Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Why is $ \frac{5}{64}((161+72\sqrt{5})^{-n}+(161+72\sqrt{5})^{n}-2)$ always a perfect square? I'm working on a puzzle, and the solution requires me somehow establishing that
$$ f(n):=\frac{5}{64}\Big(\big(161+72\sqrt{5}\big)^{-n}+\big(161+72\sqrt{5}\big)^{n}-2\Big)$$
is a perfect square for $n\in \mathbb{Z}_{\geq 0}$.
... | Note that $161+72\sqrt{5} = (161-72\sqrt{5})^{-1}$ and $161+72\sqrt{5}=(9+4\sqrt{5})^2,$ and therefore
\begin{align*}
\big(161+72\sqrt{5}\big)^{-n}+\big(161+72\sqrt{5}\big)^{n}-2 &= \big( (161+72\sqrt{5})^{n/2} - (161-72\sqrt{5})^{n/2} \big)^2 \\
&= \big( (9+4\sqrt{5})^n - (9-4\sqrt{5})^n \big)^2.
\end{align*}
It there... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3747774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
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Find $L=\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^{n}\left\lfloor 2\sqrt{\frac{n}{k}} \right\rfloor -2\left\lfloor \sqrt{\frac{n}{k}} \right\rfloor$ Question:- Find Limit $$L=\lim_{n\to \infty }\frac{1}{n}\sum_{k=1}^{n}\left\lfloor 2\sqrt{\frac{n}{k}} \right\rfloor -2\left\lfloor\sqrt{\frac{n}{k}} \right\rfloor \text , ... | As this is a Riemann sum, you can convert it into an integral.
This becomes:
$$\int_0^1 \left \lfloor \frac2{\sqrt x} \right \rfloor -2\left \lfloor\frac1{\sqrt x} \right \rfloor\,dx$$
Put $\sqrt x \rightarrow 1/t$ to get:
$$ = 2\int_1^\infty \frac{\left \lfloor 2t \right \rfloor}{t^3} -2\frac{\left \lfloor t \right \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}$ given that $\sin\alpha-\cos\alpha=\frac12$ Given that $\sin\alpha-\cos\alpha=\frac12$. What is the value of $$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}?$$
My work:
$$\sin\alpha-\cos\alpha=\frac12$$
$$\sin\alpha\frac1{\sqrt2}-\cos\alpha\frac1{\sqrt2}=... | You can easily evaluate it without calculator as follows
$$\frac{1}{\sin^3\alpha}-\frac{1}{\cos^3\alpha}=\frac{\cos^3\alpha-\sin^3\alpha}{\sin^3\alpha\cos^3\alpha}$$
$$=\frac{(\cos\alpha-\sin\alpha)(\cos^2\alpha+\sin^2\alpha+\cos\alpha\sin\alpha)}{\sin^3\alpha\cos^3\alpha}$$
$$=\frac{-(\sin\alpha-\cos\alpha)(1+\cos\alp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3750756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Another way to solve $\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$ without the substitution $y=\tan\left(\frac{x}{2}\right)$? Is there another way to solve an integral $$\int \frac{\sin^4(x)}{1+\cos^2(x)}\ dx$$ without the substitution $y=\tan\left(\frac{x}{2}\right)$?
$\large \int \frac{\sin^3(x)}{1+\cos^2(x)}\ dx$ is easi... | Hint:
Bioche's rules suggest to use the substitution
$$t=\tan x,\quad \mathrm d x=\frac{\mathrm dt}{1+t^2}.$$
Indeed, as $\cos^2x=\frac 1{1+t^2}$, $\:\sin ^2x=\frac{t^2}{1+t^2}$, one obtains
$$\int\frac{\sin^4(x)}{1+\cos^2(x)}\,\mathrm dx=\int\frac{\frac{t^4}{(1+t^2)^2}}{1+\frac1{1+t^2}}\,\frac{\mathrm dt}{1+t^2}= \int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3751405",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 5
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How to evaluate $\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}$ directly? The rational zeta series
$$
\sum_{n=1}^{\infty}\frac{\zeta (2n)-1}{n+1}=\frac{3}{2}-\ln \pi \tag1
$$
can be derived from other well known rational zeta series.
$$
\sum_{n=2}^{\infty}\frac{\left ( -1 \right )^{n}\left ( \zeta (n)-1 \right )}{n+1}=\f... | We write
$$ S := \sum_{n=1}^{\infty} \frac{\zeta(2n)-1}{n+1} $$
for the sum to be computed.
1st Solution. We have
\begin{align*}
S
= \sum_{n=1}^{\infty} \frac{1}{n+1} \sum_{k=2}^{\infty} \frac{1}{k^{2n}}
= \sum_{k=2}^{\infty} \sum_{n=1}^{\infty} \frac{1}{n+1} \frac{1}{k^{2n}}
= \sum_{k=2}^{\infty} \left( - k^2 \log ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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How can I integrate $\int \frac{u^3}{(u^2+1)^3}du?$ How to integrate following
$$\int\frac{u^3}{(u^2+1)^3}du\,?$$
What I did is here:
Used partial fractions
$$\dfrac{u^3}{(u^2+1)^3}=\dfrac{Au+B}{(u^2+1)}+\dfrac{Cu+D}{(u^2+1)^2}+\dfrac{Au+B}{(u^2+1)^3}$$
After solving I got
$A=0, B=0, C=1, D=0, E=-1, F=0$
$$\dfrac{u^3}{... | Hint: $\frac {u^{3}} {(u^{2}+1)^{3}}=u\frac {(u^{2}+1)-1} {(u^{2}+1)^{3}}=\frac u {(u^{2}+1)^{2}}-\frac u {(u^{2}+1)^{3}}$. Split then integral into two parts and use the substitution $x=1+u^{2}$ in both . The answer is $-\frac 1 {2(u^{2}+1)} -\frac 1 {4(u^{2}+1)^{2}}+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3753883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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For $x≠y$ and $2005(x+y) = 1$; Show that $\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$ Problem:
Let $x$ and $y$ two real numbers such that $x≠0$ ; $y≠0$ ; $x≠y$ and $2005(x+y) = 1$
*
*Show that $$\frac{1}{xy} = 2005\left(\frac{1}{x} + \frac{1}{y}\right)$$
*Calculate $l$:
$$l = \frac{y}{y-x} - \frac{y-... | *
*$$\frac{1}{xy} = 2005(\frac{1}{x}+\frac{1}{y}) \iff \frac{1}{xy}=\frac{2005(x+y)}{xy}$$
which follows immediately from the condition
*$$l = \frac{y}{y-x} - \frac{y-x}{y} - \frac{x}{y-x} - \frac{y-x}{x} + \frac{y}{x} - \frac{x}{y} +2=$$$$= \frac{y}{y-x}-({1}-\frac{x}{y})-\frac{x}{y-x}-(-1+\frac{y}{x})+\frac{y}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3756456",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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Evaluate $\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x}$ What I attempted thus far:
Multiply by conjugate
$$\lim_{x \to 0} \frac{\sqrt{1 + x\sin x} - \sqrt{\cos x}}{x\tan x} \cdot \frac{\sqrt{1 + x\sin x} + \sqrt{\cos x}}{\sqrt{1 + x\sin x} + \sqrt{\cos x}} = \lim_{x \to 0} \frac{1 + x\sin x - \cos ... | Similar to Varun Vejalla's answer, but without L'Hopital: Multiplying by the conjugate and replacing $\tan x$ by $\sin x/\cos x$, we have
$$\begin{align}
{\sqrt{1+x\sin x}-\sqrt\cos x\over x\tan x}
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos x}\cdot{1+x\sin x-\cos x\over x\sin x}\\
&={\cos x\over\sqrt{1+x\sin x}+\sqrt\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3758133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Finding units of $\mathbb{Z}[\delta]$ with $\delta=\frac{1+\sqrt{d}}{2}$ and $d\equiv 1\pmod{4}.$ Let $d$ be a square free integer congruent to 1 modulo 4. Let
$$\delta=\frac{1+\sqrt{d}}{2}$$
(a) Prove that the subset $\mathbb{Z}[\delta]$ of the complex numbers defined here
$$\mathbb{Z}[\delta]:=\left\{a+b\delta:a,b\in... | Hint for (a): $$\delta^2={d+1+2\sqrt d\over4}={d-1\over4}+\delta$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759473",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to prove that if $x>0$ and $y>0$, then $\sqrt{x}+\sqrt{y}>\sqrt{x+y}$, using the relation of arithmetic and geometric means?
How to prove that if $x>0$ and $y>0$, then $$\sqrt{x}+\sqrt{y}>\sqrt{x+y}\,,$$ using the relation of arithmetic and geometric means?
I started by showing that if $x>0$ and $y>0$, based on t... |
Theorem (A Baby Version of the Triangle Inequality). Let $a$, $b$, $c$, and $d$ be real numbers. We have
$$\sqrt{a^2+b^2}+\sqrt{c^2+d^2}\geq \sqrt{(a+c)^2+(b+d)^2}\,.$$
The equality holds if and only if
*
*$(a,b)=(0,0)$, or
*there exists $\lambda\geq 0$ such that $(c,d)=(\lambda a,\lambda b)$.
Let $x$ and $y$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3759816",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Integral: $\int \dfrac{dx}{(x^2-4x+13)^2}$?
How can I integrate $$\int \dfrac{dx}{(x^2-4x+13)^2}?$$
Here is my attempt:
$$\int \dfrac{dx}{(x^2-4x+13)^2}=\int \dfrac{dx}{((x-2)^2+9)^2}$$
Substitute $x-2=3\tan\theta$, $\ dx=3\sec^2\theta d\theta$
\begin{align*}
&=\int \dfrac{3\sec^2\theta d\theta}{(9\tan^2\theta+9)^2}\... | substitute $\theta=\tan^{-1}\left(\frac{x-2}{3}\right)$ & $$\sin2\theta=\frac{2\tan\theta}{1+\tan^2\theta}=\frac{2\left(\frac{x-2}{3}\right)}{1+\left(\frac{x-2}{3}\right)^2}=\frac{6(x-2)}{x^2-4x+13}$$
After substituting $\theta$ and $\sin2\theta$, you will get final answer
$$I=\frac{1}{54}\tan^{-1}\left(\frac{x-2}{3}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3761986",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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How to solve polynomial rational relations for $y$ (e.g $\sqrt{4-3y-y^2} = x(y+4)$)? From time to time, I struggle to solve polynomial relations for $y$.
A trivial example is :
$$ \frac{y}{x} = x \iff y = x^2$$
Easy.
