Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
|---|---|---|
Telescoping exercise with iterations? For every $x>0$, consider the sequence $(x_n)$ defined by $x_0=x$ and, for every $n\geqslant0$, $$x_{n+1} = \sqrt{x_n + \frac12}$$
Then $x_n\to x_*=\frac{1+\sqrt3}2\ne0$ hence the sequence $$S_n(x)=\sum_{k=1}^n(-1)^kx_k^4$$ diverges. Consider its Cesàro sums, defined by $$C_n(x)=\f... | Here is the complete proof. I had wanted to post this before but had to wait for all close voters to rescind their votes after trying to save this question.
First off -- I think the key step in the proof is made a little clearer by using the iterated function notation for the question instead of the other that has been... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluating $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$
I have to evaluate $\displaystyle\lim_{x\rightarrow y}\frac{\sin^2x-\sin^2y}{x^2-y^2}$
I tried several ways like replacing $x^2-y^2$ by $z$ and then solving and breaking the denominator into $(x+y)(x-y)$ and then replacing $(x-y)$ by $z$ a... | $$\lim _{ x\rightarrow y } \frac { \sin ^{ 2 } x-\sin ^{ 2 } y }{ x^{ 2 }-y^{ 2 } } =\lim _{ x\rightarrow y } \frac { \left( \sin { x-\sin { y } } \right) \left( \sin { x+\sin { y } } \right) }{ \left( x-y \right) \left( x+y \right) } =\\ =\lim _{ x\rightarrow y } \frac { \left( 2\sin { \frac { x-y }{ 2 } } \cos... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How do I solve $18^{{20}^{19}}$ divided by 7? I was re-practicing some number theory and I came across this question that I had previously solved and it seems like... I can't solve it anymore. My thoughts are that $18^6 \equiv 1 \text{ mod } 7$ by Euler's formula so I am looking to express the power $20^{19}$ as some $... | Since $\gcd(18,7)=1$, by Euler's theorem, $18^{\phi(7)}\bmod{7}=1$.
Since $7$ is prime, $\phi(7)=7-1=6$.
Therefore $18^{6}\bmod{7}=1$.
There exists a positive integer $n$ such that:
$20^{19}=$
$6n+(20^{19}\bmod{6})=$
$6n+((20\bmod{6})^{19}\bmod{6})=$
$6n+(2^{19}\bmod{6})$
Therefore:
$18^{20^{19}}\bmod{7}=$
$18^{6n+(2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2059715",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3,xyz=4$ Evaluate $\frac{1}{zx+y-1}+\frac{1}{zy+x-1}+\frac{1}{xy+z-1}$ if $x+y+z=2,x^2+y^2+z^2=3, xyz=4$
The first thing that I notice is that it is symetric to $a,b,c$ but it can't help me .The other idea is finding the numbers but g... | Replacing $x=2-y-z$ and similar, the sum becomes:
$$
S = \sum_{cyc}\frac{1}{x+yz-1} = \sum_{cyc}\frac{1}{1-y-z+yz} = \sum_{cyc}\frac{1}{(y-1)(z-1)}=\cfrac{1}{\prod_{cyc} (x-1)} \sum_{cyc} (x-1)
$$
From the given relations $\sum_{cyc} xy = \frac{1}{2}\left((\sum_{cyc} x)^2-\sum_{cyc} x^2\right)=\cfrac{1}{2}\,$, so:
$$
... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Proving inequality using Lagrange multipliers. I started learing about Lagrange Multipliers and I got the following question:
Prove that $\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}≥\frac{3}{2}$, for each $a,b,c>0$.
I'm not sure how to use Lagrange multipliers for it... any ideas?
| Let the objective function be
$$f = a(c+a)(a+b) + b(b+c)(a+b) + c(b+c)(c+a)$$
and the constraint be
$$g = (a+b)(b+c)(c+a) - k.$$
Note that if $g=0$, then
$$\frac{f}{k} = \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}.$$
Solving $\nabla f-\lambda g = 0$ and $g=0$ we find $f$ is minimized for
$a=b=c=k^{1/3}/2$, with value... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 3
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p-adic number for polynomial I have three polynomials:
$$f=x^5+x^4+4x^3+3x^2+3x$$
$$g=x^5+4x^3+2x^2+3x+6$$
$$p=x^2+3$$
Question: what is $v_p(fg)$?
First, I have multiplied f and g:
$h :=f \cdot g = x^{10}+x^9+8x^8+9x^7+24x^6+29x^5+36x^4+39x^3+27x^2+18x $
Then I have divided that $h$ with $x^2+3$ to get a factor to k... | Because $x^2+3$ is irreducible in $\Bbb{Z}[x]$ (Eisenstein) it is the minimal polynomial of $i\sqrt3$. You can then check that
*
*$f(i\sqrt3)=0$, so $f(x)$ must be divisible by $x^2+3$. A long division shows that $f(x)/(x^2+3)=x+x^2+x^3$, and you can check this is not a multiple of $x^2+3$.
*$g(i\sqrt3)=0$, so $g(x... | {
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Find arc length of implicit function Result of integrating I've been trying to find the arc length of the following function:
$e^y=\frac{e^x+1}{e^x-1}$
from $x = a$ to $x = b$.
I've tried designating $y$, but it doesn't seem to work. I would really appreciate your help
| We have $y = $ ln$(\frac{e^x + 1}{e^x-1})$. Then $\frac{dy}{dx} = \frac{e^x -1}{e^x + 1} * \frac{{e^{2x} - e^x -e^{2x} - e^x}}{(e^x -1)^2} = -2\frac{e^x}{(e^x -1)(e^x +1)} = -2\frac{e^x}{e^{2x} -1}$.
Now, letting $y(x)$ be our curve and $l$ its length, $l(y) = \int_a^b \sqrt{1 + (\frac{dy}{dx})^2} dx$.
Then we have now... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Why doesn't this property work on the following composite function? Given $f(x),g(x)$ find $(f \circ g)(x)$.
$$f(x) = x + \frac 1x \\g(x) = \frac{x+1}{x+2}\\(fog)(x) = \frac{x+1}{x+2}+\frac{1}{\frac{x+1}{x+2}} = \frac{x+1}{x+2}+\frac{x+2}{x+1}$$
Using the property $\frac{a}{b}+\frac{c}{d}=\frac{ad+bc}{bd}$ I now get: $... | The property works.
$$\frac{(x+1)^2+(x+2)^2}{(x+1)(x+2)}=\frac{x^2+2x+1+x^2+4x+4}{(x+1)(x+2)}=\frac{2x^2+6x+5}{(x+1)(x+2)}$$
I believe your careless mistake is $(x+1)^2+(x+2)^2 \neq (x+1)^2(x+2)^2$
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the number of integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$? What is the number of non negative integer solutions of $2{x_{1}}+2{x_{2}}+{x_{3}}+{x_{4}}={12}$ ?
I tried it as :
${x_{3}}+{x_{4}}={12} - 2{x_{1}}-2{x_{2}}$
Now, finding the solutions of ${x_{1}}+{x_{2}}$
*
*${x_{1}}+{x_{2}} = 0 ... | Sorry for making this an answer - it's a bit too long for a comment on Elliot's answer. As Elliot showed we only need to find the coefficient of $x^{12}$ for $(1-x^2)^{-2}(1-x)^{-2}$. This is quite simple without computer aid with a few tricks. First notice that $1/(1-x)^2 = 1+2x+3x^2+\cdots$. Since we are only int... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Periodicity of an inhomogeneous system of ODEs
Consider the linear $2 \times 2$-system: $x'=Ax+b(t),$ where $b \in C(\mathbb{R}, \mathbb{R}^2)$ is a $2 \pi /{\omega}$-periodic and
$$A=
\begin{pmatrix}
0 & \omega \\ -\omega & 0
\end{pmatrix}$$
for some $\omega >0.$
(a). Find the general solution (depends on two ... | Suppose $x(t)$ is periodic. Then $x(0)=x(p)$ for some $p>0$.
Direct computation shows that $x(0)=\begin{pmatrix}c_1\\c_2\end{pmatrix}$. And
$$x(p)=\begin{pmatrix}
\cos \omega p & \sin \omega p \\ -\sin \omega p & \cos \omega p
\end{pmatrix} \cdot \begin{pmatrix}
c_1 \\ c_2
\end{pmatrix}+\begin{pmatrix}
\cos \omega p &... | {
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How to calculate the determinant of a $4 \times 4$ matrix with multiple variables? What is the determinant: $$ \begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix} $$
Someone gave me the following hint
Replace $d$ by a variable $x$; make use of the fact that the s... | $$\text{Det}(A)=\begin{vmatrix}1& a & a^2 & a^4 \\ 1 & b & b^2 & b^4 \\ 1 & c & c^2 & c^4 \\1 & d & d^2 &d^4 \end{vmatrix}=\begin{vmatrix}1& a & a^2 & a^4 \\ 0 & b-a & b^2-a^2 & b^4-a^4 \\ 0 & c-a & c^2-a^2 & c^4-a^4 \\0 & d-a & d^2-a^2 &d^4-a^4 \end{vmatrix}=\begin{vmatrix} b-a & b^2-a^2 & b^4-a^4 \\ c-a & c^2-a^2 &... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2068179",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "26",
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A.P. terms in a Quadratic equation.
The terms $a,b,c$ of quadratic equation $ax^{2}+bx+c=0$ are in A.P. and positive. Let this equation have integral root $\alpha,\ \beta$. Then find the value of $\alpha+ \beta + \alpha \cdot \beta$ ?
please point where I'm wrong:
Let common difference be $d$
$\implies \alpha+ \beta ... | We have, $$p(x)=ax^2+bx+c=ax^2+(a+p)x+(a+2p)$$
and thus, $$t+r+r\cdot t=-\frac{a+p}{a}+\frac{a+2p}{a}=\frac{p}{a}$$
then $p=a\cdot k$, $k=(t+r+t\cdot r) \in \Bbb Z$.
Then our polynomial becomes,$$p(x)=ax^2+a(1+k)x+a(1+2k)$$
$$t=\frac{-(1+k) \pm \sqrt{k^2-6k-3}}{2} \quad...(1)$$
So
$$k^2-6k-3=q^2 \Rightarrow (k-3)^2-12... | {
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"timestamp": "2023-03-29T00:00:00",
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Multiplicative inverse of $2x + 3 + I$ in $\mathbb{Z}_5[x]/I $? In $\mathbb{Z}_5[x]$, let $ I = \langle x^2 + x + 2\rangle$. Find the multiplicative inverse of $2x + 3 + I$ in $\mathbb{Z}_5[x]/I$.
