Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Does $\sum^\infty_{s=1}\delta^{s-1}\frac{1}{1 + cs}$ have a closed-form expression?
I'm wondering whether
$$
\sum^\infty_{s=1}\delta^{s-1}\frac{1}{1 + c \cdot s},
$$
with $\delta \in (0,1)$ and $c\ge 0$ has a closed-form expression.
While I don't know of a general closed-form I can think if a special case:
Setting $c... | For $|x^c|< 1$, consider the geometric series $\sum\limits_{n=0}^{\infty}x^{nc}$. Then we know
$$\frac{1}{1-x^c}=\sum\limits_{n=0}^{\infty}x^{nc}.$$
So
\begin{align*}
\frac{x^c}{1-x^c}& =\sum_{n=0}^{\infty}x^{nc+c}\\
\int\frac{x^c}{1-x^c} \, dx& =\sum_{n=0}^{\infty}\frac{x^{nc+c+1}}{1+c(n+1)}\\
\int\frac{x^c}{1-x^c} \,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4187156",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integer solution of $abc=a+b+c+2$. Let $a,b,c$ be integers greater than $1$. I am trying to prove that $$abc\geq a+b+c+2$$
with equality if and only if $a=b=c=2$.
I can prove the inequality by using the fact that $ab\geq a+b$. Since $ab\geq 4$ and $c\geq 2$, it follows that $abc\geq 4c$ and $abc\geq 2ab$. Therefore $ab... | WLOG we assume $c \geq b \geq a \geq 2$
Then $3c \geq a + b + c$
We have,
$(ab-3) c \geq 2$
$abc \geq 2 + 3c \geq 2 + a + b + c$
For the equality case,
we are given, $abc = a + b + c +2 \leq 2 + 3 c$
$ \implies (ab-3) c \leq 2$
$ab - 3 \leq \dfrac{2}{c} \leq 1 \ ( \text {as } c \geq 2)$
That leads to $ab = 4, c = 2$.
i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4189755",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Is $R$ a linear transformation from $\Bbb R^2 $ to $\Bbb R^2$? Question :
Let $R$ be the rotation around the origin clockwise by $\frac{ \pi}{3}$. Is $R$ a linear transformation from $\Bbb R^2 $ to $\Bbb R^2$? Write down the corresponding matrix $Au = R(u)$. Compute $R(u)$ with $u=[1,3]$
My approach:
$$R(u)=\underbrac... | Your approach is correct for the last two questions. However, you have not proven that $R$ is a linear transformation. I will answer your question with detail.
A linear transformation from a $K$-vector space $V$ to another $K$-vector space $W$ is a function $R:V\to W$ such that the following two properties hold:
*
*$... | {
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"timestamp": "2023-03-29T00:00:00",
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Are there positive integers $a,b,c$ such that both $2a(b^c+1),2a(b^c-1)$ are perfect powers? Are there positive intgers $a,b,c$ such that both $2a(b^c+1),2a(b^c-1)$ are perfect powers?
My research
We have $2a(b^c+1)-2a(b^c-1)=4a$, so I looked at perfect powers which have a multiple of $4$ as a difference. I found this... | If $b$ is even then $b^c+1$ and $b^c-1$ are coprime, and so we have
$$b^c+1=\prod p^{u_p}\qquad\text{ and }\qquad b^c-1=\prod q^{v_q},$$
where the $p$ and $q$ are all distinct prime numbers, and the $u_p$ and $v_q$ are distinct positive integers. To find a positive integer $a$ such that $2a(b^c+1)$ and $2a(b^c-1)$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4199200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find a matrix for a unitary transform between matrices or prove that there is none I have hermitian matrices $A,\,B$ and would like to find a unitary matrix $U$ such that
$$UAU^\dagger=B$$
or show that there is no such matrix.
Example:
For
$$
A=\begin{pmatrix}
0 & 1 & 0 & 0\\
1 & 0 & 0 & 0\\
0 & 0 & 0 & -1\\
0 & 0 & -1... | In the second problem, the two matrices are unitarily similar when $abcd=0$. When $abcd\ne0$, one can verify that
\begin{aligned}
\operatorname{tr}(A^4)&=4\left[(a^2+b^2+c^2+d^2)^2+4(ab-cd)^2\right],\\
\operatorname{tr}(B^4)&=4\left[(a^2+b^2+c^2+d^2)^2+4(ab+cd)^2\right].
\end{aligned}
Hence $\operatorname{tr}(A^4)\ne\o... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Min and max of $f(x,y)=e^{-xy}$ where $x^2+4y^2 \leq 5$ I am trying to use Lagrange multipliers to find the maximum and minimum values of the function $$f(x,y)=e^{-xy}$$
constrained as $$x^2+4y^2=5$$
I began this problem by setting up the Lagrangian:
$$f(x,y) = e^{-xy}$$
$$g(x,y) = x^2+4y^2-5$$
$$L(x,y) = f(x,y) - \lam... | The function $ \ f(x \ , \ y) \ = \ e^{\ -xy} \ $ to be extremized has "diagonal symmetry", which is to say that $ \ f(-x \ , \ -y) \ = \ f(x \ , \ y) \ \ ; $ it also has an anti-symmetric property in that $ \ f(-x \ , \ y) \ = \ f(x \ , \ -y) \ = \ \frac{1}{f(x \ , \ y)} \ \ . $ The elliptical region within which we... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Finding the surface area of a sphere within a cylinder and above the $xy$-plane using double integral I'm trying to find the surface area of the part of the sphere $x^2+y^2+z^2=a^2$ that lies within the cylinder $x^2+y^2=ax$ and above the xy-plane.
I want to do this using $$A(S)=\iint_D \sqrt{[f_{x}(x,y)]^2+[f_{y}(x,y)... | Yes it is easier to evaluate using polar coordinates. Based on your working,
$\displaystyle \sqrt{1+z_x^2 + z_y^2} = \cfrac{a}{\sqrt{a^2-x^2-y^2}}$
$\displaystyle dS = \cfrac{a}{\sqrt{a^2-x^2-y^2}} \ dx \ dy$
So integral to find surface area of sphere inside the cylinder above xy-plane is,
$\displaystyle \int_{x^2+y^2 ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$p$-adic structure of Pell-type equations As an example, consider an integer solution of $ x ^ 2-3y ^ 2 = 13 $.
$ y $ that satisfies this equation is
$y_k = \frac {(4+ \sqrt {3}) (2+ \sqrt {3}) ^ k-(4- \sqrt {3}) (2- \sqrt {3}) ^ k} { 2 \sqrt {3}} $
for any integer $ k $.
By calculation
$ 3 | y_k \Leftrightarrow k \eq... | As $d$ gets larger and $p^d|y_k$, this can be thought of as taking the limit as $y \to 0$ in the p-adics, since $\lim_{d \to \infty}|p^d|_p=0$. So in your equation we're looking at solving just $x^2=N$.
Since in your specific problem you're interested in $k$, we can set $y_k=0$ and solve for $k$,
$$0 = \frac {(4+ \sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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To prove $f=x^4+x^3+x^2+x^1+1$ is irreducible over the $\mathbb{Q}$ My line of proof is as follows:
*
*$\pm1$ are the only candidates for being rational roots (Rational Root Theorem)
*Since $f(1)=5$ and $f(-1)=1$, none is a root
And hence the given polynomial is irreducible over the rationals $\mathbb{Q}$.
Is this ... | (Alternative proof, not using Eisenstein's criterion.)
The polynomial is reciprocal and can be easily factored over the reals:
$$
\begin{align}
x^4+x^3+x^2+x+1 &= x^2 \left(\left(x^2+\frac{1}{x^2}\right)+\left(x+\frac{1}{x}\right)+1\right)
\\ &= x^2 \left(\left(x+\frac{1}{x}\right)^2+\left(x+\frac{1}{x}\right)-1\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4209504",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solutions to $2^a3^b+1=2^c+3^d$
Find all $a,b,c,d$ positive integer such that:
$2^a3^b+1=2^c+3^d$
My progress:
One solution satisfying is $$\boxed{a=1,b=1,c=2,d=1} $$
We first take $\mod 3$ which gives $$ L.H.S\equiv 1\mod 3,~~R.H.S\equiv 2^c\mod 3$$
Hence we get $c$ even. So let $c=2k.$
We get $$2^a3^b-3^d=2^{2k}-1=... | There might not be an easy elementary solution to this equation. It was solved (there are $12$ nontrivial solutions in integers, where trivial solutions have at least $2$ of the exponents equal to $0$) by Tijdeman and Wang in a paper in the Pacific Journal in 1988; their argument uses bounds for linear forms in logarit... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4211699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Conditional inequality $2a^3+b^3≥3$
Non-negative $a$ and $b$ such that $a^5+a^5b^5=2$. How then do I prove
the following inequality $2a^3+b^3≥3$?
So, we can try using the Lagrange multiplier method:
Let $f(a, b)=2 a^{3}+b^{3}+\lambda(a^{5}+a^{5} b^{5}-2), \quad a, b \geq 0$
$$\tag1 \frac{\partial f}{\partial a}=6 a^{... | Remarks: Without calculus.
We have $a^5 = \frac{2}{1 + b^5}$.
We need to prove that $2a^3 \ge 3 - b^3$.
WLOG, assume that $3 - b^3 \ge 0$.
It suffices to prove that
$$2^5 a^{15} \ge (3 - b^3)^5$$
or
$$2^5 \left(\frac{2}{1 + b^5}\right)^3 \ge (3 - b^3)^5$$
or
$$2^9 \ge (1 + b^5)^3 \cdot (3 - b^3)^3 \cdot 2(3 - b^3)^2.$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4213233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $(a+b)^c\cdot(b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot (a+b+c)\right]^{a+b+c}$ where $a,b,c\in \mathbb{Q}^{+}$ unless $a=b=c$.
If $a,b,c$ be positive rational numbers, prove that
$$(a+b)^c\cdot (b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot
(a+b+c)\right]^{a+b+c}$$ unless $a=b=c$
My try :
Consider $... | We need to prove that:
$$\sum_{cyc}\frac{c}{a+b+c}\ln(a+b)\leq\ln\left(\frac{2}{3}(a+b+c)\right).$$
Now, $\sum\limits_{cyc}\frac{c}{a+b+c}=1$ and we can use Jensen for the concave function $\ln$.
Indeed, $$\sum_{cyc}\frac{c}{a+b+c}\ln(a+b)\leq\ln\left(\sum_{cyc}\frac{c(a+b)}{a+b+c}\right)$$ and it's enough to prove tha... | {
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"question_score": "1",
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Why is AM-GM giving 2 contradictory results for the same problem? The problem is the following:
given x and y such that $x+y=1$ and $x,y \geq 0$, find the maximum value of $x^2y$.