But consider this relation:
$$ \sqrt{4-3y-y^2} = x(y+4)$$
No matter how much I mess around it, seems i... | $$ \sqrt{4-3y-y^2} = x(y+4)$$
$$ {4-3y-y^2} = x^2(y+4)^2$$
$${ y^2+3y-4} = -x^2(y+4)^2$$
$$(y+4)(y-1) = -x^2(y+4)^2$$
$$\frac{y-1}{y+4} = -x^2$$
$$\frac{1-y}{y+4} = x^2$$
$$\sqrt{\frac{1-y}{y+4}} = x$$
$$ \frac{1-y}{y+4} = x^2$$
$$1-y = x^2y + 4x^2$$
$$1-y = x^2y + 4x^2$$
$$-y = x^2y + 4x^2-1$$
$$y = 1-x^2y - 4x^2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3764098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Newton's evaluation of $1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$ How might have Newton evaluated the following series?
$$\sqrt{2} \, \frac{\pi}{4} = 1 + \frac{1}{3} - \frac{1}{5} - \frac{1}{7} + \frac{1}{9} + \frac{1}{11} - \cdots$$
The method of the this thread applies by sett... | Nick Mackinnon gives what appears to be the story in an article that appeared in the Mathematical Gazette in March 1992 (Vol. 76, No. 475), entitled "Newton's Teaser." He writes that Newton conjured up the poser, in fits and starts, in response to Leibniz's series
$$
1 - \frac13 + \frac15 - \frac17 + \cdots = \frac\pi... | {
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"url": "https://math.stackexchange.com/questions/3764826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving $\frac1{2\pi} \int_0^{2\pi} \frac{R^2-r^2}{R^2-2Rr\cos\theta+r^2} d\theta =1$ by integrating $\frac{R+z}{z(R-z)}$ without residue theorem. I was given the function:
$$ \frac{R+z}{z(R-z)} $$
And I was asked to integrate it around a closed contour to prove:
$$\frac1{2\pi} \int_0^{2\pi} \frac{R^2-r^2}{R^2-2Rr\cos\... | We could avoid complex analysis altogether
$$I = \frac{1}{\pi}\int_0^\pi \frac{R^2-r^2}{R^2+r^2 - 2Rr\cos\theta}\:d\theta$$
$$ = \frac{1}{\pi}\int_0^\pi \frac{R^2-r^2}{R^2+r^2\left(\cos^2\frac{\theta}{2}+\sin^2\frac{\theta}{2}\right) - 2Rr\left(\cos^2\frac{\theta}{2}-\sin^2\frac{\theta}{2}\right)}\:d\theta$$
$$\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3767266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Rolling Dice Game, Probability of Ending on an Even Roll The game is described as follows. $A$ and $B$ take turns rolling a fair six sided die. Say $A$ rolls first. Then if $A$ rolls {1,2} they win. If not, then $B$ rolls. If $B$ rolls {3,4,5,6} then they win. This process repeats until $A$ or $B$ wins, and the game st... | The problem is not clear as stated.
Interpretation $\#1$: If you interpret it as "find the probability that the game end in an evenly numbered round" you can reason recursively.
Let $P$ denote the answer. The probability that the game ends in the first round is $\frac 26+\frac 46\times \frac 46=\frac 79$. If you don't... | {
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"url": "https://math.stackexchange.com/questions/3768049",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\iint dx\,dy$ over the region bounded by $y^2=x$ and $x^2+y^2=2x$ in the first quadrant
Identify the region bounded by the curves $y^2=x$ and $x^2+y^2=2x$, that lies in the first quadrant and evaluate $\iint dx\,dy$ over this region.
In my book the solution is like:
$$\begin{align}\\
\iint dx\,dy &=\int_... | $$\int\limits_0^1\left(\sqrt{2x-x^2}-\sqrt{x}\right)\,dx =\\=
\int_\limits0^1\sqrt{2x-x^2}\,dx-\int\limits_0^1\sqrt{x}\,dx = \frac{\pi}{4} - \frac{2}{3}
$$
For first one:
$$\int_\limits0^1\sqrt{2x-x^2}\,dx = \int_\limits0^1\sqrt{1-(x-1)^2}\,dx=\\=\int\limits_{-1}^{0}\sqrt{1-y^2}\,dy=\int\limits_{-\frac{\pi}{2}}^{0}\cos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768368",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$
Evaluate:
$$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx$$
I could find the integral by setting it equal to $$\frac{ax+b}{(1+x^3)^{1/2}}$$
and differentiating both sides w.r.t.$x$ as
$$\frac{2-x^3}{(1+x^3)^{3/2}}=\frac{a(1+x^3)^{3/2}-(1/2)(ax+b)3x^2(1+x^3)^{-1/2}}{(1+x^3)}$$$$=\f... | $$\int \frac{2-x^3}{(1+x^3)^{3/2}} dx=\int \frac{\frac{1}{x^3}(2-x^3)}{\frac1{x^3}\left(1+x^3\right)^{3/2}} \ dx$$
$$=\int \frac{\left(\frac{2}{x^3}-1\right)dx}{\left(\frac{1}{x^2}+x\right)^{3/2}}$$
$$=-\int \frac{d\left(\frac{1}{x^2}+x\right)}{\left(\frac{1}{x^2}+x\right)^{3/2}}$$
$$=- \frac{\left(\frac{1}{x^2}+x\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2 (using proof by contradiction).
The question is "Show that the solution of the equation $x^5-2x^3-3=0$ are all less than 2."
I have attempted to answer this question using proof by contradiction and I think my answer is either wrong or not a we... | Here is another way of doing this:
If $x\gt 2$ then $x^5\gt 4x^3\left(=2x^3+2x^3\right) \gt 2x^3+16\gt 2x^3+3$
I have added this because it clicked into my head immediately as a way of estimating and thinking though different approaches gives you a range of tools to solve such problems.
| {
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"url": "https://math.stackexchange.com/questions/3768647",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If the largest positive integer is n such that $\sqrt{n - 100} + \sqrt{n + 100}$ is a rational no. , find the value of $\sqrt{n - 1}$ . So here is the Problem :-
If the largest positive integer is n such that $\sqrt{n - 100} + \sqrt{n + 100}$ is a rational no. , find the value of $\sqrt{n - 1}$ .
What I tried :- I thin... | Let $\sqrt{n-100} + \sqrt{n+100} = p$, where $p$ is rational.
$$\implies 2n + 2\sqrt{n^2 - 10000} = p^2$$
But that must mean that $2\sqrt{n^2 - 10000}$ is rational.
Which must mean that $\sqrt{n^2 - 10000}$ is rational.
$$\implies n^2 - 10000 = k^2$$
$$\implies (n+k)(n-k) = 10000$$ The problem requires us to maximize $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3768870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Proofs by Induction: are my two proofs correct? I have been trying to understand how proof by mathematical induction works, and I am struggling a bit. But, I think I am understanding it and I just want to verify that what I am doing is correct (and if not, why?)
I have attached a screenshot (as a link) of my problem (b... | Yes.... it is okay but it needs a bit of exposition-- that is an explanation of what you are doing and why you are doing it and why it proves what you want.
I would add after you write the word "Induction:" I would expound words to the effect
Induction step: Supose $P(k)$.
Suppose that for some $k$ that $1+ 3+ 6 + ...... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\nabla_x\left\langle L,Z-\begin{bmatrix}Tu&x\\ x^H&t\end{bmatrix}\right\rangle+p/2\left\|Z-\begin{bmatrix}Tu&x\\ x^H&t\end{bmatrix}\right\|^2$ I am currently reading Atomic norm denoising with applications to line spectral estimation by Bhaskar et al.
In appendix E, an ADMM algorithm is presented to solve the SDP... | The error is in the derivative of the last term. That should be $\rho/2 \cdot4(x-z_1) = 2\rho(x-z_1)$.
The easiest way to see this is by writing the Frobenius norm as the sum of the squared components (or squared lenghts of the matrix component for the complex plane). The only places where $x$ occurs are the first colu... | {
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"url": "https://math.stackexchange.com/questions/3771652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving the system $\sqrt{x} + y = 7$, $x + \sqrt{y} = 11$ I want to solve the following nonlinear system of algebraic equations. Indeed, I am curious about a step by step solution for pedagogical purposes. I am wondering if you can come up with anything. I tried but to no avail.
\begin{align*}
\sqrt{x} + y &= 7 \\
x +... | Here is another elegant solution that one of my friends suggested. Although it is so simple and straight forward, it just works for these specific numbers $7$ and $11$! Let us start by rewriting the equations in the following forms
\begin{align*}
(\sqrt{x}-3)+(y-4)&=0,\\
(x-9)+(\sqrt{y}-2)&=0.
\end{align*}
Next, using ... | {
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"url": "https://math.stackexchange.com/questions/3772635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Show that $(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$ Let $a,b,c\ge 0$ be such that $a^2+b^2+c^2=3$. Show that
$$(a^3+a+1)(b^3+b+1)(c^3+c+1)\le 27$$
I want to consider the function
$$f(x)=\ln{(x^{3/2}+x^{1/2}+1)}$$ Maybe it isn't the case $f''(x)\le 0$, so I can't use Jensen's inequality.
| Your idea works, but in another writing:
$$3\ln3-\prod_{cyc}(a^3+a+1)=\sum_{cyc}\left(\ln3-\ln(a^3+a+1)\right)=$$
$$=\sum_{cyc}\left(\ln3-\ln(a^3+a+1)+\frac{2}{3}(a^2-1)\right)\geq0$$
because easy to see that $f(x)\geq0$ for any $x\geq0$, where $$f(x)=\ln3-\ln(x^3+x+1)+\frac{2}{3}(x^2-1).$$
Indeed, $$f'(x)=\frac{4(x-1)... | {
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"url": "https://math.stackexchange.com/questions/3774774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving a result for $\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)$ $$\prod_{k=0}^{\infty}\Bigl(1-\frac{4}{(4k+a)^2}\Bigr)=\frac{(a^2-4)\Gamma^2\bigl(\frac{a+4}{4}\bigr)}{a^2\Gamma\bigl(\frac{a+2}{4}\bigr)\Gamma\bigl(\frac{a+6}{4}\bigr)}$$
According to WA. I attempted using
$$\prod_{k=0}^{\infty}\Bigl(1-\frac{x... | Use (proved by induction)
$$
\prod_{k=0}^{n-1} (a+4k) = \frac{4^n\Gamma(n+\frac{a}{4})}{\Gamma(\frac{a}{4})}
$$
together with
$$
1-\frac{4}{(4\,k+a)^2}={\frac { \left( a+4\,k+2 \right)
\left( a+4\,k-2 \right) }{ \left( 4\,k+a \right) ^{2}}}
$$
to get
$$
\prod _{k=0}^{n-1}{\frac { \left( a+4\,k+2 \right) \left( a+4\,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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Is this a valid proof for $I(n^2) \geq \frac{5}{3}$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $\sigma(x)$ denote the sum of divisors of the positive integer $x$, and let $I(x)=\sigma(x)/x$ be the abundancy index of $x$.
