Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
What are different ways to compute $\int_{0}^{+\infty}\frac{\cos x}{a^2+x^2}dx$? I am interested to compute the following integral
$$I=\int_{0}^{+\infty}\frac{\cos x}{a^2+x^2}dx$$
where $a\in\mathbb{R}^+$. Let me explain my first idea. As the integrand is an even function of $x$ then
$$2I=\int_{-\infty}^{+\infty}\frac{... | A rather exotic approach leading to a known functional equation, which has an exponential solution.
Consider a function ($a>0$):
$$f(a)=a \int_{-\infty}^\infty \frac{\cos x}{a^2+x^2} dx=\int_{-\infty}^\infty \frac{\cos a x}{1+x^2} dx=\pi e^{-a}$$
Let's square it and change the dummy variable:
$$f^2(a)=\int_{-\infty}^\i... | {
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"url": "https://math.stackexchange.com/questions/2734148",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 1
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Evaluate $\int \frac{x\tan(x) \sec(x)}{(\tan(x)-x)^2}$ Find value of following integral $$\int \frac{x\tan(x) \sec(x)}{(\tan(x)-x)^2}\text{dx}$$
the numerator is $\text{d[sec(x)]}$ but that isnt work due to $x$ in denominator. First we can simplify as $$\int\frac{x\sin(x)}{(\sin(x)-x\cos(x))^2} \text{dx} = \int \frac{x... | \begin{array}{rcl}
\displaystyle \int \dfrac{x \sec x \tan x \,dx}{(\tan x - x)^2} &=& \displaystyle\int\color{red}{\left(-x\csc x\right)}\cdot \color{blue}{\left(\dfrac{-\tan^2 x}{\left(\tan x - x\right)^2}\right)\,dx}\\
&=&\displaystyle \int \color{red}{(-x\csc x)\cdot \color{blue}{d\left(\dfrac{1}{\tan x - x}\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does Tom Apostol deduce the inequality $\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}$ in section I 1.3 of his proof by induction example? In his book Calculus, Vol. 1, Tom Apostol writes the following proof for proving the following predicate by induction:
$$A(n): 1^2 + 2^2 + \cdots + (n - 1)^2 \lt \frac{n^3}{3}.$$
... | If $a<b, b<c; a<c$
So, as $1^2 + 2^2 + \cdots + k^2 \lt \dfrac{k^3}3+k^2,$
and if we can show $\dfrac{k^3}3+k^2<\dfrac{(k+1)^3}3$
we can safely conclude,
$$1^2 + 2^2 + \cdots + k^2 \lt\dfrac{(k+1)^3}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Equations involving $2 \times 2$ Matrices There are two possible values of A in the solution of the matrix equation
$\left[\begin{matrix}2A+1&-5\\-4&A\end{matrix}\right]^{-1}\cdot \left[\begin{matrix}A-5&B\\2A-2&C\end{matrix}\right]= \left[\begin{matrix}14&D\\E&F\end{matrix}\right]$ where $A,B,C,D,E,F \space all \spac... | First keep in mind that if $x_{1,2}$ are the roots of a quadratic $ax^2+bx +c=0,\,a\neq 0$ then
$$|x_1-x_2|={\sqrt{\Delta}\over a}$$
Second you do not need to invert the first matrix just multiply both sides with it so the matrix equation is equivalent to
$$\begin{bmatrix} A-5 & B\\2A-2 & C\end{bmatrix}=\begin{bmatrix}... | {
"language": "en",
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How to solve the cubic $x^3-3x+1=0$? This was a multiple choice question with options being
$$(A)-\cos\frac{2\pi}{9},\cos\frac{8\pi}{9},\cos\frac{14\pi}{9} \\
(B)-2\cos\frac{2\pi}{9},2\cos\frac{8\pi}{9},2\cos\frac{14\pi}{9} \\
(C)-\cos\frac{2\pi}{11},\cos\frac{8\pi}{11},\cos\frac{14\pi}{11} \\
(D)-2\cos\frac{... | To follow on from your method, $\cos(a) + \cos(b) = 2 \cos({a+b\over2}) \cos({a-b\over2})$, so $$\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}=2 \cos\frac{5\pi}{9}\cos\frac{3\pi}9=\cos\frac{5\pi}{9}$$
but $\cos\frac{5\pi}{9} = -\cos\frac{14\pi}{9}$, so the sum $\cos\frac{2\pi}{9}+\cos\frac{8\pi}{9}+\cos\frac{14\pi}{9}=0$.
| {
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"source": "stackexchange",
"question_score": "1",
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I want to verify my proofs. 1).Prove: $\lim_{n \to \infty} \frac{3n^2-4}{n^2-4} = 3$.
A proof: there exists an $\epsilon>0$. We define $N=\lceil{\sqrt{\frac{8}{\epsilon}+4}}\rceil$ (notice $N>2$). So:
$$
\begin{aligned}
|\frac{3n^2-4}{n^2-4}-3| < \epsilon \Leftrightarrow |\frac{3n^2-4-3n^2+12}{n^2-4}|<\epsilon \Leftr... | As an alternative to simplify note that
$$\lim_{n \to \infty} \frac{3n^2-4}{n^2-4} = \lim_{n \to \infty} \frac{3n^2-12+8}{n^2-4} =\lim_{n \to \infty} 3+\frac{8}{n^2-4}$$
and we can show that eventually $0\le\frac1{n^2-4}\le\frac1n$ and $\frac1n\to 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Short way for upper triangularization
We are given a matrix $$A =
\begin{bmatrix}
3 & 0 & 1 \\
-1 & 4 & -3 \\
-1 & 0 & 5 \\
\end{bmatrix}$$
and we are asked to find a matrix $P$ such that $P^{-1}AP$ is upper triangular.
Here, we first find one eigenvalue as $\lambda= 4$. Then the matrix $$
A-4I =
\begin{bmatrix}
... | Suppose we begin with the matrix $$A =
\begin{bmatrix}
3 & 0 & 1 \\
-1 & 4 & -3 \\
-1 & 0 & 5 \\
\end{bmatrix}$$
and use "double Gaussian elimination". That is, whenever we multiply by an elementary row operation $E$ on the left, we must also multiply by the corresponding column operation $E^{-1}$ on the right. We beg... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2744829",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Use of principle of inclusion and exclusion in counting Here is the question
Q:- How many bit strings of length 8 contain three consecutive zeros?
I tried to solve this by making the cases that three consecutive zeros start from bit number 1, 2 and so on. I approached to my answer correctly. I obtained 107 as my answe... | As N. F. Taussig shows, Inclusion-Exclusion is not easy to apply to this problem. One other method that can be used is generating functions. To count the number of strings with no more than two zeros in succession, we will represent the atomic strings
$$
\begin{array}{}
1&x\\
10&x^2\\
100&x^3
\end{array}\tag1
$$
Since ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747128",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to find mean and standard deviation? An urn contains a large number of cards of which 1/4 have the number 1,1/4 have the number 2 and 1/2 have the number 3.
a) Let $X$ be the number of the card when a card is taken from the urn. Find the mean and standard deviation of $X$.
b) Let $X$ be the sample mean when card sa... | Your probability mass function (pmf) $p$ is
$$
p(x)=P(X=x)=
\begin{cases}
\frac{1}{4} &\mbox{ if } x=1, \\
\frac{1}{4} &\mbox{ if } x=2, \\
\frac{1}{2} &\mbox{ if } x=3. \\
\end{cases}
$$
a.) The expected value (mean) of $X$ is
$$
\mu=\mu_X = E[X] = \sum_{x=1}^{3} xp(x) = 1\left(\frac{1}{4}\right)+ 2\left(\frac{1}{... | {
"language": "en",
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Find the value of $ab$ $$5^a = 16, 8^b = 25$$
*
*Find the value of $ab$.
So, this question seems simple to solve. However, the most important thing is to know where to start. That's why I couldn't solve this problem. I've been looking for a method/strategy to solve all kinda questions which involve exponential te... | $5^a = 16$ so $a =\log_5 16= \log_5 2^4 = 4\log_5 2$
$8^b = 25$ so $b = \log_8 25=\log_8 5^2 = 2\log_8 5$
So $ab =4\log_5 2*2\log_8 5=8\log_5 2*\log_8 5$.
That may (or may not) be simplified.
Use $\log_a b = \frac {\log_M b}{\log_M a}$ to convert to a common base.
$\log_5 2 = \frac {\ln 2}{\ln 5}$ and $\log_8 5 = \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2752147",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to maximise this bivariate function: $\frac{x}{x + y} + \frac{50 - x}{100 - x - y}$ I wish to maximise
$ g(x,y) = \frac{x}{x + y} + \frac{50 - x}{100 - x - y}$
subject to the constraints $x , y \in \mathbb{N}$. $ 1≤ x≤ 50 ,0≤y≤50$
I know $g(x,y) = f(x,y) + f(50-x,50-y)$ if that helps.
I know it is maximised at $... | We have $$g(x,y) = \frac{x}{x + y} + \frac{50 - x}{100 - x - y}=2-\frac y{x+y}+\frac{y-50}{100-x-y}$$ so $$\frac{\partial g}{\partial x}=\frac y{(x+y)^2}+\frac{y-50}{(100-x-y)^2}=0$$ gives $$\color{red}{y(100-(x+y))^2=-(y-50)(x+y)^2}\implies200y-4y(x+y)=(x+y)^2\tag{1}$$ and since $$g(x,y) = \frac{x}{x + y} + \frac{50 -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Real roots of an equation A friend gave me this equation which I have trouble finding the real roots. $$x^9+3x^6+3x^3-16x+9=0$$ One can easily see that 1 is a root then with the help of Horner's method this can be simplified. However I am looking for an elegant solution if possible not just to use a computer to do that... | The polynomial can be factored by considering the product of
$$(x^3 + a x^2 + b x + c) \cdot (x^6 + \alpha x^5 + \beta x^4 + \gamma x^3 + \delta x^2 + \eta x + \mu).$$
With a little bit of effort it can be found that
$$(x - 1) (x^2 + x - 1) (x^6 + 2 x^4 + 2 x^3 + 4 x^2 + 2 x + 9) = 0.$$
The real roots are then obtain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2753587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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the volume of the expanding cube is increasing at the rate of $24 cm^3/min$, how fast is the surface area increasing when surface area is $216cm^2$? If the volume of the expanding cube is increasing at the rate of $24 cm^3/min$, how fast is the surface area increasing when surface area is $216cm^2$?
