using Schroders iteration. How to start? I am new to matlab

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Part 2 Implement a MATLAB function schroderbisection.m of the form
function [r, h] = schroderbisection(a, b, f, fp, fpp, t)
% a: Beginning of interval [a, b]
% b: End of interval [a, b]
% f: function handle f(x) to find a zero for
% fp: function handle f’(x)
% fpp: function handle f’’(x)
% t: User-provided tolerance for interval width
which combines the fast convergence of the Schro ̈der iteration for multiple roots
g(x)=((x - f(x)/f'(x) * 1 / ( 1 - (f(x) - f'(x))/f'(x)^2))
= x - (f(x) f'(x)/ f'(x)^2 - f(x)f''(x))
with the bracketing guarantee of bisection. At each step j = 1 to n, carefully choose p as in geometric mean bisection (watch out for zeroes!). Define
ε = min(|f (b) f (a)|/8, |f ′′ (p)||b a|2 )
Apply the Schro ̈der iteration function g(x) to two equations f±(x) = f(x) ± ε = 0, yielding two candidates x = q± = g±(p). Replace [a,b] by the smallest interval with endpoints 1 chosen from a, p, q+, q− and b which keeps the root bracketed. Repeat until a f value exactly vanishes, b − a ≤ t, or b and a are adjacent floating point numbers, whichever comes first. Return the final approximation to the root r = (a + b)/2 and a 6 × n history matrix h[1:6,1:n] with column h[1:6,j] = (a,p,q−,q+,b,f(p)) recorded at step j.

Answers (1)

Arnab Sen
Arnab Sen on 24 Feb 2016
Hi Cleven ,
I think you are looking for a way to represent the function 'f' and take it's differential. You can use the anonymous function for this purpose like below:
>> f = @(p)600*p^4-550*p^3+200*p^2-20*p-1;
Now, the function is created and the handle is help by the variable 'f'. For example, f(3) will evaluate the function at x=3 and output 35489.
For more detail about the 'Anonymous Function', refer to the following link:
For differentiation, you may refer to the following link:
You can refer to this link for an illustration on how to implement bisection and secant method which you can easily extend to your required algorithm.


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