3 Differentiation of Maclaurin series

We have already noted that, by the binomial series,

1 1 − x = 1 + x + x 2 + x 3 + ⋯ x < 1

Thus, with x replaced by − x

1 1 + x = 1 − x + x 2 − x 3 + ⋯ x < 1

We have previously obtained the Maclaurin expansion of  ln ( 1 + x ) :

ln ( 1 + x ) = x − x 2 2 + x 3 3 − x 4 4 + ⋯ − 1 < x ≤ 1

Now, we differentiate both sides with respect to x :

1 1 + x = 1 − x + x 2 − x 3 + ⋯

This result matches that found from the binomial series and demonstrates that the Maclaurin expansion of a function f ( x ) may be differentiated term by term to give a series which will be the Maclaurin expansion of d f d x .

As we noted in Section 16.4 the derived series will have the same radius of convergence as the original series.

Task!

Find the Maclaurin expansion of ( 1 − x ) − 3 and state its radius of convergence.

First write down the expansion of ( 1 − x ) − 1 :

1 1 − x = 1 + x + x 2 + ⋯ x < 1

Now, by differentiation, obtain the expansion of 1 ( 1 − x ) 2 :

1 ( 1 − x ) 2 = d d x 1 + x + x 2 + ⋯ = 1 + 2 x + 3 x 2 + 4 x 3

Differentiate again to obtain the expansion of ( 1 − x ) − 3 :

1 ( 1 − x ) 3 = 1 2 d d x 1 ( 1 − x ) 2 = 1 2 2 + 6 x + 12 x 2 + 20 x 3 + ⋯ = 1 + 3 x + 6 x 2 + 10 x 3 + ⋯ Finally state its radius of convergence:

The final series: 1 + 3 x + 6 x 2 + 10 x 3 + ⋯ has radius of convergence R = 1 since the original series has this radius of convergence. This can also be found directly using the formula R = lim n → ∞ b n b n + 1 and using the fact that the coefficient of the n t h term is b n = 1 2 n ( n + 1 ) .