Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Prove $abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}$ for $a+b+c+d=1$
Let $a,b,c$, and $d$ be four positive reals satisfying $a+b+c+d=1$. Show that
$$abc+abd+acd+bcd\le\frac{1}{27}+\frac{176abcd}{27}.$$
I tried the inequality between $27abc$ and $(a+b+c)^3$ but it didn't help me
| We rewrite the inequality as :
$$f(a,b,c,d)=-\left(\ln\left(a\right)+\ln\left(b\right)+\ln\left(c\right)+\ln\left(d\right)+\ln\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}-\frac{176}{27}\right)\right)\geq -\ln\left(1/27\right)$$
We use the equal variable method corollary 1.5 as :
$$a+b+c+d=1$$
$$\frac{1}{a}+\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521003",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
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If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
\dfrac{5}{2} & \dfrac{3}{2}\\
-\dfrac{3}{2} & -\dfrac{1}{2}
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it ... | COMMENT.-$A$ has a double eigenvalue $\lambda=1$, is not diagonalizable and its Jordan matrix decomposition is $$A=\begin{pmatrix}-1&-\frac23\\1&0\end{pmatrix}*\begin{pmatrix}1&1\\0&1\end{pmatrix}*\begin{pmatrix}0&1\\-\frac32&-\frac32\end{pmatrix}^{-1}$$ so we have
$$A^{2022}=\begin{pmatrix}-1&-\frac23\\1&0\end{pmatrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4521130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
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Let $r,s,t$ are roots of the cubic equation $x^3+bx^2+cx+d=f(x)$ then write down $D=((r-s)(r-t)(s-t))^2$ in terms of $b, c, d$ Let $r,s,t$ are roots of the cubic equation $f(x) = x^3+bx^2+cx+d$ then write down $D=((r-s)(r-t)(s-t))^2$ in terms of $b,c,d$.
Is there any clever way to solve it?
We know $r+s+t=-b, rs+st+rt=... | Remark: A solution using middle school skill.
We have
$$(r-s)(r-t)(s-t) = r^2s + s^2t + t^2r - r^2t - t^2s - s^2 r = A - B$$
where
$$A := r^2s + s^2t + t^2r, \quad B := r^2t + t^2s + s^2 r.$$
We have
$$[(r-s)(r-t)(s-t)]^2 = (A - B)^2 = (A + B)^2 - 4AB.$$
We have
\begin{align*}
A + B &= r^2(s + t) + t^2(r + s) + s^2(t ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Determine the convergence of $\sum_{n=1}^\infty (1-cos(\frac{1}{n}))(\sqrt{n+1}-\sqrt{n})$ I'm having trouble determining the convergence of the series:
$\sum_{n=1}^\infty (1-cos(\frac{1}{n}))(\sqrt{n+1}-\sqrt{n})$
I used $\sqrt{n+1}-\sqrt{n} = \frac{1}{\sqrt{n+1}+\sqrt{n}}$ and then tried the comparison test for $0 \l... | Use :
$$1-\cos(\frac{1}{n}) \sim_{+\infty} \frac{1}{2n^2} $$
and
\begin{align}
\sqrt{n+1} - \sqrt{n} &= \sqrt{n} (\sqrt{1+\frac{1}{n}} -1) \\
&\sim_{+\infty} \sqrt{n} \times \frac{1}{2n} \\
&= \frac{1}{2\sqrt{n}}\\
\end{align}
Thus
$$ (1-\cos(\frac{1}{n}))(\sqrt{n+1}-\sqrt{n}) \sim_{+\infty} \frac{1}{4 n^{\frac{5}{2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4524724",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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help with mathematical induction exercise the instructions say: "Consider n and $a_1<a_2<...<a_n$ natural numbers, $n\ge1$. Prove that $$(\sum_{k=1}^n a_k)^2 \le \sum_{k=1}^n a_k^3$$"
this is how I proceeded:
induction base: n = 1 $\implies a_1^2 \le a_1^3$ which is always true, since $a_1$ is a natural number
inductiv... | As you pointed out, you need to prove that $2\sum_{k=1}^n a_k \leq a_{n+1}^2-a_{n+1}$. And you have the hypothesis $a_1<\dots<a_{n+1}$.
Then we have that:
$2\sum_{k=1}^n a_k\leq 2\sum_{k=1}^n (a_{n+1}-k) = 2na_{n+1}-(n+1)n$, since $a_{n+1-k} \leq a_{n+1}-k$.
So, it suffices to prove that $2na_{n+1}-(n+1)n\leq a_{n+1}^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526840",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solving the integral $\int_{0}^{1} \frac{s \ln(1+s) - t \ln(1+t)}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s$ For $t \in [0,1]$, let
$$f(t) = \int_{0}^{1} \frac{s \ln(1+s) - t \ln(1+t)}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s.$$
Is it possible to evaluate this integral to obtain an explicit expression for $f(t)$? ... | Continue with
\begin{align}
& \int_{0}^{1} \dfrac{s \ln(1+s) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s\\
= &\ \underset{=J}{\int_{0}^{1} \dfrac{s \ln(1+s) - s \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s} + {\int_{0}^{1} \dfrac{s \ln(1+t) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4527259",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Find all natural numbers $n$ such that the sum of the squares of the positive divisors of $n$ is equal to $n^2+2n+37$. If $f(n)$ is the sum of the squares of the positive divisors of n,then find all natural numbers $n$ such that $f(n)=n^2+2n+37$. I tried it first by substituting few values of $n$ but that doesn't work ... | Let us call the function $f$ as $\sigma_2$ since that's standard notation. Then, we have $\mathbb N\ni n\mapsto \sigma_2(n)=\sum_{d\mid n}d^2$. We start with a few lemmas and propositions some of which are easy enough so I will leave the proofs to you.
Lemma 1 $\sigma_2$ is multiplicative, that is, if $\gcd(a,b)=1$ th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$ $$\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}dx$$
I am given solution for this definite integral is $\frac{\pi}{2}\left(\ln\frac{\sqrt2+3}{\sqrt2+1}\right)$. Any idea or approach you would use to solve this?
| \begin{align}
&\int_{0}^{\infty}\tan^{-1}\left(\frac{2x}{1+x^2}\right)\frac{x}{x^2+4}\ dx\\
=& \int_{0}^{\infty}\int_{\sqrt2-1}^{\sqrt2+1}\frac{x}{x^2+y^2}\frac{x}{x^2+4}\ dy \ dx\\
=& \ \frac\pi2 \int_{\sqrt2-1}^{\sqrt2+1}\frac1{y+2}dy= \frac{\pi}{2}\ln\frac{\sqrt2+3}{\sqrt2+1}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4529185",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 0
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If there are only two lattice points $(n, n+1),(n+1, n)$ on the first circle quadrant with radius $r$, then $r^2$ is prime. Conjecture
For lattice points $(x,y) \in \mathbb{Z}^{+}$, define a circle at the origin
$x^2+y^2=r^2$
Let $r = \sqrt{2 n^2 + 2 n + 1}$
where $n$ is odd, $n>3$.
If there are only two lattice points... | We have the equation
$$n^2 + (n+1)^2 = m \tag{1}\label{eq1A}$$
for positive integers $n$ and $m$. Since $\gcd(n,n+1) = 1$, then $\gcd(n,m)=\gcd(n+1,m)=1$ as well. Next, consider if there's a prime factor $p \equiv 3 \pmod{4}$ which divides $m$. Then $n$ has a multiplicative inverse modulo $p$, call it $n^{-1}$, so \eqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4531551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Proving using Mathematical Induction from my discrete math class This is a practice exercise from our class about proving inequalities using mathematical induction. I've been stuck on the last step for quite a while now.
This is the Question. "Prove that $\sum_{k=1}^n\frac{1}{k^2}=\frac{1}{1^2}+\frac{1}{2^2}+\cdots+\fr... | It is very straightforward induction. You just add
$\dfrac{1}{(n+1)^{2}}$ to both sides of $P(n)$ and you get
$2-\dfrac{1}{n}+\dfrac{1}{(n+1)^{2}}$ must be less than $2-\dfrac{1}{n+1}$ .
Making all the simplifications we get
$\dfrac{n+2}{(n+1)^{2}}<\dfrac{1}{n}$ which is obviously true. So we get
$P(n+1)$ and we are do... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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How do I show $\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$ for $n \ge 1$? So this was part of a bigger induction proof problem, but I'm stuck at what I think is the last step. I need to show that
$\frac{2n+1}{2n+2} \le \frac{\sqrt{n+1}}{\sqrt{n+2}}$, for $n \ge 1$
Would it be enough to show that the limit as i... | Although this is a little indirect, it keeps calculation to a
minimum.
For any positive integer $m,$ we have
$$
(m - 1)(m + 1) = m^2 - 1 < m^2,
$$
therefore
$$
\frac{m - 1}{m^2} < \frac1{m + 1},
$$
therefore
$$
\left(\frac{m - 1}m\right)^2 = \frac{(m - 1)^2}{m^2} \leqslant
\frac{m - 1}{m + 1} = 1 - \frac2{m + 1} < 1 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 3
} |
Reducing $ax^6-x^5+x^4+x^3-2x^2+1=0$ to a cubic equation using algebraic substitutions
Use algebraic substitutions and reduce the sextic equation to the cubic equation, where $a$ is a real number:
$$ax^6-x^5+x^4+x^3-2x^2+1=0$$
My attempts.
First, I tried to use the Rational root theorem, when $a$ is an integer $x=\pm... | The coefficient $a$ should be something difficult to deal with, so isolating $a$ might help.
We know that $x=0$ is not a root, so we can divide by $x^6$ to have
$$a=\frac{x^5-x^4-x^3+2x^2-1}{x^6}$$
$$a=\frac 1x-\frac{1}{x^2}-\frac{1}{x^3}+\frac{\color{red}2}{x^4}-\frac{1}{x^6}$$
Now, we can focus on the RHS which has n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4538322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 1
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Diophantine Equation $a^3+b^3=22c^3$ Found this equation in a book on the Riemann Hypothesis. I've found solution without proof on MathStack asking for an algorithm for all the solutions. I'm trying to find a proof for the smallest positive integer solutions. I think I'm close, but not sure how to proceed.
