Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Proving $\int_{0}^{\pi/2} \ln \sin x ~ \ln \cos x~ dx=-\frac{\pi^3}{48}+\frac{\pi}{2}\ln^22$ Interestingly the integral $$I=\int_{0}^{\pi/2} \ln \sin x ~ \ln \cos x~ dx~~~~~~~~~(1)$$
is doable by hand by using Fourier series: $$ \ln \sin x=-\sum_{j=1}^{\infty} \frac{\cos 2j x}{j}-\ln 2,\quad \ln \cos x=\sum_{k=1}^{\inf... | Solution without using beta function:
Let $a=\ln(\sin x)$ and $b=\ln(\cos x)$ in the algebraic identity
$$ab=\frac12a^2+\frac12b^2-\frac12(a-b)^2,$$
we have
$$\ln(\sin x)\ln(\cos x)=\frac12\ln^2(\sin x)+\frac12\ln^2(\cos x)-\frac12\ln^2(\tan x).$$
Integrate both sides from $x=0$ to $\pi/2$,
\begin{gather*}
\int_0^{\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4184823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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Alternative approaches to maximize $y=x\sqrt{100-x^2}$ I could find three good approaches to find maximum of the function $y=x\sqrt{100-x^2}$. I will explain them briefly :
First: Finding $x$ satisfies $y'=0$ then plugging it in the function.
Second: Using the substitution $x=10\sin\theta$ (or $x=10\cos\theta)$ for
$\t... | Square both sides to get $y^2 = x^2 (100-x^2) \implies -x^4+100x^2-y^2=0$. This is a quadratic in $x^2$: when $\Delta = 0$, $100^2 - 4(-1)(-y^2) = 0 \implies y = ±50$.
The maximum value is $y = 50$ as considering the negative branch, $f(-x) = -f(x)$, hence the maximum of the positive branch is the same as the negative ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4185702",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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What insight is there on negative number bases? I don't really have much to say, other than that I was curious about how negative number bases work, and I was wondering if I could get some insight on how to understand them.
I know how to convert negative base numbers to base 10, but have no clue as to how to write a ba... |
I know how to convert negative base numbers to base 10, but have no clue as to how to write a base 10 number to an arbitrary negative base, at least not easily.
Write it to the positive absolute value base first. subtract the odd terms from $|b|$ and carry to even terms.
Example: To write $1234$ in base $-6$.
$1234... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4188029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving $(x^3-4)^3=(\sqrt[3]{(x^2+4)^2}+4)^2$ I'm trying to solve the equation below for $x$: $$(x^3-4)^3=(\sqrt[3]{(x^2+4)^2}+4)^2$$
I tried to solve it by doing some algebraic operations such as expanding both sides and then try to remove the third roots by rising both sides of the equation to power of 3 but it gets ... | As mentioned by @Player0, the equation can be solved using monotonicity noticing that $x=2$ is a solution. But in this answer, I will be solving the equation algebraically (and for real numbers).
We first remove the bizarre cube roots from our equation. To do that, we consider $y=\sqrt{x^3-4}$ and $z=\sqrt[3]{x^2+4}$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4191706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Does $ a_{\lfloor n + \sqrt{n/2}\rfloor} \le a_n\left(1 + \frac{b}{n^{\frac{1}{2}+\varepsilon}}\right) $ imply that $a_n$ is bounded? Let $(a_n)_{n\ge 1}$ be an increasing sequence of positive real numbers and suppose there is a constant $b > 0$ and $\varepsilon > 0$ such that
$$
a_{\lfloor n + \sqrt{n/2}\rfloor} \le a... | As the sequence $(a_n)_n$ increases, it tends either to a finite limit (the sequence converges), either to $+\infty$ (the sequence diverges).
We will prove that the second case can't happen. Suppose by contradiction that the sequence tends to $+\infty$, then, there exists $n_0$ such that $a_{n_0} >2$.
Let us define the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Using generating functions to count number of ways to plan a semester This is the question:
The semester of a college consists of n days. In how many ways can we separate the semester into sessions if each session has to consist of at least five days?
My work:
If $A(x)$ is the generating function for making subinterval... | You want to count ordered partitions of $n$ into parts of size at least $5$. You have $$A(x)=x^5+x^6+\dots= \frac{x^5}{1-x}$$
and
\begin{align}
B(x) &= \frac{1}{1-A(x)}
= \frac{1}{1-\frac{x^5}{1-x}}
= \frac{1-x}{1-x-x^5} \\
&= 1x^0 + 1x^5 + 1x^6 + 1x^7 + 1x^8 + 1x^9 + 2 x^{10} + 3 x^{11} + 4 x^{12} + 5 x^{13} + 6 x^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4196997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Vector quadruple product How did the author arrive at step $(3)$ from step $(1)$ and step $(2)$ in the following definition of vector quadruple product?
Many combinations of vector and scalar products are possible, but we consider only one more, namely the vector quadruple product $(\mathbf{a} \times \mathbf{b}) \time... | We have: $[(a\times b)\cdot d]c - [(a\times b)\cdot c]d = [(c\times d)\cdot a]b - [(c\times d)\cdot b]a$
isolating the intended left side we get $[(a\times b)\cdot c]d = [(c\times d)\cdot b]a - [(c\times d)\cdot a]b + [(a\times b)\cdot d]c$
Recognizing the things in the brackets as the scalar triple product, we can r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4198269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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if $a+b=2$, finding minimum value of $\frac{a^2+b^2}{\sqrt{a^2+1} \sqrt{b^2+4}}$ if $a+b=2$, Finding minimum value of $\frac{a^2+b^2}{\sqrt{a^2+1} \sqrt{b^2+4}}$.
I only thought of the Lagrangian multiplier method, it’s a lot of calculation, I can’t do it.
$\frac{a^2+b^2}{\sqrt{a^2+1} \sqrt{b^2+4}}+\lambda (a+b-2)$
(1... | $$ b=2-a$$
$$ f(a) = \frac{a^2+(2-a)^2}{\sqrt{a^2+1} \sqrt{(2-a)^2+4}} $$
$$ f'(a) = \dfrac{2(a^3+6a^2+2a-12)}{\left(\left(2-a\right)^2+4\right)^\frac{3}{2}\left(a^2+1\right)^\frac{3}{2}}$$
The denominator of the derivative is always positive. It is sufficient to analyse the numerator. Observe that $a=-2$ is a root of ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4198913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Min and max of $f(x,y)=e^{-xy}$ where $x^2+4y^2 \leq 5$ I am trying to use Lagrange multipliers to find the maximum and minimum values of the function $$f(x,y)=e^{-xy}$$
constrained as $$x^2+4y^2=5$$
I began this problem by setting up the Lagrangian:
$$f(x,y) = e^{-xy}$$
$$g(x,y) = x^2+4y^2-5$$
$$L(x,y) = f(x,y) - \lam... | $\min f$ is at $\max (xy)$
If $g = x y$ is defined on $x^2 + 4 y^2 <= 5$
Then maximum of $g$ occurs at the boundary of the elliptical region,
namely the ellipse $x^2 + 4 y^2 = 5$
The semi-major and semi-minor axes are:
$a = \sqrt{5}$
$b = \sqrt{5}/2$
The parameteric equation of the boundary is
$p(t) = ( \sqrt{5} \cos t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200342",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the smallest odd value of $k$ such that $\displaystyle\int_{10}^{19}{\sin x\over1+x^k}dx<\frac 19$
Find the smallest odd value of $k$ such that $\displaystyle\int_{10}^{19}{\sin x\over1+x^k}dx<\frac 19$
$$\displaystyle\begin{align*}\int_{10}^{19}{\sin x\over1+x^k}dx&<\int_{10}^{19}{\mid\sin x\mid\over1+x^k}dx\\&... | You want to show that
$$\int_{10}^{19}\frac{\sin x dx}{1+x}\leq\frac19.$$
If this is true, it's because the "oscillations" of $\sin x$ cancel out, since the integral of the absolute value is a good bit larger than $1/9$. Since $1/(1+x)$ doesn't vary much in the interval $[10,19]$, you can subtract something close to it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4200936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Find the minimum of $P(X=0)$ when $E[X]=1,E[X^2]=2,E[X^3]=5$ using the probability generating function I was given the following exercise.
Let $X$ be a random variable that takes non-negative natural number
values such that $E[X]=1,E[X^2]=2,E[X^3]=5$. Find the minimum value of
$P(X=0)$ using the taylor expansion of th... | We have absolute convergence on the (complex) unit ball around $0$.
Using the Taylor expansion of order three with an explicit form of the rest
we have for some $c=c(z)\in(z,1)$:
$$
\begin{aligned}
G(z) &=
\frac 1{0!}G(1)
+ \frac 1{1!}G'(1)(z-1)
\\
&\qquad\qquad
+ \frac 1{2!}G''(1)(z-1)^2
+ \frac 1{3!}G'''(1)(z-1)^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
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Solving a first order homogenous differential equation Problem:
Solve the following differential equation.
$$ x^2 \, dy + (y^2 -xy) \, dx = 0 $$
Answer:
$$ \left( \dfrac{x^2}{y^2} \right) \, dy
+ \left( \dfrac{y^2}{x^2} - \dfrac{y}{x} \right) \, dx = 0 $$
Hence we have a homogeneous differential equation. Let $y = vx$... | Let $y = vx$
$\cfrac{dy}{dx} = v + x\dfrac{dv}{dx}$
$x^2 \, dy + (y^2 -xy) \, dx = 0$
$ \cfrac{dy}{dx} + \left(\cfrac{y^2}{x^2} - \cfrac{y}{x}\right) = 0 \ $ (this is where you have a mistake)
$v + x\dfrac{dv}{dx} = v - v^2$
$\cfrac{dx}{x} = - \cfrac{dv}{v^2}$
$ \ln |x| + C = \cfrac{1}{v} = \cfrac{x}{y}$
$y = \cfrac{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4201875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to derive this series for $\gamma$ that is only involving odd integer values of $\zeta(s)$? With $\gamma$ being the Euler Mascheroni constant, this series is well known:
$$1- \sum_{n=2}^{\infty} \frac{\zeta(n)-1}{n} = \gamma \tag{1}$$
The following series involving $\zeta(2n+1)$ also seems to converge to the same v... | Note that
$$\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}=2\frac{\zeta(2n+1)-1}{2n+1}-\frac{\zeta(2n+1)-1}{n+1}+\frac{2}{2n+1}-\frac{2}{2n+2}.$$
Hence
$$\sum_{n=1}^{\infty}\frac{\zeta(2n+1)}{(n+1)\,(2n+1)}=
2\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{2n+1}-\sum_{n=1}^{\infty}\frac{\zeta(2n+1)-1}{n+1}+2\sum_{n=3}^{\infty}\frac{(-1)^{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4203294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find the number of solutions of the equation $e^z = 2z+1$ in the open unit disc $\{z \in \Bbb C : |z| < 1\}$ Find the number of solutions of the equation $e^z = 2z+1$ in the open unit disc $\{z \in \Bbb C : |z| < 1\}$.
