MATLAB Optimizing Taylor Series Approximations in Matlab for Trigonometric Functions

[NYK]

I have been working on writing g a script file that will:

1. Calculate f(x)=5sin(3x) using the Taylor series with the number of terms n=2, 5, 50, without using the built-in sum function. 
2. Plot the three approximations along with the exact function for x=[-2π 2π]. 
3. Plot the relative true error for each of the approximations 
4. Calculate the value of sin(x) and the error for x=π and x=3π/2 for each of the approximations 
5. How many terms are necessary for an error E<.000001?

I have been able to get as far as the third part of the question, any advice, tips or pointers are greatly appreciated!

I pasted the script I have so far bellow:

clear, clc, close all
%Define the limits, the original function and the Taylor series.
syms x

a = -2*pi:2*pi;

g = (5*sin(3*x));

T_2 = taylor(g, 'Order', 2);

T_5 = taylor(g, 'Order', 5);

T_50 = taylor(g, 'Order', 50);

z = (5*sin(3*a));%plot the original function and the three Taylor series.

fg=figure;
ax=axes;
ez1=plot(a,z, 'r--');
hold on
ez2=ezplot(char(T_2),[-2*pi, 2*pi]);
ez3=ezplot(char(T_5),[-2*pi, 2*pi]);
ez4=ezplot(char(T_50),[-2*pi, 2*pi]);

legend('5sin(3x)','T2','T5','T50')

set(ez2, 'color', [0 1 0])
set(ez3, 'color', [0 0 1])
set(ez4, 'color', [1 0 1])title(ax,['Graph of 5sin(3x) and taylor expansions T2, T5 and T50'])

 

[RUber]

for part 4, you should just evaluate the taylor series at the points mentioned and subtract that from the true value.
Once you have done that, you can set up a script to try different values for 'order' until you reach the accuracy needed.

 

[kreil]

3. "Relative true error" to me just sounds like you're plotting the absolute difference between the approximations and the true values.

4. Pretty self explanatory, I'm with RUber on this one ^^.

5. Use the remainder term to figure this out. See the following links:

http://en.wikipedia.org/wiki/Taylor's_theorem#Motivation

http://www.millersville.edu/~bikenaga/calculus/remainder-term/remainder-term.html

 

[RUber]

If you wanted to do this without the built in Taylor functions, you could define the derivatives of g:
g(x)= 5sin(3x)
d(n,g(x))=5*3^n*sin(3x+n*pi/2)
and the taylor series is
(x-x_0)^n/n!*d(n,g(x_0)).
Often, the series is evaluated at x_0 = 0...which produces a pretty simple result for the sine function.

 

[NYK]

Thanks RUber, you are compltely correct, I read the problem statement a little more closely and releaized i wasnt do it correct.

So I've tried working with this:

clear;clc

n =[2 5 50]
do=linspace(-2*pi,2*pi,720);
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
endhavent had any luck with it so far though