Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b) Let $f(x)=(x+1)(x+2)(x+3)(x+4)+5$ where $x\in[-6,6]$. If the range of the function is [a,b] where $a,b\in N$, then find the value of (a+b)
My attempt is as follows:-
Rewrite $f(x)=g... | Continuing discussion in comments, use first derivative test.
$$
f’(x)=2x^3+15x^2+35x+25=(2x+5)(x^2+5x+5)
$$
Now the roots for the quadratic are
$x = \frac{-5+\sqrt{5}}{2}, \frac{-5-\sqrt{5}}{2}$. So to check for maximum and minimum value, check $f(-6)=120$, $f(\frac{-5+\sqrt{5}}{2})=-1$, $f(\frac{5}{2})>0$ cannot be m... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove convergence of $n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ I am working on some old analysis exams and i got stuck on this exercise :
Using the epsilon definition show that $a_{n} = n \cdot \left(\sqrt{1+ \frac{1}{n}} -1\right)$ converges and determine its limit.
Knowing that the limit is 1/2, I know need ... | $$n\left(\sqrt{1+\frac1n}-1\right)<n\left(\sqrt{1+\frac1n+\frac1{4n^2}}-1\right)=\frac12$$
and
$$n\left(\sqrt{1+\frac1n}-1\right)=\frac1{\sqrt{1+\dfrac1n}+1}>\frac1{\sqrt{1+\dfrac1n+\dfrac1{4n^2}}+1}=\frac12-\frac1{4n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3432944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Showing that solutions to $x^3y^{\prime\prime\prime}+2x^2y^{\prime\prime}-4xy^\prime+4y=0$ are linearly independent
Find solutions for $x^3y^{\prime\prime\prime}+2x^2y^{\prime\prime}-4xy^\prime+4y=0$ which have the form $y(x)=x^r$ and then show that they are linearly independent.
So my method for solving this was to ... | Then solving $ r(r−1)(r−2)+2⋅r(r−1)−4r+4=0$ I guessed the solution $r=1$
You need to solve completely your equation :
$r(r−1)(r−2)+2⋅r(r−1)−4r+4=0$
$r(r−1)(r−2)+2⋅r(r−1)−4(r-1)=0$
$(r-1)(r(r-2)+2r-4)$
$(r-1)(r^2-4)=0 \implies (r-1)=0 \text{ or } (r^2-4)=0$
$r^2-4 =0 \implies r=2,r=-2$
$r-1 =0 \implies r=1$
$$S_r=\{1,2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3435877",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Proving the Sum to Product Identities (without using the product to sum identities) I got this project for my students from another teacher, and it did not come with a key. There are four versions, each one is proving a different Sum to Product identity, but they can't use the product to sum identity. The instruction... | Let's correct your calculation.
\begin{align*}
-2&\sin\left(\frac{\alpha + \beta}{2}\right)\sin\left(\frac{\alpha - \beta}{2}\right)\\
& = -2\sin\left(\frac{\alpha}{2} + \frac{\beta}{2}\right)\sin\left(\frac{\alpha}{2} - \frac{\beta}{2}\right)\\
& = -2\left[\sin\left(\frac{\alpha}{2}\right)\cos\left(\frac{\beta}{2}\ri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
} |
Convergence of $\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\cdots$ Does the following series converge? If yes, what is its value in simplest form?
$$\left( \frac{1}{1} \right)^2+\left( \frac{1}{2}+\frac{1}{3} \right)^2+\left( \frac{1}{4}+\frac{1}{5}+\frac{1}{6} \right)^2+\left( \frac{1}{7}+\f... | Suggestion.
This is merely a suggestion that is too large to fit in a comment. Euler's integral formula for the harmonic numbers, $H_n=\int_0^1\frac{1-x^n}{1-x}\ \mathrm{d}x$, gives us a formula for the series as $a\to1^-$:
$$S=4\sum_{n=1}^{\infty}\left[f_a(n(n+1))-f_a(n(n-1))\right]^2$$
where $f_a(x)=\int_{0}^{a}\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3436804",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 2
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Why does this process map every fraction to the golden ratio? Start with any positive fraction $\frac{a}{b}$.
First add the denominator to the numerator: $$\frac{a}{b} \rightarrow \frac{a+b}{b}$$
Then add the (new) numerator to the denominator:
$$\frac{a+b}{b} \rightarrow \frac{a+b}{a+2b}$$
So $\frac{2}{5} \rightarrow ... | First consider the sequence of every second fraction:
$$
\frac{a_{2n+2}}{b_{2n+2}} = \frac{a_{2n}+b_{2n}}{a_{2n}+2b_{2n}} = \frac{\frac{a_{2n}}{b_{2n}} +1}{\frac{a_{2n}}{b_{2n}} + 2} = f(\frac{a_{2n}}{b_{2n}})
$$
where $f(x)$ is defined as
$$
f(x) = \frac{x+1}{x+2} = 1 - \frac{1}{x+2}
$$
for $x \ge 0$.
Use the monoto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3440647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "40",
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"answer_id": 0
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Solve the equation $\cos^{-1}\frac{x^2-1}{x^2+1}+\tan^{-1}\frac{2x}{x^2-1}=\frac{2\pi}{3}$ $\cos^{-1}\dfrac{x^2-1}{x^2+1}+\tan^{-1}\dfrac{2x}{x^2-1}=\dfrac{2\pi}{3}$
Let's first find the domain
$$-1<=\dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1}{x^2+1}>=-1 \text { and } \dfrac{x^2-1}{x^2+1}<=1$$
$$\dfrac{x^2-1+x^2+1}{x^2+1... | Following your approach, you are letting $x=\tan(\theta)$ for some $\theta \in (-\pi/2,\pi/2)$, and arriving at $\frac{\pi}{3}=\arccos(\cos(2\theta))+\arctan(\tan(2\theta))$. The important thing is that $\arccos$ maps to $[0,\pi]$ while $\arctan$ maps to $(-\pi/2,\pi/2)$. Thus:
*
*if $2 \theta \not \in [0,\pi]$, the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3442053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then Find $6xyz$. If $x+y+xy=3$ , $y+z+yz=8$ and $z+x+xz=15$ then
Find $6xyz$.
How can I approach this problem?
I need some hints. Thanks.
| For convenience, we just need to look at one equation, $$x+y+xy=3 \implies (x+1)(y+1)=4$$
hence we define
$a=x+1,b=y+1,c=z+1$
and we obtain
$ab=4,bc=9,ac=16$
You solve easily by that hint then.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3444729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Stuck on a probability law problem I'm currently trying to solve a problem, I completed the first question but I am stuck at the second one, here is the problem:
One person roll a die until the result is a $1$, a second person toss a coin until he gets $3$ tails.
*
*How many tries are they going to make on average.
... | To show that $$\sum_{k=3}^\infty\left[\left(\frac56\right)^{k-1}\cdot\frac16\binom{k-1}2\left(\frac12\right)^{k-3}\left(\frac12\right)^3\right]= \frac{25}{343},\tag1$$ it suffices to show that $$\sum_{k=3}^\infty\left(\frac5{12}\right)^k\binom{k-1}2= \frac{125}{343}\tag2$$ after combining constants. The binomial term c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Solve $2 \cos^2 x+ \sin x=1$ for all possible $x$ $2\cos^2 x+\sin x=1$
$\Rightarrow 2(1-\sin^2 x)+\sin x=1$
$\Rightarrow 2-2 \sin^2 x+\sin x=1$
$\Rightarrow 0=2 \sin^2 x- \sin x-1$
And so:
$0 = (2 \sin x+1)(\sin x-1)$
So we have to find the solutions of each of these factors separately:
$2 \sin x+1=0$
$\Rightarrow \sin... | Your method's fine, the answer's right. The only improvement I can suggest is to make the definition of "base solution" clear upfront. Each "and so" acts as if a specific value of $\sin x$ has finitely many solutions rather than finitely many per period, so before you obtain them you should mention a restriction to $[0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3445884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 1
} |
Evaluate $\lim\limits_{n \to \infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}$
$$\lim_{n \to
\infty}\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}$$
Maybe we can believe with assurance that
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}}\approx\sum_{k=1}^{n}\frac{1}{\sqrt{n-k}\cdot\sqrt{k}}=\frac{1}{n... | You can turn the sum into two Riemann sums each of which converges to $\frac \pi 2$:
$$\sum_{k=1}^{n}\frac{1}{\sqrt{n+1-k}\cdot\sqrt{k}} = \sum_{k=1}^{n}\frac{n+1-k + k}{\sqrt{n+1-k}\cdot\sqrt{k}}\cdot \frac 1{n+1}$$
$$=\frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{n+1-k}{k}} + \frac 1{n+1}\sum_{k=1}^{n}\sqrt{\frac{k}{n+1-k}}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3446462",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$ Let $ m_a, m_b $ and $ m_c $ be the medians relative to the $ a, b, c $ sides of a triangle, show that:
$$\frac{(m_a)^4+(m_b)^4+(m_c)^4}{a^4+b^4+c^4} = \frac{9}{16}$$
What i tried:
ust use Stewart’s theorem. We have $m=n=a/2,d=m_a$ so
$$\frac{a^3}{... | Note,
$$m_a^4+m_b^4+m_c^4 = (m_a^2+m_b^2+m_c^2 )^2 - 2(m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2) $$
Then, evaluate,
$$m_a^2+m_b^2+m_c^2 = \frac{2b^2+2c^2-a^2}{4} +\frac{2c^2+2a^2-b^2}{4} +\frac{2a^2+2b^2-c^2}{4} = \frac34 (a^2+b^2+c^2)$$
$$m_a^2m_b^2+m_b^2m_c^2+m_c^2m_a^2 = \frac9{16}(a^2b^2+b^2c^2+c^2a^2)$$
Thus,
$$\frac{m_a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3453618",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Mathematical Induction Involving The Floor Function I need to prove the identity $\lfloor \sqrt{n}+\sqrt{n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$ for all natural numbers $n$.
I wanted to use mathematical induction. The identity is true for $n=1$. Then, I assume $$\lfloor \sqrt{k}+\sqrt{k+1}\rfloor=\lfloor\sqrt{4k+2}\rf... | Here is an (maybe) easier approach.
To prove $\lfloor\sqrt n+\sqrt {n+1}\rfloor=\lfloor\sqrt{4n+2}\rfloor$, and it's easy to notice that $\lfloor\sqrt{4n+2}\rfloor>\lfloor\sqrt n+\sqrt {n+1}\rfloor$ and that $\lfloor\sqrt{4n+2}\rfloor-\lfloor\sqrt n+\sqrt {n+1}\rfloor<1$, it suffices to prove that there's no integer be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3456333",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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How to remove square roots from denominator in $\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}}$? I have this math problem: Remove all square roots in the denominator of
$\frac{\sqrt{x}+\sqrt{y}}{\sqrt{x}-\sqrt{y}} = ?$
The obvious solution is to multiply the fraction with $\frac{\sqrt{x}-\sqrt{y}}{\sqrt{x}-\sqrt{y}}$.