The point is, I tried to solve it in 2 different ways, which are shown below, but the second method I used gives a wrong result, and I can't... | Your mistake is in this assumption in your second approach:
It follows that $x^2y$ is maximum when it is equal to the other side of the inequality
This is not true; while at $x = \frac12$,
$$x^2y = \frac{(x+1)^3}{27},$$
the right-hand side is not a constant upper bound, but increasing.
Then it turns out (from the fir... | {
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How many different ordered triples are there such that $a+b+c = 50$ and $a\geq b\geq c\geq 0$? What if $a>b>c>0$? How many different ordered triples are there such that $a+b+c = 50$ and $a\geq b\geq c\geq 0$? What if $a>b>c>0$?
a,b, and c are all integers
I found this question from a textbook, but the author didn't giv... | Using generating functions:
First rewrite the problem by using the substitutions
$$d = a-b, \qquad e = b-c, \qquad f = c.$$
so we have $d,e,f \geq 0$ are integers such that $d + 2e + 3f = 50$.
The number of such triplets can be found as the coefficient of $x^{50}$ in the generating function
$$\frac{1}{(1-x)(1-x^2)(1-x^... | {
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Exercise on limit of indeterminate form: $\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt2$ I'm not able to show that
$\lim_{x \rightarrow 1} \frac{\sqrt{x+3} - \sqrt{5-x}}{\sqrt{1+x} - \sqrt{2}} = \sqrt{2}$
I proceeded as follows:
Substituting 1 to $x$ gives the indeterminate form ... | If you rationalize both the numerator and the denominator you will end up with $\frac {2(\sqrt {1+x}+\sqrt 2)} {\sqrt {x+3}+\sqrt {5-x}}$ so the limit is $\frac {2(\sqrt2 +\sqrt 2)} {\sqrt 4 +\sqrt 4}=\sqrt {2}$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Function corresponds to $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+\alpha^2}$ using the Fourier expansion
Find the function corresponds to $\sum_{n=1}^\infty \frac{(-1)^n}{n^2+\alpha^2}$ using the Fourier expansion
$$
f(x)=\frac{a_0}{2}+\sum_{n=1}^\infty (a_n\cos(nx)+b_n\sin(nx))
$$
where $$\frac{a_0}{2}=\frac{1}{2\pi}\int... | You may use the Poisson summation formula:
If $f\in L_1(\mathbb{R})$ and $\widehat{f}\in L_1(\mathbb{R})$, then $Pf(x)=\sum_{n\in\mathbb{Z}}f(x+n)$ converges uniformly, $f\in\mathcal{C}(\mathbb{S}^1)$, and
$$Pf(x)=\sum_{n\in\mathbb{Z}} \widehat{f}(n) e^{2\pi kx}$$
where $\widehat{f}$ is the Fourier transform of $f$.
U... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$. I want to find limit of $a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}$ as $n\to \infty$.
$\displaystyle a_{n} =\frac{\left( 2n^{\frac{1}{n}} -1\right)^{n}}{n^{2}} =\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right)^{n}$
$\displaystyle \begin{array}{{>{\displaystyl... | Use these two results:
We can use that for $|\alpha x|>>1$, $$(1+\alpha)^x \sim e^{\alpha x}
\tag{1}$$
We have the basic limit for logarithm as; $$\lim_{n \rightarrow
\infty}n\left(a^{1/n}-1\right)=\log a \hspace{5px}\text{, for all
positive } a$$ for $a =n$ itself we have: $$\lim_{n \rightarrow
\infty}n\left(n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 3
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Complicated Integration involving Bessel function and exponential function Sir,
while studying the quantum scattering for various cases, I have got the following integrals:
$$\int_{t_0=-\infty}^{t-R/c}\Big[\frac{\exp{(-i\omega t_0)}}{(t-t_0)}\Big]J_0\big(\sqrt{(t-t_0)^2-(R/c)^2}\big)\,dt_0$$ and similarly,
$$\int_{t_0=... | $$
\int_{-\infty}^{t-a}{\frac{e^{-i\omega u}}{t-u}J_0\left( \sqrt{\left( t-u \right) ^2-a^2} \right) du}
\\
u=t-r,=e^{-i\omega t}\int_a^{\infty}{\frac{e^{i\omega r}}{r}J_0\left( \sqrt{r^2-a^2} \right) dr}
\\
I\left( \omega \right) =\int_a^{\infty}{\frac{e^{i\omega r}}{r}J_0\left( \sqrt{r^2-a^2} \right) dr}
\\
I'\left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4218760",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$ The question is
Find all primes $(p,q)$ such that $pq$ divides $p^3 + q^3 +1$.
My attempt:
This reduces to finding primes $p,q$ such that p divides $q^3+1=(q+1)(q^2-q+1)$ and q divides $p^3+1=(p+1)(p^2-p+1)$.
Now we have 4 cases: If p divides $q+1$ and q d... | If $p \mid q+1$ and $q \mid p^2-p+1$, then $\frac{p^2-p+1}q$ must be $-1$ modulo $p$, since $q\equiv -1\pmod p$. As a result, there exist positive $m,n$ for which
$$(mp-1)(np-1)=p^2-p+1.$$
We clearly can't have $m=n=1$, so we must have
$$(p-1)(2p-1)\leq (mp-1)(np-1)=p^2-p+1\implies p\leq 2,$$
and thus $(2,3)$ is the on... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4219435",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Determining convergence of sequence $a_n=a_{n-1}^{-1}+a_{n-2}^{-1}$ I'm trying to prove convergence for the sequence
$$a_n=\frac{1}{a_{n-1}}+\frac{1}{a_{n-2}}$$
with $a_0=a_1=1$. If it does converge then it converges to $\sqrt{2}$, which agrees with numerical tests. $$$$I've managed to prove two facts about the sequenc... | By induction, we can easily show that $1 \leq a_n \leq 2$ for all $n$. Next, note that the recurrence relation can be recast as
$$ a_n - \sqrt{2} = -\frac{a_{n-1} - \sqrt{2}}{\sqrt{2}a_{n-1}} - \frac{a_{n-2} - \sqrt{2}}{\sqrt{2}a_{n-2}}. $$
Applying this relation to the factor $a_{n-1} - \sqrt{2}$ on the right-hand sid... | {
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"timestamp": "2023-03-29T00:00:00",
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Evaluate $\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2\sin^2(\theta) + b^2\cos^2(\theta)} \ d\theta$ Evaluate
$$\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2\sin^2(\theta) + b^2\cos^2(\theta)} \ d\theta$$
I transformed it into:
$$\int_{0}^{2\pi} b|\sin(\theta)| \sqrt{a^2(1-\cos^2(\theta)) + b^2\cos^2(\theta)} \ d\theta\\ \rig... | Let $$F(a,b)=\int_0^{2\pi}|\sin(x)|\sqrt{a^2\sin^2(x)+b^2\cos^2(x)}\,dx.$$
Symmetry of the integrand gives
$$F(a,b)=4\int_0^{\pi/2}|\sin(x)|\sqrt{a^2\sin^2(x)+b^2\cos^2(x)}\,dx.$$
For $x\in[0,\pi/2]$, $|\sin x|=\sin x$, so the integral is
$$F(a,b)=4\int_0^{\pi/2}\sin(x)\sqrt{a^2\sin^2(x)+b^2\cos^2(x)}\,dx.$$
Then the t... | {
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Or/Not Conditions in Number Theory How many positive integers, not exceeding 100, are multiples of 2 or 3 but not 4?
Hello, I was wondering how to solve problems like these. My thoughts are to count the number that are multiples of 2 but not 4, multiples of 3 but not 4, and then multiples of 6 but not 4, then use Princ... | What you did is not quite correct, although it does give the correct result. This is because each floor function in inclusion-exclusion is for just one set of divisions only, not multiple ones combined as you did, although combining them still sometimes gives the same values anyway, as it does in your case. An example ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Given a polynomial $W(x)$. Find all pairs of integers $a,b$ that satisfy $W(a)=W(b)$ Given a polynomial $W(x) = x^4-3x^3+5x^2-9x$. Find all pairs of distinct integers $a,b$ that satisfy $W(a)=W(b)$. My approach was to factor the polynomial.
$$\begin{align*}
x^4-3x^3+5x^2-9x &= (x^4-3x^3+2x^2)+(3x^2-9x+6)-6\\
&= x^2(x^... | Some calculation of $W(a)-W(b) = 0$ give us :
$$(a+b)\Big((a+b)^2-2ab\Big) - 3\Big((a+b)^2-ab\Big) +5(a+b)-9=0$$
Let $x=a+b$ and $y=ab$ and we get $$(2x-3)y = x^3-3x^2+5x-9$$
Let $n=2x-3$ so $n$ is odd, then we have $ x\equiv_n {3\over 2}$ so $$0 \equiv_n x^3-3x^2+5x-9 \implies 39\equiv_n 0 $$ so $$2x-1\mid 39 \implie... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4222908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Are there infinitely many positive integer solutions to $(xz+1)(yz+1)=P(z)$? Let $P(z)\equiv 1($ mod $ \ z) $ be a polynomial of degree $n>3$ with integer coefficients. Are there infinitely many positive integers $x, y, z$ such that $(xz+1)(yz+1)=P(z)$?
If $P(z) = a_nz^n+1$, it has be proven that the Diophantine equati... | your first recipe, 5,20, 51, 104.. works
$$ z = n^3 + 2n^2 + 2n $$
$$x = n+1 $$
$$ 1+xz = n^2 + 3 n^3 + 4 n^2 +2n+1 $$
$$ y = n^5 +3n^4 +5n^3 + 4 n^2 + 1 $$
$$ 1 + yz = n^8 +5n^7 +13n^6 +20n^5 + 19n^4 + 10n^3 + 2 n^2 + 1$$
$$ ( 1+xz)(1+yz) = n^{12} + 8n^{11} + 32n^{10} + 81n^9 + 142n^8 + 178n^7 + 161n^6 + 104n^5 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4226419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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Solving $m^3-n^3=2mn+8$ We have to find integer solutions to the given equation, this is what i tried :-
For ease, let us denote $x=-m, \enspace n=-y$, and then we are basically considering $x,m,y,n$ as nonnegative wherever they occur below :-
We have four cases :-
CASE $1$ : $mn<0$ and $x^3+n^3=-(8+2xn)$
This case is... | With a little help from ultralegend5385's hint.