My Approach :
$V=l^... | The volume: $V=x^3$. The surface: $S=6x^2$. Given the surface is $216$, we can find the side:
$$6x^2=216 \Rightarrow x=6.$$
Given the volume increases at the rate $24$, we can find at what rate the side is increases:
$$\frac{dV}{dt}=3x^2\cdot \frac{dx}{dt}=24 \Rightarrow \frac{dx}{dt}=\frac{24}{3\cdot 6^2}=\frac29.$$
G... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solution of system of equations involving $x_{1},x_{2},x_{3}$
Solve for $x_{1},x_{2},x_{3}$, given
$ax^2_{1}+bx_{1}+c=x_{2}$
$ax^2_{2}+bx_{2}+c=x_{3}$
$ax^2_{3}+bx_{3}+c=x_{1}$
Try: from $(1)$ and $(2)$
$a(x^2_{1}-x^2_{2})+b(x_{1}-x_{2})=(x_{2} - x_{3})$
And from $(2)$ and $(3)$
$a(x^2_{2}-x^2_{3})+b(x_{2}-x_{3})=(x_... | I would start with solving $$ax^2_{1}+bx_{1}+c=x_{2}$$
Via the quadratic formula,
$$x_1=\frac{-b\pm\sqrt{b^2-4a(c-x_2)}}{2a}$$
Repeating the same process, we have
$$x_2=\frac{-b\pm\sqrt{b^2-4a(c-x_3)}}{2a}$$
and
$$x_3=\frac{-b\pm\sqrt{b^2-4a(c-x_1)}}{2a}$$
Hence,
\begin{align}
x_1&=\frac{-b\pm\sqrt{b^2-4a(c-x_2)}}{2a}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2754246",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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$\sqrt[5]{2^4\sqrt[3]{16}} = ? $ $$\sqrt[5]{2^4\sqrt[3]{16}} = ? $$
Rewriting this and we have
$$\sqrt[5\cdot 3]{2^{4\cdot 3}4}$$
$$\sqrt[15]{2^{12}2^2}$$
Finally we get
$$\sqrt[15]{2^{12}2^2} = \sqrt[15]{2^{14}} = 2^{\frac{15}{14}}$$
Am I right?
| $$2^4\sqrt[3]{16}=2^4\cdot2^{4/3}=2^{16/3}\implies\sqrt[5]{2^4\sqrt{16}}=\left(2^{16/3}\right)^{1/5}=2^{16/15}$$
so no, you are not right...but almost.
| {
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"timestamp": "2023-03-29T00:00:00",
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A Problem Dealing with putting balls in bin and Expected Value - possible wrong answer Please consider the problem below. Is my answer correct. If is not correct then where did I go wrong?
Problem:
You keep tossing balls into $n$ bins until one of the bins has two balls. For each toss there is a $\frac{1}{n}$ probabili... | The number $n$ of bins is given, and fixed. Denote by $E(j)$ $(0\leq j\leq n)$ the expected number of additional tosses when there are $j$ empty bins and the game is not yet over. Then we have the following recursion:
$$E(j)=1+{j\over n}E(j-1)\quad(n\geq j\geq 1),\qquad E(0)=1\ .$$
This gives
$$\eqalign{
E(n)&=1+{n\ove... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2758092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 1
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Infinite positive integer solutions of the equation: $x^2+x+1=(y^2+y+1)(n^2+n+1)$ Could anybody help me some ideas on the below problem:
Let $n$ be a positive integer. Prove that there are infinite pairs of positive integer $(x\, y)$ such that
$$x^2+x+1=(y^2+y+1)(n^2+n+1).$$
Thanks in advance.
| HINT:
Fix $n$. You get a quadratic equation in $x$, $y$. You should reduce it to a Pell equation as follows:
$$(4 x^2 + 4 x + 4 ) = (n^2 + n+1)(4 y^2 + 4 y + 1)\\
(2x+1)^2 + 3 = N \cdot ((2y+1)^2 + 3)\\
(2x+1)^2 - N (2y+1)^2 = 3(N-1)\\
a^2 - N b^2 = 3(N-1)$$
You have a solution for the last equation, $a=2n+1$, $b=1$, c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2761139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How to solve this complex limits at infinity with trig? Please consider this limit question
$$\lim_{x\rightarrow\infty}\frac{a\sin\left(\frac{a(x+1)}{2x}\right)\cdot \sin\left(\frac{a}{2}\right)}{x\cdot \sin\left(\frac{a}{2x}\right)}$$
How should I solve this? I have no idea where to start please help.
| $$\lim_{x\rightarrow\infty}\frac{a\color{#f00}{\sin(\frac{a(x+1)}{2x})}\cdot \sin(\frac{a}{2})}{\color{#00ff}{x\cdot \sin(\frac{a}{2x})}}\tag{1}$$
$\color{#f00}{\sin\left(\frac{a(x+1)}{2x}\right)} = \sin\left(\frac{a}{2}+\frac{a}{2x}\right);\quad$ so $x \to \infty \implies \sin\left(\frac{a}{2}+0\right) = \sin\left(\f... | {
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"source": "stackexchange",
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EC Point calculation Following the Guide to Elliptic Curve Cryptography, it provides the following elliptic curve on $E(\mathbb{F}_p)$ with $p=29$ on page 80:
$E: y^2 = x^3 + 4x + 20$
Page 81 provides a list of the points on the curve. For $x=2$ it includes the points $(2,6)$ and $(2,23)$. Let's do the calculation:
\be... | This is arithmetic in finite fields. You have to take modular square roots. Here the examples with Pari/GP
? x=Mod(36,29)
%1 = Mod(7, 29)
? x^(1/2)
%2 = Mod(6, 29)
? x=Mod(59,29)
%3 = Mod(1, 29)
? x^(1/2)
%4 = Mod(1, 29)
So yes, there is there a special sqrt-operation for mod?
See e.g. Modular square root or th... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Inequality about the most common divergent series We know that:
$1+\frac{1}{2}+\frac{1}{3}+....$ is a divergent series.
I have a small problem about this series.
Show that: $1+\frac{1}{2}+\frac{1}{3}+....+\frac{1}{2^{2006}} >50$
How to prove this one?
Can you use a basic knowledge to explain for a 14 years old student... | Notice $1+\frac{1}{2}+\dots + \frac{1}{n} =\int\limits_{i=1}^{n+1} \frac{1}{\lfloor x \rfloor } \geq \int\limits_{i=1}^{n+1} \frac{1}{ x } = \ln(n+1)$
We conclude your sum is at least $\ln(2^ {2006} +1)$ and the rest is easy.
A basic way to prove it is by noticing
$\frac{1}{2^ n} + \frac{1}{2^n + 1} + \dots + \frac{1... | {
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"timestamp": "2023-03-29T00:00:00",
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British Maths Olympiad (BMO) 2004 Round 1 Question 1 alternative approaches? The questions states:
Solve the simultaneous equations (which I respectively label as $
> \ref{1}, \ref{2}, \ref{3}, \ref{4}$)
$$\begin{align} ab + c + d &= 3 \tag{1} \label{1} \\ bc + d + a &= 5
\tag{2} \label{2} \\ cd + a + b &= 2 \tag{3... | Here's a way after you get $a+c=2$.
Take equation $2$ subtract equation $1$ to get $(b-1)(c-a)=2$.
Likewise, take equation $3$ subtract equation $4$ to get $(d-1)(c-a) = -4$.
Finally, we see that
\begin{align}
\frac{b-1}{d-1}=\frac{(b-1)(c-a)}{(d-1)(c-a)}= -\frac{1}{2} \ \ \implies \ \ 2b+d =3.
\end{align}
Next, take... | {
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Finding the range of $f(x)=\cos x\left(\sin x+\sqrt{\smash[b]{\frac12+\sin^2x}}\right)$
How do I find the number of integers in the range of
$$f(x)= \cos x\left( \sin x + \sqrt{\dfrac 12 +\sin^2 x} \right)?$$
I set the derivative equal to $0$ but the method isn't efficient here because it gives a very complicated ... | Since you are asking about integers in the range, we do not need to fully know the end points of the range, and we don't need calculus.
Clearly the integer $0$ is in the range, considering $x=\pi/2$.
Since $f(\pi/4)=\frac{\sqrt{2}}{2}\left(\frac{\sqrt{2}}{2}+1\right)=\frac{1+\sqrt{2}}{2}>1$, the integer $1$ is in the r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2771734",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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A functional equation problem: $ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $
Let $ \mathbb R ^ + $ denote the set of the positive real numbers. Find all functions $ f : \mathbb R ^ + \to \mathbb R ^ + $ satisfying
$$ \frac { f \left( f ( y ) ^ 2 + x y \right) } { f ( y ) } = f ( x ) + y $$... | $ \def \R {\mathbb R ^ +} $
You can show that the only $ f : \R \to \R $ satisfying
$$ f \left( x y + f ( y ) ^ 2 \right) = \big( f ( x ) + y \big) f ( y ) \tag 0 \label 0 $$
for all $ x , y \in \R $ is the identity function. It's straightforward to verify that the identity function is a solution. We try to prove the c... | {
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Finding the residue of $\frac{1}{z^3 \sin z}$ at z = 0 Given the function $f(z) = \frac{1}{z^3 \sin{(z)}}$, what is the residue of $f(z)$ at $z = 0$?
I want to find $b_1$ from the Laurent expansion. So I did the following:
\begin{align*}
\frac{1}{z^3 \sin{(z)}}
&= ( \dots \frac{b_5}{z^5} + \frac{b_4}{z^4} + \frac{b_... | The function is even so the residue at $z = 0$ is zero, since the residue is the coefficient of ${1 \over z}$ in the Laurent expansion.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Generating function for the Catalan numbers I know that generating function $f(x)$ for the Catalan numbers is
\begin{equation}
f(x)=\cfrac{1\pm \sqrt{1-4x}}{2x}\ .
\end{equation}
It is often said that we should choose
\begin{equation}
f(x)=\cfrac{1- \sqrt{1-4x}}{2x}
\end{equation}
because $f(x)$ should be continuous a... | We choose the negative sign in
\begin{align*}
f(x)=\frac{1\color{blue}{\pm} \sqrt{1-4x}}{2x}
\end{align*}
since we want to expand $f$ in a power series.
According to the binomial series expansion we have for $|x|<\frac{1}{4}$ the following representation at $x=0$
\begin{align*}
\sqrt{1-4x}&=\sum_{n=0}^\infty \bi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2778800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find integer triples $(x,y,z)$ such that $2018^x=y^2+z^2+1$ To find the triples, I had some tries.
First , consider $x\ge 2$, then $2018^x=0\pmod 4$. But for the right side $y^2+z^2+1$. Consider 3 situations.
*
*case 1. Both $y$ and $z$ are even. Then $y^2+z^2+1=1\pmod 4$.
