Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Calculate $\frac{1}{\sin(x)} +\frac{1}{\cos(x)}$ if $\sin(x)+\cos(x)=\frac{7}{5}$ If
\begin{equation}
\sin(x) + \cos(x) = \frac{7}{5},
\end{equation}
then what's the value of
\begin{equation}
\frac{1}{\sin(x)} + \frac{1}{\cos(x)}\text{?}
\end{equation}
Meaning the value of $\sin(x)$, $\cos(x)$ (the denominator) wi... | If
$$\sin x+\cos x=a\qquad\text{and}\qquad{\frac{1}{\sin x}}+{\frac{1}{\cos x}}=b$$ then
$$a=b\sin x\cos x\qquad\text{and}\qquad a^2=1+2\sin x\cos x=1+\frac{2a}{b}$$
from which it follows that
$$b={\frac{2a}{a^2-1}}$$
Letting $a=\frac{7}{5}$, we have
$$b={\frac{\frac{14}{5}}{\frac{49}{25}-1}}={\frac{70}{24}}={\frac{3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1649606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "29",
"answer_count": 10,
"answer_id": 2
} |
Solving a determinant using properties of a determinant
$$\begin{vmatrix}
y+z & x & x\\
y & z+x & y\\
z & z & x+y
\end{vmatrix}=k(xyz)$$
Find the value of $k$.
I solved this question by substituting $x=y=z=1$ and then expanding the determinant to get $k=4$ which is the correct answer.
Is there any method ot... | Use the property that adding multiples of rows does not affect the determinant. Subtracting row 2 and 3 from row 1, we get
$$
\,\,\,\,\,\,\begin{vmatrix}
y+z & x & x\\
y & z+x & y\\
z & z & x+y
\end{vmatrix}\\=
\begin{vmatrix}
0 & -2z & -2y\\
y & z+x &y\\
z & z & x+y
\end{vmatrix}
$$
Add half row 1 to row... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1650016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Factor the matrix (scalar $\times A$) into permutations of $A$ Here's an example of $A . B = scalar \times C$, done with magic squares. The last square does not have a consecutive range of digits.
Drop the magic square requirement. In $2\times2$ matrices we have the following, where a matrix times a scalar is factore... | Try
$$ \pmatrix{7 & -3 & 6 & 8 \cr
1 & -5 & 4 & -1 \cr
-7 & 3 & -4 & -6 \cr
0 & -2 & 5 & 2 \cr}
\pmatrix{
-5 & 3 & 7 & -1 \cr
-6 & 4 & 8 & -2 \cr
-4 & 2 & 6 & 0 \cr
5 & -3 & -7 & 1 \cr}
=\pmatrix{
-1 & -3 & 5 & 7 \cr
4 & -6 & -2 & 8 \cr
3 & 1 & -7 & -5 \cr
2 & -4 & 0 & 6 \cr}$$
and
$$
\pmatrix{
4 & -6 & 0 & 8 \cr
-2 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651157",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Solution Set of System of ODE. I am trying to find the solution of the system
$$\begin{bmatrix}x_1\\x_2\end{bmatrix}'= \begin{bmatrix}1&3\\3&1\end{bmatrix}\begin{bmatrix}x_1\\x_2\end{bmatrix}$$.
I am given that \begin{bmatrix} e^{4t} \\ e^{4t} \end{bmatrix} and \begin{bmatrix} e^{-2t} \\ -e^{-2t} \end{bmatrix} are sol... | \begin{equation*}
\mathbf{x}^{\prime }=\mathbf{W\cdot x\Rightarrow x}(t)=\exp [\mathbf{W}%
t]\cdot \mathbf{x}(0),\;\mathbf{W}=\left(
\begin{array}{cc}
1 & 3 \\
3 & 1
\end{array}
\right)
\end{equation*}
We can diagonalise $\mathbf{W}$. Thus
\begin{equation*}
\mathbf{W}=\left(
\begin{array}{cc}
1 & 3 \\
3 & 1
\end{array}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651752",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
integrate $\int \frac{dx}{x\sqrt{1-x}}$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$$\int \frac{dx}{x\sqrt{1-x}}$$
$u=1-x$
$du=-dx$
$$-\int \frac{du}{(1-u)\sqrt{u}}$$
$a(1-u)+b\sqrt{u}=1\Rightarrow a-au+b\sqrt{u}=1$
$a=1\Rightarrow b\sqrt{u}-u=0\Rightarrow b=\sqrt{u}$
$$\int \frac{\sqrt{u}}{1-u}du-\int \frac{du}{\sqrt{u}}=\int ... | HINT:
Let $ x = u^2, \, dx= 2 u du ; \, \int \dfrac{2 du }{u\sqrt{1-u^2} }$
again let $$ u = \sin(t) ; \ du = \cos (t) dt ...$$
leading to
$$ =2 log \dfrac{u}{1+\sqrt{1-u^2}} = 2 log \dfrac{\sqrt x}{1+\sqrt{1-x} }$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1651932",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 4
} |
Clever way of calculating the integral $ \int \frac{dt}{t^2\sqrt{t-2} } $
$$ \int \frac{\text{d}t}{t^2\sqrt{t-2} } $$
I know it can be calculated using somewhat complicated substitutions, but is there possibly some clever way of solving that type of integral? I don't even expect full solution, just ideas.
| \begin{align}
u & = \sqrt{t-2} \\[8pt]
u^2 & = t-2 \\[8pt]
u^2 + 2 & = t \\[8pt]
2u\,du & = dt
\end{align}
\begin{align}
\int \frac{dt}{t^2\sqrt{t-2}} & = \int \frac{2u\,du}{(u^2+2)^2 u} = \int\frac{2\,du}{(u^2+2)^2}
\end{align}
Then let $u = \sqrt 2 \tan \theta$, $du= \sqrt 2 \sec^2\theta\,d\theta$, $u^2+2=2\sec^2\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652299",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
To show the $\epsilon-\delta$ definition for limits holds. Question: Check if the following limit exists, if so show that the $\epsilon$ $\delta$ definition for limits holds.
$$\lim_{(x,y) \to (1,2)} \frac{(x-1)^2(y-2)^2}{x^2+y^2-2xy-4y+5}$$
My answer: So clearly the limit is zero, you can write $\frac{(x-1)^2(y-2)^2}{... | Hint:
Note we can factorize the denominator $D$ as
$x^2+y^2-2xy-4y+5=(x-\frac{y}{2})^2+\frac{3}{4}(y-2)^2-(y-2)-xy$
If $(x,y)$ closed to $(1,2)$, we know $(x-\frac{y}{2})^2,\frac{3}{4}(y-2)^2,(y-2)$ are very closed to $0$ and $-xy$ is closed to $-2$, hence we know the denominator $D$ satisfies
$$|D|\ge |xy|-|(x-\frac{y... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1652711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\cdots \left(\frac{2}{n-1}\right)^{2}\frac{1}{n}$ If $n$ be a positive integer $>1$, prove that
$$2^{n(n+1)}>(n+1)^{n+1}\left(\frac{n}{1}\right)^n\left(\frac{n-1}{2}\right)^{n-1}\left(\frac{n-2}{3}\right)^{n-2}\cdots \left(\frac... | Take $\log$ of both sides and verify that the inequality becomes equality for $n=1$. Take backward difference:
$$\nabla RHS=(n+1)\log (n+1)- n\log n+n\log n-\sum_{i=1}^n \log i\le 2n\log 2 =\nabla LHS \tag 1$$
holds for $n=1$. Take another backward difference $$\nabla^2 RHS=(n+1)\log(1+\frac 1 n)\le 2\log 2 =\nabla^2 L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654070",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 4,
"answer_id": 3
} |
Power series of $\frac{1+x}{(1-x)^2}$ This question is continuing from the previous question here:
Power Series representation of $\frac{1+x}{(1-x)^2}$
I am trying to calculate the power series representation of the equation:
$$
\begin{align}
f(x) = \frac{1+x}{(1-x)^2}
\end{align}
$$
My workout is as follow:
$$
\begin{... | Hint: In order to find the power series expansion around $x=0$ you could also use the binomial series representation
\begin{align*}
(1+x)^\alpha=\sum_{n=0}^{\infty}\binom{\alpha}{n}x^n\qquad\qquad \alpha\in\mathbb{C}, |x|<1
\end{align*}
We obtain
\begin{align*}
\frac{1+x}{(1-x)^2}&=(1+x)\sum_{n=0}^{\infty}\binom{-2}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1654261",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that $|\mathbb{N} \times \mathbb{N}| \leq |\mathbb{N}|$ Claim: $|\mathbb{N} \times \mathbb{N}| \leq |\mathbb{N}|$
Proof:
For all $ x \in \mathbb{N}, y \in \mathbb{N}$, we can define a function $f(x,y) = x + \frac{(x+y-1)(x+y-2)}{2}$
We now prove that $f$ is an injection. We will show that if $f(x,y) = f(u,v)$ t... | Here's a way of proving that result that's easier to understand:
$$
\begin{array}{cccccccccc}
1,1 & & 1,2 & \rightarrow & 1,3 & & 1,4 & \rightarrow & 1,5 & & 1,6 & \rightarrow \\
\downarrow & \nearrow & & \swarrow & & \nearrow & & \swarrow & & \nearrow \\
2,1 & & 1,1, & & 2,3 & & 2,4 & & 2,5 & & \cdots \\
& \swarrow & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1657399",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$ Prove that $\int\frac{3x^2+1}{(x^2-1)^3}dx=\frac{-x}{(x^2-1)^2}+c$
My Try:
$\int\frac{3x^2+1}{(x^2-1)^3}dx$
Put $x=\sec\theta$
$\int\frac{3\sec^2\theta+1}{\tan^6\theta}\sec\theta\tan\theta d\theta$
Converting to $\sin$ and $\cos$ we get
$=\int\frac{(3+... | $\dfrac{d\left(\dfrac{ax+b}{(x^2-1)^n}\right)}{dx}=-\dfrac a{(x^2-1)^n}-\dfrac{2a}{(x^2-1)^{n+1}}-b\cdot\dfrac{2x}{(x^2-1)^{n+1}}$
Integrating both sides wrt $x,$
$$-\dfrac{ax+b}{(x^2-1)^n}=a\int\dfrac{dx}{(x^2-1)^n}+2a\int\dfrac{dx}{(x^2-1)^{n+1}}+b\int\dfrac{2x\ dx}{(x^2-1)^{n+1}}$$
$$\implies2a\int\dfrac{dx}{(x^2-1)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1658367",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Find the value of $\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \frac{1}{90}$ Find the value of $p+q$, where $p$ and $q$ are two positive integers such that $p$ and $q$ have no common factor larger than $1$ and
$$\frac{1}{20} + \frac{1}{30} + \frac{1}{42} + \frac{1}{56} + \frac{1}{72} + \... | We can observe that
\begin{align}
20&=4\cdot 5\\
30&=5\cdot6\\
42&=6\cdot7\\
56&=7\cdot8\\
72&=8\cdot9\\
90&=9\cdot10.