Let $m, n \in \mathbb{Z}$. Factoring the LHS, we obtain $$m^3-n^3 = (m-n)(m^2+mn+n^2)$$
Then our equation becomes
$$ m^3-n^3=(m-n)(m^2+mn+n^2)=2mn+8=2(mn+4)$$
$$ (m-n)(m^2+mn+n^2)=2(mn+4)$$
We can split this equation into two cases:
Case 1: $$(m-n) = 2 \tag{1}\label{eq1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Solve: $y''=(1+(y')^2)^{3/2}$ / $y(1)=\frac 15, y'(1)=\frac 34$. My attempt:
$y''=(1+(y')^2)^{3/2}$
$\displaystyle\implies \frac {y''}{(1+(y')^2)^{3/2}}=1$
$\displaystyle\implies \int_{}^{} \frac {y''}{(1+(y')^2)^{3/2}} dx=x+c$
$\displaystyle\implies \int\limits_{}^{}\frac{1}{(1+u^2)\sqrt{1+u^2}}du=x+c,$ where $ u=y'$... | You have found that
$\displaystyle\int\frac{du}{(1+u^2)^{\frac{3}{2}}}=x+c$
Let $u=\tan\theta$ , therefore $du=\sec^2\theta d\theta$
$\displaystyle\int\frac{\sec^2\theta d\theta}{\sec^3\theta}=x+c$
$\displaystyle\int\cos\theta=x+c$
$\sin\theta=x+c$
$\displaystyle \frac{u}{\sqrt{1+u^2}}=x+c$
where $u=y'$
We can find $c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4228956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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What is the value of the PL segment if QL = 12? For reference:
My progress...
the idea would be to demonstrate that $triangle AQL \cong \triangle LQC$
$\triangle PAB \sim QCB (AA)\\
\measuredangle PBA = \measuredangle QCB=90-\theta$
Could I say that $\triangle ABC$ is isoscelels without showing that BL is perpendicul... | This is an interesting question. Please note your mistake is in trying to prove that $\triangle ABC$ is isosceles. $PL = QL$ holds regardless of whether $\triangle ABC$ is isosceles or not.
All we are given is that $\angle BAP = \angle BCQ = \theta$, and $BP$ and $BQ$ are perpendicular drawn to $AP$ and $CQ$ respective... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4229808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Simplifiying $A \cos(x-\phi)$ I have to proof that we can transform this expression : $$a \cos x + b \sin x$$
to this one : $$A \cos (x-\phi)$$
Indeed, I can do the reverse path using the formula of $\cos(x-y)= \cos x \cos y + \sin x \sin y$
but I get stuck on the initial path.
Which trig formula should I use?
Thanks
| You can proceed with your initial path as follows
$$a\cos x +b\sin x$$
Multiply and divide by $\displaystyle\frac{1}{\sqrt{a^2+b^2}}$
$$\sqrt{a^2+b^2}(\frac{a}{\sqrt{a^2+b^2}}\cos x +\frac{b}{\sqrt{a^2+b^2}}\sin x)$$
Now $(\frac{a}{\sqrt{a^2+b^2}})^2+(\frac{b}{\sqrt{a^2+b^2}})^2=1$ and both terms are less than $1$
Th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4236594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Why is $\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac n4$? I found the relation for $n\geq3$ $$\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac n4$$But despite my best efforts, I still have no idea as to how to prove it.
Thin... | Following your way, by geometric series we have that for $n=2N$
$$\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\exp\left((2k-1)\frac{2i\pi}n\right)=\sum\limits_{k=1}^{N}\exp\left((2k-1)\frac{i\pi}N\right)=\\=e^{-\frac{i\pi }N}\sum\limits_{k=1}^{N}\left(e^{\frac{i2\pi}N}\right)^k=e^{-\frac{i\pi }N}\frac{e^{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Integrating inverse of a polynomial using valid method gives wrong result The integral is:
$$\int \frac{1}{x^2+(x-1)^2}dx$$
One way to integrate inverses of polynomials is to find their roots and find the fractions which when added together, give the original fraction (with the roots being the denominators). For exampl... | You did it as if$$\int\frac1{x-i}\,\mathrm dx=\log|x-i|.$$Actually, not only this is not true as it doesn't make sense, since $\log|x-i|$ is not differentiable, in the sense of Complex Analysis. You should work with$$\int\frac1{x-i}\,\mathrm dx=\log(x-i),$$with a carefully chosen branch of the logarithm.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241899",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Showing a solution to 3rd order differential equations forms a subspace Let $S$ denote the set of all solutions of the following differential equation defined on $C^3[0,\infty)$;
$$
\begin{align}
\frac{d^3x}{dt^3} + b \frac{d^2x}{dt^2} + c \frac{dx}{dt} + dx = 0
\end{align}
$$
Show that $S$ is a linear subspace of $C... | It is easy to see that $S\subset \mathcal{C}^3[0,\infty)$.
All you need do is show that if $f$ and $g$ are in $S$ and $\alpha$ is a scalar then $\alpha f+g$ is a solution of the ODE. In other words, $\alpha f+g \in S$.
Clearly $\frac{d^3}{dt^3}(\alpha f+g) = \alpha \frac{d^3f}{dt^3} + \frac{d^3g}{dt^3}$, $\frac{d^2}{dt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Elementary solutions of the equation of a quadratic formula We know that when $A=0$, the quadratic equation $Ax^2+Bx+C=0$ has one solution, $-\frac{C}{B}$.
Since for every quadratic equation, a quadratic equation has at most two solutions: $\frac{-B + \sqrt{B^2 - 4AC}}{2A}$ and $\frac{-B - \sqrt{B^2 - 4AC}}{2A}$, I thi... | If $A=0$, then the equation $Ax^2+Bx+C=0$ is a linear equation, not a quadratic equation. Hence, it only has one root. Since $Bx+C=0$ is not a quadratic equation, there is no reason to think that applying the quadratic formula would produce the correct roots.
If you consider the derivation of the quadratic formula, in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4244828",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
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What is the value of side $AC$ in the triangle below? For reference:The angle B of a triangle ABC measures 60°. The AN and CM medians are plotted. The radius of the circle inscribed in the MBNG quadrilateral
(G is centroid(barycenter) of ABC) measures $\sqrt3$ . Calculate AC.
My progress
$\triangle BED: \\sen30 = \fra... |
Given $ \small MBNG$ is a tangential quadrilateral, it is easy to see that,
$ \small BN + MG = BM + GN$
[How? $ \small BE + EN = BF + NH, MK + KG = MF + GH$. In fact there is a theorem called Pitot Theorem that states the same.]
So if $ \small AN = d, CM = e$, we have $ \small MG = \frac{e}{3}, \small GN = \frac{d}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4245493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can pmf of discrete random variables vary? Can probability mass function be in different form even though when we plug in $k =1,2,3,\cdots$ for $P(X=k)$, the probability for each $k$ between pmf A and pmf B would still be the same. For example :
$1.6.2$. Let a bowl contain $10$ chips of the same size and shape. One and... | Note that \begin{align*}
\frac{\binom9{x-1}}{\binom{10}{x-1}}\times \frac{1}{11-x}&=\frac{\frac{9!}{(x-1)!(10-x)!}}{\frac{10!}{(x-1)!(11-x)!}}\times \frac{1}{11-x}=\frac{9!}{10!}\times \frac{(11-x)!}{(10-x)!}\times \frac{1}{11-x}\\
&= \frac{1}{10}\times \frac{(10-x)!}{(10-x)!}=\frac{1}{10}
\end{align*}
and your answer ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that: $\sum\limits_{cyc}\frac{1}{\sqrt{2a^2+5ab+2b^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$ \dfrac{1}{\sqrt{2a^2+5ab+2b^2}}+\dfrac{1}{\sqrt{2b^2+5bc+2c^2}}+\dfrac{1}{\sqrt{2c^2+5ca+2a^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}.$$
I solved thi... | My attempt for this very hard inequality as conjecture :
Conjecture :
Let $x\geq z$ and $z\leq y\leq 1.5z$ and $z\geq 1$ then it seems we have :
$$0\le f\left(x\right)+f\left(y\right)-2f\left(\frac{2}{\frac{1}{x}+\frac{1}{y}}\right)$$
Where : $$f\left(a\right)=\frac{1}{\sqrt{2a^{2}+z\left(a-z\right)}}$$
The trick her... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Integrate $\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {{x^2} - 8x + 16} \right]}} + \left[ {{x^2}} \right]}}dx} = $ Solve $\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {{x^2} - 8x + 16} \right]}} + \left[ {{x^2}} \right]}}dx} = \_\_\_\_\_$ where [.] represent greatest integer function.
My... | $$I=\int\limits_1^3 {\frac{{\left[ {{x^2}} \right]}}{{{{\left[ {(x-4)^2} \right]}} + \left[ {{x^2}} \right]}}dx}!$$
Use $$\int_{a}^{b} f(x) dx=\int_{a}^{b} f(a+b-x) dx$$, then
$$I=\int\limits_1^3 {\frac{{\left[ {{(x-4)^2}} \right]}}{{{{\left[ {x^2} \right]}} + \left[ {{(x-4)^2}} \right]}}dx}$$
Adding the two we get $$2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4248914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Finding all solutions to quartic Diophantine equation Consider the Diophantine equation $6x^2 = y^2(2y-1)(y-1)$. I am interested in finding all solutions to this such that $y$ is a positive integer -- or at the very least knowing whether there are infinitely many. Certainly there are some; the smallest being $(x,y) = (... | *
*Using the change of variables $ z = \frac{x}{y}$, we have $ 6z^2 = (2y-1)(y-1)$.
*Completing the Square on RHS gives us $ 48z^2+1 = (4y-3)^2 $.
This is a Pell's equation with solutions $$ z = \frac{ ( 7+ 4\sqrt{3}
)^n - ( 7 - 4 \sqrt{3} ) ^n } { 8 \sqrt{3} }, 4y-3 =
\frac{1}{2} [ ( 7+ 4\sqrt{3} )^n - ( 7 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4250587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$ Let $a,b,c>0$, find the minimum value of: $$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3}{4}$ occur when $a=b=c$
$\bullet \dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}= \dfrac{a^2}... | Another way.
We know that the expression is closed to $\frac{1}{3}$, but by C-S
$$\sum_{cyc}\frac{a}{b+3a}=\frac{a^2}{ab+3a^2}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(3a^2+ab)}\geq\frac{(a+b+c)^2}{\sum\limits_{cyc}(3a^2+6ab)}=\frac{1}{3},$$ which says that $\frac{1}{3}$ is an infimum.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4251200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Solving Inequality $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$ If someone could help with solving the inequality above, that would be awsome!
