Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS My thinking:
Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$
By the AGM (Arithmetic-Geometric Mean Inequality):
We have
$x_1\cdot x_2\le \left(\frac{x_1\cdo... | $x_1^2+x_2^2=(x_1+x_2)^2-2(x_1x_2)=a^2-2(a-1)$
The last equility is Vieta or if you wish $(x-x_1)(x-x_2)=x^2-(x_1+x_2)x+(x_1x_2).$
So the min is when $a=1$ (because of usual calculus).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Can we re-write Newton's Binomial formula as a power series in $\ r\ $ without any problems? Newton's Generalised Binomial theorem states that if $\ x\ $ and $\ y\ $ are real numbers with $\ \vert x \vert > \vert y \vert\ (\text{note that } \left\vert \frac{y}{x} \right\vert < 1),\ $ and $\ r\ $ is any complex number, ... | $$ (x+y)^r = \sum_{k=0}^{\infty} \binom{r}{k} x^{r-k} y^k = x^r + \frac{r}{1!} x^{r-1} y^1 + \frac{r(r-1)}{2!} x^{r-2} y^2 + \frac{r(r-1)(r-2)}{3!} x^{r-3} y^3 + \ldots\ .$$
Dividing through by the constant $\ x^r,\ $ we see that the above series converges if and only if
$$ \left(1+\frac{y}{x}\right)^r = \sum_{k=0}^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360296",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integral $\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx$ I am trying to compute the integral
$$\int_0^{\pi/3}\frac{x}{\cos(x)}\,dx \tag{1}$$
Context: Originally I was trying to prove the following result:
$$\sum_{n=0}^\infty\frac{1}{(2n+1)^2\binom{2n}{n}}=\frac83\beta(2)-\frac{\pi}3\ln(2+\sqrt{3})\tag{2}$$
Where $\beta(2)$ is th... | $$I=\int_0^\frac{\pi}{3}\frac{x}{\cos(x)}\,dx $$
Making the substitution $x=\frac{\pi}{2}-t$
$$I=\frac{\pi}{2}\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{dt}{\sin t}-\int_\frac{\pi}{6}^\frac{\pi}{2}\frac{t\,dt}{\sin t}=I_1-I_2$$
Next, we make the substitution $\frac{dt}{\sin t}=d\big(\ln(\tan\frac{t}{2})\big)$
$$I_1=\frac{\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4360729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Show that $f(x) = x^3 + 3\sin x + 2\cos x$ is one-to-one. How would I show that $f(x) = x^3 + 3\sin x + 2\cos x$ is one to one?
Showing that the function is strictly increasing seemed to be the way to go, but then I need to show that $f'(x) = 3x^2 + 3\cos x - 2\sin x > 0$. I graphed the derivative function and it is in... | To show that $f(x) = x^3 + 3 \sin x + 2 \cos x $ is one-to-one, it is enough to show that $ f'(x) = 3 x^2 + 3 \cos x - 2 \sin x > 0 $ for all $ x $. Notice that $ (3 \cos x - 2 \sin x) \ge -2 $ for all $ x $. So, whenever $ 3 x^2 > 2 $, $ f'(x) $ must be positive. Note that for $ x \le -\pi/2 $ or $ x \ge \pi/2 $, $ 3 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Determinant of a special block matrix $\big(\begin{smallmatrix}A&B\\B&A\end{smallmatrix}\big)$ Let be $ A,B\in \mathbb{K}^{n,n} $ arbitrary matrices. Then
$$ \det\Bigg(\underbrace{\begin{pmatrix}A&B\\B&A\end{pmatrix}}_{=:L}\Bigg)=\det(A+B)\cdot \det(A-B) $$
My idea: I consider
$$ \begin{align}&\det(A+B)\cdot \det(A-B)\... | You can use this decomposition:
$$
\begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix}^{-1} \begin{pmatrix} A &B \\ B &A \end{pmatrix} \begin{pmatrix} I_n &-I_n \\ I_n &I_n \end{pmatrix} = \begin{pmatrix} A+B &0 \\ 0 &A-B \end{pmatrix}
$$ You can see the motivation for this by thinking of $A, B$ as real numbers and $I_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many ways are there to prove that $ \int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x=2 G?$ In my post, I had found the values of the couple of integrals,
$$
\int_{0}^{\frac{\pi}{4}} x \tan x d x=-\frac{\pi}{8} \ln 2+\frac{G}{2}\tag*{(1)}
$$
and
$$
\int_{0}^{\frac{\pi}{4}} x \cot x d x=\frac{\pi}{8} \ln 2+\frac{G}{2}\... | Inspired by Mr Svyatoslav,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{x}{\sin x} d x &=4 \int_{0}^{\frac{\pi}{2}} x d\left(\ln \left(\tan \frac{x}{2}\right)\right) \\
&=4\left[x \ln \left(\tan \frac{x}{2}\right)\right]_{0}^{\frac{\pi}{2}}-4 \int_{0}^{\frac{\pi}{2}} \ln \left(\tan \frac{x}{2}\right) d x \\\\
&=-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ improve $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the abundancy index of the positive integer... | This is a partial answer.
Note that
$$I(n^2) = \dfrac{2}{I(q^k)} \leq \dfrac{2}{I(q)} = \dfrac{2q}{q+1},$$
since $k \equiv 1 \pmod 4$ implies that $k \geq 1$.
Following exactly the same steps in the original post, we obtain
$$I(q^k) < I(n^2) \leq \dfrac{2q}{q+1}.$$
After working out the details, we get
$$I(q^k) + I(n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to find exact value of integral $\int_{0}^{\infty} \frac{1}{\left(x^{4}-x^{2}+1\right)^{n}}dx$? When I first encountered the integral $\displaystyle \int_{0}^{\infty} \frac{1}{x^{4}-x^{2}+1} d x$, I am very reluctant to solve it by partial fractions and search for any easier methods. Then I learnt a very useful tr... | Note
$$
\int_0^\infty \frac1{(x^4-x^2+1)^{n+1}}dx=\frac{(-1)^{n}}{n!} \frac{d^{n} J(a)}{da^{n}}\bigg|_{a=1}
$$
where
$$J(a)=\int_0^\infty \frac{dx}{x^4-x^2+a}=\frac1{2\sqrt a}
\int_0^\infty \frac{d(x-\frac{\sqrt a}x)}{(x-\frac{\sqrt a}x)^2+2\sqrt a-1}
= \frac\pi{2\sqrt{a(2\sqrt a-1)} }
$$
Then
\begin{align}
& \int_0^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4367988",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
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Approximating factorial using identity $\frac{1}{x}!\frac{2}{x}!\cdots\cdot\frac{x}{x}!=\frac{ {x}!\cdot(2\pi)^{\frac{x-1}{2}} }{ x^x\cdot\sqrt{x} }$ I created a function that describes the product of the inverse multiples of a factorial
$$ m(x) = \frac{1}{x}!\cdot\frac{2}{x}!\cdot\frac{3}{x}!\cdots\frac{x-1}{x}!\cdot\... | The Gauss multiplication formula states that
$$
\prod\limits_{k = 0}^{n - 1} {\Gamma\! \left( {z + \frac{k}{n}} \right)} = (2\pi )^{(n - 1)/2} n^{1/2 - nz} \Gamma (nz)
$$
for $n\geq 1$ and for all complex $z$ for which both sides are defined. Taking $z=1$ and using $r!=\Gamma(1+r)$ gives
$$
\prod\limits_{k = 1}^n {\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4371422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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Show using induction that for every $n\in \mathbb{N}$ the number $P_n=2^{2^{n+1}}+2^{2^n}+1$ is divisible by $21$. Show using induction that for every $n\in \mathbb{N}$ the number $P_n=2^{2^{n+1}}+2^{2^n}+1$ is divisible by $21$.
So the first step is to check for $n=1$ $$P_1=2^{2^2}+2^2+1=16+5=21,$$ which is divisible ... | Let $A_n=2^{2^n}$. Then $P_n=2^{2^{n+1}}+2^{2^n}+1=A_n^2+A_n+1$ and...
$$P_{n+1}=2^{2^{n+2}}+2^{2^{n+1}}+1=2^{4\cdot2^n}+2^{2\cdot2^n}+1=A_n^4+A_n^2+1\\=(A_n^2+A_n+1)(A_n^2-A_n+1)=P_n(A_n^2-A_n+1)$$
And $P_1=21$. So... the answer is clear! If $21\mid P_n$, then automatically $21\mid P_{n+1}$ holds!
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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If x and y are real numbers such that $4x^2+2xy+9y^2=100$ then what are all possible values of $x^2+2xy+3y^2$ If $x$ and $y$ are real numbers and $4x^2+2xy+9y^2=100$ then what are all possible values of $x^2+2xy+3y^2$.
The first thing I did with this equation was to try and factor the first equation, but that didn't wo... | From $4 x^2 + 2 x y + 9 y^2 = 100 $, by defining $ r= [x, y]^T $ we can write
$ r^T Q r = 100 $
where $ Q = \begin{bmatrix} 4 && 1 \\ 1 && 9 \end{bmatrix} $
Diagonlizing $Q$, we can factor it into $ Q = R D R^T $
We get
$D = \begin{bmatrix} \frac{1}{2} ( 13 - \sqrt{29} ) && 0 \\ 0 && \frac{1}{2} ( 13 + \sqrt{29}) \end{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4376993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Can the value $\lim \frac{\sin x}{x} = 1$ be used even if $x$ is not approaching $0$? I was going through a solved example that involved calculating the coefficient $a$ for a piecewise function to be continuous at $x = 0$:
$f(x) = a \sin{\frac{\pi}{2} (x + 1)}$ if $x \le 0$
$f(x) = \frac{\tan{x} - \sin{x}}{x^3}$ if $x ... | Let $f(x)$ a function such that $f(x)\to0$ when $x\to x_0\,\in\,\bar{\mathbb{R}}$, then:
$$\sin(f(x))\,\,\sim\,\,f(x)$$
This results can be shown using composite function limit theorem.
