Numerical Methods Course Notes Version 0.1 (UCSD Math 174, Fall ...

4.3. SECANT METHOD 57
relation between e k+1 and the previous two errors, e k , e k−1 .
Indeed, this is the case. Omitting all the nasty details (see Cheney & Kincaid [7]), we arrive at
the imprecise equation:
e k+1 ≈ − f ′′ (r)
2f ′ (r) e ke k−1 = Ce k e k−1 .
Again, the proof relies on finding some “attractor” region and using continuity.
We now postulate that the error terms for the secant method follow some power law of the
following type:
|e k+1 | ∼ A |e k | α .
Recall that this held true for Newton’s method, with α = 2. We try to find the α for the secant
method. Note that
|e k | = A |e k−1 | α ,
so
Then we have
|e k−1 | = A − 1 α |ek | 1 α .
A |e k | α = |e k+1 | = C |e k | |e k−1 | = CA − 1 α |e k | 1+ 1 α ,
Since this equation is to hold for all e k , we must have
α = 1 + 1 α .
This is solved by α = 1 ( √ )
2 1 + 5 ≈ 1.62. Thus we say that the secant method enjoys superlinear
convergence; This is somewhere between the convergence rates for bisection and Newton’s method.