Here is my thinking of using AGM
*
*$\sqrt{\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)}\le \frac{\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}}{2}$... | Let $u = \sqrt{x/y}$. The problem is equivalent to showing
$$
u^3 + 1 \ge u^2 + u,
$$
which follows immediately from
$$
(u-1)^2(u+1) \ge 0
$$
for all $u \ge 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4252299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \rig... | $$\int_0^\infty \frac{1-x^2}{(1+x^2)^2}\operatorname{sech}^2 \left(\frac{\pi x}{2}\right) dx\overset{IBP}=\pi\int_{0}^\infty \frac{x}{1+x^2}\operatorname{sech}^2 \left(\frac{\pi x}{2}\right)\tanh\left(\frac{\pi x}{2}\right)dx$$
$$=2\pi\int_{-\infty}^\infty \frac{x}{1+x^2}\color{blue}{\left(\frac{e^{\pi x/2}}{\left(e^{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
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Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\... | Edited for clarity.
Expanding on the comment of sonicsid, when you are faced with the necessity of clearing the denominator of $(x + 4)$, since $x + 4 = 0$ is disallowed, you must break the analysis into two cases:
Case 1: $(x + 4) > 0.$
$-1 < \frac{7}{x+4} < 1.$
$ (-1) (x+4) < 7 < (x+4).$
$(x + 4) > -7$ and $(x + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Error in my approach to prove that $(m-1)^{m-1} \equiv m - 1 \pmod m$ using a concrete example I read somewhere that if $m$ is composite then $(m - 1)^{m - 1} \equiv m - 1 \pmod m$ and I was curious to try to prove it myself.
So I took as $m = 6$.
Now I can see that $5^5 \equiv 5 \pmod 6$
I was thinking along the follo... | A most immediate thing to note is that $(m-1)\equiv -1 \pmod m$ always.
So $(m-1)^{m-1} \equiv (-1)^{m-1}\pmod m$. And $(-1)^{m-1} = -1$ if $m-1$ is odd and $(-1)^{m-1} = 1$ if $m-1$ is even. And as $-1 \equiv m-1\pmod m$ and $1\equiv m+ 1\pmod m$ we realize what you probably heard was
$(m-1)^{m-1} \equiv (-1)^{m-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Choosing the "right order" of approximation in Taylor Series I was evaluating the following limit using Taylor series, $\lim_{x \rightarrow 0}(2 \frac{e^x-1-x}{x^3}-1/x)$, and I wrote the exponential function as $e^x=1+x+\frac{x^2}{2}+ \frac{x^3}{6}+o(x^4)$ and got that the limit is $\frac13$. But, if I had written $e^... | $$\lim_{x\to0}2\frac{e^x-1-x}{x^3}-\frac{1}{x}=\lim_{x\to0}\frac{1}{x}+\frac{1}{3}+\frac{1}{12}x+\cdots-\frac{1}{x}=\frac{1}{3}$$
However, if you choose to use little-o notation, recall that the definition of little-0 is that:
$$f(x)\in\omicron(g(x)),\,x\to0\iff\lim_{x\to0}\frac{f(x)}{g(x)}=0$$
Whereas the definition o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4254532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that cube root of product of 2 prime numbers are irrational I am stuck with this problem from my son's homework:
Given $p$ and $q$ are prime numbers, prove that $\sqrt[3]{pq}$ is irrational
Could someone please shed some light? Thanks!
| (You don't explicitly state $p \neq q$. I assume this here. A similar argument can be made for the $p = q$ case.)
Suppose $\sqrt[3]{pq} = \frac{a}{b}$ is rational in lowest terms (so $a$ and $b$ are integers, $b > 0$, and $\gcd(a,b) = 1$). Then \begin{align*}
pq &= \frac{a^3}{b^3} \\
b^3pq &= a^3
\end{align*}
Since... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4255086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Solve integral $\int_0^{2\pi}\frac{\sin^2\frac{(N+1)x}{2}}{2\pi(N+1)\sin^2\frac x2} dx $ My question is to solve
$$\int_0^{2\pi}\frac{\sin^2\left(\frac{N+1}{2}x\right)}{2\pi(N+1)\sin^2(x/2)} dx $$
Up to $1000$ or more, Wolfram-Alpha solves this integral and gives the value $1$ for every integer. But, I couldn't prove ... | \begin{align}
\frac{1}{2\pi(N+1)}\int_0^{2\pi}\frac{\sin^2\left(\frac{N+1}{2}x\right)}{\sin^2\left(\frac{x}{2}\right)} dx &\overset{\color{blue}{u = \frac{x}{2}}, \, \color{green}{n=N+1}}{=}\frac{1}{2\pi n}\color{blue}{2}\int_0^{\color{blue}{\pi}}\frac{\sin^2(nu)}{\sin^2(u)} du\\
& \overset{(\color{purple}{\text{Even})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Isn't the book wrongly taking $\sin^{-1}$ on both sides here? Question:
If $\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$, then show that $\theta=\pm\frac{1}{2}\sin^{-1}\frac{3}{4}$.
My book's solution:
$$\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$$
$$\sin(\pi\cos\theta)=\sin(\frac{\pi}{2}\pm\pi\sin\theta)\ [\text{Formula:}\co... | The reasoning shown in the book is incorrect. Indeed, were it correct, then we could make the following deduction.
$$\sin(\pi\cos\theta)=\cos(\pi\sin\theta)$$
$$\sin(\pi\cos\theta)=\sin(\frac{\pi}{2}+\pi\sin\theta)$$
This is exactly the same as your step (i), but with a $+$, rather than a $\pm$. That's perfectly fine, ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
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Prove that $\lim\limits_{n\to\infty} \frac{\sin\left(\log n + n^5 + e^{n^2}\right)}{n} = 0$. Prove that $\lim\limits_{n\to\infty}x_n = 0$, where $x_n = \frac{\sin(\log n + n^5 + e^{n^2})}{n}$.
Demostración: Suppose $\epsilon > 0$. We first note the $\left|\sin(x)\right| \leq 1$ for $x\in\mathbb{R}$. Thus, $\left|\sin\l... | HINT
We can use
$$-\frac1n \le \frac{\sin\left(\log n + n^5 + e^{n^2}\right)}{n}\le \frac1n$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Is $g(x, y) = \frac{f(x, y)^{y+1}-1}{(f(x,y)-1)(xf(x,y)+1)} $ always an integer? Let \begin{equation} f(x, y) = \frac{x^y-1}{x+1} \end{equation}
And \begin{equation}
g(x, y) = \frac{f(x, y)^{y+1}-1}{(f(x,y)-1)(xf(x,y)+1)}
\end{equation}
where $x, y $ are positive integers with $y$ even, $f(x, y) \not = 1$.
It appears ... | Here is an elementary proof:
Let us change the notation so that
$$ f(x,y):= \frac{x^{2y}-1}{x+1}$$ and
$$ g(x,y):= \frac{f(x,y)^{2y+1}-1}{(f(x,y)-1)(xf(x,y)+1)}.$$
We first show that $f(x,y)-1$ and $xf(x,y)+1$ are coprime. Suppose that there exist $p$ a prime which divides both $f(x,y)-1$ and $xf(x,y)+1$ then $p$ divi... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Euler substitution inconsistency in evaluating $\int_{-1}^1 \frac{\mathrm dx}{\sqrt{1-x}+2+\sqrt{1+x}}$
$$I = \int_{-1}^1 \frac{\mathrm dx}{\sqrt{1-x}+2+\sqrt{1+x}} = 4\sqrt2-2-\pi$$
Euler's substitutions immediately come to mind. But as is, the integrand is free of $\sqrt{ax^2+bx+c}$ (with $a\neq0$). So, I considere... | The alternative substitution $x=\sin2t$ may be simpler
$$\begin{align}
& \hspace{5mm}\int_{-1}^1 \frac{1}{\sqrt{1-x}+2+\sqrt{1+x}} \ dx\\
& = \int_0^{\pi/4}\frac{2\cos2t}{1+\cos t}dt =\int_0^{\pi/4}\left( \sec^2 \frac t2-8\sin^2\frac t2\right)dt = 4\sqrt2-2-\pi
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Is this matrix of maximal rank? Let $(A_0,A_1)\in (M_{p\times r}(\mathbb{R}))^2$, $(B_0,B_1)\in (M_{p\times q}(\mathbb{R}))^2$, and suppose that the matrix
$$\begin{pmatrix}A_1&B_0\\A_0&B_1\end{pmatrix}$$
has rank $2p$. Has the matrix
$$\begin{pmatrix}A_1&B_0&&&\\
&B_1&A_0&&\\
&&A_1&B_0&\\
&&&B_1&A_0\end{pmatrix},$$
wh... | For simplicity, set $$M = \begin{pmatrix} A_{1} & B_{0}\\ A_{0} & B_{1} \end{pmatrix} \in M_{2 p, q +r}(\mathbb{R}) \quad \text{and} \quad N = \begin{pmatrix} A_{1} & B_{0} & & & \\ & B_{1} & A_{0} & & \\ & & A_{1} & B_{0} & \\ & & & B_{1} & A_{0} \end{pmatrix} \in M_{4 p, 2 q +3 r}(\mathbb{R}) \, \text{.}$$ To prove t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Integrating $\int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$ I found the following integral and wanted to know if there is a nice closed form solution in terms of elementary or some special functions (Polylogarithm, Clausen, etc).
$$\displaystyle \int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$$
I know that the ... | Beside @Zacky's solution, using partial fraction, you can write
$$\frac{\tan ^{-1}(x)}{x^4+1}=\frac i 2 \left(\frac{\tan ^{-1}(x)}{x^2+i}-\frac{\tan ^{-1}(x)}{x^2-i}\right)$$ and face two integrals
$$I(k)=\int \frac{\tan ^{-1}(x)}{x^2+k}\,dx$$ which can be integrated (have a look here).
The problem is that the result f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Solve below somewhat symmetric equations Solve $x,y,z$ subject to
$$x^2+y^2 - xy = 3$$
$$(x-z)^2+(y-z)^2 - (y-z)(x-z) = 4$$
$$(x-z)^2+y^2 - y(x-z) = 1$$
$$x,y,z \in R^{+}$$
My attempts:
$(x-z)^2+(y-z)^2 - (y-z)(x-z) - ((x-z)^2+y^2 - y(x-z)) = xz - 2yz = (x-2y)z = 3$
$x^2+y^2 - xy - ((x-z)^2+y^2 - y(x-z)) = 2xz - yz - z... | We can take a quadrilateral $ABCD$, for which:
$AC\cap BD=\{E\},$ $AC=BD=z$,$ED=x$, $EC=y$, $\measuredangle AEB=60^{\circ},$$AB=2$, $BC=1$ and $CD=\sqrt3$.
