Trying to do central difference method

%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.

Answer 1

[John D'Errico]

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.

[Joe Bennett]

Thank you, its all sorted now.

Answer 2

[Alan Stevens]

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.

[Joe Bennett]

Thank you, its all sorted now.