Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Find the first three Laurent expansion terms of $\frac{1-z}{z^2} e^z$ I've this function $f(z)=\frac{1-z}{z^2} e^z$ and I've to find the first three Laurent expansion terms in $z=0$.
I've proceeded in this way:
First of all I've considered the expansion series of $e^z = \sum_{n=0}\frac{z^n}{n!}$ and I found the first t... | First, you can either write
$$
e^z=1+z+\frac{z^2}{2}+o(z^2)
$$
with “little-oh,” or
$$
e^z=1+z+\frac{z^2}{2}+O(z^3)
$$
with “big-oh.” Let's assume that you meant the latter. Then your calculation gives
$$
(1-z)e^z = 1+z+\frac{z^2}{2}+O(z^3) - (z+z^2+\frac{z^3}{2}+O(z^4)) \\
= 1 - \frac{z^2}{2}- \frac{z^3}{2}+O(z^3)+O... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479009",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Simplest proof of $|I_n+J_n|=n+1$ Answering another question, I realised that I "know" that the determinant of the sum of an identity matrix $I_n$ and an all-ones square matrix $J_n$ is $n+1$. I.e.
$$|I_n+J_n|=\left|\begin{pmatrix}
1 & 0 & 0 &\cdots & 0 \\
0 & 1 & 0 &\cdots & 0 \\
0 & 0 & 1 &\cdots & 0 \\
\vdot... | I'm going to elaborate on Exodd's comment. I'm inclined to agree that this is the simplest proof, provided you're comfortable with the basic constructions of linear algebra.
Lemma: If $A \in \mathbf{C}^{n \times n}$ has eigenvalues (counting multiplicity) $(\lambda_i)_{i=1}^n$, then $A + I$ has eigenvalues $(\lambda_i ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ For $n \in \mathbb{N}$, evaluate
$$\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$$
I could not use wolframalpha, I do not know the reason.
For $... | We seek to calculate $$\begin{align}\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx&= \frac{1}{2}+\frac{1}{4}-\frac{1}{6}-\frac{1}{8}+\frac{1}{10}+\frac{1}{12}-\frac{1}{14}-\frac{1}{16}+\cdots \\&=\frac{1}{2}\left(\left(\frac{1}{1}-\frac{1}{3}+\frac{1}{5}-\frac{1}{7}+\cdots\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4481299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
limit of function near a given point Find the limit of the given function as $x\to -2$ without using L’Hôpital’s Rule:
$$\displaystyle \lim_{x \to -2} \frac{\sqrt[3]{x-6} + 2}{x^3 + 8}$$
I used the identity: $x^3 + 2^3 = (x+2)(x^2 -2x +4)$ at the denominator. Then i tried to use the same identity at the numerator and d... | As you have noticed that $x^3 + 2^3 = (x+2)(x^2 -2x +4)$, we first calculate the limit
$$\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}.$$
Let $t=\sqrt[3]{x-6}$, then $t\to-2$ as $x\to-2$ and $x+2=t^3+8=(t+2)(t^2-2t+4)$, so
$$\lim_{x\to-2}\frac{\sqrt[3]{x-6} + 2}{x+2}=\lim_{t\to-2}\frac{t + 2}{(t+2)(t^2-2t+4)}=\lim_{t\to-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Integral $\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt$ $$I=\int_0^1 \frac{\ln(1-\sqrt{3}t+t^2)}{t} dt=-\frac{13}{72}\pi^2$$
Here is what I tried: $1-\sqrt{3}t+t^2=(z_1-t)(z_2-t)$ where $z_1=e^{\frac{\pi}{6}i}, z_2=e^{-\frac{\pi}{6}i}$
$$I=\int_0^1 \frac{\ln(z_1)+\ln(1-\frac{t}{z_1})}{t}+\frac{\ln(z_2)+\ln(1-\frac{t}{z_2... | As @Frank W pointed out in comment...(although I saw the comment after I came up with the proof). Actually, I found the integral $I(a) = \pi a - \frac{a^2}{2} - \pi^2/3$ in the book "Table of Integrals, Series, and Products"
by I.S. Gradshteyn and I.M. Ryzhik. Then I came up with the proof.
For $a \in (0, \pi/2]$, let
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Similar sums $\sum _{n=1}^\infty \frac{(-1)^{n-1}}{n^3}$ and $\sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi ^3}{32}$ Is it possible to find $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^3}$ if we know that $\displaystyle \sum_{n=1}^\infty \frac{(-1)^{n-1}}{(2n-1)^3}=\frac{\pi^3}{32}$?
Any help is welco... | The short answer is no, but nevertheless they are some how related as we will shortly see below.
Lets start by the first series which is an instance of the Dirichlet eta function $\eta(s)=\sum_{n=1}^\infty \frac{(-1)^{n-1}}{n^s}$, which is related to the Riemann Zeta function by the equation
\begin{align*}
\eta(s)=(1-2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4486809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$. If $a > 0$, $b > 0$ and $b^{3} - b^{2} = a^{3} - a^{2}$, where $a\neq b$, then prove that $a + b < 4/3$.
Now what I thought is to manipulate given result somehow to get something in the form of $a + b$:
\begin{align... | We have:
\begin{align}
a^2+ab+b^2 = a+b &\implies 4(a+b)^2 - 4(a+b) = 4ab < (a+b)^2\\
&\implies 3(a+b)^2 - 4(a+b) < 0\\
&\implies (a+b)(3(a+b)-4) < 0\\
&\implies 3(a+b) - 4 < 0\\
&\implies (a+b) < \dfrac{4}{3}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487764",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Irrational equation $\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$ I saw the problem from one math competition:
$$\left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0$$.
I tried to solve it this way:
\begin{align*}
& \left(x^{3}-3x+1\right)\sqrt{x^{2}-1}+x^{4}-3x^{2}+x+1=0\\
\Leftrightarrow\,\, & \left... | $$(x^3-3x+1)\sqrt{x^2-1}+x^4-3x^2+x+1=0$$
$$(x^3-3x+1)\sqrt{x^2-1}=-(x^4-3x^2+x+1)$$
$$(x^3-3x+1)^2(x^2-1)=(x^4-3x^2+x+1)^2$$
If you expand everything, and then subtract the two sides of the equation, you get
$$x^6-4x^4+2x^3+3x^2-4x+2 = 0$$
By the Rational Root Theorem, potential rational roots are $\pm 1$ and $\pm 2$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490883",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 1
} |
Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {2-x}}4\right)=\sqrt 2$ The following question appeared in a JEE Mock exam, held two days ago.
Question:
Find the number of solutions to the equation $\sin\left(\frac{\pi\sqrt x}4\right)+\cos\left(\frac{\pi\sqrt {... | Let $x = 2 \cos^2 \theta, \theta \in \left[0, \dfrac{\pi}{2}\right]$
Then we have $\sin \left(\dfrac{\pi \cos \theta}{2\sqrt 2}\right) +\cos \left(\dfrac{\pi \sin \theta}{2\sqrt 2}\right) = \sqrt 2 $
Now if $\sin \theta > \cos \theta $ we have $\cos \left(\dfrac{\pi \sin \theta}{2\sqrt 2}\right)<\cos \left(\dfrac{\pi \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find $\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $ Let $a \in (0,1)$, I would like to integrate
$$\int_{t}^1 \frac{\sin(\pi x)^2}{(\sin(\pi x/2)^2-a^2)^2} \ dx. $$
Now $\sin(\pi x/2)^2$ is a monotonically increasing function from $0$ to $1$, therefore there exists a unique $t^*$ such that $\sin(\pi ... | With $x\to 1-x$
\begin{align}
&\int_{t}^1 \frac{\sin^2\pi x}{(\sin^2\frac{\pi x}2-a^2)^2} \ dx\\
=&\ 4\int_0^{1-t}-1
+\frac{(1-2a^2)\cos^2\frac{\pi x}2+a^4}{(\cos^2\frac{\pi x}2-a^2)^2}\ dx\\
=& \ 4(t-1)+\frac2a\frac d{da}\int_0^{1-t} \frac{a^2-a^4}{\cos^2\frac{\pi x}2-a^2}\ dx\\
=& \ 4(t-1)+\frac2a\frac d{da}\bigg(
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluating $\lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}$ $$
\begin{align*}
&\text { Let } \mathrm{x}=2 \mathrm{y} \quad \because x \rightarrow 0 \quad \therefore \mathrm{y} \rightarrow 0\\
&\therefore \lim _{x \rightarrow 0} \frac{12-6 x^{2}-12 \cos x}{x^{4}}\\
&=\lim _{y \rightarrow 0} \frac{12-6(2 y)^{... | $5th$ $step$
$1-\cos2y=\sin^2y$
$6th$ $step$
$expansion\ of\ \sin = y−\frac{y^3}{3!}+\frac{y^5}{5!}−⋯∞$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
$\int_0^1 \frac{\arcsin x\arccos x}{x}dx$ Someone on Youtube posted a video solving this integral.
I can't find on Math.stack.exchange this integral using search engine https://approach0.xyz
It is related to $\displaystyle \int_0^\infty \frac{\arctan x\ln x}{1+x^2}dx$
Following is a solution that is not requiring the u... | Just for the fun (it is to long for a comment)
Using @Quanto's solution
$$I=\frac18\int_0^\pi\frac{x(\pi-x)}{\sin (x)}dx$$ and the $\large 1,400^+$ years old approximation
$$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$
$$I \simeq \frac{1}{128}\int_0^\pi \left(5 \pi ^2-4 (\pi -x) x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4500073",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
Conjectured closed form of $\int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x$ Consider
$$ I = \int_0^{\infty} \frac{1}{(x^2+a^2)^{n}} \, \mathrm d x $$
where $a > 0$ is a constant. We can evaluate this using the Leibnitz theorem for specific values of $n$. Mathematica can solve it for specific values of $n$ as we... | Letting $x=a\tan \theta$ transform the integral into a Beta function
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{a \sec ^{2} \theta d \theta}{\left(a^{2} \tan ^{2} \theta+a^{2}\right)^{n}}
&=\int_{0}^{\frac{\pi}{2}} \cos ^{2 n-2} \theta d \theta =\frac{1}{2 a^{2 n-1}} B\left(\frac{1}{2}, n-\frac{1}{2}\right)
\end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Prove that $a^{4/a} + b^{4/b} + c^{4/c} \ge 3$
Let $a, b, c > 0$ with $a + b + c = 3$. Prove that
$$a^{4/a} + b^{4/b} + c^{4/c} \ge 3.$$
This question was posted recently, closed and then deleted, due to missing of contexts etc.
By https://approach0.xyz/, the problem was proposed by Grotex@AoPS.