My Attempt:
Let $f(z) = e^z-2z-1$. Let $A \subseteq \{z \in \Bbb C : |z| < 1\}$ be the solution set. ... | You already figured out that $z=0$ is one solution of the equation $e^z=2z+1$. Using the Taylor series of the exponential function and simple estimates one can show that there are no other solutions in the unit disk:
For $0 < |z| < 1$ is
$$
e^z - (1+2z) = -z + \frac{z^2}{2!}++ \frac{z^3}{3!} + \cdots
$$
and therefore
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4204635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Find the value of $\int_0^1{4dx\over 4x^2-8x+3}$
Find the value of $\displaystyle\int_0^1{4dx\over 4x^2-8x+3}$
$$\begin{align*}\int_0^1{4dx\over 4x^2-8x+3}&=\int_0^1{dx\over (x-1)^2-(\frac 12)^2}
\\&=\int_0^1{dx\over (x)^2-(\frac 12)^2}
\\&=\int_{0}^{1/2}{dx\over (x)^2-(\frac 12)^2}+\int_{1/2}^{1}{dx\over (x)^2-(\fra... | Find the value of $\int_{0}^{1} \frac{4}{4x^2-8x+3} dx$
$$\int_{0}^{1} \frac{4}{4x^2-8x+3} dx = 4 \int_{0}^{1} \frac{1}{4x^2-8x+3} dx $$
Splitting the denominator we get,
$$4 \int_{0}^{1} \frac{1}{4x^2-8x+3} dx = 4 \int_{0}^{1} \frac{1}{(2x-3)(2x-1)} dx$$
Now using the method f partial fractions we get,
$$ 4 \int_{0}^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Converting $\intop_{0}^{a}\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}dx$ to elliptic integral Tried using $x=a\sin\left(\theta\right)$ $$\rightarrow \intop_{0}^{\pi/2}\sqrt{\frac{a^{2}-\left(a\sin\left(\theta\right)\right)^{2}}{1-\left(a\sin\left(\theta\right)\right)^{2}}}a\cos{\left(\theta\right)d\theta}$$
$$\iff \intop_{0}^{\... | Approximating the elliptic integral seems to be difficult.
What you could do is to expand the integrand around $a=1$
$$\sqrt{\frac{a^{2}-x^{2}}{1-x^{2}}}=1+\frac{a-1}{1-x^2}-\frac{(a-1)^2 x^2}{2 \left(x^2-1\right)^2}-\frac{(a-1)^3 x^2}{2
\left(x^2-1\right)^3}-\frac{(a-1)^4 \left(x^2 \left(x^2+4\right)\right)}{8
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206635",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Conditional Distribution Brownian process $X(t) = W^{2}(t)$ This is a homework problem. Please tell me if my solution is correct, if I am not correct please point out my error. If there is a better way to solve the problem, please let me know.
Problem
W(t) is a standard Brownian process $X(t) = W^{2}(t)$
a) find $f_{X}... | (a)
$$P(W_t^2\leq x)=P(-\sqrt{x}\leq W_t\leq \sqrt{x})=\int_{[-\sqrt{x},\sqrt{x}]}f_{W_t}(y,t)dy=\Phi\bigg(\sqrt{\frac{x}{t}}\bigg)-\Phi \bigg(-\sqrt{\frac{x}{t}}\bigg) $$
where $\Phi$ is the standard normal cdf. Thus
$$f_{X_t}(x,t)=\frac{1}{2\sqrt{xt}}\phi \bigg(\sqrt{\frac{x}{t}}\bigg)+ \frac{1}{2\sqrt{xt}}\phi \bigg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4206973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving by Lagrange's auxiliary equations $ \frac {dx}{z} =\frac {dy}{z} =\frac {(3z+1)(dz)}{x+y} $ By taking 1st and 2nd equations 1st solution is $$ x=y+c_{0}$$
but i am getting two different solutions slightly differing :
Solution 1 : using lagrange multiplier 1,1,0 we get
$$ \frac{dx+dy}{2z}=\frac{(3z+1)dz}{x+y}$$... | There is no mistake in your calculus.
Both :
$$ (x+y)^{2}= 4z^{3}+2z^{2}+c_{1'} \to(1)$$
$$ 2(x^{2}+y^{2})=4z^{3}+2z^{2}+c_{2''} \to(2)$$
are correct, each one on different characteristic curves (because different constants of integrations).
This is not contradictory (on the characteristic curves, but... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4207285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $a^2 + b^2 + c^2 + ab + bc +ca \ge 6$ given $a+b+c = 3$ for $a,b,c$ non-negative real. I want to solve this problem using only the AM-GM inequality. Can someone give me the softest possible hint? Thanks.
Useless fact: from equality we can conclude $abc \le 1$.
Attempt 1:
Adding $(ab + bc + ca)$ to both sides of i... | By the Cauchy-Schwarz, we have
$$a^2+b^2+c^2+ab+bc+ca=\frac{(a+b)^2+(b+c)^2+(c+a)^2}{2} \geqslant \frac{[(a+b)+(b+c)+(c+a)]^2}{2 \cdot 3} $$
$$=\frac{2}{3}(a+b+c)^2 = 6.$$
Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4208010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 5
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Conditional inequality $2a^3+b^3≥3$
Non-negative $a$ and $b$ such that $a^5+a^5b^5=2$. How then do I prove
the following inequality $2a^3+b^3≥3$?
So, we can try using the Lagrange multiplier method:
Let $f(a, b)=2 a^{3}+b^{3}+\lambda(a^{5}+a^{5} b^{5}-2), \quad a, b \geq 0$
$$\tag1 \frac{\partial f}{\partial a}=6 a^{... | Proof by contradiction. Suppose $2a^3 + b^3 < 3$.
Then $ 0 \leq a \leq \sqrt[3]{3/2}$ and $a^5 + a^5b^5 < a^5 + a^5 (3-2a^3)^{5/3} $.
Let $ f(a) = a^5 ( 1 + (3-2a^3)^{5/3})$.
Verify that $f'(a) = 5a^4 ( 1 + (3-4a^3) (3-2a^3)^{2/3}) $.
Verify that $f'(a)$ is $0$ on $\{ 0, 1 \}$, positive on $(0, 1)$, and negative on $... | {
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"url": "https://math.stackexchange.com/questions/4213233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Prove that $(a+b)^c\cdot(b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot (a+b+c)\right]^{a+b+c}$ where $a,b,c\in \mathbb{Q}^{+}$ unless $a=b=c$.
If $a,b,c$ be positive rational numbers, prove that
$$(a+b)^c\cdot (b+c)^a\cdot (c+a)^b\leq \left[\frac{2}{3}\cdot
(a+b+c)\right]^{a+b+c}$$ unless $a=b=c$
My try :
Consider $... | $\displaystyle\left[\frac{2(ab+bc+ca)}{a+b+c}\right]^{a+b+c}\gt (a+b)^c\cdot (b+c)^a\cdot (c+a)^b-----(1)$
unless $a=b=c$
You can continue from here as follows
Note that $a^2+b^2+c^2> ab+bc+ca$
It follows that $\displaystyle\frac{(a+b+c)^2}{3}>ab+bc+ca$ and equivalently
$\implies\displaystyle\frac{2(a+b+c)^2}{3}>2(ab+b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4214783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Calculate the angle measure $TBA$ for reference:
On the internal bisector of $\measuredangle C$ of an isosceles triangle $ABC (AB = BC)$ if
externally marks point $T$, in such a way that $\measuredangle CTA = \measuredangle TAB = 30^o$.
Calculate $\measuredangle TBA$ (Answer $20^o$)
My progress:
$\triangle ABC(Isóscel... | Longer method with the use of calculator!
Drop altitude $TD$ to the side $AB$. Then we have:
$$\sin 30^\circ = \frac{DT}{AT} \Rightarrow DT=\frac12AT\\
\begin{cases}\tan x=\frac{DT}{BD}\\ \tan 30^\circ =\frac{DT}{AD}\end{cases}\Rightarrow AD+BD=\frac{DT}{\tan x}+\sqrt3\cdot DT\Rightarrow AB=\frac12AT\left(\frac1{\tan x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4216537",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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Finding $\lim \frac{(2n^{\frac 1n}-1)^n}{n^2}$. I want to find limit of $a_n= \frac{(2n^{\frac 1n}-1)^n}{n^2}$ as $n\to \infty$.
$\displaystyle a_{n} =\frac{\left( 2n^{\frac{1}{n}} -1\right)^{n}}{n^{2}} =\left(\frac{2}{n^{\frac{1}{n}}} -\frac{1}{n^{\frac{2}{n}}}\right)^{n}$
$\displaystyle \begin{array}{{>{\displaystyl... | Note that $b_n=n^{1/n}-1\to 0$ and by Taylor series we have $$nb_n= \log n+\frac{(\log n) ^2}{2n}+ o\left(\frac{(\log n) ^2}{n}\right)=\log n+o(1)$$ and hence we have $$\log a_n=n\log (1+ 2b_n) -2\log n$$ Next we have via Taylor series $$\log a_n=2nb_n-2\log n-2nb_n^2+o(nb_n^2)=o(1)$$ (as $nb_n-\log n=o(1)$ and $nb_n^2... | {
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"url": "https://math.stackexchange.com/questions/4218136",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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calculate cubic equation discriminant Let $p(x)=x^3+px+q$ a real polynomial.
Let $a,b,c$ the complex root of $p(x)$.
What is the easiest way to calculate $\Delta=(a-b)^2(b-c)^2(a-c)^2$ in function of $p,q$?
The result is $\Delta=-4p^3-27q^2$
Ps. $f=(x-a)(x-b)(x-c)$ and so $a+b+c=0$; $ab+bc+ac=p$ and $abc=-q$; how to ... | Let me offer you a different method which can be useful when calculating other symmetric functions of the roots.
$\Delta$ is symmetric in the roots, of total degree $6$.
In general then, where the cubic is $X^3-e_1X^2+e_2 X-e_3$, $\Delta$ must be a sum of monomials $e_1^{k_1}e_2^{k_2}e_3^{k_3}$ with $k_1+2k_2+3k_3=6$. ... | {
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"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
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Find the largest volume of a cuboid inside a sphere with radius $3m$ I've solved this one by using geometry, but is there a way of finding the max by using derivatives?
My work:
So, because of geometry it has to be a cube.
The diagonal of a cube is equal to $a\sqrt{3}$ and also to $2r$
So: $a\sqrt{3}=2r$
$a=\dfrac{2r}{... | Consider a sphere with the equation $x^2 + y^2 + z^2 = 9m^2$ with volume generated by $V = 8|x||y||z|$.
Consider $0 \leq A \leq 3m$ such that $x^2 + y^2 = A^2$
To maximize $xy$, it is clear that since $xy =\frac{(x+y)^2 - A^2}{2}$ that a maximum occurs when $x + y$ is maximized.
Since $\frac{d}{dx} (x+y) = \frac{d}{dx}... | {
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Relation of coefficient of Cubic polynomial If $a.b $ & $c$ are rational number satisfying the equation $x^3+ax^2+bx+c=0$, then which of the following can also be true .
