Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Stuck solving $z^2$ + $zw^*$= $18$ and $2z^*$=$w^*(1−i)$ as a system of equations Let $z$ and $w$ be complex numbers that satisfy
$z^2$ + $z\overline{w}$= $18$
and $2\overline{z}$=$\overline{w}(1−i)$
with $\Re(z)>0$. Find $w$.
I tried to find $z$ by subbing $\overline{w}$ = $\frac{18-z^2}z$ into $2\overline{z}$=$\overl... | Preliminaries
$z = |z|e^{i\theta}\\
\bar z = |z|e^{-i\theta}\\
w = |w|e^{i\phi}\\
\bar w = |w|e^{-i\phi}\\
(1-i) = \sqrt 2 e^{-\frac {\pi}{4} i}$
Let's use the second equation to express $w$ in terms of $z.$
$2\bar z = \bar w(1-i)\\
2|z|e^{-i\theta} = (|w| e^{-i\phi})\sqrt 2 e^{-\frac {\pi}{4} i}\\
2|z|e^{-i\theta} = (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3802347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Is my calculation of the integral $\int \tan^{-1} x \, dx$ correct? Compute $\int \tan^{-1}x \,dx$.
First, set $u = \arctan(x)$ and $dv = dx$. We want to find $du$ and we already have $v = x$.
We start by taking the tangent of both sides, leaving us with
$$\tan(u) = x.$$
Next, using implicit differentiation, we get $\f... | Your explanation of the integration by parts, which you combine with a change of variable, is a little tedious.
Using $x=\tan(u)$ which implies $dx=(\tan^2(u)+1)\,du$,
$$\int\arctan(x)\,dx=\int u(\tan^2(u)+1)\,du=u\tan(u)-\int\tan(u)\,du$$
and back to the original variable with $u=\arctan(x)$,
$$\int\arctan(x)\,dx=x\ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805484",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $(a-b^2)b>0$, then $\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}+\sqrt[3]{a-\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}{3b}}}$ is rational From Hardy´s "A course of pure mathematics" 10th edition, problem 31 miscellaneous problems of chapter I.
If $(a-b^2)b>0$, then
$$
\sqrt[3]{a+\frac{9b^3+a}{3b}\sqrt{\frac{a-b^3}... | Hint . Put $$a=y^3$$
We get :
$$b\Bigg(\sqrt[3]{x^3+\Big(3+\frac{x^3}{3}\Big)\sqrt{\frac{x^3-1}{3}}}+\sqrt[3]{x^3-\Big(3+\frac{x^3}{3}\Big)\sqrt{\frac{x^3-1}{3}}}\Bigg)\quad (1)$$
Where $x=\frac{y}{b}$
Now if $b$ is rational the expression in front of $b$ must be a rational but I don't see how ...
In fact if $x=1$ the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3805796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Finding remainder of $123^{456}$ divided by 88 using Chinese Remainder Theorem I tried using Chinese remainder theorem but I kept getting 19 instead of 9.
Here are my steps
$$
\begin{split}
M &= 88 = 8 \times 11 \\
x_1 &= 123^{456}\equiv 2^{456} \equiv 2^{6} \equiv 64 \equiv 9 \pmod{11} \\
y_1 &= 9^{-1} \equiv 9^9 \equ... | By Euler's theorem, we first get $123^{40}\cong1\pmod{88}$, since $\varphi(88)=40$. This results in $35^{16}\pmod{88}$, easily.
Now we use CRT: $\begin{cases}x\cong 35^{16}\pmod8\\x\cong35^{16}\pmod{11}\end{cases}$.
So, $x\cong3^{16}\pmod8\implies x\cong1\pmod8$, and $x\cong2^{16}\pmod{11}\implies x\cong5^4\pmod{11}\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3806122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
In quadrilateral $ABCD$, $\angle BAC=\angle CAD=2\,\angle ACD=40^\circ$ and $\angle ACB=70^\circ$. Find $\angle ADB$.
Let quadrilateral $ABCD$ satisfy $\angle BAC = \angle CAD = 2\,\angle ACD = 40^\circ$ and $\angle ACB = 70^\circ$. Find $\angle ADB$.
What I tried
*
*Ceva’s Theorem (Trigonometry version)
*Try to c... |
Since $\angle ACB=\angle ABC=70^\circ$, the triangle $ABC$ is isosceles and $\;\overline{AB}=\overline{AC}$.
By applying the law of sines to the triangle $ACD$, we get that:
$\overline{AD}=\overline{AC}\cdot\cfrac{\sin\angle ACD}{\sin\angle ADC}=\overline{AC}\cdot\cfrac{\sin 20^\circ}{\sin 120^\circ}=\cfrac{2\overline... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3807102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Maximum value of $abc$ for $a+b+c=5$ and $a^2+b^2+c^2=11$
$a,b,c$ are three real numbers such that $a+b+c=5$ and $a^2+b^2+c^2=11$,
what's the maximum value of $abc$?
I thought of a way, $ab+bc+ca$ is not hard to find, $a,b,c$ satisfy the cubic equation $x^3 - 5 x^2 + 7 x - abc = 0$ , then use the discriminant of the ... | From condition we get $ab+bc+ca=7.$ Using the Cauchy-Schwarz inequality, we have
$$11 \geqslant a^2 + \frac{(b+c)^2}{2} = a^2 + \frac{(5-a)^2}{2},$$
so
$$a^2 + \frac{(5-a)^2}{2} \leqslant 11 \Rightarrow \frac 13 \leqslant a \leqslant 3.$$
Similar we get $\frac 13 \leqslant b,\,c \leqslant 3.$ Therefore
$$(a-3)(b-3)(c-3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$
If $a > b > 1$ and $\frac{1}{log_a(b)} + \frac{1}{log_b(a)} = \sqrt{1229}$ , find the value of :- $\frac{1}{log_{ab}(b)} - \frac{1}{log_{ab}(a)}$ .
What I Tried :- I tried ... | Instead of trying to find the values of $\log b$ and $\log a$, just look at the equation $\log_b a + \frac 1{\log_b a} = 1229$.
Now, if $(ab)^x = a$, then $a^xb^x = a$ so $b^x = a^{1-x}$, then $a = b^{\frac x{1-x}}$, therefore $\frac x{1-x} = \log_b a$, therefore $x = \frac{\log_b a}{\log_b a + 1} = \log_{ab} a$.
Simil... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3809407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$ without a calculator? Is there a way to get this value without calculator?
$$\sum_{a=1}^6\sum_{b=1}^6\sum_{c=1}^6\frac{ab(3a+c)}{2^a2^b2^c(a+b)(b+c)(c+a)}$$
I'm currently studying AIME.
| Let's consider the sum
$$
S(n) = \sum_{a = 1}^{n} \sum_{b = 1}^{n} \sum_{c = 1}^{n} \frac{ab(3a + c)}{2^a 2^b 2^c (a + b)(b + c)(c + a)}.
$$
The key is to look at what happens if we exchange the roles of $a$, $b$, and $c$. The sum is, for example, equal to
$$
\sum_{a = 1}^{n} \sum_{c = 1}^{n} \sum_{b = 1}^{n} \frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3810306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Complete $\int \frac{x^2}{\sqrt{9x^2-1}}dx$ I am trying to solve the integral $$\int \frac{x^2}{\sqrt{9x^2-1}}dx,$$ but I am not sure how to solve it.
I have thought substitute $x$ by $\frac{\sec(u)}{3}$. Then $dx = \frac{1}{3} \tan(u)\sec(u)$,
$\sqrt{9x^2 -1} = \sqrt{\sec^2(u) - 1} = \tan(u)$ and $u = \sec^{-1}$(3x)
$... | First Way
Break down the $\sec^3{x}$ and integrate by parts
$$ I = \int \sec^3{x} dx = \int \sec^2{x} \sec{x} dx $$
Recalling:
$$ \int u dv = uv - \int vdu$$
Useful tip: Always try to recognize famous derivatives. Note that $ \sec^2{x}$ seems familiar so lets
use $dv = \sec^2{x} dx$ and $ u = \sec{x}$ then $v = \tan{x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3812332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
I have to solve this irrational equation on $\mathbb{R}$ :
$$ \sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$$
I tried to do a substitution with $u=1-x$ but the only things I manage to reach is the following equation by squaring and using $(a-b)(a+b)=a^2 -b^2$:
$$ (\sqrt{1-x}-2x\sqrt{1-x^... | Let $x=\cos\alpha,$ where $\alpha\in[0,\pi]$.
Thus, we need to solve
$$\sqrt{1-\cos\alpha}=2\cos^2\alpha-1+2\cos\alpha\sqrt{1-\cos^2\alpha}$$ or
$$\sqrt{2\sin^2\frac{\alpha}{2}}=\cos2\alpha+2\cos\alpha\sin\alpha$$ $$\sqrt2\sin\frac{\alpha}{2}=\cos2\alpha+\sin2\alpha$$ or
$$\sin\frac{\alpha}{2}=\sin\left(\frac{\pi}{4}+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3815246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Is there a simpler expression for this piecewise-defined function? As a math-for-fun exercise, I challenged myself to find a globally-defined, everywhere-differentiable antiderivative of $\sqrt{1-\sin(x)}$. Because of the fundamental theorem of calculus, the problem boils down to evaluating the integral
$$f(x)=\int_{-\... | $$\sqrt{1-\sin x}=\sqrt{\cos^2\frac x2-2\cos\frac x2\sin\frac x2+\sin^2\frac x2}=\left|\cos\frac x2-\sin\frac x2\right|
\\=\sqrt2\left|\sin\left(\frac x2-\frac\pi4\right)\right|$$ and we can focus on the antiderivative of $|\sin t|=\pm\sin t$, where the sign alternates with period $\pi$.
