Average value of recurrent function.

Question

Given a function f(x) = a*f(x-1) where a is a number between 0 and 1, what is the average value of f(x) for x >= 0?

Clarification:

f(1) is a constant, say 1
f is only defined for integer inputs

Disclaimer: I'm still in high school, so it's possible that I'm asking/saying something very stupid.

Answer 1

Ok, so let f(1)=k. Then for any n we have that the sum of $f(1)+f(2)...f(n)= k(\frac{1-a^{n+1}}{1-a})$. then the average value of $f(m)$ for $1\leq m \leq n$ is $\frac{k(\frac{1-a^{n+1}}{1-a})}{n}$.

So it makes sense to say that the average value is going to be $\lim_{n \to \infty} \frac{k(\frac{1-a^{n+1}}{1-a})}{n}=k\lim_{n \to \infty} \frac {1-a^{n+1}}{n-a}=0 $for $k\neq 1 $and $k$ for $ a=1 $

Answer 2

If $f(1) = 1$ then $f(2) = af(1) = a$, $f(3) = af(2) = a^2$ ... $f(n) = a^{n-1}$.

For $0 < a < 1$, as $n$ grows, $a^{n-1}$ approaches $0$, so that's your average.

Answer 3

It's not too difficult to show (by induction) that if $f(0)=1$ and $f(n)=af(n-1)$ for all integers $n\ge 1,$ then $f(n)=a^n$ for all integers $n\ge 0.$

Another thing that isn't too tricky to show is that for any integer $k\ge 0,$ we have $$f(0)+\cdots+f(k)=\frac{1-a^{k+1}}{1-a}.$$ $($*Hint*: Use the explicit formula for $f(n)$ and multiply both sides of the above equation by $1-a$.$)$

From this, and the fact that $0\le a<1,$ we can see that $$f(0)+f(1)+f(2)+\cdots=\frac1{1-a}.$$

The problem is this: how would we define the average of a countably infinite set of numbers? There isn't really a notion of "dividing by infinity" that's well-defined. Arguably the best we can do is look at the tendency of $$\frac{f(0)+\cdots+f(k)}{k+1}=\frac{1-a^{k+1}}{(1-a)(k+1)}$$ as $k$ grows without bound. But $0<\frac{1-a^{k+1}}{1-a}<\frac{1}{1-a}$ for all $k\ge 0,$ so $$0<\frac{1-a^{k+1}}{(1-a)(k+1)}<\frac{1}{(1-a)(k+1)}$$ for all $k\ge 0,$ so since the quantity on the far right tends to $0$ as $k$ grows without bound, then the quantity in the middle also tends to $0$. Hence, though $f(n)$ is positive for all $n\ge 0,$ the "average value" of all the $f(n)$s is most sensibly said to be $0$. Strange, yes, but that sort of thing can happen when we delve into the realms of the infinite.

A less surprising case occurs when $a=1,$ since then we're taking the average of an infinite list of exactly the same number, and we get exactly what you'd expect.