My strategy is to spl... | I show the inequality (almost algebraically speaking) for $x\in[0,4/3]$ :
$$4\left(x-1\right)+1\leq x^{\frac{4}{x}}$$
In fact we have a stronger inequality for $x\in[3/4,4/3]$:
$$\left(x^{-\frac{1}{4}}\left(1+\left(\frac{1}{x}+\frac{1}{4}\right)\left(x-1\right)\right)\right)^{4}\leq x^{\frac{4}{x}}\tag{I}$$
To show it ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4502186",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 2
} |
Limit of $n^2$ and a recurrence relation with ceiling function For all positive integer $n$ we define a finite sequence in the following way:
$n_0 = n$, then $n_1\geq n_0$ and has the property that $n_1$ is a multiple of $n_0-1$ such that the difference $n_1 - n_0$ is minimal among all multiple of $n_0 -1 $ that are bi... | Under the substitution
$$t_k = \frac kn, \quad a_k = \frac{n_k}{n(n - k)}, \quad Δt = \frac1n,$$
we have
$$t_0 = 0, \quad a_0 = Δt, \quad t_k = t_{k - 1} + Δt, \quad a_k = a_{k - 1} + \left\lceil\frac{a_{k - 1}}{1 - t_k}\right\rceil Δt,$$
which approximates* the differential equation
$$a(0) = 0, \quad a'(t) = \left\lfl... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4503801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 0
} |
$\varepsilon-\delta$ proof of this multivariable limit We've to prove that
$$
\lim_{(x,y)\to(0,0)} \frac{x^3+y^4}{x^2+y^2} =0
$$
Kindly check if my proof below is correct.
Proof
We need to show there exists $\delta>0$ for an $\varepsilon>0$ such that
$$
\left| \frac{x^3+y^4}{x^2+y^2} \right| < \varepsilon \implies \sqr... | That is not correct. You are supposed to find, for each $\varepsilon>0$, some $\delta>0$ such that$$\sqrt{x^2+y^2}<\delta\implies\left|\frac{x^3+y^4}{x^2+y^2}\right|<\varepsilon,$$and that's not what you did.
Note that$$\left|\frac{x^3}{x^2+y^2}\right|=|x|\frac{x^2}{x^2+y^2}\leqslant|x|$$and that$$\left|\frac{y^4}{x^2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504768",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Is it possible for Riemann Sum and Standard Integration to have different answers? I have a specific question for this equation :
$$\frac{1}{x+1}
$$
Using Standard Integration,
$$\int_{0}^{1} \frac{1}{x+1}
$$
which is approximately 0.69
Using Riemann Sum (right end point), however, I get this
$$ \lim _{n \to \infty } ... | Yes, it's possible to show both expressions equal $\ln(2)$. Just like how others pointed out, we can't just "plug in $\infty$" all the time and get the right answer because $\infty$ is not a real number. To show our Riemann Sum converges to $\ln{(2)}$, we do
$$\eqalign{
\lim_{n\to\infty}\sum_{i=1}^{n}\frac{1}{n+i} &= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
} |
Summation of reciprocal products When studying summation of reciprocal products I found some interesting patterns.
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)}=\frac{1}{1\cdot1!}-\frac{1}{1\cdot(N+1)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdot(k+1)(k+2)}=\frac{1}{2\cdot2!}-\frac{1}{2\cdot(N+1)(N+2)}$$
$$\sum_{k=1}^{N} \frac{1}{k\cdo... | Starting the summations at $k=1$ and using Pochhammer symbols
$$\prod_{n=0}^i(k+n)=k (k+1)_i$$
$$\sum_{k=1}^N\frac 1{k (k+1)_i}=\frac{1}{i \,\Gamma (i+1)}-\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}$$ Now, using Stirling approximation for large values of $N$
$$\log\Bigg[\frac{\Gamma (N+1)}{i\, \Gamma (N+1+i)}\Bigg]=-(i \lo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Proving that a quadrilateral is an isosceles trapezoid if and only if the diagonals are congruent. I would like a proof to a Theorem (via OnlineMathLearning.com) I found:
A quadrilateral is an isosceles trapezoid if and only if the diagonals are congruent.
And more specifically, Wikipedia's "Isosceles trapezoid" entr... | Label the trapezoid as the shown below. The base of the quadrilateral is $\overline{AD}$. The lengths of the sides $\overline{AB},\overline{BC},\overline{CD},\overline{AD}$ are respectively $b,c,d$, and $a$. The diagonals $\overline{AC}$ and $\overline{BD}$ both have length $r$. Angles $\angle BAC$ and $\angle CAD$ are... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4506778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Calculate the sum $\sum_{n=0}^{\infty }\frac{1}{n!!}$ \begin{align*}
S =\sum_{n=0}^{\infty }\frac{1}{n!!}=\sum_{n=0}^{\infty }\left ( \frac{1}{(2n)!!}+\frac{1}{(2n+1)!!} \right )
&= \sum_{n=0}^{\infty }\left ( \frac{1}{(2n)!!}+\frac{(2n)!!}{(2n+1)!} \right )=\sum_{n=0}^{\infty }\left ( \frac{1}{2^nn!}+\frac{2^nn!}{(2n+... | This is not an answer but it is too long for a comment.
Without integration, the partial sums can be computed (thanks to Mathematica).
$$S_m=\sum_{n=0}^{m }\frac{1}{n!!}$$
$$S_{2p}=\sqrt{\frac{e \pi }{2}} \,\frac{\Gamma
\left(p+\frac{1}{2},\frac{1}{2}\right)}{\Gamma
\left(p+\frac{1}{2}\right)}+\sqrt{e}\,\frac{ \G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
How many methods to tackle the integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x ?$ $ \text{We are going to evaluate the integral}$
$\displaystyle \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x \tag*{} \\$
by letting $ y=\frac{\pi}{4}-x. $ Then
$$\begin{aligned} \displaystyle ... | Same solution, made slightly more general.
Let $a,b,c$ be real numbers such that
$$
c^2 = a^2 + b^2\ , \qquad a,c>0\ .
$$
Consider now the integral
$$
\begin{aligned}
J = J(a,b,c)
&:=
\int_0^{\pi/4}
\frac{\sin x+\cos x}{a^2 + b^2 \sin 2x}\; dx
=
\int_0^{\pi/4}
\frac{\sqrt 2\cos y}{a^2 + b^2 \cos 2y}\; dy
=
\int_0^{\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508889",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
prove that this expression equals the area of a specific triangle Prove that the area of a triangle with one of its angles $\frac{\pi}4$ rad, and the side opposite to this angle is $2$cm equals $\sin(2\theta)-\cos(2\theta)+1$, where $\theta$ is the angle adjacent to the $2$cm side.
and here is a picture to make things ... | Let $a = 2, A = \frac {\pi} 4, B = \theta$
Then,
$$
\frac{b}{\sin \theta} = \frac{2}{\frac{\sqrt{2}}2}
\implies b = 2\sqrt 2 \sin \theta
$$
Then,
$$
\sin C = \sin(A+B) = \frac {\sqrt{2}}2 (\sin \theta + \cos \theta)
$$
Then,
$$
S = \frac 12 ab\sin C =\frac 12 \times 2\times 2\sqrt 2\sin \theta\times \frac {\sqrt{2}}2 (... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511797",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the sum of radicals without squaring, Is that impossible?
Find the summation:
$$\sqrt {3-\sqrt 5}+\sqrt {3+\sqrt 5}$$
My attempts:
\begin{align*}
&A = \sqrt{3-\sqrt{5}}+ \sqrt{3+\sqrt{5}}\\
\implies &A^2 = 3-\sqrt{5} + 3 + \sqrt{5} + 2\sqrt{9-5}\\
\implies &A^2 = 6+4 = 10\\
\implies &A = \sqrt{10}
\end{align*}
S... | In this answer I tried to generalize the answer given above.
Generalization:
Let,
$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n,\,\,\,m\ge n$$
and
$$A=\sqrt {a+\sqrt b}+\sqrt {a-\sqrt b}=2\sqrt m$$
Then we have
$$\sqrt {a\pm\sqrt b}=\sqrt m\pm\sqrt n\\
\begin{cases} m+n=a\\mn =\frac {b}{4}\end{cases}\\
t^2-at+\frac b4=0\\
4t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 3
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Graph Transformation Knowing dealing with graph transformations come handy MANY times. I searched on google to get a comprehensive graph transformation list but couldn't find one. Some good while back I learned them all with great enthusiasm but have now all forgotten them and like before, I'm not finding a great teach... | \begin{array}{|c|c|c|c|}
\hline
\text{New function}(c>0,\lambda>1)& \text{Change of coordinate}& \text{Geometry of transformation}
\\ \hline
f(x) +c & (a, b) \mapsto(a,b+c)& \text{ shift up by} c\\ \hline
f(x) -c & (a, b) \mapsto(a,b-c)&\text{shift down by} c\\ \hline
\lambda f(x) & (a, b) \mapsto(a,\lambda b)&\text{ve... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Finding the equation of a plane given three points Below is a problem I did from a Calculus text book. My answer matches the
back of the book and I believe my answer is right. However, the method
I used is something I made up. That is, it is not the method described
in the text book.
Is my method correct?
Problem:
Find... | A plane can be defined by its normal $n$ and a point $p$ of the plane. The equation of the plane is, for any point $x$ in the plane, $x\cdot n=d$, where $d=n\cdot p$.
If we have three points in the plane, $a,b,c$ then $a.n=b.n=c.n=d$. So we can solve for $n$ by forming a matrix $$Q=\begin{pmatrix}a&b&c\end{pmatrix}^T$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Given $a^4+8b=4(a^3-1)-16\sqrt 3$ and $b^4+8a=4(b^3-1)+16\sqrt 3$, find $a^4+b^4$ Given $a^4+8b=4(a^3-1)-16\sqrt 3$ and $b^4+8a=4(b^3-1)+16\sqrt 3$, find $a^4+b^4$
I tried adding and subtracting both equations, but didn't get anywhere. Would appreciated any ideas. Thanks!
| Adding both equation gives:
$$a^4+b^4+8(a+b)=4(a^3+b^3)-8$$
i.e: $$(a^2-2a)^2-4(a^2-2a)+b^4-4b^3+8b+8=0$$
Consider this equation is a quadratic equation with variable $a^2-2a$
So the discriminant is:$(-2)^2-(b^4-4b^3+8b+8)=-b^4+4b^3-8b-4=-(b^2-2b-2)^2$
Then:$b^2-2b-2=0$, so $b=1-\sqrt{3}$ or $b=1+\sqrt{3}$ then you can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4517697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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How to find this indefinite integral? $\int\frac{1+x^4}{(1-x^4)\cdot \sqrt{1+x^4}}dx$ I am thinking of a trig sub of $x^2 = \tan{t}$ but its not leading to a nice trigonmetric form, which i can integrate. Our teacher said that it can be computed using elementary methods, but I'm unable to think of the manipualtion.
| With parts of the calculation left as an exercise:
The surd-in-the-denominator expression I commented on is actually a subtle hint. Rewriting the numerator as$$1+x^4=\frac12(1-x^2)^2+\frac12(1+x^2)^2,$$we want to separately evaluate$$\frac12\int\frac{(1-x^2)^2}{(1-x^4)\sqrt{1+x^4}}dx=\frac12\int\frac{(1-x^2)}{(1+x^2)\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
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Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$ Question:
Show that for any two positive real numbers $a, b: \frac a{a + 2b} + \frac b{b + 2a} ≥ \frac12$.