For the antiderivative, we can ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3817952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Evaluating the limit $\lim_{h\to 0} \frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}$ Find the limit
$\lim_{h\to 0} \frac{\sqrt {1 - (x+h) ^2} - \sqrt {1-x^2}} {h}$
I tried substituting "h" and also multiplying $\frac{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}{\sqrt {1 - (x+h) ^2} + \sqrt {1-x^2}}$
| Let $f(x)=\sqrt{1-x^2}$
$$f'(x)=\lim_{h \to 0}\frac{\sqrt{1-({x+h})^2}-\sqrt{1-x^2}}{h}=\frac{-x}{\sqrt{1-x^2}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3818468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
finding a relation in $p:p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$
if $$p=\frac{1}{3}+\frac{1}{3}\frac{3}{6}+\frac{1}{3}\frac{3}{6}\frac{5}{9}+\cdots$$ and $$p^2+ap+c=0.$$ Find $a,c$ also $|c|=2$
My progress:The general term $$T_{m+1}=\frac{(1)(3)\cdots(2m+1)}{(3)(6)\cdots(3m+3)}$... | Since $p=\tfrac13\sum_{m\ge0}\tfrac{\binom{2m+1}{m}}{6^m}$ and $\binom{n}{k}=\oint_{|z|=1}\frac{(1+z)^ndz}{2\pi iz^{k+1}}$, simplifying a geometric-series integrand eventually gives$$p=-2\oint_{|z|=1}\frac{1+z}{1-4z+z^2}\frac{dz}{2\pi i}.$$The denominator's roots are $2\pm\sqrt{3}$, but only $2-\sqrt{3}$ is enclosed. S... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3823606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How does one go about solving $arg(\frac{z-2i}{z-6}) = \frac{1}{2}\pi$ This should give $$\frac{z-2i}{z-6} = bi$$
but solving that gives me $$z = \frac{-2b +6b^2-6bi +2i}{1+b^2}$$
and substituting $z$ for $x + yi$ gives me $x = \frac{-2b +6b^2}{1+b^2}$ and $y=\frac{-6b +2}{1+b^2}$
And I have no clue how to continue now... | You have not made a mistake. Observe that $x$ is a multiple of $y$:
$$x = \frac{-2b + 6b^2}{1 + b^2} = -b \cdot \frac{2 - 6b}{1 + b^2} = -by$$
which is what the answer states.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3826976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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If $a^2+b^2+c^2+d^2=4$ then $(a+2)(b+2)\geq cd$ Let $a,b,c,d$ be real numbers with $a^2+b^2+c^2+d^2=4$. Prove that $(a+2)(b+2)\geq cd$.
My approach: I have considered an expression $$\begin{aligned}(a+2)(b+2)-cd=&4+2(a+b)+(ab-cd)\\=&(a^2+b^2+c^2+d^2)+2a+2b+(ab-cd)\end{aligned}$$ I was trying to write it as the sum of s... | Because by AM-GM $$(a+2)(b+2)=ab+2(a+b)+4=$$
$$=\frac{1}{2}(2ab+4a+4b+4+a^2+b^2)+2-\frac{1}{2}(a^2+b^2)=$$
$$=\frac{1}{2}(a+b+2)^2+2-\frac{1}{2}(a^2+b^2)\geq$$
$$\geq2-\frac{1}{2}(a^2+b^2)=\frac{1}{2}(c^2+d^2)\geq |cd|\geq cd.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3830865",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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What is the solution to $\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$ $$\frac{{x}^{2}}{({x+1-\sqrt{x+1}})^2} < \frac{{x}^{2}+3x+18}{({x+1})^{2}}$$
I've already tried this many times but still not get the answer, this was one of my solution which I can't go further anymore.
$\frac{{x}^{2}}{(... | Note that $$x=(x+1)-1=(\sqrt{x+1}-1)(\sqrt{x+1}+1)$$ Thus the left-hand side simplifies to $$(\sqrt{x+1}+1)^2<\frac{x^2+3x+18}{x+1}$$ This doesn't look promising, but let's see if we can get an approximate solution by pulling out the leading term on the right-hand side. Well, $x^2+3x+18=(x+1)(x+2)+16$. Thus partial... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833035",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
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How to integrate hyperbolic function $\frac{1-\sin(x)}{1+\sin(x)}$ I'm trying to find a primitive of $ \frac{1-\sin(x)}{1+\sin(x)} $. By changing the variable to $t=\tan(\frac{x}{2})$ and letting $\sin(x) = \frac{2t}{1+t^2}$ I get the following integral:
\begin{align}
& \int \frac{1-\sin(x)}{1+\sin(x)} \, dx \\[8pt] = ... | $$\begin{align}\int\frac{1-\sin x}{1+\sin x}dx &= \int\left(\frac2{1+\sin x}-1\right)\,dx
\\&= \int\frac2{1+\frac{2t}{1+t^2}}\cdot\frac2{1+t^2}\,dt-x
\\&= \int\frac4{1+2t+t^2}\,dt-x
\\&= \int\frac4{(t+1)^2}\,d(t+1)-x
\\&= -\frac4{t+1}-x
\\&= -\frac4{\tan\frac x2+1}-x
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3833386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
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Can a triangle ABC be made if $\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$
I would like to know the simplest approach to find out whether a triangle ABC will be made if $$\frac{\cos A}{2}=\frac{\cos B}{3}=\frac{\cos C}{7}$$
The counterpart questions for sine and tangent can be handled as follows:
*
*If $\df... | Show for yourself that if $A,B,C$ are the angles of a triangle then
$$
\cos^2A + \cos^2B + \cos^2C = 1-2\cos A \cos B \cos C
$$
This is not very difficult, use the fact that $A+B+C = 180$ along with double angle identities.
Therefore, if each of those ratios equaled $k$ , we get $62k^2 = 1-84k^3$, which can be solved ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
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Need help with this proof via induction
Let $x_1,...,x_n > 0$.
I'm having troubles proving this formula via induction:
$$
(x_1 + \ldots + x_n)\left(\frac1{x_1} + \ldots + \frac1{x_n}\right) \ge n^2
$$
So far, I've managed to rewrite it like this:
$$
\sum_{k=1}^n x_k \sum_{k=1}^n \frac{1}{x_k} \ge n^2
$$
Also the base... | $$\tag{*} \underbrace{(x_1 + \ldots + x_n)}_{A}~\underbrace{\left(\frac1{x_1} + \ldots + \frac1{x_n}\right)}_B \ge n^2$$
$$
\begin{split}
(x_1 + &\ldots + x_n + x_{n+1})
\left(\frac1{x_1} + \ldots + \frac1{x_n} + \frac{1}{x_{n+1}}\right) \\
&= \left(A+x_{n+1}\right)\left(B+\frac{1}{x_{n+1}}\right) \\
&= AB + \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3834682",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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A nice way to remember trigonometric integrals? Is there a "nice" way to remember trigonometric integrals, beyond what is typically taught in a standard calculus class? I'm currently in Calculus II, and up to now I've found calculus rather accessible. I love that, at least in my classes, we learn the "how" and the "why... | The pythagorean theorem, as applied to trigonometry says
$\sin^2 \theta + \cos^2 \theta = 1$
This is the key piece of knowledge for these integrals.
The implications are:
$\cos \theta = \pm \sqrt {1-\sin^2 \theta}\\
\sin \theta = \pm \sqrt {1-\cos^2 \theta}\\
\tan^2 \theta + 1 = \sec^2 \theta$
How does this relate to t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3835081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Sketch the graph of the function $f(x) =\frac{(x+1)^3} {x^2 - x+1}$ I tried to follow the rules of sketching a graph, I found the intersection with the ox and oy in (0,1)and (-1,0) and I think that the oblique asymptote is y=x+4. What about the vertical ones? The roots of the denominator are complex ones. Can we say th... | You have $$f(x) =\frac{(x+1)^3} {x^2 - x+1} = x + 4 + \frac{6x - 3}{x^2-x+1}.$$
As $\lim\limits_{x \to \pm \infty} \frac{6x - 3}{x^2-x+1} = 0$, $y=x+4$ is indeed an asymptote of $f$. Also the sign of $\frac{6x - 3}{x^2-x+1}$ is the one of $6x-3$ as $x^2-x+1$ is always positive. So $f$ crosses its asymptote at $x = 1/2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3837386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Find the area bounded by the curve $x^4+y^4=x^2+y^2$ I am stuck with this problem which deals with evaluating an Area
The problem reads :
Find the area bounded by the curve $x^4+y^4=x^2+y^2$.
I tried factorizing the expression and expressing $y$ in terms of $x$, not able to proceed with that idea. Someone please help... | Expressing the curve in polar coordinates, we have\begin{align}x^4+y^4=x^2+y^2&\iff\rho^4\cos^4\theta+\rho^4\sin^4\theta=\rho^2\\&\iff\rho=\frac1{\sqrt{\cos^4\theta+\sin^4\theta}}.\end{align}Hence, the area that you're after is\begin{align}\int_0^{2\pi}\int_0^{1/\sqrt{\cos^4\theta+\sin^4\theta}}\rho\,\mathrm d\rho\,\ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3840121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Messy Gaussian Integral I am trying to understand how to better perform the following integral.
$$\int^{\infty}_{0} x^4 e^{\frac{-x^2}{\beta^2}}\mathrm{d}x$$
I've done a little research and found that $e^{-x^2}$ doesn't integrate easily, for it is the Gaussian integral. Many sources are pointing me to use polar coordin... | I just wanted to expand on Claude's answer. First a transformation:
$$\int_0^\infty x^n\exp\left(\frac{-x^2}{a^2}\right)\mathrm{d}x=a^{n+1}\int_0^\infty x^n\exp(-x^2)\mathrm{d}x$$
Let
$$I_n=\int_0^\infty x^n\exp(-x^2)\mathrm{d}x$$
Integration by parts. Let $u=\exp(-x^2)$, $\mathrm{d}u=-2x\exp(-x^2)\mathrm{d}x$, $\mathr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3843441",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Meaning of drawing any number of balls from an urn containing $n$ balls There is a question as follows:
From an urn containing $n$ balls any number of balls are drawn. Show that the probability of drawing an even number of balls is $\frac{2^{n-1}-1}{2^n-1}$
Firstly what does it mean to draw any number of balls from the... | since we need to find the probability of choosing even number of balls it means the probability is
$$ \frac{\binom{n}{2}+\binom{n}{4} + \ .... \ }{\binom{n}{1} + \binom{n}{2} + \binom{n}{3} + \ .... \ \binom{n}{n}}
$$
which equals $$ \frac{2^{n-1} \ - \ \binom{n}{0}}{2^n - \binom{n}{0}} $$
which immediatly yeilds you... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3845866",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Let ABCD be a trapezoid; find point M on AB, such that $DM+MC$ is minimal possible Let ABCD be a trapezoid with $A=B=90$, $AD=a$ and $AB=BC=2a$.
Find point M on AB, such that $DM+MC$ is minimal possible.
I have been trying to do this question, but without success. I have succeeded in proving that $DA+AC<DB+BC$ somethi... | Another way to solve -
Choose a point $M$ on $AB$ such that $AM = x, \,$ then BM = $(2a-x)$
$DM = \sqrt{AD^2 + AM^2} = \sqrt{a^2+x^2}$
Similarly $MC = \sqrt{BC^2 + BM^2} = \sqrt{4a^2+(2a-x)^2}$
$L = DM + MC = \sqrt{a^2+x^2} + \sqrt{8a^2+ x^2 - 4ax}$
To find min value of $L$,
$ \displaystyle \frac{dL}{dx} = \frac {x} {\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3846563",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Remainder of dividing polynomial of the $n$th degree
What is the remainder when dividing the polynomial
$$P(x)=x^n+x^{n-1}+\cdots+x+1$$ with the polynomial
$$x^3-x$$ if $n$ is a natural odd number?
