Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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How to convert $\frac{1}{(1-x)(1-x^3)}$ into a sum of multiple fractions? What I mean is, that one can convert $\frac{1}{(1-x)^2(1-x^2)}$ into the following sum:
$\frac{1}{8}(\frac{1}{1+x}+\frac{1}{1-x} +\frac{2}{(1-x)^2} + \frac{4}{(1-x^3)})$
But I can't seem to do the same here, because when I try to simplify $1-x^3$... | To get the exact answer requested first use a difference of two squares;
$$\frac1{(1-x)^2(1-x^2)}=\frac1{(1+x)(1-x)^3}
$$The standard partial fraction treatment for repeated factors gives
$$\frac1{(1+x)(1-x)^3}=\frac{A}{(1+x)}+\frac{B}{(1-x)}+\frac{C}{(1-x)^2}+\frac{D}{(1-x)^3}
$$
$$1=A(1-x)^3+B(1+x)(1-x)^2+C(1+x)(1-x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087102",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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series convergence lang page 26 exercise 7 Show that the series
$$\sum_n \frac{z^{n-1}}{(1-z^n)(1-z^{n+1})},\quad z\in\mathbb{C},$$
converges to $\frac{1}{(1-z)^2}$ for $|z|<1$ and to $\frac{1}{z(1-z)^2}$ for $|z|>1$...
I try prove it so;
I know that $\sum z^n = \frac{1}{1-z}$ when $|z|<1$ thus $\sum (n+1)z^n = \frac{1... | Note $$\frac{z^{n-1}}{(1-z^n)(1-z^{n+1})} = \frac{1}{z(1-z)}\left[\frac{1}{1-z^n} - \frac{1}{1-z^{n+1}}\right]$$ and $$\lim_{n\to \infty} \frac{1}{1-z^{n+1}} = \begin{cases}1&\text{if $\lvert z\rvert < 1$}\\ 0 & \text{if $\lvert z\rvert > 1$}\end{cases}$$
So the series telescopes to $$\frac{1}{z(1-z)}\left[\frac{1}{1-z... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3087882",
"timestamp": "2023-03-29T00:00:00",
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Calculate $\lim_{n\rightarrow \infty }\left(\frac{n^{2}+1}{n+1}\right)^{\tfrac{n+1}{n^{2}+1}}$
$$\lim_{n\rightarrow \infty }\left(\frac{n^{2}+1}{n+1}\right)^{\tfrac{n+1}{n^{2}+1}}$$
I tried to use $f^{g}=e^{g \ln f}$ and I got $e^{\tfrac{n+1}{n^{2}+1}\ln \left(\tfrac{n^2+1}{n+1} \right)}$. How to continue ?
| Making the problem more general, consider
$$a_n=\left(\frac{n^{2}+a}{n+b}\right)^{\tfrac{n+c}{n^{2}+d}}\implies \log(a_n)={\tfrac{n+c}{n^{2}+d}}\log\left(\frac{n^{2}+a}{n+b}\right)$$ Now, use Taylor expansions for large $n$
$$\log\left(\frac{n^{2}+a}{n+b}\right)=\log
\left({n}\right)-\frac{b}{n}+\frac{a+\frac{b^2}{2... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Solve $z^2 + 4|z| + 4 = 0$ in $\mathbb{C}$
How can I solve this equation:
$z^2 + 4|z| + 4 = 0$?
If I rewrite $z$ as $a+ib$, I get 2 real solutions. But they are wrong.
Any help?
My attempt. By letting $z=a+ib$, we have that
$$(a+ib)^2 + 4 \sqrt{a^2+b^2} + 4 = 0 $$
$$ a^2 + 2iab - b^2 + 4 \sqrt{a^2+b^2} + 4 = 0 $$
$... | Hint: $z^2 + 4|z| + 4 = 0$ iff $z^2 =-4 |z| - 4 \le 0$ and so $z=bi$ for some $b$.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Proving irrationality of $\sqrt[3]{3}+\sqrt[3]{9}$ I need to prove $$\sqrt[3]{3}+\sqrt[3]{9}$$
is irrational, I assumed
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m}{n}$$
I cubed both sides and got
$$\sqrt[3]{3}+\sqrt[3]{9} = \frac{m^3-12n^2}{9n^3}$$
I tried setting $$\frac{m^3-12n^2}{9n^3} = \frac{m}{n}$$ but that led me nowhe... | Let $\sqrt[3]3+\sqrt[3]9=r$.
Thus, since for all reals $a$, $b$ and $c$ we have:
$$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-ac-bc),$$ we obtain:
$$3+9-r^3+9r=0.$$
Now, let $r=\frac{m}{n},$ where $m$ and $n$ are naturals with $gcd=1.$
Thus, $$m^3-9n^2m-12n^3=0,$$ which says that $m$ is divisible by $3$.
Let $m=3m'$, whe... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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"answer_id": 2
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The greatest value of $|z|$ if $\Big|z+\frac{1}{z}\Big|=3$ where $z\in\mathbb{C}$
$\bigg|z+\dfrac{1}{z}\bigg|=3$ then the greatest value of $|z|$ is ___________
My Attempt
$$
\bigg|z+\frac{1}{z}\bigg|=\bigg|\dfrac{z^2+1}{z}\bigg|=\frac{|z^2+1|}{|z|}=3\\
\bigg|z+\frac{1}{z}\bigg|=3\leq|z|+\frac{1}{|z|}\implies |z|^2-3... | (First, assume that $|z|>1$; our range for $|z|$ will be mirrored with its reciprocal, but the maximum we're looking for is definitely greater than $1$.)
Your inequality $\left|z+\frac1z\right|\le |z|+\left|\frac1z\right|$ is true, but it's only half the picture; we get equality when $z$ and $\frac1z$ are positive real... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can I simplify this fraction problem? I have the problem $\frac{x^2}{x^2-4} - \frac{x+1}{x+2}$ which should simplify to $\frac{1}{x-2}$
I have simplified $x^2-4$, which becomes:
$\frac{x^2}{(x-2)(x+2)} - \frac{x+1}{x+2}$
However, if I combine the fractions I get, $x^2-x-1$ for the numerator, which can't be factored... | You need to put them over a common denominator,
$$\frac {x+1}{x+2}=\frac {(x+1)(x-2)}{(x+2)(x-2)}=\frac {x^2-x-2}{x^2-4}$$
Now you can subtract the numerators
$$x^2-(x^2-x-2)=x+2$$
and finally divide out the $x+2$ from numerator and denominator
Your $-1$ should be $-2$. You didn't show your work, so I can't see why ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer? For the equation $y = 4x^2 + 8x + 5$, what are the integer values of x such that y/13 is an integer?
For example, if x = 3, $y = 4(3^2) + 8(3) + 5$ = 65 which is divisible by 13
if x = 8, y = 325 which is divisible by 1... | Let us try with this approach:
$\begin{align} 4x^2+8x+5 &= 4(x+1)^2+1 &\equiv 0 &\quad(\text{mod} 13)\\
&\Rightarrow 4(x+1)^2 &\equiv 12 &\quad(\text{mod} 13)\\
&\Rightarrow (x+1)^2 &\equiv 3 &\quad(\text{mod} 13)\end{align}\\$
Substituting for $x+1 = f$ we are looking to take the square-root of 3 modulo 13. The follow... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Simplify $\frac{x^3-x}{x^2+xy+x+y}$ $$\frac{x^3-x}{x^2+xy+x+y}$$
What I did:
$$\frac{x}{xy+x+y}$$
through simplifying the $x$'s.
But it's not right.
What did I do wrong?
| You can only simplify factors of the numerator with factors of the denominator. Thus first you have to decompose the numerator and the denominator, and then simplify the fraction.
Start with
$$
\frac{x^3-x}{x^2+xy+x+y} = \frac{x(x^2-1)}{x(x+y)+x+y} = \frac{x(x+1)(x-1)}{(x+1)(x+y)},
$$
and now simplify the only common ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096139",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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value of $x$ in Trigonometric equation Find real $x(0<x<180^\circ)$ in
$\tan(x+100^\circ)=\tan(x-50^\circ)+\tan(x)+\tan(x+50^\circ)$
what i try
$\displaystyle \tan(x+100^\circ)-\tan(x)=\tan(x+50^\circ)+\tan(x-50^\circ)$
$\displaystyle \frac{\sin(100^\circ)}{\cos(x+100^\circ)\cos x}=\frac{\sin(2x)}{\cos(x+50^\circ)\co... | Using the addition theorem for $\tan $ you get
$$
\frac{\tan(x) +\tan(100^\circ)}{1 - \tan(x) \tan(100^\circ)}=\frac{\tan(x) - \tan(50^\circ)}{1 + \tan(x) \tan(50^\circ)}+\tan(x)+\frac{\tan(x) + \tan(50^\circ)}{1 -\tan(x) \tan(50^\circ)}
$$
Let $y= \tan(x)$, then this is
$$
\frac{y +\tan(100^\circ)}{1 - y\tan(100^\ci... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3096262",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Subgroup of general linear group over $\mathbb{F}_{3}$ generated by two matrices I have two matrices: $A=\begin{bmatrix}0 & 2 &1\\1 & 2&1\\1 &2&0\end{bmatrix}$
and $B=\begin{bmatrix}2 & 1 &0\\0 & 1&0\\0 &1&2\end{bmatrix}$. I need to describe subgroup generated by these matrices.I know that group generated by A and B, s... | We have $A^2=\begin{bmatrix}0&0&2\\ 0&2&0\\ 2&0&0\end{bmatrix}$, $A^4=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}$, $B^2=\begin{bmatrix}1&0&0\\ 0&1&0\\ 0&0&1\end{bmatrix}$ and $BAB=A^{-1}$ since $A^3=A^{-1}=\begin{bmatrix}2&1&0\\ 2&1&2\\\ 0&1&2\end{bmatrix}$ and $BAB=\begin{bmatrix}2&1&0\\ 2&1&2\\\ 0&1&2\end{bma... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find all values of a for which the system has trivial solutions $$\begin{pmatrix}1+a & 1 & 1 & 1\\\ 1 & 1-a & 1 & 1\\1 & 1 & 1 & a\\1 & 2 & 2 & 1\\\end{pmatrix}$$
As the title says, find all values of a for which the system has trivial solutions.
