1. ## graphs and points

Suppose $S = (a,b)$ with $0 < b < a$. We generate points as follows: $x_0 = a, y_0 = b$, $x_{n+1} = \frac{x_n+y_n}{2}$ and $y_{n+1} = \frac{2x_{n}y_{n}}{x_n+y_n}$.

So $0 < x_{n+1}-y_{n+1} = \frac{x_n-y_n}{x_n+y_n} \cdot \frac{x_n-y_n}{2} < \frac{x_n-y_n}{2}$.

Does the $\frac{x_n-y_n}{2}$ act similarly to $\epsilon/2$? And so $\lim x_n = \lim y_n = x$?

2. Originally Posted by Sampras
Suppose $S = (a,b)$ with $0 < b < a$. We generate points as follows: $x_0 = a, y_0 = b$, $x_{n+1} = \frac{x_n+y_n}{2}$ and $y_{n+1} = \frac{2x_{n}y_{n}}{x_n+y_n}$.

So $0 < x_{n+1}-y_{n+1} = \frac{x_n-y_n}{x_n+y_n} \cdot \frac{x_n-y_n}{2} < \frac{x_n-y_n}{2}$.

Does the $\frac{x_n-y_n}{2}$ act similarly to $\epsilon/2$?
the later is a constant, the former varies depending on n.
And so $\lim x_n = \lim y_n = x$?
how did you get x here?

3. Originally Posted by Jhevon
the later is a constant, the former varies depending on n. how did you get x here?
because $x_{n+1}y_{n+1} = x_ny_n$.