Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
System of congruent equations Given the following system, how do you prove that there only is one solution congruent 8?
$$3x+7y \equiv 2 (\text{mod } 8)$$
$$4x+5y \equiv 7 (\text{mod } 8)$$
My main idea have been to solve the equation, then to come to the solution that there only is one solution, however I get flawed r... | Write this linear system in matrix form:
$$\begin{pmatrix}3&7\\4&5\end{pmatrix}\begin{pmatrix}x\\y\end{pmatrix}\equiv\begin{pmatrix}2\\7\end{pmatrix}$$
and multiply both sides by the adjugate matrix, which is invertible mod. $8$:
\begin{align}
&\begin{pmatrix}5&1\\4&3\end{pmatrix}\begin{pmatrix}3&7\\4&5\end{pmatrix}\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2573706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Determine the value of $\cos(2)$ I'm trying to prove that the function $f(x)=\cos(x)$ has an $x$-intercept in the interval $[0,2]$. To do that I showed that $f(0)=\cos(0)=1>0$. Then I would have to show that $\cos(2)<0$, however how would I workout the value of $\cos(2)$? My professor mentioned something about using th... | Just for the fun of it.
Using the same idea as Felix Marin, let us build the $[n,n]$ Padé approximant of $\cos \left(x+\frac{2\pi }{3}\right)$ around $x=0$ and evaluate it for $x=2-\frac{2\pi }{3}$.
For $n=1$
$$-\frac{\frac{1}{2}+\frac{7 }{4 \sqrt{3}} x}{1+\frac{1}{2 \sqrt{3}}x }\implies \cos(2)\approx -\frac{9+21 \sqr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2574738",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
If $c_n=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{i(n-i)x}dx $ ,then find $\sum_{n\in \mathbb{Z}}|c_n|^2$ Let $$c_n=\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{i(n-i)x}dx .$$ Then what will be the value of $$\sum_{n\in \mathbb{Z}}|c_n|^2.$$
My attempt:
\begin{align*}
c_n & =\frac{1}{\sqrt{2\pi}}\int_{-\pi}^\pi e^{i(n-i)x}dx... | from here David Cardon
$$\sum_{k=-\infty}^{\infty}\frac1{1+k^2}=\pi\coth (\pi)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2575332",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
I don't know how to solve this integral
Could you please help me compute the following integral: $$\int \frac{(x^2+x-4)}{x(x^2+4)}dx$$
What I've done so far is:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{A}{x}dx+\int \frac{B}{x^2+4}dx$$
So
$$x^2+x-4 = A(x^2-4)+Bx$$
And I did this so I could figure out the v... | Hint:
It must be
$$\dfrac{x^2+x-4}{x(x^2+4)}=\dfrac Ax+\frac{Bx+C}{x^2+4}$$
then
\begin{align*}
x^2+x-4&=A(x^2+4)+(Bx+C)x\\
&=(A+B)x^2+Cx+4A
\end{align*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577273",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Factorising $x^2-x+1$ I attempted to factorise $x^2-x+1$, which seemed fairly straightforward:
$$
\begin{align}
x^2-x+1 &= (x+1)^2-3x \\
&= (x+1-\sqrt{3x})(x+1+\sqrt{3x}) \\
&= (x-\sqrt{3x}+1)(x+\sqrt{3x}+1)
\end{align}$$
Technically neither $(x-\sqrt{3x}+1)$ nor $(x+\sqrt{3x}+1)$ are linear functions,... | The function $(x+1-\sqrt{3x})(x+1+\sqrt{3x})$ is only defined for $x \ge 0$, therefore it is not the same function as $x^2-x-1$, which is defined on $\forall x \in \mathbb{R}$. Why? Think about how the square root is defined.
For $x < 0$, you have $|x|=-x$, hence this factorization becomes
$$ x^2 - x + 1 = (x+1)^2 + 3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2577376",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 1
} |
Minimum value of rational expression
If $x,y,z$ are distinct real number, Then minimum value of $\displaystyle \left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2$
Try: $$\left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2\geq \frac{x^2+y^2+z^2}{... | we have to prove that $$ \left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2\geq 2$$
this is true since this inequality is equivalent to $${\frac { \left( {x}^{3}-y{x}^{2}-z{x}^{2}-{y}^{2}x+3\,zyx-{z}^{2}x+{y}
^{3}-z{y}^{2}-{z}^{2}y+{z}^{3} \right) ^{2}}{ \left( y-z \right) ^{2}
\lef... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2578651",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solving $\sin x - \cos x = 1$ I have been trying to solve $\sin x - \cos x = 1$ by squaring both sides but has not been able to obtain the solution. Here is what I did:
$$\begin{align}(\sin x - \cos x)^2 &=1^2\\\sin^2x-2\sin x\cos x+\cos^2x&=1\\1-\sin 2x&=1\\\sin2x&=0\end{align}$$
Obviously $x=0$ is not a solution. May... | $$
\begin{aligned}
\sin x-\cos x &=1 \\
-\frac{1}{\sqrt{2}} \sin x+\frac{1}{\sqrt{2}} \cos x &=-\frac{1}{\sqrt{2}} \\
\cos \left(x+\frac{\pi}{4}\right) &=-\frac{1}{\sqrt{2}} \\
x+\frac{\pi}{4} &=2 n \pi \pm \frac{3 \pi}{4} \\
x &=\frac{\pi}{2}(4 n+1) \text { or }(2 n-1) \pi, \text{ where }n\in Z.
\end{aligned}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2580343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 8,
"answer_id": 7
} |
Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$ Got these problems as separate sections of a question in a book's chapter on 'Divisibility & primes'.
*
*Show that if $x^2 + 5y^2 =1$, then $x = \pm 1$.
Can state it in terms of two factors as : $(x + y\sqrt{-5})(x-y\sqrt{-5}) =x^2 + 5y^... | Both $x$ and $y$ are drawn from $\mathbb Z$, right?
*
*If $x = \pm 1$, then $x^2 = 1$. Then $5y^2 = 0$, so obviously $y = 0$. Okay, I kinda took that backwards. Going forward: if $y \in \mathbb R$, then $y^2 \geq 0$. Therefore, if $y \neq 0$, then $5y^2 > 4 > 1$.
*Unless there is some restriction on $x$ and/or $y$,... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581390",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Extremal problem with positive integer numbers Let $a,b$ be two positive integer numbers such that $a\sqrt{3}-b\sqrt{7}>0$. Find the minimum value of
$$
S=(a\sqrt{3}-b\sqrt{7})(a+b).
$$
Attempt I have tried and guess that the minimum value of $S$ is $(55+36)(55\sqrt{3}-36\sqrt{7})$, where $(55,36)$ is the integer solut... | Here is a sketch proof that the greatest lower bound for $S$ is $(3/2)(1/\sqrt3 + 1/\sqrt7) = 1.43297\ldots$. Call this $m$. Note that $(3 - \sqrt3m)/\sqrt3 = (\sqrt7m - 3)\sqrt7 = k$, say.
For integers $a,b$ such that $\sqrt3a - \sqrt7b > 0$ define
$$\delta(a,b) = (\sqrt3a - \sqrt7b)(a + b) - m.$$
As noted in the link... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2581498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Show that $4^{3x+1} + 2^{3x+1}+ 1$ is divisible by 7 I want to show that $4^{3x+1} + 2^{3x+1} + 1$ is divisible by 7, I am trying to show this with modular arithmetic.
If I break up each part of the equation, I can see that
$4^{3x+1} = 4$ x $2^{6x}$
which implies that $4^{3x+1}mod(7) = 4$
I can't quite find a nice f... | If $x^2+x+1=0,x=w$ where $w$ is a complex cube root of unity.
$$x^{2m}+x^m+1=w^{2m}+w^m+1= \begin{cases}(w^3)^{2n}+(w^3)^n+1=3 &\mbox{if }m=3n\\
w^{2(3n+1)}+w^{3n+1}+1=w^2+w+1=0 & \mbox{if }m=3n+1\\=\cdot=0 & \mbox{if }m=3n+2 \end{cases} $$
$\implies x^2+x+1$ will divide $x^{2m}+x^m+1$ if $3\nmid m$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583756",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 4
} |
How to prove that 2 is not a Gaussian prime? To prove an algebraic number is not a Gaussian prime, need find factors of it.
Let us assume, $(a + bi)(x +yi) = 2, \exists a,b,x,y \in \mathbb{Z}$; so $(ax -by) + i(ay + bx) =2$.
As imaginary part is null in $2$, so $ax - by =2, ay + bx =0$.
Now, how to reach from here, the... | If $(a+bi)(x+yi)=2$, $(a-bi)(x-yi)=2$ and therefore $(a^2+b^2)(x^2+y^2)=4$. In how many ways can $4$ be written as the product of two integers, each of which is a sum of $2$ squares? Well, $4=4\times1=(2^2+0^2)\times(1^2+0^2)$ and $4=2\times2=(1^2+1^2)\times(1^2+1^2)$. Therefore, there are few possibilities for you to ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2583963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
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-
$$\Rightarrow \sin A = \sqrt2 - \cos A$$
$$\Rightarrow \tan A = \frac{\sqrt2 - \cos A}{\cos A}$$
$$\Rightarrow \tan A = \frac{ \sqrt 2 }{\co... | Your second attempt is actually what helps here. Note:
$$\begin{align}\tan A+\cot A&=\frac{\sin A}{\cos A}+\frac{\cos A}{\sin A}\\&=\frac{\sin^2 A+\cos^2 A}{\sin A\cos A}\\&=\frac{1}{\sin A\cos A}\\&=\frac{1}{1/2}\\&=2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586230",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Jordan decomposition - help with calculation of transformationmatrices? Let $$A=\begin{pmatrix}5&1&1 \\-1&3&1\\0&0&4\end{pmatrix}$$
The Jordan-decomposition is $$A=\begin{pmatrix}-1&-1&-1/2 \\1&0&0\\0&0&-1/2\end{pmatrix}\begin{pmatrix}4&1&0 \\0&4&1\\0&0&4\end{pmatrix}\begin{pmatrix}0&1&0 \\-1&-1&1\\0&0&-2\end{pmatrix}$... | We are looking for P such that $P^{-1}AP=J$ thus you need to solve the following systems $AP=PJ$ that is
$Av_1=4v_1$
$Av_2=v_1+4v_2$
$Av_3=v_2+4v_3$
where $v_1$,$v_2$,$v_3$ are the columns of P.
