Question

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

Answer 1

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

Answer 2

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

Answer 3

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.

Answer 4

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.

Answer 5

\(C_{k} = [x,x,x\cdots,( \text{ktimes})] = \dfrac{p_k}{q_k}\)

where \(p_k\) and \(q_k\) are defined recursively by :
\(p_k=x p_{k-1}+p_{k-2}\)
\(q_k=xq_{k-1}+q_{k-2}\)

Answer 6

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

Answer 7

\(p_0 = x\\p_1=x^2+1\\p_2 = x(x^2+1)+x = x^3+2x\\p_3=x(x^3+2x)+x^2+1=x^4+3x^2+1\\
p_4=x(x^4+3x^2+1)+x^3+2x=x^5+4x^3+3x
\)


see any pattern ?

Answer 8

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.

Answer 9

could you please give initial conditions as \(p[0]=p[1]=k\)

@thomas5267

Answer 10

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

Answer 11

i think it should be

p[k]==x*p[k-1]+p[k-2], p[0]=p[1]=x

Answer 12

that "x" has nothing to do with the index

Answer 13

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

Answer 14

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

Answer 15

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.

Answer 16

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

Answer 17

idk did I mess up my recurrence relation?

Answer 18

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

Answer 19

Small error I made when integrating, should have put at the beginning oh well lol

\[\int_0^1 x dx = \frac{1}{2} \ne 1\]

Answer 20

\[
g(x,n)=[x,x,g(x,n-1)]\,,g(x,0)=x
\]

\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

Answer 21

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 \(r^2-rx-1=0\)

Answer 22

@thomas5267 http://mathb.in/40411

Answer 23

Ok easy to integrate now!

Answer 24

Can we exchange limits and integration?

Answer 25

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

Answer 26

\[\int_0^1 \frac{xa}{xb+c}dx\]

\[xb+c=u\]\[dx=\frac{1}{b}du\]

\[\int_c^{b+c} \frac{a\frac{1}{b} u - \frac{c}{b}}{u}\frac{1}{b}du = \int_c^{b+c}\frac{a}{b^2} -\frac{c}{b^2 u} du\]
\[\frac{a}{b}+\frac{c}{b^2} \ln\frac{c}{b+c} \]

now plug in those \(\phi\) expressions for a,b,c.

Answer 27

hmm \(\phi_1,\phi_2\) are functions of \(x\) right

Answer 28

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

Answer 29

replacing \(n\) by \(2n\) might simplify something, im gonna try after eating something...

Answer 30

\(2n+1\) 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 \(r^2-rx-1=0\)

Answer 31

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.

Answer 32

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

Answer 33

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

Answer 34

You guys lost me at hello

Answer 35

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

Answer 36

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

Answer 37

.

Answer 38

nice

Answer 39

$$f(x,n+1)=x+\frac1{f(x,n)}\\\implies f_{n+1}=x+\frac1{f_n}$$
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 $$f_n=\frac12\frac{(x-\sqrt{4-x^2})^{n+1}+(x+\sqrt{4-x^2})^{n+1}}{(x-\sqrt{4-x^2})^{n}+(x+\sqrt{4-x^2})^{n}}$$

Answer 40

oops, that should be \(\sqrt{4+x^2}\) throughout

Answer 41

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 \(x\)

Answer 42

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.