Q stringlengths 70 13.7k | A stringlengths 28 13.2k | meta dict |
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Sum of series $\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$
The problem is to find the sum of series $$\frac{x}{1-x^2} + \frac{x^2}{1-x^4} + \frac{x^4}{1-x^8} +\ldots$$
to infinite terms, if $|x|<1$
I tried taking $\frac{x}{1-x^2}$ outside but could not solve it.
How to proceed further?
| If $|x|<1$, for any $k\in\mathbb{N}$ we have
$$ \frac{x^{2^k}}{1-x^{2^{k+1}}} = x^{2^k}+x^{3\cdot 2^k}+x^{5\cdot 2^k}+\ldots\tag{A}$$
but given any $n\in\mathbb{N}^+$, there is a unique representation of $n$ as $2^k\cdot d$, with $k\in\mathbb{N}$ and $d\in 2\mathbb{N}+1$. It follows that:
$$ \sum_{k\geq 0}\frac{x^{2^k}}{1-x^{2^{k+1}}} = \sum_{n\geq 1} x^n = \color{red}{\frac{x}{1-x}}.\tag{B}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Let $a,b,c$ be positive real numbers then prove the inequality $\frac{abc}{(a+1)(a+b)(b+c)(c+16)}≤\frac1{81}$ For positive real numbers $a, b, c$ prove that
$$\frac{abc}{(a+1)(a+b)(b+c)(c+16)}≤\frac1{81}$$
I subjected $a+1$, $1+\frac{b}{a}$, $1+\frac{c}{b}$ and $1+\frac{16}{c}$ to AM-GM inequality and obtained the following equations
$$a+1≥2\sqrt{a}$$
$$1+\frac{b}{a}≥2\sqrt{\frac{b}{a}}$$
$$1+\frac{c}{b}≥2\sqrt{\frac{c}{b}}$$
$$1+\frac{16}{c}≥2\sqrt{\frac{16}{c}}$$
Multiplying, we get,
$$(a+1)(1+\frac{b}{a})(1+\frac{c}{b})(1+\frac{16}{c})≥16\sqrt{16}$$
or $$\frac{(a+1)(a+b)(b+c)(c+16)}{abc}≥64$$
i.e.$$\frac{abc}{(a+1)(a+b)(b+c)(c+16)}≤\frac1{64}$$
I am not arriving at the desired result and have realised that the question should be solved in another way. Please let me know how to solve this and also mention where I missed out.
| Take $a=2x$, $b=4y$ and $c=8z$ and use AM-GM.
Indeed, we need to prove that
$$(1+a)(a+b)(b+c)(c+16)\geq81abc$$ or
$$(1+2x)(2x+4y)(4y+8z)(8z+16)\geq81\cdot64xyz$$ or
$$(1+2x)(x+2y)(y+2z)(z+2)\geq81xyz,$$
which is AM-GM:
$$(1+2x)(x+2y)(y+2z)(z+2)\geq3\sqrt[3]{x^2}\cdot3\sqrt[3]{xy^2}\cdot3\sqrt[3]{yz^2}\cdot3\sqrt[3]{z}=81xyz.$$
Done!
| {
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If $ab+bc+ca+abc=4$, then $\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c$ Let $a,b,c$ be positive reals such that $ab+bc+ca+abc=4$.
Then prove
$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq 3\leq a+b+c $
So high guys im a high schooler trying to solve this inequality. I did a few things such as try to set a> b>c and then try to generalize it but i couldn't make much progress at all. I would appreciate a clear solution to help me understand the problem in depth.
I did manage to generalize that
$\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\leq a+b+c $
By using the AM-GM inequality on each of a b and c however havent made much progress in proving that the LHS is less than 3 and the RHS is greater than 3.
| AM-GM on $ab$, $bc$, $ca$ and $abc$ gives:
$$1=\frac{ab+bc+ca+abc}{4}\ge\sqrt[4]{(abc)^3}\implies 1\ge abc$$
So, still using the first equation:
$$ab+bc+ca\ge 3$$
Now, by AM-GM
$$a^2+b^2\ge 2ab$$
$$b^2+c^2\ge 2bc$$
$$c^2+a^2\ge 2ac$$
Adding them all up and dividing by $2$ gives:
$$a^2+b^2+c^2\ge ab+bc+ca$$
So now:
$$(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)\ge 3(ab+bc+ca)\ge 9$$
From which it follows that:
$$a+b+c\ge 3$$
I'll work on the other inequality when I have time
| {
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Value of $(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$ if $z^4-2z^3+z^2+z-7=0$ for $z=\alpha$, $\beta$, $\gamma$, $\delta$ Let $\alpha$, $\beta$, $\gamma$, $\delta$ be the roots of $$z^4-2z^3+z^2+z-7=0$$ then find value of $$(\alpha^2+1)(\beta^2+1)(\gamma^2+1)(\delta^2+1)$$
Are Vieta's formulas appropriate?
| transform the equation so that you get the desired roots.
so the given equation has roosts $\alpha, \ \beta,\ \gamma, \ \delta $
first transform $x \to \sqrt{x}$ so the new equation ha roots $\alpha^2, \ \beta^2,\ \gamma^2, \ \delta^2 $ then transform $x \to x-1$ to get the equation whose roots are $\alpha^2+1, \ \beta^2+1,\ \gamma^2+1, \ \delta^2+1 $
now the answer is product of roots of new equation.
| {
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Determine the solution of $x^4y''=(y')^2+3x^6$ Determine the solution of $x^4y''=(y')^2+3x^6$
My idea:
suppose $(y')^2=t \to 2y'y''=\frac{dt}{dx}\to y''=\frac{1}{2\sqrt{t}}\frac{dt}{dx}$
then equation reduced into $x^4\frac{1}{2\sqrt{t}}\frac{dt}{dx}=t+3x^6$
how to solve from here ig ot no idea
| First put $u = \dfrac{y'}{x^2}\implies y'' = 2xu+x^2u' = u^2+3x^2\implies x^2u'=u^2-2xu+x^2+2x^2 = (u-x)^2+2x^2\implies u' = \left(\dfrac{u-x}{x}\right)^2+2 $ . Next put $v = \dfrac{u-x}{x} \implies u = vx+x \implies u' = xv'+v + 1\implies xv'+v+1 = v^2+2\implies xv' = v^2-v+1\implies \dfrac{dv}{v^2-v+1} = \dfrac{dx}{x}$ . Can you take it from here to solve for $v$, then $u$, and finally then $y$ ? I hope it helps...
| {
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Calculate the triple integral $\iiint_D \sqrt{x^2+y^2+z^2}\, dV$.
Calculate the triple integral of
$$\iiint_D \sqrt{x^2+y^2+z^2}\, dV$$ where $D$ is bounded by
(1) $x^2+y^2+z^2=2ay$ and (2) $y=\sqrt{x^2+z^2}$.
So far I was thinking that if I made $y$ turn to $z$ and $z$ turn to $y$ I could use cilyndrical coordinates
So (1)$a^2=r^2+(z-a)^2$ and (2) $z=r$ and both intersect at $z=a$
It is correct to propose the integral$$\int_0^{2\pi}\int_0^a\int_0^{\sqrt{a^2-(z-a)^2}}r\sqrt{z^2+r^2} \,dr\,dz\,d\theta$$?
And if so I evaluate and get $a^3\pi$ but Im suspicious of this result
| The domain $D$ is the area inside the intersection between the sphere $$S=\{(x,y,z)\in\Bbb R^3:x^2+(y-a)^2+z^2=a^2,\;a\in \Bbb R\}$$
and the half cone $$C=\{(x,y,z)\in\Bbb R^3:y=x^2+z^2\}$$
The function to integrate is $f(x,y,z)=\sqrt{x^2+y^2+z^2}$.
In spherical coordinates, with radial distance $r$, azimuthal angle $\theta$, and polar angle $\phi$, we have $x=r\sin\phi\cos\theta$, $y=r\sin\phi\cos\theta$, $z=r\cos\phi$ and $\mathrm dV=r^2\sin\phi\,\mathrm d r\,\mathrm d\theta\,\mathrm d\phi$.
For $x=0$, we have $y\le\sqrt{z^2}=|z|$ and then $\phi\in\left[\frac\pi 4,\frac{3\pi}4\right]$. For $z=0$, we have $y\le\sqrt{x^2}=|x|$ and then $\theta\in\left[\frac\pi 4,\frac{3\pi}4\right]$.
Thaat is we have to integrate the function $f(r\sin\phi\cos\theta,r\sin\phi\cos\theta,r\cos\phi)=\tilde f(r,\phi,\theta)=r$ over the region inside the spherical wedge
$$
W=\left\{(r,\phi,\theta)\in\Bbb R^3,\, a\in\Bbb R: r\le |a|,\,\frac\pi 4\le\phi\le\frac{3\pi}4,\,\frac\pi 4\le\theta\le\frac{3\pi}4\right\}
$$
So the integral becomes
$$
\iiint_D f\mathrm dV=\iiint_W \tilde f\mathrm dW=\int_{\pi/4}^{3\pi/4}\int_{\pi/4}^{3\pi/4}\int_{0}^{|a|}r^3\sin\phi\,\mathrm d r\,\mathrm d\theta\,\mathrm d\phi=\frac{a^4}{4}\frac{\pi}{2}{\sqrt 2}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Summation of $\dfrac{1}{1+1^2+1^4}+ \dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}...$
Find the sum of $\dfrac{1}{1+1^2+1^4}+ \dfrac{2}{1+2^2+2^4}+\dfrac{3}{1+3^2+3^4}...$ till n terms.
The $i^{th}$ term is given by $\dfrac{i}{1+i^2+i^4}$. I did a similar problem previously but there I was able to decompose the fraction into partial fractions. Here, I am unable to do so. What would be an efficeinent way to solve it then? (Just a hint would suffice)
| HINT: $1+i^2+i^4=(1+i^2)^2-i^2$
| {
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Closed form $\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx=12\log A-\frac{4}{3}\log 2$ Evaluate
$$\int_0^\infty\left(\frac{\tanh x}{x^2}-\frac{1}{xe^{2x}}\right)dx$$
I haven't been able to find references to the indefinite integral of the $\tanh$ term except for some similar forms that had solutions. see here and here.
Edit:
Following Random Variable's result we have the form
$$12\log A-\frac{4}{3}\log 2$$
| Let $$I(z) =\frac{z}{a}\int_0^\infty \left(\frac{1}{2}-\frac{z}{at}+\frac{1}{e^{at/z}-1}\right) \frac{1-e^{-at}}{t^2} dt$$
Then Binet's first formula says
$$I(z) = \int_0^z \left[\ln\Gamma(x) - (z-\frac{1}{2})\ln z + z - \frac{\ln(2\pi)}{2}\right] dx $$
Letting $a=2,z=1/2$ gives
$$4I(\frac{1}{2}) =\int_0^\infty \left(\frac{1}{2}-\frac{1}{4t}+\frac{1}{e^{4t}-1}\right) \frac{1-e^{-2t}}{t^2} dt$$
and $a=4,z=1$ gives
$$4I(1) =\int_0^\infty \left(\frac{1}{2}-\frac{1}{4t}+\frac{1}{e^{4t}-1}\right) \frac{1-e^{-4t}}{t^2} dt$$
Some algebraic manipulation yields
$$4I(1)-4I(\frac{1}{2}) = \underbrace{\int_0^\infty \left[\frac{1}{2t^2}-\frac{e^{-2t}}{2t} - (\frac{1}{2}-\frac{1}{4t})\left(\frac{e^{-2t}-e^{-4t}}{t^2}\right)\right] dt}_{J} - \frac{I}{2}$$
With $I$ your desired integral. Surprisingly, $J$ has elementary primitive (even it were not, we still have systematic way to crash it), with value $5/8$.
Hence it remains to evalute $$\int_0^1 \ln \Gamma(x) dx \quad \quad \int_0^{1/2} \ln \Gamma(x) dx$$
The former is just $\ln(2\pi)/2$, for the latter, we can use the integral representation of Barnes G function:
$$\int_0^z \ln \Gamma(x) dx = \frac{z(1-z)}{2}+\frac{z}{2}\ln(2\pi) + z \ln\Gamma(z) - \log G(1+z)$$
and the special value $$\ln G(\frac{3}{2}) = -\frac{3}{2}\ln A + \frac{\ln \pi}{4}+\frac{1}{8}+\frac{\ln 2}{24}$$
with $A$ being the Glaisher-Kinkelin constant.
Alternatively, use Fourier expansion of $\ln \Gamma(x)$, integrate termwise, and remember the relation between $A$ and $\zeta'(2)$ also gives the value of the integral $$\int_0^{1/2} \ln\Gamma(x)dx = \frac{3}{2}\ln A + \frac{5}{24}\ln 2 + \frac{\ln \pi}{4}$$
| {
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"timestamp": "2023-03-29T00:00:00",
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minimize $x^4 - 6x^2 y^2 + y^4$ given $x^2 + y^2 \leq 1$ I have a constrained optimization problem. Can we maximize / minimize this function on the unit sphere?
$$ f(x,y,z) = x^4 - 6 x^2 y^2 + y^4 \quad\text{given that}\quad x^2 + y^2 + z^2 = 1$$
One idea could be to use the Cauchy-Schwartz inequality. Since I forget the proof:
$$ (x^2 + y^2)^2 \geq 0 \text{ so that }x^4 + y^4 \geq 2 x^2 y^2 \text{ and }f(x,y,z) \geq - 4 x^2 y^2 \geq - 4 $$
I could try other rearrangements as well. This one gives me an upper bound of $3$.
$$ x^4 - 6x^2 y^2 + y^4 \leq x^4 + 6x^2 y^2 + y^4
= (x^2 + y^2)^2 + 4x^2 y^2 \leq 3\, \big( x^2 + y^2 \big)^2 \leq
3\, \big( x^2 + y^2 + z^2 \big)^2 = 3$$
If I use some real analysis we know that the sphere as a subset of Euclidean space is compact, so that:
$$ -\infty < -4 \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) \leq \min_{x^2 +y^2 + z^2 = 1} f(x,y,z) < 3 < +\infty$$
I'm trying to avoid Lagrange multipliers unless the're really natural here. Observer also that:
$$ \left[ x^2 + y^2 + z^2 = 1 \right] \to \left[ x^2 + y^2 \leq 1\right] $$
as the original problem was defined on the unit sphere but the $z$ is extraneous.
They might not be extraneous we could set spherical coordinates:
$$ (x,y,z ) = \big(\cos \theta \, \cos \varphi, \;\cos \theta \sin \varphi, \;\cos \varphi\big)$$
and we could put into our inequality:
\begin{eqnarray*} x^4 - 6x^2 y^2 + y^4 &=& \cos^4 \theta \cos^4 \varphi - 6 \cos^4 \theta \sin^2 \varphi \cos^2 \varphi+ \cos^4 \theta \sin^4 \varphi \\ \\
&=& \cos^4 \theta \,\big( \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi \big) \\ \\
&\leq & \cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
\end{eqnarray*}
This looks promising as I have reduced a three-dimensional problem to a problem with only an angle $\varphi$, but I may have lost something with the final "$\leq$" sign.
Just a tiny bit more:
$$
\cos^4 \varphi - 6 \cos^2 \varphi \sin^2 \varphi + \sin^4 \varphi
= (\cos^2 \varphi - \sin^2 \varphi)^2 - 4 \cos^2 \varphi \sin^2 \varphi
= \cos^2 2\varphi - \sin^2 2\varphi
$$
and if we use the double-angle identity.
$$ 1 \geq \cos^2 2\varphi - \sin^2 2\varphi
= \cos^2 2\varphi - (1 -\cos^2 2\varphi)
= 2\, \cos^2 2\varphi - 1 \geq - 1$$
This is very similar to what I obtained before.
| Let $x=r \cos \theta, y = r \sin \theta$, then the cost reduces to
$r^4(\cos^4 \theta -6 \cos^2 \theta \sin ^2 \theta + \sin^4 \theta) = r^4 \cos(4 \theta)$.
Hence we can write an equivalent problem as extremising
$r^4 \cos(4 \theta)$ subject to $r \in [0,1]$.
| {
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"timestamp": "2023-03-29T00:00:00",
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Calculus simplification Could someone explain how this:
$\int(\frac{x}{\sqrt{x+1}})dx$
becomes:
$\int(\sqrt{x+1})dx - \int(\frac{1}{\sqrt{x+1}})dx$
| It’s got nothing to do with the integral:
$$\begin{align}
\frac{x}{\sqrt{x+1}} &= \frac{x+1-1}{\sqrt{x+1}}\\
&= \frac{x+1}{\sqrt{x+1}} - \frac{1}{\sqrt{x+1}}\\
&= \sqrt{x+1} - \frac{1}{\sqrt{x+1}}
\end{align}$$
Does that help?
| {
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Show that $\sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $ I wonder why this is true
$$ \sum_{m,n = - \infty}^{\infty} \frac{(-1)^m}{m^2 + 58 n^2} = - \frac{\pi \ln( 27 + 5 \sqrt {29})}{\sqrt {58}} $$
Where the sum omits the case $n = m = 0$ ofcourse.
| Finally I managed to sum this series using Ramanujan's class invariants. We have the definition $$g(q) = 2^{-1/4}q^{-1/24}\prod_{n = 1}^{\infty}(1 - q^{2n - 1}), \, g_{p} = g(e^{-\pi\sqrt{p}})\tag{1}$$ Ramanujan established that if $p$ is a positive rational number then $g_{p}$ is an algebraic real number. Moreover we have the non-trivial identity $$g_{4/p} = 1/g_{p}\tag{2}$$ Another ingredient we need is the formula $$\sin \pi z = \pi z\prod_{n = 1}^{\infty}\left(1 - \frac{z^{2}}{n^{2}}\right)\tag{3}$$ Taking logs and differentiating with respect to $z$ we get $$\pi\cot\pi z = \frac{1}{z} -2z \sum_{n = 1}^{\infty}\frac{1}{n^{2} - z^{2}}$$ Replacing $z$ by $iz$ we get $$\pi\coth\pi z = \frac{1}{z} + 2z\sum_{n = 1}^{\infty}\frac{1}{n^{2} + z^{2}}$$ The above sum can be written as $$\sum_{n = 1}^{\infty}\frac{1}{n^{2} + z^{2}} = \frac{\pi}{2z} - \frac{1}{2z^{2}} + \frac{\pi e^{-2\pi z}}{z(1 - e^{-2\pi z})}\tag{4}$$ Consider the sum in question
\begin{align}
S(p) &= \sum_{m, n\in\mathbb{Z}}'\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\
&= 2\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2}} + 2\sum_{n = 1}^{\infty}\sum_{m\in\mathbb{Z}}\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\
&= 2\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2}} + \frac{2}{p}\sum_{n = 1}^{\infty}\frac{1}{n^{2}}+ 4\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\
&=\frac{(2- p)\pi^{2}}{6p} + 4\sum_{n = 1}^{\infty}\sum_{m = 1}^{\infty}\frac{(-1)^{m}}{m^{2} + pn^{2}}\notag\\
&=\frac{(2- p)\pi^{2}}{6p} + 4\sum_{m = 1}^{\infty}(-1)^{m}\sum_{n = 1}^{\infty}\frac{1}{m^{2} + pn^{2}}\notag\\
&= \frac{(2- p)\pi^{2}}{6p} + \frac{4}{p}\sum_{m = 1}^{\infty}(-1)^{m}\sum_{n = 1}^{\infty}\frac{1}{m^{2}(1/p) + n^{2}}\notag\\
&= \frac{(2- p)\pi^{2}}{6p} + \frac{4}{p}\sum_{m = 1}^{\infty}(-1)^{m}\left(\frac{\pi\sqrt{p}}{2m} - \frac{p}{2m^{2}} + \frac{\pi\sqrt{p} e^{-2\pi m/\sqrt{p}}}{m(1 - e^{-2\pi m/\sqrt{p}})}\right)\text{ (using equation }(4)) \notag\\
&=\frac{\pi^{2}}{3p}-\frac{2\pi\log 2}{\sqrt{p}}+\frac{4\pi}{\sqrt{p}}\sum_{m=1}^{\infty} \frac{(-1)^{m}q^{m}}{m(1-q^{m})},\,q=e^{-2\pi/\sqrt{p}}\notag\\
&=\frac{\pi^{2}}{3p}-\frac{\pi\log 4}{\sqrt{p}}+\frac{4\pi}{\sqrt{p}}(a(q^{2})-a(q))\notag\\
&\, \, \, \, \,\,\,\,\text{where }a(q) =\sum_{n=1}^{\infty} \frac{q^{n}} {n(1-q^{n})}=-\log\prod_{n=1}^{\infty}(1-q^{n})\notag\\
&=\frac{\pi^{2}}{3p}-\frac{\pi\log 4} {\sqrt{p}}+\frac{4\pi}{\sqrt{p}}\log\prod_{n=1}^{\infty}(1-q^{2n-1}) \notag\\
&=\frac{\pi^{2}}{3p}-\frac{\pi}{\sqrt {p}} \log\frac{2}{q^{1/6}g^{4}(q)}\text{ (via equation }(1)) \notag\\
&= -\frac{\pi\log(2/g_{4/p}^{4})}{\sqrt{p}}\notag\\
&= -\frac{\pi\log (2g_{p}^{4})}{\sqrt{p}}\text{ (using equation }(2))\notag
\end{align}
It is well known that $g_{58} = \sqrt{(5 + \sqrt{29})/2}$ and this gives the desired closed form for $S(58)$. The above technique can also be used (with some more effort) to prove the Kronecker's second limit formula.
The function $a(q) $ used above is related to work of Simon Plouffe. See this answer for details.
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Prove that $x^2 y+ y^2 z + z^2x ≥ 2(x + y + z) − 3$
Let $x$, $y$ and $z$ be positive real numbers such that $xy + yz + zx = 3xyz$. Prove that $x^2 y+ y^2 z + z^2x ≥ 2(x + y + z) − 3$
This question is from BMO 2014.
My working:
By AM-GM inequaluity,
$$x^2 y+ y^2 z + z^2x ≥ 3xyz=xy+yz+zx$$
So now I need to prove that
$$xy+yz+zx\ge2(x+y+z)-3$$
To prove this do I need the relation $xy + yz + zx = 3xyz$?
Need help.
| Let $x=\frac{1}{a}$, $y=\frac{1}{b}$ and $z=\frac{1}{c}.$
Thus, $a+b+c=3$ and we need to prove that
$$a^2c+b^2a+c^2b\geq2abc(ab+ac+bc)-3a^2b^2c^2$$ opr
$$\frac{(a+b+c)^3(a^2c+b^2a+c^2b)}{27}\geq\frac{2abc(ab+ac+bc)(a+b+c)}{3}-3a^2b^2c^2$$ or
$$\sum_{cyc}(a^5c+a^4b^2+3a^4c^2+3a^3b^3+3a^4bc-12a^3b^2c-11a^3c^2b+12a^2b^2c^2)\geq0,$$
which is true by Schur and AM-GM.
The Schur inequality is the following.
Let $x$, $y$ and $z$ be positive numbers.
Prove that:
$$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0.$$
Now, let $x=bc$, $y=ac$ and $z=ab$.
Thus, $$\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0$$ or
$$3\sum_{cyc}(a^3b^3-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0.$$
Also, by AM-GM $$\sum_{cyc}a^5c\geq\sum_{cyc}a^4bc$$,
$$\sum_{cyc}(a^4b^2+a^4c^2)\geq2\sum_{cyc}a^4bc.$$
Thus, by Schur again
$$6\sum_{cyc}(a^4bc-a^3b^2c-a^3c^2b+a^2b^2c^2)\geq0.$$
Hence, it remains to prove that
$$\sum_{cyc}(2a^4c^2-3a^3b^2c-2a^3c^2b+3a^2b^2c^2)\geq0,$$
which is obvious by the $uvw$ method.
Another way.
By C-S
$$\sum_{cyc}x^2y=\frac{1}{3}\sum_{cyc}\frac{1}{y}\sum_{cyc}x^2y\geq\frac{1}{3}(x+y+z)^2\geq2(x+y+z)-3,$$
where the last inequality it's just $$(x+y+z-3)^2\geq0.$$
Done!
