Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Geometry Question Points on a Parabola We have the curve with equation $y^2=x+4$. Point $A$ is on the curve and has coordinates $(0,2)$, $B$ and $C$ are other points lying on the curve where the line segment $AB$ is perpendicular to $BC$. What values can the $y$-coordinate of the point $C$ be?
| Let $B(b^2-4,b)$ and $C(c^2-4,c)$, thus:
$$m_{AB}=\frac{2-b}{4-b^2}=\frac{1}{2-b}$$
$$m_{BC}=\frac{b-c}{b^2-c^2}=\frac{1}{b-c}$$
Perpendicularity condition:
$$m_{AB}=-\frac{1}{m_{BC}}\implies c-b=\frac{1}{2-b}\implies c=b+\frac{1}{2-b}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577853",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the minimum of expression: $\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$ If $x+y+z=1$ and $x,y,z$ are positive numbers, Find the minimum of expression:
$$\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z}$$
My solution:
$$\left[\frac{2-x}{3+x}+\frac{2-y}{3+y}+\frac{2-z}{3+z} \right]_{min}\Rightarrow \left[\frac 5... |
My solution:
$$[...]\quad \implies \quad\left[ \frac 1{x+3}+\frac 1{y+3}+\frac 1{z+3}\right]_{min}$$
And we have $a+b+c≥3\sqrt[3]{abc}$
Which that $\color{red}{\left[ a+b+c\right]_{min}\Rightarrow a=b=c}$
$\color{red}{\style{font-family:inherit}{\text{That step}}}$ does not follow.
You presumably mean to use $a=\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578231",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluate: $-T_1^{2k}+T_2^{2k}-T_3^{2k}+T_4^{2k}-T_5^{2k}+T_6^{2k}-\cdots+T_{2n}^{2k}$ $T_n={n(n+1)\over 2}$, observing only the even terms of $T_n$ such as the series below
$$\color{red}{-1^2+3^2=8\times1^2}$$
$$\color{blue}{-1^2+3^2-6^2+10^2=8\times3^2}$$
$$\color{Red}{-1^2+3^2-6^2+10^2-15^2+21^2=8\times6^2}$$
$$\colo... | The formula for $k=2$ is
$$
4 n^2 (n + 1)^2 (2 n^4 + 4 n^3 - 2 n + 1)
\\= 16 T_n^2 (8 T_n^2- 4 T_n + 1)
$$
The formula for $k=3$ is
$$
n^2 (n + 1)^2 (32 n^8 + 128 n^7 + 96 n^6 - 160 n^5 - 61 n^4 + 294 n^3 - 33 n^2 - 228 n + 114)
\\= 8 T_n^2 (256 T_n^4 - 384 T_n^3 + 390 T_n^2 - 228 T_n + 57)
$$
Thus, an interesting ques... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578915",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
On $\sum_{n=0}^\infty \frac{1}{n^2+bn+c} = \frac{\pi \tan\big(\frac{\pi}2\sqrt{b^2-4c}\big)}{\sqrt{b^2-4c}}+x$ and solvable Galois groups? In this comment, I hastily assumed that,
$$\sum_{\color{blue}{n=0}}^\infty \frac{1}{n^2+bn+c} = \frac{\pi \tan\big(\frac{\pi}2\sqrt{b^2-4c}\big)}{\sqrt{b^2-4c}}$$
In fact, this is... | Well, if a quadratic polynomial $p(x)$ has a negative discriminant, $\sum_{n\in\mathbb{Z}}\frac{1}{p(n)}$ can be found through the Poisson summation formula or just by considering $\frac{d}{dx}\log(\cdot)$ applied to the Weierstrass product for the cosine function. For instance
$$\sum_{n\in\mathbb{Z}}\frac{1}{n^2+9n+c}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580695",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
check convergence $\int _{-\infty }^0\:\frac{e^{3x}}{1+x^2} $
check convergence of $\int _{-\infty }^0\:\frac{e^{3x}}{1+x^2} $
so for $x\le0$ : $e^{3x}\le1$ then we get $0\le\frac{e^{3x}}{1+x^2}\le\frac{1}{1+x^2}\le\frac{1}{x^2}$
and because $\int _{-\infty }^{-1}\:\frac{1}{x^2}$ is convergence we get $\int _{-\inft... | Well, you are correct, but your arguments could be a lot more precise.
Note that when: $x \in (-\infty,0), e^x < 1$. Hence, $$f(x) = \frac{e^x}{1+x^2} < \frac{1}{1+x^2} = g(x)$$
Hence, by the comparison test as: $$\int_{-\infty}^{0} \frac{1}{1+x^2}\, dx = \arctan x \bigg \lvert_{-\infty}^{0} = \frac{\pi}{2}$$ which con... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581648",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find the equation from the given equation in which leading term has c0-efficient $1$
Find the equation obtained from $5x^3-7x^2+8x-9=0$ into one in which the co-efficients of leading term is unity and the others are integers.
Below I am showing my attempt:
Applying the transformation $x\mapsto x+h$,then $f(x)=5x^3-... | Then, divide it by $5$ to obtain: $$f’(x+h) = x^3+x^2(3h-\frac75) + x(3h^2-\frac{14h}{5}+\frac85) +(h^3-\frac{7h^2}{5}+\frac{8h}{5}-\frac95)$$
Now, note that $f’(x+h)$ is monic and the coefficients of every power of $x$ should $\in \mathbb{Z}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581964",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Let $f_n(x) = \frac{x^2}{(1+x^2)^n}$ If $\sum_{n=0}^\infty$ converges pointwise, find it's limit.
Let $f_n(x) = \frac{x^2}{(1+x^2)^n}$ If $\sum_{n=0}^\infty f_n(x)$ converges pointwise, find it's limit. Determine if it converges uniformly.
It is clear that $0\le \frac{x^2}{(1+x^2)^n} < 1, \forall n \ge 1$.
With the a... | Note that we have
$$\sum_{n=0}^N \frac{x^2}{(1+x^2)^n}=x^2\left(\frac{1-\frac{1}{(1+x^2)^{N+1}}}{1-\frac{1}{1+x^2}}\right)=1+x^2-\frac{1}{(1+x^2)^N}$$
For $x>0$, we have $\lim_{N\to \infty }\sum_{n=0}^N \frac{x^2}{(1+x^2)^n}=1+x^2$.
Now, take $\epsilon=\frac1e$. Then,
$$\left|\sum_{n=0}^N \frac{x^2}{(1+x^2)^n}-\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2584895",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
limit of a trigonometric function $ \lim\limits_{x\to\pi/3} \frac{1 - 2\cos (x)}{\sin (3x)} $ compute the limit of
$$ \lim_{x\to \frac{\pi }{3} } \frac{1 - 2 \cos (x)}{\sin (3x)} $$
I would like to not do a translation with the change of variable $ t = x - \frac{\pi }{3} $
| Taking into account "I would like to not do a translation with the change of variable ...", let us use the Taylor series around $x=a$.
$$\cos(x)=\cos (a)-(x-a) \sin (a)-\frac{1}{2} (x-a)^2 \cos (a)+O\left((x-a)^3\right)$$
$$\sin(x)=sin (a)+(x-a) \cos (a)-\frac{1}{2} (x-a)^2 \sin (a)+O\left((x-a)^3\right)$$ So, using $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2585787",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 2
} |
To find the Value of $\tan A + \cot A$, if the value of $\sin A + \cos A$ is given To Find -
$$\tan A + \cot A$$
Given,
$$\sin A + \cos A = \sqrt2$$
My progress as far -
1st way-
$$\Rightarrow \sin A = \sqrt2 - \cos A$$
$$\Rightarrow \tan A = \frac{\sqrt2 - \cos A}{\cos A}$$
$$\Rightarrow \tan A = \frac{ \sqrt 2 }{\co... | Let's use $\sin^2A=\sqrt{1-\cos^2A}$
Equation transform into$$\sin A+\sqrt{1-\sin^2A}=\sqrt2$$
Which is quadratic equation and easy to solve
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Evaluating $ \prod_{i=1}^n 3^{2^i} + \sum_{i=1}^n 3^{2^{i+1}-1} \prod_{j=i+1}^n 3^{2^j}$ I'm working on difference equations and got stuck on the algebra with this one:
$$ u_n = \prod_{i=1}^n 3^{2^i} + \sum_{i=1}^n 3^{2^{i+1}-1} \prod_{j=i+1}^n 3^{2^j}$$
I need to simplify this to just a function of n. Any help would ... |
We obtain
\begin{align*}
\color{blue}{\prod_{i=1}^n3^{2^i}}&\color{blue}{+\sum_{i=1}^n3^{2^{i+1}-1}\prod_{j=i+1}^n3^{2^j}}\\
&=3^{2^1+2^2+\cdots+2^n}+\sum_{i=1}^n3^{2^{i+1}-1}\cdot 3^{2^{i+1}+2^{i+2}+\cdots+2^{n}}\\
&=3^{2\left(2^n-1\right)}+\sum_{i=1}^n3^{2^{i+1}-1}\cdot 3^{2^{i+1}\left(2^{n-i}-1\right)}\\
&=3^{2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589326",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Given matrix times vector, find the inverse of the matrix times the same vector. Given $AV= \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$, where A is a matrix and V is a vector, find $A^{-1}V$.
My thought are the following:
Step 1:
Times $A^{-1}$ for both sides.
$A^{-1}AV=A^{-1} \begin{bmatrix}8\\8 \\ 8 \\8\end{bmatrix}$... | In general, no you can't assume $AV = \lambda V$. Not every vector is an eigenvector. Even if $V$ were an eigenvector, then this would still not give you a well-defined answer.
For one example, take $A = I$, the $4 \times 4$ identity matrix. Then $AV = V$, hence $V = \begin{bmatrix} 8 \\ 8 \\ 8 \\ 8 \end{bmatrix} \neq ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2589677",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve $(y+ x^3y + 2x^2)dx + (x + 4xy^4+ 8y^3)dy = 0$ Basically I tried to group the terms
ydx + xdy is a perfect differential but I couldn't think a method for ($x^3$y)dx + (4x$y^4$)dy . Also dividing by xy didn't help me.
| $$(y+ x^3y + 2x^2)\text{d}x + (x + 4xy^4+ 8y^3)\text{d}y = 0$$
$$\Rightarrow (y\text{d}x + x\text{d}y)+ x^3y\text{d}x + (2x^2\text{d}x) + 4xy^4\text{d}y+ (8y^3\text{d}y) = 0$$
$$\Rightarrow \text{d}(xy)+ (x^3y\text{d}x + 4xy^4\text{d}y) + \frac{2}{3}\text{d}(x^3) + 2\text{d}(y^4) = 0$$
$$\Rightarrow \text{d}(xy)+ xy(x^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590017",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Find $f(f(f(f(f(f(\cdots f(x)))))))$ $2018$ times
I was given a problem to calculate $f(f(f(\dots f(x))))$ $2018$ times if $f(x)=\frac{x}{3} + 3$ and $x=4$.