Note that both $\sigma$ and $I$ are multiplicative functions.
A number $m$ ... | I think that the answer for your initial question is yes. I've found no errors in the argument.
I think that the answer for your final question is no since under the condition that $q=5$, we see that $I(n^2)\ge \dfrac 53$ is equivalent to $k=1$ as follows :
$$\begin{align}I(n^2)\ge\frac 53&\iff \frac{8\cdot 5^k}{5^{k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3776712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Special value of hypergeometric function $\, _2F_1\left(a,a+\frac{1}{3};\frac{4}{3}-a;-\frac{1}{8}\right)$ How can one prove $$\, _2F_1\left(a,a+\frac{1}{3};\frac{4}{3}-a;-\frac{1}{8}\right)=\frac{\left(\frac{2}{3}\right)^{3 a} \Gamma \left(\frac{2}{3}-a\right) \Gamma \left(\frac{4}{3}-a\right)}{\Gamma \left(\frac{2}{3... | From the transformation ${_2F_1}(a,b,c,z) = (1-z)^{c-a-b}{_2F_1}(c-a,c-b,c,z)$, it is equivalent to find
$${_2F_1}(\frac{4}{3}-2a,1-2a,\frac{4}{3}-a,-\frac{1}{8}) = {_2F_1}(\frac{4b}{3},\frac{4b+1}{3},\frac{4b+5}{6},-\frac{1}{8})\qquad b=\frac{3}{4}-\frac{3a}{2}$$
We have the quartic transformation:
$$\tag{1}{_2F_1}(\f... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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Inverse Laplace Transform, need help with this $\frac{s^2}{s^2+\sqrt{2}s+1}$ Inverse Laplace Transform of $\frac{s^2}{s^2+\sqrt{2}s+1}$
I transformed the denominator as $(s+\frac{\sqrt{2}}{2})^{2}$ + $\frac{1}{2}$
$\frac{s^2}{(s+\frac{\sqrt{2}}{2})^{2} + \frac{1}{2}}$ and I have no idea how to move forward because part... | Substitute $u=s+\dfrac{\sqrt{2}}{2}$:
$$\dfrac{s^2}{(s+\dfrac{\sqrt{2}}{2})^{2} + \frac{1}{2}}=\dfrac{(u-\dfrac{\sqrt{2}}{2})^2}{u^{2} + \dfrac{1}{2}}$$
Then expand...And apply the first Shifting Theorem.
$$\mathcal{L^{-1}}\dfrac {\left(s+\frac{\sqrt2}2\right)}{{\left(s+\frac{\sqrt2}2\right)^2+\frac12}}=e^{-\frac {\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3778940",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evalution of a function where $t = x + \frac{1}{x}$ Consider a function $$y=(x^3+\frac{1}{x^3})-6(x^2+\frac{1}{x^2})+3(x+\frac{1}{x})$$ defined for real $x>0$. Letting $t=x+\frac{1}{x}$ gives: $$y=t^3-6t^2+12$$
Here it holds that $$t=x+\frac{1}{x}\geq2$$
My question is: how do I know that $t=x+\frac{1}{x}\geq2$ ?
I w... | We have that for $x>0$
$$x+\frac{1}{x}\geq2 \iff x\cdot x+x\cdot \frac{1}{x}\geq x\cdot 2 \iff x^2-2x+1\ge 0 \iff (x-1)^2 \ge 0$$
and the equality holds if and only if $x=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3779787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$ Question:
Show that the transformation $w=\frac{2z+3}{z-4}$ maps the circle $x^2+y^2-4x=0$ onto the straight line $4u+3=0$.
My try:
$$\begin{align}\\
&x^2+y^2-4x=0\\
&\implies (x-2)^2+y^2=4\\
&\implies |z-... | This is equivalent to showing that $4w+3$ is pure imaginary i.e. $\arg(4w+3)=\pm\frac{\pi}{2}$.
Since $0$ and $4$ are end points of the circle diameter and $4w+3=11\frac{z-0}{z-4}$, then $\arg(4w+3)=\pm\frac{\pi}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3781558",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Let $\frac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$ Let $\dfrac{\tan A}{1-\tan^2A}=\sin^220^\circ-\sin160^\circ\sin220^\circ+\sin^2320^\circ$, find $\tan6A$
My attempt :
\begin{align*}
\dfrac{\tan2A}{2}=\sin^220^\circ-\sin20^\circ\sin40^\circ+\sin^240^\circ\\
\tan2A=2... | Note that:
\begin{array}{}
\dfrac{\tan2A}{2}&=&\sin^2160^\circ-\sin160^\circ\sin220^\circ+\sin^2220^\circ\\
\dfrac{\tan2A}{2}&=&(\sin160^\circ-\sin220^\circ)^2+\sin160^\circ\sin220^\circ\\
\dfrac{\tan2A}{2}&=&(\sin20^\circ+\sin40^\circ)^2-\sin20^\circ\sin40^\circ\end{array}
Apply product-sum and sum-product
\begin{arra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3783658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Is $g(x) = \frac{x^3+9}{x^2}$ one-to-one? Let $g(x) =\frac{x^3+9}{x^2}$ restricted to $D(g) = (-\infty,0)$. Is it 1-1?
My approach
$g(x) \text{ 1-1 }: g(x_1)=g(x_2) \iff x_1 = x_2$
Let $x_1, x_2 \in D(g)$ then,
$$\frac{x_1^3+9}{x_1^2} = \frac{x_2^3+9}{x_2^2} \iff x_2^2 \cdot x_1^3 + 9x_2^2 - x_1^2 \cdot x_2^3 + 9x_1^... | The step "In order for (1) to hold..." is incorrect. Just before it, you had an expression of the form $A-B=0$ ($A$ happens to be $x_2^2(x_1^3+9)$ and $B$ happens to be $x_1^2(x_2^3+9)$.) From it you have concluded that $A=0$ and $B=0$. This is not correct - the difference of two numbers can be $0$ even if neither of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3784121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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If $a^2+a b+b^2=40$ and $a^2-\sqrt{a b}+b=5$, then find $a^2+\sqrt{a b}+b$ I was given this problem to solve with elementary methods (High School level).
Knowing that
$$\begin{align}
a^2+a b+b^2 &=40 \\
a^2-\sqrt{a b}+b &=\phantom{0}5
\end{align}$$
find
$$a^2+\sqrt{a b}+b$$
I tried to look for $\sqrt{a b}$, since the... | We could try to "brute force" it at the high school level. Start with the second equation
$$a^2 - \sqrt{ab} + b = 5$$
$$a^2 + b -5 = \sqrt{ab}$$
$$a^4 + b^2 + 25 + 2a^2 b -10a^2 -10b = ab$$
$$b^2 + ( 2a^2 -10 -a) b + (a^4 -10a^2 +25) =0$$
$$b = \frac{-(2a^2 -10 -a) \pm \sqrt{( 2a^2 -10 -a)^2 - 4(a^4 -10a^2 +25))}}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 2,
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Sum of Series S =?
Proving:
$$\displaystyle\sum_{n=0}^{\infty}\frac{(n^2+4n-1)(\pi-\phi)^{3n}}{n!(n+2)}=(\pi-\phi)^3e^{(\pi-\phi)^3}+2\left(e^{(\pi-\phi)^3}-1\right)-5\left(\frac{e^{(\pi-\phi)^3}((\pi-\phi)^3-1)+(\pi-\phi)^4-(\pi-\phi)^3+1}{(\pi-\phi)^6}\right) $$
Where .$ \phi $ is golden ration
Note:this series is... | Note that for $n\geq 1$,
$$
\frac{{n^2 + 4n - 1}}{{n!(n + 2)}}x^{3n} = \frac{{x^{3n} }}{{(n - 1)!}} + 2\frac{{x^{3n} }}{{n!}} - 5\frac{{x^{3n} }}{{n!(n + 2)}}.
$$
Hence,
$$
\sum\limits_{n = 0}^\infty {\frac{{n^2 + 4n - 1}}{{n!(n + 2)}}x^{3n} } = e^{x^3 } (x^3 + 2) - 5\sum\limits_{n = 0}^\infty {\frac{{x^{3n} }}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Quasilinear PDE $u_t + (u^2)_x = 0$ cauchy problem The problem I am trying to solve is:
\begin{equation}\label{eq:3.1}
\begin{cases}
\partial_t u + \partial_x(u^2)=0 & x\in \mathbb{R}, t \in (0,\infty]\\
u(x,0)=
\begin{cases}
0 & x\leq 0\\
x & 0<x\leq 1\\
1 & x>1
\end{cases}
\end{cases}
\... | The main part that you did seems correct. Except the limits at the end.
$$\begin{equation}
\frac{\partial u}{\partial x} +2u\frac{\partial u}{\partial t} = 0
\end{equation}$$
Your Charpit-lagrange characteristic ODEs are correct :
$$\frac{dx}{2u}=\frac{dt}{1}=\frac{du}{0}$$
A first characteristic equation comes from $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3785527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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What kind of curve do these lines make? Our teacher has given us a question to be solved.
What curve does the intersection points of the given lines make? A parabola, hyperbola, or none of them? (please look at the image I've posted. I am not speaking about all of the intersection points. Just the tangent curve beneath... | Inspection shows that the given lines can be produced as follows:
$$y_k(x)={\sqrt{3}\over10}\bigl(2kx+(k^2+25)\bigr)\qquad(-5\leq k\leq 5)\ .\tag{1}$$One now should look for the envelope of the family $\bigl(\ell_c\bigr)_{c\in{\mathbb R}}$ of lines
$$\ell_c:\quad y-y_c(x)=0\ ,$$where the integer $k$ in $(1)$ has been r... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluate $\frac{1+3}{3}+\frac{1+3+5}{3^2}+\frac{1+3+5+7}{3^3}+\cdots$ It can be rewritten as
$$S = \frac{2^2}{3}+\frac{3^2}{3^2}+\frac{4^2}{3^3}+\cdots$$
When $k$ approaches infinity, the term $\frac{(k+1)^2}{3^k}$ approaches zero. But, i wonder if it can be used to determine the value of $S$. Any idea?
Note: By using ... | Hint: Let $f(t)=\sum t^{k}$. Then $tf'(t)=\sum kt^{k}$ and $t^{2}f''(t)=\sum (k^{2}-k)t^{k}$. This gives $\sum kt^{k}$ and $\sum k^{2}t^{k}$ in terms of $f,f'$ and $f''$. Put $t=1/3$ and use the fact that $(k+1)^{2}=k^{2}+2k+1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3788078",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Show $ \prod_{k=1}^{n} 4^k = 2^{n*(n+1)}$, where did I go wrong in my induction step? Can someone help me out with the induction step?
