Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
find the minimum of $A=\frac{x^3}{3y+1}+\frac{y^3}{3z+1}+\frac{z^3}{3x+1}$ with $x^3+y^3+z^3=3$ With $x,y,z \ge 0, x^3+y^3+z^3=3$: find the minimum of $A=\dfrac{x^3}{3y+1}+\dfrac{y^3}{3z+1}+\dfrac{z^3}{3x+1}$
My attempts: $A=\dfrac{x^6}{3yx^3+x^3}+\dfrac{y^6}{3zy^3+y^3}+\dfrac{z^6}{3xz^3+z^3} \ge \dfrac{(x^3+y^3+z^3)^2... | The key observation is the term $3x,3y,3z$ at the denominators. This suggests the application of the AM-GM inequality somehow. Indeed, by AM-GM inequality: for $a \ge 0 \implies a^3 + 2 = a^3 + 1 +1 \ge 3a\implies a^3+ 3 \ge 3a+1$. Using this fact for $a = x, y, z$ in the denominators of $A$. Thus: $A \ge \dfrac{x^3}{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Show that there is no pair of coprime positive integers $(x,y)$ such that $(x+y)^3 \mid (x^n+y^n)$
Let $n$ be a square-free integer. Show that there is no pair of coprime positive integers $(x,y)$ such that $$(x+y)^3 \Big| (x^n+y^n)$$
The problem can be apparently solved by LTE, but I don't know what are the cases to... |
$(x+y)^3 | (x^n+y^n), (x, y)=1.$
\begin{align}
\text{Case 1. } \; & n=1: \\
&(x+y)^2|1, x+y=\pm1. \Rightarrow \text{cannot be positive.} \\
\ \\
\text{Case 2. } \; & n=2: \\
&(x+y)^3 | x^2+y^2. \\
&\text{We can easily find that $(x+y)^3$ will be absolutely bigger than $x^2+y^2$ when $x, y$ increases.} \\
& \therefore... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4474935",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find inverse of rational function $f(x) = \frac{x^3+2}{x-3}$ when x not equal to 3 Question:
I'm trying to find inverse of $f(x) = \frac{x^3+2}{x-3}$ for an assignment.
$y = \frac{x^3+2}{x-3}, x\neq3$
Replace $x$ and $y$.
$x = \frac{y^3+2}{y-3}$
$x(y-3) = y^3 + 2$
$xy - 3x = y^3 + 2$
I don't know how to proceed after ... | If $x\neq 3$ $$y= \frac{x^3+2}{x-3}\quad \implies \quad x^3-y \,x+(3y+2)=0$$ For $y>0$, following the steps given here, we have
$$\Delta=4 y^3-27 (3 y+2)^2$$ which is negative; so, only one real root.
Using the hyperbolic solution for one real root, we end with
$$x=-\frac{2 \sqrt{y} }{\sqrt{3}}\cosh \left(\frac{1}{3} ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Why am I getting extraneous solutions solving $\frac{1+\sqrt{1-x}}{x-\sqrt{1-x^{2}}}=2 x$? I came across a beautiful problem:
Solve $$\frac{1+\sqrt{1-x}}{x-\sqrt{1-x^{2}}}=2 x$$
My approach:
Obviously $x=0$ is not a solution.
Now we have:
$$1+\sqrt{1-x}=2x^2-2x\sqrt{1-x^2} \tag1$$
Squaring the above equation both sid... |
we get four values of $x:$
$$x=\frac{\pm \sqrt{3}}{2}, \frac{\pm \sqrt{10+2 \sqrt{5}}}{4} \tag7$$
out of which only $$x=\frac{-\sqrt{3}}{2}, \frac{-\sqrt{10+2 \sqrt{5}}}{4}$$ will satisfy.
Actually, among these four solutions, only $x=\frac{\sqrt{3}}{2}$ is extraneous; it was introduced, together with the other extra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find the maximum of $\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$,where$a,b,c>0$ $a,b,c>0$, find the maximum of :
$$\frac{abc}{(4a+1)(9a+b)(4b+c)(9c+1)}$$
I try to find the minimum of $\frac{(4a+1)(9a+b)(4b+c)(9c+1)}{abc}=\frac{4a+1}{\sqrt{a}}\cdot\frac{9a+b}{\sqrt{ab}}\cdot\frac{4b+c}{\sqrt{bc}}\cdot\frac{9c+1}{\sqrt{c}}
=\... | Hint
Try instead to minimize
$$\Phi=\frac{(4a+1)(9a+b)(4b+c)(9c+1)}{abc}$$ Compute the partial derivatives
All of them being equal to $0$, the solution is immediate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478545",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find minimum of $\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+ab+bc+ca$ Let $a, b,c$ be the lengths of the sides of a triangle such that $a+b+c=2$, find the minimum value of $$\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}+ab+bc+ca$$
I don't have many ideas for this problem, my attemps:
The minimum value is $2\sqrt{2}+\d... | Denote the expression by $f$.
Using Cauchy-Bunyakovsky-Schwarz inequality, we have
\begin{align*}
&(\sqrt{a^2 + b^2} + \sqrt{b^2 + c^2} + \sqrt{c^2 + a^2})^2\\
=\,& 2a^2 + 2b^2 + 2c^2
+ 2\sqrt{(a^2 + b^2)(c^2 + b^2)}
+ 2\sqrt{(c^2 + b^2)(c^2 + a^2)}\\
&\quad
+ 2\sqrt{(c^2 + a^2)(b^2 + a^2)}\\
\ge\,& 2a^2 + 2b^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Proving $(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$ with various solutions.
$(ac+bd)^2+(ad-bc)^2=(a^2+b^2)(c^2+d^2)$
Solutions in the answers.
$\ \\ \ \\ \ \\ \ \\$
Edit) Since this question is closed, I'll add more contexts for this question.
This identity is called "Brahmagupta-Fibonacci identity", which the comment... | \begin{align}& (a^2+b^2)(c^2+d^2) \\= \; & (ac)^2+(bc)^2+(ad)^2+(bd)^2 \\= \; & \Big((ac)^2+(bd)^2\Big)+\Big((ad)^2+(bc)^2\Big) \\= \; & \Big((ac)^2+2(ac)(bd)+(bd)^2\Big)+\Big((ad)^2-2(ad)(bc)+(bc)^2\Big) \\= \; & (ac+bd)^2+(ad-bc)^2\end{align}
| {
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Calculating complex integral with Residue - where is my fault? \begin{align}\int_0^{2\pi} \frac{\cos(x)}{13+12\cos(x)} dx & = \displaystyle\int_0^{2\pi} \frac{(z+1/z)\frac{1}{2}}{13+12(z+1/z)\frac{1}{2}}\frac{1}{iz} dz \\
& = \cdots \\
&= -i\displaystyle\int_0^{2\pi} \frac{z^2+1}{\left(z+\frac{2}{3}\right)\left(z+\frac... | On the right-hand side of the first line, the integral should no longer be over the real interval $[0,2\pi]$; rather, it should be over the unit circle in the complex plane. In addition to correcting that calculational step and subsequent ones, that also explains why the residue at $z=-\frac23$ is relevant: it's inside... | {
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"url": "https://math.stackexchange.com/questions/4480941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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limits of 2 lebesgue integrals
Compute the following limits, with justification, where the integrals denote Lebesgue integrals:
*
*$\lim\limits_{m\to\infty} \int_0^\infty \dfrac{m\sin (y/m)}{y(1+y^2)} dy$.
*$\lim\limits_{m\to\infty} \int_0^1 \dfrac{1+my^2}{(1+y^2)^m}dy.$
For 1), I think it's useful to use the Tayl... | For 1) use the fact that $|\sin t| \leq t$ for all $t >0$ so a dominating function is $\frac 1 {1+y^{2}}$.
For 2) use the fact that $(1+y^{2})^{m} \geq 1+my^{2}$ so a dominating function is the constant function $1$.
[The function $(1+x)^{m}-1-mx$ vanishes when $x=0$ and its dervative is $m[(1+x)^{m-1}-1]$ which is non... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481034",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof for a prime-generating sequence Let $k$ be a positive integer.
Let $n$ be an integer such that $n=6k-1$
Let $r$ be the remainder of the division of $(n-1)!-n$ by $(n+2)$
Conjecture: if $6k+1$ is prime $r=3k+2$
For example the first 25 values of $r$ are:
${5,8,11,2,17,20,23,2,2,32,35,38,41,2,2,50,53,56,2,2,65,2,71... | $$n+2=6k+1$$ by Wilson's theorem $$(n+1)!\equiv -1\pmod {n+2}$$ it follows that $$n!\equiv 1\pmod{n+2}$$ and that $$(n-1)!\equiv -(2^{-1})\equiv 3k\pmod {n+2}$$ and $$-n\equiv 2\pmod {n+2}$$ so together we have $$3k+2$$ as the remainder $\blacksquare$
As to the addition chains $$(3k+2)+(3(k+1)+2)=6(k+1)+1$$ so what.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Evaluating $\lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}$
Problem statement:
If
$$ A = \lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}, $$
Find $A^2$.
Solution:1)
\begin{align*}
\prod_{k=0}^{n} \binom{n}{k}
&= \prod_{k=1}^{n} \frac{n!}{k!(n-k)!}
= \frac{(n!)^{n+... | Let $A_n = \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}$ denote the expression before the limit is taken. Then
\begin{align*}
\log A_n
&= \frac{1}{n(n+1)} \log \left[\prod_{r=0}^n \binom{n}{r}\right] \tag{1} \\
&= \frac{1}{n(n+1)} \sum_{r=0}^{n} \log \binom{n}{r} \tag{2} \\
&= \frac{1}{n(n+1)} \sum_{r=0}^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Probability of rolling exactly 1 number exactly 3 times in 6 rolls of a fair die I'm trying to calculate the probability based on the size of the event space divided by the size of the sample space $P=\frac{|E|}{|S|}$
I know that $|S|=6^6$, but am not sure what exactly the event space consists. Currently my thoughts ar... | Since there are six possible values for each of the six rolls, there are indeed $6^6$ elements in the sample space.
For the favorable cases, since we want to calculate the number of cases in which exactly one number appears three times, there are two possibilities:
*
*One number appears three times and three other nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485857",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Finding the value of $\sqrt[4]{(4+\sqrt7)^{-1}}\sqrt{1+\sqrt7}$ with other approaches It is a problem from a timed exam,
What is the value of $\sqrt[4]{(4+\sqrt7)^{-1}}\sqrt{1+\sqrt7}$ ?
$1)1\qquad\qquad2)\sqrt[4]2\qquad\qquad3)2\qquad\qquad4)2\sqrt[4]2$
I solved it with two approaches.
