Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Solving $\cos x + \cos 2x - \cos 3x = 1$ with the substitution $z = \cos x + i \sin x$ I need to solve
$$\cos x+\cos 2x-\cos 3x=1$$
using the substitution$$z= \cos x + i \sin x $$
I fiddled around with the first equation using the double angle formula and addition formula to get
$$\cos^2 x+4 \sin^2x\cos x-\sin^2 x=1$... | Another idea ,maybe
$$cos(x)+cos(2x)-cos(3x)=1\\
cos(x)-cos(3x)=1-\cos(2x)\\\cos
(2x-x)-\cos(2x+x)=1-\cos(2x)\\\cos(2x)\cos(x)+\sin(2x)\sin(x)-(\cos(2x)\cos(x)-\sin(2x)\sin(x))=1-\cos(2x)\\2\sin(2x)\sin(x)=1-\cos(2x)\\
2\sin(2x)\sin(x)=2\sin^2(\frac{2x}{2})\\
2\sin(2x)\sin(x)=2\sin^2(x)\\
\sin(2x)\sin(x)-\sin^2(x)=0\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586848",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 4
} |
Given $3\cos x - 4 \sin x = 2$, find $3 \sin x + 4 \cos x$ without first solving for $x$ If $$3\cos{x}-4\sin{x}=2$$
find $$3\sin{x} +4\cos{x} $$
I have solved the equation for $x$, then calculated the required value, but I think there is a direct solution without solving the equation.
| $$3\sin x+4\cos x=A$$
$$3\cos x-4\sin x=2$$
$$9\sin^2x+24\sin x\cos x+16\cos^2x=A^2$$
$$9\cos^2x-24\sin x\cos x+16\sin^2x=4$$
add both equations
$$9+16=A^2+4$$
$$A^2=21$$
$$A=\pm \sqrt{21}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588061",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\int (2x+3) \sqrt {3x+1} dx$ Evaluate $\int (2x+3) \sqrt {3x+1} dx$
My Attempt:
Let $u=\sqrt {3x+1}$
$$\dfrac {du}{dx}= \dfrac {d(3x+1)^\dfrac {1}{2}}{dx}$$
$$\dfrac {du}{dx}=\dfrac {3}{2\sqrt {3x+1}}$$
$$du=\dfrac {3}{2\sqrt {3x+1}} dx$$
| Would you not just introduce substitution $u=\sqrt{3x+1}$, $x=\frac{u^2-1}{3}$, $dx=\frac23 udu$ so:
$$\int (2x+3) \sqrt {3x+1} dx=\int \left(\frac23 (u^2-1)+3\right)u\cdot \frac23 udu$$
which is an integral of a polynomial in $u$ - easy to solve.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2588847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Find $\sum_{k=1}^\infty\frac{1} {k(k+1)(k+2)(k+3)}$ I have to solve this series transforming it into a telescopic sequence
$$\sum_{k=1}^\infty\frac{1}
{k(k+1)(k+2)(k+3)}$$
But I'm lost in the calculation!
| You can also go the hard way: write
$$
\frac{1}{x(x+1)(x+2)(x+3)}=
\frac{A}{x}+\frac{B}{x+1}+\frac{C}{x+2}+\frac{D}{x+3}
$$
so
$$
A(x+1)(x+2)(x+3)+Bx(x+2)(x+3)+Cx(x+1)(x+3)+Dx(x+1)(x+2)=1
$$
Now,
*
*for $x=0$: $6A=1$
*for $x=-1$: $-2B=1$
*for $x=-2$: $2C=1$
*for $x=-3$: $-6D=1$
Thus
$$
\frac{1}{k(k+1)(k+2)(k+3)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589176",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
using $x^9 - 1$ as a product of linear and quadratic factors with real coefficients to solve trig So here's where I'm at I know i can find the ninth roots of unity by using the nth roots of unity formula $$ \zeta_n = e^{\frac{i2\pi k}{n}} ,$$ when $k = 1,2,3,...,n-1$
so the roots of unity where $e^{i\frac{2\pi}{9}},e... | You should have the four pairs of complex roots you have accounted for plus a factor $x-1$ as $x=1$ is a root of your polynomial. That gives you the linear term.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Is this the right solution for an unsolvable system? I have to find the values of $k$ such that this system $\left[
\begin{array}{ccc|cc}
1&0&-3&-3\\
2&k&-1&-2\\
1&2&k&1
\end{array}
\right]$ is:
a) unsolvable b) solvable with infinite solutions c) one solution.
After some steps we can see in this m... | from the first equation we get
$$x_1=-3+3x_3$$
plugging this in the second equation and solving for $x_3$ we have
$$x_3=\frac{4}{5}-\frac{k}{5}x_2$$
and for the third equation we get
$$x_2\left(2-\frac{k}{3}(3+k)\right)=4-\frac{4}{5}(3+k)$$
can you finish?
from the last equation we get
$$x_2(6-3k-k^2)=\frac {3}{5}(8-4k... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Behaviour of $\sum\limits_{n=1}^\infty \frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n$ for $|x|=2$ I want to determine the radius of convergence for the power series
$$\sum_{n=1}^\infty \frac{(\sqrt{n^2+n} - \sqrt{n^2+1})^n}{n^2} x^n$$
and determine what happens at the boundaries. I determined the ratio of c... | Note that
$$\sqrt{n^2+n}=n\left(1+\frac1n\right)^{\frac12}\sim n+\frac12 \quad \sqrt{n^2+1}=n\left(1+\frac1{n^2}\right)^{\frac12}\sim n$$
$$\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n\sim\frac{1}{2^n}$$
thus
$$\frac1{n^2}\left(\sqrt{n^2+n} - \sqrt{n^2+1}\right)^n x^n \sim \frac{x^n}{2^nn^2}$$
which diverges when $|x|>2$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598350",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find $\lim\limits_{n\to+\infty}\sum\limits_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}$
Compute
$$\lim_{n\to+\infty}\sum_{k=1}^{n}\frac{k^{3}+6k^{2}+11k+5}{\left(k+3\right)!}.$$
My Approach
Since $k^{3}+6k^{2}+11k+5= \left(k+1\right)\left(k+2\right)\left(k+3\right)-1$
$$\lim_{n\rightarrow\infty}\sum_{k=1}... | $\displaystyle
\begin{align}
\sum_{k=1}^\infty\left[\frac{1}{k!} - \frac{1}{(k+3)!}\right]
&=
\left\{\begin{array}{c}
\dfrac{1}{1!} &+\dfrac{1}{2!} &+\dfrac{1}{3!} &+\dfrac{1}{4!} &+\dfrac{1}{5!}
&+\dfrac{1}{6!} &+\dfrac{1}{7!} &+\dfrac{1}{8!} &+\dfrac{1}{9!} &+\cdots \\
& & &-\dfrac{1}{4!} &-\dfrac{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Monotonicity of $\frac{n}{\sqrt[n]{(n!)}}$ It is known that when $n\rightarrow\infty$ the sequence $$\frac{n}{\sqrt[n]{(n!)}}$$ has limit $e$ but I don't know how to prove it's monotonicity. After a short calculus using WolframAlpha I found that this sequence is actually increasing. I tried to compare $2$ consecutive m... | We have that $\frac{n}{(n!)^{\frac{1}{n}}}\leq\frac{n+1}{((n+1)!)^{\frac{1}{n+1}}}\Leftrightarrow\frac{n^{n(n+1)}}{(n!)^{n+1}}\leq\frac{(n+1)^{n(n+1)}}{((n+1)!)^{n}}\Leftrightarrow$ simplifying, $\frac{n^{n^2+n}}{n!}\leq(n+1)^{n^2}\Leftrightarrow\frac{n^{n^2+n}}{(n!)(n+1)^{n^2}}\leq1$;let us proceed by induction. For n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 6,
"answer_id": 2
} |
If $k$ is composite, then $2^{k} -1$ is composite for $k \geq 2$ I am trying to prove that if $k$ is composite, then $2^{k} -1$ is composite, $k \geq 2$.
I have already established the equality $$n^k - 1 = (n-1)(n^{k-1} + ... + 1) \tag{*}$$ If I let $n=2$ and $k=ab$, I don't really get anything useful. The hint says t... | We have in general
$$ 2^n-1 = \underset {n \text { times}} { \underbrace {1 + 2^1 + 2^2 + ... + 2^{n-1}}}$$
If $n$ is composite, say $a \cdot b$ then
$$\begin{array}{lll} 2^n-1 &= \underset { a \cdot b \text { times}} { \underbrace {1 + 2^1 + 2^2 + ... + 2^{n-1}}} \\ \phantom{X} \\
&=\underset {a \text{ times}} { \unde... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Beginner troubleshooting an eigenvector calculation I am having some difficulty identifying the error in my eigenvector calculation. I am trying to calculate the final eigenvector for $\lambda_3 = 1$ and am expecting the result $ X_3 = \left(\begin{smallmatrix}-2\\17\\7\end{smallmatrix}\right)$
To begin with, I set up ... | Uh, that's just bad algebra- your fractions are upside down! You have $x_1= -\frac{2}{7}x_3$ and $x1= -\frac{2}{17}x_2$. Setting $x_1= 1$ gives $-\frac{2}{7}x_3= 1$ so $x_3= -\frac{7}{2}$ and $-\frac{2}{17}x_2= 1$ so $x_2= -\frac{17}{2}$, not what you have. that gives $X_3= \begin{pmatrix} 1 \\ -\frac{17}{2} \\ -\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2601685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Proof of (complicated?) summation equality This is a simplified case of something I'm trying to prove.
Suppose that $N,h$ are even. I want to show that
$$ \sum_{k=1}^{(N-h)/2} \frac{2^{2k}}{kN^{\underline{2k}}}\left(\frac{N}{2}\right)^{\underline{k}}\left(\frac{N-h}{2}\right)^{\underline{k}} = \sum_{j=1}^{(N-h)/2} \fr... | Consider the sum
$$
S^n_m=\sum_{k=1}^m4^k\frac{\binom{n}{k}\binom{m}{k}}{\binom{2n}{2k}\binom{2k}{k}};\quad n\ge m\ge1.\tag{1}
$$
We are going to prove:
$$
S^n_m=\frac{2m}{2n-2m+1}.\tag{2}
$$
It is easy to check that for $m=1$ and arbitrary $n\ge1$ the expression (2) is valid:
$$
S^{n}_1=4^1\frac{\binom{n}{1}\binom{1}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2608686",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Find the domain of convergence of the series
Find the domain of convergence of the series :
$(1)$ $$\sum\limits_{n=1}^{\infty}\frac{1\cdot3\cdot5\cdots(2n-1)}{n!}\Big(\frac{1-z}{z}\Big)^n$$
$(2)$ $$\frac{1}{2}z+\frac{1\cdot3}{2\cdot5}z^2+\frac{1\cdot3\cdot5}{2\cdot5\cdot8}z^3 +\dots$$
I was reading Rudin and I know t... | Let us first write
$$\frac{1\cdot3\cdot5\cdot\ldots\cdot(2n-1)}{n!}=\frac{1\cdot2\cdot3\cdot4\cdot\ldots\cdot(2n)}{2^nn!n!}=\frac{(2n)!}{2^n(n!)^2}$$
and now apply the quotient test:
$$\left|\frac{a_{n+1}}{a_n}\right|=\left|\frac{(2n+2)!}{2^{n+1}\left((n+1)!\right)^2}\frac{(1-z)^{n+1}}{z^{n+1}}\cdot\frac{2^n(n!)^2}{(2n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2608796",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$ One of my friend had just given me an inequality to solve which is stated below.
