Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Ring theory - cosets problem; prove specific result Let $F=\mathbb{Z}_2[x]/I$ where $f(x)=x^3+x^2+[1]\in\mathbb{Z}_2[x]$ and $I=(x^3+x^2+[1])\mathbb{Z}_2[x]$. With $X=x+I\in F$ and writing $a$ for $a+I$ with $a\in\mathbb{Z}_2$ show that the following hold:
(i) $X^3=X^2+[1]$
(ii) $X^4=X^2+X+[1]$
The way I've been taught... | You could use the congruence notation, in the same way as you would write $13 \equiv 3 \bmod 5$. In your case, your question amounts to show that
(i) $X^3 \equiv X^2+1 \bmod X^3 + X^2 + 1$
(ii) $X^4 \equiv X^2+X+1 \bmod X^3 + X^2 + 1$
You already observed that, in your ring of characteristic $2$, the conditions $g = h$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4418605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Showing $\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$ for $x,y,z>0$ and $xyz=1$
Let $x,y,z>0$ with $ xyz=1$ then prove that,
$$\frac{xy+yz+xz}{x+y+z}>5-\sqrt{4(x^2+y^2+z^2)+6}$$
Let $$5≤\sqrt{4(x^2+y^2+z^2)+6}\implies x^2+y^2+z^2≥\frac {25-6}{4}=\frac {19}{4}$$ then the inequality is always true.
This also shows... | In fact, we can prove a stronger inequality
$$
\frac{xy+yz+zx}{x+y+z}+\sqrt{4(x^2+y^2+z^2)+6}\geq 1+3\sqrt{2} >5
$$
for all $x,y,z>0$ with $xyz=1$.
To prove the above inequality, we use the Cauchy-Schwarz Inequality to deduce that
$$
\left( 1+1+1+\frac{3}{2} \right)\left( 4x^2+4y^2+4z^2 +6 \right)\geq \left( 2x+2y+2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4423904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$
Prove the inequality $9(a+b)(b+c)(c+a) \geq 8(a+b+c)(ab+bc+ca)$ for $a, b, c \in \mathbb{R_{>0}}$
I tried by first using AM-HM inequality on $a, b, c$ to get the following result.
$\frac{a+b+c}3 \geq \frac 3{\frac 1a+\frac1b+\frac1c}$
$\implies (a+b+c)(\... | $$9(a+b)(b+c)(c+a) - 8(a+b+c)(ab+bc+ca)$$
$$ = (a+b)(b+c)(c+a) - 8abc$$
$$\ge0$$
because $(a+b)(b+c)(c+a) = (a+b+c)(ab+bc+ca)-abc$
and $(a+b)(b+c)(c+a) \ge 8abc$ by AM-GM: $(a+b)(b+c)(c+a)=abc+\dots+bca\ge 8\sqrt[8]{a^8b^8c^8}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4427378",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Demonstration of an imaginary ellipse I'm studying the classfication of non-degenerate conics based on the determinant of the matrix associated to it $A_Q$.
I didn't understand, in the case of a non-degenerate ellipse (det $A_{33}>0$, minor of $A_Q$, and det $A_Q \neq 0$) why we have an imaginary ellipse if $(A+C)detA_... | When $D=b^2-4ac\neq 0$, you can translate the center $(\frac{2cd-be}{b^2-4ac},-\frac{bd-2ae}{b^2-4ac})$ to the origin without changing the form of the conic:
$$ax^2+bxy+cy^2+f+\frac{-ae^2+bde-cd^2}{4ac-b^2}=0\tag1$$ or
$$ax^2+bxy+cy^2+\frac{\begin{vmatrix}a&b/2&d/2\\b/2&c&e/2\\d/2&e/2&f\end{vmatrix}}{\begin{vmatrix}a&b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4431864",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$ 2 \cos ^{2021} x+\sin ^{2022} x=2 $ Someone sent me this task, it is meant to be solved quickly, nothing to think too much about.
At first I tried representing $\sin ^{2022}x$ as $(\sin ^{2}x)^{1011}$
$$
2 \cdot \frac{\left(\cos ^{1011} x\right)^{2}}{\cos (x)}+\left(1-\cos ^{2} x\right)^{1011}=2.
$$
Then using using ... | One elegant solution has been given in the comments. Here is an alternative one, using the Cauchy-Schwarz inequality. For any positive integer $n$ is
$$
\begin{align}
f(x) &= 2 \cos^{2n+1}(x)+\sin^{2n+2}(x) \\
&= \cos^{2n}(x) \cdot 2 \cos(x) + \sin^{2n}(x) \cdot \sin^2(x) \\
&\le \sqrt{\cos^{4n}(x) + \sin^{4n}(x)} \cd... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4433498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Conjecture: Every $n \geq 20 \in \mathbb{N}$ can be written as a sum of three integers $(\geq 2)$ that are pairwise coprime This question on the sum of pairwise prime numbers piqued my interest, and I started looking at what numbers can be written as the sum of three pairwise coprime numbers (excluding $1$):
$$
\begin{... | Assume that $n$ is not the power of sum prime, i.e $n$ cannot be written as $n=p^{\alpha}$ for some prime $p$ and exponent $\alpha$. In this case, we can find two distinct primes $p,q$ that both divide $n$. Clearly $p+q<pq<n$ and so $n-(p+q)$ is a positive integer. This means that we can write
$$n=p+q+(n-(p+q)).$$
The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4434707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solution of the recurrence relation $y_n = \frac{1}{2} + \frac{1}{2}y_{n-1}$ $y_0 = 0$ and $y_n = \frac{1}{2} + \frac{1}{2}y_{n-1}$. Solution of this reccurent equation is $y_n = 1 - \frac{1}{2^n}$, accordingly with the software. But I do not understand the minus sign since it would be $y_n = \frac{1}{2} + \frac{1}{2}\... | Repeat the formula $n$ times: $y_n = \dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\cdots + \dfrac{1}{2^{n-1}} + \dfrac{1}{2^n}+\dfrac{1}{2^n}y_0= 1 -\dfrac{1}{2^n}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4437950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How many squares of each colour are in a generalized checkerboard $C$-coloured $m \times n$ rectangle?
How many squares of each colour are in a generalized checkerboard $C$-coloured $m \times n$ rectangle?
Assume an $m\times n$ rectangle has been been divided into a grid of $mn$ unit squares, and the squares have bee... | Let $a = m \bmod C$ and $b = n \bmod C$ i.e. the remainders of $m,n$ divided by $C$.
As noted in the comments above, as long as we have a square or a rectangle with at least one side divisible by $C$, all the colors will be present in that square or rectangle and their occurrences will be all equal.
We note that we can... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the sum: $\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+2)\ n!}$
How to evaluate the below sum?
$$\sum_{n=0}^{\infty}\frac{x^{n+2}}{(n+2)\ n!}$$
I was trying to find the integral of $x\ e^x$ without using Integral By Parts.
Here's what I got so far:
$$\begin{aligned}e^x &= \sum_{n=0}^\infty \dfrac{x^n}{n!}\\x\ e^x &... | You have\begin{align}\frac1{(n+1)!}-\frac1{(n+2)!}&=\frac{n+2-1}{(n+2)!}\\&=\frac{\cancel{n+1}}{(n+2)\cancel{(n+1)}n!}\require{cancel}\\&=\frac1{(n+2)n!}\end{align}
and therefore you have\begin{align}\sum_{n=0}^\infty\frac{x^{n+2}}{(n+2)n!}&=\sum_{n=0}^\infty\left(\frac1{(n+1)!}-\frac1{(n+2)!}\right)x^{n+2}\\&=\sum_{n=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4439878",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
The solutions of the equation $ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $ are? I tried this:
$$ \sin{x} + \sin{3x} = \frac{8}{3\sqrt{3}} $$
$$ 2\sin{2x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ 4\sin{x}\cos{x}\cos{x} = \frac{8}{3\sqrt{3}} $$
$$ \sin{x}(1-\sin^2{x}) = \frac{2}{3\sqrt{3}} $$
Here, I tried to set $\sin x = t$
... | HINT
To begin with, notice that $\sin(3x) = 3\sin(x) - 4\sin^{3}(x)$. Hence it results that:
\begin{align*}
\sin(x) + \sin(3x) = \frac{8}{3\sqrt{3}} & \Longleftrightarrow 4\sin(x) - 4\sin^{3}(x) = \frac{8}{3\sqrt{3}}\\\\
& \Longleftrightarrow \sin(x) - \sin^{3}(x) = \frac{2\sqrt{3}}{9}
\end{align*}
By inspection, one c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4440902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Finding Laurent Series Expansion for $f(z) = \frac{z^3}{z^2-3z+2}$ for different domains So I wanted to find the Laurent series expansion for $f(z) = \frac{z^3}{z^2-3z+2}$. So I guess more specifically, my question pertains how the expansion may change for different domains: $|z| < 1; 1 < |z| < 2; |z| > 2; 0 < |z − 1| ... | If you want to find the Laurent expansion of $\frac{1}{z-b}$ in an annulus centred at $a$ with some given radii, you may rewrite the fraction as
$$
\frac{1}{z-b}=\begin{cases}
\frac{1}{a-b}\frac{1}{1-w}
&\text{when $b\ne a$ and $|z-a|<|b-a|$, where $w=\frac{z-a}{b-a}$,}\\
\frac{1}{b-a}\frac{w}{1-w}
&\text{when $b\ne a$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4441782",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $\binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m$ for $m\in \mathbb{N}\backslash \{0,1\}$ Prove the following inequality for every $m\in \mathbb{N}\backslash \{0,1\}$:
$$ \binom{13+m}{m}-(m+1)\binom{6+m}{m}\geq m.$$
By some computational arguments, the inequality seems to be true and in particular the difference b... | Using the identity (well-known, easy to prove)
$$\binom{n + 1}{k} = \binom{n}{k} + \binom{n}{k - 1}, $$
we have
\begin{align*}
\binom{13 + m}{m} &= \binom{12 + m}{m} + \binom{12 + m}{m - 1}\\
&= \binom{11 + m}{m} + \binom{11 + m}{m - 1} + \binom{11 + m}{m - 1} + \binom{11 + m}{m - 2}\\
&\ge \binom{11 + m}{m} + 2\bin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4449254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Prove that $c^3 + a^3 − 2abc = 6b − 11$ Let $a, b, c$ be real numbers such that $(a + b + c)^2 = 3(ab + bc + ca + 1).$ Given,
$$a^3 + b^3 − 2abc = 6c + 2,$$
$$b^3 + c^3 − 2abc = 6a + 9,$$
Prove that $$c^3 + a^3 − 2abc = 6b − 11$$
Note that, subtracting the two equations, we get $$a^3-c^3=6(c-a)+7. $$
Moreover, we have ... | Let $s=a+b+c$ and $t=ab+bc+ca$, so that $s^2=3t+3$ is the first given equation. Note that
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=s(s^2-3t)=3s.$$
So,
$$(a^3+b^3-2abc)+(b^3+c^3-2abc)+(c^3+a^3-2abc)=6s.$$
Now simply subtract the other two given equations.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4450018",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Computing Correlation cofficient from joint pdf X and Y have the joint pdf:
$f(x,y)= \frac{2}{x} e^{-2x}$ for $0<x<\infty , 0<y<x$
I'm using the formula: $E(h(x,y))=\int_{0}^{\infty} \int_{0}^{x} h(x,y)\frac{2}{x} e^{-2x} dydx $ . And I'm simply setting $x$,$x^2$, etc. as h(x,y).
