6.2 Properties of power series  (Page 4/10)

Multiplication of power series

We can also create new power series by multiplying power series. Being able to multiply two power series provides another way of finding power series representations for functions.

The way we multiply them is similar to how we multiply polynomials. For example, suppose we want to multiply

and

It appears that the product should satisfy

In [link] , we state the main result regarding multiplying power series, showing that if ${\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n}$ and ${\displaystyle \sum _{n=0}^{\infty }{d}_n{x}^n}$ converge on a common interval I , then we can multiply the series in this way, and the resulting series also converges on the interval I .

We omit the proof of this theorem, as it is beyond the level of this text and is typically covered in a more advanced course. We now provide an example of this theorem by finding the power series representation for

using the power series representations for

Differentiating and integrating power series

Consider a power series ${\displaystyle \sum _{n=0}^{\infty }{c}_n{x}^n}=c_0+c_{1}x+c_2{x}^2+\text{⋯}$ that converges on some interval I , and let $f$ be the function defined by this series. Here we address two questions about $f.$

• Is $f$ differentiable, and if so, how do we determine the derivative $f^{\prime }?$
• How do we evaluate the indefinite integral ${\displaystyle \int f\left(x\right)dx}?$

We know that, for a polynomial with a finite number of terms, we can evaluate the derivative by differentiating each term separately. Similarly, we can evaluate the indefinite integral by integrating each term separately. Here we show that we can do the same thing for convergent power series. That is, if

converges on some interval I , then

and

Evaluating the derivative and indefinite integral in this way is called term-by-term differentiation of a power series    and term-by-term integration of a power series    , respectively. The ability to differentiate and integrate power series term-by-term also allows us to use known power series representations to find power series representations for other functions. For example, given the power series for $f\left(x\right)=\frac{1}{1-x},$ we can differentiate term-by-term to find the power series for $f^{\prime }\left(x\right)=\frac{1}{{\left(1-x\right)}^2}.$ Similarly, using the power series for $g\left(x\right)=\frac{1}{1+x},$ we can integrate term-by-term to find the power series for $G\left(x\right)=\text{ln}\left(1+x\right),$ an antiderivative of g . We show how to do this in [link] and [link] . First, we state [link] , which provides the main result regarding differentiation and integration of power series.