Consider the three positive reals $a, b, c$ then prove that
$$a+b+c\le \frac {a^3}{bc} + \frac {b^3}{ac} + \frac {c^3}{ab}$$
I have solved this inequality very e... | Using strictly familiar Angel form of the CS inequality and the traditional CS inequality itself we have: $\displaystyle \sum_{\text{cyclic}} \dfrac{a^3}{bc}= \displaystyle \sum\dfrac{(a^2)^2}{abc}\ge\dfrac{(a^2+b^2+c^2)^2}{abc+abc+abc}=\dfrac{(a^2+b^2+c^2)^2}{3abc}\ge\dfrac{(ab+bc+ca)^2}{3abc}= \dfrac{(ab)^2+(bc)^2+(c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610126",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Show that if $x+y+z=x^2+y^2+z^2=x^3+y^3+z^3=1$ then $xyz=0$ For any real numbers $x,y,z$ such that $x+y+z = x^2+y^2+z^2 = x^3+y^3+z^3 =1\\$ show that $x \cdot y \cdot z=0$.
I think that because $x \cdot y \cdot z = 0$ at least one of the three number should be equal to zero, but I'm stuck into relating the other things... | Let $q(t)=(t-x)(t-y)(t-z)$ and $p_n=x^n+y^n+z^n$. Power sums give a base of the ring of symmetric polynomials in the variables $x,y,z$, and by Newton's identities
$$ e_1 = x+y+z = p_1=1, $$
$$ 2e_2 = 2xy+2yz+2xz = p_1^2-p_2 = 0, $$
$$3e_3 = 3xyz = e_2 p_1 - e_1 p_2 + p_3 = 0, $$
so $q(t) = t^3-t^2$ and $x,y,z$ are give... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612380",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 8,
"answer_id": 1
} |
How can I calculate $\lim\limits_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$?
How can I calculate this limit?
$$\lim_{x\rightarrow 0} \frac{(1+x)^{\frac{1}{x}}-e}{x}$$
I thought about L'Hospital because case of $\frac{0}{0}$, but I don't know how to contiune from this point..
| let $n = \frac 1x$
$\lim_\limits{n\to \infty} n((1+\frac 1n)^n - e)$
Binomial expansion:
$n(1 + 1 + \frac 12 (1-\frac 1n) + \frac 1{3!} (1-\frac 1n)(1-\frac 2{n})+\cdots +\frac {1}{n!} (1-\frac 1n)\cdots(1-\frac {n-1}{n}) - e)$
$e = 1 + 1 + \frac 1{2!} + \frac {1}{3!}\cdots$
$n( (-\frac 12 -\frac 1{3!} {3\choose 2} - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2617346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 1
} |
Evaluate $\int_{1}^{2} \ \frac{\arctan (x)}{\arctan \left (\frac{1}{x^2-3x+3} \right )} dx$ In this exercise, we are given two integrals $I$ and $J$:
$$I = \int_{0}^{\pi/2}\frac{\sin^nx}{\sin^nx+\cos^nx} dx, \quad n\in \mathbb{N^*},$$
and
$$J = \int_{1}^{2} \frac{\arctan(x)}{\arctan \left (\frac{1}{x^2-3x+3} \right )} ... | As you suspected, there may be one more typo in the textbook.
As I commented, I was wondering if approximation would be require. So, for the fun of it, I tried to.
$$I=\int_1^2\frac{\tan ^{-1}(x)}{\tan ^{-1}(2-x)-\tan ^{-1}(1-x)}\,dx=\int_0^1 \frac{\tan ^{-1}(y+1)}{\tan ^{-1}(y)-\tan ^{-1}(y-1)}\,dy$$
Using Taylor expa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Proving that the sequence $\{\frac{3n+5}{2n+6}\}$ is Cauchy. I'm not quite sure how to tackle these kinds of questions in general, but I tried something that I thought could be right. Hoping to be steered in the right direction here!
Let $\{\frac{3n+5}{2n+6}\}$ be a sequence of real numbers. Prove that this sequence i... | Let $ \epsilon >0$ be an arbitrary small number.
Choose a positive integer $n_0$ such that $$\frac {2}{n_0 +3} <\frac {\epsilon }{2}$$
Note that if $m \ge n_0$, and $n\ge n_0$ are integers, then $$ |f(n)-f(m)| = |\frac{3n+5}{2n+6}-\frac{3m+5}{2m+6}|= $$ $$|\frac {2(m-n)}{(n+3)(m+3)} |\le $$ $$\frac {2(m+n)}{(n+3)(m+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2620493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 4
} |
Sum of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $ What is the limit of series $\frac{1}{2 \cdot 3 \cdot 4} + \frac{1}{4 \cdot 5 \cdot 6} + \frac{1}{6 \cdot 7 \cdot 8} + \cdots $?
The $n$th summand is $\frac{1}{(2n)(2n + 1)(2n+2)} = \frac{1}{4} \frac{1}{n(2... | $$
\begin{align}
\sum_{k=1}^\infty\frac1{2k(2k+1)(2k+2)}
&=\frac12\sum_{k=1}^\infty\left(\frac1{2k(2k+1)}-\frac1{(2k+1)(2k+2)}\right)\\
&=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac1{2k}-\frac2{2k+1}+\frac1{2k+2}\right)\\
&=\lim_{n\to\infty}\frac12\sum_{k=1}^n\left(\frac2{2k}-\frac2{2k+1}\right)-\lim_{n\to\infty}\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2621718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Compute $\int\frac{x^2}{\tan{x}-x}dx$ for $x \in (0, \frac{\pi}{2})$
Compute the indefinite integral $\displaystyle{\int\frac{x^2}{\tan{x}-x}dx, x \in (0, \frac{\pi}{2})}$.
I have tried substituting $\tan{x} = t$ or write $\tan{x} = \displaystyle{\frac{\sin{x}}{\cos{x}}}$ and then try to solve it from there, but I di... | $$\int\frac{x^2}{\tan{x}-x} =\int\frac{x^2 \cos x +x \sin x - x \sin x}{\sin{x}-x \cos x}=\int \left(\frac{x \sin x}{\sin{x}-x \cos x} - x \right) $$
Now notice that $$\sin x - x \cos x=t \implies x\sin x = dt$$
Hence $$\int \left(\frac{x \sin x}{\sin{x}-x \cos x} - x \right) = \ln |(\sin{x}-x \cos x) | - \frac{x^2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find lengths of tangents drawn from $(3,-5)$ to the Ellipse Find lengths of tangents drawn from $A(3,-5)$ to the Ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$
My try: I assumed the point of tangency of ellipse as $P(5\cos a, 4 \sin a)$
Now Equation of tangent at $P$ is given by
$$\frac{x \cos a}{5}+\frac{ y \sin a}{4}=1$$... | let $$y=mx+n$$ then the equation of the line through $P(3;-5)$ is given by
$$y=m(x-3)-5$$ plug this in the equation of the Ellipse and solve this equation for $x$, since it should be a tangentline, the discriminante must be Zero
you will get $$16m^2-30m-9=0$$ solve this equation for $m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623243",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Why is $\int_{-\infty}^{\infty}\left\lvert\log\frac{a^2+x^2}{b^2+x^2}\right\rvert dx=2\pi(b-a)$?
If $a,b\in\mathbb R_{>0}$ and $ b>a$, then why is $\int_{-\infty}^{\infty}\left\lvert\log\frac{a^2+x^2}{b^2+x^2}\right\rvert dx=2\pi(b-a)$ ?
Usual change of variables doesnt bring anything, I think. Is there a special fun... |
METHODOLOGY $1$:
Note that we can write
$$\begin{align}
\int_{-\infty}^\infty\log\left(\frac{b^2+x^2}{a^2+x^2}\right)\,dx&=2\int_0^\infty \int_{a^2}^{b^2} \frac{1}{y+x^2}\,dy\,dx\\\\
&\overbrace{=}^{\text{Fubini}}2\int_{a^2}^{b^2}\int_0^\infty \frac{1}{y+x^2}\,dx\,dy\\\\
&=2\int_{a^2}^{b^2}\frac{\pi/2}{\sqrt y}\,dy\\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625357",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
how to get the limit of $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right)$ How to get the limit $\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) = \frac{1}{2}$ ?
$\begin{align}
\lim_{n \to \infty}\left( \sqrt{n+\sqrt{n}}-\sqrt{n} \right) &=
\lim_{n \to \infty}\left( \frac{(\sqrt{n+\sqrt{n}}-\sq... | $$\begin{align}
\sqrt{n+\sqrt{n}}-\sqrt{n} &=
\frac{(\color{red}{\sqrt{n+\sqrt{n}}}-\color{blue}{\sqrt{n}})(\color{red}{\sqrt{n+\sqrt{n}}}+\color{blue}{\sqrt{n}})}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{(\color{red}{\sqrt{n+\sqrt{n}}})^2-(\color{blue}{\sqrt{n}})^2}{\sqrt{n+\sqrt{n}}+\sqrt{n}} \\
&= \frac{(n+\sqrt{n})-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626347",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Why is this proof false? $\sqrt 2 + \sqrt 6 < \sqrt 15$ What is wrong with the given proof?
$$
\sqrt 2 + \sqrt 6 < \sqrt 15 \\
(\sqrt 2 + \sqrt 6)^2 < 15 \\
2 + 6 + 2\sqrt2\sqrt6 < 15 \\
2\sqrt2\sqrt6 < 7\\
2^2 \cdot 2 \cdot 6 < 49 \\
48 < 49
$$
| As it stands, the proof is flawed, but not for any error of reasoning (really), but because it is argued in a slightly nonsensical manner. Typically, when you write a sequence statements, you want to connect them to each other. For example, "$A$ implies $B$" or "$A$ if and only if $B$". If no such connecting verbs a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2630542",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
how to maximize $f(x,y)=\frac{x+y-2}{xy}$? How to maximize $f(x,y)=\frac{x+y-2}{xy}$ where $x,y \in \{1,2,\ldots,n\}$?
