Power Series

Definition of a Power Series

We now investigate a generalization of polynomials.  The polynomial below is a fifth degree polynomial

        3x5 - 2x4 + 5x3 + 4x - 1

if instead of a fifth degree polynomial, we consider a polynomial of infinite degree.


          Definition of a Power Series

Let f(x) be the function represented by the series


Then f(x) is called a power series function.

More generally, if f(x) is represented by the series


Then we call f(x) a power series centered at x = c.  



The power series


is centered at 0.

The power series 


is centered at -2.


The Radius of Convergence

Next we want to investigate the domain of power series.  Recall to find the domain, we ask what values of x can the function handle?  This is particularly important with power series, since infinite series often do not converge.  It would be an insurmountable task to plug in each value of x and see if the series converges for that value.  Fortunately, we have the following theorem.





is a power series centered at c then only the following three are possibilities for the domain of f.

  1. The domain is the value c only.

  2. There domain is all real numbers.

  3. There exists a real number R such that all values of x that satisfy
       | x - c |  <  R

    are in the domain and values that satisfy

            | x - c |  >  R

    are not in the domain.

R is called the radius of convergence of f.  

To compute the radius of convergence, we use the ratio test.

 Find the radius of convergence of



We use the Ratio Test:


We solve     
        |x - 3| < 2 

so that  

        1 < x < 5


        1/2(5 - 1) = 2

the radius of convergence is 2



Find the radius of convergence of


Taylor and Maclaurin Series

Since power series are functions, a natural question to ask is, "Can our everyday functions be represented as power series?"  Also, "Given a power series, can we find an everyday function that is equivalent to the power series?"

The following definition helps to answer these questions.


The Taylor Series for f(x) centered at x = c is

If c = 0, then the series is called the Maclaurin series for f.

We use the notation f (n) to denote the nth derivative of f.



Find the McLaurin Series  for

        f(x) = cos(x)


We compute:

        f(0) = 1           f '(0) = 0        f ''(0) = -1        f (3)(0) = 0

       f (4)(0) = 1      f (5)(0) = 0       f (6)(0) = -1        f (7)(0) = 0

       f (8)(0) = 1      f (9)(0) = 0        f (10)(0) = -1        f (11)(0) = 0

Hence we have the series

        1 - x2 /2+ x4/4! - x6/6!  + x8/8! - x8/8!  + ...

We see that the series is


Exercises  Find the Taylor series expansion for

  1. sin(x) centered at x = 0

  2. ln(x) centered at x = 0


Differentiation  and Integration of Power Series

Since a power series is a function, it is natural to ask if the function is continuous, differentiable or integrable.  The following theorem answers this question.


Suppose that a function is given by the power series

           f(x) =San(x - c)n

and that the interval of convergence is 

          (c - R,c + R) (plus possible endpoints)

then f(x) is continuous, differentiable, and integrable on that interval (not necessarily including the endpoints).  To obtain the derivative or the integral of f(x) we can pass the derivative or integral through the S.  In other words



Furthermore, the radius of convergence for the derivative and integral is R.


Consider the series

        f(x) = Sxn

by the GST this series converges for |x| < 1,  hence the center of convergence is 0 and the radius is 1.  By the  above theorem

        f '(x) = Snxn-1

has center of convergence 0 and radius of convergence 1 also.  We can also say that


also has center of convergence 0 and radius of convergence 1.


Show that


satisfies the differential equation

        y'' + xy' - y = 0


We can also use substitution to find power series.


Find the Maclaurin series for 

      1 - x2


Substituting x2 for x in


 we have 



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