Based on this, I got:
$E(XY)=\frac{1}{4... | Here is a useful trick. Instead of individually computing each moment, consider the generalized expression for nonnegative integers $a, b$
$$\begin{align}
\operatorname{E}[X^a Y^b] &= \int_{x=0}^\infty \int_{y=0}^x x^a y^b \frac{2}{x} e^{-2x} \, dy \, dx \\
&= \int_{x=0}^\infty 2x^{a-1} e^{-2x} \left[ \frac{y^{b+1}}{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4451416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve the PDE $(xz-y)p+(yz-x)q=xy-z$ using lagrange method I have the following PDE,
$$(xz-y)p+(yz-x)q=xy-z$$where $p=z_x,\quad q=z_y$
Now having a hard time to get two solution from,
$$\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{xy-z}$$
I can't think of any multipliers trick which can help me to get one. In fact, $dx-dy... | This is not a complete answer rather expanding the @ShengtongZhang's one as the OP commented there,
characteristic equation of that PDE,
$$
\begin{align}
&\frac{dx}{xz-y}=\frac{dy}{yz-x}=\frac{dz}{xy-z}=dt\\
&x'=xz-y,\quad y'=yz-x, \quad z'=xy-z
\end{align}
$$
Now, inspecting the $x'y-y'x$ term,
$$
\begin{align}
x'y-y'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4454310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Is it possible to compute $-\log\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right)$ without a calculator? The following question is part of a chemistry problem that came in the Dhaka University admission exam 2013-14.
What is $-\log\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right)$?
(a) 2.672
(b) 2.772
(c) 2.872... | Working with whole numbers
$$1.8\times 10^{-5}\times 0.1= \frac 9 5 \times 10^{-6}$$
$$\sqrt{1.8\times 10^{-5}\times 0.1}=\frac{3}{ \sqrt{5}}\times 10^{-3}$$
$$\frac{3}{ \sqrt{5}}=\frac{3}{ \sqrt{4+1}}=\frac 3 2 \times \frac 1{\sqrt{1+\frac 14}}\sim\frac 3 2 \times \left(1-\frac 18\right)=\frac{21}{16}=1+\frac 5 {16}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4455969",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 2
} |
How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ How to show that $\frac{\pi}{2} \le \sum_{n=0}^\infty \frac{1}{n^2+1} \le \frac{3\pi}{4}$ ?
My Attempt : I was using Integral Test of a Series. I got $\int_0^\infty \frac{1}{1+x^2} \le \sum_{n=0}^\infty \frac{1}{1+n^2} \le \f... | This can be computed using the result of this answer:
$$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+1}
&=-\frac1{2i}\sum_{n=0}^\infty\left(\frac1{i-n}+\frac1{i+n}\right)\\
&=\frac12-\frac1{2i}\sum_{n\in\mathbb{Z}}\frac1{i+n}\\
&=\frac12-\frac1{2i}\pi\cot(\pi i)\\[3pt]
&=\frac12+\frac\pi2\coth(\pi)\\[9pt]
&\approx2.07667... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4456882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to find the closed form of $\int_{-\infty}^{\infty} \frac{x^{2 n+1} \sin x}{\left(1+x^{2}\right)^{n+1}} d x, \textrm{ where }n=0,1,2,3,…?$ In my post, I had found the exact value of the integral
$$\displaystyle I:=\int_{-\infty}^{\infty} \frac{x^{3} \sin x}{\left(1+x^{2}\right)^{2}}dx= \frac{\pi}{2 e}\tag*{} $$
by ... | This not an answer.
Computing
$$I_{n}=\left.\frac{(-1)^{n}}{n !} J^{(n)}(1) = \pi \frac{d^{n}}{d a^{n}}\left(a^{-1} e^{-\frac{1}{\sqrt{a}}}\right)\right|_{a=1}$$
$${I_n}= \frac \pi{2^n \,e }\, b_n$$ and the $b_n$ generate the sequence
$$\{-1,2,-5,5,176,-3881,70337,-1285390,24806665,-512336809,\cdots\}$$ which is not id... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4459778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Algebraic Manipulation involving square roots I have the following expressions:
$$E(g) := \frac{Ug^2 - 4tg}{1+g^2} \hspace{2cm} (1)$$
$$ g_0 := \frac{1}{4t} \left[ -U + \sqrt{U^2 + (4t)^2} \right] \hspace{2cm} (2) $$
By plugging (2) in (1) I should arrive at:
$$ E(g_0) = \frac{1}{2} \left( U - \sqrt{U^2 + (4t)^2} \ri... | It follows from $\,(2)\,$ that $\require{cancel}\,(4 t g_0+U)^2 = U^2 + 16t^2\,$, or $\,16t^2g_0^2 + 8 Utg_0 + \cancel{U^2} = \cancel{U^2} + 16t^2\,$, and assuming $\,t \ne 0\,$ it follows that $\,2tg_0^2 + Ug_0-2 t = 0 \iff \color{blue}{2tg_0^2} = -Ug_0+2t\,$. Then:
$$
\begin{align}
E(g_0) &= \frac{Ug_0^2 - 4tg_0}{1+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4466905",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Calculate double integral $\iint_D\frac{\sqrt{x^2+y^2 - a^2}x}{(x^2+y^2)^2}dxdy$ over unbounded region $D$ Calculate
$$I = \iint\limits_D\frac{\sqrt{x^2+y^2 - a^2}x}{(x^2+y^2)^2}\,dx\,dy$$
where the region of integration is:
$$D = \{(x,y) \in \mathbb{R}^2 \mid x+y - a\sqrt{2} \geq 0, -x+y +a\sqrt{2} \geq 0, a > 0\}$$
I... | With the variable changes $x=au$ and $y=av$
\begin{align}
&\iint_D\frac{\sqrt{x^2+y^2 - a^2}x}{(x^2+y^2)^2}\,dx\,dy\\
=&\int_0^\infty \int_{-v+\sqrt2}^{v+\sqrt2} \frac{\sqrt{u^2+v^2 - 1}\ u}{(u^2+v^2)^2}\,du\,dv\\
=& \int_0^\infty \frac12\bigg(\tan^{-1}{\sqrt{u^2+v^2 - 1}}-\frac{\sqrt{u^2+v^2 - 1}}{u^2+v^2}\bigg)\bigg|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluate the line integral along a parabola
Evaluate the line integral of $f(x,y)=-y+x$ along part of the parabola $y=2(x+1)^2$ from the point $(0,2)$ to the point $(-1,0)$
I need help trying to find a good parameterization for this because what I've done just lands me in a mess.
My work so far:
Let $x=t, y=2(t+1)^2$... | Using the substitution $4(x+1)=y$ on the latter integrand, we get:
\begin{align}\frac{1}{16}\int_{0}^{1\over4}(\frac{y^2}{2}-y-4)\sqrt{y^2+1}\,dy\end{align}
The integral$\int_{0}^{1\over4}(y+4)\sqrt{y^2+1}\,dy$ is quite straightforwardly evaluated.
For the $4\int_{0}^{1\over4}\sqrt{y^2+1}\,dy$, substituting $y=\sinh u$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 1
} |
Calculate a triple integral where the integrand contains cubic terms Calculate
$$I = \iiint\limits_Q(x^3 + y^3 + z^3)\,dx\,dy\,dz$$
where $Q: x^2 + y^2 + z^2 - 2ax-2ay-2az + 2a^2 = 0$ is a sphere
If I try using spherical coordinates I don't go anywhere. Any hint about
how to calculate it?
| Due to symmetry
$$I = \iiint_Q(x^3 + y^3 + z^3)\,dV
=3 \iiint_Q z^3\,dV$$
Then, recenter the sphere at origin and integrate in spherical coordinates
\begin{align}
I =& \ 3 \iiint_{x^2+y^2+z^2<a^2} (z+a)^3\,dV
=\ 3 \iiint_{x^2+y^2+z^2<a^2} (3az^2 +a^3)\,dV\\
=& \ 3a \iiint_{r<a} r^2 \ r^2\sin\phi dr d\theta d\phi + 3a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4467916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\tan\frac{\alpha}{2},$ if $\sin\alpha+\cos\alpha=\frac{\sqrt7}{2}$ and $\alpha\in(0^\circ;90^\circ)$. Find $\tan\dfrac{\alpha}{2},$ if $\sin\alpha+\cos\alpha=\dfrac{\sqrt7}{2}$ and $\alpha\in(0^\circ;90^\circ)$.
So we have $$\sin\alpha+\cos\alpha=\dfrac{\sqrt7}{2}$$ Then $$\sin^2\alpha+\cos^2\alpha+2\sin\alpha\co... | Let $\beta=\alpha +45^o.$ Then $\sin \beta=\sin \alpha \cos 45^o+\cos \alpha \sin 45^o=(1/\sqrt 2)(\sin \alpha +\cos \alpha)=(\sqrt 7)/(2\sqrt 2).$ And so $\cos \beta=\pm \sqrt {1-\sin^2\beta}=\pm \sqrt {1-7/8}=\pm 1/(2\sqrt 2).$
Now $\cos \alpha =\cos (\beta -45^o)=\cos \beta \cos 45^o+\sin \beta \cos 45^o=(\cos \beta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4468299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $x^2-3-\operatorname{frac}(x)=0$
Problem. Solve the equation
$$ x^2 - 3 - \operatorname{frac}(x) = 0 $$
I tried to solve this question by moving $3$ to the other side:
$$x^2 - \operatorname{frac}(x) = 3$$
Since the fractional part of R.H.S. is $0$,
$$
\operatorname{frac}\left(x^2\right) = \operatorname{frac}(... | We have
$$
0 = x^2 - 3 - {\rm frac}(x) = x^2 - 3 - \left\{ x \right\}
$$
where
$$
x = \left\lfloor x \right\rfloor + \left\{ x \right\}\quad \left| {\,0 \le \left\{ x \right\} < 1} \right.
$$
So we can put first of all
$$
\eqalign{
& 3 \le x^2 = 3 + \left\{ x \right\} < 4\quad \Rightarrow \quad \sqrt 3 \le \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4469822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Compute $f^{(2020)}(0)$ Problem :
Let
$$f(x)=\frac{x}{(x+1)(1-x^2)}$$
Then find $f^{(2020)}(0)$.