So, what I know so far is:
$$P(x)=Q(x)D(x)+R(x)$$
In this case I'll call $Q(x) = x^3-x$
$$Q(x) = 0 \iff x=\pm1$$
So from... | Hint:
Experimenting the first few values of $n$: $\:n= 3,5,7,9,11$, you may conjecture the remainder for the general polynomial $P_n(x)=x^n+x^{n-1}+\dots+x^3+x^2+x+1\:$ is
$$R_n(x)=kx^2+(k+1)x+1,\quad\text{ where } k=\left\lfloor \frac n2 \right\rfloor.$$
and try to prove it by induction on $n$ (odd).
You can prove th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3849311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, what is $|a + b + c + d|$? From the Pre-Regional Mathematics Olympiad, 2019:
If $x = \sqrt{2} + \sqrt{3} + \sqrt{6}$ is a root of $x^4 + ax^3 + bx^2 + cx + d$, where $a, b, c$ and $d$ are integers, ... | The only part missing in the existing answer is the fact that the minimal polynomial of $a=\sqrt{2}+\sqrt{3}+\sqrt{6}$ has degree $4$, i.e. the degree of the extension $|\Bbb{Q}[a]:\Bbb{Q}|$ is $4$. Since $a$ is a root of an integer polynomial of degree $4$, the degree of the extension is $1, 2$ or $4$. It cannot be $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3851795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
What is $\tan \alpha$, if $(a+2)\sin\alpha +(2a - 1)\cos\alpha =2a + 1$? I tried the following:
$$\begin{aligned}a\sin\alpha +2\sin\alpha + 2a\cos\alpha - \cos\alpha &= 2a+1\\
a(\sin\alpha +2\cos\alpha)+(2\sin\alpha-\cos\alpha)&=2a+1\end{aligned}$$
Therefore,
$$\sin\alpha +2 \cos\alpha=2$$
$$2\sin\alpha - \cos\alpha=1$... | Hint:
Use Weierstrass substitution to form a quadratic equation $$\tan\dfrac\alpha2=t$$
$$(a+2)\cdot\dfrac{2t}{1+t^2}+(2a-1)\cdot\dfrac{1-t^2}{1+t^2}=2a+1$$
$$\iff-t^2(2a+1+2a-1)+2t(a+2) +2a-1-(2a+1)=0$$
$$\iff2at^2-t(a+2)+1=0$$
Now use $\tan2A=\dfrac{2\tan A}{1-\tan^2A}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3852291",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Consider a function $f(x)= \arcsin (\frac {2x}{1+x^2}) + \arccos (\frac{1-x^2}{1+x^2}) +\arctan (\frac{2x}{1-x^2})-a\arctan x$
Consider a function $f(x)= \arcsin (\frac {2x}{1+x^2}) + \arccos (\frac{1-x^2}{1+x^2}) +\arctan (\frac{2x}{1-x^2})-a\arctan x$, where $a$ is any real constant. Find the value of $a$ if $f(x)=0... | For any $\;x\in\left]-\infty,-1\right[\;,\;$ it results that
$\arcsin\left(\dfrac{2x}{1+x^2}\right)=-\pi-2\arctan x\;,$
$\arccos\left(\dfrac{1-x^2}{1+x^2}\right)=-2\arctan x\;,$
$\arctan\left(\dfrac{2x}{1-x^2}\right)=\pi+2\arctan x\;.$
Hence, for all $\;x\in\left]-\infty,-1\right[,\;$ it results that
$\arcsin\left(\dfr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3855490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How to solve this geometric+arithmetic recurrence I've been stuck on this recurrence closed form question for a while now:
$S(n)=9S(n-1)+4n, n > 1$
$S(1) = 4$
After expanding a couple iterations to find a pattern, I came up with this:
$4*9^{n-1}+4*(n*\sum_{k=0}^{n-2}9^k-\sum_{k=0}^{n-2}k*9^k)$
$=4*9^{n-1}+4*(n*\frac{9^... | The observed pattern is fine. We have
\begin{align*}
S(n)&=4\cdot 9^{n-1}+4\left(n\frac{9^{n-1}-1}{9-1}-\sum_{k=0}^{n-2}k\,9^k\right)\\
&=\frac{1}{2}9^{n-1}\left(n+8\right)-\frac{1}{2}n-4\sum_{k=1}^{n-2}k\,9^{k}\tag{1}
\end{align*}
In (1) we have collected terms and we start the index $k$ with $1$, since the term with ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3857650",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Prove $\forall z\in\mathbb C-\{-1\},\ \left|(z-1)/(z+1)\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$ I'm trying to prove
$$\forall z\in\mathbb C-\{-1\},\ \left|\frac{z-1}{z+1}\right|=\sqrt2\iff\left|z+3\right|=\sqrt8$$
thus showing that the solutions to $\left|(z-1)/(z+1)\right|=\sqrt2$ form the circle of center $-3$ and ... | We have that
$$\frac{z-1}{z+1}=\sqrt 2 e^{i\theta} \implies z=\frac{1+\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}}$$
and $$z+3= \frac{4-2\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}} \implies |z+3|^2=\frac{4-2\sqrt 2 e^{i\theta}}{1-\sqrt 2 e^{i\theta}} \frac{4-2\sqrt 2 e^{-i\theta}}{1-\sqrt 2 e^{-i\theta}} =\frac{24-16\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3860623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Where did I go wrong in applying the remainder theorem?
When the expression $3x^3 + px^2 + qx + 8$ is divided by $x^2 - 3x + 2$, the remainder is $5x + 6$. Find the values of $p$ and $q$.
I tried to answer this question using the factor theorem (remainder theorem in this case), but got the answer wrong:
$$ \text{Let}... | It's difficult to keep track of all the lines, but you have the correct idea. In one case we have
$$f(2)=5\cdot 2+6\Rightarrow 24+4p+2q+8=16\Rightarrow2p+q=-8$$
and in the other case we have
$$f(1)=5\cdot(1)+6\Rightarrow 3+p+q+8=11\Rightarrow p+q=0$$
From the second equation we get $q=-p$ so by plugging that in into th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3864529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Remainder Theorem: $f(x) = x^3+1$ divided by $\operatorname {d}(x)= x$ Take for example $\operatorname {f}(x) = x^3+1$ divided by $\operatorname {d}(x)= x-1$ $$\frac {x^3+1}{x-1}$$
The Remainder Theorem tells us that the remainder will be $$\operatorname {f}(zero-of-\operatorname {d}(x))$$
In this case the remainder sh... | Leave fractions aside. The remainder theorem tells you that, if $f(x)$ is a polynomial with integer coefficients and $a$ is an integer, then
$$
f(x)=(x-a)q(x)+f(a)
$$
so $f(a)$ is the remainder of the division by $x-a$ in the ring of polynomials with integer coefficients.
What you're conjecturing is that $f(a)$ should ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3864990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Solving limit - $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$ $\lim_{x\to0,y\to0}(x^2+y^2)^{x^2y^2}$
Since $x$ approaches $0$ and $y$ also approaches $0$ we can suspect that $0<x^2 + y^2<1$. For every $x,y\in\Bbb R$, we have that $\frac{1}{4}(x^2 + y^2)^2\geq x^2y^2$.
Now, $1\geq (x^2+y^2)^{x^2y^2}\geq (x^2+y^2)^{\frac{1}{4}(... | Here's another approach to this limit. Notice for any $(x,y)\neq (0,0)$ that $$
x^2y^2 \ln(x^2+y^2)=f(x,y)(x^2+y^2)\ln(x^2+y^2)$$ where $f(x,y)=\frac{x^2y^2}{x^2+y^2}$. Clearly $(x^2+y^2)\ln(x^2+y^2)\rightarrow 0$ as $(x,y)\rightarrow (0,0)$ while $f(x,y)$ is bounded on the punctured disc $x^2+y^2<1$, $(x,y)\neq (0,0)$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3865972",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve $\cos\theta-3\cos2\theta+\cos3\theta=\sin\theta-3\sin2\theta+\sin3\theta$ My attempt:
\begin{align*}
\cos\theta-3\cos2\theta+\cos3\theta&=\sin\theta-3\sin2\theta+\sin3\theta\\
\cos\theta-3\cos2\theta+4\cos^3\theta-3\cos\theta&=\sin\theta-3\sin2\theta+3\sin\theta-4\sin^3\theta\\
-2\cos\theta-3\cos2\theta+4\cos^3\t... |
$\cos\theta+\cos3\theta = 2\cos\theta\cos2\theta$
$\sin\theta+\sin3\theta = 2\sin2\theta\cos\theta$
So,
$\cos2\theta( 2\cos\theta-3) = \sin2\theta(2\cos\theta -3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869446",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Application of the Cauchy-Schwarz Inequality Need to prove the following:
$(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{1}{2}a^2+\frac{1}{3}b^2+\frac{1}{6}c^2$
using the Cauchy-Schwarz inequality however direct application yields $(\frac{1}{2}a+\frac{1}{3}b+\frac{1}{6}c)^2\leq \frac{7}{18}(a^2+b^2+c^2)$ which i... | Hint: $\frac 12a+\frac 13b+\frac16c=\frac16a+\frac16a+\frac16a+\frac16b+\frac16b+\frac16c$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3869873",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve for closed form $a_j$ given $a_j = 1 + pa_{j+1} + (1-p)a_{j-1}$ where $a_1 = 0$.
Solve for a closed form expression for $a_j$ given constant $p \in (0,1)$ and $q=1-p$ and:
\begin{align*}
a_j &= 1 + pa_{j+1} + qa_{j-1} \\
\end{align*}
Given that $a_1 = 0$.