First approach,
$$\sqrt[4]{\fr... | If we call $ \ u \ = \ 4 + \sqrt7 \ \ , $ then we have $ \ 1 + \sqrt7 \ = \ u - 3 \ $ and $ \ \frac{1}{u} \ = \ \frac{4 - \sqrt7}{9} \ \ . $ Consequently,
$$\sqrt[4]{(4+\sqrt7)^{-1}} \ · \ \sqrt{1+\sqrt7} \ \ = \ \ \sqrt[4]{\frac{(u \ - \ 3)^2}{u}} \ \ = \ \ \sqrt[4]{ \ u \ - \ 6 \ + \ \frac{9}{u} \ }$$
$$ = \ \ \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Find $\lim_{n \to \infty}\frac{x_n}{\sqrt{n}}$ where $x_{n+1}=x_n+\frac{n}{x_1+x_2+\cdots+x_n}$ Assume a positive sequence $\{x_n\} $ satisfies $$x_{n+1}=x_n+\frac{n}{x_1+x_2+\cdots+x_n}.$$ Find $\lim\limits_{n \to \infty}\dfrac{x_n}{\sqrt{n}}$.
Assume the limit we want is $L$. Then by Stolz theorem, one can obtain
\be... | This is a community-wiki answer illustrating Iosif Pinelis's solution (with some simplifications).
*
*If you like this, please give kudos to his original solution as well.
*Also, feel free to improve this answer as you please!
Setting. We will write $s_n = \sum_{k=1}^{n} x_k$, so that the recurrence relation takes... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
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$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$ is independent of $n$
If $n$ is a positive integer, prove that
$$\left\lfloor\frac{8n+13}{25}\right\rfloor-\left\lfloor\frac{n-12-\left\lfloor\frac{n-17}{25}\right\rfloor}{3}\right\rfloor$$
is i... | This is an amendment of my solution inspired by fleablood's comment:
To simplify the expression $\left \lfloor \frac{n-17}{25} \right \rfloor$, let $n-17=25k+r$ i.e. $n=25k+r+17$ where $0 \le r \le 24$.
Then
$$\left\lfloor \frac{8n+13}{25}\right \rfloor=\left\lfloor \frac{8(25k+r+17)+13}{25} \right\rfloor = 8k+5+\left\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489162",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is $\frac{x^2-3x-4}{-3x-15}\Big/\frac{x^2-16}{x^2-x-30}$ defined at $x=6$ or not? If a number is in the denominator of a rational expression that itself is the denominator of another rational expression, such that the number causes the latter denominator to be undefined, then isn't the entire expression undefined?
For ... |
$$\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}} \implies\quad\frac{(x-4)(x+1)(x-6)(x+5)}{-3(x+5)(x+4)(x-4)},x\neq{-5},{-4},4,6\quad?$$
Rather: $$f(x)=\frac{\quad\frac{x^2-3x-4}{-3x-15}\quad}{\frac{x^2-16}{x^2-x-30}}\iff\bigg(\color\red{x\neq-5,4,6}\quad\text{and}\quad f(x)=\frac{(x+1)(x-6)}{-3(x+4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving the equation $x^2+(\frac{x}{x+1})^2=\frac54$
Solve the equation$$x^2+\left(\frac{x}{x+1}\right)^2=\frac54$$
I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sur... | Let $t=x+1$
$\displaystyle (t-1)^2 + \left( \frac{t-1}{t} \right)^2 = \frac{5}{4}$
$(t^2 + 2 + \frac{1}{t^2}) - 2\left(t + \frac{1}{t} \right) = \frac{5}{4}$
Let $s = \left(t + \frac{1}{t} \right)$
$s^2 - 2s - \frac{5}{4} = (s+\frac{1}{2})(s-\frac{5}{2}) = 0$
We reduced solving quartic for x to quadratic for s, then qu... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $P(P(x))=Q(x)^2$, then $P(x)=R(x)^2$? Let $P$ and $Q$ are polynomials from $\mathbb{R}[x]$. Suppose that $P(P(x))=Q(x)^2$. Does it mean that for any such $P(x)$ exists some $R(x)\in \mathbb{R}[x]$ such that $P(x)=R(x)^2$?
| Yes, $P(x)$ must be a square. Suppose we have a counterexample $(P(x), Q(x))$, and let $d$ be the degree of $P(x)$. Then $P(P(x))$ has degree $d^2$, so in order for it to be a square $d$ must be even. Write $d = 2n$ and $P(x) = a_{2n} x^{2n} + a_{2n-1} x^{2n-1} + \cdots + a_1 x + a_0$. The leading coefficient $a_{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show that $\sum_{\ell=0}^k \binom{2k+1}{2\ell} = 2^{2k}$ In acquainting myself with the Cauchy product, I am trying to derive the power series of $\sin(x)\cos(x)$. Of course, $\sin(x)\cos(x)=\frac12\sin(2x)$ is easy to work with so I know what to expect.
We have
$$\sin(x) = \sum_{n=0}^\infty \frac{(-1)^n}{(2n+1)!} x^{2... | Hint: Since $\dbinom{2k+1}{2\ell} = \dbinom{2k+1}{2k+1-2\ell}$, the sum (which we'll denote by $S$) satisfies:
\begin{align*}
S &= \dbinom{2k+1}{0}+\dbinom{2k+1}{2}+\cdots+\dbinom{2k+1}{2k-2}+\dbinom{2k+1}{2k}
\\
S &= \dbinom{2k+1}{2k+1}+\dbinom{2k+1}{2k-1}+\cdots+\dbinom{2k+1}{3}+\dbinom{2k+1}{1}
\end{align*}
Now, add... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does one show this complex expression equals a natural number? We have:
$$\left(\frac{10 }{3^{3/2}}i-3\right)^{1/3}+ \frac{7}{3 \left(\frac{10}{3^{3/2}}i-3\right)^{1/3}}=2$$
This comes from solving the cubic equation of $x^3-7x+6=0$ which factors as $(x-2)(x-1)(x+3)=0$
We can simplify the problem into finding just ... | Let $\,u = \left(-3 + \dfrac{10}{3^{3/2}}i\right)^{1/3}\,$ and $\,v = \left(-3 -\dfrac{10}{3^{3/2}}i\right)^{1/3}\,$ be the principal values of the cube roots. Being the principal value, $\,u\,$ is the root with the greatest real part of $\,
z^3 = -3 + \dfrac{10}{3^{3/2}} i\,$.
Taking complex conjugates, $\,\overline{u... | {
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"url": "https://math.stackexchange.com/questions/4497637",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Pythagorean Quadruples (Integer Question) We have $x^2+y^2+z^2=m^2$ with $x,y,z,m$ integers, $\gcd(x,y,z)=1$, $z$ is odd, $x$ and $y$ are even.
Set $x_1:=\frac{x}{2}$ und $y_1:=\frac{y}{2}$. We get
\begin{gather*}
x_1^2+y_1^2=\frac{1}{4}(x^2+y^2)=\frac{1}{4}(m^2-z^2)=\left(\frac{m+z}{2}\right)\left(\frac{m-z}{2}\right)... | Since $f_1|x_1,y_1$, we have
$$f_1^2\,|\,x_1^2+y_1^2\ =\ \left(\frac{m+z}2\right)\left(\frac{m-z}2\right)$$
and $f_1$ is coprime to $f_2$, hence also to $\frac{m-z}2$, so $f_1^2\,|\,\frac{m+z}2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$8x^4+27y^4+64z^4$$
where $$x+y+z=\frac{13}{4}$$
I tried to apply C-S inequality but it has little difference,
The form what I know is :
$$(a^4+b^4+c^4)(x^4+y^4+z^4)\ge (ax+by+cz)^4$$
But in this problem, coefficient is form of $()^3$, not a $... | Using Lagrange multiplier method:
$$\frac{\partial}{\partial x}\left(8x^4+27y^4+64z^4-\lambda(x+y+z-\frac{13}4)\right)=0\\4\cdot8x^3=\lambda$$
Similarly for $y$ and $z$ you get:
$$4\cdot 27y^3=\lambda\\4\cdot 64z^3=\lambda$$
If we use $$\frac\lambda4=\mu^3$$
one can take the cubic root and get $$x=\frac\mu2\\y=\frac\mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving that $\frac {(n+1)n(n-1)...(n-i)}{i+1}$ can be used as the summ of $k^h$ I have previously proved that $$\sum_{k=i}^{n} k(k-1)(k-2)\cdots(k-i+1) = \frac{(n+1)\cdot n\cdot (n-1)\cdots(n-i+1)}{i+1}$$
From here I am supposed to show that the formula above can be used to compute the sum of $k^h$, for example, $\sum... | Yes, it can be used.
Example with $\sum_{k=1}^{n}k^2$
$\sum_{k=1}^{n}k^2$
$=\sum_{k=1}^{n}k(k-1+1)$
$=(\sum_{k=1}^{n}k(k-1)) + \sum_{k=1}^{n}k$
$=(\sum_{k=2}^{n}k(k-1)) + \sum_{k=1}^{n}k$ ( means that $i = 2$)
$=\frac{(n+1)n(n-1)}{3} + \frac{n(n+1)}{2}$
$=\frac{(n+1)n(2(n-1) + 3)}{6}$
$=\frac{(n+1)n(2n+1)}{6}$
Example... | {
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Find $f:\mathbb{R} \to [-1, 1]$ which satisfies these two conditions:
Find $f:\mathbb{R} \to [-1, 1]$ which satisfies these two conditions:
$f(x+y)=f(x)f(a-y)+f(y)f(a-x)$, $f(a-x-y)=f(a-x)f(a-y)-f(x)f(y).$ ($a$: constant.)
My expectation of $f$ is $f(x)=\sin(\frac {\pi x} {2a} +2\pi n).$ But, how can we prove this?... | If we let $g(x) = f(a-x) + i.f(x)$, the two equations are equivalent to $g(x+y) = g(x).g(y)$. This resembles Cauchy exponential functional equation from $\mathbb{R}$ to $\mathbb{C}$, and there are infinitely many solutions via Hamel's basis.
