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anonymous
 one year ago
Is there a closed form for the limit?
\[\lim_{n\to\infty}\int_0^1f(x,n)\,dx\]where
\[f(x,n)=[\underbrace{x,x,\cdots,x,x}_{2n+1\text{ times}}]=x+\frac{1}{x+\dfrac{1}{x+\cdots}}\quad(2n+1\text{ times})\]
anonymous
 one year ago
Is there a closed form for the limit? \[\lim_{n\to\infty}\int_0^1f(x,n)\,dx\]where \[f(x,n)=[\underbrace{x,x,\cdots,x,x}_{2n+1\text{ times}}]=x+\frac{1}{x+\dfrac{1}{x+\cdots}}\quad(2n+1\text{ times})\]

This Question is Closed

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0A halfbaked conjecture: something in terms of the golden ration \(\phi\approx1.618\).

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0isn't it same as \[\frac{1}{2}\int_0^1 x+\sqrt{x^2+4}\, dx\]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0I don't think so... We have \(f(0,n)=0\) and \(f(1,n)=\phi\) as \(n\to\infty\), whereas \[x+\sqrt{x^2+4}=\begin{cases}1&\text{x=0}\\[1ex]\dfrac{1+\sqrt5}{2}&\text{x=1}\end{cases}\] Some comparative plots on WA shows that your function is greater over the entire interval.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0One thing we could do is approximate the limit pretty well by determining a sufficient \(p(x)\) such that \(\lim\limits_{x\to\infty} (f(x,n)p(x))=0\), essentially a polynomial asymptote. Not sure what degree, though. Degree 1 is as good a place to start as any I suppose.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(C_{k} = [x,x,x\cdots,( \text{ktimes})] = \dfrac{p_k}{q_k}\) where \(p_k\) and \(q_k\) are defined recursively by : \(p_k=x p_{k1}+p_{k2}\) \(q_k=xq_{k1}+q_{k2}\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Sure, take your time. Meanwhile, here's at least an upper bound for the limit determined with geometric methods. (See pic, in which I'm using \(n=25\).) As \(n\to\infty\), it appears that \(f(0,n)\le1\), so I'll use \(p(x)=(\phi1)x+1\) as my asymptote. Then \[\lim_{n\to\infty}\int_0^1f(x,n)\,dx\le\int_0^1p(x)\,dx=\frac{1+\phi}{2}\]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(p_0 = x\\p_1=x^2+1\\p_2 = x(x^2+1)+x = x^3+2x\\p_3=x(x^3+2x)+x^2+1=x^4+3x^2+1\\ p_4=x(x^4+3x^2+1)+x^3+2x=x^5+4x^3+3x \) see any pattern ?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Mathematica output: p[k]==k p[k1]+p[k2] p[k]==BesselI[1+k,2] C[1]+BesselK[1+k,2] C[2] C[1] and C[2] are constants.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0could you please give initial conditions as \(p[0]=p[1]=k\) @thomas5267

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0p[k]==2 k BesselI[1 + k, 2] BesselK[0, 2] + 2 k BesselI[0, 2] BesselK[1 + k, 2] with p[0]==p[1]==k

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0i think it should be p[k]==x*p[k1]+p[k2], p[0]=p[1]=x

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0that "x" has nothing to do with the index

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0\[ \begin{align*} p(k)&=x p(k1)+p(k2),\quad p(0)=p(1)=x\\ p(k)&=2^{k1} x \left(x\sqrt{x^2+4}\right)^k+\frac{2^{k1} x^2 \left(x\sqrt{x^2+4}\right)^k}{\sqrt{x^2+4}}\\ &\frac{2^{k} x \left(x\sqrt{x^2+4}\right)^k}{\sqrt{x^2+4}}+2^{k1} x \left(\sqrt{x^2+4}+x\right)^k\\ &+\frac{2^{k} x \left(\sqrt{x^2+4}+x\right)^k}{\sqrt{x^2+4}}\frac{2^{k1} x^2 \left(\sqrt{x^2+4}+x\right)^k}{\sqrt{x^2+4}} \end{align*} \]

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0I think we can solve the recurrence relation : \[[x,x,\ldots, (\text{k+1 terms})] = \dfrac{p_k}{q_k}\] \(p_k = xp_{k1}+p_{k2} \) characteristic equation is \(r^2xr1=0 \implies r = \dfrac{x\pm\sqrt{x^2+4}}{2}\) so the general solution is \[p_k = c_1 \left(\dfrac{x+\sqrt{x^2+4}}{2}\right)^k + c_2 \left(\dfrac{x\sqrt{x^2+4}}{2}\right)^k\] not sure if this goes anywhere

