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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 \[f(x,n)=[\underbrace{x,x,\cdots,x,x}_{2n+1\text{ times}}]=x+\frac{1}{x+\dfrac{1}{x+\cdots}}\quad(2n+1\text{ times})\]

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  1. anonymous
    • one year ago
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    A half-baked conjecture: something in terms of the golden ration \(\phi\approx1.618\).

  2. ganeshie8
    • one year ago
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    isn't it same as \[\frac{1}{2}\int_0^1 x+\sqrt{x^2+4}\, dx\]

  3. anonymous
    • one year ago
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    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.

  4. anonymous
    • one year ago
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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.

  5. ganeshie8
    • one year ago
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    \(C_{k} = [x,x,x\cdots,( \text{ktimes})] = \dfrac{p_k}{q_k}\) where \(p_k\) and \(q_k\) are defined recursively by : \(p_k=x p_{k-1}+p_{k-2}\) \(q_k=xq_{k-1}+q_{k-2}\)

  6. anonymous
    • one year ago
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    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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  7. ganeshie8
    • one year ago
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    \(p_0 = x\\p_1=x^2+1\\p_2 = x(x^2+1)+x = x^3+2x\\p_3=x(x^3+2x)+x^2+1=x^4+3x^2+1\\ p_4=x(x^4+3x^2+1)+x^3+2x=x^5+4x^3+3x \) see any pattern ?

  8. thomas5267
    • one year ago
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    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.

  9. ganeshie8
    • one year ago
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    could you please give initial conditions as \(p[0]=p[1]=k\) @thomas5267

  10. thomas5267
    • one year ago
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    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

  11. ganeshie8
    • one year ago
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    i think it should be p[k]==x*p[k-1]+p[k-2], p[0]=p[1]=x

  12. ganeshie8
    • one year ago
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    that "x" has nothing to do with the index

  13. thomas5267
    • one year ago
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    \[ \begin{align*} p(k)&=x p(k-1)+p(k-2),\quad p(0)=p(1)=x\\ p(k)&=2^{-k-1} x \left(x-\sqrt{x^2+4}\right)^k+\frac{2^{-k-1} x^2 \left(x-\sqrt{x^2+4}\right)^k}{\sqrt{x^2+4}}\\ &-\frac{2^{-k} x \left(x-\sqrt{x^2+4}\right)^k}{\sqrt{x^2+4}}+2^{-k-1} x \left(\sqrt{x^2+4}+x\right)^k\\ &+\frac{2^{-k} x \left(\sqrt{x^2+4}+x\right)^k}{\sqrt{x^2+4}}-\frac{2^{-k-1} x^2 \left(\sqrt{x^2+4}+x\right)^k}{\sqrt{x^2+4}} \end{align*} \]

  14. ganeshie8
    • one year ago
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    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

  15. Kainui
    • one year ago
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    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.

  16. thomas5267
    • one year ago
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    Is this true? \[ f(x,n) = x+\cfrac{1}{x+\cfrac{1}{f(x,n-1)}}=x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{f(x,n-3)}}}} \]

  17. Kainui
    • one year ago
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    idk did I mess up my recurrence relation?

  18. thomas5267
    • one year ago
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    No I messed it up. The second continued fraction should be f(x,n-2).

  19. Kainui
    • one year ago
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    Small error I made when integrating, should have put at the beginning oh well lol \[\int_0^1 x dx = \frac{1}{2} \ne 1\]

  20. thomas5267
    • one year ago
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    \[ g(x,n)=[x,x,g(x,n-1)]\,,g(x,0)=x \] \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

  21. ganeshie8
    • one year ago
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    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 \(r^2-rx-1=0\)

  22. Kainui
    • one year ago
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    @thomas5267 http://mathb.in/40411

  23. Kainui
    • one year ago
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    Ok easy to integrate now!

  24. thomas5267
    • one year ago
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    Can we exchange limits and integration?

  25. ganeshie8
    • one year ago
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    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

  26. Kainui
    • one year ago
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    \[\int_0^1 \frac{xa}{xb+c}dx\] \[xb+c=u\]\[dx=\frac{1}{b}du\] \[\int_c^{b+c} \frac{a\frac{1}{b} u - \frac{c}{b}}{u}\frac{1}{b}du = \int_c^{b+c}\frac{a}{b^2} -\frac{c}{b^2 u} du\] \[\frac{a}{b}+\frac{c}{b^2} \ln\frac{c}{b+c} \] now plug in those \(\phi\) expressions for a,b,c.

  27. ganeshie8
    • one year ago
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    hmm \(\phi_1,\phi_2\) are functions of \(x\) right

  28. Kainui
    • one year ago
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    ohhhhh they are? Yeah then I guess that doesn't really help much lol.

  29. ganeshie8
    • one year ago
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    replacing \(n\) by \(2n\) might simplify something, im gonna try after eating something...

  30. ganeshie8
    • one year ago
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    \(2n+1\) 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 \(r^2-rx-1=0\)

  31. thomas5267
    • one year ago
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    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.

  32. Kainui
    • one year ago
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    What if this is all we have to do: \[\frac{1}{2}\int_0^1 x \color{red}{-} \sqrt{x^2+4}\, dx\]

  33. Kainui
    • one year ago
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    Wait I guess that won't work will it. Oh well lol.

  34. zzr0ck3r
    • one year ago
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    You guys lost me at hello

  35. thomas5267
    • one year ago
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    \[ \begin{align*} f(x,n)&=\frac{r_1(x)^n\left(x \left(\sqrt{x^2+4}-x\right)-2\right)+r_2(x)^n\left(x \left(\sqrt{x^2+4}+x\right)+2\right) }{r_1(x)^n\left(\sqrt{x^2+4}-x\right) +r_2(x)^n\left(\sqrt{x^2+4}+x\right) }\\ r_1(x)&=-x \left(x^2-x\sqrt{x^2+4}+2\right)\\ r_2(x)&=-x \left(x^2 +x\sqrt{x^2+4}+2\right) \end{align*} \]

  36. thomas5267
    • one year ago
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    \[ r_2(x)\leq-1\text{ for }0.352201\leq x\leq1.\\ \lim_{n\to\infty}r_2(x)^n\text{ does not exist for all }x,\,x\in [0,1] \]

  37. ikram002p
    • one year ago
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    .

  38. anonymous
    • one year ago
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    nice

  39. anonymous
    • one year ago
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    $$f(x,n+1)=x+\frac1{f(x,n)}\\\implies f_{n+1}=x+\frac1{f_n}$$ 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 $$f_n=\frac12\frac{(x-\sqrt{4-x^2})^{n+1}+(x+\sqrt{4-x^2})^{n+1}}{(x-\sqrt{4-x^2})^{n}+(x+\sqrt{4-x^2})^{n}}$$

  40. anonymous
    • one year ago
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    oops, that should be \(\sqrt{4+x^2}\) throughout

  41. anonymous
    • one year ago
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    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 \(x\)

  42. anonymous
    • 11 months ago
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    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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