Trying to do central difference method

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Joe Bennett
Joe Bennett 2020-11-13
评论： Joe Bennett 2020-11-13
%QUESTION3 Implements the central differnce method to find the gradient of
%y=e^x-2sin^2(2x) at the point where x=1.
clear all
clc
format long
x=1; % defining x
y=1; % defining y, (y=x)
h=0.1 % first value of h
x=x+h; % getting x+h within f(x)
y=y-h; % getting x-h within f(x)
f = @(x) exp(x)-2*(sin(2*x)).^2; % defining f(x)
g = @(y) exp(y)-2*(sin(2*y)).^2; % defining f(y) to later use as f(x-h)
Grad=(f(x)-g(y))/2*h % Central difference method, Grad = Gradient
for n=1:2 % 2 iterations after h=0.1
h=h/10 % new value for h
x=x+h; % to replace f(x) with f(x+h)
y=y-h; % to replace f(x) with f(x-h)
Grad=(f(x)-g(y))/2*h % Central difference method
end
Here is my code, but my output is incorrect. I did the calculations on paper and they're quite different to my answers. My first value of "Grad" is correct but off by a factor of 1/100. My other two answers are completely wrong.

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采纳的回答

John D'Errico
John D'Errico 2020-11-13
编辑：John D'Errico 2020-11-13
Ran in:
You need to understand this:
6/2*3
ans = 9
Why does it return 9, and not 1?
I told matlab to divide 6 by 2, and then to multiply the result by 3. So the answer is 9. Do you see why the following result is different?
6/(2*3)
ans = 1
Why am I pointing this out? What did you write?
Grad=(f(x)-g(y))/2*h % Central differnce method, Grad = Gradient
h needs to be in the DENOMINATOR. You want to divde by h, yet you multiplied by h. Ergo, the factor of 100, because h was 0.1.

更多回答（1 个）

Alan Stevens
Alan Stevens 2020-11-13
You should probably use the following to calculate the gradient
Grad = (f(1+h) - f(1-h))/(2*h);
though a much smaller value for h will be needed for an accurate result.

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