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1I thought it might be nice to find a reduction formula starting with the recurrence relation: \[f(x,n) = x+\frac{1}{x+\frac{1}{f(x,n1)}} \] \[\int_0^1f(x,n) dx = 1+\int_0^1\frac{f(x,n1)}{xf(x,n1)+1}dx \] Unfortunately it sorta stops there although I recognized immediately that this has the same form as another integral I'm familiar with, the lambert product log. For refresher, it's the inverse of \(xe^x\). \[W(xe^x)=W(x)e^{W(x)}=x\] Actually if we take that second form and write it like this and differentiate it we get: \[W'(x) = \frac{e^{W(x)}}{xe^{W(x)}+1}\] So we have this fact: \[W(x) = \int \frac{e^{W(x)}}{xe^{W(x)}+1} dx\] Maybe there's some kind of substitution we can do in solving the integral that relates the two? \[\int_0^1\frac{f(x,n1)}{xf(x,n1)+1}dx \] Part that looks interestingly similar: \(e^{W(x)}\) and \(f(x,n1)\) I'm a little tired to be playing around with that so I'm going to bed and thought I'd share, someone can probably figure out a way to make these two similar things work out.

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Is this true? \[ f(x,n) = x+\cfrac{1}{x+\cfrac{1}{f(x,n1)}}=x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{f(x,n3)}}}} \]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1idk did I mess up my recurrence relation?

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0No I messed it up. The second continued fraction should be f(x,n2).

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Small error I made when integrating, should have put at the beginning oh well lol \[\int_0^1 x dx = \frac{1}{2} \ne 1\]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0\[ g(x,n)=[x,x,g(x,n1)]\,,g(x,0)=x \] \frac{\left(x^2+2\right) \left(x \left(x^2\sqrt{x^2+4} x+2\right)\right)^n+x \sqrt{x^2+4} \left(x \left(x^2\sqrt{x^2+4} x+2\right)\right)^n+\left(x^2+2\right) \left(x \left(x \left(\sqrt{x^2+4}+x\right)+2\right)\right)^n+x \sqrt{x^2+4} \left(x \left(x \left(\sqrt{x^2+4}+x\right)+2\right)\right)^n}{x \left(\left(x \left(x \left(\sqrt{x^2+4}+x\right)+2\right)\right)^n\left(x \left(x^2\sqrt{x^2+4} x+2\right)\right)^n\right)+\sqrt{x^2+4} \left(\left(x \left(x^2\sqrt{x^2+4} x+2\right)\right)^n+\left(x \left(x \left(\sqrt{x^2+4}+x\right)+2\right)\right)^n\right)} Too long. Paste the code into this website. https://www.codecogs.com/latex/eqneditor.php

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0that can be simplified into nice form, but i don't see an easy way to integrate : \[f(x,n) = \dfrac{x(\phi_1^n\phi_2^n+\phi_1^{n1}\phi_2^{n1})}{x(\phi_1^n\phi_2^n)+\phi_1^{n1}\phi_2^{n1}}\] where \(\phi_1, \phi_2\) are the roots of \(r^2rx1=0\)

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1@thomas5267 http://mathb.in/40411

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Ok easy to integrate now!

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0Can we exchange limits and integration?

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0Nope, exchanging is not working as per second reply of S&A http://www.wolframalpha.com/input/?i=%5Cint_0%5E1+%28x%2Bsqrt%28x%5E2%2B4%29%29%2F2+dx

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1\[\int_0^1 \frac{xa}{xb+c}dx\] \[xb+c=u\]\[dx=\frac{1}{b}du\] \[\int_c^{b+c} \frac{a\frac{1}{b} u  \frac{c}{b}}{u}\frac{1}{b}du = \int_c^{b+c}\frac{a}{b^2} \frac{c}{b^2 u} du\] \[\frac{a}{b}+\frac{c}{b^2} \ln\frac{c}{b+c} \] now plug in those \(\phi\) expressions for a,b,c.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0hmm \(\phi_1,\phi_2\) are functions of \(x\) right

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1ohhhhh they are? Yeah then I guess that doesn't really help much lol.

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0replacing \(n\) by \(2n\) might simplify something, im gonna try after eating something...

ganeshie8
 one year ago
Best ResponseYou've already chosen the best response.0\(2n+1\) terms in the continued fraction, so i think the exponent in \(f(x,n)\) has to be \(2n\) since it is starting form \(0\), correct integrand : \[f(x,n) = \dfrac{x(\phi_1^{2n}\phi_2^{2n}+\phi_1^{2n1}\phi_2^{2n1})}{x(\phi_1^{2n}\phi_2^{2n})+\phi_1^{2n1}\phi_2^{2n1}}\] where \(\phi_1, \phi_2\) are the roots of \(r^2rx1=0\)

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0I do believe that we can take the limit first before integrating as f(1,n) is bounded above by the golden ratio. By the monotone convergence theorem, exchanging limit and integration is valid.