\end{align}
Can you use partial fraction?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1659227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solving for x with matrices $$
\begin{bmatrix}
-5 & -9 \\
-6 & -2 \\
\end{bmatrix} X + \begin{bmatrix}
-3 & 2 \\
-9 & -3 \\
\end{bmatrix} = \begin{bmatrix}
-9 & -7 \\
-2 & -3 \\
\end{bmatrix} X
$$
I am asked to solve for mat... | You have formally
$$ AX +B = CX$$
You should proceed as follows
$$ \begin{array}{rll}
AX + B &=& CX \\
B &=& CX - AX \\
B &=& (C- A)X \\
(C-A)^{-1} B &=& (C-A)^{-1} (C- A)X \\
(C-A)^{-1} B &=& X \\
\end{array} $$
Now $$ C-A = -\begin{bmatrix} 9 & 7 \\ 2 &3 \end{bmatrix}
- (-\begin{bmatrix} 5 & 9 \\ 6 &2 \end... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1661443",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
integrate $\int_0^{\frac{\pi}{4}}\frac{dx}{2+\tan x}$
$$\int^{\frac{\pi}{4}}_{0}\frac{dx}{2+\tan x}$$
$v=\tan(\frac{x}{2})$
$\tan x=\frac{2v}{1-v^2}$
$dx=\frac{2\,dv}{1+v^2}$
$$\int^{\frac{\pi}{4}}_0 \frac{dx}{2+\tan x}=\int^{\frac{\pi}{8}}_0 \frac{\frac{2\,dv}{1+v^2}}{2+\frac{2v}{1-v^2}}=\int^{\frac{\pi}{8}}_0 \frac... | Universal trigonometric substitution is done in order to get the tangents instead of sines and cosines. In this case it is no longer necessary, so:
$$t= \tan x,\quad x = \arctan t,\quad dx = \dfrac1{t^2+1},$$
$$J =\int\limits_0^{\pi/4}\frac{dx}{2+\tan x} = \int_0^1\dfrac{dt}{(t+2)(t^2+1)}.$$
Let
$$R(t) = \dfrac1{(t+2)(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662272",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Find the quadratic equation, given three points on the parabola
Given that $f(x)=a(x-b)^2+c$, and that $f(1)=f(5)=-1$ and $f(0)=-11$, find the values of $a,b$,and $c$.
I found the axis of symmetry is $3$, because $f$ has the same value at $1$ and $5$, and $(1+5)/2=3$. Therefore $b=3$. How to find $a$ and $c$?
| You have found $b$ correctly as 3. I.e. $f(x)=a(x-3)^2+c=0$.
From this point you could solve simultaneously to get $a$ and $c$. I.e. solving:
$$f(1)=a(1-3)^2+c=-1$$
$$f(0)=a(0-3)^2+c=-11$$
But there is an even easier way thinking about it graphically:
As you have two points with the same y-value we could also express t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1662363",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that $(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)$ This is Problem 6 of the 2007 Indian National Math Olympiad (INMO).
If $x, y, z$ are positive real numbers, prove that
$(x+y+z)^2(yz+zx+xy)^2 \leq 3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2).$
My best idea was to expand this and simplify. Althoug... | $(x+y+z)^2=x^2+y^2+z^2+2xy+2yz+2xz \\ \le x^2+xy+y^2+y^2+yz+z^2+x^2+xz+z^2$
$3(y^2+yz+z^2)(z^2+zx+x^2)(x^2+xy+y^2)-(x^2+xy+y^2+y^2+yz+z^2+x^2+xz+z^2)(yz+zx+xy)^2=(x+y+z)^2(y^2z^2-xyz^2+x^2z^2-xy^2z-x^2yz+x^2y^2) \ge0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1663294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solve $I=\iiint\limits_\Omega z^2 dv$ using spherical coordinate system, $\Omega: x^2 +y^2 + z^2 \le R^2 \cap x^2 +y^2 + z^2 \le 2Rz$ Question: Solve $I=\iiint\limits_\Omega z^2 dv$ using spherical coordinate system. $\Omega$ is the common part of $x^2 +y^2 + z^2 \le R^2 $ and $ x^2 +y^2 + z^2 \le 2Rz$.
My attempt:Beca... | Your second equation gives $0\leq r\leq 2R\cos(\varphi)$, so $0\leq r\leq \min{(R,2R\cos{\varphi})}$ is the correct bound.
You now need to break up the integral into the regions where each upper bound applies, integrate, and add them.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1665521",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Calculate the limit $\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$ Calculate the limit
$$\lim_{x \to 2} \frac{x^2\sqrt{x+2}-8}{4-x^2}$$
I tried to factorise and to simplify, but I can't find anything good.
$$\lim_{x \to 2} \frac{\frac{x^2(x+2)-8\sqrt{x+2}}{\sqrt{x+2}}}{(4-x^2)}$$
| Substitute $t=\sqrt{x+2}$; if $x\to2$, then $t\to2$, so, after doing $x=t^2-2$, you get
\begin{align}
\lim_{t\to2}\frac{t(t^2-2)^2-8}{4-(t^2-2)^2}
&=\lim_{t\to2}\frac{t^5-4t^3+4t-8}{4t^2-t^4} \\[4px]
&=\lim_{t\to2}\frac{t^3(t^2-4)+4(t-2)}{t^2(4-t^2)} \\[4px]
&=\lim_{t\to2}\left(-t-\frac{4}{t^2(2+t)}\right)
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1666082",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 1
} |
Find $\int \frac {x^2}{x^3+1} dx$. What is my mistake? $\int \frac {x^2}{x^3+1} dx$
$ u = 3x+1, du=3x^2 dx$
$\int \frac{3 du}{u} $
Am I wrong something? Why the answer is $\int \frac{du}{3u}$ instead of $\int \frac{3 du}{u} $ ?
Thank you.
| Set $t=x^3+1$ and $dt=3x^2dx$
Therefore $\color{red}{ x^2dx=\frac{du}{3}}$
$$=\frac 1 3\int \frac 1 t dt =\frac{\ln t}{3}+\mathcal C=\boxed{\color{blue}{\frac 1 3 \ln (x^3+1)+\mathcal C}} $$
hope it helps
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1667729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Prove that if $0 \le k \le \frac {n-1}{2}$, then ${n \choose k} \le {n \choose k+1}$, with equality holding if and only if $k = \frac{n - 1}{2}$ Prove that if $0 \le k \le \frac {n-1}{2}$, then ${n \choose k} \le {n \choose k+1}$. Further, prove that equality is met if and only if $k = \frac {n-1}{2}$
I tried to use th... | $$\binom n{k+1}=\frac{n^\underline{k+1}}{(k+1)!}=\frac{n^\underline{k}(n-k)}{(k+1)k!}=\frac{n-k}{k+1}\cdot \frac{n^\underline{k}}{k!}=\frac{n-k}{k+1}\cdot \binom nk\\
\binom n{k+1}\ge\binom nk \;\;\;\text{ if }\frac{n-k}{k+1}\ge 1\;\;\;\Rightarrow n\ge 2k+1 \;\;\;\Rightarrow k\le\frac {n+1}2\qquad\blacksquare$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1668085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Fraction of length-$n$ binary numbers satisfying a constraint Problem: Derive an expression for the fraction of length-$n$ binary numbers that do not contain the subsequence $010$.
Background: The total number of $n$-bit binary numbers is of course $2^n$. Some of these numbers have the sequence $010$ within it; ot... | I upvoted the answer by @Joriki which is exemplary.
This can also be done using generating functions.
We get
$$G(z) = \frac{1}{1-z} + \frac{1}{1-z} \frac{z}{1-z}
\sum_{q\ge 0} \left(\frac{z^2}{1-z}\frac{z}{1-z}\right)^q
\frac{1}{1-z}.$$
Explanation: first term a string of ones. Remaining term: admissible
sequences c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1669371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Integer solutions for $x^4 + 4xy^3 = z^2$ Find all triplets $(x,y,z)$ of integers so that $$ x^4 + 4xy^3 = z^2. $$
What I've done:
Suppose $x=0$. Then we see $z=0$ and hence $(0,y,0)$ is a solution.
Suppose $y=0$. Then we see $x^4 = z^2$ and hence $(x,0,\pm x^2)$ is a solution.
Suppose $z=0$. Then we see $x^3=-4y^3$, w... | I found an answer based on what MXYMXY posted earlier.
The crucial step is to observe that:
$$
(2x^2)(z-x^2)(z+x^2) = 2x^2(z^2-x^4) = 2x^2(4xy^3) = (2xy)^3.
$$
Note that from my analysis above we may conclude that all factors on the LHS are non-zero.
Also note that $$(2x^2)+(z-x^2) = z+x^2.$$
Let $d = \gcd(2x^2,z-x^2)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1675668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Integral using partial fractions I am having trouble using partial fractions to evaluate
$$\int \frac{6x}{x^3-8} dx.$$ I can find the denominator but using equations to find the numerator is difficult.
| First, factor $x^3-8$. $x^3-8=(x^2+2x+4)(x-2)$
$$\frac{6x}{x^3-8}=\frac{ax+b}{x^2+2x+4}+\frac{c}{x-2}$$
Then multiply the lhs denominator out on the right side:
$$6x=(x-2)(ax+b)+c(x^2+2x+4)$$
When $x=2$, $(x-2)(ax+b)$ vanishes and $$6(2)=c(4+4+4)\Rightarrow c=1$$
Now, we can use $c$ to solve for $ax+b$.
$$6x-(1)(x^2+2x... | {
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Solving $\int {\frac{1}{(-x^{2}+6x-5)^{1/2}}}\;dx$ such that $(1\lt x\lt5)$ using trigonometric substitutions and pythagorean identities First post so I'll get right to the question;
$$\int {\frac{1}{(-x^{2}+6x-5)^{1/2}}}\;dx,\qquad(1\lt x\lt5)$$
To begin with I completed the square which yields: $$-(x-3)^{2}+4$$
subst... | You are on the right track. You should proceed as follows:
$$\begin{align} &\int {\left(\frac{1}{-x^{2}+6x-5}\right)}^{1/2} dx
\\ &=\int {\left(\frac{1}{4-x^{2}+6x-9}\right)}^{1/2} dx
\\ &=\int \frac{dx}{\sqrt{2^2-(x-3)^2}}
\\ &=\int \frac{d(x-3)}{\sqrt{2^2-(x-3)^2}}
\\ &=\color{blue}{\arcsin \frac{x-3}{2} + c}\end{a... | {
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How does one project the gradient at a point on a surface into a plane? I am studying Multivariable Calculus and have come to the following excerpt in my book:
I can see clearly how they get from the given function to
$$ y'(x)\ =\ \frac{3y}{x} $$
And understand that this slope passes through the given point. The foll... | Thanks to user @Moo for helping out in the comments on the original post.