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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$\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}$ For any vectors $\mathbf{a},\mathbf{b},\mathbf{c}$, if $$\mathbf{a}\times\mathbf{b}=\mathbf{b}\times\mathbf{c}=\mathbf{c}\times\mathbf{a}$$ where $\mid{\mathbf{a}\times\mathbf{b}}\mid\neq0$
Show that $\mathbf{a}+\mathbf{b}+\mathbf{c}=0$, i.e. a triangle
What I have done so far:
Subtract each term from another, change the sign of cross product when we switch elements s.t. $$\mathbf{a}\times\mathbf{b}+\mathbf{c}\times\mathbf{b}=0, \mathbf{a}\times\mathbf{b}+\mathbf{a}\times\mathbf{c}=0,\mathbf{b}\times\mathbf{c}+\mathbf{a}\times\mathbf{c}=0$$
Which gives $$\mathbf{a}\times\mathbf{(b+c)}=\mathbf{b}\times\mathbf{(a+c)}=\mathbf{c}\times\mathbf{(a+b)}=0$$
Any hints for what to do next? And I think there should be a geometric proof somewhere as well.
| An alternative argument: as $a\times b$ is nonzero, $a$ and $b$ are
nonzero and non-parallel. Also as $(a+b+c)\times a=0$ then $a+b+c$
is a multiple of $a$. Similarly $a+b+c$ is a multiple of $b$. A vector
can only be a multiple of $a$ and of $b$ if it is zero.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Recurrence relation $a_n = -a_{n-1}+n -1 , a_0=7$ $a_n = -a_{n-1} +n -1 , a_0=7$
I am trying to find an explicit formula for this recurrence relation by using backward iteration.
So far I've ended up with this and I don't know what to do to find an explicit formula for this.
$a_n = -a_{n-1}+n -1 , a_0=7$
$= -a_{n-1} +n -1$
$= a_{n-2} +1$
$= -a_{n-3} +n -2$
$= a_{n-4} +2$
$= - a_{n-5} +n-3$
$= a_{n-6} +3$
......
| Subtracting the given equations for $n$ and $n-1$, gives
$$
\begin{align}
a_n+a_{n-1}&=n-1\tag{1a}\\
a_{n-1}+a_{n-2}&=n-2\tag{1b}\\
a_n-a_{n-2}&=1\tag{1c}
\end{align}
$$
Therefore,
$$
a_{2n}=n+a_0\tag2
$$
The equation for $n=1$ gives $a_0+a_1=0$, and therefore
$$
\begin{align}
a_{2n+1}
&=n+a_1\\
&=n-a_0\tag3
\end{align}
$$
Putting these together gives
$$
a_n=\left\lfloor\frac n2\right\rfloor+(-1)^na_0\tag4
$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2472277",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Find, $ \lim_{n\to\infty}\cos\frac{x}{2}\cos\frac{x}{4}\dotsm\cos\frac{x}{2^{n}} $ What's the answer is this limit and how is it solved?
$$
\lim_{n\to\infty}\cos\frac{x}{2}\cos\frac{x}{4}\dotsm\cos\frac{x}{2^{n}}
$$
| Check by induction that
or $$ \sin(x)=2\sin(\frac {x}{2^{}})\cos(\frac {x}{2^{}})=2^{2}\sin(\frac {x}{2^{2}})\cos(\frac {x}{2^{2}})\cos(\frac {x}{2^{}})\\\\ =2^{3}\sin(\frac {x}{2^{3}})\cos(\frac {x}{2^{3}}) \cos(\frac {x}{2^{2}})\cos(\frac {x}{2^{}}) =....=2^{n}\sin(\frac {x}{2^{n}}) \prod_{j=1}^n \cos(\frac {x}{2^{j}} x)$$
that is
$$\sin(x)=2^{n}\sin(\frac {x}{2^{n}}) \prod_{j=1}^n \cos(\frac {x}{2^{j}} x) $$
So $$\prod_{j=1}^n \cos(\frac {x}{2^{j}} x) = \frac{\sin x}{2^{n}\sin(\frac {x}{2^{n}}) } $$
Let set $h =\frac {x}{2^{n}}\to 0~~as~~n\to \infty $ then
$$\lim_{n\to \infty }\prod_{j=1}^n \cos(\frac {x}{2^{j}} x) = \lim_{n\to \infty }\frac{\sin x}{x } \frac{x}{2^{n}\sin(\frac {x}{2^{n}}) } \\=\frac{\sin x}{x } \lim_{h\to 0}\frac{h}{\sin(h) } =\frac{\sin x}{x }$$
since $$ \lim_{h\to 0}\frac{\sin(h) }{h} = 1$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Understanding $a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$ by means of determinants. The identity comes from expanding the determinant
$$\begin{vmatrix}
a & b & c \\
c & a & b \\
b & c & a \\
\end{vmatrix}$$
in two ways.
The LHS comes from expanding the determinant by Sarrus' rule. The RHS comes from adding up all columns to the first, factoring (a+b+c), and then expanding the remaining determinant.
The derivation of the RHS is what I don't understand. I'm guessing that there is some elementary determinant operation at work that I'm unfamiliar with.
| In more detail, the thing that's being suggested to get the RHS is the following reasoning:
\begin{align}
\begin{vmatrix}a & b & c \\ c & a & b \\ b & c & a\end{vmatrix} &= \begin{vmatrix}a+c & a+b & b+c \\ c & a & b \\ b & c & a\end{vmatrix} & \text{(Add row 2 to row 1)} \\
&= \begin{vmatrix}a+b+c & a+b+c & a+b+c \\ c& a & b \\ b & c & a\end{vmatrix} & \text{(Add row 3 to row 1)} \\
&= (a+b+c) \begin{vmatrix}1 & 1 & 1\\c & a & b\\ b & c & a\end{vmatrix} & \text{(Factor out $a+b+c$)} \\
&= (a+b+c) \left(\begin{vmatrix}a & b \\ c & a\end{vmatrix} - \begin{vmatrix}c & b \\ b & a\end{vmatrix} + \begin{vmatrix}c & a \\ b & c\end{vmatrix}\right) & \text{(Expand on $1^{\text{st}}$ row)} \\
&= (a+b+c)(a^2-bc - (ca-b^2) + c^2-ab) \\
&= (a+b+c)(a^2+b^2+c^2 - ab - bc - ca).
\end{align}
| {
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Definite integral over a semicircular area $\int_0^{2a}\int_0^{\sqrt{2ax-x^2}}\frac{\phi'(y)(x^2+y^2)x}{\sqrt{4a^2x^2-(x^2+y^2)^2}}dy\,dx$. change the order of integration in
$$\int_0^{2a}\int_0^{\sqrt{2ax-x^2}}\frac{\phi'(y)(x^2+y^2)x}{\sqrt{4a^2x^2-(x^2+y^2)^2}}dy\,dx$$
I was able to change the order of integration here to
$$\int_0^a\int_{a-\sqrt{a^2-y^2}}^{a+\sqrt{a^2-y^2}}\frac{\phi'(y)(x^2+y^2)x}{\sqrt{4a^2x^2-(x^2+y^2)^2}}dx\,dy$$
Now i am stuck with the integration here w.r.t $x$. I tried substituting $(x^2+y^2)^2$ with $t$ but then replacing value of $4a^2x^2$ becomes a problem. I was thinking of substituting both $x$ and $y$ as $k\cos\theta$ and $k\sin\theta$ but then $y$ won't be constant w.r.t $x$, so that too I believe is out of the question!
| Next thing you want to do is re-scale
$$\int_0^a\int_{a-\sqrt{a^2-y^2}}^{a+\sqrt{a^2-y^2}}\frac{\phi'(y)(x^2+y^2)x}{\sqrt{4a^2x^2-(x^2+y^2)^2}}dx\,dy =$$ $$ a^3 \int_0^1 \phi'(a y) \,dy \int_{1-\sqrt{1-y^2}}^{1+\sqrt{1-y^2}}\frac{(x^2+y^2)x}{\sqrt{4 x^2-(x^2+y^2)^2}}dx$$
After this you can prove the second integral is constant. From there the solution is easy.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Complex Roots of the Characteristic Equation I have the following IVP: $y"+6y'+18y=0$, $y(\frac{\pi}{2})=0$, $y'(\frac{\pi}{2})=6$.
Converting this into a characteristic equation:
$$r^2+6r+18=0$$
Now, solving for r:
$$r=-3\pm 3i$$
When I try to solve for the constants of integration, I get $C_1=0$, $C_2=-2e^{\frac{3\pi}{2}}$.
Plugging all of this in, I get: $y=-2e^{\frac{3\pi}{2}}*e^{-3t}*sin(3t)$, but my homework program says I'm incorrect. Anyone have any idea where I might be going wrong?
| First, the form of $y(x)$ is the following:
$$y(x)=e^{-3x}\left(c_{1}\cos(3x)+c_{2}\sin(3x)\right),$$
for two constants $c_{1}$ and $c_{2}$ to be computed using your initial considitons. Then
$0=y\left(\frac{\pi}{2}\right)=e^{-\frac{3\pi}{2}}\left(c_{1}\cdot\cos\left(\frac{3\pi}{2}\right)+c_{2}\sin\left(\frac{3\pi}{2}\right)\right)=-c_{2}e^{-\frac{3\pi}{2}}$ and this implies $c_{2}=0$ because:
$$\cos\left(\frac{3\pi}{2}\right)=0 \qquad \text{and} \qquad \sin\left(\frac{3\pi}{2}\right)=-1$$
So, the solution reduces to:
$$y(x)=c_{1}e^{-3x}\cos\left(3x\right)$$
Now take the derivative of $y(x)$ :
$$ y^{\prime}(x)=(-3)c_{1}e^{-3x}\cos(3x)-3c_{1}e^{-3x}\sin(3x) $$
and we evaluate at $x=\frac{\pi}{2}$ to get:
$$6=y^{\prime}\left(\frac{3\pi}{2}\right)=-3c_{1}e^{-\frac{3\pi}{2}}(-1)$$
then $c_{1}=2e^{-\frac{3\pi}{2}}$ and this finally gives:
$$ y(x)= 2e^{\frac{3\pi}{2}}e^{-3x}\cos(3x),$$
or $$ y(x)=2e^{\left(\frac{3\pi}{2}-3x\right)}\cos(3x).$$
| {
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If $a,b,c$ are rational roots for $x^3+a x^2+b x+c=0$, find the value of a, b, c
Given: $a,b,c$ are rational roots for the equation $x^3+a x^2+b x+c=0$
Find: all possible values of $a,b,c$.
Question used within the preparation for an entrance exam.
My attempt: from Vieta's formulas, we know that (1) $a+b+c=-a$, (2) $ab+bc+ac=b$ and (3) $abc=-c$. Developing this last equation, we get $c(ab+1)=0$. Let's assume that $c=0$. Replacing $c=0$ in (2) we get $ab=b$, or $a=1$. Replacing $a=1$, $c=0$ in (1) we get $b=-2$. Therefore $a=1,b=-2,c=0$ is a possible set of values. By inspection $a=b=c=0$ is another set.
Questions: are there other (rational) values for $a,b,c$ beyond these 2 ones I found? Are there other approaches to address this question?
| You already have $c(ab+1)=0$. But you only considered one case $c=0$.
Let consider the case $ab=-1$. One then has, from (2), $c(a+b) = b+1$
Or $c(a^2-1)=a-1$.
Subcase 1: $a=1$, then $b=-1$. From (1), one has $c=-2a-b=-1$.
Subcase 2: $c(a+1)=1$, then from (1), one has $$a-\frac{1}{a} + \frac{1}{a+1}=-a$$, then $$2a-\frac{1}{a(a+1)}=0$$ $$2a^2(a+1)=1.$$ Solve this equation, you don't have $a$ is rational (for example, link).
So you have one more tuple $(1,-1,-1)$.
| {
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Find $\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}$ I am just trying to calculate
$$\lim_{n\to\infty} \frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}.$$
To do this I apply formula for sum of fourth powers of $n$ number. My result: $$\lim_{n\to\infty}\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}=1$$. I'm intrested in finding other method to solve the following problem.
| Hint:
$$\frac{1^4+2^4+\dots+n^4}{1^4+2^4+\dots+n^4+(n+1)^4}
=\frac{1^4+2^4+\dots+n^4+(n+1)^4-(n+1)^4}{1^4+2^4+\dots+n^4+(n+1)^4}
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Maximum and minimum value of an inequality Hello during a problem I have to solve this with $a,b,c$ positiv real number:
$$6\geq\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}+\frac{c^2 + c b + 2 b + b^2 + 3}{c^2 + c b - 2 b + b^2 + 3}+\frac{a^2 + a c + 2 c + c^2 + 3}{a^2 + a c - 2 c + c^2 + 3}\geq 3$$
Edit:Wolfram alpha says that the minimum value is 3.
Second edit :This is what I want to solve :
$$\sum_{cyc}\frac{a^2 + a b + 2 a + b^2 + 3}{a^2 + a b - 2 a + b^2 + 3}\leq \sum_{cyc}\frac{a + \sqrt{a b} + 2 \sqrt{a} + b + 3}{a + \sqrt{a b}- 2 \sqrt{a} + b + 3}$$
I try many classical inequalities , but without success .
Thanks a lot .
| Your first inequality is wrong take $a=\frac{1}{1000}$ and $a=b=1.5$
It's more than 6 .
| {
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Minimize $a+b+c$ where $a!=b! \cdot c^2$ and $c \geq 10$, $a,b,c \in \Bbb N^+$ For $c=10$, we get $a=100$ and $b=99$.
However, I am wondering if it is possible to write a square of a number as a multiplication of consecutive integers. So, if
$c^2 = a (a-1) (a-2) \cdots (a-n)$
then we would possibly get a smaller $a+b+c$.
$b=1$, $\ \ $ ($n=a-1$) is not possible because the only square factorials are $0!$ and $1!$.
These are my thoughts on the problem. Thanks for your help.
| This is a variant of Xpw's answer, intended to reduce the number of separate cases that need to be examined in order to show that $(a,b,c)=(100,99,10)$ minimizes the sum $a+b+c$ if $c\ge10$.
Suppose $a!=b!\cdot c^2$ with $c\ge10$ and $a+b+c\lt209$. Let's write $a=b+n$. Then we have
$$c^2=(b+1)(b+2)\cdots(b+n)\le(208-2b-n)^2$$
We cannot have $n=2$, since any two consecutive integers are relatively prime, hence each would have to be a square in order for their product to be a square, but positive squares differ by at least $3$.
For $n\ge3$ we have the following crude inequality:
$$b^n\le205^2\lt50000$$
(You can be less crude if you like, but there's little to be gained from it.) Since $40^3=64000$ and $20^4=160000$, we see that $b\lt40$ if $n=3$ and $b\lt20$ if $n\ge4$. Let's do the $n\ge4$ case first.
Any string of $4$ or more consecutive numbers starting at a number less than or equal to $20$ contains at least one prime within the final $4$ numbers in the string (the first prime gap greater than $4$ occurs at $29-23$). The product of such a string contains that prime to the first power only, hence cannot be a square.
Any string of three consecutive numbers contains exactly one number divisible by $3$. In order for their product to be a square, the number divisible by $3$ must be divisible by an even power of $3$. For strings beginning at $40$ or less, the only possibilities are $9$, $18$, and $36$. Since $7$, $11$, $17$, $19$, and $37$ are primes, the only strings to check are $8\cdot9\cdot10$ and $34\cdot35\cdot36$, neither of which is a square.
| {
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Prove for all n∈N $1^2+3^2+5^2+...+(2n-1)^2=\frac{4n^3-n}{3}$
Prove for all n∈N $1^2+3^2+5^2+...+(2n-1)^2=\frac{4n^3-n}{3}$
This is what I got, but I dont think it is right.
| The fact that you are dealing with a sum of squares seems to have completely disappeared in the step.
Here's what you need:
$$1^2+3^2+...+(2k-1)^2+(2k+1)^2= \text{ (Inductive Hypothesis)}$$
$$\frac{4k^3-k}{3}+(2k+1)^2= $$
$$\frac{4k^3-k}{3}+\frac{3(2k+1)^2}{3}= $$
$$\frac{4k^3-k+3(2k+1)^2}{3}= $$
$$\frac{4k^3-k+3(4k^2+4k+1)}{3}= $$
$$\frac{4k^3-k+12k^2+12k+3}{3}= $$
$$\frac{4k^3+12k^2+12k+3-k+1-1}{3}= $$
$$\frac{4k^3+12k^2+12k+4-(k+1)}{3}$$
$$\frac{4(k^3+3k^2+3k+1)-(k+1)}{3}$$
$$\frac{4(k+1)^3-(k+1)}{3}$$
| {
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Deriving $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ When I first saw that $x^n-1=(x-1)(x^{n-1}+x^{n-2}+...+x+1)$ I wondered how does one find/derive something like that (not just prove it by induction), now that I have some very basic knowledge about polynomials I gave it a try.
Im asking if there are mistakes in the following. Thanks in advance to anyone who will bother checking that.
Let $n$ be positive integer, then:
$x^n-1$ is polynomial of degree $n$ where the coefficients of the terms between $x^n$ and $-1$ (the constant term) are equal to $0$.
$$f(x)=a_nx^n+a_{n-1}x^{n-1}+...+a_1x+a_0$$
where $a_n=1, a_0=-1$ $\,$ and $\,$ $a_i=0$ for every $i=n-1,...,1$
If $n$ is even we have two roots, $1$ and $-1$
If $n$ is odd, $1$ is the only root.
In both cases we can rewrite it as
$$f(x)=x^n-1=(x-1)(b_{n-1}x^{n-1}+...+b_1x+b_0)=\\ x(b_{n-1}x^{n-1}+...+b_0)-(b_{n-1}x^{n-1}+...+b_0)=\\ (b_{n-1}x^n+b_{n-2}x^{n-1}+...+b_1x^2+b_0x)-b_{n-1}x^{n-1}-b_{n-2}x^{n-2}...-b_1x-b_0\tag{A}$$
After rearranging we get:
$$f(x)=b_{n-1}x^n+(b_{n-2}-b_{n-1})x^{n-1}+(b_{n-3}-b_{n-2})x^{n-2}+...+(b_0-b_1)x+(-b_0)$$
The highest term, namely $a_nx^n$ comes from multiplying $x$ with $b_{n-1}x^{n-1}$, that is
$$a_nx^n=x(b_{n-1}x^{n-1})=b_{n-1}x^n$$
We can conclude that $a_n=b_{n-1}=1$
$f(0)=-1=(-1)(0+...+0+b_0)$, hence $b_0=1$.
Then the constant term $(-b_0)=a_0=-1$
Then the coefficients $(b_{i-1}-b_{i})=a_{i}=0$ for every $i=n-1,...,1$
We observe the following, $\, b_{n-2}-b_{n-1}=b_{n-2}-1=0,\,$ then $\, b_{n-2}=1$
the next coefficient, $b_{n-3}-b_{n-2}=b_{n-3}-1=0,\,$ then $\,b_{n-3}=1$
We can keep going like that with the other coefficients. From that we conclude that $b_i=1$ for every $i=n-1,...,1$
Hence $$f(x)=x^n-1=(x-1)(x^{n-1}+...+x+1)$$
At $(A)$ I use the Polynomial remainder theorem to get $(x-1)$ as factor.
I wasnt able to find something like that for $x^n+1$ when $n$ is even, there are no real roots.
Prabably could be writen like that:
$x^n+c=(x^n-1)+c+1\;$ where $c$ is some integer.
| You are making it much too complicated.
By multiplication of the polynomials,
$$(x-1)(x^{n-1}+x^{n-2}+x^{n-3}+\cdots x^2+x+1)=\\
x^n+x^{n-1}+x^{n-2}+\cdots\ \ \ \ x^3+x^2+x\\
\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ -x^{n-1}-x^{n-2}-x^{n-3}\cdots-x^2-x-1=\\x^n-1.$$
| {
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Find the Center and Radius of a Circle with a Given Circle Equation There is a question that is asking the center and radius of the circle that is defined by the equation $x^2 + y^2 - 2ax + 4by =0$.
This was my work (completing the square):
$$x^2 - 2ax + (\frac{-2a}{2})^2 + y^2 + 4by + (\frac{4b}{2})^2 = (\frac{-2a}{2})^2 + (\frac{4b}{2})^2$$
$$x^2 - 2ax + a^2 + y^2 + 4by + 4b^2 = a^2 + 4b^2$$
$$(x-a)^2 + (y-2b)^2 = a^2 + 4b^2$$
I got the center of the circle to be $(a, 2b)$, and the radius of the circle to be $\sqrt{a^2 + 4b^2}$ (although I don't know how to simplify the radius any further than that).
Am I wrong in any step?
Thanks in advance.
| $$x^2+y^2−2ax+4by=0$$
Using method of completing the square
$$x^2-2ax+(-a)^2-(-a)^2+y^2+4by+(2b)^2-(2b)^2=0$$
$$(x-a)^2+(y+2b)^2=a^2+4b^2$$
The center of the circle is given as
$$(a,-2b)$$
The radius of the circle is
$$r^2=a^2+4b^2$$
$$r=\sqrt{a^2+4b^2}$$
Consider the general form of equation for object circle,
$$(x-x_0)^2+(y-y_0)^2=r^2$$
Center is then
$$(x_0,y_0)$$
The radius is then
$$r$$
There is a mistake of the sign in your work.
| {
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Equation System with 3 variables "Find all the possible solutions $(x,y,z) \in$ $\Bbb R$ that satisfies the following equation system".
My teacher gave me this problem and i got some clues in how to solve it. I evaluated $1$ and $-1$ and i got these solutions $(1,-1,1)$ and $(-1,-1,-1)$. I think these are the only ones but i can't prove it.
Am i wrong?, are there more solutions?.
Thanks
$$
\left\{
\begin{array}{c}
x+y-z=-1 \\
x^2-y^2+z^2=1\\
-x^3+y^3+z^3=-1
\end{array}
\right.
$$
| Use the first equation to eliminate $z$: $z=x+y+1$.
Substituting into the second equation gives
$$x^2-y^2+(x+y+1)^2=1 \Longleftrightarrow (x+y)(x+1)=0\,.$$
Therefore, either $x=-1$ or $x=-y$.
If $x=-1$, then the first equation gives $y=z$, and the third equation becomes $2y^3=-2$. The only solution to this equation is $y=-1$. Hence $(x,y,z)=(-1,-1,-1)$ is a solution.
If $x=-y$, then the first equation becomes $-z=-1$ so $z=1$. The third equation is now $2y^3=-2$ so $y=-1$ and $x=-y=1$. Hence $(x,y,z)=(1,-1,1)$ is a solution.
There are no other solutions.
| {
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Induction proof: $2^n + 3^n ≡ 5^n (mod 6)$ I'm trying to prove that $2^n + 3^n ≡ 5^n\ (mod\ 6)$ using induction.
$n=1$:
$2+3≡5\ (mod\ 6)$
$n=k$:
$2^k + 3^k ≡ 5^k\ (mod\ 6)$
$n=k+1$:
$2^{k+1} + 3^{k+1} ≡ 5^{k+1}\ (mod\ 6)$
$2*2^k + 3*3^k ≡ 5*5^k\ (mod\ 6)$
$6\ |\ 5*5^k - 2*2^k - 3*3^k$
$6\ |\ (2+3)*5^k - 2*2^k - 3*3^k$
$6\ |\ 2*5^k + 3*5^k - 2*2^k - 3*3^k$
$6\ |\ 2*(5^k - 2^k) + 3*(5^k - 3^k)$
Not quite sure if I'm going in the right direction or where to go from here...
| To do the inductive case:
$5^k \equiv 2^k + 3^k \mod 6$ is true. So now we must look at $5^{k+1}$, which we know is $5 \times 5^k$, therefore is congruent to $5(2^k + 3^k) \mod 6$. Note that:
$$
2^{k+1} + 2^k + 3^{k+1} + 3^k = 3(2^k + 3^k + 3^{k-1})
$$
therefore, $2^{k+1} + 3^{k+1} \equiv -(2^k+3^k) \equiv 5(2^k+3^k) \mod 6$. Combining what we have, $5^{k+1} \equiv 2^{k+1} + 3^{k+1} \mod 6$, completing the inductive step.
Alternately, note that $5^k -2^k-3^k= \sum_{k=1}^{n-1} \binom nk 2^k3^{n-k}$ is a multiple of $6$ as each term on the summation is a multiple of $6$.
| {
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How to see the order relation for $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}$ and $\frac{n^4}{n^7}$ as $n$ very large? I have two sequences $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}$ and $\frac{n^4}{n^7}$. How do I check whether these two sequence have a fixed order relation when $n$ getting larger? Say, for all $n$ that is large than $N$, $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}\leq\frac{n^4}{n^7}$.
I know that $\lim_{n\to\infty}\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}=0$, but this doesn't help to see the relation for these two sequence(only helpful to see that the nominator$<$denominator). Also, even I know $$\lim_{n\to\infty}\frac{\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}}{\frac{n^4}{n^7}}=1$$
it appears that we can't deduce which is necessarily bigger than the other for large $n$.