The text says to use the hint that $a_j = c(1-j)$, but... | Let $b_j = a_{j+1} - a_j$, $b_j$ satisfies
$$-pb_{j} + qb_{j-1} = 1$$
Let $c_j = b_j - \frac{1}{q-p}$, $c_j$ satisfies
$$c_j = \frac{q}{p}c_{j-1}$$
So
$$c_{j} = c_1\left(\frac{q}{p}\right)^{j-1}$$
We deduce
$$a_{j} = a_1 + \frac{j-1}{q-p} + c_1\sum_{k=0}^{j-2}\left(\frac{q}{p}\right)^{k}$$
with the convention
$$\sum_{k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3871293",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Follow-up 'Diophantine' question: extrapolating to the general case I came across a problem involving a certain Diophantine equation a few days ago. I learnt quite a few extremely helpful things about them on this thread here, which I started: A model that can be followed when solving Diophantine equations - ideas? -
F... | $$ 3xy - 2px - py = 0$$
$$ 9xy - 6px - 3py = 0$$
$$ 9xy - 6px - 3py + 2p^2 = 2p^2$$
$$ (3x-p)(3y-2p) = 2p^2$$
\begin{array}{|c|c|c|c|c|c|}
\hline
&\color{red}{(3x-p)}\color{blue}{(3y-2p)} & x & y\\ \hline
A&\color{red}{1}\cdot\color{blue}{2p^2} & (p+1)/3 & 2p(p+1)/3 & \color{green}{?}\\ \hline
B&\color{red}{2p^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3875009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Finding expectation of minimum of $(X,Y)$ where $(X,Y)$ is bivariate normal distribution. Let $(X,Z)$ be bivariate normal with parameters
$\mu_X := E(X) = 1, \mu_Z := E(Z ) = 1, \sigma_X^2 := Var(X) = 1$,
$ \sigma_Z^2
:= Var(Z ) = 1$,
and the correlation coefficient of (X, Z ) is $\rho$ with density
$$f(x,z)=\dfrac{1}{... | You can use the following
$$\min(X,Z) = \frac{(X+Z)}{2}-\frac{|X-Z|}{2} $$
the distribution of $W =(X -Z)$ is normal with mean $0$ and variance $2(1-\rho)$
Now this is a random variable and we can take out the distribution of $|W|$ as
$$\mathbb{P}(|W|\leq t) = \mathbb{P}(W\leq t)-\mathbb{P}( W \leq -t)=F_W(t) - F_W(-t)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3876934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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General tribonacci explicit formula I have been thinking about finding an explicit formula for the tribonacci numbers, where, namely:
$$a_n = a_{n-1}+a_{n-2}+a_{n-3}$$
and $a_1 = 0, a_1 = 1, a_2 = 1.$ Obviously, these beginning terms can be shifted, but we'll leave them as such for now.
This has proven difficult, and I... | Please see https://www.fq.math.ca/Scanned/36-2/wolfram.pdf in The Fibonacci Quarterly, sub-section 6.2.2. That approach also considers having initial functions instead of any given initial values.
Changing the constant coefficients to $x$, $y$ and $z$ from 1 can be solved similarly.
Specifically, we are seeking a solut... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3877319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find the value of $x^5 + \frac{1}{x^5}$ - question about correctness of method The task is: if $x+ \frac{1}{x}= 1$ find $x^5 + \frac{1}{x^5} $.
I used the binomial formula and proved that $x^5 + \frac{1}{x^5} = 1$, but I have a question about following method, I am not sure if it's correct. If I take square of the th... | $x+\frac{1}{x} =1$
$\Rightarrow $ $ (x+\frac{1}{x}) ^5=1$
$\Rightarrow $ $x^5+(\frac{1}{x})^5 +5(x^3+(\frac{1}{x})^3)+10
(x+\frac{1}{x})=1$
$\Rightarrow $ $x^5+(\frac{1}{x})^5=-9-5(x^3+(\frac{1}{x})^3)(*)$
Let's find :$ x^3+(\frac{1}{x})^3$
$ (x+(\frac{1}{x}))^3 =1$
$\Rightarrow $ $x^3+(\frac{1}{x})^3+3(x+\frac{1}{x})... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3878620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
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Maximising volume of a cylinder when surface area fixed I know how to start off this problem but get bogged down when it comes to differentiating at the end.
A right circular cylinder is of radius r cm. and height pr cm. The total surface area of the cylinder is $S cm^2$ and its volume is $V cm^3$.
Find an expression f... | The first term of your derivative is wrong.$$\begin{align*}&\frac{dV}{dp}=\frac d{dp}\left[\frac{S\sqrt S}{2\sqrt{2\pi}}\frac p{(1+p)^{3/2}}\right]=\frac{S\sqrt S}{2\sqrt{2\pi}}\left[\frac1{(1+p)^{3/2}}-\frac{3p/2}{(1+p)^{5/2}}\right]\\&\frac{dV}{dp}=\frac{S\sqrt S}{4\sqrt{2\pi}}\left[\frac{2-p}{(1+p)^{5/2}}\right]\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3882241",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find positive definite $X$ such that $AXA^T=\alpha X$. Given an invertible real matrix $A\in\mathbb{R}^n$, is it possible to obtain a (real) positive definite matrix $X$ (and scalar $\alpha>0$) such that
$$
AXA^T=\alpha X?
$$
Attempt: The previous equation can be written as $AXA^T-X +Q=0$ with $Q=(1-\alpha)X$. So, if $... | Consider $A = \pmatrix{2 & 1 \\ 0 & 1}$. Then you seek a matrix
$$
X = \pmatrix{a & b \\ c & d}
$$
and a constant $K$ such that
$$
\pmatrix{2 & 1 \\ 0 & 1}\pmatrix{a & b \\ c & d}\pmatrix{2 & 0 \\ 1 & 1} = K \pmatrix{a & b \\ c & d}
$$
i.e.,
$$
\pmatrix{2a+c & 2b+d \\ c & d}\pmatrix{2 & 0 \\ 1 & 1} = K \pmatrix{a & b \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3883198",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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An integer $n \geq 2$ is called square-positive- proof? An integer $n \geq 2$ is called square-positive if there are $n$ consecutive positive integers whose sum is a square. Determine the first four square-positive integers.
So I have found the first four square-positive numbers, but I need to prove that why it $4$ is ... | We prove the following result:
If $n=2^b \cdot d, d \text{ is odd, then } n \text{ is square-positive if and only if } b=0 \text{ or } b \text{ is odd}$.
$\underline{\text{Case 1}}$: $b$ is even and $b>0$. Let $b=2c$. We prove by contradiction $n$ cannot be square-positive. If it were, then there exist $n$ consecutiv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3885279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Prove by induction that $\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$ What would be the right way to solve this by induction proof?
$$\frac{1}{2n}\leq\frac{1\text{·}3\cdot5\text{·}\ldots\text{·}(2n-1)}{2+4+6+\ldots+2n}$$
This is what I've done (reference https://www.slader.com/d... | Let $\varphi, \psi:\mathbb N\to\mathbb N$ given by
\begin{aligned}
\varphi(n) &= \sum_{k=1}^n 2k = n(n+1)\\
\psi(n) &= \prod_{k=1}^n (2k-1)\\
\end{aligned}
and notice that
\begin{aligned}
\varphi(n+1) &= \frac{n+2}n\varphi(n)\\
\psi(n+1) &= (2n+1)\psi(n)\\
\end{aligned}
If you assume that, for a certain $n\in\mathbb N$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3886912",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Why using fewer terms of Taylor series doesn't give $0/0$ but gives a wrong answer? I was reading a Calculus book and I saw this problem which looks easy:
$$\lim _{x \rightarrow 0} \frac{2 x \cos x- \sin 2x}{x^3} = ?$$
It's a 0/0 limit and it's using some of the Taylor series of $\sin$ and $\cos$ expressions to solve t... | In all the cases we should use remainder to proceed properly as follows
$$\frac{2 x \cos x- \sin 2x}{x^3}=\frac{2 x \left(1-\frac12 x^2+O(x^3)\right)- \left(2x-\frac16 (2x)^3+O(x^4)\right)}{x^3}=$$
$$=\frac{2x-x^3-2x+\frac43x^3+O(x^4)}{x^3}=\frac13+O(x) \to \frac13$$
without remainder we can easily get wrong with the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3890092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Finding $\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+ \sin(x))} \right)$ I am trying to evaluate the following limit:
$$L=\lim_{x \to 0} \left( \frac{\sin(x)-3\sinh(x)+2x}{x^2(\tanh(2x)+\sin(x))} \right)$$
Begin by rewriting the limit as:
$$L=\frac{\lim\limits_{x \to 0}\left(\cfrac{\sin(x)-3\sinh(x)... | $$\sinh x=x+\frac{x^3}{6}+O\left(x^4\right)$$
$$\tanh(2x)=2 x-\frac{8 x^3}{3}+O\left(x^4\right)$$
$$\sin x=x-\frac{x^3}{6}+O\left(x^4\right)$$
The limit can be rewritten as
$$\frac{x-\frac{x^3}{6}-3 \left(\frac{x^3}{6}+x\right)+2 x}{x^2 \left(2x-\frac{8 x^3}{3}+x-\frac{x^3}{6}\right)}\to -\frac{2}{9}\text{ as }x\to 0$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3891527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
$(ab+bc+ca)^3=abc(a+b+c)^3$, prove that $a,b,c$ are in $G.P.$ Suppose $a,b,c$ are non-zero real numbers such that
$$(ab+bc+ca)^3=abc.(a+b+c)^3$$
Prove that $a,b,c$ must be terms of a $G.P.$
I simplified this equation too
$$(ab)^3+(bc)^3+(ca)^3=abc.(a^3+b^3+c^3)$$
I tried to subtract $3(abc)^2$ from both sides and it g... | This one is a long-way of solving (With a very obvious result).
\begin{gather*}
\implies(ab+bc+ca)^3-abc(a+b+c)^3=0\\
\end{gather*}
\begin{multline*}
\implies (a^3b^3+b^3c^3+c^3a^3+3a^2b^3c+3a^2bc^3+3a^3b^2c+3ab^2c^3+3a^3bc^2+3ab^3c^2)-\\abc(a^3+b^3+c^3+3a^2b+3ab^2+3b^2c+3bc^2+3c^2a+3ca^2)=0
\end{multline*}
After simpl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3893615",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Why is $\sqrt{n} + \sqrt{n + 1} > 2\sqrt{n}$? I am trying to prove an inequality, but that itself is not the question here. Looking around for ideas how to proceed I found solutions for the same problem.
But in several occasions I couldn't follow the answers because of this step.
In both of these excerpts (taken from a... | We have that
$$\sqrt{n} + \sqrt{n + 1} >\sqrt{n}+\sqrt{n}= 2\sqrt{n}$$
since $\sqrt{n + 1}>\sqrt{n}$ or also dividing by $\sqrt n$
$$\sqrt{n} + \sqrt{n + 1}> 2\sqrt{n} \iff 1+\sqrt{1+\frac1n}>2$$
which is true since $\sqrt{1+\frac1n}>1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3895572",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $n\leq a_1+a_2+...+a_n \leq n+1$ If $a_1=2,a_{n+1}=\sqrt{a_n+8}-\sqrt{a_n+3}$. Prove that $n\leq a_1+a_2+...+a_n \leq n+1$ for every $n\ge1$ and $\lim a_n=1$.
I have showed that by squaring and inequality techniques:
*
*$a_i<\sqrt{3}$ for every $i>1$.
*If $a_i>1$ then $a_{i+1}<1$ for every $i\ge 1$
I thi... | We have
$
a_{n+1}-1 = (a_n-1) \left( \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right)
$
and
$(a_{n+1}-1)+(a_n-1) = (a_n-1) \left( 1+ \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right)$,
a) $ \left| \cfrac{1}{3+\sqrt{8+a_n}} - \cfrac{1}{2+\sqrt{3+a_n}} \right| < \left| \cfrac{1}{3+\sqrt{8+a_n}} +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3896187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Which values of $a$ and $b$ will give $x^{2}+2$ as a factor of $x^{17}+ax+b$ over $\mathbb{Z}_{3}$? The question:
Let $x^{17}+ax+b$ be a polynomial over $\mathbb{Z}_{3}$. For which values of $a$ and $b$ will $x^{2}+2$ be a factor?