In your case, we are to study the continuity of:
$$f(x)=\begin{cases}
a \sin{\frac{\pi}{2} (x + 1)} & \text{ if } x\leq 0 \\
\frac{\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4377698",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Find two 2 permutations of the numbers of set $\{\frac 11,\frac 12,\dots,\frac 1n\}$ with certain properties. The problem is: Let $n>1$ be a natural numbers, $(a_1,a_2,\dots,a_n)$ and $(b_1,b_2,…,b_n)$ are $2$ permutations of the numbers in set $\{\frac 11,\frac 12,\dots,\frac 1n\}$, satisfying $a_i+b_i\ge a_j+b_j$ for... | Consider the counterexample:
\begin{align}
(a_1,a_2,\ldots,a_{12}) &= \bigg(\frac{1}{1},\frac{1}{4},\frac{1}{2},\frac{1}{3}, \frac{1}{5}, \frac{1}{9}, \frac{1}{6}, \frac{1}{10}, \frac{1}{7}, \frac{1}{11}, \frac{1}{8}, \frac{1}{12} \bigg) \\
(b_1,b_2,\ldots,b_{12}) &= \bigg(\frac{1}{4},\frac{1}{1},\frac{1}{3},\frac{1}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4379227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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A typical inequality: $\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}$
For $x, y, z\in (0, \infty)$ prove that: $$\frac{8}{27}\leq\frac{(x^2+yz)(y^2+zx)(z^2+xy)}{(xy+yz+zx)^3}.$$
My attempts to apply media inequality or other inequalities have been unsuccessful. In desperation I did the calculations an... | Another solution based on what you have already obtained and AM-GM:
It is easy to check that;
$$x^2y^2z^2+x^4yz+x^3y^3 \geq3x^3y^2z.$$
So we need to show that;
$$17(x^3y^3+x^3z^3+z^3y^3)+25(x^4yz+xy^4z+xyz^4)\geq21(x^3y^2z+x^3yz^2+x^2y^3z+x^2yz^3+xy^3z^2+xy^2z^3).$$
Now realize that we simply get;
$$x^3y^3+x^3y^3+x^3y^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4381204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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What is the decomposition into even and odd function of a function equal to $x$ when $x \geq 0$ and equal to zero when $x <0$? Every function $f$ with domain in $\mathbb{R}$ can be written
$$f=E+O$$
where $E$ is an even function and $O$ is an odd function.
Proof
Assume $f(x) = E(x) + O(x)$.
Then
$$f(-x)=E(-x) + O(-x)=E... |
True, I felt a bit weird about it too. I should have made the proof something like as follows. Let $f$ be any function. Let $E(x)=\frac{f(x)+f(-x)}{2}$ and let $O(x)=\frac{f(x)-f(-x)}{2}$. Then, I easily show that $E$ is even, $O$ is odd, and $f=E+O$. Thus we can say $f=E+O$ with $E$ even and $O$ odd, for any function... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4381416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\sum n(n+1)$ According to WolframAlpha,
$$\sum_{n=1}^k n(n+1)=\frac{1}{3}k(k+1)(k+2)$$
and it is easy to verify this if we use induction.
However, I would like to know how one can actually come up with this, other than by thinking about how to force the terms to cancel out.
I tried: Since $k(k+1)/2=1+2+\cdots... | Note that $n(n+1) = 2\bigg(\frac{n(n+1)}{2!}\bigg)=2\binom{n+1}{2}$. Then the sum becomes ....
$$\begin{align}
S(k) &= \sum_{n=1}^k{n(n+1)}\\
&= 2\sum_{n=1}^k{\binom{n+1}{2}}\\
(1) &= 2\binom{k+2}{3}\\
&= 2\cdot\frac{k(k+1)(k+2)}{3!}\\
&= \frac{k(k+1)(k+2)}{3}\\
\end{align}$$
Where $(1)$ is just the hockey-stick identi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382770",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 4
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Computing explicit Riesz projection Consider the matrix:
$ A = \begin{pmatrix} 1 & 1 \\ 0 & 1 \end{pmatrix} $
which has eigenvalue $1$ with algebraic multiplicity $2$ and geometric multiplicity $1$.
I am trying to explictly construct the Riesz projection
\begin{align*}
P = - \frac{1}{2 \pi i} \int_{\gamma} \frac{1}{A -... | It seems easiest to pick the contour $\gamma: \lbrack 0, 2 \pi \rbrack \to \mathbb{C} $ defined by
\begin{align*}
\gamma( \theta) = 1+ e^{i \theta}
\end{align*}
which is the circle of unit radius around $1$ and $\gamma'(\theta) = i e^{i \theta} $.
Now, to evaluate the contour integral
\begin{align*}
- \frac{1}{2 \pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4382900",
"timestamp": "2023-03-29T00:00:00",
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Finding the greatest common divisor of $(n+1)(n+2)^2(n+3)^3(n+4)^4$ Determine the largest positive integer that divides $(n+1)(n+2)^2(n+3)^3(n+4)^4$ for all positive integers n.
First, I noticed that out of $n+1, n+2, n+3,$ and $n+4,$ there must be one multiple of $4$, at least one multiple of $3$, and a multiple of $... | Following the comment of lulu, look at numbers of the form $n=19+24k$.
For $k=0$, $$(n+1)(n+2)^2(n+3)^3(n+4)^4=20\cdot 21^2\cdot 22^3\cdot 23^4$$ which has the factorization $2^5\cdot 3^2\cdot 5\cdot 7^2\cdot 11^3\cdot 23^4$ and is not divisible by any number containing more than five factors of $2$ and two factors of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4385677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Distributing 6 marbles among 3 containers What's the probability of all containers being non-empty?
Interpretation: All configurations are equiprobable.
Each container can contain min. 0 and max. 6
marbles.
Total number of configurations is implied by both the condition of equiprobability
and the saturation condition: ... | Your numerator is larger than your denominator, so your result cannot represent a probability. It is best to avoid choosing representatives of a set, then selecting additional members of a set since that leads to over counting.
Method 1: There are three choices for each marble, so there are $3^6$ ways to distribute th... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Maximize $\lambda$ over $a^2 + b^2 + c^2 + d^2 \ge ab + \lambda bc + cd$ and $a,b,c,d\geq 0$. Find the largest real number $\lambda$ such that
$$a^2 + b^2 + c^2 + d^2 \ge ab + \lambda bc + cd$$ for all nonnegative real numbers $a,$ $b,$ $c,$ $d.$
I tried using AM-GM on like $a^2+\dfrac{b^2}{4}\geq ab$, and I tried com... | You're very much on the right track. You can use
\begin{align*}
a^2+b^2+c^2+d^2
&=\left(a^2+\frac{b^2}4\right)+\left(\frac{c^2}4+d^2\right)+\frac{3(b^2+c^2)}4\\
&\geq ab+cd+\frac{3\cdot 2bc}4
\end{align*}
to get that $\lambda=3/2$ works. Now, you just need to determine if any larger $\lambda$ will work. Can you look at... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Given diagonals in a parallelogram. Find sides. Given a parallelogram with $d_1 = AC = 26$ cm, $d_2 = BD = 18$ cm and $\sin \displaystyle \angle AOD = \frac{12}{13}$.
Find $AB = a$ and $AD = b$.
What I did in order to solve it
*
*Using the formula for the area
$S = \frac{d_1d_2\sin \displaystyle \Phi}{2} = \frac{26... | $\cos AOD =\pm \sqrt{1-\sin^2 AOD}=\pm \sqrt{1-(12/13)^2}=\pm 5/13$.
Since $AOD < 90^\circ$, $\cos AOD = 5/13$
So $b^2=AO^2+OD^2-2 AO\cdot OD \cos AOD = 13^2+9^2-2\cdot 13\cdot 9\cdot (5/13)=160$.
Since $\sin DOC=\sin(180^\circ-AOD)=\sin AOD=12/13$, we also have $\cos DOC=\pm 5/13$, but this time we choose the minus si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4395766",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
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Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order Show that $\frac{3}{5} + \frac{4}{5}i$ number in multiplicative complex numbers field(apart from $0)$ has infinite order and prove that $\frac{1}{\pi}\arctan(\frac{4}{3})$ is irrational.
by contradic... | For the second part of the question: O.P. assumed that $\frac{1}{\pi}\arctan(\frac{4}{3})=\frac{m}{n}$ is rational. Then, $\cos(\frac{m}{n}\pi)=\frac{3}{5}$.
Let $U_{n-1}(x)$ be the Chebyshev polynomial of second kind of order $n-1$. Then $U_{n-1}(\frac{3}{5})=\frac{\sin m\pi)}{\sin(\frac{m}{n}\pi)}=0.$ We will show th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4398533",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to solve a difference equation $p(a,b)=[p(a-1,b)+p(a,b-1)]/2$ from the gambler's ruin problem with several boundary conditions Recently I read a probability book and it is going to deduce pascal distribution from the famous gambler's ruin problem,where person A need to win for $a$ times or person B to win for $b$ t... | Boundary conditions. If we assume $p(a,0)=0$ and $p(0,b)=1$ for $a,b>0$, together with the equation $2p(a,b)=p(a-1,b)+p(a,b-1)$ as in $(1)$, then we have $p(a,b)+p(b,a)=1$ for $(a,b)\neq(0,0)$, shown using induction. Thus, $p(a,a)=1/2$ holds for $a>0$; it need not be given. Assume $p(0,0)=q$ is also given (it doesn't p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4399086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Galois group of splitting field of $x^4-16x^2+4$ isomorphic to $\mathbb{Z}/(2)\times \mathbb{Z}/(2)$ I've already shown irreducible. I know I need to get intermediate fields since they correspond to subgroups (which I can use for isomorphisms). But how do I get the intermediate fields here?
edit: I'm thinking perhaps t... | $\dfrac{1}{\sqrt{5}-\sqrt{3}}=\dfrac{\sqrt{5}+\sqrt{3}}{2}$.
Therefore, any root generates all three other roots. Therefore, the splitting field is of a degree-$4$ extension of $\mathbb{Q}$.
Now, since you want three distinct intermediate subfields of degree $2$,
$(\sqrt{5}+\sqrt{3})+(\sqrt{5}-\sqrt{3})=2\sqrt{5}$
$(\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4400642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continued fraction for $\sqrt 2$ I discovered this continued fraction for $\sqrt 2$ but could not find any sources in which it appeared. It goes as follows:
\begin{equation}
\sqrt{2} = 1 + \frac{1 + \frac{1 + \frac{1 + \dots}{3 + \dots}}{3 + \frac{1 + \dots}{3 + \dots}}}{3 + \frac{1 + \frac{1 + \dots}{3 + \dots}}{3 + \... | Let $x = \frac {1 + \frac {1 + \cdots}{3+\cdots}}{3 + \frac {1 + \cdots}{3+\cdots}}$
$x = \frac {1+x}{3+x}\\
x^2+ 3x = 1+x\\
x^2 + 2x - 1 = 0\\
x = -1 \pm \sqrt {2}$
We know that $x>0$ so we can discard the negative value.
