Q
stringlengths 70
13.7k
| A
stringlengths 28
13.2k
| meta
dict |
|---|---|---|
Multivariable limit which should be simple ! How to calculate the following limit WITHOUT using spherical coordinates?
$$
\lim _{(x,y,z)\to (0,0,0) } \frac{x^3+y^3+z^3}{x^2+y^2+z^2}
$$
?
Thanks in advance
| Since $x^2 \le x^2+y^2+z^2$, we have $|x| \le \sqrt{x^2+y^2+z^2}$, and similarly for $|y|$ and $|z|$. This gives
$$
\left|\frac{x^3+y^3+z^3}{x^2+y^2+z^2}\right|
\le \frac{|x|^3+|y|^3+|z|^3}{x^2+y^2+z^2}
\le \frac{(x^2+y^2+z^2)^{3/2}+(x^2+y^2+z^2)^{3/2}+(x^2+y^2+z^2)^{3/2}}{x^2+y^2+z^2}
.
$$
Can you take it from there?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1282418",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
converge value of series $\sum_{n=0}^{\infty} \left( \frac{1}{n+d+1} - \frac{1}{n+5d+1} \right) $ \begin{align}
\sum_{n=0}^{\infty} \left( \frac{1}{n+d+1} - \frac{1}{n+5d+1} \right) = \sum_{n=0}^{\infty}\frac{4d}{(n+d+1)(n+5d+1)}=
?
\end{align}
I know from the $p$-test, ($i.e$ $\sum \frac{1}{n^p}$ : $p>1$ series converges)
The above series converges.
I want to know the exact value(or function) in terms of $d$.
| One may recall the following series representation of the digamma function $\displaystyle \psi : = \Gamma'/\Gamma$,
$$
\psi(u+1) = -\gamma + \sum_{n=1}^{\infty} \left( \frac{1}{n} - \frac{1}{u+n}
\right), \quad u >-1, \tag1
$$ where $\gamma$ is the Euler-Mascheroni. From $(1)$ you get
$$
\sum_{n= 1}^{N}\frac{1}{n+u}=\psi(N+u+1) -\psi(u+1) .
$$
Assume $d$ is any real number such that $d>-1/5$. You may write, for $N\geq1$,
$$
\begin{align}
\sum_{n= 1}^{N}\left(\frac{1}{n+d+1} - \frac{1}{n+5d+1}\right)&=\sum_{n= 1}^{N}\frac{1}{n+d+1} -\sum_{n= 1}^{N} \frac{1}{n+5d+1}\\\\
&=\left(\psi(N+d+2)-\psi(d+2)\right)-(\psi(N+5d+2)-\psi(5d+2))
\end{align}
$$ Then letting $N \to \infty$, using $\displaystyle \psi(M)=\log M-O(1/M)$ as $M \to +\infty$, gives
$$
\begin{align}
\sum_{n= 1}^{\infty}\left(\frac{1}{n+d+1} - \frac{1}{n+5d+1}\right)&=\psi(5d+2)-\psi(d+2).
\end{align}
$$
Many special values of $\psi$ are known, for example
$$
\begin{align}
\psi \left(\frac12\right) & = -\gamma - 2\ln 2, \\
\psi \left(\frac13\right) & = -\gamma + \frac\pi6\sqrt{3}- \frac32\ln 3.
\end{align}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1283313",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
how to evaluate the product $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$? Evaluating the infinite product of $\prod _{n=2}^\infty (1+ \frac{1}{n^2}+\frac{1}{n^4}+\frac{1}{n^6}+\cdots )$. Please Help.
| Step $1$: Notice that $$\prod_{n=2}^{\infty} \bigg(\sum_{k=0}^{\infty} \frac{1}{n^{2k}}\bigg) = \prod_{n=2}^{\infty} \frac{1}{1-\frac{1}{n^2}} = \prod_{n=2}^{\infty}\frac{n^2}{n^2-1}$$
Step $2$: Use induction on $N$ to show that $$\prod_{n=2}^N \frac{n^2}{n^2-1} = \frac{2N}{N+1}$$
Step 3: Conclude that $$\prod_{n=2}^{\infty} \frac{n^2}{n^2-1} = \lim_{N \to \infty} \frac{2N}{N+1} = 2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284512",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 0
} |
$ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $
$$ \lim_{x\to o} \frac{(1+x)^{\frac1x}-e+\frac{ex}{2}}{ex^2} $$
(can this be duplicate? I think not)
I tried it using many methods
$1.$ Solve this conventionally taking $1^\infty$ form in no luck
$2.$ Did this, expand $ {(1+x)^{\frac1x}}$ using binomial theorem got $\frac13$ then grouped coefficients of $x^0$ and it cancelled with $e$ then took coefficient of $x$ cancelled with $\frac{ex}{2}$ and so on very messy right ? at last I got $\frac13$ but that's not the expected answer!
I must have went wrong somewhere can anyone help me with this.
| We can proceed as follows
\begin{align}
L &= \lim_{x \to 0}\dfrac{(1 + x)^{1/x} - e + \dfrac{ex}{2}}{ex^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x}\right) - e + \dfrac{ex}{2}}{ex^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(\dfrac{\log(1 + x)}{x} - 1\right) - 1 + \dfrac{x}{2}}{x^{2}}\text{ (cancelling }e\text{ from Nr and Dr)}\notag\\
&= \lim_{x \to 0}\dfrac{\exp\left(-\dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right) - 1 + \dfrac{x}{2}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{1 +\left(-\dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right) + \dfrac{1}{2}\left(-\dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right)^{2} + o(x^{2}) - 1 + \dfrac{x}{2}}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{x^{2}}{3} + \dfrac{1}{2}\left(-\dfrac{x}{2} + \dfrac{x^{2}}{3} + o(x^{2})\right)^{2} + o(x^{2})}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{x^{2}}{3} + \dfrac{x^{2}}{8} + o(x^{2})}{x^{2}}\notag\\
&= \lim_{x \to 0}\dfrac{\dfrac{11x^{2}}{24} + o(x^{2})}{x^{2}}\notag\\
&= \frac{11}{24}
\end{align}
Note that using Taylor series seems to be the only option as L'Hospital will only complicate the limit expression.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1284801",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 4,
"answer_id": 2
} |
How to find cotangent? Need to find a $3\cot(x+y)$ if $\tan(x)$ and $\tan(y)$ are the solutions of $x^2-3\sqrt{5}\,x +2 = 0$.
I tried to solve this and got $3\sqrt{5}\cdot1/2$, but the answer is $-\sqrt{5}/5$
| \begin{align}
& x^2 - 3\sqrt{5}\,x+ 2 = (x-a)(x-b) \\[6pt]
= {} & x^2 - (a+b) x + ab.
\end{align}
Therefore $3\sqrt 5= a+b$ and $2=ab$.
Hence $3\sqrt 5 = \tan x + \tan y$ and $2 = \tan x\tan y$.
So
$$
\tan(x+y) = \frac{\tan x+\tan y}{1-\tan x\tan y} = \frac{a+b}{1-ab} = \frac{3\sqrt 5}{1-2},
$$
and finally,
$$
3\cot(x+y) = \frac{-1}{\sqrt 5} = \frac{-\sqrt 5}{5}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285409",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 3,
"answer_id": 2
} |
Finding the surface area of the spheroid $\frac{x^2}{3} + \frac{y^2}{3} + \frac{z^2}{4} = 1$ I'm asked to evaluate this:
What is the surface area of the surface defined by $\frac{x^2}{3} + \frac{y^2}{3} + \frac{z^2}{4} = 1$?
I first parameterized it with spherical coordinates and then I took the cross product and then the magnitude of the cross product of $r(\phi) \times r(\theta)$.
\begin{equation}
\left.
\begin{aligned}
x &= \sqrt{3} \cos\theta \sin\phi, \\
y &= \sqrt{3} \sin\theta \sin\phi, \\
z &= 2\cos\phi;
\end{aligned}\right\}\qquad
0 < \theta < 2\pi,\quad
0 < \phi < \pi.
\end{equation}
I took the partials with respect to $\phi$ and $\theta$ and arrived at
\begin{align*}
r(\theta) &= (-\sqrt{3} \sin\theta \sin\phi, \sqrt{3} \cos\theta \sin\phi, 0), \\
r(\phi) &= (\sqrt{3} \cos\theta \cos\phi, \sqrt{3} \sin\theta \cos\phi, -2 \sin\phi).
\end{align*}
Once I found the partials, I took the cross product between them and then took its magnitude to get
$$
\int \sin\phi \sqrt{12\sin^2\phi + 9\cos^2\phi}\, d\phi.
$$
Now, I've tried plenty of trig things but I just can't solve this integral. Can someone help me?
| Note that two of the axes of the ellipsoid are the same, and the third is longer. Thus it is actually a prolate spheroid. The formula for the surface area of the prolate spheroid $\frac{x^2 + y^2}{a^2} + \frac{z^2}{c^2}=1$ is:
$$ S=2\pi a^2 + 2\pi\frac{ac}{e}\sin^{-1}e $$
where $e$ is the ellipticity $\sqrt{1-\frac{a^2}{c^2}}$.
This formula is derived by considering the spheroid as a surface of revolution about the $z$-axis and doing the usual integration to find its area:
$$S = 2\pi \int_{-c}^{c}{r(z) \sqrt{1+(r'(z))^2}\,dz}$$
where $r(z) = a\sqrt{1-\frac{z^2}{c^2}}$ is the radius of the circular cross section at height $z$.
Source:
*
*Wolfram MathWorld - Prolate Spheroid
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1285879",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$
Let $a$ and $b$ be two integers. Show that $\gcd(a^2, b^2) = \gcd(a,b)^2$.
This is what I have done so far:
Let $d = \gcd(a,b)$. Then $d=ax+by$ for some $x,y$. Then $d^2 =(ax+by)^2 = a^2x^2 + 2axby+b^2y^2$.
I am trying to create a linear combination of $a^2$ and $b^2$ but do not know what to do with the middle term.
EDIT: I would be interested in seeing a proof using the Fundamental Theorem of Arithmetic as well. I simply thought it would be easiest to use Bezout's Identity.
| Suppose that $(a,b)=d$. Then Bezout's Identity says that we have some $x,y$ so that $ax+by=d$, and therefore
$$
a^3x^3+3a^2bx^2y+3ab^2xy^2+b^3y^3=d^3\tag{1}
$$
Dividing $(1)$ by $d$, remembering that both $d\mid a$ and $d\mid b$, we get
$$
a^2\left(\frac adx^3+3\frac bdx^2y\right)+b^2\left(3\frac adxy^2+\frac bdy^3\right)=d^2\tag{2}
$$
$(2)$ says that $\left.\left(a^2,b^2\right)\middle|\,d^2\right.$. Since $d\mid a$ and $d\mid b$, we also have $\left.d^2\,\middle|\left(a^2,b^2\right)\right.$.
Therefore,
$$
\left(a^2,b^2\right)=d^2=(a,b)^2
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1286904",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
"answer_count": 3,
"answer_id": 0
} |
show that $\frac{1}{F_{1}}+\frac{2}{F_{2}}+\cdots+\frac{n}{F_{n}}<13$ Let $F_{n}$ is Fibonacci number,ie.($F_{n}=F_{n-1}+F_{n-2},F_{1}=F_{2}=1$)
show that
$$\dfrac{1}{F_{1}}+\dfrac{2}{F_{2}}+\cdots+\dfrac{n}{F_{n}}<13$$
if we use
Closed-form expression
$$F_{n}=\dfrac{1}{\sqrt{5}}\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right)$$
$$\dfrac{n}{F_{n}}=\dfrac{\sqrt{5}n}{\left(\left(\dfrac{1+\sqrt{5}}{2}\right)^n-\left(\dfrac{1-\sqrt{5}}{2}\right)^n\right)}$$
Well and now I'm stuck and don't know how to proceed
| Lemma
$$F_{n}\ge\dfrac{n(n+1)(n+2)}{42},n\ge 5$$
proof:use induction, since
$$\dfrac{1}{42}[k(k+1)(k+2)+(k+1)(k+2)(k+3)]=\dfrac{1}{42}(k+1)(k+2)(2k+3)$$
and
$$(k+1)(2k+3)\ge (k+3)(k+4)$$
so
$$\dfrac{n}{F_{n}}\le \dfrac{42}{(n+1)(n+2)}$$
so
$$\sum_{k=1}^{n}\dfrac{k}{F_{k}}\le 1+2+\dfrac{3}{2}+\dfrac{4}{3}+42\left(\dfrac{1}{6}-\dfrac{1}{n+2}\right)<13$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1287613",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "14",
"answer_count": 6,
"answer_id": 2
} |
For which values of $\alpha \in \mathbb{R}$, does the series $\sum_{n=1}^\infty n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1})$ converge? How do I study for which values of $\alpha \in \mathbb{R}$ the following series converges?
(I have some troubles because of the form [$\infty - \infty$] that arises when taking the limit.)
$$\sum_{n=1}^\infty n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1})$$
| Using Taylor expansion $\sqrt{1+x} =_{x\to 0} 1 + \frac x 2 - \frac {x^2}{8} + O(x^3)$, we obtain for $n \to \infty$:
$$\begin{eqnarray}
n^\alpha(\sqrt{n+1} - 2 \sqrt{n} + \sqrt{n-1}) & = & n^\alpha \sqrt{n} \left(\sqrt{1+\frac 1 n} - 2 + \sqrt{1-\frac 1 n}\right) \\
& = & n^{\alpha + 1/2} \left(1 + \frac 1 {2n} - \frac 1 {8n^2} - 2 + 1 - \frac {1}{2n} - \frac 1 {8n^2} + O(\frac 1 {n^3}) \right) \\
& = & -n^{\alpha + 1/2} \left(\frac 1 {4n^2} + O(\frac 1 {n^3})\right) \\
& \sim_{n\to\infty} & \frac{-1}{4 n^{3/2 - \alpha}}
\end{eqnarray}$$
Therefore, the series converges if and only if $\frac 3 2 - \alpha > 1$, ie $\alpha < \frac 1 2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290025",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Solving a special Quartic Equation.
Solve for $x$
$$(x^2-4)(x^2-2x)=2$$
I have tried the Rational Root Theorem and found that there are no rational roots. Further, the polynomial $p(x)=(x^2-4)(x^2-2x)-2$ is irreducible since when I tried expanding it and writing it as a product of two quadratics, there were no integer solutions for the coefficients. I also depressed the quartic polynomial $p(x)$ hoping that the coefficient of $x$ would also vanish along with the coefficient of $x^3$, giving me a biquadratic. But that didn't happen. I also tried using substitutions, but none of them worked so far.
Any help will be appreciated.
Thanks.
| $$(x^2-4)(x^2-2x)=2$$
$$\Rightarrow x^4-2x^3-4x^2+8x-2=0$$
$$\Rightarrow (x^2-x-1)^2-3(x-1)^2=0$$
$$\Rightarrow (x^2-x-1+\sqrt 3\ (x-1))(x^2-x-1-\sqrt 3\ (x-1))=0$$
$$\Rightarrow x^2+(\sqrt 3-1)x-1-\sqrt 3=0\ \ \text{or}\ \ x^2-(\sqrt 3+1)x-1+\sqrt 3=0$$
$$\Rightarrow x=\frac{-\sqrt 3+1\pm\sqrt{8+2\sqrt 3}}{2},\frac{\sqrt 3+1\pm\sqrt{8-2\sqrt 3}}{2}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1290127",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 1,
"answer_id": 0
} |
How to prove $\int_0^1 \ln\left(\frac{1+x}{1-x}\right) \frac{dx}{x} = \frac{\pi^2}{4}$? Can anyone suggest the method of computing
$$\int_0^1 \ln\left(\frac{1+x}{1-x}\right) \frac{dx}{x} = \frac{\pi^2}{4}\quad ?$$
My trial is following
first set $t =\frac{1-x}{1+x}$ which gives $x=\frac{1-t}{t+1}$
Then
\begin{align}
dx = \frac{2}{(1+t)^2} dt, \quad [x,0,1] \rightarrow [t,1, 0]
\end{align}
[Thanks to @Alexey Burdin, i found what i do wrong in substitution]
then the integral reduces to
\begin{align}
\int^{0}_1 \ln(t)\frac{2}{1-t^2} dt
\end{align}
How one can obtain above integral?
Please post an answer if you know the answer to this integral or the other methods to evaluate above integral.
Thanks!
| Setting the convergence issue aside, I'll compute the integral formally using power series.
Note that
\begin{eqnarray}\frac{\ln\frac{1+x}{1-x}}{x}=\frac{\ln(1+x)-\ln(1-x)}{x}=\sum_{n=1}^\infty\frac{2x^{2n-2}}{2n-1}\end{eqnarray}
Integrating termwise,
\begin{eqnarray}
\int_0^1\sum_{n=1}^\infty\frac{2x^{2n-2}}{2n-1}=2\sum_{n=1}^\infty\frac{1}{(2n-1)^2}
\end{eqnarray}
Using the fact that $\displaystyle\sum_{m=1}^\infty\frac{1}{m^2}=\frac{\pi^2}{6}$ and $\sum_{n=1}^\infty\frac{1}{(2n)^2}=\frac{\pi^2}{24}$, we have the required sum is $\displaystyle2\left(\frac{\pi^2}{6}-\frac{\pi^2}{24}\right)=\frac{\pi^2}{4}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1292295",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
$3^x + 4^y = 5^z$ This is an advanced high-school problem.
Find all natural $x,y$, and $z$ such that
$3^x + 4^y = 5^z$.
The only obvious solution I can see is $x=y=z=2$. Are there any other solutions?
| Note
$$2^z\equiv (3+2)^z=5^z=3^x+4^y\equiv 0+1=1\pmod 3$$
so
$z$ is even number,let $z=2z_{1}$,then we have
$$(5^{z_{1}}+2^y)(5^{z_{1}}-2^y)=3^x$$
so we have
$$5^{z_{1}}+2^y=3^x,5^{z_{1}}-2^y=1$$
then we have
$$(-1)^{z_{1}}+(-1)^y\equiv 0\pmod 3,(-1)^{z_{1}}-(-1)^y\equiv 1\pmod 3$$
so we have $z_{1}$ is odd,$y$ is even,let $y=2y_{1}$
we have
$$(4+1)^{z_{1}}+4^{y_{1}}=(4-1)^x\Longrightarrow (-1)^x\equiv 1\pmod 4$$
so $x$ is also even.
if $y>2$, since
$$5^{z_{1}}+2^y=3^x\Longrightarrow 5\equiv 1\pmod 8$$
that's not impossible.so
$y=2\Longrightarrow z_{1}=1,z=2,x=2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1300730",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to solve the difference equation $u_n = u_{n-1} + u_{n-2}+1$ Given that:
$$
\begin{equation}
u_n=\begin{cases}
1, & \text{if $0\leq n\leq1$}\\
u_{n-1} + u_{n-2}+1, & \text{if $n>1$}
\end{cases}
\end{equation}
$$
How do you solve this difference equation?
Thanks
EDIT:
From @marwalix's answer:
$$
u_n=v_n-1
$$
$$
\begin{equation}
v_n=\begin{cases}
2, & \text{if $0\leq n\leq1$}\\
v_{n-1} + v_{n-2}, & \text{if $n>1$}
\end{cases}
\end{equation}
$$
Characteristic equation of $v_n$ is
$$
r^2=r+1
$$
Therefore,
$$
r=\frac{1\pm\sqrt{5}}{2}
$$
Therefore, the general solution for $v_n$ is
$$
v_n=A\left(\frac{1+\sqrt{5}}{2}\right)^n+B\left(\frac{1-\sqrt{5}}{2}\right)^n
$$
When $n=0$,
$$
2=A+B
$$
When $n=1$,
$$
2=A\left(\frac{1+\sqrt{5}}{2}\right)+B\left(\frac{1-\sqrt{5}}{2}\right)
$$
Therefore,
$$
A=\frac{5+\sqrt{5}}{5}
$$
$$
B=\frac{5-\sqrt{5}}{5}
$$
Therefore,
$$
u_n=\frac{5+\sqrt{5}}{5}\left(\frac{1+\sqrt{5}}{2}\right)^n+\frac{5-\sqrt{5}}{5}\left(\frac{1-\sqrt{5}}{2}\right)^n-1
$$
| Same (essentially) as marwalix ... Can you do the homogeneous equation? Can you find a particular solution to the inhomogeneous equation? Put them together to get the general solution. Then use your initial values to get your solution.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1302599",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Solve Inequality for $ |x| $ Given $$\big|\frac{(x-2)}{(x+3)}\big| < 4,$$ solve for $x.$
\ My solution
$$|x - 2| < 4|x + 3|$$
Since,
$ |x - 2| \ge |x| - |2| $ and
$ |x + 3| \le |x| + |3| $ according to triangle inequality;
$|x| - |2| < 4|x| + 4|3| $
$-14 < 3|x|$
$|x| > \frac{-14}{3}$
Is this the final answer?
| $$\left|\frac{x-2}{x+3}\right|<4$$
$$-4<\frac{x-2}{x+3}<4~~ |\cdot (x+3)^2\ne 0$$
$$-4(x+3)^2<(x-2)(x+3)<4(x+3)^2$$
$$\begin{cases}
-4(x+3)^2<(x-2)(x+3)\\
(x-2)(x+3)<4(x+3)^2
\end{cases}$$
$$\begin{cases}
0<(x+3)((x-2)+4(x+3))\\
0<(x+3)(4(x+3)-x+2)
\end{cases}$$
$$\begin{cases}
0<(x+3)(5x+10)\\
0<(x+3)(3x+14)
\end{cases}$$
$$\begin{cases}
\left[\begin{array}{}x<-3\\ x>-2\end{array}\right.\\
\left[\begin{array}{}x<-\frac{14}{3}\\ x>-3\end{array}\right.
\end{cases}$$
So the final answer is $x\in(\infty;-\frac{14}{3})\cup(-2;\infty)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1303632",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
Determine whether or not the limit exists: $\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$
Determine whether or not the limit $$\lim_{(x,y)\to(0,0)}\frac{(x+y)^2}{x^2+y^2}$$
exists. If it does, then calculate its value.
My attempt:
$$\begin{align}\lim \frac{(x+y)^2}{x^2+y^2} &= \lim \frac{x^2+y^2}{x^2+y^2} + \lim \frac {2xy}{x^2+y^2} =\\&= 1 + \lim \frac 2{xy^{-1}+yx^{-1}} = 1+ 2\cdot\lim \frac 1{xy^{-1}+yx^{-1}}\end{align}$$
But $\lim_{x\to 0^+} x^{-1} = +\infty$ and $\lim_{x\to 0^-} x^{-1} = -\infty$
Likewise, $\lim_{y\to 0^+} y^{-1} = +\infty$ and $\lim_{y\to 0^-} y^{-1} = -\infty$
So the left hand and right hand limits cannot be equal, and therefore the limit does not exist.
| If we approach $y=mx $ then $\lim_{(x,y)\to (0,0)}\frac{(x+y)^2}{x^2+y^2}
=\lim_{x\to 0}\frac{(x+xm)^2}{x^2+m^2x^2}=\lim_{x\to 0}\frac{(1+m)^2}{1+m^2}=\frac{(1+m)^2}{1+m^2} $ which is depends on $m$ so limit is not unique . So limit does not exist .
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1307149",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 6,
"answer_id": 5
} |
Solution of a recurrence equations $T(1) = 1$
$T(n) = 2T(\frac{n}{3}) + n + 1$
How do you solve this equzione recurrence? I arrived at this point and then I don't know how to proceed...
