4.4. TAYLOR SERIES 157
Problem 4.76 Find a power series for:
q.x/ D
1
x
2
3x C 2
Problem 4.77 Find a power series for:
q.x/ D x
2
ln
1 Cx
3
Problem 4.78 Find a power series for:
q.x/ D
x
2
C 9
x
2
9
4.4 TAYLOR SERIES
In the last section, we managed to create power series for several of the standard transcendental
functions. Notably absent were sin.x/, cos.x/, and e
x
. e key to these is Taylor series. We
need a little added notation to build Taylor series. We will denote the nth derivative of f .x/ by
f
.n/
.x/. Notice that this means that f
.0/
.x/ D f .x/.
Knowledge Box 4.23
Taylor series
If f .x/ is a function that can be differentiated any number of times,
f .x/ D
1
X
nD0
f
.n/
.c/.x c/
n
nŠ
:
is formula is called the Taylor series expansion of f .x/ at c. e constant c is
called the center of the expansion.
Example 4.79 Use Taylor’s formula to find a power series centered at c D 0 for f .x/ D e
x
and
find its radius of convergence.
158 4. SEQUENCES, SERIES, AND FUNCTION APPROXIMATION
Solution:
e function f .x/ D e
x
is a very good choice for a first demonstration of the Taylor
expansion. is is because every derivative of of e
x
is e
x
. In other words,
f
.n/
.x/ D e
x
and so f
.n/
.0/ D 1
Applying the formula we get:
e
x
D
1
X
nD0
f
.n/
.0/.x 0/
n
nŠ
D
1
X
nD0
1 x
n
nŠ
D
1
X
nD0
x
n
nŠ
e radius of convergence of this series was computed in Example 4.56 – it is r D 1. is
means that the expansion of e
x
converges everywhere.
˙
Here is an interesting calculation. Recall Euler’s identity from Fast Start Differential Calculus:
e
ix
D i sin.x/ C cos.x/. Let’s look at the Taylor series for e
ix
:
e
ix
D
1
X
nD0
.ix/
n
nŠ
D
1
X
nD0
i
n
x
n
nŠ
D 1 C ix
1
2
x
2
i
1
3Š
x
3
C
1
4Š
x
4
C i
1
5Š
x
5
1
6Š
x
6
i
1
7Š
x
7
C
D
1
X
nD0
.1/
n
x
2n
.2n/Š
C i
1
X
nD0
.1/
n
x
2nC1
.2n C1/Š
From Euler’s identity, we get that the real part of the expression above is cosine, and the imag-
inary part is sine. is gives us power series for sin.x/ and cos.x/.
4.4. TAYLOR SERIES 159
Knowledge Box 4.24
Taylor series for sin.x/ and cos.x/ and e
x
e
x
D
1
X
nD0
x
n
nŠ
sin.x/ D
1
X
nD0
.1/
n
x
2nC1
.2n C 1/Š
cos.x/ D
1
X
nD0
.1/
n
x
2n
.2n/Š
with all three expansions having an interval of convergence of .1; 1/.
e calculations above show that we can use algebraic manipulation to create power series based
on the power series we get from Taylor expansions. Let’s do a couple more examples.
Example 4.80 Find a power series for h.x/ D
e
2x
.
Solution:
e
x
D
1
X
nD0
x
n
nŠ
e
2x
D
1
X
nD0
.2x/
n
nŠ
D
1
X
nD0
2
n
x
n
nŠ
D
1
X
nD0
2
n
nŠ
x
n
˙
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