I have tried row reduction but I haven't gotten anywhere with it. So wha... | Your determinant is correct, so we get some values to check
$$\det A = a \left(2 a^2-1\right) \implies a = 0, \pm \dfrac{1}{\sqrt{2}}$$
If $a = 0$, the RREF is
$$\begin{bmatrix}
1 & 0 & 0 & 0 \\
0 & 1 & 1 & 0 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end{bmatrix}$$
If $a = \dfrac{1}{\sqrt{2}}$, the RREF is
$$\begin{bma... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098042",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find a Jordan's matrix and basis I have a matrix: $A={\begin{bmatrix}3&-1&0&1\\0&3&4&4\\0&0&-5&-8\\0&0&4&7\end{bmatrix}}$.1) I calculate characteristic polynomial. It is: $p_{A}(\lambda)=(\lambda-3)^{3}(\lambda+1)$ 2) So exist Jordan's matrix: $J={\begin{bmatrix}3&?&0&0\\0&3&?&0\\0&0&3&0\\0&0&0&-1\end{bmatrix}}$ 3)I ... | I'll write down an algorithm to find a basis with eigenvectors / generalized eigenvectors, when the wanted vector will be denoted by $\;w\;$ .
First, we must find out what the eigenvalues are. You already did that, they are $\;-1,3\;$, with the second one of geometric multiplicity $2$ . Now
$$(A-(-1)I)w=0\iff \begin{pm... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3098501",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluating $\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}$ What is the value of $$\frac{2013^3-2\cdot 2013^2\cdot 2014+3\cdot 2013\cdot 2014^2-2014^3+1}{2013\cdot 2014}?$$
What I have tried:
$$\implies\frac{2013^2(2013-2\cdot2014)+2014^2(3\cdot 2013-2014)+1}{2013\cdot 2014}$$
$... | Hint
Choose $2013=a$
$$\dfrac{a^3-2a^2(a+1)+3a(a+1)^2-(a+1)^3+1}{a(a+1)}$$
$$=\dfrac{-3a(a+1)+a(a+1)\{3(a+1)-2a\}}{a(a+1)}$$
$$=-3+3+a$$ as $a(a+1)\ne0$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How can i find a inverse of a polynomial in a quotient ring? I am asked to find the inverse of $\widehat {x^3+1} $ in Q/[x]/I where I = $x^2-2$.
I am used to find the inverse by keep doing the division until a find a irreducible elements but the degree of the polynomial i asked to find the inverse is larger than the d... | extended Euc:
$$ \left( x^{3} + 1 \right) $$
$$ \left( x^{2} - 2 \right) $$
$$ \left( x^{3} + 1 \right) = \left( x^{2} - 2 \right) \cdot \color{magenta}{ \left( x \right) } + \left( 2 x + 1 \right) $$
$$ \left( x^{2} - 2 \right) = \left( 2 x + 1 \right) \cdot \color{magenta}{ \le... | {
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"url": "https://math.stackexchange.com/questions/3100289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Binomial Theorem expansion and proving an interesting identity? In the identity
$$\frac {n!}{x(x+1)(x+2)...(x+n)} = \sum ^n_{k=0}\frac {A_k}{x+k} $$
Prove that $$A_k =(-1)^{k}\:\cdot\: ^{n}C_k$$
Also from this deduce that,
$$ \;^{n}C_0\frac 1{1.2} - \:^{n}C_1\frac1{2.3} +\; ^{n}C_2\frac1{3.4} \; ... \;{(-1)^n}\; ^{n... |
We obtain
\begin{align*}
\frac{n!}{x(x+1)\cdots(x+n)}&=\sum_{k=0}^n\frac{A_k}{x+k}\\
n!&=\sum_{k=0}^nA_k\frac{x(x+1)\cdots(x+n)}{x+k}\tag{1}
\end{align*}
Substituting $x=-j,0\leq j\leq n$ in (1) we get
\begin{align*}
n!&=A_j(-j)(-j+1)\cdots (-1)\cdots1\cdot 2\cdots(n-j)\\
&=A_j(-1)^j j!(n-j)!\\
\end{ali... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3100400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof by induction: $a_n > n + \frac{1}{3} \space \forall n \in N,\space n\geq 4$ How do I prove this by induction?
Let $(a_n)_{n\in N}$ be the sequence defined by:
$$a_1=1,\space a_2=\frac{3}{2},\space a_{n+2}=a_{n+1}+\frac{2n+1}{n+2}a_n \space \space (n\in N)$$
Prove that $a_n > n + \frac{1}{3} \space \forall n \in ... | What you did is correct. A shortcut would be to notice that
$$a_{n+2}=a_{n+1}+\frac{2n+1}{n+2}a_n > n+\frac{4}{3}+ \underbrace{\frac{2n+1}{n+2}a_n}_{\geq 1} > n+\frac{4}{3}+1 = n + \frac{7}{3}$$
since we have that $a_n > 4 + \frac{1}{3} > 1$ and $\frac{2n+1}{n+2} > 1$ are true for all $n \geq 4$. Hence, $\frac{2n+1}{n... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3101063",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Proving $\left(1+\frac{1}{m}\right)^m < \left(1+\frac{1}{n}\right)^n$ Let $m,n\in \mathbb{N}$. If $m > n$ show that
$$\left(1+\frac{1}{m}\right)^m > \left(1+\frac{1}{n}\right)^n$$
My works: I tried to show if $g(x)=\left(1+\frac{1}{x}\right)^x$ then $g'(x) > 0$.
\begin{align}
g'(x) &= \frac{d e^ { x \ln(1+\frac{1}{x}) ... | Normally the proof goes something like, let $a_n=(1+\frac{1}{n})^n$ for all $n\in\mathbb Z^+$. Then, $$\frac{a_{n+1}}{a_n}=\frac{(1+\frac{1}{n+1})^{n+1}}{(1+\frac{1}{n})^n}=\frac{(\frac{n+2}{n+1})^{n+1}}{(\frac{n+1}{n})^n}=\frac{(n+2)^{n+1}n^n}{(n+1)^n(n+1)^{n+1}}=\frac{(n^2+2n)^n}{(n+1)^{2n}}\frac{n+2}{n+1}=\frac{((n+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3105024",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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How do we factor out $x^2 - x -2$ in this expression? Suppose we're given that
$$x^4 - 2x^3 +x-2$$
How do we factor out $x^2 - x -2$ in this expression?
$$(x^4 -x^3- 2x^2)-(x^3-x^2-2x)+(x^2-x-2) = x^2(x^2 -x-2)-x(x^2-x-2)+(x^2-x-2) = (x^2-x-2)(x^2-x+1)$$
This satisfies with what we want to get. However, I do not seem... | Another way: complete the square in $\rm\color{#c00}{lead\ 2\ terms}$ yielding
$$ \underbrace{\color{#c00}{(x^2\!-\!x)^2} - (\color{#c00}{x^2}\!-\!x)-2}_{\Large \color{#c00}{x^4\ -\ 2x^3}\ +\ x\ -\ 2\!\!\!\!\!\!\!\!\!\!\!\! } = X^2\!-\!X\!-\!2 = (X\!+\!1)(X\!-\!2) = (x^2\!-\!x\!+\!1)\underbrace{(x^2\!-\!x\!-\!2)}_{\L... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3106729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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If If $x^2+ax+b+1=0$ has roots which are positive integers, then $(a^2+b^2)$ can be which of the given choices?
If $x^2+ax+b+1=0$ has roots which are positive integers, then $(a^2+b^2)$ can be
(A) 50
(B) 37
(C) 19
(D) 61
My approach: I first took roots $\alpha$, $\beta$ and then applied sum and product of roots accor... | We need that $a$ and $b$ will be naturals and there is a natural $n$ for which $$a^2-4(b+1)=n^2.$$
Thus, $$a^2+b^2=n^2+b^2+4b+4=n^2+(b+2)^2.$$
Can you end it now?
| {
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To find the inverse of a special kind of matrix. In a matrix analysis problem, I encountered the following special kind of matrix
$$
\begin{bmatrix}
0 & 1 & a & a & a & a \\
1 & 0 & a& a& a& a \\
a& a &0 & 1& a& a \\
a& a &1 & 0 & a& a \\
a & a & a & a &0 & 1\\
a & a & a & a &1 & 0
\en... | You can write it as a sum:
$$ \begin{bmatrix}
0 & 1 & a & a & a & a \\
1 & 0 & a & a & a & a \\
a & a & 0 & 1 & a & a \\
a & a & 1 & 0 & a & a \\
a & a & a & a & 0 & 1 \\
a & a & a & a & 1 & 0
\end{bmatrix} =
\begin{bmatrix}
-a & 1-a & 0 & 0 & 0 & 0 \\
1-a & -a & 0 & 0 & 0 & 0 \\
0 & 0 & -a & 1-a & 0 & 0 \\
0... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3109913",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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Is $8n^3+12n^2-2n-3$ divisible by 5, with $n$ congruent with 1,2,3 mod 5? I make some test using python and I find that this is the case for $n = 5k +p$ with $k$ an integer and $p =$1,2,or 3.
I was able to prove for $p = 1$ and $p=2$ but not for $p = 3$.
What I'm doing wrong? This is my work:
Let's assume 5 is a diviso... | OP changed the question:
So, another solution:
$$8n^3+12n^2-2n-3 \equiv_5 3n^2 + 2n^2 -2 n - 3 \stackrel{n \equiv_5 k}{\equiv_5} 3k^3 + 2k^2 - 2k -3$$
Now:
*
*$k = 1$: $3+2-2-3 \equiv_5 0$
*$k = 2$: $3\cdot 8 +2\cdot 4 -2\cdot2 -3 \equiv_5 -1 -2-4-3 \equiv_5 0$
*$k = 3$: $3\cdot 27 +2\cdot 9 -2\cdot3 -3 \equiv_5 ... | {
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"timestamp": "2023-03-29T00:00:00",
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A poker hand contains five cards. Find the probability that a poker hand can be.... a) Four of a kind (Contains four cards of equal face value)
So for this one, we want four cards that have the same face value, different suit. And the last card can be any remaining card.
There are 13 ranks, (A, 2, 3, 4, 5, 6, 7, 8, 9, ... | You correctly calculated the probabilities of four of a kind and a full house, not the total number of ways. The number of ways these hands can be obtained is given by the numerators of your probabilities.
What is the probability of three of a kind?
