Advanced Calculus Single Variable

8.4 The Binomial Theorem

The following is a very important example known as the binomial series.

Use Theorem 8.2.1 to do this. First note that if y

≡^α, then y is a solution of the following initial value problem.

y′ −---α--y = 0,y(0) = 1. (8.14) (1+ x)

(8.14)

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 8.14 by

^−α. When this is done, one obtains

d ( −α ) −α ( α ) dx- (1+ x) y = (1+ x) y′ − (1+-x)y = 0. (8.15)

(8.15)

Therefore, from 8.15, there must exist a constant, C, such that

−α (1+ x) y = C.

However, y

= 1 and so it must be that C = 1. Therefore, there is exactly one solution to the initial value problem in 8.14 and it is y = ^α.

The strategy for finding the Taylor series of this function consists of finding a series which solves the initial value problem above. Let

∞∑ y(x) ≡ anxn (8.16) n=0

(8.16)

be a solution to 8.14. 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 8.2.1 and the initial value problem,

∞∑ n−1 ∑∞ n (1 +x ) annx − αanx = 0 n=0 n=0

and so

∞∑ n−1 ∞∑ n annx + an(n− α) x = 0 n=1 n=0

Changing the variable of summation in the first sum,

∑∞ ∑∞ an+1(n + 1)xn + an (n − α)xn = 0 n=0 n=0

and from Corollary 8.2.2 and the initial condition for 8.14 this requires

a (α− n) an+1 = -n------,a0 = 1. (8.17) n+ 1

(8.17)

Therefore, from 8.17 and letting n = 0, a₁ = α, then using 8.17 again along with this information,

a2 = α-(α-−-1). 2

Using the same process,

( α(α−1)) ----2----(α-−-2) α(α−-1)(α-−-2) a3 = 3 = 3! .

By now you can spot the pattern. In general,

◜---n-of thes◞e◟ factors-◝ α-(α-−-1)⋅⋅⋅(α-−-n+-1) an = n! .

Therefore, the candidate for the Taylor series is

∑∞ α-(α−-1)⋅⋅⋅(α−-n-+-1) n y(x) = n! x . n=0

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

^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 ^st term to the absolute value of the n^th term is

| | ||α(α−1)⋅⋅⋅(α−-n+1)(α−n)|||x|n+1 ----|--(n+1)n!----|-------= |x| |α−-n| → |x| ||α(α−1)⋅⋅n⋅(!α−n+1)|||x|n n+ 1

showing that the radius of convergence is 1 since the series converges if

< 1 and diverges if > 1.

The expression,

is often denoted as . With this notation, the following theorem has been established.

There is a very interesting issue related to the above theorem which illustrates the limitation of power series. The function f

= ^α makes sense for all x > −1 but one is only able to describe it with a power series on the interval . 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.