$2^kT(\frac{n}{3^k}) + \frac{2^{k-1}n}{3^{k-1}} + 2^{k-1} + \frac{2^{k-2}n}{3^{k-2}} + 2^{k-2} + \frac{2n}{3} + 2 + 1$
Thanks a lot!
| Use the Master Method:
$T(n) = 2T(\frac{n}3) + n + 1$ falls into Case 3, because $c > \log_{b}(a)$, with $c=1$, $a=2$, $b=3$, $f(n) = n+1$. Additionally $af(\frac{n}{b}) \leq kf(n)$ is trivially satisfied with $k = \frac{2}{3}$.
Thus, $T(n) = \Theta(f(n)) = \Theta(n)$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309319",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Finding $\lim\limits_{n\to\infty }\frac{1+\frac12+\frac13+\cdots+\frac1n}{1+\frac13+\frac15+\cdots+\frac1{2n+1}}$ I need to compute: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}$.
My Attempt: $\displaystyle\lim_{n\rightarrow \infty }\frac{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\cdots+\frac{1}{n}}{1+\frac{1}{3}+\frac{1}{5}+\frac{1}{7}+\cdots+\frac{1}{2n+1}}=\lim_{n\rightarrow \infty }\frac{2s}{s}=2$.
Is that ok?
Thanks.
| If the numerator is always twice the denominator, then this works provided you include a proof of that. Let's try it when $n=2$:
$$
\frac{1+\frac12}{1+\frac13+\frac15} = \frac{30+15}{30+10+6}= \frac{45}{46}\ne 2.
$$
Let's try it when $n=3$:
$$
\frac{1+\frac12+\frac13}{1+\frac 13+\frac15+\frac17}= \frac{210+105+70}{210+70+42+30} = \frac{385}{352} \ne 2.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1309606",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
"answer_count": 5,
"answer_id": 2
} |
What is the most unusual proof you know that $\sqrt{2}$ is irrational? What is the most unusual proof you know that $\sqrt{2}$ is irrational?
Here is my favorite:
Theorem: $\sqrt{2}$ is irrational.
Proof:
$3^2-2\cdot 2^2 = 1$.
(That's it)
That is a corollary of
this result:
Theorem:
If $n$ is a positive integer
and there are positive integers
$x$ and $y$ such that
$x^2-ny^2 = 1$,
then
$\sqrt{n}$ is irrational.
The proof is in two parts,
each of which
has a one line proof.
Part 1:
Lemma: If
$x^2-ny^2 = 1$,
then there are arbitrarily large integers
$u$ and $v$ such that
$u^2-nv^2 = 1$.
Proof of part 1:
Apply the identity
$(x^2+ny^2)^2-n(2xy)^2
=(x^2-ny^2)^2
$
as many times as needed.
Part 2:
Lemma: If
$x^2-ny^2 = 1$
and
$\sqrt{n} = \frac{a}{b}$
then
$x < b$.
Proof of part 2:
$1
= x^2-ny^2
= x^2-\frac{a^2}{b^2}y^2
= \frac{x^2b^2-y^2a^2}{b^2}
$
or
$b^2
= x^2b^2-y^2a^2
= (xb-ya)(xb+ya)
\ge xb+ya
> xb
$
so
$x < b$.
These two parts
are contradictory,
so
$\sqrt{n}$
must be irrational.
Two things to note about
this proof.
First,
this does not need
Lagrange's theorem
that for every
non-square positive integer $n$
there are
positive integers $x$ and $y$
such that
$x^2-ny^2 = 1$.
Second,
the key property of
positive integers needed
is that
if $n > 0$
then
$n \ge 1$.
| I just thought of this one:
Consider the equation $x^2-n=0$ for natural $n$. Evidently, $\sqrt n$ is a solution to the equation. Now the rational root theorem implies that for a root to be rational for that equation, it must be a factor of $n$ (up to sign). If $\sqrt n$ is a factor of $n$ ($n$ is a perfect square), then $\sqrt n$ is rational. If $\sqrt n$ is not a factor of $n$ ($n$ is not a perfect square), then $\sqrt n$ must be irrational.
Now just plug in $n=2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311228",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "114",
"answer_count": 19,
"answer_id": 15
} |
Proving a formula using another formula These questions are from the book "What is Mathematics":
Prove
formula 1: $$1 + 3^2 + \cdots + (2n+1)^2 = \frac{(n+1)(2n+1)(2n+3)}{3}$$
formula 2: $$1^3 + 3^3 + \cdots + (2n+1)^3 = (n+1)^2(2n^2+4n+1)$$
Using formulas 4 and 5;
formula 4: $$1^2 + 2^2 + 3^2 + \cdots + n^2 = \frac{n(n+1)(2n+1)}{6}$$
formula 5: $$1^3 + 2^3 + \cdots + n^3 = \left(\frac{n(n+1)}{2}\right)^2$$
My approach for proving the first one was to do something like a subtraction between formula 4 and 1, substituting the value of $n $ in formula 4 for $2n + 1$. I was left with a formula which I proved by mathematical induction gives me $2^2 + 4^2 + \cdots + (2n)^2$. I am wondering if this is the correct approach before doing the second proof? Is there a better and simpler way of doing this?
| I think you're basically on the right track. Notice that
$$2^2 + 4^2 + 6^2 + \cdots + (2n)^2 = 4(1^2 + 2^2 + 3^2 + ... + n^2).$$
That should make things relatively easy.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1311480",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Sum with Generating Functions Find the sum
$$\sum_{n=2}^{\infty} \frac{\binom n2}{4^n} ~~=~~ \frac{\binom 22}{16}+\frac{\binom 32}{64}+\frac{\binom 42}{256}+\cdots$$
How can I use generating functions to solve this?
| For an alternative technique, let
$$ S = \sum_{n\geq 2}\frac{\binom{n}{2}}{4^n}.$$
We have:
$$\begin{eqnarray*} 3S = 4S-S &=& \sum_{n\geq 2}\frac{\binom{n}{2}}{4^{n-1}}-\sum_{n\geq 2}\frac{\binom{n}{2}}{4^{n}}=\frac{1}{4}+\sum_{n\geq 2}\frac{\binom{n+1}{2}-\binom{n}{2}}{4^n}\\&=&\frac{1}{4}+\sum_{n\geq 2}\frac{n}{4^n}.\end{eqnarray*}$$
If we set:
$$ T = \sum_{n\geq 2}\frac{n}{4^n}$$
we have:
$$ \begin{eqnarray*}3T = 4T-T = \sum_{n\geq 2}\frac{n}{4^{n-1}}-\sum_{n\geq 2}\frac{n}{4^n}=\frac{1}{2}+\sum_{n\geq 2}\frac{1}{4^n}=\frac{7}{12}\end{eqnarray*} $$
hence $T=\frac{7}{36}$ and $3S=\frac{4}{9}$, so $\color{red}{S=\frac{4}{27}}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1313644",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Integral of Trigonometric Identities $$\int(\sin(x))^3(\cos(2x))^2dx$$
I can write $$\sin^3(x)=\sin(x)(1-\cos^2(x)=\sin(x)-\sin(x)\cos^2(x)$$
for $$\cos^2(2x)=(1-\sin^2(x))^2=1-4\sin^2(x)+4\sin^4(x)$$
after simplifying the Trig identities i get:
$$\int(sin^3(x)-4sin^5(x)+4sin^7(x))dx$$
so i need to know how to go further :)
| Note $f(x)=\sin^3 x \cos^2 2x$. You have
$$\cos 2x = 2 \cos^2 x -1$$
Hence
$$(\sin^3 x)(\cos^2 2x)=\sin x \sin^2 x (2 \cos^2 x -1)=\sin x (1- \cos^2 x) (2 \cos^2 x -1)$$
As $(\cos x)^\prime = -\sin x$, we get by expending the right hand side $$f(x)=-(\cos x)^\prime (- 2 \cos^4 x +3 \cos^2 x -1)$$ and finally
$$\int f(x)dx = \frac{2}{5} \cos^5 x - \cos^3 x + \cos x + a$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1315757",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 0
} |
Finding the partial fraction decomposition of $\frac{4s^2 - 5s + 2}{s^2(s^2 +9)}$ I am trying to find the partial fraction decomposition of $\dfrac{4s^2 - 5s + 2}{s^2(s^2 +9)}$ into something of the form $A\dfrac{1}{s} + B\dfrac{1}{s^2} + C\dfrac{1}{s^2+9} + D\dfrac{s}{s^2 + 9}$.
I am unable to factor the numerator into linear terms and how to factor the denominator into linear terms in a useful way is unclear to me since if I just break $s^2$ into $s \cdot s$ then it doesn't seem very helpful. Thank you in advance.
| $$\dfrac{4s^2 - 5s + 2}{s^2(s^2 +9)}=\frac{As+B}{s^2}+\frac{Cs+D}{s^2+9}$$
Multiply both sides by $s^2(s^2-9):$
$${4s^2 - 5s + 2}=({As+B})({s^2}+9)+{(Cs+D)}s^2$$
Combine terms:
$$4s^2-5s+2=(A+C)s^3+(B+D)s^2+(9A)s+9B$$
Solve for coefficients:
$$9B=2\implies B=\frac 29$$
$$\left(\frac 29+D\right)=4\implies D=\frac{34}9$$
$$9A=-5\implies A=-\frac{5}9$$
$$-\frac 59+C=0\implies C=\frac 59$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316289",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
with inequality $\frac{y}{xy+2y+1}+\frac{z}{yz+2z+1}+\frac{x}{zx+2x+1}\le\frac{3}{4}$ Let $x,y,z\ge 0$, show that
$$\dfrac{y}{xy+2y+1}+\dfrac{z}{yz+2z+1}+\dfrac{x}{zx+2x+1}\le\dfrac{3}{4}$$
I had solve
$$\sum_{cyc}\dfrac{y}{xy+y+1}\le 1$$
becasuse After some simple computations, it is equivalent to
$$(1-xyz)^2\ge 0$$
| Note that by CS inequality,$$\frac{4y^2}{xy^2+2y^2+y} =\frac{(y+y)^2}{(xy^2+y^2)+(y^2+y)}\le \frac{y^2}{xy^2+y^2}+\frac{y^2}{y^2+y}=\frac{1}{x+1}+\frac{y}{y+1}$$
Summing three such terms we get
$$4\sum_{cyc}\frac{y}{xy+2y+1}\le \sum_{cyc}\frac1{x+1}+\sum_{cyc}\frac{x}{x+1}=3$$
P.S. You will need to handle the case $xyz=0$ separately, shouldnt be too difficult.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1316870",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 0
} |
How to find $\lim_{x \to 0}\frac{\cos(ax)-\cos(bx) \cos(cx)}{\sin(bx) \sin(cx)}$ How to find $$\lim_\limits{x \to 0}\frac{\cos (ax)-\cos (bx) \cos(cx)}{\sin(bx) \sin(cx)}$$
I tried using L Hospital's rule but its not working!Help please!
| One can use only the following standard limits (without L'Hospital's rule nor Taylor series)
\begin{equation*}
\lim_{u\rightarrow 0}\frac{1-\cos u}{u^{2}}=\frac{1}{2}\ \ \ \ \ \ \ \ ,\
\lim_{u\rightarrow 0}\cos u=\cos 0=1\ \ \ \ \ \ \ \ \lim_{u\rightarrow 0}%
\frac{\sin u}{u}=1
\end{equation*}
To this end, transform the original expression as follows
\begin{equation*}
\frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx}=\frac{\cos ax-\cos bx\cos cx}{%
x^{2}}\frac{bx}{\sin bx}\frac{cx}{\sin cx}\frac{1}{bc}
\end{equation*}
Note that
\begin{equation*}
\cos bx\cos cx=(1-\cos bx)(1-\cos cx)+\cos cx+\cos bx-1
\end{equation*}
then
\begin{eqnarray*}
\frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx} &=&\left( \frac{\cos
ax-(1-\cos bx)(1-\cos cx)-\cos cx-\cos bx+1}{x^{2}}\right) \times \frac{bx}{\sin bx}\frac{cx}{\sin cx}\frac{1}{bc} \\
&=&\left( \frac{\cos ax-1}{(ax)^{2}}a^{2}+\frac{1-\cos bx}{(bx)^{2}}b^{2}+%
\frac{1-\cos cx}{(cx)^{2}}c^{2}+\frac{1-\cos cx}{(cx)^{2}}(\cos
bx-1)c^{2}\right) \times \frac{bx}{\sin bx} \frac{cx}{\sin cx} \frac{1}{bc}.
\end{eqnarray*}
Passing to the limit as $x$ tends to $0$ and using standard limits cited
above one gets
\begin{eqnarray*}
\lim_{x\rightarrow 0}\frac{\cos ax-\cos bx\cos cx}{\sin bx\sin cx} &=&\left(
\frac{-1}{2}a^{2}+\frac{1}{2}b^{2}+\frac{1}{2}c^{2}+\frac{1}{2}%
(1-1)c^{2}\right) \times 1\cdot 1\cdot \frac{1}{bc} \\
&=&\frac{-a^{2}+b^{2}+c^{2}}{bc}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1317215",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 5
} |
Proof by induction that $3^n - 1$ is an even number How to demonstrate that $3^n - 1$ is an even number using the principle of induction?
I tried taking that $3^k - 1$ is an even number and as a thesis I must demonstrate that $3^{k+1} - 1$ is an even number, but I can't make a logical argument.
So I think it's wrong the assumption...
Thanks
| First, $ 3^0 - 1 = 0 $ which is even so $3^n - 1 $ is even for $ n = 0 $
Suppose that
$3^k - 1 $ is even. We need to show that $3^{k+1}- 1 $ is even. Well,
$ 3^{k+1}-1 = 3(3^{k}) -1 = 3(3^k - 1) + 2 $
By our assumption, $ 3^k-1 $ is even so $ 3^{k} - 1 = 2m $, for some integer $m$
Thus, $ 3(3^k - 1) + 2 = 3(2m) + 2 = 2(3m + 1) $ Note that, $3m + 1 \in Z $ since addition and multiplication are closed in $Z$. Thus, $2(3m + 1) = 3^{k+1} -1 $ is even which is want we need to show.
Therefore, $3^n -1$ is even for all $n > 0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1319074",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 9,
"answer_id": 3
} |
Show that $30 \mid (n^9 - n)$ I am trying to show that $30 \mid (n^9 - n)$. I thought about using induction but I'm stuck at the induction step.
Base Case: $n = 1 \implies 1^ 9 - 1 = 0$ and $30 \mid 0$.
Induction Step: Assuming $30 \mid k^9 - k$ for some $k \in \mathbb{N}$, we have $(k+1)^9 - (k+1) = [9k^8 + 36k^7 + 84k^6 + 126k^5 + 126k^4 + 84k^3 + 36k^2 + 9k] - (k+1)$.
However I'm not sure where to go from here.
| Here's an alternative way of doing it:
Since $\phi(30)=8$, we get that $m^9 \equiv m \pmod{30}$ whenever $(m,30)=1$.
Since the only factors of $30$ are $2$,$3$, and $5$, we write:
$$ n=2^a \cdot 3^b \cdot 5^c \cdot m$$
Where $(m,30)=1$. Now:
$$ n^9 = 2^{9a} \cdot 3^{9b}\cdot 5^{9c} \cdot m^9 \equiv (2^9)^a \cdot (3^9)^b \cdot (5^9)^c \cdot m \pmod{30}$$
Now note:
$$2^9 \equiv 2 \pmod{30}$$
$$3^9 \equiv 3 \pmod{30}$$
$$5^9 \equiv 5 \pmod{30}$$
So therefore: $$n^9 \equiv 2^a \cdot 3^b \cdot 5^c \cdot m \equiv n \pmod{30}$$
And therefore $30 \mid n^9-n$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320452",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
trigonometric inequality bound? Suppose that one wants to determine an upper bound of the trigonometric expression
$$
a\sin(x)+b\cos(x),
$$
where $a,b\in\mathbb{R}$. My instinct is to proceed as follows:
$$
a\sin(x)+b\cos(x)\leq |a|+|b|,
$$
which is correct. If one were to drop the modulus signs, would that make the bound incorrect?
| the conservative(naive) upper bound $|a| + |b|$ is too big because $\sin x$ and $\cos x$ cannot be both max or both min simultaneously. the reason is the ever present constraint $$\sin^2 x +\cos^2 x = 1.$$
what happens is $$\begin{align}a\sin x + b \cos x &= \sqrt{a^2 + b^2}\left(\frac a{\sqrt{a^2 + b^2}}\sin x + \frac b{\sqrt{a^2 + b^2}}\cos x\right) \\
&=\sqrt{a^2+b^2}\cos(x-t), \cos t = \frac b{\sqrt{a^2 + b^2}}, \sin t = \frac a{\sqrt{a^2 + b^2}}\end{align}$$
therefore $$\lvert a\cos x + b\sin x \rvert \le \sqrt{a^2 + b^2.}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1320788",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Evaluate $\int_{\pi/6}^{\pi/2} (\frac{1}{2} \tan\frac{x}{2}+\frac{1}{4} \tan\frac{x}{4}+ \cdots+\frac{1}{2^n} \tan\frac{x}{2^n}+\cdots) dx$ $$f(x)=\frac{1}{2} \tan\frac{x}{2}+\frac{1}{4} \tan\frac{x}{4}+....+\frac{1}{2^n} \tan\frac{x}{2^n}+...$$
Check the function $f(x)$ is continuous on $[\frac{\pi}{6},\frac{\pi}{2}]$ and evaluate $\displaystyle\int_{\pi/6}^{\pi/2} f(x)$.
| $$\sum_{k=1}^{n}\dfrac{1}{2^k}\tan{\dfrac{x}{2^k}}=-\left(\sum_{k=1}^{n}\ln{\left|\cos{\dfrac{x}{2^k}}\right|}\right)'=-\left(\ln{\left|\dfrac{\sin{x}}{2^n\sin{(x/2^n)}}\right|}\right)'=\dfrac{1}{2^n}\cot{\dfrac{x}{2^n}}-\cot{x}$$
so
$$\sum_{k=1}^{+\infty}\dfrac{1}{2^k}\tan{\dfrac{x}{2^k}}=\dfrac{1}{x}-\cot{x}$$
then it is easy
$-\displaystyle\int_{\pi/6}^{\pi/2}\dfrac{1}{x}- \cot(x) = \left(\ln{x}-\log(\sin(x))\right)_{\pi/6}^{\pi/2} =\ln{3}-\log(2)$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 0
} |
square of a permutation cycle $$\sigma = \begin{bmatrix}
1 &2 &3 &4 &5 &6 &7 &8 &9 \\
1&5 &7 &4 &6 &9 &3 &2 &8
\end{bmatrix}$$
$$\sigma^{2} = \begin{bmatrix}
1 &2 &3 &4 &5 &6 &7 &8 &9 \\
1& 6&3 &4 &9 &8 &7 &5 & 2
\end{bmatrix}$$
Given $$\sigma$$ I found $$\sigma^2.$$
I want to find $$\sigma^{-2}$$ but I'm being told that my permutation cycle for $\sigma^2$ is wrong.
Very frustrating.
Please help...
Thanks in advance..
| Permutations are represented in two ways. One is a verbose description of the mapping: $$\sigma = \begin{bmatrix}
1 &2 &3 &4 &5 &6 &7 &8 &9 \\
1&5 &7 &4 &6 &9 &3 &2 &8
\end{bmatrix}$$ means that $1$ goes to $1$, $2$ goes to $5$, and so on.
We abbreviate this to cycle notation, which is more compact and also more revealing of important structure. This permutation abbreviates to $$(1)(25698)(37)(4)$$ which means the same thing as before, only more compactly. The $(25698)$ means that $2$ goes to $5$, $5$ goes to $6$, $6$ goes to $9$, $9$ goes to $8$, and $8$ goes back to $2$. Then we abbreviate further by dropping the $(1)$ and $(4)$, which can be inferred even if not written explicitly.
In the cycle notation, $\sigma^2$ is indeed written $(26859)$, as you should check. You are being asked to write $\sigma^{-2}$ in the compact cycle notation.
The inverse of a permutation $p$ is another permutation that un-does the effect of $p$. If $p$ takes $3$ to $7$, then $p^{-1}$ should take $7$ to $3$. Let's take a simple example:
$$p = \begin{bmatrix}
1 &2 &3 &4 &5 \\
4&5 &2 &1 &3
\end{bmatrix}$$
The inverse of $p$, in mapping notation, is
$$p^{-1} = \begin{bmatrix}
4&5 &2 &1 &3 \\
1 &2 &3 &4 &5 \\
\end{bmatrix}$$
because where $p$ took $2$ to $5$, $p^{-1}$ takes $5$ to $2$. As commented earlier, we have “flipped the notation upside down”. We usually arrange the columns in order:
$$p^{-1} = \begin{bmatrix}
1 &2 &3 &4 &5 \\
4 & 3& 5 & 1 & 2 \\
\end{bmatrix}$$
In cycle notation, $p$ is written $(14)(253)$. Again, this says that $2$ goes to $5$. As you observed in the comments, you can find $p^{-1}$ by writing the cycle notation for $p$ backwards: $$p^{-1} = (352)(41)$$ which we would usually write as $$p^{-1} = (14)(235).$$ Again, notice that this says that $p^{-1}$ takes $5$ to $2$, as before.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 1,
"answer_id": 0
} |
Puiseux Series? WolframAlpha says that $$\sqrt{x^2-1}$$ expanded in Puiseux series near 1 is $\sqrt 2 \sqrt{x-1}$.
I don't know what a Puiseux series is; I have searched on the net but I haven't understood much... can you briefly explain it to me and how I can obtain this result?
| If you are familiar with Taylor series, in this case you easily can get the same expansion "for free," without having to sweat too much. Set $y=x-1$, so that you are looking at
$$
\sqrt{x^2-1} = \sqrt{(x+1)(x-1)} = \sqrt{(y+2)y} = \sqrt{2y}\sqrt{1+\frac{y}{2}}
$$
when $y\to 0$ (i.e., $x\to 1$). Recalling the Taylor expansion of $t\mapsto \sqrt{1+t}$ around $0$, you get
$$
\sqrt{1+\frac{y}{2}} = 1+\frac{y}{4}-\frac{y^2}{32} + o(y^3)
$$
(I only went to order $3$, but you can go much further) so that
$$\begin{align}
\sqrt{x^2-1} &= \sqrt{2y}\left(1+\frac{y}{4}-\frac{y^2}{32} + o(y^3)\right) = \sqrt{2y}+\frac{\sqrt{2}}{4}y^{3/2}-\frac{\sqrt{2}}{32}y^{5/2} + o\!\left(y^{7/2}\right) \\
&=\sqrt{2}\sqrt{x-1}+\frac{\sqrt{2}}{4}(x-1)^{3/2}-\frac{\sqrt{2}}{32}(x-1)^{5/2} + o\!\left((x-1)^{7/2}\right).
\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1321992",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "16",
"answer_count": 3,
"answer_id": 0
} |
Simplifying sum equation. (Solving max integer encoded by n bits) Probably a lack of understanding of basic algebra. I can't get my head around why this sum to N equation simplifies to this much simpler form.
$$
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = 2^n - 2
$$
Background
To give you some background I am trying to derive $MaxInt(n) = 2^n-1$ which describes that the maximum integer which can be created using the two's complement where $n$ is the number of bits the integer is encoded by.