Solve $a^3+b... | COMMENT.-Apropos of Tomita's comment, I want to add that it is not necessary to go to the (more popularly known) Weierstrass form and that the curve $x^3+y^3=nz^3$ is also in their own right an elliptic curve in which the formulas of sum are given as follows:
If $A=(X_1,Y_1,Z_1)\ne B=(X_2,Y_2,Z_2)$ then $A+B=(X_3,Y_3,Z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4540644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Rotating a Matrix into a Pauli Spin Matrix Suppose we are given the matrix
$$
M=\begin{pmatrix}
z & x-iy\\
x+iy & -z
\end{pmatrix}
$$
constrained by $x^2+y^2+z^2=1.$ Note that by the constraints, the eigenvalues of $M$ are given by $\pm1$. My question is, how do we find a unitary matrix $R$ such that
$$
R^{\dagger}MR=\... | The eigenvalues of $M$ are found by calculating the characteristic polynomial
$$ \det(M-\lambda I_2)=\det\begin{pmatrix}z-\lambda & x-yi \\ x+yi & -z-\lambda\end{pmatrix}=\lambda^2-(x^2+y^2+z^2). $$
Thus, $\lambda=\pm1$ since $x^2+y^2+z^2=1$. Now, what about the eigenvectors? For $\lambda=1$,
$$ \underbrace{\begin{pmat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4541371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Trignometry problem: If $\sin^2\theta + 3\cos\theta = 2$ then find $\cos^3\theta + \frac{1}{\cos^3\theta}$ If $\sin^2\theta + 3\cos\theta = 2$ then find $\cos^3\theta + \frac{1}{\cos^3\theta}$
What I did:
$\sin^2\theta + 3\cos\theta = 2$
$3\cos\theta - 1 = 1 - \sin^2\theta$
$3\cos\theta - 1 = \cos^2\theta$
$\cos^3\... | Let $c=\cos\theta$. From what you found, we have
$$c+\frac{1}{c}=3.$$
Then,
$$c^3+\frac{1}{c^3}=(c+\frac{1}{c})^3-3(c+\frac{1}{c})=3^3-3\times 3=18.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542328",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Find $\lim \limits_{x\to 0} \frac{\sin({\pi \sqrt{\cos x})}}{x} $ So I am having trouble finding this limit:
$$\lim \limits_{x\to 0} \frac{\sin({\pi \sqrt{\cos x})}}{x}$$
The problem is I can't use the derivative of the composition of two functions nor can I use other techniques like l'Hôpital's theorem.
I tried numero... | In this answer I will use the fact that $\lim_{x \to 0} \frac{\sin x}{x} = 1$ to derive the limit
\begin{align}
\lim_{x \to 0} \frac{\sin \left( \pi\sqrt{\cos x} \right)}{x}&=\lim_{x \to 0} \frac{\sin \left(\pi - \pi \sqrt{\cos x} \right)}{x}\\
&= \lim_{x \to 0} \frac{\sin \left(\pi\left(1 - \sqrt{\cos x}\right) \righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4542673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$. The following question is taken from JEE practice set.
The equation $x^2-3ax+b=0$ does not have distinct real roots, find the least possible value of $\frac{b}{a-2}$, where $a\gt2$.
My Attemp... | We get from $9a^2\le4b$ that $a,b$ and $\frac b{a-2}$ are all positive. It is clear that raising $b$ does not violate the inequality but increases $\frac b{a-2}$, so for any $a$ we minimise $\frac b{a-2}$ by setting $b=\frac94a^2$.
Once that is fixed we then rewrite
$$\frac b{a-2}=\frac94\cdot\frac{a^2}{a-2}=\frac94\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to factorize $\frac{\cos(3x)-\cos(x)}{\tan(2x)-\tan(x)}$? How to factorize $\dfrac{\cos(3x)-\cos(x)}{\tan(2x)-\tan(x)}$?
Which trigonometric identities to use?
I'm stuck when it comes to $\tan(2x)+\tan(x)$. I don't know which identity to use to turn it into the product.
I was thinking of just transforming them to s... | First note the following identities:
\begin{align}
\cos (3x) &= 4 \cos ^3 (x) - 3 \cos (x) \\
\sin(2x) &= 2 \sin(x) \cos(x) \\
\cos(2x) &= 2 \cos^2(x) - 1
\end{align}
Substituting into the expression yields
\begin{multline}
\frac{\cos(3x) - \cos(x)}{\tan(2x) - \tan(x)} = \frac{4 \cos(x)(\cos^2(x) - 1)}{\frac{2 \sin(x) ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
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How to solve the equation for $x$ where $x^{(x-1)^2} = 2x +1 $ How to solve for $x$ in the following equation
$$ x^{(x-1)^2} = 2x +1$$
Guessing $(x-1)^2=2$ is a pretty good guess and it works giving $x = \sqrt{2} +1 $
But I want a non-guess solution. I tried thinking in terms of logarithm and binomial expansion but cou... | I'm assuming $x$ is non negative, since otherwise $x^{(x-1)^2}$ might not be well defined over the reals.
If $x<1$, then $x^{(x-1)^2}<1$, but $2x+1\ge 1$, so that can't yield a solution. Now suppose $x\ge 1$.
Let $y = (x-1)^2$, so $x^y = 2x+1 =$ $x^2+2-y$, therefore
$$x^y - x^2 = 2-y$$
Because $x \ge 1$, we have $x^y \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548239",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Prove that it converges to zero. Let $a_n$ be a sequence of positive real numbers satisfying $a_n$ < $\frac{1}{2}a_{n-1}$ for all $n \ge 2$. Use an appropriate theorem and the fact that $\lim_{n\to\infty} (\fra... | Testing
Note that for all $n\ge 1$ we have
\begin{align*}
a_n\le \frac{1}{2}a_{n-1}\le \dots \le \left(\frac{1}{2}\right)^{n-1}a_{1}
\end{align*}
Also note that for all $x>0$ exists $m\in \mathbb{N}$ such that $\frac{1}{m}<x$, with this we can conclude that for all $\varepsilon>0$ exists $N\in \mathbb{N}$ such that
\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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Find the first derivative of $y=(x^4-1)\sqrt[3]{x^2-1}$
Find the first derivative of $$y=(x^4-1)\sqrt[3]{x^2-1}$$
We can write the function as $$y=(x^4-1)\left(x^2-1\right)^\frac13$$ For the derivative we have $$y'=4x^3\left(x^2-1\right)^\frac13+\dfrac13\left(x^2-1\right)^{-\frac23}2x(x^4-1)\\=4x^3\left(x^2-1\right)^... | $$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}}\left(7x^2+1\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}}\left(7x^2+1\right)\left(1\right)$$
$$=\frac{2}{3}x\left(x^2-1\right)^{\frac{1}{3}}\left(7x^2+1\right)\left(\frac{\left(x^2-1\right)^{\frac{2}{3}}}{\left(x^2-1\right)^{\frac{2}{3}}}\right)$$
$$=\frac{2}{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548869",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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A smarter (not bashy) way to solve this roots of unity problem? (Mandelbrot) Let $\xi = \cos \frac{2\pi}{5} + i \sin \frac{2pi}{5}$ be a complex fifth root of unity. Set $a = 20\xi^2 + 13 \xi, b = 20\xi^4 + 13\xi^2, c = 20\xi^3 + 13\xi^4, \text{and } d = 20\xi + 13\xi^3$. Find $a^3 + b^3 + c^3 + d^3$
Immediately what c... | Let $\omega_1, \ldots, \omega_p$ be the $p$th roots of unity (including $1$), where $p$ is prime. Note that:
$$\omega_1^k + \ldots + \omega_p^k = \begin{cases} p & \text{if } p \mid k \\ 0 & \text{otherwise.} \end{cases}$$
Why? The group of $p$th roots of unity are isomorphic to $\Bbb{Z} / p\Bbb{Z}$, and the map $x \ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549205",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
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A Question on which solution is false and where (with logarithms) This question was put to me yesterday during a meeting with one of my highschool school teacher who taught me mathematics.
The question:
Find the domain of the function $f:x \to$ $log_\frac{1}{10} \frac{2x+1}{x+2}$.From this, solve the inequality $f(x) ... | The second solution is wrong because $f$ is only decreasing on each of the two intervals $(-\infty,-2)$ and $(-\frac12,\infty)$ but not on their union.
So, a large forgotten part of the solution is when $y<-2$ (and $x>0$, and $y=\frac{2-x}{2x-1}$), i.e. $0<x<\frac12.$
Moreover, the value $x=\frac12$ is also forgotten b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4549387",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show a function is continuous on $\mathbb{R}$ by any method Let $f(x)=\frac{Kx}{K^2+x^2}$ where K is some constant, show this is continuous on $\mathbb{R}$.
Here are my scratch work in looking for a delta. let $x,y\in \mathbb{R} $ WTS $|\frac{Kx}{K^2+x^2}-\frac{Ky}{K^2+y^2}|<\epsilon,\forall \epsilon>0$ whenever $|x-y|... | Since you started your proof attempting to use the $\epsilon$-$\delta$ definition here is a hint. One can proceed in the following way:
*
*If $K=0$, then $f(x)=0$ for all $x$ and the proof is trivial.
*If $K\neq 0$, then we can express:
$$f(x)=\frac{x/K}{1+x^2/K^2}$$.
The $\epsilon$-$\delta$ proof is a bit laboriou... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding a relationship between coefficients to solve a cubic polynomial From E.J Barbeau's Polynomials, the question states:
Find the relationship between $p$ and $q$ in order that the equation $x^3 + px + q$ may be put into the form $x^4 = (x^2 + ax + b)^2$. Hence, solve the equation $8x^3 - 36x + 27 = 0$.
I have no c... | You want $x^3+px+q=0$ to be equivalent to $x^4=(x^2+ax+b)^2$. Notice that:
$$x^4=(x^2+ax+b)^2 \iff (x^2)^2-(x^2+ax+b)^2=0$$
$$\iff (x^2-(x^2+ax+b))(x^2+(x^2+ax+b))=0 \iff (ax+b)(2x^2+ax+b)=0$$
$$\iff 2ax^3+a^2x^2+abx+2bx^2+abx+b^2=0$$
$$\iff 2ax^3+(a^2+2b)x^2+2abx+b^2=0 \iff x^3+\frac{a^2+2b}{2a}x^2+bx+\frac{b^2}{2a}=0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4550359",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
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Prove that $\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln\big(\frac mn \frac{n+p}{m+p} \frac{m+2p}{n+2p} \frac{n+3p}{m+3p}\dots\big)$ page 371 in ‘Synopsis Of Elementary Results In Pure Mathematics’ contains the following result,
$$\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p) \ln(x)}\,dx = \ln \left(\frac{m}{n} ... | Let $a\geq 0$ and $$f(a)=\int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p)\ln x}x^a dx$$
You're really looking for the value of $f(0)$.