Need help.
| If you imagine writing $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}-\frac {n^4}{n^7}$ and putting them over a common denominator, the denominator will be positive. The first terms in the numerator will be $n^{11}$ and $-n^{11},$ which add to zero. The next terms will be $99999n^3 \cdot n^7$ and $-9999n^5 \cdot n^4$. The first is clearly larger, so for large $n$ we have $\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}\gt\frac {n^4}{n^7}$
Added: another way to look at it, of course with the same result, is to write $$\frac{n^4+99999n^3-99n^2+3}{n^7+9999n^5-9999n^2}=\frac 1{n^3} \cdot\frac{1+99999n^{-1}-99n^{-2}+3n^{-4}}{1+9999n^{-2}-9999n^{-5}}$$ and use the fact that for $n$ large $\frac 1{1+n^{-1}}\approx 1-n^{-1}$ to make the right side $\frac 1{n^3}(1+99999n^{-1}-9999n^{-2}-99n^{-2}+$ a few other terms) The leading term is enough to conclude that the fraction goes to zero like $n^{-3}$ but the next term tells you it is greater than $n^{-3}$ for $n$ large enough. It is just like a Taylor series. The comparison with $n^{-3}$ subtracts off the leading term, so the next one dominates.
| {
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Explicit formula from recursive formula I'm failing to find any reasonable solution to this given problem.
Find explicit formula of $n$-th element from given sequence:
$$\begin{cases}
a_1 = 1\\
a_2 = 2\\
a_{n+2} = 4 a_{n+1} + 4 a_n + 2^n
\end{cases}$$
I have tried finding some clues whilst typing out few first expressions to no avail. I also tried inserting the recursion formula instead of $a_{n+1}$ or $a_n$ though it got very messy in no time and all I have gotten was something like this:
$$a_{n+2} = 4^3 a_{n-3} + \sum\limits_{j=0}^3 (2^{n+j} + 4^{j+1}a_{n-j})$$
Which I believe would get me to:
$$a_{n+2} = 4^{n-1} a_{1} + \sum\limits_{j=0}^{n-1} (2^{n+j} + 4^{j+1}a_{n-j})$$
Any tips on how can I solve this one?
| Solving recurrences can be tricky. A systematic approach is to try to solve for the generating function of the sequence and then work from there, but to get the general sequence term from the generating function can also be hard as it involves differentiating $n$ times. This kind of problem is usually solved nicely by a CAS like Mathematica. You can also try to look up the sequence terms in the Online Encyclopedia of Integer Sequences, however for this sequence there doesn't seem to be an entry.
Think about the sequence generating function, that is the formal power series
$$
G(x) = \sum_{n=1}^\infty a_{n+2} x^n
$$
we know the general term for $a_{n+2}$ giving
$$
a_{n+2} = 4 a_{n+1} + 4 a_{n} +2^{n}
$$
Then we know the generating function is
$$
G(x) = \sum_{n=1}^\infty (4 a_{n+1} + 4 a_{n} +2^{n}) x^n \\
G(x) = 4\sum_{n=1}^\infty a_{n+1}x^n + 4\sum_{n=1}^\infty a_{n}x^n +\sum_{n=1}^\infty 2^{n} x^n
$$
consider what these terms mean in relation to our definition of the generating function
$$
G(x) = \sum_{n=1}^\infty a_{n+2} x^n = a_3x^1+a_4x^2+a_5x^3+\cdots\\
4\sum_{n=1}^\infty a_{n+1}x^n =4(a_2x^1+a_3x^2+a_4x^3+\cdots)=4a_2x+4xG(x)\\
4\sum_{n=1}^\infty a_{n}x^n = 4(a_1x^1+a_2x^2+a_3x^3+\cdots)=4a_1x+4a_2x^2+4x^2G(x)
$$
we also know that (when it converges) from the geometric series
$$
\sum_{n=1}^\infty 2^n x^n = \frac{2x}{1-2x}
$$
Altogether we have
$$
G(x) = 8x+4x G(x) + 4x+8x^2+4x^2G(x) + \frac{2x}{1-2x}
$$
$$
G(x)(1-4x-4x^2) = 12x+8x^2 + \frac{2x}{1-2x}\\
G(x) = -\frac{2x(8x^2+8x-7)}{1-6x+4x^2+8x^3} = 14x + 68x^2 + 336x^3 + \cdots
$$
which fits with our original definition. Now we need to extract the sequence of coefficients from this. We can define the full sequence generating function to include the first two terms
$$
g(x) = 1 + 2x + x G(x) = \sum_{n=0}^\infty a_0 x^n = \frac{1-4x+6x^2}{1-6x+4x^2+8x^3}
$$
from this we can differentiate as many times as necessary then set the generating function to zero to get the desired coefficient multiplied by some factor, this can formally be seen as the inverse Z transform of g(1/x).
InverseZTransform[(1-4x+6x^2)/(1-6x+4x^2+8x^3)/.x->1/x,x,n]
Which in Mathematica gives the answer
$$
a_n = -2^{n-2} + \frac{5}{8}(2-2\sqrt{2})^n + 5\cdot2^{n-3}(1+\sqrt{2})^n
$$
which can be seen to recreate the relabelled coefficients $a_0=1, a_1=2, a_3=14, \cdots$. Other ways to get the $n^{th}$ derivative are the Cauchy differentiation formula or by spotting patterns for simple generating functions.
| {
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Prove that $\operatorname{ord}_{3^{2n}+3^n+1}2 \equiv 0 \pmod{4}$
Prove that $\operatorname{ord}_{3^{2n}+3^n+1}2 \equiv 0 \pmod{4}$ where $n$ is a positive integer.
For $n = 1$ we have $3^{2n}+3^n+1 = 3^2+3+1 = 13$. In this case, since $12$ is prime we know that the order must divide $12$. We find that $12$ is the order.
For $n = 2$ we have $3^{2n}+3^n+1 = 3^4+3^2+1 = 91$. In this case, $91$ is not prime but we find that the order is $12$.
In both of these cases, the order was $12$ and is divisible by $4$.
For the general case, I thought about first proving that $\operatorname{ord}_{3^{2n}+3^n+1}2$ is even. We have $$3^{2n}+3^n+1 = (3^n+1)^2-3^n.$$ How can we prove it is even?
| Clearly, it's sufficient to find divisor $d$ of $3^{2n}+3^n+1$ such that $\operatorname{ord}_d 2 \equiv 0 \pmod 4$.
If $n$ is even, say $n=2k$, then
$$
3^{2n} + 3^n + 1 = (3^{k})^4 + (3^k)^2 + 1 =
\bigl([3^k]^2+3^k+1\bigr)\bigl([3^k]^2-3^k+1\bigr),
$$
it is divisible by $3^{2k}+3^k+1$.
And when $n$ is odd, then
$$
3^{2n} + 3^n + 1 = (3^n - 1)^2 + 3^{n+1}
$$
is sum of two coprime squares, hence all its prime divisors are congruent to $1$ modulo 4 (that's obvious if you're familiar with gaussian integers). Moreover,
$$
3^{2n} + 3^n + 1 \equiv 1 + 3 + 1 = 5 \pmod 8,
$$
it means that it has prime divisor $p \equiv 5 \pmod 8$.
And $2$ is quadratic non-residue modulo that $p$, it means that
$$
2^{(p-1)/2} \equiv -1 \pmod p.
$$
Put for clarity $p = 4k+1$. Now it's easy to see that $\operatorname{ord}_p 2 \equiv 0 \pmod 4$: $\operatorname{ord}_p 2 $ divides $(p-1) = 4k$ but not divides $(p-1)/2 = 2k$.
| {
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How to solve $a^3 + 39 ab^2 - 18 = 0$, $3a^2 b + 13 b^3 - 5 = 0$ In this answer, the user @123 has claimed that by solving the system
$$\begin{cases}a^3 + 39 ab^2 - 18 = 0 \\ 3a^2 b + 13 b^3 - 5 = 0\end{cases}$$
we give $ a = \dfrac 32$ and $ b = \dfrac12$. Could anyone explain for me that how one can solve such a system, please?
| You can try eliminating the $a^3$ term, by multiplying the first equation by $b$ and the second equation by $\frac{a}{3}$ and subtracting the two.
$$ \begin{cases}
b \left(a^3 + 39 ab^2 - 18 = 0 \right) \\
\frac{a}{3} \left( 3a^2 b + 13 b^3 - 5 = 0 \right)
\end{cases} \Rightarrow $$
$$ \begin{cases}
b a^3 + 39 ab^3 - 18b = 0 \\
b a^3 + \frac{13}{3} a b^3 - \frac{5}{3} a = 0
\end{cases} \Rightarrow \frac{104}{3} a b^3 + \frac{5}{3} a - 18 b = 0 $$
$$ a= \frac{54 b}{104 b^3 + 5} $$
Substituting $a$ into the second equation yields
$$ \frac{8748 b^3}{(104 b^3+5)^2} + 13 b^3 -5 =0$$ which is solved with $$b^3 = \frac{1}{8} \Rightarrow b = \frac{1}{2} $$
finally you get
$$ \begin{cases} a = \frac{3}{2} \\ b = \frac{1}{2} \end{cases} $$
| {
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prove that 15 divides $n^7+2n^5+4n^3+8n$ for any integer n
Prove that for all $n\in\mathbb{Z}$, $15$ divides $n^7+2n^5+4n^3+8n$.
I know that I have to show that 5 divides $n^7+2n^5+4n^3+8n$ and that
3 divides $n^7+2n^5+4n^3+8n$, I also know that by FLT $n^5 \equiv n \pmod{5}$ and that $n^3 \equiv n \pmod{3}$, but I'm not sure on how to further proceed with this. Any help would be appreciated.
Thanks For the Help guys, I figured it out on my own though!
| $$n^3-n=n(n^2-1), n^5-n=n(n^2-1)(n^2+1)\implies n^5-n$$ is divisible by $(3,5)=15$
Now $$n^7+2n^5+4n^3+8n=n^5(n^2+2)+4n^3+8n\equiv 5n^3+10n\pmod{15}$$
We need $15|(5n^3+10n)\iff3|(n^3+2n)$
But $n^3+2n=n^3-n+3n$
| {
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Using EGFs to solve the recurrence relation $a_n=n a_{n-1}+(n+1)!$ I have to solve the recurrence relation
$$a_n = na_{n-1} + (n + 1)!,\qquad a_0 = 1.$$
I am struggling to finish the problem. I have attached my work. Can you please help me finish?
$$\begin{eqnarray*}A(x)&=&\sum_{n\geq 0}\frac{a_n}{n!}x^n = a_0+\sum_{n\geq 1}\frac{a_n}{n!}x^n=a_0+\sum_{n\geq 1}\left(na_{n-1}+(n+1)!\right)\frac{x^n}{n!}\\&=&a_0+\sum_{n\geq 1}a_{n-1}\frac{x^n}{(n-1)!}+\sum_{n\geq 1}(n+1)x^n = a_0 + x \sum_{n\geq 1}a_{n-1} \frac{x^{n-1}}{(n-1)!}+\sum_{n\geq 1}(n+1)x^n\\&=&a_0+ x\,A(x)+\underbrace{\sum_{n\geq 1}(n+1)x^n}_{\text{I know that }(n+1)x^n = \frac{d}{dx}x^{n+1}.}\end{eqnarray*}$$
| Define the exponential generating function:
$\begin{equation*}
\hat{A}(z) = \sum_{n \ge 0} \frac{a_n}{n!} z^n
\end{equation*}$
Write the recurrence shifted:
$\begin{equation*}
a_{n + 1} = (n + 1) a_n + (n + 2)!
\end{equation*}$
Multiply by $z^n / n!$, sum over $n \ge 0$, recognize some sums:
$\begin{align*}
\sum_{n \ge 0} a_{n + 1} \frac{z^n}{n!}
&= \sum_{n \ge 0} (n + 1) a_n \frac{z^n}{n!}
+ \sum_{n \ge 0} (n + 2)! \frac{z^n}{n!} \\
\hat{A}'(z)
&= z \hat{A}'(z) + \hat{A}(z)
+ \sum_{n \ge 0} n (n + 1) z^n
\end{align*}$
Now:
$\begin{align*}
\frac{1}{1 - z}
&= \sum_{n \ge 0} z^n \\
\frac{d^2}{d z^2} \frac{1}{1 - z}
&= \frac{2}{(1 - z)^3} \\
&= \sum_{n \ge 0} (n + 2) (n + 1) z^n
\end{align*}$
thus:
$\begin{equation*}
\sum_{n \ge 0} n (n + 1) z^n
= \frac{2 z}{(1 - z)^3}
\end{equation*}$
Thus you get the differential equation:
$\begin{equation*}
(1 - z) \hat{A}'(z) = \hat{A}(z) + \frac{2 z}{(1 - z)^3}
\end{equation*}$
This one is linear of the first order. For initial condition you know that $\hat{A}(0) = a_0 = 1$, so (written as partial fractions):
$\begin{equation*}
\hat{A}(z)
= \frac{2}{1 - z}
- \frac{2}{(1 - z)^2}
+ \frac{1}{(1 - z)^3}
\end{equation*}$
Expanding by the binomial theorem:
$\begin{align*}
\hat{A}(z)
&= \sum_{n \ge 0} 2 z^n
- \sum_{n \ge 0} 2 \binom{-2}{n} (-z)^n
+ \sum_{n \ge 0} \binom{-3}{n} (-z)^n \\
&= \sum_{n \ge 0} 2 z^n
- \sum_{n \ge 0} 2 \binom{n + 1}{1} z^n
+ \sum_{n \ge 0} \binom{n + 2}{2} n^n \\
&= \sum_{n \ge 0}
\left(
2 - 2 (n + 1) + \frac{(n + 2) (n + 1)}{2}
\right) z^n \\
&= \sum_{n \ge 0} \frac{n^2 - n + 2}{2} z^n
\end{align*}$
The coefficients are:
$\begin{align*}
a_n
&= n! [z^n] \hat{A}(z) \\
&= n! (n^2 - n + 2)
\end{align*}$
But note that this whole mess could be avoided by noting that the recurrence is linear, first order. Dividing through by $n!$ gives a simple recurrence in $a_n / n!$.
| {
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setting of this induction proof I would like to see if this is a correct induction proof and whether or not this is a good setting out of it
A sequence is defined by
$$a_n = a_{n-1} + a_{n-2} + a_{n-3}$$
for $n\geq 3, a_0 = 1, a_1 = 2, a_2 = 4$.
Prove that $a_n \leq 4^n$ for all $n\in\mathbb{N}$.
Let $P(n)$ be the proposition that
$$''a_n\leq 4^n{''}.$$
Now since we have $a_0 = 1 = 4^0 \leq 4^0$ then $a_0 \leq 4^0$.
Also, $a_1 = 2 < 4 \leq 4 = 4^1$ then $a_1 \leq 4^1$.
Also, $a_2 = 4<16 = 4^2 \leq 4^2$ so $a_2 \leq 4^2$.
Hence, $P(0),P(1),P(2)$ are true.
Now, let $k-3\in\mathbb{N}$ and suppose $P(k-3),P(k-2),P(k-1)$ is true.
We must show that $P(k)$ is true.
Now by definition
$$\begin{align}a_n &= a_{n-1} + a_{n-2} + a_{n-3} \\ &\leq 4^{n-1} + 4^{n-2} + 4^{n-3} \qquad \text{by the inductive hypothesis}\\ &= 21\times 4^{k-3} \\ &\leq 64\times 4^{k-3} \\ &= 4^{k}.\end{align}$$
Hence, $P(k)$ is true if $P(k-3),P(k-2),P(k-1)$ is true for $k\geq 3$.
So by induction, $P(n)$ is true for all $n\in\mathbb{N}$.
Would appreciate any advice.
| Suppose that $P(n)$ is true and prove it for $P(n+1)$
$
a_n = a_{n-1} + a_{n-2} + a_{n-3}\quad $ by the inductive hypothesis
if we suppose that $P(k)$ is true for any $1\leq k\leq n$ we are using strong induction
We must prove that
$a_{n+1}\leq 4^{n+1}$
$a_{n+1}=a_n + a_{n-1} + a_{n-2} \leq 4^n+4^{n-1}+4^{n-2}=4^{n-2}\left(4^2+4+1\right)=21\cdot 4^{n-2}\leq 64\cdot 4^{n-2}=4^{n+1}$
proved
Hope this helps
| {
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How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$? How do I prove that $1^4+2^4+3^4\cdots\ + n^4 = \frac{1}{5}n^5 + \frac{1}{2}n^4 + \frac{1}{3}n^3-\frac{1}{30}n$?
I've spent quite some time on this problem. So far, I've simplified the right-hand side to $\frac{1}{30}(n+1)[(2n+3)(3n^3)+n(n-1)]$. But then, the algebra becomes very complicated when I add $(n+1)^4$ to both sides of the inductive hypothesis.
| The usual trick is to evaluate the sum $$\sum_{k=1}^n k(k+1)(k+2)(k+3)$$ using $$k(k+1)(k+2)(k+3) = \dfrac{1}{5}k(k+1)(k+2)(k+3)(k+4) - \dfrac{1}{5}(k-1)k(k+1)(k+2)(k+3).$$
Then simply expand the summands to obtain $\sum_{k=1}^n k^4$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2515600",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
"answer_count": 4,
"answer_id": 1
} |
Finding average of two variables in two equation The given equations are:
$$7x+3y=17$$
and
$$3x+7y=19$$
How can i find the average of two variables $x$ and $y$ with simple calculation steps?
| Since we got two (differing from each other) formulas with only 2 unknown numbers, we can quite easily calculate both $x$ and $y$. I'll explain my steps, without using matrices, just algebra (just like you asked).
So, we got $7x + 3y = 17$ and $3x + 7y = 19$. Because we can say $17 = 19 - 2$ we can state that
$$7x + 3y = 3x + 7y - 2$$
We take away $3x$ and $3y$ on both sides and we get
$$4x = 4y - 2$$
We divide both sides by $4$ and we get
$$x = y - \frac{1}{2}$$
Now, if we use that in our first equation, it becomes
$$7(y - \frac{1}{2}) + 3y = 17$$
$$7y - 3\frac{1}{2} + 3y = 17$$
We add $ 3\frac{1}{2}$ on both sides and it says
$$10y = 20\frac{1}{2}$$
Now we divide both sides by 10, and we calculated $y$!
$$y = 2\frac{1}{20}$$
And since we already know that $x = y - \frac{1}{2}$ we calculated that
$$x = y - \frac{1}{2} = 2\frac{1}{20} - \frac{10}{20} = 1\frac{11}{20}$$
Now, if you still want the average, it is
$$\frac{x + y}{2} = \frac{2\frac{1}{20} + 1\frac{11}{20}}{2} = \frac{72}{40} = 1\frac{32}{40} = 1\frac{4}{5}$$
There you go
EDT: Whoops, looks like I made a mistake somewhere since the answer is not correct, but you get the idea.
EDT2: Corrected the mistake (typo) thanks to the comments :) and this is in accordance with the answer placed by the question asker himself, so that's entirely correct!
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522689",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$ $$\left\lfloor \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}\right\rfloor =?$$
I can't find $$? \leq \frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{n}} \leq ?$$ I remeber there was a sharp bound , Am i right ?
I am thankful if someone help me to find the bound , or give another idea
Last question : Is it possible to find $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{4}}+...+\frac{1}{\sqrt{1024}}$ by integral ?
| By the EML formula
$$ H_n^{(1/2)} = 2\sqrt{n}+\zeta\left(\tfrac{1}{2}\right)+\frac{1}{2\sqrt{n}}+O\left(\frac{1}{n^{3/2}}\right) $$
hence $H_{1024}^{(1/2)}\approx \color{red}{62}+\frac{1}{2}$. There are many questions on MSE asking for approximations of $H_n^{(1/2)}$: the Hermite-Hadamard inequality/the trapezoid method for the estimation of $\int_{1}^{1024}\frac{dx}{\sqrt{x}}$ provide straightforward answers for the task at hand.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2522888",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "9",
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Verify the Divergence Theorem The vector field $ \ \large \vec F=(xy,yz,zx) \ $ , on the closed cylinder $ \ x^2+y^2 \leq 1 , \ \ z=0 , \ \ z \leq 1$.
Verify the Divergence Theorem.
Answer:
$ (i) $ we have to calculate here $ \iint_S F \cdot n \ dS$
Let $ z=1-x^2-y^2 \ $
Then $ \vec F=(xy,y(1-x^2-y^2) , x(1-x^2-y^2)) \ $
Now ,
$$ F_x=(y,-2xy, 1-3x^2) , \ \ F_y=(x, 1-x^2-3y^2, -2xy) $$
Then,
$$ F_x \times F_y=(13x^2y^2+3x^4+2x^2-3y^2, \ 2xy^2+x-3x^3, \ x^2y+y-3y^3) $$
Then,
$$ \iint_S F \cdot n \, dS= \int_{-1}^1 \int_{-1}^1 \vec F \cdot (F_x \times F_y) \, dx\,dy$$
But this becomes complicated .
I am confused .
Am I right so far .
Help me out
| You should have started with choosing the paramaterization of your surface. We have three parts. We have the upper part the side part and the bottom part. I only do the side part. A paramterization for the side part is:
\begin{align}
\phi(t,\theta) =
\begin{pmatrix}
\cos \theta\\
\sin\theta\\
t
\end{pmatrix}
\end{align}
$t\in [0,1]$ and $\theta \in[0,2\pi]$. Then the normal vector is:
\begin{align}
\partial_\theta\phi(t,\theta) \times \partial_t\phi(t,\theta) =
\begin{pmatrix}
-\sin\theta \\
\cos\theta\\
0\\
\end{pmatrix} \times
\begin{pmatrix}
0\\
0\\
1\\
\end{pmatrix}
=
\begin{pmatrix}
\cos\theta \\
\sin\theta \\
0
\end{pmatrix}
\end{align}
So:
\begin{align}
\iint_S \textbf{F}\cdot \textbf{n} dS &= \int^{2\pi}_0 \int^1_0 \textbf{F}(\phi(t,\theta))\cdot( \partial_\theta\phi(t,\theta) \times \partial_t\phi(t,\theta) ) dt d\theta\\
&= \int^{2\pi}_0 \int^1_0 \cos^2\theta \sin\theta + t\sin^2\theta dtd\theta\\
&= \frac{\pi}{2}
\end{align}
For the last integral you need to use the trig identities as usual. As I already said I'll leave the top part and the bottom part for you.
The divergence:
\begin{align}
\iiint_K \text{div} \textbf{F} dV
\end{align}
Where $K=\{ (x,y,z) | x^2+y^2\leq 1 , 0\leq z\leq 1\} $. Using polar coordinates:
\begin{align}
\iiint_K \text{div}\textbf{F} dV &=\iiint_K y+z+x dV\\
&=\int^1_0\int^1_0 \int^{2\pi}_0 (r\cos\theta+r\sin\theta + z)r d\theta dr dz\\
&= \frac{\pi}{2}
\end{align}
I assume integrating that is not a problem. Now do the bottom part and the top part of the surface integral to verify the Divergence Theorem. Let me know if there is something unclear.
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Simplify $\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$
Which one is equivalent to: $$\frac{1-\sin(x)+\cos(x)}{1+\sin(x)+\cos(x)}$$
1)$\dfrac{\cos(x)}{1+\sin(x)}$
2)$\dfrac{\cos(x)}{1-\sin(x)}$
3)$\dfrac{1-\sin(x)}{\cos(x)}$
4)$\dfrac{1+\sin(x)}{\cos(x)}$
Of course we can find the correct choice by testing each of them,but I'm looking for an analytical solution assuming we don't know the final expression.
I tried the following techniques,all failed: multiplying by $\frac{\sin(x)}{\sin(x)}$ , multiplying by $\frac{\cos(x)}{\cos(x)}$ , using $\sin^2(x)+\cos^2(x)=1$
|
$$\frac { 1-\sin { x } +\cos { x } }{ 1+\sin { x } +\cos { x } } =\frac { \sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) +\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } -2\sin { \left( \frac { x }{ 2 } \right) } \cos { \left( \frac { x }{ 2 } \right) } +\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } -\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } } }{ \sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) +\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } +2\sin { \left( \frac { x }{ 2 } \right) } \cos { \left( \frac { x }{ 2 } \right) } +\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } -\sin ^{ 2 }{ \left( \frac { x }{ 2 } \right) } } } =\\\ =\frac { 2\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } -2\sin { \left( \frac { x }{ 2 } \right) } \cos { \left( \frac { x }{ 2 } \right) } }{ 2\cos ^{ 2 }{ \left( \frac { x }{ 2 } \right) } +2\sin { \left( \frac { x }{ 2 } \right) } \cos { \left( \frac { x }{ 2 } \right) } } =\\=\frac { \cos { \left( \frac { x }{ 2 } \right) } -\sin { \left( \frac { x }{ 2 } \right) } }{ \cos { \left( \frac { x }{ 2 } \right) } +\sin { \left( \frac { x }{ 2 } \right) } } =\frac { { \left( \cos { \left( \frac { x }{ 2 } \right) } -\sin { \left( \frac { x }{ 2 } \right) } \right) }^{ 2 } }{ \left( \cos { \left( \frac { x }{ 2 } \right) } +\sin { \left( \frac { x }{ 2 } \right) } \right) \left( \cos { \left( \frac { x }{ 2 } \right) } +\sin { \left( \frac { x }{ 2 } \right) } \right) } =\\ =\frac { 1-\sin { x } }{ \cos { x } } $$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2526453",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using the divergence theorem to calculate the surface integral of a sphere Hi so I have the question to use the divergence theorem to calculate the surface integral of the sphere
Let $S=\{(x,y,z):(x-a)^2 + (y-b)^2 + (z-c)^2 = R^2\}$ and $f = (x^2,y^2,z^2)$. Evaluate
$$I=\iint_S fdS.$$
By the divergence theorem, I get that
$$I=\iiint_B \text{div}(f)dV$$
where $B=\{(x,y,z):(x-a)^2 + (y-b)^2 + (z-c)^2 \leq R^2\}$.