A) $a+b^2+c^3=0$
B) $a+b^2+c^3=5$
C) $a+b^2+c^3=1$
D) None of these
If it is a cubic function the equation has at least one real root. So for any valu... | Since no relation between $a, b$ and $c$ need to hold in order to satisfy $$x^3+ax^2+bx+c=0 \tag{i}$$ as there always exist some real $x$ which is one of the roots of $\text{(i)}$, I'm assuming that $a, b$ and $c$ satisfy $\text{(i)}$ for all $x \in \mathbb R$
Thus, for $x=0$, equation $\text{(i)}$ becomes
$0+0+0+c=0 \... | {
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Divisibility rules with prime number Let $n \in \mathbb{Z}$ with the property that
$$
7 \mid\left(n^{3}+1\right)
$$
but $7$ does not divide $\left(n^{2}-2 n-3\right) .$ Prove that $7 \mid(4 n+1)$.
So far I got:
$n^3+1 = (n+1)\cdot(n^2-n+1)$, since $7$ is a prime, so
$7\mid(n+1)$ or $7\mid(n^2-n+1)$.
And since $7$ does ... | Further factor $n^2 - n + 1 \equiv (n - 3)(n - 5) \mod 7$. Since $n - 3 \neq 0 \mod 7$, we see that $n - 5 \equiv 0 \mod 7$. So $n \equiv 5 \mod 7$. So $4n + 1 \equiv 0 \mod 7$.
| {
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"question_score": "2",
"answer_count": 3,
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Solving $m^3-n^3=2mn+8$ We have to find integer solutions to the given equation, this is what i tried :-
For ease, let us denote $x=-m, \enspace n=-y$, and then we are basically considering $x,m,y,n$ as nonnegative wherever they occur below :-
We have four cases :-
CASE $1$ : $mn<0$ and $x^3+n^3=-(8+2xn)$
This case is... | Another way to solve this is to introduce $x=m-n$. Then you have quadratic equation on $n$:
$$(3x-2)n^2+(3x^2-2x)n +x^3-8=0$$
This equation has integer solution if it discriminant is perfect square:
$$(3x^2-2)^2-4(3x-2)(x^3-8) = d^2$$
In particular $$3x^4-8x^3+12x^2-96x+60\leq 0$$ which can not occurre many times...
| {
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"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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The number of solutions of $a+b+c+d=n,\ a\geq b,\ d,\ {\rm and}\ c\geq b+1,\ d$ Define $P(n)$ to be the number of solutions : $a,\ b,\ c,\ d$ are nonnegative integers s.t.
$$a+b+c+d=n $$ and $$ a\geq b,\ d,\ {\rm and}\ c\geq b+1,\ d$$
Define $Q(n)$ to be the number of solutions : $x,\ y,\ z,\ w $ are nonnegative int... | Let
\begin{align*}
P_n &= \{(a, b, c, d) \in \mathbb{N}_0^4 : a + b + c + d = n, a \geq b, d, c \geq b + 1, d\}\\
Q_n &=\{(x, y, z, w) \in \mathbb{N}_0^4 : x + y + 2z + 3w = n - 1\}
\end{align*}
so $|P_n| = P(n)$ and $|Q_n| = Q(n)$.
We will proceed by induction to show $P(n) = Q(n)$ for all $n \geq 0$.
Note that the ba... | {
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"source": "stackexchange",
"question_score": "2",
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Trigonometric elimination between two variables Eliminate $\theta$ and $\phi$ between the following equations: $$\begin{cases}\sin \theta + \sin \phi = x \\ \cos \theta + \cos \phi = y \\ \tan \frac {\theta}{2} \tan \frac {\phi}{2} = z\end{cases}$$
What I've done so far
I've established that $$\tan \left(\frac {\theta+... | Write $u = \tan \theta/2$, $v = \tan \phi/2$. Then we get
$$
\begin{cases}
2(u+v)(1 + uv) = x(1+u^2)(1+v^2) \\
2 -2u^2v^2 = y(1+u^2)(1+v^2) \\
uv = z.
\end{cases}
$$
Now setting $s = u + v$ and using the third relation, we obtain
$$
\begin{cases}
2s(1+z) = x(1 - 2z + z^2 + s^2) \\
2(1-z^2) = y(1 - 2z + z^2 + s^2).
\end... | {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Calculating the Image of $A.$ Where did I err? $$A=\begin{pmatrix}
4 &-1& 1\\
8&-2&2\\
-6&1&-2\\
\end{pmatrix}$$
I have to show $p=\begin{pmatrix}
1 \\
2\\
-2 \\
\end{pmatrix} \in \mathrm{Im}A=\left\{Ax \mid x\in \mathbb{R^3} \right\}$
If I let $x=\begin{pmatrix}
0 \\
0 \\
1 \\
\end{pmatrix},$ then $Ax=\begin{pmatrix}
... | The image is the span of linearly independent vectors in the column of the matrix. It suffices to show that $p$ is redundant and can be written as a linear combination of the other columns. After your row reduction, notice that only the first two columns have free variables whereas the third doesn't. This automatically... | {
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"source": "stackexchange",
"question_score": "1",
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Wrong answer in Thomas Calculus 14th Edition textbook There is this question on derivatives to which the answer is given as $\frac{43}{75}$rad/sec in the answers section. This answer appears to be wrong.
My Solution
$$
\theta+\tan^{-1}\left(\frac{6}{4-x}\right)+\tan^{-1}\left(\frac{3}{x}\right)=\pi
$$
which gets reduc... | Here's an alternative explicit calculation.
Let $D$ be the end of the $6$-cm segment opposite $C.$
Let $E$ be the end of the $3$-cm segment opposite $A.$
The angle $\theta$ is the sum of the angles $\angle BDC$ and $\angle BEA.$
We can see that when $x=4,$ then $B$ has just reached $C$ and is moving perpendicular to th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4235161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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For an acute angled triangle $ABC,$ if $p=\frac{\sqrt3+\sin A+\sin B+\sin C}{2\sin A\sin B\sin C}$, find the range of $p$ $$ p=\frac{\sqrt3+\sin A+\sin B+\sin C}{2\sin A\sin B\sin C}$$
$\displaystyle \sin A+\sin B+\sin C=4\cos\frac{A}{2}\cos\frac{B}{2}\cos\frac{C}{2}$ and
$\displaystyle \sin A\sin B\sin C=8\cos\frac{A}... | The lower bound.
For $\alpha=\beta=\gamma=60^{\circ}$ we obtain a value $\frac{10}{3}$.
We'll prove that it's a minimal value.
Indeed, in the standard notation we need to prove that $$\sqrt3+\sum_{cyc}\frac{2S}{bc}\geq\frac{20}{3}\cdot\frac{8S^3}{a^2b^2c^2}$$ or
$$\sqrt3+\frac{2S(a+b+c)}{abc}\geq\frac{160S^3}{3a^2b^2c^... | {
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"source": "stackexchange",
"question_score": "4",
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Why is $\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac n4$? I found the relation for $n\geq3$ $$\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)=\frac n4$$But despite my best efforts, I still have no idea as to how to prove it.
Thin... | Let$$T=\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\cos\left((2k-1)\frac{2\pi} n\right)$$$$S=\sum\limits_{k=1}^{\left\lfloor\frac n2\right\rfloor}\sin^2\left((2k-1)\frac\pi n\right)$$
We can get the relation $T+2S=\lfloor\frac n2\rfloor$
If $n$ is even then using summation of cosines whose angles are in AP we... | {
"language": "en",
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"source": "stackexchange",
"question_score": "7",
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Is it possible to prove equality between complicated algebraic expression and integer without mathematical programs or websites? Prove:
$\sqrt[3]{3+10i\sqrt{\frac{1}{27}}}+\sqrt[3]{3-10i\sqrt{\frac{1}{27}}}=3$
From WolframAlpha:
First:
by
https://www.wolframalpha.com/input/?i=cubrt%283%2B10isqrt%281%2F27%29%29
$\sqrt[3... | Recall the algebraic identity
$$a^3+b^3 + c^3 - 3abc = \frac12(a+b+c)((a-b)^2 + (b-c)^2+(c-a)^2)$$
Whenever $a + b + c = 0$, we have $a^3 + b^3 + c^3 = 3abc$.
Let $\rho = 3 \pm \frac{10}{\sqrt{27}}i$ and $u = \sqrt[3]{\rho} + \sqrt[3]{\bar{\rho}} = 2\Re(\sqrt[3]{\rho})$. Substitute $a,b,c$ by $\rho$, $\bar{\rho}$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4241752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Is the answer of a problem about the equation of a straight line given by my book wrong? This is the problem and the answer given by my book:
My solution:
$3x+\sqrt{3}y+2=0...(i)$
$x\cos\alpha+y\sin\alpha=p...(ii)$
Since (i) & (ii) are equations of the same straight line,
$$\frac{3}{\cos\alpha}=\frac{\sqrt{3}}{\sin\al... | You are correct that $p$ needn't be negative: $$3x+\sqrt{3}y+2=0,$$ $$x\cos\left(\frac{\pi}{6}\right)+y\sin\left(\frac{\pi}{6}\right)=-\frac{1}{\sqrt{3}},$$ and $$x\cos\left(\frac{7\pi}{6}\right)+y\sin\left(\frac{7\pi}{6}\right)=\frac{1}{\sqrt{3}}$$ are all represented by the same line.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4242184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Norm of the trimming map Let $M_n$ denote the linear space of $n\times n$ matrices over $\mathbb{C}$. Consider $M_n$ as a normed space with operator norm. Is it true that the trimming map
$$
T:M_n\to M_n:
\begin{pmatrix}
a_{1,1} & a_{1,2} & \ldots & a_{1,n} \\
a_{2,1} & a_{2,2} & \ldots & a_{2,n} \\
\ldots & \ldots &... | *
*$T$ is a contraction for the operator $\left\|\cdot\right\|_p$ norms for $p=1$ and $p = \infty$ since those \begin{align}
\left\|A\right\|_1 &= \max_{j}\sum_{i=1}^n \left|a_{i,j}\right| &\text{and}\quad \left\|A\right\|_\infty &= \max_{i}\sum_{j=1}^n \left|a_{i,j}\right|.
\end{align}
*However $T$ is not a contra... | {
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"source": "stackexchange",
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Find an ellipse from two points and the tangent at a third point I am searching an ellipse in the form $\frac{\left(x - x_\circ\right)^2}{a^2} + \tfrac{\left(y - y_\circ\right)^2}{b^2} = 1$ (its axis parallel to the coordinate axes). What I know are three points, and for one of them I additionally know the direction ϑ... | You have $4$ unknowns, which are, $a, b, x_0, y_0$, and you can write down $4$ equations, so from dimensionality the problem should be solvable. Let the three points be $P_1(x_1, y_1), P_2(x_2, y_2), P_3 (x_3, y_3)$. The equation of the ellipse to be identified is
$ A x^2 + B xy + C y^2 + D x + E y + F = 0 $
Since, t... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If $ \frac{X_n}{1024^n} $ is an odd integer, find the smallest possible value of $n$, where $n\ge2$ is an integer.
