{4}$$
now set $$x=\tan(\alpha/2),y=\tan(\beta/2),z=\tan(\gamma/2)$$
after simplification we get
$$\cos(\gamma)+\cos(\alpha)+\sin(\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1588051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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An infinite nested radical problem From this link, problem 36, I found that
$$\sqrt{4+\sqrt{4+\sqrt{4-\sqrt{4+\sqrt{4+\sqrt{4-...}}}}}}=2\left(\cos{\dfrac{4\pi}{19}}+\cos{\dfrac{6\pi}{19}}+\cos{\dfrac{10\pi}{19}}\right).$$
The signs : + + - + + - + + - ... .
How to prove it?
Furthermore, how to represent $\sqrt{7+2\sqr... | I can´t solve in detail this question, however I want to write a remark about because I think it might be of interest to some people (we disregard the convergence whose existence is implicit by the statement assuming to be true).
(1) Let $E$ be equal to the RHS; we have the challenging equality $$E^3+E^2-6E-7=0$$ (you ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Better proof for $\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$ It's required to prove that
$$\frac{1+\cos x + \sin x}{1 - \cos x + \sin x} \equiv \frac{1+\cos x}{\sin x}$$
I managed to go about out it two ways:
*
*Show it is equivalent to $\mathsf{true}$:
$$\frac{1+\cos x + \sin x}{1... | Observe $$(1 - \cos x + \sin x)(1 + \cos x) = (1 - \cos^2 x) + (1 + \cos x)\sin x = \sin^2 x + (1 + \cos x)\sin x = (1 + \cos x + \sin x)\sin x,$$ from which the result immediately follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1590059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 5,
"answer_id": 1
} |
Integral $\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$ It's a follow-up to my previous question.
Can we find an anti-derivative
$$\int\arcsin x\cdot\ln^3x\,dx$$
or, at least, evaluate the definite integral
$$\int_0^{1/2}\arcsin x\cdot\ln^3x\,dx$$
in a closed form (ideally, as a combination of elementary functions and polyloga... | As can be checked by differentiation, there is an antiderivative continuous on $(0,1)$:
$$\begin{align}&\int\arcsin x\cdot\ln^3x\,dx=\\
&\hspace{1cm}\frac32\left[\operatorname{Li}_3\left(\frac{\alpha}2\right)-\operatorname{Li}_3\left(\frac\beta2\right)\right]+3\,(2-\ln x)\cdot\operatorname{Li}_2\left(\frac\alpha2\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1591155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
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Groups with Complex Numbers I'm working on a quick proof related to the complex numbers.
Take $G$ to be the group of all nonzero complex numbers under
multiplication ($\circ$). Let $H$ be the set of of complex
numbers such that the sum of squares of the two real parts of the
complex number is equal to 1. I want to sho... | Remember that the identity under multiplication is 1. So the inverse of $z \in \mathbb C$ is the reciprocal $\frac{1}{z}$. Since $|z| = 1$, we know $\left|\frac{1}{z}\right| = \frac{1}{1} = 1$.
We can compute this explicitly if you are not comfortable with the absolute value of a complex number. Let $z = x+iy$ such tha... | {
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"url": "https://math.stackexchange.com/questions/1592210",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $(1+2+...+k)^2 = 1^3 + ... + k^3$ using induction I need to prove that
$$(1+2+{...}+k)^2 = 1^3 + {...} + k^3$$
using induction.
So the base case holds for $0$ because $0 = 0$ (and also for $1$: $1^2 = 1^3 = 1$)
I can't prove it for $k+1$ no matter what I try! Can you give me a hint?
| We have:
$$(1+2+...+k+(k+1))^2=(1+2+...+k)^2+(k+1)^2+2(k+1)(1+2+...+k)$$
So by induction hypothesis:
$$(1+2+...+k+(k+1))^2=1^3+2^3+...+k^3+(k+1)^2+2(k+1)(1+2+...+k)$$
We know $1+2+...+k = \frac{k(k+1)}{2}$.(You can prove it by induction easily). Therefore:
$$(1+2+...+k+(k+1))^2=1^3+2^3+...+k^3+(k+1)^2+k(k+1)^2=1^3+2^3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
} |
How to compute $\lim_{x \to 0} (\frac{x^5 e^{-1/x^2}+x/2 - \sin(x/2))}{x^3})$? I have a problem with this limit. I have no idea where is the problem.
Can you correct my mistake? Thanks
$$\lim\limits_{x \to 0} \left(\frac{x^5 e^\frac{-1}{x^2}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}\right)$$
I used the developments of McLa... | One may write, as $x \to 0$,
$$
\begin{align}
\frac{x^5 e^{-\frac1{x^2}}+\frac{x}{2} - \sin(\frac{x}{2})}{x^3}&=\frac{x^5 e^{-\frac1{x^2}}+\frac{x}{2} - \left(\frac{x}2-\frac{x^3}{48}+O(x^5)\right)}{x^3}\\\\
&=\frac{x^5 e^{-\frac1{x^2}} + \frac{x^3}{48}+O(x^5)}{x^3}\\\\
&=\frac{x^2 e^{-\frac1{x^2}} + \frac{1}{48}+O(x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1592872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Find $a,b$ for which $xyz+z=a,\quad xyz^2+z=b,\quad x^2+y^2+z^2=4$ has unique solution Find all values of $a,b$ for which the system of equations $xyz+z=a,\quad xyz^2+z=b,\quad x^2+y^2+z^2=4$ has only one real solution.
$xyz+z=a$
$xyz^2+z=b$
So,$\frac{xy+1}{xyz+1}=\frac{a}{b}$
I can think no method by so as this syste... | If $(x, y, z)$ is a solution, then $(y, x, z)$ is a solution. Hence if there is a unique solution, we need $x=y$. This reduces the problem to $2$ variables.
$$\begin{align*}
x^2z+z&=a\\
x^2z^2+z&=b\\
2x^2+z^2&=4\\
\end{align*}$$
If $(x, x, z)$ is a solution, note that $(-x, -x, z)$ is also a solution. Hence, $x=0$ if t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1593466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Evaluate the following integral I got stuck with the integral below. I have tried to make it look like the derivative of arctan.
$$\int \frac{2-x}{x^2-x+1}\,dx$$
Thank you!
| HINT: $$\int \frac{2-x}{x^2-x+1}\ dx=\frac{1}{2}\int \frac{3-(2x-1)}{x^2-x+1}$$
$$=\frac{1}{2}\int \frac{3}{x^2-x+1}\ dx-\frac 12\int \frac{(2x-1)}{x^2-x+1}\ dx$$
$$=\frac{3}{2}\int \frac{1}{\left(x-\frac 12\right)^2+\frac{3}{4}} dx-\frac 12\int \frac{d(x^2-x+1)}{x^2-x+1}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1598638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Help finding the determinant of a 4x4 matrix? Sorry for the lack of notation but the work should be easy to follow if you know what you are doing. Okay my problem is that the book says it can be done by expanding across any column or row but the only way to get what the book does in their practice example is to choose ... | You are all wrong. The determinant of this matrix is 380.
You can check it here:
https://www.wolframalpha.com/input/?i=matrix+determinant&assumption=%7B%22F%22,+%22Determinant%22,+%22detmatrix%22%7D+-%3E%22%7B+%7B+5,-7,2,2%7D,%7B0,3,0,-4%7D,+%7B+-5,-8,0,3%7D,%7B+0,5,0,6%7D%7D%22&assumption=%7B%22C%22,+%22matrix+determi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1599082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 7,
"answer_id": 6
} |
Prove that $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$
Let $a,b,c$ be three nonnegative real numbers. Prove that $$a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2} \geq 2(ab+bc+ca).$$
It seems that the inequality $a^2+b^2+c^2 \geq ab+bc+ca$ will be of use here. If I use that then I will get $a^2+b^2+c^2+3\sqrt[3]{a^2b^2c^2}... | Yet another way in which this can be shown using Schur's inequality in tandem with the AM-GM inequality is as follows:
$$
a^2+b^2+c^2+3(a^2b^2c^2)^{1/3}\geqslant a^{2/3}b^{4/3} + a^{4/3}b^{2/3} +b^{2/3}c^{4/3} + b^{4/3}c^{2/3} + a^{2/3}c^{4/3} + a^{4/3}c^{2/3} \\[2ex]= 2\left({a^{2/3}b^{4/3} + a^{4/3}b^{2/3}\over 2} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1601427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Not Understanding a specific substitution rule I was given the question,
If $f(3x+5) = x^2-1$, what is $f(2)$? I am trying to understand the reasoning why $3x+5$ is set equal to $2$.