The answer is supposed to be $3x + 1 + I$. But I don't understand how you find this. I divided $x^2 + x + 2$ by $2x+3$ to ... | Your original approach was on the right track. By dividing, you found that
$$
x^2 + x + 2 = (2x + 3)(3x + 1) + 4 \, .
$$
Rearranging, we have
\begin{align*}
(2x + 3)(3x + 1) = -4 + x^2 + x + 2 = 1 + x^2 + x + 2 \equiv 1 \pmod{x^2 + x + 2}
\end{align*}
so $3x + 1$ is the inverse of $2x+3$.
In general, if you have the e... | {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Find the GCD of $10$ integers
Find the greatest common divisor of the following ten integers:
\begin{align}
& 2000^3+3\cdot2000^2+2\cdot2000,\ 2001^3+3\cdot2001^2+2\cdot2001,\ \dots, \\[5pt]
& \dots, \ 2008^3+3\cdot2008^2+2\cdot2008,\ 2009^3+3\cdot2009^2+2\cdot2009.
\end{align}
How are these numbers related to... | You've got $n(n+1)(n+2)$ for $n=2000,\ldots,2009$. You want divisors that they all share in common.
Given three consecutive integers, exactly one of them is divisible by $3$; therefore their product is divisible by $3$. That is true of all $10$ of the products, so $3$ is a common divisor. Among the ten products, there ... | {
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"url": "https://math.stackexchange.com/questions/2070076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve $\dfrac{\mathrm dx(t)}{\mathrm dt}=(2x(t)+6)(-2t^5+4t^4+2t^2)$ if $x(0)=-3$? How do I solve the following differential equation?
$$\frac{\mathrm dx(t)}{\mathrm dt}=(2x(t)+6)(-2t^5+4t^4+2t^2),\quad x(0)=-3$$
I've searched on Wolfram Alpha, and it shows: expand the $(2x(t)+6)(-2t^5+4t^4+2t^2)$, distribute $-... | $\dfrac{dx}{dt}=(2x+6)(-2t^5+4t^4+2t^2)$
$\implies \displaystyle \int \dfrac{dx}{2x+6}=\int -2t^5+4t^4+2t^2\,dt$
$\implies \dfrac{1}{2}\ln|2x+6|=-\dfrac{1}{3}t^6+\dfrac{4}{5}t^5+\dfrac{2}{3}t^3+\ln C_1$
$\implies \ln|2x+6|=-\dfrac{2}{3}t^6+\dfrac{8}{5}t^5+\dfrac{4}{3}t^3+\ln C$
$\implies \ln\left|\dfrac{2x+6}{C}\right|... | {
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"timestamp": "2023-03-29T00:00:00",
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How can I isolate $x$ in this equation? Simultaneous equations - $7^{x-1} = 49 \hspace{1 mm} (7^{y})$ --------- (i)
$\hspace{46 mm} 3^x+3^y = 84$ ---------- (ii)
The first I make x subject of equation in this way:
$\Longrightarrow 7^{x-1} = 49 \hspace{1 mm} (7^{y})$
$\Longrightarrow 7 \times 7^x = 7^2 \times 7^y$
$\Lon... | First multiply equation $(i)$ by $7$ to get rid of the confusing $-1$ in the exponent and obtain $$7^x=7^{y+3}$$
Then for the second part, note that $$3^{y+3}+3^y=3^y\left(3^3+3^0\right)$$
These two key steps should make a solution rather easy.
| {
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If $a+b+c = 6$ and $a$, $b$, $c$ are nonnegative then $a^2+b^2+c^2 \geq 12$ Let $a,b,c$ be three positive real numbers such that $a+b+c = 6$. Prove that $a^2+b^2+c^2 \geq 12$.
I tried using the AM-GM inequality to solve the same, however I wasn't able to make any considerable progress.
| We need to prove that
$$a^2+b^2+c^2\geq12\left(\frac{a+b+c}{6}\right)^2$$ or
$$\sum\limits_{cyc}(a-b)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2074513",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Diophantine Equation $x^2y^2-4x-4y=z^2$ Is it true that the Diophantine equation
$$
x^2y^2-4x-4y=z^2\; ; \; x,y>0
$$
has the only solutions $(x,y)=(2,2),(2,3),(1,5),(3,2),(5,1)$?
What is the general solution in all integers?
| Note that the perfect squares next to $x^2y^2$ are
$$x^2y^2 \pm 2xy +1.$$
Therefore if $(x , y, z)$ with $x, y > 0$ is an integral point on $z^2 = x^2y^2 - 4(x + y)$, then
$$-4(x + y) \leq 1 - 2xy.$$
It follows that any solution must be of the form $(1, y)$, $(2, y)$, $(x, 1)$, $(x, 2)$, $(x, 3)$, $(x, 4)$, $(x, 5)$ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2076952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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solving logarithmic equation $\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x}$ Solve for $x$
$$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x} $$
| Assume that $\log$ is the natural logarithm and $x > 1$ (Otherwise, it would not make sense the power in LHS). Since $\log \sqrt{x} = \dfrac{\log x}{2}$, we can simplify the equation as $$\left(\frac{\log x}{2}\right)^{(\log^2x + \log x^2 - 2)} = \log \sqrt{x} \implies \left(\frac{\log x}{2}\right)^{(\log^2x + \log x^... | {
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If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$. I have an inequality problem which is as follow:
If $a,b,c$ are positive integers, with $a^2+b^2-ab=c^2$ prove that $(a-b)(b-c)\le0$.
I am not so good in inequalities. So, please give me some hints so that I can proceed.
Thanks.
| From given equality we have $c=\sqrt {a^2+b^2-ab}$, then you can plug this in your inequality and you get $(a-b)(b-\sqrt{a^2+b^2-ab})$, now you have 2 cases to look at:
First: $a>b, a=b+x$ with this your first term is positive, so we have to show second one is negative, thus we have $b-\sqrt{b^2+bx+x^2};\sqrt{b^2+bx+x^... | {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "2",
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Prove that: $\csc 50° + \cot 100° = \cot 25° -\csc 100°$ Prove that: $$\csc 50° + \cot 100° = \cot 25° -\csc 100°$$
My Attempt:
$$L.H.S=\csc 50° + \cot 100°=\frac {1}{\sin 50°} + \frac {\cos 100°}{\sin 100°}$$
$$=\frac {1}{\sin 50°} + \frac {\cos 100°}{2\sin 50°\cos 50°}$$
$$=\frac {2\cos 50° + \cos 100°}{2\sin 50°\cos... | This would work better may be :
$$\csc 50° + \cot 100° = \cot 25° -\csc 100°$$
$$\csc 50° + \cot 100° +\csc 100° = \cot 25° $$
$$\frac{1}{\sin 50°} + \frac{\cos 100°}{\sin 100°} +\frac{1}{\sin 100°} = \cot 25° $$
$$\frac{1}{\sin 50°} + \frac{\cos 100° + 1}{\sin 100°} = \cot 25° $$
$$\frac{1}{\sin 50°} + \frac{... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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How to find $3^{3^{3^{\dots}}}\pmod{100}$? I can show that $3^{3^{3^n}}\equiv7\pmod{10}$ since
$3^1\equiv3\pmod{10}$
$3^2\equiv9\pmod{10}$
$3^3\equiv7\pmod{10}$
$3^4\equiv1\pmod{10}$
Thus, it reduces to $3^{(3^{3^n}\mod4)}$. I can then notice that
$3^1\equiv3\pmod4$
$3^2\equiv1\pmod4$
Reducing it down to $3^{(3^{(3^n\... | As $3\equiv-1\pmod4,3^{2n+1}\equiv-1\equiv3$ for any integer $n\ge0$
So, $3^{3^{3^{\cdots}}}$ can be written as $\displaystyle3^{4k+3}$ or $\displaystyle3^{3^{(4b+3)}}$
Now,$\displaystyle3^{4k+3}=27(10-1)^{2k}=27(1-10)^{2k}$
and $\displaystyle(1-10)^{2k}\equiv1-\binom{2k}110\pmod{10^2}\equiv1-20k$
So, it is sufficient ... | {
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Taylor theorical question: solving an inequality Prove by using Taylor formula that in a neighbourhood of zero this inequality holds:
$\cosh x \le 12 \frac{x^2+2}{24-x^4}$
Please explicitate everything you do :)
Thanks in advance
| The hint pretty much tells you everything you need to do: take the Taylor expansion of the difference of the two sides, until you see a non-zero term. Then check the sign of that term near $0$.
In particular, the theorem that we will use is the following:
Theorem. Let $f,g$ be two functions defined on a (pointed) neigh... | {
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"answer_id": 0
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How to integrate $\frac{x}{e^x - 1}$ w.r.t. x? A friend of mine and I wanted to solve the following indefinite integral but got stuck:
$$
\int \frac{x}{e^x - 1} dx.
$$
My approach:
Let
$$
I = \int \frac{x}{e^x - 1} dx.\\
\implies I = x\int \frac{dx}{e^x - 1} - \int \left ( \int \frac{dx}{e^x - 1} \right ) dx.
$$
Now, ... | \begin{equation}
I = \int \frac{x}{\mathrm{e}^{x}-1} dx = -I_{1} = -\int \frac{x}{1-\mathrm{e}^{x}} dx
\end{equation}
Integrate by parts
\begin{align}
I_{1} &= \int \frac{x}{1-\mathrm{e}^{x}} dx \\
\tag{1}
&= x^{2} - x \ln(1-\mathrm{e}^{x}) - \int x dx + \int \ln(1-\mathrm{e}^{x}) dx \\
\tag{2}
&= x^{2} - x \ln(1-\math... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2082941",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Solutions for $a^2+b^2+c^2=d^2$ I have to find an infinite set for $a,b,c,d$ such that $a^2+b^2+c^2=d^2$ and $a,b,c,d\in\mathbb N$
I have found one possible set which is $x,2x,2x,3x$ and $x\in\mathbb N$
But I want another infinite sets and how to reach them because I have found my solution by hit and trial .
| As you noticed, given a solution you can multiply it by any natural greater than $1$ to get another solution, so we can concentrate on solutions that do not have a common divisor. You can just keep finding solutions by hand. A more systematic approach is to write $a^2+b^2=d^2-c^2=(d+c)(d-c)$ The two terms on the rig... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2088960",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 5,
"answer_id": 2
} |
Formula for calculating moduli of roots of complex numbers from determinants - not working Determinants are used here as scaling factors for the areas of the polygons generated by connecting the individual roots of $z$.
*
*The roots of complex numbers are given by the formula $$z_k=\sqrt[n]{r}\Big(\cos\frac{\theta+2... | Let me write some words, from your statements. Maybe it works:
Suppose all vertices of a regular polygon are on a circle with radius $r$.