Show $ \prod_{k=1}^{n} 4^k = 2^{n*(n+1)}$
Base case n=1: $$4^1 = 2^{1*(1+1)} = 2^2$$
Induction step (to show: $2^{(n+1)*(n+2)} = 2^{n^2+3n+2}$ ) :
$$ \prod_{k=1}^{n+1} 4^k = 4^{n+1} *... | Your mistake was writing $2^{n(n+1)}$ as$\tfrac124^{n(n+1)}$ rather than $4^{\tfrac12n(n+1)}$. The inductive step should be$$4^{\tfrac12n(n+1)}4^{n+1}=4^{\tfrac12n(n+1)+n+1}=4^{(\tfrac12n+1)(n+1)}=4^{(n+1)(n+2)}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3791672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to prove this equality of the determinant of matrix?
Prove that
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end{pmatrix}
=(a^2-bc)(b^2-ca)(c^2-ab)\end{equation*}
My attempt:
\begin{equation*}
\det\begin{pmatrix}
a^2 & b^2 & c^2 \\
ab & bc & ca \\
b^2 & c^2 & a^2
\end... | It has to be a degree-$6$ homogeneous polynomial in $a,\,b,\,c$ that vanishes if $a^2=bc$, because if $a/b=b/c=k$ the determinant is $c^6$ times$$\left|\begin{array}{ccc} k^{4} & k^{2} & 1\\ k^{3} & k & k^{2}\\ k^{2} & 1 & k^{4} \end{array}\right|=k^{4}\left(k^{5}-k^{2}\right)-k^{2}\left(k^{7}-k^{4}\right)=0.$$Repeatin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3793112",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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The ratio of the area of two regular polygons The polygons in the figure below are all regular polygons(regular heptagon), share a vertex and the orange line crosses the three vertices of the two regular polygons, the area of the small regular polygon and the large regular polygon is denoted as $S_1$, $S_2$, what is... |
Won't go through the calculation, but this is the idea.
First since $\triangle ADE$ and $\triangle BDF$ are similar, we know $AE$ pass through $G$.
Now we can calculate $DG$,$GC$,$AG$ based on the left heptagon and since $AD\parallel CE$ we can calculate $GE=GC\cdot {AD\over DG}$. Also we know $\angle DGE=180^{\circ}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3794914",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
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Solve for $a,b,c$ :- $a + \frac{1}{b} = \frac{7}{3}$ , $b + \frac{1}{c} = 4$ , $c + \frac{1}{a} = 1$
Solve for $a,b,c$ :- $a + \frac{1}{b} = \frac{7}{3}$ , $b + \frac{1}{c} = 4$ , $c + \frac{1}{a} = 1$
What I Tried :- We have $3$ equations , $3$ variables, should be easy .
I did everything, solved these equations but... | From the last equation we have $c=1-\frac{1}{a}=\frac{a-1}{a}$ and from the second $b=4-\frac{1}{c}=4-\frac{a}{a-1}==\frac{3a-4}{a-1}$ so the first equation becomes $a+\frac{1}{b}=a+\frac{a-1}{3a-4}=\frac{7}{3}$ or $3a^{2}-4a+a-1=\frac{7}{3}(3a-4)$ so $9a^{2}-9a-3=21a-28$ and hence we have $9a^{2}-30a+25=(3a-5)^{2}=0.$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3795057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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$(a+1)(b+1)(c+1)\leq4$ for triangle sides $a,b,c$ with $ab+bc+ac=1$
Given that $a,b,c$ are the lengths of the three sides of a triangle, and $ab+bc+ac=1$, the question is to prove $$(a+1)(b+1)(c+1)\leq4\,.$$
Any idea or hint would be appreciated.
This is Problem 6 of Round 1 of the BMO (British Mathematical Olympiad)... | We need to prove $$abc+a+b+c\leq2$$ or
$$(abc+(a+b+c)(ab+ac+bc))^2\leq4(ab+ac+bc)^3$$ or
$$\prod_{cyc}(a(b+c-a)+bc)\geq0$$ and we are done!
We can get a last factoring by the following way.
For $ab+ac+bc=a^2$ we obtain:
$$(abc+(a+b+c)(ab+ac+bc))^2=(abc+(a+b+c)a^2)^2=a^2(a^2+ab+ac+bc)^2=$$
$$=(ab+ac+bc)(2(ab+ac+bc))^2=4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Check the convergence of the series $\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$ I want to check if the following series converge or not.
*
*$\displaystyle{\sum_{n=1}^{+\infty}\frac{\left (n!\right )^2}{\left (2n+1\right )!}4^n}$
I suppose we have to find here an upper boun... | The Stirling approximation implies $\binom{2n}{n}\sim\frac{4^n}{\sqrt{n\pi}}$, so $\frac{n!^24^n}{(2n+1)!}\sim\frac{\sqrt{\pi}/2}{\sqrt{n}}$, so the series diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3797499",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Two inequalities with parameters $a,b,c>0$ such that $ca+ab+bc+abc\leq 4$
Let $a,b,c>0$ be such that $bc+ca+ab+abc\leq 4$. Prove the following inequalities:
(a) $8(a^2+b^2+c^2)\geq 3(b+c)(c+a)(a+b)$, and
(b) $\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}+\dfrac{2}{a^2b}+\dfrac{2}{b^2c}+\dfrac{2}{c^2a}\geq 9$.
Prove al... | The first inequality.
Let $a=kx$, $b=ky$ and $c=kz$ such that $k>0$ and $xy+xz+yz+xyz=4.$
Thus, the condition gives $$k^2(xy+xz+yz)+k^3xyz\leq xy+xz+yz+xyz$$ or
$$(k-1)((k+1)(xy+xz+yz)+(k^2+k+1)xyz)\leq0$$ or $$k\leq1.$$
Thus, we need to prove that $$8(x^2+y^2+z^2)\geq3k(x+y)(x+z)(y+z)$$ and since $0<k\leq1$, it's eno... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3798343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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If $f(x)=\sin^{-1} (\frac{2x}{1+x^2})+\tan^{-1} (\frac{2x}{1-x^2})$, then find $f(-10)$ Let $x=\tan y$, then
$$
\begin{align*}\sin^{-1} (\sin 2y )+\tan^{-1} \tan 2y
&=4y\\
&=4\tan^{-1} (-10)\\\end{align*}$$
Given answer is $0$
What’s wrong here?
| $$\arcsin(\sin(x))=\left\{
\begin{array}{ll}
x-2\pi n&\hbox{for }-\frac{\pi}{2}+2\pi n\le x\le \frac{\pi}{2}+2\pi n\\
\pi-x-2\pi n&\hbox{for }\frac{\pi}{2}+2\pi n\le x\le \frac{3\pi}{2}+2\pi n\\
\end{array}\right.,$$
$$\arctan(\tan(x))=x-\left\lfloor\frac{x+\pi/2}{\pi}\right\rfloor\pi$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3800521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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solve for $x$, $(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a$ Find the value of $x$ when $$(\sqrt{a+ \sqrt{a^2-1}})^x+(\sqrt{a- \sqrt{a^2-1}})^x=2a.$$See, by hit and trial method it is clear that $x=2$ is a solution. But I failed to solve this explicitly to get the solutions.
My Attempt: \begin{align*} &(\s... | put $u=(\sqrt{a+ \sqrt{a^2-1}})^x$
then taking conjugate,$\frac{1}{u}$=$(\sqrt{a- \sqrt{a^2-1}})^x$
or we have to solve
$u+\frac{1}{u}=2a$ which can be easily solved.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3801818",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find all $x\in \mathbb{R}$ for which $\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$
Find all $x\in \mathbb{R}$ for which $$\frac{8^x+27^x}{12^x+18^x}=\frac{7}{6}$$
Letting $a=2^x$ and $b=3^x$ we get $$\frac{a^3+b^3}{a^2b+ab^2} = \frac{7}{6}$$
from the numerator we have that $$a^3+b^3=(a+b)(a^2-ab+b^2)=7$$
since $7$ is a pr... | The problem here is that $a+b$ and $a^2 - ab + b^2$ are not necessarily integers.
@lab bhattacharjee gives a complete answer for the same problem. In short, observe that both $a^3+b^3$ and $a^2b + ab^2$ are homogeneous of degree $3$. Dividing by $a^3$ yields $$\frac{7}{6}=\frac{a^3+b^3}{a^2b+ab^2}=\frac{1+t^3}{t+t^2}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3804820",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Integrate $ \int \frac{1}{\sin^{4}x+\cos^{4}x}dx $ Show that$$ \int \frac{1}{\sin^{4}(x)+\cos^{4}(x)}dx \ = \frac{1}{\sqrt{2}}\arctan\left(\frac{\tan2x}{\sqrt{2}}\right)+C$$
I have tried using Weierstrass substitution but I can't seem to get to the answer... Should I be using the said method or is there another way I c... | Original post was to evaluate $$\int\frac{\,dx}{\sin^4(x)\cos^4(x)}$$ but I believe that the intended integral was $$\int\frac{\,dx}{\sin^4(x)+\cos^4(x)}$$ which I evaluate after the original request.
$$\begin{align}I=&\int\frac{\,dx}{\sin^4(x)\cos^4(x)}\\&=2^4\int\frac{\,dx}{\bigr[2\sin(x)\cos(x)\bigr]^4}\\&=2^4\int\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 1
} |
Find the value of $a^{2020}+b^{2020}+c^{2020}$.
Question: Let $f(x)=x^3+ax^2+bx+c$ and $g(x)=x^3+bx^2+cx+a$ where $a,b,c\in\mathbb{Z}$, $c\neq 0$. Suppose $f(1)=0$ and roots of $g(x)=0$ are squares of the roots of $f(x)=0$. Find the value of $a^{2020}+b^{2020}+c^{2020}$.