I get $\text{div}(f) = 2x+2y+2z$, I’m just wondering where to go from here, are the $(x-a)^2$ and $0$ the boundaries for my triple integral?
| We have to evaluate
$$\iiint_{(x-a)^2 + (y-b)^2 + (z-c)^2 \leq R^2} (2x+2y+2z)dxdydz$$
which is equal to
$$2\iiint_{X^2 + Y^2 + Z^2 \leq R^2} ((a+X)+(b+Y)+(c+Z))dXdYdZ$$
where $X=x-a$, $Y=y-b$, $Z=z-c$.
Since the ball $X^2 + Y^2 + Z^2 \leq R^2$ is symmetric with respect to the plane $X=0$, it follows that the integral of $X$ over the ball is zero. The same for $Y$ and $Z$. Therefore, we finally obtain
$$2(a+b+c)\cdot \frac{4\pi R^2}{3}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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prove that $2\sqrt5 +\sqrt{11}$ is irrational how would you prove that $2\sqrt5 +\sqrt{11}$ is irrational?
I started with a proof by contradiction that assumes that $2\sqrt5 +\sqrt{11}$ is rational and therefore there exist integers $a$ and $b$ such that $\frac{a}{b}=2\sqrt5 +\sqrt{11}$ and squaring both sides yields
$\frac{a^2}{b^2}=31 +4\sqrt5\sqrt{11}$ and from this point im stuck as i dont know how to continue to arrive at a contradiction.
| If $2\sqrt{5}+ \sqrt{11}=r$ is rational then $4*5 + 4*\sqrt{55} + 11 = r^2$ is rational. Then $\sqrt{55} = \frac {r^2 - 31}4$ is rational.
NOW do the proof by contradiction
Let $\sqrt{55} = \frac ab$ so $55b^2 = a^2$ so $5|a$ and so $25|a^2$ so $5|b^2$ so $5|b$ so $a$ and $b$ aren't in lowest terms.... yadda, yadda, yadda....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2533270",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "7",
"answer_count": 3,
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Toward Explicit Formula from Recursion : Is Generating function "the only" answer? I am trying to draw the explicit formula of $S_n$ that is defined as below:
$S_n$ is the number of words of length n using 0,1 and 2 such that no two consecutive 0's occur.
Myself, as I had learned from the basic skills in combinatorics, just easily get to the point constructing recursive relation such that:
$$S_n = 2S_{n-1} + 2S_{n-2}$$
since $S_n$ splits up to disjoint two cases: one with no 0 at the last posit, and always 0 at the last locus, then there exist bijection between the former one and 1 or 2 at the n-th posit multiplied with $S_{n-1} $ cases, and also another bijection between the latter one and 1 or 2 at the n-1-th posit multiplied with $S_{n-2}$.
So If my given recursion is correct, next step is how could I go further into the formulating with what.
I superficially knows the concept of $OGF$, and $EGF$, and their formal definition. Generating function contains its sequential information in a form of coefficients with a corresponding polynomial degrees as an index (as far as I understand).
Now if I define $S_n$ a functional form, $s(n)$, generating function would be :
$$g(x) = \sum_{n=0}^{\infty}s(n)x^n$$
Then let's little bit refer to a few terms of $S_n$:
$$1, 3, 8, 22, ...$$
And revise the $g(x)$:
$$g(x) = \sum_{n=0}^{\infty}s(n)x^n =\sum_{n=2}^{\infty}s(n)x^n+3x+1=2\sum_{n=2}^{\infty}s(n-1)x^n+2\sum_{n=2}^{\infty}s(n-2)x^n+3x+1 $$
Now, it looks like the problem has been changed into solve the functional equation(I am not sure this is right term)
1) What should be the next step?
2) Is the generating function the only approach toward the explicit formula?
| This is where signals & systems knowledge comes in handy. :-)
Eigenvectors are your friend.
\begin{align*}
S_n &= 2 S_{n-1} + 2 S_{n-2} \\
\\
\begin{bmatrix} S_n \\ S_{n - 1} \end{bmatrix} &= \begin{bmatrix} 2 & 2 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} S_{n - 1} \\ s_{n - 2} \end{bmatrix}
\end{align*}
Let $\vec{S}_{n+1} = \begin{bmatrix} S_n \\ S_{n-1} \end{bmatrix}$. We now have:
\begin{align*}
\vec{S}_{n+1} &= \begin{bmatrix} 2 & 2 \\ 1 & 0 \end{bmatrix} \vec{S}_{n} \\
\\
\implies\ \ \ \vec{S}_{n+1} &= \begin{bmatrix} 2 & 2 \\ 1 & 0 \end{bmatrix}^n \vec{S}_1
\end{align*}
We're basically done at this point, but you can compute the power explicitly by diagonalizing it:
\begin{align*}
\begin{bmatrix} 2 & 2 \\ 1 & 0 \end{bmatrix} x &= \lambda x
\implies \lambda = 1\pm\sqrt{3} \\
\begin{bmatrix} 2 & 2 \\ 1 & 0 \end{bmatrix} &= \begin{bmatrix} 1 + \sqrt{3} & 1 \\ 1 - \sqrt{3} & 1 \end{bmatrix} \begin{bmatrix} 1 + \sqrt{3} & 0 \\ 0 & 1 - \sqrt{3} \end{bmatrix} \begin{bmatrix} 1 + \sqrt{3} & 1 \\ 1 - \sqrt{3} & 1 \end{bmatrix}^{-1} \\
\end{align*}
So this means we have our answer $S_n$ in the first row of $\vec{S}_{n+1}$:
\begin{align*}
\vec{S}_{n+1} &= \begin{bmatrix} 1 + \sqrt{3} & 1 \\ 1 - \sqrt{3} & 1 \end{bmatrix} \begin{bmatrix} (1 + \sqrt{3})^n & 0 \\ 0 & (1 - \sqrt{3})^n \end{bmatrix} \begin{bmatrix} 1 + \sqrt{3} & 1 \\ 1 - \sqrt{3} & 1 \end{bmatrix}^{-1} \vec{S}_1
\end{align*}
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "10",
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How to integrate $\int_1^{\infty}\frac{\sqrt{x-1}}{(1+x)^2}dx$ using contour integration? How to integrate using contour integration?
$$\int_1^{\infty}\frac{\sqrt{x-1}}{(1+x)^2}dx$$
I was putting $y = x-1$ then $\frac{dy}{dx}$= $1-0$, ${dy}={dx} $ then i get $$\int_1^{\infty}\frac{\sqrt{y}}{(2+y)^2}dy$$
I don't know how to take it from here. I would appreciate if someone could help me and give me some hints or the solution. Thanks.
| I won't prefer a contour here, making the substitution $y=x-1$ we have,
$$ \displaystyle \int_1^\infty \dfrac{\sqrt{x-1}}{(1+x)^2}\; dx = \int_0^\infty \dfrac{\sqrt{y}}{(2+y)^2}\; dy$$
Next substitute $y=u^2$ ,
$$\displaystyle I = 2\int_0^\infty \dfrac{u^2}{(u^2+2)^2}\; du $$
Using partial fractions this will be equal to,
$$ \displaystyle \begin{align} I &= 2\int_0^\infty \dfrac{du}{u^2+2}-\int_0^\infty \dfrac{4\; du}{(u^2+2)^2} \\ &= \dfrac{\pi}{\sqrt{2}} - \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} (1+\cos 2\theta)\; d\theta \\ &=\dfrac{\pi}{\sqrt{2}}-\dfrac{\pi}{2\sqrt{2}} \\ &= \dfrac{\pi}{2\sqrt{2}} \end{align}$$
Footnotes :
1) $\displaystyle \int \dfrac{dx}{x^2+a^2} = \dfrac{1}{a} \tan^{-1}\dfrac{x}{a}$
If you integrate it then,
$\displaystyle \int_0^{+\infty} \dfrac{dx}{x^2+a^2} = \dfrac{1}{a}\times \dfrac{\pi}{2}$
2) $\displaystyle \int_0^{+\infty} \dfrac{4\; du}{(u^2+2)^2}$
We make the substitution $u=\sqrt{2}\tan \theta$ , this means $du=\sqrt{2}\sec^2 \theta d\theta$ .The new limits for $\theta$ would be according to
$ \displaystyle \theta = \tan^{-1}\dfrac{u}{\sqrt{2}}$ .The new limits are thus from $0\to\dfrac{\pi}{2}$ since $\tan^{-1}\dfrac{+\infty}{\sqrt{2}}=\dfrac{\pi}{2}$ & $\tan^{-1}\dfrac{0}{\sqrt{2}}=0$ . The integral now reads :
$$\displaystyle \begin{align}\int_0^{+\infty} \dfrac{4\; du}{(u^2+2)^2} &= 4\int_0^{\pi/2} \dfrac{\sqrt{2}\sec^2 \theta}{(2(\tan^2 \theta+1))^2} \; d\theta \\ &= 4\sqrt{2}\int_0^{\pi/2} \dfrac{\sec^2 \theta}{4(\sec^2 \theta)^2}\; d\theta \\ &= \sqrt{2} \int_0^{\pi/2} \dfrac{\sec^2 \theta}{\sec^4 \theta} \; d\theta \\ &= \sqrt{2} \int_0^{\pi/2} \cos^2 \theta \; d\theta \\ &= \dfrac{\sqrt{2}}{2}\int_0^{\pi/2} 2\cos^2 \theta \; d\theta \\ &= \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} (1+\cos 2\theta)\; d\theta \\ &= \dfrac{1}{\sqrt{2}} \int_0^{\pi/2} d\theta + \dfrac{1}{\sqrt{2}}\int_0^{\pi/2} \cos 2\theta \; d\theta \\ &= \dfrac{\pi}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}\left[\sin 2\theta\right]_0^{\pi/2} \\ &= \dfrac{\pi}{2\sqrt{2}} - \dfrac{1}{2\sqrt{2}}(\sin \pi - \sin 0) \\ &= \dfrac{\pi}{2\sqrt{2}} \end{align} $$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using Vieta's to create polynomial The quadratic $x^2+\frac{3}{2}x-1$ has the following unexpected property: the roots, which are $\frac{1}{2}$ and $-2$, are one $\textit{less}$ than the final two coefficients. Now find a quadratic with leading term $x^2$ such that the final two coefficients are both non-zero, and the roots are one $\textit{more}$ than these coefficients.
I thought of using Vieta's and designating roots p and q, but I wasn't able to go far. Here is my thought process.
Let there be 2 roots, $p$ and $q$.
$$p+q = -b$$
$$pq= a$$
$$p=-b+1$$
$$q = a+1$$
At this point, I got lost.
| Let $x^2 + ax + b$ be the wanted polynomials and $x_1,x_2$ it's roots. Then by Vieta's Formulae we have $x_1 + x_2 = -a$ and $x_1x_2 = b$. Also from the condition $x_1 = a + 1$ and $x_2 = b+1$.
Now subsituting in the first equation we have: $2a + b = -2$, so we get that $a= \frac{-2-b}{2}$. Then $x_1 = \frac{-2-b}{2} - 1 = -\frac{4+b}{2}$. Substituting in the second Vieta's Formula we have:
$$b = \left(-\frac{4+b}{2}\right)(b-1) \iff -2b = b^2 + 3b - 4 \iff b^2 + 5b - 4 =0 $$
Solving this equation will give you a solution.
| {
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"timestamp": "2023-03-29T00:00:00",
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Area of a Triangle in $3-$Dimensions After Undergoing Three Linear Transformations There is a triangle in $3-$space with vertices at the three following points:
$A=(1,1,1)$, $B=(1,1,-1)$ and $C=(-1,1,1).$
These points then undergo a sequence of three transformations to give $A'$, $B'$ and $C',$ respectively. First, a rotation of $30°$ about the $x-$axis, second, a rotation of $30°$ about the $z-$axis, and finally a contraction with a factor of $k=1/4.$
What would be the transformation matrix that represents these three transformations combined in the order stated?
What would be the area contained in the triangle whose vertices are the points $A',B',C'$?
Finally, what would be the ratio of the area of the triangle whose vertices are $A', B', C'$ to the area of the triangle whose vertices are $A, B, C$?
| Here is a sketch assuming you mean rotation wrt right hand rule. disclaimer: it is the opposite of elegant:
Apply the function:
$$\begin{pmatrix}x\\y\\z \end{pmatrix} \mapsto \begin{pmatrix}\frac{1}{4}&0&0\\0&\frac{1}{4} &0\\0&0&\frac{1}{4} \end{pmatrix}\begin{pmatrix}\frac{\sqrt{3}}{2}&-\frac{1}{2}&0\\\frac12&\frac{\sqrt{3}}{2}&0\\0&0&1\end{pmatrix}\begin{pmatrix}1 & 0 &0\\0&\frac{\sqrt{3}}{2}& -\frac{1}{2}\\0&\frac{1}{2}&\frac{\sqrt{3}}{2} \end{pmatrix} \begin{pmatrix}x\\y\\z\\ \end{pmatrix}$$
to all three points in order to obtain $A^{\prime}, B^{\prime}, C^{\prime}$. From this, form vectors $v:=A^{\prime}-C^{\prime}$ and $u:=B^{\prime}-C^{\prime}$. Take the projection $\frac{(v \cdot u)}{\|u\|}$, and take $\|v-\frac{(v \cdot u)}{\|u\|}\|$ for the height. Multiply this by $\|u\|$ which is the basis and divide by $1/2$ for the area. Or replace the last step with cross product, if you prefer.
| {
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"timestamp": "2023-03-29T00:00:00",
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What is the limit of this infinite sum of an infinite product? In a statistics problem (the Ross-Littlewood paradox) we encounter the following sum term
$$\sum_{k=1}^\infty \prod_{n=k}^\infty \left( \frac{9n}{9n+1} \right)$$
how do we evaluate such term?
Could we use the following with a limit on both sum and product together
$$\lim_{l \to \infty} \sum_{k=1}^l \prod_{n=k}^l \left( \frac{9n}{9n+1} \right) = \lim_{l \to \infty} \frac{9l}{10} = \infty$$
Or should we evaluate the terms independently
$$\lim_{l \to \infty} \sum_{k=1}^l \lim_{m \to \infty} \prod_{n=k}^m \left( \frac{9n}{9n+1} \right) = \lim_{l \to \infty} \sum_{k=1}^l 0 = 0$$
I have a visual interpretation of these integrals which is like a triangular form:
$$\begin{array}\\
\sum_{k=1}^l \prod_{n=k}^m \left( \frac{9n}{9n+1} \right) = & \frac{9 \cdot 1}{9 \cdot 1 +1} & \cdot & \frac{9 \cdot 2}{9 \cdot 2 +1} & \cdot & \frac{9 \cdot 3}{9 \cdot 3 +1} & \cdots & \cdots & \cdots & \cdots & \frac{9 \cdot m}{9 \cdot m +1} & + & & \\
& & & \frac{9 \cdot 2}{9 \cdot 2 +1} & \cdot & \frac{9 \cdot 3}{9 \cdot 3 +1} & \cdots & \cdots & \cdots& \cdots & \frac{9 \cdot m}{9 \cdot m +1} & +\\
& & & & & \frac{9 \cdot 3}{9 \cdot 3 +1} & \cdots & \cdots & \cdots & \cdots & \frac{9 \cdot m}{9 \cdot m +1} &+\\
& & & & & & \ddots & \cdots & \cdots & \cdots & \frac{9 \cdot m}{9 \cdot m +1} & +\\
& & & & & & & \ddots & \frac{9 \cdot l}{9 \cdot l +1} & \cdots & \frac{9 \cdot m}{9 \cdot m +1} \\
\end{array} $$
and when ${(l,m) \to (\infty,\infty)}$ I suspect we end up with the same triangle independent from the path.
Of particular interest is the reasoning on the difference between the two.
*
*Why it is a different result while the terms for both limits count up to the same collection?
*Does the infinite term in the second case make sense. Would it be valid to state that $lim_{k \to \infty} \prod_{n=k}^\infty \left( \frac{9n}{9n+1} \right) = 0$ while this result has only been shown for finite $k$?
*Is the product term zero or almost zero?
*Is there literature on this topic of such nested expressions?
| Note that
$$
\exp\left(\frac{-1}{9n+1}\right)\geq1-\frac{1}{9n+1}
$$
for all $n\geq 1$ so that
$$
0=\exp\left(\sum_{n=k}^{\infty}\frac{-1}{9n+1}\right)\geq \prod_{n=k}^{\infty}\left(1-\frac{1}{9n+1}\right)
$$
for each $k\geq 1$. Hence
$$
\sum_{k=1}^\infty \prod_{n=k}^\infty \left( \frac{9n}{9n+1} \right)=0.
$$
| {
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"timestamp": "2023-03-29T00:00:00",
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} |
All irreducible polynoms of degree 4 in $\mathbb{Z}_2$ I need to find all irreducible polynoms of degree at most 4 in $\mathbb{Z}_2$
This is my result:
$$ x^4 + x^1 + 1 \\
x^4 + x^2 + 1\\
x^4 + x^3 + 1 \\
x^4 + x^3 + x^2 + x^1 + 1 \\
$$
Is this correct?
| $x^4 + x^2 + 1= (x^2 + x + 1)^2$ is not irreducible.
An irreducible polynomial of degree $4$ defines a field extension of $\mathbb Z_2$ of degree $4$. Its $2^4$ elements are the roots of $x^{2^4}-x$ because of Lagrange's theorem in group theory.
Therefore, the irreducible polynomials of degree $4$ are those that divide $x^{2^4}-x$ over $\mathbb Z_2$:
$$
x^{16}-x=x (x + 1) (x^2 + x + 1) (x^4 + x + 1) (x^4 + x^3 + 1) (x^4 + x^3 + x^2 + x + 1)
$$
and so there are exactly three of them.
To factor $x^{16}-x$, factor $x^{15}-1$, starting with the factorization into cyclotomic polynomials:
$$
x^{16}-x = x(x^{15}-1)=x \Phi_1(x) \Phi_3(x) \Phi_5(x) \Phi_{15}(x)
$$
which gives
$$
x^{16}-x = x(x - 1)(x^2 + x + 1) (x^4 + x^3 + x^2 + x + 1) (x^8 - x^7 + x^5 - x^4 + x^3 - x + 1)
$$
We then need to further factor these mod 2. The last factor is the only one that is reducible.
| {
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"source": "stackexchange",
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"answer_count": 1,
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} |
Using Lagrange multiplier to find all extrema It is given that
$$f(x,y,z)=x^2-y^2+z^2$$
and the constraint is
$$g(x,y,z)=x^2+2y^2+3z^2-1=0$$
I started to take Lagrange multiplier approach, here is my work
$$
\left\{
\begin{array}{c}
2x-2xλ=0 \\
-2y-4yλ=0 \\
2z-6zλ=0 \\
x^2+2y^2+3z^2-1=0
\end{array}
\right.
$$
Which gives
$$
\left\{
\begin{array}{c}
x(1-λ)=0 \\
y(1+2λ)=0 \\
z(1-3λ)=0 \\
x^2+2y^2+3z^2-1=0
\end{array}
\right.
$$
Then, I have no idea how to plug equation (1) to (3) in the constrain since x, y and z equal to zero. If x, y and z equal to zero, then equation (4) will not work.
$$$$
If x, y and z not equal to zero, then
$$
λ=1 \ or \ λ=-1/2 \ or \ λ=1/3 \
$$
What's wrong in my approach, do I missed other constraints?
| The system of equations that you obtained, that is
$$
\begin{cases}
x(1-\lambda)=0 \\
y(1+2\lambda)=0 \\
z(1-3\lambda)=0 \\
x^2+2y^2+3z^2-1=0
\end{cases}
$$
is correct. Now looking at $x(1-\lambda)=0$, you should say that if $x\not=0$ then $\lambda=1$ and
$$
\begin{cases}
3y=0 \\
-2z=0 \\
x^2+2y^2+3z^2-1=0
\end{cases}
\implies y=z=0,\; x=\pm 1.$$
Can you take it from here and see what happens when $y\not=0$? What if $z\not=0$?
Note that the constraint $x^2+2y^2+3z^2-1=0$ implies that at least one of $x$, $y$, $z$ is different from zero.
| {
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"answer_id": 0
} |
Equation 1.15 in Jaynes Probability Theaory - The Logic of Science The equation 1.15 states:
$C ≡ (A + \overline{B})(\overline{A} + A\overline{B}) + \overline{A}B(A + B)$
Further the text it is suggested to the reader to verify that the equation is equal to C = (B ⇒ $\overline{A}$)
I can't grasp the idea behind it:(
For me it: $(A + \overline{B})(\overline{A} + A\overline{B}) + \overline{A}B(A + B) = A\overline{A} +AA\overline{B} +\overline{A}\overline{B} + A\overline{B}\overline{B} + \overline{A}AB +\overline{A}BB = A\overline{B} + \overline{A}\overline{B} + \overline{A}B =\overline{B} +\overline{A}B$
Please help me to undestend how an implication can be derived from this equation.
upd. Ok, I think I see why it's true (from Venn diagram): the only way B can be True is in conjunction with $\overline{A}$, so if $B$ is True then $\overline{A}$ is also True.
|
For me it was: $(A + \overline{B})(\overline{A} + A\overline{B}) + \overline{A}B(A + B) \\= A\overline{A} +AA\overline{B} +\overline{A}\,\overline{B} + A\overline{B}\,\overline{B} + \overline{A}AB +\overline{A}BB \\= A\overline{B} + \overline{A}\,\overline{B} + \overline{A}B \\=\overline{B} +\overline{A}B$
Yes, so far so good. From there, on to $C = \overline B+\overline A$ by the Absorption Law.
Or from the second last line, use idempotence. $C~{= A\overline{B} + \overline{A}\,\overline{B} + \overline{A}B \\ = A\overline{B} + \overline{A}\,\overline{B}~+~\overline{A}\,\overline{B}+\overline{A}B\\ =\overline B+\overline A}$
Therefore $C= (B\to\overline A)$ by Implication Equivalence.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
"answer_count": 1,
"answer_id": 0
} |
Generalizing the Pell equation $x^2-9\times89y^2 = 1$? I. The fundamental solution to the Pell equation,
$$x^2-61y^2=1$$
will stand out being "largish" as it is the $\color{blue}{6\text{th}}$ power of a fundamental unit $U_d$,
$$(U_{61})^6 = \big(\tfrac{39+5\sqrt{61}}{2}\big)^6 = x+y\sqrt{61} =1766319049+226153980\sqrt{61}$$
A necessary (but not sufficient) condition is $d=8n+5$.
II. Likewise, the fundamental solution to,
$$x^2-9\times89y^2=1$$
turns out to involve a $\color{blue}{4\text{th}}$ power,
$$(U_{89})^4 = (500+53\sqrt{89})^4 = x+3y\sqrt{89} = 500002000001+3\times 17666702000\sqrt{89}$$
More generally,
Q. Is it true that given fundamental solution $p,q$ of the negative Pell equation,
$$p^2-dq^2 = -1$$
if we define $x,y$ as,
$$(p+q\sqrt{d})^4 = x+3y\sqrt{d}$$
then $x,y$ is the fundamental solution of,
$$x^2-9dy^2 = 1$$
for all prime $d = 12n+5$? Equivalently, since
$$(p^2 - d q^2)^4 = \big(p^4 + 6 d p^2 q^2 + d^2 q^4\big)^2 - d \big(4 p q (p^2 + d q^2) \big)^2=1$$
then for such $d$, is $p^2 + d q^2$ a multiple of $3$?