(1). The n-th roots of 1 as a complex number are given by the formula $$z_k=\cos\frac{\theta+2k\pi}{n}+i\sin\frac{\theta+2k\pi}{n}$$
(2). All of these roots lie on circle $|z|=r$, so... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2089076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Strategy on factoring quartic polynomials: cross method. I found a method to factorize quartic polynomials which I don't understand how it works.
It is presented like this:
Cross method
This methodology lets to factorize ordered and completed 4th
polynomials of the form:
$$F(x) = ax^4+bx^3+cx^2+dx+e$$
Rules:
... | I might be misunderstanding, but this just looks like concealed expanding $$(x^2+ax+b)(x^2+cx+d)\tag{1}$$ and comparing coefficients.
*
*Factoring extremes
Well, what we want is to choose $b$ and $d$ such that $(x^2+b)(x^2+d)$ matches leading and constant factor. What it amounts to is finding $b$ and $d$ such that $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2090587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 1
} |
Inequality with square roots : $\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}+\frac{\sqrt{a+b}}{c}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$ Let $a, b, c$ be positive real numbers.
Prove that $\frac{\sqrt{b+c}}{a}+\frac{\sqrt{c+a}}{b}+\frac{\sqrt{a+b}}{c}\geq \frac{4(a+b+c)}{\sqrt{(a+b)(b+c)(c+a)}}$
Please check if my ... | I think this inequality is not so strong. See my solution here:
https://artofproblemsolving.com/community/c6h1446900p8274628
and here:
https://artofproblemsolving.com/community/c6h1447863p8284378
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Question about inverse function $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ This was from iranian university entrance exam .Suppose $f(x)=\frac{1}{2}(x+\sqrt{x^2+4})$ find $f^{-1}(x)+f^{-1}(\frac{1}{x}),x \neq 0$.
It is easy to find $f^{-1}$ and solve this like below ...
$$y=\frac{1}{2}(x+\sqrt{x^2+4}) \\(2y-x)^2=(\sqrt{x^2+4})... | The short version:
$$f(x)f(-x)=1\implies f^{-1}(f(x))+f^{-1}\left(\frac1{f(x)}\right)=f^{-1}(f(x))+f^{-1}\left(f(-x)\right)=0.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091552",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Calculate $\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$ $$\int_0^{2\pi} \frac{1}{2+\sin(x)} \ dx$$
$$\sin(x)=\frac{e^{ix}-e^{-ix}}{2i}$$
$$z=\alpha(x)=e^{ix}$$
$$\sin(x)=\frac{z^2-1}{2zi}$$
$$2+\sin(x)=\frac{4zi+z^2-1}{2zi}$$
$$\frac{1}{2+\sin(x)}=\frac{2zi}{4zi+z^2-1}$$
$$\int_\alpha \frac{2zi}{4zi+z^2-1} \ \ \frac{1}{zi... | Yes, absolutely correct but you may try alternative approach by converting $\sin\space x$ into $\tan\space (x/2)$ $$\sin(x) = \frac{2\cdot \tan(x/2)}{1+\tan^2(x/2)}$$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091783",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Find indefinite integral $\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$ $$\int \frac{\arcsin (x)}{x^2}\, \mathrm{d}x$$
$$\frac {1}{x^2\sqrt{1-x^2}}+2\int \frac{1}{x^3\sqrt{1-x^2}}\, \mathrm{d}x$$
I try to integrate by parts method, but its doesnt want to be solved. I try to substitue $x=\sin u \mathrm{d}x=\cos u$ but fai... | $\int \frac{\arcsin(x)}{x^2}dx=\int \arcsin(x) d(-\frac{1}{x})=-\frac{1}{x}\arcsin(x)+\int \frac{dx}{x \sqrt{1-x^2}}= -\frac{1}{x}\arcsin(x)+\int \frac{dx}{x^2 \sqrt{\frac{1}{x^2}-1}}=-\frac{1}{x}\arcsin(x)+\int \frac{d(-\frac{1}{x})}{\sqrt{\frac{1}{x^2}-1}}=-\frac{1}{x}\arcsin(x)-\int \frac{dy}{\sqrt{y^2-1}}=-\frac{1}... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2091970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Prove $\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\frac{\pi ^{2}}{48}+\frac{1}{4}\ln^22$. How to prove the following series,
$$\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\frac{\pi ^{2}}{48}+\frac{1}{4}\... | Yes! As you mentioned, it's a very useful formula.
Use
$$\sum_{j=1}^{n}\frac{\left ( -1 \right )^{j-1}}{j}=\ln 2+\left ( -1 \right )^{n-1}\int_{0}^{1}\frac{x^{n}}{1+x}\, \mathrm{d}x$$
we get
\begin{align*}
&\sum_{k=1}^{\infty }\frac{\left ( -1 \right )^{k}}{k}\sum_{j=1}^{2k}\frac{\left ( -1 \right )^{j}}{j}=\ln 2\sum_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2092814",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 2,
"answer_id": 0
} |
What is the minimum value of $\dfrac {9x^2\sin^2 x+4}{x\sin x}$
Question: What is the minimum value of $$\dfrac {9x^2\sin^2x+4}{x\sin x}\tag1$$For $0<x<\pi$.
I solved this, but somewhere, I did something wrong. My work is as follows:
First, set $y=x\sin x$. Therefore,
$$\begin{align*}\dfrac {9y^2+4}{y}=\dfrac 4y+9y\... | Hint: you can show it as $$\frac { 9x^{ 2 }\sin ^{ 2 } x+4 }{ { x\sin x } } =\frac { \left( 3x\sin { x } -2 \right) ^{ 2 }+12x\sin { x } }{ x\sin x } =\frac { \left( 3x\sin { x } -2 \right) ^{ 2 } }{ x\sin x } +12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2093819",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 0
} |
Given the quadratic equation $p(x^2 +9)= -5qx$ has two equal roots, find the ratio of $p:q$. Hence, solve the quadratic equation Given the quadratic equation $p(x^2 +9)= -5qx$ has two equal roots, find the ratio of p:q. Hence, solve the quadratic equation
so this is what i got so far :
$$px^2+9p+5qx=0$$
$$(5q)^2 - 4(p)... | Answer 1 -
When we have equal roots then discriminate equals to zero. As you are solving.
Answer 2 -
We have,
5q = 6p
$\frac{p}{q} = \frac56$
Or
You can also write
$(5^2 q^2) - (6^2 p^2) = 0$
As $(5q)^2 - (6p)^2 = 0$
$(5q - 6p)(5q + 6p) = 0$
Either 5q - 6p = 0 or 5q + 6p = 0
$\frac{p}{q} = \frac56$ or $\frac{p}{q} = \f... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097159",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Integral of $\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$ I was asked to find the following integral:
$$\int\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
What I tried to replace $\sqrt{1-\frac{1}{x^2}}$ with $u$ so that: $$du=\frac{dx}{x^3\sqrt{1-\frac{1}{x^2}}} \Rightarrow du*x=\frac{dx}{x^2\sqrt{1-\frac{1}{x^2}}}$$
And:$$x=\s... | Substitute $\text{u}=\frac{1}{x}$:
$$\mathcal{I}\left(x\right)=\int\frac{1}{x^2\sqrt{1-\frac{1}{x^2}}}\space\text{d}x=-\int\frac{1}{\sqrt{1-\text{u}^2}}\space\text{d}\text{u}$$
Now, use:
$$\frac{\text{d}\arcsin\left(x\right)}{\text{d}x}=\frac{1}{\sqrt{1-x^2}}$$
So, we get:
$$\mathcal{I}\left(x\right)=\text{C}-\arcsin... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2097549",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
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Mathematica gives: $\int_{0}^{\infty}{\cos(x^n)-\cos(x^{2n})\over x}\cdot{\ln{x}}\mathrm dx={12\gamma^2-\pi^2\over 2(4n)^2}$ How do we show that the given result by Mathematica is correct?
$$\int_{0}^{\infty}{\cos(x^n)-\cos(x^{2n})\over x}\cdot{\ln{x}}\mathrm dx={12\gamma^2-\pi^2\over 2(4n)^2}\tag1$$
$n>0$
Where $\... | Define
$$\mathcal{I}=\int_{0}^{\infty }\frac{\cos x-\cos x^2}{x}\, \ln x\, \mathrm{d}x$$
and
$$\mathcal{I}\left ( \alpha \right )=\int_{0}^{\infty }x^{\alpha-1}\left ( \cos x-\cos x^{2} \right )\mathrm{d}x$$
and $\mathcal{I}\left(0\right)=-\dfrac{\gamma}{2}
$, see a proof here.
Now let's see the integral below
$$\int... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2100775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 2,
"answer_id": 1
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Prove $2^{3n}3^n | (4n)!$ Algebraically
Prove that $2^{3n}3^n | (4n)!$
I can make a combinational prove.
Consider the set $a_1a_1a_1a_1a_2a_2a_2a_2\cdots a_na_na_na_n$ this has permutations equal to
$$\frac{(4n)!}{4!^n} \\ = \frac{(4n)!}{(2^33)^n} = \frac{(4n)!}{2^{3n}3^n}$$
Since this is a integer number thus $2^{... | In the integers from $1$ to $4n$, there will be $2n$ even numbers, of which half ($n$) will be divisible by $4$.
Also in those integers, there will be at least $n$ numbers divisible by $3$.
Since $(4n)!$ multiplies all such integers together, we can see that $2^{3n}$ divides $(4n)!$, as does $3^n$. And as these two nu... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2103565",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Known that $a + b + c = 0$ and $a^3 + b^3 + c^3 = 27$. Find ABC!
Known that
$$a + b + c = 0$$
$$a^3 + b^3 + c^3 = 27$$
What is the value of $abc?$
A.) 1
B.) 0
C.) 7
D.) 8
E.) 10
My Work:
$$a + b = -c$$
$$a + c = -b$$
$$b + c = -a$$
Then
$$(a + b + c)(a + b + c)(a + b + c) = 0$$
Expanded into:
$$a^3 + b... | No, the question is wrong, and your answer is right. However, it might be easier by recalling the identity $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$
From $a+b+c=0$, we have $abc=9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2104523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 0
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How do I continue solving this polynomial division? I had this equation as a bonus question on a quiz today. I got to a certain extend and was completely unsure how to continue.
*$$(10x^4-18x^3-94x^2-8x+2) \div (10x+2)$$
I figured the easiest first step would be to divide. (Actually, I assumed it was factorable, bu... | One idea would be to guess that $10x+2$ is a factor.