Once we find $v_1$ from the first system, we can find $v_2$ from the second and finally $v_3$ from the third.
See also this e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586921",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Smallest $n$ such that $125|2^n+3^n$
Find the smallest positive integer $n$ such that $125|2^n+3^n$
Suppose $125|2^n+3^n$. Now, $n$ must be odd because if $n$ is even then
$$0\equiv 2^n+3^n\equiv2^n+(-2)^n\equiv 2^n+2^n\equiv2^{n+1}\pmod 5,$$
which is in contrast to hypothesis. How to go on from here?
| As $3\equiv2^7\pmod{125}$
$$2^n+3^n\equiv2^n+(2^7)^n\equiv2^n(2^{6n}+1)\pmod{125}$$
So, we need $2^{6n}\equiv-1\pmod{125}$
Using Show that $r$ is a primitive root (mod $p^k$), $2$ is a primitive root $5^n$ for $n\ge1$
$$\implies6n\equiv\dfrac{\phi(125)}2\pmod{\phi(125)}$$
$$\iff3n\equiv25\pmod{50}$$
As $17\cdot3\equiv1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2587700",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
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?
| We can rewrite the problem as a first-order difference equation:
\begin{align*}
a_{n+1} &= \frac{1}{3} a_n + 3 \\ a_0 &= 4
\end{align*}
We want to know $a_{2018}$.
There are a number of standard techniques to solve these, that are covered in discrete math courses. One trick is temporarily ignore the $+3$ at the en... | {
"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": 0
} |
Verify that for $k=3/2 ,$ $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$ This is a part of a proof I am studying on:
Let $f_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\}) $
For $k=3/2 ,$ it's shown that $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$
How is it conc... | Note that in general
$$\sin x=\sum\limits_{k=0}^\infty (-1)^k \frac{x^{2k+1}}{(2k+1)!} = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \ldots.$$
thus
$$\sin 2x =2x-\frac{8x^3}{6}+\frac{32x^3}{120}...$$
and
$$\frac{\sin 2x}{2x} =1-\frac{4x^2}{6}+\frac{16x^2}{120}...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Problem in Triangles. $A{A_1}, B{B_1}, C{C_1}$ are the median of triangle ${ABC}$ whose centroid is ${G}$ . If the points ${A}, {C_1} ,{G}, {B_1}$ are concyclic, then show that ${2a^2 =c^2 +b^2} $ if $a,b,c$ are sides of triangle?
| Let $a$, $b$ and $c$ be sides-lengths of $\Delta ABC$ in the standard notation.
Thus, by the Ptolemy theorem we obtain
$$AC_1\cdot B_1G+AB_1\cdot C_1G= C_1B_1\cdot AG$$ or
$$\frac{c}{2}\cdot\frac{1}{6}\sqrt{2a^2+2c^2-b^2}+\frac{b}{2}\cdot\frac{1}{6}\sqrt{2a^2+2b^2-c^2}=\frac{a}{2}\cdot\frac{1}{3}\sqrt{2b^2+2c^2-a^2}$$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591172",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Factoring the polynomial $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$
can you help me factoring the following polynomial?
$6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$
I don't know where to start from.
| Note that $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4$ is homogenous. So set $u=b/a$. Then $6a^4 + a^3b +5a^2b^2 + ab^3 - b^4 = a^4(6 + u +5u^2 + u^3 - u^4)$.
We try integer roots of $6 + u +5u^2 + u^3 - u^4$ by considering the divisors of $6$. We find $6 + u +5u^2 + u^3 - u^4=-(u + 2) (u - 3) (u^2 + 1)$.
This gives
$
6a^4 + a^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593602",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
$n$th Term of the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots$ Prove that $n$th Term of the sequence $1,2,2,3,3,3,4,4,4,4,5,5,5,5,5,\cdots$ is given by
$$T_n =\left[\sqrt{2n+\frac{1}{2}}\right]$$ where $[.]$ is Floor function
My Try:
Its clear that $1$st Term is $1$, $3$rd Term is $2$, $6$th term is $3$ and so on
W... | Note that $$T_n=\frac{-1+\sqrt{1+8n}}{2}$$ is true only if $$n=m(m+1)/2.$$ Now you have to define $T_n$ for other values of $n$. Since terms repeat between two such triangular numbers and you know the value of $T_n$ for $n=m(m+1)/2$ the rest of the solution should be obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2593941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Let $x$ and $y$ are linearly interdependent and that $x+y+z = 0$. Show that $\theta(x,y) + \theta(y,z) + \theta(z,x) = 2\pi$ In the book of Linear Algebra by Werner Greub, at page 202 Q.11.a, it is asked that
Let $x,y,z$ be three vectors of a plane such that $x$ and $y$ are
linearly interdependent and that $x+y+z = ... | Let $X=\theta(y,z)$, $Y=\theta(z,x)$, $Z=\theta(x,y)$.
$$\cos Z=\frac{x\cdot y}{|x|\cdot|y|}$$
$$\cos X=\frac{-x\cdot y-|y|^2 }{|y|\cdot|x+y|}$$
$$\cos Y=\frac{-x\cdot y-|x|^2 }{|x|\cdot|x+y|}$$
We'll need also $\sin X$ and $\sin Y$:
$$\sin X=\sqrt{1-\frac{(x\cdot y)^2+2x\cdot y|y|^2+|y|^4}{|y|^2|x+y|^2}}=\frac{\sqrt{|... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594020",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Using induction, prove that $({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$. Question: Using the principle of mathematical induction, prove that for every integer $n\geq 1, ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$.
My attempt:
Let $P(n): ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not ... | Supose that $3^{2^n}-1$ is not divisible by $2^{n+3}$. Then there are $q\in \mathbb{N}$ and $r\in \{1,2,3,\ldots, 2^{n+3}-1\}$ such that
$$
3^{2^n}-1 = q\cdot 2^{n+3}+r
$$
Then
\begin{align}
3^{2^{n+1}}-1 =& (3^{2^{n}})^2-1 \\
=& (3^{2^{n}}-1)(3^{2^{n}}+1) \\
=& (3^{2^{n}}-1)[(3^{2^{n}}-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2594371",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 3
} |
Find $\sum_{n=1}^{\infty}$ $\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$
$$\sum_{n=1}^{\infty}\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$$
MY Approach$\sum_{n=1}^{\infty}$$\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\righ... | Another approach , applicable to this type of sums in general, is to decompose the fraction as a sum
of fractions with Rising Factorials at the denominator.
$$
\eqalign{
& {{n^{\,2} } \over {\left( {n + 1} \right)\left( {n + 2} \right)\left( {n + 3} \right)\left( {n + 4} \right)}} = {{\left( {n + 1} \right)\left( {n ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2595778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 3
} |
Find $\lim_{x\rightarrow\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right).$ This is what my prof did:
Rewrite as $$x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)=\frac{x}{e}\left(1-e\cdot e^{-x\ln{\left(1+\frac{1}{x}\right)}}\right).$$
Here he uses that
$$\ln{(1+t)}=t-\frac{t^2}{2}+O(t^3),$$
so
$$\... | The whole argument must be understood for $t=1/x\to0$, so that the lowest exponents yield the largest terms ($t^2$ exceeds $t^4$ because $1$ exceeds $t^2$; if you prefer, $1/x^2$ exceeds $1/x^4$ because $x^4$ exceeds $x^2$).
Then
$$e^{t+O(t^3)}=1+t+O(t^3)+O\left(\left(t+O(t^3)\right)^2\right)=1+t+O(t^2)$$ by keeping th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
} |
Computing volume using shell resulted in negative value? The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...
Use cylindrical shells to compute the volume of $y=x^2$ and $y=2-x^2$, revolved about $x=2$
$$x^2=2-x... | In the region $-1 \leq x \leq1$, $2-x^2 > x^2$ (the $2-x^2$ curve is above the $x^2$ curve). So the height is $(2-x^2)-x^2$, not $x^2-(2-x^2)$.
This fix should simply change the sign of your answer -- which is exactly the issue you wanted to address!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Prove that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$= $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$ QuestionProve that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$=
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$
My Approach R.H.S
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$=$\frac{x^{2^{k-1}\left(1-x^{2}\ri... | $$\begin{align}
\frac{1}{1-x^{2^{k-1}}} - \frac{1}{1-x^{2^k}}
&= \frac{(1-x^{2^k}) - (1-x^{2^{k-1}})}{(1-x^{2^{k-1}})(1-x^{2^k})}\\
&= \frac{x^{2^{k-1}}(1-x^{2^{k-1}})}{(1-x^{2^{k-1}})(1-x^{2^k})}\\
&= \frac{x^{2^{k-1}}}{1-x^{2^k}}
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2599202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Derive the conditions $xy<1$ for $\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy}$ and $xy>-1$ for $\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy}$
$$
\tan^{-1}x+\tan^{-1}y=\tan^{-1}\frac{x+y}{1-xy} \text{, }xy<1\\
\tan^{-1}x-\tan^{-1}y=\tan^{-1}\frac{x-y}{1+xy} \text{, }xy>-1
$$
But, How do I reach the conditions $... | Thanks @Rohan for the hint.
$$
\tan^{-1}:\mathbb{R}\to \Big(\frac{-\pi}{2},\frac{\pi}{2}\Big)
$$
Taking,
$$
\alpha=\tan^{-1}x\implies x=\tan\alpha\text{ , where }\tfrac{-\pi}{2}<\alpha<\tfrac{\pi}{2}\\
\beta=\tan^{-1}y\implies{y}=\tan\beta\text{ , where }\tfrac{-\pi}{2}<\beta<\tfrac{\pi}{2}\\
$$
For,
$$
\tan^{-1}x+\ta... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2604703",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
polar coordinate subtitution
$$\int\int_{(x^2+y^2) \le1} \left(\frac{x^4}{a^4}+ \frac{y^4}{b^4} \right)dxdy$$
I try to solve this multivariable integration, by substitute $x=ar \sin \theta$ and $y=br \sin \theta$
to $(\frac{x^4}{a^4}+ \frac{y^4}{b^4} )$ so it will produce $r^4 $ with Jacobin $abr$ but when I substi... | Just use plain old polar coordinates. Let $x = r\cos \theta$ and $y = r\sin \theta$, then everything boils down to
\begin{align}
\int^{2\pi}_0 \int^1_0\ \left( \frac{r^4 \cos^4\theta}{a^4}+\frac{r^4\sin^4\theta}{b^4}\right)\ rdrd\theta = \frac{1}{6} \int^{2\pi}_0 \frac{\cos^4 \theta}{a^4}+\frac{\sin^4\theta}{b^4}\ d\th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2605810",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
existence of solutions of $a^n+b^n+c^n=6^n$ $$2^2+4^2+4^2=6^2$$
$$3^3+4^3+5^3=6^3$$
There is no such thing for exponent 4. It's not hard to prove that the only integer solutions of $a^4+b^4+c^4=6^4$ are trivial ones.