P.S. There is similar behavior for
$$x^2-49dy^2 = 1$$
but now involves a $\color{blue}{24\text{th}}$ power. For example, the fundamental solution to,
$$x^2-49\times13y^2 = 1$$
is,
$$(U_{13})^{24} = \big(\tfrac{3+\sqrt{13}}{2}\big)^{24} = x+7y\sqrt{13} =1419278889601 +7\times56233877040\sqrt{13}$$
though its sequence of primes $d = 5, 13, 61, 157, 181, 397,\dots$ is harder to characterize compared to section II.
|
Equivalently, since
$$(p^2 - d q^2)^4 = \big(p^4 + 6 d p^2 q^2 + d^2 q^4\big)^2 - d \big(4 p q (p^2 + d q^2) \big)^2=1$$
then for such $d$, is $p^2 + d q^2$ a multiple of $3$?
Since $d=12n+5\equiv 2\pmod 3$, $$p^2-dq^2=-1\implies p^2+q^2\equiv 2\pmod 3\implies p^2\equiv q^2\equiv 1\pmod 3$$
It follows from this that
$$p^2+dq^2\equiv 1+2\cdot 1\equiv 0\pmod 3$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "3",
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} |
Find the sum of the double series $\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)} $ First, the original problem follows,
$$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}$$
where $$H_{k}=\sum_{j=1}^k \frac{1}{j}$$ is the $k$-th partial sum of harmonic series.
Using the following identity,
$$H_{k+1} = \sum_{j=1}^\infty (\frac{1}{j} - \frac{1}{k+j+1}).$$
I was able to get this one.
$$\sum_{k=1}^\infty \frac{H_{k+1}}{k(k+1)}=\sum_{j=1}^\infty [\frac{1}{j} - \frac{1}{j+1} + \frac{1}{j+1}\sum_{k=1}^\infty \frac{1}{(k+1)(k+1+j)}] $$
So, If I get the sum of the double series,
$$\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)}.$$
I can also find the original problem.
What method can I use at this problem?
| Going back to the original problem, note that for any positive integer $N$,
$$
\begin{align}
\sum_{k=1}^N \frac{H_{k+1}}{k(k+1)}&=\sum_{k=1}^N \frac{H_{k+1}}{k}-\sum_{k=1}^N \frac{H_{k+1}}{k+1}\\
&=\sum_{k=1}^N \frac{H_{k}}{k}+\sum_{k=1}^N \frac{1}{k(k+1)}-\sum_{k=2}^{N+1} \frac{H_{k}}{k}\\
&=1+\sum_{k=1}^N \left(\frac{1}{k}-\frac{1}{k+1}\right)- \frac{H_{N+1}}{N+1}\\
&=2-\frac{1}{N+1}- \frac{H_{N+1}}{N+1}.
\end{align}$$
Hence, by taking the limit as $N\to +\infty$, we get
$$\sum_{k=1}^{\infty} \frac{H_{k+1}}{k(k+1)}=2.$$
P.S. It follows that
$$\sum_{k=1}^\infty \sum_{j=1}^\infty \frac{1}{(k+1)(j+1)(k+1+j)}=1.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
"answer_count": 4,
"answer_id": 0
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Find $a$ if $(a+3)x^2-4x+2<0$ $\forall x\in [-2,1]$ Find $a$ if $(a+3)x^2-4x+2<0 \qquad \forall x\in [-2,1]$
My Attempt:
$a+3=0$ $\implies -4x+2<0$
$$2x-1>0$$
$$x>\dfrac {1}{2}$$
$\implies a \neq -3$.
| HINT
Let $f(x)= (a+3)x^2-4x+2$
Firstly, we discuss for $a+3$ is positive, negative, or zero.
Then for each case, we check for $x=-\frac{-4}{2(a+3)}=\frac{2}{a+3}$, is it on the left of $[-2,1]$, on the right, or in between?
For example, if $a+3 < 0$:
If $\frac{2}{a+3} \le -2$: it is decreasing, and we just want $f(-2) <0$
If $-2 < \frac{2}{a+3} <1$: we need its local maximal to be negative, $f(\frac{2}{a+3}) < 0$
If $\frac{2}{a+3} \ge 1$: it is increasing, $f(1) <0$
| {
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"answer_id": 3
} |
$\ 3\,\tan(x)-\sqrt3 $ You have to find alternate form, that has only multiplication/division signs between trigonometric functions
The form is: $\ 3\,\tan(x)-\sqrt3 $
The solution form is: $\ \left(\frac{2*\sqrt 3\sin(x-π/6)}{\cos(x)}\right) $
My try: $\ 3\,\tan(x)-\sqrt3 = 3*\left(\frac{\sin(x)}{\cos(x)}\right)-\left(\frac{\sin(π/3)}{\cos(π/3)}\right)= \left(\frac{\sin(x)\cos(π/3)-\sin(π/3)\cos(x)}{\cos(x)\cos(π/3)}\right)=\left(\frac{\sin(x)\cos(π/3)+\sin(x)\cos(π/3)-\sin(π/(3)\cos(x)}{\cos(x)\cos(π/3)}\right)$
I am certain that I have done something wrong in one of the steps, as this leads to only more complicated form, that isn't even near the solution
| Note that we can write $$3\tan x- \sqrt{3} = 2\times \frac{\sqrt{3}}{2}\frac{\sin x}{\cos x}\times \sqrt{3} - 2\times \sqrt{3}\times \frac{\cos x}{\cos x}\frac{1}{2}$$ $$=2\times \sqrt{3}\frac{1}{\cos x}[\sin x(\frac{\sqrt{3}}{2})-\cos x(\frac{1}{2})]$$ $$=\frac{2\times \sqrt{3}}{\cos x}[\sin x\cos \frac{\pi}{6}-\cos x\frac{\pi}{6}]$$ $$=\frac{2\sqrt{3}\sin(x-\frac{\pi}{6})}{\cos x}$$
| {
"language": "en",
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"source": "stackexchange",
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Fourier series proving $\sum \frac{1}{n^4} = \frac{\pi^4}{90}$ I have calculated the Fourier series of $x^4$ in $(-4,4)$ as
$$f(x) = \frac{256}{5} + \sum_{n=1}^\infty \left(\frac{2048}{\pi^2n^2} \cos(\pi n ) -\frac{12288}{\pi^4n^4} \cos(\pi n )\right)\cos (\frac{\pi n x}{4})$$
I have been asked to use
$$ \sum_{n=1}^∞ \frac{1}{n^2} = \frac{\pi^2}{6} $$
to show that
$$ \sum_{n=1}^∞ \frac{1}{n^4} = \frac{\pi^4}{90} $$
Not sure how to go about this any pointers would help.
| Since $x^4$ is $C^1$ and has equal values at the endpoints of $[-4,4]$ we have
$$
x^4= \frac{256}{5} + \sum_{n=1}^\infty \left(\frac{2048}{\pi^2n^2} \cos(\pi n ) -\frac{12288}{\pi^4n^4} \cos(\pi n )\right)\cos \left(\frac{\pi n x}{4}\right)
$$
for $x\in[-4,4]$. Hence, taking $x=4$ we get
$$
\begin{split}
256
&= \frac{256}{5} + \sum_{n=1}^\infty \left(\frac{2048}{\pi^2n^2} \cos(\pi n ) -\frac{12288}{\pi^4n^4} \cos(\pi n )\right)\cos (\pi n )\\
&= \frac{256}{5} + \sum_{n=1}^\infty \left(\frac{2048}{\pi^2n^2} -\frac{12288}{\pi^4n^4}\right)\\
&= \frac{256}{5} + \frac{2048}{\pi^2}\cdot\frac{\pi^2}{6} -\frac{12288}{\pi^4}\sum_{n=1}^\infty\frac{1}{n^4}.
\end{split}
$$
And after some simple computations we get
$$\sum_{n=1}^\infty \frac{1}{n^4} = \frac{\pi^4}{90}.
$$
| {
"language": "en",
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Evaluating $\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$ I saw this problem somewhere recently and I was having some difficulty getting started on it.
The problem is twofold. The first is to evaluate:
$$\sum_{k=0}^\infty \left(\frac{1}{5k+1} - \frac{1}{5k+2} - \frac{1}{5k+3} + \frac{1}{5k+4} \right)$$
and once this is done, to explain what this has to do with the construction of a pentagon (maybe some other polygon?) using a compass and straight edge.
In terms of evaluating the series, I tried writing each $n$ as $m \cdot 2^k$ and evaluating the summation there since $2^k$ will alternate between + and - mod 5. However, this leads to a divergent series and I think this is not a valid thing to do since the original series is not absolutely convergent so we can't rearrange terms like that.
| As a followup to Lord Shark's answer, the idea of using $\frac{1}{n+1}=\int_{0}^{1}x^n\,dx $ can be naive but it is pretty effective. Once we have
$$ L(\chi,1)=\sum_{n\geq 1}\frac{\left(\frac{n}{5}\right)}{n}=\int_{0}^{1}\frac{1-x^2}{1+x+x^2+x^3+x^4}\,dx $$
the integral can be evaluated by partial fraction decomposition, since $\int_{0}^{1}\frac{dx}{x-\xi}=\log\left(1-\frac{1}{\xi}\right) $.
Let $\omega=\exp\left(\frac{2\pi i}{5}\right)$. We have
$$\begin{eqnarray*} \operatorname*{Res}_{x=\omega^k}\frac{1-x^2}{1+x+x^2+x^3+x^4}&=&\lim_{x\to \omega^k}\frac{(1-x-x^2+x^3)(x-\omega^k)}{1-x^5}\\&\stackrel{d.H.}{=}&\lim_{x\to \omega^k}\frac{-1-2x+3x^2}{-5x^4}\\&=&\frac{1}{5}\lim_{x\to \omega^k}\left(x+2x^2-3x^3\right)\end{eqnarray*} $$
for any $k\in[1,4]$, hence
$$\begin{eqnarray*} L(\chi,1) &=& \frac{1}{5}\sum_{k=1}^{4}\left(\omega^k+2\omega^{2k}-3\omega^{3k}\right)\log(1-\omega^{-k})\\&=&\color{red}{\frac{2\log(5+\sqrt{5})-\log(20)}{\sqrt{5}}}.\end{eqnarray*}$$
In a similar way you may prove that
$$ L(\chi,2)=\sum_{k\geq 0}\left[\frac{1}{(5k+1)^2}-\frac{1}{(5k+2)^2}-\frac{1}{(5k+3)^2}+\frac{1}{(5k+4)^2}\right]=\frac{4\pi^2}{25\sqrt{5}}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "11",
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Taylor series for $\sqrt{1+x^2}$ I want to expand
$$f(x)=\sqrt{1+x^2}$$ in powers of $x-2$
I started by getting the maclaurin series
$$\sqrt{1+x}=1+\frac{1}{2}x+\frac{1}{2} \left( \frac{-1}{2} \right) \frac{x^2}{2!} + \frac{1}{2} \left( \frac{-1}{2}\right) \left(\frac{-3}{2}\right)\frac{x^3}{3!}$$
$$\sqrt{1+x^2}=1+\frac{1}{2}x^2+\frac{1}{2}\left(\frac{-1}{2}\right)\frac{x^4}{2!}+\frac{1}{2} \left(\frac{-1}{2}\right)\left(\frac{-3}{2}\right)\frac{x^6}{3!}$$
Then
$$\sqrt{1+x^2}=\sqrt{1+(x-2)^2-4+4x}=\sqrt{(x-2)^2-3+4x}=\sqrt{(x-2)^2-3+4(x-2)+8}$$
I could not complete , what can we do then ?
(I know that we can differentiate the function and substitute in the Taylor formula , but I want a shorter way)
for example :
if we want to expand $$g(x)=\frac{1}{1-x}$$ around $x=2$ we can start by
\begin{align}
g(x) & =\frac{1}{1-x}=\frac{1}{1-(x-2)-2}=\frac{-1}{1+(x+2)} \\[10pt]
& =-[1-(x-2)+(x-2)^2-(x-2)^3+\cdots]
\end{align}
So I want to get convert $f(x)$ to a form that we can write its expansion without getting derivatives, like I did with $g(x)$ above
| If you want to take the limit when $x\to 2$ it is easier to set $x=2+u$ with $u\to 0$ because you will get it easier to make the expansions in variable $u$.
Thus $f(x)=\sqrt{1+x^2}=\sqrt{1+(2+u)^2}=\sqrt{5+4u+u^2}$
The proper expansion in zero is $\displaystyle \sqrt{1+v}=1+\frac v2-\frac{v^2}8+\frac{v^3}{16}-\frac {5v^4}{128}+\cdots$
We need first to factorize $5$ out of the square root.
$f(x)=\sqrt{5}\sqrt{1+\frac 45u+\frac 15u^2}\quad$ and then substitute $v=(\frac 45u+\frac 15u^2)$ in the expansion.
For instance let's limit ourselves to $o(u^3)$
$$f(x)=\sqrt{5}\left(1+\frac 12\left(\frac 45u+\frac 15u^2\right)-\frac 18 \left(\frac 45u+\frac 15u^2\right)^2 + \frac 1{16} \left(\frac 45u+\frac 15u^2 \right)^3 + o(u^3)\right)$$
We will do the calculation while ignoring all terms smaller than $o(u^3)$ (i.e terms in $u^4, u^5, \ldots$).
\begin{align}
f(x) & =\sqrt{5}\left(1+\frac 12\left(\frac 45u+\frac 15u^2\right)-\frac 18 \left(\frac {16}{25}u^2+2\frac 45\frac 15u^3\right)+\frac 1{16} \left(\frac 45u\right)^3 + o(u^3)\right) \\[8pt]
&=\sqrt{5}\left(1+\frac 25 u+\frac 1{50}u^2-\frac 1{125}u^3+o(u^3)\right)
\end{align}
In the end replace $u$ by $(x-2)$ to have the desired expansion:
$$f(x)=\sqrt{5}\left(1+\frac 25 (x-2)+\frac 1{50}(x-2)^2-\frac 1{125}(x-2)^3+o((x-2)^3)\right)$$
| {
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"question_score": "5",
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Convert the double integral into polar $$\int^{4}_{0}\int_{0}^{\sqrt{4x-x^2}}\sqrt{x^2+y^2}\text{ dy dx}=14.22$$
In polar, we have $0\leq y\leq \sqrt{4x-x^2}$
Simplying, we get $y^2=4x-x^2$, which turns into $(x-2)^2+y^2=4$. Here is the region we are considering.
Clearly $0\leq\theta\leq\pi$, and we know that
$x^2+y^2=4x$, which means $r^2=4r\cos\theta\to r=4\cos\theta$
Therefore, $0\leq r\leq 4\cos\theta$, and the original function $\sqrt{x^2+y^2}$ simply becomes $r$. So our integral now is:
$$\int^{\pi}_{0}\int^{4\cos\theta}_{0}r\text{ r dr d$\theta$}=\int^{\pi}_{0}\int^{4\cos\theta}_{0}r^2\text{ dr d$\theta$}=\int^{\pi}_{0}\left[\dfrac{r^3}{3}\right]^{4\cos\theta}_{0}\text{ d$\theta$}$$
$$=\dfrac{64}{3}\int^{\pi}_{0}\cos^3\theta\text{ d$\theta$}=0$$
But according to WA^, the cartesian coordinate integral is different?
| You have to integrate $d\theta$ from $0$ to $\pi/2$.
This will give you
$$\int_0^{\pi/2}\cos^3\theta\ d\theta = \frac{2}{3}$$
Hence
$$\frac{128}{9} = 14.22(...)$$
| {
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Is $Log(i^{1/3})=\frac{1}{3}Log(i)$? The definition of general complex log for any non-zero complex number $z$ is
$$Log(z)=\log|z|+i[\arg(z)+2m\pi], m\in \mathbb{Z}$$
With this, if $n\in \mathbb{N}$ then
$Log(z^{1/n})=\frac{1}{n} Log(z)$ holds for all non-zero complex number $z$.
I verified this for $Log(i^{1/2})=\frac12 Log(i)$ successfully but could not make up with the following: $$Log(i^{1/3})=\frac13 Log(i)$$
Let me show what I have done and where I got stuck.
Since $i=\cos(2n\pi+\frac{\pi}{2})+i \sin(2n\pi+\frac{\pi}{2}), n\in \mathbb{Z}$ then by De-Moivre' s theorem, we have
\begin{align}
i^{1/3}= & \cos\left(\frac{2n\pi+\frac{\pi}{2}}{3}\right)+i\sin\left(\frac{2n\pi+\frac{\pi}{2}}{3}\right), n=0,1,2 \\
=& \cos\left(\frac{(4n+1)\pi}{6}\right)+i\sin\left(\frac{(4n+1)\pi}{6}\right), n=0,1,2\\
= &\begin{cases}
\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}) \\
\cos(\frac{5\pi}{6})+i\sin(\frac{5\pi}{6}) \\
\cos(\frac{9\pi}{6})+i\sin(\frac{9\pi}{6})
\end{cases}\\
= &\begin{cases}
\cos(\frac{\pi}{6})+i\sin(\frac{\pi}{6}) \\
\cos(\pi-\frac{\pi}{6})+i\sin(\pi-\frac{5\pi}{6}) \\
-\cos(\frac{\pi}{2})-i\sin(\frac{\pi}{2})
\end{cases}\\
= &\begin{cases}
\frac{\sqrt{3}}{2}+i\frac{1}{2} \\
-\frac{\sqrt{3}}{2}+i\frac{1}{2} \\
0-i
\end{cases}
\end{align}
Now LHS:
\begin{align}
\frac13 Log(i)=&\frac13[\log|i|+i\{\arg(i)+2n_1 \pi\}], n_1\in \mathbb{Z}\\
=&\frac13(2n_1\pi+\frac{\pi}{2}), n_1\in \mathbb{Z}\\
=&(4n_1+1)\frac{\pi i}{6}, n_1\in \mathbb{Z}
\end{align}
whereas we see that
\begin{align}
&Log(i^{1/3})\\
=&\begin{cases}
\log|\frac{\sqrt{3}}{2}+i\frac{1}{2}|+i[\arg(\frac{\sqrt{3}}{2}+i\frac{1}{2})+2m_1\pi]\\
\log|-\frac{\sqrt{3}}{2}+i\frac{1}{2}|+i[\arg(-\frac{\sqrt{3}}{2}+i\frac{1}{2})+2m_2\pi]\\
\log|-i|+i[\arg(-i)+2m_3\pi]
\end{cases}\\
=&\begin{cases}
i[\frac{\pi}{6}+2m_1\pi]\\
i[\frac{5\pi}{6}+2m_2\pi]\\
i[\frac{-\pi}{2}+2m_3\pi]
\end{cases}, m_1, m_2, m_3\in \mathbb{Z}
\end{align}
And here I got stuck. I don't know how to finish. Any help will be appreciated.
| You have
$$ i = \exp\left(i\frac{\pi}{2} + i2n\pi\right) $$
Taking the third power yields
$$ i^{1/3} = \exp\left(i\frac{\pi}{6} + i \frac{2n\pi}{3} i\right) $$
Since $|i| = |i^{/3}| = 1$, taking the log yields
$$ \log (i^{1/3}) = i\left(\frac{\pi}{6} + \frac{2n\pi}{3} \right) $$
On the other hand, we have
$$ \frac{1}{3}\log (i) = \frac{i}{3} \left(\frac{\pi}{2} + 2n\pi \right) = i \left( \frac{\pi}{6} + \frac{2n\pi}{3} \right) $$
Does this help?
EDIT: Continuing from your work, you can show that the 3 arguments for $\log(i^{1/3})$ are evenly spaced by an angle of $2\pi/3$. Therefore they can all be combined as
$$ \left\{\begin{aligned} i\left(-\frac{\pi}{6} + 2m_1\pi\right) \\ i\left(\frac{5\pi}{6} + 2m_2\pi\right) \\ i\left(-\frac{\pi}{2} + 2m_3\pi\right) \end{aligned}\right. = i\left(\frac{\pi}{6} + \frac{2n\pi}{3}\right) = i(4n+1)\frac{\pi}{6} $$
where $n$ is mapped by every alternating triplet of $(m_1,m_2,m_3)$. For example $m_3 = 0, m_1 = 0, m_2 = 0 \to n = -1,0,1$, etc
| {
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How do you find the values that $x$ can take by squaring? $$|x+4| \cdot |x-4| = x+4$$
How do you find the values that $x$ can take by squaring?
How I've tried
$$|x+4|^2 \cdot |x-4|^2 = (x+4)^2$$
and
$$|x^2+16| \cdot |x^2+16| = x^2+16$$
I think I went wrong.
| You can consider two cases:
If $\ x+4=0$, then $x=-4$. So
$$\vert{x+4}\vert(x-4)=0=x+4.$$
IF $\ x+4\neq 0$, then you can write
$$x-4=\frac{x+4}{x+4}=\mathrm{sign}(x+4)=\begin{cases} 1 &x+4\geq0\\ -1&x+4<0\end{cases}=\begin{cases} 1 &x\geq-4\\ -1&x<-4\end{cases}$$
from that the solutins are $x=3$ is $\ x=5$.
Or you can take squares like you said:
$$(x+4)^{2}(x-4)^{2}=(x+4)^{2}$$
and saying again if $x=-4$ both sides are equal to $0$. Otherwise you can cancel both factors $(x+4)^{2}$ (remember only when $x\neq -4$) to get:
$$ (x+4)^{2}(x-4)^{2}=(x+4)^{2} \overset{\displaystyle{x\neq-4}}{\Longrightarrow} (x-4)^2=1 \Longrightarrow x-4=\pm 1$$
so, $x-4=-1$ or $x-4=1$, so the 3 solutions are:
$$x_1=-4, x_2=3, x_3=5$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "1",
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Deriving least squares coefficients for curve of form $y=a/(x^2+b)$ I am trying to derive the coefficients $a$ and $b$ for a least squares curve of the form $y=\frac{a}{x^2+b}$
To start I have written $$S=\sum_i(y_i-y)^2=\sum_i\left(y_i-\frac{a}{x_i^2+b}\right)^2=\sum_i\left(y_i^2-\frac{2a y_i}{x_i^2+b}+\frac{a^2}{(x_i^2+b)^2}\right) \\S=\sum_iy_i^2-2a\sum_i\frac{y_i}{x_i^2+b}+a^2\sum_i\frac{1}{(x_i^2+b)^2}$$
Now the way I have seen it done for the line $y=a+bx$ is they minimize $S$ with respect to both $a$ and $b$, applying that for this curve gives the two constraint equations:
$$0=a\sum_i\frac{1}{(x_i^2+b)^2}-\sum_i\frac{y_i}{x_i^2+b}\\
0=-a\sum_i\frac{1}{(x_i^2+b)^3}+\sum_i\frac{y_i}{x_i^2+b}$$
Summing these together we find $$0=a\left(\sum_i\frac{1}{(x_i^2+b)^2}-\sum_i\frac{1}{(x_i^2+b)^3}\right)$$
Since $a\neq 0$ we must have $$\sum_i\frac{1}{(x_i^2+b)^2}=\sum_i\frac{1}{(x_i^2+b)^3}$$ Which seems to imply $$(x_i^2+b)^2=(x_i^2+b)^3$$ and unfortunately there doesnt appear to be a constant $b$ that solves this equation. Did I make a mistake somewhere or is this the result?
| We have $$\frac{\partial S}{\partial a} = -2\sum \frac{y_i}{x_i^2+b}+2a\sum \frac{1}{(x_i^2+b)^2}$$ $$\frac{\partial S}{\partial b} = 2a \sum \frac{y_i}{(x_i^2+b)^2} -2a^2 \sum \frac{1}{(x_i^2+b)^3}$$
| {
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} |
Solving the Differential equation: $y'=\frac{2}{x}y+x^3$ We have the differential equation $$y'=\frac{2}{x}y+x^3$$ and we know $x \in (0, \infty)$.
My attempt with variation of constants
\begin{align}
\phi(x) &= \exp \left(\int \frac{2}{x} dx \right) \\
&= \exp(2\ln|x|) \\
&= x^2c
\end{align}
and
\begin{align}
\psi(x) &= (x^2c) \cdot \int \frac{x^3}{x^2} dx \\
&= (x^2c) \cdot \frac{x^2}{2}
\end{align}
but this solution is wrong. Where is the mistake?
| The mistake is with the integration constant:
$$\exp\left(\int\frac{2}{x}\,dx\right) = \exp\bigr(2\log(x) + d\bigl) = cx^2$$
and not $x^2 +c$
| {
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"source": "stackexchange",
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Show that $x^3+x^2+x+1 \mid x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4s}$ in $\Bbb Z[x]$ Problem:
Show that $x^3+x^2+x+1 \mid x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4s}$ in $\Bbb Z[x]$ for all $m,n,q,s$ nonnegative integers
So my algebra sucks.
I'm reviewing some algebra and found this problem. I was able to play with it and get a trivial solution that I'm sure is missing the point of the problem.
My (dumb) solution:
We prove by induction.
For $m=n=q=s=0$ we have that $x^3+x^2+x+1 \mid x^3+x^2+x+1$.