My Attempt :
From partial fraction decomposition,
$$f(x)=\frac{1}{4(1-x)}+\frac{1}{4(x+1)}-\frac{1}{2(x+1)^2}$$
and,
$$\frac{1}{1-x}=\sum_{n=0}^\infty x^n,\quad \frac{1}{1+x}=\sum_{n=0}^\infty (-1)^nx^n, \quad\frac{1}{(1+x... | Since the other responses have dissected the OP's (i.e. original poster's) work, I can give an alternative approach.
$\underline{\text{Preliminary Results}}$
$\displaystyle f(u) = u^{-1} \implies f^{(2020)}(u) = [(2020)!] \times u^{-2021}.$
$\displaystyle f(u) = u^{-2} \implies f^{(2020)}(u) = [(-2) \times (-3) \times ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4471123",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Slight variation of Mertens' third theorem. Do we have an estimate of $\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}$? Define $f(n)$ to be:
$$
\sum_{d \mid n\#}(-1)^{\omega(d)}\dfrac{\sigma_0(d)}{d}
$$
But $\sigma_0(d) = 2^{\omega(d)}$ for any $d \mid n\#$ a primorial, so:
$$
f(n) = \prod_{p \text{ prime} \\ ... | We have
\begin{align*}
\prod\limits_{2<p \le n} {\left( {1 - \frac{2}{p}} \right)^{ - 1} } & = \exp \left( { - \sum\limits_{2<p \le n} {\log \left( {1 - \frac{2}{p}} \right)} } \right) \\ & = \exp \left( {\sum\limits_{2<p \le n} {\frac{2}{p}} - \sum\limits_{2<p \le n} {\left[ {\frac{2}{p} + \log \left( {1 - \frac{2}{p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472059",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Finding the minimum value of $(x_1-x_2)^2 +\left(\frac{9}{x_1} - \sqrt{1-x_2^2}\right)^2$ by using trigonometry
Find the minimum value of the given function by using trigonometry
$$(x_1-x_2)^2 + \left(\dfrac{9}{x_1} - \sqrt{1-x_2^2}\right)^2$$
I know the distance formula method but is there any other suitable method ... | Expand as usual, and using Cauchy-Schwarz inequality: $f(x_1,x_2) = x_1^2 - 2x_1x_2 + \dfrac{81}{x_1^2} - \dfrac{18\sqrt{1-x_2^2}}{x_1}+1 = 1+x_1^2+\dfrac{81}{x_1^2} - 2\left(x_1x_2+\dfrac{9\sqrt{1-x_2^2}}{x_1}\right)\ge 1+x_1^2+\dfrac{81}{x_1^2} - 2\sqrt{x_1^2+\dfrac{81}{x_1^2}}\sqrt{x_2^2+(1-x_2^2)} = 1 + y^2- 2y = g... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4472906",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Let $a$ and $b$ be roots of $x^2-7x+2$. Find the value of $a^6 + b^6$.
Answer:
$a+b = 7, ab = 2$
$$\begin{align}
(a+b)^6 &= a^6 + 6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5+b^6 \\[4pt]
a^6 + b^6 &= (a+b)^6 - (6a^5b+15a^4b^2+20a^3b^3+15a^2b^4+6ab^5) \... | A better way to handle this is to observe the factorization
$$a^6 + b^6 = (a^2 + b^2)(a^4 - a^2 b^2 + b^4),$$ from which we observe
$$a^2 + b^2 = (a+b)^2 - 2ab,$$ and
$$a^4 - a^2 b^2 + b^4 = (a^2 + b^2)^2 - 3a^2 b^2.$$
Since we know that $$x^2 - 7x + 2 = (x - a)(x - b) = x^2 - (a+b)x + ab,$$ it follows that $a+b = 7$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473560",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 9,
"answer_id": 3
} |
Evaluate $\sum _{n=0}^{\infty } \frac{16^n}{(2n+1)^3 \binom{2 n}{n}^2}$ In this post give the result of $$\sum _{n=1}^{\infty } \frac{16^n}{n^3 \binom{2 n}{n}^2}=8\pi\text{G}-14 \zeta (3)$$
Then i wonder if i change $n^3$ to $(2n+1)^3$ so what is the result. I check with CAS give the result: $$\sum _{n=0}^{\infty } \f... | Other than @Frank W's elegant solution, here is a different approach.
It is well-known that:
$\displaystyle \tag*{}\frac {\arcsin x}{\sqrt{1-x^2}}=\sum\limits_{n\ge 0}\frac {4^nx^{2n+1}}{(2n+1)\binom {2n}n}$
Dividing by $x$ and doing $\int_0^x \,dx$ on both sides gives:
$\displaystyle \tag{1}\int_0^x\frac {\arcsin x}{x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4473923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Given $0\leq a\leq b\leq c$ and $b > 0$, prove that: $2\sqrt{\frac{a}{c}}\leq \frac{b}{c} + \frac{a}{b}\leq 1 + \frac{a}{c}$ This is an exercise that got me stuck. It's from the book "Functions of Several Real Variables 1.1 - exercise 10" by Martin Moskowitz and Fotios Paliogiannis.
Exercise:
Given $0\leq a\leq b\leq c... | Let $a=x^2, b=y^2, c=z^2.$ For fixed $b,c$ let $f(x)=(b/c)+(a/b)-2\sqrt {a/c}\,=(y^2/z^2)+(x^2/y^2)-2(x/z)$ and let $g(a)=1+(a/c)-(b/c)-(a/b).$
Show that $f'(x)>0\iff x>y^2/z.$ So $\min f(x)=f(y^2/z)=0.$
Show that $g'(a)\le 0.$ So $\min \{g(a):a\le b\}=g(b)=0.$
The substitution $a=x^2, b=y^2, c=z^2$ is not really neces... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4475725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Laurent series on different areas? Looking for the Laurent series of
$f(z)=\frac{1}{z(z-1)(3i-z)}=\frac{i}{3 z} + \frac{\frac{1}{10} - \frac{i}{30}}{z - 3 i} - \frac{\frac{1}{10} + \frac{3i}{10}}{z - 1}$
So we look for $z_0=0,z_1=1,z_2=3i$ on
$A=\{z \in \Bbb C: |z| <1\}$
$B=\{z \in \Bbb C: 1< |z| <3\}$
$C=\{z \in \Bbb ... | Hint: for $|z|<1$ we have $$\frac{1}{1-z} = \sum_{n=0}^\infty z^n$$ but for $|z|>1$ we have $$\frac{1}{1-z} =-\frac{1}{z} \frac{1}{1-\frac{1}{z}} = -\sum_{n=1}^\infty\frac{1}{z^n}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476308",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
What will be the value of $ 2\cos^{2}\theta - 1 $ , if $ \cos^{4}\theta - \sin^{4}\theta = \frac{2}{3}$ The question was What will be the value of $ 2\cos^{2}\theta - 1 $, if $ \cos^{4}\theta - \sin^{4}\theta = \frac{2}{3}$
Here is my working:
$\cos^{4}\theta - \sin^{4}\theta = \frac{2}{3} $
$(\cos^{2}\theta)^{2}-(\sin... | I finally got the answer
$\cos2\theta = 2\cos^{2}\theta - 1$
$2\cos^{2}\theta - 1 = \frac{2}{3} $
This is our required answer
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4476844",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Integrating $\int { \arctan{t} \over \sqrt{t} } $ Integrate: $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt$$
My attempt:
I first substituted $y = \sqrt{t}$
then $$\int { \arctan{t} \over \sqrt{t} }\,\mathrm dt = 2\int { \arctan{y^2}}\,\mathrm dy$$
then I integrated by parts:
$u = \arctan{y^2}$, $v' = 1$
$$ = 2\left(... | Elaborating on the comment, the key is to note that the denominator can indeed be further factored
$$\begin{align*}y^4+1 & =y^4+2y^2+1-2y^2\\ & =\left(y^2+1\right)^2-2y^2\\ & =\left(y^2+y\sqrt2+1\right)\left(y^2-y\sqrt2+1\right)\end{align*}$$
Therefore, we can use standard partial fraction decomposition to arrive at
$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4479786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}$
If $a+b+c=3$ ,$ a,b,c>0$ and $\lambda \geq 1$,
prove that : $$\frac{a}{a^2+\lambda}+\frac{b}{b^2+\lambda}+\frac{c}{c^2+\lambda} \leq \frac{3}{\lambda +1}.$$
my attempt:
using CBC twice and AM_AG inequality
$\fr... | A comment on the tangent line method. Why does it work?
The function $f$ is not concave everywhere. What's going on?
The function $f_{\lambda}=f\colon x\mapsto \frac{x}{x^2 + \lambda}$
has derivatives
$$f'(x) = \frac{\lambda - x}{(\lambda + x^2)^2} \\
f^{''}(x) = \frac{2x(x^2 -3 \lambda) }{(x^2 + \lambda)^3}$$
Therefor... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4480984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Closed form of $\sum _{k\ge 1} \frac{(-1)^{\binom{k}{p}}}{k}$, an alternating harmonic series with the signs determined by a binomial coeffcient In a comment to Evaluating $\int_{0}^{1} \lim_{n \rightarrow \infty} \sum_{k=1}^{4n-2}(-1)^\frac{k^2+k+2}{2} x^{2k-1} dx$ for $n \in \mathbb{N}$ I proposed to study this alter... | Here, I will examine Statement 2 and 3.
Step 1. To determine the parity of $\binom{k}{2^q}$, we consider its generating function in the ring $\mathbb{F}_2[\![x]\!]$ of formal power series with coefficients in $\mathbb{F}_{2}$:
\begin{align*}
\sum_{k=0}^{\infty} \binom{k}{2^q} x^k
&= \frac{x^{2^q}}{(1-x)^{2^q + 1}}
= \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483345",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Trying to solve $|2x-15| = -x^2 - 5x -8$ My first instinct was to take the positive and negative of the right hand side, resulting in
$2x-15 = -x^2 - 5x - 8$, and $2x-15 = x^2 + 5x + 8$, which results in the first giving me two real answers using the quadratic equation, and the second being two imaginary solutions. The... | For complex numbers, the expression $ \ |2x - 15| \ $ is interpreted as the modulus of the number(s) in brackets. If we use $ \ z \ = \ a + bi \ \ , \ \ a \ , \ b \ $ real, the given equation becomes
$$ \sqrt{(2a - 15)^2 \ + \ (2b)^2 } \ \ = \ \ -(a + bi)^2 \ - \ 5(a + bi) \ - \ 8 $$
$$ \Rightarrow \ \ \sqrt{4a^2 - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4483922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Better upper bound for $ \sum_{k=2}^{\infty} \frac{1}{2^{k-1}} \sum_{n=1}^{\infty} (n! \: \text{mod} \: k) $ Since the sum $\sum_{n=1}^{\infty} (n! \: \text{mod} \: k)$ will be zero beyond $k-1$, the series could be interpreted as an finite sum of length $k-1$. Also, the max value of $n! \: \text{mod} \: k$ is naturall... | Calculate more terms we could reach more accurate bound.