$\f... | Instead multiply by $\dfrac{\sqrt{x}+\sqrt{y}}{\sqrt{x}+\sqrt{y}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3458074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the limit of $\sqrt[n]{n^2 + n}$ To find the limit I got the $\sqrt[3n]{n^2+n}$
Particularly, $\sqrt[3n]{n^2+n} \ge 1 \rightarrow \sqrt[3n]{n^2+n} = 1 + d_n$ where $d_n\ge 0$.
According to the Bernoulli's rule
$\sqrt{n^2+n} = (1+d_n)^n \ge d_n\cdot n \rightarrow d_n \le \frac{\sqrt{n^2+n}}{n}$
The $\frac{\sqrt{n^... | First, we will show that $\lim_{n\to\infty}n^{1/n}=1$. We define the sequence $x_n$ as
$$x_n=n^{1/n}-1\tag1$$
ASIDE:
Note that $n^{1/n}\ge 1$for $n\ge1$. This is true since if $0\le y\le 1$, then $0\le y^n\le 1$ for all $n\in \mathbb{N}$.
From $(1)$, it is easy to see that $(1+x_n)^n=n$. Then, using the binomial th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3460379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 1
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Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Solve $yy^\prime+x=\sqrt{x^2+y^2}$
Tried dividing by $y$ to get $$y^\prime+\frac{x}{y}=\frac{\sqrt{x^2+y^2}}{y}$$
$$(y^\prime)^2+2\frac{x}{y}+\frac{x^2}{y^2}=\frac{x^2+y^2}{y^2}$$
$$(y^\prime)^2+2\frac{x}{y}=1$$
$$y^\prime=\sqrt{1-2\frac{x}{y}}$$
Tried using $v=\frac{y}{x}$
$$y^\pri... | Let $r=\sqrt{x^2+y^2}$, then the equation becomes
$$
rr'=r
$$
which means that $r'=1$; that is, $r=x+c$ which is the horizontal parabola
$$
y^2=2cx+c^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3466125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Limit $\lim\limits_{n\to\infty}\left(\frac{n}{n^2+1}+\frac{n}{n^2+2}+\frac{n}{n^2+3}+\cdots+\frac{n}{n^2+n}\right)$ $\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}+\dfrac{n}{n^2+2}+\dfrac{n}{n^2+3}\cdots\cdots+\dfrac{n}{n^2+n}\right)$
Can we write it as following
$E=\lim\limits_{n\to\infty}\left(\dfrac{n}{n^2+1}\right)... | Observe that
$$\frac{n}{n^2+1}+\frac{n}{n^2+2}+\cdots+\frac{n}{n^2+n}$$ lies between
$$\frac{n}{n^2+n}+\frac{n}{n^2+n}+\cdots+\frac{n}{n^2+n}=\frac{n}{n+1}
$$ and
$$\frac{n}{n^2}+\frac{n}{n^2}+\cdots+\frac{n}{n^2}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3467697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 7,
"answer_id": 0
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Interval for which $x^{12}-x^9+x^4-x+1>0$
The largest interval for which $x^{12}-x^9+x^4-x+1>0$ is :
$$
x^{12}-x^9+x^4-x+1>0\\
x(x^3-1)(x^8+1)+1>0\\
x(x-1)(x^2+x+1)(x^8+1)+1>0
$$
I can see that this is satisfied for $x\in(-\infty,0]\cup[1,\infty)$. But how can I verify the above function is greater than zero when $x\... | $$x^{12}+x^4(1-x^5)+(1-x)>0$$ for $0<x<1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3470096",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate in closed form: $ \sum_{m=0}^\infty \sum_{n=0}^\infty \sum_{p=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}$
Evaluate in closed form:
$$ \sum_{m=0}^\infty \sum_{n=0}^\infty \sum_{p=0}^\infty\frac{m!n!p!}{(m+n+p+2)!}$$
I tried to use same method for similar question and two variables.Was not able to get the final ans... | Playing with Mathematica a bit and you will see that the sum is
$$
S(3)=\sum_{p=0}^\infty \sum_{m=0}^\infty \sum_{n=0}^\infty\frac{m!n!p!}{(m+n+p+3)!}
=
\frac{1}{12} \left(3 \zeta (3)+2 \pi ^2 \log (2)\right)+
\sum _{n=1}^{\infty } \left(\frac{\psi ^{(1)}(n)}{4 n (2 n-1)}-\frac{\psi ^{(1)}\left(n+\frac{1}{2}\right)}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3471368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Prove $x^2+\frac{1}{x^2}=2\cos(2\theta)$ Prove $x^2+\frac{1}{x^2}=2\cos(2\theta)$ and $x^3+\frac{1}{x^3}=2\cos(3\theta)$
knowing that there exist a number $x$ given angle $\theta$ such that $x+\frac{1}{x}=2\cos(\theta)$
Doesn't really know how to start this problem, thought that I would some how need to use the double ... | $$\left(x+\frac 1x\right)^2 = x^2+2+\frac 1{x^2} = 4\cos^2\theta \\\implies x^2+\frac{1}{x^2} = 2(2\cos^2\theta-1) = 2\cos2\theta$$
Similarly
$$\left(x+\frac 1x\right)^3 = x^3 + 3\left(x+\frac 1x\right) + \frac{1}{x^3} = x^3+\frac{1}{x^3} + 6\cos\theta = 8\cos^3\theta$$
$$ \implies x^3+\frac{1}{x^3} = 2(4\cos^3\theta-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3473553",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solve equation $(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$ Find the value of $x$ given the equation,
$$(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}=5$$
I think they are powers of $ \frac {1} {\Phi} $ or $ \Phi $. In case I think ... | There are two solutions (see below).
Here is how we can prove this fact in a rigorous way. Let :
$$f(x):=(2+ \sqrt{5})^{\frac{x}{2}}+2 \cdot \left(\frac{7 - 3 \sqrt{5}}{2}\right)^{\frac{x}{4}}\tag{1}$$
which has the form
$$f(x)=a^x+2b^x \ \text{with} \ a>1, b<1.\tag{2}$$
Its derivative being $f'(x)=\ln(a) a^x +2\ln(b... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Generating Function & Sequence Find the generating functions of the sequences
2, 1, 2, 1, 2, 1, . . .
I get $\frac{1}{1+x} + \frac{1}{1-x} = \frac{2}{1-x^2}$
But the solution ends up with $\frac{2}{1-x^2} + \frac{x}{1-x^2} = \frac{2+x}{1-x^2}$.
The solutions starts with $\sum_{n\ge 0} (2)x^{2n}+\sum_{n\ge 0} (1) x^{2n... | As an alternative to the other answers, you can simply use a variant of one function, that being
$$\frac{1}{1-x^2}=\sum_{n=0}^{\infty} x^{2n}$$
If you consider that the $2$'s are even indexed (assuming in the sequence $\{a_n\}$ the first term is $a_0$) and the $1$'s are odd indexed, you have that
$$\frac{2}{1-x^2}$$
w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3474486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Verify $JJ' = 1$, given $x = e^u \cos v$, $y = e^u \sin v$. I was trying to solve this jacobian problem but I'm not getting the required solution i.e $JJ' = 1$
I'm getting $J = e^{2u}$ and $J' = \frac{x^2 - y^2}{x^2 + y^2}$ So multiplying both
$J*J'$ will not return 1 in my case.
What am I doing wrong? Here's my solut... | You're almost there. First, a miscalculation: your $\frac{\partial v}{\partial x}$ has the wrong sign. In particular: we indeed have
$$
u = \frac 12 \log(x^2 + y^2), \quad v = \tan^{-1}(y/x).
$$
It follows that
$$
\frac{\partial u}{\partial x} = \frac{x}{x^2 + y^2}, \quad
\frac{\partial u}{\partial y} = \frac{y}{x^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3475837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$ Any suggestions how to solve: $$\begin{array}{|l}x^2-y^2=7 \\ x^2+xy+y^2=37\end{array}$$
I can simplify the system and get a homogeneous polynomial of degree $2$, but I think it must have an easier way.
| Hint Let
$$a=x+y \\ b=x-y$$
Then, the equations become
$$ab=7 \\
3a^2+b^2=4 \cdot 37$$
Thus
$$3a^2+\frac{49}{a^2}=148$$
This is a quadratic in $a^2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3476644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$ I've tried to solve the limit
$$ \lim_{n \to \infty} (1+ \frac{3}{n^2+n^4})^n$$
but I'm not sure.
$ (1+ \frac{3}{n^2+n^4})^n = \sqrt [n^3]{(1+ \frac{3}{n^2+n^4})^{n^4}} \sim \sqrt [n^3]{(1+ \frac{3}{n^4})^{n^4}} \sim \sqrt [n^3]{e^3 } \rightarrow 1$
Is it r... | You may also just use the binomial formula and squeeze by applying
*
*$(\star)$: $\frac 1{n^k}\binom nk \leq \frac 1{k!}$ for $0\leq k\leq n$
Hence,
$$1\leq \left(1 +\frac{3}{n^2+n^4}\right)^n\leq \left(1 +\frac{3}{n^4}\right)^n$$
$$\leq 1+\sum_{k=1}^n\left( \frac{3}{n^4}\right)^k\binom nk \stackrel{(\star)}{\leq} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find the inverse element of an unknown complex root. I got asked if I could solve the following task: Let $f:=X^3-3X+4 \in \mathbb{Q}[X]$ be the polynomial with $\alpha \in \mathbb{C}$, $f(\alpha)=0$ and $K=\mathbb{Q}(\alpha)$.
Furthermore, let $\beta := \alpha^2+\alpha+1$.
Find the inverse of $\beta$.
I tried solvi... | Since $\alpha$ is a root of $x^3-3x+4$, we have that $\alpha^3-3\alpha+4=0$, so
$$\alpha^3=3\alpha-4.$$
Since, $\beta=\alpha^2+\alpha+1$ we have that $\beta\alpha=\alpha^3+\alpha^2+\alpha$. Hence
$$\beta\alpha=\alpha^2+4\alpha-4.$$
Similarly, $\beta\alpha^2=\alpha^3+4\alpha^2-4\alpha.$ Hence
$$\beta\alpha^2=4\alpha^2-\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3477625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Number of ordered pairs of $A,B$ in Probability
Let $|X|$ denote the number of elements in a set $X,$
Let $S = \{1,2,3,4,5,6\}$ be a sample space,
where each element is equally likely to occur.