For each integer $k\ge2$, the decimal expansions of the numbers $1024, 1024^2, \dots, 1024^k$ are concatenated, in that order, to obtain a number $X_k$. (For example, $X_2 = 10241048576$.) If $ \frac{X_... | As $1024$ is just a little larger than $10^3, 1024^k$ will be a little larger than $10^{3k}$, so will have $3k+1$ decimal digits. According to Alpha, the number of digits does not increase above this until $k=98$. I think you can rely on $n$ being smaller than that.
| {
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"source": "stackexchange",
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Finding $x^8+y^8+z^8$, given $x+y+z=0$, $\;xy +xz+yz=-1$, $\;xyz=-1$
The system says
$$x+y+z=0$$
$$xy +xz+yz=-1$$
$$xyz=-1$$
Find
$$x^8+y^8+z^8$$
With the first equation I squared and I found that $$x^2+y^2+z^2 =2$$
trying with
$$(x + y + z)^3 =
x^3 + y^3 + z^3 + 3 x^2 y + 3 x y^2 + 3 x^2 z++ 3 y^2 z + 3 x z^2 +
... | This is an expansion of my comment. Recall Vieta's formulas (in this case, for cubics):
If $p, q, r$ are the roots of $t^3 + bt^2 + ct + d$, where $a \neq 0$, then
\begin{align*}
p + q + r &= -b \\
pq + qr + pr &= c \\
pqr &= -d.
\end{align*}
This follows easily by just expanding $(t - p)(t - q)(t - r)$ and equating ... | {
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"source": "stackexchange",
"question_score": "12",
"answer_count": 8,
"answer_id": 5
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Showing that $\int_{0}^{\infty}\frac{x^{m-1}-x^{n-1}}{1-x}dx$ is convergent if $ 0
Showing that $\int_{0}^{\infty}\frac{x^{m-1}-x^{n-1}}{1-x}dx$ is convergent if $ 0<m<1 , 0<n <1$
Let $I = \int_{0}^{\infty}\frac{x^{m-1}-x^{n-1}}{1-x}dx$ then $I = \int_{0}^{\infty}\frac{x^{m-1}}{1-x}dx - \int_{0}^{\infty}\frac{x^{n-1}}... | A simple way to show that the integral is convergent is by cutting the integral into three parts
\begin{align}
\int^{1/2}_0 + \int^{3/2}_{1/2} + \int^\infty_{3/2}\frac{x^{m-1}-x^{n-1}}{1-x}d x = I_1+I_2+I_3.
\end{align}
It is clear that $I_3$ is convergent since
\begin{align}
I_3 \le C\int^\infty_{3/2}\left(\frac{1}{x... | {
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How to solve $\sin(2\theta)$ questions Given that: $\sin\theta=\displaystyle{}\frac{12}{13}$ and $0<\theta<\displaystyle{}\frac{\pi}{2}$ the value of $\sin(2\theta)$ is:
I figured out a way to solve it, though I'm not sure if it is the best solution.
Here we will combine two different trigonemtric identities. First:
$\... | Another way to do this is to first find $\cos\theta$. This is easier if you recognise small Pythagorean triads. ;)
Let
$y/r=\sin\theta$ and $x/r=\cos\theta$, where $r^2=x^2+y^2$.
We have $\sin\theta=5/13$, so
$$x^2 = 13^2-5^2 = (13+5)(13-5) = 18\cdot8=12^2$$
thus $\cos\theta=12/13$.
Now, we know that $\sin(2\theta)=2\s... | {
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"source": "stackexchange",
"question_score": "5",
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Prove that: $\sum\limits_{cyc}\frac{1}{\sqrt{2a^2+5ab+2b^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}$ Let $a$, $b$ and $c$ be non-negative numbers such that $ab+ac+bc\neq0$. Prove that:
$$ \dfrac{1}{\sqrt{2a^2+5ab+2b^2}}+\dfrac{1}{\sqrt{2b^2+5bc+2c^2}}+\dfrac{1}{\sqrt{2c^2+5ca+2a^2}} \geq\sqrt{\frac{3}{ab+ac+bc}}.$$
I solved thi... | Remark: I give a proof using the so-called isolated fudging.
It suffices to prove that
$$\frac{\sqrt{\frac{ab + bc + ca}{3}}}{\sqrt{2a^2 + 5ab + 2b^2}}\ge \frac{8c^2 + 9(a + b)c + 8ab}{8(a^2 + b^2 + c^2) + 26(ab + bc + ca)}. \tag{1}$$
Note: Taking cyclic sum on (1), we get the desired inequality.
If $c = 0$, it is easy... | {
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"source": "stackexchange",
"question_score": "5",
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How to prove $\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}$ Prove that
$$
\int_{0}^{\pi/2}\frac{\sqrt{\cos \theta}}{1 + \cos^2\theta}d\theta = \frac{\pi}{4}
$$
My attempt :I tried to use the beta function, but I couldn't.
| We have
$$\int_{0}^{\frac{\pi}{2}} \frac{\sqrt{\cos(\theta)}}{1+\cos^{2}(\theta)}d\theta=\int_{0}^{\frac{\pi}{2}}\sum_{n=0}^{\infty}(-1)^{n}\cos^{2n+\frac{1}{2}}(\theta)d\theta$$
$$=\sum_{n=0}^{\infty}(-1)^{n}\int_{0}^{\frac{\pi}{2}}\cos^{2n+\frac{1}{2}}(\theta)d\theta$$
by Fubini/Tonelli theorems. Then using the trigo... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the minimum value of: $\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$ Let $a,b,c>0$, find the minimum value of: $$\dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}$$
I have tried:
$\bullet$ The minimum value is $\dfrac{3}{4}$ occur when $a=b=c$
$\bullet \dfrac{a}{b+3a}+\dfrac{b}{c+3b}+\dfrac{c}{a+3c}= \dfrac{a^2}... | Another way:
We know that our expression is closed to $\frac{1}{3}$ but
$$\sum_{cyc}\frac{a}{b+3a}\geq \sum_{cyc} \frac{a}{3b+3a+3c}=\frac{1}{3}$$ and we got the same result again.
| {
"language": "en",
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"source": "stackexchange",
"question_score": "5",
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Solving Inequality $\sqrt{\frac{x^2}{\:y}}+\sqrt{\frac{y^2}{x}}\ge \sqrt{x}+\sqrt{y}$ If someone could help with solving the inequality above, that would be awsome!
Here is my thinking of using AGM
*
*$\sqrt{\left(\sqrt{\frac{x^2}{y}}\sqrt{\frac{y^2}{x}}\right)}\le \frac{\sqrt{\frac{x^2}{y}}+\sqrt{\frac{y^2}{x}}}{2}$... | Note that we must have $x > 0$ and $y > 0$ for the inequality to even be defined at all.
Start by squaring both sides to get the equivalent inequalty
$$\frac{x^2}{y} + \frac{y^2}{x} + 2 \sqrt{\frac{x^2}{y} \frac{y^2}{x}} = \frac{x^2}{y} + \frac{y^2}{x} + 2 \sqrt{xy} \geq x + y + 2 \sqrt{xy}$$
Equivalently,
$$\frac{x^2}... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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} |
How to prove $\int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2}\, dx = \frac{\zeta(3)}{\pi}$? I was recently searching for interesting looking integrals. In my search, I came upon the following result:
$$ \int_{0}^{\infty} \frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \rig... | Following K.defaoite's suggestion, I've managed to get a partial solution using the Residue Theorem. Rewriting the integrand with the Weierstrass product for $\cosh$ squared in the denominator we get
$$
f(x) =\frac{(1-x^2) \, \text{sech}^2\left(\frac{\pi x}{2} \right)}{(1+x^2)^2} = \frac{(1-x^2) }{(x+i)^4(x-i)^4 \prod_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 6,
"answer_id": 4
} |
Why does $-1<\frac{7}{x+4}$ imply $x<-11$? I keep getting $x>-11$ I'm trying to solve for the $|\frac{7}{x+4}|<1$. Which becomes $-1<\frac{7}{x+4}<1$.
Evaluating the positive side is fine, $3<x,$ but for the negative side:
$-1<\frac{7}{x+4}$ should evaluate to $-11 > x$. But I keep getting $-11 < x.$
My working:
$$-1<\... | $$\frac{7}{x+4}>-1 \implies \frac{x+11}{x+4} >0$$
This means
(1): $(x+11)>0 \& x+4>0 \implies x>-4 \& x >-11 \implies x>-4$
(2): $(x+11) <0 \& x+4 <0 \implies x<-11 \& x <-4 \implies x<-11$
So the total answer is union of the two: $x\in (-\infty, -11) \cup (-4, \infty)$ or
$x<-11$ or $x >-4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 1
} |
Error in my approach to prove that $(m-1)^{m-1} \equiv m - 1 \pmod m$ using a concrete example I read somewhere that if $m$ is composite then $(m - 1)^{m - 1} \equiv m - 1 \pmod m$ and I was curious to try to prove it myself.
So I took as $m = 6$.
Now I can see that $5^5 \equiv 5 \pmod 6$
I was thinking along the follo... | The fallacy in your argument is in the last step, where you "cancel" $1\cdot2\cdot3 \cdot 4\cdot 5$ from each side of the congruence.
As in ordinary arithmetic, you can only cancel factors that aren't $0$. (In fact, you can only reliably cancel factors that are relatively prime to the modulus.)
So in this case the asse... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4253887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
problem with power series $\sum \frac{x^n}{n+3}$ I need to evaluate the following power series, and I don’t really know how to do it, this is the series $$\sum \frac{x^n}{n+3}$$
This is how I tackled this problem, but the final solution looks very dumb.
So we know that $$\frac{1}{1-x}=\sum x^n$$
But we also know that... | Denote
\begin{equation}
f(x)=\sum_{n=1}^\infty\frac{x^n}{n+k}, \quad k\in\{0,1,2,\dotsc\}.
\end{equation}
Then
\begin{equation}
x^kf(x)=\sum_{n=1}^\infty\frac{x^{n+k}}{n+k}
\end{equation}
and
\begin{equation}
\bigl[x^kf(x)\bigr]'=\sum_{n=1}^{\infty}x^{n+k-1}=x^k\sum_{n=0}^{\infty}x^n=\frac{x^k}{1-x}, \quad -1\le x<1.