For which $n$ there exist $a,b,c\in\mathbb{N}$ such that $a^n+b^n+c^n=6^n$?
(own question, a very prob... | For $n \geq 7,$
$$ \left( \frac{5}{6} \right)^n < \frac{1}{3} $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2606543",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove $a^2+b^2+c^2\gt \frac {1}{2018}$ given $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$ Let $a, b, c$ be real numbers such that $\left({3a + 28b + 35c}\right)\left({20a + 23b +33c}\right) = 1$. Prove that $a^2+b^2+c^2\gt \frac {1}{2018}$.
It looks like an easy question, but I thought for a while a... | By using C-S we have :
$$|3a+28b+35c| \le \sqrt{3^2+28^2+35^2} \cdot \sqrt{a^2+b^2+c^2} \tag 1$$
and
$$|20a+23b+33c|\le \sqrt{20^2+23^2+33^2} \cdot \sqrt{a^2+b^2+c^2} \tag 2$$
Multiplying $(1)$ and $(2)$
$$|(3a+28b+35c)(20+23b+33c)| \le \sqrt{20^2+23^2+33^2} \cdot \sqrt{a^2+b^2+c^2} \cdot \sqrt{3^2+28^2+35^2} \cdot \sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2607417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Evaluation of Integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$ Calculate integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$ This was my exam question. I've tried many online math solvers and math programs but none were able to solve. If anybody has an answer would be helpful. Thanks
| I have found a solution in terms of a hypergeometric function for positive $x$:
$$
I = \frac{2}{3\sqrt{x}}\left(\sqrt{1+\frac{3}{x^3}}(1+x^2)-4\;_2F_1\left(-\frac{1}{2},\frac{1}{6},\frac{7}{6},-\frac{3}{x^3}\right)\right) +C
$$
and verified the derivative equals the integrand (for positive $x$), and in fact for general... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2608973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Solve $15\cdot x=27\pmod{18}$ EXERCISE:
Solve $15\cdot x\equiv 27\pmod{18}$
SOLUTION:
We know that $\gcd(15,18)=3$ and $3\setminus 27=3\cdot 9$.
So from http://www.math.niu.edu/~richard/Math420/lin_cong.pdf we can conclude that we have 3 solutions!
Then we have that $15\cdot x\equiv 27\pmod{18}\implies 5\cdot x\equ... | The primary task is finding $5^{-1}$ modulo $6$, because once we find it, we multiply both sides by it to make $x$ have coefficient $1$ (therefore determining $x$). In general, here's how to find $z^{-1}$ modulo $a$.
*
*Want to find $y$ such that $zy \equiv 1 \pmod{a}$
*This is equivalent to: find $y$ and $b$ (modu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 1
} |
Tricky convergence/divergence of a sum Let be $A \subset \mathbb{N}$ composed by all the positive integers that have NO $0$ in their decimal expression (so $A=\{1,2,3,\dots 9,11,12, \dots 19,21... \}$).
Show the convergence or divergence of $\sum_{n\in A} \frac{1}{n}$
| Define $S_n = [10^{n - 1}, 10^n) \cap A$ for $n \in \mathbb{N}_+$. Because for every $(a_n \cdots a_1 a_0)_{10} \in S_{n + 1}$, there is $(a_n \cdots a_1)_{10} \in S_n$, and for every $(a_{n - 1} \cdots a_0)_{10} \in S_n$, there is $(a_{n - 1} \cdots a_0 j)_{10} \in S_{n + 1}$ $(1 \leqslant j \leqslant 9)$, then$$
\sum... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2611392",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Functional with 4th grade characteristic equation I get a functional example, which looks like this:
$\mathcal{F}(u) = \int_0^{\frac{\pi}{4}} (x^2 + 16u^2 - u''^2) dx $
with conditions: $u(0) = 1, u'(0) = 0, u(\frac{\pi}{4}) = 0, u(\frac{\pi}{4}) = -2 $
Using Euler-Lagrange equation :
$\frac{\partial F}{\partial u} - \... | Your method isn't very clear. You introduce a delta, then it goes away with no explanation. You seem to be using t where i is standard. Solving the equation $\lambda^4 + 8 = 0$ is pretty simple; just move the 8 to the other side, and then take the fourth root.
$\lambda^4 = 8^{(1/4)}\omega $
where $\omega^4 = -1$
No... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2615492",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\Bbb Z\Big(\frac{1+\sqrt{d}}{2}\Big)$ is closed under multiplication. Here, we have that $d\equiv 1\bmod4$ and $$\Bbb Z\Big(\frac{1+\sqrt{d}}{2}\Big):=\Big\{u+v\Big(\frac{1+\sqrt{d}}{2}\Big): u,v\in\Bbb Z\Big\}.$$
Here is my attempt so far:
Suppose $A:=u+v\Big(\frac{1+\sqrt{d}}{2}\Big)$ and $B:=w+x\Big(\frac... | \begin{eqnarray*}
\left(u+v \left( \frac{1+ \sqrt{d}}{2} \right) \right) \left(w+x \left( \frac{1+ \sqrt{d}}{2} \right) \right) = uw+(ux+vw) \left(\frac{1+ \sqrt{d}}{2}\right)+ vx \left(\frac{1+d +2\sqrt{d}}{4}\right)
\end{eqnarray*}
now use $d=4k+1$
\begin{eqnarray*}
\left(u+v \left( \frac{1+ \sqrt{d}}{2} \right) \rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2616697",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 1
} |
A faster way to evaluate $\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt$? I need to evaluate the integral
$$\int_1^\infty\frac{\sqrt{4+t^2}}{t^3}\,\mathrm dt\tag1$$
After some workarounds I found the change of variable $t=2\sqrt{x^2-1}$, then
$$\int_1^\infty\frac{2\sqrt{1+(t/2)^2}}{t^3}\,\mathrm dt=\frac12\int_{\sq... | Enforcing a substitution of $x \mapsto 1/x$ we have
$$I = \int_0^1 \sqrt{4 x^2 + 1} \, dx.$$
This integral can be readily found using a hyperbolic substitution of $x \mapsto \frac{1}{2} \sinh x$. Doing so yields
\begin{align*}
I &= \frac{1}{2} \int_0^{\sinh^{-1} (2)} \cosh^2 x \, dx\\
&= \frac{1}{4} \int_0^{\sinh^{-1} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2618271",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Finding $\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}$
find the :
$$\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... | Let $x=20^{\circ}$
\begin{eqnarray}\dfrac{1}{1+\tan 70^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}&=&\dfrac{1}{1+\cot 20^{\circ}}+\dfrac{1}{1+\tan 20^{\circ}}\\&=& \dfrac{\sin x}{\sin x+\cos x}+\dfrac{\cos x}{\cos x+\sin x}\\ &=&1\end{eqnarray}
| {
"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": 0
} |
Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find $\lim_{n \to \infty} (a_n -\sqrt{n})$
Let $a_{n+1}=\sqrt{a_n+n}$ with $a_1=1$ then find
$\displaystyle \lim_{n \to \infty} (a_n -\sqrt{n}).$
My Try:
$$\lim_n( \sqrt{a_{n}+n}-\sqrt{n}) \times \dfrac{( \sqrt{a_{n}+n}+\sqrt{n}) }{( \sqrt{a_{n}+n}+\sqrt{n})}=\lim_n \df... | Let $b_n=a_n-\sqrt n$ therefore:$$b_{n+1}=\sqrt{b_n+\sqrt n+n}-\sqrt{n+1}=\dfrac{b_n+\sqrt n-1}{\sqrt{b_n+\sqrt n+n}+\sqrt{n+1}}=\dfrac{\dfrac{b_n-1}{\sqrt n}+1}{\sqrt{\dfrac{n+1}{n}}+\sqrt{\dfrac{b_n}{n}+\dfrac{1}{\sqrt n}+1}}\qquad\qquad\qquad(1)$$if we assume $\lim_{n\to\infty}b_n=l<\infty$ we have:$$l=\dfrac{1}{2}$... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2623306",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 1
} |
Computing $\lim_{x \to 1}\frac{x^\frac{1}{5}-1}{x^\frac{1}{6} -1}$ I cannot figure out how to
get around the zero numerator and denominator in order to compute the limit below:
$$\lim_{x \to 1}\frac{\left(x^\frac{1}{5}\right)-1}{ \left( x^\frac{1}{6}\right) -1}$$
I tried:
$$ \lim_{x \to 1} \frac{ (x^\frac{1}{5... | Take $u=x^{\frac{1}{5}}$ and $v=x^{\frac{1}{6}}$.
$\displaystyle u-1=\frac{u^5-1}{u^4+u^3+u^2+u+1}=\frac{x-1}{u^4+u^3+u^2+u+1}$
$\displaystyle v-1=\frac{v^6-1}{v^5+v^4+v^3+v^2+v+1}=\frac{x-1}{v^5+v^4+v^3+v^2+v+1}$
$\displaystyle \frac{u-1}{v-1}=\frac{v^5+v^4+v^3+v^2+v+1}{u^4+u^3+u^2+u+1}\to\frac{6}{5}$ as $x\to 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2624926",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 9,
"answer_id": 6
} |
How do I find the quotient of the division of $x^{1993}+x^2+1$ by $x^2+x+1$? I have an exercise where I need to find the quotient of the division of polynomial $f=x^{1993}+x^2+1$ by $g=x^2+x+1$. I know to do it the long way but I think there is a trick to do it fast. By the way, I know the remainder is $0$.
| Just wrote $$X^{1993}+X^2+1=X^{1993}-X+X^2+X+1=X\left(X^{1992}-1\right)+X^2+X+1.$$
Since $$X^{1992}-1=(X^3-1)(X^{1989}+...+X^3+1)=(X-1)(X^2+X+1)(X^{1989}+...X^3+1),$$
we see that $X^{1993}+X^2+1$ is divisible by $X^2+X+1$.