Thus, we obtain your conditions:
$$x^2+y^2-xy=ED^2+EC^2-2ED\cdot EC\cos60^{\circ}=CD^2=3,$$
$$(x-z)^2+(y-z)^2-(y-z)(x-z)=(z-x)^2+(z-y)^2-(z-x)(z-y)=$$
$$=EB^2+EA^2-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Proving $\sum_{k=2}^n \frac{(k-2){n-k+2\choose k-1}+k{n-k+1\choose k-1}}{k{n\choose k}}=1$ for $n\geq 2$ My friend showed me a difficult combinatorial identity I cannot solve.
Prove that:
$$\sum_{k=2}^n \frac{(k-2){n-k+2\choose k-1}+k{n-k+1\choose k-1}}{k{n\choose k}}=1$$ for all $n\ge 2$.
How do I prove this?
Edit: ... | This can be turned into a telescoping sum.
Applying the Pascal's triangle recurrence to the first binomial coefficient in the numerator yields
$$
\binom{n-k+2}{k-1}=\binom{n-k+1}{k-2}+\binom{n-k+1}{k-1},
$$
which allows us to rewrite the identity as
$$
D_n=\sum_{k\ge2}\frac{(k-2)\binom{n-k+1}{k-2}+2(k-1)\binom{n-k+1}{k... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
"answer_id": 1
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Solve for $x$ in $\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$
Solve for $x$:
$$\sqrt{x+2\sqrt{x+2\sqrt{x+2\sqrt{3x}}}} = x$$
I tried to substitute $y=x+2$ and then I try to solve the equation by again and again squaring.
Then I got equation, $$(y-2)(3y^{14}-(y-2)^{15})=0$$
One solution is $y = 2$ and another is $y =... | It's not so hard to imagine someone guessing the solutions $0$ and $3$. The solution $0$ is something one can see from the positions of all of the "$x$". The solution $3$ might be inspired by asking what would make $\sqrt{3x}$ rational.
Now the idea is to prove that there can be no more than two solutions by showing th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
} |
Solving $x^3-7x^2+14x-8-\frac12\sin x=0$ With the given problem:
$$x^3-7x^2+14x-8-\frac12\sin x=0$$
I factorized the cubic part:
$$x^3-7x^2+14x-8=0$$
where we test the following solutions of the fraction of the last coefficient divided by the first coefficient and its even multiplicates:
$$\pm\frac{8}{1}, \pm\frac{8}{2... | Here is an approximation to the zero points.
*
*We put $f_1=x^3−7x^2+14x−8$ and $f_2=\frac{1}{2}\sin x$, so that $f_1(x)=f(x)-f_2(x)$.
We then calculate the values:
$f(0)=-8\leqslant0$, $f(1.5) = 0.1263 > 0$ , $f(3) = −2.0706 < 0$, $f(5) = 12.4795 > 0$
By the mean-value theorem of continuous functions we know that $... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Prove the binomial sum $\sum_{k=0}^n k\binom{2n}{k} = \frac{n4^n}2$. Prove $\displaystyle \sum_{k=0}^n k\binom{2n}{k} = \frac{n4^n}2$.
This is from section 0.241 of Gradshteyn and Ryzhik's table of integrals.
I have managed to find a proof for this using the identity $k\binom nk = n \binom{n-1}{k-1}$.
Hence \begin{alig... | Too complex solution based on algebra.
Consider
$$k \binom{2 n}{k} x^{k-1}=\Bigg[ \binom{2 n}{k} x^k\Bigg]'$$
$$S_n=\sum_{k=0}^n \binom{2 n}{k} x^k=(x+1)^{2 n}-\binom{2 n}{n+1} x^{n+1} \, _2F_1(1,1-n;n+2;-x)$$ where appears the gaussian hypergeometric function.
Computing the derivative with respect to $x$, we have
$$S... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove $\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1 $ Prove $$\log_{a}{(\frac{b^2}{ac}-b+ac})\cdot\log_{b}{(\frac{c^2}{ab}-c+ab})\cdot\log_{c}{(\frac{a^2}{bc}-a+bc})\geq1,
$$
where $a,b,c \in (0,1)$.
I tried to solve it in this way:
$$\log_{a}{(\frac{b^2}{... | Another attempt:
At least of logarithms must be $\geq1$. We suppose the first:
$$\frac{b^2}{ac}-b+ac=ac\left(\left(\frac{b}{ac}\right)^2-\frac{b}{ac}+1\right)$$
$$ac\left(\left(\frac{b}{ac}\right)^2-\frac{b}{ac}+1\right)\geq a\leftrightarrow c\left(\left(\frac{b}{ac}\right)^2-\frac{b}{ac}+1\right)-1\geq0$$
Let
$\frac{b... | {
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"source": "stackexchange",
"question_score": "2",
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Prove: $\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge\sqrt{3}\sum_{cyc}{\sqrt[4]{\frac{5ab}{c}+4a}}$
Prove that the following inequality :$$\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)^2\ge\sqrt{3}\left(\sqrt[4]{\frac{5ab}{c}+4a}+\sqrt[4]{\frac{5bc}{a}+4b}+\sqrt[4]{\frac{5ca}{b}+4c}\right)$$ holds for all positive real numb... | We need to prove that $$(a+b+c)^2\geq\sum_{cyc}\sqrt[4]{9a^4b^2(5b^2+4c^2)}.$$
Now, by Holder and Rearrangement we obtain:
$$\sum_{cyc}\sqrt[4]{9a^4b^2(5b^2+4c^2)}=\sum_{cyc}\sqrt[4]{9a^2b^2\cdot a(5b^2a+4c^2a)}\leq$$
$$\leq\sqrt[4]{9(ab+ac+bc)^2(a+b+c)\sum_{cyc}(5b^2a+4c^2a)}\leq$$
$$\leq\sqrt[4]{9(ab+ac+bc)^2(a+b+c)\... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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$\sqrt{6}+\sqrt{3}$ is not rational proof I want to prove $\sqrt{6}+\sqrt{3}$ is not rational; here is my attempt:
Assume for the sake of contradiction that $\sqrt{6}+\sqrt{3}$ is rational. Then $(\sqrt{6}+\sqrt{3})^2$ must also be rational. Since $$(\sqrt{6}+\sqrt{3})^2=9+2\sqrt{6}\sqrt{3}=9+2\sqrt{2}\sqrt{3}\sqrt{3}=... | Looks more or less ok, I would just add one more step.
You correctly prove that if $\sqrt 6+\sqrt 3$ is rational, then $9+6\sqrt2$ is also rational.
I would now add one more step and say that this also means that because $\sqrt{2} = \frac{9+6\sqrt{2} - 9}{6}$, that therefore, $\sqrt{2}$ is also rational, and only then ... | {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Probability of randomly assigning elements to groups
10 students are tested in an exam with 4 different versions. Each
student is randomly assigned to one of the versions. What is the
probability that there are exactly $i$ versions in which exactly > $3$ students were assigned. Answer separately for $i=2, i = 3$.
I... | For another approach and showing the mistake :
$1-)$ For $i=2$ , By your reasoning select firstly $2$ case which take exactly $3$ sutdents by $C(4,2)$ ,after that , choose $3$ students for the first and the second by $C(10,3) \times C(7,3)$ ,respectively .Now , we have $4$ students to disperse to $2$ versions , but th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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$x(t)= a\cos(t)$ , $y(t)= b\sin(t)$ in terms of the arc length $S$ I'm trying to parameterize the ellipse $x(t)= a\cos(t)$ , $y(t)= b\sin(t)$ in terms of the arc length $S$ but I don't know how to do it.
Supposing that $\gamma:[a,b]\to \mathbb{R}$ is a smooth curve with $\gamma'(t)\neq 0$ for $t\in [a,b]$ , I know th... | There is no standard closed form but series solutions only
*
*Local canonical form starting from minor axis (clockwise convention):
\begin{align}
k &= \sqrt{1-\frac{b^2}{a^2}} \\
s(t) &= \int_0^t \sqrt{a^2\cos^2 \theta+b^2\sin^2 \theta} \, d\theta \\
&= aE(t,k) \\
t &= E^{-1} \left( \frac{s}{a},k \right) \\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4284133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Integration of complicated equation I was looking at a Wolfram Mathworld article about bean curves: https://mathworld.wolfram.com/BeanCurve.html
And it states that the area enclosed by the curve of
$$x^4+x^2y^2+y^4=ax(x^2+y^2)$$ is
$$A=\sqrt{2}a^2\int_0^1\sqrt{x\left(1-x+\sqrt{1+(2-3x)x}\right)}\text{d}x=\frac{7\pi a^2... | use polar coordinates to evaluate the area.
$$r^4(\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta))=ar^3\cos(\theta)$$
$$r=\frac{a\cos(\theta)}{\sin^4(\theta)+\cos^4(\theta)+\sin^2(\theta)\cos^2(\theta)}$$
$$\frac{A}{a^2}=\frac{1}{2a^2}\int_{-\pi/2}^{\pi/2}r^2 d\theta$$
$$=\frac{1}{2}\int_{-\pi/2}^{\pi/2} \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4285709",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How do you prove by induction that $\frac{1}{2} + \frac{2}{2^2} + \ldots + \frac{n}{2^n} = 2 - \frac{n+2}{2^n}$? For $n=1$ this is true because $\frac{1}{2^{1}}=2-\frac{1+2}{2^{1}}=\frac{1}{2}$. Further, it is a little more complicated, can we now assume that this is true up to the number $n-1$? Then do the induction s... | You can write this sum using sigma notation, for clarity.
$\frac{1}{2} +\frac{2}{2^2} + \dots+\frac{n}{2^n} = \sum_{i=1}^{n} \frac{i}{2^i} = 2-\frac{n+2}{2^n}.$ (*)
Let our induction hypothesis $P(n)$ be the equation (*).
Let's apply mathematical induction now.
Base case $(n=1):$ $$\sum_{i=1}^{1} \frac{i}{2^i} = \fra... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can we approximate $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$? It seems like from the graph $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$ is somehow alike to the graph $e^{-x^2}$, the main problem is that the limitations of the software makes it hard to graph for large numbers. Is it possible to do so or... | The partial product $$P_m=\prod _{n=1}^{m } \left(1-\frac{(2 n+1) }{n^2\pi ^2 }x^2\right)$$ effectively looks like a gaussian (tested for $m=10^5$). Using Pochhammer symbols
$$P_m=\frac{\left(\frac{\pi^2-x^2-\sqrt{x^2 \left(x^2+\pi ^2\right)}}{\pi
^2}\right){}_m \left(\frac{\pi^2-x^2+\sqrt{x^2 \left(x^2+\pi ^2\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288574",
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"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
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Where is the mistake? Finding an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
Find an equation for the ellipse with foci $(1,2)$, $(3,4)$, and sum of distance to the foci equals to $5$.