We suppose
$$10x^4 - 18x^3 -94x^2 -8x +2 = (10x+2)(ax^3+bx^2+cx+d).$$
What does this tell us? We immediately know by considering the $x^4$ term, that $a=1$. By considering the $x^3$ term we have $-18 = 10b +2a = 10b+2$. This tells us that $b=-2$. The $x^2$ term tells... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105033",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Prove $\lim_{n\rightarrow \infty }\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\mathrm dt=\frac{1}{4}$ How to prove
$$\lim_{n\rightarrow \infty }\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\mathrm dt=\frac{1}{4}$$
I used $x\to \dfrac{1}{t}$ but it didn't work.
Any hint?Thank you.
| It's easy to see that when $0<t\leq 1$, we have $1-2t^2< \cos2t<1$, hence
$$\frac{1}{4t^2}-\frac{1}{2}<\frac{\cos2t}{4t^2}< \frac{1}{4t^2}$$
so
$$\frac{1}{n}\int_{\frac{1}{n}}^{1}\left ( \frac{1}{4t^2}-\frac{1}{2} \right )\mathrm{d}t<\frac{1}{n}\int_{\frac{1}{n}}^{1}\frac{\cos2t}{4t^{2}}\, \mathrm{d}t<\frac{1}{n}\int_{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2105469",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 0
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Write the quadratic function in the form $g(x)= a(x-h)^2 + k$ Write the quadratic function in the form $g(x)= a(x-h)^2 + k$
Then, give the vertex of its graph.
$g(x)= 2x^2-16x+35$
~
I tried, but still lost:
$(2x^2-16x)+35$
$2(x^2-4^2-16)+35$
$2(x^2-4^2-32+35$
| I'm not too sure what you did, one of your brackets is missing, but I believe that your error lies in $2(x^2-4^2-16)+35$.
When you complete the square for $x^2+bx$, you take half of $b$ and then square that. However, starting with $x^2+16x$, you subtracted $16$, and then "turned" the $8x$ into a $4^2$. Not possible!
In... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2106909",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
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Calculate the following indefinite integral Calculate: $$\int \frac{(2x^{2}+x+\frac{1}{2})\cos2x+(6x^{2}-7x+\frac{13}{2})\sin2x}{\sqrt{(x^{2}-x+1)^{3}}}dx.$$
I have no idea how to start it.
| $$\int \frac{(2x^{2}+x+\frac{1}{2})\cos2x+(6x^{2}-7x+\frac{13}{2})\sin2x}{\sqrt{(x^{2}-x+1)^{3}}}dx \\=\int \frac{\cos2x}{\sqrt{x^2-x+1}}\Big(2+\frac{3x-\frac{3}{2}}{x^2-x+1}\Big)+ \frac{\sin2x}{\sqrt{x^2-x+1}}\Big(6+\frac{-x+\frac{1}{2}}{x^2-x+1}\Big)dx \\=\int\Bigg(\Big(\frac{2\cos2x}{\sqrt{x^2-x+1}}+\frac{6\sin2x}{\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2110970",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
Solve $y'=-\frac{x+y-2}{x-y+4}$
$$y'=-\frac{x+y-2}{x-y+4}$$
I have used the substation:
$x=u-1$
$y=v+3$
And got $$\frac{dv}{du}=\frac{-1-\frac{v}{u}}{1-\frac{v}{u}}$$
$z=\frac{v}{u}$
And got:
$\frac{du}{u}\frac{1-z}{z^2-2z-1}$
$$u=e^{\frac{-z+2z+1}{4}+C}$$
What should I do next?
| You could approach it this way:
$$
\frac{dy}{dx} = -\frac{x + y - 2}{x - y + 4} \\
\implies x\ dy - y\ dy + 4\ dy = -x\ dx - y \ dx - 2 \ dx \\
\implies (x \ dy + y \ dx) - x \ dx - 2 \ dx - y \ dy = 0 \\
\implies \int d(xy) - \int x \ dx - 2\int dx - \int y \ dy = 0 \\
\implies x^2 + y^2 - 2xy + 4x + c = 0
$$
which is... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2112026",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$
Find $x^2+y^2$ if $x^2-\frac{2}{x}=3y^2$ and $y^2-\frac{11}{y}=3x^2$.
My try:
$$\frac{y^2-\frac{11}{y}}{3}-\frac{2}{x}=3y^2$$
then?
| Assume that $x,y\in\mathbb{R}$ and $a,b\in\mathbb{R}$ satisfy $$x^2-\frac{a}{x}=3y^2\text{ and }y^2-\frac{b}{y}=3x^2\,.$$
Then, taking $z:=x+\text{i}y$, we have
$$z^3=\left(x^3-3xy^2\right)-\text{i}\left(y^3-3x^2y\right)=a-b\text{i}\,.$$
Thus,
$$x^2+y^2=|z|^2=\sqrt[3]{\left|z^3\right|^2}=\sqrt[3]{|a-b\text{i}|^2}=\sqrt... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113062",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
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Algebraic Fractions grade 7 I am a student and I am having difficulty with answering this question. I keep getting the answer wrong. Please may I have a step by step solution to this question so that I won't have difficulties with answering these type of questions in the future.
Write as a single fraction in its simple... | The standard approach as I understand it is:
$$\begin{align}
\frac{3}{2x+1}&+\frac{8}{2x^2-7x-4} \\
&=\frac{3}{2x+1}+\frac{8}{(2x+1)(x-4)} \tag{factorise denom.}\\
&= \frac{3(x-4)}{(2x+1)(x-4)}+\frac{8}{(2x+1)(x-4)} \tag{common denom.}\\
&= \frac{3(x-4)+8}{(2x+1)(x-4)} \tag{common frac.}\\
&= \frac{3x-12+8}{(2x+1)(x-4)... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2113579",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
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Problem on Trigonometry If $ \sin x + \cos x = a$ , then find:
(i) $ \sin^6 x + \cos^6 x$
(ii) $ \lvert \sin x – \cos x \rvert $
I tried using the identity $a^3 + b^3$ for (i) but that didn't work and I ended up with an expression that was of no help. I have absolutely no idea on how to go about the problem. Any kind o... | Given $\displaystyle \sin x+\cos x = a\Rightarrow (\sin x+\cos x)^2 = a^2\Rightarrow \sin x \cos x = \frac{a^2-1}{2}$
Using $(\sin ^2 x+\cos^2 x)^3 = (\sin^6 x+\cos^6 x)+3(\sin x\cos x)^2(\sin^2 x+\cos^2 x)$
So $\displaystyle 1=\sin^6 x+\cos^6 x+3\left(\frac{a^2-1}{2}\right)$
$\displaystyle \bullet\; |\sin x-\cos x|^2... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2114711",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 0
} |
If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$ If $a^b = b^a$ and $a=2b$ then find the value of $a^2+b^2$
My Attempt,
$$a^b = b^a$$
$$a^b =b^{2b}$$
$$a^b =b^b.b^b$$.
Now, what should I do further?
| You didn't specify what are $a,b$ ... Integers ? positive real numbers ? ...
I will assume that $a,b$ are positive real numbers, such that $a^b=b^a$ and $a=2b$.
So we have $(2b)^b=b^{2b}$, which simplifies to $2^b=b^b$, hence $b=2$ and $a=4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116583",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 6,
"answer_id": 0
} |
Find the general solution of $(x + \sin{x} +\sin{y})\ \textrm{d}{x} + \cos{y}\ \textrm{d}{y}=0$ I have tried to make this equation seperable, homogeneous, exact and use integrating factors and nothing seems to be working. Could you tell me what form to use?
| if you multply through by $e^x$ you have an exact diff eq.
$(x e^x + e^x \sin x + e^x \sin y)dx + (e^x \cos y) dy = 0\\
\nabla (x e^x-e^x + \frac 12 e^x \sin x - \frac 12 e^x \cos x + e^x \sin y) = (x e^x + e^x \sin x + e^x \sin y)dx + (e^x \cos y) dy\\
x e^x-e^x + \frac 12 e^x \sin x - \frac 12 e^x \cos x + e^x \sin ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2116982",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Two consecutive numbers removed from first $n$ natural numbers If two consecutive numbers are removed from first $n$ natural numbers, and the arithmetic mean of remaining numbers is$ \frac {105}{4}$, find $n $.
Let $k $ and $k+1$ be two removed numbers, then we have :
$$ 1+2+3+\cdots +n = 2k+1+\frac{105n-210}{4}. $$
... | The sum of the first n natural number is $\frac{n(n+1)}{2}$. If the arithmetic mean of these remaining n-2 numbers is 105/4 then their sum is $\frac{(n- 2)105}{4}$. The difference in sums is the sum of those two missing numbers: $\frac{n(n+1)}{2}- \frac{105(n- 2)}{4}= \frac{2n^2+ 2n- 105n+ 210}{4}= \frac{2n^2- 103n+ ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2118140",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Any hints on how to show that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $ I am trying to prove that $$\frac{2 - \sqrt{3}}{2 \sqrt{3}} < \frac{1}{10} $$
My attempt was to suppose that $\frac{2 - \sqrt{3}}{2 \sqrt{3}} \geq \frac{1}{10}$, and show that it was absurd.
Here's what I did:
Suppose that $\frac{2 - \sqrt{... | If you multiply the numerator and denominator by $2+\sqrt3$, you get $\dfrac1{4\sqrt3+6}$. Since $\sqrt3>1$
$$4\sqrt3+6>4(1)+6$$
$$4\sqrt3+6>10$$
$$\frac1{10}>\frac1{4\sqrt3+6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2119085",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 3
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Finding the number of $4$-digit numbers so no two adjacent digits are even Find the number of $4$-digit numbers with distinct digits chosen from the set $\{0,1,2,3,4,5\}$ in which no two adjacent digits are even.
| Even digits $\{0,2,4\}$
Odd digits $\{1,3,4\}$
Numbers of the format: Even Odd Even Odd
$2 \times 3 \times 2 \times 2=24$
Numbers of the format: **Odd Even Odd Even
$3 \times 3 \times 2 \times 2=36$
Numbers of the format: Odd Odd Odd Even
$3 \times 2 \times 1 \times 3=18$
Numbers of the format: Odd Odd Even Odd ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121166",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
Evaluate$\sum_{n=0}^\infty\binom{3n}{n}x^n$ The question is:
Evaluate $$\sum_{n=0}^\infty\binom{3n}{n}x^n$$
After applying a few numbers as $x$ in Wolfram Alpha, I guess that the answer is probably:
$$2\sqrt{\frac1{4-27x}}\cos\left( \frac13\sin^{-1}\frac{3\sqrt{3x}}{2} \right)$$
that I can never prove. (Interestingly... | This is too long for a comment.
