8.6. THE BINOMIAL THEOREM 167

Example 8.6.1 Find a Taylor series for the function (1+ x)α centered at 0 valid for |x|< 1.

Use Theorem 8.2.1 to do this. First note that if y(x)≡ (1+ x)α , then y is a solution ofthe following initial value problem.

y′− α

(1+ x)y = 0, y(0) = 1. (8.16)

Next it is necessary to observe there is only one solution to this initial value problem. Tosee this, multiply both sides of the differential equation in 8.16 by (1+ x)−α . When this isdone, one obtains

ddx

((1+ x)−α y

)= (1+ x)−α

(y′− α

(1+ x)y)= 0. (8.17)

Therefore, from 8.17, 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 initialvalue problem in 8.16 and it is y(x) = (1+ x)α .

The strategy for finding the Taylor series of this function consists of finding a serieswhich solves the initial value problem above. Let y(x) ≡ ∑

∞n=0 anxn be a solution to 8.16.

Of course it is not known at this time whether such a series exists. However, the process offinding it will demonstrate its existence. From Theorem 8.2.1 and the initial value problem,(1+ x)∑

∞n=0 annxn−1−∑

∞n=0 αanxn = 0 and so

∞

∑n=1

annxn−1 +∞

∑n=0

an (n−α)xn = 0

Changing the variable of summation in the first sum,

∞

∑n=0

an+1 (n+1)xn +∞

∑n=0

an (n−α)xn = 0

and from Corollary 8.2.2 and the initial condition for 8.16 this requires

an+1 =an (α−n)

n+1,a0 = 1. (8.18)

Therefore, from 8.18 and letting n = 0, a1 = α, then using 8.18 again along with this

information, a2 =α(α−1)

2 . Using the same process, a3 =

(α(α−1)

2

)(α−2)

3 = α(α−1)(α−2)3! . By

now you can spot the pattern. In general,

an =

n of these factors︷ ︸︸ ︷α (α−1) · · ·(α−n+1)

n!.

Therefore, the candidate for the Taylor series is

y(x) =∞

∑n=0

α (α−1) · · ·(α−n+1)n!

xn.

Furthermore, the above discussion shows this series solves the initial value problem on itsinterval of convergence. It only remains to show the radius of convergence of this series