My approach: Let the other roots of $f(x)$ exc... | Another way to reach the solution is this:
Since $a=-c^2$ and $b=c^4$ you have $f(x) =x^3-c^2x^2+c^4x+c$ and since $f(1)=0$ you obtain $1-c^2+c^4+c=0$ i.e. $c^2(c^2-1)+c+1=(c+1)(c^3-c^2+1)=0$, by observing that $c^3-c^2+1=0$ can't have neither odd or even solution, you obtain $c=-1$ and your solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3808144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Inconsistent solution to differential equation $Q.$
Let f(x) be a continuous function satisfying the following differential equation-
$$f(x)=(1+x^2)(1+\int_0^x\frac{f^2(t)dt}{1+t^2})$$
$$\text{Find f(1)}$$
My Work-
1)Putting $x=0$ in the equation we get $f(0)=1$
2)Dividing by $1+x^2$ and differentiating w.r.t. $x$ we ... | $$\frac{f(x)}{1+x^2}=1+\int_{0}^{x} \frac{f^2(t)}{1+t^2}dt \implies f(0)=1$$
D. w.r.t. $x$, using Lebnitz, to get
$$f'(x)-\frac{2x}{1+x^2}f(x)=f^2(x)\implies f^{-2}f'-\frac{2x}{1+x^2}\frac{1}{f}=1$$
This is Bernoulli equation. Let $1/f=v$, then we get
$$\frac{dv}{dx}+\frac{2x}{1+x^2}v=-1$$
This is linear equation, whos... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813224",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Number of non negative integer solutions of $x+y+2z=20$
The number of non negative integer solutions of $x+y+2z=20$ is
Finding coefficient of $x^{20}$ in
$$\begin{align}
&\left(x^0+x^1+\dots+x^{20}\right)^2\left(x^0+x^1+\dots+x^{10}\right)\\
=&\left(\frac{1-x^{21}}{1-x}\right)^2\left(\frac{1-x^{11}}{1-x}\right)\\
=&\... | Found the mistake. Instead of $\left(x^0+x^1+\dots+x^{10}\right)$, it should have been $\left(x^0+x^2+\dots+x^{20}\right)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3813339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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How to deduce this factorization of $x^5+x+1$ by looking at $\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$? The question is:
$$\int\frac{3x^4+2x^3-2x+1}{x^5+x+1}dx$$
I tried a lot but couldn't solve it so I looked at the solution which is:
$$x^5+x+1=(x^2+x+1)(x^3-x^2+1)$$ and we can write $$3x^4+2x^3-2x+1=(x^3-x^2+1)+(3x^2-2... | Just it's better to see that any polynomial $x^{3k-1}+x^{3n-2}+1$ has a factor $x^2+x+1$ for any naturals $k$ and $n$.
For example, your reasoning with $\omega\neq1$ and $\omega^3=1$ helps to understand it.
In our case we can get this factoring so:
$$x^5+x+1=x^5-x^2+x^2+x+1=(x^2+x+1)(x^3-x^2+1).$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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For $n\ge 6$, can we partition the set $\{1 , 4 , 9 , ...,n^2\}$ into two subsets whose sums are equal or differ by one?
For $n\ge 6$, can we partition the set $\{1 , 4 , 9 , ...,n^2\}$ into two subsets such that the sums of the elements in the two subsets are equal or differ by one?
For example : for $n = 10$, we ca... | The difference of sums is of the from $S:=\sum_{k=1}^n s_kk^2$ where each $s_k=\pm1$. Our task is to find $s_k$ such that the sum is $0$ or $1$.
Observe that $$\tag1a^2-(a+1)^2-(a+2)^2+(a+3)^2=4.$$
Therefore, we can pick four consecutive signs such that they contribute either $+4$ or $-4$.
Hence with $8$ consecutive si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Cauchy-Schwarz Inequality problems Let $a,$ $b,$ $c,$ $d,$ $e,$ $f$ be nonnegative real numbers.
(a) Prove that
$$(a^2 + b^2)^2 (c^4 + d^4)(e^4 + f^4) \ge (ace + bdf)^4.$$
(b) Prove that
$$(a^2 + b^2)(c^2 + d^2)(e^2 + f^2) \ge (ace + bdf)^2.$$
I'm not sure how I should start approaching both problems. I believe I shoul... | For (a),
Just use Holder's Inequality.
For (b),
Expand $(a^2 + b^2) (c^2 + d^2) (e^2 + f^2) $. (It is equivalent to the LHS of my answer)
Expand $(ace + bdf)^2$. (It is equivalent to the RHS of my answer)
The problem is just proving
$$
a^2 c^2 e^2 + a^2 c^2 f^2 + a^2 d^2 e^2 + a^2 d^2 f^2 + b^2 c^2 e^2 + b^2 c^2 f^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $a^2+b^2-ab=c^2$ for positive $a$, $b$, $c$, then show that $(a-c)(b-c)\leq0$
Let $a$, $b$, $c$ be positive numbers. If $a^2+b^2-ab=c^2$. Show that
$$(a-c)(b-c)\leq0$$
I have managed to get the equation to $(a-b)^2=c^2-ab$, but I haven't been able to make any progress.
Can someone help me?
| Equation, $(a^2-ab+b^2)=(c)^2$ has solution shown below:
$a=p^2-q^2$
$b=2pq-q^2$
$c=p^2-pq+q^2$
Where, $p>q>0$
We need:
$w=(a-c)(b-c)\leq0$
Or, $w=(c-a)(c-b)\leq0$
$(c-b)=(p-q)(p-2q)$
$(c-a)=-q(p-2q)$
Hence,
$w=(c-a)(c-b)=(-)(p-q)(q)(p-2q)^2$
Since, $p>q>0$:
$(p-2q)^2$ & $(p-q)$ & $(q)$ is a positive quantity and the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3820511",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
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Solve differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$ I need to solve this differential equation: $y' = \frac{y^2}{x^3} + 2\frac{y}{x} - x$.
My attempt
I found that $y = x^2$ is a solution. Then I tried to put $y = x^2f(x)$, and solved this way:
$$2xf(x) + x^2f'(x) = xf^2(x) + 2xf(x) - x \implies x^2f'... | 1.) It is correct, if a little unusual.
2.a) The usual way for a Riccati equation with a particular solution is to set $y(x)=y_p(x)+\frac1{u(x)}$ which should result in a linear first order DE for $u$.
2.b) It is typical for Riccati equation that the solution family is similar to a circle, that is, that the real line f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find the roots of $x^3 - 6x = 4$ This exercise is from the book Complex Analysis by Joseph Bak, and it says:
"Find the three roots of $x^{3}-6x=4$ by finding the three real-valued possibilities for $\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$".
I know that these numbers were found by Cardan's method, but I don't understand why they... | Let $x=\sqrt[3]{2+2i}+\sqrt[3]{2-2i}$. Then,
$x^3=2+2i+2-2i+3(\sqrt[3]{2+2i})^2(\sqrt[3]{2-2i})+3(\sqrt[3]{2+2i})(\sqrt[3]{2-2i})^2$
$=4+3(\sqrt[3]{2+2i}+\sqrt[3]{2-2i})(\sqrt[3]{2+2i})(\sqrt[3]{2-2i})$
$=4+3x\sqrt[3]{(2+2i)(2-2i)}$
$=4+3x\sqrt[3]{8}$
$=4+6x$
$\therefore x^3-6x=4$
Therefore, solving for $x=\sqrt[3]{2+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3821909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Area of triangle under tan conditions. For an acute triangle $ABC$, the following conditions hold.
$$\frac{1}{\tan A} + \frac{1}{\tan B} + \frac{1}{\tan C } =2$$
$$ a^2 + b^2 + c^2 =50 $$
Compute the area of such a triangle.
I used a tan addition law and get,
$$\tan A + \tan B + \tan C = \tan A \tan B \tan C $$.
And le... | We have $$2=\sum_{cyc}\frac{ab\cos\gamma}{ab\sin\gamma}=\frac{\sum\limits_{cyc}(a^2+b^2-c^2)}{4S}=\frac{50}{4S},$$ which gives $$S=\frac{25}{4}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3823715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Computing $\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$ The problem is: $$\lim_{x\to-5}\frac{x^2+2x-15}{|x+5|}$$
I factored the numerator to get: $\frac{(x-3)(x+5)}{|x+5|}$
How do i solve the rest?
| After the correction in the limit $x\rightarrow -5$.
Recall that $|x+5|=\begin{cases}x+5 \space\space \text{if $x\geq-5$} \\-(x+5)\space\space \text{if $x<-5$}\end{cases}.$ Then
*
*$\lim_{x\rightarrow (-5)^{-}}\frac{(x-3)(x+5)}{|x+5|}=\lim_{x\rightarrow (-5)^{-}}\frac{(x-3)(x+5)}{-(x+5)}=\lim_{x\rightarrow (-5)^{-}}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3824902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Real roots of $x^7+5x^5+x^3−3x^2+3x−7=0$? The number of real solutions of the equation,
$$x^7+5x^5+x^3−3x^2+3x−7=0$$
is
$$(A) 5 \quad (B) 7 \quad (C) 3 \quad (D) 1.$$
Using Descartes rule we may have maximum no. of positive real roots is $3$ and negative real root is $0.$ So there can be either $3$ real roots or $1$ re... | If the Descartes rule does not help there is always the following (or similar) way: $$x^7+5x^5+x^3−3x^2+3x−7=(x-1)(x^6+x^5+6x^4+6x^3+7x^2+4x+7).$$
But $$x^6+x^5+6x^4+6x^3+7x^2+4x+7=$$
$$=\left(x^3+\frac{1}{2}x^2-1\right)^2+\frac{1}{4}(23x^4+40x^3+36x^2+16x+12)=$$
$$=\left(x^3+\frac{1}{2}x^2-1\right)^2+\frac{1}{4}(23x^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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If $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}$ ; $\frac{1}{a} + \frac{1}{x} + \frac{1}{y} = 0$.
If $a \neq 0$ , $b \neq 0$ , $c \neq 0$ and if :- $\frac{1}{a} + \frac{1}{b} + \frac{1}{a+x} = 0$ ; $\frac{1}{a} + \frac{1}{c} + \frac{1}{a+y}=0$ ; $\frac{1}{a} + \frac{1}{x... | For those who love tedious algebra, from
$$\begin{cases}
\frac1a+\frac1b+\frac1{a+x}=0\\
\frac1a+\frac1c+\frac1{a+y}=0\\
\frac1a+\frac1x+\frac1y=0
\end{cases}$$
We have:
$$a = \frac {-1}{\frac1x+\frac1y} = -\frac{xy}{x+y}$$
$$b = \frac {-1}{\frac 1a+ \frac 1{a+x}}=-\frac {a(a+x)}{2a+x}$$
$$c = \frac {-1}{\frac 1a+ \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831155",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Another upper bound for the Stirling numbers of the first kind It is shown in this question that
$${n \brack n-k}\leq\frac{n^{2k}}{2^kk!}.$$
But a sharper bound seems to be $${n \brack n-k}\leq\frac{n^{k}}{2^k}{n-1 \choose k}.$$
I don't see how to derive this inequality. Any idea?