Considering $$f_k=\sum_{n=0}^\infty\binom{kn}{n}x^n$$ where $k$ is a positive integer,we could notice nice patterns using the generalized hypergeometric function
$$f_4=\, _3F_2\left(\frac{1}{4},\frac{2}{4},\frac{3}{4};\frac{1}{3},\frac{2}{3};\frac{4^4
}{3^3}x\right)$$
$$f_5=\,
_4F_... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2121911",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
"answer_count": 3,
"answer_id": 1
} |
Solve: $(y\sqrt{1-y^2})dx+(x\sqrt{1-y^2}+y)dy=0$
$$(y\sqrt{1-y^2})dx+(x\sqrt{1-y^2}+y)dy=0$$
$$\frac{\partial M}{\partial y}=\sqrt{1-y^2}-\frac{-2y^2}{2\sqrt{1-y^2}}=\frac{1-2y^2}{\sqrt{1-y^2}}$$
$$\frac{\partial N}{\partial x}= \sqrt{1-y^2}$$
$$\frac{M_{y}-N_{x}}{M}=\frac{\frac{1-2y^2}{\sqrt{1-y^2}}-\sqrt{1-y^2}}{y\... | Apart from your question you can solve this equation by this way as well
$$(y\sqrt { 1-y^{ 2 } } )dx+(x\sqrt { 1-y^{ 2 } } +y)dy=0\\ ydx+xdy+\frac { ydy }{ \sqrt { 1-{ y }^{ 2 } } } =0\\ d\left( xy \right) -\frac { 1 }{ 2 } d\left( \sqrt { 1-{ y }^{ 2 } } \right) =0\\ d\left( xy-\frac { \sqrt { 1-{ y }^{ 2 } } }{ 2 ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2122745",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Finding $u$ for $u_t+uu_x=f$ For a river flowing on a slope, the resistive force is $R = av^2$ and gravitational force, $F = bh$. The flow speed adjusts itself to $av^2=bh \implies v^2=\frac{b}{a}h$. Given the mass conservation law,
$$h_t + (hv)_x = r.$$
If $v^2=\frac{b}{a}h$,
$$v^2=\frac{b}{a}h\rightarrow (v^2)_t =\f... | $$\frac{u_t}{3}+\frac{u^{\frac{1}{2}}u_x}{6}=r\frac{b}{a} \qquad (1)$$
Supposing $a$ , $b$ and $r$ are constants.
Solving thanks to the method of characteristics :
The system of differential equations for the characteristics curves is :
$$3dt=6\frac{dx}{ u^{\frac{1}{2} }}=\frac{a}{br}du$$
From $\quad 3dt=\frac{a}{br}du... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2123728",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solve the system ${x}^{2}+ \left( y-1 \right) ^{2}=4,z^{4}+y{z}^{2}+xz+1=0. $ Find all real solutions of the system of equation
\begin{cases}
{x}^{2}+ \left( y-1 \right) ^{2}=4,\\{z}^{4}+y{z}^{2}+xz+1=0.
\end{cases}
| let $t = y-1$
$x^2+t^2 = 4$
$tz^2 + xz = -z^4 - z^2 - 1$
Solutions (x,t) exist if and only if $\frac{z^4 +z^2 +1}{\sqrt{z^4 +z^2}} \le 2$ (geometrically)
Let $k = z^4 + z^2$
$\frac{k +1}{sqrt(k)} \le 2$ equals $k^2 +2k +1 \le 4k$ equals $k^2 -2k +1 \le 0$ equals k = 1;
So, $z^4 + z^2 = 1$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2124568",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Range of a 1-2 function $$f(x)=\frac x {x^2+1}$$ I want to find range of $f(x)$ and I do like below . If someone has different Idea please Hint me .
Thanks in advanced .
This is 1-1 function $\\f(x)=\dfrac{ax+b}{cx+d}\\$,
This is 2-2 function $\\f(x)=\dfrac{ax^2+bx+c}{a'x^2+b'x+c'}\\$,
This is 1-2 function $\\f(x)=\dfr... | Method 1
$$y=\dfrac{x}{x^2+1} \to yx^2-x+y=0 \\\Delta \geq 0 \to 1-4(y)(y) \geq0 \\y^2 \leq \dfrac14 \\ y \in \left[-\dfrac12,\dfrac12\right]$$
Method 2
It is well known that$$\left|x+\frac1x\right| \geq 2 \to y=\dfrac{x}{x^2+1} \to \dfrac1y=\dfrac{x^2+1}{x}=x+\dfrac1x \\\to |\dfrac1y|\geq 2 \\|y| \leq \dfrac12 \spa... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2126509",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 0
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Cubic system of equations modulo $n$. What would be the best method to solve the systems of equations? If solving it, please show how to solve for $x$ and $y$ because that's where it's confusing. Thanks.
$5y^3+4x^2-y^2+10x-y$ $=$ $1$ $\pmod {31}$
$2y^3+13x^3+7y^2-x^2+5x-2$ $=$ $1$ $\pmod {31}$
| Systematically eliminating $y$ by starting with the cubic term and then moving on to lower powers of $y$ leads to the equation:
$$16 x^9+9 x^8+10 x^7+14 x^6+28 x^5+28 x^4+19 x^3+13 x^2+2 x+26=0 \bmod 31$$
and
$$y = \frac{15 x^3+11 x^2+29 x+9}{x^3+6 x^2+12 x+15}\bmod 31$$
where the denominator is to be interpreted as t... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2129725",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Solution of $\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $ using Mclaurin series The problem is
$$\lim_{x\to 0} \frac{(x \sqrt{1 + \sin x} - \ln{\sqrt{(1 + x^2)}-x)}}{\tan^3{x}} $$
I know that
$$\sin x = x - \frac{x^3}{3!} + \frac{x^5}{5!}... $$
got this
$$\tan x = x + \frac{x^3... | $$\ln\sqrt{1+x^2}=\dfrac{\ln(1+x^2)}2=\dfrac12\left(x^2-\dfrac{x^4}2+\dfrac{x^6}3-\cdots\right)$$
Now $1+\sin x=\left(\cos\dfrac x2+\sin\dfrac x2\right)^2$
Now $\cos\dfrac x2+\sin\dfrac x2=\sqrt2\sin\left(\dfrac x2+\dfrac \pi4\right)>0$ for $x\to0$
$\implies\sqrt{1+\sin x}=\cos\dfrac x2+\sin\dfrac x2$
and $\cos\dfrac x... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130076",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
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A question about domains and ranges: find the domain and range of the function $f(x)=\sqrt{1-x^2}$, without any digital help If $f(x)=\sqrt{1-x^2}$, then the domain of $f$ is $[-1,1]$. Respectively, the range of $f$ is $[0,1]$.
How do I understand that the domain of $f$ is $[-1,1]$ and that the range of $f$ is $[0,1]$... | One way to determine the domain and range of $f(x) = \sqrt{1 - x^2}$ is to graph the function. If $y = \sqrt{1 - x^2}$, then $y \geq 0$. Moreover, \begin{align*}
y & = \sqrt{1 - x^2}\\
y^2 & = 1 - x^2\\
x^2 + y^2 & = 1
\end{align*}
which is the equation of the unit circle, that is, the equation of the circle with cen... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 3
} |
How to find the second derivative of $f(x)=\frac{x^2}{10+x}$ How do you find the second derivative of $f(x)=\frac{x^2}{10+x}$?
I get $f'(x)={20x+x^2}/{(10+x)^2}$
Then get stuck here:
$(10+x)^2 (20+2x) - (20x+x^2) \frac {2(10+x)}{((10+x)^2)^2}$
| First make the Euclidean division:
$$x^2=(x-10)(x+10)+100, \enspace\text{whence}\quad\frac{x^2}{x+10}=x-10+\frac{100}{x+10}$$
Thus
\begin{align}
f'(x)&=1-\frac{100}{(x+10)^2},\\
f''(x)&=\frac{200}{(x+10)^3}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2130806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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What is the sum of $100$ terms of the sequence $a_{k} = \frac{1}{k} - \frac{1}{k+1}$? In the sequence $a_{1}, a_{2}, a_{3}, ..., a_{100}$, the $k$th term is defined by $$a_{k} = \frac{1}{k} - \frac{1}{k+1}$$ for all integers $k$ from $1$ through $100$. What is the sum of $100$ terms of this sequence?
The answer given ... | Such a sum is called telescop sum. We have
$$\sum_{k=1}^{100} \left( \frac{1}{k}- \frac{1}{k+1} \right) \stackrel{\star}{=}1 - \frac{1}{101}.$$
In $(\star)$ we split the sum into two and make an index shift.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2135191",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Solving a polynomial in an easier manner I need to solve the following polynomial: $$(12x-1)(6x-1)(4x-1)(3x-1)=15$$
I tried do the multiplication and ended up with an even worse expression. There are a few other questions like this one on the list of exercises I'm trying to solve, and I just wanted an easier method, wi... | Hint: write it as:
$$12\cdot 6\cdot 4 \cdot 3 \cdot \left(x-\frac{1}{12}\right)\left(x-\frac{1}{6}\right)\left(x-\frac{1}{4}\right)\left(x-\frac{1}{3}\right)=15$$
Note that the symmetric terms have equal sums $\left(x-\frac{1}{12}\right)+\left(x-\frac{1}{3}\right)=\left(x-\frac{1}{6}\right)+\left(x-\frac{1}{4}\right)$ ... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
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Prove that $\cos(\pi/7)$ is root of equation $8x^3-4x^2-4x+1=0$
Prove that $\cos\theta$ is root of equation $8x^3-4x^2-4x+1=0$, given $\theta=\frac{\pi}{7}$.
I put $\cos\theta$ in equation, but couldn't show the left-hand side to be zero.
| We can use theory of equations (and some disguised Galois theory).
Given equation can be rewritten as $(2x)^3-(2x)^2 -2( 2x)+1=0$. So we need to show $2\cos \theta $ is a root of $x^3-x^2-2x+1=0$. We will do this by showing there is a cubic equation with integer coefficients satisfied by $2\cos\theta$, and determine th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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How to solve this system of equations in Lagrange Multiplier problem Find the maximum and minimum values of ${x^{2} + y^{2} + z^{2}}$ subject to the conditions ${\frac{x^{2}}{4} + \frac{y^{2}}{5} + \frac{z^{2}}{25} = 1}$ and ${x + y - z = 0}$.