The quotient it's $$X(X-1)(X^{1989}+...X^3+1)+1.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625086",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How to factor $2x^3 + 21x^2 +27x$ I am having trouble with factoring $2x^3 + 21x^2 +27x$. The answer is $x(x+9)(2x+3)$ but not sure how that was done. Obviously I factored out the $x$ to get $x(2x^2+21x+27)$ then from there I am lost. I tried the AC method and grouping. Can someone show the steps? Thanks!
| $$2x^3+21x^2+27x\\=x(2x^2+21x+27)\\=\dfrac{x}{2}(4x^2+42x+54)\\=\dfrac{x}{2}(2x+18)(2x+3)\\=x(x+9)(2x+3)$$for factorizing $4x^2+42x+54$ we know that it must in form of $(2x+a)(2x+b)$. So:$$4x^2+42x+54=(2x+a)(2x+b)=4x^2+(a+b)2x+ab$$which implies$$ab=54\\a+b=21$$clearly the only numbers satisfying those equalities are $a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2625763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Find extremal values of $f(x,y)=x^2+4y^2$ on the set $D$. Find abs.min and max values of $f(x,y)=x^2+4y^2$ on the set $D=\{(x,y) \in \mathbb{R}^2; x^2 + (y+1)^2 \leq 4, y \geq -1, y \leq x+1\}$.
For the interior, it is as usually $f_x=0$ and $f_y=0$, and I got $(0,0)$ as a critical point, and $f(0,0) =0$.
For the boun... | Your approach seems to be correct. However, note that you are able to in fact "talk" about extremum values since the function $f(x,y)$ is continuous on a compact set $D$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2626719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Solve $n$th order determinant Solve $n$th order determinant:
$$
\begin{pmatrix}
1 & 2 & 3 & \cdots & n-2 & n-1 &n\\
2 & 3 & 4 & \cdots & n-1& n& 1\\
3 & 4 & 5 & \cdots & n &1 &2\\
\vdots& \vdots& \vdots& \ddots& \vdots& \vdots&\vdots\\
n-2 &n-1&n &\cdots &n-5 &n-4 &n-3\\
n-1 &n&1 &\cdots &n-... | With the help of WA, we get this sequence
$$
1, -3, -18, 160, 1875, -27216, -470596, \dots
$$
oeis/A052182 tells us that the determinant is $(-1)^{n-1}n^{n-2}\dfrac{(n^2 + n)}{2}$, except for the sign.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629389",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Find all solutions to $y^2 \equiv 5x^3 \pmod {7}$
Find all solutions to $y^2 \equiv 5x^3 \pmod {7}$
So basically one can evaluate all $a^2, 5b^3 \pmod {7}$ and look for all $(a,b) \in \mathbb Z_7\times \mathbb Z _7$ such that $a^2 \equiv 5b^3 \pmod 7$
Two questions:
*
*Is that the "best" way doing it?
*If I coun... | If $7\mid x\iff7|y$
else $7\nmid xy$
In that case as $\phi(7)=6, x^3\equiv\pm1\pmod7$
If $x^3\equiv1,y^2\equiv5$
Now as $y\equiv\pm1,\pm2,\pm3\pmod7, y^2\equiv1,4,9\equiv2$
Hence, $y^2\not\equiv5$
If $x^3\equiv-1,y^2\equiv-5\equiv2\equiv3^2\implies y\equiv\pm3$
Now $x^3\equiv1\implies x\equiv3,5\equiv-2,6\equiv-1\pmod... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629505",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$9^x=5$, solve for $5\cdot27^{-x-1}$ Question:
Given $9^x=5$
Compute the value of $5\cdot27^{-x-1}$
My attempt:
$9^x=(3^x)^2=5\Rightarrow3^x=\sqrt{5}$
$5\cdot27^{-x-1}=3^{2x}\cdot(3^{3-x}\div3^{3})=3^{2x}\cdot3^{-x}=3^{x}=\sqrt{5}$
My answer is incorrect and I wonder why.
| If you want to use your result
$$5.27^{-x-1}=\frac{9^x}{27.27^{x}}=\frac{1}{27}\frac{1}{3^x}=\frac{1}{27 \sqrt{5}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2629596",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
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 $x = a\sin \theta, dx = a\cos \theta d\theta$
$$\int \frac{a cos\theta d\theta}{(a^2 + a^2 sin^2 \theta)^2} = \int \frac{a\cos \theta d\theta}{(a^2(1+si... | You can use that
$$
\int\frac{\text{d}x}{a^2+x^2}=\frac{1}{a}\text{arctan}\left(\frac{x}{a}\right)+K
$$
And then integrating by parts gives you
$$
\int\frac{\text{d}x}{a^2+x^2}=\left[\frac{x}{a^2+x^2}\right]+\int \frac{2x^2}{\left(a^2+x^2\right)^2}\text{d}x$$
$$=\left[\frac{x}{a^2+x^2}\right]+2\int \frac{\text{d}x}{a^2... | {
"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": 2
} |
Expressing complex numbers in exponential form I do not know why I am having so many issues with complex numbers. I am basically trying to teach myself, but keep doubting myself and getting very frustrated. I've no way of checking my answers so I'm just lost. So I want to write the following in polar form $re^{i\theta}... | $\frac{2+2i}{-\sqrt{3} +i} =$
$\frac {(2+2i)(-\sqrt{3} -i)}{(-\sqrt{3} +i)(-\sqrt{3} -i)}=$
$\frac {(-2\sqrt{3} +2)- (2\sqrt 3+2)i}{3+1}=$
$\frac {-\sqrt3 + 1}2 -\frac{\sqrt 3 + 1}2i=$
$r(\cos \theta + i\sin \theta)$
So we need to solve for $r\cos \theta = \frac {-\sqrt3 + 1}2$ and $r\sin \theta = -\frac{\sqrt 3 + 1}2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2637575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
A fair six-sided die is rolled repeatedly I need help with this homework problem.
A fair six-sided die is rolled repeatedly. Each time the die is rolled, the number showing is written down. Let $X$ be the number of rolls until some number shows twice. The possible values of $X$ are $2,3,4,5,6,$ or $7$. For each of th... | You are on the right track, but the last roll of the die could be any of the numbers already rolled. Then
$$\text{Pr}(2)=\frac66\times\frac16=\frac{6}{36}=\frac{1}{6}$$
$$\text{Pr}(3)=\frac66\times\frac{(6-1)}{6}\times\frac26=\frac{60}{216}=\frac{5}{18}$$
$$\text{Pr}(4)=\frac66\times\frac{(6-1)}{6}\times\frac{(6-2)}{6}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2638788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$ Prove that if $a$ and $b$ satisfy $2a+4b=1$, then $a^2+b^2\ge \frac {1}{20}$
The only idea I have is that I could apply Cauchy-Schwartz but i don't see how, any hints?
| Using coordinate geometry
The perpendicular distance of $2x+4y=1$ from the origin is given by
$$d=\frac{1}{\sqrt{2^2+4^2}}=\frac{1}{\sqrt{20}}$$
By $\sqrt{x^2+y^2} \ge d$ gives
$$x^2+y^2 \ge \frac{1}{20}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2639048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 2
} |
ODE Substitution Problem Have some practice exam questions and answers to work through, I'm almost there but get stuck.
what we have is:
$$
\begin{cases}
\dfrac{dx}{dt}=\dfrac{x^{3}-xt^{2}}{t^{3}} \\
x(1)=2 \\
\end{cases}
$$
and the (general solution) answer is:
$x(t) = \pm t\sqrt{\frac{2}{1-Dt^{4}}}$
now I figured... | Assume that $\frac{dy}{y^3-2y}=\frac{dt}{t}$ is done alright. The anti derivative of $\frac{1}{t}dt$ is indeed $ln|t|$, however the anti derivative of $\frac{dy}{y^3-2y}$ cannot be $ln|y^3-2y|$ for the very simple reason, the derivative of $ln|y^3-2y|$ is not $\frac{1}{y^3-2y}$, in fact it is $\frac{3y^2-2}{y^3-2y}$ by... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2640851",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $xy+yz+zx$ given systems of three homogenous quadratic equations for $x, y, z$ This is a question from Math Olympiad.
If $\{x,y,z\}\subset\Bbb{R}^+$ and if $$x^2 + xy + y^2 = 3 \\ y^2 + yz + z^2 = 1 \\ x^2 + xz + z^2 = 4$$ find the value of $xy+yz+zx$.
I basically do not know how to approach this question. Pleas... | To my surprise, this problem can be solved using geometry.
Identify the Euclidean plane with complex plane $\mathbb{C}$.
and let $\omega = e^{\frac{2\pi}{3}i}$ be the cubic root of unity.
Consider triangle $\triangle ABC$ with vertices at $A = x$, $B = y\omega$ and $C = z\omega^2$.
