Taylor
expansion  series experiments with Matlab
Once
you know how Maclaurin
series work, Taylor
series are easier to
understand.
Taylor
expansions are very similar to Maclaurin expansions
because
Maclaurin
series actually are Taylor series centered at x
= 0.
Thus, a Taylor series
is a more generic form of the
Maclaurin series, and it
can be centered at any xvalue.
The
goal of a Taylor
expansion is
to approximate function values.
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 addup to get the Taylor approximation
approx =
sum(seq)
%
Let's compare with the official number
real_value
= cos(x)
The resuls are:
seq =
0 0.10
0
approx =
0.10
real_value =
0.29
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 n1)
for i = 0 :
n1
% Implement the Taylor
expansion
t(i+1, :) = deriv(1) * (xc).^(i) /
factorial(i);
% Find derivative for the next
term
deriv = circshift(deriv, 1);
end
%
Addup 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 asis 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)', '4term
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.