*case 2. One of them are even, the other... | You have solved almost all cases. So let me add the solution for $x=1$, and exclude negative $x$.
Since $2017$ is a prime which is congruent $1$ modulo $4$, it is representable as the sum of two squares. The solutions are, up to sign and order, given by
$$
2017=44^2+9^2,
$$
see here:
$2017$ as the sum of two squares
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2781254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Proof Verification: Using Taylor series to arrive at a certain inequality If $x>0$ show that $|(1+x)^{1/3} - (1 + \frac{1}{3}x - \frac{1}{9}x^2)| \le \frac{5}{81}x^3$.
Proof:
Let $f(x) = (1 +x)^{1/3}$
Then, estimating $f(x)$ at the point $x_o = 0$, we have that:
$f'(x) = 1/3(1+x)^{-2/3}, f"(x) = -2/9(1+x)^{-5/3}, f^{(... | You've got $(1+x)^{1/3}- (1+ \frac{1}{3}x -\frac{1}{9}x^2 ) <\frac{5}{81}x^3$, but the value on the left side could be a very big negative number, so you actually have to show that the absolute value is smaller than $\frac{5}{81} x^3$.
Working with $P_3, R_3$ is "too much". I suggest to get $P_2,R_2$. Doing that you wi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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If $x^2-bx+c=0$ has real roots, then prove that both are greater than $1$ when $c+1>b>2$.
If $ x^2-bx+c=0$ has real roots, prove that both roots are greater than $1$, when $c+1>b>2$.
Working
I tried to prove the given inequality by taking roots greater than $1$.
Let $\alpha$, $\beta$ be the roots of the quadratic ... | Given $x^2-bx+c=0$ we need
*
*$\Delta =b^2-4c\ge0$
and
*
*$x_1=\frac{b-\sqrt{b^2-4c}}{2}>1 \iff b-2> \sqrt{b^2-4c} \stackrel{b>2}\iff b^2-4b+4\ge b^2-4c \iff 4c+4\ge4b \iff c+1\ge b$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2782758",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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Integration of $ \int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2} \mathrm dx$ Evaluate $$\int_1^2 \frac{3x^3 + 3x^2 - 5x + 4}{3x^3 - 2x^2 +3x -2}\mathrm dx$$I believe I should use the partial fractions technique to evaluate this integral, but I am not getting anywhere when I try it.
| Apologies if it's unclear - I'm not sure how to get the polynomial division in Mathjax, if someone could advise me on how to do this it would be appreciated.
From here I'll use Mathjax however:
$$5x^2-8x+6=A(x^2+1)+B(3x-2)$$
Let $A=px+q, B=rx+s$. If $p>0$, this equation transforms into a cubic so $p=0$.
Hence $qx^2+q+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2784031",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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A trigonometric integral (guessed from a combinatorics formula) In class, I defined the binomial coefficient using an integral:
$$\binom{n}{k} = \displaystyle \int_0^{2\pi}\dfrac{dt}{2\pi} e^{-ikt}(1+e^{it})^n.$$
I succeeded in demonstrating many standard properties of the binomial coefficient directly using integrat... | There is a mistake in my presentation that eludes me, but the essence is the following.
Using
$$\sum_{k=1}^{\infty} p^{k} \, \sin(k \, x) = \frac{p \, \sin(x)}{1 - 2 \, p \, \cos(x) + p^{2}}$$
then let $S_{n}$ be the desired summation to be evaluated to obtain:
\begin{align}
S_{n} &= \sum_{k=0}^{n} \binom{n}{k} \\
&= ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2789309",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 2
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solving differential equation $\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$ How would you solve this third order differential equation:
$$\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}=x^2+2x+2$$
My first thought was to take a double integral:
$$\iint\frac{d^3y}{dx^3}+\frac{d^2y}{dx^2}dxdx=\iint{x^2+2x+2}dxdx$$
so:
$$y+\frac{dy}... | Well, there are many ways you can you solve this DE. However, the simplest would be using the characteristic equation and since its a non-homogeneous DE, we need to find a particular solution as well. Finally we need to use the superposition principle to add the solution.
Now, in symbols and numbers
$$
r^3 +r^2 = 0 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2791300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
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Eccentricity of ellipse in terms of eccentric angles of extremities of focal chord?
If $A$ and $B$ are eccentric angles of the extremities of a focal chord of an ellipse then eccentricity of the ellipse is?
The answer given is $\dfrac{\sin A + \sin B}{\sin (A+B)}$.
Now I tried to do this by assuming extremities to be... |
The equation of chord joining pairs of point $[A]$ and $[B]$ is:
$$\frac{x}{a} \cos \frac{A+B}{2}+\frac{y}{b} \sin \frac{A+B}{2}
=\cos \frac{A-B}{2}$$
Put $(x,y)=(\pm ae,0)$, then
$$e=\pm \frac{\cos \frac{A-B}{2}}{\cos \frac{A+B}{2}}$$
Also,
\begin{align}
\frac{\sin A+\sin B}{\sin (A+B)} &=
\frac{2\sin \frac{A+B}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2792481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding the area between 4 curves by changing my basis to create a square and integrating The question number $3$ at hand: https://i.imgur.com/DL11Izh.jpg
My work on it: https://i.imgur.com/AzaQKHS.jpg?1
I thought I was doing everything right but it seems to be wrong, the final answer should be $$\frac{6}{35}(b^{\frac{... | Let $(x,y)=(u^3 v^3,u^2 v^4)$
\begin{array}{ccccc}
ax^2<y^3<bx^2 & \implies &
au^6v^6<u^6v^{12}<bu^6v^6 & \implies &
a<v^6<b \\
cy^3<x^4<dy^3 & \implies &
cu^6v^{12}<u^{12}v^{12}<du^6v^{12} & \implies & c<u^6<d
\end{array}
\begin{align*}
\iint_A dx\, dy &=
\int_{\sqrt[6]{a}}^{\sqrt[6]{b}}
\int_{\sqrt[6]... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793197",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Evaluation of $\int\frac{1}{(\sin x+a\sec x)^2}dx$
Evaluation of $$\int\frac{1}{(\sin x+a\sec x)^2}dx$$
Try: Let $$I=\int\frac{1}{(\sin x+a\sec x)^2}dx=\int\frac{\sec^2 x}{(\tan x+a\sec^2 x)^2}dx$$
put $\tan x=t$ and $dx=\sec^2 tdt$
So $$I=\int\frac{1}{(a+at^2+t)^2}dt$$
Could some help me how to solve above Integral ... | You can write
$$
\frac{1}{(a+t+at^2)^2}=-\frac{1-2a^2+2at+2a^2t^2}{(-1+4a^2)(a+t+at^2)^2}+\frac{2a}{(-1+4a^2)(a+t+at^2)}.
$$
Also is
$$
-\int\frac{1-2a^2+2at+2a^2t^2}{(-1+4a^2)(a+t+at^2)^2}dt =\frac{1+2at}{(-1+4a^2)(a+t+at^2)}
$$
and
$$
\int \frac{2a}{(-1+4a^2)(a+t+at^2)}dt=\frac{4a}{(-1+4a^2)^{3/2}}\arctan\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2793823",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Show that $ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$ Show that:
$$ \sqrt{2 + \sqrt{2 + \sqrt{2 + 2 \cos 8\theta}}} = 2 \cos \theta$$
My try:
As we can see that the LHS is in form of $\cos 8 \theta$ which must be converted into $\cos \theta $ in order to solve the equation.
There are several ques... | $$\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta}}}$$
Firstly, $2+2cos8\theta=2+2(2cos^24\theta-1)=4cos^24\theta$
$$\therefore\sqrt{2+\sqrt{2+\sqrt{2+2cos8\theta}}}=\sqrt{2+\sqrt{2+2cos4\theta}}$$
Then, $2+2cos4\theta=2+2(2cos^22\theta-1)=4cos^22\theta$
$$\therefore\sqrt{2+\sqrt{2+2cos4\theta}}=\sqrt{2+2cos2\theta}$$
Finally, $2+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2795571",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
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What is the closed form of :$\int\frac{x+2}{\left(x^2+x+2\right)\left(\sqrt{x^2+2x+3}\right)}\,dx$ over $\mathbb{R}$? I have took many times to get antiderivative of the below integral A in $\mathbb{R}$ , but i didn't succeed however it is a constant from $-\infty \to +\infty$ and it's equal :$1.4389$ , but i'm sure th... | Using Mathematica with Rubi 4.15.1 we can find very simple form antiderivative:
$$\int \frac{x+2}{\left(x^2+x+2\right) \sqrt{x^2+2 x+3}} \, dx=\\-\sqrt{\frac{1}{14} \left(11+8 \sqrt{2}\right)} \tan ^{-1}\left(\frac{1+2 \sqrt{2}-\left(5+3
\sqrt{2}\right) x}{\sqrt{7 \left(11+8 \sqrt{2}\right)} \sqrt{3+2 x+x^2}}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2796429",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Integral $\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$ I am tring to evaluate
$$I=\int_0^2 \frac{\arctan x}{x^2+2x+2}dx$$
The first thing I did was to notice that
$$\frac{1}{x^2+2x+2}=\frac{1}{(x+1)^2+1}=\frac{d}{dx}\arctan(x+1)$$
So I integrated by parts in order to get
$$I=\arctan 2\arctan 3-\int_0^2\frac{\arctan(x+1)}{1+x... | An elementary solution. Let $I$ denote the integral. Apply the substitution $x=\frac{2t}{t+\sqrt{5}}$ to obtain
$$ I = \int_{0}^{\infty} \frac{\arctan\left(\frac{2t}{t+\sqrt{5}}\right)}{\sqrt{5}+4t+\sqrt{5}t^2} \, dt. \tag{1} $$
Substituting $t \mapsto 1/t$, we find that
$$ I = \int_{0}^{\infty} \frac{\arctan\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2798550",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
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Integration and hyperbolic function problem According to my cosmology book:$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$where $a$ is the scale factor, the dot indicates derivative wrt time and $\varLambda$ and $K$ are constants. The author then says “Introducing a new variable $x$ by $a^{3}=x^{2}$ and integrating once ... | $$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$
$$x^2=a^3 \implies 3a^2 \dot a =2\dot x x \implies 9a^4 \dot a ^2=4x^2\dot x ^2 $$
$$\implies 9a^3(a\dot a^2)=4x^2\dot x^2 \implies (a\dot a ^2)=\frac {4\dot x^2}{9}$$
The equation becomes
$$a\dot{a}^{2}=\frac{\varLambda}{3}a^{3}+K,$$
$$\frac {4\dot x^2}{9}=\frac{\varLambda... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2799610",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate indefinite integral $\int \tan(\frac{x}{3}) \, dx$ I need help verifying why I am getting an incorrect answer for the question evaluate the integral
$$\int \tan\left(\frac{x}{3}\right) \, dx$$
I simplify the above equation using trig identities to get
$$\int \frac {\sin \left(\frac{x}{3}\right)}{\cos\left(\fr... | Just to make your life a bit simpler without this fraction $\frac{x}{3}$, just use u-sub.