Is it even possible to be done without a computer?
| Let's try to spot a pattern.
$$f(x)=3+\frac x3=\frac13(3^2+x)$$ $$f(f(x))=f\left(\frac13(9+x)\right)=3+\frac{9+x}{9}=4+\frac x9=\frac1{3^2}(3^3+3^2+x)$$ $$f(f(f(x)))=f\left(\frac19(36+x)\right)=3+\frac{36+x}{27}=\frac{13}3+\frac x{27}=\frac1{3^3}(3^4+3^3+3^2+x)$$
So iterating $2018$ times, we have $$f(f(f(\cdots f(x))... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590631",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 11,
"answer_id": 3
} |
Sum of all Fibonacci numbers $1+1+2+3+5+8+\cdots = -1$? I just found the sum of all Fibonacci numbers and I don't know if its right or not.
The Fibonacci sequence goes like this : $1,1,2,3,5,8,13,\dots$ and so on
So the Fibonacci series is this $1+1+2+3+5+8+13+\dots$
Let $1+1+2+3+5+8+\dots=x$
$$\begin{align}
1 + 1 + 2 ... | Since the terms of the series don't go to $0$, the series does not converge. Relatedly, $x=\infty$ also solves your equation $x-1=2x$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591315",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
simplification of trigonometric expression I search a simplification for
$$ \arcsin\biggl( \frac{1}{\sqrt 2} \frac{x}{\sqrt {(x-1)^2 +1}}\biggr)$$
any idea ? I tried composition, derivation and setting $x$ as a sine, cosine or a tangent in vain.
Thanks
| Let us substitute $x = 1 + \tan \theta $. We then get: $$E = \arcsin \left (\frac1 {\sqrt 2} \frac {1+\tan \theta}{\sqrt {\sec^2 \theta}} \right) $$ $$=\arcsin \left (\frac {1}{\sqrt 2} (\cos \theta + \sin \theta) \right) $$ $$=\arcsin \left (\sin \theta \cos \frac {\pi}{4} + \cos \theta \sin \frac {\pi}{4} \right) $$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Some infinite series (just for fun!) I have a few infinite series problems that I think MSE might enjoy, whose answers I already know:
$$\sum_{n=0}^\infty \frac{2^{n-2^n}}{1+2^{-2^n}}=\text{?}$$
$$\sum_{n=0}^\infty \frac{4^{n+2^n}}{(1+4^{2^n})^2}=\text{?}$$
$$\sum_{n=0}^\infty \frac{3^n(1+2\cdot3^{-3^n})}{2\cosh(3... | \begin{align}
\frac{1}{1-2}+S_1&=\lim_{N\to\infty}(\frac{1}{1-2}+\sum_{n=0}^N \frac{2^{n}}{1+2^{2^n}})\\
&=\lim_{N\to\infty}(\frac{1}{1-2}+\frac{1}{1+2}+\sum_{n=1}^N \frac{2^{n}}{1+2^{2^n}})\\
&=\lim_{N\to\infty}(\frac{2}{1-2^2}+\sum_{n=1}^N \frac{2^{n}}{1+2^{2^n}})\\
&=\lim_{N\to\infty}\frac{2^{N+1}}{1-2^{2^{N+1}}}\\
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2592425",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Find all ordered pairs such that $x + y^2 = 2$ and $y + x^2 = 2$ How can I find all ordered pairs $(x,y)$ such that $x + y^2 = 2$ and $y + x^2 = 2$?
A solution/explanation would be greatly appreciated.
If it helps, the solutions given in the textbook are
x=-2, y=-2; x=1, y=1; x= (1+root 5)/2, y=(1-root 5)/2; ... | Well $x+y^2=y+x^2$ so $x-y=x^2-y^2=(x-y)(x+y) $
If $x-y=0$ then $x=y$ and $x+x^2=2$. We can use the the quadratic equation to solve that.
If $x-y\ne 0$ then $1=(x+y) $ and $y=1-x $
So $x+y^2=x^2+y=x^2-x+1=2$ and we can use the quadratic equation to solve that.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How to prove that $\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}$ for non-negative $a,b$? If $a, b$ are non-negative real numbers, prove that
$$
\frac{a+b}{1+a+b} \leq \frac{a}{1+a} + \frac{b}{1+b}
$$
I am trying to prove this result. To that end I added $ab$ to both denominator and numerator as we know
$$
\frac... | We get $$\frac{a}{1+a}+\frac{b}{1+b}-\frac{a+b}{1+a+b}={\frac {ba \left( a+b+2 \right) }{ \left( 1+a \right) \left( 1+b
\right) \left( 1+a+b \right) }}
\geq 0$$
the numerator can be calculated as
$$a(1+b)(1+a+b)+b(1+a)(1+a+b)-(a+b)(1+a)(1+b)=...$$
we also have under the same conditions
$$\frac{a+b+c}{1+a+b+c}\le \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595966",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 6,
"answer_id": 0
} |
Find the area above the $x-$ axis included between the parabola $y^2=ax$ and the circle $x^2+y^2=2ax$.
Find the area above the $x-$ axis included between the parabola $y^2=ax$ and the circle $x^2+y^2=2ax$.
The circle $x^2+y^2=2ax$ can be represented as $(x-a)^2+y^2=a^2$ which has centre at $(a,0)$ and radius $a$.
Sin... |
The graph is drawn taking $a=2$. What you need to notice is that the second integral is just $1/4$-th the area of the circle. So required area is $$\int_0^a\sqrt{ax}dx+\frac{\pi a^2}4=\frac{a^2}{3/2}+\frac{\pi a^2}4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597776",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find values of x so that the matrix is not invertible Given
$$A=\begin{pmatrix}
5 & 6 & 6-x & 8 \\
2 & 2-x & 2 & 8 \\
6 & 6 & 2 & 8 \\
2 & 3 & 6 & 7
\end{pmatrix}\in\mathbb{R}^{4\times 4}$$
One must find all values of $x\in\mathbb{R}$ such that the matrix is not invertible.
I tried finding the determinant... | HINT
Let's calculate $\det A$ as a function of x and then set $\det A=0$.
Note that adding a scalar multiple of one column/row to another column/row does not change the value of the determinant. Thus you can simplify the matrix A before to perform the calculation.
You should obtain:
$$\det A=2(-13 x^2 + 19 x - 4)$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2598421",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $ab$ given the value of a limit If $$\lim_{x\to 0} \frac{ae^x -b}{x} = 2$$ then find $ab$.
I have tried everything that atleast I know about limits or algebra but I got not a single idea how to do this.
Please help
| We have
$$e^x = \sum_{k = 0}^\infty\frac{x^k}{k!} = 1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...$$
by Taylor Series expansion of $e^x$, so
$$\lim_{x\to 0} \frac{ae^x -b}{x} = \lim_{x\to 0}\frac{a(1+x+\frac{x^2}{2!}+\frac{x^3}{3!}+...)-b}{x} = \lim_{x\to 0}\bigg(\frac{a}{x}+a+\frac{x}{2!}+...-\frac{b}{x}\bigg) = 2$$
Now, notic... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Convergence of $\sum\limits_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $
I've got in my assignment to show if the following series converges or diverges.
$$\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right) $$
Attempt:
\begin{align*}
\sum_{n=1}^\infty \left(e^\frac{1}{n} -1 -\frac{1}{n}\right... | Just use comparison test. Using this Proving Schwarz derivative $\frac{f''(0)}{2} =\lim\limits_{x\to 0} \frac{\frac{f(x) -f(0)}{x}-f'(0)}{x}$ without Taylor expansion or L'Hopital rule? you have $$\lim_{n\to\infty}{n^2}\left(e^\frac{1}{n} -1 -\frac{1}{n}\right) =\lim_{x\to 0}\frac1{x^2}\left(e^x -1 -x\right) =\lim\lim... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599622",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
For triangle $ABC$ there are median lines $AH$ and $BG$ with $\angle CAH=\angle CBG={{30}^{0}}$ . Prove that $ABC$ is the equilateral triangle. For triangle $ABC$ there are median lines $AH$ and $BG$ with $\angle CAH=\angle CBG={{30}^{0}}$. Prove that $ABC$ is the equilateral triangle.
| With the standard notation, if $M_A$ is the midpoint of $BC$,
$$ AM_A^2 = \frac{2b^2+2c^2-a^2}{4},\quad AC^2 = b^2,\quad CM_A^2=\frac{a^2}{4} $$
hence if $\widehat{M_A A C}=30^\circ$ we have
$$ \cos 30^\circ = \frac{\sqrt{3}}{2} = \frac{AM_A^2+AC^2-CM_A^2}{2\cdot AM_A\cdot AC}=\frac{-a^2+3b^2+c^2}{4b\cdot AM_A} $$
and
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Alternating summation and subtraction of square roots I encountered a problem, to find the integer part of: $$\frac{1}{\sqrt{1} + \sqrt{2}} + \frac{1}{\sqrt{3} + \sqrt{4}} +...+\frac{1}{\sqrt{99} + \sqrt{100}}$$.
I multiplied the conjugate of each denominator. Meaning, for $\frac{1}{\sqrt{a} + \sqrt{b}}$, I multiply $... | $$\frac {1}{\sqrt n +\sqrt {n+1}} = \sqrt {n+1} - \sqrt {n}$$
Therefore the dsire sum is $ A-B$ where,
$$A= \sqrt 100+\sqrt 98 +....+\sqrt 2 = 338.047...$$ and
$$B=\sqrt 99+\sqrt 97 +....+\sqrt 1 = 333.415....$$
Thus [A-B]=4
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2600710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 2
} |
Differentiating two trig functions. Check. I have two problems that I'd like a quick checkup on. I am trying to differentiate these two functions:
*
*$f(x) = x \cdot \cos x + 2\cdot \tan x$
and
*$f(x) = e^x \cdot \cos x$
Here are my attempts:
*
*Going to use a combo of sum and product rules.
$$f'(x) = \frac{... | Perhaps extending the steps will help. We have
$$f(x) = x\cos(x) + 2\tan(x)$$
Then
\begin{align*}
f'(x) &= \frac{d}{dx}\left(x\cos(x) + 2\tan(x)\right)\\
&= \frac{d}{dx}\bigg( x \bigg) \cos(x) + x\frac{d}{dx}\bigg( \cos(x) \bigg) + \frac{d}{dx}\bigg(2\tan(x) \bigg )\\
&= \cos(x) - x\sin(x) + 2\sec^2(x)
\end{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Solve $15\cdot x=27\pmod{18}$ EXERCISE:
Solve $15\cdot x\equiv 27\pmod{18}$
SOLUTION:
We know that $\gcd(15,18)=3$ and $3\setminus 27=3\cdot 9$.