$1 + x = \sqrt 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluate $\int\left(\frac{\ln(x)-x}{x^2 -1}\right)\;\mathrm{dx}$ Question
Evaluate $$\int\left(\frac{\ln(x)-x}{x^2 -1}\right)\;\mathrm{dx}$$
Here is my work (and where I stopped)
Solution found on Internet
(I didn’t understand it at all to be honest)
| You should get $Li_{2}(x)$ function
https://mathworld.wolfram.com/Dilogarithm.html
$\int{\frac{ln(x)-x}{x^2-1}dx} = \int{\frac{ln{(x)}}{x^2-1}}-\int{\frac{x}{x^2-1}dx}\\
=-\int{\frac{ln{(x)}}{1-x^2}}dx - \frac{1}{2}\int{\frac{2x}{x^2-1}dx}\\
=-\int{\frac{ln{(x)}}{(1-x)(1+x)}}dx - \frac{1}{2}\ln{|x^2-1|} \\
-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4405201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solutions of Triangles inequalities - Duplicate A,B,C are angles of a triangle we are supposed to prove that $sin(\frac{A}{2})sin(\frac{B}{2})sin(\frac{C}{2})$ $\leq$ $\frac{1}{8}$. I used trigonometric ratios of half angles which would give $\frac{(s-a)(s-b)(s-c)}{abc}$. How can I proceed after this step?
Any help wou... | As @Esgeriath's answer mentioned, the correct inequality should be
$$
\sin \frac{A}{2} \; \sin \frac{B}{2} \; \sin \frac{C}{2} \le \frac{1}{8}.
$$
That answer used multivariable calculus methods to calculate the extremum of the left-hand side.
Here is an elementary proof of this inequality.
You have shown that
$$
\sin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4408491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Existence of $\sin\alpha=\frac{b}{\sqrt{a^2+b^2}}; \cos\alpha=\frac{a}{\sqrt{a^2+b^2}}$ Let $a,b\in\mathbb{R}$ such that $(a,b)\neq(0,0)$. Let $x$ be a real number.
We make a function such that $A(x)= a\cos(x)+b\sin(x)$. I need to show the existence of a number $\alpha\in\mathbb{R}$ such that:
$$\sin\alpha=\frac{b}{\sq... | Note that $\sin(x)$ is continuous in $x$, and that there is an $x'$ such that $\sin(x')=1$, and there is an $y$ such that $\sin(y)=-1$. So for any $c \in [-1,1]$, there is an $x$ such that $\sin(x)=c$. Now, for any such $a,b$, note that $\frac{b}{\sqrt{a^2+b^2}}$ is in $[-1,1]$, so there indeed
is an $\alpha'$ such tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4409503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Find $\exp\bigg[t\left ( \begin{matrix} -1& 1 \\ 0 & -1 \\ \end{matrix} \right )\bigg]$ by definition Find $\exp\bigg[t\left ( \begin{matrix}
-1& 1 \\
0 & -1 \\
\end{matrix} \right )\bigg]$ by definition
Denote $A=\left ( \begin{matrix}
-1& 1 \\
0 & -1 \\
\end{matrix} \right )$
I find out that
$$A... | From your insight, it may be more helpful to rewrite the powers of $A$ (you made a slight error) as
$$A^n = (-1)^n \begin{pmatrix}
1 & -n \\ 0 & 1 \end{pmatrix}$$
So,
$$\sum_{n=0}^{\infty} \frac{t^n}{n!}A^n = \sum_{n=0}^\infty \frac{(-t)^n}{n!} \begin{pmatrix}
1 & -n \\ 0 & 1 \end{pmatrix} = \sum_{n=0}^\infty \begin{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4410167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to integrate $\frac{\cos2x}{ \sin x + \sin 3x}$? I thought of using trig identities to get rid of $\cos2x$ and $\sin3x$, then use Weierstrass substitution, but I got myself into big trouble as the expression got too complicated.
Is there a simpler way to tackle this problem?
| First, note that
$$\sin(x) + \sin(3x) = 4 \sin(x) \cos^2(x)$$
by the uncommonly-used triple angle identity for sine and the Pythagorean identity:
$$\begin{align*}
\sin(3x)
&= 3 \sin(x) - 4 \sin^3(x) \\
&= \sin(x) \Big( 3 - 4 \sin^2(x) \Big) \\
&= \sin(x) \Big( 3 - 4 \Big( 1 - \cos^2(x) \Big) \Big) \\
&= \sin(x) \Big( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4411141",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Maximizing $\sum_{i=1}^{10} \cos(3x_i)$, for real $x_i$ such that $\sum_{i=1}^{10} \cos(x_i)=0$
Determine the greatest possible value of $\sum_{i=1}^{10} \cos(3x_i)$ for real $x_1,x_2,\cdots,x_{10}$, where $\sum_{i=1}^{10} \cos(x_i)=0$.
My approach was to somehow get a equation solely in terms of one of the $\cos x_... | We apply Lagrange multipliers.
Let $f(x_1,\ldots, x_{10})=\sum_{i=1}^{10} \cos 3x_i$ be the objective function and $g(x_1,\ldots, x_{10})=\sum_{i=1}^{10} \cos x_i=0$ be the constraint.
For each $1\leq i\leq 10$, we have by the triple angle formula,
$\lambda \sin x_i = 3(3\sin x_i - 4\sin^3 x_i)$
which gives
$\sin x_i =... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4414200",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving recursive equation without Master's Theorem Cheers, I am asked to find the time complexity of a recursive algorithm, which splits the starting problem on $\sqrt{n}$ subproblems, of size $\sqrt{n}$ each, and then combines the solution in linear time.
My try for creating the equation is: $T(n) = \sqrt{n}T(\sqrt{n... | $$T(n) = n ^ {\frac{1}{2}}T(n ^ {\frac{1}{2}}) + n$$
$$T(n ^ {\frac{1}{2}}) = n ^ {\frac{1}{4}}T(n ^ {\frac{1}{4}}) + n ^ {\frac{1}{2}}$$
Substitute:
$$T(n) = n ^ {\frac{1}{2}}n ^ {\frac{1}{4}}T(n ^ {\frac{1}{4}}) + n + n$$
Now,
$$T(n ^ {\frac{1}{4}}) = n ^ {\frac{1}{8}}T(n ^ {\frac{1}{8}}) + n ^ {\frac{1}{4}}$$
Substi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4416132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Proving the identity of $\sum_{k = 0}^n{4n \choose 4k} = 2^{4n - 2} + (-1)^n2^{2n - 1}$ combinatorially I want to prove the following identity combinatorially:
$$\sum_{k = 0}^n{4n \choose 4k} = 2^{4n - 2} + (-1)^n2^{2n - 1}$$
Here's my attempt so far: The left hand side is counting the number of teams among $4n$ people... | Here's an indirect approach that works.
First prove combinatorially that
$$(x-1)(1+x+x^2+\dots+x^{n-1}) = x^n - 1 \quad \text{for $n \ge 1$},$$
which is Identity 216 in Proofs That Really Count: The Art of Combinatorial Proof.
Then take $n=4$ and $x=i^k$ to yield
$$\frac{(i^k)^0+(i^k)^1+(i^k)^2+(i^k)^3}{4} = \begin{cas... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4417155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Is it hard to tackle the integral $\int_{0}^{\infty} \frac{x^{2}}{\left(1+x^{4}\right)^{2}} d x?$ Putting $ \displaystyle=4, \alpha=2, n=2 $ in my post,
$$\int_{0}^{\infty} \frac{x^{n}}{\left(1+x^{m}\right)^{\alpha}}dx=\frac{\pi}{m(\alpha-1) !} \csc\frac{(n+1) \pi}{m}\prod_{k=1}^{\alpha-1}\left(\alpha-k-\frac{n+1}{m}\r... | Letting $x\mapsto \frac{1}{x} $ yields
$$
I=\int_{0}^{\infty} \frac{x^{4}}{\left(x^{4}+1\right)^{2}} d x
$$
Adding them together gives $$
\begin{aligned}
2 I &=\int_{0}^{\infty} \frac{x^{2}+x^{4}}{\left(x^{4}+1\right)^{2}} d x \\
&=\int_{0}^{\infty} \frac{1+\frac{1}{x^{2}}}{\left(x^{2}+\frac{1}{x^{2}}\right)^{2}} d x \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4419223",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Is this nice-looking inequality actually trivial?
Let, $x,y,z>0$ such that $ xyz=1$, then prove that
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz)≥2(x+y+z)^2 $$
I tried to use the inequality
$$x^2+y^2+z^2≥xy+yz+xz$$
Then I got,
$$(xy+yz+xz)(x^2+y^2+z^2+xy+yz+xz) ≥2(xy+yz+xz)^2≥2(x+y+z)^2\implies xy+yz+xz≥x+y+z$$
Which is corre... | Your solution is wrong, a value of a polynomial of smaller degree can be bigger than the corresponding value of a polynomial of bigger degree. For example $x+5$ has degree $1$, $x^9$ has degree $9$ but $1+5>1^9$.
Here is a correct solution. I would be glad to see a simpler solution, but I do not see a substantially dif... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4421129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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ab+bc+ca=3 then $\sum\frac{\sqrt{a+3}}{a+\sqrt{bc}}\ge\frac{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{\sqrt{a+b+c+1}}$ Let $a,b,c≥0:ab+bc+ca=3$. Prove that:
$$\frac{\sqrt{a+3}}{a+\sqrt{bc}}+\frac{\sqrt{b+3}}{b+\sqrt{ca}}+\frac{\sqrt{c+3}}{c+\sqrt{ab}}\ge\frac{2(\sqrt{a}+\sqrt{b}+\sqrt{c})}{\sqrt{a+b+c+1}}$$
This is really tough... | *Partial solution?