Since $\sum_{k=n}^{\infty}\frac{(k-1)^2}{2^{k-1}}=
\frac{(n-1)^2}{2^{n-4}}-\frac{2n^2-6n+3}{2^{n-3}}+\frac{(n-2)^2}{2^{n-2}}$
So we have the sum is between $\sum_{k=2}^{h-1}\frac1{2^{k-1}}\sum_{n=1}^{\infty}n!\pmod k$ and $\left(\sum_{k=2}^{h-1}\frac1{2^{k-1}}\su... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4484476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Derivative of $\tan^{-1} \sqrt{\frac{1-\cos x}{1-\sin x}}$ Is there any way to differentiate $\tan^{-1} \sqrt{\frac{1-\cos x}{1-\sin x}}$ without any messy calculations? Here are my thoughts:
We write $\cos x$ and $\sin x$ in terms of $\tan{\frac{x}{2}}$. Then our function becomes $\tan^{-1} \frac{\sqrt{2}\tan \frac{x}... | $\newcommand{\d}{\mathrm{d}}$Using some algebra and trigonometric identities, we can save ourselves a lot of work: $$\frac{1-\cos x}{1-\sin x}=\frac{(1-\cos x)(1+\sin x)}{\cos^2 x}=(\sec x-1)(\tan x+\sec x)$$
Using the chain rule: $$\frac{\d}{\d x}\arctan\left(\sqrt{\frac{1-\cos x}{1+\sin x}}\right)=\frac{1}{2}\cdot\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4485625",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Compute $\int_0^1 \frac{\sqrt{t(1-t)}}{a+(t-b)^2} \ dt$ for $a,b>0$ I would like to compute the integral
$$\int_0^1 \frac{\sqrt{t(1-t)}}{a+(t-b)^2} \ dt$$
where $a$ and $b>0$ are positive parameters.
Wolfram Alpha is able to provide an answer for the indefinite integral, but is struggeling with computing the definite o... | Hint Here's a sketch of one method: The substitution $$t = \frac{u^2}{(1 + u^2)} , \qquad dt = \frac{2 u \,du}{(1 + u^2)^2}$$ transforms the original integral in $t$ to the improper integral
$$2 \int_0^\infty \frac{u^2 \,du}{(1 + u^2) (A u^4 + B u^2 + C)},$$
where
$$A := a + (b - 1)^2, \quad B := 2 [a + b(b - 1)], \qua... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4487327",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
Inverse Triple Laplace Transform of $\frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}}$ I want to find the inverse triple Laplace transform of $L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right]$. I did
\begin{align*}
L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^... | I tried another way to solve the problem by using the Taylor series of three variables and I am wondering whether it is correct :
\begin{align*}
L^{-1}_{x_{3}} L^{-1}_{x_{2}} L^{-1}_{x_{1}} \left[ \frac{-1}{s^2_{1} + s^2_{2} + s^2_{3}} \right] &= L^{-1}_{x_{3}} \left[ L^{-1}_{x_{2}} \left[L^{-1}_{x_{1}} \left[ \frac{-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4488585",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The number of binary strings of length $n$ without the sequence (1;0;1) in it. First i know that a similat question had been answered here: The number of binary strings of length $n$ with no three consecutive ones But i am really a numb on this subject so i want to be sure to understand it correctly.
But my question is... | Hint: First I'd like to point at the somewhat problematic part of OPs conclusion. Let's have a look at
\begin{align*}
s_{n}&=s_{n-1}0+s_{n-1, \text{valid}}1\\
&=s_{n-1}0+\color{blue}{s_{n-2}11}+s_{n-2, \text{valid}}01\\
&=s_{n-1}0+s_{n-2}11+s_{n-3}001\\
\\
&s_{n}\stackrel{?}{=}s_{n-1}+s_{n-2}+s_{n-3}
\end{align*}
The ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4490638",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
$\epsilon-N$ for $\lim\limits_{n \to \infty} \sqrt{n^{2} +3n-3} -n = \frac{3}{2}$ First, I tried to use the triangle inequality only once to find an N:
$$
\left | \sqrt{n^2+3n-3}-n-\frac{3}{2} \right | \leqslant \left | \sqrt{n^2+3n-3}-n \right | + \left | \frac{3}{2} \right | = \epsilon
$$
$$
N=\left \lfloor \frac{(... | You have
$$\sqrt{n^2+3n-3}-n = \frac{ (\sqrt{n^2+3n-3}-n)(\sqrt{n^2+3n-3}+n)}{\sqrt{n^2+3n-3}+n}$$
$$= \frac{3n-3}{\sqrt{n^2+3n-3}+n}.$$
Divide top an bottom by $n$ to get
$$\frac{3-\frac{3}{n}}{\sqrt{1+\frac{3}{n} - \frac{3}{n^2}}+1} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4491540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Profit maximization with Markov Chains
Problem : An opera singer is due to perform a long series of concerts. Having a bad temper, they are liable to pull out each night with probability $1/2$ . Once this has happened they will not sing again until the promoter convinces them of the promoter’s high regard. This the pr... | Suppose the opera singer has $n$ performance days. Let $X_i = \begin{cases} 1 & \text{if performance occurs on day $i$}\\ 0 & \text{else} \end{cases}$ for $1 \leq i \leq n$.
Then the profit is $M = \left (\frac{3}{4} + x \right )\left (X_1 + \dots + X_n\right ) - nx$ (in thousands of pounds).
We have transition matrix
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4492054",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
$\frac{dx^5}{dx^2},$ the derivative of $x^5$ with respect to $x^2$
Let $f(x)=x^5.$ Find its derivative with respect to $x^2$ , i.e., find $\frac{dx^5}{dx^2}.$
We know that $\frac{dx^5}{dx}=5x^4.$
But what should I do when it is needed to take derivative with respect to $x^2 ,x^3$ or other weird things different from ... | In your second approach you should substitute as follows:
(note:in my solution $f(x) \neq x^5$)
$\frac{dx^5}{dx^2}=\frac{d((x^2)^{\frac{5}{2}})}{dx^2}$
So $f(x^2)=(x^2)^{\frac{5}{2}}$ which implies $f(x)=x^{\frac{5}{2}} $
\begin{align*} \lim_{h\to 0}\frac{f(x^2+h)-f(x^2)}{h}&=\lim_{h\to 0}\frac{(x^2+h)^{\frac{5}{2}}-(x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4493384",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Probability of not playing an opponent in a contest with $2^n$ players, from Ross Introduction to Probability In a certain contest, the players are of equal skill
and the probability is $\frac{1}{2}$ that a specified one of
the two contestants will be the victor. In a group
of $2^n$ players, the players are paired off ... | Sorry that I still stick to counting $i$ from $1$ to $n$. This sounds more natural to me.
I think that the calculation is incorrect because
First, in your formula, $\sum_{i=1}^n P(A_i)=\sum_{i=1}^n \frac {1}{2^i}=1-\frac{1}{2^n} \neq 1 $
It means that something is wrong. Actually $P(A_n)$ should be $\frac {1}{2^{n-1}}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494459",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
What is the number of different sums of the items of a set of consecutive natural numbers? I have been playing around with sets of consecutive natural numbers like say $S=\{1,2,3,...,10\}$ and I have come up with this problem for which however I have not yet found an answer and I don't know if there is one. My problem ... | Here is an answer with the same content as Thomas Andrews' answer, but presented a bit more heuristically.
Since the sum of the first $n$ positive integers is given by $\frac{n(n+1)}{2}$, you can obviously make that sum with the $n$ integers. By removing one integer addend at a time, you can make a number which is $1$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4494779",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Getting the area between $y$-axis, $f(x)$, $f(2-x)$ when both function is given by their subtraction?
*
*For a polynomial $f(x)$, let function $g(x) = f(x) - f(2-x)$.
*$g'(x) = 24x^2 - 48x + 50$
What is the area between $y=f(x)$, $y=f(2-x)$, and $y$-axis?
My approach:
*
*$g(1) = f(1) - f(1) = 0$.
*From $g'(x)$,... | From where you left off,
$$ g(x) \ \ = \ \ f(x) \ - \ f(2 - x) $$ $$ = \ \ 2a·x^3 \ - \ 6a·x^2 \ + \ (12a + 4b + 2c)·x \ - \ (8a + 4b + 2c) \ \ , $$
which does match up with your integration of $ \ g'(x) \ $ with $ \ a \ = \ 4 \ \ , \ \ 2b + c \ = \ 1 \ \ : $
$$ g(x) \ \ = \ \ 2·4·x^3 \ - \ 6·4·x^2 \ + \ (12·4 + 2·1)·... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4496396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Quadratic formula $x = \frac{- (b +\sqrt{b^2- 4ac})}{ \pm2a}$ In the proof of the quadratic formula
$$x = \frac{- b +\sqrt{b^2- 4ac}}{2a}$$
shouldn't there be $\pm 2a$ instead of $2a$, since both can be the square root of $4a^2$?
| Another way to approach it which does not rely on the study of the sign of $a$:
\begin{align*}
ax^{2} + bx + c = 0 & \Longleftrightarrow 4a^{2}x^{2} + 4abx + 4ac = 0\\\\
& \Longleftrightarrow (4a^{2}x^{2} + 4abx + b^{2}) = b^{2} - 4ac\\\\
& \Longleftrightarrow (2ax + b)^{2} = b^{2} - 4ac\\\\
& \Longleftrightarrow 2ax +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4499745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Issues with integrating $\int x^2 \sqrt{x^3 +1}$ via integration by parts I want to integrate
$$\int x^2 \sqrt{x^3 +1}~dx$$
I tried it with integration by parts (because we have a product here), but an online calculator did it with integration by substition.
Would this still be correct?
$$\frac{1}{3}x^3 (x^3+1)^\frac{... | The first line of your calculation is missing a factor of $x^2$, which arises when using the chain rule to compute $d(\sqrt{x^3 + 1})$.