| Let's say $u=3x+5$. Then, $x^2-1=(\frac{u-5}{3})^2-1$. $\Rightarrow$ $f(u)=(\frac{u-5}{3})^2-1$ $\Rightarrow$ $f(x)=(\frac{x-5}{3})^2-1$. Now we can easily see that $f(2)=0$ . However, instead of doing all these complicated stuff, we could have made the following deduction: $3x+5=2$ $\Rightarrow$ $x=-1$ $\Rightarrow$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1602407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 2
} |
Value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Find the value of $X^4 + 9x^3+35X^2-X+4$ for $X=-5+2\sqrt{-4}$
Now the trivial method is to put $X=5+2\sqrt{-4}$ in the polynomial and calculate but this is for $2$ marks only and that takes a hell lot of time for $2$! So I was thinking may be there is some tri... | You could consider carrying out polynomial long division, dividing the given polynomial by the minimal polynomial of $-5+4i$, which is $x^2+10x+41$. Viz:
$x^4+9x^3+35x^2-x+4=(x^2-x-4)(x^2+10x+41)+160$
We know that if $x=-5+4i$, then $x^2+10x+41$ vanishes, so the given polynomial evaluates to $160$ there.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1603412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Some equations from Russian maths book. Could you please help me with solving these equations. I would like to solve them in the most sneaky way. All of the exercises in this book can be solved in some clever way which I can't often find.
$$
\frac{(x-1)(x-2)(x-3)(x-4)}{(x+1)(x+2)(x+3)(x+4)} = 1
$$
$$
\frac{6}{(x+1)(x+2... | factorizing the left minus the right-hand side we get $$-\frac{20 x \left(x^2+5\right)}{(x+1) (x+2) (x+3) (x+4)}=0$$
can you proceed from here?
making the same with the second equation we get $$-\frac{x (x+3) \left(x^2+3 x-16\right)}{(x-1) (x+1) (x+2) (x+4)}=0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1604713",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 4
} |
Pythagorean triples with the same c value $a^2 + b^2 = c^2$
There are, Primitive Pythagorean Triples, that share the same c value. For example,
$63^2 + 16^2 = 65^2$ and $33 ^2 + 56^2 = 65^2$.
I have been trying to figure out why the following theorem for finding such triples works.
Take any set of primes. Ex: $5,13,1... | 1) $(a + bi)(a - bi) = a^2 + b^2$ always.
2) $(a + bi)(c + di)(e + fi) = g + hi \ne (a + bi)(c + di)(e - fi) = j + ki$
yet $(a - bi)(c - di)(e - fi) = g - hi \ne (a - bi)(c - di)(e + fi) = j - ki$
so
3) $(a + bi)(c + di)(e + fi)(a - bi)(c - di)(e - fi) = (g+hi)(g - hi) = g^2 + h^2 =K$
and
4) $(a + bi)(c + di)(e - fi)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Closed form of $\sum\frac{1}{k}$ where $k$ has only factors of $2,3$
Consider the set containing $A$ all positive integers with no prime
factor larger than $3$, and define $B$ as
$$
B= \sum_{k\in A} \frac{1}{k}
$$
Thus, the first few terms of the sum are:
$$
\frac{1}{1} +\frac{1}{2} +\frac{1}{3} +\frac{1}{4... | So there are obviously only two primes, $2$ and $3$. Thus, we have to consider the sum
$$\sum_{j\ge0}\sum_{i\ge0}\frac{1}{2^i3^j}$$
I would break this down further into the following:
$$\sum_{i\ge0}\frac{1}{3^0\cdot2^i}
+\sum_{i\ge0}\frac{1}{3^1\cdot2^i}
+\sum_{i\ge0}\frac{1}{3^2\cdot2^i}+\cdots$$
Now the first term is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1605124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find the minimum of $\sum\limits_{cyc} {\sqrt{\frac a{2(b+c)}}}$ with $a,b,c \gt 0$ As said in the title, I have to find the minimum of the following:
$$\sum_{cyc} {\sqrt{\frac a{2(b+c)}}} $$ with $a+b+c>0$
In my very last attempt, I tried to work it out using AM-GM:
Since $$\sqrt{\frac a{2(b+c)}}= \frac 1{\sqrt 2}\f... | Stronger inequality:
With $a, b, c> 0$: $$\sqrt{\frac{a}{b+ c}}+ \sqrt{\frac{b}{c+ a}}+ \sqrt{\frac{c}{a+ b}}\geq \sqrt{2+ 2.\frac{abc}{\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )}}$$
Proof:
Thus, we have:
$$\left ( a+ b \right )\left ( b+ c \right )\left ( c+ a \right )+ abc= \left ( a+ b+ c \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1610947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the value of the expectation $E(X)$ of the following distribution.
Let $X$ be a random variable having the distribution function
$$F(x)=\begin{cases}
0 &\text{if } x<0\\
\frac{1}{4} & \text{if } 0\leq x<1\\
\frac{1}{3} & \text{if } 1\leq x<2\\
\frac{1}{2} & \text{if } 2\leq x<11/3\\
1& \text{if } x\geq 11/3\en... | This thing isn't a continuous random variable, so the formula you're using doesn't work.
I recommend drawing this thing out.
The expectation is
$$E[X] = (1-0)\left(1-\frac{1}{4}\right)+(2-1)\left(1-\frac{1}{3}\right) +\left(\frac{11}{3}-2\right)\left(1-\frac{1}{2}\right)=2.25$$
Alternatively, the distribution of $X$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1611736",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Second derivative of $x^3+y^3=1$ using implicit differentiation I need to find the $D_x^2y$ of $x^3+y^3=1$ using implicit differentiation
So,
$$
x^3 + y^3 =1 \\
3x^2+3y^2 \cdot D_xy = 0 \\
3y^2 \cdot D_xy= -3x^2 \\
D_xy = - {x^2 \over y^2}
$$
Now I need to find the $D_x^2y$.
I am pretty sure that means the second d... | $1=x^3+y^3\implies 0=3 x^2+3 y^2 y',$ so $$0=x^2+y^2 y'.$$ Differentiate this to get $$0=2 x +2 y y'^2+y^2 y''.$$ Therefore for $y\ne 0$ we have $$y''=-y^{-2}(2 x+2 y y'^2).$$ From the first differentiation we have $$y'=-x^2/y^2.$$ Therefore for $y\ne 0$ (equivalently, for $x\ne 1$),$$y''=-y^{-2}(2 x +2 y y'^2)=- y^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1612571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove that $\sqrt{x+y+z}\ge{\sqrt{x-1}+\sqrt{z-1}}$. Let $x,y,z>1$ such that $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=2$. Prove that
$$\sqrt{x+y+z}\ge{\sqrt{x-1}+\sqrt{z-1}}$$.
I took $x=\sec^2{a}$, $y=\sec^2{b}$, $z=\sec^2{c}$ but it was not useful. Then I took $x-1=a$ and similarly $b$ and $c$. On simplifying I got
$$b+... | Applying Cauchy the following is true. $$(x+y+z)(\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z})\ge (\sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1})^2$$
However, $\frac{x-1}{x}+\frac{y-1}{y}+\frac{z-1}{z}=1$ from the condition in the problem.
$\therefore$ $\sqrt{x+y+z}\ge \sqrt{x-1}+\sqrt{y-1}+\sqrt{z-1}$.
However $\sqrt{x-1}+\sqrt{y-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Evaluating the Fourier coefficients of $abs(x)$ Let's get started:
$$\hat f(n) = \frac{1}{2\pi}\int_0^{2\pi} |x|e^{-inx} dx$$
since $|x|$ is an even function:
$$= \frac{1}{\pi}\int_0^{\pi} xe^{-inx} dx$$
Integration by parts yields:
$$e^{-inx}\Big|_0^{\pi} + \frac{1}{in} \int_0^\pi e^{-inx} dx = (-1)^n - 1 + \frac{1}{i... | Since $|x|$ is positive in $[0, 2 \pi]$, then
\begin{align}
\hat f(n) &= \frac{1}{2\pi} \, \int_0^{2\pi} |x|e^{-inx} \, dx \\
&= \frac{1}{2\pi} \, \int_{0}^{2\pi} x \, e^{-i n x} \, dx \\
&= \frac{1}{2\pi} \, \left[ \int_{0}^{\pi} x \, e^{-i n x} \, dx + \int_{\pi}^{2\pi} x \, e^{-i n x} \, dx \right] \\
&= \frac{1}{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1613645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Induction proof without summation I have to prove this induction:
$\dfrac{1}{(n+1)}+\dfrac{1}{(n+2)}+\dots+\dfrac{1}{2n} = \dfrac{1}{(1\times2)}+\dfrac{1}{(3\times4)}+\dots+\dfrac{1}{(2n-1)\times2n}$
Can someone help me with it?
| If you want to use the summation symbol, note that
$$
a_n=\frac{1}{n+1}+\frac{1}{n+2}+\dots+\frac{1}{2n}=
\sum_{k=1}^n\frac{1}{n+k}
$$
Therefore
$$
a_{n+1}=\sum_{k=1}^{n+1}\frac{1}{n+1+k}
$$
The right-hand side can be written
$$
b_n=\sum_{k=1}^{n}\frac{1}{2k(2k-1)}
$$
So your task is to prove $a_n=b_n$. The case $n=1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Calculte indefinte integral of $\int \frac{dx}{\sqrt{(x+2)(3-x)}}$ I have to calculate $$\int \frac{dx}{\sqrt{(x+2)(3-x)}}$$.
