help me in trying to solve the secant method

Question

Ishmum Monjur Nilock 2020-2-10

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https://ww2.mathworks.cn/matlabcentral/answers/504714-help-me-in-trying-to-solve-the-secant-method

编辑： Jim Riggs 2020-2-10

Hello all I try to write the code for secant method below:

clear all
close all
clc
n=100;
err=0.004;
x0=0.5;
x1=1;
x(1)=x0;
x(2)=x1;
f=@(x) exp(-x)-x;
i=0;
for i=3:n
    x(i) = x(i-1) - (f(x(i-1)))*((x(i-1) - x(i-2))/(f(x(i-1)) - f(x(i-2))));
    i=i+1;
    if abs((x(i)-x(i-1))/(x(i)))*100 < err  %in this line
        root=x(i);
        break
    end
end

When i try to run the code it says that error index exceeds the maximum array elements(3) in the line no specified with green comment. How to solve the problem? Is there any way so that I can run the program from here?

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Answer 1

Jim Riggs 2020-2-10

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https://ww2.mathworks.cn/matlabcentral/answers/504714-help-me-in-trying-to-solve-the-secant-method#answer_414852

编辑：Jim Riggs 2020-2-10

Try preallocating x, right after you define n:

n = 100;
x = nan(1,n);
...

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Jim Riggs 2020-2-10

编辑：Jim Riggs 2020-2-10

I just noticed that you are incrementing your loop variable "i" inside the loop.

This is usually a bad idea.

the line "for i=3:n" controls the value of i. When i gets to a value of n, then you add 1 to i, so i is now n+1.

The way that your loop is written now, x(i) is defined on each pass of the loop. When you increment i, you are trying to reference x(i+1) which does not yet exist. If you preallocate the way I suggested, then x(i+1) will exist, but it will be defined as "nan". Either way, you will have a problem with this construct.

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