Instead, applying integration by parts with $$u = \sqrt{x^3 + 1}, \qquad dv = x^2$$ gives
$$\left(\frac{1}{3} x^3\right) \left(\sqrt{x^3 + 1}\right) - \int \left(\frac{1}{3} x^3\right)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$ $( \frac{1}{a} +\frac{1}{b} +\frac{1}{c}) +a +b +c \geq 4\sqrt{3}$
If a, b, c are positive real numbers such that $a^2+ b^2+ c^2 = 1$
Show that:
$$\frac{1}{a} +\frac{1}{b} +\frac{1}{c}+a +b +c \geq 4\sqrt{3}.$$
My attempt:
First , I used Holder's : $$... | By your argument, we already have
$$\frac1a+\frac1b+\frac1c+a+b+c\geq\frac9{a+b+c}+\frac{(a+b+c)^3}3=\frac9u+\frac{u^3}3\equiv f(u)$$
where $u=a+b+c>0$. Try to show that $f'(x)=0$ only has one root $x^*=\sqrt3$ when $x>0$ and $f''(x^*)>0$, therefore $f(u)\geq\min_{x>0}f(x)=f(x^*)=4\sqrt3$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501286",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Evaluating the limit of $\lim_{n \to \infty} \frac{4n^2 +9}{2n^2+2n+3}$ with definition I have troubles with "estimating".
We want to evaluate the limit with the definition. For all $\epsilon >0$ we have to find an $N$, such that $|a_n-a|<\epsilon$ holds for $n>N$.
The limit in question is:
$$\lim_{n \to \infty} \frac... | In your approach you need to find an upper bound for the numerator and a lower bound for the denominator.
Your calculation was ok but I would present it as below to make it a little bit more explicit:
*
*$4n-3<4n$
*$2n^2+2n+3>2n^2$
In fact your way of proceeding is totally fine, you can even skip the epsilon part an... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4501404",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
How to show that two matrices are similar without using anything about eigenvalues or eigenvectors Problem: Show that $A$ and $B$ are similar without using eigenvalues or eigenvectors:
$$
A =
\left[ {\begin{array}{ccc}
1 & 1 & 1 \\
0 & 2 & 2 \\
0 & 0 & 3 \\
\end{array} } \right]
$$
$$
B =
\left[ {... |
I also know that I could set up a system of equations to solve for the entries of $S$ like $AS=SB$, but that strikes me as crude.
Crude as it may be, it's certainly doable.
$$S = \begin{bmatrix}a & b & c \\ d & e & f \\ g & h & j\end{bmatrix}$$
$$AS = \begin{bmatrix}a + d + g & b + e + h & c + f + j \\ 2 d + 2 g & 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4504984",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What to conclude from $x y \le N$ and $x\le 1$? For real $x, y, N$, what can we conclude for $y$ from two inequalities:
$$x y \le N\qquad {\rm and} \qquad x\le 1?$$
| Let $x \ne 0$. Then you have six cases following from $x y \le N$:
*
*For $0 < x \le 1$ and $N \ge 0$ you have $0 \le y \le \frac{N}{x}$ .
*For $0 < x \le 1$ and $N < 0$ you have $ y \le \frac{N}{x} < N < 0 $ .
*For $-1 \le x <0$ and $N \ge 0$ you have $0 \ge y \ge \frac{N}{x}$ .
*For $-1 \le x <0$ and $N < ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4505470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Interesting integral $\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{2}}$ Latest Edit
Inspired by @J.G., I find a formula in general,
$$
\begin{aligned}
\int_{0}^{\frac{\pi}{2}} \frac{d x}{\left(1+\sin ^{2} x\right)^{n}} &=2 \int_{0}^{\pi} \frac{d x}{(3-\cos x)^{n}} \\
&=\left.\frac{2(-1)^{n}}{(n-1) !}... | \begin{align}
\int_{0}^{\frac{\pi}{2}} \frac{dx}{(1+\sin ^{2} x)^{2}}
= &\int_{0}^{\frac{\pi}{2}} \frac1{4\tan^3x} \ d\left(\frac{\tan^2x}{1+\csc^2x}\right)\\
\overset{ibp}=&\ \frac34 \int_{0}^{\frac{\pi}{2}} \frac{d(\tan x)}{1+2\tan^2x}
=\frac{3\pi}{8\sqrt2}
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4508178",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 4,
"answer_id": 0
} |
Find the function $f(x)$ when $f(f(x))=1-x$, for $x\in [0,1]$
The function f(x) is continuous and $f(f(x))=1-x$, for $x\in [0,1]$ then,
(A) $f(\frac{1}{8})+f(\frac{7}{8})=3$
(B) $f(\frac{2}{3})+f(\frac{1}{3})=2$
(C) $f(\frac{5}{6})+f(\frac{1}{6})=1$
(D) None of These
My approach is as follow $f\left( {f\left(... | From $f(f(x)) =1-x$ follows
$$1-f(x)=f(f(f(x)))=f(1-x) .$$
Hence,
$$f(x) +f(1-x) =1$$
for all $x\in[0, 1]$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510680",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
How to find sum of this series using GP formula??? Given G.P:- $1 + 2^{1/n} + 2^{2/n} + . . . + 2^{(N-1)/N}$
My approach:-
We know, $S_{n} = \frac{a(r^n - 1)}{r - 1}$
Hence, here as we have n-1 terms, the formula will be:-
$S_{n-1} = \frac{a(r^{n-1} - 1)}{r - 1}$
We can see, $a = 1$, $r = 2^{1/n}$
Hence, $S_{n-1} = \fr... | Answer edited.
The sum of the first $N$ terms of a geometric series with first term $a$ and common ratio $r$ is: $a + ar + \ldots + ar^{N-1} = \frac{a(1-r^N)}{1-r}.$
To see that this finite series has $N$ terms and not $N-1$ terms, write it as:
$$a + ar + \ldots + ar^{N-1} = ar^0 + ar^1 + \ldots + ar^{N-1} = ar^{-1}\u... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4510822",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\int_0^\pi x\frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx$
How am I supposed to solve the following definite integral?
$$
\mathcal{I} = \int_0^\pi x \cdot \frac{\sin{\frac{x}{2}} - \cos{\frac{x}{2}}}{\sqrt{\sin{x}}} dx
$$
This definite integral is solved if the minus sign is replaced by... | Here is a solution using real analysis. First, denote the two integrals as
\begin{align*}
J & \equiv\int\limits_0^{\pi}\frac {x\left(\sin\frac x2\color{red}-\cos\frac x2\right)}{\sqrt{\sin x}}\,\mathrm dx\\K & \equiv\int\limits_0^{\pi}\frac {x\left(\sin\frac x2\color{red}+\cos\frac x2\right)}{\sqrt{\sin x}}\,\mathrm dx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4511773",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 4,
"answer_id": 1
} |
Computing $\int_{0}^{\pi} \ln (\sin x+2) d x$ and $\int_{0}^{\pi} \ln (2-\sin x) d x$ I first encountered this integral
$$
I=\int_{0}^{\pi} \ln (\sin x+2) d x
$$
several months ago without any idea and had tried many methods such as integration by parts, substitution and Fourier series etc. but all are in vain. Today, ... | Let $ I_{\pm}=\int_0^{\pi/2}\ln(2\csc2x\pm 1)dx$ and evaluate
\begin{align}
I_+ +I_- =& \int_0^{\pi/2}\ln(4\csc^2 2x-1)dx
=\int_0^{\pi/2}\ln(\tan^4 x+\sec^2x)dx\\
=& \int_0^{\pi/2}\int_0^1 \frac{2y\sec^2 x}{\tan^4 x+y^2(1+\tan^2x)}dy\ dx\\
=&\int_0^1 \frac\pi{\sqrt{(y+1)^2-1}}dy=\pi\ln(2+\sqrt3)\\
\\
I_+ -I_- =& \int_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4513080",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
How to compute derivative of $\sin(x^3)$ by definition? I am trying to proof that derivative of $\sin(x^3) = 3x^2\cos(x^3)$ by definition.
But I don't know an identity for $\sin(x^3)$ for getting $\cos(x^3)$.
Even I try to find a quantity similar to $\frac{\sin(x)}{x}$.
| $$\frac{d\sin x^3}{dx}=\lim_{h\rightarrow 0} \frac{\sin(x+h)^3-\sin x^3}{h}$$
$(x+h)^3=x^3(1+h/x)^3$ Let us use $(1+z)^k=1+kz+\frac{k(k-1)}{2} z^2+O(z^3)$ to get
$$\frac{d\sin x^3}{dx}=\lim_{h\rightarrow 0}\frac{\sin[x^3(1+h/x)^3]-\sin x^3}{h}$$
$$=\lim_{h\rightarrow 0}\frac{\sin[x^3+3hx^2+O(h^2)]-\sin x^3}{h}$$
$$=\li... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4514793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
What is wrong with my solution to find the size of an angle of a triangle? The sides of a triangle $a$, $b$, and $c$ has the following lengths.
$$a=x^2+x+1$$
$$b=2x+1$$
$$c=x^2-1$$
Since $a,b,c>0$, $x>1$.
$$a-b=x(x-1)>0\ (\because x>1)$$
$$a-c=x+2>0$$
Therefore, $a$ is the longest side, and its opposite angle $A$ is th... | Well, the angles A,B,C are not "proportional" as you wrote.
By the cosine law, you have
$$
a^2=b^2+c^2-2bc \cos(A).
$$
Hence
$$
\cos(A)=\frac{b^2+c^2-a^2}{2bc}=\frac{(2x+1)^2+(x^2-1)^2-(x^2+x+1)^2}{2(2x+1)(x^2-1)}=-\frac{1}{2}.
$$
Therefore $A=\frac{2}{3}\pi$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4515335",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solving $\cos(2\theta)=\sin\left(\frac{\theta}{2}\right)$ Consider an acute angle $\theta$, which has the property that
$$\cos(2\theta)=\sin\left(\frac{\theta}{2}\right).$$
I am trying to find the value of $\theta$. Using the identity
$$\cos(2\theta)=\sin\left(\frac{\pi}{2}-2\theta\right),$$
I can write the first equat... | Note that the solution to $$\sin(x) = \sin(y)$$ is $$x = 2k\pi + y \text{ and } x = 2k\pi + \pi - y$$ So in this case we have $$\frac\pi 2 - 2\theta = 2k\pi + \frac\theta 2 \text{ and } \frac\pi 2 - 2\theta = 2k\pi + \pi - \frac\theta 2$$The solutions are $$\frac{5\theta}{2} = \frac \pi 2 - 2k\pi \implies \theta = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4518438",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If the matrix A is given, then find the value of $A^{2022}$ If it is given that matrix $ A = \begin{bmatrix}
\dfrac{5}{2} & \dfrac{3}{2}\\
-\dfrac{3}{2} & -\dfrac{1}{2}
\end{bmatrix}$ then find the value of $A^{2022}$.
Here is my try on it : $$\\$$
If we look at the pattern while multliplying A with itself it ... | Another option is to see that $A = I + B$, where
$B = \begin{pmatrix}
\frac 3 2 & \frac 3 2 \\
-\frac 3 2 & -\frac 3 2
\end{pmatrix}$
and $B^2 = 0$.
So $A^n = I + nB$.