How two's complement encodes numbers with 4 bits is explained by the images below:
Therefore:
$$
MaxInt(n) = \sum_{i=0}^{n-2} 2^i = (2^0 + 2^1 + ... + 2^{n-3} + 2^{n-2} )
$$
Maybe there is a way of integrating this or simplifying it but I figured that this is a similar problem to sum to N where
$$
\frac{T(n) + T(n)}{2} = T(n) = \sum_{i=1}^{n} n-i+1 = \sum_{i=1}^{n} i
$$
So following this logic $MaxInt(n)$ is also equal to:
$$
MaxInt(n) = \frac{MaxInt(n) + MaxInt(n)}{2}
$$
Since
$$
(2^0 + 2^1 + ... + 2^{n-3} + 2^{n-2}) = (2^{n-2} + 2^{n-3} + ... 2^2 + 2^1 + 2^0)
$$
Then
$$
MaxInt(n) = \sum_{i=0}^{n-2} 2^{n-2-i} = (2^{n-2} + 2^{n-3} + ... 2^2 + 2^1 + 2^0)
$$
Putting it all together:
$$
MaxInt(n) = \frac{\sum_{i=0}^{n-2} 2^{n-2-i} + \sum_{i=0}^{n-2} 2^i}{2}
$$
$$
MaxInt(n) = \frac{\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i}{2}
$$
Which is when I got stuck, cheating with wolfrom alpha I found that
$$
\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = 2^n - 2
$$
But I don't know why. If you see a better alternative way (i.e. not using sum to N method) of deriving $MaxInt(n) = 2^n-1$ please let me know.
Thanks for reading.
| First notice that for any $m$, $\sum_{i=0}^{m-1} 2^i = 2^m-1$. To see this
(and really this is the geometric series proof applied to your situation), let $$S=\sum_{i=0}^{m-1} 2^i=1+2+4+8+\cdots+2^{m-1}.$$ Then \begin{eqnarray*} 2S&=&2+4+8+\cdots+2^{m-1}+2^m\\ &=& (-1+1)+2+4+\cdots+2^{m-1}+2^m\\&=& -1+(1+2+4+\cdots+2^{m-1})+2^m\\ &=& -1 + S + 2^m.\end{eqnarray*}
So $2S=-1+S+2^m$ and so $S=2^m-1$.
Now your sum is $$\sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i = \sum_{i=0}^{n-2} 2^{(n-2)-i} + \sum_{i=0}^{n-2} 2^i.$$
But $$\sum_{i=0}^{n-2} 2^{(n-2)-i} = \sum_{i=0}^{n-2} 2^i.$$ Here, the left-side is the sum $2^{n-2}+2^{n-3}+\cdots + 2+1$ and the right-side just expresses this sum in the reverse order $1+2+\cdots+2^{n-3}+2^{n-2}$.
So putting this together you have \begin{eqnarray*} \sum_{i=0}^{n-2} 2^{-i+n-2} + 2^i &=& \sum_{i=0}^{n-2} 2^{(n-2)-i} + \sum_{i=0}^{n-2} 2^i\\ &=& \sum_{i=0}^{n-2} 2^{i} + \sum_{i=0}^{n-2} 2^i\\ &=&2\sum_{i=0}^{n-2} 2^i \\ &=& 2(2^{n-1}-1) \\ &=& 2^{n}-2\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1323931",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Summation of series in powers of x with certain combinations as coefficients How can I find the sum: $$\sum_{k=0}^{n} \binom{n-k}{k}x^{k}$$
Edit: The answer to this question is: $$\frac{{(1+\sqrt{1+4x})}^{n+1}-{(1-\sqrt{1+4x})}^{n+1}}{2^{n+1}\sqrt{1+4x}}$$ I don't know how to arrive at this answer.
| This answer is similar to this answer and this answer. That is, compute the generating function of the given sequence:
$$
\begin{align}
&\sum_{n=0}^\infty\sum_{k=0}^n\binom{n-k}{k}x^ky^n\\
&=\sum_{k=0}^\infty\sum_{n=k}^\infty\binom{n-k}{k}x^ky^n\tag{1}\\
&=\sum_{k=0}^\infty\sum_{n=0}^\infty\binom{n}{k}x^ky^{n+k}\tag{2}\\
&=\sum_{k=0}^\infty\frac{(xy^2)^k}{(1-y)^{k+1}}\tag{3}\\
&=\frac1{1-y}\frac1{1-\frac{xy^2}{1-y}}\tag{4}\\
&=\frac1{1-y-xy^2}\tag{5}\\
&=\frac1{\left(1-\frac{1-\sqrt{1+4x}}2y\right)\left(1-\frac{1+\sqrt{1+4x}}2y\right)}\tag{6}\\
&=\frac{\frac{-1+\sqrt{1+4x}}{2\sqrt{1+4x}}}{1-\frac{1-\sqrt{1+4x}}2y}+\frac{\frac{1+\sqrt{1+4x}}{2\sqrt{1+4x}}}{1-\frac{1+\sqrt{1+4x}}2y}\tag{7}\\
&=\frac1{\sqrt{1+4x}}\sum_{n=0}^\infty\left[\left(\frac{1+\sqrt{1+4x}}2\right)^{n+1}-\left(\frac{1-\sqrt{1+4x}}2\right)^{n+1}\right]y^n\tag{8}
\end{align}
$$
Explanation:
$(1)$: change order of summation
$(2)$: substitute $n\mapsto n+k$
$(3)$: $\sum\limits_{n\ge k}\binom{n}{k}y^n=\frac{y^k}{(1-y)^{k+1}}$
$(4)$: $\sum\limits_{k\ge 0}x^k=\frac1{1-x}$
$(5)$: simplify
$(6)$: quadratic formula
$(7)$: partial fractions
$(8)$: $\sum\limits_{n\ge 0}x^n=\frac1{1-x}$
Equating coefficients of $y^n$ yields
$$
\sum_{k=0}^n\binom{n-k}{k}x^k
=\frac1{\sqrt{1+4x}}\left[\left(\frac{1+\sqrt{1+4x}}2\right)^{n+1}-\left(\frac{1-\sqrt{1+4x}}2\right)^{n+1}\right]
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1324616",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 3,
"answer_id": 0
} |
What does $\sum_{k=0}^\infty \frac{k}{2^k}$ converge to? This problem comes from another equation on another question (this one).
I tried to split it in half but I found out that
$$\sum_{k=0}^\infty \frac{k}{2^k}$$
can't be divided.
Knowing that $$\sum_{k=0}^\infty x^k=\frac{1}{1-x}$$
I wrote that
$$\sum_{k=0}^\infty \frac{k}{2^k}=\sum_{k=0}^\infty \left(\frac{\sqrt[k] k}{2}\right)^k=\frac{1}{1-\frac{\sqrt k}{2}}=\frac{2}{2-\sqrt[k] k}$$
But that's not what I wanted. Could anyone help me?
| You can also view it this way, which is quite intuitive.
\begin{align*}
\sum_{k=0}^\infty \frac{k}{2^k}
&= \frac{1}{2} + \frac{2}{4} + \frac{3}{8} + \frac{4}{16} + \frac{5}{32} + \cdots \\
&= \frac12 + \Bigl(\frac14 + \frac14\Bigr) + \Bigl(\frac18 + \frac18 + \frac18\Bigr) + \Bigl(\frac1{16} + \frac1{16} + \frac1{16} + \frac1{16}\Bigr) + \cdots \\
&= \Bigl(\frac12 + \frac14 + \frac18 + \frac1{16} + \cdots\Bigr) + \Bigl(\frac14 + \frac18 + \frac1{16} + \cdots\Bigr) + \Bigl(\frac18 + \frac1{16} + \cdots\Bigr) + \cdots \\
&= 1 + \frac12 + \frac14 + \frac18 + \cdots \\
&= 2.
\end{align*}
We are able to rearrange the terms of the infinite series in this way because the series is absolutely convergent. It is important that we meet this condition, as without this we would not be able to reorder the terms of the series and group them in this way.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325254",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "18",
"answer_count": 3,
"answer_id": 2
} |
How prove $\frac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$ Let $a,b,c\in R$,and such $ab+bc+ac=0,a+b+c\neq 0$
show that
$$\dfrac{(a-b)^4+(b-c)^4+(c-a)^4}{(a+b+c)^4}=2$$
| WLOG, assume $a+b+c=1 \Rightarrow (a-b)^2+(b-c)^2+(c-a)^2 = 2(a^2+b^2+c^2)=2$, since $1=(a+b+c)^2 = a^2+b^2+c^2$, and also $2((a-b)(b-c))^2 = 2(ab-ac-b^2+bc)^2=2(-2ac-b^2)^2=2(b^4+4acb^2+4a^2c^2)$. Thus:$\displaystyle \sum_{cyclic}2((a-b)(b-c))^2=2(a^4+b^4+c^4)+8abc(a+b+c)+8(a^2b^2+b^2c^2+c^2a^2)=2(1-2(a^2b^2+b^2c^2+c^2a^2))+8abc+8(a^2b^2+b^2c^2+c^2a^2)=2+4(a^2b^2+b^2c^2+c^2a^2)+8abc=8abc+2+4(0-2abc(a+b+c))=2\Rightarrow \displaystyle \sum_{cyclic} (a-b)^4 =\left(\displaystyle \sum_{cyclic} (a-b)^2\right)^2-\displaystyle \sum_{cyclic} 2((a-b)(b-c))^2= 2^2-2 = 4-2=2$. Done.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1325944",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
If $x ^ 6 = x$, prove that $x ^ 2 = x$, in a ring I found a short and interesting problem:
Given a ring $(R, +, \cdot)$ and knowing that $x ^ 6 = x\ (\forall x\in R)$, prove that $x ^ 2 = x\ (\forall x \in R)$.
While it is short, I cannot figure out how to solve it. If it would be the reverse, then the solution were simple: $(x ^ 2) ^ 3 = x$.
Given this information, can be the problem be solved? If so, which is the simplest way to solve it?
| Consider the case where the ring has a unit (if not, then one could consider $R$ as a $\mathbb{Z}$-algebra, but the details would change in that case).
Observe $2^6=2$ and $3^6=3$. In other words, $64=2$ and $729=3$. So $62=0$ and $726=0$. Since $\gcd(62,726)=2$, it follows that $2=0$ (by repeated subtraction).
Therefore, we have a ring where $2=0$. Now, consider $(x+1)^6=(x+1)$. The LHS expands as
$$
x^6+6x^5+15x^4+20x^3+15x^2+6x+1=x+1.
$$
Simplifying the even coefficients, it follows that
$$
x^6+x^4+x^2+1=x+1.
$$
Since $x^6=x$, we know that $x^4+x^2=0$ or that $x^4=x^2$. Since $x^4=x^2$, by multiplying by $x^2$, we have $x^6=x^4$ so $x^6=x^2$, but since $x^6=x$, $x=x^2$.
(There are a few places in this calculation, where one should be careful to make sure that I'm not cheating, but the ideas should work, at least in the case where $R$ has a unit.)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1326023",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
"answer_id": 0
} |
Proving “The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.” This site
states:
Example $\boldsymbol 3$. The sum of consecutive cubes. Prove this remarkable fact of arithmetic: $$1^3 +2^3 +3^3 +\ldots +n^3 =(1 +2 +3 +\ldots +n)^2.$$
“The sum of $n$ consecutive cubes is equal to the square of the sum of the first $n$ numbers.”
In other words, according to Example $1$: $$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}.$$
Should:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^2 (n+1)^2}{4}$$
not be:
$$1^3 +2^3 +3^3 +\ldots +n^3 = \frac{n^3 +(n + 1)^3}{2^3}$$
as everything in the left-hand side is cubed?
| No! Generally speaking, one shows by induction that $\,1^r+2^r+\dots+n^r\,$ has a closed form which is a polynomial in $n$ of degree $\color{red}{r+1}$.
Examples:
*
*$1 +2 +\dots+n =\dfrac{n(n+1)}2$.
*$1^2+2^2+\dots+n^2=\dfrac{n(n+1)(2n+1)}6$
*$1^0+2^0+\dots+n^0=\underbrace{1+1++\dots+1}_{n \ \text{times}}=n$
and the formula you posted about. What you propose hasn't the required degree, so it can't be true.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1328798",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 5,
"answer_id": 0
} |
Rotation of a line by a matrix
Give the equation of the line $\ell'$ that is obtained by rotating $\ell$: $x+2y=5$ by an angle of $\theta=\frac{1}{2}\pi$ with center point $O(0,0)$.
The rotation matrix is $\left.\begin{pmatrix}\cos \alpha & -\sin \alpha \\ \sin \alpha & \cos \alpha\end{pmatrix}\right|_{\alpha=\frac{1}{2}\pi}=\begin{pmatrix}0 & -1 \\ 1&0\end{pmatrix}$. Two points on $\ell$ are $(1,2)$ and $(3,1)$, whose images are respectively $\begin{pmatrix}0 & -1 \\ 1&0\end{pmatrix}\begin{pmatrix}1\\2\end{pmatrix}=\begin{pmatrix}-2\\1\end{pmatrix}$ and $\begin{pmatrix}0 & -1 \\ 1&0\end{pmatrix}\begin{pmatrix}3\\1\end{pmatrix}=\begin{pmatrix}-1\\3\end{pmatrix}$.
Hence, the gradient of $\ell'$ is $\dfrac{\Delta y}{\Delta x}=2$. Therefore, the equation of $\ell'$ is $y=2(x+2)+1\iff \boxed{y=2x+5}$.
I have two questions: my book says the answer should be $y=2x+6$. If I am wrong, what did I wrong? Second, is there anyone who suggests a more elegant of faster method for this problem?
| A shorter way:
Given line is $x+2y=5$, so we could say $x=r$ and $y=\dfrac{5-r}{2}$.
Now we multiply $\begin{pmatrix} 0&-1\\1&0\\ \end{pmatrix}\begin{pmatrix} r\\ \dfrac{5-r}{2}\\ \end{pmatrix}$, it gives us coordinates after rotation which are
$\begin{pmatrix} \dfrac{r-5}{2}\\ r\\ \end{pmatrix}$. namely, the new $x=\dfrac{r-5}{2}$ and the new $y=r$, so the line after rotation is $y=2x+5$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1329193",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
$\frac{(-1)^n}{2\cdot 4\cdot \cdot\cdot2n}=\frac{(-1)^n}{2^n\cdot n!}$ $$\frac{(-1)^n}{2\cdot 4\cdot \cdot\cdot2n}=\frac{(-1)^n}{2^n\cdot n!}$$
$$\frac{(-1)^n}{3\cdot 5\cdot \cdot \cdot(2n+1)}=\frac{{(-2)^n} \cdot n! }{(2n+1)!}$$
can anyone tell me if these are true or false?
| Yes, both are true.
For the first, note that
$$ 2 \cdot 4 \dots 2n = (1 \cdot 2) \cdot (2 \cdot 2) \cdot (3 \cdot 2) \cdot (4 \cdot 2) \dots (n \cdot 2) = n! \cdot 2^n$$
For the second:
$$3 \cdot 5 \dots (2n + 1) = \frac{1 \cdot 2 \cdot 3 \cdot 4 \cdot 5 \dots (2n + 1)}{2 \cdot 4 \dots 2n} = \frac{(2n+1)!}{n! \cdot 2^n}$$
and
$$2^n \cdot (-1)^n = (-2)^n$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1330229",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Quick way to solve the system $\displaystyle \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} = \frac{65}{36}$, $xy-x+y=118$. Consider the system
$$\begin{aligned} \left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = \frac{65}{36}, \\ xy -x +y & = 118. \end{aligned}$$
I have solved it by performing the substitutions $x-y=u$ and $xy=v$. Then I multiplied the first equation by $6^u$ and used $a^2-b^2=(a+b)(a-b)$ to find
$$(3^u+2^u)(3^u-2^u) = 65 \cdot 6^{u-2}.$$
By inspection I found $u=2$ and $v=120$. I solved the original system in $x,y$ and got the answers. Is there another quicker way to solve this without resorting to this sort of ninja inspection? I have found a second solution by solving $a^u +1/a^u = 65/36$, which assures $u=2$ but takes much more time.
Could there be a third way faster than these?
| Let $u=x-y$ and $a=\frac23$
\begin{align}
\left( \frac{3}{2} \right)^{x-y} - \left( \frac{2}{3} \right)^{x-y} & = a^{-u} - a^{u}\\
&=e^{-u\log a}-e^{u \log a}\\
&=-2\sinh (u\log a)
\end{align}
therefore \begin{align}
\log a^u &=\text{arcsinh} \frac{-36}{72}\\
&=\log\Big(-\frac{36}{72}+\sqrt{1+(-\frac{36}{72})^2}\Big)\\
&=\log(\frac49)
\end{align}
hence $$(\frac23)^{x-y}=(\frac23)^2$$ results in $x-y=2$. The rest is as others did.
Note: $\text{arcsinh}z=\log(z+\sqrt{1+z^2})$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331134",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 3
} |
Find the cubic equation of $x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$ Find the cubic equation which has a root $$x=\sqrt[3]{2-\sqrt{3}}+\sqrt[3]{2+\sqrt{3}}$$
My attempt is
$$x^3=2-\sqrt{3}+3\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)+2-\sqrt{3}$$
$$x^3=4+\left(\sqrt[3]{(2-\sqrt{3})^2}\right)\left(\sqrt[3]{(2+\sqrt{3})}\right)+3\left(\sqrt[3]{(2-\sqrt{3})}\right)\left(\sqrt[3]{(2+\sqrt{3})^2}\right)$$
then what I will do??
| The general solution to a cubic:
$$\begin{align}
r & = \sqrt[3]{
\underbrace{ \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right) }_{p}
+ \sqrt[2]{
\left(\underbrace{ \frac{c}{3a} - \frac{b^2}{9a^2} }_{q} \right)^3 + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2
}
} \\
& + \sqrt[3]{
\left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)
- \sqrt[2]{
\left( \frac{c}{3a} - \frac{b^2}{9a^2} \right)^3 + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2
}
} \\
& - \underbrace{ \frac{b}{3a} }_{r}
\end{align}$$
The problem gives $p = 2$, $q^3 + p^2= 3$, $r = 0$.
$$\begin{cases}
0 = \frac{b}{3a} \\
2 = \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \\
3 = \left( \frac{c}{3a} - \frac{b^2}{9a^2} \right)^3 + \left( \frac{-b^3}{27a^3} + \frac{bc}{6a^2} - \frac{d}{2a} \right)^2 \\
\end{cases}$$
Since polynomials have the same roots up to a scaling factor (that is, $f(x)$ has the same roots as $k~f(x)$), we can choose $a = 1$. And solve and get $b = 0$.
$$\begin{cases}
a = 1 \\
b = 0 \\
2 = -\frac{d}{2} \\
3 = \left( \frac{c}{3} \right)^3 + \left( \frac{d}{2} \right)^2 \\
\end{cases}$$
$$\begin{cases}
a = 1 \\
b = 0 \\
3 = \left( \frac{c}{3} \right)^3 + 4 \\
d = -4 \\
\end{cases}$$
$$\begin{cases}
a = 1 \\
b = 0 \\
c = -3 \\
d = -4 \\
\end{cases}$$
$$x^3 - 3x - 4 = 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331417",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 8,
"answer_id": 7
} |
Substituting the value $x=2+\sqrt{3}$ into $x^2 + 1/x^2$ My teacher gave me a question which I am not able to solve:
If $x=2+\sqrt{3}$ then find the value of $x^2 + 1/x^2$
I tried to substitute the value of x in the expression, but that comes out to be very big.
| By brute-force:
$$(2+\sqrt3)^2+\frac1{(2+\sqrt3)^2}=\frac{(2+\sqrt3)^4+1}{(2+\sqrt3)^2}.$$
Then
$$(2+\sqrt3)^2=7+4\sqrt3,\\
(2+\sqrt3)^4=(7+4\sqrt3)^2=97+56\sqrt3,$$
$$\frac{(2+\sqrt3)^4+1}{(2+\sqrt3)^2}=\frac{98+56\sqrt3}{7+4\sqrt3}=\frac{14\cdot7+14\cdot4\sqrt3}{7+4\sqrt3}=14.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1331489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 8,
"answer_id": 6
} |
Finding the Jacobian of a system of 1st-order ODEs I am trying to find the Jacobian matrix of the following system of 1st-order ODEs.
My system is: $\dfrac{dx}{dt} = \left(x-3\right)\!\left(y+x\right) \\ \dfrac{dy}{dt} = \left(x+4\right)\!\left(y-2x\right)$
Since $(x-3)(y+x) = xy+x^2 - 3y - 3x$ and $(x+4)(y-2x) = xy-2x^2+4y-8x$, I get a matrix like this after taking the partial-derivatives. However, it's apparently incorrect and I'm not understanding where I went wrong. Any help is appreciated.
$\left[\begin{array}{cc}
y+2x-3 &x-2\cr
y-4x-8 &x+4\cr
\end{array}\right]$
| Note that
$\dfrac{\partial (xy + x^2 - 3x - 3y)}{\partial y}$
$ = x - 3, \tag{1}$
not $x- 2$; the correct matrix is thus
$\begin{bmatrix} y + 2x -3 & x - 3 \\ y - 4x - 8 & x + 4 \end{bmatrix}. \tag{2}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1332311",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding $ \lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x} $ I'm kind of stuck on this problem, I could use a hint.
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}
$$
After some algebra, I get
$$
{\lim_{x\to 2}\frac{x+2 - 2x}{x(x-2)-\sqrt{x+2}+\sqrt{2x}}}
$$
EDIT above should be:
$$
\lim_{x\to 2}\frac{x+2 - 2x}{x(x-2)(\sqrt{x+2}+\sqrt{2x)}}
$$
I'm stuck at this point, any help would be greatly appreciated.
| You can notice that the denominator is $x(x-2)$, you have
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}=
\left(\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x-2}\right)
\left(\lim_{x\to 2}\frac{1}{x}\right)
$$
provided the limit exists. Thus we can concentrate on
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x-2}
$$
which is the derivative at $2$ of the function
$$
f(x)=\sqrt{x+2}-\sqrt{2x}
$$
Since
$$
f'(x)=\frac{1}{2\sqrt{x+2}}-\frac{1}{\sqrt{2x}}
$$
we have
$$
f'(2)=\frac{1}{4}-\frac{1}{2}=-\frac{1}{4}
$$
Thus your limit is
$$
\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x^2-2x}=
\left(\lim_{x\to 2}\frac{\sqrt{x+2}-\sqrt{2x}}{x-2}\right)
\left(\lim_{x\to 2}\frac{1}{x}\right)=
\left(-\frac{1}{4}\right)\left(\frac{1}{2}\right)=-\frac{1}{8}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1333110",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Evaluate $\int^0_1 \frac{\ln(t)}{1-t^2}dt$
Evaluate: $$\int^0_1 \dfrac{\ln(t)}{1-t^2}dt$$
This actually came up while solving another integral. It was suggested that I use a binomial series, but unfortunately I do not understand how to use this. Can anyone help me out?
| First consider the operation
\begin{align}
\partial_{n} \, t^{n} = \frac{d}{dn} \, e^{n \ln(t)} = \ln(t) \, e^{n \ln(t)} = t^{n} \, \ln(t).
