Rate of Convergence
Up to Error Analysis
Exercise 3
Newton's method converges quadratically because
Exercise 4
e[n+1] is approximately c e[n]^2
To get a feel for whether or not this is a fast rate of convergence, let's look at a simplified case. Suppose c is approximately 1 and e[0] = 10^(-1). Then
e[1] @ e[0]^2 = 10^(-2)
e[2] @ e[1]^2 = 10^(-4)
e[3] @ e[2]^2 = 10^(-8)
e[4] @ e[3]^2 = 10^(-16)
Wonderful! You can see that the error diminishes very quickly.