Put: $\frac{x}{3}=u$
Then $\int tan(\frac{x}{3})dx= 3\int tan\ u\ du= 3 $ln |sec u| +C$=$3 ln|sec($\frac{x}{3})|+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2804424",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
} |
Verify that $\int_{0}^{\pi/2}\tan x\ln(\tan x)\ln(\cos x)\ln^3(\sin x)\mathrm dx=-\frac{3\zeta(6)}{2^8}$ $$\int_{0}^{\pi/2}\tan x\ln(\tan x)\ln(\cos x)\ln^3(\sin x)\mathrm dx=-\frac{3\zeta(6)}{2^8}$$
Any hints how to simplify this integral to more manageable to be solve.
| Considering $$
\begin{aligned}
J(a, b, c) &=\int_0^{\frac{\pi}{2}} \tan ^a x \cos ^b x \sin ^c x d x \\
&=\int_0^{\frac{\pi}{2}} \sin ^{a+c} x \cos ^{b-a} x d x \\
&=\frac{1}{2} B\left(\frac{a+c+1}{2}, \frac{b-a+1}{2}\right),
\end{aligned}
$$
then $$\boxed{I=\left.\frac{1}{2} \frac{\partial^3}{\partial a \partial b \pa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2805317",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How can I find the sum of $\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}$ How can I find the sum of the series:
$$\sum_{n=2}^\infty \frac{(-1)^n}{n^2 - n}$$
I did this:
$$\sum_{n=2}^\infty \left(\frac{(-1)^n}{n-1} + \frac{(-1)^{n-1}}{n}\right)$$
But I am not sure how to proceed.
| Let us consider a sequence
$$a_n = \sum_{k=n+1}^{2n} \frac{1}{k}$$
Now, let's attempt to get a recursive formula for this.
$$a_n - a_{n-1} = \sum_{k=n+1}^{2n} \frac{1}{k} - \sum_{k=n}^{2n-2} \frac{1}{k} = -\frac{1}{n} + \frac{1}{2n-1} + \frac{1}{2n} = \frac{1}{(2n-1)(2n)}$$
$$a_n = a_{n-1} + \frac{1}{(2n-1)(2n)}$$
Sinc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2808959",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Using the limit definition to find a derivative for $\frac{-5x}{2+\sqrt{x+3}}$ I am trying to find the derivative of $\frac{-5x}{2+\sqrt{x+3}}$ using the limit definition of a derivative.
$$\lim_{h \to 0} \frac {f(x+h)-f(x)}{h}$$
What I did is
$$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}\dfrac{2-\sqrt{x+h+3}}... | You better not rush to multiply by conjugates:
$$\lim_{h \to 0} \frac{\dfrac{-5(x+h)}{2+\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}}{h}=\\
\lim_{h \to 0} \frac{\left(\dfrac{-5x}{2+\sqrt{x+h+3}}-\dfrac{-5x}{2+\sqrt{x+3}}\right)+\dfrac{-5h}{2+\sqrt{x+h+3}}}{h}=\\
-5x\cdot \lim_{h \to 0} \frac{\dfrac{1}{2+\sqrt{x+h+3}}-\dfrac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2810506",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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$1^n-3^n-6^n+8^n$ is divisible by $10$ Prove that $1^n-3^n-6^n+8^n$ is divisible by $10$ for all $n\in\mathbb{N}$
It is divisible by $2$ and $5$ if we rearrange it will it be enough
$(1^n -3^n)$ and $(6^n -8^n)$ is divisible by $2$.
And
$(1^n-6^n)$ and $(8^n-3^n)$ is divisible by $5$.
Hence $\gcd(2,5)$ is $1$ and it ... | Your proof is OK, however, for completeness, you must further show why the binomials are divisible by $2$ and $5$.
If you are familiar with modular arithmetic, it is:
$$1^n-3^n-6^n+8^n\equiv 1-3-6+8\equiv 0 \pmod {10}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2813432",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
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What is the probability of a symmetric coin fall
The prompt: We have 10 coins and 1 of them is non-symmetric (with probability of head equal $\frac{1}{3}$). We toss a randomly selected coin 6 times, and obtain 3 tails. What is the probability that we tossed a symmetrical coin?
The way I went on about solving the prob... | The probability of getting exactly 3 tails out of 6 tosses is:
a) ${6\choose 3}\left(\frac{1}{2}\right)^6$ in the case of symmetric coin.
b) ${6\choose 3}\left(\frac{1}{3}\right)^3 \left(\frac{2}{3}\right)^3$ in the case of the asymmetric coin.
Then, if you picked your initial coin uniformly at random (probability $1/1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815284",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Determine conics by four points and a tangent line. I have small troubles determining conics that go through four points and have a given tangent line.
More specifically $P_1 = (0:1:0), P_2 = (0:0:1), P_3 = (1:0:1), P_4 = (1:-1:0) \in \mathbb{RP}^2$ and $ t: \{x-y+z=0\}$.
I can determine the conics as far as $ax^2+axy+... | $\det(\begin{pmatrix}x^2&xy&y^2&xz&yz&z^2\\0&0&1&0&0&0\\0&0&0&0&0&1\\1&0&0&1&0&1\\1&-1&1&0&0&0\\a^2&ab&b^2&ac&bc&c^2\end{pmatrix})=a(c-b-a)yz+bc(-xz+xy+x^2)=0$
Now assume $(a:b:c)$ is the point of tangency making $b=a+c$ and $y=x+z$, substitute and expand:
$2((ac+c^2)x^2-a^2xz-a^2z^2)$
The discriminant reduces to $a^2(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2815990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to solve the given determinant by using properties if possible Let $\alpha,\beta$ be the real roots of the equation $ax^2+bx+c=0$ and let $ {\alpha}^n+{\beta}^n=S_{n}\ for \ n \geq 1 $ then find the value of the determinant of A, where $$A = \left[ \begin{array} { l l l } {\ \ \ \ { 3 }} & { 1 + S _ { 1 } } & { 1 +... | $S _ { n } = \alpha ^ { n } + \beta ^ { n } \forall n \geq 1$
$A= \left| \begin{array} { l l l } { 1+1+1 } & { 1+\alpha+\beta } & { 1+\alpha^{2}+\beta^{2} } \\ { 1+\alpha +\beta } & { 1+\alpha^{2}+\beta^{2} } & { 1+\alpha^{3}+\beta^{3}} \\ { 1+\alpha^{2}+\beta^{2} } & { 1+\alpha^{3}+\beta^{3} } & { 1+\alpha^{4}+\beta^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2818182",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Prove this $\sin(nx)$ identity Without induction How to prove that $$\sin(2nx)=2n\sin x \cos x \prod_{k=1}^{n-1}\left(1-\frac{\sin^2(x)}{\sin^2\frac{k\pi}{2n}}\right)$$
for Natural n
I Tried by several way and the last try is to use euler formula which lead me to
$$\sin(2nx)=\sin x \cos^{2m+1} x \sum_{k=0}^{n-1} \left... | Finally it solved ^_^ :
first by use: $\sin^2x=\frac{1-\cos 2x}{2}$
$$\left(1-\frac{\sin^2(x)}{\sin^2\frac{k\pi}{2n}}\right)=\left(\frac{\cos 2x - \cos{\frac{k\pi}{n}}}{2\sin^2\frac{k\pi}{2n}}\right)$$
then by use :$\cos x - \cos y $ formula
I got
$$\left(1-\frac{\sin^2(x)}{\sin^2\frac{k\pi}{2n}}\right)=-\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2820548",
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"source": "stackexchange",
"question_score": "4",
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How to evaluate this integral $\int_{1/3}^{3} \frac{\sin(\frac1x -x)dx}{x}$?
Evaluate
$$\int_{1/3}^{3} \frac{\sin(\frac1x -x)dx}{x}.$$
I am unable to think of any way to solve this.
| $$I=\int_{\frac{1}{3}}^3 \dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx=\underbrace{\int_{\frac{1}{3}}^1\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx}_J+\underbrace{\int_1^3\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx}_H$$
$$J=\int_{\frac{1}{3}}^1\dfrac{\sin \left( \dfrac{1}{x}-x\right)}{x}dx\quad t=\dfrac{1}{x}\rig... | {
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"url": "https://math.stackexchange.com/questions/2820998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Consider the next succession and prove by induction The exercise says:
Knowing the next succession,
$$a_1=1$$
$$a_{n+1}=\frac{2a_n}{(2n+2)(2n+1)}, n>=1$$
Prove by induction that $a_n=\frac{2^n}{(2n)!}$
What I've done so far is prove for $n=1$
For $n=1$:
$$a_1=\frac{2^1}{(2\times1)!}=\frac{2}{2}=1$$
Which is correct.... | No, it is not correct. What you need to show is that, assuming
$$a_n=\frac{2^n}{(2n)!},$$
then
$$a_{n+1} = \frac{2a_n}{(2n+2)(2n+1)} = \frac{2^{n+1}}{\big(2(n+1)\big)!}.$$
So
$$\frac{2a_n}{(2n+2)(2n+1)} = \frac{2}{(2n+2)(2n+1)}\frac{2^n}{(2n)!}=\dots$$
Try to follow from there.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving the product of four consecutive integers, plus one, is a square I need some help with a Proof:
Let $m\in\mathbb{Z}$. Prove that if $m$ is the product of four consecutive integers, then $m+1$ is a perfect square.
I tried a direct proof where I said:
Assume $m$ is the product of four consecutive integers.
If $m... | To get a feel for the problem, let's work backwards, starting from a square number. Take some integer $n$.
$n^2 - 1 = (n+1)(n-1).$
OK, so it looks like $m$ has two factors whose difference is 2. Could it be that in our product of consecutive integers, the product of two of them is $n-1$ and of the other two is $n+1$?