As you know, there are $\binom{52}{5}$ possible five-card hands tha... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113091",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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Can anybody help me figure out what the author did in this task? In the expression
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg(\frac{\ln(1+xe^{-x})}{x}+1\bigg)\bigg)$$
The author claims in the next step that
$$L = \lim_{x\to+\infty}x\bigg(... | $$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}\bigg(\frac{\ln(1+xe^{-x})}{x}+1\bigg)\bigg)$$
This is equal to
$$L = \lim_{x\to+\infty}x\bigg(\sqrt[3]{1+\frac{1}{x}+\frac{1}{x^2}+\frac{1}{x^3}}-\sqrt{1+\frac{1}{x}+\frac{1}{x^2}}-\sqrt{1+\frac{1}{x}+\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3113214",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
How to find limits of integral to find volume? Find the volume generated by the plane region, in the first quadrant, bounded by the graph of the function $ y=\sqrt{9-x^2} $ sbout the y-axis.
I know how to solve it using the formula but how do I get the upper and lower limits of the integration.
| If I understood the question correctly, the volume of the solid they're asking you to find is a hemisphere. If you have to rotate it about the y-axis, use the shell method:
$$
V=2\pi\int_{0}^{3}x\sqrt{9-x^2}\,dx
$$
Your bounds of integration are from $0$ to $3$ because $\sqrt{9-x^2}$ is a circle of radius $3$ centered ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115240",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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When is a matrix equal to its own inverse? When is a matrix equal to its own inverse?
If you have a $2\times2$ matrix and one if the entries is equal to $x$, for what values of $x$ is this matrix equal to its own matrix? And why?
| The easiest way is just to write that
$$
A^2=\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)\left(\begin{matrix}a & b \\ c & d\end{matrix}\right)=\left(\begin{matrix}a^2+bc & (a+d)b\\(a+d)c & d^2+bc\end{matrix}\right)=I.
$$
Since $(a+d)b=(a+d)c=0$, either $a+d=0$ or $b=c=0$. In the latter case, we must have $a^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3115411",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
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Find the convergence radius for this power series The Problem: Find the convergence radius of $\sum_{n=0}^{\infty} \frac{n}{5^{n-1}} z^{\frac{(n)(n+1)}{2}}$
My attempts to find a solution I apply either the ratio test and end up with this expression:
$\lim_{n \to \infty} \frac{1}{5} |z|^{(n+1)} =L$
Since I need $L<1$... | Observe we have
\begin{align}
\sum^\infty_{k=0}a_k z^k
\end{align}
where
\begin{align}
a_k =
\begin{cases}
\frac{n}{5^{n-1}}&\text{ if } k = \frac{n(n+1)}{2},\\
0 & \text{ otherwise }
\end{cases}.
\end{align}
Then by Cauchy-Hadamard theorem (i.e. root test), we see that
\begin{align}
\frac{1}{R}=\limsup_{k\rightarrow ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117173",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Prove: $2^b-1\mid{}2^a-1 \iff b\mid{}a$ I figured we'd have to show $a=qb$ from $2^a-1=r(2^b-1)$ for $q,r\in\mathbb{Z}$ for the first implication and vice versa, how should I proceed from here?
| Hint $:$ $2^{qb}-1 = (2^b -1) (2^{b(q-1)} + 2^{b(q-2)} + \cdots + 1).$
The above hint is enough to prove that $b \mid a \implies 2^b-1 \mid 2^a - 1.$
To prove that $2^b - 1 \mid 2^a - 1 \implies b \mid a$ use division algorithm to write $a = bq + r,$ where $0 \leq r < b.$ Then we have
$$\begin{align}
2^a - 1 & = 2^{b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117279",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find $K=a^2b+b^2c+c^2a$ for roots $a>b>c$ of a cubic. If $a>b>c$ are the roots of the polynomial $P(x)=x^3-2x^2-x+1$ find the value of $K=a^2b+b^2c+c^2a$.
Using Vièta's formulas:
$$a+b+c=2$$
$$ab+bc+ca=-1$$
$$abc=-1$$
Using those I found that
$$a^2+b^2+c^2=6$$
$$a^3+b^3+c^3=11$$
$$a^2b+b^2c+c^2a+ab^2+bc^2+ca^2=1$$
but... | HINT: Consider the discriminant of your polynomial; what can you say about its square root given that $a>b>c$? Hover over the yellow box for a (more) complete solution.
The discriminant of your polynomial is $\Delta=49$, and because $a>b>c$ you also have
$$a^2b+b^2c+c^2a-a^2c-b^2a-c^2b=(a-b)(a-c)(b-c)=\sqrt{\Delta}=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3117967",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Why is $\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$ = a - b? I'm given the complex rational expression
$$\frac{\frac{a}{b} - \frac{b}{a}}{\frac{a+b}{ab}}$$
and asked to simplify. The solution provided is $a - b$; however I get $\frac{a^2 - b^2}{a+b}$.
My working:
Numerator first:
$$\frac{a}{b} - \frac{b}{a}$$
Le... | You did nothing wrong. Factor the numerator $a^{2} - b^{2}$ into the difference the squares:
$$a^{2} - b^{2} = (a + b)(a - b)$$
This will leave with:
$$\frac{(a+b)(a-b)}{a+b}$$
Finally, cancel the $a+b$ expression from the numerator and denominator to obtain the desired result.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3119058",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
} |
A question about INMO 2017, Question 3
Find all triples $(x,a,b)$ where $x$ is a real number, and $a,b$ are integers belonging to $\{1,2,\dots,9\}$ such that $$x^2-a\{x\}+b=0$$ Here $\{x\}$ denotes the fractional part of $x$.
My contention, which I know is wrong, is that the equation has no solutions.
Reason: If $x... | As for where in your argument you went wrong, as stated in the comments, you can't necessarily assume that $x$ is a terminating decimal.
As to determine what values will work, let
$$x = c + r \text{ where } c \in \mathbb{Z} \text{ and } 0 \le r \lt 1 \tag{1}\label{eq1}$$
This gives
$$\left(c + r\right)^2 - ar + b = 0 \... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3120830",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the number of pairs of $(a,b)$ of positive real numbers satisfying $a^4+b^4<1$ and $a^2+b^2>1$?
The number of pairs of $(a,b)$ of positive real numbers satisfying
$a^4+b^4<1$ and $a^2+b^2>1$ is -
$(i)$0
$(ii)$1
$(iii)$2
$(iv)$ more than 2
Solution:We have $a,b>0$,
According to the given situation,$0<a^4+b^... | Let $a^3+b^3=1$, there are uncountably many such pairs with $a\in (0,1)$ as we can set $a^3=\cos^2t, b^3=\sin^2t$.
Now for all these pairs, $a^4+b^4<a^3+b^3=1<a^2+b^2$ is obvious.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3122623",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Solving the linear first order differential equation? I was given the equation
$y' = -xy$, where $y(0) = 1$.
My solution was as follows:
$$\frac{dy}{dx} = -xy $$
$$dy = -xy \ dx $$
$$\int {dy} = \int {-xy dx} $$
$$y = -\frac{x^2}{2}y + c $$
$$y + \frac{x^2}{2}y = c $$
$$y(1 + \frac{x^2}{2}) = c $$
$$y = \frac{c}{1 + \f... | (Too long for a comment.)
No one else has mentioned this, so I will...
Remember: $y$ is a function of $x$, not something that is constant with respect to variation of $x$. (Forgetting this is a fairly common error when first performing implicit differentiation.) It can help to write "$y(x)$" rather than just "$y$"... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124604",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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How many four-digit integers are there that contain exactly one $8$? My book gives:
$$1 \times 9 \times 9 \times 9 + 3 \times 8 \times 9 \times 9 = 2673$$ integers
I don't understand where the $3$ comes from in the second part.
I understand the first part, $1 \times 9 \times 9 \times 9$. I believe it's saying we have ... | You made a good start. Indeed, if you fix the first spot as $8$, the second, third and fourth spot can be filled with any digit in $\{0,1,2,3,4,5,6,7,9\}$, hence you have $9^3$ possible choices.
It remains to count how many number you could have when the first spot is not $8$. Notice that, as Peter suggests above, usu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3124673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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integrate sin(x)cos(x) using trig identity. Book tells me the answer is:
$$ \int \sin(x)\cos(x) dx = \frac{1}{2} \sin^{2}(x) + C $$
however, I get the result:
$$
\sin(A)\cos(B) = \frac{1}{2} \sin(A-B)+\frac{1}{2}\sin(A+B)
$$
$$
\begin{split}
\int \sin(x)\cos(x) dx
&= \int \left(\frac{1}{2}\sin(x-x) + \frac{1}{2}\sin... | Note that
$$\cos(2x)=1-2\sin^2x$$
From here you get
$$-\frac{1}{4}\cos(2x) + C = -\frac{1}{4} + \frac{1}{2}\sin^2 x+C = \frac{1}{2}\sin^2x + C_1$$
where $C_1 = -\frac{1}{4}+C$ is a new constant.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126227",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
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What is the $n^{th}$ term derivative of $f(x) = (x^2-x-1)(\ln(1-x))$? I have the first three terms but am struggling with finding the $n^{th}$ term derivative of the function. Here is my work:
$$\\$$
$$f(x) = (x^2-x-1)(\ln(1-x)) $$
$$f'(x) = (2x-1)(\ln(1-x))-\left(\dfrac{x^2-x-1}{1-x}\right)$$
$$f''(x) = \dfrac{3x^2-5x... | It might be useful here to make a small substitution and to expand the parentheses:
*
*$y = 1-x \Rightarrow y' = -1$
*$\Rightarrow f(x) = g(y(x)) = -(1+x(1-x))\ln (1-x) = -(1+(1-y)y)\ln y$
Now differentiate $\boxed{g(y) = -\ln y - y\ln y + y^2 \ln y}$ with respect to $x$ and have in mind that $\color{blue}{y'(x) = ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3126890",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
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Vector Fields on Lie Groups in a Coordinate Parameterisation My question is about how to write out vector fields on Lie groups in terms of a coordinate basis when a coordinate parametrisation of the group is given.
Consider the group $E(2)$ of Euclidian transformations. A general element can be written in coordinates a... | Careful: you mean $SE(2)$, the group of proper rigid transformations.
If $g = \begin{bmatrix}\cos c & -\sin c & a \\ \sin c & \cos c & b \\ 0 & 0 & 1 \end{bmatrix}$ is a fixed group element and $h = \begin{bmatrix}\cos z & -\sin z & x \\ \sin z & \cos z & y \\ 0 & 0 & 1 \end{bmatrix}$ then
$$L_g(h) = \begin{bmatrix}\co... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3127453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
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What is the value of $2x+3y$ if $x+y=7$ and $x^2-y^2=21$? If $x+y=7$ and $x^2-y^2=21$, then what is $2x+3y$?