It seems that maximum will occur when $(x,y)=(1,n)$ or $(n,1).$
| Note that $f(1,1)=0$ and $f(1,2)=f(2,1)=\frac {1}{2}$
Otherwise,
$$\begin{align}
f(x,y)&=\frac{x+y-2}{xy}\\
&=\frac {1}{y} +\frac {1}{x} - \frac {2}{xy}\\
&=\frac {1}{y} \left(1-\frac {1}{x}\right) +\frac {1}{x}\left(1-\frac {1}{y}\right)\\
&\le\frac {1}{2} \left(1-\frac {1}{x}\right) +\frac {1}{2}\left(1-\frac {1}{y}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2631119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Solving $\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)$ Solve
$$
\cos\big(\tan^{-1}x\big)=\sin\big(\cot^{-1}\frac{3}{4}\big)
$$
My Attempt:
From the domain consideration,
$$
\boxed{0\leq\tan^{-1}\frac{3}{4},\tan^{-1}x\leq\frac{\pi}{2}}
$$
$$
\cos\big(\tan^{-1}x\big)=\cos\big(\frac{\pi}{2}-\cot^{-1}\frac... | Yes because $\cos$ is an even function.
It follows also from your way:
$$\arctan{x}=\pm\arctan\frac{3}{4},$$ which gives $x=\frac{3}{4}$ or $x=-\frac{3}{4}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
If $|a|\ge 2$,$|b|\ge 2$, $P(x)=x^4-(a+b)x^3+(ab+2) x^2-(a+b)x+1$, prove that all roots of $P(x)$ are real
Given: $a,b\in \mathbb R$, $|a|\ge 2$,$|b|\ge 2$, $$P(x)=x^4-(a+b)x^3+(ab+2) x^2-(a+b)x+1,$$
Prove or disprove: all roots of $P(x)$ are real.
From a math contest. The polynomial is reciprocal, but I can't see h... | Given the helpful hints and suggestions from David Quinn and Alex Francisco, I will try a full solution to my question, for sake of completeness.
First, by dividing $P(x)=x^4-(a+b)x^3+(ab+2)x^2-(a+b)x+1=0$ by $x^2$ we get
$$x^2-(a+b)x+(ab+2)-(a+b)\frac{1}{x}+\frac{1}{x^2}=0$$
$$\Leftrightarrow x^2+\frac{1}{x^2}-(a+b)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632617",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Reduction formula for $\int\frac{dx}{(x^2+a^2)^n}$ How can I use integration by parts to write $$\int\frac{dx}{(x^2+a^2)^n} = \frac{x}{2a^2(n-1)(x^2+a^2)^{n-1}}+\frac{2n-3}{2a^2(n-1)}\int\frac{dx}{(x^2+a^2)^{n-1}}?$$
I would try $$u=\frac{1}{(x^2+a^2)^n},du=\frac{-2nx}{(x^2+a^2)^{n+1}}dx; dv=dx,v=x.$$ Integration by pa... | Call your integral $I_n$ so that the result of your integration by parts is $$I_n=\frac{x}{(x^2+a^2)^n}+2nI_n-2na^2I_{n+1}$$
Now rearrange so that $I_{n+1}$ is the subject of the formula.
Then replace all $n$ with $n-1$ and you have the formula you require.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633107",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the nth term of the given series The series is $$1^3 + (2^3 +3^3)+(4^3+5^3+6^3)+...$$
i.e., $1^3$, $(2^3 +3^3)$ , $(4^3+5^3+6^3)$ , $(7^3+8^3+9^3+10^3)$ , and so on are the $1^{st}$, $2^{nd}$, $3^{rd}$ and so on terms respectively. ($(4^3+5^3+6^3)$ is the third term). So we need to find the $n^{th}$ term.
So, the ... | Consider the sequence of consecutive integers beginning with one more than the $(n-1)$th triangular number and ending with $(n-1)$ more than the first term. This sequence has $n$ terms. If you take the terms of this sequence, cube them, and add them together, you get the $n$th term of the sequence that you want, which ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Limit using Taylor series expansion
Can someone help me with this limit. I know I have to expand in Taylor series
$\sqrt{1+x}=1+\frac{1}{2}x-\frac{1}{8}x^2+\frac{1}{16}x^3+o(x^3)$
$a^x=e^{x\ln a}=1+x\ln a+\frac{1}{2!}x^2(\ln a)^2+\frac{1}{3!}x^3(\ln a)^3+o(x^3)$.
But how to proceed next?
| Note that from Taylor's expansion
$$\frac{ a^{\sqrt{x+1}}-a^{1+\frac{x^2}2-\frac{x^2}8} }{x^3}
=\left(a^{1+\frac{x^2}2-\frac{x^2}8}\right)\frac{ a^{\frac{x^3}{16}+o(x^3)} -1 }{x^3}
=\left(a^{1+\frac{x^2}2-\frac{x^2}8}\right)\frac{ a^{\frac{x^3}{16}+o(x^3)} -1 }{\frac{x^3}{16}+o(x^3)}\frac{\frac{x^3}{16}+o(x^3)}{x^3}$$
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$ Let $f(x)=\big(|x+a|-|x+2|+b|x+5| \big)$ be an odd function then find $a+b$.
My try :
$$\begin{align}f(-x)&=|-x+a|-|-x+2|+b|-x+5| \\ &=-f(x) \\&= -\big(|x+a|-|x+2|+b|x+5| \big) \end{align}$$
$$|-x+a|-|-x+2|+b|-x+5|+ \big(|x+a|-|x+2|+b|x+5|) =0... | Since, $f(-x)=-f(x),$ we obtain:
$$|x-a|-|x-2|+b|x-5|+|x+a|-|x+2|+b|x+5|=0.$$
Now, for $x=2$ we obtain:
$$10b+|a-2|+|a+2|-4=0$$ and for $x=5$ we obtain:
$$10b+|a-5|+|a+5|-10=0,$$ which gives
$$|a-2|+|a+2|=|a-5|+|a+5|-6$$ or
$$-2\leq a\leq2,$$ which gives$$b=0.$$
Thus, $$f(x)=|x+a|-|x+2|$$ and
$$|x-a|-|x-2|+|x+a|-|x+2|=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638472",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Unable to prove that $n^5-n$ is a multiple of 30. It is a problem in sec. 2.2 of the book titled by Griffin Harriet, and although have no counter-example till now as below:
$n(n-1)(n+1)(n^2+1) \implies$ if $n=4$, then $n-1=3, n+1=5, n^2+1=17$, and $4*3*5*17=30*2*17$.
Basically, the issue is how to show that :
$(n-1)(n)... | Modulo $5$ we have $$n^5-n=n(n^4-1)=n(n^2-1)(n^2+1)\equiv n(n^2-1)(n^2-4) =n(n-1)(n+1)(n-2)(n+2)$$ and the product of five successive integers is divisible by $5$, while $n(n^2-1)=n(n-1)(n+1)$ is divisible by $6$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638620",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 1
} |
Is $\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$ is divergent or not? I'm searching for a continuous function whose Fourier series diverges at x=0.
Now I've come up with an idea and don't know if it's right.
$$\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4... | $\sum_{n=0}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$=$\int_0^1 \frac{1}{\log\log\frac{4}{x}}dx$+$\sum_{n=1}^\infty \int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$
Using Riemann sum for the integral:
$\int_0^1 \frac{\cos\left(n\pi x\right)}{\log\log\frac{4}{x}}dx$=$\lim_{n -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the area of a quadrilateral in a parallelogram.
Assume quadrilateral $ABCD$ is a parallelogram and its area is $S$. And satisfy the following conditions: $AE=BE$, $BF=FC$, $AQ // PC$. Find the area of quadrilateral $APCQ$
| Assume the origin of the coordinate system is $B$. so coordinate of point $B$ is $(0,0)$. And the $x$-axes is $BC$. Assume the coordinate of point $A$ is $(2a,2b)$. The coordinate of point $C$ is $(2c,0)$. Then we can calculate other points' coordinates. $E$ is $(a,b)$, $F$ is $(c,0)$ $D$ is $(2c+2a, 2b)$.
Then the e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Is $\int \frac{\sqrt{1-x^2}}{x}\;dx = \ln\lvert x\rvert + C$? I have the integral
$$
\int \frac{\sqrt{1-x^2}}{x}\;dx
$$
I do a trig. substitution $x = \sin(\theta)$. Then I get $\sqrt{1-x^2} = \cos(\theta)$. So
$$
\int \frac{\sqrt{1-x^2}}{x}\;dx = \int \frac{\cos(\theta)}{\sin(\theta)}\; d\theta = \ln\lvert \sin(\thet... | $$x=\sin\theta\implies dx=\cos(\theta)\,d\theta$$
$$\begin{align}
\int{ \sqrt{1-x^2} \over x}\,dx
&= \int{ \sqrt{1-\sin^2\theta} \over \sin\theta} \cos(\theta)\,d\theta \\
&= \int{ \sqrt{\cos^2\theta} \over \sin\theta } \cos(\theta)\,d\theta \\
&= \int{ \cos^2\theta \over \sin\theta}\,d\theta \\
&= \int{ -\sin^2(\theta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Prove that $\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$ Prove that
$$\sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy} \le 3$$
for $x \ge 0$, $y \ge 0$, $z \ge 0$ and $x+y+z \le 2$.
My work:
\begin{align*}
&\mathrel{\phantom{=}} \sqrt{x^2+yz}+\sqrt{y^2+xz}+\sqrt{z^2+xy}\\
&\le\sqrt3 \sqrt{x^2+yz+y^2+xz+z^2+xy}\\
&\le\sq... | It is enough to show the homogeneous inequality $\sum_{cyc} \sqrt{x^2+yz} \leqslant \frac32(x+y+z)$. WLOG, let $x \geqslant y \geqslant z$.
Note by AM-GM, we have $\sqrt{x^2+yz} \leqslant \sqrt{x^2+xz} \leqslant x + \frac12z$. This type of AM-GM is motivated by noting $(1, 1, 0)$ is a solution for equality.
Further... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$ Find all the solutions $\in \Bbb R^+$ that satisfy $x(6-y)=9$, $y(6-z)=9$, $z(6-x)=9$
My try
I found that $0 \lt x,y,z \lt 6$
Multiplying the equations we got $x(6-y)y(6-z)z(6-x)=9^3$
$x(6-x)y(6-y)z(6-x)=9^3$
And here is the problem,... | If two of $x,y,z$ are equal, wlog, $x=y$ then $x(6-x)=9 \to x=3$ and $z(6-3)=9 \to z=3$.
If all are $\neq$, wlog $0<x<y<z$, then $9=x(6-y)<z(6-x)=9$, abs.