If $f$ is continuous on the other hand, plugging $x=0$ yields $g(y) = g(0)g(y)... | {
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How to show $\sum_{k=1}^{80} {1 \over \frac{k}{81} - \frac{1}{2} - \frac{\iota}{2}} + {1 \over \frac{k}{81} - \frac{1}{2} +\frac{\iota}{2}} = 0$? I came across this sum while doing the homework my teacher gave us on series. It was originally $$a_k = \frac{k}{81}, S = \sum_{k=1}^{80} {a_k^2 \over 1+2a_k^2-2a_k}$$
I then... | There is no need to use complex numbers. Here the keyword is symmetry: note that $a_{81-k}=1-a_k$ and therefore
\begin{align}
2S&=\sum_{k=1}^{80} \frac{a_k^2}{1-2a_k(1-a_k)}+\sum_{k=1}^{80} \frac{a_{81-k}^2}{1-2a_{81-k}(1-a_{81-k})}\\
&=\sum_{k=1}^{80} \frac{a_k^2}{1-2a_k(1-a_k)}+\sum_{k=1}^{80} \frac{(1-a_k)^2 }{1-2(1... | {
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Probability that a $5$ cards hand from a standard deck of $52$ contains at least $1$ ace $\frac{C(4, 1)C(51, 4)}{C(52,5)}$? I understand the direct and indirect methods for solving this problem, but I do not understand why $\frac{C(4, 1)C(51, 4)}{C(52,5)}$ is wrong. The answer that $\frac{C(4, 1)C(51, 4)}{C(52,5)}$ res... | Since there are
$$\binom{4}{k}\binom{48}{5 - k}$$
ways to select exactly $k$ aces and $5 - k$ non-aces, the correct count of favorable cases should be
$$\binom{4}{1}\binom{48}{4} + \binom{4}{2}\binom{48}{3} + \binom{4}{3}\binom{48}{2} + \binom{4}{4}\binom{48}{1}$$
Your failed attempt counts each hand with more than one... | {
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Solving a functional equation $ g(x) = \frac{1}{1-x} g\left(\frac{ax}{1-x}\right) $ Consider a following functional equation
$$ g(x) = \frac{1}{1-x} g\left(\frac{ax}{1-x}\right) $$
for a given fixed real parameter $a$.
I don't know whether the solution for $g$ is unique. However, assuming it has a Taylor series around ... | I will assume that $g$ is continuous at $0$ and $a\in (0,1)$.
Let's prove by induction the following: for $n\in \Bbb N$,
\begin{align}
g(x) &= \frac{1}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}g\left(\frac{a^{n+1}x}{1 - \left(\sum\limits_{k=0}^{n}a^k\right)x}\right)
\end{align}
For $n=0$ the statement is equivalement... | {
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How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \thet... | By Contour Integration
Considering the integral
$$
J=\int_{0}^{2 \pi} \frac{d \theta}{a+b \cos \theta}.
$$
Let $z=e^{\theta i}$, then $\cos \theta=\frac{1}{2}\left(y+\frac{1}{z}\right)$ and $d z =i e^{\theta i} d \theta=i z d \theta .$
$$
\begin{aligned}
\\
J &=\int_{K(0,1)} \frac{1}{a+\frac{b}{2}\left(z+\frac{1}{z}\ri... | {
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Prove that $\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$
Show that for any non negative real numbers $x_1,x_2,\cdots x_n,$
$$\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$$
My work:
Let$$S(n)=\f... | Another simple solution by induction, motivated by OP's questions and by the currently most voted answer (which is also by induction but a bit complicated).
\begin{equation}
\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}
\end{equation}
First, the inequality is... | {
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Can anyone explain this process of solving? (Differentiation) I'm at differentiation of algebraic functions. There's an example in the module that I couldn't quite get how it led to that.
$y=\frac {(x+1)^3}{x^2}$
It's solved by using a combination of quotient and power rules. I'll enumerate how it's solved.
$(1) y′= \f... | Here is one way to solve it. Use the fact that the derivative of $(ax+b)^n=n(ax+b)^{n-1}(ax+b)'$. This just follows from the chain rule.
Rewrite the equation as $y=(x+1)^3x^{-2}$. Then apply the product rule. The product rule says $(f(x)g(x))'=f(x)g'(x)+g(x)f'(x)$.
Thus, we get $((x+1)^3x^{-2})'=((x+1)^3\cdot-2x^{-3})+... | {
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Squeeze theorem to prove that the sequence $a_n= \frac{3^n}{n!}$ is convergent? I want to know if the following proof of the convergence of a sequence is correct.
Proof. Let $a_n= \frac{3^n}{n!}$. Firstly, it is trivial to see that $a_n \geq 0$ for all $n \in \mathbb{N}$. Secondly, see that
$$a_n = \frac{3^n}{n!}= \fra... | Alternatively replacing all $n \geq 4$ to 4: $$0\leq a_n\leq \frac{9}{2}(\frac{3}{4})^{n-3}, $$ you will get the result. This is stronger that you can see the sum of this series also converges.
| {
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Show that $4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$. The problem asks us to show that the following equation holds true. $$4\sin^2(24^{\circ})+4\sin(24^{\circ})\sin(12^{\circ}) = 1$$
This equation has been verified on my calculator.
Perhaps some basic trigonometric formulas will be enough to solve th... | We can avoid values of $\cos36^\circ,\sin18^\circ$ to derive at the identity.
Let $x=6^\circ$
using Prosthaphaeresis Formulas
$$4\sin^24x+4\sin4x\sin2x=4\sin4x(\sin4x+\sin2x)=8\sin4x\sin3xcos x$$
$$\implies\sin24^\circ\sin18^\circ\cos6^\circ=\sin24^\circ\cos(90-18)^\circ\sin84^\circ$$
Using Prove that: $\sin\beta\sin\l... | {
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$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $ Find all x's
It is given that $$\frac{1}{\log_2(x-2)^2}+\frac{1}{\log_2(x+2)^2}=\frac{5}{12} $$ Find all possible values of $x$.
What I did: $$\frac{1}{\log_2(x-2)}+\frac{1}{\log_2(x+2)}=\frac{5}{6} $$ $$\log_{(x-2)}2 +\log_{(x+2)}2 =\frac{5}{6}$$
What i... | Let $a = \log_2(x-2)$ and $b = \log_2(x+2)$. Then your equation becomes:
$$\frac{1}{a} + \frac{1}{b} = \frac{5}{6}$$
$$\frac{a + b}{ab} = \frac{5}{6}$$
$$6(a + b) = 5ab$$
$$b = \frac{6a}{5a - 6}$$
$$\log_2 (2^a + 4) = \frac{6a}{5a - 6}$$
This has the integer solution $a = 2$. Unfortunately, while this solution is eas... | {
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Equation has two distinct solutions We have $\displaystyle{ a,b,c }$ real numbers such that $\displaystyle{ a^2+b^2+c^2>0 }$. Which condition must hold so that the equation $\displaystyle{ ax^2+bx+c=0 }$ has two different solutions?
$$$$
I have done the following :
\begin{align*}&ax^2+bx+c=0 \Rightarrow x^2+\frac{b}{a}... | for any degree two polynomial in order for it to have real roots we must have :
$$\Delta = b^2-4ac > 0$$
and if $\Delta \neq 0$ it always has two distinct solutions real or complex so that I believe suffices.
| {
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Infinite summation with multiple terms in denominator and variables in exponents I need to find $\sum\limits_{n=1}^{\infty} \frac{6^n}{2^{1 + 2 n} + 3^{1 + 2 n} - 5\cdot 6^n}$, but I don't know how to do infinite summation with multiple terms in the denomiator and variables in exponents. Can anybody give me a hint? Tha... | Hint
Let $x=2^n$ and $y=3^n$ to make
$$\frac{6^n}{2^{1 + 2 n} + 3^{1 + 2 n} - 5\times 6^n}=\frac{x y}{2 x^2-5 x y+3 y^2}=\frac{x y}{(2 x-3 y) (x-y)}$$ Now, as @person commented, using partial fractions
$$\frac{x y}{(2 x-3 y) (x-y)}=\frac{x}{y-x}-\frac{2 x}{3 y-2 x}=\frac{2^n}{3^n-2^n}-\frac{2^{n+1}}{3^{n+1}-2^{n+1}}$$ ... | {
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Maximize $f(x,y)=x+y$ subject to $\sqrt{4-2 x}+\sqrt{4-2 y}=\sqrt{x y}$ Given that $x,y \in [0,2]$ find the maximum value of $x+y$ if $$\sqrt{4-2 x}+\sqrt{4-2 y}=\sqrt{x y}$$
Looking for an elementary approach. No Lagrange multipliers please.
My try:
I tried using CS inequality:
$$\sqrt{4-2x}+\sqrt{4-2y}\leq \sqrt{2}\t... | Remark: Alternatively, we can use the substitutions $\sqrt{4 - 2x} = u, \sqrt{4-2y} = v$.
From $\sqrt{4-2x} + \sqrt{4-2y} = \sqrt{xy}$, we have
$$(8 - 2x - 2y) + 2\sqrt{(4-2x)(4-2y)} \ge xy$$
or
$$4(8 - 2x - 2y) + 8\sqrt{(4-2x)(4-2y)} \ge 4xy$$
or
$$4(8 - 2x - 2y) + 8\sqrt{(4-2x)(4-2y)} \ge (4-2x)(4-2y) - 16 + 8x + 8y... | {
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Darboux sum $\int_{2}^{5}1-x+3x^2dx$ calculate the Darboux sum of
$$\int_{2}^{5}1-x+3x^2dx$$
I think i did the calculation good but I need help in getting my partition more formal please tell me if you see any errors and mis
$P=\{2,2+\frac{1}{n}....5\}$,
$\Delta x_i =\frac{1}{n} $
\begin{align}
&=\int_{2}^{5}1dx-\int_{... | There are plenty of errors.
*
*Take $x_i=2+\frac3n$, not $x_i=2+\frac1n$. Hence $\Delta x_i=\frac3n$, not $\frac1n$.
*Always keep "lim" in front of your "Darboux" (Riemann) sums because as such, they are not "equal" to your integrals.