I tried to find related term due to AM-GM: $$2\sqrt{a+3}=2\sqrt{\frac{a+ab+bc+ca}{a+\sqrt{bc}}(a+\sqrt{bc})}\le\frac{a(a+b+c+1)}{a+\sqrt{bc}}+2\sqrt{bc}$$
It means that: $$\frac{a(a+b+c+1)}{a+\sqrt{bc}}\ge\frac{2a(b+c+1)}{\sqrt{a+3}+\sqrt{bc}}\implies \frac{\sqrt{a+3}+\sqrt{bc}}{a+\sqrt{bc}}\ge\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4425033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding $(a,b)\in\mathbb{N}^2$ such that $\dfrac{a^2+b^2+1}{a+b} \in \mathbb{N}$. A pair $(a,b)\in\mathbb{N}^2$ is called good if $a < b$ and
$$\frac{a^2+b^2+1}{a+b}\in\mathbb{N}.$$
I think I've shown that there are infinitely many good pairs. However, the family of good pairs that I've found is not all of them, and th... | $a^2-b^2$ = $(a-b)(a+b)$ is divisible by a+b, so $a^2+b^2+1$ is divisible by a+b if and only if $2a^2+1$ is.
To find all solutions with a, b >= 1:
Iterate through a = 0, 1, 2, …
Let z = 2a^2+1
Determine all divisors d of z.
For each d >= a, (a, d-a) is a solution.
Since z is divisible by z, we have the obvious b = z... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solving $(x^2+4x+3)\sqrt{x+2}+(x^2+9x+18)\sqrt{x+7}\geq x^3+10x^2+33x+36$ I am trying but I cannot solve the following inequation:
$(x^2+4x+3)\sqrt{x+2}+(x^2+9x+18)\sqrt{x+7}\geq x^3+10x^2+33x+36$
My attemption.
The inequation is equvalent to
$$
(x+1)(x+3)\sqrt{x+2}+(x+3)(x+6)\sqrt{x+7}\geq (x+3)^2(x+4)
$$
Since $x+2\g... | Let us solve
$$(x+1)\sqrt{x+2}+(x+6)\sqrt{x+7}\geq (x+3)(x+4).$$
We have
\begin{align*}
\mathrm{LHS} - \mathrm{RHS}
&= (x + 1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x + 7} - 3)\\[5pt]
&\qquad + 2(x + 1) + 3(x + 6) - (x + 3)(x + 4)\\[5pt]
&= (x + 1)(\sqrt{x + 2} - 2) + (x + 6)(\sqrt{x + 7} - 3) + (x + 4)(2 - x)\\[5pt]
&=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4429521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
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Finding the conditions for $\mathbf{A}×(\mathbf{B}×\mathbf{C}) = (\mathbf{A}×\mathbf{B})×\mathbf{C}$. I wanted to find the conditions in which $\mathbf{A}×(\mathbf{B}×\mathbf{C})$ is equal to $(\mathbf{A}×\mathbf{B})×\mathbf{C}$.
On solving $\mathbf{A}×(\mathbf{B}×\mathbf{C})-(\mathbf{A}×\mathbf{B})×\mathbf{C}=\mathbf{... | The Jacobi identity is
$$
\mathbf{A}\times(\mathbf{B}\times\mathbf{C})+
\mathbf{B}\times(\mathbf{C}\times\mathbf{A})+
\mathbf{C}\times(\mathbf{A}\times\mathbf{B})=\mathbf{0}
$$
Using anticommutativity, we get
$$
\mathbf{A}\times(\mathbf{B}\times\mathbf{C})-
(\mathbf{A}\times\mathbf{B})\times\mathbf{C}=
-\mathbf{B}\time... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4430693",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Calculating a matrix-exponential Let A be the following matrix.
$$\begin{pmatrix}
1 & 1 \\
0 & 1 \\
\end{pmatrix}
$$
I have to calculate $e^A$.
My idea was to diagonalize A because then $e^A = Pe^DP^-1$ if $A = PDP^-1$.
But A cannot be diagonalized since 1 is a double eigenvalue and therefore A does not have 2 linearly... | Note that $A^k = \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix}$.
So $e^A = \sum_{k=0}^{\infty} \frac{1}{k!}A^k = \sum_{k=0}^{\infty} \frac{1}{k!} \begin{pmatrix} 1 & k \\ 0 & 1 \end{pmatrix} = \begin{pmatrix} \sum_{k=0}^{\infty} \frac{1}{k!} & \sum_{k=0}^{\infty} \frac{k}{k!} \\ 0 & \sum_{k=0}^{\infty} \frac{1}{k!} \en... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4432299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to deal with odd $m$ in integral $\int_{0}^{\frac{\pi}{4}}(\sin^{6}m x+\cos^{6}m x) \ln (1+\tan x) d x $ Latest edit
*
*Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)$$
By our resu... | Thanks to @Quanto for settling down the question by proving the odd one as:$$I_{2n+1}= \frac{5\pi}{64}\ln2+\frac3{16(2n+1)}\bigg(\frac\pi4-\sum_{j=0}^{2n}\frac{(-1)^j}{2j+1} \bigg)
$$
As asked by @Claude Leibovici for the powers other than 6, I had generalised my result to even powers below:
$$
I(m,n):=\int_{0}^{\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436587",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "7",
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Find minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$
Find the minimum of $\sqrt{a}+\sqrt{b}+\sqrt{c}$, given that $a,b,c \ge 0$ and $ab+bc+ca+abc=4$.
I can prove that $\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1,$ because:
$$ab+bc+ca+abc=4 \\ \implies (a+2)(b+2)(c+2)=(a+2)(b+2)+... | Not beautiful solution.
WLOG $a\leq b\leq c$. Suppose $a>1$, then $b>1$, $c>1$, $ab+bc+ca+abc > 4$. Therefore $a\leq 1$.
Let $\sqrt{b}+\sqrt{c}=d$, $\sqrt{bc}=e$. Then $bc=e^2$, $d^2=b+c+2e$, $b+c=d^2-2e$. $(\sqrt{b}-\sqrt{c})^2=b+c-2e=d^2-4e\geq 0$, therefore $d^2\geq 4e$.
$$ab+bc+ca+abc=4\Rightarrow a(b+c+bc)+bc=4 \R... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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Show that, if $t_{3n}=x$, $t_{3n+1} = y$ and $t_{3n+2} = z$ for all values of $n$, then $p^3+q^3+3pq-1=0$
A sequence of numbers $t_0,t_1,t_2,...$ satisfies $$t_{n+2} = pt_{n+1} + qt_n, ~~~n\geq0$$
where $p$ and $q$ are real. Throughout this question, $x,y,z$ are non-zero real numbers.
Show that, if $t_{3n}=x$, $t_{3n... | For a shortcut, note that the characteristic polynomial of the linear recurrence is $\,x^2-px-q\,$, but the additional conditions also require $\,t_{n+3}=t_n\,$ with characteristic polynomial $\,x^3-1\,$. It follows that they must have a common root, so $\,0 = \text{res}(x^3-1, x^2 - p x - q)=-p^3 - 3 p q - q^3 + 1\,$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4436910",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Lower bound for $I(a,b)=\int_0^\infty \frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} \label{elliptic integral 2}$ I am studiyng this paper which states that the integral
$$I(a,b)=\int_0^\infty \frac{dx}{\sqrt{a^2+x^2}\sqrt{b^2+x^2}} \tag{1}$$
satisfies the inequality
\begin{align}
\frac{\pi}{2a} \leq I(a,b) \leq \frac{\pi... | hint We have
$$\frac{1}{\sqrt{b^2+x^2}\sqrt{b^2+x^2}} \ge \frac1{\sqrt{a^2+x^2}\sqrt{b^2+x^2}}
\ge \frac{1}{\sqrt{a^2+x^2}\sqrt{a^2+x^2}} .$$
Therefore...
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate this limit $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $? $ \lim\limits_{x\rightarrow0}\frac{1-\sqrt{1+x^2}\cos{}x}{x^4} $
I tried L'Hopital, but it gets complicated.
I tried also this:
$$ \lim\limits_{x\rightarrow0}\frac{1-(1+x^2)\cos^2{x}}{x^4(1+\sqrt{1+x^2}\cos{x})} $$
$$ \lim\lim... | I shall use this idea of yours:$$\lim_{x\to0}\frac{1-\sqrt{1+x^2}\cos(x)}{x^4}=\lim_{x\to0}\frac{1-(1+x^2)\cos^2(x)}{x^4\left(1+\sqrt{1+x^2}\cos(x)\right)}.$$Note that$$\lim_{x\to0}\left(1+\sqrt{1+x^2}\cos(x)\right)=2.$$So, what remains is to compute\begin{align}\lim_{x\to0}\frac{1-(1+x^2)\cos^2(x)}{x^4}&=\lim_{x\to0}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4444063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that for $\alpha \leq2$ the sum $\sum_{n=1}^\infty (\sqrt{n+1}-\sqrt{n})^\alpha$ is divergent, and if $\alpha \geq 4$ the sum is convergent. Prove that for $\alpha \leq2$ the sum $\sum_{n=1}^\infty (\sqrt{n+1}-\sqrt{n})^\alpha$ is divergent, and if $\alpha \geq 4$ the sum is convergent.
Attempt:
$$\sum_{n=1}^\inf... | You could write $\sqrt{n+1} - \sqrt{n} = \sqrt{n}\left(\sqrt{1 + \frac{1}{n}} - 1\right)$ and then note that for $n\geqslant 1$,
$$
\frac{\sqrt 2 -1}{n} \leqslant \left( \sqrt{1+\frac{1}{n}} -1 \right) \leqslant\frac{1}{n}
$$
where the right hand inequality is derived from the basic property $x \geq \sqrt x$ when $x \g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\cot\left(\frac{\pi}{4}+\beta\right)+\frac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$ Show that $$\cot\left(\dfrac{\pi}{4}+\beta\right)+\dfrac{1+\cot\beta}{1-\cot\beta}=-2\tan2\beta$$
I'm supposed to solve this problem only with sum and difference formulas (identities).