Finally, the solution is $(3,3,3)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646322",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
A limit question of 3-variable-functions. $\lim\limits_{ (x,y,z) \to (0,0,0)} \frac {xyz^2}{x^2+y^4+z^6}$
$$\lim\limits_{ (x,y,z) \to (0,0,0)} \frac {xyz^2}{x^2+y^4+z^6}$$
I checked that the limit does not exist but I cannot prove that.
I tried $y=mx$, $z=nx$ and also $y=x^m$, $z=x^n$ but they gave me nothing but the... | Let consider
$$\left|\frac {xyz^2}{x^2+y^4+z^6}\right|=\frac {|x||y|z^2}{x^2+y^4+z^6}$$
and let
$$\begin{cases}
|x|=|X|\\\\
|y|=\sqrt{|Y|}\\\\
z=\sqrt[3] Z
\end{cases}$$
then
$$\frac {|x||y|z^2}{x^2+y^4+z^6}=\frac {|X|\sqrt{|Y|}\sqrt[3] {Z^2}}{X^2+Y^2+Z^2}=\frac{\rho^{1+\frac12+\frac23}f(\theta,\phi)}{\rho^2}=\frac{\rh... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Difficulty finding the integral of this rational function? I'm attempting to do the following integral
$$\int \frac{2x+1}{x^3 +2x^2 +1} \, dx$$
I wanted to try using partial fractions but I'm unsure how to factor the denominator. I've been unable to make any progress on this question because of this which is why I don'... | Hint
The denominator has three roots, only one neing real. So, write
$$x^3+2x^2+1=(x-a)(x^2+bx+c)$$ and use partial fractions
$$\frac{2x+1}{x^3 +2x^2 +1}=\frac 1 {a^+ab+c}\left(\frac{2 a+1}{x-a}-\frac{ (a+b-2 c)+(2 a+1) x}{x^2+bx+c} \right)$$
Now, work the problem of expressing $a,b,c$ which is not the most pleasant si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648372",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Any simple way for proving $\int_{0}^{\infty} \mathrm{erf(x)erfc{(x)}}\, dx = \frac{\sqrt 2-1}{\sqrt\pi}$? How to prove
$$\int_{0}^{\infty} \mathrm{erf(x)erfc{(x)}}\, dx = \frac{\sqrt 2-1}{\sqrt\pi}$$ with $\mathrm{erfc(x)} $ is the complementary error function, I have used integration by part but i don't succed
| Recalling that $\text{erf} (x) = 1 - \text{erfc} (x)$, the integral can be rewritten as
$$\int_0^\infty \text{erf}(x) \text{erfc}(x) \, dx = \int_0^\infty \text{erfc}(x) \, dx - \int_0^\infty \text{erfc}^2 (x) \, dx.$$
As
$$\frac{d}{dx} \left (\text{erfc}(x) \right ) = -\frac{2}{\sqrt{\pi}} e^{-x^2},$$
integrating by p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Using integration to solve a formula for the area of a ellipse Problem:
Set up a definite integral to find the area of an ellipse with axis lengths $a$ and $b$. Use a trigonometric substitution to find a formula for the area. What happens if $a=b$? Does this agree with a Geometry formula for a circle? Explain.
$$\frac... | The integral is pretty trivial. Spoiler:
$$\int \sqrt{a^2-t^2}\ dt = \frac{1}{2} \left(t \sqrt{a^2-t^2}+a^2 \tan ^{-1}\left(\frac{t}{\sqrt{a^2-t^2}}\right)\right)$$
This can be obtained by a simple substitution $t = a\sin(z)$, $dt = a\cos(z)\ dz$
Which turns the integral into
$$a^2\int \cos^2(z)\ dz$$
Which is trivial ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Unable to evaluate limit correctly I want to find the limit of
$$\frac{e^x + \frac1{e^x} - 2\cos x}{x\tan x}$$
as $x$ tends to $0$.
My attempt:
The limit of $\frac{e^x + 1/e^x - 2cosx}{xtanx}$ should be the same as the limit of $\frac{2 - 2\cos x}{x\tan x}$, which can evaluated using the standard limits of $x/\sin x$ ... | You cannot apply partial limits in addition and subtraction. Though, it can be used in products.
\begin{align} \lim_{x \to 0} \frac{e^x + 1/e^x - 2 \cos x}{x \tan x} &=\lim_{x \to 0} \frac{e^x + 1/e^x - 2 \cos x}{x^2} \times \frac{x}{\tan x }\\
&=\lim_{x \to 0} \frac{e^x + 1/e^x - 2 \cos x}{x^2} \times \lim_{x \to 0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
What is the value of $\delta $ if $\epsilon=0.01$?
Let $f(x,y) = \begin{cases} \frac{2x^2y+3xy^2}{x^2+y^2}, & \text{if
$(x,y)\neq(0,0)$} \\[2ex] 0, & \text{if $(x,y)=(0,0)$ } \end{cases}$
Then the condition on $\delta $ such that $\vert f(x,y)-f(0,0)
\vert<0.01$ whenever $\sqrt {x^2+y^2}<\delta $ is-
1.$\delta <0.01... | To get a larger possible value of $\delta$, use the polar coordinate transformation $x = r \cos t, y = r \sin t$ to rewrite the equality in the question.
$$\left\vert \frac{2x^2y+3xy^2}{x^2+y^2}-0\right\vert =\left\vert \frac{r^3 \cos t \sin t (2\cos t+3\sin t)}{r^2}\right\vert = r |\cos t| |\sin t| |2 \cos t + 3 \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Conjecture: $N=x_1^2+x_2^2-y^3$ has integer solutions for all $N$, with $x_1>0$,$x_2>0$ and $y>0$ Background.
I constructed some code for Algorithm for diophantine equation and decided to reuse it to investigate $N=x_1^2+x_2^2+z^3$ with $z$ integer. Negative values for $z$ seemed to produce plentiful small $N$ values, ... | This was a problem posed by Noam Elkies and Irving Kaplansky in the January 1995 M.A.A. Monthly. They did not ask that your $y$ be positive, but that is how it comes out anyway, with a very few exceptions for the variable $x$ below being, for example $0,1,2.$ These can then be fixed by a quick search.
Later, Kap and I ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2651855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
What is the probability that some people are chosen in different groups? Let's say that I have $n$ people and $m \le n$ of this people are malicious. Then I run a process that chooses 6 people randomly and assigns each of them to one of two different groups $A$ and $B$ such that each group in the end has 3 people. What... | Not sure if this is the correct answer, but maybe my reasoning can help you nonetheless.
The total number of arrangements is: ${n \choose 6} \cdot { 6 \choose 3}$.
${n \choose 6}$ represents the total ways of choosing 6 people out of the n people
${ 6 \choose 3}$ represents the number of ways of splitting the people in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
On the sides $BC$, $CA$, $AB$ of $\Delta ABC$, points $D$, $E$, $F$ are taken in such a way that $\frac{BD}{DC} = \frac{CE}{EA} = $ How to find the area ?
The Diagram and the question.
| Let $G\in EC$ such that $DG||BE$ and $GC=2x$.
Thus, by Thales $$\frac{EG}{2x}=\frac{BD}{DC}=2,$$ which gives $EG=4x$, $EC=6x$ and $AE=3.$
Thus, by Thales again we obtain:
$$\frac{AP}{PD}=\frac{AE}{EG}=\frac{3x}{4x}=\frac{3}{4}.$$
Now, let $I\in DC$ such that $EI||AD$ and $DI=y$.
Thus, by Thales $IC=2y$ and from here $D... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653843",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Area of a triangle using a matrix Find the area of a triangle whose vertices are (1,0), (2,2), and (4,3)
$$\begin{vmatrix}x_1&y_1&1\\x_2&y_2&1\\x_3&y_3&1\end{vmatrix}=\begin{vmatrix}1&0&1\\2&2&1\\4&3&1\end{vmatrix}$$
$$=1(-1)^2\begin{vmatrix}2&1\\3&1\end{vmatrix}+0(-1)^3\begin{vmatrix}2&1\\4&1\end{vmatrix}+1(-1)^4\begi... | It comes from
$$\det(A) = \sum_{j=1}^n (-1)^{i+j} a_{i,j} M_{i,j}$$
with $n=3$ and minor $M_{i,j}$, where $i=1$. So in this case
$$\det(A) = \sum_{j=1}^3 a_{1,j}(-1)^{1+j} M_{1,j}=1(-1)^2\begin{vmatrix}2&1\\3&1\end{vmatrix}+0(-1)^3\begin{vmatrix}2&1\\4&1\end{vmatrix}+1(-1)^4\begin{vmatrix}2&2\\4&3\end{vmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654841",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
For $x, y \in \mathbb{R}$, prove that $\max(x, y) = \frac{x + y + |x - y|}{2},$ and $\min(x, y) = \frac{x + y - |x - y|}{2}$. Prove that for all real numbers $x$ and $y$,
$$\max(x, y) = \dfrac{x + y + |x - y|}{2},$$
and
$$\min(x, y) = \dfrac{x + y - |x - y|}{2}.$$
For any real number $x$, the absolute value of $x$, d... | $|x-y|$ means the distance between $x$ and $y$ in the number line.
On the other hand, $\frac{x+y}{2}$ means the midpoint between $x$ and $y$.
(you can check those two statements if you didn't know them!)
Therefore, $\frac{x+y+|x-y|}{2}=\frac{x+y}{2}+\frac{|x-y|}{2}$, which means adding half the distance between $x$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656234",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Elementary Matrix and Row Operations Find an elementary matrix E such that EA = B
$$A = \begin{bmatrix} 2 & -1 \\ 5 & 3 \end{bmatrix}$$
$$B = \begin{bmatrix} -4 & 2 \\ 5 & 3 \end{bmatrix}$$
$$\begin{bmatrix} 2 & -1 \\ 5 & 3 \end{bmatrix}*E=\begin{bmatrix} -4 & 2 \\ 5 & 3 \end{bmatrix}$$
So the Elementary Matrix is an I... | Since the operation is $B=EA$ we have
\begin{equation}
\begin{pmatrix}
-2 & 0\\0 & 1
\end{pmatrix}\begin{pmatrix}
2 & -1\\5 & 3
\end{pmatrix} = \begin{pmatrix}
-4 & 2\\5 & 3
\end{pmatrix}.
\end{equation}
Your elementary matrix is correct but you meant to multiply it to $A$ on the left not on the right.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656407",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Difference Calculus of the Factorial Function for Negative Integer Powers of $x$ Under the difference calculus, we have that
$$\Delta f(x)=f(x+1)-f(x)$$.