So from http://www.math.niu.edu/~richard/Math420/lin_cong.pdf we can conclude that we have 3 solutions!
Then we have that $15\cdot x\equiv 27\pmod{18}\implies 5\cdot x\equ... | It is possible to solve this problem using algebraic symbols or augmented matrices. For example the following uses augmented matrices to solve the problem:
$$\begin{array}{c|c|c|}
\hline
\text{x} & \text{y} & \text{c} & \text{s} & \text{t} \\
\hline
\text{15} & \text{-18} & \text{27} \\
& \text{+3} & \text{+3} & \tex... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
How to calculate this exponential integral? I am trying to calculate
$\int_0^{+\infty}\log_2(1+ax)\frac{1}{a}e^{-\frac{x}{a}}dx$
by using the integration by parts method like:
\begin{equation}
\begin{aligned}
\int_0^{+\infty}\log_2(1+ax)\frac{1}{a}e^{-\frac{x}{a}}dx &= -\int_0^{+\infty}\log_2(1+ax)(e^{-\frac{x}{a}})'d... | You are almost done. Continuing from where you left: $$I =\int_{0}^{\infty} \frac{e^{-x/a-1/a^2}}{-x/a-1/a^2}\, dx$$ Substituting $u = \frac{x} {a} + \frac1{a^2} \implies du = \frac1{a} \, dx$ gives us: $$I = a \int_{\frac1{a^2}}^{\infty} - \frac{e^{-u}}{u} \, du$$ which is the exponential integral $E_1(u)$.
Hope you ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2612981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Form an equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$ If $a,b,c$ are the roots of the equation $x^3+x+1=0,$ Then the equation whose roots are $(a-b)^2,(b-c)^2,(c-a)^2.$
Try: $a+b+c=0,ab+bc+ca=1,abc=-1$
Now $(a-b)^2+(b-c)^2+(c-a)^2=2(a+b+c)^2-6(ab+bc+ca)=-6$
Could some help me to explain short way to calculate pro... | You already know the value of $(a-b)^2+(b-c)^2+(c-a)^2$. Use the fact that\begin{multline}(a-b)^2(b-c)^2(c-a)^2=-4 a b c (a+b+c)^3+(a b+c b+a c)^2 (a+b+c)^2+\\+18 a b c (a b+c b+a c) (a+b+c)-4 (a b+c b+a c)^3-27 a^2 b^2 c^2\end{multline}and that\begin{multline}(a-b)^2(b-c)^2+(b-c)^2(c-a)^2+(c-a)^2(a-b)^2=(a+b+c)^4+\\-6... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2613388",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
Proof verification: If $a \equiv b (mod$ $n)$, then $a^3 \equiv b^3 (mod$ $n)$ Would someone be willing to verify the following proof?
Theorem: Suppose $a, b \in \mathbb{Z}; n \in \mathbb{N}$. If $a \equiv b (mod$ $n)$, then $a^3 \equiv b^3 (mod$ $n)$.
Proof:
$a \equiv b (mod$ $n) \rightarrow xn = a - b; x \in \mathb... | Your proof is correct.
You could have stopped at $a^3=b^3+kn$, it doesn't matter whether $k$ has a complicated expression, if it's an integer, then $a^3\equiv b^3\pmod n$.
However, notice that you can factorize $xn$ actually, and that $xn=a-b$.
Thus $a^3-b^3\equiv(a-b)(a^2+ab+b^2)\equiv 0\pmod n$ since $(a-b)\equiv 0\p... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2614158",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Let $X$ and $Y$ be independent and identically distributed random variables, find $P(X \geq 2Y)$
Let $X$ and $Y$ be independent and identically distributed random variables with probability mass function $p(n)=\frac{1}{2^{n}}$. Find $P(X \geq 2Y)$.
My attempt: \begin{align} P(X \geq 2Y) & = \sum\limits_{y=1}^{\infty}... | We have $$\mathbb{P}(X \geq 2Y) = \sum_{y \geq 1} \mathbb{P}(X \geq 2Y, Y=y) = \sum_{y \geq 1} \sum_{x \geq 2y} \mathbb{P}(X=x, Y=y) = \sum_{y \geq 1} \sum_{x \geq 2y} \frac{1}{2^x} \frac{1}{2^y}.$$
Thus,
$$\mathbb{P}(X \geq 2Y) = \sum_{y \geq 1} \frac{1}{2^y} \frac{2}{2^{2y}} = 2 \sum_{y \geq 1} \frac{1}{2^{3y}}= \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$
find the :
$$\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$$
My Try :
$$\dfrac{1}{1+\dfrac{\sin 70^{\circ}}{\cos 70^{\circ}}}+\dfrac{1}{1+\dfrac{\sin 20^{\circ}}{\cos 20^{\circ}}}$$
$$\dfrac{\cos70^{\circ}}{\cos 70^{\circ}+\sin 70^{\cir... |
See that $$\tan (x) = \frac{\sin x}{\cos x}=\frac{\cos (90^{\circ}-x)}{\sin (90^{\circ}-x)}=\frac{1}{\tan (90^{\circ}-x)}$$
what happen if you take $x=70$
$$\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}} = \dfrac{1}{1+\frac{1}{\tan 20^{\circ}}}+\dfrac{1}{1+\tan 20^{\circ}}= \dfrac{\tan 20^{\circ}}{1+\tan... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2619711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Find the centre of a circle given the equation of a tangent to the circle and the x coordinate of the centre The line with equation $2x+y-5=0$ is a tangent to the circle with equation $(x-3)^2 + (y-p)^2=5$
Find the two possible values of $p$.
| $$y=5-2x$$
Since this is a tangent, it touches the circle.
$$\begin{align}(x-3)^2+(y-p)^2&=5\\(x-3)^2+(5-2x-p)^2&=5\\x^2-6x+9+25+4x^2+p^2+2(-10x)+2(2px)+2(-5p)&=5\\5x^2+(4p-26)x+(p^2-10p+29)&=0\end{align}$$
We require that $$\boxed{b^2-4ac=0}$$
$$\begin{align}(4p-26)^2-20(p^2-10p+29)&=0\\4(2p-13)^2-20(p^2-10p+29)&=0\\4... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2622606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Limit: $\lim_{x\to2} \frac{1}{x-2}\cdot \sin\left(\frac{x-2}{x+2}\right)$ Can someone help me understand how the solution for the following limit is $1/4$?
I've been trying to solve it but I always end up in a 'dead end' with an
indetermination. If someone could help me, that would be awesome.
$$\lim_{x\to2} \frac{1... | Note that for $x\to 2$
$$\frac{1}{x-2}\cdot \sin\left(\frac{x-2}{x+2}\right)=\frac{\sin\left(\frac{x-2}{x+2}\right)}{x-2}=\frac{\sin\left(\frac{x-2}{x+2}\right)}{\frac{x-2}{x+2}}\cdot\frac{1}{x+2}\to 1\cdot\frac14=\frac14$$
indeed
$$y=\frac{x-2}{x+2}\to 0 \quad \implies \frac{\sin\left(\frac{x-2}{x+2}\right)}{\frac{x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623777",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Proving a limit through the delta-epsillon definition of a limit, discarding the delta upper bound Prove that the limit exists:$$\lim\limits_{x \to -3} \frac{x+5}{x+2} = -2$$
My answer is:
Let $\epsilon > 0$, there exists a $\delta=\boxed{min\{\frac{\epsilon}{6},\frac{1}{2}\}}$. Let $x$ such that $0<|x + 3| < \delta$,
... | We can find the optimal value for $\delta$ in this way, let assume
$$ \left|\frac{x+5}{x+2} +2\right|<\epsilon \implies -2-\epsilon<\frac{x+5}{x+2}<-2+\epsilon \implies -2-\epsilon<1+\frac{3}{x+2}<-2+\epsilon \implies -3-\epsilon<\frac{3}{x+2}<-3+\epsilon \implies \frac{-3-\epsilon}{3}<\frac{1}{x+2}<\frac{-3+\epsilon}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623892",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
"Different form" of Ceva's theorem?
Point $P$ is inside $\triangle ABC$. Line segments $APD$, $BPE$, and $CPF$ are drawn with $D$ on $BC$, $E$ on $AC$, and $F$ on $AB$ (see the figure below). Given that $AP=6$, $BP=9$, $PD=6$, $PE=3$, and $CF=20$, find the area of $\triangle ABC$.
Now, the Art of Problem Solving sit... | One way to prove it is to use Mass Points. We assign weights $a$ to vertex $A$, $b$ to $B$, $c$ to $C$. Then the weights of $F$ is $a+b$, of $D$ is $b+c$ of $E$ is $a+c$ and of $P$ is $a+b+c$.
Then we have that
$$\frac{AP}{PD} = \frac{b+c}{a} \implies \frac{AD}{PD} = \frac{a+b+c}{a} \implies \frac{PD}{AD} = \frac{a}{a+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2627832",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Sum of $\frac{C_r}{(r+1)(r+2)}$ from $r=0$ to $r=n$ I am trying to compute the following sum: $$\sum_{r=0}^{n}\frac{C_r}{(r+1)(r+2)}$$
Integrating $(1+x)^n = \sum_{r=0}^{n}C_rx^r$ with respect to $x$,
$$\int(1+x)^n dx=\int\sum_{r=0}^{n}C_rx^rdx$$
$$\implies \frac{(1+x)^{n+1}}{n+1}=\sum_{r=0}^{n}C_r\frac{x^{r+1}}{r+1}$$... | We have to respect the integration constant when evaluating an indefinite integral.