I solved it like this:
\begin{align}
y& =7-x \\
x^2-(7-x)^2-21&=0 \\
x^2-49+14x-x^2-21&=0 \\
14x&=70 \\
x&=5
\end{align}
Then I solved for $y$ and I got $2$. I plugged in the values of $x$ and $y$ to $2x+3y$, a... | Hint: A slightly easier way.
$$x^2-y^2=(x+y)(x-y)=7(x-y)=21 \implies x-y=3$$
But your solution is correct.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3129493",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Proof of exponential law using limit definition of exponential function? For fun, I tried to prove the well-known exponential property $e^{a+b} = e^a e^b$ using the limit definition of the exponential function, below.
Definition. The exponential function is defined as follows.
$$e^x := \lim_{\epsilon \rightarrow 0} \l... | Taking for granted that $e^x$ is well-defined in this way for all real $x$, it follows that for all real $x$:
$$
e^x = \lim_{n\to\infty}\left(1+\frac{x}{n}\right)^n.
$$
For all real $a$ and $b$, and every positive integer $n$,
\begin{align*}
& \phantom{={}} \left\lvert\left(1+\frac{a}{n}+\frac{b}{n}+\frac{ab}{n^2}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3131673",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 0
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solve $\frac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$ I came across a question from another forum -
find the $x$ in the following diagram:
I managed to deduce an equation from the following diagram:
which is:
$\dfrac{\sqrt{5}(\cos \theta - \sin \theta)}{3\sqrt{2}}=\tan \theta$
and I know the answe... | Defining $c := \cos\theta$ and $s := \sin\theta$, we can write
$$c^2\sqrt{5} = s\left(3 \sqrt{2} + c\sqrt{5}\right) \tag{1}$$
Squaring, re-writing $s^2 = 1-c^2$, and re-arranging,
$$10 c^4 + 6 c^3 \sqrt{10} + 13 c^2 - 6c \sqrt{10} - 18 = 0 \tag{2}$$
At this point, if we had the presence of mind to identify $10$ as $\sq... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3132010",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 3,
"answer_id": 1
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Proving $\lim\limits_{n{\rightarrow}\infty} \int_{0}^{n} \frac{{\sqrt{x}}\ln x}{1+x^2}dx=\pi$
Show that $$\lim\limits_{n{\rightarrow}\infty} \int_{0}^{n} \frac{x^{\frac{1}{2}}\ln x}{1+x^2}dx=\pi$$
After a number of transformations I ended having $$4\int_{0}^{\frac{1}{n^2}}\frac{\ln(x)}{x^4+1}dx$$ From here on I have
... | Here is an approach that will make use of the derivative of the Beta function $\operatorname{B} (m,n)$ of the form
$$\operatorname{B} (m,n) = \int_0^\infty \frac{x^{m - 1}}{(1 + x)^{m + n}} \, dx, \quad m,n > 0.$$
Let
$$I = \int_0^\infty \frac{\sqrt{x} \, \ln x}{1 + x^2} \, dx.$$
Enforcing a substitution of $x \mapsto... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3133817",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
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How to evaluate a binomial sum with $2n$ in the exponent. The question is to evaluate the sum $$\sum_{k=0}^n\binom nk 3^{2n-k}$$
have tried fitting into the binomial form of $\binom nk \times x^k\times y^{n-k}$
but I can't seem to bring it to the correct form.
| Using @lulu's hint:
\begin{align}
\sum_{k = 0}^{n} \binom{n}{k} 3^{2n - k}
& = \sum_{k = 0}^{n} \binom{n}{k} 3^{n + n - k}
= 3^{n} \sum_{k = 0}^{n} \binom{n}{k} 3^{n - k} \\
& = 3^n \sum_{k = 0}^{n} \binom{n}{k} 1^k 3^{n - k}
= 3^n (1 + 3)^n
= 3^n 4^n
= 12^n.
\end{align}
Alternatively, one could pull out the $3^{2n}$:
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3134238",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Evaluate $\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^2+y^2} dxdy$ [Background]: I'm trying to find the volume of the region bounded by the $xy$-plane, the cone $z^2=x^2+y^2$ and the cylinder $(x-1)^2+y^2=1$.
[Attempt]: I tried to use the polar coordinate:
\begin{align*}
\iint_{\{(x,y)\mid (x-1)^2+y^2\leq1\}}\sqrt{x^... | Hint Writing the domain in polar coordinates gives
$$(r \cos \theta - 1)^2 + (r \sin \theta)^2 \leq 1 .$$
Expanding and using the Pythagorean identity gives
$$r^2 - 2 r \cos \theta + 1 \leq 1,$$
rearranging gives
$$r^2 \leq 2 r \cos \theta ,$$
and for $r \neq 0$, dividing by $r$ gives
$$r \leq 2 \cos \theta .$$
So, in ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3135745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Locus of orthocenter of triangle inscribed in ellipse While messing around with ellipses in Geogebra, I found the following interesting result:
Let $\alpha$ be an ellipse. Let $AB$ be a fixed chord, and let $P$ be a point that moves freely on $\alpha$. Then as $P$ traces the ellipse, the orthocenter of triangle $PAB$ t... | With the help of Mathematica, I was able to confirm your suspicion.
Let the ellipse be parameterized by $(a \cos\theta, b \sin\theta)$, and let $A$ and $B$ be points corresponding to $\theta = 2\alpha$ and $\theta = 2\beta$. After some symbol-crunching, we find that the orthocenter $(x,y)$ satisfies
$$\begin{align}
a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3135970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 1,
"answer_id": 0
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Showing that $\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$ Since I don't have the answer to this one, I want to make sure I've done this correctly.
Show that
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
Since
$$\arctan x = \arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right)$$
we have
$$\ta... | Let's consider the definitions of arcsine and arctangent.
Let $\arctan x = \theta$. Then $\theta$ is the unique angle in the interval $\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$ such that $\tan\theta = x$.
Let
$$\arcsin\left(\frac{x}{\sqrt{1 + x^2}}\right) = \varphi$$
Then $\varphi$ is the unique angle in the interv... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Pattern for Induction?
Find and prove a formula for the sum
$$\frac{1^3}{1^4+4} - \frac{3^3}{3^4+4}+...+\frac{(-1)^n(2n+1)^3}{(2n+1)^4 + 4}$$ where $n$ is an integer.
I tried listing out the partial sums of the sequence to see if there was a pattern, however the only thing I can make out is that the denominators o... | Extended hint:
Writing down 4-5 first partial sums reveals after reducing the fractions the general pattern:
$$
S_n=(-1)^n\frac{n+1}{4(n+1)^2+1}.
$$
It remains to apply induction or telescoping. The following facts will be useful:
$$\begin{array}{ll}
(2n+1)^4+4&=(4n^2+1)(4n^2+8n+5),\\
4n^2+8n+5&=4(n+1)^2+1,\\
(2n+1)^3... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137706",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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Eigenvalue by Gaussian elimination? Let $A=\begin{bmatrix}-1&4&0\\-2&4&-1\\2&-5&0 \end{bmatrix}$
I want to find out the eigenvalue of matrix my Gaussian elimination. My idea was to make the matrix triangular and by that being able to write an expression for the determinant that equals 0. Like this: $(aλ+a_1)(bλ+b_1)(... | Here is a simple way to obtain the characteristic polynomial:
\begin{align}
&\begin{vmatrix}
\lambda+1&-4&0\\ 2&\lambda- 4&\phantom{-}\\-2&5&\lambda
\end{vmatrix} =
\begin{vmatrix}
\lambda+1&-4&0\\ 0&\lambda+1&\lambda+1\\-2&5&\lambda
\end{vmatrix} =
(\lambda+1)\begin{vmatrix}
\lambda+1&-4&0\\ 0 & 1 & 1\\-2&5&\lambda
\e... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3137831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Help finding the derivative of $f(x) = \cos{\left(\sqrt{e^{x^5} \sin{x}}\right)}$ I am trying to find the derivative of $f(x) = \cos(\sqrt{(e^{x^5} \sin(x)})$.
I keep getting the wrong answer, and I'm not sure what I'm doing wrong.
$$\frac{d}{dx} e^{x^5} = e^{x^5} \cdot 5x^4$$
$$\frac{d}{dx} \sin(x) = \cos(x)$$
$$\fra... | I differentiated your function and did not look at your result. As you can see, they're identical:
$$\begin{align}
\frac{d}{dx}\left[\cos\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)\right]
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)}\left(\sqrt{e^{x^5}\cdot\sin{x}}\right)'\\
&=-\sin{\left(\sqrt{e^{x^5}\cdot\sin{x}}\righ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3138417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
} |
A simpler method to prove $\log_{b}{a} \times \log_{c}{b} \times \log_{a}{c} = 1?$ This is the way I think about it:
$
1 = \log_aa = \log_bb = \log_cc \\~\\
\textbf{Using the ‘change of base rule':} \\
\log_{a}{b} = \frac{\log_{b}{b}}{\log_{b}{a}}, \hspace{0.3cm} \log_{b}{c} = \frac{\log_{c}{c}}{\log_{c}{b}}, \hspace{0... | Well it is much simpler. By definition
$$\log_b(a) = \frac{\ln(a)}{\ln(b)}, \quad \log_c(b) = \frac{\ln(b)}{\ln(c)}, \quad log_a(c) = \frac{\ln(c)}{\ln(a)}$$
where $\ln$ is the logarithm in base $e$.
Your identity directly follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3143783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate integral $\int \sin^4(t)\cos^3(t)dt$ $$\int \sin^4(t)\cos^3(t)dt = \int \sin^4(t)(1-\sin^2(t))\cos(t) dt $$
$$u = \sin(t) \\ du = \cos(t)dt$$
$$ \int \sin^4(t)\cos^3(t)dt = \int u^4(1-u^2) du \\
= u^4 - u^6 = \frac{1}{5}u^5 - \frac{1}{7}u^7 + C \\
= \frac{1}{5}\sin^5(t) - \frac{1}{7}\sin^7(t) + C $$
This s... | Use the formula $$\int \cos^m(t)\sin^n(t)dt=-\frac{\cos^{m+1}(t)\sin^{n-1}(t)}{m+n}+\frac{n-1}{m+n}\int\cos^m(t)\sin^{n-2}(t)dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3146621",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Closed form representation of alternating series I consider the following series
\begin{align}
\sum_{n=0}^\infty{\frac{(-1)^{n+2}}{n+2}} \stackrel{?}{=} 1-\ln(2)
\end{align}
Wolfram tells me that it is equal to $1-\ln(2)$.