The solution is simple: you know the first derivative of the line and a point it goes through. If you take the first, indefinite integral of your derivative--you get the original line, plus some constant C. Plug your point into this line, solve f... | {
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Factor $16x^4-x^2y^2+y^4$ So basically I have the answer to this problem because it's in the book, but I have no clue how the author solved it. It's from Schaum's Precalculus. Thanks.
Factor $(16x^4)-(x^2y^2)+(y^4) = (4x^2+y^2-3xy)(4x^2+y^2+3xy)$
| Hint
$$16x^4+y^4=(4x^2+y^2)^2-8x^2y^2$$ $$16x^4-x^2y^2+y^4=(4x^2+y^2)^2-9x^2y^2=(4x^2+y^2)^2-(3xy)^2$$ Now, use $a^2-b^2=(a+b)(a-b)$
| {
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Find the solutions of the diophantine equation $(x^2-y^2)(z^2-w^2)=2xyzw$ Let $x,y,z,w$ be postive integers. Find all solutions of:
$$(x^2-y^2)(z^2-w^2)=2xyzw$$
This gives:
$$\left(\dfrac{x}{y}-\dfrac{y}{x}\right)\left(\dfrac{z}{w}-\dfrac{w}{z}\right)=2$$
$$\left(p-\dfrac{1}{p}\right)\left(q-\dfrac{1}{q}\right)=2$$
| Given,
$$\Big(p-\frac{1}{p}\Big)\,r=2\tag1$$
$$q-\frac{1}{q} = r\tag2$$
Solve for $p$ in $(1)$ and $q$ in $(2)$,
$$p=\frac{1\pm\sqrt{r^2+1}}{r}$$
$$q=\frac{r\pm\sqrt{r^2+4}}{2}$$
The discriminant should be a square. We have to find rational number $r=u/v$ such that,
$$u^2+v^2 = w_1^2\\
u^2+4v^2 = w_2^2$$
But $d=4$ is ... | {
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Positive determinant of a symmetric matrix Let $a,b,c,d,e$ positive reals for which $b<\sqrt{ac}$, $c<\sqrt{bd}$, and $d<\sqrt{ec}$. Then, consider the following (Hankel) matrix
$$
M=\begin{pmatrix} a+4b+6c+4d+e & a+3b+3c+d & a+2b+c \\ a+3b+3c+d & a+2b+c & a+b \\ a+2b+c & a+b & a\\ \end{pmatrix}
$$
Is it possible to p... | The claim is not true. The conditions on the coefficients ensure that certain submatrices of the matrix
$$
\left( \begin{array}{ccc} e & d & c \\ d & c & b \\ c & b & a \end{array} \right)
$$
are positive definite. These conditions are not sufficient to ensure that the whole matrix is positive definite.
Here is a coun... | {
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Indefinite integral of a rational function problem... Edit: Corrected my question.
If we have a rational function in general case:
$$f(x)=\frac{1}{(x^2+a^2)^n}$$
And we denote its integral as:
$$I_n=\int \frac{dx}{(x^2+a^2)^n}$$
For $n=1$ we have the integral in the lists:
$$\int \frac{dx}{x^2+a^2}=\frac{1}{a}\arctan\f... | You obviously made a mistake of putting n=2 in the formula, when you are actually trying to calculate I_2, so n+1=2 and n=1.
Putting n=1, you will get the result you got from Wolfram Alpha.
| {
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Integrate $\int \frac{2x}{x^2 +2x+1}$ Here is my attempt:
$$ \frac{2x}{x^2 +2x+1}= \frac{2x}{(x+1)^2 } = \frac{2}{x+1}-\frac{2}{(x+1)^2 }$$
Then I tried to integrate it,I got $2\ln(x+1)+\frac{2}{x+1}+C$ as my answer. Am I right? please correct me if I'm wrong.
| $$=\int\frac{2x+2}{x^2+2x+1}dx-2\int\frac{dx}{x^2+2x+1}$$
set $t=x^2+2x+1$ and $dt=(2x+2)dx$
$$=\int\frac{1}{t}-2\int\frac{dx}{(x+1)^2}dx$$
Set $\nu=x+1$ and $d\nu=dx$
$$=\int\frac{1}{t}-2\int\frac{1}{\nu ^2}d\nu=\ln|t|+\frac{2}{\nu}+\mathcal C=\color{red}{\ln|x^2+2x+1|+\frac{2}{x+1}+\mathcal C}$$
| {
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Solutions of the trigonometric equation $\sin(x+ \frac \pi 4) = \sin(\frac 3 8 \pi-3x)$ The following equation
$$\sin(x+ \frac \pi 4) = \sin(\frac 3 8 \pi-3x)$$
has these solutions:
$$x = \frac {\pi} {32} + k \frac \pi 2 \space \vee x = -\frac 3 {16} \pi + k\pi$$
Is it safe to say the equation has this solution below? ... | The equality $\sin\alpha=\sin\beta$ is equivalent to the condition
$$
\alpha=\beta+2k\pi\quad\text{or}\quad \alpha=\pi-\beta+2k\pi
$$
(where $k$ is an integer). So in your case you have
$$
x+\frac{\pi}{4}=\frac{3\pi}{8}-3x+2k\pi
$$
which gives
$$
\boxed{x=\frac{\pi}{32}+k\frac{\pi}{2}}
$$
or the other branch
$$
x+\frac... | {
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Prove that $n(n+1)(n+5)$ is a multiple of $6$ I need to prove that $n(n+1)(n+5)$ is divisible by 6. where $n$ is a natural number. I have used the method of induction. But not successful
I got the expression $(k^3+6k^2+5k)+3k^2+15k+12$ when $n=k+1$.
The term inside the bracket is divisible by 6 since we have assumed t... | One number among $n$, $n+1$, $n+2$ is a multiple of 3.
Since $5 \pmod 3 \equiv 2$, among $n$, $n+1$ and $n+5$, one is a multiple of 3.
Either $n$ or $n+1$ must be an even number (since they are consecutive numbers).
Therefore, n(n+1)(n+5) is a multiple of 2.
As $n(n+1)(n+5)$ is a has both 3 and 2 as factors, it is a mu... | {
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How to prove this inequality $\sqrt{2x^2-6x+5}+\sqrt{y^2-4y+5}+\sqrt{2x^2-2xy+y^2}\ge \sqrt{10}$ Let $x,y\in R$, show that
$$\sqrt{2x^2-6x+5}+\sqrt{y^2-4y+5}+\sqrt{2x^2-2xy+y^2}\ge \sqrt{10}$$
| Hint:
*
*$\sqrt{2x^2-6x+5}= \sqrt{(x-1)^2+(x-2)^2}$ is the distance from $A(1,2)$ to $B(x,x)$.
*$\sqrt{2x^2-2xy+y^2}=\sqrt{(x-y)^2+x^2}$ is the distance from $B(x,x)$ to $C(y,0)$.
*$\sqrt{y^2-4y+5}=\sqrt{(y-2)^2+(-1)^2 }$ is the distance from $C(y,0)$ to $D(2,-1)$.
The inequality is then equivalent to
$$AB+BC+CD... | {
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Recovering a quadratic polynomial from three values using calculus I'm asked to solve this using calculus:
Let $$ f(x) = ax^2 + bx +c .$$ If $ f(1) = 3 $, $f(2) = 7$, $f(3) = 13$, then find $a$, $b$, and $f(0)$.
I know I can solve this using solving three equations simultaneously. And I can also solve this using Gau... | Use calculus to solve this... Well, I can wedge some calculus in here...
The derivative of $f(x) = a x^2 + b x + c$ with respect to $x$ is $2ax+b$. By the Fundamental Theorem of Calculus, we have that \begin{align*}
f(x) &= f(0) + \int_0^x 2 a t + b \,\mathrm{d}t \\
&= f(0) + (a t^2 + bt)_{t=x} - (a t^2... | {
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Evaluation of $1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\dotsb$
How can we calculate sum of following infinite series
$\displaystyle \bullet\; 1+\frac{1}{3}-\frac{1}{5}-\frac{1}{7}+\frac{1}{9}+\frac{1}{11}-\cdots$
$\displaystyle \bullet\; 1-\frac{1}{7}+\frac{1}{9}-\frac{1}{15}+\frac{1}{17}-\cdots$
$\textb... | Writing $x^2=y$
$$\dfrac{1+x^2+x^4}{(1+x^2)(1+x^4)}=\dfrac{1+y+y^2}{(1+y)(1+y^2)}$$
Let $$\dfrac{1+y+y^2}{(1+y)(1+y^2)}=\dfrac A{1+y}+\dfrac{By+C}{1+y^2}$$
$$\iff1+y+y^2=A(1+y^2)+(1+y)(By+C)=A+C+y(C+B)+y^2(A+B)$$
Comparing the constants, coefficients $y,y^2$
$$A+B=B+C=C+A=1\iff A=B=C=\dfrac12$$
For $\displaystyle \dfr... | {
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Solving modulus inequality $|x - 1| + |x - 6|\le11$ geometrically Find all possible values of $x$ for which $x$ for which the inequality $$|x - 1| + |x - 6|\le11$$ is true.