So for this question, I began by expanding all terms and moving them all to one side. However, I do not know how to... | You almost have it, just use
$$2a^2-ab+2b^2 = 2(a-b)^2+3ab > 0$$
(and the denominator is positive as well).
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
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Combinatorial Identity involving Bernoulli numbers For all $n\geq 1$ and $m\geq0$, I'm trying to prove that
$\sum_{k=0}^n\sum_{l=0}^m\binom{n}{k}\binom{m}{l}\frac{(n-k)!(m-l)!}{(n+m-k-l+1)!}(-1)^l B_{k+l}=0$
where $B_n$ are the Bernoulli numbers with $B_{1}=-\frac{1}{2}$.
I made a couple of attempts using the recursive... | The instructive derivation of @RenéGy reduces OPs problem to show the nice symmetric Bernoulli number identity:
\begin{align*}
\color{blue}{(-1)^n \sum_{g=0}^m}&\color{blue}{ \binom{m}{g}\frac{B_{n+g+1}}{n+g+1}
+(-1)^m \sum_{g=0}^n \binom{n}{g}\frac{B_{m+g+1}}{m+g+1}}\\
&\color{blue}{= - \frac{1}{(n+m+1){\binom{n+m}{m}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4520057",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "17",
"answer_count": 4,
"answer_id": 0
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If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\ {g'(x)} & {x< k} \end{cases} $ , $\max (k)=?$ if $f$ is a differentiable function
If $g(x)=ax^2+bx+c$ and $f(x)= \begin{cases} {g(x)} & {x\ge k} \\
{g'(x)} & {x< k} \end{cases} $. If $f(x)$ is a differentiable
function, what is the maximum value of $k... | $$g(x) = (b + c)x^2 + bx + c$$
$$g'(x) = 2(b + c)x + b$$
$$g''(x) = 2(b + c)$$
For $f$ to be continuous at $k$, we must have $g(k) = g'(k)$,
$$(b + c)k^2 + bk + c = 2(b + c)k + b$$
$$(b + c)k^2 + (- b - 2c)k + (c - b) = 0$$
$$k = \frac{b + 2c \pm \sqrt{(-b-2c)^2 - 4(b+c)(c-b)}}{2(b+c)}$$
$$k = \frac{b + 2c \pm \sqrt{5b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4523509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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How did people come up with the formula $a^3+b^3=(a+b)(a^2-ab+b^2)$? Every resource that I've read proves the formula
$$ a^3 + b^3 = (a+b)(a^2-ab+b^2) \tag1$$
by just multiplying $(a+b)$ and $(a^2 - ab + b^2)$.
But how did people come up with that formula? Did they think like, "Oh, let's just multiply these polynomials... | As noticed by $a=-b$ we obtain $a^3+b^3=0$ therefore we can guess by homogeneity
$$a^3+b^3=(a+b)(Xa^2+Yab+Zb^2)$$
with $X$, $Y$ and $Z$ unknown, to obtain
$$(a+b)(Xa^2+Yab+Zb^2)=Xa^3+Ya^2b+Zab^2+Xa^2b+Yab^2+Zb^3=$$
$$=Xa^3+(Y+X)a^2b+(Z+Y)ab^2+Zb^3$$
which requires
*
*$X=1$
*$Z=1$
*$Y+X=0 \implies Y=-1$
*$Z+Y=0 \im... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Calculate $\sin^5\alpha-\cos^5\alpha$ if $\sin\alpha-\cos\alpha=\frac12$ Calculate $$\sin^5\alpha-\cos^5\alpha$$ if $\sin\alpha-\cos\alpha=\dfrac12$.
The main idea in problems like this is to write the expression that we need to calculate in terms of the given one (in this case we know $\sin\alpha-\cos\alpha=\frac12$).... | Alternative approach:
Let $~\displaystyle x = \sin(\alpha), y = \cos(\alpha) = \implies
(x - y) = \frac{1}{2}, ~\left(x^5 - y^5\right) = \color{red}{what?}$
Tools
*
*$T_1 ~: ~(x - y)^2 = x^2 - 2xy + y^2.$
*$T_2 ~: ~(x - y)^3 = x^3 - 3x^2y + 3xy^2 - y^3.$
*$T_3 ~: ~(x - y)^5 = x^5 - 5x^4y + 10x^3y^2 - 10x^2y^3 + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real Solutions of the expression $\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}-1=0$ I recently stumbled across the following question:
Find all the real solutions $x$ of the expression $\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}-1=0$
It was given by one of my teachers (he was teaching us a... | $\sqrt[2]{x+3-4\sqrt[2]{x-1}}+\sqrt[2]{x+8-6\sqrt[2]{x-1}}=1$
Let $x=1+t^2$, then we re-write the eq. as
$$\sqrt{(t-2)^2}+\sqrt{(t-3)^2}=1$$
$$\implies |t-2|+|t-3|=1$$
When $2\le t \le 3$. then the equation becomes
$$ (t-2)+(3-t)=1\implies 1=1$$
So all roots are $2\le t\le 3$ \implies $5 \le x \le 10.$
Hence this equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527266",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Find the all possible values of $a$, such that $4x^2-2ax+a^2-5a+4>0$ holds $\forall x\in (0,2)$
Problem: Find the all possible values of $a$, such that
$$4x^2-2ax+a^2-5a+4>0$$
holds $\forall x\in (0,2)$.
My work:
First, I rewrote the given inequality as follows:
$$
\begin{aligned}f(x)&=\left(2x-\frac a2\right)^2+\fr... | I suggest a solution different from OP; albeit simpler.
First, transform the condition $\forall x\in (0,2)$. Equivalent to this condition, introduce $x = \frac{2}{y}$ with $y >1$, so only one boundary has to be inspected and the term remains quadratic in $y$ and $a$:
The inequality $4x^2-2ax+a^2-5a+4>0$ becomes
$16-4a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528746",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 11,
"answer_id": 0
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Is this proof of $\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil$ correct? I've been practicing proving things about floor and ceiling functions, so I thought I'd try to prove this well-known identity:
$$\left\lfloor \frac{n}{m} \right\rfloor = \left\lceil \frac{n-m+1}{m} \right\rceil... | As in your previous question on floor and ceiling, I am finding the interval of both sides with inequalities.
Both sides are defined and are integers. They are respectively within these intervals:
$$\begin{array}{rrcl}
\text{LHS:}&\dfrac nm-1 <&\left\lfloor\dfrac nm\right\rfloor&\le \dfrac nm\\\hline
\text{RHS:}&\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4530070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$
Evaluate $\sin^2x = \frac{2+ \sqrt{3}}{4}$
Find value of $2x$
I worked it out as such:
$\sin^2x = \frac{2+ \sqrt{3}}{4} \implies \frac{1- \cos 2x}{2} = \frac{2+ \sqrt{3}}{4}$
$2 - 2 \cos 2x = 2 + \sqrt{3}$
$- 2 \cos 2x = + \sqrt{3}$
$ \cos 2x = \frac{- \sqrt{3}}{2}$
And $ 2x... | The reason why you have two solutions is because $\sin^2(x) = \sin^2(2\pi - x)$
therefore both solutions work.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Let $p(x)$ and $q(x)$ be the predicates: $p(x):x^2-3x+2=0$ and $q(x):(x-1)^{3}(x-2)(x-3)x=0$. Show: $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$. I would like to know if this is correct. Thanks in advance.
Let $p(x)$ and $q(x)$ be the predicates:
$p(x):x^2-3x+2=0$
$q(x):(x-1)^{3}(x-2)(x-3)x=0$
Show that: $\forall x... | I would rephrase like that:
then,
$(x-1)^{3}(x-2)(x-3)x= (x-1)^{2}(x-1)(x-2)(x-3)x=(x-1)^{2}(0)(x-3)x=0$
Then, $\forall x\in\mathbb{R}$, $p(x)\Rightarrow q(x)$.
or
To show $q(x): (x-1)^{3}(x-2)(x-3)x=0$
$L.H.S.= (x-1)^{3}(x-2)(x-3)x = (x-1)^{2}(x-1)(x-2)(x-3)x=(x-1)^{2}(0)(x-3)x = 0$
$R.H.S. = 0$
$L.H.S. = R.H.S.$
Then... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Representing $\frac{1}{x^2}$ in powers of $(x+2)$ I was asked to represent $f(x)=\frac{1}{x^2}$ in powers of $(x+2)$ using the fact that $\frac{1}{1 − x} = 1 + x + x^2 + x^3 + ...$. I am able to represent $\frac{1}{x^2}$ as a power series, but I am struggling withdoing it in powers of $(x+2)$. This is what I attempted.... | Let $u = x + 2$ so that $x = u - 2$. Then,
$$
\frac{1}{x^2} = \frac{1}{(u - 2)^2}
= \frac{1}{(2 - u)^2}
= \frac{1}{4 \bigl(1 - \tfrac12 u \bigr)^2}
= \frac{1}{4} \frac{1}{(1 - v)^2}
$$
where $v = \tfrac12 u$ or $u = 2v$.
We can derive a nice power series for $(1 - v)^{-2}$ centered at $v = 0$ by differentiating the ge... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4534269",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Is there any natural number between $(N^2 + N)$ and $(N^2 + 1)$ that can divide $(N^2 + N)\times(N^2 + 1)$ My question is fairly simple and explained in title. I'm trying to prove that there are no natural number(s) between $(n^2+n)$ and $(n^2+1)$ that can divide $(n^2+1) \times (n^2 + n)$
[EDIT]
I was trying to solve ... | A proof by contradiction. Assume there is a $n^{2}+1+k|(n^{2}+n)(n^{2}+1)$
where $1\leq\,k\,\leq\,n-1$. Then
$n^{2}+k+1|(n^{2}+n)(n^{2}+k-k+1)$ hence $n^{2}+k+1|(n^{2}+n)k$ and then
$n^{2}+k+1|(n^{2}+k+1-k-1+n)k$ hence $n^{2}+k+1|-(k+1-n)k$ and the latter
is positive ($k=n-1$ is not acceptable since it gives $n^{2}+k+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4535406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$
Let $S$ be the set of all $3\times 3$ matrices with entries in $\{0,1,-1\}$. Prove that $\{\det(A) : A\in S\} = \{-4,-3,-2,-1,0,1,2,3,4\}$.