But that's a post-hoc explanation: it is not obvious to remark that when looking at the matrix - although such a simplification can be expected, in an e... | {
"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": 1
} |
Can't understand why cosh(1/x) changes the result for this limit I have this limit:
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cosh{\frac 1x}\over x^2}-1\right)$$
If I solve this limit by hand, I get 1
$\require{cancel}$
$$\lim_{x \to -\infty} x^2\left({(x^2+1)\cosh{\cancelto{0}{\frac 1x}}\over x^2}-1\right)=$$
$$\lim_{x... | Note that the Taylor series of $\cosh(z)$ at $z=0$ is $1+\frac{z^2}{2}+o(z^2)$. Hence, as $x\to -\infty$, we find
$$\begin{align}
x^2\left({(x^2+1)\cosh\left({\frac 1x}\right)\over x^2}-1\right)
&=x^2\left({(x^2+1)(1+\frac{1}{2x^2}+o(1/x^2))\over x^2}-1\right)
\\
&=(x^2+1)\left(1+\frac{1}{2x^2}+o(1/x^2)\right)-x^2\\
&... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4522344",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Calculate the integral of $\frac{1}{x^2 +x + \sqrt x}$ How to correctly calculate the integral:
$$\int_0^\infty \frac{1}{x^2 +x + \sqrt x}dx$$
Edit: I tried to figure out if the limit exists:
Step 1: break the integral to two parts: from 0 to 1, from 1 to infinity.
Step 2: use limit comparison test for both of the inte... | $$\int \frac{dx}{x^2 +x + \sqrt x}=2\int \frac{dy}{y^3 +y + 1}$$
Write
$$\frac{1}{y^3 +y + 1}=\frac 1{(y-a)(y-b)(y-c)}$$ Use partial fraction decomposition
$$\frac{1}{y^3 +y + 1}=\frac{1}{(a-b) (a-c) (y-a)}+\frac{1}{(b-a) (b-c) (y-b)}+\frac{1}{(c-a) (c-b) (y-c)}$$ Then three logarithms to be recombined before using the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Calculate $\frac{4-5\sin\alpha}{2+3\cos\alpha}$
Calculate $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}$$ if $\cot\dfrac{\alpha}{2}=-\dfrac32$.
My first approach was to somehow write the given expression only in terms of the given $\cot\frac{\alpha}{2}$ and just put in the value $\left(-\dfrac{3}{2}\right)$. Now I don't thi... | $$\dfrac{4-5\sin\alpha}{2+3\cos\alpha}=\dfrac{4-5.2\sin\frac{\alpha}{2}\cos\frac{\alpha}{2}}{2+3\left(1-2\sin^2\frac{\alpha}2\right)}=\dfrac{4-10\sin\frac{\alpha}2\cos\frac{\alpha}2}{5-6\sin^2\frac{\alpha}2}$$Now divide numerator and denominator by $\sin^2\frac{\alpha}2$,
$$\dfrac{\dfrac{4}{\sin^2\frac{\alpha}2}-10\cot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4525454",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
What is wrong in this solution of $\sec x + \csc x = 2 \sqrt{2}$
Find number of solutions in the interval $[0,2\pi]$ of the equation - $$\csc x + \sec x = 2 \sqrt{2}.$$
$⇒ \dfrac{1}{\sin x}+ \dfrac{1}{\cos x }= 2 \sqrt2$
$⇒ \dfrac{\sin x +\cos x}{\sin x \cos x} = 2 \sqrt 2 ⇒ (\sin x+\cos x)^2=8\sin^2x\cos^2x⇒1+2\sin... | As noticed in the comments, the issue is that by squaring we can add solutions, as for the following trivial example
$$x=-1 \implies x^2 =1 \implies x=1 \lor x=-1$$
In this case to avoid this issue we can proceed as follows (since $\sin x\cos x\neq 0$):
$$\dfrac{1}{\sin x}+ \dfrac{1}{\cos x }= 2 \sqrt2 \iff \cos x + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4526494",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving a function is continuous using rigorous definition of a limit Im trying to prove that the function $$\begin{cases}f(x,y)=\dfrac{(2x^2y^4-3xy^5+x^6)}{(x^2+y^2)^2}, & (x,y)≠0\\ 0, & (x,y)=0\end{cases}$$ is continuous at point (0,0) using the rigorous defintion of a limit.
Attempting to find the upper limit of the... | Use the fact that $|xy| \le \frac{x^2 + y^2}{2}$. You can then get,
$$3|xy^5| = 3|xy|y^4 \le \frac{3}{2}(x + y)^3$$
and
$$2x^2y^4 = 2|xy|^2y^2 \le \frac{1}{2}(x + y)^3.$$
You can then proceed as you were.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4528027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
How do you compute $\int_{0}^{\pi}\sqrt{\sin^{-1}(\cos(x))\tan^{-1}(\cot(x))}dx$ I tried using the identity of $\frac{\pi}{2}-x$ and doing substitutions like $u=\cos(x)$, but I keep ending up with $0$.
Is there a concept I'm missing?
| $$\int_{0}^{\pi}\sqrt{\sin^{-1}(\cos(x))\tan^{-1}(\cot(x))}dx$$
$$= \int_{0}^{\pi} \sqrt{\left(\frac{\pi}{2} - x\right)^2} dx$$
$$= \int_{0}^{\pi} \left|\frac{\pi}{2} - x\right| dx$$
$$= \int_{0}^{\pi/2} \left(\frac{\pi}{2} - x\right) ~dx + \int_{\pi/2}^\pi \left(x-\frac{\pi}{2}\right) ~dx$$
$$= \left[\frac{\pi}{2}x - ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4532858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Density of hitting time for a two-sided barrior for Brownian motion with drift For the following Brownian motion with drift $X_t = X_0 + \mu t + \sigma B_t$ where $\mu \in \mathbb{R}$, $ \sigma > 0$ and $X_0 \in \mathbb{R}$ which is a solution to the stochastic differential equation:
\begin{align*}
dX_t = \mu dt + \sig... | I have now found all mistakes, so here is the final answer.
We use separation of variables to find a solution we define $u(x,t) = f(x)g(t)$ then we get the two equations:
\begin{align*}
\begin{cases}
D\partial_x^2 f - \mu \partial_x f + \lambda f = 0 \\
\partial_t g = - \lambda g
\end{cases}
\end{align*}
where ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4533948",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
If $x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$, show that $\frac{201}{403}
If $$x=\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{2015^2}$$
show that
$$\frac{201}{403}<x<\frac{2014}{2015}$$
So, I manage to do the RH inequality using that $$x<\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+\cdots+\frac{1}{2014\cdot2015}=1-\... | We have
$$\frac{1}{k}-\frac{1}{k+1} < \frac{1}{k^2} < \frac{1}{k-1} - \frac{1}{k}$$
and so (telescoping sum)
$$\frac{1}{2} - \frac{1}{2016} < \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{1}{1} - \frac{1}{2015}$$
that is
$$\frac{1007}{2016}< \sum_{k=2}^{2015} \frac{1}{k^2} < \frac{2014}{2015}$$
Now, LHS $= \frac{2014}{4032}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4539222",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Finding all $x \in \mathbb{R}$ with $|x+1| + |x-1| < 4$ We want to find all $x \in \mathbb{R}$ with
$$|x+1| + |x-1| < 4$$
I tried the following, but I'm uncertain if this is correct:
Critical points: $x= -3$ and $x = 1$.
Case $1$:
$$x \in (-\infty, -3) \\
|x+1| = -(x+1) = -x-1 \\
|x-1| = -(x-1) = -x+1 \\
|x+1| + |x-1|... | Consider a cord of length $4$ with ends attached to the points $-1$ and $1.$ When the cord is maximaly extended to the right or left it reaches the points $2$ and $-2,$ respectively. Therefore the answer is $(-2,2).$
Similarly let the cord of length $2L$ be attached to the points $a<b$ such that $2L>b-a.$ Then the solu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\int_{0}^{\infty} \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-1}\:dx$ Find $$I=\int_{0}^{\infty} \frac{1-\frac{1}{x^2}}{\left(x+\frac{1}{x}\right)^2-1}dx$$
I have a very small query. I used the substitution $x+\frac{1}{x}=v$, then the limits of integration are both infinity. Can we say the integral is zer... | If $I$ exists, then $I = I_1 + I_2$ where $$I_1 = \int_{x=0}^1 f(x) \, dx + \int_{x=1}^\infty f(x) \, dx, \quad f(x) = \frac{1-x^{-2}}{(x+x^{-1})^2-1}.$$
Since the antiderivative of $f$ is
$$F(x) = \int f(x) \, dx = \int \frac{dv}{v^2-1} = \frac{1}{2} \int \frac{1}{v-1} - \frac{1}{v+1} \, du = \frac{1}{2} \log \left|\f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4543285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the value of given integral The given integral is :$$\displaystyle\int_0^1 \dfrac{\sin{\theta}(\cos^2{\theta} -\cos^2{\frac{\pi}{5}})(\cos^2{\theta} -\cos^2{\frac{2\pi}{5}})}{\sin{5\theta}} \, d\theta$$
I tried solving it with some trigonometric identities but comes out it does not work here.
However solution g... | Note that
$$\begin{aligned}
z^5-\frac{1}{z^5}
=&\ \left(z-\frac{1}{z}\right)\left(z-2 \cos \frac{\pi}{5}+\frac{1}{z}\right)\left(z-2 \cos \frac{2 \pi}{5}+\frac{1}{z}\right) \\
& \>\>\> \times\left(z+2 \cos \frac{2 \pi}{5}+\frac{1}{z}\right)\left(z+2 \cos \frac{\pi}{5}+\frac{1}{z}\right)\\
\end{aligned}$$
Then, set $z=e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
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... | Hint: $0 < a_n < a_1\cdot \left(\dfrac{1}{2}\right)^{n-1}, n \ge 2$ .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4548483",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Find the smallest possible value for $x^2-3x+2y^2+4y+2$ Find the smallest possible value for $x^2-3x+2y^2+4y+2$.
I know this: $x^2-3x+2y^2+4y+2=(x-\frac{3}{2})^2+2(y+1)^2-\frac{9}{4}$ so the smallest value is $-\frac{9}{4}$.
Now we also know this: $x^2-3x+2y^2+4y+2=(x-1)^2-1-x+2(y+1)^2=(*)(x-1)^2+2(y+1)^2-(1+x)$
My que... | If you (think you) have completed the square, but your left-over term isn't a constant, then you haven't completed the square.
The fact that $x=1$ minimises the expression $(x-1)^2$ has nothing to do with minimising $g_1(x) = (x-1)^2 + 3x\ $ or $\ g_2(x) =(x-1)^2 - 26x^3 + 45,\ $ or $\ g_3(x) =(x-1)^2 + f(x),\ $ unless... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4554540",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Find the integers $n\geq 2$ that satisfy $0 \leq 1 +\frac{n}{n-1}\cos\big(\pi\cdot\frac{n}{n+1}\big)$ I want to find the integers $n\geq 2$ that satisfy the inequality
$$
x_1x_2+x_2x_3+\cdots+x_{n-1}x_n\leq\frac{n-1}{n}\Big(x_1^2+x_2^2+\cdots+x_n^2\Big)
$$
for all $x_i>0$.