I tried to use -
$\int u'v = uv - \int v'u$, but im pretty stuck.
Thanks.
| \begin{align}
\int \frac{dx}{\sqrt{(x+2)(3-x)}}
=& \ \text{sgn} (\frac12-x)\int \frac{d\left(\sqrt{(x+2)(3-x)}\right)}{\sqrt{\frac{25}4- (x+2)(3-x)}}\\
=& \ \text{sgn} (\frac12-x) \sin^{-1}\bigg(\frac25 \sqrt{(x+2)(3-x)}\bigg)+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1615788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Linear algebra: Solving a system of equation matrix with a variable as coefficient. Let's consider this augmented matrix
$$\left(\begin{array}{ccc|c}
3 &-6 &6 &15\\
-2 &7 &a &-25\\
2 &-6 &6 & 20
\end{array}\right)$$
I'm trying to figure how to solve a matrix like this when there is $a$ as one of the coeffic... | Procede with the usual row reduction, trying to avoid pivoting on the element with $a$ as long as possible:
$$\left(\begin{array}{ccc|c}
3 &-6 &6 &15\\
-2 &7 &a &-25\\
2 &-6 &6 & 20
\end{array}\right)
\xrightarrow{\begin{matrix}R1~/~3\\R3~/~2\end{matrix}}
\left(\begin{array}{ccc|c}
1 &-2 &2 &5\\
-2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then $\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$
Prove that if $x,y,z$ are positive real numbers and $ xy+xz+yz = 1$ then $$\sqrt{x}+\sqrt{y}+\sqrt{z} > 2$$
I am having a hard time relating the square roots in the inequality to the given condition. I was think... | My attempt (not a full solution). The best I was able to achieve is this inequality:
$$\sqrt{x}+\sqrt{y}+\sqrt{z}>\sqrt{2+\sqrt{3}}=1.93185\dots$$
We will use the mean inequalities:
$$\sqrt{\frac{x^2+y^2+z^2}{3}} \geq \frac{x+y+z}{3} \geq \sqrt[3]{xyz} \geq \frac{3}{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}$$
By squaring ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1616429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Probability using indicator function There are 10 pairs of different socks in a drawer. You take 5 pairs out of it. What is the expected value of matching socks?
Is it OK to use the indicator function as follows?
Let $X$ be the number of pairs that match. Let $I_i$ be the indicator function that takes $1$ if the the $i... | I am not familiar with the method of indicator functions, but we may check your solution using basic counting methods. We count cases for each specific number of matching pairs. We will use derangements in our solution, which is notated by $!n.$
The total number of ways to draw the socks is $\frac{\dbinom{10}{2}\dbinom... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617169",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the least value of $4\csc^{2} x+9\sin^{2} x$
Find the least value of $4\csc^{2} x+9\sin^{2} x$
$a.)\ 14 \ \ \ \ \ \ \ \ \ b.)\ 10 \\
c.)\ 11 \ \ \ \ \ \ \ \ \ \color{green}{d.)\ 12} $
$4\csc^{2} x+9\sin^{2} x \\
= \dfrac{4}{\sin^{2} x} +9\sin^{2} x \\
= \dfrac{4+9\sin^{4} x}{\sin^{2} x} \\
= 13 \ \ \ \ \ \ \ \... | Let $y=4\csc^2x+9\sin^2 x$
$$y'=-8\csc^2x\cot x+9\sin2 x$$
$$y''=-8\csc^2x(-\csc^2 x)-8\cot x(-2\csc^2 x\cot x)+18\cos2 x$$
$$y''=8\csc^4x+16\csc^2\cot^2 x+18\cos2 x$$
for maxima or minima, $y'=0$ hence, $$-8\csc^2x\cot x+9\sin2 x=0$$
$$2\cos x\left(9\sin x-\frac{4}{\sin^3 x}\right)=0$$
$$\frac{\cos x}{\sin^3 x}\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1617363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to show $\frac{1}{2\sqrt{2} + \sqrt{3}} = \frac{2\sqrt{2} - \sqrt{3}}{5}$? Show that:
$$ \dfrac{1}{2\sqrt2+\sqrt3}=\dfrac{2\sqrt2-\sqrt3}{5}$$
So I multiplied everything by $\sqrt3$
Then I got
$$\frac{\sqrt{3}}{2\sqrt{2}+3}$$
Then multiply it by $\sqrt2$ to obtain
$$\frac{\sqrt{2}\sqrt{3}}{2 \cdot 3+3}$$
Which is ... | \begin{align}
\frac {1}{2\sqrt 2 + \sqrt 3}= \frac {2\sqrt 2-\sqrt 3}{5}&\iff \\\frac {1}{2\sqrt 2 + \sqrt 3}\cdot {2\sqrt 2 + \sqrt 3} = \frac {2\sqrt 2-\sqrt 3}{5}\cdot {(2\sqrt 2 + \sqrt 3)} &\iff\\
1=\frac {(2\sqrt2)^2-(\sqrt 3)^2}{5}&\iff\\
1=\frac {4\cdot 2-3}{5}&\iff 1=\frac 5 5.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1618339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Prove that if $(a^2+b^2+c^2+d^2)^2 > 3(a^4+b^4+c^4+d^4)$, then, using any three of them we can construct a triangle.
Prove that if $a,b,c,$ and $d$ are positive numbers and satisfy $(a^2+b^2+c^2+d^2)^2 > 3(a^4+b^4+c^4+d^4)$, then, using any three of them we can construct a triangle.
I find it hard to go from the give... | We'll prove that $a$, $b$ and $c$ are sides-lengths of a triangle.
Indeed, by C-S
$$3(a^4+b^4+c^4+d^4)=(2+1)(a^4+b^4+c^4+d^4)\geq\left(\sqrt{2(a^4+b^4+c^4)}+d^2\right)^2$$
which gives $a^2+b^2+c^2>\sqrt{2(a^4+b^4+c^4)}$ or
$$(a+b+c)(a+b-c)(a+c-b)(b+c-a)>0$$
and we are done because $a+b-c<0$ and $a+c-b<0$ impossible. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1620424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
What is the coefficient of $x^4$ in the expansion of $\sqrt[3]{1+x}$ Here's what I tried:
$$\sum_{n \ge0} {\frac{1}{3} \choose n} x^n= \sum_{n \ge0} = \frac{\frac{1}{3}!}{n!(n-\frac{1}{3})!}x^n=\sum_{n \ge0} \frac{(\frac{1}{3}-1)(\frac{1}{3}-2)\cdot ...\cdot(\frac{1}{3}-(n-1)) }{n!}x^n$$
What to do more, or is this all... | Hint: Supplement to the already given answer
If a series
\begin{align*}
A(x)=\sum_{n\geq 0}a_nx^n
\end{align*}
is stated, the coefficient of $x^k$ is $a_k$ without any notion of $x$. A convenient notation for the coefficient of $x^k$ is $[x^k]$.
Since the coefficient of $x^4$ should be obtained, an answer could be sta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1621435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Using contour integrals to evaluate sum - Problem calculating residues
"Compute $$\sum_1^{\infty} \frac{1}{{n^2}{(n+1)^2}}$$ using contour integration"
I have used the function $F(z) = \frac {\pi cot\pi z}{z^2(z+1)^2}$
Which has double poles at $z=0$ and $z=-1$
For the pole at $z=0$, if I calculate the residue by tak... | Consider
$$f(a) = \sum_{n=-\infty}^{\infty} \frac1{(n^2+a^2)[(n+1)^2+a^2]} $$
The sum we want will be $\frac12 \lim_{a \to 0} [f(a) - \frac{2}{a^2 (1+a^2)}]$. (These represent the $n=0$ and $n=-1$ terms in the sum that become singular as $a \to 0$.) We may evaluate this sum using the residue theorem. Recall that, f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Looking for "an easy to understand" proof for following Power series I'm looking for proof for the following Power series
$exp(X) = \sum_{k=0}^{n} \frac{X^{k}}{k!}$
If X is $A_{nxn}$ matrix, then prove the series is converge
Given
$\exp(X) = \sum_{k=0}^{n} \frac{X^{k}}{k!}$
$\sin(X) = \sum_{k=0}^{n} (-1)^{k}\frac{X^{2... | Better to proceed as follows.