We consider the foci in the coordinate system $XY$ such that $X=x-2$ and $Y=y... | You just need to write it like what it is, the sum of the distance to the foci is 5:
$$\sqrt{(x-1)^2+(y-2)^2} + \sqrt{(x-3)^2+(y-4)^2} = 5$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now... | Treat $4x^2 - 2xy - 4x + 3y - 3$ as a quadratic in $x$.
Set $4x^2 - 2x(y + 2) + 3(y - 1) = 0$, then solve for $x$ (using the Quadratic Formula), to factor the LHS:
$$x = \dfrac{2(y+2) \pm \sqrt{(2(y+2))^2 - 4(4)(3)(y-1)}}{8} = \dfrac{y + 2 \pm \sqrt{y^2 + 4y + 4 - 12y + 12}}{4} = \dfrac{y + 2 \pm \sqrt{y^2 - 8y + 16}}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 3
} |
Minimize the ratio involving the ellipse
Let $P$ be any point on the curve $\dfrac{x^2}{4}+\dfrac{y^2}{3}=1$,
and $A,B$ be two fixed points $\left(\dfrac{1}{2},0\right)$ and
$(1,1)$ respectively. Find the minimum value of
$\dfrac{|PA|^2}{|PB|}$.
Assume $x=2\cos\theta,y=\sqrt{3}\sin\theta$, then
$$\dfrac{|PA|^2}{|PB|}... | Denote $c = \cos \theta$ and $s = \sin\theta$.
We have
$$|\mathrm{PA}|^2 = (2c - 1/2)^2 + (s\sqrt 3 - 0)^2= c^2 - 2c + \frac{13}{4}$$
and
$$|\mathrm{PB}|^2 = (2c - 1)^2 + (s\sqrt 3 - 1)^2
= c^2 - 4c + 5 - 2\sqrt 3\, s.$$
We have
\begin{align*}
|\mathrm{PA}|^4 - |\mathrm{PB}|^2
&= (c^2 - 2c + 13/4)^2 - (c^2 - 4c + 5 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290875",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Let $p, q$ be different primes. Then $p^2+q^2-pq$ is not a perfect square. This question originally comes from the following problem:
Let $ABC$ be a triangle with integer side lengths with $\angle ABC = 60^{\circ}$. Suppose length $\overline{AB}$ and $\overline{BC}$ are prime numbers. Determine and prove what kind of ... | We have: $$(p-q)^2<p^2-pq+q^2<(p+q)^2.$$
If $p>q>0$ and $p^2-pq+q^2=c^2,$ for $c>0,$ then then $q^2\equiv c^2\pmod p,$ so $$c\equiv \pm q\pmod p$$ Now, since $0<p-q< c<p+q,$ this means our options are $c=q$ or $c=2p-q.$ In both cases, $c$ is in the range only if $p<2q.$
If $c=q,$ then $p^2-qp=0,$ so $p=0$ or $p=q.$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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How should I evaluate $\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$? $$\int_0^{\infty}\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3]{x}}e^{-y\pi\left(x^2+1/x^2+1\right)}\ \mathrm{d}y\ \mathrm{d}x$$
$$\int_0^{\infty}\frac{\sqrt[3]{x}}{1+\sqrt[3... | If, as @RAHUL commented, you use partial fraction decomposition, you should be able to prove that the result is the smallest positive root of
$$19683 y^6-94041 y^4+105786 y^2-2809=0$$
Solving the cubic equation in $y^2$ and taking the square root of it, the result is
$$\sqrt{\frac{1}{27} \left(43-2 \sqrt{543} \cos \lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4291631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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$\lim_{n \rightarrow \infty } \int_0^1 \frac{nf(x)}{1+ n^{2} x^{2} }dx = \frac{ \pi }{2} f(0)$ for continuous $f$. If f is a continuous function on $[0,1]$, then show that $$\lim_{n \rightarrow \infty } \int_0^1 \frac{nf(x)}{1+ n^{2} x^{2} }dx = \frac{ \pi }{2} f(0)$$
Can anybody help me to solve this? I tried bu... | Put $I = \int_{0}^1 \frac{nf(x)}{1+n^2x^2} dx = I_1 + I_2$ where
$I_1 = \int_{0}^{n^{-\frac13} } \frac{nf(x)}{1+n^2x^2} dx$ and $I_2 = \int_{n^{-\frac13} }^1 \frac{nf(x)}{1+n^2x^2} dx$.
Further $|f(x)| \le M$ because $f$ is continious. We have
$$|I_2| \le \int_{n^{-\frac13} }^1 \bigg|\frac{nf(x)}{1+n^2x^2} \bigg| dx \... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Integral of $x^2dx$ by definition - what am I doing wrong? $$\int_a^bx^2dx$$
By definition
$$b-a=nh$$
$$\lim_{h\to0}a^2h+(a+h)^2h+(a+2h)^2h+...+(a+(n-1)h)^2h$$
$$\lim_{h\to0}h(a^2+(a+h)^2+(a+2h)^2+...+(a+(n-1)h)^2)$$
$$\lim_{h\to0}h(a^2+(a^2+1^2h^2+2*1h)+(a^2+2^2h^2+2*2h)+...+(a^2+(n-1)^2h^2+2*(n-1)h))$$
$$\lim_{h\to0}... |
Please tell me where did I go wrong
$$\lim_{h\to0}a^2h+(a+h)^2h+(a+2h)^2h+...+(a+(n-1)h)^2h$$
$$\lim_{h\to0}h(a^2+(a+h)^2+(a+2h)^2+...+(a+(n-1)h)^2)$$
$$\lim_{h\to0}h(a^2+(a^2+1^2h^2+2*1h\color{red}{*a})+(a^2+2^2h^2+2*2h\color{red}{*a})+...+(a^2+(n-1)^2h^2+2*(n-1)h\color{red}{*a}))$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Convergence or Divergence of $ 1 - 1 + \frac{1}{2} - . . .$ and another series. I'm testing ideas about the convergence and divergence of series:
Say I have a series $$ 1 - 1 + \frac{1}{2} - \frac{1}{2} + \frac{1}{3} - \frac{1}{3} + . . .$$
I believe that this series would converge to $0$ as every odd term in the serie... | $$1 - \frac{1}{2} + \frac{1}{2} - \frac{1}{2^2} + \frac{1}{3} - \frac{1}{2^3} + \frac{1}{4} - \frac{1}{2^4} + . . .$$
can be coded as $\sum_{n\geq1}\Big(\frac{1}{n}-\frac{1}{2^n}\Big)$. If this series would converge, as the sum of convergent series is a convergent series, then $\sum_{n\geq1}\Big(\frac{1}{n}-\frac{1}{2... | {
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"source": "stackexchange",
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Prove that the sequence $(a_n)$ is Cauchy and find the limit. Let us define a sequence $(a_n)$ as follows:
$$a_1 = 1, a_2 = 2 \text{ and } a_{n} = \frac14 a_{n-2} + \frac34 a_{n-1}$$
Prove that the sequence $(a_n)$ is Cauchy and find the limit.
I have proved that the sequence $(a_n)$ is Cauchy. But unable to find the... | Rewrite $a_n$ as $$a_1=1,\ a_2=2,\ a_{n+2}=\dfrac{3}{4}a_{n+1}+\dfrac{1}{4}a_n \mathrm{\ for\ } n\geqq 1.$$
We can get
\begin{align}
&a_{n+2}-a_{n+1}=-\dfrac{1}{4}(a_{n+1}-a_{n}) \cdots (A)\\
&a_{n+2}+\dfrac{1}{4}a_{n+1}=a_{n+1}+\dfrac{1}{4}a_{n} \cdots (B)
\end{align}
Letting $b_n=a_{n+1}-a_n$, we get $b_{n+1}=-\dfrac... | {
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"source": "stackexchange",
"question_score": "4",
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The Operator Norm of Basis Change Matrices for a more explicit Gelfand Formula Gelfand's formula states that the operator norm of a matrix exponent $A^n$ is in some sense equal to the exponent of the spectral radius $\rho(A)^n$ in the sense that
$$
\lim_{n\to\infty} \sqrt[n]{\|A^n\|} =
\inf_{n\in\mathbb{N}} \sqrt[n]{\... | Instead of using the Jordan decomposition one should use the Schur Decomposition in this case as Unitary matrices are isometries in euclidean norms and therefore isometries on the induced operator norms. This means we do not have to estimate them at all.
Lemma: Explicit Schur Decomposition for Specific Matrix
For a rea... | {
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"source": "stackexchange",
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What's the measure of the segment $EF$ in the triangle below? For reference:
The triangle $ABC$ where $AB = 7$, $BC = 8$ and $AC = 9$ is inscribed in a circle. Calculate the measure of the arrow ($EF$) of the side $AC$. (Answer: $\frac{3\sqrt5}{2}$)
My progress:
Here are the relations I found:
$EF = R - OE \\
\trian... | Draw altitude $AD$ and call $BD=x$.
From Pythagoras' theorem, $$7^2-x^2=9^2-(8-x)^2\implies x=2$$
It follows that $AD=3\sqrt5$.
We see $\triangle ABD\sim\triangle AOE$. Therefore, $$\frac{AO}7=\frac{OE}2=\frac{9/2}{3\sqrt5}\\\implies AO=\frac{3\cdot7}{2\sqrt5},\: OE=\frac{3\cdot2}{2\sqrt5}$$
Thus $$EF=\underbrace{OF}_... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the value of $ \frac{1}{{2(2^2 - 1)}} + \frac{1}{{3(3^2 - 1)}} + \frac{1}{{4(4^2 - 1)}} + \cdots \;. $ To find the value of $$
\frac{1}{{2(2^2 - 1)}} + \frac{1}{{3(3^2 - 1)}} + \frac{1}{{4(4^2 - 1)}} + \cdots \;.
$$
I presented it as $$
\sum\limits_{n = 2}^\infty {\frac{1}{{n(n^2 - 1)}}} .
$$
Then using part... |
Note : $$\frac{1}{a.b.c}=\frac{1}{c-a}\left(\frac{1}{a.b}-\frac{1}{b.c}\right)$$
and $$\frac{1}{a.b}=\frac{1}{b-a}\left(\frac{1}{a}-\frac{1}{b}\right)$$
Use this in $$\sum_{n\geq2}\frac{1}{(n-1)(n)(n+1)}$$
Here's what you'll get : $$\sum_{n\geq2}\frac{1}{2}\left(\frac{1}{(n-1)(n)}-\frac{1}{(n)(n+1)}\right)\implies\fr... | {
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What's the the value of $x$ in the circumference below? For reference:
In the figure; calculate $x$, if $r =\sqrt2$.