Note that
$$\begin{split}
f^\prime(a) &= \int_0^1 \frac{x^{m-1}-x^{n-1}}{(1+x^p)}x^a dx\\
&= \int_0^1 (x^{m-1}-x^{n-1})\left( \sum_{k\geq 0} (-1)^k x^{kp}\right)x^a dx\\
&= \sum_{k\geq 0} (-1)^k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4553487",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
How do I calculate ln(12) approximate value to two decimals? (edit) The problem I've been assigned literally says just to "Calculate ln(12) without a calculator to two decimals using ln(1+x) series." I've been trying to figure out what series I'm dealing with here and don't seem to find an answer anywhere.
| When $x> 1/2$ we can use the following series for the logarithm:
$$\ln(x) = \sum_{k = 1}^{+\infty} \dfrac{1}{k} \left(\dfrac{x-1}{x}\right)^k = \dfrac{x-1}{x} + \dfrac{1}{2}\left(\dfrac{x-1}{x}\right)^2 + \dfrac{1}{3} \left(\dfrac{x-1}{x}\right)^3 + \ldots$$
Plugging $x = 12$ you get
$$\ln(12) \approx \dfrac{11}{12} + ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554897",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
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... | Integrate by parts as follows
\begin{align}
\int \frac{3}{(x^2 +5)^2}dx &=\int \frac3{10x}d\left(\frac{x^2}{x^2+5}\right)\\
&= \frac{3x}{10(x^2+5)}+\frac3{10}\int\frac{1}{x^2+5}dx\\
&= \frac{3x}{10(x^2+5)}+\frac3{10\sqrt5}\tan^{-1}\frac x{\sqrt5}+C
\end{align}
| {
"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": 0
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Proving a combinatorial identity having sums of fractions of binomial coefficients Recently I was solving a probability question, and I encountered a summation that I was unable to figure out. I put it on Wolfram Alpha, and it returned an unexpectedly simple solution. The answer I mention is here, and it is correct as ... | We seek to evaluate
$$\sum_{q=1}^{r+1} {r\choose q-1} {b+r\choose q}^{-1}
= \sum_{q=0}^r {r\choose q} {b+r\choose q+1}^{-1}.$$
Recall from MSE
4316307 the
following identity which was proved there: with $1\le k\le n$
$$\frac{1}{k} {n\choose k}^{-1}
= [z^n] \log\frac{1}{1-z} (z-1)^{n-k}.$$
We obtain
$$\sum_{q=0}^r {r\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 3
} |
A discrete difference computation I am attempting to compute the following discrete difference:
$$\Delta[(3x+2)^2]$$ where $\Delta[f(x)] = f(x+1)-f(x)$
Method 1:
\begin{align*}
\Delta[(3x+2)^2] &= [3(x+1)+2]^2 - (3x+2)^2 \\
&=(3x+5)^2 - (3x+2)^2 \\
&= (9x^2+30x+25)-(9x^2+12x+4) = 18x+21
\end{align*}
Method 2:
Define $f... | Using
$$ (a x + b)^2 = (a x + b)(a (x-1) + b) + a \, (a x + b) = (a x + b)^{(2)} + a \, (a x + b)^{(1)}$$
then
\begin{align}
\Delta (a x + b)^{2} &= \Delta (a x + b)^{(2)} + a \, \Delta (a x + b)^{(1)} \\
&= 2 \, a \, (a x + b)^{(1)} + a \, a \, (a x + b)^{(0)} \\
&= 2 \, a \, (a x + b) + a^2 \\
&= a \, (2 a \, x + 2 b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4557905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Convert $z=-1$ to polar form? I have an issue when wanting to convert $z=-1$ to the polar form.
Indeed, assuming that $z$ is of the form $z=a+ib$, I start by finding $z=(-1)+(0)i$, thus $a=-1$ and $b=0$.
I know that the polar form of $z$ is $z=re^{i\theta}$, where $r=\sqrt{a^2+b^2}$, thus $r=\sqrt{(-1)^2}=1$.
Then, I ... | Your attempt is good, but your formula for the argument $\theta=\operatorname{arg}(z)=\arctan\left(\frac{b}{a}\right)=\arctan\left(\frac {0}{-1}\right)=0$ only works in special cases (when
$a > 0$). Actually the formulas are:
$$
\begin{align*}
z &= a + b \cdot \mathrm{i}\\
z &= r \cdot \operatorname{cis}(\theta) = r \c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4558242",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"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... | Have you ever seen Archemedes' proof of the quadrature of a parabola?
https://en.wikipedia.org/wiki/Quadrature_of_the_Parabola
Archimedes proved that any secant through a parabola intersecting at points $\alpha, \beta$ form a region with area $\frac{a(\alpha - \beta)^3}{6}$
But using calculus...
$y = ax^2 + bx + c$
let... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4562637",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Evaluating $\int_0^{2 \pi} \frac{1}{5-3 \sin z}dz$ So I need to evaluate $\int_0^{2 \pi} \frac{1}{5-3 \sin z}dz$. Here's what I have thus far:
\begin{align*}
\int_0^{2 \pi} \frac{1}{5-3 \sin z}dz&= \frac{1}{3} \int_0^{2 \pi} \frac{1}{\frac{5}{3}- \sin z}dz\\
&=\frac{1}{3} \int_0^{2 \pi} \frac{1}{\frac{10i-3e^{i z}+3e^{... | Let $t=\tan(\frac{z}{2})$, $\sin z=\frac{2t}{t^2+1}$ and $dz=\frac{2dt}{t^2+1}$ and thinking the integral from $-\pi$ to $\pi$, we get
$$\int_{-\infty}^{\infty}\frac{2}{5}\frac{1}{(t-\frac{3}{5})^2+(\frac{4}{5})^2}dt=\left.\frac{1}{2}\arctan\left(\frac{5t-3}{4}\right)\right\rvert_{-\infty}^{\infty}=\frac{\pi}{2}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566936",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $xyz = x+y+z$ for $x,y,z > 0$, then $\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$
If $x,y,z \in \Bbb R_{>0}$ satisfy $xyz = x+y+z$, prove that $$\sqrt{1+x^2} + \sqrt{1+y^2} + \sqrt{1+z^2} \ge 6$$
We can express $1+x^2$ as
$$1+x^2 = (1-xy)(1-xz) = \frac{(y-xyz)(z-xyz)}{yz} = \frac{(x+z)(x+y)}{yz}$$
since $x + ... | Your idea gives a simple solution.
Indeed, by AM-GM twice we obtain: $$\sum_{cyc}\sqrt{\frac{(x+y)(x+z)}{yz}}\geq3\sqrt[3]{\frac{\prod\limits_{cyc}(x+y)}{xyz}}\geq3\sqrt[3]{\frac{8xyz}{xyz}}=6.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4575712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Exponential equations: Why is this method wrong? While solving exponential equations with my Y10 students, they found that you can break down a number into powers with the same base and add one $a^0$ and just use exponents, but never more than one power of one. They tried this out of laziness to do variable change and ... | It works because the equation is actually involving a variation of the hyperbolic sine function $\sinh(x)=\frac{e^x-e^{-x}}{2}$ which is known to be invertible.
If $a$ is a positive real number different from $1$, let $\sinh_a(x)=\frac{a^x-a^{-x}}{2}$ (I will call it the "hyperbolic sine of base $a$"). Your equation is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
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... | Let $$x^{6m+2}-x^{6m+1}+1=\lambda(x^2-x+1)$$
$$x^{6m+8}-x^{6m+7}+1=S$$
$$S=x^6(x^{6m+2}-x^{6m+1})+1$$
$$S=x^6(\lambda(x^2-x+1)-1)+1$$
$$S=x^6\lambda(x^2-x+1)-x^6+1$$
Now dividing $S$ by $x^2-x+1$ yields
$$\frac{x^6\lambda(x^2-x+1)}{x^2-x+1}+\frac{1-x^6}{x^2-x+1}$$
$$\frac{x^6\lambda(x^2-x+1)}{x^2-x+1}+\frac{(1+x-x^3-x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584894",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
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Prove that $2\cdot 3^x +1= p^y$ has no solution Prove that the Diophantine equation of $3$ variables $(x,y,p)$
$$2\cdot 3^x +1= p^y$$
has no solution where $x,y\in\mathbb{N}_+$, $x\ge2, y\ge2$ and $p$ is a prime number.