Consider for example the series $\sum_{k=0}^\infty\frac1{k!}t^kA^k$, say with $t$ complex or if you prefer take $t\in\mathbb R$. You can verify that the all entries of the series are power series.
So, all that matters for convergence is the radius of convergence. It is easy to show that th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1622954",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to compute the integral of $ \frac{\sqrt{(x^2+1)}}{x^2}$? II have spent hours trying to find the right substitution but I have had no results.
I have tried using $x=\tan u$ but it did not give the result shown in my book.
| $$\int \frac{\sqrt{x^2+1}}{x^2}dx$$
Apply Integration By Parts:$\:\int \:uv'=uv-\int \:u'v$
$u=\sqrt{x^2+1},\:u'=\frac{x}{\sqrt{x^2+1}},\:\:v'=\frac{1}{x^2},\:\:v=-\frac{1}{x}$
Then
$$\int \frac{\sqrt{x^2+1}}{x^2}dx=-\frac{\sqrt{x^2+1}}{x}-\int \:-\frac{1}{\sqrt{x^2+1}}dx$$
$$=-\frac{\sqrt{x^2+1}}{x}-\int \:-\frac{1}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1623501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the maximum value of $(12\sin x-9\sin^{2} x)$
The maximum value of $(12\sin x-9\sin^{2} x)$
is equal to
$a.)\ 3 \\
\color{green}{b.)\ 4} \\
c.)\ 5 \\
d.)\ \text{none of these}$
As
$-1\leq \sin x\leq 1 ,\\
12\sin x-9\sin^{2} x \\
=12-9=3 \\
$
But the answer given is $4.$
I am looking for a short and simple w... | Hint: If you complete the square, you will get:
$$12\sin x - 9\sin^2 x$$
$$= 4-4 + 12\sin x - 9\sin^2 x $$
$$= 4-2^2 + 2 \cdot 2 \cdot 3\sin x - (3\sin x)^2 $$
$$= 4-(3\sin x - 2)^2.$$
Now, it is easy to determine the maximum value of this expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1625692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Proving Trig Identities (Complex Numbers)
Question: Prove that if $z = \cos (\theta) + i\sin(\theta)$, then
$$ z^n + {1\over z^n} = 2\cos(n\theta) $$
Hence prove that $\cos^6(\theta)$ $=$ $\frac 1{32}$$(\cos(6\theta)$ + $6\cos(4\theta)$ + $15\cos(2\theta)$ + $10$)
I learnt to prove the first part in another post l... | $n=1\implies2\cos x=z+\dfrac1z$
$$\cos^6x=\left(\dfrac{z+\dfrac1z}2\right)^6$$
$$64\cos^6x=\left(z+\dfrac1z\right)^6=z^6+\dfrac1{z^6}+\binom61\left(z^4+\dfrac1{z^4}\right)+\binom62\left(z^2+\dfrac1{z^2}\right)+\binom63$$
Can you take it from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1627410",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find $\lim\limits_{n\to\infty}\left(\frac{1}{\sqrt{n}}\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}\right)$ I don't know how to find the sum of $\sum\limits_{k=1}^{n}\frac{1}{\sqrt{2k}+\sqrt{2k+2}}$. After rationalization we have
$\left(\frac{1}{\sqrt{2}+\sqrt{4}}+\frac{1}{\sqrt{4}+\sqrt{6}}+...+\frac{1}{\sqrt{2... | Hint rationlize you will get a telescoping series for eg after rationalizing first two terms we get $\frac{\sqrt{4}-\sqrt{2}+\sqrt{6}-\sqrt{4}}{2}=\frac{\sqrt{6}-\sqrt{2}}{2}$ on rationalizing till $n$ see which terms remain then you will get your steps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1633010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find all functions $F(x)$ for which $F (x) + F ((x − 1)/x) = 1 + x$
Let $F (x)$ be the real-valued function defined for all real $x$ except for $x = 0$ and $x = 1$ and satisfying the functional equation $F (x) + F ((x − 1)/x) = 1 + x$. Find $F (x)$.
This functional equation looks like I could do an inverse substituti... | Solve the system of questions:
$$
\begin{align}
F(x) + F\left(\frac{x-1}{x}\right) &= 1+x \\
F\left(\frac{x-1}{x}\right) + F\left(\frac{1}{1-x}\right) &= \frac{2x-1}{x} \\
F\left(\frac{1}{1-x}\right) + F(x) &= \frac{2-x}{1-x}
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642298",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Show if $A^TA = I$ and $\det A = 1$ , then $A$ is a rotational matrix Show if $A^TA = I$ and $\det A = 1$ where
$ A =
\begin{bmatrix}
a & b \\
c & d
\end{bmatrix}
$, then $A =\begin{bmatrix}
\cos\theta & -\sin\theta \\
\sin\theta & \cos\theta
\end{bmatrix}$.
attempt:
Suppose $ A^TA =\begin{bmatri... | from $A^TA = I$, we konw $(a, c)$ and $(b, d)$ is orthogonal, for some $\theta$, we know $(a, c) = \sqrt{(a^2 + c^2)}(\cos\theta, \sin\theta) = (\cos\theta, \sin\theta)$, the unit vector orthogonal to $(a, c)$ is $(-\sin\theta, \cos\theta)$, so we know, for some scalar $k$, $(b, d) = k(-\sin\theta, \cos\theta)$. $k$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1642625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find parallel line value I've an homework problem that i'm unable to find the right answer.
The problem is:
The line $tx + sy = 2$ goes through point $(2,1)$ and is parallel to line $y = 8 -3x$, find the value of $t^2 + s^2$.
$ A. {32\over49}$ $B.{18\over49}$ $C.{36\over49}$ $D.{25\over49} $ $E.{40\over49} $
I was abl... | Rewrite $y = -3x + 7$ in the form $tx + sy = 2$: $$\begin{align*}y & = -3x + 7 \\ 3x + y & = 7 \\ \frac{6}{7} x + \frac{2}{7} y & = 2.\end{align*}$$ By comparison of the form of $\frac{6}{7} x + \frac{2}{7} y = 2$ to that of $tx + sy = 2$, we find $t = \frac{6}{7}$ and $s = \frac{2}{7}$. Therefore, $$t^2 + s^2 = {\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1643579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Can an infinite sum of irrational numbers be rational? Let $S = \sum_ {k=1}^\infty a_k $ where each $a_k$ is positive and irrational.
Is it possible for $S$ to be rational, considering the additional restriction that none of the $a_k$'s is a linear combination of the other ?
By linear combination, we mean there exists ... | Taking
$a_k = \frac{1}{\sqrt{k}} - \frac{1}{\sqrt{k+1}}$
we have
$\sum_{k=1}^\infty a_k = (1-\frac{1}{\sqrt{2}}) + (\frac{1}{\sqrt{2}} - \frac{1}{\sqrt{3}})+(\frac{1}{\sqrt{3}}-\frac{1}{\sqrt{4}}) +\dots = 1$
As per Mario Carneiro's suggestion in the comments, let us instead take
$a_k = \frac{1}{\sqrt{p_{k-1}}} - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1647409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "113",
"answer_count": 23,
"answer_id": 13
} |
Calculation of area in 2 definite integrals given function $y=x^2$ Here is a graph for $y=x^2$
Given that the area in blue is equal to the area in pink, find a in terms of b and solve for a.