I know the following
\begin{align}
&\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k}} = \ln(2)\\
&\sum_{n=0... | \begin{align}
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k}}
&=
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=0}^\infty{\frac{(-1)^{k}}{k+1}}
\\ &=
\sum_{k=1}^\infty{\frac{(-1)^{k+1}}{k+1}} + \sum_{k=1}^\infty{\frac{(-1)^{k}}{k+1}}+1
\\ &=1
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149130",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Closed form of the sum of two binomial expansions I would like to know if there is a simple closed form for the following expression:
$(x+y)^n + (x+z)^n$
Expanding the above I get
$(y^n + z^n) + nx(y^{n-1}+z^{n-1}) + \frac{(n-1)n}{2}x^2(y^{n-2} + z^{n-2}) + \mathcal{O}(x^3) $,
but it isn't clear to me if this can facto... | Maybe you are looking for a symmetric formulation like this
$$
\eqalign{
& \left( {x + y} \right)^{\,n} + \left( {x + z} \right)^{\,n} = \cr
& = \left( {x + \left( {{{z + y} \over 2}} \right) - \left( {{{z - y} \over 2}} \right)} \right)^{\,n}
+ \left( {x + \left( {{{z + y} \over 2}} \right) + \left( {{{z -... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find the number of roots of the equation, $x^3 + x^2 +2x +\sin x = 0$ in $[-2\pi , 2\pi]$.
Find the number of roots of the equation,
$$x^3 + x^2 +2x +\sin x = 0$$
in $[-2\pi , 2\pi]$.
What I have tried:
$$x^3 + x^2 +2x = -\sin x$$
$$x^2 +x +2 = \frac{-\sin x }{x}$$
$$(x + \frac{1}{2})^2 + \frac{7}{4} = \frac{-\sin ... | For $x<0, \sin x=x-\frac{x^3}{3!}+O(x^5)>x$:
$$x^3+x^2+2x-\sin x <x^3+x^2+2x-x=x(x^2+x+1)<0,$$
for $x>0, \sin x<x$:
$$x^3+x^2+2x-\sin x>x^3+x^2+2x-x=x^3+x^2+x>0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3149963",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
How is the partial sum of a geometric sequence calculated? I have been asked to calculate a general formula for $1 + b + 2b^2 + 3b^3 + ... + Nb^N$
I think that a part of this is a geometric sequence, and I have rewritten this as
$f(x) = 1 + \sum_{i=1}^n i\cdot x^i$
(I couldn't figure out a way to make the first term b... | Start with $$S=1 + b + 2b^2 + 3b^3 + \cdots + Nb^N$$
Multiply by $b$ and then subtract to give
$$bS=b + b^2 + 2b^3 + 3b^4 + \cdots +(N-1)b^N + Nb^{N+1}$$
$$(b-1)S = -1 - b^2 - b^3 - b^4 - \cdots -b^N + Nb^{N+1}$$
Do that again
$$b(b-1)S = -b - b^3 - b^4 - b^5 - \cdots -b^N-b^{N+1} - Nb^{N+2}$$
$$(b-1)^2S = 1-b +b^2 -(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3151564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
} |
How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$? How to prove that function $f(p)= \frac{1}{(p^2 - 1)} \sum_{q=3}^p \frac{q^2-3}{q}[q\text{ is prime}]$ is less than 0.3 for all $p$?
Let's suppose that we have a function defined as follo... | $$f(p)=\frac{1}{p^2-1}\sum_{q=3}^p\frac{q^2-3}{q}$$
If we use the fact that:
$$\sum_{q=3}^p\frac{q^2-3}{q}=\sum_{q=3}^pq-3\sum_{q=3}^p\frac1q$$
Now we know that:
$$\sum_{q=3}^pq=\sum_{q=1}^pq-\sum_{q=1}^2q=\frac{p(p+1)}{2}-3$$
$$-3\sum_{q=3}^p\frac1q=-3\left[\sum_{q=1}^p\frac1q-\sum_{q=1}^2\frac1q\right]=-3\left[\ln(p)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3153287",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 1
} |
Find the number of $k$ satisfying $\frac{n}{3} - 1 < k \leq \frac{n}{2}$ for $n > 9$. Show that the number of $k \in \mathbb{N}$ satisfying $\frac{n}{3} - 1 < k \leq \frac{n}{2}$ for $n \in \mathbb{N}, n> 9$ is greater than $\frac{n}{6}$.
I looked at the distance $|\frac{n}{3} - 1 - (\frac{n}{2})| = 1 + \frac{n}{6}$ wh... | Let $n = 6m + r$ for some non-negative integer $m$ and integer $0 \le r \le 5$. Thus, the question asks to prove there are more than $\frac{n}{6} = m + \frac{r}{6} \lt m + 1$ values of $k$, i.e., that there are at least $m + 1$ such values.
The inequality becomes
$$2m + \frac{r}{3} - 1 \lt k \le 3m + \frac{r}{2} \tag{1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3154858",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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$\iint_{\Sigma} \frac{d\sigma}{\sqrt{x^2+y^2+(z+R)^2}}$ with $\Sigma$ the upper half of the sphere $x^2+y^2+z^2=R^2$ My attempt:
$K = \{(r,\theta): 0 \le r \le R, 0 \le \theta \le 2\pi \}$. I chose following parameterization: $$ \vec{\varphi}(r,\theta)=(r\cos\theta, r\sin\theta,\sqrt{R^2-r^2}).$$ And after further cal... | When you wrote $\int_0^R \frac{r^2}{\sqrt{(2 R^2+2 R\sqrt{R^2-r^2} ) (R^2-r^2)}}$ the numerator should have been $rR$ rather than $r^2$. You can evaluate the integral using your substitution $t=\sqrt{R^2-r^2}$ as follows:
$$\int_0^{2 \pi}d\theta\int_0^R \frac{rR}{\sqrt{(2 R^2+2 R\sqrt{R^2-r^2} ) (R^2-r^2)}}dr$$
$$
= 2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3155541",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
On the transport equation $x \cdot \nabla u = |x|^2$ I have no clue about how to solve the following system
\begin{equation*}
\begin{cases}
x\cdot\nabla u=|x|^2, \quad x\in\mathbb{R}^n, \\
u|_{x_1=1}=3x_n.
\end{cases}
\end{equation*}
Study the domain where the function is defined.
Can you help me with some hints?... | Use the method of characteristics to decompose the problem. Observe, we have the string of equations:
$$\frac{dx_1}{x_1} = \ldots = \frac{dx_n}{x_n} = \frac{dz}{x_1^2 + \ldots x_n^2}$$
Solving for the constant characteristics for the first n-1 equations, we get:
$$\phi_i = \frac{x_{i-1}}{x_i}$$
So the last equation b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3157133",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Using Cardano's method to find an algebraic equation whose root is $\sqrt{2} +\sqrt[3]{3}$ $\sqrt{2} +\sqrt[3]{3}$ solution of an algebraic equation $ a_{0}x^{n}+a_{1}x^{n-1}+a_{2}x^{n-2}+\cdots+a_{0}=0 $
How to find this equation?
I tried Cardano's method, noting that
$$\sqrt{2} +\sqrt[3]{3} = \sqrt[3]{\sqrt{8}} +\sq... | Let $x=\sqrt 2 + \sqrt[3]{3}$. Then,
$$
x-\sqrt 2 = \sqrt[3]{3}
$$
Cubing both sides and rearranging we have,
$$
x^3+6x-3=\sqrt 2 (3x^2+2)
$$
Squaring both sides and rearranging we have,
$$
x^6-6x^4-6x^3+12x^2-36x+1=0
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3159254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Sum of all values that satisfy $\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$.
What is the sum of all values of $x$ that satisfy the equation
$\frac{x^{x-3}}{x}=\frac{x}{x^\frac{4}{x}}$?
I start off by cross multiplying.
$$x^2=x^{x-3+\frac{4}{x}}$$
Taking the square root of both sides gives me:
$$x=\pm x^{\frac{1}{2}x... | In your second case, after you render
$x^{\frac{x^2-5x+4}{2x}}=-1$
forget about logarithms. Taking absolute values of the equation $a^b=-1$ and remembering that anything to a zero power can only be $+1$, conclude that the only way to get a value of $-1$ using real variables is to have $a=x=-1$. So put $x=-1$ into the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160152",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
nth element in the given sequence Given two constants, a and b, and a sequence constructed as follows:
1, 1, 1, 1...(a - 1 times), 2, 2, 2, 2,...(a-2) times,......a-1
For example, for a = 6,
1, 1, 1, 1, 1, 2, 2, 2, 2, 3, 3, 3, 4, 4, 5
Ques: Determine the bth term in the sequence
*
*This is opposite to the sequence ... | Let's flip the sequence to give,
5, 4, 4, 3, 3, 3, 2, 2, 2, 2, 1, 1, 1, 1
Now if we perform, $a-t_n$ for all the terms it gives us,
1, 2, 2, 3, 3, 3.....5
We know the term $t_n = \lfloor \sqrt{2n} + \frac{1}{2} \rfloor$. Our term for the flipped sequence is just $t_n = a-\lfloor \sqrt{2n} + \frac{1}{2} \rfloor$.
The in... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3160845",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Prove that the conic $x^2 - 4xy + y^2 -2x -20y -11 = 0$ is a hyperbola and find the centre $(h,k)$
I have to prove that the conic $$x^2 - 4xy + y^2 -2x -20y -11 = 0$$ is a hyperbola and find the centre $(h,k)$.
I proved it is a hyperbola using discriminant $b^2-4ac $ and the answer was greater than zero hence a hype... | Write first the terms containing $x$ as the beginning of the square of an affine function in $x$ and $y$:
$$x^2-4xy-2x= (x-2y-1)^2-(4y+4y^2+1),$$
so that the equation becomes
\begin{align}
x^2 - 4xy + y^2 -2x -20y -11 &= (x-2y-1)^2-(4y+4y^2+1)+y^2-20y-11 \\
&= (x-2y-1)^2-3(y^2+8y+4)\\
&= (x-2y-1)^2-3\bigl((y+4)^2-16\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3163098",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
Isn't there any divisor $k$ of $n^4$ such that $n^2-nI did some experiment with my Python script to find a number which could divide $n^4$ in this interval ($n^2-n$, $n^2$).
