# extracting arrays from a taylor series and plotting

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Max on 11 Feb 2015
Commented: Max on 11 Feb 2015
I have been working on a script that calculates a Taylor series without using the built-in function.
I am having a lot of trouble extracting the proper array of numbers for the three iteration values I need to plot together with the error (I need to plot N = 2 5 & 50 with the exact function of 5sin(3x)).
This is what I have:
clear;clc
n =[2 5 50];
do=linspace(-2*pi,2*pi,720);
T = zeros(51);
for i =1:720
for k=1:1:50;
ns=2*k+1;
T(i)=T(i)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);
end
end
approx2 = T(:,2);
approx5 = T(:,5);
aprrox50 = T(:,50);
exact = 5*sin(3*do);
plot(do,exact, '-r')
hold on
ez1=ezplot('approx2');
ez2=ezplot('approx5');
ez3=ezplot('approx50');
legend('5sin(3x)','T2','T5','T50')
set(ez1, 'color', [0 1 0])
set(ez2, 'color', [0 0 1])
set(ez3, 'color', [1 0 1])
title('Graph of 5sin(3x) and taylor expansions T2, T5 and T50')
I cannot get it to graph properly though......
I have looked through the matlab help file and searched through google.
I am thinking it might have to do with not defning the Taylor iterations properly.
This is a homework assignment so I AM NOT LOOKING FOR SOMEONE TO DO THIS FOR ME.
Any hints or tips to put me in the right direction will be greatly appreciated!
Thank You

per isakson on 11 Feb 2015
Edited: per isakson on 11 Feb 2015
Hints:
• aprrox50 &nbsp is a typo
• read the doc on ezplot, note that approx.. are double vectors
• use plot to plot approx..
• I find it easier to work with functions than with scripts when it comes to debugging - it used to be anyhow.
• try &nbsp figure, imagesc(T), colorbar
• ALWAYSONTOP, by Elmar Tarajan makes it possible to see the effect on the diagram of each line of code; step through the code and watch - or dock the figure

Max on 11 Feb 2015
hence the reason why k was always definded as k = 1:n
per isakson on 11 Feb 2015
Yes, Matlab's indexing is "one-based", but in one way or other you have to include the first term in the series!
Max on 11 Feb 2015
well the first term is always 15x
so would defining the series as
(for example)
approx2(i) = 15*x + 5*(((-1)^k)*(3*x(i))^(ns))/factorial(ns);
i want to say that takes care of the 0 term

Roger Stafford on 11 Feb 2015
It looks as though your T is incorrectly computed. For each of three different counts, n = 2, 5, and 50, you need to compute n terms of the Taylor series for each of 720 different values of the angle. That would imply three, not two, nested for-loops or their equivalent, and the resulting T should be two-dimensional, not one-dimensional.
for n = [2,5,50]
for i = 1:720
for k = 1:n
etc.
(Of course, with the appropriate use of matlab's 'sum' function you could eliminate at least one of these loops.)

Show 1 older comment
Roger Stafford on 11 Feb 2015
I really should have written the loops in my suggestion this way:
T = zeros(720,3);
n = [2,5,50];
for j = 1:3
for i = 1:720
for k = 1:n(j)
T(i,j) = T(i,j) + ....
.......
approx2 = T(:,1);
approx5 = T(:,2);
approx50 = T(:,3);
Max on 11 Feb 2015
Yes this does help, but for the three loops would i define the j values as
j = 1:2
j = 1:5
J = 1:50
I know they would go at the begging of each loop for that specific number of iterations.
Max on 11 Feb 2015
I tried this:
T = zeros(720,3);
n = [2 5 50];
do=linspace(-2*pi,2*pi,720);
for j = 1:2
for i = 1:720
for k = 1:n(j)
ns=2*k+1;
T(i,j)=T(i,j)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);
end
end
end
for j = 1:5
for i = 1:720
for k = 1:n(j)
ns=2*k+1;
T(i,j)=T(i,j)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);
end
end
end
for j = 1:50
for i = 1:720
for k = 1:n(j)
ns=2*k+1;
T(i,j)=T(i,j)+5*(((-1)^k)*(3*do(i))^(ns))/factorial(ns);
end
end
end
approx2 = T(:,1);
approx5 = T(:,5);
approx50 = T(:,50);
But matlab complains about n(4) being out of bonds because numel(n)=3
which makes no sense to me because its defined at the begging of the script that n = [2 5 50]...

daniel on 11 Feb 2015
The first problem I see is that "exact" is a 1x720 vector and your approximations are 51x1 vectors; my first suggestion would be to get them all in the same domain. Once they all span the same domain and are of the same size, your plot should look better.

#### 1 Comment

Max on 11 Feb 2015
Thats an excellent point, I could just transpose the approx values but then it still wouldnt match size wise