anonymous
  • anonymous
Is there a closed form for the limit? \[\lim_{n\to\infty}\int_0^1f(x,n)\,dx\]where \[f(x,n)=[\underbrace{x,x,\cdots,x,x}_{2n+1\text{ times}}]=x+\frac{1}{x+\dfrac{1}{x+\cdots}}\quad(2n+1\text{ times})\]
Mathematics
  • Stacey Warren - Expert brainly.com
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
A half-baked conjecture: something in terms of the golden ration \(\phi\approx1.618\).
ganeshie8
  • ganeshie8
isn't it same as \[\frac{1}{2}\int_0^1 x+\sqrt{x^2+4}\, dx\]
anonymous
  • anonymous
I 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.

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anonymous
  • anonymous
One 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
  • ganeshie8
\(C_{k} = [x,x,x\cdots,( \text{ktimes})] = \dfrac{p_k}{q_k}\) where \(p_k\) and \(q_k\) are defined recursively by : \(p_k=x p_{k-1}+p_{k-2}\) \(q_k=xq_{k-1}+q_{k-2}\)
anonymous
  • anonymous
Sure, 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 \(p(x)=(\phi-1)x+1\) as my asymptote. Then \[\lim_{n\to\infty}\int_0^1f(x,n)\,dx\le\int_0^1p(x)\,dx=\frac{1+\phi}{2}\]
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ganeshie8
  • ganeshie8
\(p_0 = x\\p_1=x^2+1\\p_2 = x(x^2+1)+x = x^3+2x\\p_3=x(x^3+2x)+x^2+1=x^4+3x^2+1\\ p_4=x(x^4+3x^2+1)+x^3+2x=x^5+4x^3+3x \) see any pattern ?
thomas5267
  • thomas5267
Mathematica output: p[k]==k p[k-1]+p[k-2] p[k]==BesselI[1+k,-2] C[1]+BesselK[1+k,2] C[2] C[1] and C[2] are constants.
ganeshie8
  • ganeshie8
could you please give initial conditions as \(p[0]=p[1]=k\) @thomas5267
thomas5267
  • thomas5267
p[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
  • ganeshie8
i think it should be p[k]==x*p[k-1]+p[k-2], p[0]=p[1]=x
ganeshie8
  • ganeshie8
that "x" has nothing to do with the index
thomas5267
  • thomas5267
\[ \begin{align*} p(k)&=x p(k-1)+p(k-2),\quad p(0)=p(1)=x\\ p(k)&=2^{-k-1} x \left(x-\sqrt{x^2+4}\right)^k+\frac{2^{-k-1} x^2 \left(x-\sqrt{x^2+4}\right)^k}{\sqrt{x^2+4}}\\ &-\frac{2^{-k} x \left(x-\sqrt{x^2+4}\right)^k}{\sqrt{x^2+4}}+2^{-k-1} x \left(\sqrt{x^2+4}+x\right)^k\\ &+\frac{2^{-k} x \left(\sqrt{x^2+4}+x\right)^k}{\sqrt{x^2+4}}-\frac{2^{-k-1} x^2 \left(\sqrt{x^2+4}+x\right)^k}{\sqrt{x^2+4}} \end{align*} \]
ganeshie8
  • ganeshie8
I think we can solve the recurrence relation : \[[x,x,\ldots, (\text{k+1 terms})] = \dfrac{p_k}{q_k}\] \(p_k = xp_{k-1}+p_{k-2} \) characteristic equation is \(r^2-xr-1=0 \implies r = \dfrac{x\pm\sqrt{x^2+4}}{2}\) 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
  • Kainui
I thought it might be nice to find a reduction formula starting with the recurrence relation: \[f(x,n) = x+\frac{1}{x+\frac{1}{f(x,n-1)}} \] \[\int_0^1f(x,n) dx = 1+\int_0^1\frac{f(x,n-1)}{xf(x,n-1)+1}dx \] 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? \[\int_0^1\frac{f(x,n-1)}{xf(x,n-1)+1}dx \] Part that looks interestingly similar: \(e^{-W(x)}\) and \(f(x,n-1)\) 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
  • thomas5267
Is this true? \[ f(x,n) = x+\cfrac{1}{x+\cfrac{1}{f(x,n-1)}}=x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{f(x,n-3)}}}} \]
Kainui
  • Kainui
idk did I mess up my recurrence relation?
thomas5267
  • thomas5267
No I messed it up. The second continued fraction should be f(x,n-2).
Kainui
  • Kainui
Small 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
  • thomas5267
\[ g(x,n)=[x,x,g(x,n-1)]\,,g(x,0)=x \] \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
  • ganeshie8
that can be simplified into nice form, but i don't see an easy way to integrate : \[f(x,n) = \dfrac{x(\phi_1^n-\phi_2^n+\phi_1^{n-1}-\phi_2^{n-1})}{x(\phi_1^n-\phi_2^n)+\phi_1^{n-1}-\phi_2^{n-1}}\] where \(\phi_1, \phi_2\) are the roots of \(r^2-rx-1=0\)
Kainui
  • Kainui
@thomas5267 http://mathb.in/40411
Kainui
  • Kainui
Ok easy to integrate now!
thomas5267
  • thomas5267
Can we exchange limits and integration?