L... | {
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"source": "stackexchange",
"question_score": "18",
"answer_count": 13,
"answer_id": 4
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Different way $\int_{0}^{\infty}e^{-2x}\frac{\ln \cosh(x)}{1+\cosh(x)}dx$ I would like to evaluate this integral, $$I=\int_{0}^{\infty}e^{-2x}\frac{\ln \cosh(x)}{1+\cosh(x)}\mathrm dx$$
An approach might would be:
*
*Remove the $\cosh(x)$
$$I=2\int_{0}^{\infty}e^{-x}\ln\left(\frac{e^x+e^{-x}}{2}\right)\frac{dx}{(e... | Let $I$ denote the integral. Then applying the substitution $u = e^{-x}$ followed by integration by parts,
\begin{align*}
I
&= \int_{0}^{1} \frac{2u^2}{(1+u)^2} \log\left(\frac{1+u^2}{2u}\right)\,du \\
&= \int_{0}^{1} 2\left( 1+u - \frac{1}{1+u} - 2\log(1+u)\right)\frac{1-u^2}{u(1+u^2)} \, du \\
&= \underbrace{ \int_{0... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What tools one should use for inequalities? If $a,b,c>0$ prove that:
$$\frac{1}{a+4b+4c}+\frac{1}{4a+b+4c}+\frac{1}{4a+4b+c}\leq \frac{1}{3\sqrt[3]{abc}}.$$
My first try was the following:
$$\sum_{cyc}\frac{1}{a+4b+4c}\leq\sum_{cyc}\frac{1}{\sqrt[3]{16abc}}=\frac{1}{\sqrt[3]{16abc}}$$
But $\frac{1}{\sqrt[3]{16abc}}\geq... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that $f(v^2)\geq0,$ where
$$f(v^2)=16u^3+12uv^2-w^3-24u^2w-3u^2w,$$ which is a linear function.
But a linear function gets a minimal value for an extreme value of $v^2$, which happens for equality case of two variables.
Sines our inequality is homog... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solving Quasi-Linear Transport Equation with two shockwaves. I want to solve the following Partial Differential Equation by using the method of characteristics. This is the transport equation.
\begin{align}u_t - (1-2u)u_{x}&=0, &-\infty < x < \infty, t >0, \end{align}
\begin{align}u(x,0) &= \begin{cases} \frac{1}{4}... | Here, we are dealing with the macroscopic traffic-flow model by Lighthill-Witham-Richards (LWR), where $0≤u≤1$ represents the number of cars per unit length. To get an insight about the problem, here is a plot of the characteristic curves $x(t) = (1-2u(x_0,0))\, t + x_0$ in the $x$-$t$ plane:
As written in OP, two sho... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Find $B=A^2+A$ knowing that $A^3=\begin{bmatrix}4&3\\-3&-2\end{bmatrix}$
Is there a way to solve this rather than just declaring a matrix $$A=\begin{bmatrix}a&b\\c&d\end{bmatrix}$$ and then trying to solve a system of cubic equations? My attempt... | Denote the third root of unity as $w=e^{i2\pi/3}$.
Then any of the quantities $\rho=\{w,w^2,w^3=1\}\,$ satisfy $\,\rho^3=1,\,$ i.e. is a cube root of one.
Note that the matrix $X=A^3$ has a single eigenvalue of $\{\lambda=1\}$ with multiplicity two.
Since it is a $2\times 2$ matrix, any analytic function of $X$ (and it... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2838190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
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show this inequality $ab+bc+ac+\sin{(a-1)}+\sin{(b-1)}+\sin{(c-1)}\ge 3$
let $a,b,c>0$ and such $a+b+c=3abc$, show that
$$ab+bc+ac+\sin{(a-1)}+\sin{(b-1)}+\sin{(c-1)}\ge 3$$
Proposed by wang yong xi
since
$$\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ac}=3$$
so use Cauchy-Schwarz inequality we ha... | Hint : Prove with Jensen's inequality (apply to the function $f(x)=0.5x^2+\sin(x-1)$ ) this with the obvious restriction $ab+bc+ca\leq6$:
$$ab+bc+ca+\sin(a-1)+\sin(b-1)+\sin(c-1)$$$$\geq$$$$ ab+bc+ca-0.5(a^2+b^2+c^2)+3\sin(\frac{a+b+c}{3}-1)+1.5(\frac{a+b+c}{3})^2$$
And now a last hint prove this (with the condition ... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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The logic in radical symplification I'm having troubles while studying radicals, namely with converting expressions with the form $$\sqrt{a+b\sqrt{c}}$$ to $$a+b\sqrt{c}$$ and vice versa. When I'm dealing with these kind of problem, I usually add and subtract $\sqrt{c}^2$. In example: $$\sqrt{27+10\sqrt{2}}$$ I factor ... | Do not expect integer solutions for problems like that.
For $$\sqrt{27+10\sqrt{2}}= a+b\sqrt{2}$$
We square both sides to get $$27+10\sqrt{2}=a^2+2b^2 +2ab\sqrt 2$$
Thus we solve a system $$ 2ab=10, a^2+2b^2=27$$ and we are lucky to find $$a=5, b=1$$
Therefore $$\sqrt{27+10\sqrt{2}}= 5+\sqrt{2}$$
On the other hand fo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2840858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using Method of Undetermined coefficients to find maximum value Find the maximum value of ($1$+$x$)$^5$($1$$-$$x$)($1$$-$$2$$x$)$^2$ given that $1/2$$<$$x$$<$$1$.
The solution states the follows
Consider the maximum value of
($A$($1$$+$$x$))$^5$ $*$ ($B$($1$$-$$x$)) $*$ ($C$($2$$x$$-$$1$)$^2$), where $A$, $B$, $C$... | Since ${\large{\frac{1}{2}}} < x < 1$, it follows that $1+x,\;1-x,\;2x-1$ are positive.
Let $A,B,C$ be fixed positive real numbers, not yet selected.
Applying $\text{AM-GM}$, to $5$ copies of $A(1+x)$, $1$ copy of $B(1-x)$, and $2$ copies of $C(2x-1)$, yields
$$\left(\frac{\;5\bigl(A(1+x)\bigr)+B(1-x)+2\bigl(C(2x-1)\... | {
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"source": "stackexchange",
"question_score": "2",
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Prove that $2^{2^5} +1 \equiv 0 \mod 641$
$2^{(2^5)} +1 \equiv 0 \mod 641$
From the totient funcion we have:
$2^{32^{20}} \equiv 1$
Thus:
Either $2^{32} \equiv 1$ or $2^{32} \equiv -1$
But how do I prove that $2^{32} \equiv -1$?
| You only need $$641=2^4+5^4$$
and $$641=2^7\cdot 5+1$$
This gives you $$2^{32}=(2^7)^4\cdot 2^4\equiv -(2^7)^4\cdot 5^4=-(2^7\cdot 5)^4\equiv -1\mod 641$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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For $x^3+px+q=0$, one of the solutions is $\sqrt 3-1$. Find the value of rational number $p$ and $q$, and other two solutions of the equation.
I cannot find any error from my answer. What’s wrong?
| If $\sqrt{3} -1$ is a root of $x^3+px+q$, it doesn't necessarily follow that $\sqrt{3}+1$ is also a root.
In fact, if $\sqrt{3} -1$ is a root then $\sqrt{3}+1$ cannot be a root, assuming $p,q \in \mathbb{Q}$.
Indeed
$$\alpha = \sqrt{3} + 1 \implies \alpha-1 = \sqrt{3} \implies (\alpha-1)^2 = 3 \implies \alpha^2-2\alpha... | {
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"source": "stackexchange",
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Is $\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=1$?
Evaluate $$\lim_{x\to-\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}$$
My attempt:
$$\lim_{x\to -\infty} \frac{5x+9}{3x+2-\sqrt{4x^2-7}}=\lim_{x\to-\infty}\frac{5x}{-\sqrt{4x^2}}=\frac{-5}{2}$$ According to the answer key, it actually equals $1$.
Thanks in advance.... | Here's my way :
$$ \lim_{x\to -\infty} \frac {5x+9}{3x+2-\sqrt{4x^2-7}} = \lim_{x\to -\infty} \frac {x (5+ \frac {9}{x})}{3x+2-\sqrt{4x^2(1-\frac{7}{4x^2})}}=\lim_{x\to -\infty}\frac {x (5+ \frac {9}{x})}{3x+2-2|x|\sqrt{1-\frac{7}{4x^2}}}=\lim_{x\to -\infty} \frac {x (5+ \frac {9}{x})}{3x+2+2x\sqrt{1-\frac{7}{4x^2}}} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2849212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove $ \min \left(a+b+\frac1a+\frac1b \right) = 3\sqrt{2}\:$ given $a^2+b^2=1$ Prove that
$$ \min\left(a+b+\frac1a+\frac1b\right) = 3\sqrt{2}$$
Given $$a^2+b^2=1 \quad(a,b \in \mathbb R^+)$$
Without using calculus.
$\mathbf {My Attempt}$
I tried the AM-GM, but this gives $\min = 4 $.
I used Cauchy-Schwarz to get $\q... | Solution 1
By $H_n \leq G_n \leq A_n \leq Q_n$,
$$a+b+\frac{1}{a}+\frac{1}{b} \ge a+b+\frac{4}{a+b}=(a+b+\frac{2}{a+b})+\frac{2}{a+b} \geq 2\sqrt{2}+\sqrt{\frac{2}{a^2+b^2}} =3\sqrt{2}.$$
Solution 2
$$a+b+\frac{1}{a}+\frac{1}{b} \geq 2(\sqrt{ab}+\frac{1}{2\sqrt{ab}})+\frac{1}{\sqrt{ab}}\geq 2\sqrt{2}+ \sqrt{2}=3\sqrt{2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2850000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 9,
"answer_id": 8
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Convergence of the series $\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$.
Show that $$\sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{x^n}$$ converges for every $x>1$.
let $a(x)$ be the sum of the series. does $a$ continious at $x=2$? differentiable?
I guess the first part is with leibniz but I am not sure about it.
| Let's look at the
partial sums,
and let $y = -1/x$
so
$-1 < y < 0$..
$\begin{array}\\
s_m(y)
&=\sum_{n=1}^{m} (-1)^{n-1}(-y)^n\\
&=\sum_{n=1}^{m} (-1)^{n-1}(-1)^ny^n\\
&=-\sum_{n=1}^{m} y^n\\
&=-y\sum_{n=0}^{m-1} y^n\\
&=-y\dfrac{1-y^m}{1-y}\\
&=\dfrac{-y}{1-y}-\dfrac{-y^{m+1}}{1-y}\\
\end{array}
$
so
$\begin{array}\\
... | {
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"url": "https://math.stackexchange.com/questions/2853668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
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$\tan(a) = 3/4$ and $\tan (b) = 5/12$, what is $\cos(a+b)$ It is known that
$$\tan(a) = \frac{3}{4}, \:\:\: \tan(b) = \frac{5}{12} $$
with $a,b < \frac{\pi}{2}$.
What is $\cos(a+b)$?