The sides of the triangle equal to
$$... | {
"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": 1
} |
Derivative of natural log with chain rule. Is there a better way? I'm a bit stuck on taking these two derivatives:
$$h(x) = \ln(x + \sqrt{x^2-1})$$
\begin{align}
h'(x) &= \frac{1}{x + \sqrt{x^2 - 1}} \cdot \frac{d }{dx} (x + \sqrt{x^2 -1} )\\
&= \frac{1}{x + \sqrt{x^2 - 1}} \biggl(1 + \frac{2x}{2\sqrt{x^2 - 1}}\biggr)
... | $$\frac{1}{x + \sqrt{x^2 - 1}} \cdot (1 + \frac{2x}{2 \cdot \sqrt{x^2 - 1}})$$
$$=\frac{1}{x + \sqrt{x^2 - 1}}\frac{x + \sqrt{x^2 - 1}}{\sqrt{x^2 - 1}}$$
$$=\frac{1}{\sqrt{x^2 - 1}}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2646373",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 3
} |
On the convergence of the series $ \sum_n \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)} $ I've got some problem with this series
$$
\sum_{n=1}^{\infty} \sqrt[3]{(n^{3}+n)} - \sqrt[3]{(n^3-n)}
$$
I know that the series diverges by a comparison test, but I've got a lot of trouble to prove it. I've tried some algebraic tricks l... | Using asymptotic calculus:
$$\sqrt[3]{n^3+n}-\sqrt[3]{n^3-n}=n\left(\sqrt[3]{1+\frac{1}{n^2}}-\sqrt[3]{1-\frac{1}{n^2}}\right)=n\left(\left(1+\frac{1}{n^2}\right)^{1/3}-\left(1-\frac{1}{n^2}\right)^{1/3}\right)=n\left(\left(1+\frac{1}{3n^2}+o\left(\frac{1}{n^2}\right)\right)-\left(1-\frac{1}{3n^2}+o\left(\frac{1}{n^2}\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2648836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Use Lagrange multiply to find the local extremes of $f$ Use Lagrange multiply to find the local extremes of $f$
Let $f(x,y)=y^2-4xy+4x^2$ and $x^2+y^2=1$
My work:
Let $f:\mathbb{R}^2\rightarrow\mathbb{R}$ defined by $f(x,y)=y^2-4xy+4x^2$ and $g:\mathbb{R}^2\rightarrow\mathbb{R}$ defined by $g(x,y)=x^2+y^2-1$
We need de... | If you solve the first two equations in your system of three equations for $\alpha$ you get
$$ \alpha=-4+2\frac{y}{x}=-1+2\frac{x}{y} $$
Substituting $u=\dfrac{y}{x}$ into $$ -4+2\frac{y}{x}=-1+2\frac{x}{y} $$
gives the equation
$$ 2u^2-3u-2=0 $$
with solutions $u=-\dfrac{1}{2}$ or $u=2$ leading to the result that $y=-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2649348",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
If $x \cos\theta+y\sin\theta=a$ and $x\sin\theta-y\cos\theta=b$, then $\tan\theta=\frac{bx+ay}{ax-by}$. (Math Olympiad) I tried to solve it but I can’t get the answer. Please help me in proving this trig identity:
If
$$x \cos\theta+y\sin\theta=a$$
$$x\sin\theta-y\cos\theta=b$$
then $$\tan\theta=\frac{bx+ay}{a... | We first try to find $\sin \theta $ in terms of $a, b, x, y$
Multiply equation first by $y$ and the equation $2^{nd}$ by $x$ and add the resultant equations to get $$\sin \theta= \frac {bx+ay}{x^2+y^2}$$
Now multiply the first equation by $x$ and second equation by $y$ to get following equations
$$x^2\cos \theta+xy\si... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2652527",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Let $a,b,c$ form a primitive Pythagorean triple, with $a$ odd. Prove that $(c-b)$ and $(c+b)$ share no common prime factors. Let $a,b,c$ form a primitive Pythagorean triple (meaning $(a,b,c) \in \mathbb{Z}^{3}$). Let $a$ be odd. Use a proof by contradiction to show that $(c-b)$ and $(c+b)$ share no common prime factors... | Since $a$ is odd, $b$ is even . Because if $a=2a'+1$ is odd and if $b=2b'+1$ is also odd then $c^2=(2a'+1)^2+(2b'+1)^2$ has a remainder of $2$ modulo $4,$ which is impossible for a square.
So $c^2$ is odd, so $c$ is odd . So $c+b$ and $c-b$ are both odd.
Let $d$ be a common divisor of $c+b$ and $c-b$. Note that $d$ mus... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2653670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 4
} |
Compute $\iint_D(-2x-y)\cos((x-y)(-2x-y))dxdy$
Compute $$\iint_D(-2x-y)\cos((x-y)(-2x-y)) \ dxdy$$
Where $D$ has vertices at the points
$$P_1=\left(-1 + \frac{\pi}{6} , 2+\frac{\pi}{6} \right)$$
$$P_2=\left(-2 + \frac{\pi}{6} , 4+\frac{\pi}{6} \right)$$
$$P_3=\left(-2 + \frac{\pi}{4} , 4+\frac{\pi}{4} \right)$$
$$P_4... | (Edited in response to OP's comment.)
Exact evaluation of the edited integral
You have correctly set up the relationship between $x,y$ and $u,v$.
\begin{align}
\begin{bmatrix}u \\ v\end{bmatrix} &= \begin{bmatrix}1 & -1 \\ -2 & -1 \end{bmatrix} \begin{bmatrix} x \\ y \end{bmatrix} \\
\begin{bmatrix} x \\ y \end{bmatri... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654244",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Prove $15\mid 4^{2n}-1$ without induction
Prove without induction that $\forall n \in\mathbb Z$, $15\mid4^{2n}-1$.
$4^{2n} = (4^2)^n = 16^n$.
If $n=1,$ then $(4^2)^n-1=15$, and for $n=2$, it is $255$, which is divisible by $15$.
Using the congruence arithmetic, any $n$ in $4^{2n}-1$ can be expressed as a product of p... | note that $$4^{2n}-1=(2^n-1)(2^n+1)(2^{2n}+1)$$
also possible: $$4^{2n}=16^n$$ and $$16\equiv 1 \mod 15$$ so $$16^n\equiv 1^n\equiv 1 \mod 15$$therefore $$16^n-1\equiv 0 \mod 15$$ so no factorization needed
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2654354",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
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... | hint: $\sin x = -\dfrac{\sqrt{3}}{2}$ would lead to the other two angles.
| {
"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": 1
} |
How to win in this game You select 10 numbers from the set $\{2,3,\dots,12\}$.
You then continually roll 2 fair dice and sum them up, until your selection of 10 numbers come up.
For example if your selection was 7,7,7,7,8,8,8,6,6,6 (4 7's, 3 8's and 3 6's), and you roll the dice repeatedly and get 7,7,6,5,8,7,7,9,8,... | Using a brute-force exhaustive search, and a recursion for each case, I get the following answer:
There are two equivalent optimal strategies, namely
$$4,5,6,6,7,7,7,8,8,9$$
$$5,6,6,7,7,7,8,8,9,10$$
yielding, for the expected number of rounds,
$$
e=\frac{a}{b} \approx 28.26676327
$$
where
$$
\begin{align*}
a&=71526610... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2657427",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Number of solution set such that $a \geq b \geq c \geq d \geq 0$; $a, b, c, d \in \mathbb{Z}$ and $a + b + c + d = 10$ Given $(a, b, c, d)$ is a set of integers and $a \geq b \geq c \geq d \geq 0$. Find the number of solution sets for $a + b + c + d = 10$.
(This is problem from a competition, the answer key says its $2... | $a \geq b \geq c \geq d \geq 0$ give easily as follows
$$\begin{cases}a=10\text{ gives } 1\\a=9\text{ gives } 1\\a=8\text{ gives }2\\a=7\text{ gives }3\\a=6\text{ gives }4\\a=5\text{ gives }5\\a=4\text{ gives }5\\a=3\text{ gives }2
\end{cases}$$
examples: $8200,8110\text{ are the two solutions for a=8 }\\ 5500,5410,532... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2658875",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral $\int \sqrt{1+x^2}dx$ I was trying to do this integral
$$\int \sqrt{1+x^2}dx$$
I saw this question and its' use of hyperbolic functions. I did it with binomial differential method since the given integral is in a form of $\int x^m(a+bx^n)^p\,dx$ and I spent a lot of time on it so I would like to see if it can ... | Solutions are same. Observe that $\sqrt{1+x^2}-x=\frac{1+x^2-x^2}{\sqrt{1+x^2}+x}=\frac{1}{\sqrt{1+x^2}+x}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2660140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 1
} |
$\sum_{k=1}^n\frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2n}\right)}$ How to calculate this sum
$$S_n = \sum_{k=1}^n\frac{1}{\sin^2\left(\frac{(2k+1)\pi}{2n}\right)} ?$$
Note :
1)
$$S_n = n+\sum_{k=1}^n{\rm cotan}^2\left(\frac{(2k+1)\pi}{2n}\right)$$
2) $S_n$ seems to be equal to $\frac{n^2}{4}$.
3) I kwow how to calcula... | Hint 1: If you know the answer for $\sum\limits_{k=1}^{2n-1}\csc^2\left(\frac{k}{2n}\pi\right)$, then use
$$
\sum_{k=1}^n\csc^2\left(\frac{2k-1}{2n}\pi\right)+\sum_{k=1}^{n-1}\csc^2\left(\frac{k}{n}\pi\right)=\sum_{k=1}^{2n-1}\csc^2\left(\frac{k}{2n}\pi\right)
$$
Hint 2: $\frac{2n/z}{z^{2n}-1}$ has residue $1$ at $z=e^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2660233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
irrationality of $e$ and series I'm trying to understand a passage in the proof that e is an irrationale number.
We assume that $e$ is a rational number, $e= \frac{p}{q}$ in which $p$ and $q$ are two coprime numbers.
$$e= \sum_{k=0}^\infty \frac{1}{k!} \implies e-\sum_{k=0}^\infty \frac{1}{k!}=\frac{p}{q}-\sum_{k=0}^\... | \begin{align}\sum_{k=q+1}^\infty\frac1{k!}&=\frac1{q!}\left(\frac1{q+1}+\frac1{(q+1)(q+2)}+\cdots\right)\\&<\frac1{q!}\left(\frac1{q+1}+\frac1{(q+1)^2}+\cdots\right)\\&=\frac1{q!}\times\frac{\frac1{q+1}}{1-\frac1{q+1}}\\&=\frac1{q!}\times\frac1q.\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663131",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
The 4 digit base $6$ number $abcd$ with $a>0$ and $d$ odd is a perfect square.
The 4 digit base $6$ number $abcd$ with $a>0$ and $d$ odd is a perfect square. List all possible values of $c$. (The letters are the digits of the base 6 number.)
I've rewritten $abcd_6$ into $216a + 36b + 6c + d$, and I know that $d$ ca... | Work modulo $36$, test only the odd numbers $15$ through $35$:
$^\text{(we may omit $1$ through $13$, since these would not satisfy $a>0$)}$
$$\,$$
$$(15)^2 \equiv 9 \pmod{36} $$
$$(17)^2 \equiv 1 \pmod{36} $$
$$(19)^2 \equiv 1 \pmod{36}$$
$$(21)^2 \equiv 9 \pmod{36} $$
$$(23)^2 \equiv 25 \pmod{36}$$
$$(25)^2 \equiv 13... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663336",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Calculate the determinant of a special matrix I am having some difficulties to calculate a determinant. The matrix is an $n\times n$ one, but I will show the particular case $5\times 5$ in order to simplify the notation.
