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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

A half-baked conjecture: something in terms of the golden ration $$\phi\approx1.618$$.

2. ganeshie8

isn't it same as $\frac{1}{2}\int_0^1 x+\sqrt{x^2+4}\, dx$

3. 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.

4. 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.

5. ganeshie8

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

6. 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 $\lim_{n\to\infty}\int_0^1f(x,n)\,dx\le\int_0^1p(x)\,dx=\frac{1+\phi}{2}$

7. 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 ?

8. 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.

9. ganeshie8

could you please give initial conditions as $$p[0]=p[1]=k$$ @thomas5267

10. 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

11. ganeshie8

i think it should be p[k]==x*p[k-1]+p[k-2], p[0]=p[1]=x

12. ganeshie8

that "x" has nothing to do with the index

13. 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*}

14. 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

15. 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.

16. thomas5267

Is this true? $f(x,n) = x+\cfrac{1}{x+\cfrac{1}{f(x,n-1)}}=x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{x+\cfrac{1}{f(x,n-3)}}}}$

17. Kainui

idk did I mess up my recurrence relation?

18. thomas5267

No I messed it up. The second continued fraction should be f(x,n-2).

19. Kainui

Small error I made when integrating, should have put at the beginning oh well lol $\int_0^1 x dx = \frac{1}{2} \ne 1$

20. 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

21. 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 $$r^2-rx-1=0$$

22. Kainui

@thomas5267 http://mathb.in/40411

23. Kainui

Ok easy to integrate now!

24. thomas5267

Can we exchange limits and integration?

25. 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

26. 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.

27. ganeshie8

hmm $$\phi_1,\phi_2$$ are functions of $$x$$ right

28. Kainui

ohhhhh they are? Yeah then I guess that doesn't really help much lol.

29. ganeshie8

replacing $$n$$ by $$2n$$ might simplify something, im gonna try after eating something...

30. 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 $$r^2-rx-1=0$$

31. 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.

32. Kainui

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

Wait I guess that won't work will it. Oh well lol.

34. zzr0ck3r

You guys lost me at hello

35. 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*}

36. thomas5267

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

37. ikram002p

.

38. anonymous

nice

39. 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}} 40. anonymous oops, that should be \(\sqrt{4+x^2}$$ throughout 41. anonymous now imagine even $$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 $$x$$

42. 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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