So the LHS is $$\dfrac{\cot\dfrac{\pi}{4}... | $$\frac{4 \sin \beta \cos \beta}{\sin ^{2} \beta-\cos ^{2} \beta} = \frac{2 \sin2\beta}{-(\cos ^{2} \beta-\sin ^{2} \beta)}=-2\frac{\sin 2\beta}{\cos 2\beta}=-2 \tan(2\beta)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4447522",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $c$ for $P(\frac{5}{12}-c \le Y \le \frac{5}{12}+c)=\frac{1}{2}$ Find $c$ for $P(\frac{5}{12}-c \le Y \le \frac{5}{12}+c)=\frac{1}{2}$
We also have that
$$f_Y(y) = k\sum_{i=0}^{\infty}y^i, y\in(1/3, 1/2) \\ \implies k\int_{1/3}^{1/2}\frac{1}{1-y}dy \implies k=\frac{1}{\log(\frac{4}{3})}$$
What I have tried:
I had ... | If the distribution is not "normal" then why would you think of applying tricks from normal distribution? . Secondly it is Chebycheff's Inequality. So applying it won't yield you any "equality" for $c$.
$$P(\frac{5}{12}-c<Y<\frac{5}{12}+c)=k\int_{\frac{5}{12}-c}^{\frac{5}{12}+c}\frac{1}{1-y}\,dy = k\ln\bigg(\frac{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $|1 - (1+\frac{z-1}{n})^m| \leq 1$ where $|z|\leq 1$ and $mHow to prove
$$\Big|1 - \big(1+\frac{z-1}{n}\big)^m\Big| \leq 1$$
where $z\in\mathbb{C}$, $|z|\leq 1$ and $m,n\in\mathbb{N}$ and $m<n$.
I have run numerical experiments and believe the inequality is correct.
| By Maximum modulus principle, the maximum of $\displaystyle |1-(1+\frac{z-1}{n})^m|$ cannot be reached when $|z| < 1$. So we can suppose $|z|=1$.
Let $z=e^{i\theta}$, and $\displaystyle w = 1 + \frac{z-1}{n} = \rho e^{i \varphi}$. Then:
$$
\begin{align}
|1-(1+\frac{z-1}{n})^m | &= |1-w^m| \\
&=|1-w|\cdot|1+w+\cdots+w^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove the product of first n consecutive odd numbers is a square less than another square? I have observed for first few values of consecutive odd numbers, the result is always of the form:
$m^2 - n^2$, where $m$ and $n$ are two distinct positive integers. That is:
$$1\cdot 3\cdot 5 \cdot 7\cdots (2k-1) = m^2 - ... | You might note that the difference sequence of the sequence of squares is exactly the odd numbers. Here's the sequence of squares:
$$0, 1, 4, 9, 16, 25, 36, \ldots$$
Here's the sequence of the differences:
$$1, 3, 5, 7, 9, 11, \ldots$$
That means every odd number is the difference of two consecutive squares. And your... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find volume of solid bounded by given surfaces. $z=a+x,z=-a-x,x^2+y^2=a^2$ Find volume of solid bounded by given surfaces.
$$z=a+x, \qquad z=-a-x, \qquad x^2+y^2=a^2$$
This is the solid. We can find volume of solid that has positive $z$ value and multiply by $2$. And for finding volume of that solid I thought doing do... | Intersection of planes is a line given by, $z = a + x = - a - x \implies x = - a, z = 0$
At intersection with the cylinder, $(-a)^2 + y^2 = a^2 \implies y = 0$
So, both planes and the cylinder intersect at a single point $(-a, 0, 0)$. That means the projection of the region in xy-plane is circle $x^2 + y^2 \leq a^2, z ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let $a_n$ be a sequence such that $a_1 >0$ and for all $n$ $a_{n+1}=a_n+ \frac{1}{a_n^3}$ then $\lim _\limits {n \to \infty}a_n=\infty$
Let $a_n$ be a sequence such that $a_1 >0$ and for all $n$ $a_{n+1}=a_n+ \frac{1}{a_n^3}$ then $\lim _\limits {n \to \infty}a_n=\infty$
I am not sure how to formally prove this , t... | Squaring both sides,
\begin{align*}
a_{n+1}^2 = a_n^2 + 2\frac{a_n}{a_n^3} + \frac{1}{a_n^6} > a_n^2 + 2\frac{1}{a_n^2}
\end{align*}
and squaring again,
\begin{align*}
a_{n+1}^4 > a_n^4 + 4 + \frac{4}{a_n^4} > a_n^4 + 4
\end{align*}
and so
\begin{align*}
a_{n+1}^4 = a_1^4 + \sum_{k=1}^{n}(a_{k+1}^4 - a_k^4) > 4n + a_1^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to do the following trigonometric simplification: $ \frac{1- \cos (3\alpha) }{1- \cos (\alpha)} = (1 + 2\cos (\alpha)^2) $ This is probably a trivial question but I don't understand it. Somehow I can't seem to understand how to simplify this expression:
$$ \frac{1- \cos (3\alpha) }{1- \cos (\alpha)} = (1 + 2\cos (\... | I prefer using capital $A$, so bare with me.
We will prove that $$1-\cos3A=(1-\cos A)(1+2\cos A)^2$$
First we will show that $$\cos3A=4\cos^3A-3\cos A$$
Indeed, we know that $$\cos(x+y)=\cos x\cos y-\sin x\sin y$$
Thus, for $x=2A, y=A$ we obtain: $$(*):\cos 3A=\cos 2A\cos A-\sin 2A\sin A$$
but from $(*)$ we see that: $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Other ways to factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ To factorize $xy\left(x+y\right)+yz\left(y-z\right)-xz\left(x+z\right)$ I used the fact that $x=-y$ and $y=z$ and $x=-z$ make the expression zero. Hence it factorize to $\lambda (x+y)(y-z)(x+z)$ and we can check that the number $\lambda$... | You can observe that, if $x=-y$ or (not "and") $x=-z$ or (not "and") $y=z$ , then the expression equals to zero. Thus, we have
$$xy(x+y)+yz(y-z)-xz(x+z)=A(x + y) (x + z) (y - z)$$
Here, $A$ must be a real number, not an expression. Because, it is enough to see that the coefficient of $x^2y$ equals to $1$. This implies,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4458244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Calculate $C=\sin3\alpha\cos\alpha$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$. Calculate $$C=\sin3\alpha\cos\alpha$$ if $\tan2\alpha=2$ and $\alpha\in(0^\circ;45^\circ)$.
My idea was to find $\sin\alpha$ and $\cos\alpha$. Then we have $\sin3\alpha=3\sin\alpha-4\sin^3\alpha$. So $$\tan2\alpha=2=\dfrac{2\tan\a... | Hint:
Writing $A$ for $\alpha$
$$2C=2\sin3A\cos A=\sin4A+\sin2A$$
Now Weierstrass substitution to find $\sin4A$
$$\tan2A=2\implies\dfrac{\sin2A}2=\dfrac{\cos2A}1=\pm\sqrt{\dfrac1{1^2+2^2}}$$
As $0^\circ\le2A\le90^\circ, \sin2A\ge0$
Can you take it from here?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Group theory, when are elements conjugate to each other? In my book "Group Theory and Chemistry" by Bishop, it is written that if $$R = Q^{-1}PQ$$ $R$ is the transform of $P$ by $Q.$ But isn't $R$ not just equal to $P (Q^{-1} Q = E )$?
If I transform $P$ by $Q$ wouldn't I have $R = PQ$?
| I will give two examples where this fails. The first is the smallest possible group that is non-abelian; that is, where the elements do not necessarily commute, so there exist $a,b$ in the group such that $ab\neq ba$. The second is usually one of the first examples one studies where commutativity fails.
Consider the g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4465517",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How many real solutions are there for the equation $x^{3}-300x=3000$ How many real solutions are there for the equation $$x^{3}-300x=3000$$
Attempt
There is a solution here I found, without using any algebraic factorization: https://youtu.be/y845hT7aYAQ
and this seems to be most efficient.
But how about one that uses ... | By Descartes rule of signs, we have one positive root. Zero isn’t a root. Now suppose there are negative roots, then using $x \mapsto -x$, these are positive roots of $x^3+3000=300x$.
However by AM-GM,
$x^3+2000+1000> 300\sqrt[3]2\,x>300x$, hence there are no negative roots.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4466146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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find an equation satisfied by the three bisectors of a right triangle let $a,b,c$ be the sides of a generic triangle and $\alpha, \beta, \gamma$ the bisectors of the respective opposite angle.
It is well know that, for example, $\alpha$ divides the side $a$ in two segments, $x$ and $a-x$, and that we have $(a-x)c=bx$, ... | (Too long for a comment.)
In the case of a right triangle with $\,b^2+c^2=a^2\,$ the bisector formulas simplify to:
*
*$\require{cancel}(b+c)^2\alpha^2 = bc\left((b+c)^2-a^2\right)=bc\left(\cancel{b^2+c^2}+2bc-\cancel{a^2}\right)=2b^2c^2\,$;
*$(c+a)^2\beta^2=ca\left(c^2+2ac+\underbrace{a^2-b^2}_{=\,c^2}\right)=2ac^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468226",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve long integration by partial fraction decomposition problems faster? Some problems are just too time consuming for short exam times what is the fastest way to solve problems like this one for example $$\int \frac{5x^4-21x^3+40x^2-37x+14}{(x-2)(x^2-2x+2)^2}dx$$
| There are a few answers in here already, but I hope mine can contribute something. Except for very simple cases, there's no fastest partial fractions decomposition method. You have to decide which method(s) is(are) best for the case at hand. Although the algebraic method (matching coefficients) always works, the cover-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Doubt regarding validity of answer in a mod equation question
$$|x^2-2x+2|-|2x^2-5x+2|=|x^2-3x|$$
Find the set of values of $x$.
The answer given is $[0,\frac12]\cup [2,3]$.
What I don't get is how is the solution a range? I'm getting the solution as $0,\frac 12, 2 ,3, \frac 25$. That makes sense to me as the first e... | As you've mentioned, $|x^2-2x+2| = |(x-1)^2+1|$ is always positive.
So $|x^2-2x+2| =x^2-2x+2$.
$|2x^2-5x+2| = \begin{cases}2x^2-5x+2 & \text{for } x \in (-\infty,0.5] \cup [2,\infty)\\ -(2x^{2}-5x+2) &\text{for } x \in [0.5,2]\end{cases}$
So,
$|x^2-2x+2| - |2x^2-5x+2| = \begin{cases}-x^2+3x &\text{for } x \in (-\inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4470083",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $X/1430$ when $X=(^{10}C_1)^2+2(^{10}C_2)^2+3(^{10}C_3)^2+ ...+10(^{10}C_{10})^2$ Let $X=(^{10}C_1)^2+2(^{10}C_2)^2+3(^{10}C_3)^2+ ...+10(^{10}C_{10})^2$, then what's the value of $X\over1430$?