ganeshie8
  • ganeshie8
Nope, 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
  • Kainui
\[\int_0^1 \frac{xa}{xb+c}dx\] \[xb+c=u\]\[dx=\frac{1}{b}du\] \[\int_c^{b+c} \frac{a\frac{1}{b} u - \frac{c}{b}}{u}\frac{1}{b}du = \int_c^{b+c}\frac{a}{b^2} -\frac{c}{b^2 u} du\] \[\frac{a}{b}+\frac{c}{b^2} \ln\frac{c}{b+c} \] now plug in those \(\phi\) expressions for a,b,c.
ganeshie8
  • ganeshie8
hmm \(\phi_1,\phi_2\) are functions of \(x\) right
Kainui
  • Kainui
ohhhhh they are? Yeah then I guess that doesn't really help much lol.
ganeshie8
  • ganeshie8
replacing \(n\) by \(2n\) might simplify something, im gonna try after eating something...
ganeshie8
  • ganeshie8
\(2n+1\) 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 : \[f(x,n) = \dfrac{x(\phi_1^{2n}-\phi_2^{2n}+\phi_1^{2n-1}-\phi_2^{2n-1})}{x(\phi_1^{2n}-\phi_2^{2n})+\phi_1^{2n-1}-\phi_2^{2n-1}}\] where \(\phi_1, \phi_2\) are the roots of \(r^2-rx-1=0\)
thomas5267
  • thomas5267
I 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
  • Kainui
What if this is all we have to do: \[\frac{1}{2}\int_0^1 x \color{red}{-} \sqrt{x^2+4}\, dx\]
Kainui
  • Kainui
Wait I guess that won't work will it. Oh well lol.
zzr0ck3r
  • zzr0ck3r
You guys lost me at hello
thomas5267
  • thomas5267
\[ \begin{align*} f(x,n)&=\frac{r_1(x)^n\left(x \left(\sqrt{x^2+4}-x\right)-2\right)+r_2(x)^n\left(x \left(\sqrt{x^2+4}+x\right)+2\right) }{r_1(x)^n\left(\sqrt{x^2+4}-x\right) +r_2(x)^n\left(\sqrt{x^2+4}+x\right) }\\ r_1(x)&=-x \left(x^2-x\sqrt{x^2+4}+2\right)\\ r_2(x)&=-x \left(x^2 +x\sqrt{x^2+4}+2\right) \end{align*} \]
thomas5267
  • thomas5267
\[ r_2(x)\leq-1\text{ for }0.352201\leq x\leq1.\\ \lim_{n\to\infty}r_2(x)^n\text{ does not exist for all }x,\,x\in [0,1] \]
ikram002p
  • ikram002p
.
anonymous
  • anonymous
nice
anonymous
  • anonymous
$$f(x,n+1)=x+\frac1{f(x,n)}\\\implies f_{n+1}=x+\frac1{f_n}$$ 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 $$k^2-xk-1=0\\(k-x/2)^2=1-x^2/4\\k=\frac{x}2\pm\sqrt{1-\frac{x^2}4}=\frac12\left(x\pm\sqrt{4-x^2}\right)$$ ... giving $$g_n=\frac1{2^n}\left((x-\sqrt{4-x^2})^n+(x+\sqrt{4-x^2})^n\right)$$ so $$f_n=\frac12\frac{(x-\sqrt{4-x^2})^{n+1}+(x+\sqrt{4-x^2})^{n+1}}{(x-\sqrt{4-x^2})^{n}+(x+\sqrt{4-x^2})^{n}}$$
anonymous
  • anonymous
oops, that should be \(\sqrt{4+x^2}\) throughout
anonymous
  • anonymous
now imagine even \(2n+1\), we observe $$(x-\sqrt{4+x^2})^{2n}+(x+\sqrt{4+x^2})^n\\=\sum_{k=0}^{2n}\binom{2n}k \left(-1\right)^{2n-k}x^k\left(\sqrt{4+x^2}\right)^{2n-k}+\sum_{k=0}^{2n}\binom{2n}k x^k\left(\sqrt{4+x^2}\right)^{2n-k}\\=2\sum_{k=0}^n\binom{2n}{2k}x^{2k}\left(\sqrt{4+x^2}\right)^{2k}\\=2\sum_{k=0}^n\binom{2n}{2k}x^{2k}(4+x^2)^k$$ similarly, for odd \(2n+1\) we find: $$(x-\sqrt{4+x^2})^{2n+1}+(x+\sqrt{4+x^2})^{2n+1}\\=\sum_{k=0}^{2n+1}\binom{2n+1}k \left(-1\right)^{2n+1-k}x^k\left(\sqrt{4+x^2}\right)^{2n+1-k}+\sum_{k=0}^{2n+1}\binom{2n+1}k x^k\left(\sqrt{4+x^2}\right)^{2n+1-k}\\=2\sum_{k=0}^{n}\binom{2n+1}{2k+1}x^{2k+1}\left(\sqrt{4+x^2}\right)^{2n+1-(2k+1)}\\=2\sum_{k=0}^n\binom{2n+1}{2k+1} x^{2k+1} (4+x^2)^{n-k}$$ so the functions \(f_n(x)=f(x,n)\) are plain rational functions in \(x\)
anonymous
  • anonymous
It 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 mumbo-jumbo", 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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