I found that $y$ cannot be even, if $y =2k$ then $4| (p^k -1)(p^k+1) =2\cdot3^x$ which is a contradi... | We have, as the O.P. noted, the exponent of the prime $p$ is odd so
$$2\cdot3^x=(p-1)(p^{2y}+p^{2y-1}+\cdots+p+1)$$ the second factor having $y+1$ terms is odd so $2$ divides exactly $p-1=2m$, with $m$ odd. It follows
$$3^x=m(p^{2y}+p^{2y-1}+\cdots+p+1)\Rightarrow m=3^a \text { and } \boxed{p=2\cdot 3^a+1} $$ then
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4586098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
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$\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x=\frac{\pi^2}{2}$ with one-variable calculus solution Prove that $$\int_{0}^{\pi}(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x=\frac{\pi^2}{2}$$
My textbook says that we need to note that the integral can be tran... | Using the integration by parts, you have
\begin{eqnarray}
&&\int(\sqrt{x^2+\sin^2x}+\cos x\sin x\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}x\\
&=&\int\sqrt{x^2+\sin^2x}dx+\frac12\int_{0}^{\pi}\ln(x+\sqrt{x^2+\sin^2x}))\mathrm{d}\sin^2x\\
&=&\int\sqrt{x^2+\sin^2x}dx+\frac12\bigg(\sin^2x\ln(x+\sqrt{x^2+\sin^2x})-\int\frac{\sin^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4590069",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to evaluate this integral containing a Bessel function: $\int_1^2 x^{-2} J_2(x)\,dx$? $$\int_1^2 x^{-2} J_2(x)\,dx$$
I can't solve this integration problem. I want to develop the expression using Derivation of the Bessel function and Recurrence relations between the Bessel functions and get the answer.
| You can evaluate in terms of hypergeometric functions. Recall that
$$J_{\nu}(z)=\frac{(z/2)^\nu}{\Gamma(\nu+1)}~{}_0F_1\left(;\nu+1~;~\frac{-z^2}{4}\right)$$
Therefore
$$x^{-2}J_2(x)=\frac{1}{x^2}\frac{(x/2)^2}{\Gamma(2+1)}~{}_0F_1\left(;2+1~;~\frac{-x^2}{4}\right) \\ =\frac{1}{8}~{}_0F_1\left(;3~;~\frac{-x^2}{4}\right... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4592050",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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How to solve $\int\frac{x\arctan x}{x^4+1}dx$ in a practical way I need to evaluate the following indefinite integral for some other definite integral
$$\int\frac{x\arctan x}{x^4+1}dx$$
I found that
$$\int_o^\infty\arctan{(e^{-x})}\arctan{(e^{-2x})}dx=\frac{\pi G}{4}-2I(1)$$
where
$$I(t)=\int_0^1\frac{x\operatorname{Ti... | I am not sure that you could avoid dilogarithms except if you accept an infinite summation of Gaussian hypergeometric functions.
$$I=\int \frac{x }{x^4+1}\tan ^{-1}(x)\,dx$$ Using integration by parts
$$I=\frac{1}{2} \tan ^{-1}(x) \tan ^{-1}\left(x^2\right)-\frac 12\int \frac{\tan ^{-1}\left(x^2\right)}{x^2+1}\,dx$$ l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4600858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to find the constant $C$ such that $f(x)\geq Cx$ Problem :
Define for strictly positive $x$ :
$$f\left(x\right)=\left(\prod_{k=1}^{\operatorname{floor}\left(x\right)}\left(1+\sum_{n=1}^{k}\frac{1}{k\cdot2^{n}}\right)\right)$$
Does there exists a constant $C$ such that :
$$f(x)\geq Cx$$
I think definitely yes and $C... | The claim is that $f(x)\ge Cx$ for any $x>0$, where
$$
C= \prod_{k=1}^\infty\left( {1 - \frac{1}{{2^k (k + 1)}}} \right) =0.651649356319290\ldots,
$$
as given by @metamorphy in the comments. Indeed
\begin{align*}
\log f(x) & = \sum\limits_{k = 1}^{\left\lfloor x \right\rfloor } {\log \left( {1 + \sum\limits_{n = 1}^k {... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4601174",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
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Probability on Number Theory Problem: Suppose that $a,b,c \in \{1,2,3,\cdots,1000\}$ are randomly selected with replacement. Find the probability that $abc+ab+2a$ is divisible by $5$.
Answer given from the worksheet: $33/125$
My answer: $\frac{641}{3125}$
Attempt: Since $abc+ab+2a = a(bc+b+2)$, either $a \equiv 0 \pmo... | You've made a good attempt. The only mistake I can determine is that, modulo $5$, there are only $5$ possibilities for each of $b$ and $c$, so there are $5^2 = 25$ total potential combinations, not $5^4$. Thus, your $4 \cdot\frac{1}{5^4} = \frac{4}{625}$ should be $4 \cdot\frac{1}{5^2} = \frac{4}{25}$ instead. The fina... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4602923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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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
$$\begin{align}
s&:=\sin\theta\\
c&:=\cos\theta\\
I&=\int_{0}^{2\pi}{\cos(\theta)^2-\sin(\the... | Graph the integrand and observe that there is "as much area above as below the x-axis". This leads to the conjecture that the value is zero, and also gives a strategy for proving it using a symmetry argument.
First observe that the function is periodic with period $\pi$. You can confirm this by showing that $f(x + \p... | {
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"url": "https://math.stackexchange.com/questions/4603090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 2
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Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$ Solving the homogeneous first order ode $y' = \frac{2xy}{y^2-x^2}$
substituting $y=ux$ so that $y' = u + x\frac{du}{dx}$"
$u + x\frac{du}{dx} = \frac{2x^2u}{ux^2-x^2} = \frac{2u}{u^2-1}$
$\rightarrow x\frac{du}{dx} = \frac{2u}{u^2-1} - u = \frac{2u}{u^2... | HINT
According to the substitution $y = ux$, it results that
\begin{align*}
y' = \frac{2xy}{y^{2} - x^{2}} & \Longleftrightarrow y' =\frac{2(y/x)}{(y/x)^{2} - 1}\\\\
& \Longleftrightarrow xu' + u = \frac{2u}{u^{2} - 1}\\\\
& \Longleftrightarrow xu' = \frac{2u}{u^{2} - 1} - u\\\\
& \Longleftrightarrow xu' = \frac{3u - u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4603324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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How to merge odd series and even series of hypergeometric function of Legendre polynomials into one hypergeometric function? On the Wolfram MathWorld page of Legendre Differential Equation, Legendre polynomials are represented as
$$
P_l(x) = c_n
\begin{cases}\begin{align*}
&_2F_1\left(-\frac{1}{2}(l), \frac{1}{2}(l + 1... | A quadratic transform for the Gaussian hypergeometric function exists which links (1) and (2):
\begin{align}
{ }_2 F_1\left(a, b ; \frac{a+b+1}{2} ; z\right)&= \\
&\frac{\sqrt{\pi} \Gamma\left(\frac{a+b+1}{2}\right)}{\Gamma\left(\frac{a+1}{2}\right) \Gamma\left(\frac{b+1}{2}\right)}{ }_2 F_1\left(\frac{a}{2}, \frac{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4605042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Grouping the real and imaginary terms in the proof of Euler's formula using the MacLaurin series The MacLaurin series:
\begin{align}
\sin x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n+1)!}x^{2n+1}=x-\frac{x^3}{3!}+\frac{x^5}{5!}-\cdots
\\\\
\cos x&=\sum_{n=0}^{\infty}\frac{(-1)^n}{(2n)!}x^{2n}=1-\frac{x^2}{2!}+\frac{x^4}{4!}... | If $\sum_{n=1}^{\infty}a_n = L$, then so is $\sum_{n=1}^{\infty}b_n=L$ for $b_n = a_{n/2}$ if $n$ is even, and $0$ if $n$ is odd. This is because $$\sum_{n=1}^{2N}b_n = \sum_{n=1}^{2N+1}b_n = \sum_{n=1}^{N}a_n\to L$$ as $N\to\infty$. Hence from the convergence of series for cosine and sines, we have \begin{align*}&\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4608202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Contest Math Question: simplifying logarithm expression further I am working on AoPS Vol. 2 exercises in Chapter 1 and attempting to solve the below problem:
Given that $\log_{4n} 40\sqrt{3} = \log_{3n} 45$, find $n^3$ (MA$\Theta$ 1991).
My approach is to isolate $n$ and then cube it. Observe:
\begin{align*}
\frac{\l... | Alternative approach:
Let $~r~$ denote $\displaystyle \log_{4n} 40\sqrt{3}.$
Then
$$(4n)^r = 40\sqrt{3}, ~~(3n)^r = 45 \implies $$
$$\left[\frac{4}{3}\right]^r = \left[\frac{4n}{3n}\right]^r = \frac{40\sqrt{3}}{45} = \frac{8\sqrt{3}}{9} \implies $$
$$9 \times 4^r = 8\sqrt{3} \times 3^r \implies $$
$$\frac{2^{2r}}{8} = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Properties of eigenvectors $$A :=\begin{pmatrix}
5 &−6 &0\\
2 &−2& 0 \\
0 &0 &2 \\
\end{pmatrix} $$
Determine an eigenvector of the matrix that is perpendicular to the vector $\mathbf v = (1, −4, 1).$
I think I need to first determine the eigenvalues and eigenvectors of $A$. Then I can ... | From the definition of the eigenvector $\mathbf v$ corresponding to the eigenvalue $λ$ we have
$$A\mathbf v=λ\mathbf v$$
Then:
$$A\mathbf v-λ\mathbf v=(A-λI)\mathbf v=0$$
Equation has a nonzero solution if and only if
$$\det(A-λI)=0$$
$$\det(A-λI)=\begin{vmatrix} 5-λ& -6& 0 \\ 2& -2-λ& 0\\ 0& 0& 2-λ\end{vmatrix} $$$$=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4610667",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Show positivity of a function of two variables in the unit square. Let $$
f(x,y) = x^3 (1 + y + y^2) + y^2 \Big[x^2 (5 + 2 y) + x (-6 - 4 y + y^2) + (1 + 3 y + y^2)\Big]
$$
Show that $f(x,y) \ge 0 $ for $0\le x \le 1$ and $0\le y \le 1$.
Numerical evaluations seem to support the claim. This is a cubic function in... | We have
\begin{align*}
f &= (x+1)y^4 + (2x^2 - 4x + 3)y^3 + (x^3 + 5x^2 - 6x + 1)y^2 + x^3y + x^3\\
&\ge (x+1)y^4 + (2x^2 - 4x + 3)y^4 + (x^3 + 5x^2 - 6x + 1)y^2 + x^3y^2 + x^4 \tag{1}\\
&= (2x^2 - 3x + 4)y^4 + (2x^3 + 5x^2 - 6x + 1)y^2 + x^4\\
&\ge (2-x)^2 y^4 + (2x^3 + 5x^2 - 6x + 1)y^2 + x^4\tag{2}\\
&\ge 2\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4611054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
A $n \times n$ linear system Solve the linear system
$
\begin{array}{ c c c c c c c c c c c c c c }
& & x_{2} & + & x_{3} & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 2 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ( 1)\\
x_{1} & & & + & x_{3} & + & \dotsc & + & x_{n-1} & + & x_{n} & = & 4 & \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (... | When $n=1$, you have just a single equation, but the left side is an empty sum ($0$) and the right side is $2$. Thus for $n=1$ your system devolves to a contradiction, and with that you should expect the division to break down.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4615660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $
Assume that positive numbers a, b, c, x, y, z satisfy $cy+bz =a;
az + cx = b$$bx + ay = c$.