I know that I can find a solution by a brute force technique, but is there a more efficie... | If $x^2 + 2$ is a factor of
$p(x) = x^{17} + ax + b, \tag 0$
any zero of $x^2 + 2$ must also satisfy $p(x)$;. the zeroes of $x^2 + 2$ in $\Bbb Z_3$ are $1$ and $2$:
$1^2 + 2 = 3 \equiv 0 \mod 3, \tag 1$
$2^2+ 2 = 4 + 2 = 6 \equiv 0 \mod 3; \tag 2$
indeed
$(x - 1)(x - 2) = x^2 - 3x + 2 = x^2 + 2 \mod 3; \tag 3$
if $1$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3897132",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Volume of solid rotated about the x-axis I am to find the volume of the area $R$ bounded by the curve $x=y^2+2$, $y=x-4$ and $y=0$.
I have already found the points of intersection by first setting the lines equal to each other and used the quadratic formula:
\begin{align*}
y^2+2=y+4
\\-y^2+y+2=0
\\
\\y_... | Assuming $R$ is the region in the first quadrant (i.e. with $x\ge0$ and $y\ge0$), then the volume - using cylindrical shells - is
$$\begin{align}
V&=2\pi\int_0^2y((y+4)-(y^2+2))\,\mathrm dy\\[1ex]
&=2\pi\int_0^2(2y+y^2-y^3)\,\mathrm dy\\[1ex]
&=2\pi\left[y^2+\frac{y^3}3-\frac{y^4}4\right]_0^2\\[1ex]
&=2\pi\left(2^2+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3900585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How to solve this equation with matrices can you please give me some hints to solve the following? I really don't know how to start.
$$X^2= \begin{pmatrix}
6 & 2 \\ 3 & 7
\end{pmatrix}.$$
I tried to express this matrix as $4\cdot I + \begin{pmatrix}
2 & 2 \\ 3 & 3
\end{pmatrix}$ And somehow solve it, but I really have ... | Let $$X=\begin{bmatrix} a & b \\ c & d \end{bmatrix} \implies X^2=\begin{bmatrix} a^2+bc & b(a+d) \\ c(a+d) & d^2+bc\end{bmatrix}=\begin{bmatrix} 6 & 2\\ 3 & 7 \end{bmatrix}.$$
$$\implies a^2+bc=6~~~(1), b(a+d)=2~~~(2), c(a+d)=3~~~(3), d^2+bc=7~~~(4)$$
From (2) and (3) $b/c=2/3$, let $b=2k, c=3k$. From (1) and (4), we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Point $M$ lies inside $\triangle ABC$, $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC)$
In $\Delta ABC, \angle CAB = 30^\circ$ and $\angle ABC = 80^\circ$. Point $M$ lies inside the triangle such that $\angle MAC = 10^\circ$ and $\angle MCA = 30^\circ$. Find $(180^\circ - \angle BMC... | Let $\angle MBA = \theta$, $\angle MBC = 80 - \theta$
Using trig form of Ceva's theorem,
$$ \sin MAC \sin MCB \sin MBA = \sin MCA \sin MBC \sin MAB$$
$$ \sin 10 \sin 40 \sin \theta = \sin 30 \sin (80-\theta) \sin 20 $$
$$ 4\sin 10 \cos 20 =\dfrac{\sin (80-\theta)}{\sin \theta} $$
$$ 2(\sin 30 - \sin 10) =\sin 80 \cot \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3901672",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How to use matrix language to describe the solution process? I see that the linear equations $\left\{\begin{array}{l}
2 x+y-t=-2 \\
3 y+z+2 t=3
\end{array}\right.$ can be solved in the following way:
I have uploaded a picture to give the details of the algorithm for the mathematical solution. It is obvious that this m... | I don't understand your approach in the picture for to solve this problem. However, this problem in the matricix language is easy. Note that if $$(S.E.L): \left\{\begin{aligned} 2x+y+0z-t=-2 \\ 0x+3y+z+2t=3 \end{aligned} \right.$$
so, we can write the S.E.L in the form $$AX=B \quad \text{and} \quad (A|B)$$
Taking the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Perfect square involving the exponential law If $n$ is a natural number, and $2^{10} + 2^{13} + 2^n$ is a perfect square, what is the value of $n$?
I've attempted to factor out $2^{10}$ and got $2^{10}(1 + 2^3 + 2^{n-10})$. How can I move further?
| We can factor out the $2^{10}$ first:
$$2^{10}(1+2^3+2^{n−10})$$
We can simplify $2^3 = 2\cdot{2}\cdot{2} = 8$. Also, because $2^{10}$ is a perfect square, we can remove it from the equation.
$$1+8+2^{n-10}=9+2^{n-10}$$
Now we can add a new variable, $m$, where $m$ can be any positive integer. We can rewrite our equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3908331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
complex numbers exponential form I wish to show that $\cos^2(\frac{\pi}{5})+\cos^2(\frac{3\pi}{5})=\frac{3}{4}$
I know the solutions to $z^5+1=0$ are $-1$, $e^{i\frac{\pi}{5}}$, $e^{-i\frac{\pi}{5}}$, $e^{i\frac{3\pi}{5}}$, $e^{i\frac{-3\pi}{5}}$ and that
$z^5+1=(z+1)(z^4-z^3+z^2-z+1)$
which means that the solutions to... | Like Quadratic substitution question: applying substitution $p=x+\frac1x$ to $2x^4+x^3-6x^2+x+2=0$
Divide both sides by $z^2$ and replace $2u=z+\dfrac1z$
The roots of $$(2u)^2-2-2u+1=0$$ are $$\cos\frac\pi5,\cos\frac{3\pi}5$$
Now use https://mathworld.wolfram.com/VietasFormulas.html
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3909477",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Prove that the plane through the point $(\alpha, \beta, \gamma)$ and the line $x = py + q = rz + s$ Prove that the plane through the point $(\alpha, \beta, \gamma)$ and the line $x = py + q = rz + s$ is
$${\begin{vmatrix} x& py+q& rz+s\\
\alpha& p\beta+q& r\gamma+s\\
1& 1& 1\\
\end{vmatrix}} = 0$$
My Attempt:
The equat... | Hint:
$\frac { x - 0}{1} = \frac {y + \frac {q}{p}}{\frac {1}{p}} = \frac {z + \frac {s}{r}}{\frac {1}{r}}$
One of the points on the line is $P \,$ $(0,-\frac{q}{p}, -\frac{s}{r}$).
Direction vector from this point to $Q \,$ $(\alpha, \beta, \gamma)$ is $\vec{PQ} \, (\alpha, \beta + \frac{q}{p}, \gamma + \frac{s}{r})$.... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3910905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Compute E(sin(X)|X+Y) Okay, let $X,Y$ be independent random variables with the same exponential distribution. Compute $E(X|X+Y)$ and $E(\sin(X)|X+Y)$.
Solution:
$X+Y=E(X+Y|X+Y)=E(X|X+Y)+E(Y|X+Y)=2\cdot E(X|X+Y)$
Hence $E(X|X+Y)=\frac{X+Y}{2}$.
$\sin(\frac{X+Y}{2})=E(\sin(\frac{X+Y}{2})|X+Y)=E(\sin \frac{X}{2}\cos\frac{... | Set $Z=X+Y$.
First we'll show $X|Z=z\sim \mathcal{U}(0,z)$ for $z>0$ fixed. For $x\in(0,z)$ we have
$$f_{X|Z=z}(x|z)=\frac{f_{XY}(x,z-x)\sqrt{2}}{\int_0^zf_{XY}(x,z-x)\sqrt{2}dx}=\frac{1}{z}$$ Here we're using the fact that $f_{XY}(x,y)=\lambda^2e^{-\lambda(x+y)}$ for $(x,y)\in (0,\infty)^2$ and $f_{XY}(x,y)=0$ elsewhe... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914415",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to use the Fourier transform for show that $ \int_{0}^{\infty} \frac{sin^2(x)}{x^2} dx = \frac{\pi}{2} $? Suppose
$$ \int_{0}^{\infty} f(x)\cos(\alpha x)dx = F(\alpha) = \left\{ \begin{array}{lcc}
1-\alpha & if & 0\le \alpha \le 1 \\
\\ \hspace{0.6cm} 0 & if & \alpha>1 \\
... | $$ \mathcal{F}(F(\alpha) = \frac{1-\cos(x)}{x^2} = \frac{4\sin^2(\frac{x}{2})}{\pi x^2} $$
Let $\alpha = 0$
$$ \int_{0}^{\infty} \frac{4\sin^2(\frac{x}{2})}{\pi x^2} \ cos(0) dx = 1 $$
$$ t=x/2 \Rightarrow x=2t \Rightarrow \frac{dx}{dt} = 2 $$
$$ 2 \int_{0}^{\infty} \frac{4\sin^2(t)}{4 \pi t^2} dt = 1 \iff \frac{2}{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3914884",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Prove $\sum_1^n 1/i^2 \le 2 - 1/n$ for all natural $n$. I'm trying to do this by induction. It works for $n=1$ because we get $1 \le 2 - 1 = 1$. Now suppose for some natural $k \ge 1$, we have $\sum_{i=1}^k \le 2 - 1/k$. I must show $\sum_{i=1}^{k+1} \le 2 - 1/(k+1)$. I started out from the hypothesis and added $1... | Hint: prove that LHS "grows" slower than RHS
Prove that $$\frac{1}{n^2} < (2-\frac 1n) - (2-\frac{1}{n-1}), n>1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3916502",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given $T_1 = 0$ and for $i > 1$: $T_i = 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j$ Prove via induction that $T_i = \sum\limits_{j=1}^{i-1} \frac{1}{j}$
Given $T_1 = 0$ and for $i \in \mathbb{N}, i > 1$:
\begin{align*}
T_i &= 1 + \frac{1}{i-1} \sum_{j=1}^{i-1} T_j \\
\end{align*}
Manually computing terms $2,3,4$:
\begin... | Let's say it is true for $i$, and we want to prove for $i+1$ ($i>1$):
I think you can develop the left hand:
$$T_{i+1} = 1 + \frac{1}{i}\sum_{j=1}^{i}T_j = 1 + \frac{1}{i}\sum_{j=1}^{i-1}T_j + \frac{T_i}{i}$$
$$iT_{i+1} = i + \sum_{j=1}^{i-1}T_j + T_i$$
$$\frac{i}{i-1}T_{i+1} = \frac{1}{i-1} + 1 + \frac{1}{i-1}\sum_{j=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918603",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Prove inverse trigonometric equation $2\tan^{-1}2=\pi-\cos^{-1}\frac{3}{5}$ The question is:
Prove that: $2tan^{-1}2=\pi-cos^{-1}\frac{3}{5}$ Hint: use the fact that $tan(\pi-x)=-tanx$
I did the question without using the Hint, but I don't know how to do it using the hint.