I know this can be easily solved by taking $3$ cases for$x$ and then taking the intersection of those $3$ cases. The solution will be $-2\ge x\le9$... | $$|1-x| + |x-6| \le 11$$
Let $A, X, B \in \mathbb R^n$ (Euclidean $n$-space). If $X\in \overline{AB}$ ( $X$ is on the line segment $\overline{AB}$ ), we say that $X$ is between $A$ and $B$, and we write this as $A-X-B$.
$$A-X-B \; \text{ if and only if } \; \|A-X\| + \|X-B\| = \|A-B\|$$
On $\mathbb R^1$ (the real numbe... | {
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What is the fast way to evaluate the following integral: $\int{\frac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$? I am trying to evaluate the following integral:
$$\int{\dfrac{\sqrt{x^2+1}}{x^4}\mathrm{d}x}$$
I tried the trigonometric substitution: $u = \tan(x)$. Generally, The whole integral needs two substitutions: $u = \tan(x)$... | We can simplify the expression $\frac{\sqrt{x^2+1}}{x^4}$, assuming that $x>0$, as follows:
$$\frac{\sqrt{x^2+1}}{x^4} = \frac{\sqrt{1+\frac1{x^2}}}{x^3} = \frac1{x^3} \cdot \sqrt{1+\frac1{x^2}}.$$
(We have multiplied both numerator and denominator by $\frac1x$.)
After this simplification $t=\frac1x$ seems as a reasona... | {
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Inequality using Cauchy-Schwarz Let $a,b,c\in\mathbb{R}^+$, prove that $$\sqrt{\frac{a}{b+c}}+\sqrt{\frac{b}{c+a}}+\sqrt{\frac{c}{a+b}}\leq \sqrt{\frac{3}{2}\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)} $$
Hi everyone, I've been trying to do this exercise but any method that I tried has failed. First I tried to use... | Just a hint. You can denote:
$x=\frac{a}{b}, y=\frac{b}{c}, z=\frac{c}{a}$
Then we have: $xyz=1$ i.e. $z = \frac{1}{xy}$
Then use that: $\sqrt \frac{a}{b+c} = \sqrt \frac{1}{b/a+c/a} = \sqrt \frac{1}{1/x + z} $ and do analogically for the other terms in the LHS.
Then if you use that $z = \frac{1}{xy}$ it boil... | {
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Does a closed form solution exist for $x$? $$ \sqrt{x} + x^2 = \sqrt{2} $$
If so, how would one find it?
| Let $u = \sqrt{x}$: $$u + u^4 = \sqrt{2}$$
There is a general formula for the quartic, though it is rather disgusting.
$$u = \frac{1}{2} \sqrt{-\frac{\sqrt[3]{\frac{1}{2} \left(9+\sqrt{3 \left(27+512
\sqrt{2}\right)}\right)}}{3^{2/3}}+\frac{4\ 2^{5/6}}{\sqrt[3]{3 \left(9+\sqrt{3
\left(27+512 \sqrt{2}\right)}\righ... | {
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The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is The number of real roots of $x^5 + 2x^3 + x^2 + 2 = 0 $ is
A. 0;
B. 3;
C. 5;
D. 1.
I don't know how to solve this.
| Observe by trial and error that $-1$ is a root. Then,
$$x^5+2x^3+x^2+2=(x+1)(x^4-x^3+3x^2-2x+2)=(x+1)(x^4-x^3+x^2+2x^2-2x+2)=(x+1)(x^2(x^2-x+1)+2(x^2-x+1))=(x+1)(x^2+2)(x^2-x+1)$$
Thus, there is only one real root.
| {
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Prove that $1+\frac12+\frac13+\cdots+\frac1{2^n} \ge 1+\frac{n}{2}$ Is it true that $1+\dfrac12+\dfrac13+ \dots +\dfrac{1}{2^n} \geq 1+\dfrac{n}{2}$?
If so, a proof would be great!
Thank you!
| Hint: $$\frac{1}{3}+\frac{1}{4}\ge \frac{1}{4}+\frac{1}{4}=\frac{1}{2}$$
$$\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\ge \frac{1}{8}+\frac{1}{8}+\frac{1}{8}+\frac{1}{8}=\frac{1}{2}$$
| {
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Dice probability with limited reroll What is the probability of getting 2 successes when 3 dice are rolled once, and then one die is optionally rerolled?
$$
n = \text{Number of dice} = 3
$$
$$
p_s = \text{Single die success probability (when the face is 3 or more)} = \frac{1}{2}
$$
$$
k = \text{Number of successful die... | It seems to me now that there was a misunderstanding regarding your first question. In that question, you didn't state that you stop if you get the desired number of successes on the first attempt ("What is the probability of having exactly $k$ successes from $n$ coins if all $n$ coins are flipped and there are $x$ suc... | {
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There are $5$ women, $3$ men. How many ways to form a committee of $3$ with at least $1$ member of the opposite sex? I have looked through several topics for similar solutions and I have attempted an answer to the question. Unfortunately, the sample question itself does not have an answer.
From $5$ women and $3$ men, ... | Your first method is conceptually sound, but computationally flawed. With no restrictions on gender we just need to select $3$ people from the available $8$, hence $\binom 83=56$ possible choices. Now, exactly one of these is all male but there are $\binom 53=10$ ways to assemble an all female group. Hence $$56-10-1... | {
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Solve the system of equations $\frac{x}{3} + \frac{y}{2} = \frac{2x}{3} - \frac{y}{6} = 7$ for $x$ and $y$. Here is the problem. The answer is suppose to be x = 12, y = 6
$$\frac{ x }{3} + \frac{y}{2} = \frac{ 2x }{3} - \frac{ y }{6} = 7$$
I've tried this. I'm having trouble figuring out what do on the last step I will... | $6·\left[\frac{x}3+\frac{y}2=7\right]$
$18·\left[\frac{2x}3-\frac{y}6=7\right]$
$====================$
then ...
| {
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How to prove this integral problem? For $0<k<1$ put $$K(k)=\int\limits_{0}^{\pi/2}\frac{\text{d}t}{\sqrt{1-k^2\sin^2 t}} \quad \textbf{and} \quad k'=\sqrt{1-k^2}\in(0,1)$$
*
*Prove that $$K(k)=\frac{1}{1+k}K\left(\frac{2\sqrt{k}}{1+k}\right)$$
*Prove that $$K(k)=\frac{2}{1+k'}K\left(\frac{1-k'}{1+k'}\right)$$
I h... | I will show the first result using a substitution of
$$\tan t = \dfrac{\sin 2 \theta}{k + \cos 2 \theta}.$$
Under this substitution it can be shown (it is a bit of a slog) that
$$dt = \frac{2(1 + k \cos 2\theta)}{1 + 2k \cos 2\theta + k^2} \, d\theta,$$
while the term appearing in the denominator of the integral becom... | {
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An integral for $2\pi+e-9$ Motivation
Lucian asked about the almost-integer $2\pi+e\approx9$ in a comment to a partially answered why question about $e\approx H_8$. This is more involved than approximations to $\pi$ and logarithms because two transcendental constants are included, as in $e^\pi-\pi\approx20$.
Tried so f... | Approximations $e\approx \frac{163}{60} $ and $ \pi \approx \frac{377}{120}$ are related to the integrals
$$\frac{1}{2}\int_0^1 (1-x)^2\left(e^x-1-x-\frac{x^2}{2}\right)dx = e-\frac{163}{60}$$
and
$$\frac{1}{2}\int_0^1 \frac{x^5(1-x)^6}{1+x^2}dx = \frac{377}{120}-\pi.$$
Combining both, we can build
$$\frac{1}{2} \int_0... | {
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An alternative to integration by trigonometric substitution? When I was a Calculus II student many (about 4800) moons ago, our professor taught us an alternative to trig sub. For example, if we have
$$
\int \frac{dx}{x^2\sqrt{x^2 - 9}},
$$
we would evaluate with trig sub by letting $x = 3\sec\theta$ and we'd get:
\b... | set $$\sqrt{x^2-9}=t+x$$ then you will get $$x=\frac{t^2+9}{2t}$$ then we will have $$dx=\frac{1}{2}\frac{t^2-9}{t^2}dt$$
| {
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"url": "https://math.stackexchange.com/questions/1711918",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Simplifying the integral $\int\frac{dx}{(3 + 2\sin x - \cos x)}$ by an easy approach $$I=\displaystyle\int\frac{dx}{(3 + 2\sin x - \cos x)}$$
If $$\tan\left(\frac{x}{2}\right)=u$$
or $$x=2\cdot\tan^{-1}(u)$$
Then,
$$\sin{x}=\dfrac{2u}{1+u^2}$$
$$\cos{x}=\dfrac{1-u^2}{1+u^2}$$
$$dx=\dfrac{2}{1+u^2}$$
Substitute $$\tan\l... | You could write $2\sin(x)-\cos(x) = \sqrt{5} (\frac{2}{\sqrt{5}} \sin x - \frac{1}{\sqrt{5}} \cos x) = \sqrt{5} \sin (x - \alpha)$, with $\tan \alpha = \frac{1}{2}$, so that the required integral becomes $\int \dfrac{dx}{3 + \sqrt{5} \sin(x-\alpha)}$ and then change the integration variable from $x$ to $y = x - \alpha$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712481",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
Find $\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$ $$\lim_{x \to 0}\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}$$
What I have done is to take the $\ln$
$$e^{\lim_{x \to 0}\ln\left(\left(\frac{\cos x}{\cos(2x)}\right)^{\frac{1}{x^2}}\right)}$$
$$y={\lim_{x \to 0}{\frac{1}{x^2}}\cdot \ln\f... | If you start from $g(x) = \frac{\cos x}{\cos(2x)}^{\frac{1}{x^2}}$ and first take logs you have $\ln g(x) = {\frac{1}{x^2}} (\ln \frac{\cos x}{\cos(2x)})$
Expanding $\frac{\cos(x)}{\cos(2x)} \approx 1 + 3/2 x^2$ and $\ln (1+x) \approx x$ you have $\ln g(x) \approx \frac{3/2 x^2}{x^2} = 3/2$, so that the limit is $e^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1712973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Let $a,b,c$ be positive reals. Prove that $ab(a+b)+bc(b+c)+ca(c+a)\geq \sum_{cyc} ab\sqrt {{a\over b}(c+a)(c+b)}$
Let $a,b,c$ be positive reals. Prove that
$$ab(a+b)+bc(b+c)+ca(c+a)\geq \sum_{cyc} ab\sqrt {{a\over b}(c+a)(c+b)}$$
I have no clue and I am not good at inequalities. Please help.
| Denote $A = a^2b+b^2c+c^2a,B=ab^2+bc^2+ca^2$, we have $LHS=A+B$.