My attempt:
From the graph I can see that :$a^2=b$ and $a=\sqrt b$
Since area blue is equal to area in pink:
$$\int^a_1 x^2 dx=\... | Note$$\bigg(\int_{1}^{1+\sqrt{3}} x^2 dx\bigg) -\bigg(\int_{1}^{(1+\sqrt{3})^2} \sqrt{y} dy \bigg) = -(3+2\sqrt{3}) \neq 0$$
So your teacher is wrong in this case, and you are right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1648784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) wi... | \begin{align}
\sin(x)+\cos(x) &= \frac 75 \\
\left(\sin(x)+\cos(x)\right)^2 &= \left( \frac 75 \right)^2 \\
1 + 2 \sin(x) \cos(x) &= \frac{49}{25} \\
\sin(x) \cos(x) &= \frac{12}{25}
\end{align}
\begin{align}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}
&= \frac{\sin(x) + \cos(x)}{\sin(x) \cos(x)}\\
&= \frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 6
} |
Solve equation $\frac{1}{x}+\frac{1}{y}=\frac{2}{101}$ in naturals My try was $$\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}=\frac{2}{101}\\x+y=2k,xy=101k\\x=2k-y\\y(2k-y)=101k\\2ky-y^2=101k\\y^2-2ky+101k=0\\y=k+\sqrt{k^2-101k}\\x=k-\sqrt{k^2-101k}$$
Now $\sqrt{k^2-101k}$ has to be either integer or rational,if it's an integ... | $1/x + 1/y = 2/101$
$(x+y)/xy = 2/101$
So $101|xy$. But 101 is prime so 101|x or 101|y or both.
Wolog symmetry assume 101|x.
Let $x = 101x'$.
$(101x' + y)/101x'y = 2/101$
$(101x' + y)/x'y = 2$
So $x'|y$. Let y = x'y'.
$(101 + y')/x'y' = 2$
So $y'|101$.
So $y' = 1$ of $y' = 101$.
So $102/x' = 2$ and $x'=51; y'=1$ or ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 3
} |
To prove $\sum_{n=0}^\infty \binom{r}{x}\binom{N-r}{n-x}=\binom{N}{n}.$ To prove $$\sum_{x=0}^n \binom{r}{x}\cdot \binom{N-r}{n-x}=\binom{N}{n}.$$
I tried comparing the coefficients of
$(1+x)^{(n+k)} = (1+x)^n(1+x)^k$
but couldn't reach the answer.
|
We obtain for $0\leq r\leq N$
\begin{align*}
(1+x)^{N}&=(1+x)^r(1+x)^{N-r}\\
&=\sum_{k=0}^r\binom{r}{k}x^k\sum_{l=0}^{N-r}\binom{N-r}{l}x^l\\
&=\sum_{n=0}^{N}\left(\sum_{{k+l=n}\atop{k,l\geq 0}}\binom{r}{k}\binom{N-r}{l}\right)x^n\\
&=\sum_{n=0}^{N}\left(\sum_{k=0}^n\binom{r}{k}\binom{N-r}{n-k}\right)x^n\\
\end{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1653104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Calculus, Finding Integral of a quotient Seems I have forgotten some basic integrating rules, but how do I go about finding the primitive function of $f(R) = \frac {R^2}{B+R^2}$?
| You have to consider three cases: $B=0$, $B<0$ and $B>0$.
If $B=0$, then for $R\ne 0$, the function becomes $f(R)=1$, and anti-derivative is given by
$$
F(R)=R+C,
$$
where $C$ is a constant.
If $B\ne 0$, you have to write
$$
f(R)=\frac{R^2}{B+R^2}=\frac{B+R^2-B}{B+R^2}=1-\frac{B}{B+R^2}.
$$
If $B<0$, then $B=-A^2$, and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657208",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $x,y,z>0$ and $x+y+z=1$ Then prove that $xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$
If $x,y,z$ are positive real number and $x+y+z=1\,$ Then prove that
$xy(x+y)^2+yz(y+z)^2+zx(z+x)^2\geq 4xyz$
Let $$f(x,y,z)=xy(x+y)^2+yz(y+z)^2+zx(z+x)^2$$
Then $$\frac{f(x,y,z)}{xyz} = \frac{(x+y)^2}{z}+\frac{(y+z)^2}{x}+\frac{(z+x)^... | First I edited your proof, you missed the factor $4$ in the answer. To use the AM-GM you can still use what you have there in the proof.
$$\dfrac{f(x,y,z)}{xyz} = \sum_{\text{cyclic}} \dfrac{(1-x)^2}{x}=\sum_{\text{cyclic}} \dfrac{1-2x+x^2}{x} = \sum_{\text{cyclic}} \dfrac{1}{x} - 6 + 1\geq \dfrac{9}{x+y+z} -6+1 = 4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Value of the product: $ \sqrt{2} \sqrt{2 - \sqrt{2}} \sqrt{2 - \sqrt{2 - \sqrt{2}}} \sqrt{2 - \sqrt{2 - \sqrt{2-\sqrt{2}}}} \cdots $ =? Let the recursive sequence
$$ a_0 = 0, \qquad a_{n+1} = \sqrt{2-a_n},\,\,n\in\mathbb N.
$$
T
Can we find the value of the product
$$
\prod_{n=1}^{\infty}{a_n}?
$$
Well, from here I don... | Squaring the infinite product we observe that
$$
2(2-\sqrt{2})\big(2-\sqrt{2-\sqrt{2}}\big)\Big(2-\sqrt{2-\sqrt{2-\sqrt{2}}}\Big)\cdots\\ =
2\cdot\frac{2}{2+\sqrt{2}}\cdot\frac{2+\sqrt{2}}{2+\sqrt{2-\sqrt{2}}}
\cdot\frac{2+\sqrt{2-\sqrt{2}}}{2+\sqrt{2-\sqrt{2-\sqrt{2}}}}\cdots \\
=\frac{4}{2+\sqrt{2-\sqrt{2-\sqrt{2-\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "31",
"answer_count": 2,
"answer_id": 0
} |
Value of $a^3+b^3+c^3$ when values of $a+b+c$, $abc$ and $ab+bc+ca$ are known. Is there a way to to find out what $a^3+b^3+c^3$ evaluates to, when the values of $abc$, $ab+bc+ca$ and $a+b+c$ are given?
Alternatively, is there a way to express $a^3+b^3+c^3$ in terms of the aforementioned expressions?
| We have
$$
(a + b + c)^3 = a^3 + b^3 + c^3 + 3(a+b+c)(ab + bc + ca) - 3abc
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
integrate $\int \frac{dx}{1+cos^2x}$
$$\int \frac{dx}{1+\cos^2x}$$
I used $\cos x=\frac{1-v^2}{1+v^2}$ and $dx=\frac{2dv}{1+v^2}$
and got $$2\int \frac{dv}{v^4-v^2+1}=2\int \frac{dv}{(v^2-\frac{1}{2})^2+(\frac{\sqrt{3}}{2})^2}=\frac{4}{\sqrt{3}}arctan(\frac{2v^2-1}{\sqrt{3}})+c$$
How to continue?
| An alternative method:
Multiplying through by $sec^2(x)$:
$$\int \frac{dx}{1+\cos^2x} = \int \frac{sec^2x}{1+sec^2x}dx$$
Defining $u = tanx$, $du = sec^2x$ $dx$, and using the identity $1+tan^2x = sec^2x$:
$$\int \frac{sec^2x}{1+sec^2x}dx = \int \frac{1}{u^2+2}du$$
The integral is now of the form of a very well-known... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1660682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Random vector with density on triangle and trapezoid I have a random vector $(X,Y)$ with the density $f_{XY}(x,y)=\tfrac{1}{5}$ on the trapezoid $T_1$ with vertex $(0,0),(0,1),(2,1),(3,0)$ and $f_{XY}(x,y)=\tfrac{3}{2}x$ on the triangle $T_2$ with vertex $(0,4),(1,4),(1,3)$.
I need to find:
*
*support and probabilit... | *
*I think your marginal pdf for $X$ is right.
*I think your cdf for it is wrong, although the question doesn't require it. It should be:
$$ F_X(x) =
\begin{cases}
\int_{0}^{x} (3u^2/2+ 1/5)\;du &= \dfrac{x^3}{2} + \dfrac{x}{5}, & \text{if $0\leq x\leq 1$} \\
\\
F(1) + \int_{1}^{x} (1/5)\;du &= \dfrac{x}{5} + \dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac... | You have made a few mistakes in your calculation.
First of all, since $v=\tan\frac{x}{2}$ with $\tan\frac{\frac{\pi}{4}}{2}=\tan\frac{\pi}{8}=\sqrt 2-1$, you should have
$$\int_{0}^{\frac{\pi}{4}}\frac{dx}{2+\tan x}=\int_{0}^{\color{red}{\sqrt 2-1}}\frac{\frac{2dv}{1+v^2}}{2+\frac{2v}{1-v^2}}$$
Also, you should have
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 4
} |
Evaluating the integral $\int \sqrt{1 + \frac{1}{x^2}} dx$
$$\int \sqrt{1 + \frac{1}{x^2}} dx$$
This is from the problem calculating the arc length of $y=\log{x}$.