\end{align}
Now consider the integral, where the operation just presented will be used,
\begin{align}
I_{n} = \int_{0}^{1} \ln(t) \, t^{n} \, dt = \partial_{n} \, \int_{0}^{1} t^{n} \, dt = \partial_{n} \left[ \frac{t^{n+1}}{n+1} \right]_{0}^{1} = \partial_{n} \left(\frac{1}{n+1}\right) = - \frac{1}{(n+1)^{2}}
\end{align}
Now letting $n \to 2n$ and then summing over $n$ it is seen that:
\begin{align}
\sum_{n=0}^{\infty} I_{2n} = \int_{0}^{1} \frac{\ln(t) \, dt}{1-t^{2}} &= - \sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} \\
&= - \left(\sum_{n=0}^{\infty} \frac{1}{(2n+1)^{2}} + \sum_{n=1}^{\infty} \frac{1}{(2n)^{2}} \right) + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{2}} \\
&= - \sum_{n=1}^{\infty} \frac{1}{n^{2}} + \frac{1}{4} \sum_{n=1}^{\infty} \frac{1}{n^{2}} = - \zeta(2) + \frac{1}{4} \, \zeta(2) \\
&= - \frac{3}{4} \zeta(2) = - \frac{\pi^{2}}{8}.
\end{align}
The integral desired is:
\begin{align}
\int_{0}^{1} \frac{\ln(t) \, dt}{1-t^{2}} = - \frac{\pi^{2}}{8}.
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1334808",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 0
} |
Derivative of a lemniscate at the left hand side How does $${d \over{dx}}(3(x^2+y^2)^2)$$ turn into $$12y(x^2+y^2){{dy} \over{dx}}+12x(x^2+y^2)$$? I'm having a hard time solving it algebraically without it turning into a huge polynomial.
| Use the chain rule.
$$3\left({d\over{dx}}(x^2+y^2)^2\right)=3\cdot2{d\over{dx}}(x^2+y^2)(x^2+y^2)=6\left(2x+2{d\over{dx}}(y)y\right)(x^2+y^2)=6\left(2x+2y{dy\over{dx}}\right)(x^2+y^2)=12(x^2+y^2)\left(y{dy\over{dx}}+x\right)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335731",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
Solve the system of equations with $x=y$ Solve the system of equations: $\left\{\begin{array}{l}\sqrt{x^2+(y-2)(x-y)}+\sqrt{xy}=2y\\\sqrt{xy+x+5}-\dfrac{6x-5}{4}=\dfrac{1}{4}\left(\sqrt{2y+1}-2\right)^2\end{array}\right.$
I used wolframalpha.com and got the only solution: $(x;y)=(4;4)$
And I guess that we can get $x=y$ from first equation.
And this is my try
We have $\sqrt{x^2+(y-2)(x-y)}-y+\sqrt{xy}-y=0$
$\Leftrightarrow (x-y)\left(\dfrac{x+2y-2}{\sqrt{x^2+(y-2)(x-y)}+y}+\dfrac{y}{\sqrt{xy}+y}\right)=0$
But I can't prove that $\dfrac{x+2y-2}{\sqrt{x^2+(y-2)(x-y)}+y}+\dfrac{y}{\sqrt{xy}+y}\ne0$.
So who can help me?
| If we allow the non-principal square root, you can actually have more than one solution to the system,
$$\left\{\begin{array}{l}\sqrt{x^2+(y-2)(x-y)}\color{red}\pm\sqrt{xy}=2y\\\sqrt{xy+x+5}-\dfrac{6x-5}{4}=\dfrac{1}{4}\left(\sqrt{2y+1}-2\right)^2\end{array}\right.$$
The positive case is solved by $x=y = 4$, but the negative case can be solved by,
$$x=2.771913334036360207047659037932620797468\dots\\ y=0.4518271741040497609916929241427121641268\dots$$
which are roots of $12$-deg equations and can be found by using resultants.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1335998",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Finding the basis of a subset of polynomials
Let $W$ be a subspace of the polynomials with maximum degree of $3$ and $p(1) = p(2) = 0$. Find the basis and the dimension of the subspace. The field is the real numbers.
My attempt:
My first thought is that it is the usual basis of $\mathbb{P}_3 = \{1, x, x^2, x^3\}$. However, on second thought, is it a trick question? We will never have $p(1) = p(2)$, will we? So would the basis be the empty set and the dimension be $0$?
| Note that $p(x)=ax^3+bx^2+cx+d$ satisfies $p(1)=p(2)=0$ if and only if
\begin{array}{rcrcrcrcrcrcrc}
a&+&b&+&c&+&d&=&0 \\
8\,a&+&4\,b&+&2\,c&+&d&=&0
\end{array}
But
$$
\DeclareMathOperator{rref}{rref}\rref
\begin{bmatrix}
1&1&1&1&0\\8&4&2&1&0
\end{bmatrix}
=
\begin{bmatrix}
1&0&-1/2&-3/4&0\\ 0&1&3/2&7/4&0
\end{bmatrix}
$$
so
\begin{align*}
p(x)
&= ax^3+bx^2+cx+d \\
&= \left(\frac{1}{2}c+\frac{3}{4}d\right)x^3+\left(-\frac{3}{2}c-\frac{7}{4}d\right)x^2+cx+d \\
&= c\left(\frac{1}{2}x^3-\frac{3}{2}x^2+x\right)+d\left(\frac{3}{4}x^3-\frac{7}{4}x^2+1\right)
\end{align*}
This proves that
$$
\left\{\frac{1}{2}x^3-\frac{3}{2}x^2+x,\frac{3}{4}x^3-\frac{7}{4}x^2+1\right\}
$$
form a basis for $W$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1337160",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
How to integrate $\int^{\infty}_{-\infty} e^{-2\pi^2/x^2} dx$? I am wondering how can i integrate this quantity above?
Here it is again,
$$\int^{\infty}_{-\infty} e^{-2\pi^2/x^2}dx.$$
Thanks a lot.
| Since the function is even then
\begin{align}
I = \int_{ - \infty }^\infty {\exp \left( { - \frac{{2\pi ^2 }}{{x^2 }}} \right)dx} = 2\int_0^\infty {\exp \left( { - \frac{{2\pi ^2 }}{{x^2 }}} \right)dx}
\end{align}
Setting
\begin{align}
x = \frac{{\sqrt 2 }}{{\sqrt u }}\pi ,\,\,\,\text{i.e.,}\,\,\,u = \frac{{2\pi ^2 }}{{x^2 }} \Rightarrow du = - \frac{{4\pi ^2 }}{{x^3 }}dx \Rightarrow dx &= - \frac{1}{{4\pi ^2 }}\left( {\frac{{\sqrt 8 }}{{\sqrt {u^3 } }}\pi ^3 } \right)du \\
dx &= - \frac{\pi }{{\sqrt 2 \sqrt {u^3 } }}du,
\end{align}
$x = 0,u \to \infty ;x = \infty ,u = 0$
\begin{align}
I = 2\int_\infty ^0 {\exp \left( { - u} \right)\frac{{ - \pi }}{{\sqrt 2 \sqrt {u^3 } }}du} = \pi \sqrt 2 \int_0^\infty {u^{-3/2} \exp \left( { - u} \right)du}
\end{align}
But since $ \exp \left( { - u} \right) \ge 1-u$ then $ \int_0^\infty {u^{-3/2} \exp \left( { - u} \right)du} \ge \int_0^\infty {(1-u)u^{-3/2} du} $ and this is diverge
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1338795",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Finding the following definite and indefinite integrals I want to calculate the integral
$$\int_0^{\frac{\pi}{2}} e ^{ \sin t}\, dt.$$
Can we find a primitive function for $f(t) = e ^{\sin t}$?
| For $\int e^{\sin t}~dt$ ,
$\int e^{\sin t}~dt$
$=\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n}t}{(2n)!}dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$
$=\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n)!}\right)dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$
For $n$ is any natural number,
$\int\sin^{2n}t~dt=\dfrac{(2n)!t}{4^n(n!)^2}-\sum\limits_{k=1}^n\dfrac{(2n)!((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}+C$
This result can be done by successive integration by parts.
For $n$ is any non-negative integer,
$\int\sin^{2n+1}t~dt$
$=-\int\sin^{2n}t~d(\cos t)$
$=-\int(1-\cos^2t)^n~d(\cos t)$
$=-\int\sum\limits_{k=0}^nC_k^n(-1)^k\cos^{2k}t~d(\cos t)$
$=-\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{k!(n-k)!(2k+1)}+C$
$\therefore\int\left(1+\sum\limits_{n=1}^\infty\dfrac{\sin^{2n}t}{(2n)!}\right)dt+\int\sum\limits_{n=0}^\infty\dfrac{\sin^{2n+1}t}{(2n+1)!}dt$
$=t+\sum\limits_{n=1}^\infty\dfrac{t}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+1)!k!(n-k)!(2k+1)}+C$
$=\sum\limits_{n=0}^\infty\dfrac{t}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+1)!k!(n-k)!(2k+1)}+C$
$\therefore$ For $\int_0^\frac{\pi}{2}e^{\sin t}~dt$ ,
$\int_0^\frac{\pi}{2}e^{\sin t}~dt$
$=\left[\sum\limits_{n=0}^\infty\dfrac{t}{4^n(n!)^2}-\sum\limits_{n=1}^\infty\sum\limits_{k=1}^n\dfrac{((k-1)!)^2\sin^{2k-1}t\cos t}{4^{n-k+1}(n!)^2(2k-1)!}-\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!\cos^{2k+1}t}{(2n+1)!k!(n-k)!(2k+1)}\right]_0^\frac{\pi}{2}$
$=\sum\limits_{n=0}^\infty\dfrac{\pi}{2^{2n+1}(n!)^2}+\sum\limits_{n=0}^\infty\sum\limits_{k=0}^n\dfrac{(-1)^kn!}{(2n+1)!k!(n-k)!(2k+1)}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343075",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 0
} |
$ \lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\frac{1}{\sqrt{n}}$
Knowing that :
$$I_n=\int_0^{\frac{\pi}{2}}\cos^n(t) \, dt$$
$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1$$
$$I_{n}\sim \sqrt{\dfrac{\pi}{2n}}$$
Calculate:
$$
\lim_{n\to+\infty} \frac{1\times 3\times \ldots \times (2n+1)}{2\times 4\times \ldots\times 2n}\times\dfrac{1}{\sqrt{n}}$$
Indeed,
$$I_{2n}=\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}\times\dfrac{\pi}{2}\quad \forall n\geq 1 \\
\frac{1\times 3\times \ldots \times (2n-1)}{2\times 4\times \ldots\times 2n}=\dfrac{2}{\pi}\times I_{2n}$$
then
$$
\begin{align*}
\frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&=\dfrac{2}{\pi}\times I_{2n}\times (2n+1)\times\dfrac{1}{\sqrt{n}}\\
&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{\sqrt{n}} \times \sqrt{ \dfrac{2n\times I^{2}_{2n}}{2n} } \\
&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ 2n\times I^{2}_{2n} } \\
\end{align*}
$$
or $$(2n)I^{2}_{2n}\sim \dfrac{\pi}{2}$$
then
\begin{align*}
\frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&=\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ 2n\times I^{2}_{2n} } \\
&\sim\dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ \dfrac{\pi}{2} } \\
\end{align*}
$$
I'm stuck here
i think that i can go ahead
\begin{align*}
\frac{1\times 3\times \cdots \times (2n+1)}{2\times 4\times \cdots\times 2n}\times\dfrac{1}{\sqrt{n}}&\sim \dfrac{2}{\pi}\times (2n+1)\times\dfrac{1}{ \sqrt{2}\times n} \times \sqrt{ \dfrac{\pi}{2} } \\
&\sim \dfrac{2}{\pi}\times\dfrac{2n}{ \sqrt{2}\times n} \times \sqrt{ \dfrac{\pi}{2} } \\
&\sim \frac{ 2\sqrt{\pi} }{\pi}=\dfrac{2}{\sqrt{\pi}}
\end{align*}
am i right ? if that so is there any other way
| So we know that:
$$ I_{2n}=\int_{0}^{\pi/2}\cos^{2n}(\theta)\,d\theta = \frac{\pi}{2}\cdot\frac{(2n-1)!!}{(2n)!!}=\frac{\pi}{2}\cdot\frac{1}{4^n}\binom{2n}{n}.\tag{1}$$
If, in the same way, we prove:
$$ I_{2n+1}=\int_{0}^{\pi/2}\cos^{2n+1}(\theta)\,d\theta = \frac{(2n)!!}{(2n+1)!!}\tag{2} $$
then we have:
$$\lim_{n\to +\infty} n\,I_{2n}\, I_{2n+1} = \lim_{n\to +\infty}\frac{\pi n}{4n+2}=\frac{\pi}{4}.\tag{3}$$
Since $\{I_n\}_{n\in\mathbb{N}}$ is a decreasing sequence, we have $\frac{I_{2n+1}}{I_{2n}}\leq 1$ as well as:
$$ \frac{I_{2n+1}}{I_{2n}}\geq \frac{I_{2n+2}}{I_{2n}} = \frac{2n+1}{2n+2}\geq 1-\frac{1}{2n}\tag{4}$$
hence, by squeezing:
$$ \lim_{n\to +\infty} n I_{2n}^2 = \frac{\pi}{4}, \tag{5}$$
then:
$$ \lim_{n\to +\infty}\frac{(2n+1)!!}{(2n)!!}\cdot\frac{1}{\sqrt{n}}=\color{red}{\frac{2}{\sqrt{\pi}}}.\tag{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343290",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 2,
"answer_id": 0
} |
Find the equation of base of Isosceles Traingle Given the two Legs $AB$ and $AC$ of an Isosceles Traingle as $7x-y=3$ and $x-y+3=0$ Respectively. if area of $\Delta ABC$ is $5$ Square units, Find the Equation of the base $BC$
My Try:
The coordinates of $A$ is $(1,4)$. Let the Slope of $BC$ is $m$. Since angle between $AB$ and $BC$ is same as Angle between $AC$ and $BC$ we have
$$\left|\frac{m-7}{1+7m}\right|=\left|\frac{m-1}{1+m}\right|$$ solving which we get $m=2$ or $m=\frac{-1}{2}$
Can any one help me further
| The problem can be solved using parametric equations. Since you know that both equations pass through the point, $(1,4)^T$, convenient parameterizations for the lines are:
$\left(\begin{array}{c}x_1\\y_1\end{array}\right) = \left(\begin{array}{c}1\\7\end{array}\right)s+\left(\begin{array}{c}1\\4\end{array}\right)$
and
$\left(\begin{array}{c}x_2\\y_2\end{array}\right) = \left(\begin{array}{c}1\\1\end{array}\right)t+\left(\begin{array}{c}1\\4\end{array}\right)$
The lengths of the lines from the point $(1,4)^T$ must be equal and are given by
$\left|\left|\left(\begin{array}{c}x_1\\y_1\end{array}\right)-\left(\begin{array}{c}1\\4\end{array}\right)\right|\right| = \left|\left|\left(\begin{array}{c}x_2\\y_2\end{array}\right)-\left(\begin{array}{c}1\\4\end{array}\right)\right|\right|$
or
$\left|\left|\left(\begin{array}{c}1\\7\end{array}\right)s\right|\right| = \left|\left|\left(\begin{array}{c}1\\1\end{array}\right)t\right|\right|$
which reduces to $5s = \pm t$.
The area gives another equation. The area of a parallelogram is given as the cross product of two vectors that meet at one corner, so the area of the triangle is half that much.
$\left|\left|\left[\left(\begin{array}{c}x_1\\y_1\end{array}\right)-\left(\begin{array}{c}1\\4\end{array}\right)\right]\times\left[\left(\begin{array}{c}x_2\\y_2\end{array}\right)-\left(\begin{array}{c}1\\4\end{array}\right)\right]\right|\right|=\left|\left|\left(\begin{array}{c}1\\7\end{array}\right)s\times\left(\begin{array}{c}1\\1\end{array}\right)t\right|\right|=|st-7ts|=|6st|$
Therefore, $\frac{1}{2}6st=\pm 5$ or $st=\pm\frac{5}{3}$. Combining the equations yields $s = \pm\frac{1}{\sqrt{3}}$, $t=\pm\frac{5}{\sqrt{3}}$.
Finally, the equation of the line defining the third edge can be given as a parametric equation passing through the points $(x_1,y_1)^T$ and $(x_2,y_2)^T$.
$\left(\begin{array}{c}x\\y\end{array}\right) = \left[\left(\begin{array}{c}x_2\\y_2\end{array}\right)-\left(\begin{array}{c}x_1\\y_1\end{array}\right)\right]u+\left(\begin{array}{c}x_1\\y_1\end{array}\right)=\left[\left(\begin{array}{c}1\\7\end{array}\right)s-\left(\begin{array}{c}1\\1\end{array}\right)t\right]u+\left(\begin{array}{c}1\\7\end{array}\right)s+\left(\begin{array}{c}1\\4\end{array}\right)$
Substituting in the values of $s$ and $t$ found above yields the equation. Note that there are four solutions.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343573",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the value of this series what is the value of this series $$\sum_{n=1}^\infty \frac{n^2}{2^n} = \frac{1}{2}+\frac{4}{4}+\frac{9}{8}+\frac{16}{16}+\frac{25}{32}+\cdots$$
I really tried, but I couldn't, help guys?
| Assuming as @Soke does, your series is $$\sum_{n=1}^\infty \frac{n^2}{2^n}$$ we can use the geometric series in order to find a closed form solution. The trick is to take derivatives and multiply by $x$.
Note that $$f(x) = \sum_{n=1}^\infty x^n = \frac{x}{1-x}$$ yields a derivative of $$\sum_{n=1}^\infty n x^{n-1} = \frac{d}{dx} \frac{x}{1-x} = \frac{1}{(1-x)^2}.$$ We multiply it by $x$ to get: $$\sum_{n=1}^\infty nx^n = \frac{x}{(1-x)^2}.$$
If we take another derivative we find $$\sum_{n=1}^\infty n^2 x^{n-1} = \frac{d}{dx} \frac{x}{(1-x)^2} = \frac{(1-x)^2+2(1-x)x}{(1-x)^4}= \frac{1+x}{(1-x)^3}.$$
Finally we multiply by $x$ again: $$\sum_{n=1}^\infty n^2 x^n = \frac{x(1+x)}{(1-x)^3}$$ and the answer you are looking for corresponds to $x=1/2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343721",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 5,
"answer_id": 1
} |
Exponentiation of Pascal's Triangle(in matrix form) I want to find a pattern in subsequent exponentiations of the pascal triangle shown in the form below
Matrix P[K+1][K+1]:
$$
\begin{matrix}
\binom{0}{0} & 0 & 0 & 0\cdots &0\\
\binom{1}{0} & \binom{1}{1} & 0 & 0\cdots &0 \\
\binom{2}{0} & \binom{2}{1} & \binom{2}{2} & 0\cdots &0 \\
\binom{3}{0} & \binom{3}{1} & \binom{3}{2} & \binom{3}{3}\cdots &0 \\
\vdots & \vdots &\vdots &\vdots\ddots& \vdots\\
\binom{k}{0} & \binom{k}{1}& \binom{k}{2} & \binom{k}{3}\cdots &\binom{k}{k} \\
\end{matrix}
$$
My motive is to compute $P^N$ in $O(KlogN)$ complexity ,which is definately not possible for any ordinary matrix(which would take $O(K^3logN)$).
But I am pretty much sure it is possible for this special matrix as I have done it in $O(K^2logN)$ but the problem(here),which i am solving,suggests
it to be done in $O(KlogN)$,Definately there is a secret which I want to find.
Thanks in Advanced.
| You can use a binomial expansion and rearrangement to show
$$(1+N)^{i-j}{i\choose j}=\sum_{n=0}^{i-j} N^{i-j-n}{i-j \choose n} {i\choose j} =\sum_{m=j}^i N^{i-m}{i\choose m} {m\choose j} $$
and then induction to show
$$P^N=\left(\begin{matrix}
N^{0-0}\binom{0}{0} & 0 & 0 & 0&\cdots &0\\
N^{1-0}\binom{1}{0} & N^{1-1}\binom{1}{1} & 0 & 0&\cdots &0 \\
N^{2-0}\binom{2}{0} & N^{2-1}\binom{2}{1} & N^{2-2}\binom{2}{2} & 0&\cdots &0 \\
N^{3-0}\binom{3}{0} & N^{3-1}\binom{3}{1} & N^{3-2}\binom{3}{2} & N^{3-3}\binom{3}{3}&\cdots &0 \\
\vdots & \vdots &\vdots &\vdots&\ddots& \vdots\\
N^{k-0}\binom{k}{0} & N^{k-1}\binom{k}{1}& N^{k-2}\binom{k}{2} & N^{k-3}\binom{k}{3}&\cdots &N^{k-k}\binom{k}{k} \\
\end{matrix}\right)$$
which will give you a fast calculation of any power of $P$ without having to find the intermediate powers.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1343863",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Approximating $\tan61^\circ$ using a Taylor polynomial centered at $\frac \pi 3$ : how to proceed? Here's what I have so far...
I wrote a general approximation of $f(x)=\tan(x)$ , which then simplified a bit to this:
$$\tan \left(\frac{61π}{180}\right) + \sec^2\left(\frac{61π}{180}\right)\left(\frac{π}{180}\right) + \tan\left(\frac{61π}{180}\right) \sec^2\left(\frac{61π}{180}\right)\left(\frac{π}{180}\right)^2 $$
Thing is, I'm not seeing anything obvious to do next... any hints/suggestions on how to proceed in my approximation?
Thanks in advance!
| $$f(x)=\tan(x)\\x=60^\circ=\frac{\pi}{3} \\h=\Delta x=1^\circ=\frac{\pi}{180}$$now
$$ f(x+h)=f(x)+hf'(x)+\frac{h^2}{2!}f''(x)+\frac{h^3}{3!}f'''(x)+...\\f(\frac{\pi}{3}+\frac{\pi}{180})=f(\frac{\pi}{3})+hf'(\frac{\pi}{3})+\frac{h^2}{2!}f''(\frac{\pi}{3})+\frac{h^3}{3!}f'''(\frac{\pi}{3})+...\\f(\frac{\pi}{3})+(\frac{\pi}{180})f'(\frac{\pi}{3})+\frac{(\frac{\pi}{180})^2}{2!}f''(\frac{\pi}{3})+\frac{(\frac{\pi}{180})^3}{3!}f'''(\frac{\pi}{3})+...$$
$$=\sqrt{3} +\frac{\pi}{180}(1+(\sqrt{3})^2)+ \frac{(\frac{\pi}{180})^2}{2!}2\tan(\frac{\pi}{3})\sec^2(\frac{\pi}{3})+...$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1345016",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 2,
"answer_id": 1
} |
Calculating $\sum_{k=0}^{n}\sin(k\theta)$ I'm given the task of calculating the sum $\sum_{i=0}^{n}\sin(i\theta)$.
So far, I've tried converting each $\sin(i\theta)$ in the sum into its taylor series form to get:
$\sin(\theta)=\theta-\frac{\theta^3}{3!}+\frac{\theta^5}{5!}-\frac{\theta^7}{7!}...$
$\sin(2\theta)=2\theta-\frac{(2\theta)^3}{3!}+\frac{(2\theta)^5}{5!}-\frac{(2\theta)^7}{7!}...$
$\sin(3\theta)=3\theta-\frac{(3\theta)^3}{3!}+\frac{(3\theta)^5}{5!}-\frac{(3\theta)^7}{7!}...$
...
$\sin(n\theta)=n\theta-\frac{(n\theta)^3}{3!}+\frac{(n\theta)^5}{5!}-\frac{(n\theta)^7}{7!}...$
Therefore the sum becomes,
$\theta(1+...+n)-\frac{\theta^3}{3!}(1^3+...+n^3)+\frac{\theta^5}{5!}(1^5+...+n^5)-\frac{\theta^7}{7!}(1^7+...+n^7)...$
But it's not immediately obvious what the next step should be.