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4257106",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve integral $\int_0^{2\pi}\frac{\sin^2\frac{(N+1)x}{2}}{2\pi(N+1)\sin^2\frac x2} dx $ My question is to solve
$$\int_0^{2\pi}\frac{\sin^2\left(\frac{N+1}{2}x\right)}{2\pi(N+1)\sin^2(x/2)} dx $$
Up to $1000$ or more, Wolfram-Alpha solves this integral and gives the value $1$ for every integer. But, I couldn't prove ... | Apply
$$\frac{\sin \frac{(N+1)x}2- \sin\frac{(N-1)x}2}{\sin \frac x2}=2\cos\frac{Nx}2
$$
to the integral
\begin{align}
I_{N+1}&
= \int_{0}^{\pi}{\frac{\sin^2\frac{(N+1)x}2}{\sin^2\frac x2}}dx
=\int_{0}^{\pi}\left({\frac{\sin\frac{(N-1)x}2}{\sin\frac x2}}+2\cos\frac{Nx}2 \right)^2dx\tag1
\end{align}
Note that the inte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4258621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
The limit of $n^2 f \left(a-\frac{b}{n}, b+\frac{a}{n} \right)$. Let $f : \mathbb{R}^2\to \mathbb R$ be $C^2$ class function s.t. $$f(x,y)=0 \ \mathrm{ and } \ (f_x(x,y))^2+(f_y(x,y))^2=1$$ for $(x.y)\in C:=\{ (x,y) \mid x^2+y^2=1 \},$ and suppose $f$ is monotonically increasing as getting away from $(0,0)$ on the half... | Hint: If you choose $f(x, y) = \frac{1}{2}x^2+\frac{1}{2}y^2-\frac{1}{2}$, then you see it satisfies all the required conditions. Observe we have that
\begin{align}
\lim_{n\rightarrow \infty} n^2f(a-\frac{b}{n}, b+\frac{a}{n})=&\ \lim_{n\rightarrow \infty} \frac{1}{2}n^2\left((a-\frac{b}{n})^2+(b+\frac{a}{n})^2 -1\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Sum $\sum_{r=1}^n \cos(2.(\frac {3^rx}{3})).\csc (3^rx)$ Prove that $$\sum_{r=1}^n \cos(2.(\frac {3^rx}{3})).\csc (3^rx) = \frac{1}{2\sin x}-\frac{1}{2\sin (3^nx)}$$
My attempt:
$$\Sigma \frac{\cos(\frac {2.3^rx}{3})}{\sin(3^rx)} = \Sigma \frac{1-2\sin^2(\frac {3^rx}{3})}{\sin(3^rx)}$$
I am trying to convert it to a fo... | \begin{align}
\frac{1-2\sin^2(3^{r-1}x)}{\sin(3^rx)} &= \frac{1-2\sin^2(3^{r-1}x)}{\sin(3\cdot3^{r-1}x)} \\
&= \frac{1-2\sin^2(3^{r-1}x)}{3\sin(3^{r-1}x)-4\sin^3(3^{r-1}x)} \\
&= \frac{(3-4\sin^2(3^{r-1}x))-1}{2\sin(3^{r-1}x)\left(3-4\sin^2(3^{r-1}x)\right)} \\
&= \frac{1}{2\sin(3^{r-1}x)}-\frac{1}{2\sin(3\cdot3^{r-1}x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4259884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum value of $a^8+b^8+c^8+2(a-1)(b-1)(c-1)$ Let $a,b,c$ be the lengths of the three sides of the triangle, $a+b+c=3$. Find the minimum value of $$a^8+b^8+c^8+2(a-1)(b-1)(c-1)$$
My attempts:
$\bullet$ The minimum value is $3$, equality holds iff $a=b=c=1$ so by AM-GM, we have: $$a^8+b^8+c^8\ge 8(a+b+c)-21=3... | Another way.
We'll prove that the inequality
$$a^8+b^8+c^8+2(a-1)(b-1)(c-1)\geq3$$ is true for any positives $a$, $b$ and $c$ such that $a+b+c=3$.
Indeed, since $$\prod_{cyc}((a-1)(b-1))=\prod_{cyc}(a-1)^2\geq0,$$ we can assume $a(b-1)(c-1)\geq0$ and by AM-GM and C-S we obtain:
$$\sum_{cyc}a^8+2\prod_{cyc}(a-1)-3\geq\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4260844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Euler substitution inconsistency in evaluating $\int_{-1}^1 \frac{\mathrm dx}{\sqrt{1-x}+2+\sqrt{1+x}}$
$$I = \int_{-1}^1 \frac{\mathrm dx}{\sqrt{1-x}+2+\sqrt{1+x}} = 4\sqrt2-2-\pi$$
Euler's substitutions immediately come to mind. But as is, the integrand is free of $\sqrt{ax^2+bx+c}$ (with $a\neq0$). So, I considere... | Using instead $\sqrt{1-x^2} = xt\color{red}{+}1$ is indeed the correct move, as this ensures $xt+1\ge0$ for $(x,t)\in[-1,1]^2$. Then
$$\begin{cases}x = -\frac{2t}{1+t^2} \implies \mathrm dx = -\frac{2(1-t^2)}{(1+t^2)^2}\\\\ \sqrt{1-x^2} = \frac{1-t^2}{1+t^2} \\\\ \sqrt{1\pm x} = \frac{1\mp t}{\sqrt{1+t^2}}\end{cases}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4263562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How could I rewrite $\dfrac{6 + 4 {i}} { -9 - 4 {i}}$ in a+bi form? Peace to all. When I solve the problem I get $\dfrac{70} { 97}$ - $\dfrac{12{i}} {97}$ and it's the wrong answer. How exactly do you go about solving this problem?
This is my work: I received that answer by multiplying both the numerator and denominato... | Another approach that can be taken is through the application of the definition of division. If $ \ \frac{6 + 4 i}{ -9 - 4 i} \ = \ a + bi \ \ , $ then
$$ 6 \ + \ 4i \ \ = \ \ (-9 \ - \ 4i)·(a \ + \ bi) \ \ = \ \ -9a \ - \ 4ai \ - \ 9bi \ - \ 4b·i^2 \ \ . $$
Equating the real and imaginary parts produces the pair of e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4268082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Integrating $\int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$ I found the following integral and wanted to know if there is a nice closed form solution in terms of elementary or some special functions (Polylogarithm, Clausen, etc).
$$\displaystyle \int_{0}^{1} \frac{\arctan(x)\arctan(x^2)}{x^2} dx$$
I know that the ... | By continuing the OP's work and using Cornel's closed-form of $\displaystyle \int_0^1\frac{\arctan(x)}{1+x^4}\textrm{d}x$, we get that
$$\int_0^1\frac{\arctan(x)\arctan(x^2)}{x^2}\textrm{d}x$$
$$=G+\frac{11}{192}\sqrt{2}\pi^2-\frac{1}{16}\log(2)\pi-\frac{3}{16}\sqrt{2}\log(\sqrt{2}-1)\pi-\frac{1}{4}\sqrt{2}\operatorna... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4270576",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
What is the proof for variance of triangular distribution? In Wikipedia, the formula for the variance of the triangular distribution is given here. However, I don't know how to find it. I have tried a brute force method but the formula is quite complicated (polynomial of degree 5 in a, b, c) and I can't simplify it (I ... | The computation is best performed on the location-transformed variable such that the mode is at $0$. In particular, suppose $$X \sim \operatorname{Triangular}(a,b,c), \\ f_X(x) = \begin{cases} 0, & x < a \\ \frac{2(x-a)}{(b-a)(c-a)}, & a \le x \le c \\ \frac{2(b-x)}{(b-a)(b-c)}, & c < x \le b, \\ 0, & x > b. \end{case... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271314",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to calculate a permutation to the $-100$th power? I have a this permutation element $(1,2)(3,4,6,5,8)(7,9,10)$ and I need find the $-100$th power of this element.
I think firstly we need find $-1$ power, and I have this element: $(1,2)(3,8,5,6,4)(7,10,9)$.
Here $(1,2)$ have order $2$, so after $100$ permutation we ... | Your calculations are fine.
An alternative approach would be to notice that the order of the element is the least common multiple of the orders of its disjoint cyclic components, so the order is ${\rm lcm}(2,5,3)=30$. Now note that
$$\begin{align}
((1,2)(3,4,6,5,8)(7,9,10))^{-100}&=((1,2)(3,4,6,5,8)(7,9,10))^{-90}\\
&\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4271691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Probability of having an all-heterogeneous genotype
Find the probability that the genotype of an offspring will be $\mathrm{AaBbCcDdEeFfGgHhIiJj}$ from the cross $\mathrm{AaBbCcDdEeFfGgHhIiJj \times AaBbCcDdEeFfGgHhIiJj}$.
Making a Punnett square for this problem will be too tedious. However, I noticed that this prob... | There are four possible outcomes for each of the ten traits. For instance, for the first trait, they are $AA, Aa, aA, aa$, where we first list the allele from the father, then the allele from the mother. Since we are not told the traits are linked in some way, it is reasonable to assume independence. Thus, there are ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4273663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does this integral have an analytical solution? (The residue theorem method seems to fail here) The integral is the following:
\begin{equation}
\int_{-\infty}^{+\infty}dx \frac{e^{-{\sigma}^2(x+b)^2}}{x^2+a^2}
\end{equation}
I know there is an analytical solution when $b=0$. But what about $b\neq0$ ?
| May be, we could try to expand the exponential as
$$e^{-\sigma ^2 (x+b)^2}=e^{-\sigma ^2 x^2}\sum_{n=0}^\infty (-1)^n P_n(x)\, b^n$$ and face integrals
$$I_n=\int_{-\infty}^\infty \frac {x^n} {x^2+a^2}e^{-\sigma ^2 x^2}\, dx=a^{n-1}\int_{-\infty}^\infty \frac{t^n}{t^2+1}e^{-k t^2}\,dt\qquad \text{where}\qquad k=a^2\sig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4277638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Proving $\lim_{x \to +\infty}\frac{\ln(x)}{2\cdot \ln(x+1)} = \frac{1}{2}$ with $\epsilon- \delta$ Today, I spent some hours trying to solve this problem but I can't reach the conclusion:
Show with $\epsilon-\delta $ definition the following limit:
$$\lim_{x \to +\infty}\frac{\ln(x)}{2\cdot \ln(x+1)} = \frac{1}{2}$$
... | Fix $x>e^2-1$. Using
$$ \ln(1+\frac1x)<\frac1x $$
we have
$$ \left|\frac{\ln(x)}{2\cdot \ln(x+1)}-\frac{1}{2}\right|=\frac12\frac{\ln(1+\frac1x)}{\ln(x+1)}<\frac{1}{4x}. $$
Then for $\forall \varepsilon>0$, denote $N$ by $N=\max\{e^2-1,\frac1{4\varepsilon}\}$. Then when $x>N$,
$$ \left|\frac{\ln(x)}{2\cdot \ln(x+1)}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4283115",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Given $a+b+c =1$ prove $\displaystyle \sum_{\text{cyclic}}\sqrt{abc+4ab+4ac}\ge 8(ab+bc+ca)+\sqrt{abc}$
If $a,b,c$ are non-negative real numbers such that: $a+b+c=1.$ Prove that: $$\sqrt{abc+4ab+4ac}+\sqrt{abc+4bc+4ba}+\sqrt{abc+4ca+4cb}\ge8(ab+bc+ca)+\sqrt{abc}$$
Anyone can help me give a hint to get proof?