Now assume $x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4s}=p(x)(x^3+x^2+x+1)$.
Then: $$(p(x)+x^{4m+3}(x-1))(x^3+x^2+x+1)=x^{4(m+1)+3}+x^{4n+2}+x^{4q+1}+x^{4s}$$
$$(p(x)+x^{4n+2}(x-1))(x^3+x^2+x+1)=x^{4m+3}+x^{4(n+1)+2}+x^{4q+1}+x^{4s}$$
$$(p(x)+x^{4q+1}(x-1))(x^3+x^2+x+1)=x^{4m+3}+x^{4n+2}+x^{4(q+1)+1}+x^{4s}$$
$$(p(x)+x^{4s}(x-1))(x^3+x^2+x+1)=x^{4m+3}+x^{4n+2}+x^{4q+1}+x^{4(s+1)}$$
However this is very unenlightening. What is the point of the problem? Is there more enlightening solution?
| $i,-1,-i$ are all roots of your polynomial. It is divisible by $(x-i)(x+1)(x+i)=x^3 + x^2 + x + 1.$ I suppose I should add that it is so divisible in $\mathbb Q [x],$ while Gauss theorem on content says it is then divisible in $\mathbb Z [x] \; .$
| {
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} |
Prove: $(\frac{1+i\sqrt{7}}{2})^4+(\frac{1-i\sqrt{7}}{2})^4=1$ $$\left(\frac{1+i\sqrt{7}}{2}\right)^4+\left(\frac{1-i\sqrt{7}}{2}\right)^4=1$$
I tried moving the left exponent to the RHS to then make difference of squares exp. $(x^2)^2$. Didn't get the same on both sides though. Any help?
| We need to solve for $ab=2$ and $a+b=1$
$\implies a,b=?$
Use $x^2+y^2=(x+y)^2-2xy$ twice to reach at $$a^4+b^4=1$$
| {
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"source": "stackexchange",
"question_score": "4",
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System of congruent equations Given the following system, how do you prove that there only is one solution congruent 8?
$$3x+7y \equiv 2 (\text{mod } 8)$$
$$4x+5y \equiv 7 (\text{mod } 8)$$
My main idea have been to solve the equation, then to come to the solution that there only is one solution, however I get flawed results to say the least. Here is my method:
$$\left\{
\begin{array}{ll}
3x+7y \equiv 2 (\text{mod } 8)
\\4x+5y \equiv 7 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+14y \equiv 4 (\text{mod } 8)
\\8x+10y \equiv 14 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+6y \equiv 4 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
$$
$$
\Leftrightarrow
\left\{
\begin{array}{ll}
6x+18 \equiv 4 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
-2x\equiv -14 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
\Leftrightarrow
\left\{
\begin{array}{ll}
2x\equiv 6 (\text{mod } 8)
\\2y \equiv 6 (\text{mod } 8)
\end{array}
\right.
$$
Which then gives:
$$
\left\{
\begin{array}{ll}
x=4t+3
\\y=4s+3
\end{array}
\right.$$
$t$ and $s$ being arbitary whole numbers
So solution given congruent 8 is:
$$x\equiv 3 (\text{mod } 8) \text{ or } x\equiv 7 (\text{mod } 8)$$
$$y\equiv 3 (\text{mod } 8) \text{ or } y\equiv 7 (\text{mod } 8)$$
However when you put that into the original equation, the answer is wrong. So where did i do wrong...
| Another method, that's specific to this question:
If x is even, then 4x is 0, so y = 3. Then 3x+21=2. The LHS is odd but the RHS is even, so "x is even" leads to a contradiction. So x is odd, 4x = 4, y =7, and x =3.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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Evaluating a value from two given equations If $x=\frac{4\lambda}{1+\lambda^2}$ and $y=\frac{2-2\lambda^2}{1+\lambda^2}$, where $\lambda$ is a real parameter and Z= $x^2$+$y^2$-xy, then what are the possible values of Z ?
How do I solve this problem? Are there any shortcuts or any theorems to be used?
I tried solving using a direct substitution, but could not proceed beyond step 2
$$ Z = x^2 + y^2 - xy$$ $$= (\frac{4\lambda}{1+\lambda^2})^2+ (\frac{2-2\lambda^2}{1+\lambda^2})^2 - (\frac{4\lambda}{1+\lambda^2})(\frac{2-2\lambda^2}{1+\lambda^2})$$ $$=\frac{4\lambda^4+8\lambda^3+8\lambda^2-8\lambda+4}{(1+\lambda^2)^2} $$
| We can write $\lambda=\tan (t/2)$. Then $x=2\sin t$ and $y=2\cos t$,
so $x^2+y^2=4$. Also $xy=4\sin t\cos t=2\sin(2t)$. Which values can $4-2\sin 2t$ take?
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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} |
Calculate the limit: $\lim_{x\to+\infty}(\frac{x^2 -x +1}{x^2})^{\frac{-3x^3}{2x^2-1}}$ without de l'Hôpital rule I was wondering how can I calculate the limit:
$$\lim_{x\to+\infty}\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}$$
without de l'Hôpital rule.
I tried to reconduct the limit at the well known one:
$$\lim_{x\to+\infty}\left(1+\frac1x\right)^x = e$$
Now I'm concentrating only on the part of the limit with still the Indeterminate Form, I reached this form elevating $e$ to the neperian logarithm of the function, trying to get rid of the $1^\infty$ I.F.
but, at the end of the day, I could only obtain:
$$\begin{align}\lim_{x\to+\infty}\ln\left(\frac{1}{x^2}(1+x^2-x)\right) &= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} \ln\left(1 + \frac{1}{\frac{1}{x^2-x}}\right) \\&= \lim_{x\to+\infty} \ln\left(\frac{1}{x^2}\right) + \lim_{x\to+\infty} (x^2-x) \ln\left(\left(1 + \frac{1}{\frac{1}{x^2-x}}\right)^{\frac{1}{x^2-x}}\right)\end{align}$$
But then defining
$$t= \frac{1}{x^2-x}$$
the limit
$$\lim_{t\to 0} \ln\left(\left(1 + \frac{1}{t}\right)^{t}\right)$$
goes no more to
$$ \ln(e)$$
because now $$t \to 0$$
Can you please give me some help?
| $$
\left(\frac{x^2 -x +1}{x^2}\right)^{\frac{-3x^3}{2x^2-1}}=\exp\left[\frac{-3x^3}{2x^2-1}\ln\left(\frac{x^2 -x +1}{x^2}\right)\right]=\exp\left[\frac{-3x^3}{2x^2-1}\ln\left(1-\frac{x-1}{x^2}\right)\right]
$$
The exponent
$$
\left[\frac{-3x^3}{2x^2-1}\ln\left(1-\frac{x-1}{x^2}\right)\right]\sim-\frac{3}{2}x\left(-\frac{x-1}{x^2}\right)\to 3/2\ ,
$$
therefore the sought limit is $e^{3/2}$. The above asymptotic estimate is obtained using the Maclaurin expansion of $\ln$.
| {
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"question_score": "2",
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I don't know how to solve this integral
Could you please help me compute the following integral: $$\int \frac{(x^2+x-4)}{x(x^2+4)}dx$$
What I've done so far is:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{A}{x}dx+\int \frac{B}{x^2+4}dx$$
So
$$x^2+x-4 = A(x^2-4)+Bx$$
And I did this so I could figure out the value of A and B but I am having a rough time trying to calculate A because of the $x^2$.
I've also seen that the solution solves this exercise by separating the fraction's numerator like this:
$$\int \frac{(x^2+x-4)}{x(x^2+4)}dx = \int \frac{x^2}{x(x^2+4)}dx+\int \frac{x}{x(x^2+4)}dx + \int \frac{-4}{x(x^2+4)}dx $$
When do I know that I have to use this? Is what I though correct? If yes, how do I continue it?
Thank you very much. Please, let me know if something is not very clear in my question.
Agapita.
| To make the finding of partial fractions slightly easier you may care to write
$$\frac{x^2 + x - 4}{x(x^2 + 4)} = \frac{(x^2 + 4) + x - 8}{x(x^2 + 4)} = \frac{1}{x} + \frac{1}{x^2 + 4} - \frac{8}{x(x^2 + 4)},$$
and then it is prehaps easier to guess/see that
$$\frac{8}{x(x^2 + 4)} = \frac{2}{x} + \frac{-2x}{x^2 + 4}.$$
| {
"language": "en",
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"source": "stackexchange",
"question_score": "2",
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Minimum value of rational expression
If $x,y,z$ are distinct real number, Then minimum value of $\displaystyle \left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2$
Try: $$\left(\frac{x}{y-z}\right)^2+\left(\frac{y}{z-x}\right)^2+\left(\frac{z}{x-y}\right)^2\geq \frac{x^2+y^2+z^2}{(x-y)^2+(y-z)^2+(z-x)^2}$$
could some help me to solve this, thanks
| If $z=0$ and $x=-y$ then we get a value $2$.
We'll prove that it's a minimal value.
Indeed, we need to prove that
$$\sum_{cyc}\frac{x^2}{(y-z)^2}\geq2.$$
Let $y=x+u$ and $w=x+v$.
Hence, we need to prove that
$$\frac{x^2}{(u-v)^2}+\frac{(x+u)^2}{v^2}+\frac{(x+v)^2}{u^2}\geq2$$ or
$$\left(\frac{1}{(u-v)^2}+\frac{1}{u^2}+\frac{1}{v^2}\right)x^2+2\left(\frac{u}{v^2}+\frac{v}{u^2}\right)x+\frac{u^2}{v^2}+\frac{v^2}{u^2}-2\geq0,$$ for which it's enough to prove that
$$\left(\frac{u}{v^2}+\frac{v}{u^2}\right)^2-\left(\frac{1}{(u-v)^2}+\frac{1}{u^2}+\frac{1}{v^2}\right)\left(\frac{u^2}{v^2}+\frac{v^2}{u^2}-2\right)\leq0$$ or
$$\frac{(u+v)^2(u^2-uv+v^2)^2}{u^4v^4}\leq\frac{(u^2v^2+(u-v)^2(u^2+v^2))(u^2-v^2)^2}{(u-v)^2u^4v^4}$$ or
$$(u^2-uv+v^2)^2\leq u^2v^2+(u-v)^2(u^2+v^2)$$ or
$$(u^2-uv+v^2)^2-u^2v^2\leq(u-v)^2(u^2+v^2)$$ or
$$(u-v)^2(u^2+v^2)\leq(u-v)^2(u^2+v^2).$$
Done!
| {
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$x^3 + 2x^2 + 5x + 2\cos x = 0$
$x^3 + 2x^2 + 5x + 2\cos x = 0$
How do I find the number of solutions of this equation (in $[0, 2\pi]$) without a graph?
Attempt:
The equation simplifies to $x(x^2 + 2x + 5)=- 2\cos x $
Minima of the quadratic occurs at $x= -1$ and it's value is $4$
Minima of $-2\cos x$ is $-2$
|
Let $$f(x)=x^3 + 2x^2 + 5x + 2\cos x$$
Then $$f'(x)=3x^2+4x-2\sin x+5$$ Since $$3x^2+4x+5>2\ge 2\sin x$$ for all $x\in\mathbb{R}$, we have that $$f'(x)>0$$ for all $x\in\mathbb{R}$. Thus $f$ is monotonically increasing for all real $x$.
Now as $$f(-1)=-1+2-5+1.08...=-2.91...<0,$$ $$f(0)=0+0+0+2=2>0$$ and $f$ is continuous for all values of $x$, by the Intermediate Value Theorem the equation $$f(x)=0$$ has only one solution.
P.S. If you are interested, the value of $x$ such that $f$ crosses the $x$-axis is around $-0.421$: https://www.desmos.com/calculator/2vnhsxqgz2
| {
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"source": "stackexchange",
"question_score": "4",
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Sums and harmonic series I found the solution of series on Wolfram Alpha
http://www.wolframalpha.com/input/?i=sum+1%2F(2k%2B1)%2F(2k%2B2)+from+1+to+n
$ \sum\limits_{k=1}^{n} \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right) = \sum\limits_{k=1}^{n} \frac{1}{(2k+1)(2k+2)} = \frac{1}{2} \left(H_{n+\frac{1}{2}} - H_{n+1} -1 + \text{ln}(4)\right)$
Can someone tell how to prove this in the form of Harmonic numbers?
| If you use polygamma functions
$$S_1=\sum\limits_{k=1}^{n} \frac{1}{2k+1}=\frac{1}{2} \left(\psi ^{(0)}\left(n+\frac{3}{2}\right)-\psi
^{(0)}\left(\frac{3}{2}\right)\right)$$
$$S_2=\sum\limits_{k=1}^{n} \frac{1}{2k+2}=\frac{1}2\sum\limits_{k=1}^{n} \frac{1}{k+1}=\frac{1}{2} (\psi ^{(0)}(n+2)+\gamma -1)$$ making
$$\sum\limits_{k=1}^{n} \left(\frac{1}{2k+1} - \frac{1}{2k+2}\right) = S_1-S_2$$ $$S_1-S_2=\frac{1}{2} \left(\psi ^{(0)}\left(n+\frac{3}{2}\right)-\psi
^{(0)}\left(\frac{3}{2}\right)\right)-\frac{1}{2} (\psi ^{(0)}(n+2)+\gamma -1)$$
If you look here $$H_n=\gamma +\psi ^{(0)}(n+1)$$ making
$$S_1-S_2=\frac{1}{2} \left(H_{n+\frac{1}{2}}-H_{n+1}-H_{\frac{1}{2}}+1\right)$$
and $1-H_{\frac{1}{2}}=\log (4)-1$ and then the result.
| {
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Solving $\sin x - \cos x = 1$ I have been trying to solve $\sin x - \cos x = 1$ by squaring both sides but has not been able to obtain the solution. Here is what I did:
$$\begin{align}(\sin x - \cos x)^2 &=1^2\\\sin^2x-2\sin x\cos x+\cos^2x&=1\\1-\sin 2x&=1\\\sin2x&=0\end{align}$$
Obviously $x=0$ is not a solution. May I ask why this is the case or where did things go wrong?
Thank You
| As squaring will yield more answers than enough, I am now going to solve the equation using $\sin ^{2} x+\cos ^{2} x=1$.
Let $a=\sin x \text { and } b=\cos x,$ then
$$
\begin{aligned}
&\left\{\begin{array}{l}
a-b=1 \quad \cdots(1) \\
a^{2}+b^{2}=1 \quad \cdots(2)
\end{array}\right.\\
\Rightarrow \quad & a^{2}+(a-1)^{2}=1 \\
\Rightarrow \quad & 2 a(a-1)=0 \\
\Rightarrow \quad & a=0 \text { or } 1
\end{aligned}
$$
A. When $a=0, b=-1$
$$
\left\{\begin{array}{l}
\sin x=0 \\
\cos x=-1
\end{array} \Rightarrow x=(2 n+1) \pi\right. \text {, where }n\in Z.
$$
B. When $a=1, b=0$
$$
\left\{\begin{array}{l}
\sin x=1 \\
\cos x=0
\end{array} \quad \Rightarrow x=\frac{(4 n+1) \pi}{2} \text {, where } n \in \mathbb{Z}\right.
$$
Therefore the solutions are $x=(2 n+1) \pi \text{ and } \dfrac{(4 n+1) \pi}{2}, \text {where } n\in Z.$
| {
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Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$ Got these problems as separate sections of a question in a book's chapter on 'Divisibility & primes'.
*
*Show that if $x^2 + 5y^2 =1$, then $x = \pm 1$.
Can state it in terms of two factors as : $(x + y\sqrt{-5})(x-y\sqrt{-5}) =x^2 + 5y^2$, with
(i) $(x + y\sqrt{-5}) = 1$, (ii) $(x - y\sqrt{-5}) = 1$
Adding both (i) & (ii), get: $x = 1$, Subtracting (ii) from (i), get: $(y\sqrt{-5}) = 0$.
Unable to pursue after that, as $x=-1$ is not possible.
*Show that $(x + y\sqrt{-5})$ must be a prime in $\mathbb{Z}[\sqrt{-5}]$
The hint given is to use the unique factorization theorem for the integers, by supposing $(x + y\sqrt{-5}) = (a + b\sqrt{-5})(c + d\sqrt{-5})$. The hint asks to show that: $(x^2 + 5y^2) = (a^2 + 5b^2)(c^2 + 5d^2)$. I need some more hint or help to pursue, as squaring $(x + y\sqrt{-5}) = (a + b\sqrt{-5})(c + d\sqrt{-5})$ does not lead to $(x^2 + 5y^2) = (a^2 + 5b^2)(c^2 + 5d^2)$. My attempt is stated for squaring both sides below:L.H.S.: $(x + y\sqrt{-5})(x + y\sqrt{-5}) => x^2 + 2xy(-5) -5y^2$ R.H.S.: $(a + b\sqrt{-5})^2(c + d\sqrt{-5})^2 => (a + b\sqrt{-5})(a + b\sqrt{-5})(c + d\sqrt{-5})(c + d\sqrt{-5}) => (a^2 -5b^2 +2ab\sqrt{-5})(c^2 -5d^2 +2cd\sqrt{-5})$
*Find all primes less than 50 in integers that can be written in the form $x^2 + 5y^2$.
No clue except to first find the primes: $2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37, 41, 43, 47$.
Then trying to see if the factorization works, so starting with $2 = x^2 + 5y^2$, but cannot think further. Do I need to have $y$ as a imaginary number only, as $\sqrt{-5}$, or anything will work.
| *
*You're overthinking this one. We have two integers, $x$ and $y$, with $x^2 + 5y^2=1$. If $y$ is anything other than $0$, then $5y^2$ is greater than $1$, and since $x^2$ is non-negative, the equation is impossible. Having decided $y=0$, we are left with $x^2=1$, which has two solutions: $x=\pm 1$
*This question seems to be missing something. It is not true that every number of the form $x+y\sqrt{-5}$ is prime in $\Bbb Z[\sqrt{-5}]$. For example, we can take $x=1, y=5$, and note that $1+5\sqrt{-5}$ factors as $(3+\sqrt{-5})(2+\sqrt{-5})$.
*An easier way to do this is to pick values of $y$, and then for each one, try values of $x$ until you pass $50$.
| {
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Show that $4^{3x+1} + 2^{3x+1}+ 1$ is divisible by 7 I want to show that $4^{3x+1} + 2^{3x+1} + 1$ is divisible by 7, I am trying to show this with modular arithmetic.
If I break up each part of the equation, I can see that
$4^{3x+1} = 4$ x $2^{6x}$
which implies that $4^{3x+1}mod(7) = 4$
I can't quite find a nice factorization of $2^{3x+1}$
Any help specifically on how to treat $2^{3x+1}$ would be appreciated.
| Attempt at induction.
$\begin{array}\\
4^{3(x+1)+1} + 2^{3(x+1)+1}
&=4^{3x+1+3} + 2^{3x+1+3}\\
&=4^34^{3x+1} + 2^32^{3x+1}\\
&=64\cdot 4^{3x+1} + 8\cdot 2^{3x+1}\\
&=(63+1)\cdot 4^{3x+1} + (7+1)\cdot 2^{3x+1}\\
&=63\cdot 4^{3x+1} + 7\cdot 2^{3x+1}+4^{3x+1}+2^{3x+1}\\
&=7(9\cdot 4^{3x+1} + 2^{3x+1})+4^{3x+1}+2^{3x+1}\\
&\equiv 4^{3x+1}+2^{3x+1}\pmod{7}\\
\end{array}
$
Therefore, if
$4^{3x+1} + 2^{3x+1} + 1
\equiv 0 \pmod{7}
$
then
$4^{3(x+1)+1} + 2^{3(x+1)+1} + 1
\equiv 0 \pmod{7}
$.
Since
$4^{3x+1} + 2^{3x+1} + 1
\equiv 0 \pmod{7}
$
for $x = 0$,
it is true for all $x$.
| {
"language": "en",
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Find the values of $k$ that satisfy $x_1> 1$ and $x_2 <1.$ Here I found a question that caught my attention. But then I did not see it again. The question was:
The quadratic equation is given as follows:
$$(2k+1)x^2-kx+k-2=0$$
Find the values of $k$ that satisfy $x_1> 1$ and $x_2 <1.$
My way:
$$x_1=\frac{k+\sqrt{k^2-4(2k+1)(k-2)}}{2(2k+1)} \\
x_2=\frac{k-\sqrt{k^2-4(2k+1)(k-2)}}{2(2k+1)}\\ \begin{cases} x_1>1 & \\ x_2<1 & \end{cases} \Rightarrow \begin{cases} \frac{k+\sqrt{k^2-4(2k+1)(k-2)}-2(2k+1)}{2(2k+1)} >0 & \\ \frac{k-\sqrt{k^2-4(2k+1)(k-2)}-2(2k+1)}{2(2k+1)}<0 & \end{cases} \Rightarrow \begin{cases} 2(2k+1)\left(\sqrt{-7k^2+12k+8}-(3k+2) \right)>0 & \\2(2k+1)\left(\sqrt{-7k^2+12k+8}+(3k+2) \right)>0 & \end{cases} \Rightarrow 4(2k+1)^2×\left( -7k^2+12k+8-(3k+2)^2\right)>0 \Rightarrow -16k^2+4>0 \Rightarrow (2k-1)(2k+1)<0 \Rightarrow k\in \left(-\frac12 ; \frac 12 \right)$$
Actually, I doubt my solution. Can you confirm that the solution is right or wrong?
Thank you!
| To solve
$$x_1=\frac{k+\sqrt{k^2-4(2k+1)(k-2)}}{2(2k+1)}>1$$
you need to set for the existence of real solutions
$$k^2-4(2k+1)(k-2)\geq0$$
and distinguish two cases
$$2(2k+1)>0\implies \sqrt{k^2-4(2k+1)(k-2)}>2(2k+1)-k$$
$$2(2k+1)<0\implies \sqrt{k^2-4(2k+1)(k-2)}<2(2k+1)-k$$
Similarly for
$$x_2=\frac{k-\sqrt{k^2-4(2k+1)(k-2)}}{2(2k+1)}<1$$
| {
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To find the Value of $\tan A + \cot A$, if the value of $\sin A + \cos A$ is given To Find -
$$\tan A + \cot A$$
Given,
$$\sin A + \cos A = \sqrt2$$
My progress as far -
1st way-
$$\Rightarrow \sin A = \sqrt2 - \cos A$$
$$\Rightarrow \tan A = \frac{\sqrt2 - \cos A}{\cos A}$$
$$\Rightarrow \tan A = \frac{ \sqrt 2 }{\cos A } - 1$$
$$\Rightarrow \tan A + \cot A =\frac{ \sqrt 2 }{\cos A} -1 + \cot A $$
and the 2nd way as -
$$(\sin A + \cos A)^2 = 2$$
$$\sin ^2 A + \cos ^2 A + 2\sin A\cos A = 2 $$
$$\Rightarrow 2\sin A\cos A=1$$
$$\Rightarrow \sin A\cos A=\frac12$$
As we can see the first way is unable to give an answer in absolute Real Number, and the second way doesn't go even near to what is required to proof.
I know few trigonometry identities as per my textbook, those are
*
*$\sin^2 A + \cos^2 A = 1$
*$1 + \cot^2 A = \csc^2 A$
*$\tan^2A + 1 = \sec^2 A$
| $$\tan{A}+\cot{A}=\frac{1}{\sin{A}\cos{A}}=\frac{2}{(\sin{A}+\cos{A})^2-1}=2$$
| {
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Find: $\lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
Find: $\displaystyle \lim_{x\to -\infty} f(x)= \frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}$ (no L'Hopital)
After developing the expression, by multiplying the fraction by the conjugates and rearranging, I found:
$$f(x)=x\times \frac{1+\sqrt{1+1/x^6}}{1+\sqrt{1+1/x^4}}$$
I can solve it for the limit when $x\to \infty$, with answer $\infty$ which is correct (by the book and Wolfram Alpha). But when the limit is $x\to -\infty$ the answer is zero, but from this expression, I get $-\infty$ as the answer.
I'm certainly missing something.
Hints and answers appreciated. Sorry if this is a duplicate.
| $$\frac{x^2-\sqrt{x^4+1}}{x^3-\sqrt{x^6+1}}=\frac{x^4-x^4-1}{x^6-x^6-1}\cdot \frac{x^3+\sqrt{x^6+1}}{x^2+\sqrt{x^4+1}}\to \frac{x^3+|x^3|}{x^2+|x^2|}=\frac{x^3-x^3}{x^2+x^2}= 0$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2586512",
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"source": "stackexchange",
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Jordan decomposition - help with calculation of transformationmatrices? Let $$A=\begin{pmatrix}5&1&1 \\-1&3&1\\0&0&4\end{pmatrix}$$
The Jordan-decomposition is $$A=\begin{pmatrix}-1&-1&-1/2 \\1&0&0\\0&0&-1/2\end{pmatrix}\begin{pmatrix}4&1&0 \\0&4&1\\0&0&4\end{pmatrix}\begin{pmatrix}0&1&0 \\-1&-1&1\\0&0&-2\end{pmatrix}$$
And it drives me crazy that I don't get it !