I also considered expanding each $\sin(i\theta)$ using the trigonemetry identity $\sin(A+B)$, however I don't see a general form for $\sin(i\theta)$ to work with.
| $$
2\sin \frac{\theta}2 \sin k\theta = \cos \frac{(2k-1)\theta}2 -\cos \frac{(2k+1)\theta}2
$$
so (telescoping sum)
$$
2 \sin \frac{\theta}2 \sum_{k=1}^n\sin k\theta = \cos (\frac{\theta}2) -\cos \frac{(2n+1)\theta}2 \\
= 2 \sin \frac{n\theta}2 \sin \frac{(n+1)\theta}2
$$
giving
$$
\sum_{k=1}^n\sin k\theta = \frac{\sin \frac{n\theta}2 \sin \frac{(n+1)\theta}2}{\sin \frac{\theta}2}
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1349466",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 2
} |
Show that in any triangle, we have $\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$ Show that in any triangle, we have $$\frac{a\sin A+b\sin B+c\sin C}{a\cos A+b\cos B+c\cos C}=R\left(\frac{a^2+b^2+c^2}{abc}\right),$$
where $R$ is the circumradius of the triangle.
Here is my work:
We know that $A+B+C=180^\circ$, so $C=180^\circ -(A+B)$. Plugging this in, we get that $\sin C=\sin (A+B)$ and $\cos C = -\cos (A+B)$. When we plug this into the equation we get,
$$\frac{a\sin A+b\sin B+c\sin (A+B)}{a\cos A+b\cos B-c\cos (A+B)}.$$
If we expand out $c\sin (A+B)$ and $c\cos (A+B)$, we get
$$\frac{\sin A+b \sin B+c \cos A\cos B - c\sin A\sin B}{a\cos A+b\cos B-c\cos A\cos B+c\sin A\sin B}.$$
Using the Extended Law of Sines, we can use $\sin A=\frac{a}{2R}$, $\sin B=\frac{b}{2R}$, and $\sin C=\frac{c}{2R}$.
How can I continue on?
| Use Law of Sines for the numerator, $\displaystyle\sum_\text{cyc}a\sin A=\dfrac{\sum_\text{cyc} a^2}{2R}$
For the denominator, $\displaystyle\sum_\text{cyc}a\cos A=\sum_\text{cyc}(2R\sin A\cos A)=R\sum_\text{cyc}\sin2A$
Using Prove that $\sin(2A)+\sin(2B)+\sin(2C)=4\sin(A)\sin(B)\sin(C)$ when $A,B,C$ are angles of a triangle ,
$\displaystyle\implies\sum_\text{cyc}a\cos A=R(4\sin A\sin B\sin C)=4R\prod_\text{cyc}\left(\dfrac a{2R}\right)=\cdots$
Can you take it home from here?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1350412",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 2,
"answer_id": 1
} |
Product of cosines: $ \prod_{r=1}^{7} \cos \left(\frac{r\pi}{15}\right) $
Evaluate
$$ \prod_{r=1}^{7} \cos \left({\dfrac{r\pi}{15}}\right) $$
I tried trigonometric identities of product of cosines, i.e, $$\cos\text{A}\cdot\cos\text{B} = \dfrac{1}{2}[ \cos(A+B)+\cos(A-B)] $$
but I couldn't find the product.
Any help will be appreciated.
Thanks.
| Note: Here's another variation inspired by an answer to this question.
We consider the roots of unity $e^{\frac{2\pi i k}{15}}, 0\leq k < 15$ of the polynomial
$$p(z)=z^{15}-1=\prod_{k=0}^{14}(z-e^{\frac{2\pi i k}{15}})$$
We obtain
\begin{align*}
-p(-z)=z^{15}+1&=\prod_{k=0}^{14}(z+e^{\frac{2\pi i k}{15}})\\
&=(z+1)\prod_{k=1}^{7}\left[(z+e^{\frac{2\pi i k}{15}})(z+e^{-\frac{2\pi i k}{15}})\right]\\
\end{align*}
Evaluating the polynomial $-p(-z)$ at $z=1$ gives
\begin{align*}
1&=\prod_{k=1}^{7}\left[(1+e^{\frac{2\pi i k}{15}})(1+e^{-\frac{2\pi i k}{15}})\right]\\
&=\prod_{k=1}^{7}\left[(e^{-\frac{\pi i k}{15}}+e^{\frac{\pi i k}{15}})e^{\frac{\pi i k}{15}}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})e^{-\frac{\pi i k}{15}}\right]\\
&=\prod_{k=1}^{7}(e^{\frac{\pi i k}{15}}+e^{-\frac{\pi i k}{15}})^2\tag{1}\\
&=\prod_{k=1}^{7}\left(2\cos\left(\frac{k \pi}{15}\right)\right)^2\tag{2}\\
\end{align*}
In (1) we use the formula $\cos(z)=\frac{1}{2}\left(e^{iz}+e^{-iz}\right)$.
We conclude from (2)
\begin{align*}
\prod_{k=1}^{7}\cos\left(\frac{k\pi}{15}\right)=\frac{1}{2^7}
\end{align*}
Note: Writing $-p(-z)$ as
\begin{align*}
-p(-z)=\prod_{k=1}^{7}\left[z^2+\left(e^{\frac{2\pi i k}{15}}+e^{-\frac{2\pi i k}{15}}\right)z+1\right]\\
\end{align*}
and evaluating the polynomial $-p(-z)$ at $z=i$ we obtain the related formula
\begin{align*}
\prod_{k=1}^{7}\cos\left(\frac{2k\pi}{15}\right)=\frac{1}{2^7}
\end{align*}
Doubling the argument does not change the value of the product.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351337",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "36",
"answer_count": 5,
"answer_id": 3
} |
this inequality $\prod_{cyc} (x^2+x+1)\ge 9\sum_{cyc} xy$ Let $x,y,z\in R$,and $x+y+z=3$
show that:
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 9(xy+yz+xz)$$
Things I have tried so far:$$9(xy+yz+xz)\le 3(x+y+z)^2=27$$
so it suffices to prove that
$$(x^2+x+1)(y^2+y+1)(z^2+z+1)\ge 27$$
then the problem is solved. I stuck in here
| It's enough to prove that
$$\prod_{cyc}\left(x^2+\frac{x(x+y+z)}{3}+\frac{(x+y+z)^2}{9}\right)-\frac{(xy+xz+yz)(x+y+z)^4}{9}\geq$$
$$\geq\frac{1}{2916}\left(\sum_{cyc}(5x^3+6x^2y+6x^2z-17xyz)\right)^2.$$
Let $x+y+z=3u$, $xy+xz+yz=3v^2$, where $v^2$ can be negative, and $xyz=w^3$.
Since $$\sum_{cyc}(5x^3+6x^2y+6x^2z-17xyz)=$$
$$=5(27u^3-27uv^2+3w^3)+6(9uv^2-3w^3)-51w^3=135u^3-81uv^2-54w^3$$ and $$54^2=2916,$$
we see that the last inequality is a linear inequality of $w^3$, which says that
it's enough to prove this inequality for an extremal value of $w^3$, which happens
for equality case of two variables.
Since the last inequality is even degree, we can assume $y=z=1$, which gives
$$(x-1)^4(x+2)^2\geq0.$$
Done!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1351422",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 3,
"answer_id": 2
} |
Determine whether $\sum \frac{2^n + n^2 3^n}{6^n}$ converges For the series $$\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n},$$ I was thinking of using the root test? so then I would get $(2+n^2/n+3)/6$ but how do I find the limit of this?
| Hint: $\lim_{n \rightarrow \infty} \dfrac {n^4}{2^n} = 0$
$\implies $ After a particular value of $m \in \mathbb N : n^4 << 2^n ~\forall~ n \ge m$ or
$\dfrac {n^4}{2^n}<1 ~\forall~ n \ge m$.
Hence, $\dfrac {n^2}{2^n} < \dfrac {1}{n^2} \implies \sum \dfrac {n^2}{2^n} < \sum \dfrac {1}{n^2} $
and $\sum \dfrac {1}{n^2} $ is a convergent series.
Hence, $\sum_{n=1}^{\infty}\dfrac{2^n+n^23^n}{6^n} = \sum_{n=1}^{\infty} \dfrac {1}{3^n} + \dfrac {n^2}{2^n}$ is convergent
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1352763",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 5,
"answer_id": 2
} |
If $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$, then $\frac{a+e}{b+f} < \frac{c+g}{d+h}$.
If $a, b, c, d, e, f, g, h$ are positive numbers satisfying $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$ and $b+f>d+h$, then $\frac{a+e}{b+f} < \frac{c+g}{d+h}$.
I thought it is easy to prove. But I could not. How to prove this? Thank you.
The question is a part of a bigger proof I am working on. There are two strictly concave, positive valued, strictly increasing functions $f_1$ and $f_2$ (See Figure 1). Given 4 points $x_1$, $x_2$, $x_3$ and $x_4$ such that $x_1< x_i$, $i=2, 3,4$ and $x_4> x_i$, $i=1, 2, 3$, let $d=x_2-x_1$, $b=x_4-x_3$ $c=f_1(x_2)-f_1(x_1)$, $a=f_1(x_4)-f_1(x_3)$. And given 4 points $y_1$, $y_2$, $y_3$ and $y_4$ such that $y_1< y_i$, $i=2, 3,4$ and $y_4> y_i$, $i=1, 2, 3$, let $h=y_2-y_1$, $f=y_4-y_3$ $g=f_2(y_2)-f_2(y_1)$, $e=f_2(y_4)-f_2(y_3)$.
Since the functions are concave, we have $\frac{a}{b}<\frac{c}{d}$ and $\frac{e}{f}<\frac{g}{h}$. And I thought in this setting, it is true that $\frac{a+e}{b+f} < \frac{c+g}{d+h}$ even without the restriction $b+f>d+h$.
| The updated question (with the additional constraint $b+f>d+h$) is also false. For example,
$\frac{1}{1}<\frac{3}{2}$ and $\frac{9}{4}<\frac{5}{2}$, but $\frac{1+9}{1+4} = \frac{10}{5} = \frac{8}{4} = \frac{3+5}{2+2}$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354015",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
How to rewrite $\pi - \arccos(x)$ as $2\arctan(y)$? I get the following results after solving the equation $\sqrt[4]{1 - \frac{4}{3}\cos(2x) - \sin^4(x)} = -\,\cos(x)$, :
$$
x_{1} = \pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}\\
x_{2} = \pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi n: n \in \mathbb{Z}
$$
Wolfram Alpha, here the link, instead, gives the following results:
$
x_{1} = 2\pi n - 2\arctan(\sqrt{5 + 2\sqrt{6}}) : n \in \mathbb{Z}\\
x_{2} = 2\pi n + 2\arctan(\sqrt{5 + 2\sqrt{6}}) : n \in \mathbb{Z}
$
Now, supposing that my solutions are correct, this means that there must be a relation between:
$\pi - \arccos(\frac{\sqrt{6}}{3})$ and $- 2\arctan(\sqrt{5 + 2\sqrt{6}})$
or between:
$\pi + \arccos(\frac{\sqrt{6}}{3})$ and $+ 2\arctan(\sqrt{5 + 2\sqrt{6}})$
or viceversa. But, given the solutions I have found, how can I prove that they are effectively the same as the solutions Wolfram found? Mathematically?
P.S.: I have found out, by looking at the graphs of $y_{1} = \pi - \arccos(\frac{\sqrt{6}}{3})$ and $y_{2} = 2\arctan(\sqrt{5 + 2\sqrt{6}})$ e.g., that:
$\pi - \arccos(\frac{\sqrt{6}}{3}) = 2\arctan(\sqrt{5 + 2\sqrt{6}})$
Then, of course:
$\pi - \arccos(\frac{\sqrt{6}}{3}) + 2\pi = 2\arctan(\sqrt{5 + 2\sqrt{6}})+ 2\pi \\
\pi - \arccos(\frac{\sqrt{6}}{3}) + 4\pi = 2\arctan(\sqrt{5 + 2\sqrt{6}})+ 4\pi \\
... every\,\,360°n, n \in \mathbb{Z}$
And that:
$\pi + \arccos(\frac{\sqrt{6}}{3}) = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 2\pi\\
\pi + \arccos(\frac{\sqrt{6}}{3}) + 2\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 4\pi\\
\pi + \arccos(\frac{\sqrt{6}}{3}) + 4\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ 6\pi\\
...
\pi + \arccos(\frac{\sqrt{6}}{3}) + (n - 2)\pi = -2\arctan(\sqrt{5 + 2\sqrt{6}})+ n\pi\\
$
So we can say that $\pi + \arccos(\frac{\sqrt{6}}{3}) + (n - 2)\pi$ differs from $-2\arctan(\sqrt{5 + 2\sqrt{6}}) + n\pi$ by just one lap ($2\pi$), otherwise they can be safely considered the same (for all integers).
So how to rewrite $\arccos$ in terms of $\arctan$?
Thanks for the attention!
| Let $\arctan x=y\implies x=\tan y$
Using this, $0\le y\le\dfrac\pi2\iff0\le2y\le\pi$ and $\arccos$ lies in $[0,\pi]$
$\cos2y=\dfrac{1-\tan^2y}{1+\tan^2y}=\dfrac{1-x^2}{1+x^2}$
$\implies\arccos\dfrac{1-x^2}{1+x^2}=2y=2\arctan x$ if $0\le2\arctan x\le\pi\iff0\le\arctan x\le\dfrac\pi2\implies0\le x\le\infty$
For $x<0\implies-\dfrac\pi2\le\arctan x<0\implies-\pi\le2y<0\iff0<-2y\le\pi$
As $\cos(2y)=\cos(-2y),$
$\implies\arccos\dfrac{1-x^2}{1+x^2}=-2y=-2\arctan x$
$\iff2\arctan x=-\arccos\dfrac{1-x^2}{1+x^2}$ if $x<0$
See also, How do I prove that $\arccos(x) + \arccos(-x)=\pi$ when $x \in [-1,1]$?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1354993",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 1
} |
Finding period of $f$ from the functional equation $f(x)+f(x+4)=f(x+2)+f(x+6)$ How can I find the period of real valued function satisfying $f(x)+f(x+4)=f(x+2)+f(x+6)$?
Note: Use of recurrence relations not allowed. Use of elementary algebraic manipulations is better!
| You are given that $f(x)+f(x+4) = f(x+2)+f(x+6)$ for all $x \in \mathbb{R}$.
Replace $x$ with $x+2$ to get $f(x+2)+f(x+6) = f(x+4)+f(x+8)$ for all $x \in \mathbb{R}$.
Thus, $f(x)+f(x+4) = f(x+2)+f(x+6) = f(x+4)+f(x+8)$ for all $x \in \mathbb{R}$.
Subtract $f(x+4)$ from the left and right side of the last equation to get: $f(x) = f(x+8)$ for all $x \in \mathbb{R}$.
Important note: This tells you that $f$ is $8$-periodic, but $8$ may not necessarily be the minimum period. For instance, $f(x) = \sin(\pi x)$ satisfies $f(x)+f(x+4) = f(x+2)+f(x+6)$ for all $x \in \mathbb{R}$, but has period $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1357922",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Find the roots of the summed polynomial
Find the roots of: $$x^7 + x^5 + x^4 + x^3 + x^2 + 1 = 0$$
I got that:
$$\frac{1 - x^8}{1-x} - x^6 - x = 0$$
But that doesnt make it any easier.
| $\dfrac{x^7 + x^5 + x^4 + x^3 + x^2 + 1}{x+1}=x^6-x^5+2x^4-x^3+2x^2-x+1$
$=x^6-x^5+x^4+(x^4-x^3+x^2)+(x^2-x+1)$
$=x^4(x^2-x+1)+x^2(x^2-x+1)+(x^2-x+1)$
$=(x^2-x+1)[x^4+x^2+1]$
Now $x^4+x^2+1=(x^2+1)^2-(x)^2=\cdots$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358551",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 1
} |
Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. The Statement of the Problem:
Let $X$ have pdf
$$f_X(x) =
\begin{cases}
\frac{1}{4} & 0<x<1 \\
\frac{3}{8} & 3<x<5 \\
0 & \text{otherwise}
\end{cases}$$
(a) Find the cumulative distribution function of $X.$
(b) Let $Y=1/X$. Find the pdf $f_Y(y)$ for $Y$. Hint: Consider three cases: $1/5 \le y \le 1/3, 1/3 \le y \le 1,$ and $ y \ge 1.$
Where I Am:
I think I did part (a) correctly. I did the following:
$$F_X(x) =
\begin{cases}
\frac{1}{4}x +c_1 & 0<x<1 \\
\frac{3}{8}x + c_2 & 3<x<5 \\
0 & \text{otherwise}
\end{cases}$$
$$F(0)=0=\frac{1}{4}(0)+c_1 \implies c_1 = 0 $$
$$F(5)=1=\frac{3}{8}(5)+c_2 \implies c_2 = -\frac{7}{8} $$
Therefore:
$$F_X(x) =
\begin{cases}
\frac{1}{4}x & 0<x<1 \\
\frac{1}{4} & 1<x<3 \\
\frac{3}{8}x - \frac{7}{8} & 3<x<5 \\
1 & x > 5
\end{cases}$$
If that's not right, however, please let me know.
Now, for part (b), I got a little lost. Here's what I did:
$$ \text{Let } g(x) = \frac{1}{x} \implies g'(x) = -\frac{1}{x^2} $$
Then:
$$ f_Y(y)=\frac{f_X(x)}{\lvert g'(x) \rvert} = \frac{f_X(\frac{1}{y})}{\lvert g'(\frac{1}{y})\rvert} $$
Therefore:
$$f_Y(y) =
\begin{cases}
\frac{1}{4} & 0<\frac{1}{y}<1 \\
\frac{3}{8} & 3<\frac{1}{y}<5 \\
0 & \text{otherwise}
\end{cases}$$
and taking reciprocals and flipping inequalities...
$$f_Y(y) =
\begin{cases}
\frac{1}{4} & y \ge 1 \\
\frac{3}{8} & \frac{1}{5} \le y \le \frac{1}{3} \\
0 & \frac{1}{3} \le y \le 1
\end{cases}$$
This, however... doesn't seem right. For example, what is $f_Y(y)$ when $y \in [0, \frac{1}{5}]$? Is it just $0$? I know I did something wrong here, but I can't quite figure out what exactly. If anybody could help me out, I'd appreciate it.
EDIT: "Second" Attempt...
$$F_X \left( \frac{1}{y} \right) =
\begin{cases}
\frac{1}{4}\left( \frac{1}{y} \right) & y \ge 1 \\
\frac{1}{4} & \frac{1}{3} \le y \le 1 \\
\frac{3}{8}\left( \frac{1}{y} \right) - \frac{7}{8} & \frac{1}{5} \le y \le \frac{1}{3} \\
0 & \text{otherwise}
\end{cases}$$
Therefore:
$$ f_Y(y)=\frac{d}{dy}F_X \left( \frac{1}{y} \right)= \begin{cases}
-\frac{1}{4}\left( \frac{1}{y^2} \right) & y \ge 1 \\
-\frac{3}{8}\left( \frac{1}{y^2} \right) & \frac{1}{5} \le y \le \frac{1}{3} \\
0 & \text{otherwise}
\end{cases} $$
| $$
f_X(x) =
\begin{cases}
\frac{1}{4} & 0<x<1 \\
\frac{3}{8} & 3<x<5 \\
0 & \text{otherwise}
\end{cases}
$$
\begin{align}
f_Y(y) & = \frac d {dy} F_Y(y) = \frac d {dy} \Pr(Y\le y) = \frac d {dy} \Pr\left( \frac 1 X \le y \right) \\[10pt]
& = \frac d {dy} \Pr\left( X\ge \frac 1 y \right) \text{ (if }y>0) \\[10pt]
& = \frac d {dy} F_X\left(\frac 1 y \right).
\end{align}
Now notice that when $3<x<5$ then $\dfrac 1 5 <\dfrac 1 x < \dfrac 1 3$, or $\dfrac 1 5 < y < \dfrac 1 3$ and similarly for other intervals.
In part $(a)$ I'd use definite integrals. Since $f_X(x)=0$ when $x<0$, you have for $x\ge 0$,
$$
F_X(x) = \int_0^x f_X(w) \, dw.
$$
If $x>1$, this becomes
$$
F_X(x) = \int_0^x f_X(w) \, dw = \int_0^1 f_X(w)\,dw + \int_1^x f_X(w)\,dw
$$
and if $x>3$ then
$$
F_X(x) = \int_0^x f_X(w) \, dw = \int_0^1 f_X(w)\,dw + \int_1^3 f_X(w)\,dw + \int_3^x f_X(w)\,dw.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1358786",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Prove that $\sum_{i=1}^{i=n} \frac{1}{i(n+1-i)} \le1$ $$f(n)=\sum_{i=1}^{i=n} \dfrac{1}{i(n+1-i)} \le 1$$
For example, we have $f(3)=\dfrac{1}{1\cdot3}+\dfrac{1}{2\cdot2}+\dfrac{1}{3\cdot1}=\dfrac{11}{12}\lt 1$
If true, it can be used to prove:
Proving $x\ln^2x−(x−1)^2<0$ for all $x∈(0,1)$
Also, can you prove $f(n)\ge f(n+1)$?
| $\begin{array}\\
i(n+1-i)
&=i(n+1)-i^2\\
&=(n+1)^2/4-(n+1)^2/4+i(n+1)-i^2\\
&=(n+1)^2/4-((n+1)/2-i)^2\\
\end{array}
$
Therefore
$$i(n+1-i)
=(n+1)^2/4-((n+1)/2-i)^2
\le (n+1)^2/4
$$
and,
since $1 \le i \le n$,
so that
$(n-1)/2 \ge (n+1)/2-i
\ge -(n-1)/2$,
$$i(n+1-i)
=(n+1)^2/4-((n+1)/2-i)^2
\ge (n+1)^2/4-((n-1)/2)^2
=n
$$
so
$$\frac1{n}
\ge \frac1{i(n+1-i)}
\ge \frac{4}{(n+1)^2}
.$$
Therefore
$$1
\ge \sum_{i=1}^n \frac1{i(n+1-i)}
\ge \frac{4n}{(n+1)^2}
.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1359030",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Summation stuck under radical sign I am trying to evaluate the following sum, but I'm unable to solve it in any general way.
$$S=\sum_{k=1}^n\sqrt{1+\frac{1}{(k)^2}+\frac{1}{(k+1)^2} }$$
How can I do it?
| A different method:
\begin{align}1 + \frac{1}{k^2} + \frac{1}{(k+1)^2} &= 1 + \frac{2}{k(k+1)} + \left[\frac{1}{k^2} - \frac{2}{k(k+1)} + \frac{1}{(k+1)^2}\right]\\
&= 1 + 2\left(\frac{1}{k} - \frac{1}{k+1}\right) + \left(\frac{1}{k} - \frac{1}{k+1}\right)^2\\
&= \left(1 + \frac{1}{k} - \frac{1}{k+1}\right)^2,
\end{align}
and thus
$$\sum_{k = 1}^n \sqrt{1 + \frac{1}{k^2} + \frac{1}{(k+1)^2}} = \sum_{k = 1}^n \left(1 + \frac{1}{k} - \frac{1}{k+1}\right) = n + 1 - \frac{1}{n+1}.$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1360643",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
How to find the integral $\int \frac{\sqrt{1+x^{2n}}\left(\log(1+x^{2n}) -2n \log x\right)}{x^{3n+1}}dx$? How to evaluate the integral :
$$\int \frac{\sqrt{1+x^{2n}} \, \left(\ln(1+x^{2n}) -2n \, \ln x \right) \, dx}{x^{3n+1}}$$
I have attempted an evaluation, but I am at a loss as to a useful result. Thanks for any and all help.
| Let
\begin{align}
I = \int \frac{\sqrt{1+x^{2n}} \, \left(\ln(1+x^{2n}) -2n \, \ln x \right) \, dx}{x^{3n+1}}
\end{align}
and make the transformation $x = t^{1/2n}$ to obtain
\begin{align}
I = \frac{1}{2n} \, \int t^{-5/2} \, \sqrt{1+t} \, \left(\ln(1+t) - \ln(t)\right) \, dt.