I guess ... | Hint:
Observe,
$$\begin{align*}
8(ab+bc+ca)+\sqrt{abc}&=4a(b+c)+4b(c+a)+4c(a+b)+\sqrt{abc} \\
(b+c-a)^2&=(1-2a)^2=1-4a(1-a)=1-4a(b+c)
\end{align*}$$
Using the Cauchy-Schwarz inequality,
$$\sum_{\text{cyc}}\sqrt{\left((b+c-a)^{2}+4a(b+c)\right)\left(abc+4a(b+c)\right)}\geq \sum_{\text{cyc}} \left(\sqrt{abc}(b+c-a)+4a(b+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4284289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How do you prove by induction that $\frac{1}{2} + \frac{2}{2^2} + \ldots + \frac{n}{2^n} = 2 - \frac{n+2}{2^n}$? For $n=1$ this is true because $\frac{1}{2^{1}}=2-\frac{1+2}{2^{1}}=\frac{1}{2}$. Further, it is a little more complicated, can we now assume that this is true up to the number $n-1$? Then do the induction s... | Another solution : for sequences $\{a_{n}\}, \{b_{n}\}$
*
*Prove $\sum_{k=1}^{n} a_{k} b_{k} = \sum_{k=1}^{n-1} S_{k} (b_{k} - b_{k+1}) + S_{n} b_{n}$ where $S_n = \sum_{k=1}^{n} a_{k}.$ ($n \in \mathbb{N}$)
$pf)$ $(RHS) = b_{n} (S_{n} - S_{n-1}) + b_{n-1} (S_{n-1} - S_{n-2}) + \cdots + b_{1} S_{1} = \sum_{k=1}^{n} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4286400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Can we approximate $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$? It seems like from the graph $\prod_{n=1}^\infty (1-\frac{(2n+1)x^2}{n^2\pi^2})$ is somehow alike to the graph $e^{-x^2}$, the main problem is that the limitations of the software makes it hard to graph for large numbers. Is it possible to do so or... | Calling $\Pi$ the partial product up to $N$, for $0<|x|<\pi/\sqrt{3}$ we have
$$ \begin{array}{ll}
-\ln\Pi & =\displaystyle \sum_{n=1}^N \ln\Bigg(\frac{1}{1-\frac{(2n+1)x^2}{n^2\pi^2}}\Bigg) \\[5pt] & > \displaystyle \sum_{n=1}^N \ln\Big(1+\frac{(2n+1)x^2}{n^2\pi^2}\Big) \\[5pt]
& >\displaystyle \sum_{n=1}^N \left[\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4288574",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Factoring $4x^2-2xy-4x+3y-3$. Why isn't it working? I want to factorise the following expression.
$$4x^2-2xy-4x+3y-3$$
Here are the ways I tried
$$4x^2-2xy-4x+3y-3=\left(2x-\frac y2\right)^2+3y-3-\frac{y^2}{4}-4x=\left(2x-\frac y2\right)^2+\frac{12y-12-y^2}{4}-4x=\left(2x-\frac y2\right)^2-\frac 14(y^2-12y+12)-4x$$
Now... | Probably not the fastest, but I would like to suggest you use the general factorization method:
We have,
$$
\begin{align}
& 4 x^{2}-2 x y-4 x+3 y-3=4 x^{2}-x(2 y+4)+(3 y-3) \\
\implies & \Delta=(y+2)^{2}-4(3 y-3)=(y-4)^{2} \\
\implies & x_{1}=\frac{y+2+(y-4)}{4}=\frac{y-1}{2} \\
\implies & x_{2}=\frac{y+2-(y-4)}{4}=\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 8,
"answer_id": 2
} |
Prove $\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}=1$ Use $f(n) = \frac{1}{n}$ and $f(n) = \frac{1}{n^2}$ when:
*
*$\sum_{n=1}^{\infty}[f(n+1) - f(n)] = \lim_{n \to \infty} f(n) - f(1)$
to prove $\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2}=1$
What I have tried:
$$\sum_{n=1}^{\infty} \frac{2n+1}{n^2(n+1)^2} =\sum_{n=1... | $f(n)-f(1)=\frac 1 {n^{2}}-1 \to -1$ so $\sum_{n=1}^{\infty}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2} \right)=-\sum_{n=1}^{\infty}\left(\frac{1}{(n+1)^2}-\frac{1}{n^2} \right) =-\lim [f(n)-1]=1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4290869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....?$ $$\frac {\zeta(2)}{e^2} + \frac {\zeta(3)}{e^3} + \frac {\zeta(4)}{e^4} + \frac {\zeta(5)}{e^5}.....?$$
I tried to solve it by using this product formula,
$$\frac 1{\Gamma (x)}=xe^{\gamma x} \prod_{n=1}^{\infty} \left(... | Moreover,
\begin{align*}
\sum_{n=2}^{\infty} \frac{\zeta(n)}{k^{n}}&= \frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{n!k^n} \Gamma(n+1)\zeta(n+1) \\&= \frac{1}{k} \sum_{n=1}^{\infty} \frac{1}{n!k^n} \int_{0}^{\infty} \frac{x^n}{e^{x}-1} \, dx \\&= \frac{1}{k}\int_{0}^{\infty} \frac{e^{x/k}-1}{e^{x}-1} \, dx \\&= - \frac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4295777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Mistake computing $\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}$ I am looking to evaluate the integral
$$\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}} \right)\frac{dx}{x}=\ln\left(\frac{\pi}{\Gamma^2\left(\frac{1+a}{2b}\right)\cos\left(\frac{a\pi}{2}\right)}\righ... | For this specific integral We can evaluate as follows:
$$
\begin{aligned}
I(a)&=\int_0^\infty \left(\frac{\sinh(ax)}{\sinh(x)}-\frac{a}{e^{2x}}\right)\frac{dx}{x}\\
& \\
I^\prime(a)&=\int_0^\infty \left(\frac{x\cosh(ax)}{\sinh(x)}-\frac{1}{e^{2x}}\right)\frac{dx}{x}\\
&=\int_0^\infty \left(\frac{\cosh(ax)}{\sinh(x)}-\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4296809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
determining the cubic function from a graph This question may be a bit below the regular level for this forum, but I really struggle finding the solution. All of my attempts seem to end up describing $y = 1$, probably because all the points I can read from the graph are on that line. While this was given as a homework ... | The derivative of the cubic function has two zeroes at $x = 1 $ and $x = 3$. Therefore,
$f'(x) = A (x - 1)(x - 3) = A (x^2 - 4 x + 3) $
Integrate that from $0$ to $x$,
$f(x) = f(0) + A (\dfrac{x^3}{3} - 2 x^2 + 3 x ) $
Now from the graph, $f(0) = 1$ and $f(3) = 1$ , hence,
$f(x) = 1 + A (\dfrac{x^3}{3} - 2 x^2 + 3 x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4299684",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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FUN with f̶̶l̶̶a̶̶g̶̶s̶ Newton Cotes Quadrature formula and Bernoulli polynomials of the second kind I was told to phrase my question in a more exciting way when I asked it last time. The following is a preliminary consideration. If you don't need it, just scroll down to START HERE. Here we go then:
Imagine you have a ... | According to Wikipedia:
\begin{align}
\psi_{2k}(k-1+y) = \psi_{2k}(k-1-y)
\end{align}
If we now choose $y = k - 2$, it follows:
\begin{align}
&\psi_{2k}(k-1+(k-2)) = \psi_{2k}(k-1-(k-2)) \color{white}{\frac{1}{2}}\\
& \Longrightarrow \hspace{10pt} \psi_{2k}(2k-3) = \psi_{2k}(1) \color{white}{\frac{1}{2}} \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What's the measure of the segment $EF$ in the triangle below? For reference:
The triangle $ABC$ where $AB = 7$, $BC = 8$ and $AC = 9$ is inscribed in a circle. Calculate the measure of the arrow ($EF$) of the side $AC$. (Answer: $\frac{3\sqrt5}{2}$)
My progress:
Here are the relations I found:
$EF = R - OE \\
\trian... | *
*$\angle ABC=\frac12\angle AOC=\angle AOE$
*$\sin B=\sin\angle AOE=\frac{AC/2}{R}=\frac{b}{2R}$
*Similarly
$$2R=\frac a{\sin A}=\frac b{\sin B}=\frac c{\sin C}$$
*Area of $\triangle ABC$ is $$K=\frac12ab\sin C=\frac{abc}{4R}$$
*Use Heron's formula to calculate area $K$
*Get $R$
*Use Pythagoras to get $OE$
*Ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4302567",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Find the value of $ \frac{1}{{2(2^2 - 1)}} + \frac{1}{{3(3^2 - 1)}} + \frac{1}{{4(4^2 - 1)}} + \cdots \;. $ To find the value of $$
\frac{1}{{2(2^2 - 1)}} + \frac{1}{{3(3^2 - 1)}} + \frac{1}{{4(4^2 - 1)}} + \cdots \;.
$$
I presented it as $$
\sum\limits_{n = 2}^\infty {\frac{1}{{n(n^2 - 1)}}} .
$$
Then using part... | You can continue with the partial sums as
\begin{align*}
\sum\limits_{n = 2}^N {\left( { - \frac{1}{n} + \frac{1}{{2(n + 1)}} + \frac{1}{{2(n - 1)}}} \right)} & = \sum\limits_{n = 2}^N {\left( {\left( {\frac{1}{{2(n - 1)}} - \frac{1}{{2n}}} \right) + \left( {\frac{1}{{2(n + 1)}} - \frac{1}{{2n}}} \right)} \right)} \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4303104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
find derivative of $\sqrt{x}+\sqrt{y}=\sqrt{a}$ I have an Implicit Function $\sqrt{x}+\sqrt{y}=\sqrt{a}$
the graph of the function is
I need to prove that $p+q=a$
and I need to find $\frac{d}{dx}$ to find the the slop to prove that.
result:
after rearranged the function I got
$a=x+2\sqrt{xy}+y$
and the derivative o... | Let $y'$ denote $\displaystyle\frac{dy}{dx}.$
Given: $\sqrt{x} + \sqrt{y} = \sqrt{a}.$
Using implicit differentiation, you have that
$\displaystyle \frac{1}{2\sqrt{x}} + \frac{y'}{2\sqrt{y}} = 0 \implies y' = \frac{-\sqrt{y}}{x} =
\frac{-\left(\sqrt{a} - \sqrt{x}\right)}{\sqrt{x}}.$
The distance $Q$ may be represente... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What's the the value of $x$ in the circumference below? For reference:
In the figure; calculate $x$, if $r =\sqrt2$.
(Answer: $x = \sqrt2$)
My progress:
Draw $PO_1\perp HG\:(O_1 \in HG).$
Let $O$ be the center of the largest circle.
Using Euclid's Theorem:
$\triangle OPF:OP^2 = OQ^2+PQ^2-2\cdot OQ\cdot FQ$
$\implie... |
$PG = r_1 - r, PR = r_1 + r$ and so by Pythagoras, it is easy to see that $RG = 2 \sqrt{r r_1}$
Now if $O$ is the center of the circle with radius $R$,
$AG = 2r_1 - r ~$ and
$ |OG| = |AG - AO| = |2r_1 - r - R| ~ $
By Pythagoras, $OG^2 = OR^2 - RG^2$
$\implies (2r_1 - r - R)^2 = (R-r)^2 - 4 r r_1$
Solving, $R r = R r_1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4304972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Prove, for every nonnegative integer $n$, that $5^{2n} + 2^{2n} ≡ 2^{2n+1}\pmod{21}$. Is my proof correct? Thanks in advance.