I don't even know where to begin on this question. All solutions I've seen on various sites start by writing this as a su... | Here is a variation using the binomial theorem and a bit of algebra. We use the coefficient of operator $[x^n]$ to denote the coefficient of $x^n$ in a series. This way we can write for instance
\begin{align*}
\binom{r}{k}=[x^k](1+x)^r\tag{1}
\end{align*}
We obtain
\begin{align*}
\color{blue}{X}&\color{blue}{=\sum_{r=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471784",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Is there any method other than Feynman’s Integration Technique to find $ \int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x?$ We are going to find the formula, by Feynman’s Integration Technique, for
$$\int_{0}^{\frac{\pi}{2}} \ln \left(a \cos ^{2} x+b \sin ^{2} x+c\right) d x,$$
where $a+c$ $\... | Letting $t=\tan x$ transform the integral into
$$
\begin{aligned}
I &= 2 \int_{0}^{\frac{\pi}{2}} \ln (\cos x) d x+\int_{0}^{\frac{\pi}{2}} \frac{\ln \left((a+c)+(b+c) t^{2}\right)}{1+t^{2}}dt\\&=-\pi\ln 2+\int_{0}^{\frac{\pi}{2} } \frac{\ln \left((a+c)+(b+c) t^{2}\right)}{1+t^{2}} d t
\end{aligned}
$$
By my post
$$
\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $a,b,c\in\mathbb{R},$ such that $a\geq 2$, $b\geq 5$, $c\geq 5$ and $2a^2+b^2+c^2=69$. Prove that $12a+13b+11c \geq 155.$ So,I am trying to prove the following inequality without using the Lagrange multipliers:
If $a,b,c\in\mathbb{R},$ such that $a\geq 2$, $b\geq 5$, $c\geq 5$ and $2a^2+b^2+c^2=69$. Prove that
$$12a... | Let $a = 2 + x, b = 5 + y, c = 5 + z$. Then $x, y, z \ge 0$.
We have $2a^2 + b^2 + c^2 - 69
= 2x^2 + y^2 + z^2 + 8x + 10y + 10z - 11 = 0$.
We want to prove that $12x + 13y + 11z \ge 11$.
We split into two cases:
(1) If $x > 1$ or $y > 1$ or $z > 1$, we have $12x + 13y + 11z \ge 11$.
(2) If $x, y, z \in [0, 1]$, we have... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can a cylinder be rotated at any angle $\theta$ about the origin using cartesian coordinates? How can a cylinder be rotated at any angle $\theta$ at origin using cartesian coordinates?
What would be the equation for a cylinder (first figure) rotated about the origin?
| The equation of a cylinder in the upright position (i.e. with its axis along the vertical $z$ axis, is given by
$ \mathbf{p}^T Q \mathbf{p} = r^2 $
where $\mathbf{p} = [x, y, z]^T $ , and $r$ is the radius of the cylinder, and the $3 \times 3$ matrix is given by
$ Q = \begin{bmatrix} 1 && 0 && 0 \\ 0 && 1 && 0 \\ 0 && ... | {
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For all $n,k$ in positive integers, there exists $m_1,...,m_k$ such that the following holds Let $n,k$ be two positive integers. Prove that there exists $m_1,...,m_k$ in the positive integers such that
$$1+\frac{2^k-1}{n}=\prod_{i=1}^k\left(1+\frac{1}{m_i}\right)$$
Here is my attempt
We're going to construct a solution... | The $m_i$'s in your construction are not necessarily integers.
To illustrate the problem, consider the simple case of $k = 2$. You want to write $\frac{n + 3}n$ as a product $\frac{m_1 + 1}{m_1} \cdot \frac{m_2 + 1}{m_2}$.
Your strategy goes for $m_1 = n$ and $m_2 = \frac{n + 1}2$. However we see that $\frac{n + 1}2$ i... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find expression of $c_n$, where $c_n = a_n + b_n$ Given the recurrence relation $a_{n+2} = 3a_{n+1} + 6a_n$ and $b_{n+2} = b_{n+1} + b_n$ I am supposed to find an expression of the recurrence relation for $c_n := a_n + b_n$. I tried to find some form of linear dependence to obtain a recurrence relation for $c_n$ but th... | If I define the generating functions
$$ f(x) = \sum_{n=0}^\infty a_n x^n $$
$$ g(x) = \sum_{n=0}^\infty b_n x^n $$
then
$$ \begin{align*}
(6x^2 + 3x - 1) f(x) &= \sum_{n=0}^\infty (6 a_n x^{n+2} + 3 a_n x^{n+1} - a_n x^n)\\
&= -a_0 + (3 a_0 - a_1) x + \sum_{n=2}^\infty (6 a_{n-2} + 3 a_{n-1} - a_n) x^n \\
&= -a_0 + ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Evaluate $\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}$
$$\int \dfrac{dx}{(1+x)\sqrt{1+2x-x^2}}$$
I completed the square:
$\int \dfrac{dx}{(1+x)\sqrt{2-(x-1)^2}}$
And then substituted $\sqrt 2\sin θ = x-1$ which gives
$\int \dfrac{dθ}{\sqrt2 \sinθ + 2}$
But now I'm stuck. Can someone please help?
| If you have a rational function involving trigonometric functions, try a Weierstrass substitution.
$$t=\tan\left(\frac {\theta}2\right)\qquad\mathrm d\theta=\frac {2}{1+t^2}\,\mathrm dt\qquad\sin\theta=\frac {2t}{1+t^2}$$
Hence
$$\begin{align*}\int\frac {\mathrm d\theta}{2+\sqrt{2}\sin\theta} & =\int\frac {\mathrm dt}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4478268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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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... | My second proof:
Denote the expression by $f$.
Clearly, we have $a, b, c < 1$. We have
$$\sqrt{a^2 + b^2} - \frac{a + b}{\sqrt 2}
= \frac{\frac{(a - b)^2}{2}}{\sqrt{a^2 + b^2} + \frac{a + b}{\sqrt 2}}
\ge \frac{\frac{(a - b)^2}{2}}{\sqrt{1^2 + 1^2} + \frac{1 + 1}{\sqrt 2}} \ge \frac{(a - b)^2}{6}$$
which results in
$$\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Proving inequality using root of unity Let $\omega$ be a complex cube root of unity. It can be shown that if $a,b,c \in \mathbb{R}$, then $$(a+b+c)(a+b\omega+c\omega^2)(a+b \omega^2 + c \omega)= a^3+b^3+c^3-3abc$$
I was wondering if this could be extended to prove the AM/GM inequality for $n=3$. All that needs to be do... |
Leading the proof of $n=3$-AM-GM Ineq. from:
$$(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega)=a^3+b^3+c^3-3abc.$$
Enough to show that: $(a+b+c)(a+b\omega+c\omega^2)(a+b\omega^2+c\omega) \ge 0.$
Let's make all of the conditions of equality the same in the proof of the above.
\begin{align}
\newcommand{w}{\omega}
& \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Evaluating $\lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}$
Problem statement:
If
$$ A = \lim_{n\to\infty} \left(\prod_{r=0}^n \binom{n}{r}\right)^\frac{1}{n(n+1)}, $$
Find $A^2$.
Solution:1)
\begin{align*}
\prod_{k=0}^{n} \binom{n}{k}
&= \prod_{k=1}^{n} \frac{n!}{k!(n-k)!}
= \frac{(n!)^{n+... | We have the limit
$$\lim_{n\to\infty}\frac{\sum_{r=0}^{n} \log \binom{n}{r}}{n(n+1)} $$
Applying Stolz Cesaro, the limit is equivalent to
$$\lim_{n\to\infty}\frac{\sum_{r=0}^{n} \log \binom{n}{r}-\sum_{r=0}^{n+1} \log \binom{n+1}{r}}{n(n+1)-(n+1)(n+2)} $$
Since $\frac{\binom{n}{r}}{\binom{n+1}{r}}=1-\frac{r}{n+1}$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483861",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Proving $\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$ using $\epsilon$-$\delta$ definition I want to prove that $$\lim_{x \to 2^{+}} \frac{x - 2}{\sqrt{x^{2} - 4}} + 1 = 1$$ but I am not sure if my proof is valid because I did some algebraic manipulation. Can you please verify my proof?
Given $ \epsilon ... | HINT
A slightly different approach. Suppose that $0 < x - 2 < \delta_{\varepsilon}$. Then it results that
\begin{align*}
\left|\frac{x - 2}{\sqrt{x^{2} - 4}}\right| = \frac{\sqrt{(x - 2)^{2}}}{\sqrt{(x - 2)(x + 2)}} = \frac{\sqrt{x - 2}}{\sqrt{x + 2}} < \frac{\sqrt{x - 2}}{2} < \frac{\sqrt{\delta_{\varepsilon}}}{2} := ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Find the image of the parabola $x = \frac{9}{4} − \frac{y^2}{9}$ under the principal square root mapping $w = z^{1/2}$ with $z \in \mathbb{C}$. Find the image of the parabola $x = \frac{9}{4} − \frac{y^2}{9}$ under the principal square root
mapping $w = z^{1/2}$ with $z \in \mathbb{C}$.
Definition of principal square ... | Suppose $w=(u,v)$ is in the image of the parabola under the map principal branch of square root of $z$.
Then $\exists z=(x, y) $ on the parabola such that $√z=w$
Then $z=w^2$ implies
$x+iy=u^2-v^2+2iuv$
Hence $x=u^2-v^2 , y=2uv$
$(x, y) $ lies on the parabola $x=\frac{9}{4}-\frac{y^2}{9}$ implies
$u^2-v^2=\frac{9}{4}-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Calculation of Residue. Let $c=\cos \dfrac{\pi}{5}, f(z)=\dfrac{z^2-2cz+1}{z^4-z^3+z^2-z+1}$.
$e^{\frac{3\pi}{5}i}$ is one of the pole of $f$. (This is because $f$ can be written as $f(z)=\frac{(z+1)(z^2-2cz+1)}{z^5+1}$.)
Then, calculate the Residue of $f$ at $e^{\frac{3\pi}{5}i}=:a$.
I calculated using the formula of... | We can considerably simplify the residue calculation when expanding numerator and denominator with $z+1$. This way we can effectively get rid of the denominator. We consider
\begin{align*}
f(z)=\frac{(z+1)\left(z^2-2cz+1\right)}{z^5+1}
\end{align*}
and obtain
\begin{align*}
\mathrm{Res}&(f,a)
=\displaystyle\lim_{z\to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4489465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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According to the figure, what is the area of the triangle $BCD$ of type $x$?
