Does $x = \left(x + \frac {(y - x)} 2\right)$ go on forever?

Question

Basically, if you have x which lets say is 50 and y is 100. Now if I were to keep doing $x = \left(x + \dfrac {(y - x)} 2\right)$ until x = y would it be going forever? And if not, when will it stop? I'd think this can't go on forever because logically going on for infinity it can't, but it also lowers the value each time making it seem impossible.

Answer 1

If $y \neq x$, then $y-x \neq 0$, so $x \neq (x + ((y-x)/2))$, so the process will never stop unless $y = x$ to begin with.

Note that you're just taking $x_n$ to be the mean of $x_{n-1}$ and $y$, over and over; the arithmetic mean of two numbers is always distinct from both of the numbers unless they are identical. They will converge, but not become identical, by this process.

Answer 2

The sequence defined by $x_0=50$ and $x_{n+1}=x_n+\frac12(y-x_n)$ for every $n\geqslant0$, for $y=100$, is such that $x_n=(1-\frac1{2^n})y+\frac1{2^n}x_0$ for every $n\geqslant0$. Thus, $x_n-y=\frac1{2^n}(x_0-y)$. Hence yes, $x_n\to y$ when $n\to\infty$ but no, there is no $n\geqslant0$ such that $x_n=y$ except if $x_0=y$, in which case $x_n=y$ for every $n\geqslant0$.