# Integral over recurrence relationship

I'm interested in evaluating the following definite integral

$$$$I_n = \int_0^{\gamma} F_n(x)\:dx$$$$

Where $$\gamma \gt 0$$ and $$F_n(x)$$ is based on the recurrence relationship:

$$$$F_{n + 1}(x) = \frac{1}{1 + F_n(x)}$$$$

Here $$F_0(x) = f(x)$$ where $$f$$ is a continuous function on $$\left[0,\gamma\right]$$. My first task was to find a general solution for $$F_n(x)$$ and this is where I've become unstuck.

I started by letting $$F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)}$$. Applying it to the recurrence relationship we have:

$$$$F_{n + 1}(x) = \frac{\alpha_{n+1}(x)}{\beta_{n+1}(x)} = \frac{\beta_n(x)}{\alpha_n(x) + \beta_n(x)}$$$$

And so we have a recurrence relationship over both $$\alpha_n(x)$$ and $$\beta_n(x)$$ with $$F_0(x) = f(x) = \frac{\alpha_0(x)}{\beta_0(x)}$$.

To begin with I'm focused on $$f(x) = \sec(x)$$ with $$\alpha_0(x) = 1$$ and $$\beta_0(x) = \cos(x)$$ and $$\gamma = \frac{\pi}{2}$$.

With a few iterations via Wolframalpha the pattern that emerges is that $$F_n(x)$$ takes the form:

$$$$F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)} = \frac{a_n\cos(x) + b_n}{c_n \cos(x) + d_n}$$$$

Where $$a_n, b_n, c_n, d_n \in \mathbb{N}$$ Here I would like to be able to solve for each but I'm not sure how to start.

Does anyone have any good starting points/references that I can use to begin?

Edit: Changed the definition of $$I_n$$ to have generalised upper limit of $$\gamma$$.

Update Thanks to hypernova's comment's below, it can be seen that $$\beta_n(x)$$ follows a Fibonacci Sequence:

$$$$\beta_{n + 1}(x) = \beta_n(x) + \alpha_n(x) = \beta_n(x) + \beta_{n - 1}(x)$$$$

And so we can now represent $$F_n(x)$$ as:

$$$$F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)} = \frac{\beta_{n-1}(x)}{\beta_n(x)}$$$$ for $$n \geq 2$$.

For the specific example above we have:

$$$$F_n(x) = \frac{\alpha_n(x)}{\beta_n(x)} = \frac{a_{n-1}\cos(x) + b_{n-1}}{a_n \cos(x) + b_n}$$$$ Where $$a_n$$ and $$b_n$$ are Fibonacci Sequences with $$b_0 = 0$$, $$b_1 = 1$$ and $$a_0 = 1$$, $$a_1 = 1$$. We see that $$a_n \gt b_n$$ (this will be important later)

So, we may now evaluate the integral

$$$$I_n = \int_0^{\tfrac{\pi}{2}}\frac{a_{n-1}\cos(x) + b_{n-1}}{a_n \cos(x) + b_n}\:dx$$$$

I will here employ the Weierstrass substitution $$t = \tan\left(\frac{x}{2} \right)$$:

\begin{align} I_n &= \int_0^{\tfrac{\pi}{2}}\frac{a_{n-1}\cos(x) + b_{n-1}}{a_n \cos(x) + b_n}\:dx = \int_0^1 \frac{a_{n-1}\frac{1 - t^2}{1 + t^2} + b_{n-1}}{a_n \frac{1 - t^2}{1 + t^2} + b_n}\frac{2\:dt}{1 + t^2} \\ &= 2\int_0^1 \frac{a_{n - 1}\left(1 - t^2\right) + b_{n - 1}\left(1 + t^2\right)}{\left(1 + t^2\right)\left(a_n\left(1 - t^2\right) + b_n\left(1 + t^2\right)\right)}\:dt \\ &= 2\int_0^1 \frac{\left(b_{n - 1} - a_{n-1}\right)t^2 + \left(b_{n-1} + a_{n-1}\right)}{\left(1 + t^2\right)\left(\left(b_n - a_n\right)t^2 + \left(b_n + a_n\right)\right)}\:dt \\ &= 2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right) \int_0^1 \frac{t^2 - \theta_{n - 1}}{\left(1 + t^2\right)\left(t^2 + \theta_n\right)}\:dt \end{align}