$Q:$ In a circle $\omega \ $ with a center $A$, $E\in \omega \ $ is the tangent point, $[BA]\bot[AC] \ $, $\angle ABD= \angle DCB \ $, $\angle DBC= \angle DCA \ $, $D\in\omega \ $, $|EC|=x \ $ What is the area of the triangle $BCD$ of type $x... |
First of all note that $\angle ABC=\alpha+\beta=\angle ACB$, that means triangle ABC isosceles, so :
$\angle ABC=\angle ACB=45^o\Rightarrow \alpha=\beta=22.5^o$
Triangle AEC is right angled and we have:
$AC^2=r^2+x^2\Rightarrow AC=\sqrt{r^2+x^2}$
in triangle ABC:
$BC=AC\sqrt 2=\sqrt{2(r^2+x^2)}$
$AH=AC \sin 45^o=\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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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$$
I noticed that for $0< x$, both $x^2$ and $\left(\dfrac x{x+1}\right)^2$ are increasing functions so their sum is also increasing and it has only one root which is $x=1$ (by inspection). But I'm not sur... | To solve $4x^3+12x^2+15x+5=0$, start by applying the rational root theorem. In this case, potential rational roots are $\pm \lbrace \frac{1}{4}, \frac{1}{2}, 1, \frac{5}{4}, \frac{5}{2}, 5 \rbrace$. Trying them out, we see that $x=\frac{-1}{2}$ satsfies the equation. So $x + \frac{1}{2}$ (or for the sake of avoiding... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Minimum of a function defined by a sum I have the following function, for a fixed integer $n\in \mathbb N$, $\displaystyle\varphi(x) = \sum_{k=0}^{n} \frac{(-1)^k}{x-k}$, for $x\neq l$ for all $l\in\{0,\ldots,n\}$. I am suspecting that $\min\limits_{x\in (0,n)} \left|\varphi(x)\right|\ge 2$ and want to prove it. So my... | I think I have found an answer for the even case (when $n$ is even). Indeed, let $n=2p$ and since $\varphi(x) = -\varphi(2p-x)$ it is enough to study the following case:
*
*$x\in(2l-1, 2l)$ for $l\in\{1,\ldots,p\}$
\begin{align}
\left|\varphi(x)\right| &= \left|\sum_{k=0}^{2p} \frac{(-1)^k}{x-k}\right|\\
&= \left|\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
pseudo C-S inequality? Problem :
For $x,y,z\in\mathbb{R}$,
Find minimum of $$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$$
But in this problem, coefficient is form of $()^3$, not a $... | Let $\,u = 2x, v=3y, w=4z\,$ then the problem is to minimize $u^4 / 2 + v^4 / 3 + w^4 / 4$ under the constraint $u / 2 + v / 3 + w / 4 = 13/4$. By the weighted power means inequality:
$$
\sqrt[4]{\frac{u^4 / 2 + v^4 / 3 + w^4 / 4}{1/2+1/3+1/4}} \;\ge\; \frac{|u| / 2 + |v| / 3 + |w| / 4}{1/2+1/3+1/4} \;\ge\; \frac{13/4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 0
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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... | It is convenient to use the falling factorial notation
\begin{align*}
k^{\underline{i}}:=k(k-1)\cdots (k-i+1)
\end{align*}
With this notation OPs identity can be written as
\begin{align*}
\sum_{k=i}^nk^{\underline{i}}=\frac{(n+1)^{\underline{i+1}}}{i+1}
\end{align*}
and since summands with $1\leq k<i$ are zero we can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to find this two-variables limit of $f$ as $(x,y)$ approaches to $(0,0)$? I am trying to evaluate the limit $$\lim_{(x,y) \to (0,0)}\frac{2x^6-y^{10}}{x^2-y^3}$$
I have seen that both the reiterated limits are zero. Moreover, I have evaluated the limit by different curves such as $y=x^2$ and the limit equals zero a... | Consider the sequence
$(\frac{1}{ n},0)\to(0,0)$
Then $f(\frac{1}{n},0)=\frac{1}{n^4}\to 0 $
Again the sequence $(\sqrt{\frac{n^2+1}{ n^3}},\frac{1}{ \sqrt[3]n})\to(0,0)$
But $\begin{align}f(\sqrt{\frac{n^2+1}{ n^3}},\frac{1}{ \sqrt[3]n})&=\frac{\frac{2(n^2+1)^3}{n^6}-\frac{1}{n^3\sqrt[3]n{}}}{\frac{1}{n^3}}\\&=2(1+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How to find the roots of Martrix Norm?
Problem: Compute $||A||$ where $A = \begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}$
Attempt:
$$
\begin{align*}
\det(A^TA-λI_2) &=
\det \left( \begin{pmatrix} 1 & 0 \\ 1 & 1\end{pmatrix}
\begin{pmatrix} 1 & 1 \\ 0 & 1\end{pmatrix}
- \begin{pmatrix} λ & 0 \\ 0 & λ\end{pmatrix} \r... | Your calculation of the determinant is correct. To find the roots, just use the quadratic formula,
$$ax^2 + bx +c =0 \implies x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$
Here,
$$\begin{align*}
&\lambda^2 - 3 \lambda + 1 = 0 \\
&\implies \lambda = \frac{-(-3) \pm \sqrt{(-3)^2 - 4(1)(1)}}{2(1)} = \frac{3 \pm \sqrt{9-4}}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502683",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$x^{10}+x^{11}+\dots+x^{20}$ divided by $x^3+x$. Remainder? Question:
If $x^{10}+x^{11}+\dots+x^{20}$ is divided by $x^3+x$, then what is the remainder?
Options: (A) $x\qquad\quad$ (B)$-x\qquad\quad$ (C)$x^2\qquad\quad$ (D)$-x^2$
In these types of questions generally I follow the following approach:
Since divisor is... | We want to find the remainder from the division:$$\frac{x^{20} + x^{19} + \ldots + x^{11} + x^{10}}{x^3 + x}$$
Now note that $$ (x^3 + x)(x^{n+1} + x^n ) = x^{n+4} + x^{n+3} + x^{n+2} + x^{n+1}, $$
so that
$$ \frac{x^{20} + x^{19} + \ldots + x^{11} + x^{10}}{x^3 + x} = x^{17} + x^{16} + x^{13} + x^{12} + \frac{x^{12} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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"answer_id": 7
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How does one algebraically find the total number of intersections between $\sin(x)$ and a linear equation? Take the function $f(x) = \dfrac{x}{8}$ and $\sin(x)$. By plotting, we can see that the graphs intersect at 7 distinct points, indicated by the grey dots. However, how would one approach such a problem algebraical... | At $x=\dfrac{5\pi}{2}$, compare the y-values of $y=\dfrac{x}{8}$ and $y=\sin{x}$:
$\dfrac{\frac{5\pi}{2}}{8}<\dfrac{5(3.2)}{16}=\sin{\frac{5\pi}{2}}$. So there are at least $7$ intersections.
If $x=3\pi$ then $\dfrac{x}{8}=\dfrac{3\pi}{8}>1$, and $1$ is the maximum value of $\sin{x}$. So for $x\ge 3\pi$ there are no in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How many methods are there to find $\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}, \textrm{ where }a>b>0$ When I deal with this simple integral, I found there are several methods. Now I share one of them.
Letting $x\mapsto \pi-x $ yields $$I=\int_{0}^{\pi} \frac{d \theta}{a+b \cos \theta}=\int_{0}^{\pi} \frac{d \thet... | By Weierstrass substitution
Letting $t=\tan \frac{\theta}{2}$ transforms the integral
$$
\begin{aligned}
I&=\int_{0}^{\pi} \frac{1}{a+\frac{b\left(1-t^{2}\right)}{1+t^{2}}} \frac{2 d t}{1+t^{2}}\\
&=2 \int_{0}^{\infty} \frac{d t}{(a+b)+(a-b) t^{2}}\\
&=\frac{2}{\sqrt{a-b} \sqrt{a+b}} \left[\tan ^{-1}\left(\frac{\sqrt{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Prove that $\frac{{x_1}^2+{x_2}^2+\cdots+{x_n}^2}{n}x_1x_2\cdots x_n\le\left(\frac{x_1+x_2+\cdots+x_n}{n}\right)^{n+2}$
Show that for any non negative real numbers $x_1,x_2,\cdots x_n,$
$$\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... | theorem :
Let $x_1,x_2,x_3,y_1,y_2,y_3\in(0,\infty)$ then if we have :
$$x_1+x_2+x_3\geq y_1+y_2+y_3$$
And for $i\neq j,1\leq i\leq 3,1\leq j\leq 3$:
$$|x_i-x_j|\leq |y_i-y_j|$$
Then we have :
$$x_1x_2x_3\geq y_1y_2y_3$$
Case $n=3$ :
Let $y_1=\left(3-\frac{6\left(ab+bx+xa\right)}{\left(a+b+x\right)^{2}}\right)^{\frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 3
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Find the limit of $\frac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$ Find the limit $$\lim_{n\to\infty}\dfrac{1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}}{1+\frac15+\frac{1}{5^2}+...+\frac{1}{5^n}}$$
For the numerator we have $1+\frac13+\frac{1}{3^2}+...+\frac{1}{3^n}=\dfrac{1-\... | Sum of IGP is $$S_\infty=1+r+r^2+r^3+\ldots=\frac{1}{1-r}.$$
So the required limit $$=\frac{1/(1-1/3)}{1/(1-1/5)}=\frac{6}{5}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4509654",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof for lemma for existence of Taylor's Polynomial There is lemma that says, for every $a \in \mathbb{R}$, for every polynomial $P_n(x)$ of degree at most $n$ is possible to write in this form:
$$P_n(x) = \sum_{k=0}^{n}c_{k}(x - a)^k$$
where $c_k = \frac{P_n^{(k)}(a)}{k!}$.
There is proof using induction:
$$P_{n+1}(x... | I write the proof step by step. I hope that it would be helpful.