(Answer: $x = \sqrt2$)
My progress:
Draw $PO_1\perp HG\:(O_1 \in HG).$
Let $O$ be the center of the largest circle.
Using Euclid's Theorem:
$\triangle OPF:OP^2 = OQ^2+PQ^2-2\cdot OQ\cdot FQ$
$\implie... | In fact, the two small circles here are called 'Twin circles'. As the name suggests, they are congruent. Here's a proof:
Following the image, $EH\parallel AB$ leads to $FEA,$ $FHB,$ $EDC,$ $ADH$ are straight lines. Also $AF$ is extended to meet $CH$ at $G$.
Since $\angle AFB=\angle ACG=90^\circ$, we can say that $H$ i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the sum of the squares of the medians of triangle APC below For reference: The diameter of a circle measures 8m. On this diameter are located the points A and B equidistant 1 m from the center; through B draw any chord PC , determine the sum of the squares of the medians of triangle APC.(Answer: $73,5$)
My progres... | Here's a proof for the general case where $\small PC$ is not necessarily perpendicular to $\small AB$:
Given that we have already proven $$\mathsf{4.(sum\ of\ squares\ of\ medians)=3.(sum\ of\ squares\ of\ sides)}$$ using Appollonius theorem, continue from there.
So let's calculate $\small AP^2+AC^2+PC^2\tag{*}$
Drop t... | {
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"timestamp": "2023-03-29T00:00:00",
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Can a ratio of sinusoidal functions with the same frequency always be written as a tangent function? In general, two sinusoidal functions $y_1$ and $y_2$ with angular frequency $\omega$ can be written as
*
*$y_1 = A\sin(\omega x) + B \cos(\omega x)$,
*$y_2 = C\sin(\omega x) + D \cos(\omega x)$.
Where $A, B, C, D$ a... | The answer is yes.
Claim 1 : If $D=0$, then
$$\dfrac{y_1}{y_2}=-\frac BC \tan\bigg(\omega x -\frac{\pi}{2}\bigg) + \frac AC$$
Claim 2 : If $D\not=0$, then $$\frac {y_1}{y_2}=\frac{DA-BC}{C^2+D^2}\tan\bigg(\omega x-\arctan\bigg(\frac CD\bigg)\bigg)+\frac{CA+BD}{C^2+D^2}$$
Claim 1 : If $D=0$, then
$$\dfrac{y_1}{y_2}=-\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x... | Multiply by $\sqrt{3x+2}$, assuming $3x+2\ge 0$ i.e. $x\ge-2/3$:
$$\left(\sqrt{3x+2}\right)^2-2x\sqrt{3x+2}+x^2=0$$
$$\left(\sqrt{3x+2}-x\right)^2=0$$
$$\sqrt{3x+2}=x$$
$$3x+2=x^2$$
Solve the quadratic equation to find $x=\frac{3\pm\sqrt{17}}{2}$. As the final equation is a consequence of the original one, but not nece... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309941",
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"source": "stackexchange",
"question_score": "1",
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For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that...
For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$$
Here's what I've done so far:... | The key is to write:
$$\frac ab+\frac ba+\frac bc+\frac cb+\frac ca+\frac ac=\frac{b+c}a+\frac{c+a}b+\frac{a+b}c=\frac{1-a}a+\frac{1-b}b+\frac{1-c}c$$
and
$$\frac{1-a}a+2\ge2\sqrt2\sqrt{\frac{1-a}a}$$
by AM-GM.
Summing similar inequalities yields the result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310082",
"timestamp": "2023-03-29T00:00:00",
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For positive real numbers $a,b,c$ prove that $2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$ For positive real numbers $a,b,c$ prove that $2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$
I think the $A.M-G.M$ inequality needs to be used here. Pretty sure we have to use the fact that
$\frac{a^6+b^6+c^6}{3}\ge a^2b^2c^2$
and that
$\frac{x^2+... | $2+\frac{a^6+b^6+c^6}{3}\ge ab+ac+bc$
$12+2a^6+2b^6+2c^6\ge 6ab+6ac+6bc$
We can write this as
$a^6+b^6+1+1+1+1+a^6+c^6+1+1+1+1+b^6+c^6+1+1+1+1\ge 6ab+6ac+6bc$
We also know that, from the $A.M-G.M$ inequality
$a^6+b^6+1+1+1+1\ge6\sqrt[6]{a^6b^6}=6ab$
$a^6+c^6+1+1+1+1\ge6\sqrt[6]{a^6c^6}=6ac$
$b^6+c^6+1+1+1+1\ge6\sqrt[6]... | {
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Let $A,B$ be $2\times2$ matrices. Given that we know $\text{Tr}(A)$, $\text{Tr}(B),\text{Tr}(AB)$, how do I find $A$ and $B$ that have those traces? Let $A,B$ be $2\times 2$ matrices. Given that we know $\operatorname{Tr}(A)$, $\operatorname{Tr}(B),\operatorname{Tr}(AB)$ how do I find $A$ and $B$ that have those trace... | Edit: I've revamped the answer to include the determinate of $1$ condition.
For an elementary approach: Let $Tr(A)=x$, $Tr(B)=y$, and $Tr(AB)=z$. For some $a,b$, we have
\begin{align*}
A&=\begin{pmatrix}a&a_1\\a_2&x-a\end{pmatrix}&
B&=\begin{pmatrix}b&b_1\\b_2&y-b\end{pmatrix}
\end{align*}
This gives us
$$AB=\begin{pma... | {
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"timestamp": "2023-03-29T00:00:00",
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Find the minimum value of $2a+ (1/a) + (1/2b) + b$, where a, b > 0 Find the minimum value of $2a+ (1/a) + (1/2b) + b$, where a, b > 0
My approach:-
Since a and b are positive numbers, I applied AM-GM inequality
$(2a+(1/a)+(1/2b)+b) /4$ ≥ $(2a\cdot (1/a)\cdot (1/2b)\cdot b) ^ {1/4} $
Giving me the answer as $4$
but the... | AM-GM inequality applied to $(2a,1/a,1/2b,b)$ gives
$$2a+\frac{1}{a}+\frac{1}{2b}+b \ge 4 \tag{1}$$
where equality is to be achieved for $2a=1/a=1/2b=b$. Solving this gives $a=1/\sqrt{2}$ but does not give a unique/consistent value for $b$.
Conclusion is, AM-GM cannot be applied to all four quantities at once. However,... | {
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"timestamp": "2023-03-29T00:00:00",
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Conver the Equation $r=a \sin x+b\cos x$ into Cartesian Form Hello :)) Wanted to convert the equation $r=a \sin x+b\cos x$, where $a, b \in \mathbb{R}$. Knowing that $x = r \cos x$ and $y= = r \sin x$, I can make $x = (a \sin x + b \cos x) \cos x$. But this is kinda a dead end because I can't simplify this further. Als... | If you need to solve fpr $x$ you can easily divide both side by $\sqrt{a^2+b^2}$. Then, we have:
$$\frac{r}{\sqrt{a^2+b^2}}=\frac{a}{\sqrt{a^2+b^2}}\sin(x)+\frac{b}{\sqrt{a^2+b^2}}\cos(x)$$
Let:
$$\theta=\cos^{-1}\left(\frac{a}{\sqrt{a^2+b^2}}\right)$$
We have:
$$\sin(x+\theta)=\frac{r}{\sqrt{a^2+b^2}}$$
Can you finish... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4315748",
"timestamp": "2023-03-29T00:00:00",
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Coefficient extraction I want to show:
\begin{equation*}
[z^n]\frac{1}{(1-z)^{\alpha + 1}} \log \frac{1}{1-z} = \binom{n + \alpha }{n} (H_{n+\alpha} - H_{\alpha}).
\end{equation*}
where $[z^n]$ means the $n$-th coefficient of the power series and
\begin{equation}
H_{n+\alpha} - H_{\alpha} = \sum^{n}_{k=1}{\frac{1}{\a... | We will need an auxiliary identity before we can move on to the subject
by OP, which is
$$[z^n] \frac{1}{(1-z)^{\alpha+1}} \log\frac{1}{1-z}
= {n+\alpha\choose n} (H_{n+\alpha} - H_\alpha).$$
with $\alpha$ a non-negative integer.
A binomial identity
Introduce with $q\ge 1$
$$f(z) = n! (-1)^n \frac{1}{z+q} \prod_{p=0}... | {
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How do we calculate the expected value of $Y$? Let $\Omega = \{−1, 0, 1, 3\}$ with the probability function $p$ given by
I want to calculate the expected value and the variance of $X(\omega)=\omega$, $Y(\omega)=5\omega-3$, $Z(\omega)=(\omega-1)^2$.
For $X$ I have done the following :
We have that $X\in \{-1,0,1,3\}$.
... | Your calculation of $E[X]$ is correct but the calculation of $\text{Var}[X]$ is not correct.
$ \text {Var}[X] = \sum(x - E[X])^2 \cdot P(X = x)$ which translates to,
$ \text {Var}[X] = E[X^2] - (E[X])^2 $
Now, $E[X^2] = (-1)^2 \cdot \dfrac{1}{10} + 1^2 \cdot \dfrac{2}{10} + 3^2 \cdot \dfrac{4}{10} = \dfrac{39}{10}$
So,... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Finding the minimum value of $|z-3+i|$ given $z$ satisfies $\arg{(z-2i)}=\frac{\pi}{6}$ I need to find the the minimum value of $|z-3+i|$ given $z$ satisfies $\arg{(z-2i)}=\frac{\pi}{6}$. My issue is that just looking at the graph of $\arg{(z-2i)}=\frac{\pi}{6}$ (which is a ray from $(0,2)$ on the Argand diagram) and $... | Alternative approach:
$\displaystyle \tan(\pi/6) = \frac{1}{\sqrt{3}}$.
Therefore, with $z = x + iy$, two constraints must be satisfied:
*
*$\displaystyle \frac{y-2}{x} = \frac{1}{\sqrt{3}}.$
*Both $(x)$ and $(y-2)$ must be positive (i.e. in the 1st quadrant), rather than both being negative.