If $A$ and $B$ are independent events associated with
$S,$
Then the number of ordered pairs $(A,B)$
such that $1 \leq |... | The condition should be $1\le y\lt x$, not $\le$.
Since $z=xy/6$, $x$ or $y$ must be divisible by $2$ and by $3$, and $xy\le6$. With $1\le y\lt x$, that allows only two solutions: $x=6$, $y=1$ and $x=3$, $y=2$.
In the first case, we have $6$ choices for the element in $y$.
In the second case we have $z=3\cdot2/6=1$, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3478135",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Find the area bounded by the curve $\left(\frac{x}a+\frac{y}b\right)^5=\frac{x^2y^2}{c^4}$ Find the area bounded by the curve $\left(\dfrac{x}a+\dfrac{y}b\right)^5=\dfrac{x^2y^2}{c^4}$.
Let $x=ar\cos\varphi$, $y=br\sin\varphi$, then
$$r=\dfrac{a^2b^2\cos^2\varphi\sin^2\varphi}{c^4(\cos\varphi+\sin\varphi)^5}.$$
But ho... | It's probably better to use the substitution
$$\begin{cases} x = ar\cos^2\theta \\ y = br\sin^2\theta \\ \end{cases} $$
This has a Jacobian of $abr\sin(2\theta)$. From looking at the equation, it only encloses a loop in the first quadrant. Then plugging in we get that
$$ r^5 = \frac{a^2b^2}{c^4}r^4\sin^4\theta\cos^4\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3482064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim_{x\to0}\frac{1-\cos x\cos2x\cos3x}{x^2}$
Find $$\lim_{x\to0}\dfrac{1-\cos x\cos2x\cos3x}{x^2}$$
My attempt is as follows:-
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(2\cos x\cos2x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}{2}\left(\cos3x+\cos x\right)\cos3x}{x^2}$$
$$\lim_{x\to0}\dfrac{1-\dfrac{1}... | Hint:
Method$\#1:$
For $x\to0,$
$$1-\cos2ax\cos2bx\cos2cx\cdots=1-(1-2\sin^2ax)(1-2\sin^2bx)(1-2\sin^2cx)\cdots$$
$$\approx2(ax)^2+2(bx)^2++2(cx)^2+\cdots+O(x^4)$$ as $\lim_{x\to0}\dfrac{\sin px}{px}=1$
Method$\#2:$
$$1-\cos2ax\cos2bx\cos2cx\cdots=\dfrac{1-(1-\sin^22ax)(1-\sin^22bx)(1-\sin^22cx)}{1+\cos2ax\cos2bx\cos2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Functional equation $ f(x+yf(x^2))=f(x)+xf(xy) $ I have trouble solving the following functional equation: $f:\mathbb{R}\to\mathbb{R}$ such that $f(x+yf(x^2))=f(x)+xf(xy)$.
I think $f(x)=x$ is the unique solution, especially, $f(x+y)=f(x)+f(y)$. But how do we show that?
| The beginning steps are the same as in the answer of @MohsenShahriari, which I repeat for sake of completeness.
Our equation is $$f(x + yf(x^2)) = f(x) + xf(xy).\tag 0$$
Setting $y = 0$ in $(0)$, we get $xf(0) = 0$. Setting again $x = 1$ gives us $f(0) = 0$.
Setting $x = 1$ in $(0)$, we get $$f(1 + yf(1)) = f(1) + f(y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3483989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Sum of coefficients of $x^i$ (Multinomial theorem application)
A polynomial in $x$ is defined by
$$a_0+a_1x+a_2x^2+ \cdots + a_{2n}x^{2n}=(x+2x^2+ \cdots +nx^n)^2.$$
Show that the sum of all $a_i$, for $i\in\{n+1,n+2, \ldots , 2n\}$, is
$$ \frac {n(n+1)(5n^2+5n+2)} {24}.$$
I don't know how to proceed. I know th... | Let $S_n$ be the required sum. By expanding the right hand side of $$a_0 + a_1 x + \ldots + a_{2n} x^{2n} = (x + 2x^2 + \ldots + nx^n)^2$$ we have
$$a_{n+i} = n \cdot i + (n-1) \cdot (i+1) + (n-2) \cdot (i+2) + \cdots \ + i \cdot n$$ for $i=1, 2, \ldots, n.$
Summing over $i=1,2,\ldots, n$ gives
$$ S_n = n(1 + 2 + 3+ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
$\frac{x}{y}-\frac{y}{x}=\frac56$ and $x^2-y^2=5$
Solve the system: $$\begin{array}{|l}
\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \\ x^2-y^2=5 \end{array}$$
First, we have $x,y \ne 0$. Let's write the first equation as:
$$\dfrac{x}{y}-\dfrac{y}{x}=\dfrac{5}{6} \Leftrightarrow \dfrac{x^2-y^2}{xy}=\dfrac{5}{6}$$
We have ... | The given problem is equivalent to finding the intersections between a rectangular hyperbola and two lines through the origin, since the first equation gives $\frac{y}{x}\in\left\{\frac{2}{3},-\frac{3}{2}\right\}$. These lines are orthogonal, so it is pretty simple to locate the solutions $(3,2)$ and $(-3,-2)$. The lin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3484688",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
How to evaluate $\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$ I am trying to calculate this integral, but I find it is very challenging
$$\int_{0}^{1}\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$$
but somehow I have managed to local its closed form to be $$(-1)^n\left(\frac{... | In $0\le x<1$
$$\frac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}=\dfrac{x^n}{(1+x)^2}$$
If $\displaystyle I_n=\int_{0}^{1}\dfrac{1-x}{1+x}\cdot \frac{x^n}{\sqrt{x^4-2x^2+1}}\mathrm dx$
$$I_n+2I_{n-1}+I_{n-2}=\int_0^1x^{n-2} dx$$
Now $I_0=?,I_1=?$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3486268",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Determinant of the matrix $(\omega^i \theta^j)_{i,j = 0,1,2}$ Given that $\omega=\frac{-1+\sqrt{-3}}{2}$ is a complex cube root of 1, $\theta$ is a real cube root of 2 and $\omega^{2}+\omega+1=0$, I am trying to maipulate the matrix
\begin{pmatrix}
1 & \theta &\theta^{2} \\
\omega & \omega \theta & \omega \theta^{2} \... | From properties of determinants we get, by adding row 2 and row 3 to the first row
$$\det\begin{pmatrix}
1 & \theta &\theta^{2} \\
\omega & \omega \theta & \omega \theta^{2} \\
\omega^{2} & \omega^{2}\theta & \omega^{2}\theta^{2}
\end{pmatrix}=\det\begin{pmatrix}
1+\omega+\omega^2 & \theta+\omega\theta+\omega^2\theta &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find polynomials $P(x)P(x-3) = P(x^2)$
Find all polynomials $P \in \Bbb R[x]$ such that $P(x)P(x-3) = P(x^2) \quad \forall x \in \Bbb R$
I have a solution but I'm not sure about that. Please check it for me.
It is easy to see that $P(x) = 0, P(x) = 1$ satisfied.
Consider $P(x) \neq c$ :
We have $P(x+3)P(x) = P((x+3)... | Considering
$$
P(x)P(x-b)=P(x^2),\ \ \ b \ne 0\ \ \ \ (1)
$$
As long as $P(x^2)$ is even then $P(x)P(x-b)$ is also even then
$$
P(b)P(0) = P(b^2)
$$
because making $x = b$ in $(1)$ we have $P(b)P(0) = P(b^2)$, As this is for a generic $b$ we have $P(x)P(0) = P(x^2)$
or
$$
P(x)P(0) = P(x^2) = P(-x)P(-(x+b))
$$
which im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3488111",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
How to prove this formula for the determinant of a $4 \times 4$ tridiagonal matrix? This following is a problem from B. S. Grewal's Higher Engineering Mathematics.
Show
$$\begin{vmatrix} 2\cos(\theta) & 1 & 0 & 0 \\ 1 & 2 \cos(\theta) & 1
& 0 \\ 0 & 1 & 2 \cos(\theta) & 1 \\ 0 & 0 & 1 & 2 \cos(\theta)
\end{vmatrix}... | https://en.wikipedia.org/wiki/Tridiagonal_matrix
$f_n = \left|
\begin{array}{llll}
a_1 & b_1 &0 &0 &0 \\
c_1 & a_2 & b_2 &0 & 0\\
0 & c_2 & \ddots & \ddots & 0\\
0 &0 & \ddots & \ddots & b_{n-1}\\
0 &0 &0 & c_{n-1} & a_n
\end{array}
\right|
$
with $f_0 = 1, f_{-1} = 0$.
$f_n
= a_n f_{n-1}-c_{n-1}b_{n-1}f_{n-2}
$.
I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3489325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
If $\alpha, \beta, \gamma, \delta$ are distinct roots of equation $x^4 + x^2 + 1 = 0$ then $\alpha^6 + \beta^6 + \gamma^6 + \delta^6$ is I tried to find discriminant first and got two roots $$\frac{-1 + \sqrt{3}i}{2}$$ and $$\frac{-1 - \sqrt{3}i}{2}$$
I tried taking $x^2 = t$ and solving equation for root $w$ and $w^2$... | If you are going the hard way of finding explicitly the roots, then the roots and their exponential forms are
$$\alpha=-\frac{1}{2}-\frac{i\sqrt{3}}{2}=-e^{\frac{i\pi}{3}} \Rightarrow\alpha^6=e^{i2\pi}=1$$
$$\beta=\frac{1}{2}+\frac{i\sqrt{3}}{2}=e^{\frac{i\pi}{3}}\Rightarrow\beta^6=e^{i2\pi}=1$$
$$\gamma=\frac{1}{2}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 2
} |
How many pairs of integers have a fixed GCD? Let $k \ge 0, a \ge 0$ be a fixed integer and $f(x,k, a)$ be the number of positive integers pairs $(m,n), m < n \le x$ such that $\gcd(km-n, m + kn) = a$.
Question: Is there a closed form of expression $\lim_{x \to \infty}\dfrac{f(x,k,a)}{x^2}$ in terms of $a$ and $k$?
My e... | $f(n,1,1) = \sum_{m < n, \gcd(m-n,m+n) = 1} 1 = \sum_{m < n} \sum_{d \mid m-n, d \mid m+n} \mu(d) = \sum_{d \mid 2n} \mu(d)\sum_{m < n, m \equiv n \mod d} 1 = \sum_{d \mid 2n} \mu(d)\left[\frac{n}{d}+O(1)\right] = n\sum_{d \mid 2n} \frac{\mu(d)}{d}+O(\tau(2n))$.