Quick working out of what I've done:
\begin{... | It isn't necessary to use the hint. What you have done is correct and valid! Nonetheless, here's the intended method using hint.
Let $\theta=2\tan^{-1}2$ and $\alpha=\pi-\cos^{-1}\frac 3 5$. Then,
$$\tan\theta=\frac{2\tan(\tan^{-1}2)}{1-\tan^2(\tan^{-1}2)}$$
$$\tan\theta=-\frac{4}{3}\quad (*)$$
and using the hint,
$$\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3918968",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
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Infinite sum of inverse of products of successive primes If we iteratively remove the multiples of the succesive prime numbers from the natural numbers, starting from 2, i.e.,
$1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, ... \rightarrow$
$1, 3, 5, 7, 9, 11, 13, 15, 17, 19, 21, ... \rightarrow$
$1, 5, 7, 11, 13, 17, 19, ... | This is not the case:
$$\frac12+\frac1{2\cdot 3}+\frac1{2\cdot 3\cdot 5}+\frac1{2\cdot 3\cdot 5\cdot 7}+\frac1{2\cdot 3\cdot 5\cdot 7\cdot11}+\cdots \le \frac12\left(1+\frac1{3}+\frac1{3^2}+\frac1{3^3}+\frac1{3^4}+\cdots \right)=\frac34$$
It seems that your series converges to
$$0.7052301717918009651474316828882485137... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3924934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Evaluating $\lim_{x\to 0}\dfrac{(1+x)^{\tfrac1x}-e}{x}$ Since we have $\frac00$,I Applied L'Hopital rule :
$$\lim_{x\to 0} (1+x)^{\tfrac1x}\times\left(\cfrac{-\ln(1+x)}{x^2}+\cfrac{1}{x(x+1)}\right)$$$$=\lim_{x\to 0}\cfrac{x^2(x+1)(1+x)^{\tfrac1x}-(x+1)\ln(1+x)+x}{x^2(x+1)}$$
But as you can see it is getting very ugly.... | $$y=\dfrac{(1+x)^{\tfrac1x}-e}{x}$$
$$z=(1+x)^{\tfrac1x}\implies \log(z)=\tfrac1x\log(1+x)$$ Now Taylor
$$\log(z)=\tfrac1x\left(x-\frac{x^2}{2}+\frac{x^3}{3}+O\left(x^4\right) \right)=1-\frac{x}{2}+\frac{x^2}{3}+O\left(x^3\right)$$
$$z=e^{\log(z)}=e-\frac{e x}{2}+\frac{11 e x^2}{24}+O\left(x^3\right)$$
$$y=-\frac{e}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3928465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
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Series of characteristic polynomials Consider the sequence of symmetric matrices with diagonal 2 and second-diagonal s $-1$, e.g.
$$
M_4= \begin{pmatrix}
2 & -1 & 0 & 0 \\
-1 & 2 & -1 & 0\\
0 & -1 & 2 & -1\\
0 & 0 & -1 & 2\\
\end{pmatrix}
$$
I'v... | The Chebyshev polynomials of the second kind satisfy the recurrence relation
$$
\begin{cases}
U_0(y) = 1 \\
U_1(y)=x\\
U_n(y) = 2y U_{n-1}(y)-U_{n-2}(y)
\end{cases}
$$
so that $Q_n(y) = U_n(y/2)$ and $P_n(x) = U_n(1-x/2)$.
The zeros of $U_n$ are
$$
y_k = \cos\left( \pi \frac{k+1}{n+1}\right) \, , \, k = 0, \ldots, n
$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934547",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
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Prove that $(1+ \frac{1}{1^3})(1+\frac{1}{2^3})...(1+\frac{1}{n^3})<3$ I have tried to use induction, but after I assume that P(n) is true, I can't go further to prove that P(n+1) is true as well. I also have tried to find an intermediate inequality, but I can't figure out which inequality I should start from.
Somethin... | Using the fact $1+x\le e^x$ for all real $x,$ we have $$\left(1+ \frac{1}{1^3}\right)\left(1+\frac{1}{2^3}\right)\cdots\left(1+\frac{1}{n^3}\right)\le \dfrac{9}{4}\exp\left(\sum_{k=3}^{n}\dfrac{1}{k^3}\right).$$ Now use the fact that $$\sum_{k=1}^{\infty}\dfrac{1}{k^3}\lt\dfrac{\pi^2}{7}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3934847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Integrating $\int{\frac{x\,dx}{x^2 + 3x -4}}$ I saw this "Beat the Integral" problem and wanted to be sure I was approaching it correctly.
The integral is $$\int{\frac{x}{x^2 + 3x -4}dx}$$
So I decide I want to decompose this, because it's a degree-2 polynomial in the denominator and a degree-1 in the numerator, so tha... | Since the denominator has two linear factors, you can set up your partial fraction decomposition as
$$\frac{x}{(x+4)(x-1)} = \frac{A}{x+4} + \frac{B}{x-1}$$
This implies
$$ x = A(x-1) + B(x+4) $$
So $A+B=1$ and $4B-A=0$ $\Rightarrow$ $(A,B) = (\frac{4}{5},\frac{1}{5})$.
This means that
$$\int \frac{x}{(x-4)(x+1)} dx = ... | {
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Show that $x^\frac{1}{n}$ is continuous at all $a \in [0,\infty)$ I have to show $x^\frac{1}{n}$ is continuous using:
$|x-a| = |(x^\frac{1}{n})^n - (a^\frac{1}{n})^n| = |x^\frac{1}{n} - a^\frac{1}{n}||\displaystyle \sum_{k=1}^{n-1} (x^\frac{1}{n})^{n-1-k} + (a^\frac{1}{n})^{k} |$
Here's what I did:
$a=0$:
$|x^\frac{1}{... | First you need the correct factorization
$$x- a = (x^{\frac{1}{n}} - a^{\frac{1}{n}})\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}}$$
Whence,
$$|x^{\frac{1}{n}} - a^{\frac{1}{n}}|= \frac{|x-a|}{\left|\sum_{k=0}^{n-1}x^{\frac{n-1-k}{n}} a^{\frac{k}{n}} \right|} $$
Note that by the reverse triangle inequality $|a|- ... | {
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Are functions: $1$; $\cos(x)$; $\cos^2(\frac{x}{2})$ linearly independent? I used the definition.
$1$; $\cos(x)$; $\cos^2(\frac{x}{2})$
$c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$
I tried converting $\cos^2(\frac{x}{2})$ into something better: $\frac{1+\cos(x)}{2}$
$c_1+c_2\cdot\cos(x)+c_3\cdot\frac{1+\... | From $\cos^2(\tfrac{x}{2}) = \tfrac12+\tfrac12\cos(x)$ we get $$\tfrac12 + \tfrac12\cos(x) - \cos^2(\tfrac{x}{2}) = 0$$ that is, $c_1\cdot1+c_2\cdot\cos(x)+c_3\cdot\cos^2(\frac{x}{2}) = 0$ where $c_1=c_2=\tfrac12$ and $c_3=-1$.
| {
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standard deviation can not be $6$. For the frequency distribution :
Variate x : $x_1 , x_2 , x_3 , x_ 4 ........, x_{15}$
Frequency f : $f_1 , f_2 , f_3 , f_ 4 ........, f_{15}$ where $0 < x_1 < x_2 < x_3 < x_ 4 ........< x_{15} = 10$ and $\sum f_i >0$.
Then the standard deviation can not be $6$.
Can anyone... | There is a standard result that $$\sigma_x^2 \leq \frac{R_x^2}{4}$$ where $\sigma_x^2$ is the variance of $X$ and $R_x$ is the range of $X$
Note that here $R_x = 10 - 0 = 10$
Using this result, $$\sigma_x^2 \leq \frac{10^2}{4} = 25 $$
Or, $$\sigma_x \leq 5$$
Proof:
Let $\min_{1\leq i \leq n} x_i = a$ and $\max_{1\leq i... | {
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Surface area of $z=4-x^2-y^2$ over a square region
Find the surface area of the paraboloid $z=4-x^2-y^2$ over the square region $-2\le x\le 2 $ and $-2\le y\le 2$.
I can parametrize this surface with $x=u,\ y=v,\text{and} \>z=4-u^2-v^2$,
where $-2\le u\le 2$ and $-2\le v\le 2$. Then I can parametrize the surface wit... | Let $\sinh t= {\frac{4}{\sqrt{17}}}\tan\theta$ to evaluate
\begin{align}
I &= \int_0^{\pi/4}(16\sec^2\theta+1)^{3/2}\,d\theta\\
&=\frac{17^2}4\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\frac{\cosh^4 t}{1+\frac{17}{16}\sinh^2t }dt
=34\int_0^{\sinh^{-1}\frac{4}{\sqrt{17}}}
\frac{(1+\cosh 2t)^2}{\frac{15}{17}+\cosh2t}dt\\
&=... | {
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$a^2 + b^2 + c^2 + 6\ge 3(a + b + c), abc = 1$ I have not solved inequalities in a while, so I am a little rusty. Could you help me with this inequality I have found?
$$a^2 + b^2 + c^2 + 6 \ge 3(a + b + c),$$ where $a, b, c > 0$ and $abc = 1$
My initial idea was $a ^ 2 + 2 \ge 2\sqrt 2a$ and the inequalities with $b $ ... | My second solution:
WLOG, assume $c = \min(a,b,c)$.
