Why is my for loop not iterating in the way I want?

I have a piece of code

 clear;
x=2;
expapprox=0;
expapproxarray=zeros(1,12);
for i=0:12
    expapprox=expapprox+x^i/factorial(i);
    error=abs(expapprox-exp(x))
    for j=1:size(expapproxarray)
        expapproxarray(j)=error
    end
end    
plot(expapproxarray)

that compares the exponential function to an approximation of that function using the Taylor series. I want to plot the error found between the two functions for an increasing number of terms in the Taylor expansion. I thought this could be done using the code above but there is a problem with how the second loop works. I think the problem is that the full iteration of the second loop is being done within the first loop but changing the order of the loops gives another wrong answer. How can I get the value of error for each value of I to be outputted into some array that I have called expapproxarray that can then be plotted as a function of i?

Answer 1

[Stephen23]

N = 2; % the value to use.
% Approximation:
X1 = 1:12; % the number of terms.
Y1 = nan(size(X1));
for k = X1
    T = 0:k-1;
    Y1(k) = sum(N.^T./factorial(T));
end
% Inbuilt function:
X2 = X1([1,end]);
Y2 = exp([N,N]);
% Plot:
plot(X1,Y1,'*',X2,Y2,'-')
xlabel('Number of terms')
ylabel('Output value')

Plots this:

You can plot the error simply by subtracting exp(N):

plot(X1,Y1-exp(N))

Answer 2

[madhan ravi]

x=2;
expapprox=cell(1);
expapprox(1)={0};
expapproxarray=zeros(1,12);
error1=cell(1);

expapproxarray=cell(1);
for i=2:12                 %your loop didn't iterate because it should start from 2 because you already defined the first element 
    expapprox{i}=expapprox{i-1}+x^i/factorial(i);
    error1{i}=abs(expapprox{i}-exp(x));   %your second loop as no meaning because 1:1 is still one doesn't mean anything here

end    
plot(cell2mat(error),'*')     %plot function plots the function according to the index so this represents I vs error
hold on
plot(cell2mat(expapprox))    %this represents i vs expapprox

[madhan ravi]

plus I also suspect

error1{i}=abs(expapprox{i}-exp(x));

should be

error1{i}=(expapprox{i}-exp(x));

because the abs function would simply ignore the negative values which is not what we want right?