So, $\sum_{n \le x} f(n,1,1) = \sum_{n \le x} \left[n\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3494730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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A complex Banach space satisfying the parallelogram law is Hilbert Let $H$ be a Banach space with associated norm $\|-\|.$ Suppose that for any $x,y\in H,$ we have:
$$\|x+y\|^2+\|x-y\|^2=2\left(\|x\|^2+\|y\|^2\right),$$
which we call the parallelogram law.
Then it is a well-known standard fact that $H$ becomes a Hilber... | Let $\|\cdot\|: H\to \mathbb{R}$ be a norm in a complex Banach space $H$ which fulfills for all $x,y\in H$ the parallelogram law
\begin{align*}
\|x+y\|^2+\|x-y\|^2=2\left(\|x\|^2+\|y\|^2\right)\tag{1}
\end{align*}
We show the map $\alpha:H\times H\to\mathbb{C}$ defined by
\begin{align*}
\alpha(x,y)= \frac{1}{4} \left\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Evaluating $\lim_{n\to\infty}\sum_{k=0}^n\frac{1}{2^k}\tan\frac{x}{2^k}$ So I was solving a previous year question paper and stumbled upon the following question-
$${\lim_{n\to {\infty}}}\biggl(\tan x+\frac{1}{2}\tan \frac{x}{2}+\frac{1}{2^2}\tan \frac{x}{2^2}+\frac{1}{2^3}\tan \frac{x}{2^3}+{\cdots}+\frac{1}{2^n}\tan ... | Hints:
*
*$\ln\prod\cos\frac{x}{2^n}=\sum\ln\cos\frac{x}{2^n}$
*$\left(\sum\ln\cos\frac{x}{2^n}\right)'=-\sum\frac{1}{2^n}\tan\frac{x}{2^n}$
*$\prod\cos\frac{x}{2^n}$ is something well known.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3496972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Area between 2 curves and $X$-axis I have trouble doing this. I have to find the area of the region bounded by the parabola ($4X-X^2$), $y=X$ and the $X$-axis. I found the interception points but don't understand what the common area is between all of them. The answer is $\frac{37}{6}$
Thanks guys.
| It is helpful to visualize the intersection between the two curves and $x$-axis by a graphing program such as Wolfram Alpha or Desmos.
Labeling $f(x)=4x-x^2, g(x)=x, h(x)=0$, we see that $g(x) < f(x)$ when $0< x < 3$. So, for $0\le x \le 3$ the area of the region bounded by the curves and the $x$-axis is the triangle ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3497159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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$\begin{cases} x^4+x^2y^2+y^4=21 \\ x^2+xy+y^2=3 \end{cases}$
I should solve the following system: $$\begin{cases} x^4+x^2y^2+y^4=21 \\
x^2+xy+y^2=3 \end{cases}$$
by reducing the system to a system of second degree.
What can I look for in such situations? What is the way to solve this kind of systems? The only th... | Hint:
$$x^4+y^4+x^2y^2=(x^2+y^2)^2-(xy)^2=?$$
So, we know $x^2+y^2,xy=9>0$
So, $x,y$ will have the same sign
Hope you can take it from here using $y=\dfrac9x$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3500677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Help with inequality problem
Given $a$ , $b$ , $c \ge 0$ show that
$$\frac{a^2}{(a+b)(a+c)} + \frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)} \ge \frac{3}{4}.$$
I tried using Titu's lemma on it, resulting in
$$\frac{a^2}{(a+b)(a+c)}+\frac{b^2}{(b+a)(b+c)}+ \frac{c^2}{(c+a)(c+b)}\ge \frac{(a+b+c)^2}{a^2+b^2+c^2 + 3(... | Another way.
We need to prove that $$4\sum_{cyc}(a^2b+a^2c)\geq3\prod_{cyc}(a+b)$$ or
$$\sum_{cyc}c(a-b)^2\geq0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$\sum_{n=1}^\infty \frac{n^k}{1+2^n+(-2)^{nk}}$ $S = \sum_{n=1}^\infty \frac{n^k}{1+2^n+(-2)^{nk}}$
Find $k \in N: S$ coverages.
I started with checking the Cauchy's theorem for $k = 2m, m \in N$ and at this point everything is fine. But for $k = 2m - 1, m \in N$ things start to get complicated and I don't really have ... | Let be
$$
a_n \left( k \right) = \frac{{n^k }}
{{1 + 2^n + \left( { - {\text{2}}} \right)^{nk} }}
$$
If $k=1$ then the series is not convergent. Indeed, you have that
$$
a_n(1) = \frac{n}
{{1 + 2^n + \left( { - {\text{1}}} \right)^n 2^n }}
$$
Thus, when $n$ is odd you have that $a_n=n$. And the necessary condition f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3501692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solve $y' + y^2 = \frac{1}{x^2}$ by introducing $z = xy$ as a new function. Question:
Solve the equation:
$$y' + y^2 = \frac{1}{x^2},~~~~~~~~~x > 0$$
by introducing $z = xy$
Attempted answer:
$z = xy \Rightarrow y = \frac{z}{x}$
Taking the derivative of $y$ with respect to $x$ using he product rule:
$$y' = \frac{z'}{x}... | When you divide by $1-z^2+z$, you're actually excluding the possibility of this term to be $0$. But note that if it is $0$, you get the desired solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3505612",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve $2^m=7n^2+1$ Solve $2^m=7n^2+1$ with $(m,n)\in \mathbb{N}^2$
Here is what I did:
First try, I have seen first that the obvious solutions are $n=1$ and $m=3$ , and $n=3$ and $m=6$, then I proved by simple congruences that $m$ must be divisible by $3$ so $m=3k$, If we add $27$ to the equation we will have $2^{3k}+3... | If $2^m=7n^2+1$ with $m=3k$, as the OP found must be the case (since $2^m\equiv1$ mod $7$), we have
$$7n^2=2^{3k}-1=(2^k-1)(2^{2k}+2^k+1)$$
Now for $k\ge1$ we have
$$\begin{align}
\gcd(2^k-1,2^{2k}+2^k+1)
&=\gcd(2^k-1,2^{2k}+2^{k+1})\\
&=\gcd(2^k-1,2^{k+1}(2^{k-1}+1))\\
&=\gcd(2^k-1,2^{k-1}+1)\\
&=\gcd(2^k+2^{k-1},2^{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3506091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\sqrt{x^5\sqrt{x^6\sqrt{x^7\sqrt{x^8\ldots}}}}}}}\,dx$ I was attempting to solve an MIT integration bee problem (1) when I misread the integral and wrote (2) instead.
$$\int\sqrt{x\cdot \sqrt[3]{x\cdot \sqrt[4]{x\cdot\sqrt[5]{x\ldots } }}}\,dx\tag{1}$$
$$\int\sqrt{x^2\sqrt{x^3\sqrt{x^4\... | Consider
$$\sum_{n=1}^\infty n x^n=x\sum_{n=1}^\infty n x^{n-1}=x\left(\sum_{n=1}^\infty x^{n}\right)'$$
When finished, make $x=\frac 12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3507554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Natural numbers equal to the sum of the squares of their four smallest divisors I was in the process of answering this question when I fell asleep; when I woke up, I found that the question has been closed for being too vague:
Find all positive integers
Anyhow, the mathematical problem is as follows:
Given a natural n... | $n$ must be even, because if it were odd the four divisors would be odd and the sum of squares would be even.
The two smallest divisors are $1$ and $2$. $n$ cannot be a multiple of $4$ because if the fourth divisor is $p$ we have $n=1^2+2^2+4^2+p^2=21+p^2$ and squares $\bmod 4$ are $0,1$ so the right cannot be a mul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Infinite product $\prod_{n=1}^{\infty}(1+z^n)(1-z^{2n-1}) = 1$ I have to show that if $|z| < 1$, $z \in \mathbb{C}$, $$\prod_{n=1}^{\infty}(1+z^n)(1-z^{2n-1}) = 1$$
This exercise if taken from Remmert, Classical Topics in Complex Function Theory.
I don't know where to start. Could anyone give a hint ?
| The finite product is
$$
\prod_{k=1}^n\left(1+z^k\right)\left(1-z^{2k-1}\right)
=\prod_{k=n+1}^{2n}\left(1-z^k\right)\tag1
$$
This can be proven by induction. Suppose $(1)$ is true for $n$
$$
\begin{align}
\prod_{k=1}^{n+1}\left(1+z^k\right)\left(1-z^{2k-1}\right)
&=\left(1+z^{n+1}\right)\left(1-z^{2n+1}\right)\prod_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3510656",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Convex cyclic hexagon $ABCDEF$. Prove $AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA$
Convex hexagon $ABCDEF$ inscribed within a circle. Prove that
$$AC \cdot BD \cdot CE \cdot DF \cdot AE \cdot BF \geq 27 AB \cdot BC \cdot CD \cdot DE \cdot EF \cdot FA\,.$$
I... | This problem had been posted here, but was deleted by the owner for an unspecified reason. I flagged the moderators about this, but they didn't do anything. Here is the same solution I gave in that link.
Ptolemy's theorem with $\square ABCD$ yields
$$AB\cdot CD+AD\cdot BC=AC\cdot BD.$$
Ptolemy's theorem with $\square... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3513053",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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Prove $\int^{n+1}_1\frac{1}{x} dx \leq 1 + 1/2 + \cdots + 1/n$.
Source: https://math.mit.edu/~choiks/Pset%202%20solutions.pdf
I don't understand the third line $\int^{n+1}_1\frac{1}{x}dx \leq 1 + 1/2 + \cdots + 1/n$.
I think it should rather be $\int^{n+1}_1\frac{1}{x}dx \geq 1 + 1/2 + \cdots + 1/n$ because
$\int^... | There are simple diagrams that compare sums and integrals.... Here what I call $f(x)$ is real valued and continuous, also positive. Things change (order of the inequalities) depending on $f$ increasing or decreasing. I guess $a < b$ are fixed integers.
if we have $f(x) > 0$ and $f'(x) > 0,$ then
$$ \int_{a-1}^{b} \; f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Solution verification:$\lim_{x\to 2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$
Evaluate without L'Hospital:$$\lim_{x\to
2}\frac{\ln(x-1)}{3^{x-2}-5^{-x+2}}$$
My attempt:
I used: $$\lim_{f(x)\to 0}\frac{\ln(1+f(x))}{f(x)}=1\;\&\;\lim_{f(x)\to 0}\frac{a^{f(x)}-1}{f(x)}=\ln a$$
$$
\begin{split}
L &= \lim_{x\to 2} \frac{\ln(x-1... | Your answer a bit rephrased.