When calculating $\int(1+x)^n\,dx = \int\sum_{r=0}^nC_rx^r\,dx$ we obtain
\begin{align*}
\int(1+x)^n\,dx&=\frac{(1+x)^{n+1}}{n+1}+K_1\\
\int\sum_{r=0}^nC_rx^r\,dx&=\sum_{r=0}^nC_r\frac{x^{r+1}}{r+1}+K_2
\end{align*}
with $K_1,K_2$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2628346",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Prove there exists no rational solutions $a,b,c$, and $d$ such that $a + b\sqrt2 + c\sqrt3 + d\sqrt6 = \sqrt5$ I want to prove there exists no rational solutions $a,b,c$, and $d$ such that $$a + b\sqrt2 + c\sqrt3 + d\sqrt6 = \sqrt5$$I have proven there exists no rational solutions $a,b$, and $c$ such that $$a+b\sqrt2=... | $(b\sqrt{2}+c\sqrt{3})^2 = (\sqrt{5} - a - d\sqrt{6})^2\implies 2b^2+2bc\sqrt{6}+3c^2= 5+a^2+6d^2 - 2a\sqrt{5}-2d\sqrt{30}+2ad\sqrt{6}\implies m+r\sqrt{6}= q\sqrt{5}+t\sqrt{30}\implies (r\sqrt{6} - q\sqrt{5})^2 = (t\sqrt{30}-m)^2\implies 6r^2-2qr\sqrt{30}+5q^2= 30t^2-2tm\sqrt{30}+m^2\implies \sqrt{30} \in \mathbb{Q}$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How to calculate $\int \frac{dx}{(a^2 + x^2)^2}$? I'm trying to use a trig substitution but I'm stuck. Here's what I did so far:
$$\int \frac{dx}{(a^2 + x^2)^2}$$
Let $x = a\sin \theta, dx = a\cos \theta d\theta$
$$\int \frac{a cos\theta d\theta}{(a^2 + a^2 sin^2 \theta)^2} = \int \frac{a\cos \theta d\theta}{(a^2(1+si... | This could be done in a "general" way.
Denoting
$$
I_k=\int \frac{dx}{(x^2+a^2)^k},
$$
one has (for $k>1$)
$$
\begin{align}
I_k&=\frac{1}{a^2}\int\frac{(x^2+a^2)-x^2}{(x^2+a^2)^k}\ dx\\
&=\frac{1}{a^2} I_{k-1}-\frac{1}{a^2}\int\frac{x^2}{(x^2+a^2)^k}\ dx\\
&=\frac{1}{a^2} I_{k-1}+\frac{1}{2a^2(k-1)}\int x\ d\left(\fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2632980",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 5
} |
Question related to beta and gamma function I'm trying to derive the following integral.
$$\int_0^\infty \frac{x^8(1-x^6)}{(1+x)^{24}} \, dx.$$
What transformations can I use?
| In general
$$\int^{\infty}_{0}\frac{x^{u-1} }{(1+bx)^{m+1}}dx=\frac{1}{b^{u}}\beta (u,m+1-u).$$
Proof:
We make the transformation $y=\frac{bx}{1+bx}$ so
$$y-1 =\frac{bx}{1+bx} -1 = \frac{bx-1-bx}{1+bx} = \frac{-1}{1+bx}\Rightarrow $$ $$bx+1=\frac{1}{1-y} \Rightarrow (1-y)^{m+1}=\frac{1}{(1+bx)^{m+1}}.$$
We have,
$$bx... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solving an exponential with three different bases $$2^x+4^x=8^x$$ Solve for $x$. I reduced the bases to $2$ and try to use logarithms but could not get passed the logarithm of an expression. When typing into online calculators they say there is no solution but graphing shows an answer.
| Hint: $4 = 2^2$ and $8 = 2^3$.
Let $z = 2^x$. Then, using the hint, we can rewrite the original equation as \begin{align} 2^x + 4^x = 8^x &\implies 2^x + (2^2)^x = (2^3)^x \\ &\implies 2^x + (2^{x})^2 = (2^{x})^3 \\ &\implies z + z^2 = z^3. \end{align} Bringing all of the terms to one side of the equation, we obtain... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2633685",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Find the natural solutions of $a^3-b^3=999$ I want to find the natural solutions of $a^3-b^3=999$.
I got $a^3-b^3=(a-b)\cdot(a^2+ab+b^2)$, so if we consider the equation in $\mathbb{Z}/3\mathbb{Z}$ we get
$$(a-b)\cdot(a^2+ab+b^2) \equiv0 \text{ mod }3$$
and because $\mathbb{Z}/3\mathbb{Z}$ is a domain, we get
$$a\equ... | $$999=a^3-b^3=(a-b)^3+3ab(a-b)\geq(a-b)^3,$$ which says $a-b$ is divided by $3$ and $a-b\leq9$ and we get not so many cases:
$a-b=3$ or $a-b=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find all solutions of the equation $z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$ Find all solutions of the equation:
$z^4+(-2+2\sqrt{2}i)z^3+(3-4\sqrt{2}i)z^2+(4+10\sqrt{2}i)z-10=0$
I don't have any other idea than to guess them.
| HINT:
You can write the same equation as:
$$(z^2-2z+5)(z^2+2i\sqrt2 z-2)=0$$
Then you can find the roots of each brackets separately...
SOLUTION:
$$z=1-2i$$
$$z = 1+2i$$
$$z = -i\sqrt2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2634633",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How to calculate $\int _\frac{3}{4}^{\frac{3}{2}} \sqrt{9 - 4x^2}$ $$\int _\frac{3}{4}^{\frac{3}{2}} \sqrt{9 - 4x^2}$$
First I set $x = \frac{3}{2}\sin\theta, dx = \frac{3}{2}\cos \theta d\theta$:
$$\int \sqrt{9 - 4x^2}dx = \int \sqrt{4\left(\frac{9}{4} - x^2\right)}dx = 2\int \sqrt{\left(\frac{3}{2}\right)^2 - \left(\... | When you substituted $x = \dfrac 32 \sin \theta$, you could have changed the limits of integral then, according to your new variable, i.e. $\theta$.
The new limits will be $$ \text{Upper limit :} \quad x = \frac{3}{4} =\frac 32 \sin \theta \implies \theta =\dfrac{\pi}{6}$$ and $$\text{Lower limit :} \quad x = \frac{3}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2635826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
What is $\lim_{x\to 2} \frac{\sqrt{x+2}-2}{x-2}$? I tried multiplying by the conjugate which gave:
$$\frac{x-2}{(x-2)\sqrt{x+2}+2x-4}$$
But i'm still gettting $\frac{0}{0}$. According to my textbook the answer should be $\frac{1}{4}$, but how do I get there?
| For fun:
$\dfrac{\sqrt{x+2} -2}{(\sqrt{x+2})^2-4}=$
$\dfrac{\sqrt{x+2}-2}{(\sqrt{x+2}-2)(\sqrt{x+2}+2)}=$
$\dfrac{1}{\sqrt{x+2}+2}.$
The limit $x \rightarrow 2$ is?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 7,
"answer_id": 3
} |
Number of solution is twice $(x,y)$ Problem: Count the number of $2 \times 2$ matrices $A$ with $A^TA=-I$ in $Z_p$ for $p>2$.
Answer: if $p$ is an odd prime, the number of such matrices $A$ is twice the number of solutions $(x,y)$ to the congruence $x^2+y^2 \equiv -1 \pmod p$.
What's the reason behind "twice"?
| $$A^TA=-I$$
is equivalent to $A^{-1}=-A^T$ and hence implies $AA^T=-I$.
If $A=\begin{bmatrix}
a&b \\
c&d
\end{bmatrix}$ Then
$$A^TA=-I \Leftrightarrow AA^T=-I \mbox{ and }A^TA=-I \\
\begin{bmatrix}
a&b \\
c&d
\end{bmatrix}\begin{bmatrix}
a&c\\
b&d
\end{bmatrix}= \begin{bmatrix}
-1&0 \\
0&-1
\end{bmatrix} \mbox{ and } ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2636348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Identity involving binomial coefficients I might have written this in a needlessly cumbersome way, but I want to prove that for odd positive integers $n$, $$\sum_{k\ odd}^{n}\binom{2n+1}{2k}=\begin{cases}
\binom{2^n+1}{2}, & \text{if}\ n\ \text{mod}\ 4 =1\\
\binom{2^n}{2}, & \text{if}\ n\ \text{mod}\ 4 =3
... | Preliminaries
Note that if $n\mid m$, then
$$
\begin{align}
\frac1n\sum_{k=0}^{n-1}e^{2\pi ikm/n}
&=\frac1n\sum_{k=0}^{n-1}1\\[6pt]
&=1
\end{align}
$$
and that if $n\nmid m$, then
$$
\begin{align}
\frac1n\sum_{k=0}^{n-1}e^{2\pi ikm/n}
&=\frac1n\frac{e^{2\pi im}-1}{e^{2\pi im/n}-1}\\[6pt]
&=0
\end{align}
$$
Thus,
$$
\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637365",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
check if $\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k$ converges How to check if $\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k$ converges?.
$\begin{align}
\sum_{k=1}^{\infty}{ \left( \frac{7k-2}{8k-3 \sqrt{k}}\right)}^k &=
\lim_{n \to \infty} \sum_{k=1}^{n}{ \left( \frac{7k-2}{8... | It should be
\begin{align*}
\lim_{k\rightarrow\infty}\left(\left(\dfrac{7k-2}{8k-3\sqrt{k}}\right)^{k}\right)^{1/k}=\dfrac{7}{8}<1,
\end{align*}
by root test, it is convergent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Addition of $2$ Events
Let $X$ and $Y$ be independent, each uniformly distributed on $\{1, 2, ..., n\}$. Find $P(X + Y = k)$ for $2 \le k \le 2n$.
\begin{align}P(X + Y = k) &= \sum_{(x,y)\,:\,x+y=k} P(x, y) \\
&= \sum_{(x,y)\,:\,x+y=k} \frac{1}{n^2} \\
&= (k - 1)\frac{1}{n^2} \\
&= \frac{k-1}{n^2} \end{align}
When $... |
Why is there 2 intervals being considered?
Hint: Look what happens if you go "backwards" from $k=2n$ "down":
*
*$k=2n$: $(n,n)$
*$k=2n-1$: $(n,n-1)$, $(n-1, n)$
*$k=2n-2$: $(n,n-2)$, $(n-1, n-1)$, $(n-2, n)$
etc. What you've got will keep growing with growing $k$, but at some point the number of ways to get th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639312",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Nilpotent $4 × 4$ matrix
Find a nilpotent $4 × 4$ matrix with only $-1$ and $3$ as entries.
$\begin{bmatrix}
-1 & -1 & 3 & 3 \\
3 & -1 & 3 & -1 \\
3 & -1 & 3& -1 \\
-1 & -1 & 3 & 3
\end{bmatrix}$ is the matrix I came up with, but I don't think it works. Anyone have a better one?
| $$
\left(
\begin{array}{cccc}
3 & 3 & 3 & 3 \\
-1 & -1 & -1 & -1 \\
-1 & -1 & -1 & -1 \\
-1 & -1 & -1 & -1 \\
\end{array}
\right)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639471",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How can we show that $y_n=z_n$?
