Prove that the bisection method converges linearly. Solution For the bisection you simply have that $\\epsilon_{i+1}/\\epsilon_i = 1/2$, so, by definition the order of convergence is 1 (linearly) Note that x(sub n) converges to the exact root r with an order of convergence p if: lim(n-> (|r - x(n + 1)|) / (|r - x(n)|^p) = lim(n->(|e(n + 1)|) / (|e(n)|^p) = K I can rearrange the equation above as follows (K = 1/2): |e(n+1)| = 1/2|e(n)|^p and I know that the bisection method is a linear operation, reducing the interval by 1/2 each time and converging on the real root, so p = 1, but I\'m not sure how to show this.

1. Prove that the bisection method converges linearly.
Solution
For the bisection you simply have that $epsilon_{i+1}/epsilon_i = 1/2$, so, by definition the
order of convergence is 1 (linearly)
Note that x(sub n) converges to the exact root r with an order of convergence p if:
lim(n-> (|r - x(n + 1)|) / (|r - x(n)|^p) = lim(n->(|e(n + 1)|) / (|e(n)|^p) = K
I can rearrange the equation above as follows (K = 1/2):
|e(n+1)| = 1/2|e(n)|^p
and I know that the bisection method is a linear operation, reducing the interval by 1/2 each time
and converging on the real root, so p = 1, but I'm not sure how to show this