I tried $x = \sinh{t}$ or $\frac{1}{x} = \tan{t}$ but all failed.
| Let $x=\sinh t$, then $dx=\cosh t \, dt$
\begin{align*}
\int \frac{\sqrt{1+x^{2}}}{x} \,dx &=
\int \frac{\cosh t}{\sinh t} \cosh t \, dt \\ &=
\int \frac{1+\sinh^{2} t}{\sinh t} \, dt \\ &=
\int (\sinh t+\operatorname{csch} t) \, dt \\ &=
\cosh t+\int \frac{dt}{2\sinh \frac{t}{2} \cosh \frac{t}{2}} \\ &=
\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
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Show that there exist no $a, b, c \in \mathbb Z^+$ such that $a^3 + 2b^3 = 4c^3$
Find all positive integer solutions of $a^3 + 2b^3 = 4c^3$.
Proof: There don't exist any integer solutions for the give equation.
Proof by the Well Ordering Principle.
Let $d$ be the set of all such possible combinations in form $(a, b, ... | There is another proof using the well ordering principle.
We use the following fact:
Third powers are can only be congruent to $0 $, $1 $ or $8 \mod 9 $.
Now, we see that the left hand side is congruent to $0$, $1$, $2$, $3$, $6$, $7$ or $8 \mod 9$, while the right hand side is congruent to $0, 4$ or $5 \mod 9$. S... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
| For the sake of being concise, I have chosen to omit most of the intermediary algebraic simplifications. Letting $u = x+2$, we get
$$\lim_{u \to 4} \frac{{u^{\frac 52} - 4u^{\frac 32} + 4\sqrt{u} - 8}}{4 - u^2 + 4u - 4}$$
By L'hopital, this is equal to
$$\lim_{u \to 4} \frac{\frac {5u^2 - 12u+4}{2 \sqrt{u}}}{4 - 2u} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Find all possible values of rank(A) as 'a' varies For an assignment I have to solve the following problem:
Find all possible values of rank(A) as 'a' varies:
\begin{bmatrix}a&2&-1\\3&3&-2\\-2&-1&a\end{bmatrix}
I came across several answers on google and stack (Find all possible values of rank(A) as a varies?, and simil... | Your first step is correct. Now multiply the second row by $\frac{2}{3}(\frac{1}{2}a-2)$ and sum to the third line, so you find an upper triangular matrix:
$$
\begin {bmatrix}
-2&-1&a\\
0&\frac{3}{2}&-2+\frac{3}{2}a\\
0&0&a^2-\frac{8}{3}a+\frac{5}{3}
\end{bmatrix}
$$
If $a^2-\frac{8}{3}a+\frac{5}{3} \ne 0$ the matrix ... | {
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"url": "https://math.stackexchange.com/questions/1667120",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $(n + 1)a^n < \frac{b^{n + 1} - a^{n + 1}}{(b-a)} < (n + 1)b^n$ $(b-a)(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n)$
$= (b^{n + 1} - ab^n) + (ab^n - a^2b^{n - 1}) + (a^2b^{n - 1} - a^3b^{n - 2}) + \ldots + (a^nb - a^{n + 1})$
$= b^{n + 1} - a^{n + 1}$, so
$(b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n) =... | If you look at $b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n$, you have $n+1$ terms. We have
$$
a^n + a^n +\cdots + a^n < b^n + b^{n - 1}a + b^{n - 2}a^2 + \ldots + a^n < b^n + b^n + \cdots + b^n
$$
because $a^n < a^kb^l < b^n$ if $k+l = n$ (and $k, l$ are both positive). Use the definition of multiplication to rewri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Find $\int \frac {x^2}{x^3+1} dx$. What is my mistake? $\int \frac {x^2}{x^3+1} dx$
$ u = 3x+1, du=3x^2 dx$
$\int \frac{3 du}{u} $
Am I wrong something? Why the answer is $\int \frac{du}{3u}$ instead of $\int \frac{3 du}{u} $ ?
Thank you.
| A more rigoures approach.
$$ \int \frac{x^2}{x^3+1} dx $$
$$ u = x^3 +1 $$
$$ \frac{du}{dx} = 3x^2 $$
so:
$$ \frac{1}{3} \int \frac{3x^2}{x^3+1} dx = \frac{1}{3} Ln |x^3 +1| + K$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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finding the maxima and minima for $f(x) = \sin\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right)$ the actual question is to show that the function$f(x) = \sin\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right)$ for all x belongs to $R$ has a minimum at $x=\left(\frac{\pi}{6}\right)$ and maxima at $x=... | Expanding first the original formula you can see better what is happening:
$$
f(x) = \sin\left(x-\frac{\pi}{3}\right)\sin\left(x+\frac{\pi}{3}\right) =
\frac14-\cos^2 x.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solving recurrence relation? Consider the recurrence relation $a_1=8,a_n=6n^2+2n+a_{n−1}$. Let $a_{99}=K\times10^4$ .The value of $K$ is______ .
My attempt :
$a_n=6n^2+2n+a_{n−1}$
$=6n^2+2n+6(n−1)^2+2(n−1)+a_{n−2}$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+a_1$
$=6n^2+2n+6(n−1)^2+2(n−1)+6(n−2)^2+2(n−2)+......+... | Let $b_{n}=\frac{a_{n}}{2}$.
Then $b_{n}=3n^2+n+b_{n-1}$.
For any $n$th degree polynomial $Q(x)$, one can find a $n+1$th degree polynomial $P(x)$ such that $P(x)-P(x-1)=Q(x)$.
Then, use this to note that $b_{n}-n(n+1)^2=b_{n-1}-(n-1)n^2$, since $n(n+1)^2-(n-1)n^2=3n^2+n$.
Thus $b_{n}-n(n+1)^2$ is a constant. But si... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Prove $\frac {1 + a + a^2 + \ldots +a^{n - 1} }{n} < \frac {1 + a + a^2 + \ldots +a^{n - 1} + a^n}{n + 1}$ if $1 < a$
Tried induction. Not sure where my mistake is, but what I did doesn't seem to make... | Another solution:
Your inequality is equivalent to $\frac{a^n-1}{n}<\frac{a^{n+1}-1}{n+1}$. But this is $\int_1^a x^{n-1}dx< \int_1^a x^n dx$, which is true because $x^{n-1} < x^n, \forall x \in (1, a]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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$\lim_{x \to 2} \frac{x^{2n}-4^n}{x^2-3x+2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^{2n}-4^n}{x^2-3x+2}$$
I tried to use
$$\lim_{x \to 2} \frac{(x^2)^n-4^n}{x^2-3x+2}$$ but i can't find anything special
| Using $a^n - b^n = (a-b)(a^{n-1}+a^{n-2}b+\cdots + b^{n-1})$, we get
$$
x^{2n}-4^n = x^{2n}-2^{2n}=(x-2)(x^{2n-1}+\cdots + 2^{2n-1}).
$$
Then divide numerator and denominator through $x-2$, then
\begin{align}
\lim_{x \to 2} \frac{x^{2n}-4^n}{x^2-3x+2}&=\lim_{x\to 2}\frac{(x-2)(x^{2n-1}+\cdots + 2^{2n-1})}{x^2-3x+2}\\
&... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Prove that $6$ divides $n^3+11n$? How can i show that
$$6\mid (n^3+11n)$$
My thoughts:
I show that
$$2\mid (n^3+11n)$$
$$3\mid (n^3+11n)$$
And
$$n^3+11n=n\cdot (n^2+11)$$
And if $n=x\cdot 3$ for all $x \in \mathbb{N}$ then:
$$3\mid (n^3+11n)$$
And if not:
The cross sum of$$n^2+11$$
is multiple of 3.
Can this be right o... | Here is a proof by induction,
*
*setting $n=1$, one should get
$$n^3+11n=1^3+11\cdot 1=12$$
obviously, the above number $12$ is divisible by $6$ hence statement is true for $n=1$.