\end{align}
Let $t = \sinh^{2}\theta$ to obtain
\begin{align}
\frac{n}{2} \, I &= \int \coth^{2}\theta \, csch^{2}\theta \, \ln(\cosh\theta) \, d\theta - \int \coth^{2}\theta \, csch^{2}\theta \, \ln(\sinh\theta) \, d\theta = J_{1} - J_{2}
\end{align}
\begin{align}
J_{1} &= \int \coth^{2}\theta \, csch^{2}\theta \, \ln(\cosh\theta) \, d\theta \\
&= \int \ln(\cosh\theta) \, \partial_{\theta}\left(\frac{-1}{3} \, \coth^{3}\theta \right) \, d\theta \\
&= - \frac{1}{3} \, \coth^{3}\theta \, \ln(\cosh\theta) + \frac{1}{3} \, \int \coth^{2}\theta \, d\theta \\
&= \frac{1}{3} \left[ \theta - \coth\theta - \coth^{3}\theta \, \ln(\cosh\theta) \right]
\end{align}
In a similar manor
\begin{align}
J_{2} = \frac{1}{3} \left[ \theta - \frac{1}{3} \, \coth\theta \, (csch^{2}\theta + 4) - \coth^{3}\theta \, \ln(\sinh\theta) \right].
\end{align}
Now,
\begin{align}
I = \frac{2}{9 \, n} \left[ \coth\theta \, (1 + csch^{2}\theta) - \coth^{3}\theta \, \ln(\coth\theta) \right].
\end{align}
By reversing the substitutions the result
\begin{align}
I = \frac{2}{9n} \left(1+\frac{1}{x^{2n}}\right)^{3/2} - \frac{1}{3n} \left(1+\frac1{x^{2n}}\right)^{3/2} \, \ln\left(1+\frac1{x^{2n}}\right) + c_{0}
\end{align}
is obtained.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361575",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 1
} |
Are the real parts of the vectors constituting the discrete Fourier transform matrix linearly independent? Let W denote the n- dimensional symmetric discrete Fourier transform matrix and $W_{i}$ denote its column vectors. Then, is the set { Re($W_{i}$) | i= 1... n } linearly independent? Or similarly, find det( Re ( W ) ).
| The short answer is: No.
For example, consider $n=3$. You have $\omega=e^{-2\pi i/ 3} = -\frac{1}{2} - i \frac{\sqrt{3} }{2}$.
Then; the corresponging DTF matrix, $W$, becomes
$$ W=
\begin{pmatrix}
1 & 1 & 1 \\
1 & \omega &\omega^2 \\
1 & \omega^2 & \omega^4\\
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 & 1 \\
1 & e^{-2\pi i/ 3} &e^{-4\pi i/ 3} \\
1 & e^{-4\pi i/ 3} & e^{-8\pi i/ 3}\\
\end{pmatrix}
=
\begin{pmatrix}
1 & 1 & 1 \\
1 & -\frac{1}{2} - i \frac{\sqrt{3}}{2} &-\frac{1}{2} + i \frac{\sqrt{3}}{2} \\
1 & -\frac{1}{2} + i \frac{\sqrt{3}}{2} & -\frac{1}{2} - i \frac{\sqrt{3}}{2}\\
\end{pmatrix}.
$$
Giving: $$\Re(W)=
\begin{pmatrix}
1 & 1 & 1 \\
1 & -1/2 &-1/2 \\
1 & -1/2 & -1/2\\
\end{pmatrix}. $$
And the set $ \{ \, \Re(W_i) \mid i= 1,2,3 \, \}
=
\Bigg\{
\begin{pmatrix}
1 \\
1 \\
1 \\
\end{pmatrix},
\begin{pmatrix}
1 \\
-1/2 \\
-1/2 \\
\end{pmatrix},
\begin{pmatrix}
1 \\
-1/2 \\
-1/2 \\
\end{pmatrix}
\Bigg\}
$
is not linearly independent since the last two vectors are equal.
Moreover; we can prove that for each $n\geq3$, the column vectors of the real part of the corresponding DFT matrix $W$ are in fact linearly dependent! A simple proof is as follows:
$$ \Re(W)_2 =\Re(W_2) = \Re \Big(\begin{pmatrix}
1 \\
\omega \\
\omega^2 \\
\vdots\\
\omega^{n-1}
\end{pmatrix}\Big) = \Re \Big(\begin{pmatrix}
1 \\
\omega^{-1} \\
\omega^{-2} \\
\vdots\\
\omega^{-(n-1)}
\end{pmatrix}\Big) = \Re \Big(\begin{pmatrix}
1 \\
\omega^{-1} \times \omega^n\\
\omega^{-2} \times \omega^{2n} \\
\vdots\\
\omega^{-(n-1)} \times \omega^{(n-1)n}
\end{pmatrix} \Big) = \Re \Big(\begin{pmatrix}
1 \\
\omega^{n-1} \\
\omega^{2(n-1)} \\
\vdots\\
\omega^{(n-1)(n-1)}
\end{pmatrix} \Big) =\Re (W_n) =\Re (W)_n
$$
since $\omega^{nk}=1$ for any integer $k$.
Thus, the second and n-th columns of $\Re(W)$ are equal if $n\geq3$; which, in the end shows that the column vectors are not linearly independent.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1361793",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Prove that if $a^2+b^2$ is a multiple of three, then a and b are multiples of three I have attempted to prove the above. I am uncertain about the correctness of my proof:
Both numbers have to be multiples of three, i.e. $3a+3b=3n$, $\ 3(a+b)=3n$
It is not possible to arrive at an integer that is a multiple of three without adding two integers that are multiples of three.
Assumption: Suppose that $b$ is not a multiple of three, then it can be expressed as $(3v \pm 1)$, therefore we have:
\begin{align*} a^2+(3v \pm 1)^2=3n\\ a^2= 3n-9v^2 \mp 6v -1\\ a=\sqrt{3(n-3v^2\mp2v-\frac{1}{3})}
\end{align*}
which is not a multiple of three? (or is it).
As mentioned before, $(a+b) \ne 3m \ $ if either $a$ or $b$ is not a multiple of 3, in which case assumption that $b$ is not a multiple of 3 is false. And hence it is a multiple of three, so is $a$.
| For any integer $n$, we have $n^2 \equiv 0 \bmod{3}$ or $n^2 \equiv 1 \bmod{3}$.
Since $a^2+b^2 \equiv 0 \bmod{3}$, by the above fact we must have $a^2\equiv 0 \bmod{3}$ and $b^2 \equiv 0 \bmod{3}$.
Since $3$ is a prime dividing $a^2$, $3$ divides $a$. Similarly $3$ divides $b$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1362554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 0
} |
Find lim$_{x \to 0}\left(\frac{1}{x} - \frac{\cos x}{\sin x}\right).$ $$\lim_{x \to 0} \left(\frac{1}{x} - \frac{\cos x}{\sin x}\right) = \frac{\sin x - x\cos x}{x\sin x}= \frac{0}{0}.$$
L'Hopital's: $$\lim_{x \to 0} \frac{f'(x)}{g'(x)} = -\frac{1}{x^2} + \frac{1}{\sin ^2x} = \frac{0}{0}.$$
Once again, using L'Hopital's: $$\lim_{x \to 0} \frac{f''(x)}{g''(x)} = \frac{2}{x^3}- \frac{2\cos x}{\sin ^3x} = \frac{0}{0}\,\ldots$$ The terms are getting endless here. Any help? Thanks.
| Since no one mentioned it, I will go for the overkill. $\cot z$ is a meromorphic function on the complex plane and $\frac{1}{z}$ is exactly its singular part in the origin, since:
$$ \sin z = z \prod_{n\geq 1}\left(1-\frac{z^2}{n^2\pi^2}\right)\tag{1} $$
implies:
$$ \log \sin z = \log z + \sum_{n\geq 1}\log\left(1-\frac{z^2}{n^2\pi^2}\right)\tag{2} $$
then:
$$ \cot z = \frac{1}{z}-\sum_{n\geq 1}\frac{2z}{n^2\pi^2-z^2}\tag{3} $$
as well as:
$$ \frac{1}{z}-\cot z = 2z\sum_{n\geq 1}\frac{1}{\pi^2 n^2-z^2}=2z\left(\frac{\zeta(2)}{\pi^2}+\frac{\zeta(4)}{\pi^4}z^2+\frac{\zeta(6)}{\pi^6}z^6+\ldots\right)\tag{4}$$
so the wanted limit is just zero. By the way, the last formula allows us to compute the values of the zeta function over the positive even integers in terms of the derivatives of $z\cot z$ at $z=0$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363719",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 7,
"answer_id": 6
} |
Bounds for $\frac{x-y}{x+y}$ How can I find upper and lower bounds for $\displaystyle\frac{x-y}{x+y}$? So I do see that
$$\frac{x-y}{x+y} = \frac{1}{x+y}\cdot(x-y) = \frac{x}{x+y} - \frac{y}{x+y} > \frac{1}{x+y} - \frac{1}{x+y} = \frac{0}{x+y} = 0$$
(is it correct?) but I don't get how to find the upper bound.
| so we have $f(x,y)=\frac{x-y}{x+y}$, we want to show that $f(x,y)$ is unbounded, therefore we fix $y=y_0>0$ and then have
$$
f(x,y_0)=f(x)=\frac{x-y_0}{x+y_0}
$$
now choose $x=x^*-y_0$ which gives us
$$
f(x)=\frac{x^*-2y_0}{x^*}
$$
and now observe that for $\lim_{x^*\to0+}\frac{x^*-2y_0}{x^*}=-\infty$ and for $\lim_{x^*\to0-}\frac{x^*-2y_0}{x^*}=\infty$.
So $f(x,y)$ is indeed unbounded as we thought.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363835",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
$A$ and $B$ similar if $A^2=B^2=0$ and dimension of range $A$ and $B$ are equal Suppose $A$ and $B$ are linear transformations on finite dimensional vector space $V$,s.t. $A,B\neq 0$ and $A^2=B^2=0$. Suppose the dimension of range $A$ and $B$ are equal, can $A$ and $B$ be similar?
| If $A^2 = B^2 = 0$, then all the eigenvalues of $A$ and $B$ are identically equal to $0$. Now, if we put $A$ into Jordan normal form, the blocks can only be $1 \times 1$ or $2 \times 2$, else $A^2 \neq 0$; thus, $A$ is similar to $$ Q \left(\begin{array}{ccccccc}
0 & 1 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & \cdots & 0 & 0\\
0 & 0 & 0 & 1 & \cdots & 0 & 0\\
0 & 0 & 0 & 0 & \cdots & 0 & 0 \\
\vdots & \vdots & \vdots & \vdots & \ddots & \vdots & \vdots \\
0 & 0 & 0 & 0 & \cdots & 0 & 0 \\
0 & 0 & 0 & 0 & \cdots & 0 & 0
\end{array} \right) Q^{-1}$$
for some $Q$, where the matrix in the middle consists of $r$ $2 \times 2$ blocks, with $r = \text{rank}(A)$; this is because the above is thethe Jordan Normal form for $A$. If the rank of $B$ is also equal to $r$, then we have that $B = PDP^{-1}$ where $D$ is the same matrix above. We then have $$A = QP^{-1} BPQ^{-1}.$$
Thus, if $r > 0$, the two matrices must be similar.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1363902",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 4,
"answer_id": 2
} |
finding a conditional expectation It is an old exam problem about conditional expectation:
Let $(\xi_1,\xi_2)$ be a Gaussian vector with zero mean and covariance matrix A with $A_{11}=A_{22}=1, A_{12}=A_{21}=1/2.$ What is $E(\xi_1^2\xi_2|2\xi_1-\xi_2)$?
With the condition I know $E(\xi_1\xi_2)=1/2$ and $E(\xi_1^2)=E(\xi_2^2)=1$. And I tried to represent $\xi_1^2\xi_2$ with $2\xi_1-\xi_2$, like
$$
E(\xi_1^2\xi_2|2\xi_1-\xi_2)=\frac{1}{2}E(\xi_1\xi_2^2+\xi_1\xi_2(2\xi_1-\xi_2)|2\xi_1-\xi_2)
\\=\frac{1}{2}[E(\xi_1\xi_2^2|2\xi_1-\xi_2)+(2\xi_1-\xi_2)E(\xi_1\xi_2|2\xi_1-\xi_2)].
$$
However, I don't know how to continue. Is it a right path? Thanks for any help.
| Make the transformation $\eta_1 = \xi_1, \eta_2 = 2\xi_1 - \xi_2$, since
$$\begin{pmatrix}
\xi_1 \\
\xi_2 \end{pmatrix} \sim \mathcal{N}\left(\begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 & \frac{1}{2} \\ \frac{1}{2} & 1\end{pmatrix}\right),$$
it follows that
$$\begin{pmatrix}
\eta_1 \\
\eta_2 \end{pmatrix} \sim \mathcal{N}\left(\begin{pmatrix} 0 \\ 0 \end{pmatrix}, \begin{pmatrix} 1 & \frac{3}{2} \\ \frac{3}{2} & 3\end{pmatrix}\right).$$
By the theorem of conditional distribution of Gaussian random vector, we have
$$\eta_1 \mid \eta_2 \sim \mathcal{N}\left(0 + \frac{3}{2}\times\frac{1}{3}(\eta_2 - 0), 1 - \frac{3}{2}\times\frac{1}{3}\times\frac{3}{2}\right) = \mathcal{N}\left(\frac{1}{2}\eta_2, \frac{1}{4}\right). \tag{1}$$
Now
$$E(\xi_1^2 \xi_2 \mid 2\xi_1 - \xi_2) = E(\eta_1^2(2\eta_1 - \eta_2) \mid \eta_2) = 2E(\eta_1^3 \mid \eta_2) - \eta_2E(\eta_1^2 \mid \eta_2).\tag{2}$$
Can you proceed based on $(1)$ and $(2)$?
(After some algebra, you should get
\begin{align*}
& E(\eta_1^2 \mid \eta_2) = \frac{1}{4}(1 + \eta_2^2), \\
& E(\eta_1^3 \mid \eta_2) = \frac{1}{8}(3\eta_2 + \eta_2^3).
\end{align*}
You can get the final result by substituting $\eta_2$ by $\xi_1, \xi_2$.
)
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364002",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 1,
"answer_id": 0
} |
Generalizing the Fibonacci sum $\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$ Given the Fibonacci, tribonacci, and tetranacci numbers,
$$F_n = 0,1,1,2,3,5,8\dots$$
$$T_n = 0, 1, 1, 2, 4, 7, 13, 24,\dots$$
$$U_n = 0, 1, 1, 2, 4, 8, 15, 29, \dots$$
and so on, how do we show that,
$$\sum_{n=0}^{\infty}\frac{F_n}{10^n} = \frac{10}{89}$$
$$\sum_{n=0}^{\infty}\frac{T_n}{10^n} = \frac{100}{889}$$
$$\sum_{n=0}^{\infty}\frac{U_n}{10^n} = \frac{1000}{8889}$$
or, in general,
$$\sum_{n=0}^{\infty}\frac{S_n}{p^n} = \frac{(1-p)p^{k-1}}{(2-p)p^k-1}$$
where the above were just the cases $k=2,3,4$, and $p=10$?
P.S. Related post.
| That's because
$\sum_{n=0}^{\infty} F_nx^n
=\frac1{1-x-x^2}
$.
Putting
$x = \frac1{10^k}$
gives
$\sum_{n=0}^{\infty} \frac{F_n}{10^{kn}}
=\frac1{1-10^{-k}-10^{-2k}}
=\frac{10^{2k}}{10^{2k}-10^{k}-1}
$.
For the others,
the generating function is
$\sum_{n=0}^{\infty} G_n x^n
=\frac1{1-x-x^2-...-x^m}
$
where
$m=2$ for Fib,
$m=3$ for Trib,
and $m=4$ for Tetra.
For each of these,
$\sum_{n=0}^{\infty} \frac{G_n}{ 10^n}
=\frac1{1-\frac1{10}-\frac1{100}-...-\frac1{10^{m}}}
=\frac{10^m}{10^m-10^{m-1}-...-1}
=\frac{10^m}{8...(m-1 \ 8s)9}
$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364092",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 3,
"answer_id": 2
} |
The maximum and minimum values of the expression Here is the question:find the difference between maximum and minimum values of $u^2$ where $$u=\sqrt{a^2\cos^2x+b^2\sin^2x} + \sqrt{a^2\sin^2x+b^2\cos^2x}$$
My try:I have just normally squared the expression and got
$u^2=a^2\cos^2x+b^2\sin^2x + a^2\sin^2x+b^2\cos^2x +2\sqrt{a^2\cos^2x+b^2\sin^2x} \sqrt{a^2\sin^2x+b^2\cos^2x}$
$u^2=a^2+b^2 +2\sqrt{a^2\cos^2x+b^2\sin^2x} .\sqrt{a^2\sin^2x+b^2\cos^2x}$
I am not getting how to solve the irrational part,so how should we do it.Is there some general way to solve such questions?
| Expanding $u^2$ more:$$u^2=a^2+b^2 +2\sqrt{\sin^2x\cos^2x(a^4+b^4)+a^2b^2(\sin^4x+\cos^4x)}$$
Using trigonometric identity $\sin^2x+\cos^2x=1$ we can derive that:$$\sin^4x+\cos^4x=1-2\sin^2x\cos^2x$$
Rewrite $u^2$ again:$$u^2=a^2+b^2 +2\sqrt{\sin^2x\cos^2x(a^2-b^2)^2+a^2b^2}$$
The minimum value of $\sin^2x\cos^2x$ is $0$ and its maximum value is (using AM-GM) $$\frac{\sin^2x+\cos^2x}{2}=\frac{1}{2}\ge\sin x\cos x$$
$$\frac{1}{4}\ge\sin^2x\cos^2x$$
Also you can find it this way using trigonometric identities $$\sin^2x\cos^2x = \frac{\sin^2(2x)}{4}\Rightarrow \max(\sin^2x\cos^2x)=\max \left(\frac{\sin^2(2x)}{4}\right)=\frac{1}{4}$$
So $$u^2_{min}=(\left |a\right |+\left |b\right |)^2$$
$$u^2_{max}=2(a^2+b^2)$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1364170",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 1
} |
Trigonometric Integrals $\int \frac{1}{1+\sin^2(x)}\mathrm{d}x$ and $\int \frac{1-\tan(x)}{1+\tan(x)} \mathrm{d}x$ Any idea of calculating this two integrals $\int \frac{1}{1+\sin^2(x)}\,dx$ and $\int \frac{1-\tan(x)}{1+\tan(x)} \mathrm{d}x$?
I found a solution online for the first one but it requires complex numbers which have not been taught by the professor.
| Notice, $$I_1=\int \frac{dx}{1+\sin^2 x}$$
$$=\int \frac{\sec^2 xdx}{\sec^2 x+\sec^2 x\sin^2 x}$$
$$=\int \frac{\sec^2 xdx}{1+\tan^2 x+\tan^2 x}$$
$$=\int \frac{\sec^2 xdx}{1+2\tan^2 x}$$
$$=\frac{1}{2}\int \frac{\sec^2 x dx}{\left(\frac{1}{\sqrt{2}}\right)^2+\tan^2 x}$$
Now, let $\tan x=t\implies \sec^2x dx=dt$
$$=\frac{1}{2}\int \frac{dt}{\left(\frac{1}{\sqrt{2}}\right)^2+t^2 }$$
$$=\frac{1}{2}\sqrt{2}\tan^{-1}\left(t\sqrt{2}\right)$$
$$\implies I_1=\frac{1}{\sqrt{2}}\tan^{-1}\left(\sqrt{2}\tan x\right)+C_1$$
Again notice, $$I_2=\int \frac{1-\tan x}{1+\tan x}dx$$
$$=\int \frac{(1-\tan x)(1-\tan x)}{(1+\tan x)(1-\tan x)}dx$$
$$=\int \frac{1+\tan^2 x-2\tan x}{1-\tan^2 x}dx=\int \frac{1+\tan^2 x}{1-\tan^2 x}dx-\int\frac{2\tan x}{1-\tan^2 x}dx$$ $$=\int \sec 2x dx-\int\tan 2xdx$$
$$=\frac{1}{2}\ln(\sec 2x+\tan 2x)-\frac{1}{2}\ln\sec 2x+C_2$$
$$=\frac{1}{2}\ln\left(\frac{\sec 2x+\tan 2x}{\sec 2x}\right)+C_2$$
$$\implies I_2=\frac{1}{2}\ln\left(1+\sin 2x\right)+C_2$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365416",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
Value of an expression with cube root radical What is the value of the following expression?
$$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}$$
| $$\sqrt[3]{\ 17\sqrt{5}+38} - \sqrt[3]{17\sqrt{5}-38}=$$
$$\sqrt[3]{8+12\sqrt{5}+30+5\sqrt{5}} - \sqrt[3]{-8+12\sqrt{5}-30+5\sqrt{5}}=$$
$$\sqrt[3]{8+12\sqrt{5}+6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3} - \sqrt[3]{-8+12\sqrt{5}-6\left(\sqrt{5}\right)^2+\left(\sqrt{5}\right)^3}=$$
$$\sqrt[3]{\left(\sqrt{5}+2\right)^3} - \sqrt[3]{\left(\sqrt{5}-2\right)^3}=$$
$$\left({\left(\sqrt{5}+2\right)^3}\right)^{\frac{1}{3}} - \left({\left(\sqrt{5}-2\right)^3}\right)^{\frac{1}{3}}=$$
$$\left(\sqrt{5}+2\right) -\left(\sqrt{5}-2\right)=$$
$$\left(\sqrt{5}+2\right) +\left(2-\sqrt{5}\right)=$$
$$2+2+\sqrt{5}-\sqrt{5}=2+2=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365489",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 7,
"answer_id": 4
} |
Finding $F(x)$ from $F(kx),$ where $F(x)$ is the antiderivative of the function $f(x)$. I have that $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1$, and I would like to find $F(x)$.