Attempt :
$$ \cos(a+b) = \cos(a) \cos(b) - \sin(a) \sin(b) $$
And we can write $\tan(a) = \sin(a)/\cos(a) = 0.3/0.4 $
and $ \sin(b)/\co... | Solve for sin(a), cos(a), sin(b) and cos(b) (like I said in comments):
$sin(a)=\frac{3}{5}$
$cos(a)=\frac{4}{5}$
$sin(b)=\frac{5}{13}$
$cos(b)=\frac{12}{13}$
Thus $$cos(a+b)=\frac{4}{5}\times\frac{12}{13}-\frac{3}{5}\times\frac{5}{13}=\frac{33}{65}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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If $abc=1$, prove that $\frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$. If $a, b, c \in \mathbb{N}$, and $abc = 1$, prove that:
$$S = \frac{2}{(a+1)^2+b^2+1} + \frac{2}{(b+1)^2+c^2+1} + \frac{2}{(c+1)^2+a^2+1} \le 1$$
Here is my try:
$$\begin{align}\frac{2}{(a+1)^2+(b^2+1)} & \le \fr... | This is a problem from Olympiad Inequalities, Thomas J. Mildorf, 2005 (short but wonderful reading with many different ideas, a true gem). The idea presented here (replacing $a,b,c$ with $x/y,y/z,z/x$) is often used to normalize the inequality in case of $abc=1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2854448",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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permutations/combinations with repeated symbols I've been learning some about counting and basic combinatorics. But some scenarios were not explained in my class...
Example problem: You are given 6 tiles. 1 is labeled "1", 2 are labeled "2", and 3 are labeled "3".
Problem 1: How many different ways can you arrange grou... | Not sure this is the best method, but the way I would actually solve such a set of questions would be the following:
Use generating function methods for the second question.
The explicit collection of submultisets of the six tiles is given by the terms of $(1+x)(1+y+y^2)(1+z+z^2+z^3)$ (the $x$ terms correspond to tile ... | {
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"url": "https://math.stackexchange.com/questions/2856180",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
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If $AB=BA$, prove that $ A=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix} $
Let $A$ a $2\times2$ matrix, if $AB=BA$ for every $B$ of the size $2\times2$, Prove that:
$$
A=\begin{bmatrix} a & 0 \\ 0 & a \end{bmatrix}
$$
$a \in \mathbb{R}$
My attempt:
Let $$
A=\begin{bmatrix} a_1 & b_1 \\ c_1 & d_1 \end{bmatrix}
$$
$$B=... | Note:
$$\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix}
\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}=
\begin{pmatrix} b_{11} & b_{12} \\ b_{21} & b_{22} \end{pmatrix}
\begin{pmatrix} a_{11} & a_{12} \\ a_{21} & a_{22} \end{pmatrix} \Rightarrow \\
\begin{cases}\require{cancel}\cancel... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
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Solve $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}$ The question:
Without the use of a calculator, solve for all values of $x$ if $\tan {(x-\frac{\pi}{4}})=-\tan{(x+\frac{\pi}{2})}.$
Using the compound angle formula for solving equations is normally easy, but I stumbled across this problem.
The $LHS$ is easy t... | $$\tan \left(x - \frac{\pi}{4}\right) = - \tan \left(x + \frac{\pi}{2}\right)$$
$$\tan \left(x - \frac{\pi}{4}\right) = \tan \left(-x - \frac{\pi}{2}\right)$$
$$x-\frac{\pi}{4}=-x-\frac{\pi}{2}+k\pi,\quad(k\in Z)$$
$$2x=-\frac{\pi}{4}+k\pi$$
$$x=-\frac{\pi}{8}+\frac{k\pi}{2}$$
Valid for any $k\in Z$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
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What is the next limit equal to? $$\lim_{n\to\infty}(\sqrt{n^2+n}-\sqrt[3]{n^3+1})=?$$
I tried amplifying the hole substraction to form the formula $$a^3-b^3$$ but didn't worked out. Can you help me figure it out?
| Note that\begin{align}\sqrt{n^2+n}-\sqrt[3]{n^3+1}&=\sqrt[6]{(n^2+n)^3}-\sqrt[6]{(n^3+1)^2}\\&=\frac{(n^2+n)^3-(n^3+1)^2}{\sqrt[6]{(n^2+n)^3}^5+\sqrt[6]{(n^2+n)^3}^4\sqrt[6]{(n^3+1)^2}+\cdots+\sqrt[6]{(n^3+1)^2}^5}\\&=\frac{3n^5+3n^4-n^3-1}{\sqrt[6]{(n^2+n)^3}^5+\sqrt[6]{(n^2+n)^3}^4\sqrt[6]{(n^3+1)^2}+\cdots+\sqrt[6]{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2856819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Prove: $\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$
Prove: $$\sin\frac{\pi}{20}+\cos\frac{\pi}{20}+\sin\frac{3\pi}{20}-\cos\frac{3\pi}{20}=\frac{\sqrt2}{2}$$
ok, what I saw instantly is that:
$$\sin\frac{\pi}{20}+\sin\frac{3\pi}{20}=2\sin\frac{2\pi}{20}\cos\frac{\pi... | $$\sin9^\circ+\cos9^\circ+\sin27^\circ-\cos27^\circ$$
$$=\sin9^\circ+\sin(90-9)^\circ+\sin27^\circ+\sin(27-90)^\circ$$
Observe that $\sin5x=\sin45^\circ$ for $x=9^\circ,81^\circ,27^\circ,-63^\circ$
Now if $\sin5x=\sin45^\circ,$
$5x=180^\circ m+(-1)^m45^\circ$ where $m$ is any integer
$\implies x=36^\circ m+(-1)^m9^\ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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The diophantine equation $5\times 2^{x-4}=3^y-1$ I have this question: can we deduce directly using the Catalan conjecture that the equation
$$5\times 2^{x-4}-3^y=-1$$
has or no solutions, or I must look for a method to solve it. Thank you.
| From $3^y\equiv1$ mod $5$, we see that $4\mid y$, so writing $y=4z$, we have
$$5\cdot2^{x-4}=3^{4z}-1=(3^{2z}+1)(3^{2z}-1)$$
Since $3^{2z}+1\equiv2$ mod $8$, we can only have $3^{2z}+1=2$ or $10$. We can dismiss the first possibility (since it implies $z=0$, which gives $3^{2z}-1=0$). Thus $3^{2z}+1=10$, so $z=1$ and ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2858769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Verifying that $\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx$ without Cauchy-Schwarz
For real numbers $a,b,p$, verify that $$\left[\int_a^b(x-p)\,dx\right]^2\le(b-a)\int_a^b(x-p)^2\,dx\tag{1}$$ without using the Cauchy-Schwarz inequality.
Note that this is obviously an immediate consequence of C-S, and... | Or note that, if $M:=\frac{1}{b-a}\,\int_a^b\,(x-p)^2\,\text{d}x$, then
$$\int_a^b\,(x-p)^2\,\text{d}x-(b-a)\,M^2=\int_a^b\,\big((x-p)-M\big)^2\,\text{d}x\geq0\,.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2862491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Prove, using first principles, that $\lim\limits_{(x,y)\to(0,0)}{x.y^2=0}$ My try is:
$\left|x.y^2-0\right|<\left|\sqrt{x^2+y^2}.(x^2+y^2)\right|< \delta.\delta^2=\epsilon$
where $0<\sqrt{x^2+y^2}< \delta$ what did I miss here?
| Writing $f(x,y) = x\cdot y^2$. Let $\varepsilon > 0$, setting $\delta = \varepsilon^{1/3}$, you have that $\sqrt{x^2 + y^2} \leq \delta \implies |f(x,y)- 0| \leq (\sqrt{x^2 + y^2})(\sqrt{x^2 + y^2})^2 \leq \varepsilon^{1/3} \cdot \varepsilon^{2/3} = \varepsilon$.
So, you have proven by constructing it that $\exists ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864450",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots$ using sigma notation This question can be solved by method of difference
but I want to solve solve it using sigma notation:
$$\frac{1^2}{2^1}+\frac{3^2}{2^2}+\frac{5^2}{2^3}+\cdots+\frac{(2r +1)^2}{2^r}+\cdots$$
I used the geometric progression summati... | A simple answer to the question
Note that
\begin{multline*}
S\equiv\sum_{n\geq1}\frac{(2n-1)^{2}}{2^{n}}=\sum_{n\geq1}\frac{4n^{2}-4n+1}{2^{n}}=4\sum_{n\geq1}\frac{n^{2}}{2^{n}}-4\sum_{n\geq1}\frac{n}{2^{n}}+\sum_{n\geq1}\frac{1}{2^{n}}\equiv4I_2-4I_1+I_0.
\end{multline*}
since each of the series on the right hand side... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2864643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Show that $\lim_{m \to \infty} \frac{2^{2m}m!(m+1)!}{(2m+1)!(2m+1)}=0$ After a long way to solve the differential equation $y''+xy'+3y=0$, I arrived at the solution $$y(x) = A_0 \sum_{m=0}^{\infty} (-1)^m \dfrac{2m+1}{2^mm!}x^{2m}+A_1 \sum_{m=0}^{\infty} (-1)^m \dfrac{2^m(m+1)!}{(2m+1)!}x^{2m+1}$$ and now I need to pro... | $y(x) = A_0 \sum_{m=0}^{\infty} (-1)^m \dfrac{2m+1}{2^mm!}x^{2m}+A_1 \sum_{m=0}^{\infty} (-1)^m \dfrac{2^m(m+1)!}{(2m+1)!}x^{2m+1}
$
Use the root test
and Stirling
($n!\approx \sqrt{2\pi n}(n/e)^n$).
$\begin{array}\\
\dfrac{2^m(m+1)!}{(2m+1)!}
&=\dfrac{m+1}{2m+1}\dfrac{2^mm!}{(2m)!}\\
&\approx\dfrac12\dfrac{2^m\sqrt{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2872409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
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How to prove $1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\left(2n-1\right)^{1/4} $
Prove that $$1 + \frac{1}{\sqrt{2}} +... +\frac{1}{\sqrt{n}} <\sqrt{n}\ .\Bigl(2n-1\Bigr)^{\frac{1}{4}} $$
My Approach :
I tried by applying Tchebychev's Inequality for two same sets of numbers;
$$1 , \frac{1}{\sqrt... | The strict inequality (which only holds for $n\gt1$) can be proved by induction, by showing that
$$\sqrt n\cdot(2n-1)^{1/4}+{1\over\sqrt{n+1}}\lt\sqrt{n+1}\cdot(2n+1)^{1/4}$$
for $n\ge N$ for some $N$ and then checking the base cases up to $N$.