*Assume that the number $n^3+11n$ is divisible by $6$ for $n=k$ then
$$k^3+11k=6\lambda$$or $$k^3=6\lambda-11k\tag 1$$
*Now, setting... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1673037",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 6
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Integrating a function of tan inverse How would I carry out the following integration?
$$\int_0^12\arctan x^2$$
I tried to substitute $x^2 = \tan\theta$
From there I wrote the integral as:
$$\int_0^{\pi/4}\theta\cdot \frac{\sec^2\theta}{\sqrt{\tan\theta}}\ d\theta$$
Now, is my only option to use integration by parts? O... | By using the Taylor series of the arctangent function, $\arctan(x)=\sum_{n\geq 0}\frac{(-1)^n x^{2n+1}}{2n+1}$, we get:
$$ \int_{0}^{1}\arctan(x^2)\,dx = 2\sum_{n\geq 0}\frac{(-1)^n}{(4n+2)(4n+3)}=2\sum_{n\geq 0}(-1)^n\left(\frac{1}{4n+2}-\frac{1}{4n+3}\right),$$
where the series
$$\sum_{n\geq 0}\frac{(-1)^n}{4n+3} = ... | {
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"url": "https://math.stackexchange.com/questions/1673170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 2
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Describe the set of all complex numbers $z$ such that $|z-a |+| z-b |=c$
Describe the set of all complex numbers $z$ such that :
$$|z-a |+| z-b |=c$$
where $a,b, c$ are real
At a simple look I immediately recognized that this is some ellipse because it's the same with the definition of the ellipse with two foci, ... | Hint:
$$|z-a|+|z-b|=c$$
Square
$$|z-a|^2+2|z-a||z-b|+|z-b|^2=c^2.$$
Move the squares from the LHS to the RHS and square again
$$4|z-a|^2|z-b|^2=(c^2-|z-a|^2-|z-b|^2)^2\\
=c^4+|z-a|^4+|z-b|^4+2|z-a|^2|z-b|^2-2c^2|z-a|^2-2c^2|z-b|^2.$$
Then,
$$0=c^4+|z-a|^4+|z-b|^4\color{red}{-2|z-a|^2|z-b|^2}-2c^2|z-a|^2-2c^2|z-b|^2\\
=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1674177",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Methods for efficiently factoring the cubic polynomial $x^3 + 1$ $$x^3 + 1$$
factors as
$$(x^2 - x + 1)(x + 1) .$$
It would have taken me a few minutes to identify this. What are the various approaches to determining rapidly that it is factorable, and factoring it?
| Here is one approach.
Note that $$x^3+1=x^3+3x^2+3x+1 -3x^2-3x=(x+1)^3-3x(x+1)=(x+1)((x+1)^2-3x)=(x+1)(x^2-x+1)$$
Another approach would be noticing that $x=-1$ is a solution to $x^3+1=0$. Then you can divide $x^3+1$ by $x+1$ with polynomial long division.
The third approach would be to set $x^3+1=(x+a)(x^2+bx+c)$ an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Find $d$ when $(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$ There exist constants $a$, $b$, $c$, and $d$ such that
$(\sin x)^7 = a \sin 7x + b \sin 5x + c \sin 3x + d \sin x$
for all angles $x$. Find $d$.
| This is the Fourier series for $f(x) = \sin^7 x$ on the domain $[0,2 \pi]$.
You have $\int_0^{2 \pi} \sin^7 x \sin x dx = d\int_0^{2 \pi} \sin x \sin x dx$.
Here is the answer:
This gives ${35 \over 64 } \pi = d \pi$.
To compute the integral,
\begin{eqnarray}
\int_0^{2 \pi} \sin^8 x &=& { 1\over (2i)^8} \int_0^{2 \p... | {
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"url": "https://math.stackexchange.com/questions/1675940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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triple integral $\iiint e^{-x^2-y^2-z^2+xy+yz+xz} \,dx\,dy \,dz$ I need to solve this
triple integral $$\iiint e^{-x^2-y^2-z^2+xy+yz+xz} \,dx\,dy \,dz$$
where $V$ is all $\Bbb R^3$
I've spent on this task a few hours,
ok firstly
$$e^{-x^2-y^2-z^2+xy+yz+xz}=e^{-1/2[(x-y)^2+(y-z)^2+(z-x)^2]}$$
I tried to use theorm of ... | To pursue the change-of-variables method further, note that with $a$ and $b$ defined as originally done, we have
$$
x^2 + y^2 + z^2 - xy - xz - yz = \frac{1}{2} \left[ (x - y)^2 + (y - z)^2 + (z - x)^2 \right] = \frac{1}{2} \left[ a^2 + b^2 + (a + b)^2 \right],
$$
since $a + b = x - z$. Rearranging, we then get
$$
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
evaluate $\sum^{\infty}_{n=2} \frac{3}{10^n}$
evaluate $$\sum^{\infty}_{n=2} \frac{3}{10^n}$$
I know I can factor out $$\sum^{\infty}_{n=2} \frac{3}{10^n}=3\sum^{\infty}_{n=2} \frac{1}{10^n}$$
And I know that the sequence converges $${{\large \frac{1}{10^{n+1}}}\over{\large \frac{1}{10^n}}}=\frac{1}{10}<1$$
But how d... | Notice, $$\sum_{n=2}^{\infty}\frac{3}{10^n}$$
$$=3\sum_{n=2}^{\infty}\frac{1}{10^n}$$
$$=3\left(\underbrace{\frac{1}{10^2}+\frac{1}{10^3}+\frac{1}{10^4}+\ldots}_{\text{sum of an infinite G.P.}}\right)$$
$$=3\left(\frac{\frac{1}{10^2}}{1-\frac{1}{10}}\right)$$
$$=3\left(\frac{1}{90}\right)=\color{red}{\frac 1{30}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1680556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
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How to find bases $\alpha$ and $\beta$ for the following $P_3(\Bbb{R})$ and $P_2(\Bbb{R})$? Suppose a linear transformation $T:P_3(\Bbb{R})\rightarrow P_2(\Bbb{R})$ has the matrix $A=\begin{bmatrix}
1 &2 & 0 & 0 \\
0 & 1 & 2& 1 \\
1 & 1 & 1 & 1 \\
\end{bmatrix} $ relative to the standard bases of $P_3(\Bbb{... | Note that for any matrix $A$ there is an invertible matrix $E$ such that
$\DeclareMathOperator{rref}{rref}EA=\rref A$. One computes $E$ by keeping track of the row reductions. In our case, we have
\begin{align*}
A &=
\left[\begin{array}{rrrr}
1 & 2 & 0 & 0 \\
0 & 1 & 2 & 1 \\
1 & 1 & 1 & 1
\end{array}\right] &
E &=
\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1681834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Calculate the sum $x+y+z$ If positive real numbers $x,y,z$ satisfying the conditions
$$x^2+xy+y^2=25, y^2+yz+z^2=49, z^2+zx+x^2=64$$
then calculate the sum $x+y+z.$
My attempt:
By subtracting equalities are obtained $$(z-x)(x+y+z)=24, (x-y)(x+y+z)=15.$$
And here I stopped. )-:
|
$\angle ADB=\angle CDB= \angle ADC=120^{\circ}$
$D -$ Torricelli Point
$$S_{ADB}+S_{CDB}+S_{ADC}=S_{ABC}$$
$$\frac 12 xy \sin 120^{\circ}+\frac 12 zy \sin 120^{\circ}+\frac 12 xz \sin 120^{\circ}=\sqrt{10 \cdot 3 \cdot 3 \cdot 2}$$
$$\frac {\sqrt 3}{4} xy +\frac {\sqrt 3}{4} zy +\frac {\sqrt 3}{4} xz=6\sqrt{5}$$
$$\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1682985",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $\sum_{n=-2}^{\infty}\cos^n x=8$, then find $x.$ Let $0<2x<\pi$. If $$\sum_{n=-2}^{\infty}\cos^n x=8,$$ then please find $x.$
I tried $\sum_{n=-2}^{\infty}\cos^n x=\frac{1}{1-\cos x}+\frac{1}{\cos x}+\frac{1}{\cos^2 x}=\frac{1+\cos x}{\cos^2 x \sin^2 x}=8.$ But I cant find $x.$
| Substituting y to cos(x), we have $\sum_{n=-2}^\infty y^n=-\frac{1}{y^2(y-1)}=8.$
The solutions are $1/2$, $\frac{1-\sqrt{5}}{4}$, $\frac{1+\sqrt{5}}{4}$
Then we have $\cos(x)=1/2$ so $x = \pi/3+2k\pi$ or $x=-\pi/3+2k\pi$ ($k \in \mathbb{Z}$)
the two other solutions give four values for x.
$\pi/5+2k\pi$, $-\pi/5+2k\pi$... | {
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"timestamp": "2023-03-29T00:00:00",
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Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.
The term inside the bracket is divisible by 6 since we have assumed t... | You just needed one more step.
$$\begin{align}n(n+1)(n+5) = & (n+1)(n+2)(n+6) \\ = & n^3+9 n^2+20 n+12 \\ = & (n^3+6n^2+5n)+3n^2+15n+12 \\ = & n(n+1)(n+5)+\underbrace{3n(n+5)}_{\star}+12\end{align}$$
$\bigstar$ If $n$ is even, then $3n$ is divisible by 6, otherwise $n$ is odd and $3(n+5)$ is divisible by 6.