Attempt
Since $F(e^{x}x) = e^{x}x^{2} - e^{x}x + e^{x} - 1,$ $F(t) = \alpha_{1}t^{\beta_{1}} + \alpha_{2}t^{\beta_{2}} + \alpha_{3}t^{\beta_{3}} + \alpha_{4}t^{\beta_{4}}.$
Let $t = e^{x}x,$ which means that $F(t) = \alpha_{1}(e^{x}x)^{\beta_{1}} + \alpha_{2}(e^{x}x)^{\beta_{2}} + \alpha_{3}(e^{x}x)^{\beta_{3}} + \alpha_{4}(e^{x}x)^{\beta_{4}} = e^{x}x^{2} - e^{x}x + e^{x} - 1.$
Therefore, $\alpha_{2} = -1, \alpha_{4} = -1, \beta_{2} = 1,$ and $\beta_{4} = 0.$
$F(t) = \alpha_{1}t^{\beta_{1}} - t + \alpha_{3}t^{\beta_{3}} - 1 = e^{x}x^{2} - e^{x}x + e^{x} - 1.$
| Ok, let's summarize the comments in an answer. Differentiating expression for $F(xe^x)$ we get $$(e^x+xe^x)f(xe^x) = x(e^x + xe^x)$$ and thus, for $x\neq -1$ we have $$f(xe^x) = x \implies f(x) = W(x)$$ and consequently, $$F(x) = \displaystyle\int_a^x W(t)\ dt$$ Now, to determine $a$:
$$\begin{align*}
F(xe^x) &= \displaystyle\int_a^{xe^x} W(t)\ dt \\
&= \left[ t = ue^u,\ dt = (u+1)e^u \right] \\
&= \displaystyle\int_{W(a)}^x u(u+1)e^u\ du\\
&= (x^2 - x + 1)e^x - (W(a)^2 - W(a) + 1)e^{W(a)} \implies W(a) = 0 \implies a = 0
\end{align*}$$ so, $$F(x) = \displaystyle\int_0^x W(t)\ dt$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1365587",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Find product limit of this recursively-defined sequence?
Problem： if $a_1=3$, $a_n=2a_{n-1}^2-1$, $n\ge2$, find the limit of this expression:
$$\lim\limits_{n \to ∞} \prod\limits_{k=1}^{n-1} (1+\frac {1}{a_k})$$
The original problem asks to find this:
$$\lim\limits_{n \to ∞} \frac {a_n}{2^na_1a_2.....a_{n-1}}$$
The solution uses relation $a_n^2-1=2a_{n-1}^2×2(a_{n-1}^2-1)$ recursively and find out that the limit is $\sqrt2$;
while I believe there got to be some other ways to do this.
So I start by finding a simpler form of relation from expression $a_n=2a_{n-1}^2-1$ which is $a_n-1=2(a_{n-1}+1)(a_{n-1}-1)$, and finally I come to the product expression and obviously got stuck, I just don't have enough weaponry to deal with problems like this, trust me I always stumble on this type...
How should one correctly use the recursive relation to gradually approach the result? Is it mostly intuition or experience? And what are the first thing you tried when you first see this problem?
| Nice question, this following is my answer,(if I can't some wrong)
Let $a_{1}=\dfrac{1}{2}(x+\dfrac{1}{x})(x>1)\Longrightarrow x=3+2\sqrt{2}$
since
$$a_{2}=2a^2_{1}-1=\dfrac{1}{2}\left(x^2+\dfrac{1}{x^2}\right)$$
$$a_{3}=2a^2_{2}-1=\dfrac{1}{2}\left(x^4+\dfrac{1}{x^4}\right)$$
by indution you have
$$a_{n}=\dfrac{1}{2}\left(x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}\right)$$
and
$$\prod_{k=1}^{n}\left(1+\dfrac{1}{a_{k}}\right)=\prod_{k=1}^{n}\dfrac{a_{k}+1}{a_{k}}=\prod_{k=1}^{n}\dfrac{2a^2_{k-1}}{a_{k}}=\dfrac{2^{n-1}(a_{1}+1)}{a_{n}}\prod_{k=1}^{n-1}a_{k}$$
since
\begin{align*}\prod_{k=1}^{n-1}a_{k}&=\dfrac{1}{2^{n-1}}\left(x+\dfrac{1}{x}\right)\left(x^2+\dfrac{1}{x^2}\right)\cdots \left(x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}\right)\\
&=\dfrac{1}{2^{n-1}}\dfrac{x^{2^n}-\dfrac{1}{x^{2^n}}}{x-\dfrac{1}{x}}
\end{align*}
so
$$\prod_{k=1}^{n}\left(1+\dfrac{1}{a_{k}}\right)=\dfrac{x+1/x+2}{x-1/x}\cdot\dfrac{x^{2^n}-\dfrac{1}{x^{2^n}}}{x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}}\to\dfrac{(x+1)}{x-1}=\sqrt{2},n\to+\infty$$
becasuse
$$\lim_{n\to\infty}\dfrac{x^{2^n}-\dfrac{1}{x^{2^n}}}{x^{2^{n-1}}+\dfrac{1}{x^{2^{n-1}}}}=1,x>1$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366343",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 1,
"answer_id": 0
} |
Nature of the roots of quadratic equation Here is the problem that I need to prove:
If $x$ is real and $\displaystyle{\ p = \frac{3(x^2+1)}{(2x-1)}}$, prove that $\ p^2-3(p+3) \geq 0$
Here is what I did:
\begin{align*}
p(2x-1)=3(x^2+1) \\
3x^2 - 2px + (p+3)=0 \\
b^2 - 4ac = 4(p^2-3(p+3))
\end{align*}
By inspection I can see that $p^2 > -3(p+3)$ for almost all values of $p \ $, therefore
$p^2-3(p+3) > 0 $. However, the question asks to show that $\ p^2-3(p+3) \geq 0$
If I make $p^2 = 3(p+3)$ I can find roots and so $\ p^2-3(p+3) = 0$, when $\displaystyle{p = \frac{3 \pm \sqrt{45}}{2}}$. Therefore $\ p^2-3(p+3) \geq 0$
Having done this, how can I mathematically show that $p^2$ is never $<$ than $3(p+3)$? Because I am not satisfied with just saying that by inspection $p^2$ is greater than $3(p+3)$.
Thank you
| $$3x^2-2px+(p+3) = 0 \\
9x^2-6px+3(p+3) = 0 \\
(3x-p)^2+(3(p+3)-p^2)=0 \\
p^2-3(p+3) = (3x-p)^2 \ge 0$$
Furthermore you can see that $p^2-3(p+3) = 0$ when $p=3x$ and $x^2 = x+1$, i.e. $x = \frac{1 \pm \sqrt 5}2$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366447",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 4,
"answer_id": 1
} |
How to solve $\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ$ Question:
$ \sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6° $
I have partially solved this:-
$$ \sin78^\circ-\sin42^\circ +\sin6^\circ-\sin66^\circ $$
$$ 2\cos\left(\frac{78^\circ+42^\circ}{2}\right) \sin\left(\frac{78^\circ-42^\circ}{2}\right) + 2\cos\left(\frac{6^\circ+66^\circ}{2}\right)\sin\left(\frac{6^\circ-66^\circ}{2}\right) $$
$$ 2\cos(60^\circ)\sin(18^\circ) + 2\cos(36^\circ)\sin(-30^\circ) $$
$$ 2\frac{1}{2}\sin(18^\circ) - 2\cos(36^\circ)\cdot\frac{1}{2} $$
$$ \sin(18^\circ) - \cos(36^\circ) $$
At this point I had to use a calculator. Does anyone know a way to solve it without a calculator.Thanks in advance.
| As $\sin(-A)=-\sin A,\sin(180^\circ-B)=\sin B$
$S=\sin78^\circ-\sin66^\circ-\sin42^\circ+\sin6^\circ $
$=\sin(-138^\circ)+\sin(-66^\circ)+\sin6°+\sin78^\circ $
Observe that the angles are in Arithmetic Progression with common difference $=72^\circ$
Using How can we sum up $\sin$ and $\cos$ series when the angles are in arithmetic progression?
and Werner Formula: $2\sin A\sin B=\cos(A-B)-\cos(A+B),$
and finally Prosthaphaeresis Formula, $\cos C+\cos D=2\cos\dfrac{C+D}2\cos\dfrac{C-D}2$
$2\sin36^\circ\cdot S=-2\cos54^\circ\cos60^\circ\implies S=-\dfrac12$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366530",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 5,
"answer_id": 4
} |
Find the number of sets of $(a,b,c)$ for $\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{29}{72}$
If $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{29}{72},\ \ c<b<a<60,\ \ \{a,b,c\}\in\mathbb{N} $.
How many sets of $(a,b,c)$ exists ?
Options
$a.)\ 3 \quad \quad \quad \quad \quad b.)\ 4 \\
c.)\ 5 \quad \quad \quad \quad \quad \color{green}{d.)\ 6} \\ $
by trial and error i found
$$\begin{array}{c|c}
2 & 72 \\ \hline
2 &36 \\ \hline
2 &18 \\ \hline
3 &9 \\ \hline
3 &3 \\ \hline
&1\\
\end{array}$$
$\\~\\~\\$
$$\dfrac{29}{72}=\dfrac{18}{72}+\dfrac{9}{72}+\dfrac{2}{72}= \dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}\\~\\
\implies (a,b,c)=(36,8,4)$$
This question is from chapter quadratic equations.
I look for a short and simple way.
I have studied maths up to $12$th grade.
| In this type of problem,
you have to go through cases,
usually on the extreme variables.
Initially,
$\dfrac{1}{c} < \dfrac{29}{72}$,
so
$c > \dfrac{72}{29}
=2+\dfrac{14}{29}
$,
so $c \ge 3$.
In the other direction,
since
$\dfrac{1}{c}+\dfrac{1}{b}+\dfrac{1}{a}
< \dfrac{3}{c}
$,
$\dfrac{3}{c} > \dfrac{29}{72}$
or
$c < \dfrac{216}{29}
$
or $c \le 7$.
Looking at $a$,
$\dfrac{3}{a} < \dfrac{29}{72}$,
or $a \ge 8$.
For each $3 \le c \le 7$,
compute
$d
=\dfrac{72}{29}-\dfrac{1}{c}
$.
Then
$\dfrac{1}{a}+\dfrac{1}{b} = d$,
so
$\dfrac{2}{a} < d < \dfrac{2}{b}$
or
$c+1 \le b < \dfrac{2}{d}$.
For each $c+1 \le b < \dfrac{2}{d}$,
compute
$\dfrac{1}{d}-\dfrac{1}{b}$
and see if that
is of the form
$\dfrac{1}{a}$
for $a > b$.
I'll leave the actual computation
to you.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1366775",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "8",
"answer_count": 2,
"answer_id": 0
} |
Integrate area of the shadow? Today I found an interesting article here. It computes (approximately) area of the shadow.
I was wondering what is exact value of the area. My first thought was to use integrals but it doesn't seem to be easy. How to compute this using integrals?
Only thing I know is that $$S\approx 2.92$$
| There is of course a much easier approach that does not require any integrals whatsoever. Let the points of intersection of the circles be $A$ and $B$, the center of the small circle be $C=(1,1)$ and the center of the large circle, i.e., the lower left vertex of the square be $O=(0,0)$. (We will scale the resulting area to the original area of the square; in this case, we will multiply our result by $5$.)
The shaded area $K$ is equal to $K = K_1-K_2 + 2 K_3$, where $K_1$ is the area of the sector $CAB$, $K_2$ is the area of the sector $OAB$, and $K_3$ is the area of the triangle $\Delta OAC$. We have that
$$K_1 = \frac12 \cdot 1^2 \cdot \theta$$
$$K_2 = \frac12 \cdot 2^2 \cdot \phi$$
$$K_3 = \sqrt{s (s-2)(s-1)(s-\sqrt{2})} $$
where $\theta$ is the angle subtended by the sector $CAB$, $\phi$ is the angle subtended by the sector $OAB$, and $s$ is the semiperimeter of triangle $\Delta OAC$. It should be clear, then, that $K_3 = \sqrt{7}/4$.
We just need to find the angles $\theta$ and $\phi$. We do this by finding the points of intersection $A$ and $B$ and applying simple trigonometry. The points of intersection are found by solving the equations
$$(x-1)^2+(y-1)^2=1$$
$$x^2+y^2=4$$
The result is that $A = \left (\frac{5-\sqrt{7}}{4},\frac{5+\sqrt{7}}{4} \right )$ and $B = \left (\frac{5+\sqrt{7}}{4},\frac{5-\sqrt{7}}{4} \right )$. (The symmetry is expected.) Finding the angles is a relatively simple affair, so long as one has the proper respect for the range of our inverse trig functions. The result is that
$$\theta = \frac{\pi}{2}+2 \arctan{\left (\frac{\sqrt{7}-1}{\sqrt{7}+1} \right )} $$
$$\phi = \frac{\pi}{2}-2 \arctan{\left (\frac{5-\sqrt{7}}{5+\sqrt{7}} \right )} $$
The final result for the area is then
$$K = 4 \arctan{\left (\frac{5-\sqrt{7}}{5+\sqrt{7}} \right )} + \arctan{\left (\frac{\sqrt{7}-1}{\sqrt{7}+1} \right )} + \frac{\sqrt{7}}{2} - \frac{3 \pi}{4} $$
To compare with the OP's monte carlo result, I get from WA that $5 K \approx 2.927625 \dots$
ADDENDUM
Actually, there is a far more elegant and simpler method of attack. We don't need to find the points $A$ and $B$. Rather, we just find the angles $\theta$ and $\phi$ from the law of cosines and sines, respectively, using $\Delta OAC$. Using the law of cosines, I find that $\cos{(\angle OCA)} = -1/(2 \sqrt{2})$, so that $ \sin{(\angle OCA)} = \sqrt{7}/(2 \sqrt{2}) $. Thus, $\angle OCA = \pi - \arcsin{(\sqrt{7}/(2 \sqrt{2}))}$ and therefore $\theta = 2 \arcsin{(\sqrt{7}/(2 \sqrt{2}))}$. Using the law of sines, we find that $\phi = 2 \arcsin{(\sqrt{7}/(4 \sqrt{2}))}$. Thus,
$$K = \arcsin{\left (\sqrt{\frac{7}{8}} \right )} - 4 \arcsin{\left (\frac12 \sqrt{\frac{7}{8}} \right )}+ \frac{\sqrt{7}}{2}$$
This produces the exact same result as above, and much more elegantly.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1368691",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 1,
"answer_id": 0
} |
Finding $\frac {a}{b} + \frac {b}{c} + \frac {c}{a}$ where $a, b, c$ are the roots of a cubic equation, without solving the cubic equation itself Suppose that we have a equation of third degree as follows:
$$
x^3-3x+1=0
$$
Let $a, b, c$ be the roots of the above equation, such that $a < b < c$ holds. How can we find the answer of the following expression, without solving the original equation?
$$
\frac {a}{b} + \frac {b}{c} + \frac {c}{a}
$$
| We know that every symmetric function of the roots $a,b,c$ can be evaluated in terms of the elementary symmetric functions:
$$ e_1=a+b+c=0,\quad e_2=ab+ac+bc=-3,\quad e_3=abc=-1$$
or the power sums:
$$ p_1=e_1=0,\quad p_2=a^2+b^2+c^2 = 6,\quad p_3=a^3+b^3+c^3=3e_1-3=-3.$$
Now:
$$ g(a,b,c)=\frac{a}{b}+\frac{b}{c}+\frac{c}{a}=-(a^2 c+b^2 a+c^2 b)$$
is not a symmetric function of $a,b,c$, and neither it is:
$$ h(a,b,c)=\frac{a}{c}+\frac{b}{a}+\frac{c}{b}=-(a^2 b+b^2 c+c^2 a),$$
but both $g+h$ and $g\cdot h$ are. So the strategy is just to find $g+h$ and $g\cdot h$ in terms of $e_1,e_2,e_3$, then solve a quadratic equation to find $\{g,h\}$ and recognize $g$ from the constraint $a<b<c$.
We have:
$$\begin{eqnarray*} g+h &=& -(a^2(b+c)+b^2(a+c)+c^2(a+b))\\ &=& (a^3+b^3+c^3)-(a^2+b^2+c^2)(a+b+c)\\&=&p_3-p_2 p_1=-3,\end{eqnarray*}$$
$$\begin{eqnarray*} g\cdot h &=& e_3 p_3 + 3a^2b^2c^2 + e_3^3\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\\&=&6-\left(\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}-3\right)\\&=&9-3\left(\frac{e_2^2}{e_3^2}-2\frac{e_1}{e_3}\right)=-18,\end{eqnarray*}$$
hence $g,h$ are the roots of $z^2+3z-18$, and $\{g,h\}=\{-6,3\}$. Since $e_3<0$, we have $a<0<b<c$, from which:
$$ -g = a^2 c+b^2 a+ c^2 b = (b+c)^2 c-b^2(b+c)+c^2 b = c^3-b^3+3bc^2 > 0 $$
and $\color{red}{g=-6}$ follows.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1370364",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 6,
"answer_id": 3
} |
The expression $(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ where $q\ne 1$, equals The expression
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$
where $q\ne 1$, equals
(A) $\frac{1-q^{128}}{1-q}$
(B) $\frac{1-q^{64}}{1-q}$
(C) $\frac{1-q^{2^{1+2+\dots +6}}}{1-q}$
(D) none of the foregoing expressions
What I have done
$(1+q)(1+q^2)(1+q^4)(1+q^8)(1+q^{16})(1+q^{32})(1+q^{64})$ is a polynomial of degree $127$. Now the highest degree of the polynomial in option (A), (B) and (C) is $127$, $63$ and $41$ respectively. And therefore (A) is the correct answer.
I get to the correct answer but I don't think that my way of doing is correct. I mean what if, if option (B) was $\frac{1+q^{128}}{1-q}$.
Please show how should I approach to the problem to get to the correct answer without any confusion.
| $$(1+q)(1+q^2)=1+q+q^2+q^3=\frac{1-q^4}{1-q}$$
So,
$$(1+q)(1+q^2)(1+q^4)=\frac{(1-q^4)(1+q^4)}{1-q}=\frac{1-q^8}{1-q}$$
So,
$$(1+q)(1+q^2)(1+q^4)(1+q^8)=\frac{(1-q^8)(1+q^8)}{1-q}=\frac{1-q^{16}}{1-q}$$
Can you see the pattern?
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371153",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 0
} |
Find the $\int \frac{(1-y^2)}{(1+y^2)}dy$ $\int \frac{(1-y^2)}{(1+y^2)}dy$ first I tried to divide then I got 1-$\frac{2y^2}{1+y^2}$ and i still can't integrate it.
| Let $y = \tan \theta \Rightarrow I = \displaystyle \int \dfrac{\cos^2\theta - \sin^2 \theta}{\cos^2 \theta}d\theta= \displaystyle \int (1-\tan^2 \theta)d\theta = \displaystyle \int (2-\sec^2\theta)d\theta= 2\theta - \tan \theta + C=2\tan^{-1}y-y+C$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1371981",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 3,
"answer_id": 2
} |
Range of an inverse trigonometric function Find the range of $f(x)=\arccos\sqrt {x^2+3x+1}+\arccos\sqrt {x^2+3x}$
My attempt is:I first found domain,
$x^2+3x\geq0$
$x\leq-3$ or $x\geq0$...........(1)
$x^2+3x+1\geq0$
$x\leq\frac{-3-\sqrt5}{2}$ or $x\geq \frac{-3+\sqrt5}{2}$...........(2)
From (1) and (2),
domain is $x\leq-3$ or $x\geq0$
but could not solve further..Any help will be greatly appreciated.
| Notice,
for the defined function $\cos^{-1}\sqrt{x^2+3x+1}$$$\implies -1\leq \sqrt{x^2+3x+1}\leq 1$$ $$\implies x^2+3x+1\geq 1$$ $$\implies x^2+3x\geq 0$$ $$\implies x(x+3)\geq 0$$ The above inequality holds for all $x$ such that $$x\in (-\infty, -3]\cup [0, \infty)\tag 1$$
Again, for the defined function $\cos^{-1}\sqrt{x^2+3x}$$$\implies -1\leq \sqrt{x^2+3x}\leq 1$$ $$\implies x^2+3x\geq 1$$ $$\implies x^2+3x-1 \geq 0$$
Solving the quadratic equation $x^2+3x-1=0$ for $x$, we have $$\left(x-\frac{-3+\sqrt{13}}{2}\right)\left(x-\frac{-3-\sqrt{13}}{2}\right)\geq 0$$ The above inequality holds for all $x$ such that $$x\in\left(-\infty, \frac{-3-\sqrt{13}}{2}\right]\cup \left[\frac{-3+\sqrt{13}}{2}, \infty\right)\tag 2$$ Form (1) & (2), we get domain of function $f(x)$ $$\color{blue}{x\in (-\infty, -3]\cup [0, \infty)}$$
Now, setting extreme point $x=-3$ in $f(x)$, we get
$$f(-3)=\cos^{-1}\sqrt{(-3)^2+3(-3)+1}+\cos^{-1}\sqrt{(-3)^2+3(-3)}$$ $$=\cos^{-1}(1)+\cos^{-1}(0)=\frac{\pi}{2}$$ Setting extreme point $x=0$ & $x=0$ in $f(x)$, we get $$f(0)=\cos^{-1}\sqrt{(0)^2+3(0)+1}+\cos^{-1}\sqrt{(0)^2+3(0)}$$ $$=\cos^{-1}(1)+\cos^{-1}(0)=\frac{\pi}{2}$$ In both cases we get equal values hence the range of $f(x)$ is $\color{blue}{\frac{\pi}{2}}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373523",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 2,
"answer_id": 1
} |
Power serie of $f'/f$ It seems that I'm [censored] blind in searching the power series expansion of $$f(x):=\frac{2x-2}{x^2-2x+4}$$ in $x=0$.
I've tried a lot, e.g., partiell fraction decomposition, or regarding $f(x)=\left(\log((x+1)^2+3)\right)'$ -- without success.
I' sure that I'm overseeing a tiny little missing link; dear colleagues, please give me a hint.
| Since
$$
x^2-2x+4=(x-1)^2+3=(x-2u)(x-2\bar{u}),
$$
with
$$
u=e^{i\frac\pi3},
$$
we have
\begin{eqnarray}
f(x)&=&\frac{2x-2}{(x-2u)(x-2\bar{u})}=\frac{1}{x-2u}+\frac{1}{x-2\bar{u}}=-\frac{\bar{u}}{2}\cdot\frac{1}{1-\frac{\bar{u}}{2}x}-\frac{u}{2}\cdot\frac{1}{1-\frac{u}{2}x}\\
&=&-\frac{\bar{u}}{2}\sum_{n=0}^\infty\left(\frac{\bar{u}}{2}\right)^nx^n-\frac{u}{2}\sum_{n=0}^\infty\left(\frac{u}{2}\right)^nx^n
=-\sum_{n=0}^\infty\left(\frac{\bar{u}}{2}\right)^{n+1}x^n-\sum_{n=0}^\infty\left(\frac{u}{2}\right)^{n+1}x^n\\
&=&-\sum_{n=0}^\infty\frac{u^{n+1}+\bar{u}^{n+1}}{2^{n+1}}x^n=-\sum_{n=0}^\infty\frac{1}{2^n}\cos\left(\frac{n+1}{3}\pi\right)x^n.
\end{eqnarray}
Also
\begin{eqnarray}
\frac{f'(x)}{f(x)}&=&\left[\ln |f(x)|\right]'=[\ln2+\ln|x-1|-\ln|x^2-2x+4|]'\\
&=&\frac{1}{x-1}-\frac{2x-2}{x^2-2x+4}=-\frac{1}{1-x}-f(x)\\
&=&-\sum_{n=0}^\infty x^n-\sum_{n=0}^\infty\frac{1}{2^n}\cos\left(\frac{n+1}{3}\pi\right)x^n\\
&=&-\sum_{n=0}^\infty\left[1+\frac{1}{2^n}\cos\left(\frac{n+1}{3}\pi\right)\right]x^n.