Prove that $ \frac{x^2}{1+x} + \frac{y^2}{1+y} + \frac{z^2}{1+z} \geq \frac{1}{2} $
I've tried appling A.M.-G.M. inequality but it didn't ... | All things are same as that of answer given by Calvin Lin upto $x= \frac{b^2+c^2-a^2}{2bc}$ and similarly for $y,z$.
Clearly we can see that $b^2+c^2>a^2$ ($x$ is positive) and same for $b,c$.
We get an acute angled triangle of sides $a,b,c$ with $x=\cos A, y=\cos B, z=\cos C$. Let the function which to prove is $f(\De... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4616087",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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AM-GM & Minimization Proof I want to prove that for all $x, y > 0$, $$\cfrac{x+y}{2} \geq \sqrt{xy}$$
Particularly, I want to show that the minimum of $(x+y)/2$ is exactly $\sqrt{xy}$.
This is my attempt:
$\textbf{Proof}$ (Contradiction). Assume if $x, y > 0$, then $(x+y)/2 < \sqrt{xy}$. But
\begin{align*}
x+y &< 2\sqr... | Instead of talking about minimum, we may say that the inequality
$$\frac{x+y}{2} \geq \sqrt{xy}$$
holds for any $x, y>0$ and the equality holds for $x=y.$
We may prove the inequality directly by starting with $$
\begin{aligned}
& x+y-2 \sqrt{x y}=(\sqrt{x}-\sqrt{y})^2 \geqslant 0 \\
\Rightarrow \quad & x+y \geqslant 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4617393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
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Find $x$ such that $\sqrt{x+1} - \sqrt{1-x} = 1$ To solve this equation, I started by putting the condition $x\in [-1, 1]$, then squared a few times: $\sqrt{x+1} - \sqrt{1-x} = 1 \iff x + 1 +1-x-2\sqrt{1-x^2} =1 \iff 2\sqrt{1-x^2}=1 \iff 4(1-x^2)=1 \iff 4x^2=3 \iff x=\pm \frac{\sqrt{3}}{2}$
This, however, is not the ri... | Extraneous roots can be introduced by squaring. We can avoid squaring by multiplying by the conjugate.
The equation
$$\sqrt{x + 1} - \sqrt{1 - x} = 1$$
imposes the restrictions that $x + 1 \ge 0 \implies x \geq -1$ and $1 - x \ge 0 \implies x \leq 1$. Therefore, we know that any valid solution must satisfy $-1 \leq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4618757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 7,
"answer_id": 6
} |
How do we prove $x^6+3x^3+2x^2+x+1 \geq 0$ Question
How do we prove
$$x^6 + 3x^3+2x^2+x+1\geq0$$
My progress
$$x^6+3x^3+2x^2+x+1=(x+1)^2(x^4-2x^3+3x^2-x+1)$$
I appreciate your interest
| Again focusing on $x^4-2x^3+3x^2-x+1$, write this as the sum of $f(x) = x^4 - 2x^3 + 3x^2$ and $g(x) = -x + 1$. $f(x) = x^2(x^2 - 2x + 3) = x^2 ((x - 1)^2 + 2)$ which is always nonnegative.
That leaves $g(x)$ which is negative for $x \ge 1$. However, we just need to show that $f(x) > x$ in this domain, which can be sho... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4619243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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How to solve a system of equations over a finite field? I need to solve a system of equations over $\mathbb{Z}_{11}$.
My system is:
$$ \left\{
\begin{array}{l}
2x + 5y + z = 8 \\
7x + 6y + 8z = 10 \\
10x + 3y + 4z = 6
\end{array}
\right.
$$
In matrix form:
$$
\begin{bmatrix}
2 & 5 & 1\\
7 & 6 & 8\\
10 & 3 & 4\\
\end{b... | Since the determinant of your matrix $A$ is $221$, it is non-zero (actually equal to $1$) in the field $\Bbb F_{11}$. Hence the inverse exists, so that $Ax=b$ has the unique solution
$$
b=A^{-1}x=\begin{pmatrix} 0 & 5 & 1\cr 8 & 9 & 2\cr 5 & 0 & 10 \end{pmatrix}\cdot \begin{pmatrix} 8 \cr 10 \cr 6\end{pmatrix} =
\begin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4624324",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Deduce limit of sequence from another sequence We have a sequence $(a_n)$ defined by the formula $$a_n = \int_0^1 (1 - x^2)^{\frac n2} \mathrm dx$$ We are also asked to prove, which I managed to do, the following recursion formula $$a_n = \frac{n + 3}{n + 2} \cdot a_{n + 2}$$ We are then asked to deduce with this infor... | Hint. Note that $0\le (1-x^2)^{(n+2)/2} \le (1-x^2)^{(n+1)/2} \le (1-x^2)^{n/2}\le 1$ when $0\le x \le1$. This gives $0 \le a_{n+2} \le a_{n+1} \le a_n \le 1$. So, $\dfrac{a_{n+2}}{a_n} \le \dfrac{a_{n+1}}{a_n} \le 1$. Can you simplify $\dfrac{a_{n+2}}{a_n}$? What can you say about the limit $\lim_{n\to \infty} \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4626282",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Construct equation of ellipse, if focuses are $F_1 (-1, -1), F_2 (1, 1)$, height (perpendicular to $F_1 F_2$) $b = 1$ First of all, I know the right answer. It's $2x^2 - 2xy + 2y^2 - 3 = 0$.
What is unknown to me, is how to obtain that answer.
I've tried to make equation like $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$
We ... | If $\left(\begin{matrix}X\\Y\end{matrix}\right)$ is the image of the vector $\left(\begin{matrix}x\\y\end{matrix}\right)$ after rotation by $45^o$, then $$\left(\begin{matrix}X\\Y\end{matrix}\right)=M\left(\begin{matrix}x\\y\end{matrix}\right)$$
Where $$M=\left(\begin{matrix}\cos 45&-\sin 45\\\sin 45&\cos 45\end{matrix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4630590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the number of squares modulo a composite number $143$ I want to know how many classes of modulo $143$ are squares. This is equivalent to find how many $a$ can make $$\begin{align*}
x^2 \equiv a \pmod{143} && (1)
\end{align*}$$ has solutions
On the other hand, I know that $143=11 \times 13$ and $(11,13)=1$, so $$... | So there should be $6\cdot 7=42.$ These can be gotten under the isomorphism between $$\Bbb Z_{11}×\Bbb Z_{13}$$ and $$\Bbb Z_{143}.$$
For instance, using Bezout, we can map $(x,y)$ to $-7×11×y+6×13×x.$
You can now check that for instance, $(3,12)\mapsto -924+234\equiv-690\equiv-118\equiv25.$ And $(4,0)\mapsto312\eq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633047",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Evaluating $\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}-\cos x}{x^3\sin x}$ I'm trying to evaluate $$\lim\limits_{x\rightarrow0} \frac{e^{-\frac{x^2}{2}}-\cos x}{x^3\sin x}.$$
I can see the $\frac{0}{0}$ form, so I'll use L'Hôpital's rule. However, I'll eliminate the sine function in the denominator by multip... | The following steps use De l'Hopital rule.
$$\lim _{x\to 0}\left(\frac{e^{-\frac{x^2}{2}}-\cos \left(x\right)}{x^3\sin \left(x\right)}\right)=\lim _{x\to \:0}\left(\frac{-e^{-\frac{x^2}{2}}x+\sin \left(x\right)}{3x^2\sin \left(x\right)+\cos \left(x\right)x^3}\right)=\lim _{x\to \:0}\left(\frac{e^{-\frac{x^2}{2}}x^2-e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4633510",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Asymptotic behavior of the function $f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$ In one of my analysis course, we considered the function
$$f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}dy.$$
Then my teacher told us that $f$ had the following behavior
$$
f(x)\sim\begin{cases}
\frac{-1}{4}... | $$f(x) = x \int_x^\infty \frac{1 - e^{-\frac{1}{4}y^2}}{y^3}\,dy$$
Let $y=2 \sqrt{t}$ to make
$$f(x)=\frac 18 x \int^\infty_{\frac {x^2}4} \frac{1-e^{-t}}{t^2}\,dt$$ Using the incomplete gamma function
$$\int t^{-n} e^{-t}\,dt=-\Gamma (1-n,t)$$
wich makes
$$f(x)=\frac{1}{2 x}-\frac{x}{8}\, \Gamma \left(-1,\frac{x^2}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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How do I solve the equation $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$? Problem: $(x - 1)(x - 2)(x - 3) = (x - 2)(x - 3)(x - 4)$
Heres my question with this problem: why do I end up with a wrong answer when I divide both sides by $(x-2)(x-3)$ to cancel out the $(x-2)(x-3)$ on both sides. Is this not allowed and wh... | Obviously, both $x=2$ and $x=3$ are solutions.
If $x$ is neither above, then we can cancel $(x-2)(x-3)$ on both sides of the equation, so we get
$$
x-1 = x-4, \text{or} -1=-3,
$$
which is a contradiction.
Therefore we only have two solutions: $x=2$ and $x=3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4634658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Evaluating $\int \sqrt{\frac{2x+3}{2x-3}}dx$ I was evaluating $\int \sqrt{\frac{2x+3}{2x-3}}dx$ and got an answer which, I think, is not correct as it is different from wolframalpha's answer.