$y=x-2;$
$\dfrac{\log (1+y)}{3^y-5^{-y}}$;
Numerator :
$f(y):=y\dfrac{\log (1+y)-\log 1}{y}$
Denominator:
$g(x)=\dfrac{15^y-1}{5^y}=$
$5^{-y}(15^y-1)=$
$5^{-y}(y\log 15)\dfrac{e^{y\log 15}-1}{y\log 15}$;
$\dfrac{f(x)}{g(x)}=$
$[\dfrac{\log (y+1)-\log 1}{y}]\cdot$
$[\dfrac{5^y}{\log 15}]$
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3514628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 1
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A generalization of the (in)famous IMO 1988 problem 6: If $\frac{a^2 + b^2 - abc}{ab + 1}$ is a positive integer then it is a square. This question is motivated by the famous IMO $1988$ problem $6$. Is the following true?
Let $a,b$ be positive integers and $c \ge 0$ be a non-negative integer. If $\dfrac{a^2 + b^2 - a... | Yes, this fact is true. We have:
$$\frac{a^2+b^2-abc}{ab+1}=k \implies a^2-b(c+k)a+(b^2-k)=0$$
Following the standard procedure for Vieta jumping, assume that $(a,b)$ is the smallest solution with respect to the sum of solutions $a+b$, such that $k$ is not a perfect square. If one root of $$x^2-b(c+k)x+(b^2-k)$$ is $a$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3515786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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How to multiply $\sqrt{5x^2+2x+1}$ with $\frac{1}{x}$? I believe the answer says, $\sqrt{5+\frac{2}{x}+\frac{1}{x^2}}$, but I don't see how they obtained it. I'm quite rusty with radicals.
| $f(x) = \sqrt{5 x^2 + 2x +1} $ is the positive function whose square $f^2$ is $f^2=5 x^2 + 2x +1$.
Hence the square $(\frac f x)^2$ is $(\frac f x)^2 = 5 +\frac 2 x + \frac 1 {x^2}$, which implies that
$$\Big | \frac{\sqrt{ 5x^2 +2x+1}}{x}\Big|= \sqrt { 5 +\frac 2 x + \frac 1 {x^2}}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Why does't quadratic formula work to factor polynomial when $a \ne 1$? $$2x^2 + 3x + 1$$
applying quadratic formula:
$$x = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$$
$$a=2, b=3, c=1$$
$$x = \frac{-3 \pm \sqrt{3^2 - 4 \cdot 2 \cdot1}}{2\cdot2}$$
$$x = \frac{-3 \pm \sqrt{9-8}}{4}$$
$$x = \frac{1}{4}[-3 + 1],~~~x=\frac{1}{4}[-3-1... | Your error is in forgetting about the leading coefficient. If a monic polynomial $p(x) $ has roots $r_1,\ldots, r_n$, with multiplicity, then
$$p(x) =(x-r_1) \cdots(x-r_n) $$
If it is not monic but instead has a leading coefficient of $a$, then
$$p(x) =a(x-r_1) \cdots(x-r_n)$$
Note that for this to work you need to inc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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Prove that if $x$ and $y$ are both not $0$ Prove that if $x$ and $y$ are both not $0,$ then
$$x^4+x^3y+x^2y^2+xy^3+y^4>0$$
I know this seems fairly easy but I'm fairly new to calculus and need some help proving that this is true. Appreciate the help!
| \begin{align}
& x^4 + x^3y + x^2y^2 + xy^3 + y^4 \\[8pt]
= {} & \left( \left( \tfrac x y \right)^4 + \left( \tfrac x y \right)^3 + \left( \tfrac x y \right)^2 + \left( \tfrac x y \right) + 1 \right) y^4 \\[8pt]
= {} & \Big(u^4 + u^3 + u^2 + u + 1\Big) y^4.
\end{align}
Since $y^4>0$ except when $y=0,$ it is enough to lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 3
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Prove that $A^2=a_1^2+a_2^2-a_3^2-a_4^2$ for all integers $A$.
Prove that:
For every $A\in\mathbb Z$, there exist infinitely many $\{a_1,a_2,a_3,a_4\}\subset\mathbb Z$ given $a_m\neq a_n $ such that $$A^2=a_1^2+a_2^2-a_3^2-a_4^2$$
After many hours, I found a general formula satisfying the statement for all $A$ and ... | $a^2=a1^2+a2^2-a3^2-a4^2\tag{1}$
Let assume $p^2+q^2-r^2-s^2 = 1\tag{2}$
Equation $(2)$ has many parametric solutions.
We use one of the solutions, $(p,q,r,s)=(2n+1, n-1, n+1, 2n).$
( By consider two identies, $(n+1)^2 - (n-1)^2 = 4n, (2n+1)^2 - (2n)^2 = 4n+1$ )
$n$ is arbitrary.
Substitute $a1=pt+c, a2=qt+d, a3=rt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3516908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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The equation $a^{4n}+b^{4n}+c^{4n}=2d^2$ Recently, I found that if $a+b=c$, then $a^4+b^4+c^4=2d^2$ for some positive integer $d$. The parametric equation is:
$$m^4+n^4+(m+n)^4=2(m^2+mn+n^2)^2$$
The condition $a+b=c$ (assuming $c \geqslant a,b$) isn't necessary. For example:
$$7^4+7^4+12^4=2 \cdot 113^2$$
We can note t... | Problem 3
This is a scheme to generate the solutions which, like your example of $(7,7,12,113)$, have two of $a,b,c$ equal.
Consider the following system of three closely related equations.
E: $2x^4-y^4=z^2$
F: $x^4+8y^4=z^2$
G: $x^4-2y^4=z^2$
A 'base solution' $(x,y,z)$ of E can be used to generate a solution $(z,xy,2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517077",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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"answer_id": 0
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First order PDE problem $yu_{x} - xu_{y} = x^2$ I am a beginner to PDE, trying to solve $$yu_{x} - xu_{y} = x^2$$ using characteristic line, first I have \begin{align*}
\frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2}
\end{align*}
Then I get the characteristic line is given by $$C = \frac{1}{2}y^2 - \frac{1}{2}x^2$$
N... | \begin{align*}
\frac{dx}{y} &= -\frac{dy}{x} = \frac{du}{x^2}
\end{align*}
From the first two ratios,
$$\frac{dx}{y} = -\frac{dy}{x} \implies x~dx~+~y~dy~=0 $$
Integrating, $~x^2+y^2=c^2~\tag1$where $~c~$ is integrating constant.
From the last two ratios, $$ -\frac{dy}{x} = \frac{du}{x^2} \implies du=-x~dy\implies ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3517794",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Identity involving falling factorial Let $(x)_n$ denote the falling factorial,
$$(x)_n = x(x-1) \dots (x - n + 1).$$
I came across the following identity in trying to solve another problem.
$$\sum_{k=1}^m \frac{(a)_k}{(x)_k} = \frac{a}{x - a + 1}\left( 1 - \frac{(a-1)_m}{(x)_m}\right).$$
This identity allows you to fin... | We prove
\begin{align*}
\sum_{k=m}^n\frac{\binom{a}{k}}{\binom{b}{k}}=\frac{b+1}{b-a+1}\left(\frac{\binom{a}{m}}{\binom{b+1}{m}}-\frac{\binom{a}{n+1}}{\binom{b+1}{n+1}}\right)\tag{1}
\end{align*}
We start with the right-hand side of (1) and obtain
\begin{align*}
\color{blue}{\frac{b+1}{b-a+1}}&\color{blue}{\left(\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3518397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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What is the sum of the coefficients of the terms containing $x$ (Problem 105, Algebra, Gelfand) In problem 105 described below, the solution provided is:
The sum of the terms containing $x$ is the sum of the terms not
containing y less the sum of the constant terms. There can be only one
constant term which is the... | You are right, the Problem 105 must read: "What is the sum of the coefficients of the terms containing $\color{red}{\text{only}}$ $x$?"
Note that $(1+x-y)^3$ is a trinomial and its expansion is:
$$f(x,y)=(1+x-y)^3=\sum_{i,j,k\\ i+j+k=3} {3\choose i,j,k}1^ix^j(-y)^k=\\
{3\choose 3,0,0}+{3\choose 2,1,0}x+{3\choose 2,0,1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3519190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Can diagonalization of a positive definite matrix be used to compute its determinant and inverse? I have a positive definite matrix $A= E^T DE$. $A\in \mathcal{R}^{n \times n} $.
$E$ is a orthonormal matrix $E \in \mathcal{R}^{m \times n}$ consisting of n orthonormal columns where $m >n $. And, $E^T E=I_{n \times n}$ ... | I don't see any simple method for computing the determinant. What I can confirm is that there is no function for $\det A$ purely in terms of $D$; the entries in $E$ must also be considered in any formula. To prove this, all we need is a diagonal matrix, and two orthogonal $m \times n$ matrices $E_1, E_2$ such that $\de... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3522695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Summation and factorial. If $$2^{n}a_{n}=\sum_{k=0}^{n}{a_{k}a_{n-k}},\phantom{a}\forall n\in\mathbb{N},$$ with $a_{0}=1,$ then $$a_{n}=\dfrac{a_{1}^{n}}{n!},\phantom{a}\forall n\in\mathbb{N}.$$
Attempt: Notice that
(1) For $ n=2t + 1, $ we have $$2^{2t+1}a_{2t+1}=2(a_{0}a_{2t+1}+a_{1}a_{2t}+a_{2}a_{2t-1}+\cdots+a_{t-1... | The sum is a convolution of $a_n$ with itself, so the generating function of the RHS is $A(z)^2$, where $A(z)=\sum_n a_n z^n$ is the generating function of $a_n$. The generating function of $2^n a_n$ is $A(2z)$, so we have $A(2z)=A(z)^2$. Hence $A(z)=\exp(c z)$ for some constant $c$, which implies that $a_n=c^n/n!$. N... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3524061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Derivative of $\dfrac{\sqrt{3-x^2}}{3+x}$ I am trying to find the derivative of this function
$f(x)=\dfrac{\sqrt{3-x^2}}{3+x}$
$f'(x)=\dfrac{\dfrac{1}{2}(3-x^2)^{-\frac{1}{2}}\frac{d}{dx}(3-x)(3+x)-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-2x(3+x)}{2\sqrt{3-x^2}}-\sqrt{3-x^2}}{(3+x)^2}$
$=\dfrac{\dfrac{-x(3+x)}{\sqrt{3-... | You just need to take the LCM of the denominators.
$$\begin{align}\dfrac{-x}{\sqrt{3-x^2}(3+x)}-\dfrac{\sqrt{3-x^2}}{(3+x)^2} &= \frac{-x(3 + x) - \sqrt{3 - x^2}\sqrt{3 - x^2}}{(3 + x)^2\sqrt{3 - x^2}} \\ &= \dfrac{-3x - x^2 - 3 + x^2}{(3 + x)^2\sqrt{3 - x^2}} \\ &= -\dfrac{3(x + 1)}{(3 + x)^2\sqrt{3 - x^2}}\end{align}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3525571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 5
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Given the points $A(2a, a)$ and $B(2b, b)$ find the coordinates of the point $M$ such that $\vec{AM} = 3 \vec{MB}$. Consider the points:
$$A(2a, a) \hspace{2cm} B(2b, b)$$
with $a \ne b$ and $a, b \in \mathbb{R}$. Find the point $M(x, y)$ such that $\vec{AM} = 3 \vec{MB}$.