I watched the form of prim factors of the numbers in this ($n^2-n$, $n^2$) interval, but I hadn't was any idea for proofing it.
Can anyone to he... | Answer to the Question
Let $k=n^2-a$. Since $n^2-a\mid n^4$ and
$$
\frac{n^4}{n^2-a}=n^2+a+\frac{a^2}{n^2-a}
$$
we must have $d=\frac{a^2}{n^2-a}\in\mathbb{Z}$. However, since $\color{#C00}{1\le a\le n-1}$ and $\frac{a^2}{n^2-a}$ is increasing in $a$,
$$
0\lt\overbrace{\ \frac1{n^2-1}\ }^{\large\color{#C00}{a=1}}\le d\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3165826",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Evaluate definite integral using limit of a sum definition
From the definition of a definite integral as the limit of a sum, evaluate
$$\int_a^b\frac{1}{\sqrt x}dx$$
My progress so far:
$\begin{align}
\int_a^b\frac{1}{\sqrt x}dx & =\lim_{n\to\infty}h\times\bigg(f(a)+f(a+h)+\cdots+f\big(a+(n-1)h\big)\bigg) \\
& = ... | We assume $0\leq a\leq b$. In order to cope with the square root function $\frac{1}{\sqrt{x}}$ it is convenient to use variable length intervals with length $j^2\frac{b-a}{n^2}$. When taking square roots we can factor out $j$ and summation is expected to become simpler. The corresponding Riemann sum is
\begin{align*}
... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3166670",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating $\sum\limits_{i=1}^8 \cos\frac{\pi t_i}{4}$ and $\sum\limits_{i=1}^8 \cos^2\frac{\pi t_i}{4}$ for $t\in\{-7,-5,-3,-1,1,3,5,7\}$ I am dealing with a series$$\sum_{i=1}^8 \cos\frac{\pi t_i}{4}$$
with $t\in\{-7,-5,-3,-1,1,3,5,7\}$. How can I determine a global solution for this summation? Would it be the same l... | You have a formula for the sum of sines or cosines of arcs in an arithmetic progression:
$$ \cos a + \cos(a + \theta )+\cos(a + 2 \theta )+ \dots + \cos(a + n \theta ) = \frac{\sin \frac{(n +1) \theta }{2}}{\sin \frac{ \theta }{2}}\,\cos\Bigl (a + \frac{n \theta }{2}\Bigr)$$
$$\sin a+ \sin(a + \theta )+ \sin(a + 2 \the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3167735",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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About the probability in rolling dice If A and B each roll two dice, how do we calculate the probability that the maximum number of A is greater than the maximum of B and that the minimum number of A is also greater than the minimum of B?
| Hint: Start by computing the probability mass function of the maximum for player $A$, i.e., compute $Pr(\max=x)$ for each $x$.
By the way, if you are able to solve the problem for the maximum, it will be easy to get the solution for the minimum.
Edit: Since it's solved, I will have some fun adding the formulas for the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3169048",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Suppose we roll a die repeatedly and let Tk be the number of rolls until some number appears k times in a row. Suppose we roll a die repeatedly and let Tk be the number of rolls until some number
appears k times in a row. (a) Find P(T2 = k) for each integer k ≥ 2. (b) Find E[T2].
I am thinking that P(T2=2) = (1/6)(1/6)... | We can answer the question for a die with $n$ faces, rolling until
some number appears $k$ times in row. We have for the probability that
we need $m$ rolls where $m\ge k$ from first principles
$$\mathrm{P}[T=m]
= \frac{1}{n^m} \frac{n}{n-1}
[z^m] (n-1) z^k \sum_{q\ge 0} (z+\cdots+z^{k-1})^q (n-1)^q
\\ = \frac{1}{n^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174475",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Long division of polynomials: $(2x^4 - 5x^3 - 15x^2 + 10x +8) \div (x^2-x-2)$ I've been self-studying from Stroud & Booth's amazing "Engineering Mathematics". I'm currently stuck on an aspect of long division of polynomials, when the denominator itself is a polynomial. So, I know how to do long division when the denimi... | Usually in these problems, I will first analyze the denominator polynomial $g(x)=x^2-x-2$.
The roots of this polynomial are: $$g(x)=(x-2)(x+1)=0$$
giving, $x=-1,2$. Now the second thing I do is I check whether the numerator polynomial is having a common root. So we check $f(x)=2x^4-5x^3-15x^2+10x+8$, for $x=-1,2$.
$f(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3174570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 3
} |
Find number of ways to split $1$ dollar into $5$, $10$, $20$, $50$ cents
Find number of ways to split $1$ dollar into $5$, $10$, $20$, $50$ cents
I am going to use generating functions:
$$n = [x^{100}] (1+x^5+x^{10}+\cdots)(1+x^{10}+x^{20}+\cdots)(1+x^{20}+x^{40}+\cdots)(1+x^{50}+x^{100}+\cdots) = \\
[x^{100}]\frac{1... | Let $$\begin{align}f(y)&=\frac{1}{1-y}\frac{1}{1-y^2}\frac{1}{1-y^4}\\&=\frac{(1+y+y^2+y^3)(1+y^2)}{(1-y^4)^3}\\&=\frac{1+y+2y^2+2y^3+y^4+y^5}{(1-y^4)^3}\end{align}$$
Letting $p(y)=1+y+2y^2+2y^3+y^4+y^5$, then $$[y^{20}]f(y)\frac{1}{1-y^{10}} = [y^0]f(y)+[y^{10}]f(y)+[y^{20}]f(y)$$
Now use that $$\frac{1}{(1-y^4)^3}=\s... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175146",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
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} |
Finding the value of $x-y$. Given two equations \begin{align} & x^4+y^4=\dfrac {-7}{9} \\ & x^3-y^3=3. \end{align} From these equations find the value of $(x-y).$
I have just factorize $x^3-y^3=(x-y)(x^2+xy+y^2)=3$. But don't know how to proceed from here. One way to find the values of $x$ and $y$ and then compute poss... | Multiply equation (2) by $x-y$, giving $(x^4-x^3 y +xy^3 + y^4) = 3(x-y)$.
Next, (1) = -7/9, so
$-x^3 y +xy^3 - 7/9 = 3(x-y)$.
Factorise.
$xy(-x^2 + y^2) - 7/9 = 3(x-y)$.
$(y-x)(y+x)xy - 7/9 = 3(x-y)$.
$(x-y)(3+(y+x)xy) = - 7/9$
Using your idea, $x^2+xy+y^2 = 3/(x-y) = - 7/3*(3+(y+x)xy)$
Congrats, you now have a quad... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3175319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Are there any other methods to apply to solving simultaneous equations? We are asked to solve for $x$ and $y$ in the following pair of simultaneous equations:
$$\begin{align}3x+2y&=36 \tag1\\ 5x+4y&=64\tag2\end{align}$$
I can multiply $(1)$ by $2$, yielding $6x + 4y = 72$, and subtracting $(2)$ from this new equation... | Another method to solve simultaneous equations in two dimensions, is by plotting graphs of the equations on a cartesian plane, and finding the point of intersection.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3180580",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "32",
"answer_count": 14,
"answer_id": 11
} |
Finding Angle using Geometry In an equilateral triangle $ ABC $ the point $ D $ and $ E $ are on sides $ AC $ and $AB$ respectively, such that $ BD $ and $ CE $ intersect at $P$ , and the area of the quadrilateral $ ADPE $ is equal to area of $ \Delta BPC $ find $ \angle BPE $.
This question when I first tried looke... |
Let $a$ be the side length of the equilateral $\triangle ABC$,
$\angle BPE=\theta$, $\angle PBC=\phi$,
and let $[\cdot]$ denote the area.
Note that
\begin{align}
[AEPD]&=[ABC]-[BCD]-[BCE]+[BCP]
,
\end{align}
and since $[AEPD]=[BCP]$,
we must have
\begin{align}
[BCD]+[BCE]&=[ABC]
.
\end{align}
Using the three-angle... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181122",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 3
} |
Is there some strategy to find the general term of a recursive sequence? Given $(a_{n})_{n \in \mathbb{N}}$,
$a_{0} = 2, a_{1} = 4, a_{n+2} = 4a_{n+1} - 3a_{n}$,
is there any way to find the general term? I reckoned that every $a_{i}$ is a factor of 2, and then the series becomes 1, 2, 5, 14, 41, ..., and obviously th... | Writing each pair of consecutive terms as a linear combination of the previous pair of consecutive terms,
$$\pmatrix{a_{n}\\a_{n-1}} = \pmatrix{4&-3\\1&0}\pmatrix{a_{n-1}\\a_{n-2}}= \cdots = \pmatrix{4&-3\\1&0}^{n-1}\pmatrix{a_{1}\\a_{0}}$$
The question becomes how to represent the matrix power in close form, which is ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181668",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$. Find $\lim_{n \to \infty} nt_n$ If $t_n=\frac{1}{2n+1}-\frac{1}{2n+2}+\frac{1}{2n+3}-\frac{1}{2n+4}+\cdots +\frac{1}{4n-1}-\frac{1}{4n}$
Find $\lim_{n \to \infty} nt_n$
First attempt: $t_n$ is positive(grouping two... | Let $$H_n = 1 + \frac{1}{2} + \frac{1}{3} + \dotsb + \frac{1}{n}$$ be the $n$th harmonic number. Then, as you noted, $$t_n = H_{4n} - H_{2n} - [H_{2n} - H_n] = H_{4n} - 2H_{2n} + H_n.$$
By the Euler-Maclaurin summation formula,
$$H_n = \log n + \gamma + \frac{1}{2n} + O\left(\frac{1}{n^2}\right)$$
so, after the smoke... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3181789",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Quickest way to find $a^5+b^5+c^5$ given that $a+b+c=1$, $a^2+b^2+c^2=2$ and $a^3+b^3+c^3=3$
$$\text{If}\ \cases{a+b+c=1 \\ a^2+b^2+c^2=2 \\a^3+b^3+c^3=3}
\text{then}\ a^5+b^5+c^5= \ ?$$
A YouTuber solved this problem recently and, though he spent some time explaining it, took over 40 minutes to solve it.