We also have factorial polynomials (sometimes referred to the rising and or falling factorials). The text I am reading defines the factorial function as
$$x^{(n)}=x... |
Everything looks fine: We obtain
\begin{align*}
\color{blue}{\Delta x^{(-n)}}&=(x+1)^{(-n)}-x^{(n)}\\
&=\cdots\\
&=\frac{(x+1)-(x+1+n)}{(x+1)(x+1+n)}\cdot\frac{1}{(x+n)^{(n-1)}}\\
&=-n\cdot\frac{1}{(x+n+1)\cdot (x+n)^{(n-1)}\cdot (x+1)}\\
&=-n\cdot\frac{1}{(x+n+1)^{(n+1)}}\\
&=\color{blue}{-nx^{(-n-1)}}
\end{align*}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2659052",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove that $n^2 - 2n + 7$ is even then $n + 1$ is even I have to use Proof by contradiction to show what if $n^2 - 2n + 7$ is even then $n + 1$ is even.
Assume $n^2 - 2n + 7$ is even then $n + 1$ is odd. By definition of odd integers, we have $n = 2k+1$.
What I have done so far:
\begin{align}
& n + 1 = (2k+1)^2 - 2(2... | If $n + 1$ is even, then $n$ must be odd. Let $n = 2k + 1$ since it is in the form of an odd number. Now substitute into the quadratic expression: $$\begin{align} n^2 - 2n + 7 &= (2k+1)^2 - 2(2k+1) + 7 \\ &= (2k)^2 + 1 + 4k - 4k - 2 + 7 \\ &= 4k^2 + 6 \\ &= 2(2k^2 + 3).\end{align}$$ Since the quadratic is an even numb... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 1
} |
How to compute $\int_{0}^{1}x\sin (bx) J_0\!\!\left(a\sqrt{1-x^2}\right)\!\mathrm dx$? I need a help with integral below,
$$\int_{0}^{1}x\sin (bx) J_0\left(a\sqrt{1-x^2}\right)\,\mathrm dx$$
where $a \geq 0$ and $b$ are constants and $J_0(x)$ is the zeroth-order of Bessel function of the first kind.
This integral has ... | Suggestion by user tired:
Integral expression below on Gradshteyn and Ryzhik's book 7th edition, section 6.677, number 6:
$$\int_0^1 \cos (b x) J_0\left(a \sqrt{1-x^2}\right) \, dx=\frac{\sin
\left(\sqrt{a^2+b^2}\right)}{\sqrt{a^2+b^2}}$$
Differentiate for variable b:
$$\int_0^1 \frac{\partial \left(\cos (b x) J_0\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Prove that $x^4 + y^4 - 3xy = 2$ is compact The exercise consists of showing that the function $f(x,y)=x^4 + y^4$ has a global minimum and maximum under the constraint $x^4 + y^4 - 2xy = 2$.
In the solution to the exercise, it it follows that the constraint is compact if we can show that $\lim_{x^2 + y^2 \rightarrow \i... | Note that
$$\eqalign{x^4+y^4-2xy&={1\over2}(x^2+y^2)^2+{1\over2}(x^2-y^2)^2-2xy\geq{1\over2}(x^2+y^2)^2-(x^2+y^2)\cr
&=(x^2+y^2)^2\left({1\over2}-{1\over x^2+y^2}\right)\geq{1\over4}(x^2+y^2)^2\geq4\ ,\cr}$$
as soon as $x^2+y^2\geq4$. It follows that the constraint defines a closed and bounded, hence compact, set in th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
>Finding range of $f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$
Finding range of $$f(x)=\frac{\sin^2 x+4\sin x+5}{2\sin^2 x+8\sin x+8}$$
Try: put $\sin x=t$ and $-1\leq t\leq 1$
So $$y=\frac{t^2+4t+5}{2t^2+8t+8}$$
$$2yt^2+8yt+8y=t^2+4t+5$$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0$$
For real roots $D\geq 0$
So $$16(2y-1)... | MrYouMath has already provided a good answer.
This answer uses your method.
You already have a quadratic equation on $t$
$$(2y-1)t^2+4(2y-1)t+(8y-5)=0\tag1$$
where $y\not=\frac 12$.
Note here that we want to find $y$ such that $(1)$ has at least one real solution $t$ satisfying $-1\le t\le 1$.
It seems that you've mis... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
Distance between two parallel lines by having linear equations I wonder where does this formula is coming from?
It is for finding the distance between two parallel lines when we have their linear equation:
First line is:$ax+by+c=0$
Second line is:$ax+by+c_1=0$
Their distance :$$\frac{|c-c_1|}{\sqrt{a^2+b^2}}$$
| Alternatively: the perpendicular line that passes through origin is: $bx-ay=0$. It crosses the two parallel lines at:
$$\left(-\frac{ac}{a^2+b^2},-\frac{bc}{a^2+b^2}\right) \ \ \text{and} \ \ \left(-\frac{ac_1}{a^2+b^2},-\frac{bc_1}{a^2+b^2}\right).$$
The distance between these points is:
$$d=\sqrt{\left(\frac{a(c-c_1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670690",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Can a parallelogram have whole-number lengths for all four sides and both diagonals?
Is it possible for a parallelogram to have whole-number lengths for all four sides and both diagonals?
One idea I had was to arrange four identical right triangles such that the right angles are adjacent. For example if we take four ... | Your hunch is wrong. A non-rectangular non-rhomboid parallelogram with integer side and diagonal lengths exists:
Suppose that the parallelogram is $ABCD$ with $AB=a$, $AD=b$ and $BD=c$. By the law of cosines:
$$\cos\angle DAB=\frac{a^2+b^2-c^2}{2ab}$$
Since $\angle CDA=\pi-\angle DAB$, $\cos\angle CDA=-\cos\angle DAB$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670781",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "20",
"answer_count": 2,
"answer_id": 1
} |
Strange Pattern in Decimal Expansion I noticed something weird when I was fooling around with my calculator.
I calculated several powers of $30$ of the form $30^{\left(\frac{10^n-1}{10^n}\right)}$ and I noticed a pattern in the fractional part:
$$30^{\left(\frac{9999}{10000}\right)}=29.9897981429$$
$$30^{\left(\frac{99... | Short answer: it's because of the approximation
$$
(1 + x)^{1/10} \approx 1 + \frac{x}{10}
$$
valid when $x$ is very close to $0$. (This approximation is the linear truncation of the Taylor series $(1 + x)^{1/10} = 1 + \binom{1/10}{1}x + \binom{1/10}{2}x^2 + \binom{1/10}{3}x^3 + \dotsb$.)
A different way of writing ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671569",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Numbers up to 1000 divisible by 2 or 3 and no other prime My task requires to find all numbers from $1-1000$ such that they are divisible by $2$ or $3$ and no other primes. I know that $2$ divides even numbers and I can use the formula $\left \lfloor{\frac{1000}{2}}\right \rfloor $ and the numbers divisible by three $\... | Such a number should be in the form of $2^m3^n$ where $m$ and $n$ are non-negative integers such that $m$ and $n$ are not both zero.
If $n=0$, $m\ge1$ and $2^m\le1000$. So $1\le m\le 9$.
If $n=1$, $2^m\le\frac{1000}{3}$. So $0\le m\le 8$.
If $n=2$, $2^m\le\frac{1000}{9}$. So $0\le m\le 6$.
If $n=3$, $2^m\le\frac{1000}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672258",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Conditional Probability: Problem with 6 persons
There are $6$ persons, let's name them $ A, B, C, D, E$ and$ F $. They are ordered according to their popularity, but without letting them know their order, so they guess all ways they can be ordered are equally possible.
Taking as a fact that those persons learn that... | I am going to make use of Bayes' Theorem:
$$P({A > C | A > B})=\frac{P(A>C \cap A>B)}{P(A>B)}$$
Consider lining up the letters as follows:
_ _ _ _ _ _
where popularity is rated from left to right.
By symmetry $$P(A>B)=\frac{1}{2}$$
We must solve for $$P(A>C \cap A>B)$$
which is the probability that $A$ is bigger than b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
proving the sum of a few determinant values as a constant Let
$$ \Delta_a = \begin{vmatrix}
a-1 & n & 6 \\ (a-1)^2 & 2n^2 & 4n - 2 \\ (a-1)^3 & 3n^2 & 3n^2 - 3n
\end{vmatrix} $$
My task is to show that
$$ \sum_{a = 1}^n \Delta_a = c
$$
where c is some constant.
I tried taking a-1 and n common from $C_1$ and $C_... | Recall that the determinant is linear in the columns of the matrix. Since $\Delta_a$ differ only in the first column, we have:
$$\sum_{a=1}^n \Delta_a = \sum_{a=1}^n\begin{vmatrix}
a-1 & n & 6 \\ (a-1)^2 & 2n^2 & 4n - 2 \\ (a-1)^3 & 3n^2 & 3n^2 - 3n
\end{vmatrix} = \begin{vmatrix}
\sum_{a=1}^n (a-1) & n & 6 \\ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Characterizing elements with square norms in quadratic integer rings Given the ring $\mathbb{Z}\left[\sqrt{D}\right]$ (where $D$ is a positive square-free integer) can we characterize all elements $\alpha$ with positive norms for which $N(\alpha)$ is a perfect square in $\mathbb{Z}$?
Obviously this is true for all int... | Seeing Rene's answer: I am talking about arranging $x^2 - Dy^2 = w^2$ with $\gcd(x,y) = 1.$
$$ 8^2 - 15 \cdot 1^2 = 7^2 $$
$$ 16^2 - 15 \cdot 3^2 = 11^2 $$
$$ 23^2 - 15 \cdot 4^2 = 17^2 $$
$$ $$
$$ 83^2 - 15 \cdot 8^2 = 77^2 $$
$$ $$
$$ \left( 5 u^2 + 3 v^2 \right)^2 - 15 \left( 2uv \right)^2 = \left( 5 u^2 - 3 v^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2677051",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Counting Principle - Dependent Events There are 7 red and 5 yellow fish in an aquarium. Three fish are randomly caught in a net. Find the probability that the fish were:
a) All red
b) Not all of the same color.
I have solved this question using the method for dependent events as follows:
a) $\frac{7}{12} \cdot \frac... | $(a)$
$$\frac{7 \choose 3}{12 \choose 3}$$
You can show that this equals what you did.
$$\begin{align*}
\frac{7 \choose 3}{12 \choose 3}
&=\frac{\frac{7!}{3!4!}}{\frac{12!}{3!9!}}\\\\
&=\frac{7\cdot6\cdot5}{\color{red}{3\cdot2\cdot1}}\frac{\color{red}{3\cdot2\cdot1}}{12\cdot11\cdot10}\\\\
&=\frac{7}{12}\cdot\frac{6}{11... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2678292",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Convolution of i.i.d. with uniform distribution I am trying to solve another exercise about convolution of continuous variables. Here is what is asked:
Let $X_1,\cdots,X_n$ be i.i.d. such that $X_1 \sim U(0,1)$. Verify that$$
P(X_1+\cdots+X_n\leq x)=\frac{1}{n!}\sum_{k=0}^{n-1}(-1)^k {n\choose k}(x-k)_+^n, \quad \fora... | $\def\d{\mathrm{d}}\def\peq{\mathrel{\phantom{=}}{}}$The base case of induction is, in fact, $n = 1$.
