Question

How do I plot the value of Approximation - Answer as s varies in the code below? If you look at my code below, you can see the method I used (I put it in a separate file).

However, it does not show me a graph from 1 to 1000. Instead the graph is from 999 to 1001 and does not have any points on it.

for s = 1:1000
    error = LaplaceTransform(s,5) - (antiderivative(1,s)-antiderivative(0,s));
end
plot(s,error);
title('Accuracy of Approximation');
xlabel('s');
ylabel('Approximation - Exact Answer');

The functions used:

function g = LaplaceTransform(s,N);
% define function parameters
a=0; 
b=1;
h=(b-a)/N;
x = 0:h:1;
% define function
g = ff(x).*exp(-s*x);

% compute the exact answer of the integral
exact_answer=antiderivative(b,s)-antiderivative(a,s)
% compute the composite trapezoid sum
If=0;
for i=1:(N-1)
    If=If+g(i).*h;
end;
If=If+g(1).*h/2+g(N).*h/2;
If

with

function fx=ff(x)
fx=x;

and

function fx=antiderivative(x,s);
fx= (-exp(-s*x)*(s*x+1))/(s^2);

Any help would be appreciated. Thanks.

Answer 1

The following

for s = 1:1000
    error = LaplaceTransform(s,5) - (antiderivative(1,s)-antiderivative(0,s));
end
plot(s,error);

already has several issues. The two main ones are that error is getting overwritten at each iteration, as @Amro has pointed out, and that s, your loop variable, is a scalar.

Thus, you need to write

difference = zeros(1000,1); %# preassignment is good for you
for s = 1:1000
    difference(s) = LaplaceTransform(s,5) - (antiderivative(1,s)-antiderivative(0,s));
end
plot(1:1000,difference);

There is another error in the LaplaceTransform function

function g = LaplaceTransform(s,N);
[...]
g = ff(x).*exp(-s*x); %# g is an array

[...]
If %# If is calculated, but not returned.

I assume you want to write

function If = LaplaceTransform(s,N);

instead, because otherwise, you try to assign the array g to the scalar difference(s).