Counting the amount of iterations to reach specific answer

Question

I have the following function $f(x) = (x + y) \cdot z$ where $y$ is any positive number and $z \in (0,1) $. The way this works is that the output feeds the input such that $x_{n+1} = (x_n + y) \cdot z$ or $f(f(f(f...$ or $(x+y)\cdot z + y)\cdot z + y) \cdot z + y) \cdot z ...$ if we try to solve $x = (x + y) \cdot z$ we get $x = \frac{y \cdot z}{1 - z}$ which will only occur after infinitely many iterations assuming that $x_{initial} < \frac{y \cdot z}{1 - z}$ where $x_{initial}$ is the first number that we start the iteration with.

Since for $x = \frac{y \cdot z}{1 - z}$ we need infinitely many iterations we can approximate this to $x > \frac{y \cdot z}{1 - z} \cdot 0.99$ the reason I say greater is because it is very rare that after an iteration I get exactly a specific number. I've already written a computer code that performs all of these iterations and stops when $x > \frac{y \cdot z}{1 - z} \cdot 0.99$ telling the user how many iterations it has performed.

What I want to do is find some function that can express this mathematically instead of a computer having to go through all of these iterations. I want to find some function $N(i)$ where $i$ is the number of iterations (doesn't have to be an integer) and $N(i)$ gives the value of $x$ after those iterations. In this case this function must include $x_{initial}, y, z$ as constants. For example if do $\lim_{i \to \infty } N(i) = \frac{y \cdot z}{1 - z}$, therefore this function will have a horizontal asymptote.

Perhaps there isn't a clear way to express this mathematically and can only be calculated via computer iteration. If anyone has any ideas I'll greatly appreciate it if you could write them down.

Thanks

Answer 1

You have stated the recurrence relation $$x_{n+1}=(x_n+y)\cdot z=z\cdot x_n+y\cdot z$$ This has the solution $$x_n=C\cdot z^n + \frac{y\cdot z(z^n-1)}{z-1}$$ where $C$ is an arbitrary constant. We can deduce the value of $C$ by using $$x_0=x_{\text{initial}}\implies C=x_{\text{initial}}$$ $$\boxed{\therefore x_n=\overbrace{f(f(f(\dots(x)))}^{n\text{ times}}=x_{\text{initial}}\cdot z^n+\frac{y\cdot z(z^n-1)}{z-1}}$$ Finally if you want to solve for the value of $n$ which gives the value of $0.99\left(\frac{y\cdot z}{1-z}\right)$ then you have to solve $$x_n=x_{\text{initial}}\cdot z^n+\frac{y\cdot z(z^n-1)}{z-1}=0.99\left(\frac{y\cdot z}{1-z}\right)$$ $$\frac{x_{\text{initial}}\cdot z^n(z-1)+y\cdot z(z^n-1)}{z-1}=0.99\left(\frac{y\cdot z}{1-z}\right)$$ $$x_{\text{initial}}\cdot z^n(z-1)+y\cdot z(z^n-1)=-0.99\cdot y\cdot z$$ $$x_{\text{initial}}\cdot z^{n+1}-x_{\text{initial}}\cdot z^n+y\cdot z^{n+1}-y\cdot z=-0.99\cdot y\cdot z$$ $$z^n(x_{\text{initial}}\cdot z-x_{\text{initial}}+y\cdot z)=0.01\cdot y\cdot z$$ $$z^n=\frac{0.01\cdot y\cdot z}{x_{\text{initial}}\cdot z-x_{\text{initial}}+y\cdot z}$$ $$n=\log_z{\left(\frac{0.01\cdot y\cdot z}{x_{\text{initial}}\cdot z-x_{\text{initial}}+y\cdot z}\right)}$$ As you want the least integer greater than this you have $$\boxed{n_{\text{min}}=\left\lceil\log_z{\left(\frac{0.01\cdot y\cdot z}{x_{\text{initial}}\cdot z-x_{\text{initial}}+y\cdot z}\right)}\right\rceil}$$