Base step: n = 0
$$5^{2 \times 0} + 2^{2 \times 0} ≡ 2^{2 \times 0+1} \pmod{21}$$
$$= 1 + 1 ≡ 2 \pmod{21} \checkmark$$
Inductive hypothesis: assume $5^{2n} + 2^{2n} ≡ 2^{2n+1} \pmod{21}$ is true... | The proof you gave is correct.
However, it can actually be done with a lot less work and without induction:
$$5^{2n}=\left(5^2\right)^n=25^n\equiv 4^n=2^{2n} (\mathrm{mod}\ 21)$$
Hence $5^{2n}+2^{2n}\equiv 2\cdot 2^{2n}=2^{2n+1} (\mathrm{mod}\ 21)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4306267",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Laurent expansion on annulus Let $F(z)= \frac{1}{(z-1)^2(z+2)}$.
I need to find a Laurent expansion for $F$ on the annulus $A = \{z: \sqrt{2}<|z-i|<\sqrt{5}$}.
So far I have managed to write $F$ as $F(z)=\frac{-1}{9(z-1)}+\frac{1}{9(z+2)}+\frac{1}{3(z-1)^2}$. Then writing $z-i = w$, I got the following expansions:
$\fr... | You are interested in the Laurent series centered at $i$, rather than $0$. So, use the fact that\begin{align}F(z)&=F\bigl((z-i)+i\bigr)\\&=\frac1{\bigl((z-i)+i-1\bigr)^2\bigl((z-i)+i+2\bigr)}\\&=\frac1{9\bigl((2+i)+(z-i)\bigr)}+\frac1{9\bigl((1-i)-(z-i)\bigr)}+\frac1{3\bigl((1-i)-(z-i)\bigr)^2}\\&=\frac1{9(2+i)}\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307562",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of the squares of the medians of triangle APC below For reference: The diameter of a circle measures 8m. On this diameter are located the points A and B equidistant 1 m from the center; through B draw any chord PC , determine the sum of the squares of the medians of triangle APC.(Answer: $73,5$)
My progres... | If $O$ is the center of the circle, $OP$ is median of $\triangle APB$. Radius of the circle is $4$.
So, $4 OP^2 = 64 = 2 AP^2 + 2 PB^2 - AB^2$
$ \implies AP^2 + PB^2 = 34$
Similarly, $AC^2 + BC^2 = 34$
Now by intersecting chords theorem, $PB \cdot BC = 5 \cdot 3 = 15$
So, $AP^2 + AC^2 + PB^2 + BC^2 + 2 PB \cdot BC = 98... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4307773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Can a ratio of sinusoidal functions with the same frequency always be written as a tangent function? In general, two sinusoidal functions $y_1$ and $y_2$ with angular frequency $\omega$ can be written as
*
*$y_1 = A\sin(\omega x) + B \cos(\omega x)$,
*$y_2 = C\sin(\omega x) + D \cos(\omega x)$.
Where $A, B, C, D$ a... | Yes: for $C,D$ not both zero, $$\frac{A\sin \omega x+B\cos \omega x}{C\sin \omega x+D\cos \omega x}=\frac{AD-BC}{C^2+D^2}\;\tan\left(\omega x-\arctan\frac CD\right)+\frac{AC+BD}{C^2+D^2}.$$ This is true even when $D=0,$ if we let $\displaystyle\arctan\frac CD=\begin{cases}\displaystyle&\frac\pi2&\text{if }C>0;
\\ \disp... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309444",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Solve the equation $\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$ Solve the equation $$\sqrt{3x+2}+\dfrac{x^2}{\sqrt{3x+2}}=2x$$
We have $DM:3x+2>0,x>-\dfrac23, x\in DM=\left(-\dfrac23;+\infty\right)$, so we can multiply the whole equation by $\sqrt{3x+2}\ne0$. Then we will have $$3x+2+x^2=2x\sqrt{3x+2}\\x^2+3x+2-2x\sqrt{3x... | There is a clever solution here: it requires using AM-GM.
Assume that $\sqrt{3x+2} \geq 0$, which clearly must be true. Then, By AM-GM, we have
$$
\sqrt{3x+2} + \frac{x^2}{\sqrt{3x+2}} \geq 2\sqrt{x^2} = 2x
$$
with equality if and only if $\sqrt{3x+2} = \frac{x^2}{\sqrt{3x+2}}$. This is true when $x^2 = 3x+2$, or $x = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4309941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that...
For positive real numbers $a,b,c$ such that $a+b+c=1$. Prove that $$\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$$
Here's what I've done so far:... | $\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)$
$=\frac{b+c}{a}+\frac{a+c}{b}+\frac{a+b}{c}+6\ge2\sqrt2\Big(\sqrt\frac{1-a}{a}+\sqrt\frac{1-b}{b}+\sqrt\frac{1-c}{c}\Big)=$ (1)
We know that $a+b+c=1$, so
$b+c=1-a$,
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4310082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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A special inequality (related to 2021) Let $x,y,z\in[0,1]$ such that: $$x+y+z\leq 1+2xyz$$
I want to prove the following inequality (for a student in high school):
$$x^{2021}+y^{2021}+z^{2021}\leq 1+2\sqrt{xyz}^{2021}$$
Edit:
This question was sent to me by a student from high school, his teacher ask them to look for ... | Alternative proof:
Let us prove that, for any integer $n \ge 2$,
$$x^n + y^n + z^n \le 1 + 2 \sqrt{xyz}^n.$$
Let
$$f(n) = 1 + 2\sqrt{xyz}^n - x^n - y^n - z^n.$$
We have
$$f(n + 1) - f(n) = 2\sqrt{xyz}^{n + 1} - x^{n + 1} - y^{n + 1} - z^{n + 1} - 2\sqrt{xyz}^n + x^n + y^n + z^n,$$
and
\begin{align*}
&[f(n + 2) - f(n +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4313282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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$\int_{-\infty}^\infty f(x)dx$ vs. $\lim_{b\rightarrow\infty}\int_{-b}^b f(x)dx$ for odd $f(x)$. I think the below one is a problem which demonstrates a beautiful subtlety (at least for someone like me who's a beginner).
Show that
$$
\int_{-\infty}^\infty \frac{2x}{1+x^2}dx
$$
diverges and
$$
\lim_{b\rightarrow\infty}\... | No. The symbols $a$ and $b$ are just variables. You could also write
$$
\int_{-\infty}^{\infty} \frac{2 x}{1 + x^2} \ \text{d}x= \left(\lim_{c \to \infty} - \ln(1 + c^2)\right)
+ \left(\lim_{c \to \infty} \ln(1 + c^2)\right).
$$
But this is not equal to
$$
\lim_{c \to \infty} \left(- \ln(1 + c^2) + \ln(1 + c^2)\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4314085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Number of solution of ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ Number of solution of the equation ${\left( {\sin x - 1} \right)^3} + {\left( {\cos x - 1} \right)^3} + {\sin ^3}x = {\left( {2\sin x + \cos x - 2} \right)^3}$ in the interv... | Hint
$$(a+b+c)^3-c^3=(a+b+c-c)((a+b+c)^2+(a+b+c)c+c^2)$$
$$a^3+b^3=?$$
Clearly $a+b$ is a common factor.
By symmetry, we can claim that the other two factors of $$(a+b+c)^3-a^3-b^3-c^3$$ will be $b+c,c+a$
Alternatively simplify
$$((a+b+c)^2+(a+b+c)c+c^2)-(a^2-ab+b^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4322038",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
How do we calculate the expected value of $Y$? Let $\Omega = \{−1, 0, 1, 3\}$ with the probability function $p$ given by
I want to calculate the expected value and the variance of $X(\omega)=\omega$, $Y(\omega)=5\omega-3$, $Z(\omega)=(\omega-1)^2$.
For $X$ I have done the following :
We have that $X\in \{-1,0,1,3\}$.
... | You can use the formula $$\mathbb{E}[aX+b] = a\mathbb{E}[X]+b$$ to see that
$$\mathbb{E}[Y] = 5\cdot \frac{13}{10} -3 = \frac{7}{2}\ .$$
However, you can also calculate it directly because you do have the probabilities for $Y$. For example
$$\Pr(Y=-8) = \Pr(5\omega-3 = -8) = \Pr(\omega=-1)\ .$$
Since $Y$ is defined in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Show that $\ln(\sin(-20)+21)>3$ by hand Hi I hope this problem is new :
Show that :
$$\ln(\sin(-20)+21)>3$$
I have tried the power series of $\ln(x)$ and $\sin(x)$ without success because it's becomes hard by hand .
You can also find the inequality due to Michael Rozenberg wich states for $x\geq 1$ :
$$\ln(x)\leq (x-1)... | Let's approximate $\sin(-20)$ using a Taylor series centered near $-20$. Since $\pi \approx 3.14$, we have $-6.5\pi \approx -20.41$, so this is what we'll use as the center of the approximation. Note that $\sin(-6.5\pi) = -1$ and $\cos(-6.5\pi) = 0$. So we have
$$\sin(x) = \sum_{k=0}^{\infty} (-1)^k \frac{\sin(-6.5\pi)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4323774",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the area of the shaded region in the triangle below? For reference: In figure $G$ is the centroid of the triangle $ABC$; if the area of the $FGC$ triangle is $9m^2$, the area of the FGB triangle is $16m^2$
Calculate the area of the shaded region. (Answer:$7m^2$)
If possible by geometry
My progress:
$S_{FGC... | $\begin{array}{} A=(p,q) & B=(a,0) & C=(0,0) & F=(x,y) \end{array}$
$S_{FGC}=|\frac{1}{2}\left| \begin{array}{} x & y & 1 \\ x_{G} & y_{G} & 1 \\ 0 & 0 & i \\ \end{array} \right| |=9 \\ S_{FGB}=|\frac{1}{2}\left| \begin{array}{} x & y & 1 \\ x_{G} & y_{G} & 1 \\ a & 0 & i \\ \end{array} \right| |=16$
$\begin{array}{} |... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4326621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
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Prove instability using Lyapunov function Consider the system:
\begin{align}
x' &= x^3 + xy \\
y' &= -y + y^2 + xy - x^3 \\
\end{align}
I want to prove the origin is an unstable point by using the Lyapunov function $V(x,y) = \tfrac{x^4}{4} - \tfrac{y^2}{2}$ (this is a hint provided in the exercise).
In order to use Che... | The dominant term is $x^6+ x^3 y+ y^2$. This term is positive (complete the square). All other terms are small when $x,y,$ are small, so they can be controlled. Here are the details.