$\begin{pmatrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & x & x & x\\
1 & x & 0 & x & x\\
1 & x & x & 0 & x\\... | An idea: subtract line two from lines 3,4,5 and cevelop by column 1:
$$\begin{vmatrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & x & x & x\\
1 & x & 0 & x & x\\
1 & x & x & 0 & x\\
1 & x & x & x & 0\end{vmatrix}=\begin{vmatrix}
0 & 1 & 1 & 1 & 1\\
1 & 0 & x & x & x\\
0 & x &-x & 0 & 0\\
0 & x & 0 &-x & 0\\
0 & x & 0 & 0 & -x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2663451",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Show that if $p(x)=a+bx+cx^2$ is a 2nd degree polynomial such that $p(1)=p(2)=p(3)=0$ then $p(x)=0$, using determinants.
Show that if $p(x)=a+bx+cx^2$ is a 2rd degree polynomial such that $p(1)=p(2)=p(3)=0$ then $p(x)=0$ (i.e. $a=b=c=0$), using the determinant of the matrix: $\left(\begin{array}{l}1&x&x^2\\1&y&y^2\\1&... | Note that\begin{align}p(1)=p(2)=p(3)=0&\iff\left\{\begin{array}{l}a+b\times1+c\times1^2=0\\a+b\times2+c\times2^2=0\\a+b\times3+c\times3^2=0\end{array}\right.\\&\iff\begin{pmatrix}1&1&1\\1&2&2^2\\1&3&3^2\end{pmatrix}.\begin{pmatrix}a\\b\\c\end{pmatrix}=0.\end{align}So, if $(a,b,c)\neq(0,0,0)$, the determinant of that ma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664406",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show the polynomial $p(x,y) =y^3+xy^2+(x^2+x)y+(4x^3+2x)$ is irreducible in $\mathbb{Z}[x, y]$ I've thought about $p(x,y)$ being in $\mathbb{Z}[x][y]$ and tried to use Eisenstein's criterion, but I haven't had any luck. The best I've got is, we can suppose $p(x,y)$ is reducible and write it as
$$p(x,y)=(y+a)(y^2+by+c) ... | In $\mathbb{Z}[x][y]$, a generalization of Eisenstein's criterion will apply, since $\mathbb{Z}[x]$ is a UFD. Now, apply this generalization using the irreducible element $x \in \mathbb{Z}[x]$.
As an alternative: if $p$ were reducible, then its terms of highest total degree $4x^3 + x^2 y + x y^2 + y^3$ would have to b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2664974",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 3
} |
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$ How do I solve this equation I found in my textbook:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
This is what I tried:
$\sqrt{x^2-16} - (x-4) = \sqrt{x^2-5x+4}$
$\mapsto \sqrt{(x+4)(x-4)} - (x-4) = \sqrt{(x-1)(x-4)}$
Dividing both sides by $\sqrt{x-4}$
$\mapsto \sqrt{x+4} - ... | At the stage where you divided by $\sqrt{x-4}$, you forgot that $\sqrt{x-4}$ could be $0$, which gives the solution $x=4$. You can only divide by $\sqrt{x-4}$ if $x-4\neq 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2665836",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 0
} |
Solve: $\sqrt{x-\frac1x}-\sqrt{1-\frac1x}=1-\frac1x$ for $ x\neq0$ $$\sqrt{x-\frac{1}{x}}-\sqrt{1-\frac{1}{x}}{=1-\frac{1}{x}}, x\neq0$$
So my idea is that I moved the second sqrt to the RHS and left the first on the LHS. I then squared everything and ended up with $\frac{x^2-1}{x}=2-\frac{3}{x}+\frac{1}{x^2}$. I don't... | The domain gives $-1\leq x<0$ or $x\geq1$.
*
*$-1\leq x<0$.
We obtain:
$$\sqrt{\frac{(1-x)(x+1)}{-x}}=\frac{\left(\sqrt{1-x}\right)^2}{x}+\sqrt{\frac{1-x}{-x}}$$ or
$$\sqrt{\frac{x+1}{-x}}=\frac{\sqrt{1-x}}{x}+\frac{1}{\sqrt{-x}}$$ or
$$-\sqrt{-x(x+1)}=\sqrt{1-x}-\sqrt{-x}$$ or
$$\sqrt{-x}=\sqrt{1-x}+\sqrt{-x(1+x)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2666187",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Solving this ODE near x=0 I could not get the correct solution on this ODE:
$$(1+x^2)y''-xy'-8y=0$$ around $x=0$.
I tried to solve this using MacLaurin expansion on $y$ and forming recurrence relations between the coefficients.
Can anyone teach me how to solve this by steps? Thank you.
| Let $y(x) = \sum^\infty_{n=0} a_n x^n$. We see $$\sum^\infty_{n=2} n(n-1)a_nx^{n-2} + \sum^\infty_{n=2} n(n-1)a_nx^n - \sum^\infty_{n=1} n a_nx^n - \sum^\infty_{n=0} 8a_n x^n = 0.$$ By re-indexing and running off a few terms, we see $$\sum^\infty_{n=2} n(n-1)a_nx^{n-2} = \sum^\infty_{n=0} (n+2)(n+1)a_{n+2}x^n = 2 a_2 +... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2669960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Reduction to triple integral I have an integral
$$I = \int_{V}(x^2 + y^2)\, dV, $$
where $V = \{2ax \geqslant z^2 \} \cap \{ x^2 + y^2 \leqslant ax \}$.
Am I right, that
$$I = \int_{0}^{a} \int_{-\sqrt{2ax}}^{\sqrt{2ax}} \int_{-\sqrt{a^2/4 - (x-a/2)^2}}^{\sqrt{a^2/4 - (x-a/2)^2}} (x^2 + y^2)\, dy\,dz\,dx?$$
It seems... | Your region $V$ is the intersection of the parabolic plane and the cylinder.
$\hspace{13em}$
So it is not bad to begin with cylindrical coordinates:
$$ (x, y, z) = (r \cos\theta, r\sin\theta, z) $$
Then $V$ is written as
$$ V = \{ 2ar \cos\theta \geq z^2 \} \cap \{ r \leq a \cos\theta \} $$
The above description provid... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2670740",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to solve $\int\frac{2+3x}{3x^2+4x-6}\,dx$? I am trying to solve
$$\int\frac{2+3x}{3x^2+4x-6}\,dx$$ using partial fraction decomposition but it fails, how to solve it?
| $\int\frac{2+3x}{3x^2+4x-6}=\frac{1}{2}\int\frac{6x+4}{3x^2+4x-6}=\frac{\ln(|3x^2+4x-6|)}{2}+K$
Since $6x+4$ is the derivative of $3x^2+4x-6$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2671718",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the last 3 digits of $2015^{18}+2015^{19}+2015^{20}+...+2015^{2018}$. Obvious I would find the sum of the numbers mod 100, but how would I handle this when the powers are so large? Calculating one already seemds ridiculous.
| As $2015\equiv15\pmod{1000},$ we need $\displaystyle\sum_{r=18}^{2018}15^r\pmod{1000}$
As $(15^n,1000)=5^3$ for $n\ge3,$
let us find $\displaystyle S=\sum_{r=18-3}^{2018-3}15^r\pmod{1000/5^3}$
As $\displaystyle15\equiv-1\pmod8,S\equiv\sum_{r=18-3}^{2018-3}(-1)^r\equiv-1\pmod8$
$\displaystyle15^3S\equiv15^3\cdot-1\pmo... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2672805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 4
} |
Prove that $\begin{vmatrix} xa&yb&zc\\ yc&za&xb\\ zb&xc&ya\\ \end{vmatrix}=xyz\begin{vmatrix} a&b&c\\ c&a&b\\ b&c&a\\ \end{vmatrix}$ if $x+y+z=0$
If $x+y+z=0$, then prove that
$$
\begin{vmatrix}
xa&yb&zc\\
yc&za&xb\\
zb&xc&ya\\
\end{vmatrix}=xyz\begin{vmatrix}
a&b&c\\
c&a&b\\
b&c&a\\
\end{vmatrix}
$$
I can do it by... | I don't think it is possible without expanding the determinants. The determinant on the right is $a^3+b^3+c^3-3abc = (a+b+c)(a^2+b^2+c^2-ab-ac-bc)$. The determinant on the left is $(a^3+b^3+c^3)xyz-abc(x^3+y^3+z^3)$. Subtracting the left side from the right side is equal to $abc(x^3+y^3+z^3-3xyz)=abc(x+y+z)(x^2+y^2+z^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676190",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 2,
"answer_id": 1
} |
Show that $\sum\limits_{r=1}^n\sum\limits_{k=1}^{n-r+1}(2r-1)k=\frac 13\sum\limits_{r=1}^nr\sum\limits_{s=1}^{n+1}s$
Question:
Show that
$$\sum_{r=1}^n\sum_{k=1}^{n-r+1}(2r-1)k=\frac 13 \sum_{r=1}^nr\sum_{s=1}^{n+1}s$$
purely by manipulating summation limits and summands, i.e. convert the original summation (on... | Not a solution.
Consider $n=4$
$LHS = 1(1+2+3+4) + 3(1+2+3) + 5(1+2) + 7 = 1(1+3+5+7) + 2(1+3+5) + 3(1+3)+ 4 = \sum\limits_{k=1}^n (n+1 - k)k^2 = \sum\limits_{k=1}^n (n+1)k^2 - k^3 $
Separately, $$\frac{1}{3}\sum\limits_{k=1}^n (n+1)k + k^3 = \frac{1}{3}(n+1)\sum\limits_{k=1}^n k + \frac{1}{3}(\sum\limits_{k=1}^n k)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2676416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Verifying Newton's Binomial Series for $(1+x)^{−2}$
Verify the consistency of Newton’s Binomial series for the function
$(1+x)^{−2}$ in two ways:
$(a)$ by multiplying the usual geometric series for $(1 +x)^{−1}$with
itself
$(b)$ by differentiation of the series for $(1 +x)^{−1}$
$(a)$ I have that Newton's Binom... | The binomial expansion is
$$
\left( {1 + x} \right)^{\, - 2} = \sum\limits_{0\, \le \,k} {\left( \matrix{
- 2 \cr
k \cr} \right)x^{\,k} } = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \left( \matrix{
k + 1 \cr
k \cr} \right)x^{\,k} } = \sum\limits_{0\, \le \,k} {\left( { - 1} \right)^{\,k} \l... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2680014",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Prove for any odd integer $n$, $n^2-1\equiv 1\pmod{8}$ Let, $\exists k \in \mathbb{Z}, n=2k+1\implies n^2-1= 4k(k+1) \pmod{8}$
Now, if $k$ is odd, $k+1$ is even, and vice-versa, so $4k(k+1) \pmod{8}=\forall k\, \exists k' \in \mathbb{Z},\,8k' \pmod{8}$
Unable to proceed further.
| The statement should have been$$ n^2-1\equiv 0\pmod{8}$$
Let $n=2k+1$, then $n^2 -1 =(2k+1)^2-1 = 4k^2 + 4k = 4k(k+1)$
Note that $ k(k+1)$ is always even.