Where

$$$$\theta_n = \frac{b_n + a_n}{b_n - a_n}$$$$

As $$a_n \gt b_n \geq 0$$ we see that $$\theta_n \lt 0$$. Hence:

\begin{align} I_n &= 2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right) \int_0^1 \frac{t^2 - \left|\theta_{n - 1}\right|}{\left(1 + t^2\right)\left(t^2 - \left| \theta_n\right)\right|}\:dt \\ &= 2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right)\left[ \frac{1}{\left|\theta_n\right| + 1} \left[\left(\left|\theta_{n-1}\right| + 1 \right) \arctan(x) + \frac{\left|\theta_{n-1}\right| - \left|\theta_{n}\right|}{\sqrt{\left|\theta_n\right|}}\operatorname{arctanh}\left(\frac{x}{\sqrt{\left|\theta_n\right|}} \right)\right]\right]_0^1 \\ &=2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right)\left(\frac{1}{\left|\theta_n\right| + 1} \right)\left[\left(\left|\theta_{n-1}\right| + 1 \right) \frac{\pi}{4} + \frac{\left|\theta_{n-1}\right| - \left|\theta_{n}\right|}{\sqrt{\left|\theta_n\right|}}\operatorname{arctanh}\left(\frac{1}{\sqrt{\left|\theta_n\right|}} \right)\right] \end{align}

Note $$b_n = a_{n - 1}$$ and thus:

\begin{align} I_n &=2\left(\frac{b_{n - 1} - a_{n-1}}{b_n - a_n}\right)\left(\frac{1}{\left|\theta_n\right| + 1} \right)\left[\left(\left|\theta_{n-1}\right| + 1 \right) \frac{\pi}{4} + \frac{\left|\theta_{n-1}\right| - \left|\theta_{n}\right|}{\sqrt{\left|\theta_n\right|}}\operatorname{arctanh}\left(\frac{1}{\sqrt{\left|\theta_n\right|}} \right)\right] \\ &= \frac{a_{n - 1}}{a_n}\frac{\pi}{2} + \left[1 - \frac{a_{n + 1}\left(a_{n - 1} - a_{n - 2} \right)}{a_n\left(a_n - a_{n - 1} \right)} \right]\sqrt{\frac{a_n - a_{n - 1}}{a_{n + 1}}}\operatorname{arccoth}\left(\sqrt{\frac{a_{n + 1}}{a_n - a_{n - 1}}} \right) \end{align}

Since the recurrence relation is fully nonlinear, it might be better to try the first few terms with the initial condition $$F_0=f$$. By Mathematica, it is easy to find the general expression should be of the form $$F_n=\frac{a_n+b_nf}{a_n+b_n+a_nf}$$ for $$n\ge 1$$, where $$a_n$$ and $$b_n$$ are constants. By the recurrence relation, these constants must follow $$\left( \begin{array}{c} a_{n+1}\\ b_{n+1} \end{array} \right)=\left( \begin{array}{cc} 1&1\\ 1&0 \end{array} \right)\left( \begin{array}{c} a_n\\ b_n \end{array} \right),$$ with $$\left( \begin{array}{c} a_1\\ b_1 \end{array} \right)=\left( \begin{array}{c} 1\\ 0 \end{array} \right).$$ Thanks to linear algebra, you may solve this Fibonacci-like $$a_n$$ and $$b_n$$ immediately.

Hope this could be somewhat helpful for you.