Let $$y_{n+1}=2^n\sqrt{2(1-\sqrt{1-(\frac{y_n}{2^n})^2}})$$ and $$z_{n+1}=\frac{2z_n}{\sqrt{2(1+ \sqrt{1- (\frac{z_n}{2^n})^2}})}$$
$y_1=z_1=2$
How can one shows that $z_n=y_n$?
I have assumed that $y_n=z_n$ then if that implies $y_{n+1}=z_{n+1}$ then by induction they are equal. But... | see that, $$1-\sqrt{1-(\frac{y_n}{2^n})^2} = \frac{\left(1-\sqrt{1-(\frac{y_n}{2^n})^2}\right)\left(1+\sqrt{1-(\frac{y_n}{2^n})^2}\right)}{1+\sqrt{1-(\frac{y_n}{2^n})^2}} = \frac{(\frac{y_n}{2^n})^2}{1+\sqrt{1-(\frac{y_n}{2^n})^2}} $$
Hence $$y_{n+1}=2^n\sqrt{2\left(1-\sqrt{1-(\frac{y_n}{2^n})^2}\right)}=2^n\sqrt{\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Definite Integral = $\int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta$ for $0\le a<1$ I am trying to find an expression for the following Definite Integral for the range $0 \leq a <1$:
$$D = \int_0^{2\pi} \frac{\sin^2\theta}{(1-a\cos\theta)^3}\,d\theta.$$
which gives (Wolfram Alpha)
$$D= \left[
\frac{\si... | It is not difficult to check that for any $a\in\mathbb{R}$ such that $|a|<1$ we have
$$ \int_{0}^{2\pi}\frac{d\theta}{1-a\cos\theta} = \frac{2\pi}{\sqrt{1-a^2}}\tag{1}$$
hence by differentiating both sides of $(1)$ with respect to $a$ we have
$$ \int_{0}^{2\pi}\frac{\cos(\theta)\,d\theta}{(1-a\cos\theta)^2}=\frac{2a\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2641805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 2
} |
A simple problem with a simple and a nonsimple solution I have this seemingly simple problem to solve. Its statement is straightforward:
"If $x+y+z=3,x^2+y^2+z^2=5, x^3+y^3+z^3=7$ show that $x^4+y^4+z^4=9,x^5+y^5+z^5\neq 11$".
There is a highschool approach, that is I expanded $(x+y+z)^2, (x+y+z)^3$ and managed to get... | If you are in the market for more sophisticated ways... you have found $x, y, z$ to be the roots of $P(t)=t^3-3t^2+2t+\frac23=0$. Now if $\alpha$ is a root of $P(t)$, we seek the minimal polynomial of $\alpha^m$ (for $m=4, 5$).
So if $P_4(r)$ is the polynomial with roots $x^4, y^4, z^4$, then it is given by the resul... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2642114",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$ This is a question from Math Olympiad.
If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.
I basically do not know how to approach this question. Pleas... | Suppose $x,y,z > 0$ satisfy the system
$$
\begin{cases}
x^2 + xy + y^2 = 3&\;\;\;(\text{eq}1)\\[4pt]
y^2 + yz + z^2 = 1&\;\;\;(\text{eq}2)\\[4pt]
z^2 + zx + x^2 = 4&\;\;\;(\text{eq}3)\\
\end{cases}
$$
From $(\text{eq}2)$, since $y,z > 0$, we get $y,z < 1$, hence from $(\text{eq}1)$, we must have $x > 1$.
Let $a = x +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643121",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "44",
"answer_count": 6,
"answer_id": 4
} |
Showing that $ 1 + 2 x + 3 x^2 + 4 x^3 + \cdots + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$ I was studying a polynomial and Wolfram|Alpha had the following alternate form:
$$P(x) = 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} = (1 + x + x^2 + x^3 + x^4 + x^5)^2$$
Of course, we can ... | By direct factorization:
$$
\begin{align}
P(x) &= 1 + 2 x + 3 x^2 + 4 x^3 + 5 x^4 + 6 x^5 + 5 x^6 + 4 x^7 + 3 x^8 + 2 x^9 + x^{10} \\[5px]
&= 1 + x + x^2 + x^3 + x^4 + x^5 \\
&\quad\quad + x + x^2 + x^3 + x^4 + x^5 + x^6 \\
&\quad\quad\quad\quad + x^2 + x^3 + x^4 + x^5 + x^6 + x^7\\
&\quad\quad\quad\quad\quad\quad ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2643601",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 8,
"answer_id": 0
} |
Given that $X,Y$ are independent $N(0,1)$ , show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}}$ are independent $N(0,\frac{1}{4})$ It is given that $X,Y \overset{\text{i.i.d.}}{\sim} N(0,1)$
Show that $\frac{XY}{\sqrt{X^2+Y^2}},\frac{X^2-Y^2}{2\sqrt{X^2+Y^2}} \overset{\text{i.i.d.}}{\sim} N(0,\frac{1}... | If you transform $(X,Y)\mapsto(R,\Theta)$ where $X=R\cos\Theta,Y=R\sin\Theta$,
you should end up with the joint density of $(R,\Theta)$ as $f_{R,\Theta}(r,\theta)=\dfrac{r}{2\pi}e^{-r^2/2}\mathbf1_{\{r>0,\,0<\theta<2\pi\}}$.
This implies $R$ and $\Theta$ are independent, where $R$ has the Rayleigh distribution and $\T... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2647442",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Calculus sine proof Suppose that $a, b, c$ are non-zero acute angles such that
$$\frac{\sin(a − b)}{\sin(a + b)} + \frac{\sin(b − c)}{\sin(b + c)} + \frac{\sin(c − a)}{\sin(c + a)}= 0$$
Prove that at least two of $a, b, c$ are equal.
I have no idea how to begin.
| $$\frac{\sin(a-b)}{\sin(a+b)}=\frac{\alpha/\beta-\beta/\alpha}
{\alpha\beta-1/(\alpha\beta)}=\frac{\alpha^2-\beta^2}{\alpha^2\beta^2-1}$$
where $\alpha=\exp(ia)$ and $\beta=\exp(ib)$. Set $\gamma=\exp(ic)$.
Then
$$(\alpha^2-\beta^2)(\alpha^2\gamma^2-1)(\beta^2\gamma^2-1)
+(\beta^2-\gamma^2)(\alpha^2\beta^2-1)(\alpha^2\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
How is $x(2x+7)+3$ equal to $(2x+1)(x+3)$? For some reason, $x(2x+7)+3$ seems like it should be equal to $(2x+7)(x+3)$ instead of $(2x+1)(x+3)$. How does $(2x+1)$ factor out of here?
The original equation was $2x^2+7x+3$.
Proof of this: https://www.desmos.com/calculator/2xpcqznkio
Notice how $x(2x+7)+3$ and $(2x+1)(x+3... | Compute $(2x+1)(x+3)$. We obtain $(2x^2+6x)+(x+3)=2x^2+7x+3$. So why do think it could be $(2x+7)(x+3)$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2650119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$. Given $x, y$ are acute angles such that $\sin y= 3\cos(x+y)\sin x$. Find the maximum value of $\tan y$.
Attempt at a solution:
$$(\sin y = 3(\cos x \cos y - \sin x \sin y) \sin x) \frac{1}{\cos y}$$
$$\tan y = (3 \co... | Let's play.
$\sin y
= 3\cos(x+y)\sin x
= 3(\cos x \cos y-\sin x \sin y)\sin x
$
Divide by
$\cos x \cos y$
$\dfrac{\tan y}{\cos x}
= 3(\cos x -\sin x \tan y)\tan x
= 3\cos x\tan x -3\sin x\tan x \tan y
$
$\tan y(\dfrac{1}{\cos x}+3\sin x\tan x)
= 3\cos x\tan x
=3\sin x
$
$\tan y(1+3\sin^2 x)
=3\sin x\cos x
$
$\tan y
=\d... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2655203",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Solve $3\sin^2 x - \cos^2 x - 2 =0$
Find all the angles between $0$ and $360^\circ$ that satisfy $$3\sin^2 x - \cos^2 x - 2 =0$$
My attempt -
$3\sin^2 x - (1-\sin^2x) - 2 =0$
$ 3 \sin^2 x + \sin^2 x = 3 $
$4\sin^2 x = 3 $
$ \sin x= \frac{\sqrt{3}}{2} $
I found that $x= 60,120 $
Why is the answer for this $60,12... | General form of solution is :
$x=k\pi± \pi/3= k\times 180 ± 60$
$k=0 ⇒ x=0± 60 ⇒x=60,.. x=-60=300$
$k=1 ⇒ x = 180 ± 60 ⇒ x= 180+60=240, x=180-60=120$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656463",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $ Minimum value of $h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} $
Find the minimum value of $h(\theta)$
$h(\theta)= 3 \sin \theta - 4\cos \theta + \sqrt{2} = 5 \sin (\theta + 53.13) + \sqrt{2} $
Minimum value -
$5\sin (\theta + 53.13) + \sq... | Minimum is attained when $\sin (\theta + 53.13) = -1$ that is
$$h_{min}=h(3\pi/2+k\pi)=5 \sin (3\pi/2+k\pi) + \sqrt{2}=-5+\sqrt{2}$$
for the same reason the maximum is attained when $\sin (\theta + 53.13) = 1$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2656593",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.
Find the length of the tangent to the curve $x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}$ which is intercepted between the axes.
$x^{\frac{2}{3}}+y^{\frac{2}{3}}=a^{\frac{2}{3}}\imp... | Now, if $x=0$ we obtain $y=q+\sqrt[3]{qp^2}$ and for $y=0$ we obtain $x=p+\sqrt[3]{pq^2}.$
Thus, $A\left(p+\sqrt[3]{pq^2},0\right)$ and $B\left(0,q+\sqrt[3]{qp^2}\right)$ they are intersection points of the tangent with $x$-axis and with $y$- axis respectively.
Id est, $$AB=\sqrt{\left(p+\sqrt[3]{pq^2}\right)^2+\left( ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658068",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that if $3\mid(a^2+b^2)$,then $3\mid a$ and $ 3\mid b$ I am trying to prove this by contradiction. So if $3$ doesn't divide $a$ or $3$ doesn't divide $b$, then the remainder is either $1$ or $2$. I am struggling on what to do next. How do I get the remainder of $a^2$ and $b^2$ for these cases?
Any help is greatly... |
I am trying to prove this by contradiction.
Why?
A contradiction where the remainder is either 1 or 2 and $a^2$ and $b^2$ can be most combinations ... that's a lot to check.