First thing I tried was to plot everything we ... | What you're doing looks correct. Using vectors and matrices is a valid way to proceed. However, as you requested, here is an alternate method to consider.
The line going through $A$ and $B$ has a slope of
$$m = \frac{y_2 - y_1}{x_2 - x_1} = \frac{b - a}{2b - 2a} = \frac{1}{2} \tag{1}\label{eq1A}$$
Thus, the general equ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3528642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the function $f$ continuous at $(0,0)$
Is the function $f$ continuous at $(0,0)$?
$f(x, y)$ := $\frac{xy}{|x|+|y|}$, if $(x, y)$ $\not= (0, 0)$
and $f(0, 0) := 0$
My attempt:
$f(x, y)$ = $\frac{xy}{x+y}$ if $(x,y) > (0,0)$ and
$f(x, y)$ = $\frac{xy}{-(x+y)}$ if $(x,y) < (0,0)$
So using polar coordinates:
$f(x, y)$... | $(x,y)\not =(0,0)$;
1) $x\not =0$;
$|\dfrac{xy}{|x|+|y|}| \le |y| \lt \sqrt{x^2+y^2};$
2) $y \not =0$;
$|\dfrac{xy}{|x|+|y|}| \le |x| \lt \sqrt{x^2+y^2};$
$\epsilon >0$ given.
Choose $\delta =\epsilon$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536692",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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the $\max$ of $\min$ $a,b,c>0$,
$$\begin{cases}a+b+c=8\\ab+bc+ac=16\end{cases}$$
$m=\min\{ab,bc,ca\}$ , how to get $m_{\max}$
With Mathematica
Maximize[{Min[a b, b c, c a], a + b + c == 8, a b + b c + c a == 16, a > 0, b
> 0, c > 0}, {a, b, c}]
$$\left\{\frac{16}{9},\left\{a\to \frac{4}{3},b\to \frac{4}{3},c\to \frac... | Let $a\geqslant b\geqslant c$, $b=c+u$ and $a=c+u+v$.
Thus, $u$ and $v$ are non negatives, $$3c+2u+v=8$$ and
$$(c+u)(c+u+v)+c(c+u)+c(c+u+v)=16$$ or
$$3c^2+(4u+2v)c+u(u+v)=16$$ or
$$9c^2+(4u+2v)3c+3u(u+v)=48,$$ which gives
$$(8-2u-v)^2+(4u+2v)(8-2u-v)+3u^2+3uv=48$$ or
$$u^2+v^2+uv=16.$$
Now, since $$ab\geqslant ac\geqsl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3536890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $\int_0^{2\pi} \frac{1}{3 + \cos x} dx$. I have to find the integral
$$\int_0^{2\pi} \dfrac{1}{3 + \cos x} dx$$
I tried using the Weierstrass subtitution, but replacing the bounds, I get:
$$t_1 = \tan \dfrac{0}{2} = \tan 0 = 0$$
$$t_2 = \tan \dfrac{2 \pi}{2} = \tan \pi = 0$$
Resulting in the integral:
$$\int_0^0 \... | Using the identity
$$\frac{1}{a+b\cos t}=\frac{1}{\sqrt{a^2-b^2}}+\frac{2}{\sqrt{a^2-b^2}}\sum_{n=1}^{\infty}\left(\frac{\sqrt{a^2-b^2}-a}{b}\right)^n\cos{(nt)},\ a>b$$
we have
$$\frac{1}{3+\cos x}=\frac{1}{\sqrt{8}}+\frac{1}{\sqrt{2}}\sum_{n=1}^{\infty}\left(\sqrt{8}-3\right)^n\cos{(nx)}$$
giving us
$$\int_0^{2\pi}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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I don't understand how -12 is the numerator in this difference quotient. Find the difference quotient of $$\frac{\left(\frac{1}{12(x+h)+3}\right)-\left(\frac{1}{12x+3}\right)}{h}$$
When I solve it I end up canceling h's at the end and only having -12 as the answer. However, the correct answer is -12 over the denominato... | $$
\frac{\frac{1}{12(x+h)+3}-\frac{1}{12x+3}}{h}=\frac{1}{h}\cdot \left(\frac{1}{12(x+h)+3}-\frac{1}{12x+3}\right)=\\
\frac{1}{h}\cdot \frac{12x+3-12(x+h)-3}{(12(x+h)+3)(12x+3)}=\\
\frac{1}{h}\cdot \frac{-12h}{(12(x+h)+3)(12x+3)}=\frac{-12}{(12(x+h)+3)(12x+3)}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3540839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the sequence $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$: proving that $2+2a_n$ is a perfect square Question:
Let $a_1=a_2=97$ and $a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}$ for $n>1$. Prove that
(a) $2+2a_n$ is a perfect square, and (b) $2+\sqrt{2+2a_n}$ is a perfect square.
I changed the ... | $ $(Essentially taken from Sequence problem on AoPS.)
The key point is to recognize the connection between the recurrence formula
$$
a_{n+1}=a_{n}a_{n-1}+\sqrt{(a_n^2-1)(a_{n-1}^2-1)}
$$
and the addition formula for the inverse hyperbolic cosine:
$$
\DeclareMathOperator{\arcosh}{arcosh}
\arcosh u + \arcosh v=\arcosh \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3541519",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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No triple of primes Prove that $p^5+q^5+r^5$ is not divisible by $10pqr$ where $p,q,r$ are prime numbers.
I proved using orders that if $p>q>r$ then $r=2$ and $q|p+2$.
I proved (it is easy) using orders that there are no primes $p,q,r$ for which $6pqr|p^3+q^3+r^3$ .
And I tried to solve this one, but it couldn't solv... | We use Fermat little theorem:
$P^5+q^5+r^5 ≡ (p+q+r) \ mod (5)$
⇒ $(p^5-p)+(q^5-q)+(r^5-r)≡ 0 \ mod(5)$
It can be shown that $30 \big | n^5-n$; $n∈ \mathbb N$: If n is prime, then:
$n^5-n≡ 0 \ mod(5)$
$n^5-5=n(n^2-1)(n^2+1)$
$n^2-1≡ 0 \ mod(3)$
Also $n(n+1)$ is even, that is $n^5-n$ is divisible by $30$. (Notice that t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543627",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
Find all possible integral ordered pair $(x,y)$ in $x^2+y^2=2x+2y+xy$
what i try $y^2-(x+2)y+x^2-2x=0$
$$y=\frac{x+2\pm\sqrt{(x+2)^2-4(x^2-2x)}}{2}$$
$$y=\frac{x+2\pm \sqrt{-3x^2+12x+4}}{2}$$
How do i solve it Help me please
| Look at the discriminant which should be positive definite so the $y$ is real for real $x$. This demands
$$-3x^2+12x+4 \ge 0 \implies 3x^2-12x-4 \le 0 \implies (x-4.3)(x+.3) \le 0 \implies -.3 \le x \le 4.3 ~~(roughly)$$
So the possible integral values of $x$ are $x=0,1,2,3,4$
out of these we accept the ones which gi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3543962",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to calculate $\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$ $$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}$$
someone please help i’m not sure how to compute this. i’ve tried to do it this way:$$\lim\limits_{ x \to \infty } \frac {7^{2x} +7^{-2x} }{ 3(7^{... | In that limes, $x\to\infty$ is equivalent to $7^{2x}\to\infty$, hence:
$$\lim\limits_{x \to \infty} \frac {7^{2x} +7^{-2x} }{ 3(7^{2x} - 7^{-2x})}
= \lim\limits_{z \to \infty} \frac {z +z^{-1}}{ 3(z - z^{-1})}
= \frac13\lim\limits_{z \to \infty} \frac {z^2+1}{z^2 - 1}
= \frac13
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3544652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 3
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If $h(x) = \dfrac{f(x)}{g(x)}$, then find the local minimum value of $h(x)$ Let $f(x) = x^2 + \dfrac{1}{x^2}$ and $g(x) = x – \dfrac{1}{x}$
, $x\in R – \{–1, 0, 1\}$. If $h(x) = \dfrac{f(x)}{g(x)}$, then the local minimum value
of $h(x)$ is :
My attempt is as follows:-
$$h(x)=\dfrac{x^2+\dfrac{1}{x^2}}{x-\dfrac{1}{x}}$... | With $t:=x-\dfrac1x$
$$h(x)=\left(x^2+\frac1{x^2}\right)\left(x-\frac1x\right)=(t^2-1)t=t^3-2t.$$
Then by the chain rule
$$h'(x)=0\iff (3t^2-2)\left(1-\frac1{x^2}\right)=0.$$
We have the two solutions $x=\pm1$, and the solutions of
$$x-\frac1x=\pm\sqrt{\frac23},$$
which are
$$\dfrac{\pm\sqrt 6\pm\sqrt{42}}{6}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3546598",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Trouble with trig substitution I have a few questions and a request for an explanation.
I worked this problem for a quite a while last night. I posted it here.
Here is the original question: $$\int\frac{-7 x^2}{\sqrt{4x-x^2}} dx$$
And here is the work that I did on it:
Help with trig sub integral
Sorry that the negati... | I'll try to convey general lessons by colouring some coefficients. The crux of this problem is to change the square-rooted expression $4x-x^2=\color{blue}{2}^2-(x-\color{limegreen}{2})^2$ to a squared trigonometric function with $x=\color{limegreen}{2}+\color{blue}{2}\sin t=2(1+\sin t)$, so the integral becomes$$\begin... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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Integer solutions of $\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $ (AusPol 1994) Find all integer solutions of
$$\frac{(a+b)(a+c)(b+c)}{2} + (a+b+c)^3 = 1 -abc $$
Attempt:
I noticed that $a,b,c>0$ or $a,b,c<0$ can't happen. Besides, if one of them is zero, we can find some solutions. Suppose $c=0$, we get
$$(a+b)\f... | It's $$(a+b+c)(ab+ac+bc)+abc+2(a+b+c)^3=2$$ or
$$(a+b+c)^3+\sum_{cyc}a(a+b+c)^2+\sum_{cyc}ab(a+b+c)+abc=2$$ or
$$\prod_{cyc}(a+b+c+a)=2$$ or $$(2a+b+c)(2b+a+c)(2c+a+b)=2.$$
Can you end it now?
