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 Creating Power Series From Functions 
     The  Geometric Power Series 
     
    Recall that  
     
            
    
     
    
     
   
    Substituting  x for r, we have  
   
    
            
     
     
   
    We write 
     
     
     
  
     
 Milking the Geometric Power Series
 By using substitution, we can obtain power series expansions
from the geometric series. Example 1   
 
Substituting  x2  for x, we have  
 
            
 
 
 
 Example 2   
 
Multiplying by  x we have
            
    
     
     
    
 Example 3   
 
Suppose we want to find the power series
for                           
1 
        f(x)  =                     
                     
2x - 3 
  centered at  x =
4.  We rewrite the function as                     
1                           
1 
                                   
=                      
          2(x - 4) + 8 -
3          2(x - 4) + 5 
 
        
 
 
 Example 4   
 
Substituting  -x for x, we have 
 
        
 
 
 Example 5   
 
Substituting  x2  in for  x in
the previous example, we have 
 
        
 
 Example 6   
 
Taking the integral of the previous
example, we have 
 
            
     
     
 Exercise  Find the power series that represents the
following functions: 
     
      - 
ln(1 + x) 
        
      - 
  tanh-1x 
        
      - 
  -(1 - x)-2  
        
     
 
Integrating Impossible Functions
      
     
    We can use power series to integrate functions where there are no
    standard techniques of integration available. 
     
     
    Example:   
     
Use power series to
     find the integral 
     
     
    
            
     
     
     
    Then use this integral to approximate  
     
            
 
  
     
     
    Solution:    
     
 Notice that this is a very difficult integral to solve.  We resort to power series.  First we use the
    series expansion from Example 6, replacing  x with
     x2.   
     
    
            
     
     
     
    Integrating we arrive at the solution  
     
          
 
 
     
    
    Now to solve the definite integral, notice that when we plug in 0 we get
0,
    hence the definite integral is 
     
    
            
     
    
     
    
    Using the first  5 terms to approximate this we get  0.300 
     
    Notice that the error is less than the next term (which comes from  x23/253) 
     
           
    E < 1/253  =  .004.
  
  
 
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