If $c \le \frac{1}{2}$, we have
$$a^2+b^2+c^2 + 6 - 3(a+b+c) = (a-\tfrac{3}{2})^2 + (b - \tfrac{3}{2})^2 + c^2 - 3c + \tfrac{3}{2}
\ge 0.$$
If $c > \frac{1}{2}$, noting that $x^2 + 2 - 3x + \ln x \ge 0$ for all $x > \frac{1}{2}$
(see the remark at the end),
we have
$$... | {
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Find the minimum value of $7x-24y$ LMNAS $25^{th}$ UGM, Indonesian
Suppose that $x,y\in\mathbb{R}$, so that :
$x^2+y^2+Ax+By+C=0$
with $A,B,C>2014$. Find the minimum value of $7x-24y$
$x^2+y^2+Ax+By+C=0$
can be written $\rightarrow$ $(x+\frac{A}{2})^2+(y+\frac{B}{2})^2+C-(\frac{A}{2})^2-(\frac{B}{2})^2=0$
Stuck,:>
| Using Cauchy-Schwarz inequality,
$$\left(x+\frac{A}{2}\right)^2+\left(y+\frac{B}{2}\right)^2+C-\left(\frac{A}{2}\right)^2-\left(\frac{B}{2}\right)^2=0\\
\implies 7x-24y=7\left(x+\frac A2\right)-24\left(y+\frac B2\right)+\frac{-7A+24B}{2}\\
\ge -\sqrt{7^2+(-24)^2} \cdot \sqrt{\left(x+\frac A2\right)^2+\left(y+\frac B2\r... | {
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Expressing solution of matrix equations in matrix form I want to find $x_{11}$, $x_{12}$, $x_{21}$, $x_{22}$, $a$, and $b$ that solve the following system of equations
\begin{align*}
\begin{bmatrix}x_{11} & x_{12}\\x_{21} & x_{22}\end{bmatrix}\begin{bmatrix}1\\0\end{bmatrix}& =
\begin{bmatrix}\gamma_{11} & \gamma_{12}\... | Let $G=[\gamma_{ij}]$ and $X=[x_{ij}].$
As the OP already figured out themselves, $a$ and $b$ are the roots of $1-\mathrm{tr}(G)\,z+\det(G)\,z^2=0.$ If we substitute $z=\lambda^{-1}$ and multiply with $\lambda^2,$ we get $\lambda^2-\mathrm{tr}(G)\,\lambda+\det(G)=0.$ This means that $a^{-1}$ and $b^{-1}$ are the eigenv... | {
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testing if the infinite series are converges or diverges here are two problems on testing if the infinite series are conv. or div.
*
*$\sum_{n=1}^\infty \frac{{{7\sqrt[6]{n^{13}}+2n}}}{{\sqrt[3]{27n^{9}-10n+16}}}$
*$\sum_{n=2}^\infty \frac{1}{\ln(n!)}$
the first one I couldn't do :(
but the second one here is what ... | In the first one, you have
\begin{align}
& \frac{n^{13/6} + \text{comparatively negligible things}}{n^3 + \text{comparatively negligible things}} \\[12pt]
= {} & \frac 1 {n^{5/6} + \text{comparatively negligible terms}} \ge \frac {\text{some constant}} {n^{5/6}} \\[6pt]
& \text{(and “constant” means not changing as $n$... | {
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Differents ways to evaluate the sum $\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$
Evaluate $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}$$
My approach:
Let $$x=\sqrt{12+ \sqrt{12+\sqrt{12+\cdots}}}$$
so, we have that $$\sqrt{12+\sqrt{12+\sqrt{12+\sqrt{12+\cdots}}}}\iff \sqrt{12+x}=x \implies 12+x=x^{2} \iff (x+3... | Method 1 - contraction mapping theorem.
Let $g : [0,\infty) \to [0,\infty)$ be the map $x \mapsto \sqrt{x+12}$. For any $x, y \in [0,\infty)$, we have
$$\begin{align} & g(x) - g(y) = \sqrt{x+12} - \sqrt{y+12} = \frac{x - y}{\sqrt{x+12} + \sqrt{y+12}}\\
\implies & |g(x)-g(y)| \le \frac{|x-y|}{2\sqrt{12}}
\end{align}
$$
... | {
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"source": "stackexchange",
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What is the greatest number of points in the plane such that the distance between any two of them is an odd integer? Suppose the origin is one of the points, we can take $(3,0)$ as the second point. If $(x,y)$ is some other point in the set I think we can use the fact that $x^2+y^2$, for $x$ and $y$ odd, is never an in... | The answer is $3$.
Assume the contrary, let's say we can find four points $O,A,B,C$ such that following six distances are all odd integers:
$$(a,b,c,a_1,b_1,c_1) = (BC,CA,AB,OA,OB,OC)$$
Since $O, A, B, C$ lies in the plane, they form a degenerate tetrahedron with volume $V = 0$. Express $V$ in terms of corresponding Ca... | {
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Proof that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}$ Kazakstan 2012
Suppose that $a, b \in\mathbb{R}$, and $a,b>0$. If $\frac{1}{a}+\frac{1}{b}=2$
prove that $a+b+\frac{1}{1+\sqrt{ab}}\geq\frac{5}{2}.$
My idea :
$a+b+\frac{1}{1+\sqrt{ab}}$ can be written as $ab(\frac{a+b}{ab}+\frac{1-\sqrt{ab}}{1-ab})=2ab+\frac{1-\sqr... | From the hypothesis: $a+b=2ab$
So the claim can be written as: $2ab+\frac{1}{1+\sqrt{ab}}\geqslant 2.5$
Now to make everything based on $ab$, we analyze the hypothesis:
For which real values $x$, there exist $a,b$ such that $ab=x , \frac{1}{a}+\frac{1}{b}=2$?
For solving this put $a$ as our variable. We want to find $a... | {
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show that two rational functions do not have a root in common Let $$f(x)=8-\binom{4}{0}\frac{1}{x+5}-\binom{4}{1}\frac{1}{x+3}-\binom{4}{2}\frac{1}{x+1}-\binom{4}{3}\frac{1}{x-1}-\binom{4}{4}\frac{1}{x-3}$$ and $$g(x)=8+\binom{4}{0}\frac{1}{x-5}+\binom{4}{1}\frac{1}{x-3}+\binom{4}{2}\frac{1}{x-1}+\binom{4}{3}\frac{1}{x... | Here's a mechanical way that doesn't require any cleverness, although it does require one tool that you might not have seen yet.
It suffices to show that the polynomials
\begin{align*}
p(x) &= (x-5) (x-3) (x-1) (x+1) (x+3) (x+5) f(x) \\
&= 8 x^6-16 x^5-264 x^4+480 x^3+1720 x^2-2384 x-1080 \\
q(x) &= (x-5) (x-3) (x-1) (... | {
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Summation of Cosine Series Find the summation of the series $\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} $
My approach is as follow
$\sum\limits_{k = 0}^n {{{\sin }^2}\left( {\frac{{k + 1}}{{n + 2}}\pi } \right)} \Rightarrow \sum\limits_{k = 0}^n {\left( {1 - \cos \left( {\frac{{2k ... | $\dfrac{\sin\frac{(n+1)d}{2}}{\sin \frac{d}{2}} \cdot \cos(a + \dfrac{nd}{2})$
For summation from $0$ to $n$
That will work
A very basic proof if you want would be substitute n = t -1 and then using the summation formula for cosine.
Proof of summation formula of cosine and sine up to n-1 terms
| {
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Find $\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}$ As stated in the title.
My attempt. Dividing through $(x-2)^{\frac{2}{3}}$.
$$L=\lim_{x \to \infty}{\frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}}}=\lim_{x \to \in... | Here is another way to do it. Dividing by $x^{2/3}$ yields $$ \frac{(x+1)^{\frac{2}{3}}-(x-1)^{\frac{2}{3}}}{(x+2)^{\frac{2}{3}}-(x-2)^{\frac{2}{3}}} = \frac{(1+\frac 1x)^{\frac{2}{3}}-(1-\frac 1x)^{\frac{2}{3}}}{(1+\frac 2 x)^{\frac{2}{3}}-(1-\frac 2x)^{\frac{2}{3}}}
$$
Substitute $t = 1/x$. Then $$\lim_{x \to \inft... | {
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How to prove $\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) \, dx=\ln 2$ How to prove the integral
$$\begin{align}&\int _0^\infty \operatorname{si}(x) \operatorname{Ci}(x) dx\\=&\int_0^\infty\left (\int_x^\infty\frac{\sin t}{t}dt \int_x^\infty\frac{\cos t}{t}dt\right)dx\\=&\ln 2\end{align}$$
I try to use int... | The Laplace transform is useful for providing integral representations in terms of non-oscillating functions:
$$\text{si}(x)=\int_{x}^{+\infty}\frac{\sin t}{t}\,dt=\int_{0}^{+\infty}\frac{e^{-sx}}{1+s^2}\left(\cos(x)+s\sin(x)\right)\,ds $$
$$\text{Ci}(x)=\int_{x}^{+\infty}\frac{\cos t}{t}\,dt=\int_{0}^{+\infty}\frac{e^... | {
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The number of pairs of integers $(x,y)$ satisfying $x ≥ y ≥ -20$ and $2x + 5y = 99$ is I tried to solve this question but I was unable to think how to get the number of integer pairs satisfying these conditions. Till now I have broken down this into :-
$2x+5y = 99$
==>$y=(99-2x)/5$
$y\geq-20$
==> $(99-2x)/5 \geq20$
==>... | y must be of the form $y=-(2k+1)5\geq-20$ which gives:
$(2k+1)5= (-5, -15) \geq -20$ $\rightarrow$ $2k+1=1, 3$
So for any $k\geq 0$ there will be an integer solution for x. in fact number of solutions is 2 when y is negative. Examples:
$2k+1=1$ $\rightarrow$ $y=-5>-20$, gives : $2x=25+99$ $\rightarrow$ $x=62$
$2k+1=3$ ... | {
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Interchange order of integral Can someone tell me which step in the process did wrong? My result seems wrong
$$D = \{(x, y) \mid \sqrt{2x - x^2} \leq y \leq \sqrt{2x}, 0 \leq x \leq 2 \}$$
$$\iint_D f(x,y) \, dy \, dx$$
Then I can draw the Graph
According to the graph, we can know that this is the area enclosed by $x ... | The natural way to write the iterated integral is of course $$\iint_D f \, dA = \int_{x=0}^2 \int_{y=\sqrt{2x-x^2}}^{\sqrt{2x}} f \, dy \, dx.$$ The first alternative where the order of integration is switched, is as you attempted, the area of the parabolic sector minus the semicircle. Note that the parabolic segment... | {
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"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Solve for $x$ in $ \sin x + \sin x \cos x - (\frac{1}{\sqrt{2}} + \frac{1}{2})= 0 $ How can to solve for $x$ in $$ \sin x + \sin x \cos x - \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right)= 0.$$
My try:
$$ \sin x + \sin x \cos x = \left(\frac{1}{\sqrt{2}} + \frac{1}{2} \right) $$
$$ (\sin x + \sin x \cos x)^2 = (\frac{1... | Using the tangent half-angle substitution, you end with
$$\left(1+\sqrt{2}\right) t^4+2 \left(1+\sqrt{2}\right) t^2-8 t+(1+\sqrt{2})=0$$By inspection (with all these $1$'s and $\sqrt{2}$'s), $\color{red}{t=(\sqrt{2}-1)}$ is a root.
Factoring, we are left with
$$\left(1+\sqrt{2}\right) t^3+t^2+\left(1+3 \sqrt{2}\right)... | {
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Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :- $\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
Consider the function $f(x) = \frac{1}{3^x + \sqrt{3}}.$ Find :-
$\sqrt{3}[f(-5) + f(-4) + ... + f(3) + f(4) + f(5) + f(6)]$.