*Don't put anything before $\Delta x_i$ inside your first sum (the one which corr... | {
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Show that $\frac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\frac{3\pi}{4}+\alpha\right)$ Show that $$\dfrac{1-\sin2\alpha}{1+\sin2\alpha}=\tan^2\left(\dfrac{3\pi}{4}+\alpha\right)$$
I am really confused about that $\dfrac{3\pi}{4}$ in the RHS (where it comes from and how it relates to the LHS). For the LHS:
$$\dfrac{1... | $$
\begin{aligned}
\tan^2 \left(\frac{3 \pi}{4}+\alpha\right) &=\tan ^2\left(\pi-\frac{\pi}{4}+\alpha\right) \\
&=\tan ^2\left(\frac{\pi}{4}-\alpha\right) \\
&=\left[\frac{\sin \left(\frac{\pi}{4}-\alpha\right)}{\cos \left(\frac{\pi}{4}-\alpha\right)}\right]^2 \\
&=\left(\frac{\sin \alpha-\cos \alpha}{\cos \alpha+\sin ... | {
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"answer_id": 7
} |
finding integral of $ \int \frac{1}{\sin x + \sqrt{3} \cos x}\ dx $ In $ \int\frac{1}{\sin x + \sqrt{3} \cos x}\ dx $, If I multiply and divide by $1/2$ I get $$ \int\frac{1/2}{\frac{1}{2} \sin x + \frac{\sqrt{3}}{2} \cos x}\ dx ,$$
then I can write $ 1/2=\sin (\frac{\pi}{6}) $ and $ \sqrt{3}/2=\cos (\frac{\pi}{6}) $ a... | You should not get different answers (up to a constant of integration.) In fact, this is simply a matter of a calculation error.
$$\int \sec x\, \mathrm dx = \log\left|\tan\left(\frac\pi 4 + \frac x2 \right) \right| + C$$
gives
$$\frac12 \int \sec \left(x - \frac\pi 6 \right)\mathrm dx = \frac12 \log\left|\tan\left(\fr... | {
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Derivation of the behavior of solutions to $\tan x = x$ This question is related to Chapter IV, Note IV.36 of Flajolet & Sedgewick's Analytic Combinatorics, and The question: Sum of the squares of the reciprocals of the fixed points of the tangent function as well:
Let $x_k$ be the $k^{th}$ positive root of the equatio... | You don't say precisely what contour $C$ you are using. For definiteness, I'm going to assume it's the square $C_m$ with side length $2m\pi$ ($m$ a positive integer), center 0, and sides parallel to the axes given in robjohn's answer to your linked question. Then by the residue theorem, for all nonnegative integers $... | {
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Evaluating Surface Integral with Divergence Theorem
Evaluate $\displaystyle\int_S \mathbf{F\cdot n}\ dS$ over the entire surface of the region above the $xy$ plane bounded by the cone $z^2=x^2+y^2$ and the plane $z=4$ if $\mathbf F=x\hat i+y\hat j+z^2\hat k$.
Solution: By the divergence theorem, since $\nabla \cdot\m... | Your computation is correct. Indeed, the direct computation gives
$$
\begin{align}\int_S \mathbf{F\cdot n}\ dS&=\iint_{x^2+y^2\leq 4^2}(x,y,4^2)\cdot(0,0,1) dx dy
\\&\;+\iint_{x^2+y^2\leq 4^2}(x,y,x^2+y^2)\cdot(\frac{x}{\sqrt{x^2+y^2}},\frac{y}{\sqrt{x^2+y^2}},-1) dx dy
\\&=16\cdot 16\pi+\int_{\theta=0}^{2\pi}\int_{r=0... | {
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A question on Brahmagupta Identity The Brahmagupta Identity states that
$$
\begin{align}
N & = (a^2+nb^2 )(c^2+nd^2 ) \\
& = (ac-nbd)^2+n(ad+bc)^2 \\
& = (ac+nbd)^2+n(ad-bc)^2 \\
\end{align}
$$
Knowing only $N$, is there a way to find $n$ such that the two cofactors of $N$ (not necessarily prime) can each be repres... | $N=35$ is represented as $x^2 + n y^2 $ for
$$ n = 10, 19, 26, 31, 34, 35 \; . $$
In particular, it is not represented by $x^2 + y^2, x^2 + 2 y^2, x^2 + 3 y^2, x^2 + 4 y^2, x^2 + 5 y^2$
For any of the $n$ above, $5,7$ are not represented by $x^2 + n y^2 $
Same idea for $$N = 55583= 11 \cdot 31 \cdot 163 $$... | {
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Prove $\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1$ for $x,y,z > 0 $
Let $x, y, z > 0$. Prove that
$$\frac{1+xy+xz}{(1+y+z)^2} + \frac{1+zy+zx}{(1+y+x)^2} + \frac{1+yx+yz}{(1+x+z)^2} \ge 1.$$
I encountered this problem today and thought that it was straightforward, and to b... | Hint :
Can you show ? For $a,b,c>0$
$$\frac{1+ba+bc}{(1+a+c)^{2}}-\frac{b^{2}+ba+bc}{(b+a+c)^{2}}=\frac{\left(b-1\right)^{2}\left(a+c\right)}{\left(a+c+1\right)^{2}\left(a+b+c\right)}\geq 0$$
| {
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How to solve this quadratic system of equations? The equations are:
$$ \begin{aligned} b + d &= c^2 - 6 \\ b - d &= -\frac{1}{c} \\ b d &= 6 \end{aligned} $$
and I want integer solutions for this.
I tried using various methods such as using $(a+b)^2-(a-b)^2=4ab$ or trying to solve for $b+d$ using the last two equations... | From the first two equations we get $b,d$ in terms of $c$:
$$b=\frac12\left(c^2-6-\frac{1}{c}\right)$$
$$d=\frac12\left(c^2-6+\frac{1}{c}\right)$$
Now substitute into the third:
$$\frac14\left((c^2-6)^2-\frac{1}{c^2}\right)=6$$
Let $a=c^2$:
$$(a-6)^2-\frac{1}{a}=24$$
$$a^3-12a^2+12a-1=(a-1)(a^2-11a+1)=0$$
Since $c$ is ... | {
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"answer_count": 4,
"answer_id": 1
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Evaluating $\lim\limits_{x \to 0} \frac{\cos\left(\sin^2 x\right)-1}{\sin x}$ without l'Hôpital I'm stuck trying to evaluate this limit without l'Hôpital's rule:
$$
\lim_{x \to 0} \frac{\cos\left(\sin^2x\right)-1}{\sin x}
$$
Could anyone give me a hint on what I need to do? I tried many trig identities but I can't see... | My attempt, using asymptotic estimates. As $u \to 0$,
$$
\cos u = 1 -\frac12 u^2 + O(u^4) .
$$
As $x \to 0$, also $\sin x \to 0$, so
\begin{align}
\sin x &= x + O(x^3)
\\
\frac{1}{\sin x} &= \frac{1}{x}+O(x)
\\
\sin^2 x &= x^2 + O(x^4)
\\
\cos(\sin^2 x) &= 1 -\frac12 \sin^2 x + O(\sin^4 x)
= 1 -\frac12 (x^2 + O(x^4)... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
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How to calculate $\mathbb{E}(X^2)$ when $X \sim Geo(\frac{1}{2})$ I am reading Mark Joshi's "Quant Job Interview Questions And Answers" question 3.7, and I am having trouble understanding how he proved that $\mathbb{E}(X^2)$ of $X \sim Geo(\frac{1}{2})$ is 6.
Here is his original question:
QUESTION 3.6. Suppose we toss... | Applying the trick once gives:
\begin{align*}
E(X^2) &= 0.5^1+2^2\times0.5^2 + 3^2\times0.5^3+\dots\\
\implies 0.5E(X^2) &= 0.5^2+2^2\times0.5^3 + 3^2\times0.5^4+\dots\\
\implies 0.5E(X^2) &= 0.5^1 +(2^2-1)\times0.5^2 + (3^2-2^2)\times0.5^3+\dots
\end{align*}
But $k^2-(k-1)^2$ simplifies to $2k-1$, so we can rewrite th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4546241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$? How to simplify $\ln |x-1| = 2\ln |\frac {y}{x}-1| - 3\ln |\frac{y}{x} - 2| + C; C = const$ to $(y-2x)^3 = C(y-x-1)^2$?
I'm trying to solve $(2x - 4y + 6)dx + (x+y-3)dy = 0$. The two lines intersect a... | Don't complete the square; instead from
$$-\frac1\tau\,d\tau=\frac{u+1}{u^2-3u +2}\,du$$
decompose $\frac{u+1}{u^2-3u+2}$ through partial fractions, integrate and temporarily leave the equation in terms of $\tau,u$ to get
$$K-\ln\tau=3\ln(u-2)-2\ln(u-1)$$
Then simplify:
$$K=\ln\frac{\tau(\mu/\tau-2)(\mu/\tau-2)^2}{(\mu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$
Finding solutions of $x$ for $(\sin x -1)(\sqrt{2} \cos x +1)=0$ in the given interval $0 \le x \le 2\pi$
Firstly, $\sin x - 1=0 \implies x= \frac{\pi}{2}$
Next, $(\sqrt{2} \cos x +1)=0 \implies x= \frac{3\pi}{4}... |
Can you find out where did you miss? Look at the blue graph. It's the graph of $y=\sqrt{2} \cos x + 1$. So, the root of the blue graph should be $x=\pi \pm \dfrac{\pi}{4} \Rightarrow x = \dfrac{3\pi}{4}, \dfrac{5\pi}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
You roll 5 dice. Find the probability of 2 pairs. I have seen this answered on here before. I have a slightly different form of answer, also my answer differs from the solution given.
My answer is
$${6 \choose 3} {5 \choose 2,2,1}6^{-5}.$$
${6 \choose 3}$ comes from choosing 3 faces out of 6 i.e $(2,3,1)$. We then arra... | Expanding on @user51547's comment, after choosing the 3 faces and grouping of the 5 dice, you also need to decide the assignment of the faces to the dice. In this case, we just need to decide which face is the unique one (not paired), and the remaining two must be the pair faces. So there is an extra $\begin{pmatrix} 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to prove $(1 + x)^{n} \leq 1 + 2nx$ I am currently working on a math problem, and it boils down to proving $(1 + x)^{n} \leq 1 + 2nx$, for a small $x$
By the Binomial expansion, it is clear that $$(1 + x)^n= 1+ nx + \frac{n(n-1)}{ 2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + \cdots$$
However, how can we prove that $$nx \ge... | One can continue from the binomial expansion, for $x>0$ this gives estimates against a geometric series
\begin{align}
(1+x)^n&= 1+nx\left(1+\frac{(n-1)x}{2}+\frac{(n-1)(n-2)x^2}{2·3}+...\right)
\\
&\le 1+nx\left(1+\frac{(n-1)x}{2}+\frac{(n-1)^2x^2}{2^2}+...\right)
\\
&\le 1+\frac{nx}{1-\frac{(n-1)x}{2}}
\end{align}
For... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Derive that $r=a\cos θ$ from $(x-h)^2 + (y-k)^2 = r^2$ I am a beginner here so please kindly tell me how I can do better when asking my questions so I can improve next time and if there are any complaints to this question.
I want to ask, how do I obtain the formula $r=a\cos θ$ from the circle formula $(x-h)^2 + (y-k)^2... | $$x=r \cos\theta\space; y=r \sin\theta\space; r=\frac{a}{2}\space;
C(\frac{a}{2}, 0)$$
$$ (x-h)^2 + (y-k)^2 = r^2$$
$$ (r \cos\theta-\frac{a}{2})^2 + (r \sin\theta-0)^2=(\frac{a}{2})^2$$
Using the identity $ (a-b)^2 = a^2-2ab+b^2$
$$
r^2\cos^2\theta - a r \cos\theta + \frac{a^2}{4} + r^2\sin^2\theta
=\frac{a^2}{4}
$$
S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4551643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $y'-\frac{y}{x}=\frac{(x+y)^2}{x^2}$ $$
\begin{aligned}
y'-\frac{y}{x} &=\frac{(x+y)^2}{x^2}\\\\
x^2y'-xy &=x^2+2xy+y^2\\
\end{aligned}$$
I have no idea what is the next step.
