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For loops and taylor series

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SonOfAFather
SonOfAFather on 6 Sep 2012
Commented: Dillan Masellas on 8 Apr 2016
I am having issued with my for loop taking the variable that i have set to be [1:1:n] but when i run my script it turns my answer into a scular in stead of a matrix.
here is what i have any help would be great thanks.
clc;
clear;
close all;
n = input('Please give number for the total number of terms in the taylor series: ');
x = input('Please give a value for "x": ');
approxValue = 0;
% Initial value of approxValue.
for k = [0:1:n];
approxVakue = (approxValue + k);
approxValue = sum(x.^k/factorial(k));
% Gives the approx value of e^x as a taylor series
end
disp('approxValue =')
disp((approxValue))
disp('e^x =')
disp(exp(x))
  1 Comment
Dillan Masellas
Dillan Masellas on 8 Apr 2016
How can I perform this operation without using the power function?

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Accepted Answer

Matt Fig
Matt Fig on 6 Sep 2012
Edited: Matt Fig on 6 Sep 2012
Use this loop instead:
for k = 0:n
approxValue = (approxValue + x.^k/factorial(k));
% Gives the approx value of e^x as a taylor series
end
I don't know what approxVakue is supposed to be doing in your code??
And what do you expect to be a matrix? Are you trying to save each term? If so:
approxValue = zeros(1,n+1);
for k = 0:n
approxValue(k+1) = x.^k/factorial(k);
end
approxValuesum = sum(approxValue); % Holds the estimate
This could also be done without FOR loops...
  6 Comments
SonOfAFather
SonOfAFather on 6 Sep 2012
thank you i understand now. It wasn't that i needed to know that the value of k it was that it didn't do what i thought it would do. i thought that the value of k would still stay in vector form, but your explaination corrected my thought process. thank you.
SonOfAFather
SonOfAFather on 12 Sep 2012
In the end i was told that"n" should have been "n-1" in
for k = 0:n should have read k = 0:n.
the explaination i was given was that each time to loop processed through it needed to be one less.
thanks for you help. just thought i would tell you how i was corrected.

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