Let $R = [-4,4]\cap \mathbb{Z}$, $T =\{\det(A) : A\in S\}$. To show $R\subseteq T,$ it suffices to show that $R\cap ... | Clearly $\det(A)$ is an integer. We first show that $|\det(A)|\leq4$. There are only two possibilities:
*
*Some row of $A$ is at most two nonzero elements. By Laplace expansion along that row, we see that $|\det(A)|$ is at most the sum of the absolute values of two minors in $A$.
*$A$ is entrywise nonzero. If $A$ is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538325",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Proving $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ I hope I'm allowed to ask this question here, but I have to prove that $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ using the following Fourier series:
$$
1-\frac{1}{e} + \sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n... | Hint:
Try to see what happens for $x=1$.
Explanation:
Your progress so far has indicated that splitting the sum into even and odd sums may be useful. To that extend you could define
$$A=\sum_{n=0}^\infty \frac{2}{1+(2n+1)^2 \pi^2}$$ and
$$B=\sum_{n=1}^\infty \frac{2}{1+(2n)^2 \pi^2}.$$
Now, you already have one linear ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539933",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 2,
"answer_id": 0
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Inverse of the function $f(x)=\sqrt{x-3}-\sqrt{4-x}+1$ I am trying to find the inverse of the function $$f(x)=\sqrt{x-3}-\sqrt{4-x}+1$$
First of all its domain is $[3,4]$
As far as my knowledge is concerned, since $f'(x)>0$, it is monotone increasing in $[3,4]$, so it is injective. Also the Range is $[0,2]$. So $$f:[3,... | We want to solve
$$y=\sqrt{x-3}-\sqrt{4-x}+1$$
for $x$. Start by making the substitution $x=u^2+3$ (with $u\geq0$) so that
$$y=u-\sqrt{1-u^2}+1.$$
Rearranging and squaring we get
$$u^2-2u(y-1)+(y-1)^2=1-u^2.$$
Which can be rewritten as
$$u^2-u(y-1)+\frac{(y-1)^2-1}{2}=0.$$
Applying the appropriate quadratic formula we ... | {
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"url": "https://math.stackexchange.com/questions/4541103",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Integrating Question with U-Sub: $\int \tan^2\theta \sec^4\theta\ d\theta$ Question:
Find the value of the expression $$\int \tan^2\theta \sec^4\theta\ d\theta$$ by using the substitution $u=\tan\theta$.
My Working:
If $u=\tan\theta$, then
\begin{align}
\frac{du}{d\theta}&=\sec^2\theta\\
du&=\sec^2\theta\cdot d\theta\\... | \begin{align}
\int\tan^2\theta \sec^4\theta\ d\theta&=\int \tan^2\theta\cdot \sec^2\theta\cdot(\sec^2\theta\cdot d\theta)\qquad(\text{say, }\tan\theta=u)\\
&=\int u^2\cdot (1+u^2)\cdot du\\
&=\int u^2 du +\int u^4 du\\
&=\frac{u^3}{3}+\frac{u^5}{5}+c\qquad(\text{some constant, }c)\\
&=\frac{\tan^3\theta}{3}+\frac{\tan^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4547887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving Summation $(2 + 1 \sum_{n=1}^{m} \frac{2}{n}) - (2 + \sum_{n=2}^{m-1} \frac{4}{n+1} + \frac{4}{m+1})$... I saw this in a textbook and I am not sure how to get from the LHS to the RHS:
$$\left(2 + 1 + \sum_{n=1}^{m} \frac{2}{n}\right) - \left(2 + \sum_{n=2}^{m-1} \frac{4}{n+1} + \frac{4}{m+1}\right) + \left(\sum... | The series as written:
$$\left(2 + 1 + \sum_{n=1}^{m} \frac{2}{n}\right) - \left(2 + \sum_{n=2}^{m-1} \frac{4}{n+1} + \frac{4}{m+1}\right) + \left(\sum_{n=1}^{m-2} \frac{2}{n+2} + \frac{2}{m+1} + \frac{2}{m+2}\right) $$
gives
$$ S_{m} = 1 - \frac{2}{m+1} + \frac{2}{m+2} + 2 \, f_{m} $$
where
\begin{align}
f_{m} &= \sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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The value of $(a^2+b^2)(c^2+1)$. The question is:
Given real numbers $a,b,c$ that satisfy
$$ab(c^2-1)+c(a^2-b^2)=12$$
$$(a+b)c+(a-b)=7$$
Find the value of $(a^2+b^2)(c^2+1)$
From what I've done, I got $7(3ac+3a+3bc-b)-2ab(c+1)(c-1)=(a^2+b^2)(c^2+1)$. I think I'm inching further from the actual solution. Can anyone gi... | We have
$$
(a+b)^2c^2+2(a^2-b^2)c + (a-b)^2=49
\\2ab(c^2-1) +2(a^2-b^2)c = 24
$$
Subtract them and we get
$$(a^2+b^2)c^2+2ab+(a-b)^2=25$$
Thus $$(a^2+b^2)(c^2+1) = 25.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Solve the equation $(x-1)^5+(x+3)^5=242(x+1)$ Solve the equation $$(x-1)^5+(x+3)^5=242(x+1)$$ My idea was to let $x+1=t$ and use the formula $$a^5+b^5=(a+b)(a^4-a^3b+a^2b^2-ab^3+b^4),$$ but I have troubles to implement it. The equation becomes $$(t-2)^5+(t+2)^5=242t\\(t-2+t+2)\left[(t-2)^4-(t-2)^3(t+2)+(t-2)^2(t+2)^2-\... | That's a helpful start.
Notice $(t-2)^5+(t+2)^5$ is an odd function:
$$ ((-t)-2)^5+((-t)+2)^5 = -(t+2)^5-(t-2)^5 = -\left((t-2)^5+(t+2)^5\right) $$
Therefore $A = \frac{(t-2)^{5}+(t+2)^{5}}{2t}$ is an even function, so when it's multiplied out and collected to a basic polynomial form, that form must be $A = Bt^4+Ct^2+D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4555212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Integrating $\int \frac{3}{(x^2 +5)^2}dx$ by parts Integrating $$\int \frac{3}{(x^2 +5)^2}dx$$
After removing the constant, it is basically integrating $\frac{1}{x^4+10x^2+25}$. I only have learnt up to integrating $\frac{1}{ax^2 + bx +c}$ with the highest power of $x$ is 2. And this cannot be broken up into partial fr... | $$\int\frac3{2x}\cdot\frac{2x}{(x^2+5)^2}dx\\=\frac3{2x}\cdot\frac{-1}{(x^2+5)}-\int\frac{-3}{2x^2}\cdot\frac{-1}{x^2+5}dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10}\int\frac{x^2+5-x^2}{x^2(x^2+5)}dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10}\int(\frac1{x^2}-\frac1{x^2+5})dx\\=\frac{-3}{2x(x^2+5)}-\frac3{10x}+\frac3{10\sqrt5}\tan^{-1}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4556204",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Number of strings of length $n$ with no consecutive $y$'s Suppose we have a set $S$ such that $\lvert S\rvert=k+1$. Fix an element $y$ in $S$. We want to find the recurrence relation on the number of $S$-strings of length $n$ that don't have two consecutive $y$'s, namely $yy$. Use $f(n)$ to denote the answer.
I already... | This answer is based upon the Goulden-Jackson Cluster Method. We consider the set of words of length $n\geq 0$ built from an alphabet $S$ of $k+1$ letters with $y\in S$ and the set $B=\{yy\}$ of bad words, which are not allowed to be part of the words we are looking for. We derive a generating function $F(z)$ with the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
In how many ways can points be assigned to five questions with one to four points per question so that the total is $14$? A professor must write an exam with $5$ questions. Question number i should give $p_i \in \mathbb{Z}$ points. The sum of the points must be $14$ and each question must give at least $1$ point and a ... | Our generating function is
$\begin {align*}
f(x)&=\left ( x+x^2+x^3+x^4 \right )^5\\&=x^5\left ( \frac{1-x^4}{1-x} \right )^5
\end{align*}$
So the answer is
$\begin {align*}
[x^{14}]f(x)&=[x^9]\left ( 1-5x^4+10x^8-h(x) \right )\sum_{k=0}^{\infty }\binom{-5}{k}(-x)^k\\&=\left ( [x^9]-5[x^5]+10[x] \right )\binom{k+4}{4}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How to solve $\int_0^{2}\frac{x\log(2-x)}{1+x^2} \; dx $ I have problem to find the value of the integral:
$$ \int_0^{2}\frac{x\log(2-x)}{1+x^2} \; dx $$
The first step is to split the absolute part.
I thank you in advance for help.
| Using
\begin{eqnarray}
&&\int \frac{\log(b-x)}{x+a}\;dx\\
&=&\int \log(b-x)d\log\frac{x+a}{a+b}\;dx\\
&=&\log(b-x)\log\frac{x+a}{a+b}-\int \frac1{b-x}\log\frac{x+a}{a+b}\;dx\\
&=&\log(b-x)\log\frac{x+a}{a+b}-\int \frac1{b-x}\log\bigg(1-\frac{b-x}{a+b}\bigg)\;dx\\
&=&\log(b-x)\log\frac{x+a}{a+b}+\text{Li}_2(\frac{b-x}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Area enclosed between the roots of a quadratic Let $f(x)=ax^2+bx+c$
If $f(x)$ has roots $\alpha$ and $\beta$, what is the area enclosed by $f(x)$ and the $x$-axis between $x=\alpha$ and $x=\beta$ in terms of $a,b$ and $c$?
It is also given that $\alpha>\beta.$
If $a=1$, then I thought this might be easier since you ge... | To pursue your approach, not restricting the polynomial to $ \ a \ = \ 1 \ \ , $ we have
$$ A \ \ = \ \ a \int_\beta^\alpha \ \ x^2 \ - \ (\alpha+\beta)x \ + \ \alpha \beta \ \ dx $$
[you have an extra "x" on what should be the "constant term" : I presume that was a typo]
$$ = \ \ a·\left( \ \frac13x^3 \ - \ \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove that $\frac{\cos2\theta}{\sin6\theta}+\frac{\cos6\theta}{\sin18\theta}+\frac{\cos18\theta}{\sin54\theta}=\frac12(\cot2\theta-\cot54\theta)$
Prove that $\frac{\cos2\theta}{\sin6\theta}+\frac{\cos6\theta}{\sin18\theta}+\frac{\cos18\theta}{\sin54\theta}=\frac12(\cot2\theta-\cot54\theta)$
I tried formulas for $\sin... | The triplication formulas for $\;\sin\;$ and $\;\cos\;$ are
$$ \sin(3x) \!=\! \sin(x)(3\cos(x)^2 \!-\! \sin(x)^2), \;
\cos(3x) \!=\! \cos(x)(\cos(x)^2 \!-\! 3\sin(x)^2). $$
Uisng these and $\;\cos(x)^2 + \sin(x)^2 = 1\;$ implies that
$$ 1 \!-\! \frac{\cot(3x)}{\cot(x)} \!=\!