Hereafter is some numerical evidence:... | I think this proof by induction works:
\begin{align}
{n \brack n-k}
&= (n-1){n-1 \brack n-k}+{n-1\brack n-k-1}
\\&\le (n-1)\frac{(n-1)^{k-1}}{2^{k-1}}\binom{n-2}{k-1}+\frac{(n-1)^k}{2^k}\binom{n-2}{k}
\\&=\frac{(n-1)^k}{2^k}\binom{n-1}{k}\left[\frac{2k}{n-1}+\frac{n-k+1}{n-1}\right]
\\&=\frac{n^k}{2^k}\binom{n-1}{k}\cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Where did I go wrong in my proof that for all $n \in \mathbb{Z}^+$, $\sqrt{2} < a_n$ with $(a_n)$ being a particular recursive sequence? I am in Introduction to Abstract Math, and I have taken Calculus 1, Linear Algebra, and Discrete Math. I got stuck with an apparent contradiction in my proof and wanted to know where ... | As another approach, let $f(x)=\frac x2+\frac 1x$ for $x>0$. It is easy to confirm that $$f(x)=\sqrt 2\iff x=\sqrt 2$$
As $f(x)$ is clearly continuous for all $x>0$ we see that $x>\sqrt 2$ must then imply that $f(x)>\sqrt 2$ and we are done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3831743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate $\lim_{(x,y)\to (0,0)}\frac{\arctan(x^2+y^4) }{ \sqrt{x^2+y^2+1} - 1}$ How can I calculate the limit of the following?
$$ \lim_{(x,y)\to (0,0)}\frac{\arctan(x^2+y^4) }{ \sqrt{x^2+y^2+1} - 1} $$
I've tried using the sandwich rule, but after more than 2 hours, with no success.
Would my appreciate help.
Thank yo... | We have that
$$\frac{\arctan(x^2+y^4) }{ \sqrt{x^2+y^2+1} - 1}=\frac{\arctan(x^2+y^4) }{ x^2+y^4}\cdot\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}$$
with
$$\frac{\arctan(x^2+y^4) }{ x^2+y^4}\to 1$$
and using polar coordinates we obtain
$$\frac{x^2+y^4 }{ \sqrt{x^2+y^2+1} - 1}=r^2\frac{\cos^2\theta+r^2\sin^4\theta }{ \sqrt{r^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find value of $\sin x-\frac{1}{\cot x}$ If $\sin x+\frac{1}{\cot x}=3$, calculate the value of $\sin x-\frac{1}{\cot x}$
Please kindly help me
Let $\sin x -\frac{1}{\cot x}=t$
Then, $$\sin x= \frac{3+t}{2}, \cot x= \frac{2}{3-t}$$
By using $$1+\cot ^2x= \frac{1}{\sin^2 x}$$
Then, the equation $$t^4-18t^2+48t+81=0$$
| Let $\tan{x}=y$.
Thus, $$\sin{x}+y=3,$$ which gives
$$\sin^2x=(3-y)^2$$ or
$$\frac{y^2}{1+y^2}=(3-y)^2$$ or
$$y^4-6y^3+9y^2-6y+9=0$$ or for any real $k$
$$(y^2-3y+k)^2-(2ky^2-6(k-1)y+k^2-9)=0.$$
Now, we'll choose a value of $k$, for which $k>0$ and $$2ky^2-6(k-1)y+k^2-9=(ay+b)^2,$$
for which we need $$9(k-1)^2-2k(k^2-9... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Showing the pdfs of two normal distributions multiplied together is still the pdf of a normal distribution Given $p_1(x) = \cfrac{1}{\sqrt{2\pi}\sigma_1}e^{-x^2/2\sigma_1^2}$ and $p_2(x) = \cfrac{1}{\sqrt{2\pi}\sigma_2}e^{-x^2/2\sigma_2^2}$
I want to show that $p_1(x)p_2(x)$ is normal.
My thought
$$
\begin{align}
p_1(... | $$
\frac{x^2}{\sigma_1^2} + \frac{x^2}{\sigma_2^2} = \frac{x^2}{\left( \dfrac{\sigma_1^2\sigma_2^2}{\sigma_1^2+\sigma_2^2} \right)} = \frac{x^2}{\text{a positive number}}
$$
It is not generally true that
$$ \require{cancel}
\xcancel{\sigma_1 \sigma_2 = \frac{\sigma_1\sigma_2}{\sqrt{\sigma_2^2 + \sigma_2^2}},}
$$
as may... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3836236",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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How to solve this absolute value inequality? I am new to absolute value inequalities.I was looking trough a book and I found this inequality,
I know a little bit about absolute value inequalities.
The inequality is given below:
$$
\left| \frac{n+1}{2n+3} - \frac{1}{2} \right| > \frac{1}{12}, \qquad n \in \mathbb{Z}
$$
| As it is an absolute value
Case 1)
$$\frac{n+1}{2n+3} - \frac{1}{2} > \frac{1}{12} $$
Case 2)
$$\frac{n+1}{2n+3} - \frac{1}{2} < \frac{-1}{12}$$
Now I can solve it a little for you , as in Case 1
$$\frac{n+1}{2n+3} - \frac{1}{2} - \frac{1}{12}>0 $$
$$\frac{n+1}{2n+3} - \frac{7}{12} > 0 $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
How to group people so everyone meets? I have sort of a stupid maths problem which I was thinking about recently, given the UK's covid rules about only being able to meet with 6 people max.
If I have a group of $N$ friends (e.g. 15) and everyone wants to hang out with everyone else, what's the fewest number of meet-ups... | Denote $M+1$ the max number of friends per group, $N+1$ the number of people, and $S_{k}$ the minimum number of meetings required for $k$ people. There are ${N+1\choose 2}$ pairs of people which need to meet. We eliminate at most $M+1 \choose 2$ of these pairs per meeting. Hence $S_{N+1} \geq \lceil {N+1\choose 2}{M+1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Show that the solutions to the equation $ax^2 + 2bx + c =0$ are given by $x = -\frac{b}{a} \pm \sqrt{\frac{b^2-ac}{a^2}}$
Hint: Start by dividing the whole equation by $a$
At first I have tried so... | As JCAA said, you should assume $a\ne 0$, otherwise, it is not a quadratic equation, but a linear equation. Also, division by zero is undefined. We just could divide both sides by $a$ because of that.
$ax^2+2bx+c=0 \implies x^2+2\dfrac{b}{a}x+\dfrac{a}{c}=0 \implies x^2+2\dfrac{b}{a}x=-\dfrac{c}{a}$
Completing the squa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841751",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Find the area between $r=1$ and $r=3\cos\theta$
Find the area between $r=1$ and $r=3\cos\theta$.
I squared both sides to get $r^2 = 1$, then did $r^2(\cos^2 \theta + \sin^2 \theta) = (r \cos \theta)^2 + (r \sin \theta)^2$$ = x^2+y^2 = 1$ to get $x^2+y^2=1$.
For $r = 3 \cos \theta$, I multiplied by $r$ on both sides ... | The area under a polar function $r(\theta)$ is:
$$\frac{1}{2} \int_a^b \big(r(\theta) \big)^2 \ d \theta$$
Here $r(\theta) = 3 \cos \theta - 1$. Now find the limits of integration, and use the identity $\cos^2 \theta = \frac{1}{2} (\cos 2 \theta+1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3841917",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that $ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$ By first expanding $(\cos^2x + \sin^2x)^3$, otherwise, show that
$$ \cos^6x + \sin^6x = 1 - (3/4)\sin^2(2x)$$
Here's what I've done so far (starting from after expansion):
$\cos^6x + (3\cos^4x\sin^2x) + (3\cos^2x... | You have $$(\cos^2(x)+\sin^2(x))^3=\cos^6(x)+\sin^6(x)+3\cos^4(x)\sin^2(x)+3\sin^4(x)\cos^2(x)$$
$$=\cos^6(x)+\sin^6(x)+3\sin^2(x)\cos^2(x)(\cos^2(x)+\sin^2(x))$$
$$=\cos^6(x)+\sin^6(x)+\frac{3}{4}\cdot 4\sin^2(x)\cos^2(x)$$
$$=\cos^6(x)+\sin^6(x)+\frac{3}{4}(2\sin(x)\cos(x))^2$$
$$=\cos^6(x)+\sin^6(x)+\frac{3}{4}\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3842339",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 9,
"answer_id": 6
} |
PRMO level question of functions Consider the functions $f(x)$ and $g(x)$ which are defined as $f(x)=(x+1)(x^2+1)(x^4+1)\ldots\left({x^2}^{2007}+1\right)$ and $g(x)\left({x^2}^{2008}-1\right)= f(x)-1$.
Find $g(2)$
This is a PRMO level question of functions and I tried it with substituting values also but to no avail an... | We can write
$$
g\left( x \right) = \frac{{\left( {x + 1} \right)\left( {x^2 + 1} \right) \ldots \left( {x^{2^{2007} } + 1} \right) - 1}}
{{\left( {x - 1} \right)\left( {x + 1} \right)\left( {x^2 + 1} \right) \ldots \left( {x^{2^{2007} } + 1} \right)}}$$
and then get
$$
g\left( x \right) = \frac{1}
{{x - 1}} - \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3844660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
find the largest integer $m$ such that $2^m$ divides $3^{2n+2}-8n-9$
find the largest integer $m$ such that $2^m$ divides $\space 3^{2n+2}-8n-9$ when $n$ is a natural number.
If the answer was known it will be easy induction.
I started out like this :
$\space 3^{2n+2}-8n-9=9(3^{2n}-1)-8n=9\underbrace{(3^n-1)(3^n+1)}-... | Hint: For $n=1$, it is clear which is the largest power of $2$. Now consider $(8+1)^{n+1} - [(n+1)\cdot8+1]$ and use binomial expansion to conclude it works for all larger $n$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849006",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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$\frac{x-c}{x-y} |\frac{f(x) - f(c)}{x-c} - f'(c)| + \frac{c-y}{x-y}|\frac{f(y) - f(c)}{y-c} - f'(c)| < (\frac{x-c}{x-y} + \frac{c-y}{x-y}) \epsilon$? After substituting the green expression, how do you deduce the last two inequalities below? I'm guessing Triangle Inequality? But after you substitute the green expressi... | Supposing $x<c<y$, we justify it by
$$1=\frac{x-y}{x-y}=\frac{x-c}{x-y}+\frac{c-y}{x-y}$$
and with the calculations (using the triangle inequality in the "$\leq$" step)
\begin{align*}
\left|\frac{f(x)-f(y)}{x-y}-f'(c)\right| &=\left|\frac{f(x)-f(c)}{x-y}+\frac{f(c)-f(y)}{x-y}-\left(\frac{x-c}{x-y}+\frac{c-y}{x-y}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find x so that $\left(\frac{3}{2}+ \sum_{i=1}^{x} 3^{i}\right)^2$ Find $x$ so that $$\left(\frac{3}{2}+ \sum_{i=1}^{x} 3^{i}\right)^2 = \frac{3^{22}}{4}$$
I've tried with simpler values for $x$ such as $0, 1$ and $2$. But I can't seem to find any pattern I can take advantage of. How do I solve it? Where would I learn ... | $\dfrac{3^{22}}{4} = \left(\dfrac{3^{11}}{2}\right)^2$
So we have,
$$\frac{3}{2} + \sum_{i=1}^x 3^{i} = \frac{3^{11}}{2} \Rightarrow \sum_{i=1}^x 3^{i} = \frac{3^{11}-3}{2} \Rightarrow 3\left(\frac{3^\color{red}x-1}{2}\right) = 3\left(\frac{3^\color{red}{10}-1}{2}\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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prove that $\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$
prove that $$\sum_{cyc}\frac{a}{b^2+c^2}\ge \frac{4}{5}\sum_{cyc}\frac{1}{b+c}$$ for positives $a,b,c$
Attempt: By C-S; $$\left(\sum_{cyc}\frac{a}{b^2+c^2} \right) \left(\sum_{cyc} a(b^2+c^2) \right)\ge {(a+b+c)}^2$$ .
