Power Series

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.

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

        \( f(x) = \sum_{n=0}^{\infty} a_n (x - c)^n \)

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

Examples

The power series

        \( \sum_{n=0}^{\infty} \frac{x^n}{n+1} = 1 + \frac{x}{2} + \frac{x^2}{3} + \frac{x^3}{4} + ... \)

is centered at 0.

The power series 

        \( \sum_{n=0}^{\infty} \frac{(-1)^n 2^n (x+2)^n}{n!} = 1 - 2(x+2) + 2(x+2)^2 - \frac{4(x+2)^3}{3} + ... \)

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.

Theorem

If 

        \( f(x) = \sum_{n=0}^{\infty} a_n (x - c)^n \)

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.


Example:  Find the radius of convergence of    

        \( \sum_{n=0}^{\infty} \frac{(x - 3)^n}{2^n} \)

Solution

We use the Ratio Test:

         \( = \lim\limits_{n \to \infty} | \frac{(x-3){n+1}}{2^{n+1}}\frac{2^n}{(x-3)^n}| = |\frac{x-3}{2}| \)

We solve     
         \( |\frac{x-3}{2}| < 1 \)
        
or 
        |x - 3| < 2 

so that  

        1 < x < 5

Since 

        1/2(5 - 1) = 2

the radius of convergence is 2

Exercise:  

Find the radius of convergence of

        \( \sum_{n=0}^{\infty} (x - 2)^n \)

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.

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

Example:

Find the McLaurin Series  for

        f(x) = cos(x)

Solution

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

         \( cos(x) = \sum_{n=0}^{\infty} \frac{(-1)^n x^{2n}}{(2n)!} \)

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.

Example:

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) = Snx^n-1

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

                \( \int f(x) dx = \sum_{n=0}^{\infty} \frac{x^{n+1}}{n+1} \)

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

Exercise:

Show that

                \( \sum_{n=0}^{\infty} \frac{x^{2n}}{2^n n!} \)

satisfies the differential equation

        y'' + xy' - y = 0

We can also use substitution to find power series.

Example   

Find the Maclaurin series for 

          1
                       
      1 - x²

Solution

Substituting x² for x in

       \( \frac{1}{1 - x} = \sum_{n=0}^{\infty} x^n \)

 we have 

       \( \frac{1}{1 - x^2} = \sum_{n=0}^{\infty} (x^2)^n = \sum_{n=0}^{\infty} x^{2n} = 1 + x^2 + x^4 + ... \)

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