Using Lagrange multiplier method, I got following equations:
$$ {2x = \frac... | With the Lagrange multipliers
$$
L(x,y,z,\lambda,\mu) = x^2+y^2+z^2+\lambda\left(\frac{x^2}{a}+\frac{y^2}{b}+\frac{z^2}{c}-1\right)+\mu(x+y-z)
$$
The stationary points are located at
$$
\nabla L = 0 =
\left\{
\begin{array}{l}
\frac{2\lambda x}{a}+\mu +2 x \\
\frac{2 \lambda y}{b}+\mu +2 y \\
\frac{2 \lambda z}{c... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2139323",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
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Various methods to find value of $\sin 18^\circ$ To find value of $\sin 18^\circ$. Now my textbook gives a proof in which it takes $\theta=18^\circ$ and then multiply it out by 5 and write again as sum of $2\theta+3\theta$ and then taking sin on both sides forms a quadratic equation.
Are there any other elegant ways to... | $$S=\cos A+\cos3A=2\cos A\cos2A=\dfrac{(2\cos A\sin A)\cos2A}{\sin A}=\cdots=\dfrac{\sin4A}{2\sin A}$$
Now if $S=\dfrac12,\sin4A=\sin A$ with $\sin A\ne0$
either $4A=m360^\circ+A\iff A=120^\circ m$ with $3\nmid m$ as $\sin A\ne0$ $\implies A \equiv\pm120^\circ\pmod{360^\circ}$
or $4A=(2m+1)180^\circ-A\iff A=(2m+1)36^\... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2140356",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 2
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Evaluating $\sum_{r=0}^{10}\frac{\binom{10}{r}}{\binom{50}{30+r}}$ The question is to evaluate $$\sum_{r=0}^{10}\frac{\binom{10}{r}}{\binom{50}{30+r}}$$I tried rewriting the expression as $$\sum_{r=0}^{10}\frac{\binom{20-r}{10}\binom{30+r}{r}}{\binom{50}{30}\binom{20}{10}}$$ I am facing trouble on how to further simpli... | $$\begin{align}
\sum_{r=0}^{10}\frac {\displaystyle\binom {10}r}{\displaystyle\binom{50}{30+r}}
&=\frac {\color{blue}{\displaystyle\sum_{r=0}^{10}\displaystyle\binom {20-r}{10}\binom {30+r}{30}}}{\color{}{\displaystyle\binom {50}{20}\binom {20}{10}}}\\\\
&=\frac {\color{blue}{\displaystyle\binom {51}{41}}}{\color{green... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2141908",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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How to solve for $x$ algebraically, given that $x^2 = 4- \sqrt{12}$ This is a question I have been chewing on for a couple days but haven't quite solved yet. The value of $x^2$ is given as $4-\sqrt{12}$ and then the result given as $±( 1-\sqrt{3})$. How would I solve this problem algebraically without prior knowledge o... | From $x^2=4-2\sqrt3$, we can take the square root to obtain$$x=\pm\sqrt{4-2\sqrt3}\tag1$$
However, we can factor $(1)$ into$$x^2=4-2\sqrt3\iff (x-1+\sqrt3)(x+1-\sqrt3)=0\tag2$$
Which implies the roots are $x=\pm\sqrt3\mp1$. To prove $(2)$, we can employ a general formula for $\sqrt{X\pm Y}$ for $X,Y\in\mathbb{R}$ and $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2142995",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 2
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Prove: $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) > 10^{2009}$
Show that the number $ 3^{{4}^{5}} {+} 4^{{5}^{6}}$ can be expressed as the product of two integers greater than $ 10^{2009}$
By Sophie Germain:
$ 3^{{4}^{5}} {+} 4^{{5}^{6}}=\left(3^{2^9}+3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right... | $\left(3^{2^9}-3^{2^8}\cdot 2^{\frac{5^6+1}{2}}+2^{5^6}\right) = \left(2^{\frac{5^6}{2}}-3^{2^8}\right)^2 > (2^{7811}-3^{256})^2$
$2^8>3^5$ so $(2^{7811}-3^{256})^2 > (2^{7811}-2^{410})^2 >(2^{7810})^2 = 2^{15620}$
$2^{10}>10^3$ so $2^{15620}>10^{4686}>10^{2009}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2143979",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Limit of $\lim_{x\to0}\frac{f(x)-\sqrt{x+9}}{x}$? How to find this limit?
For $|f(x)-3|\le x^2$,
$$\lim_{x\to0}\frac{f(x)-\sqrt{x+9}}{x}$$
Can I make $f(x)=x^2+3$, and then
$$\lim_{x\to0}\frac{x^2+3-\sqrt{x+9}}{x}$$
Using l'Hopital's,
$$\lim_{x\to0}\frac{2x-\frac{1}{2\sqrt{x+9}}}{1} = -\frac{1}{6}$$
I'm not confident... | Note that $ -x^2 \leq (f(x) - 3) \leq x^2$, implying that $ 3-x^2 \leq f(x) \leq 3+x^2$ for all $x$. Hence, fixing $x$, we have that:
$$
\frac{3-x^2-\sqrt{x+9}}{x} \leq \frac{f(x) - \sqrt{x+9}}{x} \leq \frac{3+x^2 - \sqrt{x+9}}{x}
$$
Now, take the limit as $x \to 0$ on both sides of this inequality. Both the left and r... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2145417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
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If $a^2+b^2+c^2=1$ so $\sum\limits_{cyc}\frac{1}{(1-ab)^2}\leq\frac{27}{4}$
Let $a$, $b$ and $c$ be real numbers such that $a^2+b^2+c^2=1$. Prove that:
$$\frac{1}{(1-ab)^2}+\frac{1}{(1-ac)^2}+\frac{1}{(1-bc)^2}\leq\frac{27}{4}$$
This inequality is stronger than $\sum\limits_{cyc}\frac{1}{1-ab}\leq\frac{9}{2}$ with ... | My second proof:
Fact 1: $\frac{1}{(1-x)^2} \le \frac{729}{40}x^4 + \frac{243}{40}x^2 + \frac{27}{20}$ for all $x\in [0, 1/2]$.
(Note: $\frac{729}{40}x^4 + \frac{243}{40}x^2 + \frac{27}{20} - \frac{1}{(1-x)^2}
= \frac{(81x^4 - 108x^3 + 27x^2 - 24x + 14)(1-3x)^2}{40(1-x)^2} \ge 0$.)
Noting that $ab, bc, ca \le 1/2$, by ... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2146712",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
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$A,B$ coefficients for maximal convergence
In order to approximate the solution to the equation $f(x) = x^2-2$ we use the following iteration:
$$x_{n+1} = x_n + A\frac{x_n^2 -2}{x^n} + B \frac{x_n^2-2}{x_n^3}$$
Find $A,B$ for maximal rate of convergence.
So first we can easily see that it is a proper function si... |
By varying the coefficients $A$ and $B$, we can achieve some mathematical convenience, this is called method of undetermined coefficients.
Let $x_{n}=\sqrt{2}+e_{n}$, then
\begin{align*}
x_{n+1} &= x_{n}+A\frac{x_{n}^{2}-2}{x_{n}}+B\frac{x_{n}^{2}-2}{x_{n}^{3}} \\
&= \sqrt{2}+e_{n}+A\frac{(\sqrt{2}+e_{n})^{2}-2}{... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2147453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $a^2+b^2+c^2+d^2=4$ what is the range of $a^3+b^3+c^3+d^3$? I tried to used AM-GM but could not do.Setting any one as $2$ or $-2$ and the others $0$ we get the range as $[-8,8]$ but what is the formal way to do this?
| It's enough to prove that $a^3+b^3+c^3+d^3\leq8$
for non-negatives $a$, $b$, $c$ and $d$ such that $a^2+b^2+c^2+d^2=4$ or
it's enough to prove that $a^{\frac{3}{2}}+b^{\frac{3}{2}}+c^{\frac{3}{2}}+d^{\frac{3}{2}}\leq8$
for non-negatives $a$, $b$, $c$ and $d$ such that $a+b+c+d=4$.
Let $f(x)=x^{\frac{3}{2}}$.
Hence, $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2150951",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Sign of integral of $\frac{ 2 ^{\frac{it}{2/3}} \Gamma ( \frac{it +1}{2/3}) }{ 2 ^{\frac{it}{1.5}} \Gamma ( \frac{it +1}{1.5}) } \frac{1}{(a+it)^k}$ Can we determine the sign of the following function
\begin{align}
f(a,k)=\int_{-\infty}^\infty \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right) }{ 2 ^{... | Not quite an answer, but a good start. Let's look at
$$f(a,k)=\int_{-\infty}^\infty \frac{ 2 ^{\frac{it}{2/3}} \Gamma \left( \frac{it +1}{2/3}\right) }{ 2 ^{\frac{it}{3/2}} \Gamma \left( \frac{it +1}{3/2}\right) } \frac{1}{(a+it)^k} dt$$
which can be rewritten as
$$f(a,k)= (-i) \int_{-i\infty}^{i\infty} 2 ^{\fr... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153479",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 1,
"answer_id": 0
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Show $9 \mid a^3 + b^3 + c^3 \Rightarrow 3 \mid a \text{ or } 3 \mid b \text{ or } 3 \mid c$ Let $a, b,c \in \mathbb{Z}: 9 \mid a^3 + b^3 + c^3$
I need to show that $3 \mid a \text{ or } 3 \mid b \text{ or } 3 \mid c$
I don´t have any clue how should I approach this problem. Could someone give my hints or name some the... | If $x\not\equiv 0\pmod{3}$, then $x^3\equiv \pm 1\pmod{9}$, since
$ (3x\pm 1)^3 = 9X\pm 1$.
The sum of three numbers in $\{-1,+1\}$ cannot be $\equiv 0\pmod{9}$, hence $a^3+b^3+c^3\equiv 0\pmod{9}$ implies that at least one number among $a,b,c$ is a multiple of three.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2153564",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Evaluate integral with sin and cos $$\int _0^{3\pi }\:\frac{\mathrm{d}x}{\sin ^4\left(x\right)+\cos ^4\left(x\right)}=?$$
I have used that $\cos^2 \left(x\right)=\frac{1+\cos \left(2x\right)}{2}$ and $\sin^2 \left(x\right)=\frac{1-\cos \left(2x\right)}{2}$, so:
$$\int _0^{3\pi }\:\frac{\mathrm{d}x}{\sin ^4\left(x\right... | Through the substitution $x=\frac{t}{2}$ and the periodicity of the $\cos^2$ function we have
$$ \int_{0}^{3\pi}\frac{dx}{3+\cos(4x)} = \frac{1}{2}\int_{0}^{6\pi}\frac{dt}{3+\frac{2\cos^2(t)-1}{2}} = 6\int_{0}^{\pi/2}\frac{dt}{\frac{5}{2}+\cos^2(t)} $$
and at last is is enough to set $t=\arctan u$, so that $\cos^2\arct... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154485",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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value of trigonometric product Finding $\displaystyle \tan \left(\frac{\pi}{14}\right)\cdot \tan \left(\frac{3\pi}{14}\right)\cdot \tan \left(\frac{5\pi}{14}\right) $
assume $\displaystyle \frac{\pi}{14} = \theta$ then $\tan (14\theta) = 0$
wan,t be able to go further, some help me
| We have $\tan \frac {\pi}{14} = \cot (\frac {\pi}{2}-\frac {\pi}{14}) = \cot \frac {3\pi}{7} $. Similarly, we have, $\tan \frac {3\pi}{14} = \cot \frac {2\pi}{7} $ and $\tan \frac {5\pi}{14} = \cot \frac {\pi}{7} $.