Here's my work:
$$\begin{align}\int \sqrt\frac{2x+3}{2x-3} dx & = \int \frac{2x + 3}{\sqrt{4x^2 - 9}} \ dx\tag{1}\\& =\int \frac... |
\begin{align}\int \sqrt\frac{2x+3}{2x-3} dx & = \int \frac{2x + 3}{\sqrt{4x^2 - 9}} \ dx\tag{1p}\\& =\int \frac{x}{\sqrt{x^2 - (3/2)^2}}\ dx + \frac32\int \frac{dx}{\sqrt{x^2 - (3/2)^2}}\tag{2p}\\&= \sqrt{x^2 - (3/2)^2 } + \frac{3}{2}\cosh^{-1}\left(\frac{2x}{3}\right) + C\tag{3p}\end{align}
*
*Your work is correc... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4637209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
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Calculate $\lim \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1}$ as $x \to 2$ I don't know the answer to the problem. None of my "tricks" work: I see no way of factoring it, rationalization on the denominator doesn't seem useful (because I still end up with a zero). The book has not gone over any fancier techniques such as... | The limit has the indeterminate form $0/0$, which can be eliminated if you rationalize both the numerator and denominator.
\begin{align*}
\lim_{x \to 2} \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1} & = \lim_{x \to 2} \frac{\sqrt{6 - x} - 2}{\sqrt{3 - x} - 1} \cdot \frac{\sqrt{6 - x} + 2}{\sqrt{6 - x} + 2} \cdot \frac{\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4642874",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Let $T:M_{n\times n}(\mathbb{C})\to M_{n\times n}(\mathbb{C})$ be defined by $T(X)=A^2X+AXA+XA^2$ for some $A\in M_{n\times n}(\mathbb{C})$. Let $T:M_{n\times n}(\mathbb{C})\to M_{n\times n}(\mathbb{C})$ be defined by $T(X)=A^2X+AXA+XA^2$ for some $A\in M_{n\times n}(\mathbb{C})$. Then $T$ is a linear operator. The pro... | Kronecker product is the right tool for analysing such issues. Here is how.
Let us begin by writing
$$T(X)=T_A(X)=A^2X+AXA+XA^2\tag{1}$$
in a different way, using the "vec" operator stacking the columns of any $3 \times 3$ matrix into a single $9 \times 1$ column vector :
$$T(\operatorname{vec}(X))=\operatorname{vec}(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
$\int \frac{x^2}{\sqrt{5-x^2}}dx$ using substitution $u = \sqrt{5-x^2}$? I was trying to solve $$\int\frac{x^2}{\sqrt{5-4x^2 }} dx$$ by using the substitution
$$
u=\sqrt{5-4x^2}\\
dx = \frac{\sqrt{5-4x^2}}{-4x}du$$ So the integral can be written as $$\int \left(-\frac{x}{4}\right) du$$ and
$$
x= \begin{cases}
\sqrt{(5-... | Integration by parts
$$
\begin{aligned}
I&=\int \frac{x^2}{\sqrt{5-x^2}} d x \\
& =-\int x d\left(\sqrt{5-x^2}\right) \\
& =-x \sqrt{5-x^2}+\int \sqrt{5-x^2} d x \\
& =-x \sqrt{5-x^2}+\int \frac{5-x^2}{\sqrt{5-x^2}} d x \\
& =-x \sqrt{5-x^2}+5 \int \frac{d x}{\sqrt{5-x^2}}-I \\
& =\frac{1}{2}\left[5 \sin ^{-1}\left(\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644624",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Foci of ellipse My question is given a ellipse of the equation : $$\frac{x^2}{a^2}+\frac{y^2}{b^2} =1$$ where $a>b$ then how we can find the coordinates of the foci. I want to find those coordinates without the presuming that the foci exists because most proofs I found online assume the properties of foci to be true an... | *
*Find the eccentricity of the ellipse.
$$e=\sqrt{1-\dfrac{b^2}{a^2}}$$
*Now you have the following expressions to find the coordinates of the foci:
$$\begin{align*}
F_1 &= \left(ae, 0\right) = \left(a\sqrt{1-\dfrac{b^2}{a^2}}, 0\right),\\
F_2 &= \left(-ae, 0\right) = \left(-a\sqrt{1-\dfrac{b^2}{a^2}}, 0\right).
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4644809",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding steady state equations of infinite-dimensional Markov Chain
Suppose we have the transition matrix
$$P=\begin{bmatrix} 0&1&0&0&0&\cdots\\\\ p&0&1-p&0&0&\cdots\\\\
0&p&0&1-p&0&\cdots\\\\ 0&0&p&0&1-p&\cdots\\\\ \vdots&\vdots&\vdots&\vdots&\vdots&\ddots\\\\ \end{bmatrix}$$
with $p\in[1/2,1]$. Show that the solu... | A steady-state solution satisfies $P^T \pi = \pi$, where $\pi$ is a column vector of probabilities. Reading off the matrix system gives
$$\begin{cases}
\pi_0 = p \pi_1 \\
\pi_1 = \pi_0 + p \pi_2 \\
\pi_i = (1 - p) \pi_{i - 1} + p \pi_{i + 1} & i \ge 2 \\
\pi_0 + \pi_1 + ... = 1
\end{cases}$$
Of course, we may rewrite t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4647329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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Divergence of vector-tensor outer product multiplication I have a material derivative of a tensor quantity $\mathbf{S}$.
$$
\frac{\partial \mathbf{S}}{\partial t} + \vec{v} \cdot \nabla \mathbf{S}
$$
I would like to know if the term $\vec{v} \cdot \nabla \mathbf{S}$ can be rewritten as:
$$
\nabla \cdot \left(\vec{v} \o... | Note that following relations hold:
Scalar $\rightarrow$ Vector:
$$
\operatorname{grad} \phi(\mathbf{x}):=\frac{\mathrm{d} \phi(\mathbf{x})}{\mathrm{d} \mathbf{x}}=: \mathbf{w}(\mathbf{x})
$$
Vector $\rightarrow$ Matrix:
$$
\operatorname{grad} \mathbf{v}(\mathbf{x}):=\frac{\mathrm{d} \mathbf{v}(\mathbf{x})}{\mathrm{d} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4648173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Is there an integral that proves $\pi > 333/106$? The following integral,
$$ \int_0^1 \frac{x^4(1-x)^4}{x^2 + 1} \mathrm{d}x = \frac{22}{7} - \pi $$
is clearly positive, which proves that $\pi < 22/7$.
Is there a similar integral which proves $\pi > 333/106$?
| Let us consider the polynomial
$$P_n(x):=1-x^2+x^4-x^6+\cdots x^{2n-2}=\frac{x^{2n}+1}{x^2+1}.$$
We have
$$0<\int_0^1\left(\frac{P_n(x)}{x^2+1}-\frac1{x^2+1}\right)^2dx<\int_0^1\left(\frac{x^{2n}}{0+1}\right)^2dx=\frac1{4n+1},$$ and the integral can be made as small as desired.
On another hand, the remainder of the lon... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1956",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "129",
"answer_count": 6,
"answer_id": 4
} |
Proof that $1+2+3+4+\cdots+n = \frac{n\times(n+1)}2$ Why is $1+2+3+4+\ldots+n = \dfrac{n\times(n+1)}2$ $\space$ ?
| I didn't see an answer using the method that I used, so I'm posting an answer here to 'spread the knowledge!'
This can be applied to higher powers such $1^2+2^2+\cdots n^2$ or $1^3+2^3+\cdots n^3$.
Solving by the use of Indeterminate Coefficients:
Assume the series$$1+2+3+4+5\ldots+n\tag1$$
Is equal to the infinite ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "136",
"answer_count": 36,
"answer_id": 20
} |
Lateral surface area by hand I need to find the exact solution for the lateral surface area of the solid generated by revolving the region bounded by $y=x^2$, $y=0$, $x=0$, and $x=\sqrt 2$ about the X axis.
I have come up with my own solution, but I'm not sure if it's right.
Before substituting values back in, I got
$$... | With $x=\frac{1}{2}\tan\theta$ and $dx=\frac{1}{2}\sec^2\theta d\theta$, $2\pi\int x^2\sqrt{1+4x^2}dx=\frac{\pi}{4}\int\tan^2\theta\sec^3\theta d\theta$. Now, let
$$\begin{align}
I_1&=\int\tan^2\theta\sec^3\theta d\theta
\\
&=\int(\sec^3\theta-\sec\theta)\sec^2\theta d\theta
\\
&=\tan\theta(\sec^3\theta-\sec\theta)-\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/5944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Funny identities Here is a funny exercise
$$\sin(x - y) \sin(x + y) = (\sin x - \sin y)(\sin x + \sin y).$$
(If you prove it don't publish it here please).
Do you have similar examples?
| $$
\frac{e}{2} = \left(\frac{2}{1}\right)^{1/2}\left(\frac{2\cdot 4}{3\cdot 3}\right)^{1/4}\left(\frac{4\cdot 6\cdot 6\cdot 8}{5\cdot 5\cdot 7\cdot 7}\right)^{1/8}\left(\frac{8\cdot 10\cdot 10\cdot 12\cdot 12\cdot 14\cdot 14\cdot 16}{9\cdot 9\cdot 11\cdot 11\cdot 13\cdot 13\cdot 15\cdot 15}\right)^{1/16}\cdots
$$
[Nick... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/8814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "281",
"answer_count": 63,
"answer_id": 22
} |
Work out the values of a and b from the identity $x^2 - ax + 144 = (x-b)^2$ How do I solve the following question?
You are given the identity $x^2-ax+144 = (x-b)^2$
Work out the values of $a$ and $b$.
Question appears in AQA 43005/1H.