Write
$$ \begin{aligned} V'(x,y) &= x^6 + x^4y + y^2 - y^3 -xy^2 +
x^3y\\ &= x^6 + (1+x)x^3y +(1 - x-y)y^2 \\ & \ge x^6 +(1+x)x^3y + \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4328531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Smith normal form of matrix over $Z$? I was wondering if someone could help me find the Smith normal form of the matrix A over $Z$ defined as follows:
$A =
\begin{bmatrix}
1 & 1 & 1 & 1 & 1\\
1 & 2 & 4 & 8 & 16\\
1 & 3 & 9 & 27 & 81\\
1 & 4 & 16 & 64 & 256\\
1 & 5 & 25 & 125 & 625\\
\end{bmatrix}$
Clearly this is the s... | Define
$$P := \begin{bmatrix} 1 & 0 & 0 & 0 & 0 \\ -1 & 1 & 0 & 0 & 0 \\ 1 & -2 & 1 & 0 & 0 \\ -1 & 3 & -3 & 1 & 0 \\ 1 & -4 & 6 & -4 & 1 \end{bmatrix}.$$
Then
$$P A = \begin{bmatrix} 1 & 1 & 1 & 1 & 1 \\ 0 & 1 & 3 & 7 & 15 \\ 0 & 0 & 2 & 12 & 50 \\ 0 & 0 & 0 & 6 & 60 \\ 0 & 0 & 0 & 0 & 24\end{bmatrix} =
\begin{bmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4329594",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Solving the trigonometric equation $\sqrt{2}\sin(2x)=-\sqrt{3\sin(x) + 3\cos(x) + 8\cos^4(x-\pi/4)}$ My problem is to solve the following equation:
$$\sqrt{2}\sin(2x)=-\sqrt{3\sin(x) + 3\cos(x) + 8\cos^4(x-\pi/4)}$$
I've narrowed it down to
$$3\sin(x) + 3\cos(x) + 2 + 8\sin(x)\cos(x) = 0.$$
That's as far as... | Using auxiliary angle gives
$$
\begin{aligned}
\sqrt{2} \sin (2 x) &=-\sqrt{3 \sin x+3 \cos x+8 \cos ^{4}\left(x-\frac{\pi}{4}\right)} \\
&=-\sqrt{3 \sqrt{2} \cos \left(x-\frac{\pi}{4}\right)+8 \cos ^{4}\left(x-\frac{\pi}{4}\right)}
\end{aligned}
$$
Let $ y=x-\dfrac{\pi}{4}$, then
$$\sqrt{2} \cos \left(2 y\right)=-\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330301",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Proving $1/(a+b) + 1/(b+c) + 1/(c+a) > 3/(a+b+c)$ for positive $a, b, c\,$? I have to prove that:
$$ \frac{1}{a+b} + \frac{1}{b+c} + \frac{1}{c+a} > \frac{3}{a+b+c},$$
where $a, b , c$ are positive real numbers.
I am thinking about using arithmetical and geometrical averages:
$$A_{3} = \frac{a_{1}+a_{2}+a_{3}}{3},$$
... | Other solution is Cauchy-Schwarz, that can be applied in one of its several versions.
*
*This is the standard one:
$[(\frac{1}{\sqrt{a+b}})^2+(\frac{1}{\sqrt{b+c}})^2+(\frac{1}{\sqrt{c+a}})^2][((\sqrt{a+b})^2+(\sqrt{b+c})^2+(\sqrt{c+a})^2]\ge 3^2$
From here the factor $9/2$, as already found, follows.
*
*Titu's in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4330873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Closed form for the sum of the series $\sum^\infty_{n=1} \left( {(-1)^n}\left( 1+n\ln(\frac{2n-1}{2n+1}) \right) \right)$ If $a>2$ then $\sum\limits^\infty_{n=1} \left( {(-1)^n}\left( 1+n\ln(\frac{an-1}{an+1}) \right) \right)$ diverges by divergent test. Does it converge if $a=2$? Is it possible to find an exact form f... | One has
\begin{align*} {(-1)^n}\left( 1+n\ln\left(\frac{2n-1}{2n+1}\right) \right) &= {(-1)^n}\left( 1+n\ln\left(1 -\frac{2}{2n+1}\right) \right)\\
&= {(-1)^n}\left( 1+n \left(- \frac{2}{2n+1} - \frac{2}{(2n+1)^2} + O \left( \frac{1}{n^3}\right)\right) \right)\\
&= \frac{(-1)^n}{2n+1}\left( \frac{1}{2n+1} + O \left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4346884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show that $\frac{1}{2^{2n+1}}\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]= \frac{1}{2}$ I have a hard time showing that that
$$ \frac{1}{2^{2n+1}}\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]= \frac{1}{2}$$
Namely, I try to show hat
$$\sum_{i=0}^n \le... | We will show that $\sum_{i=0}^n \left[\binom{n}{i} \cdot \sum_{j=i+1}^{n+1} \binom{n+1}{j} \right]=4^n$
Note that $\sum_{j=i+1}^{n+1}\binom{n+1}{j}=\sum_{j=0}^{n-i}\binom{n+1}{j}$.
Now consider the power series $(1+x)^{n+1}=\binom{n+1}{0}+\binom{n+1}{1}x+\binom{n+1}{2}x^2+\ldots+\binom{n+1}{n+1}x^{n+1}$. We have that t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4347388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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What is the minimum of $BP+\frac{1}{2}CP$?
In $Rt\triangle ABC,\ \angle A=90^{\circ},\ AB=4,\ AC=6.$ The radius of $\odot A$ is $2$. What is the minimum of $BP+\frac{1}{2}CP$?
Actually, the question should be "What is the minimum of $BP+\frac{1}{3}CP$", and then it will be simple by using similarity. But I still won... |
Here is a solution using angles. Our aim was to investigate whether we could find an answer to OP’s question without much ado. But, we could not. Our answer is as laborious as that of J.W.L. Jan Eerland.
For brevity, we denote $BP = a$, $CP = b$, and $\measuredangle PAB = \omega$. Furthermore, let $L$ be the length, w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4352347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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Comparing $\frac {9}{\sqrt{11} - \sqrt{2}}$ and $\frac {6}{3 - \sqrt{3}}$ (without calculator) We want to compare the following two numbers:
$$x = \frac {9}{\sqrt{11} - \sqrt{2}} \quad\text{and}\quad y = \frac {6}{3 - \sqrt{3}}$$
My attempts so far:
I multiply both numerator and denominator of $x$ by $\sqrt{11} + \sqrt... | Not only you can multiply number by the same positive value, you can also add or subtract the same values. The general idea how to solve similar questions is to remove roots one by one (making sure that values to be squared are positive):
$$
a+\sqrt{p}+\sqrt{q}+\ldots+\sqrt{r} \qquad ?\qquad b+\sqrt{u}+\sqrt{v}+\ldots... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4354087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Equation in integers $x y^2 + 2 y z^2 + z x^2 = 0$ The equation in integers
$$x y^2 + 2 y z^2 + z x^2 = 0$$
has some integral solution $[1\colon 0\colon 0]$, $[0\colon 1\colon 0]$, $[0\colon 0\colon 1]$, $[1\colon -1 \colon 1]$. I wonder if there are other solutions ( non-proportional to the one listed).
Notes: If we ... | The case where $xyz=0$ is easy, so I will skip it. Assume then that $xyz\neq 0$.
Your equation is then $(\frac{x}{z})(\frac{y}{z})^2+2(\frac{y}{z})+(\frac{x}{z})^2=0,$ that is $(\frac{xy}{z^2})^2+2(\frac{xy}{z})+(\frac{x}{z})^3=0$.
Set $u=-\frac{x}{z}$ and $v=\frac{xy}{z}+1$. Then we get $v^2-1-u^3=0$.
According to LMF... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4356547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find $a$ such that ${x_1}^2+{x_2}^2$ takes the minimal value where $x_1, x_2$ are solutions to $x^2-ax+(a-1)=0$ DO NOT USE CALCULUS My thinking:
Let $x_1 = \frac{a+\sqrt{a^2-4a+4}}{2}$ and $x_2 = \frac{a-\sqrt{a^2-4a+4}}{2}$
By the AGM (Arithmetic-Geometric Mean Inequality):
We have
$x_1\cdot x_2\le \left(\frac{x_1\cdo... | It looks like we are asked to notice that $ \ x^2 - ax + (a-1) \ = \ (x - 1)·( \ x - [a - 1] \ ) \ = \ 0 \ \ , $ so that one solution to the quadratic equation is always $ \ x_1 \ = \ 1 \ \ , $ with the other being $ \ x_2 \ = \ a - 1 \ \ . $ The function in question is then $ \ x_1^2 + x_2^2 \ = \ 1 + (a - 1)^2 \ \ ,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4357090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How to solve $x$ in terms of $a$ and $b$ in equation $\frac{1}{2}+\frac{a}{b-a}e^{-bx}-\frac{b}{b-a}e^{-ax}=0$ ($x>0,a>0,b>0, a≠b$)? Suppose $x>0,a>0,b>0,a\neq b$. How to solve $x$ in terms of $a$ and $b$ in equation $\frac{1}{2}+\frac{a}{b-a}e^{-bx}-\frac{b}{b-a}e^{-ax}=0$? If a closed-form solution is not possible, a... | Consider that you look for the zero of function
$$f(x)=\frac{1}{2}+\frac{a}{b-a}e^{-bx}-\frac{b}{b-a}e^{-ax}$$ for which
$$f'(x)=\frac{a b }{b-a}\left(e^{-a x}-e^{-b x}\right)\qquad \text{and} \qquad f''(x)= \frac{a b }{b-a}\left(b e^{-b x}-a e^{-a x}\right)$$
The key points are
$$f(0)=-\frac{1}{2} \qquad \qquad f'(0)=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4359856",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Orbits of the action of $SL(n,\mathbb R)$ on $\mathbb R^n$. We know that $GL(n,\mathbb R)$ acts on $\mathbb R^n$ via left multiplication. We can easily see that there are two orbits viz $\{0\}$ and $\mathbb R^n-\{0\}$. Now we also know that if $G$ acts on $X$ and $H$ is a subgroup of $G$ then $H$ acts on $X$ and the $H... | Consider the action of $SL(n,\mathbb{R})$ in $\mathbb{R}^n$ via left multiplication.
If $n=1$, then $SL(1,\mathbb{R})\cong\{1\}$ and the action is trivial.
We will study $n=2$, the case $n>2$ is equal with some dots around. Take a generic vector $\begin{pmatrix} x\\y \end{pmatrix}\ne\begin{pmatrix} 0\\0 \end{pmatrix}$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4361619",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the inequality $I(n^2) \leq 2 - \frac{5}{3q}$ improve $I(q^k) + I(n^2) < 3$, if $q^k n^2$ is an odd perfect number with special prime $q$? Let $N = q^k n^2$ be an odd perfect number with special prime $q$ satisfying $q \equiv k \equiv 1 \pmod 4$ and $\gcd(q,n)=1$.
Denote the abundancy index of the positive integer... | Too long to comment :
There is no such upper bound of the form $2−\dfrac{cq+d}{q^2+aq+b}$ better than $2−\dfrac{2}{q+1}$.
Proof :
Suppose that there is $(a,b,c,d)$ such that
$$2 - 2\cdot\dfrac{q^k - 1}{q^{k+1} - 1} \leqslant 2 - \dfrac{cq+d}{q^2 + aq + b} \leqslant 2 - \dfrac{2}{q+1}$$
Then,
$$\dfrac{2q^k - 2}{q^{k+1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4362475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
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Fractions in Questions and Answers
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