Thus $n^2-1 = 4k(k+1)$ is a multiple of 8.
That implies$$ n^2-1\equiv 0\pmod{8}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2681891",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
To find a remainder, when divided by $101$
There is a large number given in an exam. It is said to find a remainder when it is divided by $101$. The number is
$$1+11^{{(1+11+11^2+11^3+...+11^{100}~~)}^{(11^{101}~~+11^{102}~~+\cdots+11^{1000}~~)}}$$
We know $11$ and $101$ are itself prime then $(101,11)=1$ and $\phi... | You can use the geometric series formula $\displaystyle \frac{a(1-r^n)}{1-r}$, where $a$ is the first term, and $r$ is the common ratio.
For $1+11+\cdots+11^{100}$ this is $1 \cdot {{1-11^{100}} \over {1-11}}$. By Fermat's little theorem, $11^{101-1} \equiv 1 \pmod {101}$, so the whole expression is congruent to $1 \pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2682105",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Compute $I=\int(x+3)\sqrt\frac{x+2}{x-2}\mathrm dx$
Compute $$I=\int(x+3)\sqrt\frac{x+2}{x-2}\mathrm dx$$
The way I approach this problem was to:
*
*Set $u=\sqrt{x-2}$ and arrive at $$I=2\int\frac{u^2(u^2+1)}{\sqrt{u^2-4}}\mathrm du$$
*Set $u=2\sec t\implies\mathrm du=2\sec t\tan t\mathrm dt$ to get $I=8\int\sec^... | Note: This is a long, long stretch.
Final Answer:$$\dfrac{16\ln\left(\left|\sqrt{\frac{x^2}4-1}\right|\right)+(x+10)\sqrt{x^2-4}}2+C$$
After applying long division on $\dfrac{(x+3)\sqrt{x^2-4}}{x-2}$ and rewriting the integral, you get:
$$5\int\frac{\sqrt{x^2-4}}{x-2}dx+\int\sqrt{x^2-4} dx$$
Let $x=2\sec(u)\rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2684606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Mathematical Induction prove that $n^3+5n$ is divisible by $6$ Sorry, I know this will be a duplicate on the site but the other solution I found confusing and the method look completely different to what I was taught.
Prove that $n^3 + 5n$ is divisible by $6$ by using induction
The question is
Prove by mathematical Ind... | I don't follow the induction step. First the base case when $n=1$ is clear since $6\mid 6$. Now suppose that the claim holds when $n=k$ for some $k\geq 1$. This means that $k^3+5k=6m$ for some $m\in \mathbb{N}$. We want to show that $(k+1)^3+5(k+1)$ is divisible by $6$. Expand
$$
(k+1)^3+5(k+1)=k^3+5k+6+3(k^2+k)
$$
and... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2687992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Spectral Decomposition of $A=PDP^{-1}$ We just crammed spectral decomposition into our last lecture of the quarter, and I'm quite confused by it.
The following question is on my homework:
Use the matrices P and D to construct a spectral decomposition of $A=PDP^{-1}$.
$A=\begin{bmatrix}-1 & -4 & -4\\-4 & -3 & 0\\-4 & 0 ... | For example, compute
$$
u_1u_1^T = \pmatrix{- \frac 23 \\ \frac 13 \\ \frac 23 } \pmatrix{- \frac 23 & \frac 13 & \frac 23 } =\\
\pmatrix{\frac 49 & - \frac 29 & - \frac 49\\
- \frac 29 & \frac 19 & \frac 29\\
- \frac 49 & \frac 29 & \frac 49}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2691304",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Limit at odd integer for $x$
$$\lim_{x\to a}\frac{1}{(a^2-x^2)^2}\cdot\left(\frac{a^2+x^2}{ax}-2\sin\frac{a\pi}{2}\sin\frac{\pi x}2\right)=?$$ if $a$ is an odd integer.
The way I set out is first assuming $a=1$ and seeing if I can spot some pattern. Now, if I rewrite this limit out as:
$$\lim_{x\to 1}\frac{1}{(1-x^2)... | Note that for $a=1$ with $x=1+y$ and $y\to 0$
$$\frac{1}{ (1-x^2)^2 }\cdot\left(\frac{1+x^2}{x}-2\sin\frac{\pi x}2\right)
=\frac{1}{ (y^2+2y)^2 }\cdot\left(\frac{y^2+2y+2}{y+1}-2\cos\frac{\pi y}2\right)=...$$
and by Taylor's series
$$\cos \frac{\pi y}2=1-\frac{\pi^2 y^2}8+o(y^2)$$
then
$$...=\frac{1}{ (y^2+2y)^2 }\cdot... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2693470",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Reduce $\frac{A(r)}{B^2(r)}(\frac{dr}{dt})^2+\frac{J^2}{r^2}-\frac{1}{B(r)} = -E$ for a slowly moving particle in a weak field I'm working in Weinberg's Gravitation and Cosmology and I'm having trouble seeing how he reduced a certain equation. He starts a derivation on p.185 going from a general metric in the standard ... | If $B \approx 1 + 2\phi$ then $\frac{1}{B} \approx 1 - 2\phi$, hence the term before the equal sign becomes $(1 - 2\phi)$, hence
$$\frac{A}{B^2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{r^2} - (1 - 2\phi) = -E$$
$$\frac{A}{B^2}\left(\frac{dr}{dt}\right)^2 + \frac{J^2}{r^2} + 2\phi = -E + 1$$
$$\frac{A}{2B^2}\left(\frac... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2696233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Summation Recurrence Relation How to solve this Summation Recurrence Relation: $$x_n=\sum_{i=1}^n a_ix_{n-i}\,,\,\,\,n\ge1$$where, $x_0=1$ and $a_n$ is some arbitrary sequence.
The right hand side of the recurrence looks partially like a discrete convolution and also like Cauchy's Product. I tried using Generating Func... | As I understand the problem we have $A\cdot X = X$, displayed in matrix-arrangement
$$ \begin{array}{}
& * & \left [\begin{array}{} 1 \\ x_1 \\ x_2 \\ x_3
\end{array} \right] \\
\left [ \begin{array}{} 1 \\ a_1 &0 \\ a_2 & a_1 &0 \\ a_3 & a_2 & a_1 &0
\end{array} \right ]
& = &\left [\begin{array}{} 1 \\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704352",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
How does $n^4 + 6n^3 + 11n^2 + 6n + 1 = (n^2 + 3n + 1)^2?$ C++ student here, not quite familiar with these type of expressions. Can someone explain how does this work? I'm familiar with $(a+b)^2$ etc. mathematics but this seems to be like $(a+b+c)^2$ and having searched online, the opened form for this formula doesn't ... | $$\begin{align}(\color{red}{n^2}+\color{blue}{3n+1})^2&=(\color{red}{n^2})^2+2\color{red}{n^2}(\color{blue}{3n+1})+(\color{blue}{3n+1})^2\\&=n^4+2(3n^3+n^2)+(3n)^2+2(3n)(1)+1^2\\&=n^4+6n^3+2n^2+9n^2+6n+1\\&=n^4+6n^3+11n^2+6n+1\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2704997",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$ I try to find sum of $\sum_{i=n}^{2n-1} 3+4i$
$\sum_{i=n}^{2n-1} 3+ \sum_{i=n}^{2n-1} 4i$ =$3 (2n-1-n+1) + 4 \sum_{i=n}^{2n-1} i $ -> am I right?
if I change the index of $4 \sum_{i=n}^{2n-1} i $ to $4 \sum_{i=0}^{2n-1-n} i+n $ am I right?
$4 \sum_{i=0}^{2n-1-n} i+n $
$4... | Alternatively:
$$\begin{align}\sum_{i=n}^{2n-1}(3+4i)&=\sum_{i=0}^{2n-1}(3+4i)-\sum_{i=0}^{n-1}(3+4i)=\\
&=3\cdot 2n+4\cdot \frac{2n(2n-1)}{2}-3n-4\cdot \frac{n(n-1)}{2}= \\
&=6n+8n^2-4n-3n-2n^2+2n=\\
&=6n^2+n=\\
&=n(6n+1).\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2705733",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 6,
"answer_id": 1
} |
Multivariable delta-epsilon limit proofs The Question asks: Find the delta epsilon limit if it exist
$$ \lim \limits_{(x, y) \to (0, 0)} \ \frac{x^2 y^3}{2x^2 + y^2} = 1$$
What I've done so far:
$$ \mid {\frac{x^2 y^3}{2x^2 + y^2} - 1 \mid} < \epsilon \quad \quad \quad \quad0 < \sqrt{x^2 + y^2} < \delta $$
Now, since $... | For both $x,y\ne 0$, then
\begin{align*}
\dfrac{|x|^{2}|y|^{3}}{2x^{2}+y^{2}}&\leq\dfrac{|x|^{2}|y|^{3}}{2\sqrt{2}|x||y|}\\
&=\dfrac{1}{2\sqrt{2}}|x||y|^{2}\\
&\leq\dfrac{1}{2\sqrt{2}}\sqrt{|x|^{2}+|y|^{2}}(|x|^{2}+|y|^{2}),
\end{align*}
if either $x=0$ or $y=0$ (but not both), the above inequality is still true. So
\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2706833",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
} |
Why $\frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt0$? Why does the following inequality hold for $x > 0$?