We first rewrite the inductive inequality above as
$$\sqrt{n(n+1)}\cdot(2n-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2875714",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Compare $\arcsin (1)$ and $\tan (1)$
Which one is greater: $\arcsin (1)$ or $\tan (1)$?
How to find without using graph or calculator?
I tried using $\sin(\tan1)\leq1$, but how do I eliminate the possibility of an equality without using the calculator?
| Here is another method. The only real disadvantage is that it requires calculating $e^{-1}$ in decimal form, but only two digits are needed, so this can be done by hand easily. It does not require using very large fractions or remembering the series for $\tan(x)$.
As in another answers, we want to find an upper bound... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2877251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 7,
"answer_id": 2
} |
order of the splitting field of $x^5 +x^4 +1 $? what is the order of the splitting field of
$x^5 +x^4 +1 = (x^2 +x +1)( x^3 +x+1)$ over $\mathbb{Z_2}$
i thinks it will $6$ because $2.3 = 6$
Pliz help me...
| Because $x^5 +x^4 +1 = (x^2 +x +1)( x^3 +x+1)$ and the factors are irreducible, the splitting field of $x^5 +x^4 +1$ has degree at least $2 \cdot 3 = 6$.
On the other hand, the splitting field of $x^5 +x^4 +1$ is contained in $\mathbb F_{2^6}$, an extension of degree $6$. Hence this is the splitting field. Indeed, WA ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2878699",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$ I want to evaluate the following limit:
$$\lim_{x\rightarrow0}\left[x\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}\right]$$
We have
$$\left(\frac{1+3x}{1+2x}\right)^{\frac 1 x} = e^{\frac{\log\left(\frac{1+3x}{1+2x}\right)}{x}}\... | Recall that $x\to 0$ implies $1+2x\sim 1$. Hence you should have
$$\log\left(\frac{1+3x}{1+2x}\right)\sim\frac{1+3x}{1+2x}-1=\frac{x}{1+2x}\sim\frac{x}{1}=x.$$
and therefore
$$\lim_{x\rightarrow0}\left(\frac{1+3x}{1+2x}\right)^{\frac {1}{x}}=e^1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2879225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove that every natural number $n>15$ there exist natural numbers $x,y\geqslant1$ which solve the equation. Prove that every natural number $n>15$ exist Natural numbers $x,y\geqslant1$ which solve the equation $3x+5y=n$.
so i try induction.
base case is for $n=16$.
so $\gcd(5,3)=1$, after Euclidean algorithm i found:... | If $x,y$ is a solution, you have $3x+5y=n=n(-3\cdot3+2\cdot5)$ and therefore
$$3(x+3n)= 5(2n-y)$$
As $3,5$ are coprime it exists $k \in \mathbb Z$ such that $x=5k-3n$ and $y= 2n-3k$.
$x >1$means $5k> 1+3n$ and $y>1$ implies $3k<2n-1$.
Which is equivalent to
$$3+9n < 15k < 10n-5 \tag{1}$$
The difference of $10n-5$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Finding n-th term of a matrix I have to find the n-th power of the following matrix $$A=\begin{pmatrix} 2 & 0&1\\ 0 & -3& 0 \\ 1 & 0&2 \end{pmatrix}
$$ I computed a few powers $$A^2=\begin{pmatrix} 4+1& 0&4\\ 0 & 9& 0 \\ 4 & 0&1+4\end{pmatrix}$$
$$A^3 =\begin{pmatrix} 14 & 0&13\\ 0 & -27& 0 \\ 13& 0&14\end{pmatrix}$$
$... | You could use Diagonalization.
Let $A$ be a matrix then if you diagonalize $A$ then you get $A=PDP^{-1}$ with $B$ as a diagonal matrix and $P^{-1}$ an invertible matrix. If you try for $A^2$, then it is equal to $A^2=PD^2P^{-1}$. Similarly for $A^n=PD^nP^{-1}$
Here are the steps:
$1.$Find the characteristic polynomial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2882911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Differential equation: $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$
The solution of $\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}$ is given by:
a) $(x+2)^4 (1+\frac{2y}{x})= ke^{\frac{2y}{x}}$
b) $(x+2)^4 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
c) $(x+2)^3 (1+ 2\frac{(y-2)}{x+2})= ke^{\frac{2(y-2)}{x+2}}$
d... | $$\dfrac{dy}{dx}= \dfrac{(x+y)^2}{(x+2)(y-2)}=\left(\dfrac{x+y}{x+2}\right)^2\dfrac{x+2}{y-2}$$
let $w=\dfrac{x+y}{x+2}$ then $y=w(x+2)-x$ and
$$w'(x+2)+w-1=y'=w^2\dfrac{1}{w-1}$$
which is separable
$$\dfrac{w-1}{2w-1}dw=\dfrac{dx}{x+2}$$
with integration the solution will be obtained.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2883253",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
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Geometric intuition for the complex shoelace formula The complex shoelace formula for the signed area of a triangle with vertices given by the complex numbers $a, b, c$ is $$\frac{i}{4}
\begin{vmatrix}
1 & 1 & 1 \\
a & b & c \\
\overline{a} & \overline{b} & \overline{c} \\
\end{vmatrix}
$$
I have se... | Consider the case $\,a=0\,$, first, where the proposed formula reduces to:
$$
S_{OBC} = \frac{i}{4}
\begin{vmatrix}
1 & 1 & 1 \\
0 & b & c \\
\overline{0} & \overline{b} & \overline{c} \\
\end{vmatrix}
=
\frac{i}{4}
\begin{vmatrix}
b & c \\
\overline{b} & \overline{c} \\
\end{vmatrix}
= \frac{i}{4}\left(b \bar c -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2885275",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Proving radius of circle $\dfrac{\triangle}{a}\tan^2\dfrac{A}{2}$
If a circle be drawn touching the inscribed and circumscribed circles of a $\triangle ABC$ and the side $BC$ externally, prove that its radius is: $$r=\dfrac{\triangle}{a}\tan^2\dfrac{A}{2}$$
I tried using triangle formed by circumcenter, incenter and ... | The center $J$ of the wanted circle can be constructed by intersecting a line and a parabola, since by setting $s=JL$ we have $JO=R-s$.
If we take $B$ as the origin and $BC$ as the $x$-axis, the equation of the wanted parabola is $y=kx(x-a)$. Since the distance of $O$ from $BC$ is $R\cos A$, the vertex lies at $\left(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 1,
"answer_id": 0
} |
Proof: homogeneous second order differential equation This is an homogeneous second order differential equation
$Ay'' + By' + Cy = 0$
Its solution can be retrieved assuming $y = e^{rx}$: in this way, we will have
$y = e^{rx}$
$y' = re^{rx}$
$y'' = r^{2}e^{rx}$
replace:
$Ar^{2}e^{rx} + Bre^{rx} + Ce^{rx} = 0$
$(e^{rx}) ... | The_lost is right in his comment, we show $xe^{rx}$ solves the equation
$Ay''(x) + By'(x) + Cy(x) = 0, \; A \ne 0, \tag 0$
by simple substitution (here I assume $A \ne 0$ to ensure (0) is a bona-fide second order equation); however, it helps to know something about $r$ first, to wit:
According to the quadratic formula,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2886909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$ for $x\rightarrow\infty$ I have the following function:
$$f(x)=x\log{\left|\frac{x+2}{3-x}\right|}$$
I want to find the limit for $x\rightarrow+\infty$.
This is what I do. Since $x>=0$, I can remove the absolute value:
$$f(x)=x\log\left({\frac{x+2}{3-x}}\right)\sim x\... | Let $x>4$;
$I:= \log\dfrac {|x+2|}{|x-3|}= {\displaystyle \int_{x-3}^{x+2}}(1/t)dt.$
MVT:
$I = (1/r){\displaystyle \int_{x-3}^{x+2}}dt=$
$(1/r)[(x+2)-(x-3)]=5/r,$
where $r \in [x-3,x+2]$.
Hence:
$5\dfrac {x}{x+2} \le xI \le 5\dfrac{x}{x-3}$.
Squeeze :
$\lim_{x \rightarrow \infty} xI =5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2888587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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Sketching complex plane Sketch all points in the complex plane such that $Re(\frac{1}{z})=1$
I managed to solve it $(\frac{1}{z})= \frac{|z|}{z|z|}=\frac{(x - iy)}{(x^2+y^2)}$
Meaning that $Re(\frac{1}{z})= \frac{x}{(x^2+y^2)}$
How should I go afterwards?
| For an alternative approach, use that $\,\operatorname{Re}(w)= \dfrac{1}{2}\left(w + \bar w\right)\,$ for complex $\,w\,$, then (assuming $\,z \ne 0\,$, otherwise $\,\dfrac{1}{z}\,$ is not even defined):
$$
\begin{align}
1=\operatorname{Re}\left(\frac{1}{z}\right)=\frac{1}{2}\left(\frac{1}{z}+\frac{1}{\bar z}\right) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2890409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Given a triangle whose vertex $A$ is bisecting the base (median), find the angle $X$ Given a triangle whose vertex $A$ is bisecting the base (median), I have to find the angle $X$.
First Approach
From $\Delta AHD$ we can get $AH = HD$. And it will make $45^\circ$ at vertex $A^*$. From $\Delta ABH$, let's Consider angl... | $\require{cancel}$
Using completely trigonometry
Suppose $AC=q$, let us solve for all the sides in terms of $q$:
$$\frac{\sin(135^\circ)}{q}=\frac{\sin(30^\circ)}{AD}\implies AD=\frac{q}{\sqrt2}$$
$$\frac{\sin(15^\circ)}{CD}=\frac{\sin(135^\circ)}{q}\implies\,CD=\frac{q(\sqrt3-1)}{2}$$
Since $CD=BD$, then $BC=q(\sqrt3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2892345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Arrangement of $15$ letters with restrictions on which letters are permitted in which positions How many $15$ letter arrangements of $5$ As, $5$ Bs and $5$ Cs have no As in the first $5$ positions, no Bs in the next $5$ positions, and no Cs in the last $5$ positions?
I am not able to frame the equation. I presumed that... | If we choose which of the middle five positions are occupied by $A$s, the others must be occupied by $C$s (since no $B$s are permitted in the middle five positions). If we then choose which of the last five positions are occupied by the remaining $A$s, the remainder must be occupied by $B$s (since no $C$s are permitte... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2893980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the radius of the black circle tangent to all three of these circles? The red, blue, and green circles have diameters 3, 4, and 5, respectively.
What is the radius of the black circle tangent to all three of these circles?
