If a sequence of numbers is determined by the equality and the values and , prove that:
Rewrite , for and . By substituting into the recursive formula, you get . Since we are taking iterative midpoints, the gap between successive values of are decreasing by half, so [tex]|a_n-a_{n-1}|=\left(\frac12\right)^{n-1}[/MATh] . It should also be apparent that when n is odd, and when n is even, again, by virtue of iterating the midpoint formula, making the value of the gap alternate between positive and negative. Therefore, for , the limit of which as n gets large is .