Updated:
Let $y=vx$, which $y′=v+v′x$
$$
\begin{aligned}
y'-\frac{y}{x} &=\frac{(x+y)^2}{x^2}\\
xv'&=(v+1)^2\\
\frac{v'}{(v+1)^2}&=\frac{1... | $$ax^2+bx+c=0\implies x=(-b\pm \sqrt{b^2-4ac})/(2a)$$
substitute: $ y^{(1)}=\frac{-(3/x)+/- \sqrt{9/x^2-4(1/x^2)(1)}}{2(1/x^2)}$
solution: $y=\int \frac{-(3/x)+/- \sqrt{9/x^2-4(1/x^2)(1)}}{2(1/x^2)} dx$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the derivative of $y=\sin(\cos^{-1}(2x))$ Question
Find the derivative of $y=\sin(\cos^{-1}(2x))$.
My Working
So I let $u=\cos^{-1}(2x)$, so
\begin{align}
\frac{du}{dx}&=-\frac{1}{\sqrt{1-(2x)^2}}\times2\\
&=-\frac{2}{\sqrt{1-4x^2}}
\end{align}
Also,
$$\frac{dy}{du}=\cos u$$
I feel like I should use the chain r... |
This differentiation can also be accomplished in a couple of other ways by constructing the right triangle implied by the argument of the sine function. If we call $ \ \theta \ = \ \cos^{-1}(2x) \ \ , \ $ then $ \ \cos \theta \ = \ 2x \ = \ \frac{2x}{1} \ \ . \ $ With the hypotenuse of the right triangle taken as $ \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\sum^{\infty}_{n=0} (-1)^n \frac{z^{4n+1}}{(4n+1)!}$ I want to evaluate $\sum^{\infty}_{n=0} (-1)^n \frac{z^{4n+1}}{(4n+1)!}$ but while doing my research, I noticed that
\begin{align*}
&\sin(z) = \sum^{\infty}_{n=1} (-1)^n \frac{z^{2n+1}}{(2n+1)!} \\
&\sinh(z) = \sum^{\infty}_{n=1} \frac{z^{2n+1}}{(2n+1)!}
\e... | Consider the following.
\begin{align}
\frac{1}{\sqrt{2}} \, \cosh\left(\frac{x}{\sqrt{2}}\right) \, \sin\left(\frac{x}{\sqrt{2}}\right) &= \sum_{r} \frac{x^{2 r}}{2^{r} \, (2 r)!} \times \sum_{s} \frac{(-1)^s \, x^{2 s + 1}}{2^{s+1} \, (2 s + 1)!} \\
&= \sum_{n} \sum_{s=0}^{n} \frac{(-1)^s \, x^{2 n +1}}{2^{n+1} \, (2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Probability of drawing two colors after four draws You have a bin with 5 balls of different colors: Red, Orange, Yellow, Green, and Blue, all of which are equally likely to be drawn. After a ball is drawn, it is put back afterwards. What is the probability of drawing both the yellow and red balls at least once after dr... | As Flip Tack pointed out in the comments, you have calculated the probability that at least one yellow ball or at least one red ball is selected. We wish to find the probability that ball of both colors are selected.
Method 1: We use the Inclusion-Exclusion Principle.
Let $R$ be the event that a red ball is chosen; l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556852",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $ then $y\leq 1$ or $y \geq 4$ Let $x \in \mathbb{R}^*$ and $y \geq \frac{1}{4}$.
Show that if $\sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y $ then $y\leq 1$ or $y \geq 4$
I tried the following idea:
$$
2\sqrt{y-\frac{1}{4}}\leq \sqrt{x^2+x+y} + \sqrt{x^2-x+y} = y
$$
which leads to:
$$
y... | Let $y=z^2+\frac14$ and
$$y=\sqrt{\left(x+\tfrac12\right)^2+z^2} + \sqrt{\left(\tfrac12-x\right)^2+z^2}$$
Now by triangle/Minkowski inequality,
$$\implies y \geqslant \sqrt{(1)^2+(2z)^2} = \sqrt{4z^2+1}=\sqrt{4y}$$
$\implies y(y-4)\geqslant 0 \implies y\geqslant 4$
Hence among the options provided, $y\leqslant1$ clearl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556991",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluate $\int x^2\log(1-x^2)dx$, and hence prove that $\frac1{1\cdot5}+\frac1{2\cdot7}+\frac1{3\cdot9}+...=\frac23\log2-\frac89$
Evaluate $\int x^2\log(1-x^2)dx$, and hence prove that $\frac1{1\cdot5}+\frac1{2\cdot7}+\frac1{3\cdot9}+...=\frac23\log2-\frac89$
My Attempt:
Integrating by parts,
$$\log(1-x^2)\cdot\frac{... | Using the Taylor series of $\log(1-t)$:
$$\log(1-t)=-\sum_{n=1}^\infty\frac{t^n}n,\qquad t\in[0,1),$$
we have
$$\int_0^1 x^2\log(1-x^2)dx=-\int_0^1x^2\sum_{n=1}^\infty\frac{x^{2n}}n\,dx=-\sum_{n=1}^\infty\frac1n\int_0^1x^{2+2n}\,dx=-\sum_{n=1}^\infty\frac1{n(3+2n)},$$
which is exactly the required LHS (except for the m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim_{n \to \infty}\left ( \frac{\pi^4}{48}-a_nb_n \right )n$ Consider the sequences $a_n$ and $b_n$ such that $a_n=\sum_{k=1}^{n}\frac{1}{k^2}$ and $b_n=\sum_{k=1}^{n}\frac{1}{(2k-1)^2}$. Compute $$\lim_{n \to \infty}\left ( \frac{\pi^4}{48}-a_nb_n \right )n$$
When I saw this question, my first thought is tha... | Hint: Notice that
\begin{align}
(ab-a_nb_n)n = n(a-a_n)b+a_n(b-b_n)n.
\end{align}
Next, observe that
\begin{align}
n(a-a_n) = n\left(\frac{\pi^2}{6}-a_n\right) = n\sum^\infty_{k=n+1}\frac{1}{k^2}
\end{align}
and that
\begin{align}
\frac{n}{n+1}=n\int^\infty_{n+1}\frac{1}{x^2} dx\le n\sum^\infty_{k=n+1}\frac{1}{k^2}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solving differential equation with step function without using Laplace Transforms. Suppose we have the differential equation:
$$ \ddot{y} + y = H(x - \pi) - H(x - 2\pi) $$
where $ H(x)$ is the Heaviside step function with initial conditions $ y(0) = \dot{y}(0) = 0 $ as initial conditions, and $ y(x) $ and $ \dot{y}(x) ... | Without the Laplace transform the solution is as follows.
Using
$$ H(t - \pi) - H(t - 2 \pi) = \begin{cases} 0 & t < \pi \\ 1 & \pi < t < 2 \pi \\ 0 & t > 2 \pi \end{cases} $$
then
$$\begin{cases} y_{1}^{''} + a^2 \, y_{1} \\ y_{2}^{''} + a^2 \, y_{2} \\ y_{3}^{''} + a^2 \, y_{3} \end{cases} = \begin{cases} 0 & t < \pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Show ${\pm 3^i}, 1\leq i \leq 2^{n-2}$ is reduced residue system mod $2^n$ Show set ${\pm 3^i}$, $1\leq i \leq 2^{n-2}$ is reduced residue system mod $2^n$.
For this, I believe I need to prove that for any integer $a$ such that $gcd(a,2^n)=1$, there exists an element $b$ in the set ${\pm 3^i}$, $1\leq i \leq 2^{n-2}$... | You're correct regarding what you need to prove, and have made a good start to determining a solution. Note that since there are $2^{n-1}$ odd integers between $1$ and $2^n$, inclusive, i.e., $\varphi(2^n)=2^{n-1}$, and the set $S = \{\pm 3^{i}, 1 \le i \le 2^{n-2}\}$ has $2(2^{n-2}) = 2^{n-1}$ elements, each of which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561942",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Can one deduce $\lim_{n\rightarrow\infty}P(X_n>1)=0$ from $P(X_n>1+\frac{x}{2\log n})\leq e^{-x}\quad\forall x$ I want to control the tail probability of some random variable $(X_n)_n$. (In particular, $X_n = \max_{1\leq i\leq n} N_i$, where $N_i$ are independent standard normal)
I come up with the following two upper ... | First, we determine the probability function of $X_n$, we have:
$$\begin{align}
P(X_n<a)&=P(\max_{i=1,..,n}N_i<a)\\
&=P^n(N_i<a)\\
&=\Phi^n(a)
\end{align}$$
With $\Phi(\cdot)$ the cumulative distribution function of $\mathcal{N}(0,1)$.
Return back to the main problem, it suffices to prove that for all $\epsilon>0$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565641",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding $x+y$, given $xy= 1$, $x^2+y^2=5$, $x^3+y^3=8$ This problem is from a math competition, but I think is wrong:
Find the value of $x+y$ if:
$$\begin{align}
xy &= 1 \\
x^2 + y^2 &= 5 \\
x^3+y^3 &= 8
\end{align}$$
Solution (I think is wrong):
$x^3 + y^3 = (x + y)(x^2-xy+y^2) = (x+y)(5-1) = 4(x+y)$
So we have:
$... | Alternative approach:
Let $z = (x + y)$.
The following equations over-determine the value of $z$.
*
*E-1: $~x^2 + y^2 = 5.$
*E-2: $xy = 1.$
*E-3: $x^3+y^3 = 8.$
Since $~z^2 = x^2 + y^2 + 2xy,~$
E-1,E-2 collectively imply that
$\displaystyle z^2 = 7 ~\implies~ z ~\in ~\left\{+\sqrt{7}, -\sqrt{7}\right\}.$
So, $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Evaluate $\sum_{n=0}^{\infty} \frac{(2n-1)!!}{2^n n! (2n+1)^2}$.
Evaluate $\sum_{n=0}^{\infty} \frac{(2n-1)!!}{2^n n! (2n+1)^2}$.