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1What if this is all we have to do: \[\frac{1}{2}\int_0^1 x \color{red}{} \sqrt{x^2+4}\, dx\]

Kainui
 one year ago
Best ResponseYou've already chosen the best response.1Wait I guess that won't work will it. Oh well lol.

zzr0ck3r
 one year ago
Best ResponseYou've already chosen the best response.0You guys lost me at hello

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0\[ \begin{align*} f(x,n)&=\frac{r_1(x)^n\left(x \left(\sqrt{x^2+4}x\right)2\right)+r_2(x)^n\left(x \left(\sqrt{x^2+4}+x\right)+2\right) }{r_1(x)^n\left(\sqrt{x^2+4}x\right) +r_2(x)^n\left(\sqrt{x^2+4}+x\right) }\\ r_1(x)&=x \left(x^2x\sqrt{x^2+4}+2\right)\\ r_2(x)&=x \left(x^2 +x\sqrt{x^2+4}+2\right) \end{align*} \]

thomas5267
 one year ago
Best ResponseYou've already chosen the best response.0\[ r_2(x)\leq1\text{ for }0.352201\leq x\leq1.\\ \lim_{n\to\infty}r_2(x)^n\text{ does not exist for all }x,\,x\in [0,1] \]

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0$$f(x,n+1)=x+\frac1{f(x,n)}\\\implies f_{n+1}=x+\frac1{f_n}$$ now consider the substitution \(f_n=g_{n+1}/g_n\) which gives $$g_{n+2}=xg_{n+1}+g_n\\g_{n+2}xg_{n+1}g_n=0$$which we ca nsolve using the ansatz \(g_n=k^n\) so $$k^2xk1=0\\(kx/2)^2=1x^2/4\\k=\frac{x}2\pm\sqrt{1\frac{x^2}4}=\frac12\left(x\pm\sqrt{4x^2}\right)$$ ... giving $$g_n=\frac1{2^n}\left((x\sqrt{4x^2})^n+(x+\sqrt{4x^2})^n\right)$$ so $$f_n=\frac12\frac{(x\sqrt{4x^2})^{n+1}+(x+\sqrt{4x^2})^{n+1}}{(x\sqrt{4x^2})^{n}+(x+\sqrt{4x^2})^{n}}$$

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0oops, that should be \(\sqrt{4+x^2}\) throughout

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0now imagine even \(2n+1\), we observe $$(x\sqrt{4+x^2})^{2n}+(x+\sqrt{4+x^2})^n\\=\sum_{k=0}^{2n}\binom{2n}k \left(1\right)^{2nk}x^k\left(\sqrt{4+x^2}\right)^{2nk}+\sum_{k=0}^{2n}\binom{2n}k x^k\left(\sqrt{4+x^2}\right)^{2nk}\\=2\sum_{k=0}^n\binom{2n}{2k}x^{2k}\left(\sqrt{4+x^2}\right)^{2k}\\=2\sum_{k=0}^n\binom{2n}{2k}x^{2k}(4+x^2)^k$$ similarly, for odd \(2n+1\) we find: $$(x\sqrt{4+x^2})^{2n+1}+(x+\sqrt{4+x^2})^{2n+1}\\=\sum_{k=0}^{2n+1}\binom{2n+1}k \left(1\right)^{2n+1k}x^k\left(\sqrt{4+x^2}\right)^{2n+1k}+\sum_{k=0}^{2n+1}\binom{2n+1}k x^k\left(\sqrt{4+x^2}\right)^{2n+1k}\\=2\sum_{k=0}^{n}\binom{2n+1}{2k+1}x^{2k+1}\left(\sqrt{4+x^2}\right)^{2n+1(2k+1)}\\=2\sum_{k=0}^n\binom{2n+1}{2k+1} x^{2k+1} (4+x^2)^{nk}$$ so the functions \(f_n(x)=f(x,n)\) are plain rational functions in \(x\)

anonymous
 11 months ago
Best ResponseYou've already chosen the best response.0It turns out we have \[\lim_{n\to\infty}f_n(x)=\frac{1}{2}\left(x+\sqrt{x^2+4}\right)\]which is due to solving \(f(x)=x+\dfrac{1}{f(x)}\). Then some "measure theory mumbojumbo", to use a technical term, \[\lim_{n\to\infty}\int_0^1f_n(x)\,\mathrm{d}x=\frac{1}{2}\int_0^1\left(x+\sqrt{x^2+4}\right)\,\mathrm{d}x\]which is easy to compute. I don't follow the reasoning behind this equality because I've only skimmed over the major result of what's called the dominated convergence theorem (as well as for lack of being initiated in measure theory), but I find this satisfying enough. It's nice to have some closure in any case.
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