By AM-GM inequality $$ab\sqrt{{a\over b}(c+a)(c+b)} = \sqrt{a^2(b+c)\cdot ab(a+c)} \le \frac{a^2(b+c) + ab(a+c)}{2},$$
Thus
$$RHS = \sum ab\sqrt{{a\over b}(c+a)(c+b)} \le \frac{1}{2}\sum [a^2(b+c) + ab(a+c)]=\frac{1}{2}(2A+B+3abc)\quad (1)$$
By AM-GM ine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713225",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Find polynomials $f(x),g(x)$ such that $f(x)p(x) + g(x)q(x) = 2x^2 + 6x +2$ Let $$p(x) = x^4 + 7x^3 + 14x^2 + 7x+1 $$$$q(x) = x^4 +10x^3 + 23x^2 + 10x+1$$
Find polynomials $f(x),g(x)$ with rational coefficients such that $$f(x)p(x) + g(x)q(x) = 2x^2 + 6x +2$$
I totally have no idea to solve this problem... Please help... | If you factor both polynomials in $\mathbb Q[x]$ you'll get
$p(x) = (x^2+3x+1) (x^2+4x+1)$
$q(x) = (x^2+3x+1) (x^2+7x+1)$
These factors are irreducible (Why?).
Therefore $gcd(p,q) = x^2+3x+1$. Now with the Euclidean algorithm (or some guessing) you should be able to get two polynomials $r,s$ such that $$r(x^2+4 x+1)+s(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
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Prove that $\lim_{x \to 3} \sqrt{x+1} = 2$ Prove that $\displaystyle\lim_{x \to 3} \sqrt{x+1} = 2$
Attempt:
$0 < |x - 3| < \delta \Rightarrow |\sqrt{x+1} - 2| < \epsilon$
Well $|\sqrt{x+1} - 2| = |(\sqrt{x+1} - 2) \cdot \displaystyle\frac{\sqrt{x+1} + 2}{\sqrt{x+1}+2}| = |\frac{x-3}{\sqrt{x+1} + 2}| = |x-3| \cdot \fra... | Hint:
If $f(x)$ is continuous,
$$\lim_{x\to c} f(x)=f(c)$$
and $\sqrt{x+1}$ is continuous on $[-1,\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713815",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
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Find $(a,b)$ such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$ implies $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$
Let $f$ denote any function such that $\lim\limits_{x\to 0}\frac{f(x)}{x}=1$. Find the value of $a$ and $b$, assuming that $\lim\limits_{x\to 0}\frac{x(1+a\cos x)-b\sin x}{(f(x))^3}=1$.
... | Others have pointed to your mistake.
The numerator in your third line can be written as
$$x\bigl(1+a(1-{x^2\over2}+?x^4)\bigr)-b\bigl(x-{x^3\over6}+?x^5\bigr)=(1+a-b)x+\bigl(-{a\over2}+{b\over6}\bigr)x^3+?x^5\ ,$$
where each question mark represents some convergent power series. It follows that $a$ and $b$ have to sati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1713981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 2
} |
Algebraic vector proof of Lagrange's identity $$|v · w|
^2 + |v × w|
^2 = |v|^2 |w|^2 $$
Edit
Despite doing it multiple times it seems I have made a meal of the expansion see Jean-Claude's answer for a great explanation
Using $v = (v_1,v_2,v_3)$ and $w = (w_1, w_2, w_3)$ i have expanded the LHS and gotten
$$(v_2)^2(w_3... | You have
\begin{align*}
|v\cdot w|^2 &= (v_1w_1+v_2w_2+v_3w_3)^2 \\
&= v_1^2w_1^2 + v_2^2w_2^2 + v_3^2w_3^2 + 2v_1w_1v_2w_2 + 2v_2w_2v_3w_3 + 2v_3w_3v_1w_1 \text{.} \\
v\times w &= \begin{pmatrix}v_2w_3-v_3w_2 \\ v_3w_1-v_1w_3 \\ v_1w_2-v_2w_1\end{pmatrix} \text{.} \\
|v\times w|^2 &= (v_2w_3-v_3w_2)^2+ (v_3w_1-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 5,
"answer_id": 1
} |
Find the roots of quadratic polynomial given one root of another quadratic polynomial? if $a,b,c$ are Real numbers and $1$ is a root of equation $ax^2+bx+c=0$ then curve $y = 4ax^2+3bx+2c$ , (a is not zero) intersects $x$ axis at how many points?
I get a relation $a+b+c = 0$ I tried substituting this in given polynomia... | You found that $a+b+c=0$, which means that $c=-a-b$, so we have $y=4ax^2+3bx-(2a+2b)$. The discriminant is as follows:
$$(3b)^2-4(4a)(-(2a+2b))=9b^2+16a(2a+2b)=9b^2+32a^2+32ab$$
Now, if we plug in $a=1$ and $b=1$ into the discriminant, we clearly get a positive number, meaning there are two solutions. We need to prove ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715589",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 4
} |
Steps to solve $\lim_{n \to \infty} (\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} ) = 4 $ $$\lim_{n \to \infty} \left(\frac{ \sqrt{n^4+1} - \sqrt{n^4-1}}{ \frac1{(2n+1)^2}} \right) = 4 $$
I think $\sqrt{n^4+1} - \sqrt{n^4-1}$ is approaching to zero, but it is not correct. What steps can evaluate above limit t... | You could also write $\sqrt{n^4 \pm 1} = n^2 \sqrt{1 \pm \frac{1}{n^4}}$ and expand the square roots out using the Binomial Theorem (for large $n, \, \frac{1}{n^4}$ is small), and see what this leaves you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1715708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Prove that $\sigma(n^2)=\sum_{d\mid n} 2^{\omega(d)}$ Let $\omega(n)$ denote the number of distinct prime divisors of $n>1$, with $\omega(1)=0$.
(a) Show that $2^{\omega(n)}$ is a multiplicative function.
(b) Prove that $$\sigma(n^2)=\sum_{d\mid n} 2^{\omega(d)}.$$
I have done the part (a) and I am stuck by (b).
Fir... | Suppose we seek to show that
$$\tau(n^2) =
\sum_{d|n} 2^{\omega(d)}.$$
This can be done using Dirichlet series and Euler products.
We have for the RHS and
$$\sum_{n\ge 1} \frac{1}{n^s} 2^{\omega(n)}$$
the Euler product
$$\prod_p
\left(1 + \frac{2}{p^s} + \frac{2}{p^{2s}} + \frac{2}{p^{3s}}
+\cdots\right).$$
which is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to show pattern in iterated differentiation Generally, how would one go about proving general patterns of $n$th derivative?
The specific problem:
$$f(x)=(4-x)^{-0.5}$$
Show that:
$$f^n(0) = 0.5 \left( \frac{(1)(3)(5)\cdots(2n-1)}{8^n} \right)$$
| We can write this formally (using the odd factorial)
$$ f^{(n)}(0) = \frac{1}{2^{3n+1}} \prod_{k=1}^n (2k-1) = \frac{1}{2^{4n+1}} \frac{(2n)!}{n!}
$$
Using the power rule,
$$ f^{(n)}(x) = (-1)^n(4-x)^{-1/2 - n} \prod_{k=1}^{n} \left(\frac{1}{2} - k\right) = (4-x)^{-1/2 - n} \frac{1}{2^n} \prod_{k=1}^{n} \left(2k - 1\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1717923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Does the series $\sum_{n=1}^{\infty}{\frac{\sin^2(\sqrt{n})}{n}}$ converge?
Does the following series converge?
$$\sum_{n=1}^{\infty}{\frac{\sin^2(\sqrt{n})}{n}}$$
It shouldn't, but I have no idea how to prove it. I was wondering about Integral Criterion, but the assumptions are not satisfied. Or perhaps Dirichlet... | Observe that $\sin^2(\sqrt{n}) \geq 1/4$ iff $|\sin(\sqrt{n})| \geq 1/2$ iff
$$\frac{\pi}{6} + k\pi \leq \sqrt{n} \leq \frac{5\pi}{6} + k\pi$$
for some nonnegative integer $k$. This chain of inequalities is equivalent to
$$\left(\frac{\pi}{6} + k\pi\right)^2 \leq n \leq \left(\frac{5\pi}{6} + k\pi\right)^2$$
For a fix... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719005",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 1
} |
Find the minimal polynomial of $\sqrt{3} +i$ over $Q(i)$ and over $Q(\sqrt{3})$
Find the minimum polynomial of $\sqrt{3} +i$ over $Q(i)$ and $Q(\sqrt{3})$.
My solution:
*
*Over $Q(i)$ : Suppose $a = \sqrt{3} +i$, $a-\sqrt{3} =i$;
so, the minimal polynomial is $x-\sqrt{3} -i=0$.
*Over $Q(\sqrt{3})$: Suppose a=$\s... | Hint Consider the product of the monomial $$x - (\sqrt{3} + i)$$ with its conjugate, in two different senses of the word conjugate. (One of the products recovers the correct partial solution from the question: The minimal polynomial of $\sqrt{3} + i$ over $\Bbb Q (\sqrt{3})$ is $x^2 - 2\sqrt{3} x + 4 = 0$.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1719431",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Summation $\frac{3}{1^2}+\frac{5}{1^2+2^2}+\frac{7}{1^2+2^2+3^2}+....$ I came across a question today...
Q. The sum $\dfrac{3}{1^2}+\dfrac{5}{1^2+2^2}+\dfrac{7}{1^2+2^2+3^2}+....$ upto $11$ terms is?
Okay, I think it can be written as
$$\sum_{r=1}^{11}\dfrac{2r+1}{1^2+2^2+...+r^2}$$
I can't see how to simplify it.
An... | As $1^2+2^2+\cdots+r^2=\dfrac{r(r+1)(2r+1)}6$
$$\dfrac{2r+1}{1^2+2^2+\cdots+r^2}=6\left(\dfrac{r+1-r}{r(r+1)}\right)=?$$
Can you recognize the Telescoping nature?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1720166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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The number of words of length $n$ from specific alphabet with rule of creating. Determination of the number of words of length n formed from the alphabet $\{ a, b , c, d \} $, where the letters $a , b $ are not adjacent.
How to find out a recurrence and explicit formula for it ?
| If $a$ and $b$ are not adjacent in a sequence, we call it a valid sequence. Let the number of valid sequences of length $n$ ending with $a$ or $b$ be $p_n$, and let the number of valid sequences of length $n$ ending with $c$ or $d$ be $q_n$. Let the total number of valid sequences of length $n$ be $k_n$.
The first equa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1721000",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Tangent line problems Problem 1
Find common tangent to the curve:
$y+x^2=-4$ and $x^2+y^2=4$.
My idea:
Let $t1... y=ax+b$ is a tangent line to the first curve.
Let $t2... y=cx+d$ is a tangent line to the second curve.