Now, our expression simplifies to evaluating $$P = \tan \frac {\pi}{14} \tan \frac {3\pi}{14} \tan \fra... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2154605",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Prove the inequality, fractions. $\{ a,b,c \in\Bbb R_+\ \}$ If $\frac {1}{ab} +\frac {1}{bc} + \frac {1}{ac} = 3$ then prove the inequality:
$ab + bc + ac \ge 3$
How I started
*
*$ab + bc + ac \ge \frac {1}{ab} +\frac {1}{bc} + \frac {1}{ac}$
*$ab- \frac {1}{ab} + bc-\frac {1}{bc} + ac- \frac {1}{ac} \ge 0$
But n... | We need to prove that
$$ab+ac+bc\geq\frac{9}{\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}}$$ or
$$(ab+ac+bc)(a+b+c)\geq9abc$$
which is just AM-GM:
$$(ab+ac+bc)(a+b+c)\geq3\sqrt[3]{a^2b^2c^2}\cdot3\sqrt[3]{abc}=9abc$$
Or by C-S:
$$(ab+ac+bc)\left(\frac{1}{ab}+\frac{1}{ac}+\frac{1}{bc}\right)\geq(1+1+1)^2=9$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How does one show that $\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}?$ Consider
$$\int_{0}^{1}{1\over 1+\phi x^4}\cdot{\mathrm dx\over \sqrt{1-x^2}}={\pi\over 2\sqrt{2}}\tag1$$
$\phi$;Golden ratio
An attempt:
$x=\sin{y}$ then $\mathrm dx=\cos{y}\mathrm dy$
$(1)$ becomes
$$... | On the path of Aditya Narayan Sharma.
Define for $a\geq 0$,
$\displaystyle F(a)=\int_0^1 \dfrac{1}{(1+ax^4)\sqrt{1-x^2}}dx$
Perform the change of variable $y=\sin x$,
$\displaystyle F(a)=\int_0^{\tfrac{\pi}{2}} \dfrac{1}{1+a(\sin x)^4}dx$
$(\sin x)^2=\dfrac{1}{1+\dfrac{1}{(\tan x)^2}}$
Perform the change of variable $... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2157531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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find the limit: $\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=?$ fine the limit:
$$\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=?$$
my try :
$$\lim_{ n \to \infty }\sin ^2(\pi\sqrt{n^2+n})=\sin ^2(\lim_{ n \to \infty }\pi\sqrt{n^2+n})\\\sin^2(\infty)=$$Does not exist
is it right ?
| $$\begin{align}\lim_{n\rightarrow \infty}\sin^2\left(\pi\sqrt{n^2+n}\right) &=\lim_{n\rightarrow \infty}\sin^2\left(n\pi-\pi\sqrt{n^2+n}\right)\\
&=\lim_{n\rightarrow \infty}\sin^2\bigg(\frac{-\pi^2 n}{n\pi+\pi\sqrt{n^2+n}}\bigg)\end{align}$$
So $$\lim_{n\rightarrow \infty}\sin^2\left(\pi\sqrt{n^2+n}\right) = 1$$
Becau... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Problem on indefinite integration $$ f(x) = \int\frac{x^2}{(1+x^2)(1 + \sqrt{1+x^2})}dx$$
where $f(0)=0$, find $f(1)$.
Despite many tries by taking $$\sqrt{1+x^2}=t $$ or
$$ 1+\sqrt{1+x^2}$$ as $t$, I could not get it solved.
| Hint:
$$I = \int \frac{x^2}{(1+x^2)(1+\sqrt{1+x^2})} \mathrm{d}x = \frac{x^2(\sqrt{1+x^2}-1)}{(1+x^2)(1+\sqrt{1+x^2})(\sqrt{1+x^2}-1)} \mathrm{d}x =\int \frac{\sqrt{x^2+1}-1}{1+x^2}\mathrm{d}x = \int \frac{1}{\sqrt{1+x^2}}-\frac{1}{1+x^2}\mathrm{d}x$$
Hope you can take it from here.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Prove equality concerning differentiable function Let $f(x)$ be a function that is n-times differentiable. Prove the following equality(without L'Hospital):
$$(x^{n-1}\cdot f(\frac{1}{x}) )^{(n)}=\frac{(-1)^{n}}{x^{n+1}}\cdot f^{(n)}(\frac{1}{x})\\$$
So, I wanted to prove it by induction, for $n=0$ the statement is obv... | Prove by induction. Let $P_n$ be the claim for the specific value of $n$
Base Case: $$ x^{-1} \cdot f \left( \frac{1}{x} \right) = \frac{1}{x} f \left( \frac{1}{x} \right) $$
$\therefore P_0$
Inductive Step: Assume $P_k$ holds for some whole number $k$, for the sake of induction. By $P_k$ we have:
\begin{align}
\frac{d... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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How can I show that this complicated expression with square and cube roots reduces to the value 7? How would I go about reducing the complicated-looking expression $$ \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}} $$ to show that it is equal to 7?
I came across the complicated expression while calcula... |
But what if I didn't know about the cubic equation?
As long as you have a reasonable suspicion that the expression might be rational, the polynomial it satisfies can be derived rather easily. Let:
$$
x = \sqrt[3]{19\sqrt{5} + 56} + \frac{11}{\sqrt[3]{19\sqrt{5} + 56}}
$$
Then using $(a+b)^3=a^3+b^3+3ab(a+b)\,$ and $(... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2161428",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Find series representation of a function $ y=\frac{1}{1+x+x^{2}+x^{3}} $ Let $y = \frac{1}{1+x+x^{2}+x^{3}}$.
And I want to find the series representation of this function.
I've noticed that I can rewrite this like $\frac{1}{1+x+x^{2}+x^{3}}=\frac{1}{1+x}*\frac{1}{1+x^{2}}$ or I can rewite this like a sum of two fracti... | Note $$\frac{1}{1 + x + \cdots + x^{n-1}} = \frac{1-x}{1-x^n}.$$ Since we know $$\frac{1}{1-z} = 1 + z + z^2 + \cdots = \sum_{k=0}^\infty z^k, \quad |z| < 1,$$ we have $$\frac{1-x}{1-x^n} = (1-x)\sum_{k=0}^\infty (x^n)^k = (1-x)(1+x^n + x^{2n} + x^{3n} + \cdots).$$ We conclude that the desired series expansion is $$1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2164113",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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How to rewrite $2n^3+9n^2+13n+6$ as a polynomial in terms of $n+1$? I'm working on an induction proof for proving that the sum of the squares of the first n natural numbers is equal to: $$\frac{2n^3 +3n^2 + n}6$$ I understand how to get to the answer starting in the form of: $$\frac{n(n+1)(2n+1)}6$$ which just seems t... | $(n + 1)(n + 2)(2n + 3) = (n + 1)((n + 1) + 1)(2(n + 1) + 1) = ((n + 1)^2 + (n+ 1))(2(n + 1) + 1) = 2((n + 1)^2 + (n + 1))(n + 1) + (n + 1)^2 + (n + 1)= 2(n + 1)^3 + 3(n + 1)^2 + (n + 1)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2165036",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
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Improper integral $\int\limits_0^\infty x\exp (-x-x^2)\,\text dx$ Integrate $\int\limits_0^\infty
x\exp (-x-x^2)\,\text dx$
Hint: Use $\int\limits_0^\infty
\exp (-x-x^2)\,\text{d}x = 0.4965$
I don't know how to use this hint in solving the integration. Help!
| $I= \int xe^{-x-x^2}dx = \frac{1}{2} \left[ \int (2x-1+1)e^{-x-x^2}dx\right]=\frac{1}{2} \left[ \int (2x+1)e^{-x-x^2}dx - \int e^{-x-x^2}dx \right]$
Observe that $\frac{d}{dx}(-x-x^2)=-(1+2x)$
Thus the previous expression simplifies to $\frac{1}{2} \left[ -e^{-x-x^2} - \int e^{-x-x^2}dx \right]$
On applying the limit... | {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Finding maximum of $x+y$ Let x and y be real numbers satisfying $9x^{2} + 16y^{2} = 1$. Then $x + y$ is maximum when
a. $y = \frac{9x}{16}$
b. $y = −\frac{9x}{16}$
c. $y = \frac{4x}{3}$
d. $y = −\frac{4x}{3}$
| HINT:
WLOG $3x=\cos t,4y=\sin t$
$$x+y=\dfrac{\cos t}3+\dfrac{\sin t}4=\dfrac{4\sin t +3\cos t}{12}$$
Now $(4\sin t +3\cos t)^2+(3\sin t -4\cos t)^2=3^2+4^2$
$\implies (4\sin t +3\cos t)^2\le25\iff-5\le4\sin t +3\cos t\le5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2166855",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 0
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Find the equation of a circle tangent to a circle and x-axis, with center on a certain line. Find the equation of the circle, which is tangent to the circle $x^2 + y^2=4$ and $x-axis$, with center on the line $x=4$.