Denote $S_n = \sum\limits_{k = 1}^n X_k$ for any $n$. For $n = 1$,$$
P(S_1 \leqslant x) = P(X_1 \leqslant x) = x_+. \quad \forall 0 \leqslant x \leqslant 1
$$
Suppose the proposition holds for $n$. For any $0 \leqslant... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$ How can we find the value of $$\sum_{k=1}^{10} (\sin \frac{2 \pi k}{11}-i \cos \frac{2 \pi k}{11})=?$$
My Approach:
By Euler's Theorem: $$\huge \cos \frac{2 \pi k}{11}+i \sin \frac{2 \pi k}{11} =e^{
\frac{i2 \pi k}{11}}$$
$$\therefore \sin \frac... | \begin{align}-i e^{
\frac{i2 \pi }{11}} \frac{e^{
\frac{i20 \pi }{11}} -1}{e^{
\frac{i2 \pi }{11}}-1}&=-i\frac{\exp(\frac{i22\pi}{11})-\exp(\frac{i2\pi}{11})}{\exp(\frac{i2\pi}{11})-1} \\
&=-i\frac{\exp(i2\pi)-\exp(\frac{i2\pi}{11})}{\exp(\frac{i2\pi}{11})-1}
\\
&=-i\frac{1-\exp(\frac{i2\pi}{11})}{\exp(\frac{i2\pi}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to find the eigen values of the given matrix
Given the matrix
\begin{bmatrix}
5&1&1&1&1&1\\1&5&1&1&1&1\\1&1&5&1&1&1\\1&1&1&4&1&0\\1&1&1&1&4&0\\1&1&1&0&0&3
\end{bmatrix}
find its eigen values(preferably by elementary row/column operations).
Since I don't know any other method other than elementary operations to f... | The characteristic polynomial $\chi_t$ of this matrix can be factored nicely, i.e.,
$$
\chi_t(A)=(t - 2)(t - 3)(t - 4)^3(t - 9).
$$
So the six eigenvalues are $2,3,4,4,4,9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2683944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Area of largest inscribed rectangle in an ellipse. Can I square the area before taking the derivative? So say I have an ellipse defined like this:
$$\frac{x^2}{9} + \frac{y^2}{4} = 1$$
I have to find the largest possible area of an inscribed rectangle.
So the area ($A$) of a rectangle is $2x2y=4xy$. Also we can redefin... | I think you are in right track but your expression for the first derivative of $A$ need a little touch. I put it in more clear way:
$$A=4x \cdot \sqrt{4 - \frac{4x^2}{9}}=4x \left(4-\frac{4x^2}{9}\right)^{1/2}$$ Thus,
$$A' = 4 \left(4-\frac{4x^2}{9}\right)^{1/2} + 4x \left(4-\frac{4x^2}{9}\right)^{-1/2}\left(\frac{-8x}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684375",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Is this sequence a integer? For $n \in \mathbb{N}$ and $n>(q+1)/4$. Where $q>1$ and integer. With $p\in \mathbb{Z}$
$\frac{(4n-1)!}{q}p - \frac{(4n-1)!}{3} -4(4n-1)! \big(\frac{1}{4n-3}-
\frac{1}{4n-1}+\frac{1}{1}- \frac{1}{3}+\frac{1}{5}-\frac{1}{7}+... \big)$
I suspect that it's not integer. $1-1/3+1/5-...$ converge... | Let $n=1, q=2, p=3$. Then $4n>q+1$ but $$\frac{(4n-1)!}{q}p - \frac{(4n-1)!}{3} -4(4n-1)! \left(\frac{1}{4n-3}-
\frac{1}{4n-1}+\frac{1}{1}- \frac{1}{3}+\frac{1}{5}-\frac{1}{7}+... \right)$$ becomes $$9-2-24\left(1-\frac13+\frac\pi4\right)=7-24+8-6\pi=-9-6\pi\not\in\mathbb{Z}$$ so the result does not hold.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Inverse of a modular matrix I have the matrix$$
\begin{pmatrix}
1 & 5\\
3 & 4
\end{pmatrix} \pmod{26}
$$
and I need to find its inverse. I do it according to this website.
I find the modular multiplicative inverse (of the matrix determinant, which is $1×4-3×5=-11$) with the extended Euclid algorithm (it is $-7 \equiv ... | Doing aritmetic modulo $\;26\;$ all the time, we get that
$$\det A=\begin{vmatrix}1&5\\3&4\end{vmatrix}=4-15=-11=15\implies$$
$$\implies A^{-1}=\frac1{15}\begin{pmatrix}4&-5\\-3&1\end{pmatrix}=7\begin{pmatrix}4&21\\23&1\end{pmatrix}=\begin{pmatrix}2&17\\5&7\end{pmatrix}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2686150",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Solving the congruence $9x \equiv 3 \pmod{47}$ For this question $9x \equiv 3 \pmod{47}$.
I used euler algorithm and found that the inverse is $21$ as $21b-4a=1$
when $a=47$ and $b=9$
I subbed back into the given equation:
\begin{align*}
(9)(21) & \equiv 3 \pmod{47}\\
189 & \equiv 3 \pmod{47}\\
63 & \equiv 1 \pmod{47}
... | We wish to solve the congruence $9x \equiv 3 \pmod{47}$. You correctly found that $21$ is the multiplicative inverse of $9$ modulo $47$. However, you made an error in this step:
$$9 \cdot 21 \equiv \color{red}{3} \pmod{47}$$
If we multiply one side of the congruence by $21$, we must multiply the other side of the con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
Prove $\prod_{c < p \leq x} \left(1 - \frac{c}{p} \right) \ll \log ^{-c} x .$
For fixed $c>0$, show
$$\prod_{c < p \leq x} \left(1 - \frac{c}{p} \right) \ll \log ^{-c} x ,$$
where $p$ is a prime.
I am not sure how to show this result. We know that $$\sum_{p \leq x} \log \left( 1-\frac{1}{p} \right) = -\log \log x... | Note that
$$\frac{t}{1+t} < \log (1+t) < t$$
for $t> -1$ and $t \neq 0$. Then
\begin{align*}
c\log \left( 1 - \frac{1}{p} \right) & > \frac{-c/p}{1 - 1/p} \\ & =
\frac{p}{p-1} \left(- \frac{c}{p} \right) \\ & >
- \frac{c}{p} \\ &
> \log \left(1 - \frac{c}{p} \right)
\end{align*}
for $p >c$. Exponentiate both sides to... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2692795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Trigonometric identity. How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$? How can I prove that $\sin A = 2\sin\frac{A}2\cos\frac{A}2$ ?
My failed take on this matter is:
$$
\sin A =
\sin\frac{A}2 + \sin\frac{A}2 = 2\sin\Big(\frac{\frac{A}2+\frac{A}2}2\Big)\cos\Big(\frac{\frac{A}2-\frac{A}2}2\Big)=
2\sin\Big... | Hint:
Your first step is wrong because
$\sin(x)$ is not a linear function so: $$\sin x \ne \sin(\frac{x}{2})+\sin(\frac{x}{2})$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2700397",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
$\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$ $\lim_{n\rightarrow \infty} n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})$, $\forall \alpha \in R$
I can change the form of this limit saying that
$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=n^\alpha (\sqrt[5]{n^2(1+\frac{1}{n})}-\sqrt[5]{n^2... | By binomial expansion
*
*$\sqrt[5]{n^2+n}=\sqrt[5]{n^2}\sqrt[5]{1+1/n}=\sqrt[5]{n^2}(1+\frac1{5n}+o(1/n))$
*$\sqrt[5]{n^2+2n+1}=\sqrt[5]{n^2}\sqrt[5]{1+2/n+1/n^2}=\sqrt[5]{n^2}(1+\frac2{5n}+o(1/n))$
then
$$n^\alpha (\sqrt[5]{n^2+n}-\sqrt[5]{n^2+2n+1})=-\frac{n^\alpha\sqrt[5]{n^2}}{5n}+o\left(\frac{n^\alpha\sqrt[5... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find coefficient of multiple binomial expansions quickly?
Find the coefficient of $(x^{-2})$ in the expansion of $(x-1)^3(\frac1{x}+2x)^6$
How can this be done quickly without expanding the two brackets?
I tried writing the general formula in terms of $r$ for both brackets and equated $-2$ to the sum of the po... | Here is a variation which goes consecutively through the factors expanding one by one up to terms which are needed ignoring other terms. We also start with the factors with greatest exponents which keep the number of terms which are to expand small.
It is convenient to use the coefficient of operator $[x^k]$ to denote ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Proving that a set has empty interior For $n=0,1,2...$ define $$A_n = \bigcup^{3^n}_{k=1}\left(\frac{k-\frac23}{3^n},\frac{k-\frac13}{3^n}\right)$$ Let $A=\bigcup_{n=0}^{\infty}A_n$, and let $B=[0,1]\cap \overline A$. I wish to show that $B$ has an empty interior. First, I suppose it has an interior. So I let $x\in B$ ... | The direction of the proof is great but it needs some tiny adjustments. In short, we should choose $n$ after $\delta$.
Suppose $x\in B$ and assume the existence of $\delta>0$ such that $(x-\delta,x+\delta)\subset B$. Then we can choose $n$ such that $\frac{\frac 13}{3^n}<\delta$. Since $x\in B$, there exists $k$ such t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712124",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
What are the units in $\mathbb{Q}(\sqrt{2}, \sqrt{3})$? I'd like to find the four independent units in (the ring of integers of ) $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \subseteq \mathbb{R}$ We also have that $\mathbb{Q}(\sqrt{2}, \sqrt{3}) \simeq \mathbb{Q}[x,y]/(x^2 - 2, y^2 - 3)$, as a field extension.