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding the limit of $\sqrt{4x^2+x+7}+2x$ I've been working on this problem for a while now, but I can't solve it
$$\lim\limits_{x\to-\infty}{\sqrt{4x^2+x+7}+2x}$$
I've tried multiplying by
$$\frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}$$
but I didn't get it. Am I missing something really obvious? Can someone hel... | Hint: Multiplying by $\ \frac{\sqrt{4x^2+x+7}-2x}{\sqrt{4x^2+x+7}-2x}\ \left(=1\right)\ $ shows that
\begin{align}
\sqrt{4x^2+x+7}+2x&=\frac{x+7}{\sqrt{4x^2+x+7}-2x}\\
&=\frac{-1+\frac{7}{|x|}}{\sqrt{4-\frac{1}{|x|}+ \frac{7}{x^2}}+2}\ \ \text{for }\ x<0\ .
\end{align}
Can you see what the limit of this last expression... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
} |
What is the locus of $z^2+\bar{z}^2=2$? I have to prove that it's the equation of an equilateral hyperbola
$$z^2+z^{-2}=2$$
I try this
$z^2+z^{-2} +2 = 2+2$
$(z^2+1/z^2 +2 ) = 4$
$(z+1/z)^2=4$
$ z+1/z= 2 $
$ x+yi + 1/(x+yi) = 2 $
$ ((x+yi)^2+1)/(x+yi)=2$
$ x^2+2xyi-y^2+1=2x+2yi$
$ x^2-2x+1 +2xyi -y^2=2yi $
$(x-1)^2 +2... | Let $z = x+iy$. Then, $x=\frac12(z+\bar z)$, $y=-\frac i2(z-\bar z)$, and
$$x^2-y^2=\frac14(z+\bar z)^2 + \frac14(z-\bar z)^2 =\frac12(z^2+\bar z^2)= 1$$
Thus, the locus is an equal-axis hyperbola.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3556719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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How to solve $ \frac{d^2 x}{dt^2}=e^x $? I want to solve the following ODE:
$$
\frac{d^2 x}{dt^2}=e^x
$$
I can't think of any immediate available tool to help me with this. Any suggestions?
| Consider $p=\frac{dx}{dt}$ therefore $\frac{dp}{dt}=\frac{dp}{dx}\frac{dx}{dt}=p\frac{dp}{dx}$ therefore we have
$$p\frac{dp}{dx}=e^x\Rightarrow pdp=e^xdx\Rightarrow \frac{p^2}{2}=e^x+c'_1\Rightarrow p=\sqrt{2e^x+c_1}$$
Now we have
$$\frac{dx}{dt}=\sqrt{2e^x+c_1}\Rightarrow \frac{dx}{\sqrt{2e^x+c_1}}=dt \\ \Rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3559989",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculus 2: Integration by Parts Stuck on Integral of Product With ArcTan Inside I'm stuck on the following problem:
$$\int_0^{1/3} y \tan^{-1}(3y)\,dy$$
I think my last line in my work below is correct but I don't know what to do beyond that.
A step through of the problem would be appreciated.
| Let $x = 3y$ then $ 3dy = dx $,
$$\displaystyle \int_0 ^{\frac {1}{3}} y \ \tan^{-1}(3y) dy = \frac {1}{9} \int_0 ^1 x\ \tan^{-1}(x) dx$$
Use ILATE
$\displaystyle \Rightarrow \frac {1}{18} [x^2 \tan^{-1}(x) |_0 ^1 - \int_0 ^1 \frac {x^2}{1+x^2} dx ] $
$\displaystyle \Rightarrow \frac {1}{18} [x^2 \tan^{-1}(x) |_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Revisiting Ahmed Integral in $(0,\infty)$ A recent post in MSE:
Evaluate $\int_0^{\infty } \frac{\tan ^{-1}\left(a^2+x^2\right)}{\left(x^2+1\right)\sqrt{a^2+x^2}} \, dx$
re-emphasizes that when Ahmed Integral is converted to a two-dimensional integral in $x,y$, then the sameness of domain the $(0,1)$ of $x$ and $y$ mak... | $$I=\int_{0}^{\infty} \frac{\tan^{-1}\sqrt{2+x^2}}{(1+x^2)\sqrt{2+x^2}}~ dx= \int_{0}^{\infty} \frac{\pi/2-\tan^{-1}(1/\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}}~dx$$
$$\implies I=\frac{\pi}{2}\int_{0}^{\infty} \frac{1}{(1+x^2)\sqrt{2+x^2}} dx
-\int_{0}^{\infty} \frac{\tan^{-1}(1/\sqrt{2+x^2})}{(1+x^2)\sqrt{2+x^2}} dx =I_1-I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3562974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find the open intervals where $x(x-6)^3$ is concave upwards and concave downwards. Find the open intervals where $f(x)=x(x-6)^3$ is concave upwards and concave downwards.
Solution: We want to find the concavity of $f(x)=x(x-6)^3$, so we have to take the second derivative, find the critical points (where it equals zero... | This may help to confirm the basic derivatives and regions:
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{n\to\infty}\sum_{k=1}^n\frac{k^3}{\sqrt{n^8+k}}$ & $\lim_{n\to\infty}n\bigg(\sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}-\frac{1}{4}\bigg)$. This is a homework question. I have to find two limits:
i. $$\lim_{n\to \infty} \sum_{k=1}^n \frac{k^3}{\sqrt{n^8+k}}$$
ii. $$\lim_{n\to \infty} n\bigg(\sum_{k=1}^n \frac... | There is another way which would allow to answer both questions at the same time using the binomial theorem
$$\frac{k^3}{\sqrt{n^8+k}}=\sum_{n=3}^\infty (n-3)^{20-8 n} \binom{-\frac{1}{2}}{n-3} k ^n=(n-3)^{20-8 n} \binom{-\frac{1}{2}}{n-3}\sum_{n=3}^\infty k^n$$ Now, use Faulhaber's formulae and you should arrive to s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3563708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate the indefinite integral $\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$ $$\int \frac{\sin^3\frac\theta 2}{\cos\frac\theta2 \sqrt{\cos^3\theta + \cos^2\theta + \cos\theta}} d\theta$$
My Attempt:
$$I = \int \frac{\sin^2\frac\theta2\cdot\sin\frac\theta2\... | Following from your approach
\begin{align}I=\frac{1}{2}\int \frac{(t-1)(t+1)}{(t+1)^2\sqrt{t^3+t^2+t}}\,dx=\frac{1}{2}\int \frac{t^2(1-\frac{1}{t^2})}{t(t+\frac{1}{t}+2)t\sqrt{t+\frac{1}{t}+1}}\,dx\\=\frac{1}{2}\int \frac{(1-\frac{1}{t^2})}{(t+\frac{1}{t}+2)\sqrt{t+\frac{1}{t}+1}}\,dx\end{align}
Let $t+\frac{1}{t}=u; \... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
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Why does $3x^4 + 16x^3 + 20x^2 - 9x - 18$ = $(x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3$? $$3x^4 + 16x^3 + 20x^2 - 9x - 18 $$
When simplified I arrive to:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6}) $$
But the math book wrote:
$$ (x+3)(x+2)(\frac{-1 \mp \sqrt 37 }{6})3 $$
with that extra 3 at the end. The graph calculator s... | By the rational root test the polynomial has the two roots $x=-2$ and $x=-3$ so that
$$
3x^4 + 16x^3 + 20x^2 - 9x - 18=(x+2)(x+3)(3x^2+x-3)
$$
The extra $3$ is not an exponent but is the leading coefficient in $3x^2+x-3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3569506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$ solve for $(\cos x)(\cos2x)(\cos3x)=\frac{1}{4}$. what is the general solution for $x$.
I wrote $\cos2x$ as $2\cos^2x-1$ and $\cos3x$ as $4\cos^3x-3\cos x$. Then the expression gets reduced to $\cos x(2\cos^2x-1)(4\cos^3x-3\cos x)=\frac{1}{4}$. substituted $\cos x=t$ and... | $$\cos x\cdot \cos 2x\cdot\cos 3x=1/4$$
$$\implies (2\cos x\cdot\cos 3x)(2\cos 2x)=1$$
$$\implies (\cos 4x+\cos 2x)(2\cos 2x)=1$$
$$\implies 2\cos 4x\cdot\cos 2x+(2\cos^22x-1)=0$$
$$\implies 2\cos 4x\cdot\cos 2x+\cos 4x=0$$
$$\implies \cos 4x~(2\cos 2x+1)=0$$
Either, $~\cos 4x=0\implies 4x=(2n+1)\dfrac π2\implies x=(2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Modular arithmetic and inverse functions The question is shown below:
Suppose $S=\{0,1,2,3,4,5,6,7,8,9,10\}$ and that the function $f:S\rightarrow S$ is given by:
$f(x)=6x^2+3x+8$ (mod 11)
Let $T=\{0,5\}$.
Find $f^{-1}\left(T\right)$.
Alright,
so my initial approach with this question was to find the inverse function,
... | If $\ a\equiv \alpha^2\pmod{11}\ $ is a quadratic residue $\mod{11}\ $, then
$$
a^6 \equiv \alpha^{12}\equiv \alpha^2\equiv a\pmod{11}\ ,
$$
so the square roots of $\ a\pmod{11}\ $ are $\ \pm a^3\pmod{11}\ $.
If $\ y\equiv 6x^2+3x+8\pmod{11}\ $ then $\ 2y+4\equiv$$ x^2+6x+9$$\equiv(x+3)^2 \pmod{11}\ $ must be a quadrat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3570186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Uniform convergence of the ratio of a sequence of function
$f_n(x): (0,1) \to (0,1)$ is a sequence of continuous functions that converges uniformly to $f(x): (0,1) \to (0,1)$. Is it true that
$$
\sup_{x \in (0,1) }\frac{f_n(x)}{f_{n+1}(x)} \to 1
$$
I think pointwise convergence follows easily: since $\{f_n\}$ conv... | Let $f_{n}(x)=x^{\frac{n+1}{n+2}}$ and $f(x)=x$, then clearly $0<f_n(x)<1$ and $0<f(x)<1$ for all $x \in (0,1)$.