The above constraints ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the roots of the polynomial $x^4+2x^3-x-1=0$. Problem
Find all $4$ roots of the polynomial: $$f(x)=x^4+2x^3-x-1.$$
My Attempt
Observe,
$$f(-2)=1,\; f(-1)=-1,\; f(0)=-1,\; f(1)=1.$$
Therefore, $f(x)$ has a root between $-2$ and $-1$, another root between $0$ and $1$.
Since $f(x)$ does not contain a second degree t... | With the substitution $x=t-b/na$, you eliminate the second highest order term of a polynomial $ax^n + bx^{n-1} + \cdots .$
If you substitute $x=t-2/4 = t-1/2$ your polynomial becomes
$$t^4 -\frac{3}{2}t^2 -\frac{11}{16}.$$
This is quadratic in $t^2$ so you can solve to get
$$t^2 = \frac{3\pm 2\sqrt{5}}{4}.$$
So the fou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4327391",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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$p$ prime and $m_p$ a proper divisor of $p−1$. Is it $\sigma(m_p)-\nu(m_p)(This is a sharper version of this other question of mine.)
Along the problem I'm facing, I've come to the following lemma (if it is true):
Let $p$ be a prime and $m_p$ a proper divisor of $p-1$. Then the difference between the sum of all the div... | By computer search, $p = 61$ and $m_p = 30$ seems to be the smallest counterexample. We get $\sigma(m_p) = 1 + 2 + 3 + 5 + 6 + 10 + 15 + 30 = 72$ and $\nu(m_p) = 8$, so $\sigma(m_p) - \nu(m_p) = 72 - 8 = 64$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Integrating $\int \frac{dx}{x^{11}\sqrt{1+x^4}}$
Find $$I=\int \frac{dx}{x^{11}\sqrt{1+x^4}}$$
With $x^2=t$, we get $I=\int _{ }^{ }\frac{d\ \sqrt{t}}{\left(\sqrt{t}\right)^{11}.\sqrt{1+t^2}}=\frac{1}{2}\int _{ }^{ }\frac{dt}{t^6\sqrt{1+t^2}}$
Now, with $t= \tan k$ we get $I=\frac{1}{2}\int _{ }^{ }\frac{\cos ^5k.\ d... | After substituting $t = \frac{1}{x^4}$:
$$I = -\frac{1}{4} \int \frac{x^5 \ dt}{x^{11} \sqrt{1+x^4}} = -\frac{1}{4} \int \frac{t^2 x^2 \ dt}{\sqrt{x^4(t+1)}}= -\frac{1}{4}\int \frac{t^2 \ dt}{\sqrt{1+t}}.$$
Now further substitute $u = 1 + t$ and expand $t^2 = (u - 1)^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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multivariable calculus divergence theorem help I am stuck on a problem:
Use the Divergence Theorem to evaluate $\iint \mathbf{F} \cdot d\mathbf{S}$, where $$\mathbf{F}(x,y,z)=z^2x\mathbf{i}+(\frac{1}{3}y^3+\tan(z))\mathbf{j}+(x^2z+y^2)\mathbf{k}$$ and $S$ is the top half of the sphere $x^2+y^2+z^2=1$.
[Hint: Note that ... | Applying divergence theorem, you get flux over closed surface $S_2$.
In spherical coordinates, $x^2 + y^2 + z^2 = \rho^2$
where $x = \rho \cos\theta \sin\phi, y = \rho \sin\theta \sin\phi. z = \rho \cos\phi$
As it is part of the sphere above $z = 0, \phi \leq \pi/2$.
So the volume integral is,
$ \displaystyle \int_0^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Calculate the ratio based on the lengths of edges of triangle The problem
My problem is,
Let $\triangle ABC$ be an acute triangle, circumscribed in $(O)$ and has orthocenter $H$. Let $HO$ intersect $(O)$ at $E$ and $F$, as shown in the image. $AE$ cuts $BC$ at $K$ and $AF$ cuts $BC$ at $L$. $AO$ cuts $BC$ at $T$. Calcu... | triangle ABC with sides a. b, and c.
$\begin{array}{} A=(\frac{a^2+b^2-c^2}{2a},\frac{2S}{a}) & B=(a,0) & C=(0,0) & S=Δ(ABC) \end{array}$
$\begin{array}{} \text{Euler line (OH)} & l·x+m·y+n=0 & l=\frac{2a^4-(b^2-c^2)^2-a^2(b^2+c^2)}{2a} & m=\frac{2S(c^2-b^2)}{a} \end{array}$
$\begin{array}{} \left\{ D,E \right\}=OH∩Cir... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4332388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find radius that minimizes the surface of the solid I have solved the following problem:
"A solid, with a volume of $8cm^3$, consists of a cylinder and two equilateral cones, external to the cylinder and each with a base in common with the cylinder itself.
Find the base radius so that the surface of the solid is minima... | The surface area of one cone will be $2\pi r^2$ (apothem will be $2r$) so two cones will give you $4\pi r^2$. Thus, $S'(r)=8\pi r(1-\frac{1}{\sqrt{3}})-\frac{16}{r^2}=0\implies r=\sqrt[3]{\frac{2\sqrt{3}}{\pi (\sqrt{3}-1)}}=\sqrt[3]{\frac{3+\sqrt{3}}{\pi }}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4338296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
About the inequality $\sum_{i=1}^{n}\frac{\frac{1}{x_i}+x_{i+1}}{\sqrt{\frac{1}{x_i}+x_i}}\geq n\sqrt{2}$ Hi It's a generalization found on Aops starting from this question Prove: $\frac{\frac{1}{a}+b}{\sqrt{\frac{1}{a}+a}}+\frac{\frac{1}{b}+c}{\sqrt{\frac{1}{b}+b}}+\frac{\frac{1}{c}+a}{\sqrt{\frac{1}{c}+c}}\ge3\sqrt{2... | Lemma 1: For $a,b>0$,
$$2\sqrt{\frac{1}{a}+b} \geqslant \sqrt2 \left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)$$
Proof:
$$\left(\frac{1}{\sqrt{a}}-\sqrt{b}\right)^2 \geqslant 0$$
$$\frac{1}a + b \geqslant 2\frac{\sqrt{b}}{\sqrt{a}}$$
$$\left(\frac{1}{\sqrt{a}}+\sqrt{b}\right)^2= \frac{1}a + b + 2\frac{\sqrt{b}}{\sqrt{a}} \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4342977",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$ Find all integers $(x,y)$ satisfying $(x+y+11)^2 = x^2 + y^2 + 11^2$.
So far,
\begin{align*}
x^2 + y^2 + 121 + 2xy + 22x + 22y &= x^2 + y^2 + 121\\
2xy + 22x + 22y &= 0\\
(2x+22)y &= -22x\\
(x+11)y &= -11x
\end{align*}
At least 1 of $x,y$ must be a mu... | Hint: First, dividing your second line by $2$ gives
$$xy + 11x + 11y = 0 \tag{1}\label{eq1A}$$
Next, using Simon's favorite factoring trick gives
$$(x + 11)(y + 11) = xy + 11x + 11y + 121 \tag{2}\label{eq2A}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4344164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx.$ Definite Integral: $$\int_{-1}^{1} \log\left(\frac{1+x}{1-x} \right) \frac{1}{1-ax} dx,$$
where $0 < a <1$.
I tried with integration by parts taking $\log\left(\frac{1+x}{1-x} \right) $ as the first function and $\frac{1}{1-ax} $ as the second functi... | The simple Mathematica result given by Varun Vejalla in the comments is indeed correct. In order to make the final result look a bit nicer, we can define the function $f \colon (-1,1) \to \mathbb{R}$,
$$ f(a) = \int \limits_{-1}^1 \frac{\operatorname{artanh}(x)}{1-ax} \, \mathrm{d} x \, . $$
Since $\operatorname{artanh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 0
} |
Extension of Euclid's formula for Pythagorean triples to higher degrees I recently discovered that you can get all the primitive pythagorean triples with this neat formula known as Euclid's formula
$$a = m^2 - n^2, b = 2mn, c = m^2 + n^2$$
Where, m and n are co-prime and each having a different parity, that is one is o... | The fourth power case is the Jacobi Madden equation: $a^4+b^4+c^4+d^4=(a+b+c+d)^4$ Jacobi Madden equation.
It gives infinitely many ineteger solutions for $a^4+b^4+c^4+d^4=e^4$ using elliptic curve.
Unfortunately, the parametric solution is not known.
Some small solutions are
$955^4 + 1770^4 + (-2634)^4 + 5400^4 = 549... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4349199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Integrate $\sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1)$ I'm trying to learn calculus through self study and I happened upon the following exercise:
$$ \int_{0}^{2} \sqrt{4 - x^2}\cdot\operatorname{sgn}(x-1) \,dx $$
Seeing the sgn I thought: well this is easy and concluded that since $ \int_{0}^{2} \operatorname{sgn}(x-1... | To expand on @José Carlos Santos' answer, you can compute $\int \sqrt{4-x^2}\; dx$ by trigonometric substitution: let $x=2\sin(t)$ with $-\frac{\pi}{2}\le t\le \frac{\pi}{2}$ . Then $\sqrt{4-x^2}=2\cos(t)$ and $dx=2\cos(t)\; dt$ so
$$\int \sqrt{4-x^2}\; dx = \int 4\cos^2(t)\; dt = 2 \int 1+\cos(2t)\; dt
=
2\left(t + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4350015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Comparing $\frac {9}{\sqrt{11} - \sqrt{2}}$ and $\frac {6}{3 - \sqrt{3}}$ (without calculator) We want to compare the following two numbers:
$$x = \frac {9}{\sqrt{11} - \sqrt{2}} \quad\text{and}\quad y = \frac {6}{3 - \sqrt{3}}$$
My attempts so far:
I multiply both numerator and denominator of $x$ by $\sqrt{11} + \sqrt... | Render
$[(\sqrt{11}+\sqrt2)-(3+\sqrt3)][(\sqrt{11}+\sqrt2)-(3+\sqrt3)]=(\sqrt{11}+\sqrt2)^2-(3+\sqrt3)^3=(13+2\sqrt{22})-(12+6\sqrt3)=1+2\sqrt{22}-6\sqrt3$
Then
$6\sqrt3-2\sqrt{22}=2(3\sqrt3-\sqrt{22})=\dfrac{2(27-22)}{3\sqrt3+\sqrt{22}}=\dfrac{10}{3\sqrt3+\sqrt{22}}$
And finally
$(3\sqrt3+\sqrt{22})^2=49+6\sqrt{66}<49... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4354087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.