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

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

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A half-baked conjecture: something in terms of the golden ration \(\phi\approx1.618\).
isn't it same as \[\frac{1}{2}\int_0^1 x+\sqrt{x^2+4}\, dx\]
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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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.
\(C_{k} = [x,x,x\cdots,( \text{ktimes})] = \dfrac{p_k}{q_k}\) where \(p_k\) and \(q_k\) are defined recursively by : \(p_k=x p_{k-1}+p_{k-2}\) \(q_k=xq_{k-1}+q_{k-2}\)
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 \[\lim_{n\to\infty}\int_0^1f(x,n)\,dx\le\int_0^1p(x)\,dx=\frac{1+\phi}{2}\]
1 Attachment
\(p_0 = x\\p_1=x^2+1\\p_2 = x(x^2+1)+x = x^3+2x\\p_3=x(x^3+2x)+x^2+1=x^4+3x^2+1\\ p_4=x(x^4+3x^2+1)+x^3+2x=x^5+4x^3+3x \) see any pattern ?
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.
could you please give initial conditions as \(p[0]=p[1]=k\) @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
i think it should be p[k]==x*p[k-1]+p[k-2], p[0]=p[1]=x
that "x" has nothing to do with the index
\[ \begin{align*} p(k)&=x p(k-1)+p(k-2),\quad p(0)=p(1)=x\\ p(k)&=2^{-k-1} x \left(x-\sqrt{x^2+4}\right)^k+\frac{2^{-k-1} x^2 \left(x-\sqrt{x^2+4}\right)^k}{\sqrt{x^2+4}}\\ &-\frac{2^{-k} x \left(x-\sqrt{x^2+4}\right)^k}{\sqrt{x^2+4}}+2^{-k-1} x \left(\sqrt{x^2+4}+x\right)^k\\ &+\frac{2^{-k} x \left(\sqrt{x^2+4}+x\right)^k}{\sqrt{x^2+4}}-\frac{2^{-k-1} x^2 \left(\sqrt{x^2+4}+x\right)^k}{\sqrt{x^2+4}} \end{align*} \]
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
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.
Is this true? \[ f(x,n) = x+\cfrac{1}{x+\cfrac{1}{f(x,n-1)}}=x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{f(x,n-3)}}}} \]
idk did I mess up my recurrence relation?
No I messed it up. The second continued fraction should be f(x,n-2).
Small error I made when integrating, should have put at the beginning oh well lol \[\int_0^1 x dx = \frac{1}{2} \ne 1\]
\[ g(x,n)=[x,x,g(x,n-1)]\,,g(x,0)=x \] \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
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 \(r^2-rx-1=0\)
@thomas5267 http://mathb.in/40411
Ok easy to integrate now!
Can we exchange limits and integration?
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
\[\int_0^1 \frac{xa}{xb+c}dx\] \[xb+c=u\]\[dx=\frac{1}{b}du\] \[\int_c^{b+c} \frac{a\frac{1}{b} u - \frac{c}{b}}{u}\frac{1}{b}du = \int_c^{b+c}\frac{a}{b^2} -\frac{c}{b^2 u} du\] \[\frac{a}{b}+\frac{c}{b^2} \ln\frac{c}{b+c} \] now plug in those \(\phi\) expressions for a,b,c.
hmm \(\phi_1,\phi_2\) are functions of \(x\) right
ohhhhh they are? Yeah then I guess that doesn't really help much lol.
replacing \(n\) by \(2n\) might simplify something, im gonna try after eating something...
\(2n+1\) 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 \(r^2-rx-1=0\)
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.
What if this is all we have to do: \[\frac{1}{2}\int_0^1 x \color{red}{-} \sqrt{x^2+4}\, dx\]
Wait I guess that won't work will it. Oh well lol.
You guys lost me at hello
\[ \begin{align*} f(x,n)&=\frac{r_1(x)^n\left(x \left(\sqrt{x^2+4}-x\right)-2\right)+r_2(x)^n\left(x \left(\sqrt{x^2+4}+x\right)+2\right) }{r_1(x)^n\left(\sqrt{x^2+4}-x\right) +r_2(x)^n\left(\sqrt{x^2+4}+x\right) }\\ r_1(x)&=-x \left(x^2-x\sqrt{x^2+4}+2\right)\\ r_2(x)&=-x \left(x^2 +x\sqrt{x^2+4}+2\right) \end{align*} \]
\[ r_2(x)\leq-1\text{ for }0.352201\leq x\leq1.\\ \lim_{n\to\infty}r_2(x)^n\text{ does not exist for all }x,\,x\in [0,1] \]
.
nice
$$f(x,n+1)=x+\frac1{f(x,n)}\\\implies f_{n+1}=x+\frac1{f_n}$$ 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 $$f_n=\frac12\frac{(x-\sqrt{4-x^2})^{n+1}+(x+\sqrt{4-x^2})^{n+1}}{(x-\sqrt{4-x^2})^{n}+(x+\sqrt{4-x^2})^{n}}$$
oops, that should be \(\sqrt{4+x^2}\) throughout
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 \(x\)
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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