or as inequality is ... | Let $p=a+b+c=1, \; q=ab+bc+ca, \; r=abc.$ We need to prove
$$5(ab+bc+ca-abc) \geqslant 4(1+ab+bc+ca)(ab+bc+ca-3abc),$$
equivalent to
$$5(q-r) \geqslant 4(1+q)(q-3r),$$
or
$$(12q+7)r \geqslant q(4q-1).$$
If $4q-1 < 0,$ then
$$(12q+7)r > 0 > q(4q-1).$$
If $4q-1 \geqslant 0,$ from Schur inequality
$$(a+b+c)^3+9abc \geqsla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3854883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Finding the smooth inverse of a function Problem: The mapping $\phi: S^2 \longrightarrow S^2 $ by $$\phi(x,y,z)=(x\cos z+y\sin z,x\sin z-y \cos z,z)$$ is a diffeomorphism. Where $S^2$ is a unit sphere in $\mathbb{R}^3$.
I've already shown that $\phi$ is smooth and bijective. The only thing I can't find is $\phi^{-1}$... | To find $\phi^{-1}$ explicitly, let's define
$$ \begin{align*} a &= x \cos z + y \sin z \\
b &= x \sin z - y \cos z \\
c &= z \end{align*} $$
so that $\phi(x,y,z) = (a,b,c)$ and $\phi^{-1}(a,b,c) = (x,y,z)$. It just remains to write $x,y,z$ in terms of $a,b,c$.
Getting rid of $z$ is obvious:
$$ \begin{align*} a &= x \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3855974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Understand how to evaluate $\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$ We are given this limit to evaluate:
$$\lim _{x\to 2}\frac{\sqrt{6-x}-2}{\sqrt{3-x}-1}$$
In this example, if we try substitution it will lead to an indeterminate form $\frac{0}{0}$. So, in order to evaluate this limit, we can multiply this exp... | Option:
$z:=\sqrt{3-x}$; and consider $\lim z \rightarrow 1;$
$\dfrac{\sqrt{3+z^2}-2}{z-1}=$
$\dfrac{z^2-1}{(z-1)(\sqrt{3+z^2}+2)}=$
$\dfrac{z+1}{\sqrt{3+z^2}+2}.$
Take the limit.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858398",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 7,
"answer_id": 6
} |
Let $a_1,a_2,...,a_n>0$ be such that $\sum_{k=1}^n a_k=\frac{1}{2}n(n+1)$
Let $a_1,a_2,...,a_n>0$ be such that $\sum_{k=1}^n a_k=\frac{1}{2}n(n+1)$, then least value of
$$\sum_{k=1}^n\frac{(k^2-1)a_k+k^2+2k}{a_k^2+a_k+1}\text{ is,}$$
What I tried:
$$\sum_{k=1}^{n}k=\frac{1}{2}n(n+1)\implies k=a_k\tag{$\because\sum_{k... | You can use the fact that
$$ a_n=\sum_{k=1}^n a_k-\sum_{k=1}^{n-1} a_k=\frac{1}{2}n(n+1)-\frac{1}{2}(n-1)n=n. $$
The rest is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3858721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Do the series converge? Let $g \colon \mathbb{N} \rightarrow \{0,1\}$ such that $g(n)=0$, if $n \equiv 0$ or $n \equiv 1 ~(\bmod~4)$; or $g(n)=1$, if $n \equiv 2$ ou $n \equiv 3 ~(\bmod~4)$.
Do $\sum_{n=1}^\infty \frac{(-1)^{g(n)}}{\sqrt{n}}$ converge?
The series is something like this
*
*If $n=1$, then $\frac{(-1)^... | We have
$$\sum_{n=1}^\infty \frac{(-1)^{g(n)}}{\sqrt{n}}=\frac{1}{1}-\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+\cdots$$
Now, note that for for $n\equiv 2\ (\text{mod}\ 4)$ we have the terms
$$\cdots-\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}+\frac{1}{\sqrt{n+2}}+\frac{1}{\sqrt{n+3}}-\cdots$$
$$\cdots-\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3864364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Polynomial with root $α = \sqrt{2}+\sqrt{5}$ and using it to simplify $α^6$ Find a polynomial $\space P(X) \in \mathbb{Q}[X]\space$ of degree 4 such that
$$\alpha = \sqrt{2} + \sqrt{5}$$
Is a root of $P$.
Using this polynomial, find numbers $\space a, b, c, d \space$ such that
$$\alpha^{6} = a + b\alpha + c\alpha^{2} ... | We can use the following identity.
$$(a+b+c)\prod_{cyc}(a+b-c)=\sum_{cyc}(2a^2b^2-a^4).$$
We obtain:
$$2\alpha^2\cdot2+2\alpha^2\cdot5+2\cdot2\cdot5-\alpha^4-2^2-5^2=0$$ or
$$\alpha^4-14\alpha^2+9=0.$$
From here $$\alpha^6=14\alpha^4-9\alpha^2=14(14\alpha^2-9)-9\alpha^2=187\alpha^2-126,$$ which gives $$(a,b,c,d)=(-126,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3867201",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
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Solution of first order inexact differential equation I have following differential equation before me:
$(2y sin(x)+3y^4 sin(x)cos(x))dx-(4y^3 cos^2(x)+cos(x))dy=0$
It is an inexact differential equation. Its Integrating factor comes out to be $cos(x)$.
New $M= 2ysin(x)cos(x)+3y^4sin(x)cos^2(x)$
New $N= -(4y^3cos^3(x)+... | $$-\dfrac {\cos (2x) }2=\dfrac 12 (2\cos^2 x-1)=\dfrac 12(1-2\sin^2 x)$$
All the constants are absorbed by the constant on the right side of the solution:
$$F(x,y)=C$$
$$(2y \sin(x)+3y^4 \sin(x)\cos(x))dx-(4y^3 \cos^2(x)+\cos(x))dy=0$$
$$y d(\cos^2x)+y^4 d(\cos^3(x))+ \cos^3(x)dy^4+cos^2(x)dy=0$$
$$ d(y\cos^2x)+d(y^4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3867909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Prove Segner's Recurrence Relation $C_{n+1} = \sum\limits_{i=0}^n C_i C_{n-i}$ on Catalan Numbers $C_n = \frac{1}{n+1} \binom{2n}{n}$
Plugging in the Catalan Equation, we want to prove... | I think the best way to solve Segner's recurrence relation is to use generator functions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871430",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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System of equations (Problem $50$ from $101$ algebra by Titu)
Let $a$ and $b$ be given real numbers.
Solve the system of equations
$$\begin{aligned} \frac{x-y \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=a \\ \frac{y-x \sqrt{x^{2}-y^{2}}}{\sqrt{1-x^{2}+y^{2}}} &=b \end{aligned}$$
for real $x$ and $y$.
Solution -
Let $... | $$u-u\sqrt{uv} = (a+b)\sqrt{1-uv}$$
$$u = (a+b)\dfrac{\sqrt{1-uv}}{1-\sqrt{uv}}$$
Now substitute $uv=a^2-b^2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Probability when choosing beads with repetitions There are 20 different beads, 4 of which are yellow. Choosing 30 beads with repetitions, I need to calculate:
*
*The probability that exactly 5 of the chosen beads are yellow
*The probability that the first and last beads are yellow and exactly 5 yellow beads were cho... | Because we are asked about probability, and not about the number of combinations, and because we are choosing with repetitions - we can ignore the identity of the individual beads and instead treat each choice as a $\frac{4}{20} = 20\%$ probability to choose yellow and $80\%$ probability to choose non-yellow.
For any g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3872614",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
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Construct an isomorphism between $V$ and $\Bbb F^2$ and justify Consider the subspace of $M_2(\Bbb F):V = \biggl\{\begin{pmatrix}a+b&a\\0&b\end{pmatrix}\Biggm\vert a,b\in \Bbb F\biggr\}$
Construct an isomorphism between $V$ and $\Bbb F^2$ and justify
| Consider $\phi:V\to F^2$ defined as $\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}=(a,b)$.
Then
$\phi\Big(\begin{pmatrix}a+b&a\\0&b\end{pmatrix}+\begin{pmatrix}c+d&c\\0&d\end{pmatrix}\Big)=\phi\begin{pmatrix}a+b+c+d&a+c\\0&b+d\end{pmatrix}=(a+c,b+d)=(a,b)+(c,d)=\phi\begin{pmatrix}a+b&a\\0&b\end{pmatrix}+\phi\begin{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877213",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
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cyclic rational inequalities $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ when $a+b+c=1$ I've been practicing for high school olympiads and I see a lot of problems set up like this:
let $a,b,c>0$ and $a+b+c=1$. Show that
$$\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$$
Any prob... | We need to prove $\frac{1}{a^2+3}+\frac{1}{b^2+3}+\frac{1}{c^2+3}\leq\frac{27}{28}$ for $a, b, c \gt 0, a + b + c = 1$
Using tangent line method,
We consider the equation of the tangent line to $f(x) = \frac{1}{3+x^2}$ at $x = \frac{1}{3}$. Point is $(\frac{1}{3}, \frac{9}{28})$
$f'(x) = -\frac{2x}{(3+x^2)^2} = -\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
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Find $\frac{\mathrm{d} }{\mathrm{d} x}x\sin \left ( \sqrt{3x^{2}+5} \right )$ without using the chain rule. $$\frac{\mathrm{d} }{\mathrm{d} x}x\sin \left ( \sqrt{3x^{2}+5} \right )$$
I can't for the life of me differentiate this function while only using Trig Identities, Basic differentiation rules, and Limits (no L'Ho... | $$f'=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}$$ separate it into two limitation by $\pm(x)\sin\sqrt{3(x+t)^2+5}$
this means
$$f'=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}-(x)\sin\sqrt{3(x)^2+5}}{t}=\\
=\lim_{t \to 0}\frac{(x+t)\sin\sqrt{3(x+t)^2+5}\pm(x)\sin\sqrt{3(x+t)^2+5}-(x)\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3880196",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
Given that $G$ is the centroid of $\Delta ABC$, $GA = 2\sqrt{3}$ , $GB = 2\sqrt{2}$, $GC = 2$ . Find $[\Delta ABC]$.