What I Tried: I checked similar questions and answers in the Art o... | Hint: We have that
$$f(x)+f(1-x)=\frac{1}{3^{x} + \sqrt{3}}+\frac{1}{3^{1-x} + \sqrt{3}}=\frac{1}{3^{x} + \sqrt{3}}+\frac{3^{x}/\sqrt{3}}{\sqrt{3} + 3^x}=\frac{1}{\sqrt{3}}$$
| {
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"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
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The limit $\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$ $\lim_{x \to ∞} (3x^4-5x^2+1)(1-e^{1/x^2})^2$
I found +∞ but the limit is 3. How to compute this limit ?
| $\displaystyle \lim_{ x \to \infty} (3x^4-5x^2+1)(1-e^{1/x^2})^2$
$=\displaystyle \lim_{x \to \infty} \left(\frac{e^{1/x^2}-1}{\frac{1}{x^2}}\right)^2\left(\frac{3x^4-5x^2+1}{x^4}\right)$
$=1\times 3$
You know that: $$\lim_{u \to 0^+} \frac{e^u-1}{u}=1$$
Putting $u=1/x^2$ we get:$$\lim_{x \to \infty} \frac{e^{\frac{1}{... | {
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Integral of a rational expression involving quadratics. I want to solve this integral but I have some problems...
$$\int_2^3 \frac{(x^2-2x+1)}{(x^2+2x+1)}$$
I transformed both in $(x-1)^2$ and $(x+1)^2$ respectively but didn't find any answer. I tried as well to transform the rational expression into $1 -\frac{4x}{(x... | $$ \frac{x^2-2x+1}{x^2+2x+1}=1-\frac {4x+4-4}{(x+1)^2}=1-4\frac {x+1}{(x+1)^2}+\frac 4{(x+1)^2}$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\lim_{x \to 1} \frac{1 - x^{1/\pi}}{1 - x^{1/e}}$ without de l'Hospital rule Is it possible to evaluate this limit without de l'Hopital's rule?
$$\lim_{x \to 1} \frac{1 - x^{1/\pi}}{1 - x^{1/e}}$$
| Why not to use
$$1-x^a=-a (x-1)+\frac{1}{2} \left(a-a^2\right) (x-1)^2+O\left((x-1)^3\right)$$
$$\frac {1-x^a}{1-x^b}=\frac{-a (x-1)+\frac{1}{2} \left(a-a^2\right) (x-1)^2+O\left((x-1)^3\right)}{-b (x-1)+\frac{1}{2} \left(b-b^2\right) (x-1)^2+O\left((x-1)^3\right)}$$
$$\frac {1-x^a}{1-x^b}=\frac{a}{b}+\frac{a (a-b)}{2... | {
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Prove that: $ \sum_{i=0}^{\infty} \frac{\tan \frac{\theta}{2^i}}{2^i}= \frac{1}{\theta} - 2 \cot 2 \theta$ My attempt:
Consider the following series:
$$ S = \sum_{i=0}^n \ln( \sec \frac{x}{2^i} )$$
Notice that $ \lim_{n \to \infty} \frac{dS}{dx}$ is the required sum.
Simplfying S,
$$ S = - \ln \left( \cos x \cdot \cos... | Thanks to @gary,
$ \frac{1}{2^n} = u$,
$$ \lim_{u \to 0 } u \cot ( ux)= \lim_{u \to 0} \cos(ux) \frac{ux}{x\sin (ux)} = \frac{1}{x}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3978919",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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Coin betting problem. A coin bet is placed between two friends such that the person who wins four tosses first is the winner. What is total number of possible ways in which the bet can play out?
| Avoiding elegance, there will end up being $k$ tosses, where $k \in \{4,5,6,7\}.$ Therefore, you want $\sum_{k=4}^7 f(k)$, where $f(k)$ is the number of different coin toss sequences that result in the game ending after exactly $k$ coin tosses.
Clearly, $f(4) = \binom{4}{0} + \binom{4}{4} = 2$.
That is, there are $\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3980803",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How to divide two inequalities in the form $a - \epsilon < x < a + \epsilon$? I have the following inequalities
$$
a - \epsilon_x < x < a + \epsilon_x \\
b - \epsilon_y < y < b + \epsilon_y
$$
with $x,y,a,b,\epsilon_x,\epsilon_y\in {\rm I\!R}$, $|\epsilon_x|>0$ and $|\epsilon_y|>0$.
How to bound $\frac{x}{y}$?
I first... | If $0 \in (b- \epsilon_y, b + \epsilon_y)$, there is no way to bound $\frac{x}{y}$.
Otherwise you have $$\vert y \vert \gt m = \begin{cases}
b-\epsilon_y & \text{ if } b - \epsilon_y \gt 0\\
-(b + \epsilon_y) & \text{ if } b + \epsilon_y\lt 0
\end{cases}$$As $\vert x \vert \lt \vert a \vert + \epsilon_x$, you get
$$\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3982534",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find n where $(2n+1)2^{4n+5} = 3 \pmod{7}$ For $n$ normal number, the book solved it like this:
If $n$ can be divided by $3$ (which is $n = 3k$) then $n = 21L + 9$.
If $n$ can't be divided by 3(Which is either $n = 3k +1$ or $n = 3k + 2$) then $n = 21L + 1$ or $n = 21L + 2$ .
But I didn't solve it like this.
My logic i... | Here is one way to solve it:
*
*$(2n+1)\pmod 7\equiv 1,3,5,0,2,4,6,\cdots\ $ and cycling.
*$2^{4n+5}\pmod 7\equiv 4,1,2,\cdots\ $ and cycling.
Both can be proved by induction on $n$.
*
*$1\times 3\equiv 2\times 5\equiv 4\times 6\equiv 3\pmod 7\ $ verify there are no others.
Finally can you find $n$ for which the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the supremum and infimum of a set X Let the set $X=\{\frac{s}{s^2+2}, s\in\mathbb{Z}\}$ I have to find the supremum and the infimum. I have done in this way:
I can observe that $X=\{\frac{s}{s^2+2}, s\in\mathbb{N}\}\cup \{\frac{-t}{t^2+2}, t\in\mathbb{N}\}$ and now I study separetely the infimum and supremum of th... | Let $f(s) = s/(s^2 + 2)$. Then clearly for all $s \in \mathbb Z$, $$f(-s) = -s/((-s)^2 + 2) = -s/(s^2 + 2) = -f(s).$$ So if $f(n) = \sup X$, we must have $f(-n) = \inf X$.
Next, observe that $f(s) > 0$ if $s > 0$, so $f(s) < 0$ if $s < 0$, and $f(s) = 0$ if $s = 0$. So it suffices to consider $s \in \mathbb Z^+$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3984209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Continued radical of powers of 4 equals 3 Can someone explain to me why
$$\sqrt{4 + \sqrt{4^2 + \sqrt{4^3 + \sqrt{4^4 + \dots}}}} = 3???$$
I need an answer
| Let $f_0(x)=0$ and recursively $$ f_{n+1}(x)=\sqrt{x+f_n(4x)}.$$
Next, let $$ f(x)=\lim_{n\to\infty}f_n(x).$$
Our goal is to find $f(4)$.
Claim 1. For $x\ge 0$ and $n\in\Bbb N$,
$$\tag1 f_{n}(x)\le f_{n+1}(x).$$
Proof.
This is trivial for $n=0$. Assume $(1)$ holds for $n$ and all $x\ge0$.
Then
$$f_{n+1}(x)=\sqrt{x+f_n(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Primes of the form $p=X^2+3Y^2$ I'm trying to work in $\mathbb{Z}\left[\frac{1+\sqrt{-3}}{2}\right]$ to show that an odd prime $p\in\mathbb{Z}$, $p\neq 3$ is of the form $p=X^2+3Y^2$ if and only if $p\equiv1$ mod $3$. The hint is to first show that if $p=1$ mod $3$, then both $p=u^2+3v^2$ and $4p=u^2+3v^2$ where this t... | HINT.- If $p\ne3M+1$ then $p=3M-1$. But then
$$3M-1=X^2+3Y^2\Rightarrow X^2\equiv-1\pmod3$$ which is impossible because $\left(\mathbb Z/3\mathbb Z\right)^2= \{0,1\}$.
COMMENT.-Maybe this could help @Saegusa in his inquiring on $\mathbb Q(\sqrt{-3})$ mainly in $4p=u^2+3v^2$. This problem was conjectured by Fermat and p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3985550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $-\frac{7}{3}
Real numbers $a$ and $b$ satisfy $a^3+b^3-6ab=-11$. Prove that $$-\frac{7}{3}<a+b<-2$$
I have shown that $a+b<-2$. My approach: $-3=8-11=a^3+b^3-6ab+2^3=\frac{1}{2}(a+b+2)((a-b)^2+(a-2)^2+(b-2)^2)$. From this we must have that $a+b<-2$.
Please... | $$a^3-6 a b+b^3=-11$$
can be written as
$$(a+b)^3-3 a b (a+b)-6 a b=-11$$
setting $$a+b=s;\;ab=p$$
we get
$$ s^3-3ps-6p+11=0$$
The equation $$z^2-sz+p=0\tag{1}$$
Gives the values of $a,b$. In order to have real roots, discriminant must be positive
$$s^2-4p\ge 0$$
Solve
$$\begin{cases}
s^3-3ps-6p+11=0\\
s^2-4p\ge 0\\
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3988781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
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"answer_id": 2
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How to find the minimum of $xy$ from the equation $2x+y+6=xy$ , $x,y>0$ I have tried doing factoring and completing the square and also AM-GM
$(2-y)(x-1)=-8$
$xy=2x+y+6≥\sqrt{12xy} $
| AM-GM works perfectly fine with a small trick.
$2x+y+6=xy \Rightarrow y(x-1)=2x+6 \Rightarrow y = \dfrac{2x+6}{x-1} \Rightarrow xy =\dfrac{2x^2+6x}{x-1}$.
As @Yves Daoust points out $\dfrac{2x^2+6x}{x-1} = 2x+8+ \dfrac{8}{x-1} = 2(x-1)+10+\dfrac{8}{x-1} $.
But by AM-GM, $ 2(x-1)+10+\dfrac{8}{x-1} \geq 10 + 2\left(\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3991256",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all positive integers $x$ such that $x(x+2021)$ is a perfect square I completed the square as to get $(x+1010.5)^2-102110.25 = k^2$ but I don't know where to go from here.
Please help, thank you
I then got $(2x+2021)^2-4084441=4k^2$ then $(2k-2x-2021)(2k+2x+2021)=43^2*47^2$
| If the product of two relatively prime numbers is a perfect square, then both factors must themselves be perfect squares. So, let's consider that case first.
How many ways can $2021$ be expressed as a difference of two squares? The answer to that is $2$. Namely, $2021=1011^2-1010^2=45^2-2^2$. Hence, $x=2^2=4$ and $x=10... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3995529",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Subsets and Splits
Fractions in Questions and Answers
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