1 \!-\! \frac{ \cos(x)^2 \!-\! 3\sin(x)^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566665",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Better approach to evaluate the limit $\lim_{x\to0^+}(\cot x-\frac{1}{x})(\cot x+\frac{1}{x})$ I solved it by rewriting the limit as indeterminate form $0/0$, then apply L'Hopital's rule 4 times, It was really lengthy and easy to make mistakes, If anyone got a better approach, please tell me!
$$
\begin{align}
\lim_{x\t... | $\cot^2x-\frac1{x^2}=\csc^2x-1-\frac1{x^2}=\frac1{\sin^2x}-\frac1{x^2}-1=\frac{x^2-\sin^2x}{x^2\sin^2x}-1=\frac{(x-\sin x)(x+\sin x)}{x^4\frac{\sin^2x}{x^2}}-1$
In the numerator, using $\sin x=x-\frac{x^3}{3!}+\frac{x^5}{5!}...$, thus,
$\frac{(\frac{x^3}{3!}-\frac{x^5}{5!}+...)(2x-\frac{x^3}{3!}+...)}{x^4\frac{\sin^2x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show there's no integer between $x_n$ and $y_n$ For positive integers $n$, let $x_n, y_n$ be defined as:
$x_n = \sqrt{n} + \sqrt{n+1}$ and $y_n = \sqrt{4n + 2}$
Show that there is no whole number between $x_n$ and $y_n$.
Observation:
Obviously $x_n \not\in \mathbb{N}$ since $n$ and $n+1$ cannot both be perfect squares ... | A very straightforward approach can be as below;
Assume $m^2\lt n \le (m+1)^2$. Hence;
$$2m \lt \sqrt n+ \sqrt {n+1}\lt 2m+3,$$
and;
$$2m\lt \sqrt {4n+2}\lt 2m+3.$$
Therefore $k$ (the integer between $x_n$ and $y_n$) is either $2m+1$ or $2m+2$. But It is impossible for $k$ to be $2m+2$ because, in this manner, $n$ has ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Finding all the rational points of the curve $x^2 -2y^2 =1$ This question was left as an exercise in my class of number theory and I want to verify my solution.
Question: Let $C$ be a curve given by $x^2 -2 y^2=1$. Find all the rational points on $C$.
Attempt: $( 1,0 )$ is a solution and let m be the slope of the lin... | What you did is basically OK. However, there are several places that could be improved.
A detailed solution.
If $x=1$, then $y=0$.
Suppose $(x,y)$ is a rational solution $(x,y)$ with $x\not=1$. Let the slope of the line through $(1,0)$ and $(x,y)$ be $t=\frac{y-0}{x-1}=\frac y{x-1}$, which must be a rational number.
$t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How do we evaluate $\sum_{k=1}^{\infty} \frac{1}{3k-1}-\frac{1}{3k}$ We have to evaulate $$\sum_{k=1}^{\infty} \frac{1}{3k-1}-\frac{1}{3k}$$
I have tried to expand it to see some kind of pattern, but there seems to be no obvious one! Then I tried to just mindlessly integrate and after solving the integral $\int_1^{\inf... | I think @ThomasAndrews' comment is worthy of becoming an answer in its own right.
We obtain
\begin{align*}
\color{blue}{\sum_{k=1}^{\infty}}\color{blue}{\left(\frac{1}{3k-1}-\frac{1}{3k}\right)}&=\sum_{k=1}^{\infty}\int_{0}^1\left(x^{3k-2}-x^{3k-1}\right)dx\\
&=\int_0^1\sum_{k=0}^{\infty}\left(x^{3k+1}-x^{3k+2}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4569888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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Prove that $\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$ for $x > 0$
Given $x>0$ , prove that $$\frac{\ln x}{x}+\frac{1}{e^x}<\frac{1}{2}$$
I have tried to construct $F(x)=\frac{\ln x}{x}+\frac{1}{e^x}$ and find the derivative function of $F(x)$ to find the maximun value, but I can't solve the transcendental equation.
... | Update: I found a simpler proof.
It suffices to prove that
$$\frac{x}{2} - \ln x - x\mathrm{e}^{-x} > 0. $$
Let $p := \frac{23}{37} - \frac{14}{23}\mathrm{e}^{-37/23}$.
Since $p < \frac12$, it suffices to prove that
$$f(x) := px - \ln x - x\mathrm{e}^{-x} > 0. \tag{1}$$
We have
$$f'(x) = p - \frac{1}{x} + \mathrm{e}^{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4573486",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 1
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How can we describe this factorization process? We know that, there exist infinite number of algebraic expressions, so that we can not factorise these with real coefficients. For example,
*
*$a^2+b^2$
*$a^2+b^2+c^2$
The simplest example, $a^2+b^2$ can not be factored over $\Bbb R$.
But, I want to consider the facto... | Note that if
$a=c^2$ and $b=2d^2$
then
$a^2+b^2=(a+b)^2-2ab=(a+b-\sqrt {2ab})(a+b+\sqrt{2ab})$
becomes
$c^4+4d^4
=(c^2+2d^2)^2-4c^2d^2
=(c^2+2d^2-2cd)(c^2+2d^2+2cd)
$
which is an
unexpected factorization
with integer coefficients.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4577459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Sum of two odd cubes plus four times a cube is zero Let $x,y$ be odd integers and let $z$ be an integer. The question is to find all solutions to the equation,
$$x^3+y^3+4z^3=0$$
Of course we have the trivial solution $(x,-x,0)$. Are there any others?
By considering the equation modulo $4$ we see that wlog $x=4k+1$ and... | $x^3+y^3+4z^3=0$
Let $X=x/z, Y=y/z$ then we get $X^3+Y^3=-4$.
In general, $X^3+Y^3=n$ can be transformed to elliptic curve $v^2=u^3-432n^2$.
Hence we get $v^2=u^3-6912$.
According to LMFDB, this elliptic curve has rank $0$ and has no integer solution.
Hence $x^3+y^3+4z^3=0$ has no nontrivial solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Find limit of trigonometric function with indeterminacy Find limit of the given function:
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arctan(3x^2)} - 1)}{(1-\cos\tan6x)\ln(1-\sqrt{\sin x^2})}
$$
I tried putting 0 instead of x
$$\lim_{x\rightarrow0} \frac{(4^{\arcsin(x^2)} - 1)(\sqrt[10]{1 - \arct... | As the comment suggests, you shall use Taylor expansion up to $O(3)$, that is:
$$4^{\arcsin(x^2)} \approx 1+x^2 \log (4)+O\left(x^4\right)$$
$$\sqrt[10]{1- \arctan(3x^2)} \approx 1-\frac{3 x^2}{10}+O\left(x^4\right)$$
$$\cos(\tan(6x)) \approx 1-18 x^2+O\left(x^4\right)$$
$$\sqrt{\sin(x^2)} \approx x + O(x^4)$$
Plug the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4582099",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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In $\triangle ABC$, $D$ lies on $BC$, $\angle ADB=60$, $\angle ACB=45$ and $BD=2CD$. Find $\angle ABC$ The question is as the title states. In the following figure, with some given angles and two sides, the goal is to find the measure of $\angle ABC$. This problem is inspired by one which appeared in a local Math conte... | Let $\angle ABC = x $, then applying the law of sines on $\triangle ADC $:
$\dfrac{AD}{\sin 45^\circ} = \dfrac{a}{\sin 15^\circ} $
Applying the law of sines on $\triangle ADB$:
$\dfrac{AD}{\sin x} = \dfrac{2 a}{\sin(120^\circ - x) } $
Dividing the two equation to eliminate $AD$ and $a$
$ \dfrac{ \sin x }{ \sin 45^\circ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584468",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Prove $^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,... How can we prove that
$^{6n+2} − ^{6n+1} + 1$ is always divisible by $^2 − + 1$; = 1, 2, 3,...
I attempted to solve this with Mathematical Induction as follows:
Let s(n) = $x^2 - x + 1$ | $^{6n+2} − ^{6n+1} + 1$; = 1, 2, 3,..
Basic Ste... | you could use the relationship between S(m) and S(m+1) such that:
$ x^{6m+2} − x^{6m+1} + 1 $ and $ x^6 * (x^{6m+2} − x^{6m+1} )+1 $ if you use the famous trick of +1 -1 in () we get :
S(m+1) = $ x^6 * (x^{6m+2} − x^{6m+1} +1 )/(x^2 −x +1) +(1-x^6) / (x^2 −x +1) $
= $ (x^6 )* S(m) + (1-x^6) / (x^2 −x +1) $
=$ (x^6 )* ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
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The parabola with the equation $y=-x^2+4x+8$ is shifted so that it passes through the points (1,1) and (3,5). Find the equation of the new parabola. Given the points $(1,1)$ and $(3,5)$, the vertex form would be:
$1=(1-h)^2+k$ for $(1,1)$ and
$5=(3-h)^2+k$ for $(3,5)$.
With a system of equations, I obtain that $h = ... | "Shifted" here means translated.
In particular, after being shifted by $p$ units in the $y$ direction and $q$ units in the $x$ direction, the shifted parabola satisfies the equation
$$
(y-p)
= (x - q)^2 + 4(x - q) + 8.
$$
Now, the problem is to solve the above for $p$ and $q$ such that $(x, y) =(1, 1)$ and $(3, 5)$ sa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4586509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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Calculate $\int_{0}^{\infty}\frac{1}{(x^2 + t^2)^n}dx$ for $t > 0, n \ge 1$ The problem is as follows:
Show that for all $t > 0, n \ge 1$,
$$
\int_{0}^{\infty}\frac{1}{(x^2 + t^2)^n}dx = {2n-2 \choose n -1}\frac{\pi}{(2t)^{2n -1}}
$$
What I have so far:
Let $f(t) = \int_{0}^{\infty}\frac{1}{(x^2 + t^2)^n}dx$
$$
\frac{d... | Let $$ I_{n} = \int_{0}^{\infty} \frac{dx}{(x^2 + t^2)^n} $$ then by using integration by parts this becomes
\begin{align}
I_{n} &= \left[\frac{x}{(x^2 + t^2)^{n+1}}\right]_{0}^{\infty} + 2 \, n \, \int_{0}^{\infty} \frac{x^2 \, dx}{(x^2 + t^2)^{n+1}} \\
&= 2 \, n \, \int_{0}^{\infty} \frac{(x^2 + t^2) - t^2}{(x^2 + t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4588953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluate $\int\frac{1}{x-\sqrt{1-x^2}}dx$ The integral $I$ in question is defined as follows
$$
I \equiv \int\frac{1}{x-\sqrt{1-x^2}}dx
$$
To solve this, I tried the trig substitution $x = \sin\theta$, with $dx = \cos\theta d\theta$, and rewrote the integral as follows
$$
\int\frac{\cos\theta}{\sin\theta-\sqrt{1-\sin^2... | I approached this integral by trigonometric substitution and partial fraction.
