Telescoping a recursively defined series?

Question

This question is inspired by an answer of mine that I gave here. Basically the problem is as follows: you are given a recursively defined sequence $(x_n)_{n\geq0}$, where you know the value of $x_0$ and you know that $x_{n+1}=\varphi(x_n)$ for all $n\geq0$, where $\varphi$ is a known function. Let’s say that you can’t find $x_n$ explicitly because it’s too hard or whatever reason. But you still want to compute $\sum_{n=0}^{+\infty}x_n$. Let’s assume that we already checked the series converges (e.g., numerically), so we are only concerned with the actual value of the sum. We hope that we can find this value by resorting to a telescoping technique.

Now, if we had an explicit formula for $x_n$ as a function of $n$, we could try to telescope directly on that expression, and I know that there are algorithms that allow you to do that, even if you can’t come up with a clever idea yourself. For example, if $x_n$ is a hypergeometric term of $n$ (i.e., $\frac{x_n}{x_{n-1}}$ is the ratio of two polynomial functions of $n$), then Gosper’s algorithm either gives you another hypergeometric term of $n$ (say $y_n$) such that $x_n=y_n-y_{n-1}$, or tells you it’s not possible. But if you do get this $y_n$, you can compute $$\sum_{n=0}^{+\infty}x_n=x_0+\sum_{n=1}^{+\infty}x_n=x_0+\sum_{n=1}^{+\infty}(y_n-y_{n-1})\\=x_0+\lim_{n\to+\infty}\sum_{k=1}^n(y_k-y_{k-1})=x_0+\lim_{n\to+\infty}(y_n-y_0)\\=x_0-y_0+\lim_{n\to+\infty}y_n.$$

However, in our case, you can’t feed $x_n$ to Gosper’s algorithm because you simply don’t have an expression for that. That’s why, in the linked question, I found a function $f$ such that $x_n=f(x_n)-f(x_{n-1})$ (not literally this, but I’m trying to mirror the above paragraph), and then I telescoped like this: $$\sum_{n=0}^{+\infty}x_n=x_0+\sum_{n=1}^{+\infty}x_n=x_0+\sum_{n=1}^{+\infty}\big(f(x_n)-f(x_{n-1})\big)\\=x_0+\lim_{n\to+\infty}\sum_{k=1}^n\big(f(x_k)-f(x_{k-1})\big)=x_0+\lim_{n\to+\infty}\big(f(x_n)-f(x_0)\big)\\=x_0-f(x_0)+\lim_{n\to+\infty}f(x_n).$$

Why am I not satified? Because the specific $f$ I used in the linked question was basically suggested by people in the comments. So I wonder whether there exists an algorithm to determine such an $f$, or a systematic way to approach this type of problems. Without a proper method, we have to rely on intuition (which doesn’t always work, of course).

Edit

By writing $x_n=f(x_n)-f(x_{n-1})$ as $\varphi(x_{n-1})=f(\varphi(x_{n-1}))-f(x_{n-1})$, I think the problem boils down to the following question: for what type of $\varphi$ do we have methods to find $f$ satisfying $f(x)+\varphi(x)=f(\varphi(x))$? Is this a known functional equation?

Answer 1

By searching $f$ such that $x_n = f(x_n)-f(x_{n+1})$ (as in your linked answer), we have: $$x = f(x) - f(\varphi(x))$$ If we assume that $f$ is bijective: $$\varphi(x) = f^{-1}(f(x)-x)$$

We can build some classes of functions $\varphi$ by fixing classes of function $f$.

For example, suppose $f$ is of the form $\frac{1}{ax+b}$ (We can choose $f$ up to a constant, so no need for a polynomial term for the numerator)

We have: $$f^{-1}(x) = \frac{1-bx}{ax}$$ And: $$f(x)-x = \frac{1-bx-ax^2}{ax+b}$$ So $\varphi$ needs to be of the form: $$\begin{split}\varphi(x) &= f^{-1}(f(x)-x)\\ &= \frac{1-b\frac{1-bx-ax^2}{ax+b}}{a\frac{1-bx-ax^2}{ax+b}}\\ &= \frac{ax+b-b+b^2x-abx^2}{a-abx-a^2x^2}\\ &= \frac{(a+b^2)x+abx^2}{a-abx-a^2x^2} \end{split}$$

This is a bit cryptic for a class of function, with some variable changes, we can rewrite it as two separate cases.

If $b = 0$, it becomes $\varphi = \frac{x}{1-ax^2}$ with $f(x)= \frac{1}{ax}$

If $b \neq 0$, let $\varphi = \frac{\frac{AB+1}{B}x+x^2}{A-x-Bx^2}$

We have $a=\frac{B}{A}$ and $b = \frac{1}{A}$, so $f(x) = \frac{A}{Bx+1}$

In your linked answer, $\varphi(x)=\frac{-x^2}{1+x+x^2}=\frac{x^2}{-1-x-x^2}$ so $A=-1, B=1$ and $f(x) = \frac{-1}{x+1}$, which is a constant away from your suggested $f(x) = \frac{x}{x+1}$

Answer 2

As was previously mentioned, I wouldn't expect a closed form solution in the most general case. This is true in the case where we are solving the cohomological equation or more specifically the coboundary equation $g=f - f\circ \varphi$ for a dynamical system $\varphi$. For general dynamical systems, it is possible to obtain a representation of the transfer function $f$ as a limit. It's not unique, since a constant can be added, $f+c$, and it will still satisfy the coboundary equation. This limit also converges to an $f$ in the situation here. If $f(x_j)=\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} \sum_{i=0}^{k-1} x_{j+i}$, then $$f(x_j)-f(x_{j+1})=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n} \Big( \sum_{i=0}^{k-1} x_{j+i} - \sum_{i=0}^{k-1} x_{j+1+i}\Big)=\lim_{n\to\infty} \frac{1}{n} \sum_{k=1}^{n} (x_j - x_{j+k})$$ $$=x_j - \lim_{n\to\infty}\frac{1}{n} \sum_{k=1}^{n} x_{j+k}=x_j,\ \mbox{since}\ \sum_{k=1}^{\infty} x_{j+k}<\infty .$$ The formula in the comments should not have a negative (unless the coboundary equation is written as $g=f\circ \varphi - f$). This is NOT an answer to the question and I'm not expecting upvotes. Just wanted to provide more details. As @[Alp Uzman] pointed out, there are certain cases where this can be solved for dynamical systems. This paper shows when the coboundary equation can be solved for general measure preserving systems: https://tinyurl.com/4zxfmvtn .