But a direct proof requires the remainder to be exactly $0$ (one option) which requires $a^2$ and $b^2$ to have opposite (add to a multiple of t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2661125",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
Computing minimal polynomial An example in chapter 14.2 in Dummit and Foote computes the minimal polynomial for $\sqrt{2}+\sqrt{3}$ over $\mathbb{Q}$. We consider the field $\mathbb{Q}(\sqrt{2}+\sqrt{3})$, and note that this is the same field as $\mathbb{Q}(\sqrt{2},\sqrt{3})$. Now, the other roots of the minimal polyn... | if $x = \sqrt 2 + \sqrt 3,$ then $x^2 = 5 + 2 \sqrt 6.$ So $(x^2 - 5)^2 = 24,$ and $x^4 - 10 x^2 + 25 = 24,$ and
$$ x^4 - 10 x^2 + 1 = 0. $$
By the Gauss lemma, without any rational roots, the possible rational factorings are
$$ (x^2 + ax +1)(x^2 - ax + 1), $$
$$ (x^2 + ax -1)(x^2 - ax - 1). $$
Neither one gives int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2662642",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ How do I solve this equation I found in my textbook:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
This is what I tried:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
$\mapsto \sqrt{(x+4)(x-4)} - (x-4) = \sqrt{(x-1)(x-4)}$
Dividing both sides by $\sqrt{x-4}$
$\mapsto \sqrt{x+4} - ... | I think it should be like this when the equation gets divided by $\sqrt{x-4}$:
$(\sqrt{x-4})(\sqrt{x+4} - \sqrt{x-4}) - \sqrt{(x-4)(x-1)} = 0$
$\mapsto (\sqrt{x-4})(\sqrt{x+4} - \sqrt{x-4} - \sqrt{x-1}) = 0$
$\therefore$ Either $\sqrt{x-4} = 0$ OR $(\sqrt{x+4} - \sqrt{x-4} - \sqrt{x-1}) = 0$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
Problem with generalized eigenvectors in a 3x3 matrix. I have this matrix:
$$
A= \begin{pmatrix}
0 & 1 & 1 \\
0 & 1 & 0 \\
-1 & 1 & 2 \\
\end{pmatrix}
$$
I have founded the eigenvalues:
$$\lambda_{1,2,3}=1$$
So
$$\lambda=1$$$$\mu=3$$
I'm expecting to have one eigenvector plus two generalized eigenvec... | I like to keep things as matrices and column vectors, visual. Since $(A - I)^2 = 0$ but $A - I \neq 0,$ we can take the third column of $R$ (for "Right") as anything we like for which
$$ (A - I) w \neq 0. $$
I like
$$
w =
\left(
\begin{array}{c}
0 \\
0 \\
1
\end{array}
\right)
$$
The second column will be $v = (A-I... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2667609",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $f(x) = \frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$ Prove $f(x) =\frac{1}{1+x^2}$ is uniformly continuous on $\mathbb{R}$.
My attempt...
Proof
$$\left| f(x) - f(y) \right| = \left| \frac{1}{1+x^2} - \frac{1}{1+y^2}\right| = \frac{\left|x+y\right|}{\left(1+x^2\right)\left(1+y^2\right)}\left|x-y\ri... | Lipschitz-continuity implies uniform continuity, we cannot disagree on that. On the other hand there is a slightly more efficient way for proving that $|f'|$ is bounded:
$$ f'(x) = f(x)\cdot \frac{d}{dx}\log f(x) = \frac{1}{1+x^2}\cdot \frac{-2x}{1+x^2}$$
where $\left|\frac{1}{1+x^2}\right|\leq 1$ is trivial and $\left... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668151",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
How to prove $e^x\ge \left(1+\frac xn\right)^n$ for any real numbers $x, n > 0$ Can someone provide a detailed proof? I saw a proof here
$$\begin{align}
\frac{e_{n+1}(x)}{e_n(x)}&=\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n}\\\\
&=\left(1+\frac{-x}{(n+x)(n+1)}\right)^{n+1}\left(1+\frac xn\right... | \begin{align}
\frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^n} &= \frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left(1+\frac xn\right)^{n + 1}}\left(1+\frac xn\right) \\
&= \frac{\left(1+\frac x{n+1}\right)^{n+1}}{\left( \frac{n + x}{n} \right)^{n + 1}}\left(1+\frac xn\right) \\
&= \left(1+\frac x{n+1}\r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2668555",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
(Full) Solution to $\int \frac{1-x}{x^2+x+1}dx$ with steps.
*
*If we use substitution, we get:
$u=x^2+x+1$
$du=(2x+1)dx$
$(-1/2)du=(-x-1/2)dx=(-x+1-3/2)dx$
*Then the integral becomes:
$\int ((-du/2)/u) + 3/2\int dx/(x^2+x+1$)
But why isn't it just:
$\int ((-1/2du)-3/2)/u$ ?
And why do you use + 3/2 in the origi... | $\displaystyle \int\frac{1-x}{x^2+x+1}dx=\frac{3}{2}\int \frac{dx}{(x+\frac{1}{2})^2+\frac{3}{4}}-\frac{1}{2}\int\frac{2x+1}{x^2+2x+1}dx$.
Let $\displaystyle x+\frac{1}{2}=\frac{\sqrt{3}}{2}\tan\theta$. Then $\displaystyle dx=\frac{\sqrt{3}}{2}\sec^2\theta d\theta$ and $\displaystyle \left(x+\frac{1}{2}\right)^2+\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670310",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $x^4 - 8x^3 + 21x^2 - 20x + 5 = 0$ given that the sum of two of its roots is $4$ Here's what I tried:
Let the roots be $a$, $b$, $c$ and $d$, $a+b=4$. Then,
$$a + b + c + d = 8 \Longrightarrow 4 + c+ d = 8 \Longrightarrow a+b = c+d = 4$$
$$(a + b)(c + d) + ab + cd = 21$$
$$ab (c + d) + cd (a + b) = 20 \Longrighta... | Two of the roots sum to 4. All 4 roots sum to 8. This means that the other two roots must also sum to 4.
If this polynomial had rational roots, they would have to be in the set $\{\pm1,\pm5\}$
By what we have above we might try $x=-1, x = 5$ alas these do not work.
We could factor the polynomial like so:
$(x^2 - 4x +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670645",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Using summation by parts to evaluate an alternating sum I want to evaluate
$$
\sum_{k=0}^n (-1)^k \binom{n}{k} k.
$$
I tried summation by parts, i.e. the formula
$$
\sum_{k=0}^n (f(k+1) - f(k))g(k)
= f(n+1)g(n+1) - f(0)g(0) - \sum_{k=0}^n f(k+1) (g(k+1) - g(k))
$$
with $f(k+1) - f(k) = (-1)^k \binom{n}{k}$ and $g(k... | Summation by parts is definitely an overkill, differentiation a lesser overkill. For any $n\geq 1$ we have
$$ k\binom{n}{k} = n\binom{n-1}{k-1} $$
and
$$ \sum_{k=0}^{n}(-1)^k\binom{n}{k}k = \sum_{k=1}^{n}(-1)^k\binom{n}{k}k = -n\sum_{k=1}^{n}(-1)^{k-1}\binom{n-1}{k-1}=-n\sum_{j=0}^{n-1}\binom{n-1}{j}(-1)^j $$
so your s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670754",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Find the eigenvalues of block matrix $C$
Find the eigenvalues of $$C = \begin{bmatrix}\begin{array}{c|c} 0 & A\\ \hline A^T & 0\end{array}\end{bmatrix}$$ where $$A = \begin{bmatrix}
0&0&0&1&1&1&1\\0& 1& 1& 0& 1& 0& 1\\0 &1 &1 &1 &0 &1 &0\\1& 0& 1& 0& 0& 1& 1\\1 &0 &1& 1& 1& 0& 0\\1 &1 &0& 0& 1& 1& 0\\1 &1& 0& 1& 0& 0 ... | Since $\det(xI-C)=\det(x^2I-AA^T)$, the eigenvalues of $C$ are the singular values of $A$ and their negatives.
Now, note that $AA^T=2I+2ee^T$, where $e$ denotes the all-one vector. Hence the set of eigenvalues of $AA^T$ consists of a simple eigenvalue $2+2\times7=16$ and the eigenvalue $2$ of multiplicity $6$. It follo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672027",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Computing $\sum_{n=1}^ \infty n^2 \cdot \left(\frac{2}{3}\right)^n$ I've been dealing with the following series for a while now, without real progress.
$$\sum_{n=1}^ \infty n^2 \cdot \left(\frac{2}{3}\right)^n$$
After using WolframAlpha, I know it converges to $30$, but I can't see how to calculate it by myself.
Any le... | Substitute $2/3$ by a variable $x$. Then the series defined by
$$\sum_{k=1}^\infty k^2 x^k$$
converges locally uniformly for $|x|<1$, hence we can integrate and differentiate it termwise so
$$\sum_{k=1}^\infty k^2 x^k=x\sum_{k=1}^\infty k^2 x^{k-1}=x\sum_{k=1}^\infty k\frac{d}{dx}( x^k)=x\frac{d}{dx}\left(\sum_{k=1}^\i... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2673348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Factorize $\det\left[\begin{smallmatrix}yz-x^2&zx-y^2&xy-z^2\\zx-y^2&xy-z^2&yz-x^2\\xy-z^2&yz-x^2&zx-y^2\end{smallmatrix}\right]$ using factor theorem Factorize and prove that
$$
\Delta=\begin{vmatrix}
yz-x^2&zx-y^2&xy-z^2\\
zx-y^2&xy-z^2&yz-x^2\\
xy-z^2&yz-x^2&zx-y^2
\end{vmatrix}\\=\frac{1}{4}(x+y+z)^2\Big[(x-y)^2+(... | Hint: Note that your matrix has the form
$$\pmatrix{a&b&c\cr b&c&a\cr c&a&b\cr }$$
Which has the determinant
$$3abc-c^3-b^3-a^3$$
which can again be factored into
$$-\left(a+b+c\right)\,\left(a^2+b^2+c^2-ab-bc-ac\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2675951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Surface integral of sphere within a paraboloid in spherical coordinates Find the surface area of the portion of the sphere $x^2 + y^2 + z^2 =3c^2$ within the paraboloid $2cz =x^2+y^2$ using spherical coordinates. ($c$ is a positive constant)
I've found it in cartesian coordinates and then polar coordinates by taking ... | The limits for $\phi$ are wrong. Note that $\sqrt{3}>1$ and $\arccos(x)$ is defined in $[-1,1]$. Solving the equation $2cz+z^2=3c^2$ we get $z=c$ and $z=-3c$ (which is not acceptable because $x^2+y^2=2cz=-6c^2<0$). Hence $\phi$ goes from $0$ (north pole) to $\arccos(1/\sqrt{3})$. Therefore
$$S=3c^2\int_{\theta=0}^{2\pi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2679032",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$ (Solve for $k$) Question:
$1^2-2^2+3^2-4^2+…-2016^2+2017^2=2017k$
Solve for $k$
My attempt:
$$1^2-2^2+3^2-4^2+…-2016^2+2017^2\\
\begin{align}= (1-2)(1+2)+(3-4)(3+4)+…+(2015-2016)(2015+2016)+2017^2 \end{align}
$$
What should I do next?
| A very simple way of seeing this is to write the following:
$$\begin{align} \\
1^2-2^2+\dots-2016^2+2017^2 &= 1+(3^2-2^2)+\dots+(2017^2-2016^2) \\
&=1+(3-2)(3+2)+(5-4)(5+4)+\dots+(2017-2016)(2016+2017) \\
&=1+2+3+...+2016+2017 \\
&=\frac{(2018)(2017)}{2}=2017k \\
\end{align}$$
Which means that $k=\frac{2018}{2}=1009$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684608",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
When is $2^{2n+2}+2^{m+2}+1$ a perfect square? Let $m,n$ be natural numbers satisfying $m\leq 2n$. Is it true that $$2^{2n+2}+2^{m+2}+1$$ is a perfect square if and only if $m=n$?