Assume the followings:
$x=\sin\theta\ \ldots(1)\\
\text{Hence, }dx=\cos\theta\ d\theta\ \ldots(2)\\
\text{Also, }\cos\theta=\sqrt{1-x^2}\ \ldots(3)\\$
Then,
$\operatorname{\Large\int}\dfrac{1}{x-\sqrt{1-x^2}}dx=\operatorname{\Large\int}\dfra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589716",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Finding the set of values of k for which a modulus equation has exactly 4 roots In my assignment, I have the following question:
Find the set of values of k for which the equation $|x^2-1|+x=k$ has
exactly four roots.
What I've tried:
Removed the modulus and made two different equations $x^2+x-(k+1)=0$ and $x^2-x+(k-... |
Find the set of values of k for which the equation $|x^2-1|+x=k$ has exactly four roots.
What I've tried:
Removed the modulus and made two different equations $x^2+x-(k+1)=0$ and $x^2-x+(k-1)=0$.
You have two cases
*
*$(x^2 - 1) \leq 0.$
*$(x^2 - 1) \geq 0.$
Note the overlap in the two cases, where $~x^2 - 1 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to show that every non-trivial orbit of ODE is a periodic circle I am working with this system of first-order ODEs.
\begin{align}
\dot{x} & = y^2 - x^2 \ ,\\
\dot{y} & = - 2 x y\ .
\end{align}
We were told to convert the system using $z=x+iy$ where I found that $z'=-z^2$. From there we were asked to show that every... | I will ask you to criticize my decision, because I just thought of it, there may be inaccuracies.
I assumed that
$$ t, x(t),y(t) \in \mathbb{R}$$
$$ z(t) = x(t) + y(t)i , $$
$$ \frac{dz}{dt} = \frac{dx}{dt} + \frac{dy}{dt}i$$
$$ -z^{2} = (y^2 - x^2) - 2x\cdot y \cdot i $$
$$ \frac{dz}{dt} = -z^2$$
$$ \frac{1}{z} = t +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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A closed form expression of $\sum_{n \ge 0} \biggl( \sum_{k=1}^n \frac{1}{k} \biggr)z^n$
I am working on the following exercise: Use the identity $\frac{1}{1-z} = \sum_{n \ge 0} z^n$ and elementary operations on power series (addition, multiplication, integration, differentiation) to find closed form expressions of th... | Where these series converge,
\begin{align}
&\Bigl( \frac11 \Bigr) \, z
+ \Bigl( \frac11 + \frac12 \Bigr) \, z^2
+ \Bigl( \frac11 + \frac12 + \frac13 \Bigr) \, z^3
+ \cdots \\
&\qquad = \frac11 \Bigl( z + z^2 + z^3 + \cdots \Bigr)
+ \frac12 \Bigl( z^2 + z^3 + \cdots \Bigr)
+ \frac13 \Bigl( z^3 + \cdots \Bigr) \\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594671",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solve the radical equation for all reals: $x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$ Question:
Solve the radical equation for all reals: $$x\left(1+\sqrt{1-x^2}\right)=\sqrt{1+x^2}$$
My approach:
$$1+\sqrt {1-x^2}=\frac {\sqrt {1+x^2}}{x}\\1+2\sqrt {1-x^2}+1-x^2=\frac{1+x^2}{x^2}\\4(1-x^2)=\left(\frac{1+x^2}{x^2}+x^... | Firstly $x>0$
Let $\theta=\sin^{-1}{x} \Rightarrow x=\sin \theta$
We get
$$\sin \theta(1+\cos \theta)=\sqrt{1+\sin ^2 \theta} $$
Squaring both sides:
$$\begin{aligned}
& \left(1-\cos ^2 \theta\right)(1+\cos \theta)^2=2-\cos ^2 \theta \\
\Rightarrow \quad & \cos ^4 \theta+2 \cos ^3 \theta-\cos ^2 \theta-2 \cos \theta+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Plot of function $(x^\frac{1}{2} + y^\frac{1}{2})^2 = 5$ Can someone explain me how can we make a plot of function:
$$ (x^\frac{1}{2} + y^\frac{1}{2})^2 = 5 $$
I tried to find a first derivative and a second derivative to understand something about this plot of function but it was useless as I understand. What should I... | $$
\begin{align}
& (x^{\frac{1}{2}}+y^{\frac{1}{2}})^{2} = 5\\
\implies & x^{\frac{1}{2}}+y^{\frac{1}{2}} = \pm \sqrt{5}\\
\implies & \sqrt{x}+\sqrt{y} = \pm \sqrt{5}
\end{align}
$$
The inequality $0\le x,y\le 5$ holds for real $x$ and $y$, as mentioned in the previous answers.
Method 1: Plot $f(x,y)=|\sqrt{x} + \sqrt{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4595481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding integer roots of an equation
Find all negative integer solutions of x and y such that
$$
x^3 + 3y^2 = xy^2 + 27
$$
To begin, I isolated $x$ and $y$ to see if there is anything I could do and got
$$
y = \pm \sqrt{x^2+3x+9} \\
x = \dfrac{1}{2} \left(-3 \pm \sqrt{y^2-27}\right)
$$
Since $x$ and $y$ are negative ... | The equation $x^3 + 3y^2 = xy^2 + 27$ can be written as
$$x^3-27+3y^2-xy^2=0$$
$$(x-3)(x^2+3x+9)+y^2(3-x)=0$$
$$(x-3)(x^2+3x+9-y^2)=0$$
Now, $x^2+3x+9-y^2=0$ can be written as
$$\bigg(x+\frac 32\bigg)^2+\frac{27}{4}-y^2=0$$
Multiplying the both sides by $4$ gives
$$(2x+3)^2+27-(2y)^2=0$$
and so
$$(2y-2x-3)(2y+2x+3)=27$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Show that $b^2+c^2-a^2\leq bc$. Let $a,b,c>0$ such that $b<\sqrt{ac}$, $c<\frac{2ab}{a+b}$. Show that $b^2+c^2-a^2\leq bc$.
I tried to construct a triangle with $a,b,c$ and to apply The cosine rule, but I am not sure that it's possible to construct it and also I have no idea how to prove that an angle it's greater than... | Using $ac - b^2 > 0$ and $2ab - (a + b)c > 0$, we have
\begin{align*}
&bc - (b^2 + c^2 - a^2)\\
>\,&bc - (b^2 + c^2 - a^2) - (ac - b^2) - [2ab - (a + b)c]\\
=\,& a^2 - 2ab + 2bc - c^2\\
=\,&(a - c)(a + c - 2b)\\
>\,& 0
\end{align*}
where we have used $a > c$ and $a + c > 2b$.
(Explanations:
By AM-GM, we have
$2b <... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4597953",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
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Evaluation of $~\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta$ $$
I:=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}\mathrm d\theta
$$
My tries
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\the... | For the antiderivative, a little bit faster could be $${\cos(\theta)^2-\sin(\theta)^2\over\sin(\theta)^4+\cos(\theta)^4}=\frac{4 \cos (2 \theta )}{3+\cos (4 \theta )}$$
$$\theta=\tan ^{-1}(t)\quad \implies I= \int \frac{4 \cos (2 \theta )}{3+\cos (4 \theta )}\,d\theta=\int \frac{1-t^2}{t^4+1}\,dt$$
$$\frac{1-t^2}{t^4+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 5
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Finding the angles of a right triangle if $\frac{\text{area of triangle}}{\text{area of incircle}}=\frac{2\sqrt{3}+3}{\pi}$
Let $ABC$ be a right triangle with $\measuredangle ACB=90^\circ$. If $k(O;r)$ is the incircle of the triangle and
$$\dfrac{S_{ABC}}{S_k}=\dfrac{2\sqrt3+3}{\pi}$$ find the angles of the triangle.
... | Your approach is reasonable. You have correctly expressed the ratio of the areas of the triangle and the incircle in terms of the side lengths of the triangle and the radius of the incircle.
To proceed, and find find the angles you can follow these steps:
Use the formula for the sine of the angle $\alpha$ in terms of t... | {
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"url": "https://math.stackexchange.com/questions/4603334",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove or disprove that the inequality is valid if $x,y,z$ are positive numbers and $xyz=1$. Prove or disprove that the inequality
$$\dfrac{1}{\sqrt{1+x}}+\dfrac{1}{\sqrt{1+y}}+\dfrac{1}{\sqrt{1+z}} \geq 1$$
is valid if $x,y,z$ are positive numbers and $$xyz=1.$$
My solution is:
Let $$x=\dfrac{a}{b}, y=\dfrac{b}{c}, z=\... | Let $x=\frac{a}{b},$ $y=\frac{b}{c}$, where $a$, $b$ and $c$ are positives.
Thus, $z=\frac{c}{a}$ and
$$\sum_{cyc}\frac{1}{\sqrt{1+x}}=\sum_{cyc}\frac{\sqrt{b}}{\sqrt{a+b}}\geq\sum_{cyc}\frac{\sqrt{b}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Solve $\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$ using hypergeometric functions How can we solve this integral?
$$\int_0^1\frac{x(x+1)^b}{\alpha-x}dx$$
I used geometric series
$$=\int_0^1(x+1)^b\sum_{n=0}^\infty\left(\frac{x}{\alpha}\right)^{n+1}dx=\sum_{n=0}^\infty\int_0^1\left(\frac{x}{\alpha}\right)^{n+1}(x+1)^bdx$$
Usin... | Answer for $\alpha> 1$ and $b\notin\Bbb Z$
For $\alpha>1$, we can write the integral as
$$
I=\int_0^1\frac{x(x+1)^b}{\alpha-x}dx=\int_0^1\frac{x(x+1)^b}{\alpha+1-(x+1)}dx
=
\frac{1}{\alpha+1}\int_0^1\frac{x(x+1)^b}{1-\frac{x+1}{\alpha+1}}dx,
$$
where $0<\frac{1}{\alpha+1}\le\frac{x+1}{\alpha+1}\le \frac{2}{\alpha+1}<1$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the greatest integer in the expression below as a function of the given conditions Find the greatest integer less than $\sqrt{2^{100}+10^{10}}.$
Answer: $2^{50}$
I tried, but I can't finish:
$\sqrt{2^{100}+(2\cdot5)^{10}}=\sqrt{2^{10}(2^{90}+5^{10})}=2^5\sqrt{2^{90}+5^{10}}$
| It results that
$2^{50}=\sqrt{2^{100}}<\sqrt{2^{100}+10^{10}}<\sqrt{2^{100}+16^{10}}=$
$=\sqrt{2^{100}+2^{40}}<\sqrt{2^{100}+2^{51}+1}=\sqrt{\left(2^{50}+1\right)^2}=2^{50}+1$
Hence ,
$2^{50}<\sqrt{2^{100}+{10^{10}}}<2^{50}+1\;.$
Consequently,
$\left\lfloor\sqrt{2^{100}+10^{10}}\right\rfloor=2^{50}\;.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4607743",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$ Without calculator prove that $9^{\sqrt{2}} < \sqrt{2}^9$.