The characteristic polynomial is $p(\lambda)=(\lambda-4)^3$ therefore $\lambda=4$ has algebraic multiplicity $3$
Then
$(A-4I)=\begin{pmatrix}1&1&1 \\-1&-1&1\\0&0&0\end{pmatrix} \Rightarrow \dim(\text{kernel})=2<3$
$(A-4I)^2=\begin{pmatrix}0&0&2 \\0&0&-2\\0&0&0\end{pmatrix}\Rightarrow \dim(\text{kernel})=1<3$
The matrix is not diagonalizable, so we get
$$J=\begin{pmatrix}4&1&0 \\0&4&1\\0&0&4\end{pmatrix}$$
Now I need the eigenvectors/eigenspaces/generalized eigenvetors/spaces and here is the part I have trouble with.
$(A-4I)v=\begin{pmatrix}1&1&1 \\-1&-1&1\\0&0&0\end{pmatrix}v=0 \Rightarrow v_1=-v_2 \land v_3=0 \Rightarrow v=\begin{pmatrix}1\\-1\\0\end{pmatrix} $
$(A-4I)^2u=\begin{pmatrix}0&0&2 \\0&0&-2\\0&0&0\end{pmatrix}u=0 \Rightarrow u_3=0\Rightarrow u=\begin{pmatrix}1\\0\\0\end{pmatrix}, u*=\begin{pmatrix}0\\1\\0\end{pmatrix}$
I don't understand how/what wolfram alpha did -.-
What did I wrong ? Thanks for answers.
| I like to run the thing backwards. We have $(A-4I)^3 = 0$ but $(A-4I)^2 \neq 0.$ So, we can find a column vector, call it $r,$ such that $(A-4I)^2 r \neq 0.$ For example, we can take $r = (0,0,1)^T.$ Then let $q = (A-4I)r,$ so that $(A-4I)^2 q = 0,$ but $(A-4I)q \neq 0.$ Finally $p = (A-4I)q,$ so $(A-4I)p = 0$ but $p \neq 0.$ Then make the coefficient matrix on the right, call it $P,$ with columns $(p,q,r).$ Jordan form becomes $P^{-1}A P.$ I got
$$
P =
\left(
\begin{array}{rrr}
2 & 1 & 0 \\
-2 & 1 & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
$$
AP =
\left(
\begin{array}{rrr}
8 & 6 & 1 \\
-8 & 2 & 1 \\
0 & 0 & 4 \\
\end{array}
\right)
$$
For the columns $p,q,r,$ we can see that $$ Ap = 4p, \; Aq = p + 4 q, \; Ar = q + 4r. $$
You also need
$$
P^{-1} =
\left(
\begin{array}{rrr}
\frac{1}{4} & -\frac{1}{4} & 0 \\
\frac{1}{2} & \frac{1}{2} & 0 \\
0 & 0 & 1 \\
\end{array}
\right)
$$
I like this method as there is little additional work after confirming the matrices $A-4I$ and $(A-4I)^2.$
Let's see, if you use that same column vectors but order $P$ backwards as $r,q,p,$ you get a different Jordan form with the extra $1$ entries below the main diagonal, rather than above. This is, more or less, a proof that a matrix and its transpose are always similar.
| {
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invert multinomial logit link with three unknown I am attempting to invert the multinomial logit link with three variables. I can do it with two variables, but I do not know how to do it with three.
A multinomial logit function for three states, i.e., three probabilities, $a, b, c$ is written as follows:
$a = \frac{e^{x}} {(1 + e^{x} + e^{y})}$
$b = \frac{e^{y}} {(1 + e^{x} + e^{y})}$
$c = 1 - a - b$
These three probabilities are defined by the parameters $x$ and $y$. If we know $x$ and $y$ we can obtain $a$, $b$ and $c$.
However, given $a, b, c$, how do we obtain $x$ and $y$? One way is to use multinomial logistic regression. However, there should be a closed form solution in which $x$ and $y$ are obtained using basic algebra. I can obtain the closed form solution for two parameters, $x$ and $y$:
$x = \log(\frac{a (1 - b) + (a b)}{ (1 - a) (1 - b) - a b})$
$y = \log(\frac{b (1 - a) + (b a) }{ (1 - b) (1 - a) - b a})$
Which simplifies to:
$x = \log(\frac{a}{1 - a - b})$
$y = \log(\frac{b}{1 - a - b})$
How can I obtain the closed form solution when there are three parameters $x, y, z\;$?
$a = \frac{e^{x}} {(1 + e^{x} + e^{y} + e^{z})}$
$b = \frac{e^{y}} {(1 + e^{x} + e^{y} + e^{z})}$
$c = \frac{e^{z}} {(1 + e^{x} + e^{y} + e^{z})}$
$d = 1 - a - b - c$
| Solve each variable $x$, $y$ and $z$ as a function of the other two.
$x = \log(\frac{(a + a e^{y} + a e^{z})} {(1 - a)})$
$y = \log(\frac{(b + b e^{x} + b e^{z})} {(1 - b)})$
$z = \log(\frac{(c + c e^{x} + c e^{y})} {(1 - c)})$
Then:
Substitute $x$ into the equation for $y$.
Substitute $x$ into the equation for $z$.
Substitute $y$ into the equation for $x$.
Substitute $y$ into the equation for $z$.
Substitute $z$ into the equation for $x$.
And substitute $z$ into the equation for $y$.
Each parameter $x$, $y$ and $z$ is now expressed as a function of just one of the other two.
To express each parameter as a function of itself, substitute, for example, the equation expressing $x$ as a function of $y$ into the function expressing $y$ as a function of $x$.
Once these three equations are simplified we have:
$x = \log(-(a * c - a) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$
$y = \log(-(b * c - b) / (c^2 + (b+a-2)*c + (a - 1)*b - a + 1 - a * b))$
$z = \log(((1-b)*c) / (((b+a-1)*c+b^2+(a-2)*b-a+1) - (a*c)))$
| {
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"source": "stackexchange",
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How to evaluate cos(2$\pi$/5)? I understand evaluation for cos($\pi$/8), but I cannot figure it out how to solve cos(2$\pi$/5).
Can anyone please show it?
| Note that:
$$\cos\left(\frac{4\pi}5\right)=\cos\left(2\pi-\frac{4\pi}5\right)=\cos\left(\frac{6\pi}5\right)\implies\cos 2x =\cos 3x$$
$$\implies 2\cos^2x-1=4\cos^3x-3\cos x\implies 4y^3-2y^2-3y+1=0$$
$$\implies (y-1)(4y^2+2y-1)=0 \implies 4y^2+2y-1=0 \implies y=\frac{-2\pm\sqrt{20}}{8} $$
since $y>0$
$$\cos\left(\frac{4\pi}5\right)=\frac{-1+\sqrt{5}}{4}$$
| {
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Let $G(x)=\sum _{n=1}^{\infty }\frac{nx}{1+n^6x^2}$ Prove that $G(x)$ is bounded
Let $G:\mathbb{R}\to\mathbb{R}$
$$G(x)=\sum _{n=1}^{\infty }\frac{nx}{1+n^6x^2},x\in\mathbb{R}$$ Prove that $G(x)$ is bounded
by AGM inequality :
$$\frac{1+n^6x^2}{2}\ge \sqrt{n^6x^2}$$
$$\iff \frac{1}{2} \ge \frac{\sqrt{n^6x^2}}{1+n^6x^2}$$
$$ \iff \frac{1}{2n^2} \ge \frac{n|x|}{1+n^6x^2}$$
so G(x) is bounded , $M=\frac{1}{2n^2}$
is this correct ? and if i want to prove that it's Pointwise convergence i can say that because $\frac{1}{2n^2}$ is convergence we get also that $\frac{n|x|}{1+n^6x^2}$ is convergence for every $x\in R$?
thanks
| Easy trick by differentiating:
Let $$g_n(x)=\frac{nx}{1+n^6x^2},\implies g'_n(x)=\frac{n-n^7x^2}{(1+n^6x^2)^2}$$
$$g'_n(x)=0 \Longleftrightarrow x= \pm\frac{1}{n^3}$$ and
$$ \lim_{x\to\infty}g_n(x)=0=\lim_{x\to0}g_n(x)$$
Observing that $g_n(x)$ is odd function one easily get that $|g_n|$ attains its maximum at $x= \frac{1}{n^3}$ therefore for all $x$,
$$|g_n(x)|\le |g_n(\frac{1}{n^3})| = \frac{1}{n^2(1+n^3)}$$
And hence, $$|G(x)|\le\sum _{n=1}^{\infty }|g_n(x)|\le \sum _{n=1}^{\infty }\frac{1}{n^2(1+n^3)} \le \sum _{n=1}^{\infty }\frac{1}{n^2}=\frac{\pi^2}{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590194",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Verify that for $k=3/2 ,$ $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$ This is a part of a proof I am studying on:
Let $f_k(x) = 1 - \frac{x^2}k+\frac{x^4}{2! k(k+1)}-\frac{x^6}{3! k(k+1)(k+2)} + \cdots \qquad (k\notin\{0,-1,-2,\ldots\}) $
For $k=3/2 ,$ it's shown that $\quad f_{3/2}(x)=\frac{\sin(2x)}{2x}$
How is it concluded this way?
| Note that (using double factorial notation)
\begin{eqnarray*}
\frac{3}{2} \frac{5}{2} \cdots \frac{2n+1}{2} = \frac{(2n+1)!!}{2^n}.
\end{eqnarray*}
So the $n^{th}$ term can be written as
\begin{eqnarray*}
\frac{x^{2n}}{n! \frac{3}{2} \frac{5}{2} \cdots \frac{2n+1}{2}} =\frac{( 2^n)^2 x^{2n}}{(2n)!! (2n+1)!!} =\frac{(2x)^{2n}} {(2n+1)!}.
\end{eqnarray*}
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2590741",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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} |
Remainder when $27^{40}$ is divided by $12$ What is the remainder when ${27}^{40}$ is divided by $12$ ?
The answer is supposed by $9$ but i’m getting it’s as $3$ . Please correct me in my approach to the problem.
This is how I did it :-
${27}^{40} = {(3^3)}^{40} = {3}^{120}$
$\frac{{3}^{120}}{12} = \frac{3^{119}}{4}$
Thus
$\frac{{3}^{119}}{4} = \frac{(3^{4})^{29}3^3}{4}$
We can now expand ${81}^{29}$ by binomial theorem
$27(\binom{29}{0}(80)^{29} + \binom{29}{1}(80)^{28} + \binom{29}{2}(80)^{27}......+\binom{29}{29}1)$
As all the numbers is the parentheses except $\binom{29}{29}$ are multiples of $4$ we can remove it out of the parentheses
Thus we will have ,
$27(4K) + 27$
$4m + 4(6) +3$
$4n +3$
Where $k,m,n$ will be multiple of $4$ this gives remainder as 3
Please help me find the flaw . Thank you
| Everything you did so far is right.
To finish off, you have
$$
\frac{3^{120}}{12} = \frac{3^{119}}{4} = \frac{4n+3}{4} = n + \frac{3}{4} $$
where $n$ is an integer. Now note that
$$
n + \frac{3}{4} = n + \frac{9}{12}$$
and you see the remainder is $9$.
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2591628",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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Using induction, prove that $({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$. Question: Using the principle of mathematical induction, prove that for every integer $n\geq 1, ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$.
My attempt:
Let $P(n): ({3^2}^n -1)$ is divisible by $2^{n+2}$ but not by $2^{n+3}$ ($n\geq 1).$
BASE CASE:
$P(1): ({3^2}^1 -1)$ is divisible by $2^{1+2}$ but not by $2^{1+3}$ , that is, $8$ is divisible by $8$ but not by $16$, which is true.
$\therefore P(1)$ is true
INDUCTION HYPOTHESIS:
Let $P(n)$ be true for $n=k$.
$\therefore P(k)$ is true, that is, $({3^2}^k -1)$ is divisible by $2^{k+2}$ but not by $2^{k+3}$.
$\therefore$ Let ${3^2}^k -1=\lambda.2^{k+2}$
$\implies {3^2}^k=1+\lambda.2^{k+2}$
INDUCTIVE STEP:
$P(k+1):({3^2}^{k+1} -1)$ is divisible by $2^{k+1+2}$ but not by $2^{k+1+3}$
We have to prove that $P(k+1)$ is true
${3^2}^{k+1}-1={3^{{2^k}2}}-1=(1+\lambda.2^{k+2})^2-1=1+2\lambda.2^{k+2}+\lambda^2.2^{2(k+2)}-1=\lambda.2^{k+3}+\lambda^2.2^{k+3+k+1}=2^{k+3}(\lambda+\lambda^2.2^{k+1})$
$\therefore {3^2}^{k+1}-1$ is divisible by $2^{k+3}$
My problem: How do I prove that the number is not divisible by $2^{n+3}$? What should I add to the induction hypothesis part and the inductive step. I am not allowed to use congruence. I am allowed to use the properties of numbers and their division. I thought of doing it by proving that the number leaves a particular remainder on dividing by $2^{n+2}$ but then realized that it would be very cumbersome since the remainder can range from $1$ to $2^{n+3}-1$.
Please help.
| Here's a variant for the inductive step:
$$3^{2^{\scriptstyle k+1}}-1= \Bigl(3^{2^{\scriptstyle k}}\Bigl)^{\!2}-1=\bigl(3^{2^{\scriptstyle k}}-1\bigl)\bigl(3^{2^{\scriptstyle k}}+1\bigl)=\bigl(3^{2^{\scriptstyle k}}-1\bigl)\Bigl((3^{2^{\scriptstyle k}}-1)+2\Bigl).$$
Now the first factor is divisible by $2^{n+2}$ and no more. There results the second factor is divisible by $2$, not $2^r$ if $r>1$.
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Find $\sum_{n=1}^{\infty}$ $\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$
$$\sum_{n=1}^{\infty}\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$$
MY Approach$\sum_{n=1}^{\infty}$$\frac{n^{2}}{\left(n+1\right)\left(n+2\right)\left(n+3\right)\left(n+4\right)}$
= Lim$_{k\rightarrow\infty}$$\sum_{k=1}^{n}\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$
$\frac{k^{2}}{\left(k+1\right)\left(k+2\right)\left(k+3\right)\left(k+4\right)}$=
$\frac{1}{6\left(n+1\right)}$-$\frac{2}{\left(n+2\right)}$+$\frac{9}{\left(n+3\right)2}$-$\frac{8}{\left(n+4\right)3}$
I don't think i can telescope it, And i don't know any other method
| The partial fractions should come out to be $$\frac 16\cdot\frac 1{r+1}-2\cdot\frac 1{r+2}+\frac 92\cdot \frac 1{r+3}-\frac 83\cdot\frac 1{r+4}$$
Examine now what happens to the fractions with denominator $5$, which comes with $r=1,2,3,4$ and you get $$\frac 15\cdot\left(-\frac 83+\frac 92-2+\frac 16\right)=0$$So you do get telescoping. I'll leave you to fill in the remaining details.
| {
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"url": "https://math.stackexchange.com/questions/2595778",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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"answer_id": 2
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Find $\lim_{x\rightarrow\infty}x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right).$ This is what my prof did:
Rewrite as $$x\left(\frac{1}{e}-\left(\frac{x}{x+1}\right)^x\right)=\frac{x}{e}\left(1-e\cdot e^{-x\ln{\left(1+\frac{1}{x}\right)}}\right).$$
Here he uses that
$$\ln{(1+t)}=t-\frac{t^2}{2}+O(t^3),$$
so
$$\frac{x}{e}\left(1-e\cdot e^{-x\left(\frac{1}{x}-\frac{1}{2x^2}+O(\frac{1}{x^3})\right)}\right)=\frac{x}{e}\left(1-e^{\frac{1}{2x}+O\left(\frac{1}{x^3}\right)}\right).$$
So far so good. Now, he uses that $e^t=1+t+O(t^2)$ so that the expression aove becomes
$$\frac{x}{e}\left(1-\left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)\right)\right)=-\frac{1}{2e}+O\left(\frac{1}{t}\right)$$
and now he lets $x\rightarrow\infty$ to get the answer. However, when he introduces the expansion of $e^t$, how does he get $O(1/x^2)$?
It is my understanding that I should plug in $t=\frac{1}{2x}+O\left(\frac{1}{x^3}\right)$ everywhere in $1+t+O(t^2),$ doing that I get
$$1+\frac{1}{2x}+O\left(\frac{1}{x^3}\right)+O\left(\left(\frac{1}{2x}+O\left(\frac{1}{x^3}\right)\right)^2\right),$$
Squaring that last ordo-term, the largest power of $x$ is $1/x^2$ and the smallest is $1/x^6.$ Why choose $1/x^2$?
| From here you get
$$e^{-x\left(\frac{1}{x}-\frac{1}{2x^2}+O\left(\frac{1}{x^3}\right)\right)}=e^{-1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)}=\frac1e\cdot e^{\frac{1}{2x}+O\left(\frac{1}{x^2}\right)}=\frac1e \cdot \left(1+\frac{1}{2x}+O\left(\frac{1}{x^2}\right)\right)$$
Alternatively with little-o notation
$$\left(\frac{x}{x+1}\right)^x=\left(\frac{1}{1+\frac1x}\right)^x=e^{-x\log\left(1+\frac1x\right)}=e^{-x\left(\frac1x-\frac1{2x^2}+o\left(\frac1{x^2}\right)\right)}=e^{-1+\frac1{2x}+o\left(\frac1{x}\right)}=$$
$$=-\frac1e+\frac1{2ex}+o\left(\frac1{x}\right)$$
Thus
$$x\left(\frac{1}{e}-\frac1e-\frac1{2ex}+o\left(\frac1{x}\right)\right)=-\frac1{2e}+o(1)\to-\frac12$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2596302",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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Solve the Second Order Nonhomogeneous ODE I'm stuck on this problem since I can't find any material in my textbook or internet to solve this. Any help?
$$2x^2 \cdot y'' - x \cdot y' + y = x$$
$$y'' - \frac{y'}{2x} + \frac{y}{2x^2} = \frac{1}{2x}$$
| Here is another way:
$$y'' - \frac{y'}{2x} + \frac{y}{2x^2} = \frac{1}{2x}$$
$$y'' -\frac 1 2 ( \frac{y'x}{x^2} - \frac{y}{x^2})= \frac{1}{2x}$$
$$y'' -\frac 1 2 ( \frac{y'x-y}{x^2})= \frac{1}{2x}$$
$$y'' -\frac 1 2 ( \frac{y}{x})'= \frac{1}{2x}$$
Integrate:
$$y' -\frac{y}{2x}=\int \frac{dx}{2x}$$
$$y' -\frac{y}{2x}=\frac{\ln(x)}{2}+K$$
Which is easy to solve ....
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2597554",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
"answer_count": 2,
"answer_id": 1
} |
Computing volume using shell resulted in negative value? The question I have solved (not sure if correctly), as I ended up with a negative volume to which I am confused whether or not I can have a negative Volume is...
Use cylindrical shells to compute the volume of $y=x^2$ and $y=2-x^2$, revolved about $x=2$
$$x^2=2-x^2→2x^2=2→\frac{2x^2}2=\frac{2}2→x^2=1$$
$$x=1, x=-1$$
Radius is $$r=2-x$$
Height is $$x^2-(2-x^2)$$
Finding the Integral,
$$\int_{\neg1}^12π(2-x)(x^2-(2-x^2 ))dx=2π(\frac{4x^3}3-4x-\frac{2x^4}3+2x^2-x^2+\frac{x^4}6)+C$$
Finding the limits,
$$\lim_{x→-1+}2π(\frac{4x^3}3-4x-\frac{2x^4}3+2x^2-x^2+\frac{x^4}6)=2π(\frac{4(-1)^3}3-4(-1)-\frac{2(-1)^4}3+2(-1)^2-(-1)^2+\frac{(-1)^4}6)=19.897$$
$$\lim_{x→1-}2π(\frac{4x^3}3-4x-\frac{2x^4}3+2x^2-x^2+\frac{x^4}6)=2π(\frac{4(1)^3}3-4(1)-\frac{2(1)^4}3+2(1)^2-(1)^2+\frac{(1)^4}6)=-13.614$$
And finally, computing the volume from the obtained limits
$$V=-13.614-19.897=-33.510$$
So, if someone could let me know if (or where) I went wrong, that would be great.
| The expression of the volume should be
$$\int_{-1}^1 2π(2-x)((2-x^2 )-x^2)dx$$
Because，
$$2-x^2 \ge x^2\quad -1\le x \le 1$$
| {
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"url": "https://math.stackexchange.com/questions/2597762",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "5",
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Prove that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$= $\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$ QuestionProve that$\frac{x^{2^{k-1}}}{\left(1-x^{2^{k}}\right)}$=
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$
My Approach R.H.S
$\frac{1}{1-x^{2^{k-1}}}-\frac{1}{1-x^{2^{k}}}$=$\frac{x^{2^{k-1}\left(1-x^{2}\right)}}{\left(1-x^{2^{k}}\right)\left(1-x^{2^{k-1}}\right)}$
i just don't know what else i can do?
| Consider $x^{2^k}=m$
The equation reduces to
$$\frac {\sqrt m}{1-m}=\frac {1+\sqrt m -1}{1-m}$$
$$=\frac {1+\sqrt m}{1-m}-\frac {1}{1-m}$$
$$=\frac {1}{1-\sqrt m} - \frac {1}{1-m}$$
Resubstitute value of $m$ to obtain desired proof
| {
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"url": "https://math.stackexchange.com/questions/2599202",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 0
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Results with Legendre Polynomials I am studying Legendre's differential equation and trying to use it to establish various results but struggling as I am not really sure how to manipulate the Legendre polynomials. I begin with:
$$\frac{d}{dz}\bigg((1-z^{2})\frac{dP_{n}}{dz}\bigg) + n(n-1)P_{n}=0.$$
I have managed to show that
$$n(n+1) \int_{-1}^{1}P_{m} (z) P_{n}(z) \, dz = \int_{-1}^{1}(1-z^2)P_{m}'(z)P_{n}'(z) \, dz$$
I then used this to deduce the orthogonality condition and now need to use it to deduce:
$$\int_{-1}^{1}z^{k}P_{n}(z) \, dz=0.$$
for $k=0,1,...,n-1$. I have also shown that:
$$P_{n+1}'-P_{n-1}' = (2n+1)P_{n}.$$
I have to use this result to deduce that:
$$\int_{-1}^{1}z^{n}P_{n} (z) \, dz=\bigg( \frac{n}{2n+1} \bigg) \int_{-1}^{1}z^{n-1}P_{n-1} (z) \, dz.$$
From this I need to deduce that:
$$\int_{-1}^{1}z^{n}P_{n} (z) \, dz= \frac{2^{n+1}(n!)^2}{(2n+1)!}.$$
In manipulating the polynomials I am able to use the fact that $P_{n}(1)=1$, $P_{n}(-1)=(-1)^n$, and $P_{0}(z)=1$.
| I will use Rodrigues formula for the Legendre polynomials $P_n (x)$, namely
$$P_n (x) = \frac{1}{2^n n!} \frac{d^n}{dx^n} [(x^2 - 1)^n].$$
We start by considering the integral
$$I = \int_{-1}^1 f(x) P_n (x) \, dx.$$
Substituting Rodrigues formula for $P_n (x)$ into the above integral followed by integrating by parts $n$ times, recoginising the term $(x^2 - 1)$ repeatedly cancels at either of the end-points, one will be left with
$$I = \frac{(-1)^n}{2^n n!} \int_{-1}^1 f^{(n)}(x) (x^2 - 1)^n \, dx.$$
Now if $f(x) = x^k$ where $k = 0,1,2,\ldots, n - 1$, then
$$f^{(n)} (x) = \frac{d^n}{dx^n} [x^k] = 0.$$
Thus
$$\int_{-1}^1 x^k P_n (x) \, dx = 0, \qquad k = 0,1,2,\ldots,n - 1.$$
Next, if $f(x) = x^n$ then
$$f^{(n)} (x) = \frac{d^n}{dx^n} [x^n] = n!.$$
Thus
\begin{align*}
\int_{-1}^1 x^n P_n (x) \, dx &= \frac{(-1)^n}{2^n} \int_{-1}^1 (x^2 - 1) \, dx\\
&= \frac{1}{2^n} \int_{-1}^1 (1 - x^2) \, dx\\
&= \frac{1}{2^{n - 1}} \int_0^1 (1 - x^2)^n \, dx,
\end{align*}
since the integral is even.