Like the v... | Fun video!
Much time was spent on finding $abc=1/6$.
Alternative method for this:
$$\begin{align}a^2+b^2&=2-c^2 \Rightarrow \\
(a+b)^2-2ab&=2-c^2 \Rightarrow \\
(1-c)^2-2ab&=2-c^2 \Rightarrow \\
ab&=c^2-c-\frac12 \Rightarrow \\
abc&=c^3-c^2-\frac c2 \end{align}$$
Similarly:
$$abc=a^3-a^2-\frac a2\\
abc=b^3-b^2-\frac b2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182260",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 4
} |
An apparently harmless exercise concerning induction Let $b \in \mathbb{R}, b \ge 2$. Prove by induction that $$(b^n - 1)(b^n - b)(b^n -b^2)\cdots(b^n - b^{n-2}) \ge b^{n(n-1)}-b^{n(n-1)-1}$$ for all $n \in \mathbb{N}, n \ge 1$.
For the case $n = 2$, I have to show that $$b^2 - 1 \ge b^2 - b$$ which is true for $b \ge... | Change variable to $c = \frac1b$. The inequality at hand (the version for $b \ge 2)$ is equivalent to
$$\prod_{k=2}^n (1-c^k) \ge 1 - c\quad\text{ for }\;n \ge 2, c \in \left(0,\frac12\right]$$
To prove this by induction, we will use a stronger form of induction statement.
For $n \ge 1$ and $c \in (0,\frac12]$, let $\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3182976",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
What am I doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$? I can't figure out what I'm doing wrong when testing for concavity with $f(x)=5x^{2/3}-2x^{5/3}$. I can't find the right intervals for where the graph is concave-down and concave-up.
My Steps:
find $f'(x)$:
$$f'(x)=\frac{10}{3}x^{-1/3}-\fr... | It is a bit easier if you factor out the smallest fractional power of $x$ at each step.
\begin{eqnarray}
f(x)&=&5x^{2/3}-2x^{5/3}=x^{2/3}(5-2x)\\
f^\prime(x)&=&\frac{10}{3}x^{-1/3}-\frac{10}{3}x^{2/3}=\frac{10}{3}x^{-1/3}(1-x)\\
f^{\prime\prime}(x)&=&-\frac{10}{9}x^{-4/3}-\frac{20}{9}x^{-1/3}=-\frac{10}{9}x^{-4/3}(1+2x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3186799",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
System of Distributional Differential Equations
Here are the problem and my attempt to the solution. Is it correct?
| Let
$$
A = \begin{pmatrix}2 & -1 \\ 3 & -2\end{pmatrix},
\quad
y = \begin{pmatrix}y_1 \\ y_2\end{pmatrix},
\quad
b = \begin{pmatrix}0 \\ 2\end{pmatrix} \delta(x).
$$
The equation can then be written
$y' = Ay + b.$
This can be rewritten as
$y' - Ay = b.$
After multiplication from the left with $e^{-Ax}$ we get
$e^{-Ax}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3187741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
Find the third expression of Taylor's for $f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}$ around $x=0$
My try:Let $y=x+1$. Then: $$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\frac{1}{(2y-1)(4-\frac{1}{y^2})^2}=-\frac{1}{2}\cdot\frac{1}{... | Just for the fun of it !
Starting from @marty cohen's answer and continuing the process
$$f(x)=\frac{(1+x)^4}{(1+2x)^3(1-2x)^2}=\sum_{n=0}^\infty 2^{n-8} \left[27 (6 n+7)+(-1)^n (2 n^3+13n+67)\right]\,x^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3188040",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Proving ${\lim\limits_ {n\to\infty}}\frac{6n^3+5n-1}{2n^3+2n+8} = 3$ I'm trying to show that $\exists \,\varepsilon >0\mid\forall n>N\in\mathbb{N}$ such that:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| < \varepsilon$$
Let's take $\varepsilon = 1/2$:
$$\left|\frac{6n^3+5n-1}{2n^3+2n+8}-3\right| = \left|\frac{6n^3+5n-1... | I do it like this:
$\forall n \ge 1, \; \dfrac{6n^3 + 5n - 1}{2n^3 + 2n + 8} = \dfrac{6 + 5n^{-2} - n^{-3}}{2 + 2n^{-2} + 8n^{-3}}; \tag 1$
since it is easy to see that
$\displaystyle \lim_{n \to \infty} (6 + 5n^{-2} - n^{-3}) = 6, \tag 2$
and
$\displaystyle \lim_{n \to \infty} (2 + 2n^{-2} + 8n^{-3} ) = 2, \tag 3$
we ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3190400",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Сonvert trigonometric equation to quadratic equation I am struggling to know how to transform this equation as it involves more than one type of trigonometric function, I know how to do it with one repeated function.
Question:
$\sin^2 \theta/2$ + $\sin \theta$ + $1$ = $0$
In must be transform in pure quadratic form:
$... | This is what is sometime called a "trick question" since it does not require you to solve a quadratic equation.
Notice that if $\sin^2\left(\dfrac{\theta}{2}\right)+\sin\theta+1=0$, then it follows that $\sin^2\left(\dfrac{\theta}{2}\right)=-\sin\theta-1\ge0$. So it must be the case that $\sin\theta\le-1$. So $\sin\the... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3193934",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Matrixes with common parameters to result in no inverse I've been given three matrices $A, B \ \& \ C$ which are defined as follows:
$$ A = {
\left[
\begin{array}{ccc}
b & 5 & 8 \\
c & 1 & 3 \\
a & 4 & 3 \\
\end{array}
\right]
},\ B = {
\left[
\begin{array}{ccc}
7 & 4 & 2 \\
5 & 5 & -1 \\
-2 & -a & -b \\
\end{array... | Your determinants are right. Next you set them equal to $0$. You do not ignore the number 28. You put it on the RHS as mentioned in the comments.
$$\left(
\begin{array}{ccc|c}
7 & -9 & 17 & 0\\
-17 & -15 & 0 & -28\\
-7 & -7 & 7 &0 \\
\end{array} \right)$$
Now you can use the Gaussian elimination algorithm to obtain th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3197374",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Advanced Complex numbers Let $\omega$ be a nonreal root of $z^3 = 1.$ Let $a_1,$ $a_2,$ $\dots,$ $a_n$ be real numbers such that
$$\frac{1}{a_1 + \omega} + \frac{1}{a_2 + \omega} + \dots + \frac{1}{a_n + \omega} = 2 + 5i.$$Compute
$$\frac{2a_1 - 1}{a_1^2 - a_1 + 1} + \frac{2a_2 - 1}{a_2^2 - a_2 + 1} + \dots + \frac{2a_... | Hint: $\overline \omega$ is another non-real root of $z^3=1$,
$$\sum_{j=1}^n\frac1{a_j+\overline\omega} = \overline{2+5i}$$
$1,\omega, \overline\omega$ are roots of $z^3-1 = 0$, so
$$\omega\overline\omega = 1,\quad \omega+\overline\omega + 1 = 0$$
Consider the following sum,
$$\begin{align*}
\sum_{j=1}^n\frac1{a_j+\om... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3198831",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
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Central forces and conservation problem An object A moves under the influence of a planet B's gravity, a central force of magnitude $\frac{1}{r^2}$. At some point, its velocity is $u$e$_{\theta}$ and $r=a$. It is given that $au^2 < 2$. I am looking to find an upper limit to the distance A gets from B.
Integrating the f... | Note that the condition that at $r=a$ the velocity is $u$e$_{\theta}$ means that when $r=a$ we are either at a minimum distance or at a maximum distance.
This is because the condition that marks that we are at an extreme distance is $\dot{r} = 0$, i.e., no radial velocity.
So, from the equation
\begin{equation*}
\frac{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3203819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Answer to - proving $\cos 207^o$ is irrational. At first, $\cos(207^o) = \cos(180^o+27^o) = - \cos(27^o)$
We have, $(\sin 27^o + \cos 27^o)^2 = \sin^2 27^o + \cos^2 27^o + 2 \sin 27^o \cos 27^o$
$$ = 1 + \sin 2.27^o = 1 + \sin 54^o = 1 + \sin (90^o - 36^o) = 1 + \cos 36^o$$
$$(\sin 27^o + \cos 27^o)^2 = 1 + \frac{\sqr... | As you noticed we have to prove that:
$$\cos(27°)\notin \Bbb{Q}$$
First of all:
$$ \cos(54°)=2\cos^2(27°)-1$$
So: $$\cos(54°)\notin \Bbb{Q}\Rightarrow \cos(27°)\notin \Bbb{Q}$$
Notice that:
$$\cos(54°)=\sin(36°)=\sqrt{\frac{5-\sqrt{5}}{8}} \notin \Bbb{Q}$$
:)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3208252",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
If $x = 2$ is a root of $\det\left[\begin{smallmatrix}x&-6&-1\\2&-3x&x-3\\-3&2x&x+2\end{smallmatrix}\right]=0$, find other two roots
If $x = 2$ is a root of equation
$$ \begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0 $$
Then find the other two roots.
I solved it and got a cubic eq... | $\begin{vmatrix}
x & -6 & -1 \\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0$
$\begin{vmatrix}
x-2 & 3x-6 & 2-x\\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0$
$R_1 \rightarrow R_1-R_2$
$(x-2)\begin{vmatrix}
1 & 3 & -1\\
2 & -3x & x-3\\
-3 & 2x & x+2
\end{vmatrix} = 0$
$\begin{vmatrix}
1 & 3 & -1\\
0 & -3x-6 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209154",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Eliminate $\theta$ from $\lambda\cos2\theta=\cos(\theta + \alpha) \space$ and $\space \space\lambda \sin2\theta=2\sin(\theta + \alpha)$ Eliminate $\theta$ from $\lambda \cos2\theta=\cos(\theta + \alpha)$ and $\lambda\sin2\theta=2\sin(\theta + \alpha)$
My approach:
Dividing the RHS and LHS of both equations by $\lambda$... | We have
\begin{align*}
\tan2\theta&=2\tan(\theta+\alpha)\\
\frac{2\tan\theta}{1-\tan^2\theta}&=\frac{2(\tan\theta+\tan\alpha)}{1-\tan\theta\tan\alpha}\\
\tan\theta-\tan^2\theta\tan\alpha&=\tan\theta(1-\tan^2\theta)+\tan\alpha(1-\tan^2\theta)\\
\tan^3\theta&=\tan\alpha
\end{align*}
From $\lambda\sin2\theta=2\sin(\theta+... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3209847",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
If $(2^b-1+c^b)^a = (2^a-1+c^a)^b,$ then is it true that $a=b?$
Question: Assume that $a,b,c$ are real numbers such that
$$1\leq a\leq b, \quad\text{and}\quad 0<c<1.$$
If we have
$$(2^b-1+c^b)^a = (2^a-1+c^a)^b,$$
then is it true that
$$a=b?$$
I can only prove that the statement is true if $a$ is an intege... | Let
$$
f(a,b,c)=a\cdot\ln\Big(2^b-1+c^b\Big)-b\cdot\ln\Big(2^a-1+c^a\Big)
$$
Obviously your statement is equal to finding roots of $f$.