My effort: I tried using the fact $9^{\sqrt{2}}<9^{1.5}=27.$
Also We have $512 <729 \Rightarrow 2^9<27^2 \Rightarrow 2^{\frac{9}{2}}<27 \Rightarrow \sqrt{2}^9=2^{4.5}<27$. But both are below $27$.
| $(\sqrt{2})^9 \quad vs.\quad 9^\sqrt{2}\qquad$// square both side
$2^9=512 \quad vs.\quad
3^{4\sqrt{2}} \;≈\; 3^{5.657} \;<\; 3^5\,3^\frac{2}{3}$
$\displaystyle 3^5\,3^\frac{2}{3} = 243×2×\sqrt[3]{1+\frac{1}{8}} \;<\;486×\left(1+\frac{1}{3×8}\right) = 506.25$
$→ (\sqrt{2})^9 \;>\; 9^\sqrt{2}\qquad$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 6,
"answer_id": 2
} |
Prove $\displaystyle\frac{H(x^2)}{H(x)}$ increases. For $x\in[0,1]$, let $f(x):=-x\ln x$ and the two-sample entropy function $H(x)=f(x)+f(1-x)$. Prove $h(x):=\displaystyle\frac{H(x^2)}{H(x)}$ increases.
Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.
The numerator of the derivati... | Here is my proof which is a bit cumbersome. I am seeking a much more elegant approach.
The numerator of the derivative of the sought fraction is
$$\begin{align}
&H(x)^2\frac{dh(x)}{dx} \\
=&\frac d{dx}H(x^2)H(x)-H(x^2)\frac d{dx}H(x) \\
=& 2x\ln\frac{x^2}{1-x^2}\,\big(x\ln x+(1-x)\ln(1-x)\big)-\big(x^2\ln x^2+(1-x^2)\l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 5,
"answer_id": 2
} |
find $\int^6_0f(x)dx$ by definition , $f(x)=\left\lfloor\frac{x}{3}\right\rfloor$
Let $f(x)=\lfloor\frac{x}{3}\rfloor$,
Is $f(x)$ integrable at $[0,6]$?
If so, find $\int^6_0f(x)dx$ by definition.
The function is integratable because it is bounded and it has a finite number of points which are not continuous in $[0,... | by your definition $\Delta x = \frac{b-a}{n} = \frac{6}{n}$, so $x_k = \frac{k \Delta x}{n}$. $k$ starts from $0$ to $n$. so the sum would be:
$$
S_n = \sum_{k=0}^n f(x_k)\Delta x = \sum_{k=0}^n \left\lfloor \frac{6k}{3n} \right\rfloor \frac{6}{n} = \sum_{k=0}^n \left\lfloor \frac{2k}{n} \right\rfloor \frac{6}{n}
$$
$2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4613663",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the value of $\frac{a+b}{10}$
If $\sin x+\cos x+\tan x+\cot x+\sec x +\csc x=7$, then assume that $\sin(2x)=a-b\sqrt7$, where $a$ and $b$ are rational numbers. Then find the value of $\frac{a+b}{10}$.
How to solve these kind of problems. I can make substitutions and convert all of them to $\sin$ and then solve f... | Let $\displaystyle \sin(x)+\cos(x)=z$
Then $[\sin(x)+\cos(x)]^2=z^2\Longrightarrow \sin(2x)=z^2-1$
So $\displaystyle \sin(x)+\cos(x)+\frac{2(\sin^2x+\cos^2x)}{\sin(2x)}+\frac{2(\sin x+\cos(x))}{\sin(2x)}=7$
$\displaystyle z+\frac{2}{z^2-1}+\frac{2z}{z^2-1}=7$
$\displaystyle z+\frac{2}{z-1}=7\Longrightarrow z^2-8z+9=0$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615546",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove or disprove that the inequality is valid if $x,y,z,u$ are positive numbers and $x+y+z+u=2$. Prove or disprove that the inequality $$ \dfrac{x^2}{\left(x^2+1\right)^2}+\dfrac{y^2}{\left(y^2+1\right)^2}+\dfrac{z^2}{\left(z^2+1\right)^2}+\dfrac{u^2}{\left(u^2+1\right)^2} \leq \dfrac{16}{25}$$ is valid if $x,y,z,u$ a... | Partial answer :
As user @alet show the inequality as the variable are not in $[0.5,2/3]$ I complete it :
Let $a,c,d\in[0.5,1/\sqrt{3}]$ and $b\in[0,0.5]$ such that $a+b+c+d=2$ and $a\geq d\ge c\ge b$ then we have :
$$af\left(a\right)+b\left(b\right)+cf\left(c\right)+df\left(d\right)\leq \left(2-b\right)f\left(\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4630608",
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"source": "stackexchange",
"question_score": "4",
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"answer_id": 0
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How many integer solutions of $2^x+3^y-7n=0,~n \in \mathbb Z$? Consider the equation $2^x+3^y=7n,~n \in \mathbb Z$.
How many integer (positive) solutions are there for this equation? Can it have infinitely many positive integer solutions?
What is an example of a positive rational solution?
To reduce the number of va... | Rewriting the equation in modulo 7, we have
$$3^y\equiv -2^x \quad \pmod 7 $$
Noting that $2^3\equiv 1 \quad \pmod 7$, we get $3$ cases for $x$,
$\quad $ A. $x=3h$,$ \quad $ B. $x=3k+1 \quad $ and $\quad $ C. $3k+2$.
A. When $x=3h, $
$$
3^y \equiv-(2^{3}) ^h\equiv-1 \equiv 6 \quad \pmod 7
$$
By the Fermat Littl... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 0
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Solve for all $x$ such that $x^3 = 2x + 1, x^4 = 3x + 2, x^5 = 5x + 3, x^6 = 8x +5 \cdots$ Question:
Solve for all $x$ such that $\begin{cases}&{x}^{3}=2x+1\\&{x}^{4}=3x+2\\&{x}^{5}=5x+3\\&{x}^{6}=8x+5\\&\vdots\end{cases}$.
My attempt:
I sum up everything.
$$\begin{aligned}\sum_{i=1}^n x^{i+2} &= (2 + 3 + 5 + 8 + \cdo... | Hint:
As $x(2x+1)\ne0,$
$$\dfrac{x^4}{x^3}=\dfrac{3x+2}{2x+1}\iff x^2-x-1=0$$
Similarly check, $$\dfrac{x^5}{x^4}=?,\dfrac{x^6}{x^5}=?$$
Find the common root of all?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4632276",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Find the sum of $\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$
Find the sum of $$\sum_{k=1}^n k(k-1) {{n} \choose {k-1}}$$
The solution in the book is a lot different than what I tried
what they did in the book is say $k-1=t$ then they expanded from this point $\sum_{t=0}^{n-1} (t+1)(n) {{n-1} \choose {t-1}}$ and the fin... | \begin{align}
\sum_{k=1}^n k(k-1) \binom{n}{k-1}
&= \sum_{k=2}^n k(k-1) \binom{n}{k-1} \\
&= \sum_{k=2}^n k n \binom{n-1}{k-2} \\
&= n \sum_{k=2}^n (k-2+2)\binom{n-1}{k-2} \\
&= n \sum_{k=2}^n (k-2)\binom{n-1}{k-2} + 2n \sum_{k=2}^n \binom{n-1}{k-2} \\
&= n (n-1) \sum_{k=3}^n \binom{n-2}{k-3} + 2n \sum_{k=2}^n \binom{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633052",
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"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Number of distinct arrangement of $(a,b,c,d,e)$
If $a<b<c<d<e $ be
positive integer such that
$a+b+c+d+e=20$.
Then number of distinct
arrangement of $(a,b,c,d,e)$ is
Here the largest value of $e$ is $10$
like $a\ b\ c\ d\ e$
as $ \ \ 1\ 2\ 3\ 4\ 10$
And least value is $6$
like $ a\ b\ c\ d\ e$
as $\ \ 2\ 3\... | For when $e=10$, there is only one arrangement of numbers that satisfies $a+b+c+d+e=20$, which is $(1,2,3,4,10)$. I am going to call this solution our default solution.
Consider when $e=9$, there are four different ways to make sure that the numbers sum to 20. You can choose one number out of $a,b,c,d$ from the default... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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"answer_id": 1
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A magic basis of $\mathbb{C}^5$ This is a a small $n$ restriction of another question.
Find a $5\times 5$ matrix of unit vectors $\xi_{ij}\in \mathbb{C}^5$ such that:
*
*Entries along rows and columns are orthogonal, that is for $1\leq i,j,k,l\leq 5$:
$$\delta_{i,k}+\delta_{j,l}=1\implies \langle \xi_{ij},\xi_{kl}\ra... | David Roberson has pointed me towards Figure 1 from Quantum symmetry vs nonlocal symmetry.
This is a so-called quantum latin square:
The produced quantum Latin squares are indexed by group elements and have the property that the inner product of the $i,j$-entry and $k,l$-entry depends only on $i^{-1}k$ and $j^{-1}l$. ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635100",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Find rational solutions to $x^2 + y^2 = 6$ This question comes from Rational Points on Elliptic Curves (Silverman & Tate) Exercise $1.7$ (a).
Find rational solutions (if any) to $x^2 + y^2 = 6$
I think there exist no solutions and here is my proof:
Suppose that there is a rational point and write it as
$$x=\frac{X}{Z... | The ideas in the proof are correct, but there are some minor errors that need correction to make your reasoning rigorous.