I am attempting to evaluate this sum as an alternate form of an integral I was trying to calculate. Can anyone give any insight? WolframAlpha says it's $\frac{1}{2}\pi\log(2)$ but I'm not ... | Consider using $ (2 n -1)!! = 2^n \, \left(\frac{1}{2}\right)_{n}$, where $(a)_{n}$ is the Pochhammer symbol, to obtain
$$ f(x) = \sum_{n=0}^{\infty} \frac{(2 n -1)!! \, x^{2 n +1}}{2^n \, n! \, (2n + 1)^2} = \sum_{n=0}^{\infty} \frac{\left(\frac{1}{2}\right)_{n} \, x^{2 n +1}}{ n! \, (2n + 1)^2}. $$
With some manipula... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4572404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Showing that $\prod_{k=1}^{n} \left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \right)$ is the square of a Fibonacci number I was experimenting with products of the form
$$\prod_{k=1}^{n} \left( a + b\cos(ck) \right)$$
when I found that the expression
$$\prod_{k=1}^{n} \left( 3 + 2\cos\left(\frac{2\pi}{n+1}k\right) \righ... | This was a blast solving. Never in a million years would I have guessed these to be true. But the $n+1$ in the denominator really hints at what to use. I will be skipping the more well known proofs.
We start with the Chebyshev polynomials of the second kind $U_n(x)$ which satisfy the following recursion relations
$$U_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4573706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 3,
"answer_id": 0
} |
About the equivalence of $\cos(x) = \sqrt{1-\sin^2(x)}$ and $\cos(x) = \cos(2x-x)$ By the title, my questions may sound trivial or silly, but I have problems with those outputs.
I need the sine of $15$ degrees. Now I thought of: $15 = 45-30$ hence:
$$\sin(15) = \sin(45-30) = \sin(45)\cos(30) - \cos(45)\sin(30) = \dfrac... | $$
\frac{\sqrt{2+\sqrt{3}}}{2} = \frac{\sqrt{8+4\sqrt{3}}}{4}
$$
Solving
$$
8+4\sqrt{3} = (a+b\sqrt{3})^2
$$
results in
$$
\begin{cases}
a^2+3b^2=8 \\
ab = 2
\end{cases}
$$
and solving that results in
$$
(a, b)= (\sqrt2, \sqrt2), (\sqrt6, \sqrt\frac{2}{3})
$$
so
$$
8+4\sqrt3 = (\sqrt6+\sqrt2)^2
$$
and thus
$$
\frac{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Expected value of highest die roll Daniel will roll a fair, six-sided die until he gets a 4. What is the expected value of the highest number he
rolls through this process?
It seems the expected number of rolls is 6. Can we reword this question to what is the expected value of the maximum of 6 die rolls?
| Let $P(6)$ denote the probability of getting a $(6)$ before the first $(4)$.
Let $P(5)$ denote the probability of getting a $(5)$ before the first $(4)$ and simultaneously, not getting a $(6)$ before the first $(4)$.
Then, the computation for the expected value of the highest roll is
$$4 +~ \left[ ~2 \times P(6) ~\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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A Nonrectifiable Curve Do Carmo's book Page 11.
Let $\alpha : [0,1] \to \mathbb{R}^2$ be given as $$\alpha(t)=
\begin{cases}
(t,t \sin (\pi/t)) & \text{ for } t \ne 0\\
(0,0) & \text{ for } t=0 \,.
\end{cases}$$ Show that the arc length of the portion of the curve corresponding to $\dfrac{1}{n+1} \leq t \leq \dfrac{1}... | Everything you have is correct. All that's remaining is to show that your lower bound is at least as large as the desired lower bound. Equivalently, we need to show that
$$\sum_1^{N-1} \frac{1}{n+0.5} - \sum_1^N \frac{1}{n+1} > 0 \,.$$
We have the following.
\begin{align*}
\sum_1^{N-1} \frac{1}{n+0.5} - \sum_1^N \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576898",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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$\int \frac{\cos 4x}{4 \sin 2x} dx$ $\int \frac{\cos 4x}{4 \sin 2x} dx$
Let $u=2x$, $dx = 1/2 du$
$\int \frac{\cos 2u}{4 \sin u} \frac{1}{2} du = \frac{1}{8} \int \frac{1-2\sin^2 u}{\sin u}du \frac{1}{8} \int \frac{1}{\sin u} du - \frac{1}{8} \int 2 \sin u$
How do I integrate $\int \frac{1}{\sin u} du$ to get $\ln (\ta... | $$\int \frac{dx}{\sin x}=\int \frac{d(\cos x)}{\cos^2x-1}
= \frac12\ln \frac{1-\cos x}{1+\cos x}=\frac12\ln\frac{\sin^2\frac x2}{\cos^2\frac x2}=\ln\tan\frac x2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that the diagonal elements of $V$ coincide Let $V$ be an $n\times n$ diagonal positive definite matrix, and suppose that
$$[I_n-XX']VX=0$$
for all $X\in\mathbb R^{n \times k}$ satisfying $X'X=I_k$, where $k<n$ is fixed.
Can I show that the diagonal elements of $V$ are all equal in this case?
| As you said you're interested in the case $k <n$, here it is. Denote $V=\mathrm{Diag}(a_1,a_2,\cdots,a_n)$ and $(e_1,\cdots,e_n)$ be the canonical basis of $\mathbb{R}^n$. Take $X \in \mathcal{M}_{n,k}(\mathbb{R})$ the matrix whose columns are $(e_1+e_2,e_3,e_4,\cdots,e_{k+1})$ It is possible as $k+1\leq n$. This is
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $\frac{1+\sin{\theta}}{\cos{\theta}} = n$, find $\tan{\frac{1}{2}\theta}$. The question is:
If $\cfrac{1+\sin{\theta}}{\cos{\theta}} = n$, find $\tan{\frac{1}{2}\theta}$.
I'm quite confused on how to solve this problem. What trigonometric identities do I use? I tried squaring them and ended with $n^2=\dfrac{1+\sin... | If we also have the "tangent half-angle" identity $ \ \tan \frac{\theta}{2} \ = \ \frac{1 \ - \ \cos \theta}{\sin \theta} \ \ , \ $ we can approach the problem in this way. From $ \ n \ = \ \frac{1 \ + \ \sin \theta}{\cos \theta} \ = \ \sec \theta + \tan \theta \ \Rightarrow \ \tan \theta \ = \ n - \sec \theta \ \ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Improper integrals with bounds 0 to infinity $$\int_{0}^{\infty }\frac{1}{x^{1/2}+x^{3/2}}dx$$
I managed to integrate correctly and get $2\arctan(\sqrt{x})+C$, but I was wondering why my teacher split it up into two pieces, one from $0$ to $1$ with a limit at $0$ evaluating $2\arctan(\sqrt{x})+C$ and the other $1$ to $... | If not wrong: You could solve the problem but you want to know why did your teacher split the above integral as
$$\int_0^\infty = \int_0^1 + \int_1^\infty$$
or as
$$\arctan(x)|_0^\infty = \arctan(x)|_0^1 + \arctan(x)|_1^\infty$$
then Comment by @Accelerator
however, for this particular case
$$\begin{align*}\int_0^\inft... | {
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"url": "https://math.stackexchange.com/questions/4588867",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Minimize function $\sqrt{x^2-4x+5}+\sqrt{4+x^2}$ Suppose I want to find
$$
\min f(x) = \min(\sqrt{x^2-4x+5}+\sqrt{4+x^2})
$$
I'd start with computing derivative and set it to $0$
$$
f'(x) = \frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}}
$$
Then
$$
\frac{x-2}{\sqrt{x^2-4x+5}} + \frac{x}{\sqrt{4+x^2}} = 0
$$
$$
(x-... | You can square, but only after ensuring that you're equating numbers that are either both positive or both negative.
In your case you have
$$\textstyle
(2-x)\sqrt{4+x^2}=x\sqrt{x^2-4x+5}
$$
The square roots give no problem: they are both defined anywhere and positive. The sign is determined by the other factors.
If $2-... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Usual method of partial fractions decomposition over the reals seems to fail. I assumed that it would be straightforward to find the partial fraction decomposition over the reals of the rational function $$f(x) = \frac{1}{(x^2 +1)^2}.$$ However, when I try what I thought would be the usual method of writing it as $$\fr... | The method didn't "fail". Partial Fractions had already reached the point that you want it to reach, namely one in which you have to do integrals all of which are of one of the following forms:
*
*$\displaystyle \int \frac{A}{(ax+b)^k}\,dx$, $k\geq 1$. These yield to a substitution $u=ax+b$.
*$\displaystyle \int\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\sqrt[2k+1]{x+1}+\sqrt[2k+1]{x+2}+\sqrt[2k+1]{x+3}+...+\sqrt[2k+1]{x+n}=0$ I noticed that when graphing the function $f(x)=\sqrt[2k+1]{x+1}+\sqrt[2k+1]{x+2}+\sqrt[2k+1]{x+3}+...+\sqrt[2k+1]{x+n}$, its roots seemed to behave predictably, that is, $x=-\frac{n+1}{2}$ seems to always be the only real zero of $f(x)... | Since $g(x)=\sqrt[2k+1]{x}$ is a strictly increasing function, therefore $f(x)$ is also a strictly increasing function. This means that, if there exist a real solution, then there is only one real solution.
We know that, the function $g(x)$ is odd. Using this property and setting $x+1=-(x+n)$ leads to:
$$\begin{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590996",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculus limit question If $f:(0,+\infty)\rightarrow\mathbb{R}$ continuous
$$f(x)=\lim_{n\to\infty}\frac{3^n(x^3+ax^2+3x+1)+x^n(2x^3+6x^2+6x+2)}{2x^n+3^n} \hspace{0.5em} \forall x\in(0,3)\cup(3,+\infty) ,a\in \mathbb{R}$$Prove $f(x)=(x+1)^3 \hspace{0.2em},x>0 $
Ok so $$\Leftrightarrow f(x)=\lim_{n\to\infty}\frac{3^n(x... | Something is wrong here. For $a=0$ and $x=1$ we have
$$f(1)=\lim_{n\to\infty}\frac{3^n(1+3+1)+1^n(2+6+6+2)}{2\cdot 1^n+3^n}=\lim_{n\to\infty}\frac{5\cdot 3^n+16}{3^n+2}=5\neq 8=(1+1)^3$$
The formula only works if $a=3$.