What I have tried: Under the assumption $m<n$, I've tried to 'squeeze' the above number between two consecutive squares, implying it cannot ... | Suppose $m>n$ and $k$ is an integer such that $$(2^{n+1}+k)^2= 2^{2n+2}+2^{m+2}+1.$$ This means $$2^{n+2}k+k^2=2^{m+2}+1$$ so $$(k-1)(k+1)=k^2-1=2^{n+2}(2^{m-n}-k).$$ Note that clearly $k$ must be odd and $\gcd(k-1,k+1)=2$, so either $k-1$ or $k+1$ is divisible by $2^{n+1}$. Since $k>1$ (if $k=1$ we would have $m=n$)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2685950",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
Solve the equation $\cos^2x+\cos^22x+\cos^23x=1$
Solve the equation: $$\cos^2x+\cos^22x+\cos^23x=1$$
IMO 1962/4
My first attempt in solving the problem is to simplify the equation and express all terms in terms of $\cos x$. Even without an extensive knowledge about trigonometric identities, the problem is solvable.... | You can shorten the argument by noting at the outset that
$$
\cos3x=4\cos^3x-3\cos x=(4\cos^2x-3)\cos x
$$
so if we set $y=\cos^2x$ we get the equation
$$
y+(2y-1)^2+y(4y-3)^2=1
$$
When we do the simplifications, we get
$$
2y(8y^2-10y+3)=0
$$
The roots of the quadratic factor are $3/4$ and $1/2$.
A different strategy ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687769",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Finding the numbers having a particular factor,an upper limit and a specific set of digits. Question :
How many numbers less than $50000$ can be formed which are multiples $6$ using the digits $0,1,2,3,4,5$?
My attempt :
First of all, for a number to be a multiple of $6$, it should be a multiple of $2$ as well as $3$... | We put it in three cases as the last digit must be either $0;2$ or $4.$ (I assume that you can use each digit only once).
First case: The number is in the form $\overline{abcd0}$, which means the numbers $a,b,c,d$ are $4$ of the remaining numbers available $(1;2;3;4;5)$.
There are $5$ ways of picking four numbers from ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Mathematical Induction prove that $n^3+5n$ is divisible by $6$ Sorry, I know this will be a duplicate on the site but the other solution I found confusing and the method look completely different to what I was taught.
Prove that $n^3 + 5n$ is divisible by $6$ by using induction
The question is
Prove by mathematical Ind... | a one line proof $$n^3+6n-n=(n-1)n(n+1)+6n$$ ready!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Verfication of deduction made using the Cauchy-Schwarz inequality Is the following proof correct?
Show that $$16\leq(a+b+c+d)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{d}\right)$$ for all positive numbers $a,b,c,d$.
Proof. Let $\mathbf{R}^4$ be the inner product space with the inner product defined as in the ... | Yes, your proof is fine. Why did you write $|\sqrt{a}|$ ? If $a$ is positive, we always have $\sqrt{a}>0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2689137",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "12",
"answer_count": 3,
"answer_id": 0
} |
Find three rational numbers $a,b,c$ s.t. $b^2-a^2=c^2-b^2=5$. A rational numbers cannot have irrational as$\exists n \in \mathbb{Z}, \, \sqrt[n]{\frac xy}$, but the two equalities give: $b^2=\frac{(a^2+c^2)}{2} \implies b = \sqrt[2]{\frac {(a^2+c^2)}{2}}$.
To avoid this, need $4\mid a$, & $a=c$; so that if $\exists t=a... | We can rewrite these numbers with common denominator $D$:
$a=\dfrac{A}{D}$, $\;b=\dfrac{B}{D}$, $\;c=\dfrac{C}{D}$. Then we'll get diophantine equation (system of diophantine equations):
$$
B^2-A^2=C^2-B^2=5D^2.\tag{1}
$$
The smallest solution of $(1)$ is $(A,B,C,D)=(31,41,49,12)$.
So, example of such three rational nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2694706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
How to use integration by parts to solve an integral? I'm having trouble with solving this integral by parts. Here is what I have so far. Can anyone please help me out?
Solve the integral $\int x^3\sqrt{x^2-1}dx$ by parts, choosing u = $x^2$ and
$dv = x\sqrt{x^2-1} dx$
$du_1 = 2xdx$
$v_1$ = $\int x\sqrt{x^2-1}dx$ $ t =... | You used integration by parts once. And you used substitution to find $v_1$. Use substitution again:
$$I=\int 2x\cdot \frac13(x^2-1)^{3/2}dx;$$
$$t=x^2-1; \ \ dt=2xdx;$$
$$I=\frac13 \int t^{3/2}dt=\frac13\cdot \frac25 \cdot t^{5/2}+C=\frac{2}{15}(x^2-1)^{5/2}+C.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698119",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Finding the the derivative of $y=\sqrt{1-\sin x}; 0A question I'm attempting is:
Find the derivative of $ y = \sqrt {1 - \sin x} ; 0 < x <\pi/2$.
I did this:
$y = \sqrt {1 - \sin x} = \sqrt {\cos^2\frac{x}{2} + \sin^2\frac{x}{2} - 2\sin \frac{x}{2}\cos \frac{x}{2}} = \sqrt { (\sin \frac{x}{2}-\cos \frac{x}{2})^2} = \... | Hint: $y = |\sin(x/2)-\cos(x/2)|$ and $|\sin(x/2)-\cos(x/2)|=\color{red}{-}(\sin(x/2)-\cos(x/2))$ for $0 \le x \le \pi/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2698798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Evaluate $\int_{0}^{\pi/2}\frac{x}{(a^2\cos^2x+b^2\sin^2x)}dx$ The question is to evaluate this definite integral:
$$\int_{0}^{\pi/2}\frac{x}{(a^2\cos^2x+b^2\sin^2x)}dx$$
I know the definite integration with limits from $0$ to $\pi$, where we use the identity $\int_{a}^{b}f(x)dx = \int_{a}^{b}f(a+b-x)dx$, but it obvio... | Write $\alpha = a/b$ and substitute $\tan x = \alpha t$ to obtain
$$ \int_{0}^{\frac{\pi}{2}} \frac{x}{a^2\cos^2 x + b^2\sin^2 x} \, dx
= \frac{1}{ab} \int_{0}^{\infty} \frac{\arctan(\alpha t)}{1 + t^2} \, dt
=: \frac{1}{ab}I(\alpha). $$
Then
\begin{align*}
I'(\alpha)
&= \int_{0}^{\infty} \frac{t}{(1+t^2)(1+\alpha^2 t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2699046",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$ Related to solving $~~2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
I've tried some arithmetic to find something like
$a^x=b~\Longrightarrow~x=\log_{a}b$
But what I've found is that
$2^{x+3}+3^{x-5}=2^{3x-7}+3^{2x-10}$
$2^8\cdot2^{x-5}+3^{x-5}=2^8\cdot2^{3(x-5)}+3^{2(x-5)}$
$256\cdot2^{x-5}+... | $$0=2^{x+3}(2^{2x-10}-1)+3^{x-5}(3^{x-5}-1)=f(x)$$
As $2^{x+3},3^{x-5}>0$ for real $x,$
If $x-5>0$, then $f(x)>0$, and if $x-5<0$, then $f(x)<0$.
Since we wanted $f(x) = 0$, it must be true that $x - 5 \not< 0$ and $x - 5 \not> 0$. So $x - 5 = 0$, and thus $x = 5$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Calculate the following seminorm of $A$ Consider the following operators defined on $(\mathbb{C}^2,\|\cdot\|_2)$:
$$S = \begin{pmatrix} 1 & 1 \\ 1 & 1 \end{pmatrix}, \qquad A = \begin{pmatrix} 1 & 0 \\ 0 & 0 \end{pmatrix}.$$
I want to calculate $\|A\|_S$, with
$$\|A\|_S:=\inf\left\{c\geq 0; \quad\sqrt{\langle SAx,Ax... | You basically did all the work. You just showed that $\forall x \in \mathbb{C}, \left < SA \left ( \begin{matrix} x \\ x \end{matrix} \right ) \middle | A \left ( \begin{matrix} x \\ x \end{matrix} \right ) \right > = |x|^2$ and $\left < S \left ( \begin{matrix} x \\ x \end{matrix} \right ) \middle | \left ( \begin{mat... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705839",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Solving an equation in $\mathbb N$ I am trying to solve the equation $$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}.$$
I have made the following progress:
1) $x, y z$ have to be larger than $1$
2) only one of x, y, z can be $2$; rest should be larger
3) WLOG I have assummed $2\leqslant x\leqslant y\leqslant z$. Kno... | This is not an exhaustive list. But it is at least a partial solution.
$$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{3}{5}.$$
We know that $\dfrac{1}{ab}=\dfrac{1}{a(a+b)} + \dfrac{1}{b(a+b)}$
Let $x=2$ and you get $\frac{1}{y}+\frac{1}{z}=\frac{1}{10}.$
We can use $a=2$ and $b=5$ to get
$$\frac{3}{5} = \frac{1}{2}+\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2708022",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 0
} |
Partial Derivative of arctan Given that $$f(x,y)=\tan^{-1}\left(\frac{x+y}{1-xy}\right)$$
Find $f_x(x,y)$
My attempt,
$$
\begin{aligned}
f_x(x,y)&=\frac{(1-xy)(1)-(x+y)(-y)}{(1-xy)^2}\cdot\frac{1}{1+\left(\frac{x+y}{1-xy}\right)^2}\\
&=\frac{1+y^2}{(1-xy)^2+(x+y)^2}\\
&=\frac{1+y^2}{1+y^2+x^2+x^2y^2}
\end{aligned}
$$
B... | Alternatively, denote: $x=\tan a; y=\tan b$. Then:
$$z=\arctan \frac{\tan a+\tan b}{1-\tan a\tan b}=a+b.$$
So:
$$z_x=a_x=(\arctan x)'=\frac{1}{1+x^2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710497",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Minimum value of $p^2+q^2+r^2+s^2$.