I see that if $n=2$, the geometric mean and qu... | Suppose the inequality is true for some $n\in\mathbb N,\ n> 1$. For $n=2$, we know the inequality is true. We show that for $n\ge 3$, there are counterexamples to the inequality.
To that end, first let $n=3$ and $x_1=2,\ x_2=3,\ x_3=2$. Then if the inequality is satisfied, we have $$x_1x_2+x_2x_3\le \frac{3-1}{3}(x_1^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4559129",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Simplifying $\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)}$ using Lagrange’s polynomial I have a question of symplifying this expression
$$
A=\frac{a^4}{(a-b)(a-c)} + \frac{b^4}{(b-a)(b-c)} + \frac{c^4}{(c-a)(c-b)} \tag{1}
$$
where $a$, $b$ and $c$ are distinct nonzero real numbers. There is ... | For your problem,
I threw it at Wolfram Alpha and got
$a^2 + a b + a c + b^2 + b c + c^2
$.
This can be rewritten as
$(a+b+c)^2-(ab+ac+bc)$.
With 3 instead of 4 this is
$a+b+c$,
and with 5 this is
$a^3 + a^2 b + a^2 c + a b^2 + a b c + a c^2 + b^3 + b^2 c + b c^2 + c^3
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4560167",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find $\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}}$ without using derivatives This limit is proposed to be solved without using the L'Hopital's rule or Taylor series:
$$
\lim\limits_{x\to 0} \left( \frac{a^{\sin x}+b^{\sin x} }{2} \right)^{\frac1{x}},
$$
where $a>0$, $b>0$ are some c... | This post has multiple answers that explain why, if $\displaystyle\lim_{x\to a} f(x)=1$ and $\displaystyle\lim_{x\to a}g(x)=\infty$ then $$\lim_{x\to a} (f(x))^{g(x)}=e^{\lim\limits_{x\to a} (f(x)-1)g(x)}$$ Using this formula, calling the limit as L, we have $$L=e^{\lim\limits_{x\to 0}\left(\frac{a^{\sin x}+b^{\sin x}-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4561863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that there are no integer solutions to: $x^4+6x^2+1=8y^4$
Prove that there are no integer solutions to:
$$x^4+6x^2+1=8y^4$$
where $x>1$.
My attempts:
Let $x^2=u, y^2=v$
$$u^2+6u+(1-8v^2)=0$$
$$\Delta=36-4(1-8v^2)=w_0^2$$
$$32v^2+32=(4w_1)^2$$
$$2v^2+2=w_1^2=(2w_2)^2$$
$$v^2+1=2w_2^2$$
$$v=w_2=1\implies x=y=1$$
... | First, observe that $8y^4$ is even, so $x^4+6x^2+1$ is even, so $x^4+6x^2$ is odd. Therefore, $x$ is odd. Rewrite $x$ as $2n+1$, where n is an integer and $n\geq0$. So the original equation can be rewritten as $(2n+1)^4+6(2n+1)^2+1=8y^4$. By simplifying it we see that $2n^4+4n^3+6n^2+4n+1=y^4$, which can be rewritten a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4563040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Prove that $\sum\limits_{\mathrm{cyc}}\sqrt{xy^2(y+z)} \le {1 \over 2} \sqrt{(x + y + z)^4 - (x^2 + y^2 + z^2)^2}$
Prove the following inequality for all positive real numbers $x$, $y$, and $z$:
$$\sqrt{xy^2(y+z)} + \sqrt{yz^2(x+z)} + \sqrt{zx^2(x+y)} \le {1 \over 2} \sqrt{(x + y + z)^4 - (x^2 + y^2 + z^2)^2}$$
My at... | Let's raise both sides of the inequality to the second power. Indeed, we just need to show:
$$2(\sqrt {xy^2(y+z)}\sqrt {yz^2(x+z)}+\sqrt {yz^2(x+z)}\sqrt {zx^2(x+y)}+\sqrt {xy^2(y+z)}\sqrt {zx^2(x+y)})\le x^3y+xz^3+y^3z+x^2y^2+x^2z^2+y^2z^2+2x^2yz+2xy^2z+2xyz^2.$$
Now, notice that: $$2\sqrt {xy^2(y+z)}\sqrt {yz^2(x+z)}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4564490",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find slope of the tangent line of $4\sqrt x + 2e^\frac {3x-12}{x+2}$ at $ x_0$ Find the slope and the equation of the tangent line to the
graph $y = f(x)$ at $x_0=4$, $$4\sqrt x + 2e^\frac {3x-12}{x+2} $$
$$\lim_{h\to 0}\tfrac{4\sqrt {4+h} + 2e^\frac {12+3h-12}{4+h+2} - 10}{h} = \lim_{h\to 0}\frac{4\sqrt {4+h} + 2e^\f... | From where you left by students's L'hospital
$$f'(4)=\lim_{h\rightarrow 0}\frac{\frac{2}{\sqrt{4+h}}+2e^{\frac{3h}{6+h}}\frac{3(6+h)-3h}{(6+h)^2}}{1}=\frac{2}{\sqrt{4}}+2.e^0.\frac{3.6}{6^2}=1+1=2$$
and the tangent line at $(4,10)$ is
$$\frac{y-10}{x-4}=2\implies y=2x+2.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4565098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Solve for $n: \cos\left(\frac\pi4(n^2+2n)\right)=\sin\left(\frac\pi4(n^2+n+1)\right), n\in \mathbb Z$
Solve for $n: \cos\left(\frac\pi4(n^2+2n)\right)=\sin\left(\frac\pi4(n^2+n+1)\right), n\in \mathbb Z$
My Attempt:
$$\cos\left(\frac\pi4(n^2+2n)\right)=\cos\left(\frac\pi2-\frac\pi4(n^2+n+1)\right)\\\implies\frac\pi4... | $$2n^2+3n-5=(2n+5)(n-1)=8p+1-5=4(2p-1)$$
As $2n-5$ is odd, $4$ must divide $n-1=4r$(say)
$\implies(2(4r+1)-5)r=2p-1\implies r$ must be odd $2t-1$(say)
$\implies n=4r+1=4(2t-1)+1=8t-3$
$$\implies p=\dfrac{2n^2+3n-1}8=\dfrac{2(8t-3)^2+3(8t-3)-1}8=16t^2-9t-1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4566423",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find highest common factor of two polynomials Find the highest common factor of the following expressions:
$6-8a-32a^2-18a^3$, $20-35a-95a^2-40a^3$
My attempt:
The polynomial on the left is divisible by $2$, the polynomial on the right is not divisble by $2$. Conversely the polynomial on the right is divisble by $5$ an... | It looks like you're kind of drifting towards applying the Euclidean algorithm. In this case, though, the easiest way is probably to (1) prove that $1+a$ is a factor of both polynomials, (2) find the quotient of each polynomial divided by $1+a$, (3) find the GCD of the quotients.
If you do that here, then first you get... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4567921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Find tangent plane of level surface of $f(x, y, z)= \cos(x^2+2y+3z)$ at $(\frac{\pi}{2}, \pi, \pi)$ I was wondering if someone could verify my solution to this problem. As stated in the title, I was requested to find the equation for the tangent plane of the level surface of $f(x, y, z)= \cos(x^2+2y+3z)$ at $(\frac{\pi... | A bit of background:
*
*If $f: \mathbf{R}^{n}\to \mathbf{R}$ is differentiable function, $x_{0}\in {\rm Dom}(f)$ and $\nabla f(x_{0})\not=0$, then gradient vector $\nabla f(x_{0})$ is orthogonal to the level set $\{x\in \mathbf{R}^{n}: f(x)=c\}$ of passing through $x_{0}$. Then the tangent plane to surface $S$ defi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568592",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to Evaluate the Integral? $\int_{0}^{1}\frac{\ln\left( \frac{x+1}{2x^2} \right)}{\sqrt{x^2+2x}}dx=\frac{\pi^2}{2}$ I am trying to find a closed form for
$$
\int_{0}^{1}\ln\left(\frac{x + 1}{2x^{2}}\right)
{{\rm d}x \over \,\sqrt{\,{x^{2} + 2x}\,}\,}.
$$
I have done trig substitution and it results in
$$
\int_{0}^{1... | Hyperbolic substitution is very convenient for this integral. We can substitute $x+1=\cosh t$
$$I=\int_0^1 \ln{\left(\frac{x+1}{2x^2}\right)}\frac{dx}{\sqrt{x^2+2x}}=\int_0^{\ln{(2+\sqrt 3)}}\ln\left(\frac{\cosh t}{2(\cosh t-1)^2}\right)dt$$$$=\int_0^{\ln{(2+\sqrt 3)}}\ln\left(\frac{e^{-t}(1+e^{-2t})}{(1-e^{-t})^4}\rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4568778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "35",
"answer_count": 5,
"answer_id": 0
} |
how to do this using properties of definite integrals? $$\int_1^2 ln_2(x^3+1)dx$$
which is the same as
$\int_1^2 ln_2((3)-(x^3+1))dx$
which gives $\int_1^2 ln_2(2-x^3)\ dx$
which does not help.
Or integrate this by parts, but that's really tedious.
| Use $x^3 + 1 = (x+1)(1 - x + x^2)$.
$$\int\limits_{x=1}^2 \log_2 (x^3 + 1)\ dx = \int\limits_{x=1}^2 \log_2 (x + 1)\ dx +\int\limits_{x=1}^2 \log_2 (1 - x + x^2)\ dx$$
Then use (from integral tables): $\int \log_2 (x+1)\ dx = (x+1) \log (x+1)-x$
and
$\int \log_2 (1 - x + x^2)\ dx = \left(x-\frac{1}{2}\right) \log \left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4571237",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
For what $x$ the expression of the equation has the same remainder after division. For what $x$ does the expression $(50x+7)(3x+11)$ have the same remainder as $(26x − 86)x$ when divided by $111$?
$(50x+7)(3x+11) \equiv y \mod 111$
$(26x − 86)x \equiv y \mod 111$
$(50x+7)(3x+11) - (26x − 86)x \equiv 0 \mod 111 $
$124x^... | First you have $x= 3y\pm 1$ for some integer $y$.
You can reduce the coeficents mod $37$ so \begin{align}-20x^2+30x+40 &=_{37} -5(4x^2-6x-8)\\
&=_{37} -5(4x^2+68x-8) \\&=_{37}-5 (4x^2+68x +289)+5\\
&=_{37}-5 (2x+17)^2+5\\
\end{align}
So $$(2x-20)^2 =_{37} 1 \implies 2x-20 =_{37}\pm1 $$
So you have 4 possibilites.