So far I have:$$(x-h)^2 + (y-k)^2=r^2$$
Since the center is on the line $x=4$, then the center is $(4;k)... | let $$B(xB,yB)$$ the common Point of the two circles then we have
$$xB^2+yB^2=4$$
$$(xB-4)^2+(yB-yM)^2=yM^2$$
solving this System we get
$$\frac{10-4xB)^2}{yM}=yB$$
from here we get the equation
$$xB^2+\left(\frac{10-4xB}{y_M}\right)^2=4$$
the dicriminant of this equation must be Zero, thus we have
$$yM^2(yM^2-9)=0$$ a... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171707",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 1
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Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$ I am studying the book Equations and Inequalities by Herman et al, and am stuck on the following exercise:
Determine: $S = \frac{1}{2}{n \choose 0} + \frac{1}{3}{n \choose 1} + \cdots + \frac{1}{n+2}{n \choose n}$... | Hint : Consider the function $$f(x) = \sum_{k=2}^{n+2}\frac{x^k}{k}\binom{n+2}{k}.$$
Then, compute $f'(x)$ and express it in a closed form, using the binomial expansion theorem. The value you wanna find is $f(1)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171805",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
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Stokes theorem problem involving sketch The question is Let $S$ be that part of the surface of the paraboloid $z=x^2+y^2$ between the planes $z=1$ and $z=4$. Using suitable diagrams, show how Stokes' theorem may be applied to this surface in order to write
$\iint_s \nabla \times V.\hat ndS$
as the sum of two line int... | We take the following parametrisation
$${\varphi ^{ - 1}}\left( {u,v} \right) = \left( {u,v,{u^2} + {v^2}} \right)$$
That is
$$\begin{gathered}
x = u, \hfill \\
y = v, \hfill \\
z = {u^2} + {v^2} = {x^2} + {y^2} \hfill \\
\end{gathered} $$
From the given parametrisation we get a base of tangent-vectors
like
$$\b... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2171916",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
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Factorize $x^9 - 1 $ in $\mathbb F_3[x]$
I was able to factorize $x^9-x$ in $\mathbb F_3[x]$, however, with $x^9-1$ I am not sure what to do.
My first intuition was to do $(x^3-1)(x^3+1)$ but this is clearly just the first step, if not completely misplaced.
Thanks for help!
| This is a very "hands-on" answer.
There are only three elements of $\mathbb F_3$, i.e. $0$, $1$ and $2$.
The remainder theorem tells us that if $\mathrm f(a)=0$ then $(x-a)$ is a factor of $\mathrm f(x)$.
Let's try $x=0,1,2$ and see what happens:
\begin{eqnarray*}
0^9-1 &\equiv& 2 \pmod 3 \\
1^9-1&\equiv& 0 \pmod 3 \\
... | {
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"timestamp": "2023-03-29T00:00:00",
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Does this integral converge? $\int \limits^{-1}_{-\infty} \frac{x^2dx}{(4+x^3)^3}$ I'm new with this topic and I try to solve this integral
$$\int \limits^{-1}_{-\infty} \frac{x^2dx}{(4+x^3)^3}$$
There is a break point in $\displaystyle{x= -\sqrt[3]{4}\thinspace}$
I think, that I need to divide line segment into two li... | As $x\to -\infty$, $\frac{x^2}{(4+x^3)^3}\sim \frac{1}{x^4}$ and so there is no issue at the lower limit.
Next, note that
$$x^3+4=(x+4^{1/3})(x^2-4^{1/3}x+4^{2/3})$$
Hence, as $x\to -4^{1/3}$, $\frac{x^2}{(4+x^3)^3}\sim \frac{4^{2/3}}{432(x+4^{1/3})^3}$ and the integral diverges.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2182349",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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If $abc=1$ so $\sum\limits_{cyc}\frac{7-6a}{2+a^2}\geq1$
Let $a$, $b$ and $c$ be real numbers such that $abc=1$. Prove that:
$$\frac{7-6a}{2+a^2}+\frac{7-6b}{2+b^2}+\frac{7-6c}{2+c^2}\geq1$$
The equality occurs also for $a=b=2$ and $c=\frac{1}{4}$.
This inequality is a similar to the very many contest's inequalitie... | Clearly, we only need to prove the case $a, b, c > 0$.
Let $f(x) = \frac{7-6x}{x^2+2}.$
We have $f'(x) = \frac{(6x+4)(x-3)}{(x^2+2)^2}$.
Thus, $f(x)$ is strictly decreasing on $(0, 3)$
and strictly increasing on $(3,+\infty).$
Also, since $f(x) + 1 = \frac{(x-3)^2}{x^2+2}\ge 0$, we have $f(x)\ge -1.$
WLOG, assume that ... | {
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Find all points $(x,y)$ on the graph of $f(x)$ with tangent lines passing through a certain point Find all points $(x,y)$ on the graph of $f(x) = x^2$ with tangent lines passing through the point $(3,8)$.
My attempt:
$f'(x) = x^2$
I substituted a random point into $f'(x)$ to get the gradient of the tangent, so $f'(0) =... | Let $(a, a^2)$ be a point on the graph of $f(x) = x^2$ the tangent line at which passes through the point $(3, 8)$. We know that the slope of the tangent line at $(a, a^2)$ is $$ \frac{\mathrm{d} f(x) }{\mathrm{d} x} \vert_{x=a} = 2a.$$
But the tangent line passes through the points $(a, a^2)$ and $(3, 8)$. So its slop... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2185320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Derivative of $f(x) = y^4$ $(x^3 \cdot y^4)' = ?$
$a = x^3$ , $b = y^4$
$$(x^3 \cdot y^4)'= a'\cdot b + a\cdot b' = 3x^2 \cdot y^4 + x^3 \cdot (b')$$
Not sure what the derivative of $y^4$ would be, $(4y^3)$? Since the derivative of $(x\cdot y) = 1\cdot y + x \cdot y'$ and the derivative of $y' = y'$ instead of followin... | If $y$ is a function of $x$, then $y(x)^4$ is a composite function and you need to apply the chain rule.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188188",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 0
} |
Find the maximum of the value $F=x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+\cdots+x_{n-2}x_{n-1}x_{n}+x_{n-1}x_{n}x_{1}$
Let $x_{i}\ge 0$ and such $$x_{1}+x_{2}+\cdots+x_{n}=1$$
Find the maximum of the value
$$F=x_{1}x_{2}x_{3}+x_{2}x_{3}x_{4}+\cdots+x_{n-2}x_{n-1}x_{n}+x_{n-1}x_{n}x_{1}$$
case $1$. when $n=3$,then
$$F=2x_... | I think you meant that $F=x_1x_2x_3+x_2x_3x_4+...+x_nx_1x_2$,
but I am ready to think also about your $F$.
For $n=4$ we'll prove that $abc+abd+acd+bcd\leq4$,
where $a$, $b$, $c$ and $d$ are non-negatives such that $a+b+c+d=4$,
Indeed, by AM-GM
$$abc+abd+acd+bcd=ab(c+d)+cd(a+b)\leq\left(\frac{a+b}{2}\right)^2(c+d)+\le... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2188952",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Evaluating a finite sum using a double sum I want to evaluate the sum $S_n = \sum_{k=0}^n k 3^k $.
Try:
Since $ \sum_{j=0}^{k-1} 1 = k $, then
$$ S_n = \sum_{k=0}^n \sum_{j=0}^{k-1} 3^k $$
Since we are summing over $(0 \leq k \leq n )$ and $(0 \leq j \leq k -1 )$, then we are summing over $(0 \leq j < k \leq n )$. Thus... | Small additional info. The expression
$$ S_n = \sum_{k=0}^n \sum_{j=0}^{k-1} 3^k $$
has for $k=0$ an empty inner sum
\begin{align*}
\sum_{j=0}^{-1} 3^k=0
\end{align*}
since the upper limit of the sum is smaller than the lower limit.
So, we have effectively
\begin{align*}
\sum_{k=0}^n \sum_{j=0}^{k-1} 3^k&= \sum_{k=... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2191872",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Given $x^9 = e$ and $x^{11} = e$ prove $x = e$. Full Problem: Prove that for any element $x$ in a group $G$ that satisfies
$$x^9 = e \\
x^{11} = e,$$
where $e$ is the identity element, that $x$ itself must be $e$.
Is this as simple as showing that
*
*$x^{11} = x^{9} \cdot x^{2} = e \cdot x^2 \Rightarrow x^2 = e$
*... | Your reasoning is correct but here is a more direct argument.
Since $x^9 = e$ and $x^{11} = e$, the order of $x$ divides both $9$ and $11$. Therefore, the order of $x$ is $1$ so $x = e$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2192881",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 4,
"answer_id": 3
} |
Trignometric integral : $\int \frac{dx}{\sin x + \sec x}$ The integral I am trying to compute is :
$$\int \dfrac{dx}{\sin x + \sec x}$$
I have tried manipulating trignometric functions and it took me nowhere. Then finally I tried putting $\tan\dfrac{x}{2} = t$, and subsequently:
$$=\int\dfrac{1-t^2}{(1+t^2)^2-2t(t^2-1... | As Michael Wang commented $$A=(1+t^2)^2-2t(t^2-1)=t^4-2 t^3+2 t^2+2 t+1$$ Computing the roots of the quartic equation (all are complex), you end with
$$A=\left(t^2+(\sqrt{3}-1)t-\sqrt{3}+2\right)\left(t^2-(\sqrt{3}+1) t+\sqrt{3}+2\right)$$ Partial fraction decomposition then leads to
$$\dfrac{1-t^2}{(1+t^2)^2-2t(t^2-1... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193094",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 1
} |
$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$ let $$f(x)=\frac{1}{1+2\cos x}$$
prove that :
$$f(\frac{2\pi}{7})+f(\frac{4\pi}{7})+f(\frac{6\pi}{7})=1$$
My Try :
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{2\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\cos (\frac{4\pi}{7})}$$
$$f(\frac{2\pi}{7})=\frac{1}{1+2\... | The minimal polynomial of $2\cos\frac{2\pi}{7}$ over $\mathbb{Q}$ is given by $x^3+x^2-2x-1$, and the conjugated roots are $2\cos\frac{4\pi}{7}$ and $2\cos\frac{6\pi}{7}$. It follows that $1+2\cos\frac{2\pi}{7}$, $1+2\cos\frac{4\pi}{7}$ and $1+2\cos\frac{6\pi}{7}$ are roots of
$$ p(x)= x^3-2x^2-x+1 $$
and by Vieta's th... | {
"language": "en",
"url": "https://math.stackexchange.com/questions/2193379",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 2
} |
Laurent series for $f(z) = \frac{1}{(z-4)(z+2)}$ Find the Laurent series for $f(z) = \frac{1}{(z-4)(z+2)}$, valid for $2<|z|<4$.
first i did partial fraction expansion:
$\frac{1}{(z-4)(z+2)} = \frac{A}{(z-4)} + \frac{B}{(z+2)}$
The result i got : $\frac{1}{6}[\frac{1}{(z-4)} - \frac{1}{(z+2)}]$
For the annuls centered... | You need to write the term such that the region of convergence changes:
$$
\frac{1}{z+2} = \frac{1}{z} \, \frac{1}{1 + 2z^{-1}}
$$
which can be series expanded (in powers of $1/z$) for $|2z^{-1}| < 1$, i.e. for $|z| > 2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2194019",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
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Fractions in Questions and Answers
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