I just want to find... | The fundamental unit of $\mathbb{Q}(\sqrt{6})$ is $$5+2\sqrt{6}$$ Does this help ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713491",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Evaluating $\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $ $$\lim_{x\to \infty } (x +\sqrt[3]{1-{x^3}} ) $$
What method should I use to evaluate it. I can't use the ${a^3}$-${b^3}$ formula because it is positive. I also tried to separate limits and tried multiplying with $\frac {\sqrt[3]{(1-x^3)^2}}{\sqrt[3]{(1-x^3... | For $n\gt2$ we have
$$\left(x-{1\over x^2}\right)^3=x^3-3+{3\over x^3}-{1\over x^6}\lt x^3-3+{3\over4}\lt x^3-1$$
Hence, since $x^3-1\lt x^3$ is clear, we have
$$0\lt x-\sqrt[3]{x^3-1}\lt x-\left(x-{1\over x^2}\right)={1\over x^2}\to0\quad\text{as }x\to\infty$$
So by the Squeeze Theorem,
$$\lim_{x\to\infty}\left(x+\sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2716247",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 5
} |
Binomial sum involving power of $2$
Finding $\displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)(k+2)}\cdot 2^{k+2}\binom{n}{k}$
Try: $$\int^{x}_{0}(1+x)^n=\int^{x}_{0}\bigg[\binom{n}{0}+\binom{n}{1}x+\cdots\cdots +\binom{n}{n}x^n\bigg]dx$$
$$\frac{(1+x)^{n+1}-1}{n+1}=\binom{n}{0}x+\binom{n}{1}\frac{x^2}{2}+\cdots \cdots +\bi... | $$f(x)=(1+x)^n =\displaystyle \sum^{n}_{k=0}x^{k}\binom{n}{k} $$
Thus
$$g(x)=\int f(x) dx = \frac{(1+x)^{n+1}-1}{n+1}= \displaystyle \sum^{n}_{k=0}\frac{1}{(k+1)}\cdot x^{k+1}\binom{n}{k}$$
Where the constant of integration was chosen so that $g(0)=0$. Also,
$$\int g(x) dx= \frac{(1+x)^{n+2}-1}{(n+1)(n+2)}-\frac{x}{n+1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
Recognizing Patterns in Alternating Signs Matricies and their Inverses Let's say we have the matrix A with alternating-sign 1's below
A = \begin{bmatrix}1&-1&1&-1\\0&1&-1&1\\0&0&1&-1\\0&0&0&1\end{bmatrix}
If we find the inverse, we get
A^-1 = \begin{bmatrix}1&1&0&0\\0&1&1&0\\0&0&1&1\\0&0&0&1\end{bmatrix}
We get a simi... | Hmm, you can also prove it from basic understanding of inverses of triangular matrices, and a proof by induction.
$$ \left( \begin{array}{c | c}
A_{TL} & A_{TR} \\ \hline
0 & A_{BR}
\end{array} \right)^{-1} =
\left( \begin{array}{c | c}
A_{TL}^{-1} & -A_{TL}^{-1} A_{TR} A_{BR}^{-1}\\ \hline
0 & A_{BR}^{-1}
\end{array} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$ Give $a, b, c$ be positive real numbers such that $a+ b+ c= 3$. Prove that:
$$\left ( a+ b \right )^{2}\left ( b+ c \right )^{2}\left ( c+ a \right )^{2}\geq 64abc$$
My try
We have:
$$\left ( a+ b \right )^{2}\geq 4ab$$
$$\l... | Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.
Hence, we need to prove that:
$$(9uv^2-w^3)^2\geq64u^3w^3$$ or $f(w^3)\geq0$, where
$$f(w^3)=(9uv^2-w^3)^2-64u^3w^3.$$
But, $$f'(w^3)=-2(9uv^2-w^3)-64u^3<0,$$ which says that $f$ decreases.
Thus, it's enough to prove our inequality for a maximal value of $w^3$,
which happ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721251",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Prove this inequality $\sum_{cyc}\sqrt[4]{\frac{a}{b+c}}\ge \sqrt[4]{16+\frac{196abc}{(a+b)(b+c)(c+a)}}$ Let $a;b;c\ge0$. Prove the inequality $$\sqrt[4]{\dfrac{a}{b+c}}+\sqrt[4]{\dfrac{b}{c+a}}+\sqrt[4]{\dfrac{c}{a+b}}\ge\sqrt[4]{16+\dfrac{196abc}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}$$
I think we should... | We need to prove that
$$\left(\sum_{cyc}\sqrt[4]{a(a+b)(a+c)}\right)^4\geq16\prod_{cyc}(a+b)+196abc.$$
Now, by Holder, C-S,AM-GM and Schur we obtain:
$$\left(\sum_{cyc}\sqrt[4]{a(a+b)(a+c)}\right)^4=\sum_{cyc}a(a+b)(a+c)+$$
$$+4\sum_{cyc}\sqrt[4]{(a^2(a+b+c)+abc)^3(b^2(a+b+c)+abc)}+$$
$$+4\sum_{cyc}\sqrt[4]{(a^2(a+b+c)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Converting a generating function from fraction form to a power series Given a fraction $\frac{(1 + ax^{n})^{m}}{(1 + bx^{p})^{q}}$, how does one convert it to a series of the form $a_{0}x^{0} + a_{1}x^{1} + a_{2}x^{2} + a_{3}x^{3} . . .$ ?
I was unable to find instructions for this, and I could use the information to t... | Looking at the hints, we have:
$$1 + x^{3} + x^{6} + \cdots = \frac{1}{1 - x^{3}}$$
and
$$1 + x + x^{2} = \frac{1 - x^{3}}{1 - x}$$
so
$$\begin{align}f(x) &= (1 + x + x^{2})^{4} (1 + x^{3} + x^{6} + \cdots)^{2}\\[1ex]&=\frac{(1 - x^{3})^4}{(1 - x)^4(1-x^3)^2}\\[1ex]&=\frac{(1 - x^{3})^2}{(1 - x)^4}\\[1ex]&=\frac{(1 - 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Proving the trig identity $\frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$ without cross-multiplying I need to prove the following identity.
$$\frac{1-\cos\theta}{\sin\theta} = \frac{\sin\theta}{1+\cos\theta}$$
I want to prove it by deduction rather than cross multiplying.
| $$\frac{1-\cos\theta}{\sin\theta}=\frac{(1-\cos\theta)(1+\cos\theta)}{\sin\theta(1+\cos\theta)}=\frac{1-\cos^2\theta}{\sin\theta(1+\cos\theta)}=\frac{\sin^2\theta}{\sin\theta(1+\cos\theta)}$$
$$\frac{1-\cos\theta}{\sin\theta}=\frac{2\sin^2\frac{\theta}{2}}{2\sin\frac{\theta}{2}\cos\frac{\theta}{2}}=\frac{2\sin\frac{\t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2722902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 1
} |
Interesting 3 Variable Inequality for Real Numbers I recently saw an olympiad style inequality that seemed very difficult. I tried to use elementary inequalities such as AM-GM or Cauchy-Schwarz, but neither have helped in making significant progress. Could anyone provide a rigorous proof, preferably using more elementa... | The inequality is homogenous in $(x, y, z)$, therefore we can assume
that $xyz=1$. Then
$$
\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \ge \dfrac{3}{2}
$$
as for example demonstrated in
If $xyz=1$, prove $\frac{1}{y(x+y)}+\frac{1}{z(y+z)}+\frac{1}{x(z+x)} \geqslant \frac{3}{2}$ or A inequality proposed at Zhauty... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. If $8R^2=a^2+b^2+c^2$ then prove that the triangle is right angled. Where $a,b,c$ are the sides of triangle and $R$ is the circum radius
My Attempt
From sine law,
$$\dfrac {a}{\sin A}=\dfrac {b}{\sin B}=\dfrac {c}{\sin C}=2R$$
So,
$$a=2R \sin A$$
$$... | Since
$$4 = 2\sin^2 A + 2\sin^2 B + 2\sin^2 C$$
we have
$$(1-2\sin^2 A ) + (1-2\sin^2B) + 2 - 2\sin^2C=0.$$
or
$$2\cos^2 C + (\cos(2A) + \cos(2B)) = 0$$
Since $$\cos(2A)+\cos(2B) = 2\cos(A+B) \cos(A-B) = -2\cos C\cos(A-B),$$
we have
$$2\cos C(\cos C - \cos(A-B)) = 0$$
Replace $\cos C = -\cos(A+B)$, we get
$$ \cos C(\c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724255",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Integration of $x^2$ using double substitution my professor has asked me to calculate: $$\int x^2dx = \frac{1}{3}x^3 + C$$ using the substitution of $$u=x$$ $$v=x$$
meaning that $$du=dx$$ $$dv = dx$$
The next step is where I think I'm wrong but I'm unsure why $$dx = \frac{du + dv}{2}$$
$$ \int uv(\frac{du+dv}{2}) = \fr... | You can't use two substitutions in the same integral for the same variable. It dosen't make sense..
Note that
$$\int uv(\frac{du+dv}{2}) = \frac{1}{2}\int uvdu + \frac{1}{2}\int uvdv = \frac{1}{4}u^2v + \frac{1}{2}uv^2 + C$$
You have uv as variable but du...and also $u=v$ so normally
$$\int uvdu=\int u^2du=\frac {u^3}3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2725588",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\lim_{x\to-\infty}\frac{x^3+1}{x^3+\sqrt{4x^6+1}}$ $$\displaystyle \lim_{x\to-\infty}\frac{x^3+1}{x^3+\sqrt{4x^6+1}}$$
I understand that we need to divide by the highest power, but first we want to factor out the $x^6$ from the radical. This gives
$$\displaystyle \lim_{x\to-\infty}\frac{x^3+1}{x^3+\sqrt{x^6... | In general, for any real $x$, $\sqrt{x^6}=|x|^3$. As $x\to-\infty$, we may assume that $x<0$ and therefore $\sqrt{x^6}=|x|^3=-x^3$. This simplification allows us to easily divide the numerator and the denominator by $x^3$, and find that
$$\lim_{x\to-\infty}\frac{x^3+1}{x^3-x^3\sqrt{4+\frac{1}{x^6}}}=\lim_{x\to-\infty}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Why does one of the units, namely $50 - 7\sqrt{51}$, in $\mathbb{Z}[\sqrt{51}]$ have a decimal expansion that is the Catalan numbers? $50 - 7\sqrt{51}$ is one of the units in $\mathbb{Z}[\sqrt{51}]$. In fact, corresponding to $n = 51$, $x = 50$ and $y = 7$ is the smallest integer solution to $x^2 - ny^2 = 1$.
The Catal... | wikipedia names $$ c(x) = \sum_0^\infty \; C(n) \; x^n $$
and says that it satisfies
$$ c(x) = 1 + x \, c^2(x) $$
Define $$ A = c \left( \frac{1}{10000} \right) $$
Then we have
$$ A = 1 + \frac{1}{10000} \; A^2 $$
so
$$ A^2 - 10000A + 10000 = 0. $$ So
$$ A = \frac{10000 \pm \sqrt{100000000 - 40000}}{2} = \frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728765",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Writing integers as a product of as few elements of $\{\frac21, \frac32, \frac43, \frac54, \ldots\}$ as possible This question is inspired by question 2 of the 2018 European Girls' Mathematical Olympiad. Also posted on mathoverflow.
