NCSSM Online AP Calculus (BC) Distance Education & Extended Programs Distance Education & Extended Programs    Lesson  3.9  ...  Operations  with  Power  Series       INTRODUCTION So far in this unit, we have discussed how and why we generate and analyze power series. We also have found that some Taylor series are difficult or rather complicated to find directly. In this lesson, we will extend our study of power series into differentiation and integration as well as address some handy methods for finding Taylor series. DIFFERENTIATION & INTEGRATION OF POWER SERIES Recall that a power series can be written in the form

𝑓 𝑥 = 𝑎! 𝑥 − 𝑥! !!

!!!

= 𝑎! + 𝑎! 𝑥 − 𝑥! + 𝑎! 𝑥 − 𝑥! ! + 𝑎! 𝑥 − 𝑥! ! +⋯

We recognize that the terms in the series are polynomials that are added together. From our work with derivatives earlier this year, we know that the derivative of a sum of functions is the sum of the respective derivatives of those functions. As a result, it follows that

𝑓! 𝑥 =𝑑𝑑𝑥 𝑎! 𝑥 − 𝑥! !

!

!!!

=𝑑𝑑𝑥 [𝑎! 𝑥 − 𝑥!

!]!

!!!

= 𝒂𝒌 ∙ 𝒌 𝒙− 𝒙𝟎 𝒌!𝟏!

𝒌!𝟎

Similarly, we recall that the integral of a sum of functions is equivalent to the respective integrals of those functions. This gives us the following:

𝑓 𝑥 𝑑𝑥

!

!

= 𝑎! 𝑥 − 𝑥! !!

!!!

𝑑𝑥

!

!

= [𝑎! 𝑥 − 𝑥! !]𝑑𝑥

!

!

!

!!!

=𝒂𝒌𝒌+ 𝟏 𝒙− 𝒙𝟎 𝒌!𝟏

!

𝒌!𝟎

  Let us look at a couple of examples to demonstrate this. Please see the video posted for you in Canvas for this Lesson 3.9. In that video, we will use a known Taylor series for a function to confirm the Taylor series for another function. As noted earlier, some Taylor series for functions are difficult to generate from scratch. Consider the function 𝑓 𝑥 = cos  (5𝑥!) for example. The derivatives of this function are quite cumbersome! We have some efficient (and convenient) ways to generate these types of series. Let’s look at a couple of examples.

NCSSM Online AP Calculus (BC) Distance Education & Extended Programs Distance Education & Extended Programs Example #2: Consider the function 𝑔 𝑥 = 𝑒!!!. We recall that the Maclaurin series for 𝑓 𝑥 = 𝑒! is

ex = xk

k!k=0

∞∑ = 1+ x + x

2

2!+ x

3

3!+ x

4

4!+ ... In order to find the Maclaurin series for 𝑔(𝑥), we can simply

substitute −𝑥! into our series for 𝑓 𝑥 = 𝑒!. This would be

e−x3=

−x3( )kk!k=0

∞∑ = (−1)k x3k

k!k=0

∞∑ = 1− x3 + x

6

2!− ...

Example #3: Consider the function ℎ 𝑥 = !

! about 𝑥 = 1. We can view this function as the derivative

of 𝑔 𝑥 = ln  (𝑥), a function whose Taylor series about 𝑥 = 1 we are familiar with! Recall that about

𝑥 = 1 and on its interval of convergence, ln x = (−1)k (x −1)k

kk=1

∞∑ .

Differentiating, we obtain the following:

 𝑑𝑑𝑥 ln 𝑥 =

𝑑𝑑𝑥

−1 ! 𝑥 − 1 !

𝑘

!

!!!

=𝑑𝑑𝑥

𝑥 − 1 !

𝑘

!

!!!

=𝑘 𝑥 − 1 !!!

𝑘

!

!!!

= 𝑥 − 1 !!! = 1− 𝑥 − 1 + 𝑥 − 1 ! − 𝑥 − 1 ! +⋯!

!!!