Of course ... | {
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"url": "https://math.stackexchange.com/questions/1685246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$ Calculate:
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}$$
I don't know how to use L'Hôpital's Rule.
I tried to make $\tan x =\frac{\sin x}{\cos x}$ for the term ${\sqrt{\tan x}}... | You can first remove a few factors
$$\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt {\tan x}-\sin x \cdot \sqrt{\sin x}}{x^3 \sqrt{x}}\\
=\lim_{x \to 0^+}\frac{\tan x \cdot \sqrt{\tan x}}{x \sqrt{x}}\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}\\
=\lim_{x \to 0^+}\frac{1-\cos x\sqrt{\cos x}}{x^2}.$$
Then multiply by the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1687677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution,then all possible values of $a$ If $a\in R$ and the equation $-3(x-\lfloor x \rfloor)^2+2(x-\lfloor x \rfloor)+a^2=0$ has no integral solution, then all possible values of $a$ lie in the interval
$(A)(-1,0)\c... | EDIT (ELABORATION)
There is probably a typo here; it should say real solutions.
Assume it has solutions.
$$0\leq t<1, t=\frac{1\pm \sqrt{1+3a^2}}{3} \Leftrightarrow 1 \le \sqrt{1+3a^2} < 2 $$
From the fact that $1-\sqrt{1+3a^2}<0$ if $a \neq 0$. From here, we square both sides to get $$0 \le a^2 \le 1$$
This implies... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1691347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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If $tan^2 \theta = \frac{x}{y}$ how can we construct the angle $\theta$? If we are given the values of $x$ and $y$ and we know that $\tan^2 \theta = \dfrac{x}{y}$ is it possible for us to construct the angle $\theta$?
| Set $z=x/y$, for simplicity. Then
$$
z=\frac{\sin^2\theta}{\cos^2\theta}=\frac{1}{\cos^2\theta}-1
$$
so
$$
\cos^2\theta=\frac{1}{z+1}=\frac{y}{x+y}
$$
and
$$
\sin^2\theta=1-\frac{1}{z+1}=\frac{z}{z+1}=\frac{x}{x+y}
$$
Depending on which quadrant $\theta$ lives in, you can choose the signs in
$$
\cos\theta=\pm\sqrt{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1692267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Multiple radicals: $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$ $(\sqrt{10}+\sqrt{11}+\sqrt{12})(\sqrt{10}+\sqrt{11}-\sqrt{12})(\sqrt{10}-\sqrt{11}+\sqrt{12})(-\sqrt{10}+\sqrt{11}+\sqrt{12})$
I don't think multiplying these out will work... | HINT:
$$(a+b+c)(a+b-c)(b+c-a)(c+a-b)$$
$$=\{(a+b)^2-c^2\}\{c^2-(a-b)^2\}$$
$$=-(a^2+b^2+2ab-c^2)(a^2+b^2-2ab-c^2)$$
$$=-\{(a^2+b^2-c^2)^2-(2ab)^2\}$$
which is symmetric on expansion.
So, WLOG choose $\{a,b,c\}$ from $\{\sqrt{10},\sqrt{11},\sqrt{12}\}$ to find the same result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Find the Cartesian equation of the locus described by $\arg \left(\frac{z-2}{z+5} \right)= \frac{\pi}{4}$ My working:
$$ \frac{x + iy - 2}{x + iy + 5} $$
$$ \frac{(x - 2 + iy)(x+5-iy)}{(x + 5 + iy)(x+5-iy)} $$
$$ \frac{x^2+5x-ixy-2x-10+2iy+ixy+5iy+y^2}{x^2+5x-ixy+5x+25-5iy+ixy+5iy+y^2} $$
$$ \frac{x^2+3x-10+y^2+7iy}{x^... | Let me try this way -
The numerator is the line joining $Z(x,y)$ to $(2,0)$ while the denominator joins $Z(x,y)$ to $(-5,0)$ .
Geometrically, These two points are at a $45^\circ$ angle to each other.
Both the points should be on the locus. (Imagine a very small line at each point making $45^\circ$ angle with the ot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1693722",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Rationalize a surd $\frac{1}{1+\sqrt{2}-\sqrt{3}}$ How can I rationalize the following surd
$$\frac{1}{1+\sqrt{2}-\sqrt{3}}$$
What would be the conjugate of the denominator
| Use twice the conjugate:
$$\begin{aligned}
\frac{1}{1+\sqrt{2}-\sqrt{3}} &= \frac{1}{(1+\sqrt{2})-\sqrt{3}}\frac{(1+\sqrt{2})+\sqrt{3}}{(1+\sqrt{2})+\sqrt{3}}\\
&=\frac{1+\sqrt{2}+\sqrt{3}}{2 \sqrt{2}}\\
&= \frac{2+\sqrt{2}+\sqrt{6}}{4}
\end{aligned}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1695393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
The integral $\int\ln(x)\cos(1+(\ln(x))^2)\,dx$ Help with a integral calculus please!?
The equation is
$$\int\ln(x)\cos(1+(\ln(x))^2)\,dx$$
My teacher told me, i have to use substitution? but i can't still solve it.
I've been solving this last week but still i can't get the answer, please help me guys. Thanks!
| First, substitute via $u=1+\ln^2(x)$ and use the complex exponential form of cosine to get
\begin{align*}
\int \ln(x) \cos(1+\ln^2(x)) \,dx
&= \int \sqrt{u-1}\, \cos(u)\, \frac{\exp(\sqrt{u-1})}{2\sqrt{u-1}} \,du \\
&= \frac{1}{2} \int \cos(u)\,\exp(\sqrt{u-1}) \,du\\
&= \frac{1}{2} \int \frac{1}{2}[\exp(iu)+\exp(-iu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1696872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Real part of $(1+2i)^n$ Is it true that for all $n\in \mathbb{N}$, $n\ge 2$ we have
$$|\textrm{Re}((1+2i)^n)|>1?$$
I do know de Moivre's Theorem.
I do not know how to show that $|\sqrt{5}^n\cos(n\arccos\left ( \frac{1}{5} \right ))|>1$ because the value $\cos(n\arccos\left ( \frac{1}{5} \right ))$ can become (theoreti... | $a_{2n} = Re((1+2i)^{2n}) = Re((-3+4i)^n) = Re((1+4(-1+i))^n)$
Applying the Binomial expansion we get $(1+4(-1+i))^n$ as a series that converges $2$-adically :
$(1+4(-1+i))^n = 1 + 4(-1+i)\binom n1 -32i \binom n2 + \ldots + 4^k(-1+i)^k\binom nk + \ldots$
Since $v_2(4^k/k!) \ge k \to \infty$, this gives a $2$-adic power... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1697435",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 2
} |
Insights needed for the following Lagrange Multipler problem Find the point of the paraboloid $z = \frac{x^2}{4} + \frac{y^2}{25}$ that is closest to the point $(3, 0, 0)$.
So this seems like a pretty standard question of Lagrange multiplier, except I ran into some problems but cant figure out where I went wrong.
Atte... | You want to minimize $$F=(x-3)^2+y^2+z^2+\lambda \left(\frac{x^2}4+\frac{y^2}{25}-z\right)$$ So $$F'_x=\frac{\lambda x}{2}+2 (x-3)=0$$ $$F'_y=\frac{2 \lambda y}{25}+2 y=0$$ $$F'_z=2 z-\lambda=0$$ $$F'_\lambda=\frac{x^2}{4}+\frac{y^2}{25}-z=0$$ Eliminating $x,y,z$ from the first derivatives leads to $x=\frac{12}{\lam... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Quadratic Formula, nature of roots with Trigonometric Functions The original problem:
If $0\le a,b\le 3$ and the equation $$x^2+4+3\cos(ax+b)=2x$$ has at least one real solution, then find the value of $a+b$
$$$$
At first, on rearranging, I got the following expression:
$$x^2-2x+(4+3\cos(ax+b))=0$$ I thought this was... | Wih a little bit of manipulation we can rewrite your given equation as $$x^2 - 2x + 4 = -3\cos{\left( ax + b \right)}$$
Let $f(x) = x^2 - 2x + 4$. Differentiating to find the minimum, we get
$$f'(x) = 2x - 2 = 0 \implies x = 1$$
$$f''(x) = 2 > 0 \implies \text{minimum at } x = 1$$
The minimum value of LHS is thus $f(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1698449",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 2
} |
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.