Next, let's show $f_n(x)$ converges to $f(x)$ uniformly on $(0,1)$:
$\displaystyle h_n(x):= f_n(x) - f(x) = x^{\frac{n+1}{n+2}} - x = x \left( \frac{1}{\sqrt[n+2]{x}} - 1 \right)$.
$\displaystyle [h_n(x)]' ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573648",
"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$ $\frac{(a+b+c)^2}{(a^2+b^2+c^2)}=\frac{\cot(A/2)+\cot(B/2)+\cot(C/2)}{\cot(A)+\cot(B)+\cot(C)}$
I need to solve this trigonometric identity for a trianlge.
I'm not allowed to use the formula $a+b+c=s$ where 's' is per... | Replace all $\cot x = \frac{\cos x}{\sin x}$ and expand straightforwardly as follows,
$$RHS = \frac{\frac{\cos \frac A2}{\sin \frac A2}+\frac{\cos \frac B2}{\sin \frac B2}+\frac{\cos \frac c2}{\sin \frac C2}}
{\frac{\cos A}{\sin A}+\frac{\cos B}{\sin B}+\frac{\cos C}{\sin C}}$$
$$= \frac{\cos \frac A2\sin \frac B2\sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3573793",
"timestamp": "2023-03-29T00:00:00",
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Prove $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N$ Prove $\frac{2k+2}{2k+3}>\frac{2k}{2k+1}, k\in\mathbb N$
The book I am using asserts a non-trivial way of proving this inequality, but I cannot see why this cannot be proven by rearranging the statement equally as rigorously.
Let
$$\frac{2k+2}{2k+3}=\frac{2k}{2k+... | $\frac{2x+2}{2x+3}>\frac{2x}{2x+1}, X\in\mathbb R^+$
$ f: x \rightarrow \frac{2x}{2x+1} = f: x \rightarrow 1- \frac{1}{2x+1}$ is easily demonstrable to be strictly increasing
$x > y \rightarrow 2x+1 > 2y+1 \rightarrow \frac{1}{2x+1} <\frac{1}{2y+1} \rightarrow 1- \frac{1}{2x+1} > 1-\frac{1}{2y+1}$ therefore $f(k+1) >... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3575307",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Closed form for $\sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}$ Does there exist any closed form for the following sum?
$$\sum_{k=0}^{l}\binom{k}{n}\binom{k}{m}$$
Where
$l \in \mathbb N$ and $m,n \in \mathbb Z$
My try:
$$ \sum_{k=\max\left(m,n\right)}^{l}\binom{k}{n}\binom{k}{m}=\sum_{k=0}^{l}\binom{k}{k-n}\binom{k}{k-m}$... | We present a proof of the identity by @Diger, which should be considered
a starting point for additional simplification. We seek to show that
$$\sum_{k=0}^l {k\choose m} {k\choose n} =
\sum_{k=0}^n (-1)^k {l+1\choose m+k+1} {l-k\choose n-k}.$$
The RHS is
$$[z^n] \sum_{k=0}^n (-1)^k {l+1\choose m+k+1}
z^k (1+z)^{l-k}.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3577193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Multinomial expansion sum Given,
$$ \frac{x^2+x+1}{1-x}= a_0+a_1x +a_2x^2+\cdots $$
then, find the sum:
$$ \sum^{50}_{r=1}a_r $$
I knew using multinomial expansion, that
$$ (1-x)^{-1} = 1+ x+x^2+\cdots$$
Hence, $$ \frac{x^2+x+1}{1-x}=1\times(1+x+x^2+\cdots) +x\times(1+x+x^2+\cdots)+x^2\times(1+x+x^2+\cdots) $$
Hence,... | An easier way to handle this is that since $(1-x)(-x-2) = x^2 + x - 2$, you get
$$\begin{equation}\begin{aligned}
\frac{x^2 + x + 1}{1 - x} & = \frac{x^2 + x - 2 + 3}{1 - x} \\
& = -x - 2 + \frac{3}{1 - x} \\
& = -x - 2 + 3(1 + x + x^2 + x^3 + \ldots) \\
& = 1 + 2x + 3x^2 + 3x^3 + 3x^4 + \ldots \\
& = 1 + 2x + \sum_{i=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3581419",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How many subsets are there with exactly 8 elements in a set of 16 elements? Below is a problem I did which I believe I did correctly. I would like somebody to confirm that I did, or tell me where I went wrong.
Problem:
Consider a set with $16$ elements in it. How many subsets does it have with exactly $8$ elements?
Ans... | Your work appears correct.
The answer is "$16$ choose $8$":
$$\binom {16}8=\dfrac{16!}{8!8!}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Differentiate $x^{x^2}$ with respect to $x^2$ I am doing it by $x^2$ as the and differentiating both of the above separately and then deciding them
I got answer $x^{x^2} (2\log x+1))/2$
I don't know if it is correct
| $$\frac{dx^{x^2}}{dx^2}=\frac{de^{x^2\log x}}{2x\,dx}=\frac{2x\log x+x}{2x}x^{x^2}=\frac{2\log x+1}2x^{x^2}.$$
Alternatively,
$$\frac{d\sqrt{x^2}^{x^2}}{dx^2}=\frac{de^{{x^2\log x^2/2}}}{dx^2}=\frac{\log x^2+1}2e^{{x^2\log x^2/2}}=\frac{2\log x+1}2x^{x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3582959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Prove that in acute triangle : $\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$ Im going to prove this identity in acute triangle :
Let $ABC$ acute triangle , $A,B,C$ are angles then :
$$\sin A\cos (B-C)=\frac{\sin (2B)+\sin (2C)}{2}$$
I know that $A,B,C<\frac{π}{2}$
$$\sin A\cos (B-C)=\sin A(\cos B\cos C+\sin A\s... | Since $A,B,C$ are angles of a triangle, we have that $A+B+C = \pi$. Recall the following trigonometric identities
\begin{align}
\sin(\pi-\theta) & = \sin(\theta)\\
\sin(2\theta) + \sin(2\phi) & = 2 \sin(\theta + \phi) \cos(\theta-\phi)\\
\end{align}
We have that
\begin{align}
\sin(2B) + \sin(2C) & = 2 \sin(B+C) \cos(B-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3583492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Comparing $2$ infinite continued fractions
$A = 1 +\dfrac{1}{1 + \frac{1}{1 + \frac{1}{\ddots}}} \\
B = 2 +\dfrac{1}{2 + \frac{1}{2 + \frac{1}{\ddots}}}$
Given the two infinite continued fractions $A$ and $B$ above, which is larger, $2A$ or $B?$
I used the golden ratio on the $2$ and came up with:
$A = 1 + \dfrac{1}{... | Isn’t $\frac12B$ equal to
$$
1+\frac1{4+\frac1{1+\cdots}}\qquad <\qquad A
$$
?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Calculate limit of this sequence $I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$ I need to find the limit of the sequence :
$$I_n = \int_{0}^{\pi/2} (\tan x)^{1/n} dx$$
The only thing I have done is : take substitution $\tan x = z$
this gives
$$dx = \dfrac{1}{1 + z^2}\,dz$$
So, integral becomes :
$$I_n = \int_{0}^{\inft... | We first convert the integral into a beta function.
$$
\begin{aligned}
\int_0^{\frac{\pi}{2}}(\tan x)^{\frac{1}{n}} d x &=\int_0^{\frac{\pi}{2}} \sin ^{\frac{1}{n}} x \cos ^{-\frac{1}{n}} x d x \\
&=\frac{1}{2} B\left(\frac{1}{2}+\frac{1}{2 n}, \frac{1}{2}-\frac{1}{2 n}\right) \\
&=\frac{\pi}{2} \csc \left(\frac{1}{2}+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3584149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
how to calculate $\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}} \frac{dx}{1+\cos^2x}$? $$
\int_{\frac{\pi}{4}}^{\frac{3\pi}{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}=\int_{\frac{\pi}{4}}^{\frac{\pi}{2}}{\frac{1}{1+\cos ^2x}\rm{dx}+\int_{\frac{\pi }{2}}^{\frac{3\pi}{4}}{\frac{1}{1+\cos ^2x}\rm{dx}}}
$$
I want to split this integral,$
\i... | HINT.-Use first $\cos^2(x)=\dfrac{1+\cos(2x)}{2}$; second $\cos(2x)=\dfrac{1-\tan(x)}{1+\tan^2(x)}$ so you arrive to
$$\int\dfrac{d(\tan(x))}{\tan^2(x)+2}=\int\dfrac{du}{u^2+2}$$ you get finally
$$\left[\frac{1}{\sqrt2}\arctan\left(\frac{\tan(x)}{\sqrt2}\right)\right]_{\dfrac{\pi}{4}}^{\dfrac{3\pi}{4}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3586622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Evaluate: $S=\sum_{n=1}^{\infty}\frac{(2n)!}{(2n+1)!!^2}$ How to evaluate this sum?
$$S=\sum_{i=1}^{\infty}\prod_{j=1}^{i}\frac{j(j-1/2)}{(j+1/2)^2}$$
$$\frac{j(j-1/2)}{(j+1/2)^2}=\frac{2j(2j-1)}{(2j+1)^2}$$
$$S=\prod_{j=1}^{1}\frac{2j(2j-1)}{(2j+1)^2}+\prod_{j=1}^{2}\frac{2j(2j-1)}{(2j+1)^2}+\prod_{j=1}^{3}\frac{2j(2j... | Choosing one of the integral representations of Catalan's constant, $$2G=\int_0^{\pi/2}\log\cot\frac{x}{2}\,dx=\frac{1}{2}\int_0^{\pi/2}\log\frac{1+\cos x}{1-\cos x}\,dx=\int_0^{\pi/2}\sum_{n=0}^\infty\frac{\cos^{2n+1}x}{2n+1}\,dx\\=\frac12\sum_{n=0}^\infty\frac{1}{2n+1}\mathrm{B}\left(n+1,\frac12\right)=\sum_{n=0}^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3587706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
Find the sum of all positive integers $k$ for which $5x^2-2kx+1<0$ has exactly one integral solution.
My attempt is as follows:
$$\left(x-\dfrac{2k-\sqrt{4k^2-20}}{10}\right)\left(x-\dfrac{2k+\sqrt{4k^2-20}}{10}\righ... | Well one way of looking at the solution is for $n, k \in Z$ it will have only one integral solution if
$$n-1 \le \dfrac{k-\sqrt{k^2-5}}{5}<n< \dfrac{k+\sqrt{k^2-5}}{5} \le n+1$$
Now, $D \ge 0 \Rightarrow |k| \ge \sqrt5$ and $|\alpha -\beta| \le 2 \Rightarrow k\le \sqrt{30}$
Combining both the conditions we get $k \in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3588314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.