This last integral can be readily found in terms of the beta function. Enforcing a substution of $x \mapsto \sqrt{x}$ the integral becomes
$$\int_0^1 (1 - x^2)^n \, dx = \frac{1}{2} \int_0^1 x^{-1/2} (1 - x)^n \, dx = \frac{1}{2} \text{B} \left (\frac{1}{2}, n + 1 \right ).$$
As
\begin{align*}
\text{B} \left (\frac{1}{2}, n + 1 \right ) &= \frac{\Gamma (1/2) \Gamma (n + 1)}{\Gamma (n + 3/2)} = \frac{\sqrt{\pi} n!}{(n + 1/2) \Gamma (n + 1/2)}.
\end{align*}
Now as
$$\Gamma \left (n + \frac{1}{2} \right ) = \frac{(2n)!}{2^{2n} n!} \sqrt{\pi},$$
we can write
$$\text{B} \left (\frac{1}{2}, n + 1 \right ) = \frac{2^{2n + 1} (n!)^2}{(2n + 1)!}.$$
Thus
$$\int_{-1}^1 (1 - x^2)^n \, dx = \frac{2^{2n} (n!)^2}{(2n + 1)!},$$
giving
$$\int_{-1}^1 x^n P_n (x) \, dx = \frac{2^{n + 1} (n!)^2}{(2n + 1)!}.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "1",
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The value of the expression $x^4-8x^3+18x^2-8x+2$ when $x=\cot\frac{\pi}{12}$ The value of the expression $x^4-8x^3+18x^2-8x+2$ when $x=\cot\frac{\pi}{12}$
I know that $\cot\frac{\pi}{12}=2+\sqrt3$,but putting this value and evaluating is a long process.Isn't there a short and intuitive method to solve this?
|
I know that $\cot\frac{\pi}{12}=2+\sqrt3$
Following up on this line, let $a=2+\sqrt3\,$, then:
*
*$a^2=7+4\sqrt{3}$
*$\dfrac{1}{a}=2-\sqrt{3}\,$ so $\,a+\dfrac{1}{a}=4\,$, and $a^2+\dfrac{1}{a^2}=\left(a+\dfrac{1}{a}\right)^2-2=14$
Then $\;a^4-8a^3+18a^2-8a+2 = a^2 \cdot \left(a^2+\dfrac{1}{a^2}-8\left(a+\dfrac{1}{a}\right)+18\right)+1 = \cdots\,$
| {
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"url": "https://math.stackexchange.com/questions/2602084",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
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Solution by differential equation by Clairaut Form Solve the differential equation:
$$y+x \frac{dy}{dx}=x^4 \bigg(\frac{dy}{dx}\bigg)^2$$
This is given under 'Clairaut form' but I am not able to convert it to Clairaut form of type $y=px+f(p)$ where $p=dy/dx$. The general solution is given as $xy+c=c^2x$ and singular solution is $4x^2y+1=0$. Could someone give me some hint with this?
| $$y=-x \frac{dy}{dx}+x^4 \bigg(\frac{dy}{dx}\bigg)^2$$
Let $\quad x=\frac1t\quad\to\quad dx=-\frac{dt}{t^2}\quad\to\quad \frac{dy}{dx}=\frac{dy}{dt}\frac{dt}{dx}= -t^2\frac{dy}{dt}$
$y=-\frac1t \bigg(-t^2\frac{dy}{dt}\bigg)+\bigg(\frac1t\bigg)^4 \bigg(-t^2\frac{dy}{dt}\bigg)^2$
$$y=t\frac{dy}{dt}+\bigg(\frac{dy}{dt}\bigg)^2 $$
$\frac{dy}{dt}=\frac{dy}{dt}+t\frac{d^2y}{dt^2}+2\frac{dy}{dt}\frac{d^2y}{dt^2}$
$$\bigg(t+2\frac{dy}{dt}\bigg)\frac{d^2y}{dt^2}=0$$
General solution :
$\frac{d^2y}{dt^2}=0\quad\to\quad y=at+b=\frac{a}{x}+b=-x(\frac{-a}{x^2})+x^4(\frac{-a}{x^2})^2=\frac{a}{x}+a^2\quad\to\quad b=a^2$
$$y=\frac{a}{x}+a^2$$
and particular solution :
$t+2\frac{dy}{dt}=0 \quad\to\quad y=-\frac{t^2}{4}+c=-\frac{1}{4x^2}+c=-x(\frac{1}{2x^3})+x^4(\frac{1}{2x^3})^2=-\frac{1}{4x^2}\quad\to\quad c=0$
$$y=-\frac{1}{4x^2}$$
| {
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"url": "https://math.stackexchange.com/questions/2605806",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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"answer_id": 1
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Evaluation of Integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$ Calculate integral $\int \frac{x^2+1}{\sqrt{x^3+3}}dx$ This was my exam question. I've tried many online math solvers and math programs but none were able to solve. If anybody has an answer would be helpful. Thanks
| $$I = \int\frac{x^2+1}{\sqrt{x^3+3}}dx=\underbrace{\int \frac{x^2}{\sqrt{x^3+3}}dx}_{I_1}+\underbrace{\int \frac{1}{\sqrt{x^3+3}}dx}_{I_2}$$
$$I_1 = \int \frac{x^2}{\sqrt{x^3+3}}dx$$
u-substitution $u=x^3$
$$\int \frac{1}{3\sqrt{u+3}}du =\frac{1}{3}\int \frac{1}{\sqrt{u+3}}du$$
u-substitution $v=u+3$
$$I_1 = \frac{1}{3} \int \frac{1}{\sqrt{v}}dv = \frac{1}{3}\int v^{-\frac{1}{2}}dv = \frac{1}{3}\frac{v^{-\frac{1}{2}+1}}{-\frac{1}{2}+1}+C$$
Revert the substitutions $v=u+3,u=x^3$
$$I_1 = \frac{2}{3}\sqrt{x^3+3} + C$$
Now
\begin{align}
I_2 &= \int \frac{1}{\sqrt{x^3+3}}dx\\
&= \int \frac{1}{\sqrt{3\left(\frac{x^3}{3}+1\right)}}dx\\
&= \int \frac{1}{\sqrt{3\left(\left(\frac{x}{\sqrt[3]{3}}\right)^3+1\right)}}dx
\end{align}
Set $u = \frac{x}{\sqrt[3]{3}} \implies dx = \sqrt[3]{3}\, du$, then
\begin{align}
I_2&= \frac{\sqrt[3]{3}}{\sqrt{3}}\int \frac{1}{\sqrt{u^3+1}}du\\
\end{align}
See : $\int\frac{1}{\sqrt{x^3+1}}$
\begin{align}
I_2&= \frac{\sqrt[3]{3}}{\sqrt{3}}\left[\frac1{\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+u}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)\right]+C\\
\end{align}
Revert the original substitution $u = \frac{x}{\sqrt[3]{3}} $:
\begin{align}
&= \frac{\sqrt[3]{3}}{\sqrt{3}\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\frac{x}{\sqrt[3]{3}}}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)+C\\
\end{align}
And
$$I = \frac{2}{3}\sqrt{x^3+3} + \frac{\sqrt[3]{3}}{\sqrt{3}\sqrt[4]{3}}F\left(\arccos\left(\frac{2\sqrt{3}}{1+\sqrt{3}+\frac{x}{\sqrt[3]{3}}}-1\right)\mid\frac{2+\sqrt{3}}{4}\right)+C$$
| {
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"url": "https://math.stackexchange.com/questions/2608973",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "6",
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Evaluate $\lim_{ x\to \infty} x^2 \times \log \left(x \cot^{-1}x\right)$ Evaluate $$L=\lim_{ x\to \infty} x^2 \times \log \left(x \cot^{-1}x\right)$$
My Try:we have by change of variable $y=\frac{1}{x}$
$$L=\lim_{ y\to 0}\frac{\log\left(\frac{\tan^{-1}y}{y}\right)}{y^2}=\lim_{ y\to 0}\frac{\log\left(\tan^{-1}y\right)-\log y}{y^2} $$
We have for $y$ very very small$$\tan^{-1}y=y-\frac{y^3}{6}$$
$$L=\lim_{ y\to 0}\frac{\log\left(y-\frac{y^3}{6}\right)-\log y}{y^2}$$ $\implies$
$$L=\lim_{ y\to 0}\frac{\log y+\log\left(1-\frac{y^2}{6}\right)-\log y}{y^2}$$ $\implies$
$$L=\lim_{ y\to 0}\frac{\log\left(1-\frac{y^2}{6}\right)}{y^2}$$ $\implies$
$$L=\frac{-1}{6}$$
Is this approach fine?
| You need to fix the expansion for $\tan^{-1}y$ and be more rigorous using little-o notation
$$\tan^{-1}y=y-\frac{y^3}{3}+o(y^3)$$
then
$$\frac{\log\left(\frac{\tan^{-1}y}{y}\right)}{y^2}=\frac{\log\left(\frac{y-\frac{y^3}{3}+o(y^3)}{y}\right)}{y^2}=\frac{\log\left(1-\frac{y^2}{3}+o(y^2)\right)}{y^2}=\\=\frac{-\frac{y^2}{3}+o(y^2)}{y^2}=-\frac13+o(1)\to-\frac13$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $15\cdot x=27\pmod{18}$ EXERCISE:
Solve $15\cdot x\equiv 27\pmod{18}$
SOLUTION:
We know that $\gcd(15,18)=3$ and $3\setminus 27=3\cdot 9$.
So from http://www.math.niu.edu/~richard/Math420/lin_cong.pdf we can conclude that we have 3 solutions!
Then we have that $15\cdot x\equiv 27\pmod{18}\implies 5\cdot x\equiv 9\pmod{6}$
So, from now and then I can't understand how to proceed!
The book continue with this:
"We have a unique solution of $5\cdot x\equiv 9\pmod{6}$ which is":
$x=5\cdot 9 =45\equiv 3\pmod 6$ and $15\cdot 3\equiv 27\pmod{18}\equiv 9\pmod{18} $
The other solutions are:$(9,15)$. So all solutions are these: $(3,9,15)$
Can anyone explain me how the author of the book continue the exercise?I really can't understand how he ends up with these solutions!Is there another way to reach these solutions?
I would really appreciate a thorough explanation, since I've just started working on these type of exercises and I have to clear my mind on them.
Thanks for your time !
| Note that
$$15\cdot x\equiv 27\pmod{18}\iff 15\cdot x\equiv 9\pmod{18}$$
and
$$15\cdot x\equiv 9\pmod{18}\iff 15\cdot x=9+18\cdot k\\\iff 5\cdot x=3+6\cdot k \iff 5\cdot x\equiv 3\pmod{6}$$
note also that
$$5\cdot 5\equiv 25 \equiv 1\pmod{6}$$
then
$$5\cdot x\equiv 3\pmod{6}\iff 5\cdot 5\cdot x\equiv 5\cdot3\pmod{6}\iff x\equiv 15\pmod{6}\\\iff x\equiv 3\pmod{6}$$
| {
"language": "en",
"url": "https://math.stackexchange.com/questions/2610458",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "1",
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"answer_id": 0
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Calculate the sum of the series: $\sum\limits_{n=1}^{\infty}\left ( \frac{3}{n^{2}+n}-\frac{2}{4n^{2}+16n+15} \right )$ I am stuck at a part when it comes to evaluating the sum of the said series... Here is my work so far (and I am not sure if the notation and simplification is correct either):
Simplifying using partial sums:
$$\sum_{n=1}^{\infty}\left ( \frac{3}{n^{2}+n}-\frac{2}{4n^{2}+16n+15} \right )=\sum_{n=1}^{\infty}\left ( 3\left ( \frac{1}{n}-\frac{1}{n+1} \right )-\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right ) \right )$$
Now I take the limit of the Nth partial sum of the series, right?
$$\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( 3\left ( \frac{1}{N}-\frac{1}{N+1} \right )-\left ( \frac{1}{2N+3}-\frac{1}{2N+5} \right ) \right )$$
$$=3\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( \frac{1}{N}-\frac{1}{N+1} \right )-\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\left ( \frac{1}{2N+3}-\frac{1}{2N+5} \right ) $$
$$=3\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \frac{1}{N}-3\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\frac{1}{N+1} -\lim_{N\rightarrow \infty}\sum_{n=1}^{N} \frac{1}{2N+3}+\lim_{N\rightarrow \infty}\sum_{n=1}^{N}\frac{1}{2N+5} $$
I assume I doing something wrong here, because each term diverges. This is what was written in the textbook:
$$\sum_{n=1}^{\infty}\left ( \frac{1}{n}-\frac{1}{n+1} \right )=1-\lim_{N\rightarrow \infty}\frac{1}{N}=1$$
$$\sum_{n=1}^{\infty}\left ( \frac{1}{2n+3}-\frac{1}{2n+5} \right )=\frac{1}{5}-\lim_{N\rightarrow \infty}\frac{1}{2N+3}=\frac{1}{5}$$
The problem is...I am not sure how they got this! They are missing a lot of steps for me to understand, hence the messiness above.
| Hint:
The given expression can be written as
$$\sum_{n=1}^\infty
\frac {3}{n(n+1)} - \frac {2}{(2n+3)(2n+5)}$$
On partial decomposition it becomes
$$\sum_{n=1}^\infty \left[3\left(\frac {1}{n}-\frac {1}{n+1}\right) - \left(\frac {1}{2n+3}-\frac{1}{2n+5}\right) \right]$$
Can you see the series telescoping. By the way the answer I guess might be $\frac {14}{3}$
| {
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"timestamp": "2023-03-29T00:00:00",
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Solve $\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x$ and avoid extra solutions while squaring Solve the equation,
$$
\sin^{-1}x+\sin^{-1}(1-x)=\cos^{-1}x
$$
My Attempt:
$$
\cos\Big[ \sin^{-1}x+\sin^{-1}(1-x) \Big]=x\\
\cos\big(\sin^{-1}x\big)\cos\big(\sin^{-1}(1-x)\big)-\sin\big(\sin^{-1}x\big)\sin\big(\sin^{-1}(1-x)\big)=x\\
\sqrt{1-x^2}.\sqrt{2x-x^2}-x.(1-x)=x\\
\sqrt{2x-x^2-2x^3+x^4}=2x-x^2\\
\sqrt{x^4-2x^3-x^2+2x}=\sqrt{4x^2-4x^3+x^4}\\
x(2x^2-5x+2)=0\\
\implies x=0\quad or \quad x=2\quad or \quad x=\frac{1}{2}
$$
Actual solutions exclude $x=2$.ie, solutions are $x=0$ or $x=\frac{1}{2}$.
I think additional solutions are added because of the squaring of the term $2x-x^2$ in the steps.
So, how do you solve it avoiding the extra solutions in similar problems ?
Note: I dont want to substitute the solutions to find the wrong ones.
| Like Barry Cipra,
$$2\arcsin x=\arccos(1-x)$$
Now $0\le\arccos(1-x)\le\pi$ and $-\pi\le2\arcsin x\le\pi$
$\implies\arcsin x\ge0\iff x\ge0$
Now for $\arcsin x\ge0,2\arcsin x=\begin{cases}\arccos(1-2x^2)&\mbox{if }x\ge0\\
-\arccos(1-2x^2)& \mbox{if }x<0\end{cases}$
$x\ge0\implies 1-x=1-2x^2$
Can you take it from here?
| {
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"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "2",
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"answer_id": 2
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Does the series $\sum_{n=1}^{\infty} (n^2+1)^{1/2}-(n^3+1)^{1/3}$ converge? Does the series $\sum_{n=1}^{\infty} (n^2+1)^{1/2}-(n^3+1)^{1/3}$ converge?
I am not able to use the standard tests on this series. Any hints/advice will be much appreciated.
The solution hint is:
I don't understand how one get's this expression. The first fraction is clear but the last fraction is not so clear. What did the author use to rationalize the expression?
| Since 'the first fraction is clear', the question about the hint seems to be if the equality below is true,
$$ n^2+1 - \sqrt[3]{(n^3+1)^2}=\frac{3n^4 - 2n^3+3n^2}{(n^2+1)^2+(n^2+1)\sqrt[3]{(n^3+1)^2} + (n^3+1)\sqrt{n^3+1}}$$
Observe that if we define $a=\sqrt{n^2+1}>0\, ,b=\sqrt[3]{n^3+1}>0$, this is asking if
$$ a^2 - b^2 = \frac{3n^4 - 2n^3 + 3n^2}{a^4 + a^2 b^2 + b^4} $$
which is equivalent to
$$ (a^2-b^2)(a^4 + a^2b^2 + b^4) =3n^4 - 2n^3 + 3n^2$$
we rewrite the left hand side,
\begin{align}
LHS
&= (a^2-b^2)(a^4 + a^2b^2 + b^4) \\&= a^6 + \color{red}{a^4b^2}+\color{blue}{a^2b^4}-\color{red}{a^4b^2}-\color{blue}{a^2b^4}-b^6\\
&= a^6 - b^6\\
&=(n^2+1)^3 - (n^3+1)^2\\
&= (n^6+3n^4+3n^2+1) - (n^6 + 2n^3 + 1)\\
&= 3n^4 - 2n^3 + 3n^2\\
&= RHS
\end{align}
as needed. I much prefer the other answers, but there you go.
| {
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"timestamp": "2023-03-29T00:00:00",
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Specific steps in applying the Chinese Remainder Theorem to solve modular problem splitting modulus I am trying to get an idea of how the Chinese Remainder Theorem (CRT) can be used to finish up this problem, in which the problem
$$7^{30}\equiv x\pmod{ 100}$$
is attempted by splitting the modulus into relatively prime factors $25$ and $4,$ arriving at
$$\begin{align}
7^{30}&\equiv1\pmod4\\
7^{30}&\equiv-1\pmod{25}
\end{align}$$
I understand that the CRT may be called upon because $m=\prod m_i,$ and we have the same $7^{30}$ value on the LHS, but I don't know how to carry it out.
The question was touched upon in this post as the second entry:
How do I efficiently compute $a^b \pmod c$ when $b$ is less than $c.$ For instance $5^{69}\,\bmod 101.$
However, I don't see this particular point clearly worked out, perhaps because it is a multi-pronged question.
Following this presentation online, this seems to be the verbatim application of the CRT without any added concepts or shortcuts:
From @gimusi's answer (upvoted):
$$\begin{cases}
x \equiv 7^{30} \pmod4\\
x\equiv 7^{30} \pmod{25}
\end{cases}$$
rearranged into
\begin{cases}
x \equiv 1 \pmod4\\
x\equiv -1 \pmod{25}
\end{cases}
Given the general form of the equations above as $x\equiv a_i \pmod {m_i},$ the CRT states $x\equiv a_1 b_1 \frac{M}{m_1}+a_2 b_2 \frac{M}{m_2}\pmod M$ with $M=\prod m_i,$ and with
$$b_i =\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}.$$
The inverse of $\frac{M}{m_i}$ is such that $\frac{M}{m_i}\left(\frac{M}{m_i}\right)^{-1}\pmod {m_i}\equiv 1.$
Calculating the components:
$$\begin{align}
a_1&=1\\
a_2&=-1\\
M&=4\times 25 =100\\
\frac{M}{m_1} &= \frac{100}{4}=25\\
\frac{M}{m_2} &= \frac{100}{25}=4\\
b_1 &= \left(\frac{M}{m_1}\right)^{-1} \pmod 4 = (25)^{-1}\pmod 4 =1\\
b_2 &= \left(\frac{M}{m_2}\right)^{-1} \pmod {25}= (4)^{-1} \pmod{25}=19
\end{align}$$
Hence,
$$x=1\cdot 25 \cdot 1 + (-1)\cdot 4 \cdot 19 = -51 \pmod{100}\equiv 49.$$
| Chinese Remainder Theorem says:
$$\mathbb{Z}/100 \simeq \mathbb{Z}/25 \times \mathbb{Z}/4$$
where the isomorphism is given by mapping $x \pmod {100}$ to $(x \pmod {25}, x \pmod {4})$. Thus, the class of $x$ is a number from $0$ to $99$ that congruence to $1 \bmod 4$ and $24 \bmod 25$. The numbers $0 \le x \le 99$ and $x \equiv 24 \pmod {25}$ are: $24, 49, 74, 99$. Now which one is congruence to $1 \bmod 4$?
| {
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"url": "https://math.stackexchange.com/questions/2615233",
"timestamp": "2023-03-29T00:00:00",
"source": "stackexchange",
"question_score": "4",
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Geometric inequality : In a equilateral , $ 4(PD^2 +PE^2 + PF^2) \ge PA^2 + PB^2 + PC^2 $ In a equilateral $ABC$ ,
Prove
$$ 4(PD^2 + PE^2 + PF^2 ) \ge PA^2 + PB^2 + PC^2 $$
$(PD, PE, PF $ are distance to sides, and P is located same plane with $ABC$)
I know that;
$$PA + PB \ge PC , $$
$$ PB + PC \ge PA , $$
$$ PC + PA \ge PB $$
and When P is located on the circumcircle , $'D, E, F$ is colinear.' But these are not useful to solve the problem. I want some hints. Thank you.
| Following is an approach using barycentric coordinates.
WOLOG, we will assume the equilateral triangle $ABC$ has unit side length.
We will also assume $D, E, F$ lie on sides $BC, CA, AB$ respectively.
If $(x,y,z)$ is the barycentric coordinates of $P$. i.e.
$$\vec{P} = x\vec{A} + y\vec{B} + z\vec{C}\quad\text{ and }\quad x + y + z = 1$$
then on LHS, we have
$PD = \frac{\sqrt{3}}{2} |x|$,
$PE = \frac{\sqrt{3}}{2} |y|$ and $PF = \frac{\sqrt{3}}{2} |z|$. This leads to
$${\rm LHS} \stackrel{def}{=} 4(PD^2 + PE^2 + PF^2) = 3(x^2+y^2+z^2)$$
On the RHS, we will use the fact when $(x_1, y_1, z_1)$ and $(x_2, y_2, z_2)$ are the barycentric coordinates of two points $Q,R$ with respect to
a triangle of sides $a, b, c$, their distance is given by following formula:
$$QR^2 = -a^2(y_1-y_2)(z_1-z_2) - b^2(z_1-z_2)(x_1-x_2) -c^2(x_1-x_2)(z_1-z_2)$$
Apply this to segments $PA$, $PB$ and $PC$, we get
$$\begin{align}
PA^2 &= y+z - (xy+yz+zx)\\
PB^2 &= x+z - (xy+yz+zx)\\
PC^2 &= x+y - (xy+yz+zx)
\end{align}
$$
This leads to
$$\begin{align}
{\rm RHS} &\stackrel{def}{=} PA^2 + PB^2 + PC^2\\
&= 2(x+y+z) - 3(xy+yz+zx) = 2 - 3(xy+yz+xz)\\
&= 2(x+y+z)^2-3(xy+yz+zx)
= 2(x^2+y^2+z^2) + (xy+yz+zx)
\end{align}
$$
Combine these two expressions, we find
$${\rm LHS} - {\rm RHS}
= (x^2+y^2+z^2) - (xy+yz+zx)
= \frac12\left((x-y)^2 + (y-z)^2 + (z-x)^2\right) \ge 0$$
This establish the inequality:
$$4(PD^2+PE^2+PF^2) = {\rm LHS} \ge {\rm RHS} = PA^2+PB^2+PC^2$$
and the inequality is strict unless $x = y = z$, i.e. when $P$ is not the centroid.
| {
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"timestamp": "2023-03-29T00:00:00",
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"question_score": "3",
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Integral of $\,\arcsin(x)/(x^2)$
I did the integration by parts and got this expression, but then I am stuck on how to take it further. I tried substituting u=1-x^2, but then I had to do a partial fraction decomposition (which I did not take). Any hints or help would be very appreciated!
| Let $\sqrt{1-x^2}=t$.
Thus, $-\frac{2x}{2\sqrt{1-x^2}}dx=dt$ and
$$\int\frac{1}{x\sqrt{1-x^2}}dx=-\int\left(\frac{1}{x\sqrt{1-x^2}}\cdot\frac{\sqrt{1-x^2}}{x}\right)dt=$$
$$=-\int\frac{1}{x^2}dt=-\int\frac{1}{1-t^2}dt=\frac{1}{2}\int\left(\frac{1}{t-1}-\frac{1}{t+1}\right)dt=$$
$$=\frac{1}{2}\ln\left|\frac{t-1}{t+1}\right|+C=\frac{1}{2}\ln\left|\frac{\sqrt{1-x^2}-1}{\sqrt{1-x^2}+1}\right|+C.$$
| {
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"timestamp": "2023-03-29T00:00:00",
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Subsets and Splits
Fractions in Questions and Answers
The query retrieves a sample of questions and answers containing the LaTeX fraction symbol, which provides basic filtering of mathematical content but limited analytical insight.