The following is a very important example known as the binomial series.
Example 11.4.1Find a Taylor series for the function
(1+ x)
^{α}centered at 0 validfor
|x|
< 1.
Use Theorem 11.2.1 to do this. First note that if y
(x)
≡
(1 + x)
^{α}, then y is a solution
of the following initial value problem.
α
y′ −------y = 0,y(0) = 1. (11.6)
(1 + x)
(11.6)
Next it is necessary to observe there is only one solution to this initial value problem. To
see this, multiply both sides of the differential equation in 11.6 by
Therefore, from 11.7, there must exist a constant, C, such that
(1+ x)−α y = C.
However, y
(0)
= 1 and so it must be that C = 1. Therefore, there is exactly one solution
to the initial value problem in 11.6 and it is y
(x)
=
(1+ x)
^{α}.
The strategy for finding the Taylor series of this function consists of finding a series
which solves the initial value problem above. Let
∞∑ n
y (x) ≡ anx (11.8)
n=0
(11.8)
be a solution to 11.6. Of course it is not known at this time whether such a series exists.
However, the process of finding it will demonstrate its existence. From Theorem 11.2.1
and the initial value problem,
∑∞ n−1 ∞∑ n
(1 + x) annx − αanx = 0
n=0 n=0
and so
∑∞ ∞∑
annxn− 1 + an(n − α)xn = 0
n=1 n=0
Changing the variable of summation in the first sum,
∞ ∞
∑ a (n+ 1)xn +∑ a (n− α)xn = 0
n=0 n+1 n=0 n
and from Corollary 11.2.2 and the initial condition for 11.6 this requires
an(α-− n-)
an+1 = n+ 1 ,a0 = 1. (11.9)
(11.9)
Therefore, from 11.9 and letting n = 0, a_{1} = α, then using 11.9 again along with this
information,
Furthermore, the above discussion shows this series solves the initial value problem on its
interval of convergence. It only remains to show the radius of convergence of this series
equals 1. It will then follow that this series equals
(1 +x )
^{α} because of uniqueness of the
initial value problem. To find the radius of convergence, use the ratio test. Thus the ratio
of the absolute values of
(n + 1)
^{st} term to the absolute value of the n^{th} term
is
showing that the radius of convergence is 1 since the series converges if
|x|
< 1 and
diverges if
|x|
> 1.
The expression,
α(α−1)⋅⋅⋅(nα!−n+1)
is often denoted as
( )
αn
. With this notation, the
following theorem has been established.
Theorem 11.4.2Let α be a real numberand let
|x|
< 1. Then
α ∑∞ ( α) n
(1 + x) = n x .
n=0
There is a very interesting issue related to the above theorem which illustrates the
limitation of power series. The function f
(x)
=
(1+ x)
^{α} makes sense for all
x > −1 but one is only able to describe it with a power series on the interval
(− 1,1)
. Think about this. The above technique is a standard one for obtaining
solutions of differential equations and this example illustrates a deficiency in the
method.
To completely understand power series, it is necessary to take a course in complex
analysis. It turns out that the right way to consider Taylor series is through the use of
geometric series and something called the Cauchy integral formula of complex analysis.
However, these are topics for another course.