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4.6. Representation of Functions As Power Series 99

1) Power series can be used to represent functions by replacing variables in known power series expressions. For example, replacing x with -2x^2 in the power series for 1/(1-x) yields a power series for 1/(1+2x^2). 2) The derivatives and integrals of power series can be obtained by term-by-term differentiation and integration of the individual terms. This results in new power series with the same radius of convergence. 3) Several examples are given of finding power series representations for functions by taking derivatives or integrals of known power series, including power series for 1/(1-x)^2, tan^-1(x), and ln(1+x).

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50 views3 pages

4.6. Representation of Functions As Power Series 99

1) Power series can be used to represent functions by replacing variables in known power series expressions. For example, replacing x with -2x^2 in the power series for 1/(1-x) yields a power series for 1/(1+2x^2). 2) The derivatives and integrals of power series can be obtained by term-by-term differentiation and integration of the individual terms. This results in new power series with the same radius of convergence. 3) Several examples are given of finding power series representations for functions by taking derivatives or integrals of known power series, including power series for 1/(1-x)^2, tan^-1(x), and ln(1+x).

Uploaded by

FernandoFierroGonzalez
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
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Download as PDF, TXT or read online on Scribd
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4.6.

REPRESENTATION OF FUNCTIONS AS POWER SERIES 99

4.6. Representation of Functions as Power Series

We have already seen that a power series is a particular kind of


function. A slightly different matter is that sometimes a given function
can be written as a power series. We already know the example
1 X∞
2 3 n
= 1 + x + x + x + ··· + x + ··· = xn (|x| < 1)
1−x n=0

Replacing x with other expressions we may write other functions in the


same way, for instance by replacing x with −2x2 we get:
1 X∞
2 4 6 n n 2n
= 1−2x +4x −8x +· · ·+(−1) 2 x +· · · = (−1)n 2n x2n ,
1 + 2x2 n=0
√
which converges for | − 2x2 | < 1, i.e., |x| < 1/ 2.

4.6.1. Differentiation and Integration of Power Series. Since


the sum of a power series is a function we can differentiate it and in-
tegrate it. The result is another function that can also be represented
with another power series. The main related result is that the deriv-
ative or integral of a power series can be computed by term-by-term
differentiation and integration:
4.6.1.1.
P Term-By-Term Differentiation and Integration. If the power
series ∞ n=0 c n (x − a)n
has radius of convergence R > 0 then the func-
tion
X
∞
f (x) = cn (x − a)n
n=0
is differentiable on the interval (a − R, a + R) and

X
∞ X
∞
(1) f 0 (x) = {cn (x − a)n }0 = ncn (x − a)n−1
n=0 n=1
Z ∞ Z
X X
∞
(x − a)n+1
(2) f (x) dx = cn (x − a)n dx = C + cn
n=0 n=0
n+1

The radii of convergence of the series in the above equations is R.


Example: Find a power series representation for the function
1
f (x) = .
(1 − x)2
4.6. REPRESENTATION OF FUNCTIONS AS POWER SERIES 100

Answer : We have
1 d 1
=
(1 − x)2 dx 1 − x
and
1 X ∞
= xn ,
1 − x n=0
hence
d X n X d n X n−1
∞ ∞ ∞
1
= x = x = nx
(1 − x)2 dx n=0 n=0
dx n=1

X
∞
2 3
= 1 + 2x + 3x + 4x + · · · = (n + 1)xn (re-indexed)
n=0

The radius of convergence is R = 1.


Example: Find a power series representation for tan−1 x.
Answer : That function is the antiderivative of 1/(1 + x2 ), hence:
Z
−1 1
tan x = dx
1 + x2
Z X ∞
= (−1)n x2n dx
n=0
∞ Z
X
= (−1)n x2n dx
n=0
X∞
x2n+1
=C+ (−1)n
n=0
2n + 1
x3 x5 x7
=C +x− + − + ...
3 5 7
Since tan−1 0 = 0 then C = 0, hence

X
∞
x2n+1 x 3 x5 x 7
−1
tan x= (−1)n =x− + − + ...
n=0
2n + 1 3 5 7

The radius of convergence is R = 1.

Example: Find a power series representation for ln (1 + x).


4.6. REPRESENTATION OF FUNCTIONS AS POWER SERIES 101

Answer : The derivative of that function is 1/(1 + x), hence


Z
1
ln (1 + x) = dx
1+x
Z X ∞
= (−1)n xn dx
n=0
∞ Z
X xn+1
=C+ (−1)n dx
n=0
n+1
x2 x3 x4
=C +x− + − + ···
2 3 4
Since ln 1 = 0 then C = 0, so
x2 x3 x4 X ∞
xn
ln (1 + x) = x − + − + ··· = (−1)n+1
2 3 4 n=1
n

The radius of convergence is R = 1.

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