But we are seracing for common tangent so that mean that $a=c$ and $b=d$.
By using implicit different... | Problem$\#1:$
The equation of tangent of $$x^2+y+4=0$$ at $(t,-(4+t^2)),$
$$x(t)+\dfrac{y+\{-(4+t^2)\}}2+4=0\iff2t x+y+4-t^2=0$$
Now this will be a tangent of $$x^2+y^2=4$$ if the perpendicular distance of tangent to center of circle $=$ radius of circle
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1725708",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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If $\sin y+\cos y=\frac{1}{2}$ then find $\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y}$
If $$\sin y+\cos y=\frac{1}{2} \tag{1}$$ Then find $$x=\frac{\sin^3y}{\cos^2y}+\frac{\cos^3y}{\sin^2y} \tag{2}$$
Given that $$\sin y=\frac{1}{2}-\cos y$$
Squaring both sides we get
$$8\cos^2y-4\cos y-3=0$$
Hence
$$\cos y=\frac... | Write
$$
x = s (s^2/c^2) + c (c^2/s^2)
$$
then replace $c^2 = 1-s^2$ and vice versa to get
$$
x = s (1-c^2)/c^2 + c (1-s^2)/s^2 = (s/c^2) - s + (c/s^2) - c
$$
so that
$$
x + s + c = \frac{s}{c^2} + \frac{c}{s^2} = \frac{sc^2 + cs^2}{c^2s^2} = \frac{c + s}{cs}
$$
Since $c + s = 1/2$, this gives
$$
x + \frac{1}{2} = \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1728834",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
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Linear transformation of a subspace How can I calculate the image of a linear transformation of a subspace?
Example:
Given a subspace $A$ defined by $x + y + 2z=0$, and a linear transformation defined by the matrix
$$M=
\left(
\begin{matrix}
1 & 2 & -1\\
0 & 2 & 3\\
1 & -1 & 1\\
\end{matrix}\right)
$$
What is the imag... | x+ y+ 2x is more commonly written as 3x+ y so I am going to assume you mean x+ y+ 2z= 0. From that equation, x= -y- 2z so any vector satisfying that can be written <-y- 2z, y, z>. Applying M to that,
$\begin{pmatrix}1 & 2 & -1 \\ 0 & 2 & 3 \\ 1 & -1 & 1 \end{pmatrix}\begin{pmatrix}-y- 2z \\ y \\ z \end{pmatrix}= \be... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1729799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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Use Darboux sums to calculate the area of $\sqrt x$ in $[0,1]$ I know that $$\Delta x=\frac{b-a}{n}=\frac{1-0}{n}=\frac{1}{n}$$
So $$\overline{D}=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^n\sup f(x_i)\Delta x=\displaystyle\lim_{n\to\infty}\displaystyle\sum_{i=1}^nf(x_i)\Delta x=\displaystyle\lim_{n\to\infty... | It is possible to compute the area (or integral) directly as the limit of the upper Darboux sum.
As you showed
$$\overline{D} = \lim_{n\to\infty}\frac{1}{n\sqrt n} \sum_{i=1}^n \sqrt {i}.$$
We can use the binomial expansion $(1-x)^{\alpha} =1 - \alpha x + O(x^2).$
For $i > 1,$ we have
$$\begin{align}(i-1)^{3/2} &= i^{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1730273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Simple differential equation $\frac{x}{y}=\frac{y'}{x+1}$ $$\frac{x}{y}=\frac{y'}{x+1}$$
The solution to this is very easy, we just multiply both sides by
$$(x+1)y\ \mathrm dx$$
Then we have
$$(x^2+x)\ \mathrm dx=y\ \mathrm dy$$
Then we integrate both sides to get solution
$$\frac{x^3}{3}+ \frac{x^2}{2}=\frac{y^2}{2... | $$\frac{x}{y}=\frac{1}{x+1}\frac{\mathrm dy}{\mathrm dx}$$
The above equation does not hold for $x=-1$ and $y=0$ because division by zero is undefined. This implies that $-1$ is excluded from the domain of $y$ and $0$ is excluded from the codomain of $y$. Therefore we can safely divide by $x+1$ and $y$, respectively. W... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1734580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Show $\frac{\pi}{\cos \pi z}=4\sum\limits_{n=0}^\infty \frac{(-1)^n(2n+1)}{(2n+1)^2-4z^2}$ using partial fraction decomposition of $\cot$ and $\tan$ I am asked to show that $\frac{\pi}{\cos \pi z}=4\sum\limits_{n=0}^\infty \frac{(-1)^n(2n+1)}{(2n+1)^2-4z^2}$. Here is what I have got so far,
$\frac{\pi}{\cos (\pi z)}=\f... | Since
$
\frac{1}{\sin z}=\cot z+\tan\left(\frac{z}{2}\right)
$, and using the partial fraction decomposition of the tangent and cotangent
$$
\pi \tan(\pi z) = 8z\sum_{n=0}^{+\infty}\frac{1}{(2n + 1)^2 − 4z^2}\quad\text{and}\quad
\pi \cot(\pi z) = \frac{1}{z}+2z\sum_{n=1}^{+\infty}\frac{1}{z^2 − n^2}
$$
we have
\begin{a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1735250",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Define the image of the function $f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$ Question
Define the range of the function:
$$f(x)= 2 \arctan x + \arcsin \frac {2x}{1+x^2}$$
Answer attempt
I assume that the domain for $f(x)$ is in ${\rm I\!R}$.
If we draw a triangle with the sides: $1, x$ and $\sqrt{1+x^2}$, with the ... | This range is correct. While you start with the assumption that the domain for $f(x)$ is $x\in\mathbb{R}$, we know that the domain of $f(x)$ is limited by the domain of $\arcsin(x)$ which is $[-1,1]$.
So, this means that $-1\leq\dfrac{2x}{1+x^2}\leq 1$ which is precisely $x\in[-1,1]$
Since $\arctan(x)$ is an increasing... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1738168",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
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Fewest steps to show $\sin^2(s) = \frac{1}{2} - \frac{1}{2} \cos(2s)$? Is there a method for getting from this statement
$
\sin^2(s)
$
to this statement
$
\frac{1}{2}\ -\ \frac{1}{2}\cos(2s)
$
in fewer steps than the following?
$\begin{align}
\cos(2s) &= \cos^2(s) - \sin^2(s) \\
\frac{1}{2} \cos(2s) &= \frac{1}{2} \cos... | $$\frac{1-\cos(2s)}2=\frac{\cos^2(x)+\sin^2(x)-\cos^2(s)+\sin^2(x)}2=\sin^2(s).$$
What else ?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739029",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
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Prove $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$ Prove that $\frac{2\sec\theta +3\tan\theta+5\sin\theta-7\cos\theta+5}{2\tan\theta +3\sec\theta+5\cos\theta+7\sin\theta+8}=\frac{1-\cos\theta}{\sin\theta}$
$$\frac{2\sec\... | Let's simplify our notation by writing $c=\cos\theta$ and $s=\sin\theta$. First note that $$\frac{1-\cos\theta}{\sin\theta}=\frac{\sin\theta}{1+\cos\theta}.$$ So, for arbitrary $a$ and $b$, not both zero, we can write either fraction as $$\frac{a(1-c)+bs}{as+b(1+c)}.$$It's not hard to spot that setting $a=5c$ and $b=3+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1739433",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
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How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$? How to calculate the sum: $\sum_{n=1}^{\infty} \frac{1}{n(n+3)}$ ?
I know the sum converges because it is a positive sum for every $n$ and it is smaller than $\sum_{n=1}^{\infty} \frac{1}{n(n+1)}$ that converges and equals $1$. I need a direction...
| Using partial fraction decomposition
$$
\frac{1}{n(n+3)} = \frac{1}{3} \cdot \frac{1}{n} - \frac{1}{3} \cdot \frac{1}{n+3}
$$
convince yourself that for any $f$, we have
$$
\sum_{n=1}^{m} \left(f(n)-f(n+3)\right) = f(1) + f(2) + f(3) - f(m+3) - f(m+2) - f(m+1)
$$
and now take the limit of $m \to \infty$ gettin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1740832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 2
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Solving exponential equation (quadratic type) I fail trying to solve the following equation:
$9^x-6^x-2^{2x+1}=0$
Trying to write it as a quadratic equation makes my constant term exponential
$(3^x)^2-2^x3^x-2^{2x+1}=0$
How can I solve this type of problem most efficiently?
| \begin{align*}
9^x-6^x-2^{2x+1} &=0 \\
(\color{red}{3^x})^2-(\color{red}{3^x})(\color{blue}{2^x})-
2(\color{blue}{2^x})^2 &=0 \\
(\color{red}{3^x}+\color{blue}{2^x})
(\color{red}{3^x}-2\cdot \color{blue}{2^x}) &= 0
\end{align*}
Rejecting $\color{red}{3^x}+\color{blue}{2^x}=0$, we have
\begin{align*}
\left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741164",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Why is the double angle formula not used in this solution? solve $\sin 2x = 0.5$
I am looking at the answer to this question and it does not use the double angle formula to make this:
$\sin 2x = 2\sin x \cos x$
And instead use
$2x = \arcsin 0.5$
Why is the double angle formula not used in this situation?
| Using the double angle formula here ends you up with two things that must multiply together to become $0.5$, and we would need to solve a nasty quadratic to discover $\sin x$ and $\cos x$. Just using $\arcsin$ directly makes your life easier: $\arcsin 0.5$ is very easy to obtain, and we reduce the rest of the problem ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Solve the equation $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$ Solve the following equation: $\sqrt{1-x}=2x^2-1+2x\sqrt{1-x^2}$
Unfortunately I have no idea.
| Put $x=\cos \theta$, then:
$\sqrt{1-\cos \theta} = 2\cos^2 \theta -1 \pm 2\cos \theta\sin\theta = \cos 2\theta \pm \sin 2\theta$.
Squaring ,we get:
$1-\cos \theta = 1 \pm \sin 4\theta$.