EDIT: As @enzotib pointed out, the function $f(x)$ is assumed to be continuous. Really, this exerci... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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In $\triangle ABC$, if $AC=4$ and $BC=5$, and $\cos(A-B)=\frac{7}{8}$, find $\cos(C)$ The problem is as the title suggests, in the given figure below, the goal is to find the Cosine of $\angle C$. I tried multiple ways of approaching this, such as with the Law of Sines, area formula, etc but none of them seemed to lead... | We can use Law of Tangents
Let $t_1 = \tan\frac{A-B}{2}$
Let $t_2 = \tan\frac{A+B}{2}$
$\displaystyle \cos(A-B) = \frac{7}{8} = \frac{1-t_1^2}{1+t_1^2}
\quad → t_1^2 = \frac{8-7}{8+7} = \frac{1}{15}$
$\displaystyle \frac{\tan\frac{A-B}{2}}{\tan \frac{A+B}{2}}
=\frac{a-b}{a+b}$
$\displaystyle \frac{t_1}{t_2} = \frac{5-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Are there any other fraction-to-decimal conversions like $\frac{5}{2} = 2.5$? I noticed that to convert 2.5 to a fraction, it's the number after the decimal divided by the number before the decimal. I wondered if there are any others like these, and I cannot find any nor prove that they don't exist.
So the problem, exp... | I don't have a complete answer to the problem, but I have shown that $(2, 5)$ is the only coprime pair.
I'll write $d=10$ in the hopes of generalizing it to other $d\in\mathbb{Z}$ with $d\ge 2$.
Multiplying by $d^{\lfloor \log_{d}(b)\rfloor+1}a$ makes it
$$d^{\lfloor \log_{d}(b)\rfloor+1}b=a^2d^{\lfloor \log_{d}(b)\rf... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4596205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
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"answer_id": 0
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Finding the smallest $k$ such that $a^3+b^3+c^3+kabc\leq\frac16(k+3)(a^2(b+c)+b^2(a+c)+c^2(a+b))$, where $a,b,c$ are sides of a triangle It is known that $a, b, c$ are the sides of the triangle. Determine the smallest value of $k$, so that
$$a^3+b^3+c^3+kabc\leq\frac {k+3}{6} (a^2(b+c)+b^2(a+c)+c^2(a+b))$$
My working:
... | Since $a,b$ and $c$ are sides of a triangle, we are allowed to set $a=x+y$, $b=y+z$ and $c=x+z$. Now, let's define:
$$A=x^3+y^3+z^3 \\ B=x^2y+xy^2+z^2x+zx^2+y^2z+yz^2 \\C=xyz.$$
Note that by the Schur's inequality we already know that: $A+3C\ge B.$ Moreover it is not hard to see (by Muirhead's inequality) that $2A\ge B... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why are there complex numbers in the exact solution of $\sum_{n=1}^{\infty} \dfrac{1}{n^4+1}$? Knowing that
\begin{align}
\cot(z)=\frac{1}{z}-2z\cdot\sum_{n=1}^{\infty} \dfrac{1}{\pi^2n^2-z^2}
\end{align}
we can easily calculate the value of
\begin{align}
\sum_{n=1}^{\infty} \dfrac{1}{n^2+1}
\end{align}
by just plugg... | Addendum to Somos' answer: by expanding $\cot(x+iy)$ via trig identities, one can determine the real and imaginary parts of $\cot(z)$, allowing the answer to be expressed purely in terms of real-valued terms:
$$-\frac{1}{2}-\frac{\pi \sin \left(\sqrt{2} \pi \right)}{2 \sqrt{2} \left(\cos
\left(\sqrt{2} \pi \right)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove or disprove the inequality if $a,b,c>0$, $a \geq b+c$. Prove or disprove the inequality
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 7abc$$ if $$a,b,c>0, a \geq b+c.$$
I thought to use this evaluation:
$$a^2b+b^2c+c^2a \geq 3abc.$$
So we have:
$$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 3abc+3abc=6abc,$$ which is obvious that $... | Dividing by $abc > 0,$ the relation to be shown is $$ \frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} \overset?\ge 7$$
Set $a = b+c + d$ for $d \ge 0.$ We then need to argue that for $b,c>0, d\ge 0,$
$$ \frac{b+c}{b+c+d} + \frac{2c+d}{b} + \frac{2b+d}{c} \overset?\ge 5$$
The LHS is invariant under scaling, so wlog set... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$ \text{sec}^2x + 3\text{cosec}^2x =8 $ Solving this trigonometric equation. I have found two different methods of solving this trigonometric equation :
$$ \text{sec}^2x + 3\text{cosec}^2x =8
$$
But these methods give different answers.
Solution 1
$$ \text{sec}^2x + 3\text{cosec}^2x =8 $$
$$\implies \frac{1}{\text{cos... | Using $\cos^{2}x + \sin^{2}x = 1$, from your first solution method and first factor, $2\cos^{2}x - 1 = 0 \; \to \; \cos^{2}x = \frac{1}{2}$, so $\sin^{2}x = 1 - \frac{1}{2} = \frac{1}{2}$ and, thus, $\tan^{2}x = \frac{1/2}{1/2} = 1$. The second factor gives $4\cos^{2}x - 1 = 0 \; \to \; \cos^{2}x = \frac{1}{4}$, so $\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Help Solve $\int \sqrt{x^{2}+x-2}\,dx$ I have this integral $\int \sqrt{x^{2}+x-2}\,dx=\int |x-1|\sqrt{\frac{x+2}{x-1}}\,dx$.
My textbook advices to use the substitution $t^{2}=\frac{x+2}{x-1}$.
Observation: This integral should be solvable without using integral of modules as it is placed before moduled function integ... | First, reduce to a simpler square root
\begin{align*}
\int \sqrt{x^2+x-2} dx &= \int \sqrt{\left(x+\frac{1}{2}\right)^2-\frac{9}{4}} \,dx = \\
&= \frac{3}{2}\int \sqrt{\frac{4}{9}\left(x+\frac{1}{2}\right)^2-1} \,dx = \\
&= \frac{3}{2}\int \sqrt{\left(\frac{2}{3}x+\frac{1}{3}\right)^2-1} \,dx = \\
&= \frac{9}{4}\int \s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
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If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, If $α,β,γ$ are the roots of the equation $f(x)=x^3+qx+r=0$ then find the equation whose roots are, $\frac{\beta^2+\gamma^2}{\alpha^2}$,$\frac{\alpha^2+\gamma^2}{\beta^2}$,$\frac{\beta^2+\alpha^2}{\gamma^2}$.
My solution g... | HINT.-From Vieta we get $$\alpha^2+\beta^2+\gamma^2=-2q\\(\alpha\beta)^2+(\alpha\gamma)^2+(\beta\gamma)^2=q^2\\(\alpha\beta\gamma)^2=r^2$$ Now instead of $a,b,c$ we consider $a+1=\dfrac{-2q}{\alpha^2},b+1=\dfrac{-2q}{\beta^2}$ and
$c+1=\dfrac{-2q}{\gamma^2}$ so the equation in $a+1,b+1,c+1$ is
$$\left(X+\dfrac{2q}{\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Integrating $\frac{x^3}{\sqrt{x^2 + 4x + 6}}$ Question:
Evaluate $\displaystyle\int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx $.
My attempt:
$\begin{align} \int \frac{x^3}{\sqrt{x^2 + 4x + 6}}\ dx & = \int \frac{x^3}{\sqrt{(x+2)^2 + 2}}\ dx \\& \overset{(1)}= \int \frac{(\sqrt{2} \tan(t) - 2 )^3 \sqrt{2} \sec^2(t)\ }{\sqrt{2... | For indefinite integral of the form
$$I_n=\int \frac{x^n}{\sqrt{x^2 + bx + c}}\ dx
$$
it is advised that $I_n$ be reduced first to $I_0$ before any substitution. This is achieved by the reduction formula below with $f(x)=x^2+bx+c$
$$\int \frac{f’(x)^{n}}{\sqrt{f(x)}}dx= K_n = \frac2nf’(x)^{n-1}\sqrt{f(x)}+\frac{n-1}n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Find the intersection points of two circles Find the intersection points of the circles $$k_1:(x-4)^2+(y-1)^2=9\\k_2:(x-8)^2+(y+4)^2=100$$
The intersections point (if they exist) will satisfy the equations of both the circles, so we can find their coordinates by solving the system $$\begin{cases}(x-4)^2+(y-1)^2=9\\(x-... | Geometric point of view: The radical axis of two circles passes through the intersection points of the circles when they intersect. If we show that the distance of the centers of the circles to that line is larger than their radii, we're done.
You found the radical axis. It is $L:4x-5y+14=0$. Now, $h_1=d(L,O_1)=\frac{|... | {
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"url": "https://math.stackexchange.com/questions/4618115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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Proving certain properties for Babylonian method The Babylonian method for approximating square roots is divided into three steps:
*
*Guess an initial approximation $a$ of $ {\sqrt N}$, where $a$ and $N$ are postive rational numebrs and $N$ is not the square of any rational numbers.
*Let $c=\frac{N-a^2}{2a}$.
*Now ... | \begin{align*}
a+c &= a + \frac{N - a^2}{2a}
= \frac{N + a^2}{2a}
= \frac{(\sqrt{N} - a)^2 + 2a\sqrt{N}}{2a}\\
&= \frac{(\sqrt{N} - a)^2}{2a} + \sqrt{N} \geq \sqrt{N}
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4625339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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prove $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$ $(\sin x)^{-2}-x^{-2}\leq 1-\frac{4}{{\pi}^{2}},x\in(0,\pi/2]$
How to deal with this problem?
Observing that when $x=\pi/2$, the above inequality becomes equality.
Firstly, denote $f(x)=(\sin x)^{-2}-x^{-2}$ and then take derivative of $f(x)$. We have... | It's related to :
$$|\frac{1}{\sin(x)}-\frac{1}{x}|\leq 1-\frac{2}{\pi}$$
Called Prestin's inequality
Use the fact :
$$\left|\frac{1}{\sin^{2}(x)}-\frac{1}{x\sin\left(x\right)}\right|-\left(1-\frac{2}{\pi}\right)\geq (\sin x)^{-2}-x^{-2}-\left(1-\frac{4}{\pi^{2}}\right)$$
And :
$$\left|\frac{1}{\sin(x)}-\frac{1}{x}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4626032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Please explain how to solve limit. I know the answer but how to explain it? Problem:
$a@b = \frac{a+b}{ab+1}$. Solve limit:
$\lim_{n \to \infty}(2@3@...@n)$.
I've tried to solve this problem by just calculating:
$$2 @ 3 = 0.714$$
$$2 @ 3 @ 4 = 1.222$$
$$2 @ 3 @ 4 @ 5 = 0.875$$
$$2 @ 3 @ 4 @ 5 @ 6 = 1.1$$
I found the pa... | $\DeclareMathOperator{@}{\operatorname@}$
As noted in a comment, the $\@$ operator is associative.