*
*... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4574763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Integral Involving Harmonic Numbers: $\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx$ (Motivation) In an attempt to answer this question, I got stuck on evaluating a certain integral. I have made up the following conjecture:
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{... | I will answer my own question by proving
$$\int_{\sqrt{3}}^{\infty}\frac{\ln\left(x^{4}-1\right)}{x^{2}-1}dx = \frac{\pi^{2}}{4}+\frac{3}{2}\ln\left(2\right)\ln\left(2+\sqrt{3}\right).$$
Let the integral in question equal $I$. Letting $x \to \dfrac{1-x}{1+x}$, we rewrite the integral as
$$I = \int_{\sqrt{3}-2}^{-1}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
limit of a geometric-type series I came across this series while studying probability theory.
Let $$T(x)=\sum_{n=1}^\infty x^n \left(\frac{1}{n}-\frac{1}{n+1}\right),$$
where $T(x)$ converges for some range of $x$.
My question is how to prove that $T(x)=1+\frac{1-x}{x}\log(1-x)$ for all $|x|<1$.
The following is my att... | How about using the known series for on(1-x) and elementary series rearrangements?
$\ln(1-x) = -\sum_{n=1}^\infty\frac{x^n}{n}$ for $-1<x\le 1$
so
\begin{align*}
T(x)
&=
\sum_{n=1}^\infty x^n\left(\frac{1}{n} - \frac{1}{n+1}\right)
\\&=
\sum_{n=1}^\infty \frac{x^n}{n} - \sum_{n=1}^\infty\frac{x^n}{n+1}
\\&=
\sum_{n=1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4576790",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\lim\limits_{n \to \infty} \sqrt[n^2]{1! + 2! + 3! + \dots + n!}$ $$\lim_{n \to \infty} \sqrt[n^2]{1! + 2! + 3! + \dots + n!}$$
I can try at least to evaluate it from bottom:
$$1 < \sqrt[n^2]{n^2} < \sqrt[n^2]{n!} < \sqrt[n^2]{1! + 2! + \dots + n!}$$
Well, it doesn't say anything. Ok, gonna try evaluation fro... | Notice that :
$$1 \leq \sqrt[n^2]{1! + \cdots + n!} \leq \sqrt[n^2]{n! + \cdots + n!} = \sqrt[n^2]{n \, n!} \leq \sqrt[n^2]{n n^n} = \sqrt[n^2]{n^{n + 1}} \to 1$$
because :
$$\ln \sqrt[n^2]{n^{n + 1}} = \dfrac{1}{n^2} \ln n^{n + 1} = \dfrac{n + 1}{n} \, \dfrac{\ln n}{n} \to 1 \times 0 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4578184",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to attempt power series questions ? I'm confused with this topic
Use the power series representation of $\frac{1}{1-x}$ to obtain the power series representation of $\frac{2x}{4-3x}$ and the corresponding values of $x$.
Firstly, I tried to make $\frac{2x}{4-3x}$ in the form of $\frac{1}{1-x}$
$\frac{x}{2} \frac{1... | Maybe it is not necessary to note that:
Another way is Maclaurin series. $f(x)=\frac{2x}{4-3x}=-\frac{2}{3}+\frac{8}{3}\frac{1}{4-3x}$. With this form we can easily compute derivatives: $n$-th derivative is $f^{(n)}(x)=\frac{8}{3}3^{n}n!(4-3x)^{-n-1}$ and $f^{(n)}(0)=\frac{1}{2}\frac{3^{n-1}}{4^{n-1}}n!$ for $n>0$.
Not... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4579367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is there any better way of finding the required value If $$z=\cos\theta+i\sin\theta$$ find the value of $$\frac{1+z}{1-z}$$
The solution that I have is this
$$z=\cos\theta+i\sin\theta \implies$$
$$\frac{1+z}{1-z}=\frac{1+(\cos\theta+i\sin\theta)}{1-(\cos\theta+i\sin\theta)}=\frac{(1+\cos\theta)+i\sin\theta}{(1-\cos\th... | Observe that $$\dfrac{\cos t+i\sin t}{\cos t-i\sin t}=\cos2t+i\sin2t$$
If $\dfrac z1=\cos2t+i\sin2t=\dfrac{\cos t+i\sin t}{\cos t-i\sin t}$
Using Componendo and Dividendo
$$\dfrac{1+z}{1-z}=\cdots=?$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/4580285",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 1
} |
Verify that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C}{2})\le\frac{1}{8}$ in a general triangle $\triangle ABC$ So, this problem is inspired by a contest preparation problem I saw back in Japan, and it is as follows:
In a general triangle $\triangle ABC$, show that $\sin(\frac{A}{2})\sin(\frac{B}{2})\sin(\frac{C... | Being angles of the triangles, the sines are all positive so we can use AM-GM inequallity:
$$
\left(\frac{\sin\left(\frac{A}{2}\right)+\sin\left(\frac{B}{2}\right)+\sin\left(\frac{C}{2}\right)}{3}\right)^{3}
\geq
\sin\left(\frac{A}{2}\right)\sin\left(\frac{B}{2}\right)\sin\left(\frac{C}{2}\right)
$$
The sine function i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4583104",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Right triangle $\triangle ABC$, $D$ lies on $AB$, inscribed in a circle of radius $9$. Find the measure of $CD$ I saw this mathematical puzzle on an Instagram post today. As title suggests, we have a right angled triangle inscribed in a circle with radius $D$ and some angles. The goal is to find the length of $CD$. I'l... | Law of Tangents, let $t = \tan α$
$\displaystyle \frac{x+6}{x-6} =
\frac{\tan \frac{(3α-90°)+(90°-α)}{2}}{\tan \frac{(3α-90°)-(90°-α)}{2}}
= \frac{\tan α}{\tan (2α-90°)} = -(\tan α)(\tan 2α) = \frac{-2t^2}{1-t^2}$
$\displaystyle \frac{x+12}{x-12} =
\frac{\tan \frac{(180°-3α)+(α)}{2}}{\tan \frac{(180°-3α)-(α)}{2}}
= \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584232",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 5
} |
given N and M count the number of solutions to some diophantine equations, prove that $N>M.$
Let $N$ be the number of integer solutions to the equation $x^3 - y^3 = z^4-w^4$ with the property that $0\leq x,y,z,w\leq 2022^{2022}$. Let $M$ be the number of integer solutions to the equation $x^3-y^3 = z^4-w^4+1$ with $0\... | Continuing from what you've provided, first note that $a_0 = 1$ and, as you stated about how $A$ is defined, we have $a_{A+1}=0$. Next,
similar to what's done in the related AoPS thread Number of solutions of two systems, we get
$$\begin{equation}\begin{aligned}
2\sum_{i=0}^{A}a_{i}(a_{i}-a_{i+1}) & = \color{blue}{\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4584887",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How many ways to place plus and minus signs in front of numbers from 0 to 12 so that the sum is divisible by 5? This question is from an old NIMO contest.
Find the number of ways a series of + and − signs can be inserted between the numbers 0, 1, 2, · · ·, 12 such that the value of the resulting expression is divisibl... | It does matter what we put in front of the $0,5,10$. Of the $2^3\pm$ combinations you can do for those the result will have the same divisibility as any other. So It suffices to solve for $1,2,3,4,6,7,8,9,11,12$ and multiply the result by $8$.
Consider $A_1 = \pm 1 \pm 2\pm 3\pm 4$ and $A_2=\pm 6\pm 7\pm 8 \pm 9$ and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
How to solve $\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n$ Question:
$$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=?$$
My work:
$\lim_{n\to\infty}\left(\dfrac{n^2+5n+3}{n^2+n+2}\right)^n=\lim_{n\to\infty}\left(\dfrac{n^2(1+5/n+3/n^2}{n^2(1+1/n+2/n^2)}\right)^n=\lim_{n\to\infty}\left(\dfrac{1... | Another way
$$\dfrac{n^2+5n+3}{n^2+n+2}=\dfrac{n^2+5n+3}{n^2\left(1+\dfrac{1}{n}+\dfrac{2}{n^2}\right)}\sim \left(1+\dfrac{5}{n}+\dfrac{3}{n^2}\right)\left(1-\dfrac{1}{n}-\dfrac{2}{n^2}\right)\sim 1+\dfrac{4}{n}$$
$$\lim_{n\to\infty}\left( 1+\dfrac{4}{n}\right)^n=\lim_{t\to\infty}\left( 1+\dfrac{4}{4t}\right)^{4t}=\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4585631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Fair Value of a Dice Game 1 die. Up to 3 rolls. Your winnings are equal to the value of a single roll. You can stop after roll 1 or roll 2. If you proceed to the next roll (i.e the 2nd or 3rd), you forfeit the previous value (no memory). Therefore, if you roll 3 times, the value of the 3rd roll is what you win.
What is... | There is $\frac{1}{2}$ chance that we get $4, 5$ or $6$ in the first throw and $\frac{1}{2}$ chance that we get $1, 2$ or $3$.
Thus for exactly one throw , the expected value is $$\frac{1}{2} \times \frac{4+5+6}{3}=\frac{1}{2} \times 5$$
For exactly two throws , we must get $ \{1, 2 \text { }\mathrm{or} \text { ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4589195",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Can we evaluate $\int_0^1 \frac{\ln (1-x) \ln ^n x}{x} d x$ without expanding $\ln(1-x)$? For $|x|<1,$ we have
$$
\begin{aligned}
& \frac{1}{1-x}=\sum_{k=0}^{\infty} x^k \quad \Rightarrow \quad \ln (1-x)=-\sum_{k=0}^{\infty} \frac{x^{k+1}}{k+1}
\end{aligned}
$$
$$
\begin{aligned}
\int_0^1 \frac{\ln (1-x)}{x} d x & =-\... | Without expanding using series at all (taking for granted any properties of special functions whose derivations require series manipulation):
$$\begin{align*}
I &= \int_0^1 \frac{\log(1-x) \log^n(x)}x \, dx \\[1ex]
&= \int_0^1 \frac{\log(x) \log^n(1-x)}{1-x} \, dx \tag{1} \\[1ex]
&= (-1)^n \int_0^\infty x^n \log\left(1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594043",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Does $(x^2+xy+y^2)$ divide $(x+y)^n-x^n-y^n$ if and only if n has no prime factors less than 5 I noticed that for $n$ with prime factors greater than 5, that $xy(x+y)(x^2+xy+y^2)$ always seems to divide $(x+y)^n-x^n-y^n$. The $xy(x+y)$ factors seem fairly obvious, but I can't figure out where the $x^2+xy+y^2$ term come... | Here is an alternative direct approach, an induction with step six, and the "small cases" of $n$ among $0,1,2,3,4,5$ are easily checked. (We already have a quick, good, compact answer by Aphelli - i just wanted to see if an arguably more elementary proof avoiding complex numbers and derivatives can be worked out. The a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/4594636",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"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.