If $p,q,r,s>0$ and $(p+q)(q+r)(r+s)(s+p)=16$ and $pq+qr+rs+sp=2$.
Then minimum value of $(p^2+q^2+r^2+s^2)$
Try: using Cauchy Schwarz Inequality
$(p^2+q^2+r^2+s^2)(q^2+r^2+s^2+p^2)\geq (pq+qr+rs+sp)^2=4$
But answer given as $6$. Could some help me to solve it. Thanks
| Write $p^2 + q^2 + r^2 + s^2 = X$. Then, $$2X + 2(pq+qr+rs+sp) = (p+q)^2 + (q+r)^2 + (r+s)^2 + (s+p)^2 = 2X + 4$$.
Write $a = p+q, b = q+r, c= r+s, d = s+p$, then by AM-GM on $a^2,b^2,c^2,d^2$ we see that $a^2+b^2+c^2+d^2 \geq 4 \sqrt[4]{(abcd)^2} = 16$.
Therefore, $2X+4 \geq 16$, and $X \geq 6$.
Equality is attaine... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713072",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
Sum of the first integer powers of $n$ up to k Pascal's triangle has a lot of interesting patterns in it; one of which is the triangular numbers and their extensions. Mathematically:
$$\sum_{n=1}^k1=\frac{k}{1}$$
$$\sum_{n=1}^kn=\frac{k}{1}\cdot\frac{k+1}{2}$$
$$\sum_{n=1}^kn^2=\frac{k}{1}\cdot\frac{k+1}{2}\cdot\frac{2... | We are trying to find the correction polynomial of degree p $$a(k,p)$$ where
$$
a(k,p)=\sum_{n=1}^k(n^p)-\frac{k}{(p+1)!}\prod_{n=1}^{p}(nk+1)=\sum_{f=0}^p(a_fx^f)
$$
The correction polynomial was of degree three for the sum of cubes and degree four for the sum of tesseracts. Assuming that the polynomial degree is alw... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2713657",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$$
First,I tried to evaluate like this:
$$\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx=\int_{0}^{\frac{\pi}{2}}x^2\left(\frac{1+\cos x}{\sin x}\right)\frac{dx}{1+\cos x}=\int_{0}^{\frac{\pi}{2}}x^... | Observe we have
\begin{align}
I=\int^{\pi/2}_0 \frac{x^2}{\sin x}\ dx = \int^{\pi/2}_0 \frac{x^2}{\cos\left(\frac{\pi}{2}-x \right)}\ dx = \int^{\pi/2}_0 \frac{(\frac{\pi}{2}-u)^2}{\cos u}\ du.
\end{align}
Then using integration by parts, we see that
\begin{align}
I&=\left(\frac{\pi}{2}-u\right)^2\left\{\log\left|1 + \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "15",
"answer_count": 9,
"answer_id": 5
} |
On the least prime in an arithmetic progression $a + nb$ where $a,b$ are distinct primes. Dirichlet's Theorem: there are infinitely many primes in every arithmetic progression $\{a + nd: n \geq 0\}$ for coprime $a, d$. Consider just the case where $a, d$ can take on prime values or $0$.
Then define $a \oplus a = 0$, $... | Take $7\oplus 11=29$, so that $(7\oplus 11)\oplus 13=29\oplus 13=107$.
On the other hand we have $11\oplus 13=37$, and $7\oplus(11\oplus 13)=7\oplus 37=229$. So
$$
(7\oplus 11)\oplus 13=107\neq 229=7\oplus(11\oplus 13)
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$ Solve the ODE $\cot (x^2+y^2)dy+xdx+ydy=0$
i am trying to solving integrating combination
since given that
$\cot (x^2+y^2)dy+xdx+ydy=0$
then $\cot (x^2+y^2)dy+d(xy)=0$ is it correct way ? and we can apply integration from here? can any one help me this problem
| Using polar coordinates, we see that
\begin{align}
dx =&\ \cos\theta dr -r\sin\theta d\theta,\\
dy =&\ \sin\theta dr + r\cos\theta d\theta
\end{align}
then we see that
\begin{align}
\cot(x^2+y^2)dy +xdx+ydy = (\cot(r^2)\sin\theta+r) dr+r\cot(r^2)\cos\theta d\theta =0.
\end{align}
Hence it follows
\begin{align}
\frac{d\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2718212",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
How do I calculate the sum of $\sum_{n=0}^{\infty}{x^{n}\over n!}\cdot{{x^2\over n+2}}?$ How do I calculate the sum of $$\sum_{n=0}^{\infty}{x^{n+2}\over (n+2)n!}=\sum_{n=0}^{\infty}\color{red}{x^{n}\over n!}\cdot{{x^2\over n+2}}?\tag1$$
$$\sum_{n=0}^{\infty}{x^{n}\over n!}=e^x\tag2$$
| Hint:
$$\dfrac{x^{n+2}}{(n+2) n!}=\dfrac{(n+1)x^{n+2}}{(n+2)!}=\dfrac{(n+2-1)x^{n+2}}{(n+2)!}$$
$$=x\cdot\dfrac{x^{n+1}}{(n+1)!}-\dfrac{x^{n+2}}{(n+2)!}$$
$$\implies\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2) n!}=x\sum_{n=0}^\infty\dfrac{x^{n+1}}{(n+1)!}-\sum_{n=0}^\infty\dfrac{x^{n+2}}{(n+2)!}$$
Now use $e^y=\sum_{r=0}^\in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2719922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Find the minimum value of $\frac{a+b}{2} + \frac{2}{ab-b^{2}}$, with $a,b \in \mathbb{R}$, $a>b>0$ Find the minimum value of $$ \frac{a+b}{2} + \frac{2}{ab-b^{2}},$$
where $a,b \in \mathbb{R}$, $a>b>0$.
Attempt : The only method I knew was using partial derivatives.
Let
$$f(a,b) = \frac{a+b}{2} + \frac{2}{ab-b^{2}}... | Denote $c = a - b > 0$, then$$
\frac{a + b}{2} + \frac{2}{ab - b^2} = b + \frac{c}{2} + \frac{2}{bc} \geqslant 3 \sqrt[3]{b · \frac{c}{2} · \frac{2}{bc}} = 3.
$$
The equality is achieved at $b = 1$, $c = 2$, i.e. $(a, b) = (3, 1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
either inequality $\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$ is hard or I need to go back to study ASAP Prove that for every $x \in(0,\frac{\pi}{2})$, the following inequality:
$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$
holds
I don't see room to use derivatives, since it seems a little messy to calculate the ... | The Maclaurin series of the tangent function has all coefficients positive (see formula). Integrating we get that the Maclaurin series of the function $-\log(\cos x)$ has all coefficients positive. Up to order $12$ it is
$$-\log (\cos x) =\frac{x^2}{2} + \frac{x^4}{12} + \frac{x^6}{45} + \frac{17 x^8}{2520} + \frac{31... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x^2-\frac{9}{2}x+\left(\frac{9/2}{2}\right)^2-\left(\frac{... | HINT
We have
$$2\left(x^2-\frac{9}{2}x+\frac{9}{2}\right)=2\left(x-\frac{9}{4}\right)^2-\frac{81}{8}+9=2\left(x-\frac{9}{4}\right)^2-\frac98$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Calculus max rectangle Consider a rectangle with sides $2x$ and $2y$ inscribed in a given fixed circle $x^2+y^2=a^2$ , and let $n$ be a positive number. We wish to find the rectangle that maximizes the quantity $z=x^n+y^n$ . If $n=2$ , it is clear that $z$ has the constant value $a^2$ for all rectangles. If $n<2$, show... | By Lagrange multipliers: Let $g(x,y) = x^2+y^2$ and $f(x,y) = x^n+y^n$. First assume $x$ and $y$ are positive. The constraint is $g = a^2$. Since the case $n=2$ is done, assume $n \neq 2.$
We have $\nabla f = (nx^{n-1},ny^{n-1})$ and $\nabla g = (2x,2y).$ So the equation $\nabla f = \lambda \nabla g$ gives us
$$nx^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2724660",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the last three digits of $19^{100}$ Find the last three digits of $19^{100}$
$19^{100}=361^{50}=(1+360)^{50}=\binom{50}{0}+\binom{50}{1}360+\binom{50}{2}360^2+\cdots$
When we divide it by $1000$, the remainder comes out to be $001$, so the last three digits must be $001$, but in my book, the answer is given as $8... | Alternatively:
$$19^{100}=(20-1)^{100}=\underbrace{20^{100}-100\cdot 20^{99}+\cdots -\begin{pmatrix} 100\\ 3\end{pmatrix}\cdot 20^3}_{=1000A}+\begin{pmatrix} 100\\ 2\end{pmatrix}\cdot 20^2-\begin{pmatrix} 100 \\ 1\end{pmatrix}\cdot 20+\begin{pmatrix} 100 \\ 0\end{pmatrix}=$$
$$1000A+1978000+1.$$
Hence:
$$19^{100} \equi... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726161",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Help with $\int \frac{x-\sqrt x}{x+\sqrt x} \, \mathrm d x$ I need help with this integral. I tried $$\int \frac{x-\sqrt x}{x+\sqrt x}\,\mathrm dx = \int \frac{(x-\sqrt x)^2}{x^2-x} \,\mathrm dx = \int \frac {x(x-2\sqrt x+1)}{x(x-1)}\,\mathrm dx = \int \frac {(x-2\sqrt x+1)}{(x-1)}\,\mathrm dx$$ but I don't know how to... | Before substitution:
$$\int \frac{x-\sqrt x}{x+\sqrt x} dx= \int \frac{\sqrt x -1}{\sqrt x+1} dx = \int \frac{\sqrt x +1-2}{\sqrt x+1} dx = \int \left(1- \frac{2}{\sqrt x+1} \right)dx$$ $$= x-2 \int \frac{1}{\sqrt x+1}dx$$
Then $x = (u-1)^2$:
$$\int \frac{1}{\sqrt x+1}dx = \int \frac{2(u-1)}{u} du = 2\int \left( 1- \fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2726503",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
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.