By expanding $(a\@ b\@c)$ and $(a\@b\@c\@d)$:
$$\begin{align*}
a\@b &= \frac{a+b}{ab+1}\\
(a\@b)\@c &= \frac{\frac{a+b}{ab+1}+c}{\frac{a+b}{ab+1}c+1}= \frac{a+b+c(ab+1)}{(a+b)c + ab+1} = \frac{abc + a+b+c}{ab+ac+bc+1}\\
(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4628158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
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Calculate the following integral $ \int\frac{2x+1}{x^{n+2}(x+1)^{n+2}}\ln\left(\frac{2x^2+2x+1}{x^2(x+1)^2}+\frac7{16}\right)dx$ Hello I am trying to solve a pretty complicated integral. It is a from a set of problems, published in a monthly journal for high school students and they are exercises in preparation for a c... | Starting as you did, with
$$du=\frac {2x+1}{x^{n+2}(x+1)^{n+2}}\,dx\qquad \text{and}\qquad v=\log\left(\frac {2x^2+2x+1}{x^2(x+1)^2} + \frac{7}{16}\right)$$as you wrote
$$u= -\frac {1}{(n+1)x^{n+1}(x+1)^{n+1}}$$ $$ dv=-\frac{32 (2 x+1) \left(x^2+x+1\right)}{x (x+1) \left(x^2+x+4\right) (7 x^2+7x+4)}$$
$$u\,dv=\frac {32... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4631821",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Determining whether $\frac{1}{2^{2}} + \frac{2}{3{^2}} + ... +\frac{n}{(n+1)^{2}}$ is a Cauchy sequence. My attempt so far: If $n \leq m$, then
$a_{m} - a_{n} = \frac{n+1}{(n+2)^2}+\frac{n+2}{(n+3)^2}+...+\frac{m}{(m+1)^2} < \frac{n+1}{(n+2)(n+1)}+\frac{n+2}{(n+3)(n+2)}+...+\frac{m}{(m+1)m} = \frac{1}{(n+2)}+\frac{1}{(... | Note that we simply have $k\geq (k+1)/2$ for any integer, so
$$
\sum_{k=1}^n \frac{k}{(k+1)^2} \geq \sum_{k=1}^n \frac{\frac{1}{2}(k+1)}{(k+1)^2} = \sum_{k=1}^n \frac{1}{2} \frac{1}{k+1} = \frac{1}{2} \sum_{k=1}^n \frac{1}{k+1}
$$
and note that the right most expression is nothing but harmonic series starting from $k=2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equation of parabola passes through $4$ distinct points
Equation of axis of parabola which
passes through the point $(0,1)\ , \ (0,2)$
And $(2,0)\ ,\ (2,2)$ is
Let general equation of conic is
$ax^2+2hxy+by^2+2gx+2fy+c=0\cdots (1)$
And it represent parabola if $h^2=ab$
Parabola passes through $(0,1)$
Then put int... | What you've done is correct.
One way to find the axis is to write the equation as $(x+4y)^2-18x-48y+32=0,$ then you know the axis is parallel to $x+4y=0.$ The tangent at vertex is orthogonal and parallel to $4x-y.$ Now you can find where $4x-y+c$ intersects doubly to find the vertex. $$x^2+8x(4x+c)+16(4x+c)^2-18x-48(4x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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Universal substitution or Feynman trick to solve this integral I started with an integral $ \int_{0}^{2\pi} \sqrt{2[\sin^2(t) + 16\cos^2(t) - 4\sin(t)\cos(t)]} \,dt $
And I simplified it to $ \int_{0}^{2\pi} \sqrt{17 + 15\cos(2t) - 4\sin(2t)} \, dt$
My question: I know this can be simplified with some sort of substitut... | $\begin{align}
\int_0^{2\pi}\sqrt{17+15\cos2t-2\sin2t}\,dt&\overset{2t=x}{=}\frac12\int_0^{4\pi}\sqrt{17+15\cos x-2\sin x}\,dx\\
&=\int_0^{2\pi}\sqrt{17+15\cos x-2\sin x}\,dx\\
&=\int_0^{2\pi}\sqrt{17+\sqrt{229}\cos(x+\alpha)}\,dx\\
&=\int_\alpha^{2\pi+\alpha}\sqrt{17+\sqrt{229}\cos x}\,dx\\
&=\int_0^{2\pi}\sqrt{17+\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4646773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Proof that $3 \mid 10^{n+2} - 2*10^n + 7, \forall n \in \mathbb{Z}^+$. This is what I have so far.
Proof by Induction. Let $n \in \mathbb{Z}^+$ Let $P(n)$ be the statement that $10^{n+2} - 2*10^n + 7$ is divisible by 3.
($\textit{Base Case}$): Let $n = 1$.
$$
10^{1+2} - 2*10^1 + 7 = 1000 - 20 + 7 = 987
$$
$3 \mid 987$... | Note that $10^k=99\dots(k\ times)\dots 9+1$. Now $99\dots(k\ times)\dots 9$ is always divisible by $9$ and therefore it is divisible by $3$. So, dividing $10^k$ by $9$ gives a remainder of $1$.
Therefore, $10^{n+2}\equiv 1 \mod 3$ and $10^{n}\equiv 1\mod 3$. So, $10^{n+2}-2\times 10^{n}\equiv 1-2\times 1\mod 3\equiv -1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4650375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Sum of the alternating harmonic series $\sum_{k=1}^{\infty}\frac{(-1)^{k+1}}{k} = \frac{1}{1} - \frac{1}{2} + \cdots $ I know that the harmonic series $$\sum_{k=1}^{\infty}\frac{1}{k} = \frac{1}{1} + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + \frac{1}{5} + \frac{1}{6} + \cdots + \frac{1}{n} + \cdots \tag{I}$$ diverges,... | Here is another proof, based on the formula
$$\frac{1}{1+x}=\frac{(-1)^nx^n}{1+x}+\sum_{k=0}^{n-1}(-1)^kx^k.$$
Integrating both sides over $[0,t]$ gives
$$\ln(1+t)=\int_0^t\frac{(-1)^nx^n}{1+x}\,dx+\sum_{k=1}^n(-1)^{k+1}\frac{t^k}{k}.$$
Setting $t=1$ shows that the partial sums $s_n$ of the alternating harmonic series ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "60",
"answer_count": 12,
"answer_id": 7
} |
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| Let $k$ be any number. We know that $(k+1)² - k² = 2k + 1$
Let $k = 1, 2, 3, 4, ..., n$
$$(1+1)² - (1)²= 2(1) + 1$$
$$(2+1)² - (2)²= 2(2) + 1$$
$$(3+1)² - (3)²= 2(3) + 1$$
$$.$$
$$.$$
$$.$$
$$.$$
$$(n+1)² - (n)²= 2(n) + 1$$
Adding all the equations, we get:
$$- (1)² + (n+1)² = 2(1 + 2 + 3 + ... + n) + n(1)$$
$$n² + 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
"answer_count": 36,
"answer_id": 35
} |
What is $\sqrt{i}$? If $i=\sqrt{-1}$, is $\large\sqrt{i}$ imaginary?
Is it used or considered often in mathematics? How is it notated?
| More generally, if you want to compute all the $n$-th roots of a complex number $z_0$, that is, all the complex numbers $z$ such that
$$
z^n = z_0 \ , \qquad \qquad \qquad \qquad [1]
$$
you should write this equation in exponential form: $z = re^{i\theta}, \ , z_0 = r_0 e^{i\theta_0}$. Then [1] becomes
$$
\left( r e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "88",
"answer_count": 10,
"answer_id": 7
} |
Summing the series $ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \cdots \ \text{ad inf}$ How does one sum the given series: $$ \frac{1}{2n+1} + \frac{1}{2} \cdot \frac{1}{2n+3} + \frac{1 \cdot 3}{2 \cdot 4} \frac{1}{2n+5} + \frac{ 1 \cdot 3 \cdot 5}{2 \cdot 4 \cdot 6} \frac{1}{2n+7} + \cdots \ \text{ad inf}$$
G... | FWIW, Mathematica is able to sum this:
In[1]:= Sum[1/(2 n + 2 k + 1) (2 k - 1)!!/(2 k)!!, {k, 0, \[Infinity]}]
3 + 2 n
Sqrt[Pi] Gamma[-------]
2
Out[1]= -----------------------
(1 + 2 n) Gamma[1 + n]
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
Algebraic puzzle: Infer $x = \frac{y + z}{2}$ from $a^{\frac{1}{x}} = b^{\frac{1}{y}} + c^{\frac{1}{z}}$ and $a = b + c$ I have a suspicion that the following expression is true, however my algebra skills aren't brilliant, so any help would be appreciated:
Is it possible to infer $x = \frac{y + z}{2}$ from $a^{\frac{1}... | Let $a=2$, $b=c=1$. Also let $x=1$ then $a=b+c$ and $a^{\frac{1}{x}}=2$. Regardless of the values of $y$ and $z$ then $1^{\frac{1}{y}}+1^{\frac{1}{z}}=2$.
Thus we cannot infer that $x=\frac{y+z}{2}$
Counterexample:
Let $a=2$, $b=c=1$, $x=1$ and $y=z=2$ then $b+c=2=a$ and $b^{\frac{1}{y}}+c^{\frac{1}{z}}=2=a^{\frac{1}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The Basel problem As I have heard people did not trust Euler when he first discovered the formula (solution of the Basel problem)
$$\zeta(2)=\sum_{k=1}^\infty \frac{1}{k^2}=\frac{\pi^2}{6}$$
However, Euler was Euler and he gave other proofs.
I believe many of you know some nice proofs of this, can you please share it w... | OK, here's my favorite. I thought of this after reading a proof from the book "Proofs from the book" by Aigner & Ziegler, but later I found more or less the same proof as mine in a paper published a few years earlier by Josef Hofbauer. On Robin's list, the proof most similar to this is number 9
(EDIT: ...which is actua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "814",
"answer_count": 48,
"answer_id": 14
} |
Prove that $\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$ Using $\text{n}^{\text{th}}$ root of unity
$$\large\left(e^{\frac{2ki\pi}{n}}\right)^{n} = 1$$
Prove that
$$\prod_{k=1}^{n-1}\sin\frac{k \pi}{n} = \frac{n}{2^{n-1}}$$
| Here is a more "1st principles" pf. I use a hint in Marsden's book.
1st, $\cos(A-B)-\cos(A+B)=2\sin A \sin B$ (1), which follows by angle summation formulas.
Next, we use Marsden's hint to consider roots of $(1-z)^n-1$. These satisfy
$$(1-z)^n=1 \leftrightarrow (1-z) \in \left\{\cos \frac{2 \pi k}{n}+i \sin \frac{2\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8385",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "76",
"answer_count": 3,
"answer_id": 2
} |
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Fractions in Questions and Answers
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