| Find the Value of $b$
$x^2-ax+144\equiv(x-b)^2$
Expand the right-hand side:
$x^2-ax+144\equiv{x^2-2bx+b^2}$
Find the coefficient $(x=0)$:
$b^2=144$
$\Longrightarrow b=\pm{\sqrt{144}}=\pm12$
Find the Value of $a$
Substitute $b=\pm12$ back into original equation:
$x^2-ax+144\equiv{x^2\pm{24x}+144}$
$\Longrightarrow -ax\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/9058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 3
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Solve trigonometric equation: $1 = m \; \text{cos}(\alpha) + \text{sin}(\alpha)$ Dealing with a physics Problem I get the following equation to solve for $\alpha$
$1 = m \; \text{cos}(\alpha) + \text{sin}(\alpha)$
Putting this in Mathematica gives the result:
$a==2 \text{ArcTan}\left[\frac{1-m}{1+m}\right]$
However I a... | Added: As explained in the comments, certain trigonometric equations such as the linear equations in $\sin x$ and $\cos x$ can be solved by a resolvent quadratic equation. One method is to write the $\sin x$ and $\cos x$ functions in terms of the same trigonometric function. Since all [direct] trigonometric functions o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/9138",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 7,
"answer_id": 5
} |
Any tricky method to solve this one? The question :
Prove that :$$ \text{ if } y = 2x^2 - 1,\text{ then } \biggl[ \frac{1}{y} + \frac{1}{3y^3} + \frac{1}{5y^5}+ \cdots \biggr]$$ is equal to $$\frac{1}{2} \biggl[ \frac{1}{x^2} + \frac{1}{2x^4} + \frac{1}{3x^6} + \cdots \biggr]$$
Here I have modified the question, in t... | One you simplify to $\frac{1}{2}\log(\frac{y+1}{y-1})$, plug in $y = 2x^2 -1$ and simplify it to get $-\frac{1}{2} \log(1-\frac{1}{x^2})$. Now expand to get the desired answer.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/11319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
How to compute the formula $\sum \limits_{r=1}^d r \cdot 2^r$? Given $$1\cdot 2^1 + 2\cdot 2^2 + 3\cdot 2^3 + 4\cdot 2^4 + \cdots + d \cdot 2^d = \sum_{r=1}^d r \cdot 2^r,$$
how can we infer to the following solution? $$2 (d-1) \cdot 2^d + 2. $$
Thank you
| You can use iterative integration of finite differences. The method works as follows. Assume your problem statement is:
$$\sum^n_{r=1} a_r = ?$$
Assume there is a function $S_n$ such that:
$$\sum^n_{r=1} a_r = S_n - S_1 + c$$
By finite difference we have:
$$a_n = \sum^n_{r=1} a_r - \sum^{n-1}_{r=1} a_r = (S_n - S_1 + c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/11464",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 9,
"answer_id": 2
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How to sum $\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \cdots$ quickly? The Problem:
$$\frac1{1\cdot 2\cdot 3\cdot 4} + \frac4{3\cdot 4\cdot 5\cdot 6} + \frac9{5\cdot 6\cdot 7\cdot 8} + \frac{16}{7 \cdot 8 \cdot 9 \cdot 10} + \cdots$$
Any smarter way to solve thi... | Problems like these are usually amenable to partial fraction decompositions.
From my experience with exam problems like these, once we do the partial fractions (which can be done quite quickly as described below), either we can do some sort of a telescoping sum, or write the resulting as a combination of well known ser... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/13888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
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How to simplify nested cubic radicals $\sqrt[3]{a+b\sqrt c}$ While trying to answer this question, I got stuck showing that
$$\sqrt[3]{26+15\sqrt{3}}=2+\sqrt{3}$$
The identity is easy to show if you already know the $2+\sqrt{3}$ part; just cube the thing. If you don't know this, however, I am unsure how one would proce... | Apply the known denesting formula
$$\sqrt[3]{a+b \sqrt c}=\frac12\sqrt[3]{3bt-a}\left(1+\frac1t \sqrt c\right)$$
where $t$ satisfies $t^3-\frac{3a}bt^2+3c t-\frac{ac}b=0$. Take the example $\sqrt[3]{26+15\sqrt{3}}$ in question and solve
$$t^3-\frac{26}5t^2+9t-\frac{26}5=\frac15(t-2)(5t^2-16t+13)=0$$
which yields $t=2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/16331",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
How to solve $\int\tan^3(x)\,dx$? How to solve $\int\tan^3(x)\,dx$ ? Do I use substitution?
| A simple way out is to write $\tan^3(x) = \frac{\sin^3(x)}{\cos^3(x)}$. Now let $\cos(x) = t$ and rewrite $\sin^2(x) = 1-t^2$.
The integral now becomes $\int \frac{\sin^3(x)}{\cos^3(x)} dx$.
$\cos(x) = t \Rightarrow -\sin(x)dx = dt$ and $\sin^2(x) = 1-t^2$.
Note the numerator of the integral $\sin^3(x) dx$ can be writt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/20074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
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How to find solutions of linear Diophantine ax + by = c? I want to find a set of integer solutions of Diophantine equation: $ax + by = c$, and apparently $\gcd(a,b)|c$. Then by what formula can I use to find $x$ and $y$ ?
I tried to play around with it:
$x = (c - by)/a$, hence $a|(c - by)$.
$a$, $c$ and $b$ are kno... | Look can be deceiving. The integer solution to the equation $ax + by = c$ is anything but easy. Please endure a rather long derivation.
To make it more comprehensible let's first solve the equation for y:
\begin{align*}
ax + by = c\\
by = c - ax\\
y = \frac{c - ax}{b}
\end{align*}
To have an integer solution, $y$ must ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/20717",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "130",
"answer_count": 4,
"answer_id": 1
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How can I prove that $n^7 - n$ is divisible by $42$ for any integer $n$? I can see that this works for any integer $n$, but I can't figure out why this works, or why the number $42$ has this property.
| There are many equivalent ways of proving it.
First observe that $42$ divides a number iff $2,3$ and $7$ divides the number.
(Since $42 = 2 \times 3 \times 7$ and $\gcd(2,3) = \gcd(3,7) = \gcd(2,7) = 1$)
Divisibility by $2$:
Clearly, $2|(n^7-n)$ since $n^7$ and $n$ are of the same parity.
Equivalently you could argue o... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/22121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 8,
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} |
help solving an integral $\int\left(\frac {x-1}{3-x}\right)^\frac{1}{2} dx$ $\displaystyle\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2}\,\rm dx$
I am stuck on this part:
Let $u=\dfrac{x-1}{3-x}~\longrightarrow$ $~~\rm du=\dfrac {2}{(x-3)^2}\,\rm dx,$ which can be represented as
$\rm du=\dfrac{1}{3-x} - \dfrac{1-x}{(3... | First note
$$
du=\frac{2}{(x-3)^2}dx
$$
which implies
$$
\frac{(x-3)^2}{2}du=dx.
$$
Also,
$$
u+1=\frac{x-1}{3-x}+\frac{3-x}{3-x}=\frac{2}{3-x}
$$
which implies
$$
(u+1)^2=\left(\frac{2}{3-x}\right)^2=\frac{4}{(x-3)^2}.
$$
Putting it together you have
$$
\int \left(\frac {x-1}{3-x}\right)^\frac{1}{2} dx = \in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/22310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 0
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How does divisibility test using congruence work? In the book, it said:
Let $n = a_{k}10^{k} + a_{k-1}10^{k-1} + a_{k-2}10^{k-2} + ... + a_110 + a_0$
Then, because $10 \equiv 0 \pmod{2}$ it follows that $10^j \equiv 0 \pmod{2^j}$
What congruence property did they use in this case? Is that:
If $a \equiv b \pmod{k_1}$ a... | You you need to use this property $j$ times, since:
$10 \equiv 0 \mod{2} $ and $10 \equiv 0 \mod{2}$, then $10 \cdot 10 \equiv 10^2 \equiv 0 \mod{2^2}$
You know that $2|10$ so it must be that $2^2 | 10^2$ (factorization). Repeat again:
$10 \equiv 0 \mod{2}$ and $ 10^2 \equiv 0 \mod{2^2}$, then $10 \cdot 10^2 \equiv ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/23780",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Formula for sum of series $1+(1+x)+(1+x)^2+.... +(1+x)^n$ What is the formula to get sum of series
$$1+(1+x)+(1+x)^2+.... +(1+x)^n,$$
where $n$ is Integer and $x$ is Rational
| Simplified after the comment below by Yuval Filmus.
$$S=1+(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots +(1+x)^{n}$$
is the sum of a geometric progression with ratio $1+x$ and $n+1$ terms, the first of which is $1$. The usual way to derive $S$ is to multiply it by the ratio
$$(1+x)S=(1+x)+(1+x)^{2}+(1+x)^{3}+\ldots +(1+x)^{n}+(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/23819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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A congruence in a quadratic number field
If $\displaystyle L = \frac{3+\sqrt{-3}}{2}$,
and if $x\equiv 1\pmod{L}$,
show that $x^3\equiv 1\pmod{L^4}$.
I have already shown that if $x\equiv 1\pmod{L}$,
then $x^3\equiv 1\pmod{L^3}$.
Thanks.
| Let $w = \frac{-1 + \sqrt{-3}}{2}$. Your question is to show that if $x \equiv 1 \pmod{2+w}$, then $x^3 \equiv 1 \pmod{(2+w)^4}$.
By direct expansion, we see that $(2+w)^4 = 9w$. So it suffices to show $x^3 \equiv 1 \pmod{9}$. Now check the following,
*
*$x \equiv 1 \pmod{2+w}$ is equivalent to $x = a+bw$, such that... | {
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"timestamp": "2023-03-29T00:00:00",
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Find $\frac{\mathrm d^{100}}{\mathrm d x^{100}}\frac{x^2+1}{x^3-x}=$? $$f(x)=\frac{x^2+1}{x^3-x}$$
$$f^{(100)}(x)=?$$
I tried differnetiating once and twice, but did not see any pattern emerging and can't guess what the 100th derivative should be.
EDIT
so decomposing this as $$f(x)=-\frac{1}{x}+\frac{1}{x+1}+\frac{1}... | $$\frac{x^2+1}{x^3-x} = -\frac{1}{x} + \frac{1}{x+1} + \frac{1}{x-1} = -x^{-1} + (x+1)^{-1} + (x-1)^{-1}$$
$$\frac{d^n(y^{-1})}{dy^n} = \frac{(-1)^n n!}{y^{n+1}}$$
Hence, the $n^{th}$ derivative is
$$\frac{(-1)^{n+1} n!}{x^{n+1}} + \frac{(-1)^n n!}{(x-1)^{n+1}} + \frac{(-1)^n n!}{(x+1)^{n+1}} = (-1)^n n! \times \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/27652",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Nice proofs of $\zeta(4) = \frac{\pi^4}{90}$? I know some nice ways to prove that $\zeta(2) = \sum_{n=1}^{\infty} \frac{1}{n^2} = \pi^2/6$. For example, see Robin Chapman's list or the answers to the question "Different methods to compute $\sum_{n=1}^{\infty} \frac{1}{n^2}$?"
Are there any nice ways to prove that $$\z... | I would like to present you a method based on Daners' derivation of $\zeta(2)$ (which I summarised on MathStackexchange in here couple years ago).
We start defining (for all $n \in \mathbb{N}_0$)
\begin{align}
A_n & =\int_0^{\pi/2}\cos^{2n}x\;\mathrm{d}x,\\
B_n& =\int_0^{\pi/2}x^2\cos^{2n}x\;\mathrm{d}x,\\
C_n& =\int_0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/28329",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "119",
"answer_count": 15,
"answer_id": 13
} |
Inequality understanding My textbook says that:
$$
\frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{2}\right)^2\cdots\left(1+\frac{1}{n}\right)^n<e^n
$$
But I do not understand this. Can you please enlighten me?