\end{eqnarray}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1373729",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 5,
"answer_id": 4
} |
Counting with potency and simplifing So I have the question: Simplify $(6^{n+4}) / 2^{n+5} \cdot 3^{n+2}$
I tried to write the expresion as $6^{n+4-(2n+7)}/6$, but that is wrong.
So I guess I should factor it out. Perhaps $2^{2} + 2^{n+4}$ / $2^{n+5} \cdot 3^{n+2}$
Can you show me how to expand this expresion?
| $$\frac { { 6 }^{ n+4 } }{ { 2 }^{ n+5 }{ 3 }^{ n+2 } } =\frac { \left( 2\cdot 3 \right) ^{ n+4 } }{ { 2 }^{ n+5 }{ 3 }^{ n+2 } } =\frac { { 2 }^{ n+4 }{ 3 }^{ n+4 } }{ { 2 }^{ n+5 }{ 3 }^{ n+2 } } ={ 2 }^{ n+4-n-5 }{ 3 }^{ n+4-n-2 }={ 2 }^{ -1 }{ 3 }^{ 2 }=\frac { 9 }{ 2 } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375369",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
Evaluate $\frac 1{1+\sqrt2+\sqrt3} + \frac 1{1-\sqrt2+\sqrt3} + \frac 1{1+\sqrt2-\sqrt3} + \frac 1{1-\sqrt2-\sqrt3}$
How to evalute this equation without using calculator?
| Another way : $$\begin{align}\\&\frac{1}{1+\sqrt 2+\sqrt 3}+\frac{1}{1-\sqrt 2+\sqrt 3}+\frac{1}{1+\sqrt 2-\sqrt 3}+\frac{1}{1-\sqrt 2-\sqrt 3}\\&=\left(\frac{1}{1+\sqrt 2+\sqrt 3}+\frac{1}{1+\sqrt 2-\sqrt 3}\right)+\left(\frac{1}{1-\sqrt 2+\sqrt 3}+\frac{1}{1-\sqrt 2-\sqrt 3}\right)\\&=\frac{1+\sqrt 2-\sqrt 3+1+\sqrt 2+\sqrt 3}{(1+\sqrt 2+\sqrt 3)(1+\sqrt 2-\sqrt 3)}+\frac{1-\sqrt 2-\sqrt 3+1-\sqrt 2+\sqrt 3}{(1-\sqrt 2+\sqrt 3)(1-\sqrt 2-\sqrt 3)}\\&=\frac{2+2\sqrt 2}{(1+\sqrt 2)^2-3}+\frac{2-2\sqrt 2}{(1-\sqrt 2)^2-3}\\&=\frac{2+2\sqrt 2}{2\sqrt 2}+\frac{2-2\sqrt 2}{-2\sqrt 2}\\&=2\end{align}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1375531",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 6,
"answer_id": 0
} |
remainder of $a^2+3a+4$ divided by 7
If the remainder of $a$ is divided by $7$ is $6$, find the remainder when $a^2+3a+4$ is divided by 7
(A)$2$ (B)$3$ (C)$4$ (D)$5$ (E)$6$
if $a = 6$, then $6^2 + 3(6) + 4 = 58$, and $a^2+3a+4 \equiv 2 \pmod 7$
if $a = 13$, then $13^2 + 3(13) + 4 = 212$, and $a^2+3a+4 \equiv 2 \pmod 7$
thus, we can say that any number, $a$ that divided by 7 has remainder of 6, the remainder of $a^2 + 3a + 4$ is 2.
is there any other way to calculate it? (Let say it given b as the remainder of a divided by 7, not 6)
| The remainder of $a^2+3a+4$ divided by $7$ is sum of the remainder of each terms, modulo $7$.
So $a^2\equiv 1 \pmod{7}$ since $a=7k+6$ then $a^2=7l+1$; $\quad$
$3a\equiv 4 \pmod{7}$ since $3a=21k+18=21k+14+4$ and clearly $4\equiv 4 \pmod{7}$.
Finally $1+4+4 \equiv 2 \pmod{7}$ then the remainder is $2$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376498",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 2
} |
Is the inequality $| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ true? I'm having some trouble deciding whether this inequality is true or not...
$| \sqrt[3]{x^2} - \sqrt[3]{y^2} | \le \sqrt[3]{|x -y|^2}$ for $x, y \in \mathbb{R}.$
| It is shown in Is $x^t$ subadditive for $t \in [0,1]$? that we have the inequality
$$(a+b)^t \le a^t + b^t \tag{1}$$
for all nonnegative $a,b$ and all $t \in [0,1]$.
For your inequality, let us assume without loss of generality that $x \ge y$. Applying the above inequality (1) with $a=y$, $b=x-y$, and $t=2/3$ we get
$$x^{2/3} \le y^{2/3} + (x-y)^{2/3}.$$
Subtracting $y^{2/3}$ from both sides,
$$x^{2/3} - y^{2/3} \le (x-y)^{2/3}.$$
Now since $x \ge y$ we have $x-y = |x-y|$. And since the function $x \mapsto x^{2/3}$ is increasing, we have $x^{2/3} \ge y^{2/3}$, so $x^{2/3} - y^{2/3} = |x^{2/3} - y^{2/3}|$. Thus we showed
$$|x^{2/3} - y^{2/3}| \le |x-y|^{2/3}$$
which is the desired inequality.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1376742",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 3,
"answer_id": 1
} |
Find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ How to find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ in the formal way? Numerically its value is $\approx 0.0217326$ and the partial sum formula contains the first derivative of the gamma function (by WolframAlpha).
| You may first write a partial fraction decomposition, giving
$$
\frac{1}{n(n+1)^2(n+2)}=\frac12\left(\frac{1}n-\frac1{n+2}\right)-\frac{1}{(1+n)^2}
$$ then obtain your sum, by telescoping for the first two terms above, then by identifying a celebrated series for the third term. You get
$$
\sum_2^N\frac{1}{n(n+1)^2(n+2)}=\frac{5}{12}-\frac{1}{2 (1+N)}-\frac{1}{2 (2+N)}-\sum_2^N\frac{1}{(n+1)^2}
$$ and, as $N \to \infty$,
$$
\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}=\frac53-\frac{\pi ^2}{6}=\color{red}{0.021732\cdots}.
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1378465",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 3,
"answer_id": 1
} |
A basic root numbers question If $\sqrt{x^2+5} - \sqrt{x^2-3} = 2$, then what is $\sqrt{x^2+5} + \sqrt{x^2-3}$?
| $$\sqrt{x^2+5} - \sqrt{x^2-3} = 2\\ \sqrt{x^2+5} = \sqrt{x^2-3} + 2\\ (\sqrt{x^2+5})^2 = (\sqrt{x^2-3})^2 + 2^2\\x^2+5=x^2-3+4-4\sqrt{x^2-3} \\5-1=-4\sqrt{x^2-3}\\\sqrt{x^2-3}=1 \\x^2-3=1\\ \rightarrow x=\pm 2\\\sqrt{x^2+5} + \sqrt{x^2-3}=\sqrt{(\pm2)^2+5} - \sqrt{(\pm2)^2-3}=3+1=4$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1380090",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
How to prove an identity (Trigonometry Angles--Pi/13) In this page http://mathworld.wolfram.com/TrigonometryAnglesPi13.html
I found equation (11) and (12).
$$\cos^2\frac{\pi}{13}+\cos^2\frac{3\pi}{13}+\cos^2\frac{4\pi}{13}=\frac{11+\sqrt{13}}{8}$$
$$\sin\frac{\pi}{13}+\sin\frac{3\pi}{13}+\sin\frac{4\pi}{13}=\sqrt{\frac{13+3\sqrt{13}}{8}}$$
How to prove it ?
Thanks in advances
| For the first one, let $\zeta=e^{2\pi i/13}$. This is a thirteenth root of $1$, so $1+\zeta+\zeta^2+...+\zeta^{12}=0$.
\begin{align}
A &=\cos^2(\pi/13)+\cos^2(3\pi/13)+\cos^2(4\pi/13)\\
B &=2A-3=\cos(2\pi/13)+\cos(6\pi/13)+\cos(8\pi/13)\\
2B &=\zeta+\zeta^{-1}+\zeta^3+\zeta^{-3}+\zeta^4+\zeta^{-4}\\
C &=-1-2B=\zeta^2+\zeta^{-2}+\zeta^5+\zeta^{-5}+\zeta^6+\zeta^{-6}\\
2BC &=3(\zeta+\zeta^2+...+\zeta^{12})=-3\\
& 2B(2B+1) *=3
\end{align}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381294",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 2,
"answer_id": 0
} |
Area of a triangle with sides $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$,$\sqrt{z^2+x^2}$ Sides of a triangle ABC are $\sqrt{x^2+y^2}$,$\sqrt{y^2+z^2}$ and $\sqrt{z^2+x^2}$ where x,y,z are non-zero real numbers,then area of triangle ABC is
(A)$\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$
(B)$\frac{1}{2}(x^2+y^2+z^2)$
(C)$\frac{1}{2}(xy+yz+zx)$
(D)$\frac{1}{2}(x+y+z)\sqrt{x^2+y^2+z^2}$
I tried applying Heron's formula but calculations are very messy and simplification is difficult.I could not think of any other method to find this area.Can someone assist me in solving this problem.
| Use cosine rule to find say $\angle C$ then use formula of area as follows
Area of $\triangle ABC$ $$=\frac{1}{2}(a)(b)\sin C=\frac{1}{2}(a)(b)\sqrt{1-(\cos C)^2}$$ $$=\frac{1}{2}(\sqrt{x^2+y^2})(\sqrt{y^2+z^2})\sqrt{1-\left(\frac{(\sqrt{x^2+y^2})^2+\sqrt{y^2+z^2})^2-(\sqrt{x^2+z^2})^2}{2\sqrt{x^2+y^2}\sqrt{y^2+z^2}}\right)^2}$$
$$=\frac{1}{2}(\sqrt{x^2+y^2})(\sqrt{y^2+z^2})\sqrt{\frac{4(x^2+y^2)(y^2+z^2)-(x^2+y^2+y^2+z^2-x^2-z^2)^2}{4(x^2+y^2)(y^2+z^2)}}$$
$$=\frac{1}{2}\frac{(\sqrt{x^2+y^2})(\sqrt{y^2+z^2})}{2(\sqrt{x^2+y^2})(\sqrt{y^2+z^2})}\sqrt{4x^2y^2+4y^4+4z^2x^2+4y^2z^2-(2y^2)^2}$$
$$=\frac{1}{4}\sqrt{4(x^2y^2+y^2z^2+z^2x^2)}$$ $$=\frac{1}{2}\sqrt{x^2y^2+y^2z^2+z^2x^2}$$
Option (A) is correct
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1381791",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
"answer_count": 4,
"answer_id": 2
} |
Resources on Variants of the Clausen Functions I am interested in locating more information about the Clausen functions. Specifically I am looking for the closed forms of the Gl-type (or Sl-type as they are sometimes called) and the alternating analogues of the Cl and Gl-type Clausen functions.
In other words, I am looking for the closed forms of:
$$\sum_{n=1}^{\infty}{\frac{\cos(nx)}{n^m}}$$
$$\sum_{n=1}^{\infty}{(-1)^{n-1}\frac{\cos(nx)}{n^m}}$$
$$\sum_{n=1}^{\infty}{\frac{\sin(nx)}{n^m}}$$
$$\sum_{n=1}^{\infty}{(-1)^{n-1}\frac{\sin(nx)}{n^m}}$$
So far I have found a few examples, in places such as Mathworld, or Gradshteyn & Rhyzhik, or Abramowitz & Stegun. None of these sources provide anywhere near a comprehensive collection of the Clausen and Alternating Clausen Functions in closed form.
I was hoping someone might know of a paper or book that could provide me with more information, especially on the Alternating Clausen Functions. The help would be greatly appreciated!
| Here's what I found:
let's pick the case where $m\quad =\quad 2k\quad /\quad k\in Z\\ $
As you know:
$\sum _{ n=1 }^{ \infty }{ \frac { { t }^{ n } }{ { n }^{ m } } } \quad ={ Li }_{ m }(t)$
Let's substitute $t={ e }^{ ix }$:
$\sum _{ n=1 }^{ \infty }{ \frac { { { e }^{ inx } } }{ { n }^{ m } } } \quad ={ Li }_{ m }({ e }^{ ix })$
We will use the following identities aswell:
$$
{ Li }_{ m }(z)\quad =\quad { (-1) }^{ m-1 }{ Li }_{ m }\left( \frac { 1 }{ z } \right) -\frac { { (2\pi i) }^{ m } }{ m! } { B }_{ m }\left( \frac { ln(-z) }{ 2\pi i } +\frac { 1 }{ 2 } \right) /z\notin (0,1) \quad (1)\\ \Re ({ Li }_{ m }(z))\quad =\quad \frac { 1 }{ 2 } \left( { Li }_{ m }(z)\quad +\quad { Li }_{ m }\left( \bar { z } \right) \right )\\
\Im \left( { Li }_{ m }(z) \right) \quad =\quad \frac { 1 }{ 2i } \left( { Li }_{ m }(z)-{ Li }_{ m }\left( \bar { z } \right) \right)
$$
So:
$$
\sum _{ n=1 }^{ \infty }{ \frac { \cos { (nx } ) }{ { n }^{ m } } \quad =\quad \Re \left( \sum _{ n=1 }^{ \infty }{ \frac { { { e }^{ inx } } }{ { n }^{ m } } } \right) } \\ \qquad \qquad \qquad \quad \quad \quad =\quad \Re \left( { Li }_{ m }\left( { e }^{ ix } \right) \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad \frac { 1 }{ 2 } \left( { Li }_{ m }\left( { e }^{ ix } \right) +{ Li }_{ m }\left( { e }^{ -ix } \right) \right)
$$
And using that m is even as we supposed we get from $(1)$:
$$
\frac { 1 }{ 2 } \left( { Li }_{ m }\left( { e }^{ ix } \right) +{ Li }_{ m }\left( { e }^{ -ix } \right) \right) \quad =\quad -\frac { 1 }{ 2 } \frac { { (2\pi i) }^{ m } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { ln(-{ e }^{ ix }) }{ 2\pi i } \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\frac { 1 }{ 2 } \frac { { (-1) }^{ \frac { m }{ 2 } }{ (2\pi ) }^{ m } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { i\left( x- \pi \right) }{ 2\pi i } \right) \\ \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad \quad =\quad -\frac { 1 }{ 2 } \frac { { (-1) }^{ \frac { m }{ 2 } }{ (2\pi ) }^{ m } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi } \right)
$$
So we get the following:
$$
\ \sum _{ n=1 }^{ \infty }{ \frac { \cos { \left( nx \right) } }{ { n }^{ m } } } =\quad -\frac { 1 }{ 2 } \frac { { (-1) }^{ \frac { m }{ 2 } }{ (2\pi ) }^{ m } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi } \right) \quad /\quad m\quad =\quad 2k
$$
For the alternating sum:
$$
\sum _{ n=1 }^{ \infty }{ \frac { { (-1) }^{ n-1 }\cos { (nx) } }{ { n }^{ m } } } \quad =\quad -\sum _{ n=1 }^{ \infty }{ \frac { { (-1) }^{ n }\cos { (nx) } }{ { n }^{ m } } } \\ \qquad \qquad \qquad \qquad \qquad \quad \quad =\quad -\Re \left( \sum _{ n=1 }^{ \infty }{ \frac { { (-1) }^{ n }{ e }^{ inx } }{ { n }^{ m } } } \right) \\ \quad \qquad \qquad \qquad \qquad \qquad \quad =\quad -\Re \left( \sum _{ n=1 }^{ \infty }{ \frac { { \left( { -e }^{ ix } \right) }^{ n } }{ { n }^{ m } } } \right) \\ \qquad \qquad \qquad \qquad \qquad \quad \quad =-\Re \left( { Li }_{ m }\left( -{ e }^{ ix } \right) \right)
$$
We will calculate the latter just as we did:
$$
-\Re \left( { Li }_{ m }\left( -{ e }^{ ix } \right) \right) \quad =\quad -\frac { 1 }{ 2 } \left( { Li }_{ m }\left( { -e }^{ ix } \right) \quad +\quad { Li }_{ m }\left( { -e }^{ -ix } \right) \right) \\ \qquad \qquad \qquad \qquad =\quad \frac { -1 }{ 2 } \left(- \frac { { (2\pi i) }^{ m } }{ m! } { B }_{ m }\left( \frac { ln\left( { e }^{ ix } \right) }{ 2\pi i } +\frac { 1 }{ 2 } \right) \right) \\ \qquad \qquad \qquad \qquad =\quad \frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m }{ 2 } } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { x }{ 2\pi } \right) \right)
$$
We conclude that:
$$
\sum _{ n=1 }^{ \infty }{ \frac { { (-1) }^{ n-1 }\cos { (nx) } }{ { n }^{ m } } } =\quad \frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m }{ 2 } } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { x }{ 2\pi } \right) \right) / m = 2k
$$
Now let's calculate $\sum _{ n=1 }^{ \infty }{ \frac { \sin { (nx) } }{ { n }^{ m } } }$ but for odd values of m using the same identities:
$m = 2k+1$
$$
\sum _{ n=1 }^{ \infty }{ \frac { \sin { (nx) } }{ { n }^{ m } } } =\Im \left( \sum _{ n=1 }^{ \infty }{ \frac { { e }^{ inx } }{ { n }^{ m } } } \right) \\ \qquad \qquad \qquad \quad \quad =\quad \Im \left( { Li }_{ m }\left( { e }^{ ix } \right) \right) \\ \qquad \qquad \qquad \quad \quad =\quad \frac { 1 }{ 2i } \left( { Li }_{ m }\left( { e }^{ ix } \right) -{ Li }_{ m }\left( { e }^{ -ix } \right) \right)
$$
So using $(1)$, we have:
$$
\frac { 1 }{ 2i } \left( { Li }_{ m }\left( { e }^{ ix } \right) -{ Li }_{ m }\left( { e }^{ -ix } \right) \right) \quad =\quad \frac { 1 }{ 2i } \left( \frac { -{ (2\pi i) }^{ m } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { ln(-{ e }^{ ix }) }{ 2\pi i } \right) \right) \\ \qquad \qquad \qquad \qquad \qquad \qquad =\quad -\frac { 1 }{ 2i } \left( \frac { { (2\pi ) }^{ m }{ (i) }^{ m } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { i(x-\pi ) }{ 2\pi i } \right) \right) \\ \qquad \qquad \qquad \qquad \qquad \qquad =\quad -\frac { 1 }{ 2i } \left( \frac { { (2\pi ) }^{ m }{ (i) }^{ 2k+1 } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi } \right) \right) \\ \qquad \qquad \qquad \qquad \qquad \qquad =\quad -\frac { 1 }{ 2i } \left( \frac { { (2\pi ) }^{ m }i{ (-1) }^{ k } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi } \right) \right) \\ \qquad \qquad \qquad \qquad \qquad \qquad =\quad -\frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m-1 }{ 2 } } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi } \right) \right)
$$
So our sum is:
$$
\sum _{ n=1 }^{ \infty }{ \frac { \sin { (nx) } }{ { n }^{ m } } } = \quad -\frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m-1 }{ 2 } } }{ m! } { B }_{ m }\left( \frac { x }{ 2\pi } \right) \right) \quad m = 2k+1
$$
And for the alternating sum it's the same technique:
$$
\sum _{ n=1 }^{ \infty }{ \frac { { (-1) }^{ n-1 }\sin { (nx) } }{ { n }^{ m } } } =\quad \frac { 1 }{ 2 } \left( \frac { { (2\pi ) }^{ m }{ (-1) }^{ \frac { m-1 }{ 2 } } }{ m! } { B }_{ m }\left( \frac { 1 }{ 2 } +\frac { x }{ 2\pi } \right) \right)
$$
This is all I can do for the time being, I'll try to find the sum for other values as well.
Reference:
*
*On a property of polylogarithm
*http://functions.wolfram.com/ZetaFunctionsandPolylogarithms/PolyLog/17/ShowAll.html
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1384710",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Find $\lim\limits_{x\to 0}\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}$ I use L'Hospitals rule, but can't get the correct limit.
Derivative of numerator in function is
$$\frac{-3x^2+4x-\sqrt{1-x^2}+1}{\sqrt{(4-2x)(1-x^2)}}$$
and derivative of denominator is
$$\frac{-3x^2+2x-2\sqrt{4-x^2}+4}{2\sqrt{(1-x)(4-x^2)}}$$
Now, L'Hospitals rule must be applied again. Is there some easier way to compute the limit?
Limit should be $L=4$
| $\lim\limits_{x\to 0}\left(\frac{\sqrt{2(2-x)}(1-\sqrt{1-x^2})}{\sqrt{1-x}(2-\sqrt{4-x^2})}\right) = \lim\limits_{x\to 0}\left(\frac{\sqrt{2(2-x)}(2 + \sqrt{4-x^2})}{\sqrt{1-x}(1+\sqrt{1-x^2})}\right) = 4$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385320",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "3",
"answer_count": 6,
"answer_id": 2
} |
last two digits of $14^{5532}$? This is a exam question, something related to network security, I have no clue how to solve this!
Last two digits of $7^4$ and $3^{20}$ is $01$, what is the last two digits of $14^{5532}$?
| The OP quickly realizes that we can't write $14^k \equiv 1 \pmod{100}$ with $k \gt 0$, but there are still relations to be found in (multiplicative) semigroups.
If the last digit of integers $a$ and $b$ end in $6$, then the last digit of the product ends in $6$. This motivates us to write
$\quad 14^2 \equiv 96 \equiv -4 \pmod{100}$
and
$\quad 96^2 \equiv 16 \pmod{100}$
$\quad 96^3 \equiv 36 \pmod{100}$
$\quad 96^4 \equiv 56 \pmod{100}$
$\quad 96^5 \equiv 76 \pmod{100}$
$\quad 96^6 \equiv 96 \pmod{100}$
and
$\quad 96^{36} \equiv 96 \pmod{100}$
$\quad 96^{216} \equiv 96 \pmod{100}$
$\quad 96^{1296} \equiv 96 \pmod{100}$
So
$\; 14^{5532} = (14^2)^{2766} \equiv 96^{2766} \equiv (96^{1296})^2 (96^{174}) \equiv 96^{176} \equiv (96^{36})^4 (96^{32}) \equiv 96^{36} \equiv 96 \pmod{100}$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385396",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
"answer_count": 6,
"answer_id": 3
} |
Radical under Radical expression how to find the sum of $\sqrt{\frac54 + \sqrt{\frac32}} + \sqrt{\frac54 - \sqrt{\frac32}} $ ? Is there a method to solve these kind of equations ?
| The general idea is to move along:
$$\sqrt{a}+\sqrt{r^2\,a} = \sqrt{a}+r\,\sqrt{a} = (1+r)\,\sqrt{a} = \sqrt{(1+r)^2\,a}$$
I am using the following systematic method as here, to first get the quotient:
$$r^2 = \frac{5/4+\sqrt{3/2}}{5/4-\sqrt{3/2}} = 49+20\sqrt{6} = (5+2\sqrt{6})^2$$
Therefore we have:
$$\sqrt{5/4-\sqrt{3/2}}+\sqrt{5/4+\sqrt{3/2}}= $$
$$\sqrt{(1+5+2\sqrt{6})^2\,(5/4-\sqrt{3/2})} = $$
$$\sqrt{(60+24\sqrt{6})\,(5/4-\sqrt{3/2})} = \sqrt{3}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/1385570",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 3
} |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.