Integral over recurrence relationship

Question

I'm interested in evaluating the following definite integral

\begin{equation} I_n = \int_0^{\gamma} F_n(x)\:dx \end{equation}

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

\begin{equation} F_{n + 1}(x) = \frac{1}{1 + F_n(x)} \end{equation}

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:

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

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:

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

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:

\begin{equation} \beta_{n + 1}(x) = \beta_n(x) + \alpha_n(x) = \beta_n(x) + \beta_{n - 1}(x) \end{equation}

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

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

For the specific example above we have:

\begin{equation} 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} \end{equation} 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

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

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

\begin{equation} \theta_n = \frac{b_n + a_n}{b_n - a_n} \end{equation}

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}

Answer 1

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.

Answer 2

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