If the induction hypothesis would be true for $n$, then for $n + 1$ we can write:
\begin{align*}
P_{n+1}(x) = \sum_{k = 0}^{n + 1} c_kx^k = c_{n+1}\color{blue}{x^{n+1}} + \sum_{k = 0}^{n}c_kx^k
\end{align*}
But $x^{n + 1} = (x - a)^{n+1} - \sum_{k = 0}^n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $\frac{x}{y}+\frac{y}{x}\ge2$ and $\sum_{cyc}\frac{x}{y+z}\ge\frac32$, can we say $\sum_{cyc}\frac{w}{x+y+z}\ge\frac43$? We know that $$\frac{x}{y}+\frac{y}{x}\ge2$$ and $$\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}\ge\frac32$$ Can we say that
$$\frac{w}{x+y+z}+\frac{x}{w+y+z}+\frac{y}{x+w+z}+\frac{z}{x+y+w}\ge\frac43... | Applying Cauchy-Schwars inequality, we have
$$\left(\overbrace{\frac{x}{y+z}+\frac{y}{x+z}+\frac{z}{y+x}}^A\right)\left(x(y+z)+y(x+z)+z(y+x)\right)\ge(x+y+z)^2\implies A≥\frac{(x+y+z)^2}{2(xy+yz+xz)}$$
Then let's apply AM-GM inequality:
$$
\begin{cases}x^2+y^2\ge 2xy\\x^2+z^2\ge 2xz\\ y^2+z^2\ge 2yz\end{cases}
$$
\begi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 1
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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... | If it gets too messy, you can always use logarithmic differentiation. Take the natural logarithm of both sides to get:
$$\ln y = \ln((x+1)^3) - \ln(x^2)$$
$$\ln y = 3 \ln(x+1) - 2 \ln x$$
and so using the chain rule, we have that $\frac{d}{dx} (\ln y)$, where $y$ is a function of $x$), is $\frac{1}{y} \cdot y' = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4512453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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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
$$a_n = \frac{3^n}{n!}= \fra... | Observe that: $0<a_{n+1} = \dfrac{3}{n+1}a_n < \dfrac{3}{n+1}\cdot \dfrac{3}{2}= \dfrac{9}{2n+2}$ for $n \ge 2$. Thus the squeeze theorem can be applied to conclude that $a_{n+1} \to 0$ as claimed.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Solving the equation $f(x) = \int_0 ^x \sqrt{4-2f(t)} dt$ A continuous function $f(x)$ satisfies the following.
For all reals $x \le b$, $f(x)=a(x-b)^2+c$.
For all reals, $f(x) = \int_0 ^x \sqrt{4-2f(t)} dt$.
If $\int_0^6 f(x) dx = \frac{q}{p}$, where $p,q$ are relatively prime positive integers, find $p+q$
My approach... | We have
$$\frac{f'(x)}{\sqrt{4-2f(x)}}=1.$$
From here
$$-\sqrt{4-2f(x)}=x+c$$
hence
$$f(x) = 2-\frac12(x+c)^2.$$
As $f(0)=0$ we have $c=\pm2$. (As the vertex is $(\pm2,2)$ the condition $4-2f(x)\ge0$ is satisfied for all $x$.) Now integrate.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4516639",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove $0\le ab+bc+ca-abc\le2$ A question asks
Let $a,b,c$ be non-negative reals such that $$a^2+b^2+c^2+abc=4$$
Prove that$$0\le ab+bc+ca-abc\le2$$
The lower bound can be proven as follows:
If $a,b,c>1$, then $$a^2+b^2+c^2+abc>4$$
This implies that at least one of $a,b,c$ is less than or equal to $1$, and we can assu... |
$a^2+b^2+c^2+abc=4, a, b, c:$ non-negative. Prove that $0\leq ab+bc+ca-abc \leq 4.$
\begin{align}
&a, b, c: \text{non-negative.} \\
& \text{You found that }a \leq 1 \text{ or } b \leq 1 \text{ or } c \leq 1.\\
\Rightarrow \; & 1 \leq \frac{1}{a}+\frac{1}{b}+\frac{1}{c}. \\
\therefore \; & abc \leq ab+bc+ca, 0 \leq ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Solving $\frac{(\sec x+\tan x)^2-(\sec 2x+\tan x)^2}{\sin 2x-\sin x}=2$ How can we solve for the value of $x$ in the range $[-\pi,\pi]$?
$$\frac{(\sec x+\tan x)^2-(\sec 2x+\tan x)^2}{\sin 2x-\sin x}=2$$
All i could do is write everything in terms of the elementary function $\sin$ and $\cos$ which gave me the equation
$... | If there is no typo, it is more than ugly.
Assuming that the denominator cannot be zero, that is to say $x\neq 0$, $x\neq \pm\frac \pi3$ and $x\neq \pm\pi$, using the tangent half-angle substitution $x=2\tan^{-1}(t)$, the lhs write
$$-\frac{ t \left(t^2-3\right) \left(t^2+1\right)^3 \left(2 t^5-3 t^4-12 t^3-2 t^2+2
... | {
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"source": "stackexchange",
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How many ways are there to use powers of $2$, each at most $3$ times, and sum up to $100$?
For $i=0,1,2,\ldots$, there're $3$ weights with mass $2^i$ gram(s). How many ways are there to weigh $100$ grams with them?
To solve with normal methods for counting doesn't seem possible. Although any way to select weights can... | Your generating function $f(x)$ might as well be infinite:
$$\prod_{n=0}^{\infty}\left(1+x^{2^n}+x^{2(2^n)}+x^{3(2^n)}\right)$$
Now split this over even $n$ and odd $n$:
$$\prod_{n=0}^{\infty}\left(1+x^{4^n}+x^{2(4^n)}+x^{3(4^n)}\right)\prod_{n=0}^{\infty}\left(1+x^{2(4^n)}+x^{4(4^n)}+x^{6(4^n)}\right)$$
The left produ... | {
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"question_score": "2",
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For $0For $0<t\leq 1$ show
$$ \text{ln}(t)\geq \frac{t-1}{2t+2}\left(1+\sqrt{\frac{2t^2+5t+2}{t}}\right)$$
On a side note, I've been taking this summer class on inequalities. The professor told us that his main source are all sorts of math olympiads. Given that I'm gonna take Calc 3 next semester and Real Analysis aft... | My try .
here (Which is greater $\frac{13}{32}$ or $\ln \left(\frac{3}{2}\right)$) Michael Rozenberg show :
let $0<x\leq 1$ :
$$\ln{x}\geq(x-1)\sqrt[3]{\frac{2}{x^2+x}}$$
Now we need to show :
$$(x-1)\sqrt[3]{\frac{2}{x^2+x}}\geq \frac{x-1}{2x+2}\left(1+\sqrt{\frac{2x^{2}+5x+2}{x}}\right)$$
or :
$$\sqrt[3]{\frac{2}{x^... | {
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"question_score": "5",
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Sum of $\sum_0^\infty \frac1{n^4+a^4}$ How to find the following series as pretty closed form by $a$?
$$S=\sum_0^\infty \frac1{n^4+a^4}$$
I first considered applying Herglotz trick, simply because the expressions are similar. So I changed it like this...
$$2S-a^{-4}=\sum_{-\infty}^\infty \frac1{n^4+a^4}$$
However, the ... | If you write
$$n^4+a^4=(n-\alpha)(n-\beta)(n-\gamma)(n-\delta)$$
$$\alpha=-\frac{(1+i) a}{\sqrt{2}}\qquad \beta=\frac{(1+i) a}{\sqrt{2}}\qquad \gamma=-\frac{(1-i) a}{\sqrt{2}} \qquad \delta=\frac{(1-i) a}{\sqrt{2}}$$ Using partial fraction
$$\frac 1{n^4+a^4}=\frac{1}{(\alpha-\beta) (\alpha-\gamma) (\alpha-\delta) (x-\a... | {
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"source": "stackexchange",
"question_score": "4",
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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:
$$\dfrac{1... | One more (why not?):
$$ \cos \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ -\frac{\sqrt2}{2} ·(\sin \alpha \ + \ \cos \alpha) $$ $$ \Rightarrow \ \ \sec \left(\alpha \ + \ \frac{3 \pi}{4} \right) \ \ = \ \ \frac{-\sqrt2}{(\sin \alpha \ + \ \cos \alpha)} \ \ ; $$
$$ \tan^2 \left(\alpha \ + \ \frac{3 \pi}{4} \righ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 6
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Six-sided and four-sided dice question contradiction Suppose we have two fair dice, with six sides (numbered 1 to 6) and with four sides (numbered 1 to 4). Suppose we pick a die at random and throw it, the result is announced to be 2 and the die is discarded. Then we pick the remaining die and throw it. What is the exp... | Let $A$ represent the event we rolled the four-sided die in the first throw. Then $A^c$ represents the event we rolled a six-sided die in the first throw. Remember the second throw we are rolling the other die.
Let $B$ represent the event the outcome of the first roll is a $2$.
We are told $\Pr(A)=\Pr(A^c)=\frac{1}{2... | {
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"question_score": "5",
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Noncommutative Ring with only Left Identity Does there exist a noncommutative ring $R$ without an identity and an element $e\in R$ such that $ex = x$ for all $x\in R$? (i.e. $xe \neq x$ for some $x\in R$).
| $R=\left\{ \left[\begin{array}{ccc}a & b\\a & b\end{array}\right]\in Mat_2(\mathbb Q) \right\} $ with the usual operations.
It satisfy the axioms of group, it is distributive, and a direct computation proves that the product is well defined in $R$, therefore is also associative:
$\left[\begin{array}{ccc}a & b\\a & b\en... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$using implicit differentiation
Proof that $\frac{d}{dx} \tan^{-1} x = \frac{1}{1+x^2}$ using implicit differentiation
My workings:
$y=\tan^{-1} x$,
$x= \tan y$
$\frac{d}{dx} (x) = \frac{d}{dx} \tan y$
$1 = \sec^2 y \frac{dy}{dx}$
$\frac{dy}{dx} = \frac{1}{\sec^2 ... | $$x=\tan y\implies x^2=\tan^2y$$$$\implies 1+x^2=1+\tan^2y=\sec^2y$$$$\implies \frac{dy}{dx}=\frac{1}{\sec^2y}=\frac{1}{1+x^2}.$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Probability that first coin flip was heads given that 7 of 10 flips were heads? Let's say we flip a coin 10 times and we get heads 7 times. What is the probability that the first flip was heads? (My naïve estimation would be 7/10 but that's just intuition, trying to prove that).
| I would use the binomial distribution to proof it, which is not very time consuming. Let $A$ be the event, that the first flip is head. And $B$ is the event that we get 7 times head in 10 coin flips. Then it is asked for $P(A|B)=\frac{P(A\cap B)}{P(B)}$.
$P(A\cap B)$ is the probability that we get head at the first fli... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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