$$
Q(x)<e^{-x^2/2}, \tag{1}
$$
where
$$
Q(x) = \frac{1}{\sqrt{2\pi}} \int_x^\infty e^{-t^2/2} dt. \tag{2}
$$
The following is my attempt: Define
$$
\begin{align}
f(x)
& \triangleq e^{-x^2/2} - Q(x) \\
& = e^{-... | Your idea is quite good: if
$$
f(x)=e^{-x^2/2}-\frac{1}{\sqrt{2\pi}}\int_x^{\infty} e^{-t^2/2}\,dt
$$
then $f(0)=1/2$ and $\lim_{x\to\infty}f(x)=0$. On the other hand,
$$
f'(x)=-xe^{-x^2/2}+\frac{1}{\sqrt{2\pi}}e^{-x^2/2}
$$
so $f$ is increasing in the interval $[0,1/\sqrt{2\pi}]$ and decreasing in the interval $[1/\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707283",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find the rational number of a, b, c, solving $\sqrt[3]{\sqrt[3]{2}-1}=\sqrt[3]{a}+ \sqrt[3]{b}+\sqrt[3]{c}$ I try as following
let
\begin{eqnarray}
x= \sqrt[3]{a} \\
y= \sqrt[3]{b} \\
z= \sqrt[3]{c} \\
x+y+z = \sqrt[3]{\sqrt[3]{2}-1 }\\
\end{eqnarray}
We know that
\begin{equation}
x^3+y^3+z^3 = (x+y+z)^3... | Here is an alternative to denesting $(2^{1/3}-1)^{1/3}$. First, set $x^3=2$ so that$$x^3-1=1$$Factoring the left-hand side and isolating $x-1$ gives$$x-1=\frac 1{1+x+x^2}=\frac 3{3+x+3x^2}=\frac 3{(1+x)^3}$$Multiply both sides by $9$ to complete the cube$$9(x-1)=\left(\frac 3{1+x}\right)^3$$Cube root both sides and set... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2707393",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 4,
"answer_id": 1
} |
$A289B$ is divisible by $90$, then what is $A+B$? It is known that $A289B$ is divisible by $90$, are with $A$ and $B$ as digits. What is $A+B$?
My approach :
If $A289B$ is divisible by $90$, then the unit-value must be a zero, $B=0$.
So now we have $A2890$, which can be written as
$$ A2890 = A\times10000 + 2890 $$
sin... | If an integer is divisible $9$ then the sum of the digits is also divisible by $9$. So
$$A+2+8+9 +B \equiv A+B + 1 \equiv 0 \pmod{9}$$
shows $A+B \equiv_9 8$. Then since $B=0$ like you noted, $A+B=8$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2709305",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
} |
limit of a recursive sequence_1
I have this sequence
$$ \begin{cases}a_1= \alpha +2 & \alpha >0\\
a_{n+1}=\left(a_n- \frac{1}{n}\right)^n+ \frac{1}{n+1} & n \ge1\end{cases}$$
and I want to know
$\lim_\limits{n\rightarrow \infty} a_n$
I can prove by induction that the sequence is a monotone increasing.
$$
a_1= \a... | You can try unfolding the recursion:
\begin{align}
a_{n+1} &= \left(a_n - \frac1n\right)^n + \frac{1}{n+1}\\
&= \left(a_{n-1} - \frac1{n-1}\right)^{(n-1)n} + \frac{1}{n+1}\\
&= \left(a_{n-2} - \frac1{n-2}\right)^{(n-2)(n-1)n} + \frac{1}{n+1}\\
&\cdots\\
&= \left(a_{2} - \frac1{2}\right)^{2\cdot3 \cdots(n-1)n} + \frac{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2710604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
How to solve DE: $x^{3}y''' + 6x^{2}y'' + 4xy' - 4y = 1 - x - 2x^{2}$ Find particular solution for:
$x^{3}y''' + 6x^{2}y'' + 4xy' - 4y = 1 - x - 2x^{2}$
$\forall x > 0$
I believe that this problem can be solved by means of the method of undetermined coefficients. Something like this:
$x^3y_1''' + 6x^2y_1'' + 4y_1' -4y_... | Its been a while. I believe this is called an Euler equation.
Try assuming a homogeneous solution of the form $y=x^n$ and work from there. You should end up having to solve a cubic equation for $n$ that will provide you with three homogeneous solutions. Some linear combination will be in your final answer.
$x^3 y''' ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2712544",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Evaluate $\int_{0}^{\frac{\pi}{2}}\frac{x^2}{ \sin x}dx$ I want to evaluate $$\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^... | I will present an evaluation that makes use of the following two Euler sums:
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3) \qquad \text{and} \qquad \sum_{n = 1}^\infty \frac{(-1)^n H_{2n}}{n^2} = \frac{23}{16} \zeta (3) - \pi \mathbf{G}.$$
Here $\mathbf{G}$ is Catalan's constant. For a proof of ... | {
"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": 2
} |
Simplistic Odd Collatz formulas I've been experimenting with the Collatz for a couple of months.
I just want to give you some formulas and show tables that give the same results for odd inputs. I have also several new functions, but ill only provide a few of them here. I tweaked the formulas. It seems very strange tha... | Sure, there's nothing strange going on here: these formulas are all equivalent. (Can you prove it?)
Hint: if $n$ is odd then $\lfloor \frac{n}{2} \rfloor = \frac{n}{2} - \frac{1}{2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2714923",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Why is $y_{n+1}=\frac{1}{2}(y_n+\sqrt{\frac{1}{2^{2n}}+y_n^2})$ giving the inverse of $\pi$? A simple and interesting recursion:
$$y_{n+1}=\frac{1}{2}(y_n+\sqrt{\frac{1}{2^{2n}}+y_n^2})$$
has these curious solutions
$$y_1=-\infty,y_{\infty}=\frac{1}{2\pi}$$
$$y_1=-\frac{1}{2},y_{\infty}=\frac{2}{3\pi}$$
$$y_1=0,y_{\inf... | The closed form of the limit doesn't come from elliptic integral, but from half-angle formula for cotangent function.
To illustrate this result, we will only condiser the case where $y_1 > 0$.
Consider the auxillary sequence $a_n = 2^n y_n$, it satisfies
$$a_{n+1} = a_n + \sqrt{1 + a_n^2}$$
If $a_n = \cot(\theta)$ for ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2715039",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
} |
either inequality $\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$ is hard or I need to go back to study ASAP Prove that for every $x \in(0,\frac{\pi}{2})$, the following inequality:
$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1$
holds
I don't see room to use derivatives, since it seems a little messy to calculate the ... | Note that since $e^x$ is strictly increasing
$$\frac{2\ln(\cos{x})}{x^2}\lt \frac{x^2}{12}-1\iff\ln \cos x< \frac{x^4}{24}- \frac{x^2}{2}\iff \cos x<e^{\frac{x^4}{24}- \frac{x^2}{2}}$$
and since
*
*$e^x>1+x$
*$\cos x < 1-\frac{x^2}{2}+\frac{x^4}{24}$
(refer to Using Taylor's theorem show that $1 - \frac{x^2}{2} ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2720539",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
How to factorize $2x^2-9x+9$ by completing the square? I know that $x^2-bx+c=(x-k)^2=x^2-2kx+k^2$ if it is a complete square. If not we create one by adding and subtracting $\left(\frac{b}{2}\right)^2$
I tried $$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{... | $$
\begin{align*}
2x^2 - 9x + 9 &= 2\left(x^2 - \frac{9}{2}x + \frac{9}{2}\right)
\\&= 2\left(x^2 - 2\frac{9}{4}x + \frac{9}{2}\right)
\\&= 2\left(x^2 - 2\frac{9}{4}x + \frac{9}{2} + \left(\frac{9}{4}\right)^2 - \left(\frac{9}{4}\right)^2\right)
\\&= 2\left(\left(x-\frac{9}{4}\right)^2 + \frac{9}{2} - \left(\frac{9}{4}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721535",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 1
} |
Show the equivalence of norms on polynomial space Let $\|ax+b\|^2:=2a^2+3b^2+2ab$ be a norm on the space of all polynomials of degree one, where $a,b\in\mathbb{R}$.
Hence, $\|ax+b\|_{L^2} ^2:=\int_0^1(ax+b)^2\,\mathrm{d}x$ be another norm ($L^2$-norm).
I want to show equivalence, in the sense that, for all polynomials ... | Both ways will always be possible for any two norms you care to choose. This is since the space of polynomials of degree less than or equal to one is finite dimensional, hence all norms are equivalent.
In this particular case, you've shown $\lVert ax + b \rVert_{L^2} \leq \lVert ax + b \rVert$. The other direction is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2721870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Rewriting $\lfloor x \rfloor+\lfloor-x\rfloor$ How can I prove that
$$\lfloor x\rfloor + \lfloor -x\rfloor = \begin{cases} \phantom{-}0 &, x \in \mathbb{Z} \\ -1 &, x \notin \mathbb{Z} \end{cases}$$
where $x$ is a real number?
| Consider the following equation: $\lfloor x \rfloor - \lfloor x \rfloor = 0.$ Doh, of course, because $\lfloor x \rfloor = \lfloor x \rfloor$. The question is then: when does $\lfloor x \rfloor = |\lfloor -x \rfloor|$? (The straight bars here represent the absolute value function).
If $x$ is positive but not an integer... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2723209",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
How has this result been obtained?? We have the quadratic equation
$$ 2y^2 -(1+x)y + x = 0 $$
The author derives the following result from the above equation -
$$ y' = \frac{y-1}{4y-1-x} = \frac{(x-3)y-x+1}{x^2-6x+1} $$
Now, I understand the first part. It has been obtained by simply differentiating w.r.t to $x$ and th... | $$(x^2-6x+1)(y-1)-(4y-x-1)((x-3)y-x+1)\\
=-2x^2-4xy^2-4xy+6x+12y^2+2yx^2-6y\\
=(6-2x)(2y^2-(1+x)y+x)\\=0.$$
The identity is valid.
To establish it, solve the initial equation for $y$
$$y=\frac{x+1\pm\sqrt{x^2-6x+1}}4$$
and plug in the fraction, to get
$$y'=\frac{x-3\pm\sqrt{x^2-6x+1}}{\pm4\sqrt{x^2-6x+1}}=\frac{\pm(x-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2728065",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"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.