Therefore $\cos \theta = \pm\sin 4\theta$. The solutions of this equation are easy to find. Also, one must eliminate extraneous soluti... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1741924",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
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Proving $1 \cdot 2 + 2 \cdot 3 + \cdots + n(n + 1) = 2\binom{n + 2}{3}$by math induction? I am working on a problem, but I don't know whether or not to use math induction on it. Here's the problem:
Prove that for all integers $n \geq 1$,
$$1 \cdot 2 + 2 \cdot 3 + \cdots + n(n + 1) = 2\binom{n + 2}{3}$$
If I used math i... | Check carefully and algebraically each step:
$$\overbrace{1\cdot2+2\cdot3+\ldots+n(n+1)}^{=2\binom{n+2}3}+(n+1)(n+2)=2\frac{(n+2)!}{3!(n-1)!}+(n+1)(n+2)=$$
$$=\frac{2(n+2)!n+6(n+2)!}{6n!}=\frac{(n+2)!}{3n!}\left(n+3\right)=2\frac{(n+3)!}{3!n!}=2\binom{n+3}3$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1742658",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Epsilon delta proof with fractions, given what delta must be Choose $\delta = \min (1, \frac{\epsilon}{10})$
is the following statement true?
$$0 < |x − 1| < δ\text{ implies that }\left|\frac{x^2+3x}{x^2+1} − 2\right| < ε$$
Okay so this is what I have so far
$$\left|\frac{x^2+3x}{x^2+1} − 2\right| = |x-1|\frac{|-x+2|}... | An alternative answer keeping the $\delta$:
We know that $|x-1|<\delta$ and
$$\left|\frac{x^2+3x}{x^2+1}-2\right|= |x-1|\frac{|x+2|}{|x^2+1|}$$
Now, we use the bounds
$$|x-2| = |x-1-1|\leq |x-1|+1 \leq\delta +1$$
and
$$|x^2+1|=x^2+1\geq 1$$
to obtain
$$\left|\frac{x^2+3x}{x^2+1}-2\right|= |x-1|\frac{|x+2|}{|x^2+1|}\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1744776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How do I show that $\frac {a^2}b + \frac {b^2}c + \frac {c^2}a \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ca}?$ For positive real numbers $a, b, c$, show that
$$\frac {a^2}b + \frac {b^2}c + \frac {c^2}a \ge \frac {(a + b + c)(a^2 + b^2 + c^2)}{ab + bc + ca}.$$
I don't know how to solve this at all. Can you pro... | $$LHS-RHS=\frac{1}{2(ab+bc+ca)}\cdot \left[\frac{b(ab-c^2)^2}{ca}+\frac{c(bc-a^2)^2}{ab}+\frac{a(ca-b^2)^2}{bc}\right]\ge 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1746023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning? You cast a pair of dice. If you get a 6 and an 8 before 7 comes up twice, you win. What is the probability of winning?
What I tried was
\begin{align*}
P(X=6) & = \frac{5}{36}\\
P(X=8) & = \frac{5}{... | It is easier to compute the probability of first getting two $7's\;\;$ and take the complement
Totals other than $6,7,8$ don't matter, so we can simply take their odds in favor as $5:6:5$
Consider the three mutually exclusive starts $(a)\;\;7-7,\;\;(b)\;\; 7-(6\;\;or\;\; 8),\;\;(c)\;\; (6\;\; or\;\; 8)-7$
In $(a),\;\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1749045",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Let $\textbf{v} = (1, 1, 1, 1)$. Find a basis for... Let $\textbf{v}=(1,1,1,1)$. Find a basis for $\{\textbf{u}\in\Bbb{R}^4\ |\ \textbf{u}\cdot\textbf{v}=0\}$
How can I do this? In particular, I do not understand $\textbf{u}\cdot\textbf{v}=0.$
| Denote $W=\{u\in\mathbb{R}^4\mid u\cdot v=0\}$, we claim that the set
$\beta=\left\{
\begin{pmatrix}-1\\1\\0\\0\end{pmatrix},
\begin{pmatrix}-1\\0\\1\\0\end{pmatrix},\begin{pmatrix}-1\\0\\0\\1\end{pmatrix}
\right\}$ is a basis for $W$. First, given
$u=(u_1,u_2,u_3,u_4)\in W$, then we have
$$0=u\cdot v=
\begin{pmatrix}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1751386",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove this inequality: $\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$
Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that
$$\frac{b+c}{\sqrt{a}}+\frac{c+a}{\sqrt{b}}+\frac{a+b}{\sqrt{c}}\geq \sqrt{a}+\sqrt{b}+\sqrt{c}+3$$
I tried various methods.... | Id est, we need to prove that $\sum\limits_{cyc}(a^2b+a^2c-a^{\frac{4}{3}}b^{\frac{5}{6}}c^{\frac{5}{6}}-abc)\geq0$
which is AM-GM and Muirhead because $(2,1,0)\succ\left(\frac{4}{3},\frac{5}{6},\frac{5}{6}\right)$
Your another problem can be solved by the same way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752175",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 2
} |
Find the limit $\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$ Find the limit if it exists. (this exercise is taken from Calculus - The Classic Edition by Swokowski Chapter 10, section 1, no. 9, p.498)
$$\lim_{x\rightarrow 0} \frac{\sin x - x}{\tan x - x}$$
since the limit is $0/0$ therefore, we use L'Hopital's ... | $$\frac{\cos(x)-1}{\sec^2(x)-1}=\frac{\cos(x)-1}{\frac{1}{\cos^2(x)}-1}=\frac{\cos^2(x)(\cos(x)-1))}{1-\cos^2(x)}=\frac{\cos^2(x)(\cos(x)-1)}{(1-\cos(x))(1+\cos(x))}=\frac{- \cos^2(x)(1-\cos(x))}{(1-\cos(x))(1+\cos(x))}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1752715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 7,
"answer_id": 0
} |
How do I find $[T]_E$ when given this linear transformation $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$? In this question I have been asked to find $[T]_E$.
I was given that $T: R_3[x] \rightarrow R_3[x]$ is a linear transformation
defined as: $T(p(x)) = (1+2x^2)p''(x)+(1-2x)p'(x)+p(x)+p(0)$,
$p(x),p(x)',p(x)'' \... | Actually, $[T]_E = [[Te_1]_E,[Te_2]_E,[Te_3]_E].$ Compute
$$\begin{align}
Te_1 & = T(1)\\
& = (1 + 2x^2)1'' + (1 - 2x)1' + 1 + 1\\
& = 2\\
& = \color{#08F}2 \cdot 1 + \color{#08F}0 \cdot x + \color{#08F}0 \cdot x^2\\
& = \color{#08F}2 e_1 + \color{#08F}0e_2 + \color{#08F}0e_3,\\
[Te_1]_E & = {\begin{bmatrix}\color... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753028",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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For the given u(x,y) find a v(x,y) such that f(z) = u(x,y) + iv(x,y) is holomorphic in some region of $\mathbb{C}$ I used the Cauchy Riemann equations and got two different functions v(x,y) by integrating. Should I just add them up?
My u(x,y) is $\frac{x}{x^2 + y^2}$
My partial wrt x is $\frac{y^2 - x^2}{x^2 + y^2}$... | $u(x,y)$ partial wrt $x$ is $\frac{y^{2}-x^{2}}{(x^2+y^2)^2}$, partial wrt $y$ is
$\frac{-2xy}{(x^2+y^2)^2}$.
By the rule of Cauchy-Riemann Equations,we have
$v(x,y)$ partial wrt $y$ is $\frac{y^{2}-x^{2}}{(x^2+y^2)^2}$, partial wrt $x$ is
$\frac{-2xy}{(x^2+y^2)^2}$.
Then we can let $v(x,y)=\frac{-y}{x^2+y^2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1753590",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Prove that $ζ(4)=π^4/90$ knowing that $\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right)$ The question
Knowing that:
$$\sin(πz) = πz \prod_{n=1}^∞ \left( 1 - \frac{z^2}{n^2} \right) \tag{1}$$
obtain the Taylor series expansion of $\frac{\sin(πz)}{πz}$ to deduce:
$$ \sum_{1 ≤ n_1 < n_2 < … < n_k} \frac{1}{n... | Hint: $\frac{\pi^4}{90}=\left(\frac{\pi^2}{6}\right)^2-2\left(\frac{\pi^4}{120}\right)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760088",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
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Finding out a limit using Taylor series. So the limit is the following:
$$\lim_{x \to 0}{\frac{x^2-\frac{x^6}{2}-x^2 \cos (x^2)}{\sin (x^{10})}}$$
Expansions for $\sin(x)$ and $\cos(x)$ are given:
$$\sin x = x-\frac{x^3}{3!} + \frac{x^5}{5!}-...+(-1)^{n-1}\frac{x^{2n-1}}{(2n-1)!} + o(x^{2n})$$
$$\cos x = 1-\frac{x^2}{... | How about factoring by $x^{10}$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1760199",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
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Does an analytical form exist for the following integral I have an integral
$$f(n,a)=\int_0^{2\pi}\mathop{\mathrm{d}x}\frac{\cos(nx)\cos^2x}{1-a\cos^2x},$$
where $n$ is an even integer and $0<a<1$ is a real number. Does an analytical form exist for this integral?
Edit: Here is a partial answer that I discovered:
$$\int... | I'm kind of surprised that this problem hasn't been attacked with a geometric series yet. Let $0<b<1$. Then
$$\sum_{k=0}^{\infty}b^ke^{ik\theta}=\sum_{k=0}^{\infty}b^k\cos k\theta+i\sum_{k=0}^{\infty}b^k\sin k\theta=\frac1{1-be^{i\theta}}=\frac{1-b\cos\theta+ib\sin\theta}{1-2b\cos\theta+b^2}$$
Taking real parts and set... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Show $x_n := \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1}$ is increasing in n for $p \in (0,1)$ I need to show that $x_n := \frac{1-p^{n+1}}{1-p^n} \frac{n}{n+1}$ is increasing in $n$ for $p \in (0,1)$. My attempts have involved trying to show
$\begin{eqnarray*} \frac{1-p^{n+2}}{1-p^{n+1}} \frac{n+1}{n+2} &>& \frac{1-p^{n+1}}... | Writing $x_n = \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}$, we see:
$$x_n>x_{n-1}\iff \frac{1-p^{n+1}}{1-p^n}\frac{n}{n+1}>\frac{1-p^{n}}{1-p^{n-1}}\frac{n-1}{n}$$
$$\iff(1-p^{n-1})(1-p^{n+1})n^2>(1-p^n)^2(n^2-1)$$
$$\iff (1-p^n)^2 > n^2[(1-p^n)^2-(1-p^{n-1})(1-p^{n+1})]=n^2[p^{n-1}+p^{n+1}-2p^n]$$
$$\iff\frac{(1-p^n)^2}{p^n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1761033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Evaluation of $\lim_{x\rightarrow 0}\left(\frac{16^x+9^x}{2}\right)^{\frac{1}{x}}$
Evaluation of $\displaystyle \lim_{x\rightarrow 0}\left(\frac{16^x+9^x}{2}\right)^{\frac{1}{x}}$
$\bf{My\; Try::}$ I am Using above question using Sandwich Theorem
So Using $\bf{A.M\geq G.M\;,}$ We get
$$\frac{16^x+9^x}{2}\geq (16^x\cd... | Whenever the base and the exponent are variables it makes sense to take logs. If $L$ is the desired limit then
\begin{align}
\log L &= \log\left\{\lim_{x \to 0}\left(\frac{16^{x} + 9^{x}}{2}\right)^{1/x}\right\}\notag\\
&= \lim_{x \to 0}\log\left(\frac{16^{x} + 9^{x}}{2}\right)^{1/x}\text{ (via continuity of log)}\nota... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1762733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 5
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How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$? How to prove that $\sin x = x \, _{0}F_{1}(-;\frac{3}{2};-\frac{x^{2}}{4})$?
where $_{0}F_{1}$ is the hypergeometric series?
| The function $_0F_1(;a;z)$ is a solution to the differential equation
\begin{equation}\tag{1}
z \frac{d^2}{dz^2}y(z)+a\frac{d}{dz}y(z)= y(z).