First, while it is true that $X,Z$ and $Y,Z$ may be taken to be relatively prime without loss of generality, and that this directly follows from $x = X/Z$ and $y = Y/Z$, it is not immediately obviou... | {
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"timestamp": "2023-03-29T00:00:00",
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Create unique table rotation for an event I am planning to host an event with 14 people participating in it. There will be five tables in total. (4 tables with 3 people each and 1 with 2).
I want to rotate the tables 5-6 times and to ensure that everyone gets to meet as many people as possible and minimize (if not comp... | Here's a 7-round schedule with every pair covered exactly once, obtained via integer linear programming:
1 {6,8,11} {10,12,14} {4,5,7} {3,9,13} {1,2}
2 {3,7,11} {1,4,10} {5,8,13} {2,6,14} {9,12}
3 {2,7,8} {5,9,10} {1,11,12} {3,4,14} {6,13}
4 {1,3,8} {4,6,9} {2,5,12} {10,11,13} {7,14}
5 {2,9,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4635982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Show this inequality $\dfrac{\sin{\frac{B}{2}}}{\sin{A}+\sin{B}}+\dfrac{\sin{\frac{C}{2}}}{\sin{A}+\sin{C}}\le\dfrac{1}{2\sin{A}}$ For any $\triangle ABC$, prove or disprove $$\dfrac{\sin{\frac{B}{2}}}{\sin{A}+\sin{B}}+\dfrac{\sin{\frac{C}{2}}}{\sin{A}+\sin{C}}\le\dfrac{1}{2\sin{A}}$$
by $$\sin{A}+\sin{B}=2\sin{\dfrac{... | Let's assume $a,b,c$ are the sides of the triangle. We may suppose that $a=x+y$, $b=y+z$, and $c=x+z.$ If $p=\frac{a+b+c}{2}$, one can show that:
$$\sin \frac{B}{2}=\sqrt {\frac{(p-a)(p-c)}{ac}} =\sqrt{\frac{yz}{(x+y)(x+z)}}\\\sin \frac{C}{2}=\sqrt {\frac{(p-a)(p-b)}{ab}}=\sqrt{\frac{zx}{(x+y)(y+z)}}.$$
Moreover, $\fra... | {
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How to simplify this trigonometric equation I am not sure how to further simplify this expression:
$$\sec^2(\arcsin(y/r)) \times \frac{\frac{1}{r}}{\sqrt{1-(\frac{y}{r})^2}} \times \frac{r}{1000}$$
How should I further simplify the $\sec^2(\arcsin(y/r))$?
The simplified version looks like this:
$$\left(\frac r{10\sqrt{... | Starting from: $$\sec^2\left(\arcsin\left(\frac yr\right)\right) \times \frac{\frac{1}{r}}{\sqrt{1-(\frac{y}{r})^2}} \times \frac{r}{1000}.\tag 1$$
We know that $$\sec \left(\arcsin \left(x\right)\right)=\frac{\sqrt{1-x^2}}{1-x^2}\implies \sec^2 \left(\arcsin \left(x\right)\right)=\frac{1}{1-x^2}$$ then rewrite the $(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4639882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Limit of $(|x| + |y|)\ln(x^2 + y^4)$ at $(0,0)$ I want to show that $$\lim\limits_{(x,y) \to (0,0)} (\lvert x \rvert + \lvert y \rvert)\ln(x^2 + y^4) = 0$$
First I let $\lVert (x,y) \rVert = \lvert x \rvert + \lvert y \rvert\ < \delta$, and assume that $x,y < 1$ so $x^2 + y^4 < \lvert x \rvert + \lvert y \rvert$. Then ... | I will use the following inequality in this answer :
$\ln z<z\;\;$ for any $\;z\in\Bbb R^+.\quad\color{blue}{(*)}$
For any $\,(x,y)\in\Bbb R^2\!\setminus\!\{(0,0)\}\,$ such that $\,x^2+y^2<1\,,\,$ it results that
$\begin{align}\color{blue}{-16\sqrt[4]{x^2+y^2}}&=-8\sqrt[4]{x^2+y^2}-8\sqrt[4]{x^2+y^2}\leqslant\\&\leqsla... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4641244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Solve the differential equation: $(x^2-y^2)dx+2xydy=0$. Given $(x^2-y^2)dx+2xydy=0$
My solution-
Divide the differential equation by $dx$
$\Rightarrow x^2-y^2+2xy\frac{dy}{dx}=0$
$\Rightarrow 2xy\frac{dy}{dx}=y^2-x^2$
Divide both sides by $2xy$
$\Rightarrow \frac{dy}{dx}=\frac{1}{2}[\frac{y}{x}-\frac{x}{y}]$
This is a ... | Your answer is correct. Here is another way to solve the DE:
$$(x^2-y^2)dx+2xydy=0$$
Divide by $x^2dx$:
$$1+\dfrac {2xyy'-y^2}{x^2}=0$$
$$1+\left (\dfrac {y^2}{x}\right)'=0$$
Integrate:
$$x+\dfrac {y^2}{x}=C$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4642892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{d}{dx}\left(3x^2+4\sqrt{x}\right) = 6x+\frac{2}{\sqrt{x}}$ Prove that $\frac{d}{dx}\left(3x^2+4\sqrt{x}\right) = 6x+\frac{2}{\sqrt{x}}$
Note 2 things here please: I KNOW the differentiation rules, I KNOW that $\frac{d}{dx}\left(3x^2+4\sqrt{x}\right) = 6x+\frac{2}{\sqrt{x}}$, but I'm having a trouble ... | You made a mistake. There was a $- 4\sqrt{x}$ not $+4\sqrt{x}$.
$$\begin{aligned}\lim_{h \to 0}\frac {3(x+h)^2+4\sqrt{x+h}-3x^2-4\sqrt{x}}{h}& = \lim_{h \to 0}\frac {3x^2 + 6xh+3h^2+4\sqrt{x+h}-3x^2-4\sqrt{x}}{h}\\& = \lim_{h \to 0}\frac {6xh+3h^2+4\sqrt{x+h}-4\sqrt{x}}{h}\\&= \lim_{h \to 0}\frac {6xh+3h^2}{h}+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4643697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How much information one root can give us about the other roots of a polynomial Suppose $a,b \in \mathbb{Q}$ and $f(X) \in \mathbb{Q}[X]$. In order to prove if $a+\sqrt{2}b$ is a root of $f(X)$, then $a-\sqrt{2}b$ is also a root, I did as follows.
I showed if $(a + \sqrt{2}b)^n=A_n+\sqrt{2}B_n$ then $(a - \sqrt{2}b)^n=... | The minimal polynomial of $\sqrt{\sqrt{2}}$ is $X^4-2$. If $f(a + \sqrt{\sqrt{2}} b) = 0$ where $f(X) \in \mathbb Q[X]$ and $b \ne 0$, that says $\sqrt{\sqrt{2}}$ is a root of
$f(a + X b)$, so $X^4 - 2 \mid f(a + X b)$, i.e. $\left(\frac{Y-a}{b}\right)^4 - 2 \mid f(Y)$. Now $a - \sqrt{\sqrt{2}} b$ is a root of $\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Complex numbers - solving for smallest positive value of $n$ Given that $z_1=2\sqrt{3}\operatorname{cis}\left(\frac{3\pi}{2}\right)$ and $z_2=2\operatorname{cis}\left(\frac{2\pi}{3}\right)$ find the smallest positive value of $n$ such that $\left(\frac{z_1}{z_2}\right)^n \in \Bbb{R}^+$.
My attempt:
$$\frac{z_1}{z_2}=\f... | For $n\in\Bbb Z$ we have
$\left(\frac{z_1}{z_2}\right)^n=\sqrt{3}^{\,n}\operatorname{cis}\frac{5n\pi}{6}\in\Bbb R^+$
iff $\operatorname{cis}\frac{5n\pi}{6}\in\Bbb R^+$
iff $(\,\cos \frac{5n\pi}{6}>0 \, \land \,\sin \frac{5n\pi}{6}=0\,)$
iff $(\,\cos \frac{5n\pi}{6}=1 \, \land \,\sin \frac{5n\pi}{6}=0\,)$
iff $\frac {5n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4645475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How many methods are there to evaluate $\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^{2n}}$? Background
As I had found the integral
$$I=\int_0^{\infty} \frac{1}{\left(x+\frac{1}{x}\right)^2} d x =\frac{\pi}{4}, $$
by using $x\mapsto \frac{1}{x}$ yields
$\displaystyle I=\int_0^{\infty} \frac{\frac{1}{x^2}}{\left(... | Writing the integral on the form
$$
I_n = \int_0^{\infty} \frac{x^{2n}}{(x^2+1)^{2n}}dx
$$
can be seen as a Mellin transform of $f(x)=(1+x)^{-\rho}$. The Mellin transform is given by
$$
\mathcal{M}[f(x);s]=\frac{1}{\Gamma(\rho)}\Gamma[s,\rho-s]$$
Which then gives that
$$
\mathcal{M}[f(x^2);s]=\frac{1}{2\Gamma(\rho)}\G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 2
} |
Integral with cosine in the denominator and undefined boundaries I am trying to solve the integral
$$ \int_0^{\frac{3}{2}\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx $$
And I get the primitive function to be
$$ \frac{2\arctan\left(\sqrt{\frac{7}{3}}\tan{x}\right)}{\sqrt{21}} $$
But $\tan(\frac{3}{2}\pi)$ is not defined. Ma... | Since the function is $\pi$-periodic, $$\int_{\pi}^{\frac{3}{2}\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx=\int_{0}^{\frac{1}{2}\pi}\frac{1}{\frac{5}{2}+\cos(2x)}dx.$$
It is also symmetric about $x=\frac{1}{2}\pi$-line and therefore by $x\rightarrow \pi-x$ substitution
$$\int_{\frac{1}{2}\pi}^{\pi}\frac{1}{\frac{5}{2}+\cos(2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4649102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$. I have the following question:
If $x^2-kx+1=0$, find the value of $x^3+\frac{1}{x^3}$.
Using the first equation, I rearrange to get $x^2=kx-1$. Then, I multiply both sides by x to get get $x^3=(kx-1)^{1.5}$. I can’t think of any other way than to substitute $(kx... | $$x^3+\frac{1}{x^3}=(x+\frac{1}{x})(x^2+\frac{1}{x^2}-1)\\=(x+\frac{1}{x})[(x+\frac{1}{x})^2-3]$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4652280",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| Let:
$$S = 1 + 2 + \ldots + (n-1) + n.$$ Write it backwards: $$S = n + (n-1) + \ldots + 2 + 1.$$
Add the two equations term by term; each addition results in $n+1.$ So:
$$2S = (n+1) + (n+1) + \ldots + (n+1) = n(n+1).$$
Divide by 2: $$S = \frac{n(n+1)}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
"answer_count": 36,
"answer_id": 23
} |
Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.