Any integer $x \ge 2$ can be written as a product of (not necessarily distinct) element... | We can compute an upper bound to $f(x)$ by adding $\lfloor \log_2(x)\rfloor$ to the number of $1$s in the binary expansion of $x$ and subtracting $1$. This gives $f(5)=2+2-1=3,5=\frac 54 \frac 21 \frac 21, f(13)=3+3-1=5, 13=\frac {13}{12} \frac 32(\frac21)^3,f(65)=6+2-1=7,65=\frac {65}{64}(\frac 21)^6$
To prove that w... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2732868",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "21",
"answer_count": 1,
"answer_id": 0
} |
Binomial coefficients $\binom{n}{b} \pmod{n}$, $n > b$, $n$ odd, $n,b$ positive integers, AKS test I am looking at the values of Binomial coefficients $\binom{n}{b} \pmod{n}$, where $n > b$, $n$ odd, and $n,b$ positive integers. I am planning to use the results in an implementation in python of the AKS primality test.... | As a consequence of theorem 4 in
*
*Andrew D. Loveless: A Congruence for Products of Binomial Coefficients modulo a Composite, in: INTEGERS 7 (A44 10 Jan 2007)
we have $\binom{n}{b}\equiv 0\pmod{n}$ if (but not only if) $\gcd(n,b)=1$. The other theorems and corollaries in the paper may be of interest to you as well... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2733800",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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 $$\int\frac{x\sin(x)}{(\sin(x)-x\cos(x))^2} \text{dx} = \int \frac{x... | Consider the function $f(x) = \frac{1}{(\sin x - x\cos x)}$. Once differentiation gives:
$$f'(x) = \frac{-(\cos x +x\sin x-\cos x)}{(\sin x - x\cos x)^2} = \frac{-x \sin x}{(\sin x - x\cos x)^2}$$
So your integral is:
$$\int \frac{x \sin x}{(\sin x - x\cos x)^2} dx = \frac{-1}{(\sin x - x \cos x)}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2735081",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
How does Tom Apostol deduce the inequality $\frac{k^3}{3} + k^2 \lt \frac{(k + 1)^3}{3}$ in section I 1.3 of his proof by induction example? In his book Calculus, Vol. 1, Tom Apostol writes the following proof for proving the following predicate by induction:
$$A(n): 1^2 + 2^2 + \cdots + (n - 1)^2 \lt \frac{n^3}{3}.$$
... | $A(k)$ states
$$1^2 + \cdots + (k-1)^2 < \frac{k^3}{3}.$$
Adding $k^2$ to both sides yields
$$1^2 + \cdots + k^2 < \frac{k^3}{3} + k^2.$$
If we prove $\frac{k^3}{3} + k^2 < \frac{(k+1)^3}{3}$
then we can tack this onto the end of the above inequality to get
$$1^2 + \cdots + k^2 < \frac{k^3}{3} + k^2 < \frac{(k+1)^3}{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2736318",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Prove if $x\gt3$ then $1\ge\frac{3}{x(x-2)}$. I tried to prove it by contradiction.
Suppose it is not true that $1\ge\frac{3}{x(x-2)}$, so $1\lt\frac{3}{x(x-2)}$. Then $\frac{3}{x(x-2)}-1\gt0$. Multiply both sides of $\frac{3}{x(x-2)}-1\gt0$ by ${x(x-2)}$.
$(\frac{3}{x(x-2)}-1\gt0)({x(x-2)}\gt0(x(x-2)$
${3-(x(x-... | Why do you make things more complex than they are?
The function $x(x-2)$ is increasing and positive on $(2,+\infty)$, hence
$\;\dfrac2{x(x-2)}\;$ is decreasing and positive on this interval.
Furthermore $f(3)=1$…
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2737144",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Does the series $\sum_{n=2}^{\infty} \frac {\sin(n+\frac{1}{n})}{\log\log n}$ converge?
Does the series $\sum_{n=2}^{\infty} \frac {\sin(n+\frac{1}{n})}{\log\log n}$ converge?
My attempts :
$\sin(x+y) = \sin x\cos y + \cos x \sin y$
now $\sin(n +\frac{1}{n}) = \sin n \cos \frac{1}{n} + \cos n\sin\frac{1}{n}$
now $\... | \begin{align}
\sum_{n=2}^{\infty} \frac {\sin( n + 1/n) }{\log\log n}&=\sum_{n=2}^{\infty} \frac {\sin n \cos 1/n }{\log\log n} + \sum_{n=2}^{\infty}{\frac {\cos n \sin 1/n}{\log\log n}}\\
&=\sum_{n=2}^{\infty} \sin n\,\frac {1 }{\log\log n} +\sum_{n=2}^{\infty} \frac {(\sin n)(\cos 1/n-1) }{\log\log n}+ \sum_{n=2}^{\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2738014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
How to solve the cubic $x^3-3x+1=0$? This was a multiple choice question with options being
$$(A)-\cos\frac{2\pi}{9},\cos\frac{8\pi}{9},\cos\frac{14\pi}{9} \\
(B)-2\cos\frac{2\pi}{9},2\cos\frac{8\pi}{9},2\cos\frac{14\pi}{9} \\
(C)-\cos\frac{2\pi}{11},\cos\frac{8\pi}{11},\cos\frac{14\pi}{11} \\
(D)-2\cos\frac{... | Put $x=2t$
Hence $$8t^3-6t+1=0$$
Now put $t=\cos \theta$
Hence $$2\cos (3\theta)=-1$$
And the rest is simple trigonometric equation
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2739299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
For rotation matrix $A$, find $B = A^4- A^3 + A^2 - A$
Find the value of $B = A^4- A^3 + A^2 - A$ where $A$ is the matrix given below $$ A=
\left [ \begin{matrix}
\cos\alpha & \sin \alpha \\
-\sin\alpha & \cos\alpha
\end{matrix} \right ]
$$
It's actually quite simple when you look at it. Find $A$, $... | By induction,
$$ A^n = A^{n-1}A = \begin{pmatrix} \cos{(n-1)\theta} & \sin{(n-1)\theta} \\
-\sin{(n-1)\theta} & \cos{(n-1)\theta} \end{pmatrix} \begin{pmatrix} \cos{\theta} & \sin{\theta} \\
-\sin{\theta} & \cos{\theta} \end{pmatrix} = \begin{pmatrix} \cos{(n-1)\theta}\cos{\theta} - \sin{\theta}\sin{(n-1)\theta} & \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2740300",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
if $d\mid n$ then $x^d-1\mid x^n-1$ proof How would you show that if $d\mid n$ then $x^d-1\mid x^n-1$ ?
My attempt :
$dq=n$ for some $q$. $$ 1+x+\cdots+x^{d-1}\mid 1+x+\cdots+x^{n-1} \tag 1$$ in fact, $$(1+x^d+x^{2d}+\cdots+x^{(q-1)d=n-d})\cdot(1+x+\cdots+x^{d-1}) = 1+x+x^2 + \cdots + x^{n-1}$$
By multiplying both sid... | You have:
$$d\mid n \Rightarrow n=ad$$
Then:
$$x^n-1=x^{ad}-1=(x^d)^a-1$$
Setting $x^d=y$ (just for simplifying the process) we have:
$$y^a-1=(y-1)(y^{a-1}+\dots+y+1)=(x^d-1)((x^d)^{a-1}+\dots + x^d+1)$$
In other words we showed that:
$$x^n-1= (x^d-1)((x^d)^{a-1}+\dots + x^d+1) $$
Which obviously implies that:
$$x^d-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 5
} |
Suppose $ x+y+z=0 $. Show that $ \frac{x^5+y^5+z^5}{5}=\frac{x^2+y^2+z^2}{2}\times\frac{x^3+y^3+z^3}{3} $. How to show that they are equal? All I can come up with is using symmetric polynomials to express them, or using some substitution to simplify this identity since it is symmetric and homogeneous but they are still... | If $x=0$, the equality is trivial. So, it is sufficient to consider $x\ne 0$.
Let $y=ax, z=abx$. Then:
$$x+ax+abx=0 \Rightarrow ab=-1-a.$$
Then:
$$\frac{x^5+a^5x^5+(ab)^5x^5}{5}=\frac{x^2+a^2x^2+(ab)^2x^2}{2}\cdot \frac{x^3+a^3x^3+(ab)^3x^3}{3} \Rightarrow$$
$$\frac{1+a^5+(-1-a)^5}{5}=\frac{1+a^2+(-1-a)^2}{2}\cdot \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2747626",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Evaluate $\displaystyle m\left( \bigcup_{x \in C} \left( x - \frac{1}{20}, x+\frac{1}{20} \right) \right)$ where $C$ is Cantor set.
Question: Let $C$ be the Cantor set, obtained from the unit interval $[0,1].$
For each $x \in C$, let $I_x$ be the interval $\displaystyle \left(x-\frac{1}{20}, x + \frac{1}{20}\rig... | Since the whole interval $(\frac 13,\frac 23)$ is excluded from the Cantor set the interval $(\frac 13+\frac 1{20},\frac 23-\frac 1{20})$ is excluded from your union. The measure is then less than $1.1$. You need to find all the intervals that are missing from your set, add them up, and subtract from $1.1$. There ar... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Verify that $x^2 + cy^2 = 1$ is an implicit solution to $\frac{dy}{dx} = \frac{xy}{x^2 - 1}.$ When I differentiate $x^2 + cy^2 = 1$ implicitly and solve for $\dfrac{dy}{dx}$, I get $\dfrac{dy}{dx} = \dfrac{-x}{cy}.$ I thought I had to multiply by a fraction in order to make it similar to $\dfrac{dy}{dx} = \dfrac{xy}{x^... | Notice that:
$$
x^2 + cy^2 = 1 \iff cy = \frac{1 - x^2}{y}
$$
So:
$$
\frac{dy}{dx} = \frac{-x}{cy} = \frac{-x}{\left(\frac{1 - x^2}{y} \right)} = \frac{-xy}{1 - x^2} = \frac{xy}{x^2 - 1}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2748212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that in a field $F, \exists a,b,c \in F$ such that $x^3+x^2+1$ is a divisor of $x^{2018} + ax^3 + bx + c$ I have no idea how to even start this! Maybe using the Euclidean algorithm and showing that the extra term is 0?
| Apply Euclidean algorithm to $x^{2018}$ and $x^3+x^2+1$, we can find a polynomial $q(x)$ and quadratic polynomial $\alpha x^2 + \beta x + \gamma$ such that
$$x^{2018} = q(x)(x^3+x^2+1) + \alpha x^2 + \beta x + \gamma\\
\Downarrow\\
x^{2018} + ax^3 + bx + c = q(x)(x^3+x^2+1) + \underbrace{ax^3 + \alpha x^2 + (\beta+b)x ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2749000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
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.