What I Tried: Here is a picture :-
I know the centroid divides each of the medians ... | (This is not likely to be what you're looking for.)
I think in this problem you can use a simpler solution.
Construct point $H$ outside $\overline{AC}$ such that $AGCH$ forms a Parallelogram. We have
*
*$\overline{AH}=\overline{GC}=2$
*$\overline{AG}=2\sqrt3$
*$\overline{GE}=\overline{EH}\Longrightarrow \overline{G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
Assume that $ab \mid (a+b)^2.$ Show that $ab \mid (a-b)^2$.
If $ab \mid (a+b)^²$, then $ab\mid a^2+2ab+b^2 \Longrightarrow ab\mid a^2, ab\mid 2ab$ and $ab\mid b^2$ right?
So since $(a-b)^2 = a^2-2ab+b^2$ from the assumption we have that $ab \mid a^2$ and $ab... | As others have said $x \mid (y+z) \nRightarrow (x \mid y) \land (x \mid z)$
What you can say is $ab \mid (a+b)^2 \iff (a+b)^2 = kab, k \in \mathbb{Z}$
This implies $a^2 + b^2 +2ab = kab \implies a^2 + b^2 = (k-2)ab \implies a^2 + b^2 - 2ab = (k-4)ab \implies (a-b)^2 = (k-4)ab \iff ab \mid (a-b)^2$, as required.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3883065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90 ^\circ$. If the Symmedian/Lemoine point of triangle $ABC$ lies on the Altitude from vertex $C$, show that either $BC = AC$ or angle $C = 90^\circ$.
I have proved it going the other way. Whe... | The coordinates of the symmedian point $X_6$ and
the foot of the altitude $H_c$
in terms of vertices $A(A_x,A_y),\ B(B_x,B_y),\ C(C_x,C_y)$ and side lengths $a,b,c$
can be found as
\begin{align}
X_6&= \frac{a^2\,A+b^2\,B+c^2\,C}{a^2+b^2+c^2}
\tag{1}\label{1}
,\\
H_c&=
\tfrac12\,(A+B)+\frac{a^2-b^2}{2c^2}\,(A-B)
\tag{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to write a polynomial function that has the roots $-2$ and $\sqrt7$? I need to write a polynomial function with integer coefficients that has the roots $-2$ and $\sqrt7$. I'm able to do this correctly when I'm given roots like $-3+i$ & $-3-i$, in which I set the roots equal to zero and then multiply them by one ano... | $x = \sqrt 7 \implies$
$x^2 = 7\implies$
$x^2 - 7 =0$. So $x^2 -7$ has $\sqrt 7$ as a root.
And $x=-20\implies x+2 = 0$ so $x+2$ has $-2$ as a root.
If $x^2-7=0$ and $x+2=0$ then $(x^2-7)(x+2)=0$ and $(x^2-7)(x+2)=x^3+2x^2-7x-14$ has $\sqrt 7$ and $-2$ as roots.
(It also has $-\sqrt 7$ as a root.)
=====
By the way. T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Integrals identity By numerical integration I found the identity
$$ \int_{-\infty}^{0}\frac{db}{\sqrt{R_4(b,W)}} = 2\int_{(W+1)^2}^{\infty}\frac{db}{\sqrt{R_4(b,W)}} $$
where $R_4(b,W)= b \ (b-4) \ (b-(W-1)^2) \ (b-(W+1)^2) $ and $W>3$.
Now I would like to prove that formally but I can't find the right way. I started b... | Let the roots of $R_4$ be $a=(W+1)^2,b=(W-1)^2,c=4,d=0$. Byrd and Friedman 251.00 and 258.00 resolves both sides in terms of elliptic integrals as follows:
$$gF(\varphi,m)=2gF(\psi,m)$$
where $g=\frac2{\sqrt{(a-c)(b-d)}}$, $m=k^2=\frac{(b-c)(a-d)}{(a-c)(b-d)}=\frac{(W-3)(W+1)^3}{(W+3)(W-1)^3}$, $\sin^2\varphi=\frac{a-c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3887493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that three numbers form an arithmetic progression
The numbers $a,b$ and $c$ form an arithmetic progression. Show that the numbers $a^2+ab+b^2,a^2+ac+c^2,b^2+bc+c^2$ also form an arithmetic progression.
We have that $2b=a+c$ (we know that a sequence is an arithmetic progression iff $a_n=\dfrac{a_{n-1}+a_{n+1}}{2... | Since $2b=a+c$
$$a^2+ab+b^2+b^2+bc+c^2=a^2+b(a+c)+2b^2+c^2=a^2+\frac{(a+c)^2}{2}+2\times \frac{(a+c)^2}{4}+c^2=a^2+(a+c)^2+c^2=2(a^2+ac+c^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3892856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solve the equation $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$ $$\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$$
$$x=?$$
I solved this but the equation $ (2x + 1) (3x ^ 4-x ^ 3 + 2x ^ 2-2x + 1) = 0 $ is formed I answer $ x =- \frac {1} {2} $ I know there is, but I couldn't do the next expression. I need help with that,... | Just to give a different approach, note that any solution to $\frac{x^3+2x}{x^2-1}=\sqrt{x^2-\frac{1}{x}}$ must lie in $(-1,0)\cup(1,\infty)$ because the left hand side is negative on $(-\infty,-1)\cup(0,1)$ whereas the right hand side, being a square root, is always non-negative, and the expression is undefined at $x=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3895007",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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How to find a continuous path between matrices?
I have matrix $A=\begin{pmatrix} 0&-1&0\\1&0&0\\0&0&1\\\end{pmatrix}$ and matrix $B=\begin{pmatrix} 1&0&0\\0&\frac{1}{2}&-\frac{\sqrt{3}}{2}\\0&\frac{\sqrt{3}}{2}&\frac{1}{2}\\\end{pmatrix}$ and I want to find a continuous path between them. How do I do this?
I know tha... | You can check that any two dimensional rotation matrix can be written as $$R_\theta = \begin{bmatrix}\cos\theta & -\sin\theta \\ \sin\theta & \cos\theta\end{bmatrix} = e^{\theta J} \quad \text{where} \quad J = \begin{bmatrix}0 & -1 \\ 1 & 0\end{bmatrix}.$$
Using this fact, we have $$R_{\pi/2} = \begin{bmatrix}0 & -1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3895837",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrating partial Weierstrass products of $\sin(x)$ divided by $\sin(x)$ Recall the Weierstrass factorization of $\sin(z)$, valid for any complex $z$:
$$
\sin(z) = z \prod_{k=1}^{\infty}\left(1-\frac{z^2}{k^2\pi^2}\right)
$$Apropos of nothing, I thought, 'what if I integrated these partial products divided by sine ov... | Use integration by parts and induction in $k$ to find $$\int_{-\pi}^\pi x^{k}e^{-inx}dx$$
then
$$\int_{-\pi}^{\pi} \frac{x^{k+1}(1-x^2/\pi^2)}{\sin(x)}d=\lim_{r\to 1^-} \int_{-\pi}^\pi \frac{x^{k+1}(1-x^2/\pi^2) 2i e^{-ix}}{1-re^{-2ix}}dx$$ $$=\lim_{r\to 1^-} \int_{-\pi}^\pi \sum_{n=0}^\infty 2i x^{k+1}(1-x^2/\pi^2) r^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3896662",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Factorial simplification help; how did wolfram get this? in this problem n is assumed to be odd:
Here's what I am trying to simplify:
$$ \frac {(n-2)!( \frac {n-1}{2})!} { (n-1)! (\frac {n-3} {2}) !}
$$
Wolfram is telling me that this should simplify to $$\frac {1}{2} $$ but I am very confused about how it gets there. ... | First, note that
$$\frac{(n-2)!}{(n-1)!} = \frac{(n-2)(n-3)(n-4) \cdots (2)(1)}{(n-1)(n-2)(n-3)(n-4) \cdots (2)(1)} = \frac{1}{n-1}$$
Similarly, notice that
$$\dfrac{ \left( \dfrac{n-1}{2} \right)! }{\left( \dfrac{n-3}{2} \right)!} = \dfrac{ \left( \dfrac{n-1}{2} \right)\left( \dfrac{n-3}{2} \right)\left( \dfrac{n-5}{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How I can solve this limit only with algebra I tried to resolve this using properties of limits, properties of logarithms and some substitutions, but i can't figure whats is the right procedure for this.
$$\lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}}$$
first all I use the logarithms properties and rewrite l... | \begin{align*}
L &= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 + x}{1 - x}} \\
&= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 +(-x+x)+ x}{1 - x}} \\
&= \lim_{x\to 0} \frac{1}{x} \log{\sqrt\frac{1 -x +2x}{1 - x}} \\
&= \lim_{x\to 0} \frac{1}{x} \log{\sqrt{1 + \frac{2x}{1 - x}}} \\
&= \lim_{x\to 0} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3898556",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is the gcd of $61+35\sqrt{3} $ and $170+32\sqrt{3}$ is $19 + 11\sqrt{3}$? As a completion or correction to the solution of my question here:
Find gcd of $a = 170 + 32\sqrt{3}$ and $b = 61 + 35\sqrt{3}.$ Then find $f,g \in \mathbb{Z}[\sqrt{3}]$ such that $af + bg = d$ using norm function.
I was told by my professor ... | Division with remainder for number of these forms should be done as follows:
$$\frac{170 + 32 \sqrt{3}}{61 + 35 \sqrt{3}} = \frac{3505}{23} - \frac{1999 \sqrt{3}}{23}$$
This would be the fraction. Now to get the integral quotient, take the closest integer for the coefficients of the rational and irrational par
$$\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3903947",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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find all ordered pairs of natural numbers (x,y,z) that satisfies $x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$ find all ordered pairs of natural numbers (x,y,z) that satisfies $x^{x^y} \cdot y^{y^z} \cdot z^{z^x} = 1990^{1990} \cdot xyz (x\lt y \lt z)$
My approach was
$$x^{3(x^x-1)} \lt x... | First note the prime factorization of $1990 = 2 \times 5 \times 199$.
Hence at least one of $x,y,z$ must be divisible by $199$, from the equation $x^{x^y-1} \cdot y^{y^z-1} \cdot z^{z^x-1} = 1990^{1990}$.
Ignore the order for now and suppose $199 \mid x$. Then $x^{x^y-1} \ge 199^{199^y-1}$ and $z^{z^x-1} \ge z^{z^{199}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3907064",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.