I just figured out the radius is exactly $\dfrac{72}{23}$ but I don't know how to do the ... | It is simply the Apollonius Problem and solved by the following three quadratic equations using WolframAlpha for center $ (x,y) $ and radius $r$ of the biggest circle, which is tangent to other three circles;
(1) $ x^2 + (y - \frac{3}{2})^2 = (r - \frac{3}{2})^2 $, circle on side of length $ = 3 $,
(2) $ (x - 2)^2 + y ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895516",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "28",
"answer_count": 4,
"answer_id": 3
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How to calculate $\int_{0}^{\pi/2}\dfrac{1}{\sin^3 x + \cos^3 x}dx$ One way that I tried to solve an integral was to divide integrand by $\cos^3x$ para para obter
$$
\int_{0}^{\pi/2}\dfrac{\sec^3x}{1 + \tan^3x}dx
$$
Through software, I saw that the result involves $\tanh^{-1}\bigl(\frac{1}{\sqrt{2}}\bigr)$, but I do n... | Let $\tan\frac{x}{2}=t$.
Thus, $$\int\limits_0^{\frac{\pi}{2}}\frac{1}{\sin^3x+\cos^3x}dx=\int\limits_0^1\frac{2}{\left(\left(\frac{2t}{1+t^2}\right)^3+\left(\frac{1-t^2}{1+t^2)}\right)^3\right)(1+t^2)}dt=$$
$$=2\int\limits_0^1\frac{(1+t^2)^2}{(2t+1-t^2)((2t)^2-2t(1-t^2)+(1-t^2)^2)}dt.$$
Now, $$2t+1-t^2=-(t^2-2t+1-2)=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2895741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find the general solution (i) $\cot \theta =-\dfrac {1} {\sqrt {3}}$ (ii)$4\cos ^{2}\theta =1$ my attempt for
(i)
$\left. \begin{array} { l } { \cot ( \theta ) = - \frac { 1 \cdot \sqrt { 3 } } { \sqrt { 3 } \sqrt { 3 } } } \\ { 1 \cdot \sqrt { 3 } = \sqrt { 3 } } \end{array} \right.$
$\cot ( \theta ) = - \frac { \sqr... | We know that $$\cot\left(-\dfrac{\pi}{3}\right)=-\dfrac{\sqrt 3}{3}$$ and $$\cot( \theta+\pi)=\cot \theta$$similarly $$\cos\left(\dfrac{\pi}{3}\right)=\dfrac{1}{2}$$and $$\cos^2(\theta+\pi)=\cos^2\theta$$therefore $$(i)\qquad x=k\pi-\dfrac{\pi}{3}\\(ii)\qquad x=k\pi\pm\dfrac{\pi}{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2897340",
"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Finding limit of $\frac{x^2y^2}{x^3+y^3}$ as $(x,y)\to (0,0)$ Here, if we go along $x=-y$ then the function itself is not defined, and limit seems to exist for all other paths of approach
Putting in another way, if we use polar coords, limit is:
$$\lim_{r\to 0} \frac{r^4 \sin^2\theta\cos^2\theta}{r^3(\sin^3\theta + \co... | We have that
*
*$x=y=t\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4}{2t^3}=\frac t{2}\to 0$
*$x=t\to 0 \quad y=-t+t^2\to 0 \implies \frac{x^2y^2}{x^3+y^3}=\frac{t^4-2t^5+t^6}{3t^4-3t^5+t^6}=\frac{1-2t+t^2}{3-3t+t^2}\to \frac13$
therefore the limit doesn't exist.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2899532",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Integrate $\frac{x^2-4x+10}{x^2\sqrt x}$
Find the indefinate integral with respect to $x$ of $$\frac{x^2-4x+10}{x^2\sqrt x}$$
For this problem I first made each individual number in the numerator separate from each other for easier integration and then simplified
$$=\int \left(\frac{x^2}{x^2\sqrt x}-\frac{4x}{x^2\sqr... | $$\frac{x^2-4x+10}{x^2\sqrt x}$$
$$\frac{1}{\sqrt{x}}-\frac{4}{x\sqrt{x}}+\frac{10}{x^2\sqrt{x}}$$
$$x^{-1/2}-4x^{-3/2}+10x^{-5/2}$$
So it should be $x^{-5/2}$ instead of $x^{-3/2}$
$$I=\int x^{-1/2}-4x^{-3/2}+10x^{-5/2} {dx}=2x^{1/2}+8x^{-1/2}-\frac{20}{3}x^{-3/2}+C$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove that $\frac{\sin x}{x}=(\cos\frac{x}{2}) (\cos\frac{x}{4}) (\cos \frac{x}{8})...$ How do I prove this identity:
$$\frac{\sin x}{x}=\left(\cos\frac{x}{2}\right) \left(\cos\frac{x}{4}\right) \left(\cos \frac{x}{8}\right)...$$
My idea is to let
$$y=\frac{\sin x}{x}$$
and
$$xy=\sin x$$
Then use the double angle iden... | Note the fact that
$$
\cos \frac{x}{2^k} = \frac12 \cdot \frac{\sin (2^{1-k} x)}{\sin(2^{-k}x)},
$$
and we have
$$
\prod_{k = 1}^n \cos \frac{x}{2^k} = \frac{1}{2^n} \cdot \frac{\sin x}{\sin(2^{-n}x)} = \frac{2^{-n}x}{\sin(2^{-n}x)} \cdot \frac{\sin x}{x}.
$$
For all $x$, as $n \to \infty$, we have
$$
\lim_{n \to \inft... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2901407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
Did I misuse the inductive hypothesis? Note $\log$ is $\log_2$ in this problem.
I am working on the proof of a lemma that my professor used in class. He said we could verify that it works if we want, but I think I used the inductive hypothesis in such a way that I cannot continue my proof. Either that or I am missing... | If you draw some simple diagrams, given some $f(x) > 0$ and $f'(x) > 0,$ you find
$$ \int_{a-1}^b f(x) dx < \sum_{k=a}^b f(k) < \int_{a}^{b+1} f(x) dx $$
You are using $a=2$ and $b=n-1$
This is especially helpful when we know a closed form indefinite integral of $f$
In particular, using logarithm base $e \approx 2.7... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2904471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Determine all integers $x,\ y,\ z$ that satisfy $x+y+z=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}$
Determine all integers $x,\ y,\ z$ that satisfy
$$x+y+z=(x-y)^{2}+(y-z)^{2}+(z-x)^{2}$$
We expand the RHS and sort the equation by $x$, which yields a quadratic equation with the discriminant:
$$D=-12y^{2}-12z^{2}+24yz+12y+12z+1$$... | Let $y=u+x$ and $z=v+x$, then $y-z = u-v$ and hence
$$
\begin{align}
(y-u) + y + (y-u+v) &= u^2 + (u-v)^2 + v^2\\
3y &= 2u^2 + 2v^2 - 2uv +2u - v
\end{align}
$$
So we can almost freely choose $u,v$, but we need RHS to be divisible by $3$. By considering the equation modulo $3$, we see that solutions are
$$
(u,v) \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2905081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 1,
"answer_id": 0
} |
Does this matrix sequence always converge? Suppose $a_0, a_1, ... , a_{n-1}$ are real numbers from $(0; 1)$, such that $\sum_{k=0}^{n-1} a_k=1$. Suppose $A = (c_{ij})$ is a $n \times n$ matrix with entries $c_{ij} = a_{(i-j)\%n}$, where $\%$ is modulo operation. Is it always true that $\lim_{m \to \infty} A^m = \frac{1... | This follows directly from the Perron-Frobenius theorem.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907338",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How does the number $3$ help in this (probably Cauchy-based) inequality?
Given $a,b,c>0$ and $a+b+c=3$. Prove that
$$\dfrac{a^2+bc}{b+ca}+\dfrac{b^2+ca}{c+ab}+\dfrac{c^2+ab}{a+bc}\ge3$$
Attempt:
Using Cauchy inequality $(a+b\ge2\sqrt{ab})$:
$\dfrac{a^2}{b+ca}+b+ca\ge 2\sqrt{a^2}=2a$
$\dfrac{b^2}{c+ab}+c+ab\ge 2\sqr... | By C-S
$$\sum_{cyc}\frac{a^2+bc}{b+ca}=\sum_{cyc}\frac{(a^2+bc)^2}{(a^2+bc)(b+ca)}\geq\frac{\left(\sum\limits_{cyc}(a^2+bc)\right)^2}{\sum\limits_{cyc}(a^2+bc)(b+ca)}.$$
Thus, it's enough to prove that
$$\left(\sum\limits_{cyc}(a^2+ab)\right)^2\geq3\sum\limits_{cyc}(a^2+bc)(b+ca)$$ or
$$\sum_{cyc}(a^4+2a^2b^2+a^2b^2+2a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2907651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Given that the quadratic equation $x^2-px+q=0$ has two roots $r$ and $s,$ Find the quadratic equation that takes $r^3$ and $s^3$ as its roots. Here's what I've tried:
Using Vieta's formulas:
$\;rs = q\;$ and $\,r+s = p$.
Then I cubed $rs$ which is $q^3$ and $(r+s)^3 + rs$ which is $p^3 + pq.$ Thinking that $x^2 - (p^3... | The equation in question would look like
$$(x-r^3)(x-s^3)=x^2-(r^3+s^3)x+r^3s^3=0$$
As you've already identified, the last term is $(rs)^3=q^3$.
The coefficient of $-x$ is
$$\begin{split}
r^3+s^3 &= (r+s)(r^2-rs+s^2) \\
&= p(r^2+2rs+s^2-3rs) \\
&= p\bigg((r+s)^2-3rs\bigg) \\
&= p(p^2-3q)
\end{split}$$
Therefore, the eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909249",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Help: $ |\frac{a+1}{a}- (\frac{xz}{y^2})^k|\leq \frac{1}{b}$ In the paper linear forms in the logarithms of real algebraic numbers close to 1, it is written on page $9$ that
On other hand, a short calculation yields
$$ \left|\frac{a+1}{a}- \left(\frac{xz}{y^2}\right)^k\right|\leq \frac{1}{b}$$
Image of the page... | We have, using $b\ge 2$ and $a\ge 2^{49}$,
$$\begin{align}\left|\frac{a+1}{a}- \frac{(a+1)(ab^2+1)}{(ab+1)^2}\right|
&=(a+1)\left|\frac{1}{a}- \frac{(ab^2+1)}{(ab+1)^2}\right|
\\\\&=(a+1)\left|\frac{(ab+1)^2-a(ab^2+1)}{a(ab+1)^2}\right|
\\\\&=(a+1)\left|\frac{1+a(2b-1)}{a(ab+1)^2}\right|
\\\\&=(a+1)\cdot \frac{1+a(2b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2909494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
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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.