11.4. THE BINOMIAL THEOREM 249

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 tothe 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 serieswhich solves the initial value problem above. Let

y(x)≡∞

∑n=0

anxn (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 11.2.1and 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 11.2.2 and the initial condition for 11.6 this requires

an+1 =an (α −n)

n+1,a0 = 1. (11.9)

Therefore, from 11.9 and letting n = 0, a1 = α, then using 11.9 again along with thisinformation,

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.