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use newton raphson method to find roots of exp(-x)-x
clear all; clc; n = 11; x = zeros(1,numel(n)); ea = zeros(1,numel(n)); g=@(x) exp(-x); f=@(x) exp(-x)-x; true_root=0.56...
3 years 前 | 0 个回答 | 0
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simple Fixed Point Iteration
clear all; clc; x(1)=0; g=@(x) exp(-x); f=@(x) exp(-x)-x; true_root=0.56714329; disp('--------------------------------...
3 years 前 | 1 个回答 | 0