Edit: How do you show that
$$
\frac{(n+1)^n}{n!}=\left(1+\frac{1}{1}\right)\left(1+\frac{1}{... | For the first part
note that each term is
$$\begin{align}
\left(\frac{1+k}{k}\right)^k =& \frac{(1+k)^k}{k^k}\\
=&\frac{1}{k}\frac{(1+k)^k}{k^{k-1}}\\
=&\frac{1}{k}\frac{(1+k)^k}{(1+(k-1))^{k-1}}
\end{align}$$
so every denominator cancels nicely with the previous numerator:
$$ \begin{align}\frac{1}{n!}\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/29214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Uniqueness of the infinite expansion in base $b$ If $b>1$ is an integer, is well know that the numbers $x\in (0,1]$, can be written as
$$x = \sum_{k=1}^{\infty} \frac{a_k}{b^k}$$
for some integers $a_k \in \{0,1,\ldots ,b-1\} $. When $x=\frac{1}{b^n}$, for some $n\in \mathbb{N}$, there is two representations. One of th... | If I recall correctly you can prove that the partial sums of the expansion satisfy the following inequality
$$0 \leq x - \sum_{k = 1}^{n} \frac{a_k}{b^k} < \frac{1}{b^n}$$
and from this inequality you can just prove by induction that if you have two expansions for $x$ then they are actually the same by proving that all... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
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How to reduce congruence power modulo prime? If I have a congruence equation, says
$$x^{15} - x^{10} + 4x - 3 \equiv 0 \pmod{7}$$
Then can I use Fermat's little theorem like this:
$$(x^{6})^2 \cdot x^3 - x^6 \cdot x^4 + 4x - 3 \equiv 0 \pmod{7}$$
$$ x^3 - x^4 + 4x - 3 \equiv 0 \pmod{7}$$
Update
Should it be
$$x^{14}x... | Not quite. Look for example at the congruence $x^6 \equiv 0 \pmod{7}$. If one assumes that $x^6 \equiv 1$, things go bad. In this case it is easy to spot that there is a problem, but perhaps in a more complicated setting one might miss it.
I would advise using the fact that $x^7 \equiv x \pmod{7}$, basically a varian... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/33257",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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For what $a$ and $b$ is $9x^4-12x^3+28x^2+ax+b$ a perfect square? If $9x^4-12x^3+28x^2+ax+b$ is a perfect square, find the value of $a$ and $b$.
This is one of my past year examination's questions, some help on it?
(The answer for this problem is $a=-16$, $b=16$.)
| Even though it doesn't affect the final outcome in this problem, I think it's better long-term practice not to assume that the leading term is positive. This emphasizes that there are two square roots to a square polynomial, differing by sign, just like with integers.
Suppose $9x^4-12x^3+28x^3+ax+b=g(x)^2$ for some po... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/40021",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
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What are a , b and c? $$y = ax^2 + bx + c$$
which is tangent at the origin with the line $y=x$, It is also tangential with the line $y=2x + 3$. Determine the function! Draw a figure!
My main question is this solvable? I am doubtful?
| This is the Graph of $f(x)= -\frac{1}{5}x^{2}+x$ which I graphed using KmPlot. The figure should give you an intuitive idea of how to go about solving.
*
*The Green line is $y=2x+3$.
*The Blue line is $y=x$.
If the line $y=2x+3$ and the parabola $y=ax^{2}+bx+c$ are going to be tangent at a given point then their sl... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Efficient way to find $a$ in $c = 6a\mod n$ Given $c$ and $n$ in $c = 6a\mod n$, how can I find the lowest positive integer value for $a$?
I could find it iteratively by rewriting as $\displaystyle a = \frac{c + xn}{6}$ and increasing $x$ until $a$ is an integer, but I would prefer a more efficient solution. Is there ... | First, general comments.
The congruence
$$Ax\equiv c \pmod{n}$$
has a solution $x$ if and only if $\gcd(A,n)|c$. If it does, the standard method is to use the extended Euclidean algorithm to find integers $r$ and $s$ such that
$$d = Ar + ns$$
(where $d=\gcd(A,n)$); then multiply by $\frac{c}{d}$ (an integer, since $d|c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/44822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
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} |
What is the length of a sine wave from $0$ to $2\pi$? What is the length of a sine wave from $0$ to $2\pi$? Physically I would plot $$y=\sin(x),\quad 0\le x\le {2\pi}$$ and measure line length.
I think part of the answer is to integrate this:
$$
\int_0^{2\pi} \sqrt{ 1 + (\sin(x))^2} \, {\rm d}x
$$
Any ideas?
| \begin{align}
\int_0^{2\pi}\sqrt{1+\cos^2(x)} dx
&= 4 \int_0^{\pi/2}\sqrt{1+\cos^2(x)}dx \\
&= 4 \int_0^{\pi/2}\sqrt{1+\dfrac{1+\cos(2x)}2 }dx\\
&= 4 \sqrt{\dfrac{3}2} \int_0^{\pi/2} \sqrt{1+\dfrac{\cos(2x)}3} dx \\
&= 2\sqrt6 \int_0^{\pi/2} \sum_{n=0}^\infty a_n (\dfrac{\cos(2x)}3)^n dx \tag1
\end{align}
where $$\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/45089",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "38",
"answer_count": 9,
"answer_id": 3
} |
Bounded Sequences I came across the following problems during the course of my self-study of real analysis:
Show that the sequence $(x_n)$ defined by $x_n = 1+ \frac{1}{2} + \frac{1}{3} + \dots + \frac{1}{n}$ is unbounded.
I know a sequence $(x_n)$ is bounded if there exists a positive number $K$ such that $|x_n| \l... | *
*The simplest way to show that a sequence is unbounded is to show that for any $K\gt 0$ you can find $n$ (which may depend on $K$) such that $x_n\geq K$.
The simplest proof I know for this particular sequence is due to one of the Bernoulli brothers Oresme. I'll get you started with the relevant observations and you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/46978",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Sum of First $n$ Squares Equals $\frac{n(n+1)(2n+1)}{6}$ I am just starting into calculus and I have a question about the following statement I encountered while learning about definite integrals:
$$\sum_{k=1}^n k^2 = \frac{n(n+1)(2n+1)}{6}$$
I really have no idea why this statement is true. Can someone please explain ... | Another method by using telescoping sum :-
We know $(a+b)^3-a^3-b^3=3ab(a+b)$ , take
$a=k-1 , b=2$ , then $a+b=k+1$ and $(k+1)^3-(k-1)^3-2^3=6(k-1)(k+1)=6k^2-6$ ,
hence $(k+1)^3-(k-1)^3-8+6=(k+1)^3-k^3+k^3-(k-1)^3-2=6k^2$ , taking sum over $k$
from $1$ to $n$ we get , $\sum_{k=1}^n [(k+1)^3-k^3] + \sum_{k=1}^n [k^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/48080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "145",
"answer_count": 32,
"answer_id": 30
} |
Inequality: $(x + y + z)^3 \geq 27 xyz$ Edit: $a,b,c$ and $x,y,z$ are positive, real numbers.
Since $(a-b)^2 \geq 0~$, $a^2 + b^2 - 2ab\geq0~$ and $a^2 + b^2 \geq 2ab~$. Similarly, $a^2 + c^2 \geq 2ac~$ and $b^2 + c^2 \geq 2bc~$.
Adding these inequalities together, $2(a^2 +b^2 + c^2) \geq 2(ab + ac +bc)~$ and accord... | Since the inequality is homogeneous we may WLOG assume that $xyz=1$ and we have to prove the inequality $(x+y+z)^3 \geq 27$ or $x+y+z \geq 3.$ But the last inequality we obtain immediately from $AM \geq GM$ under the assumption $xyz=1.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/48621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 8,
"answer_id": 6
} |
Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$ Let $a,b,c$ be positive, not equal. Prove that $3(a^3+b^3+c^3) > (a+b+c)(a^2+b^2+c^2)$.
I know the proof by subtracting LHS by RHS and then doing some arrangement.
But isn't there any inequality which can be used in $(1+1+1)(a^3+b^3+c^3) >(a+b+c)(a^2+b^2+c^2)$, or an e... | The result is also a consequence of the Chebyshev (sum) Inequality.
This is a quite useful result, not the least for contest problems! It can be proved by using the Rearrangement Inequality, which is also useful to know. For completeness, we state the full result, though only half of it is needed here.
Theorem: (Cheby... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/49211",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 5,
"answer_id": 2
} |
finding remainder by dividing a sum Suppose I am dividing 4^30-2^50 by 5.
I do understand that 4^30 will get converted to 2^60
I could also find the series of remainders for 2^x. It comes out to be 2 4 3 1
By that, I could find (4^30) % 5 = 1 and (2^50) % 5 = 4. But how do I find the combined mod?
i.e. [(4^30)-(2^50)]... | If $a\equiv b\pmod m$ and $c\equiv d\pmod m$, then $a\pm c\equiv b\pm d\pmod m$.
Since $4^{2k+1}\equiv 4\pmod 5$ and $4^{2k}\equiv 1\pmod 5$, $4^{30}=4^{2\cdot 15}\equiv 1\pmod 5$.
Also, $2^{4k}\equiv 1\pmod 5$, $2^{4k+1}\equiv 2\pmod 5$, $2^{4k+2}\equiv 4\pmod 5$ and $2^{4k+3}\equiv 3\pmod 5$. Hence $2^{50}=2^{4\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/49413",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Trigonometric equality: $\frac{1 + \sin A - \cos A}{1 + \sin A + \cos A} = \tan \frac{A}{2}$ Can you guys give me a hint on how to proceed with proving this trigonometric equality? I have a feeling I need to use the half angle identity for $\tan \frac{\theta}{2}$. The stuff I have tried so far(multiplying numerator and... | There is a theorem$^1$ that is worth knowing (I used it in this answer to this question) that states:
All direct trigonometric functions of $A$ ($2\alpha$) can be expressed rationally in terms of the tangent of $\frac{A}{2}$ ($\alpha$).
Combining $\sin A=2\sin \frac{A}{2}\cdot \cos \frac{A}{2}$ and $\cos ^{2}
\frac{A... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/50093",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "13",
"answer_count": 8,
"answer_id": 1
} |
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.