\end{equation}
Knowing this, consider the function $u(x)=x y(-x^2/4)$ and take its second derivative. We obtain
$$\tag{2} \frac{d^2}{dx^2}u(x)=x\left[\frac{x^2}{4}y''\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1764235",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 1
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Determine the maximum and the minimum of an expression Let $x,y,z \in \Bbb R, x,y,z \gt 0$ such that $x^2+y^2+z^2=1$. Determine tha maximum and the minimum possible values of the expression
$$\frac {x^3+y^3+z^3} {x+y+z}.$$
| The rearrangement inequality implies that
$$3(x^3+y^3+z^3)\ge (x+y+z)(x^2+y^2+z^2).$$
So the minimum is $\frac13$, attained when $x=y=z$.
Since obviously $x,y,z\in(0,1)$,
$$\frac{x^3+y^3+z^3}{x+y+z}<1\tag1.$$
On the other hand, if we set $y=z=\varepsilon$ and $x=\sqrt{1-2\varepsilon^2}$, then
$$\lim_{\varepsilon\to 0}\... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\frac {2 \cdot 3^{m + 1}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+$, find all possible values of $k, m$. If $$\frac {2 \cdot 3^{m + 1}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+$$ and $$\frac {2 \cdot 3^{m}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \le 1$$ where $k, m \in \Bbb N_+$ and $k \ge 2$, find all possible... | \begin{align*}
\frac {2 \cdot 3^{m + 1}}{k} - \frac {2 \cdot 3^{m}}{k + 1} \in \Bbb N_+ \\
\implies 2\cdot 3^m \Bigg(\frac {3}{k} - \frac {1}{k + 1}\Bigg) \in \Bbb N_+ \\
\implies 2\cdot 3^m\frac {2k+3}{k(k+1)} \in \Bbb N_+ \\
\implies 2\cdot 3^m\frac {2t+1}{t(t-1)} \in \Bbb N_+ \tag{$t\to k+1$} \\
\end{align*}
Clearly... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$A^2+B^2=AB$ and $BA-AB$ is non-singular The question is:
Are there square matrices $A,B$ over $\mathbb{C}$ s.t. $A^2+B^2=AB$ and $BA-AB$ is non-singular?
From $A^2+B^2=AB$ one could obtain $A^3+B^3=0$. Can we get something from this?
Edit:
$$A^2+B^2=AB\implies\\
A(A^2+B^2)=A^2B \implies\\ A^3+AB^2=A^2B \implies \\
... | The answer is yes. We shall construct a concrete example below. First, suppose $A$ and $B$ are invertible and let $C=AB^{-1}$. From $A^2+B^2=AB$ we get $C + A^{-1}C^{-1}A = I$ and $A^{-1}CA=(I-C)^{-1}$. Therefore the spectrum of $C$ is invariant under the transformation $z\mapsto \frac1{1-z}$.
Let $w$ be the cubic root... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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"answer_id": 0
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Prove $(1+x)^p+(1-x)^p \ge 2(1+x^p)$ for $0\le x\le1$ and real number $p\ge2$. I don't know how to prove the following questions: If $p\ge2$ is real, then
$$
(1+x)^p+(1-x)^p \ge 2(1+x^p) \quad \text{for } 0\le x\le1;
$$
if $1\le p<2$, then opposite direction of the inequality holds.
I try to find a convex function but ... |
Lemma. for $f(x)=x^k$. if $k\geq 1$ then for $a,b>0$,then $a^k+b^k\leq (a+b)^k$. if $0\leq k\leq 1$, then $a^k+b^k\geq (a+b)^k$.
proof.when $k\geq 1$, then we have $(\frac{a}{a+b})^k\leq\frac{a}{a+b}$ and $(\frac{b}{a+b})^k\leq\frac{b}{a+b}$ for $\frac{a}{a+b}\leq 1$, $\frac{b}{a+b}\leq 1. $
Hence:$(\frac{a}{a+b})^k+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1769761",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Is this integration question correctly answered? So we have:
$$y = (x^3 + 2)^7$$
And we want to find:
$$I = \int y \space dx \\$$
From the chain rule, we know:
$${d\over dx} g \Big( f(x) \Big) = g' \Big( f(x) \Big) \cdot f'(x)$$
Hence:
$$\int g' \Big( f(x) \Big) \cdot f'(x) \space dx = g \Big( f(x) \Big) \\ $$
Let's c... | You can apply binomial theorem to integrate $(x^3+2)^7$ directly:
\begin{align}
\int (x^3+2)^7 dx &= \int \left(\sum_{k=0}^7\binom{7}{k} 2^{7-k}x^{3k}\right) dx\\
&=\sum_{k=0}^7 \binom{7}{k}\frac{2^{7-k}}{3k+1}x^{3k+1}+C
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
use little $o$ notation. out from apostol's book.
$f(x)=o(g(x))$ if $\frac{f}{g}\rightarrow 0$ when $x\rightarrow a$ and gives some properties.
1)$o(g(x))+o(g(x))=o(g(x))$
2)$o(cg(x))=o(g(x))$
3)$f(x).o(g(x))=o(f(x).g(x))$
4)$o(o(g(x)))=o(g(x))$
5)$\frac{1}{1+g(x)}=1-g(x)+o(g(x))$ when $g(x)\rightarrow 0$ when $x\righ... | I'm not sure about this but $\text o(x^3)=\text o(x^2)$ from $(7)$ and, by $(3)$, $$\text o\left(-\dfrac{x^2}{2}+\text o(x^3)\right)=\text o\left(x^2\cdot\left(-\dfrac{1}{2}+\text o(x)\right)\right)=\left(-\dfrac{1}{2}+\text o(x)\right)\cdot\text o(x^2)$$
then the above expression becomes $-\dfrac{1}{2}\text o(x^2)+\te... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1772606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
$\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x²}-1}{\sqrt{1+x}-\sqrt{1-x}}$ First I need to prove that this limit; $\displaystyle\lim _{x\to 0}\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}$ converges, then I have to find its limit. Now I don't know how to prove that it converges (these epsilon proofs are still somethin... | $$\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}=\frac{x+x^2}{(\sqrt{1+x}-\sqrt{1-x})(\sqrt{1+x+x^2}+1)}=$$ $$=\frac{x+x^2}{\sqrt{1+x}-\sqrt{1-x}}*\frac{1}{\sqrt{1+x+x^2}+1}$$ It follows
$$\lim _{x\to 0}\frac{\sqrt{1+x+x^2}-1}{\sqrt{1+x}-\sqrt{1-x}}=\lim _{x\to 0}\frac{x+x^2}{\sqrt{1+x}-\sqrt{1-x}}*\lim _{x\to 0}\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774792",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Evaluation of $\cos\left(\frac{\pi}{2n}\right)\cdot \cos\left(\frac{2\pi}{2n} \right)....\cos\left(\frac{(n-1)\pi}{2n}\right)$
Evaluation of $$\lim_{n\rightarrow \infty}\left(\tan \frac{\pi}{2n}\cdot \tan \frac{2\pi}{2n}\cdot \tan \frac{\pi}{3n}\cdot ...............\tan \frac{(n-1)\pi}{2n}\right)^{\frac{1}{n}} = $$ wi... | For fixed $n$, call the expression $A$, and take its $\log[.] $ to get:
$$ \log (A)=\frac{1}{n} \left(\log\tan \frac{\pi}{2n}+ \tan \frac{2\pi}{2n}+ .............. + \tan \frac{(n-1)\pi}{2n}\right) $$
$$ =\frac{1}{n} \sum_{k=1}^{n-1}\log\tan (\frac{\pi k}{2n})$$
Doesn't it look familiar?! Yes, Riemann sum as in the def... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1774980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Splitting fractions with a linear denominator: $\frac{2x-1}{x+2}$ How can $$\frac{2x-1}{x+2}$$ be split to give $$A-\frac{B}{x+2}$$
where $A$ and $B$ are integers?
The solution is $$2-\frac{5}{x+2}.$$
| Imagine that you divide polynomials as usual numbers.
For example, let's divide $2x-1$ by $x+2$. Firstly, find the number such that if you multiply the divisor by it and substract from a dividend, you vanish the first term. Here $a=2$, because $2x-1 - 2(x+2) = -5$, which doesn't contain a term with $x$. -5 has the degr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775116",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
$ \sum_{n = 1}^{\infty} \frac{1}{n^4}$? Using Fourier series I have managed to show that
$$ \frac{x^4}{12} = \frac{\pi^2 x^2}{6} + 4 \sum_{n = 1}^{\infty} \frac{(-1)^n}{n^4}(1-\cos(nx)) , x \in [-\pi,\pi]$$
From here apparently one need just one step to find an expression for $ \sum_{n = 1}^{\infty} \frac{1}{n^4}$. I... | For $\;x=\pi\;$:
$$\frac{\pi^4}{12}=\frac{\pi^4}6+4\sum_{n=1}^\infty\frac{(-1)^n}{n^4}\left(1-(-1)^n\right)\stackrel{\text{Only odd index matters}}\implies\frac{\pi^4}{96}=\sum_{n=1}^\infty\frac1{(2n-1)^4}$$
and from here:
$$I:=\sum_{n=1}^\infty\frac1{n^4}=\frac1{16}\sum_{n=1}^\infty\frac1{n^4}+\sum_{n=1}^\infty\frac1{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/1775879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"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.