A closed form for the recursion?

Question

Let $x$ and $y$ be real numbers and $x < y$ Given the recursion: $m_0 = \frac{x+y}{2}$ and $m_1 =\frac{m_0+ y}{2}$, so in general, $$m_i = \frac{m_{i-1} + y}{2}$$..

What is $m_{\infty}$? thanks

Answer 1

Why not trying a few terms and see if any pattern becomes obvious. You have $$m(0)=\frac{1}{2} (x+ y)$$ $$m(1)=\frac{1}{4} (x+3 y)$$ $$m(2)=\frac{1}{8} (x+7 y)$$ $$m(3)=\frac{1}{16} (x+15 y)$$ $$m(4)=\frac{1}{32} (x+31 y)$$ I am sure you see something.

Answer 2

$$y-m_i=\frac{y-m_{i-1}}2\qquad y-m_0=\frac{y-x}2$$

Answer 3

Show that the sequence is bounded from above (by $y$) and is strictly increasing, hence convergent. Show that for the limit of the sequence, which you seem to denote $m_\infty$, we have $m_\infty=\frac{m_\infty+y}{2}$ and solve for $m_\infty$.