• This is great! thank you so much!
– user150203
Jan 7, 2019 at 1:49
• @DavidG: You are welcome! Jan 7, 2019 at 1:50
• Would this be a pseudo styled Markov Chain?
– user150203
Jan 7, 2019 at 1:53
• @DavidG: For me, it is hard to try using results from Markov chains to your original question. While there might be useful results to use, it might be hard to avoid calculating $A^n$. You may try this built-in function "JordanDecomposition" in Mathematica to figure out $P$ and $J$ such that $A=PJP^{-1}$, where $J$ is the Jordan Normal Form of $A$. Then $A^n=PJ^nP^{-1}$. $J^n$ is very easy to calculate for your question, because $J$ is 2-by-2 and upper-triangular. Jan 7, 2019 at 3:15
• @DavidG: Sure! Thanks for your message. marty cohen's solution is really beautiful. I seems to have misunderstood your expectations. While the coefficients $a_n$ and $b_n$ diverge, their resulted $F_n$ converges. marty cohen's solution definitely clarifies. Thank you! Jan 8, 2019 at 10:35

This is copied from another answer of mine:

Let $$f_1 = \frac{1}{1 + g(x) }$$ where $$g(x) > 0,$$, and let $$f_n(x) =\frac{1}{1+f_{n-1}(x)}$$.

Then $$f_n(x) \to \dfrac{\sqrt{5}-1}{2}$$.

Note: I doubt that any of this is original, but this was all done just now by me.

Proof.

$$\begin{array}\\ f_n(x) &=\frac{1}{1+\frac{1}{1+f_{n-2}(x)}}\\ &=\frac{1+f_{n-2}(x)}{1+f_{n-2}(x)+1}\\ &=\frac{1+f_{n-2}(x)}{2+f_{n-2}(x)}\\ \end{array}$$

Therefore, if $$f_{n-2}(x) > 0$$ then $$\frac12 < f_n(x) \lt 1$$.

Similarly, if $$f_{n-1}(x) > 0$$ then $$0 < f_n(x) \lt 1$$.

$$\begin{array}\\ f_n(x)-f_{n-2}(x) &=\dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}-f_{n-2}(x)\\ &=\dfrac{1+f_{n-2}(x)-f_{n-2}(x)(2+f_{n-2}(x))}{2+f_{n-2}(x)}\\ &=\dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{2+f_{n-2}(x)}\\ \end{array}$$

$$\begin{array}\\ f_n(x)+f_n^2(x) &=\dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)}+(\dfrac{1+f_{n-2}(x)}{2+f_{n-2}(x)})^2\\ &=\dfrac{(1+f_{n-2}(x))(2+f_{n-2}(x))}{(2+f_{n-2}(x))^2}+\dfrac{1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\ &=\dfrac{2+3f_{n-2}(x)+f_{n-2}^2(x)+1+2f_{n-2}(x)+f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\ &=\dfrac{3+5f_{n-2}(x)+2f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\ \text{so}\\ 1-f_n(x)-f_n^2(x) &=\dfrac{4+4f_{n-2}(x)+f_{n-2}^2(x)-(3+5f_{n-2}(x)+2f_{n-2}^2(x))}{(2+f_{n-2}(x))^2}\\ &=\dfrac{1-f_{n-2}(x)-f_{n-2}^2(x)}{(2+f_{n-2}(x))^2}\\ \end{array}$$

Therefore $$1-f_n(x)-f_n^2(x)$$ has the same sign as $$1-f_{n-2}(x)-f_{n-2}^2(x)$$. Also, $$|1-f_n(x)-f_n^2(x)| \lt \frac14|1-f_{n-2}(x)-f_{n-2}^2(x)|$$ so $$|1-f_n(x)-f_n^2(x)| \to 0$$.

Let $$p(x) = 1-x-x^2$$ and $$x_0 = \frac{\sqrt{5}-1}{2}$$ so $$p(x_0) = 0$$, $$p'(x) < 0$$ for $$x \ge 0$$.

Since $$f_n(x) > 0$$, $$f_n(x) \to x_0$$.

• Thanks for your post - I actually was playing around with a few different cases last night and noticed that it converged as you spoke to. I wasn't sure how to go about proving it, so thanks heaps for your post!
– user150203
Jan 8, 2019 at 1:01