Now, its easy to see that $f(a,b,1)=0$, so we take the derivative:
$$
\frac{\partial f}{\partial c}= \frac{ab}{c(2^b-1+c^b)(2^a-1+c^a)}(2^ac^b-2^bc^a+c^a-c^b)
$$
Notice that the last t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210495",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
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Solving limit without L'Hôpital's rule: $\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}$ How can I solve this limit without L'Hôpital's rule?
$$\begin{align}\lim\limits_{x \to 0} \frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}&=\lim\limits_{x \to 0} \frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x... | $$\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}
=\frac{\tan x-\sin x}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}.$$
From Maclaurin series $\sin x=x-x^3/6+O(x^5)$ and $\tan x=x+x^3/3+O(x^5)$.
Therefore
$$\frac{\sqrt{1+\tan x}-\sqrt{1+\sin x}}{x^3}
\sim\frac{x^3/2}{x^3(\sqrt{1+\tan x}+\sqrt{1+\sin x})}$$
as $x\to0$, and so the ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210646",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Integration by Partial Fractions $\int \sec^3xdx$ I have looked at multiple ways to do partial fractions to integrate $\sec^3x$, but there is a part where I keep getting stuck when I see the partial fractions get split up.
$$\int \frac{\cos x}{\cos^4 x} = \int \frac{1}{(1-y^2)^2}$$
The next step I have seen is shown li... |
And for the denominator, why are there two additional positive
fraction decompositions
Because $1-u^2 = (1-u)(1+u).$ It's the "difference of squares" factorization. And so, squaring both sides, we also have $(1-u^2)^2 = (1-u)^2(1+u)^2$.
In partial fractions decomposition, all irreducible factors must appear as den... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3210927",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Compute $\sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}$ How can Compute in closed form this double summation :
$$\sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}$$
I think here can use harmonic series
Actually I don't have any ideas to approach it
| The idea is to replace the double sum by a double integral which then hopefully can be solved. The hope is justified and we find the following
Result
The closed form is
$$s = \sum_{k≥1,n≥1}\frac{(-1)^{n+k}}{k(k+1+n)^2}\\=\frac{1}{2}\zeta(2) - 2 (1-\log(2))-\frac{1}{8} \zeta(3)\simeq 0.0585043\tag{1}$$
Derivation
The re... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3213711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
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Find the equation of the plane through the point $(2,1,4)$ Find the equation of the plane through the point $(2,1,4)$ and perpendicular to each of the planes $9x-7y+6z+48=0$ and $x+y+z=0$.
My attempt:
The equation of the plane passing through the point $(2,1,4)$ is given by
$$A(x-2)+B(y-1)+C(z-4)=0$$
Here, $A,B,C$ rep... | By solving the second and third equations, we have
\begin{align}
B&=\frac{3}{13}A\\
C&=-\frac{16}{13}A
\end{align}
So the plane is
\begin{align}
A(x-2)+3(y-1)A/13-16(z-4)A/13 &=0\\
13(x-2)+3(y-1)-16(z-4) &=0
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215476",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Decomposing a quartic polynomial into a difference of squares Given quartic polynomial $1 + x + x^2 +x^3 + x^4$, I want to find polynomials $p, q \in \mathbb R[x]$ such that
$$1 + x + x^2 +x^3 + x^4 = p^2 (x) - q^2 (x)$$
| $$1 + x + x^2 +x^3 + x^4 = \frac 12 \begin{bmatrix} 1\\ x\\ x^2\end{bmatrix}^\top \begin{bmatrix} 2 & 1 & t\\ 1 & 2-2t & 1\\ t & 1 & 2\end{bmatrix} \begin{bmatrix} 1\\ x\\ x^2\end{bmatrix}$$
Since we want the matrix to be rank-$2$,
$$\det \begin{bmatrix} 2 & 1 & t\\ 1 & 2-2t & 1\\ t & 1 & 2\end{bmatrix} = 2 (t - 2) (t^... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215595",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Find $P$ where $P^{-1}AP$ for a given matrix $A$ I am doing a past paper and I have been given a matrix A:
\begin{bmatrix}
4 & -1 & -3 & 2 \\
4 & -2 & -4 & 4 \\
-4 & 4 & 6 & -4 \\
-6 & 5 & 7 & -4 \\
\end{bmatrix}
and I need to find a matrix $P$ such that $P^{-1}AP$ =
\begin{... | From the given information, we know that matrix A had eigenvalues 2, -2, and 2 is an eigenvalue of multiplicity.
We want to find eigenvectors that are associated with each of these eigenvalues.
find $v$ such that
$(A-\lambda I) v = 0$
$\lambda = 2$
$\begin{bmatrix}
2 & -1 & -3 & 2 \\
4 & -4 & -4 & ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3215979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Differential equation $y'' - y + 2\sin(x)=0$ I need help and explanation with this differential equation. Actually I really don't know how to solve just this type of equations. So the problem: $$y'' - y + 2\sin(x)=0$$
In my opinion first of all we solve homogeneous equation $y''-y=0$ and the solution of this is $y=c_1e... | The given differential equation is
$y'' - y + 2\sin\ x=0$
$\implies y'' - y =- 2\sin x \implies (D^2 -1)y=- 2\sin x$
where $D \equiv \frac{d}{dx} $
I think you have an idea about how to find the Complementary Function (i.e., C.F.), (for your case, which is nothing but the solution of the homogeneous differential equati... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3216900",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Proof that $ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $ for $p>5$ Proof that $$ p^2 \equiv 1 \mod{30} \vee p^2 \equiv 19 \mod{30} $$
for $p>5$ and $p$ is prime.
$\newcommand{\Mod}[1]{\ (\mathrm{mod}\ #1)}$
My try
Let show that
$$p^2 - 1 \equiv 0 \Mod{30} \vee p^2 - 1\equiv 18 \Mod{30}$$
Let check $$p^2 -1 = (p... | $p^2-1$ is divisible by $2$ and $3$ and leaves remainder $0$ or $3$ when divided by $5$
[since if $p\not\equiv0\pmod{5}$ then $p^2\equiv1$ or $4\pmod5$].
In the former case, $p^2-1\equiv0\pmod{30}$; in the latter, $p^2-1\equiv18\pmod {30}.$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3219882",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
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Evaluate $\lim\limits_{x \to 0}\frac{e^{(1+x)^{\frac{1}{x}}}-(1+x)^{\frac{e}{x}}}{x^2}$ Solution
Expanding $(1+x)^{\frac{1}{x}}$ at $x=0$ by Taylor's Formula，we obtain
\begin{align*}
(1+x)^{\frac{1}{x}}&=\exp\left[\frac{\ln(1+x)}{x}\right]=\exp \left(\frac{x-\frac{x^2}{2}+\frac{x^3}{3}+\cdots }{x}\right)\\
&=\exp \left... | Your result is correct. A simpler method is to decompose this limit into a product of simpler limits:
Let $f(x)=(1+x)^{1/x}$. We have $\lim_{x\rightarrow 0} f(x) = e$. Note that your expression can be written as
$$ \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{x^2} = \lim_{x\rightarrow 0} \frac{e^{f(x)}- f(x)^e}{(f(x)-... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3220990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
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To find minimum distance between two curves Let $P(x, y, 1)$ and $Q(x, y, z)$ lie on the curves $$\frac{x^2}{9}+\frac{y^2}{4}=4$$ and $$\frac{x+2}{1}=\frac{y-\sqrt{3}}{\sqrt{3}}=\frac{z-1}{2}$$ respectively. Then find the square of the minimum distance between $P$ and $Q$.
My Attempt is:
I tried to find minimum distanc... | Starting from @Christian Blatter's answer, using $s=2 \tan ^{-1}(x)$ and expanding, we end with
$$2 \sqrt{3}\, x^4+70 \,x^3+72 \sqrt{3} \,x^2-274\, x-26 \sqrt{3}=0$$
Let $x=t-\frac{35}{4 \sqrt{3}}$ to get the depressed quartic
$$t^4-\frac{937 }{8}t^2+\frac{24467}{24 \sqrt{3}} t-\frac{166043}{256}=0$$ which can be exact... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222871",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Arrange $x,y,z$ in ascending order. $c>1, c \in \mathbb{R}$
$x = \frac{\sqrt{c+2} - \sqrt{c+1}}{\sqrt{c} - \sqrt{c-1}}
= (\sqrt{c+2} - \sqrt{c+1})(\sqrt{c} + \sqrt{c-1})$
$y = \frac{\sqrt{c+2} - \sqrt{c+1}}{\sqrt{c + 1} - \sqrt{c}}
= (\sqrt{c+2} - \sqrt{c+1})(\sqrt{c+1} + \sqrt{c})$
$z = \frac{\sqrt{c} - \sq... | Hint One way to look at it is that $t\mapsto \sqrt t$ is concave, so successive secants have lower slope. Now can you relate $z,x$ by considering them as ratios of slopes?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/3222990",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Three ways to find the normal of a hyperboloid? In the hyperboloid $x^2 + y^2 − z^2 = 4$, where $z \ge 0$, I have found three ways to get the normal vector, but my problem is they do not seem equivalent.
The first is as I have been taught that you can find the gradient of the function, and that will be the normal:
$$\l... | Ignoring normalization ...
*
*Your first vector, divided-through by $2$, is $(x,y,−z)$.
*Your second vector, multiplied-through by $\sqrt{x^2+y^2−4}$, is $(x,y,−\sqrt{x^2+y^2−4})=(x,y,−z)$
So, those vectors are equivalent.
Your third vector is problematic, which isn't surprising, since your parameterization doesn'... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/3228455",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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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.