Taylor expansion - series experiments with Matlab

You’ll have a good approximation only if you’re close to the series’ center.

Taylor series look almost identical to Maclaurin series: 

Note:
- The derivatives in Taylor series are evaluated at x = c, the center of the approximation, not at x = 0.
- You raise the quantity (x – c) to the n power, not just x. 

Other than that, the formulas (Maclaurin’s and Taylor’s) are identical.

As an example, let’s use a cosine series to approximate cos(1.6), using just three terms in the  Taylor polynomial for f(x) = cos(x) centered at x = pi/2 
(x is approx. 1.57). In this problem, f(x) = cos(x) and c = pi/2.

 

And we’ll need to take three derivatives (n = 0 to 2) of the function and then evaluate them at x = pi/2.
 

Definition 1: f⁽⁰⁾(x) is the function itself, in this case cos(x). It's not a derivative!
Definition 2: 0! = 1.

f⁽⁰⁾(x) = cos(x)   f⁽⁰⁾(pi/2) = cos(pi/2) = 0
f⁽¹⁾(x) = -sin(x)  f⁽¹⁾(pi/2) = -sin(pi/2) = -1
f⁽²⁾(x) = -cos(x)  f⁽²⁾(pi/2) = -cos(pi/2) = 0

 
Let’s create a fast script in Matlab to evaluate the series just with 3 terms (0 to 2):
 

clear, clc 

% Let's see more decimals
format long 

% We go from n = 0 to n = 2
n = 0 : 2; 

% Center and point to evaluate
c = pi/2;
x = 1.6; 

% These are the derivatives for each term
d = [0 -1 0]; 

% We form the sequence, following the Taylor formula
seq = d .* (x - c).^n ./(factorial(n)) 

% We add-up to get the Taylor approximation
approx = sum(seq) 

% Let's compare with the official number
real_value = cos(x)

 
The resuls are:

seq =       0  -0.02920367320510                  0
approx =       -0.02920367320510
real_value =   -0.02919952230129

We’ve got almost the exact value by using only 3 terms of the series (and two values are zero because the derivative is zero for those terms!!). If we had used n = 10, it would have been an even better approximation.

Now, let’s develop an automated series to express the cosine function (centered at pi/2) using the Taylor expansion and let’s compare the results with different number of terms included.

function stp = taylor_cosine(c, x, n)
% c = center of the function
% x = a vector with values around c
% n = number of terms in the series 

% Start the series with 0
smp = 0;

% Consider all the possible derivatives
deriv = [0 -1 0 1]'; 

% Iterate n times (from 0 to n-1)
for i = 0 : n-1

    % Implement the Taylor expansion
    t(i+1, :) = deriv(1) * (x-c).^(i) / factorial(i);

    % Find derivative for the next term
    deriv = circshift(deriv, -1);

end

% Add-up the calculated terms
stp = sum(t);

Note: the above code is not very efficient, because we’re calculating and adding terms even when their derivative is zero, which is a waste of 50% of the time, more or less. We’ll leave it as-is just to keep and show the natural flow of the formula...
 

Now, let’s create a script to test and use the function above.
 

clear, clc, close all 

% Work with values around center c
c = pi/2;
x = -4 : .1 : 6;
y = cos(x); 

% Plot the goal
plot(x, y, 'g', 'Linewidth', 3)
title('Study of Taylor series')
xlabel('x')
ylabel('cos(x) with different number of terms')
axis([-4 6 -3 3])
grid on
hold on 

% Consider 4 terms in the series
smp = taylor_cosine(c, x, 4);
plot(x, smp, 'ro') 

% Consider 6 terms
smp = taylor_cosine(c, x, 6);
plot(x, smp, 'b-.') 

% Consider 10 terms
smp = taylor_cosine(c, x, 10);
plot(x, smp, 'k', 'linewidth', 3) 

% Label the calculated lines
legend('cos(x)', '4-term series', ...
       '6 terms', '10 terms')

 
The results are:

We see that all of the Taylor expansions work well when we are close to pi/2 (x approx. 1.57). More terms approximate better a larger portion of the cosine curve.