Solution to Exercise 8.4-4


© 1999 by Karl Hahn 		KCT logo

The problem was to construct a Maclaurin series for 

            ex + e-x
   f(x)  =          
                2
that is, the hyperbolic cosine function.

There are two approaches you could have taken on this. One would be simply to observe that we already have a Maclaurin series for ex. 

                       1         1         1
   ex  =  1  +  x  +     x2  +     x3  +     x4  +  ...
                      2!        3!        4!
If you put -x in for x into that, you get 
                        1         1         1
   e-x  =  1  -  x  +     x2  -     x3  +     x4  +  ...
                       2!        3!        4!
where all the even powered terms are positive and all the odd powered terms are negative. If you add those two series together, you find that all the odd powered terms cancel and all the even powered terms double up 
                       2         2         2
   ex + e-x  =  2  +     x2  +     x4  +     x6  +  ...
                      2!        4!        6!
Now simply divide the 2 out of each term and you have the Maclaurin series for hyperbolic cosine.

The other way to do it is to run through the steps as you learned them in the main text.

Step 1: Find the derivatives of the function. 

             ex - e-x
   f'(x)  =          
                 2

             ex + e-x
   f"(x)  =          
                 2
Since the second derivative brings this one back to the original function, you can expect that this pattern of these two functions will continue into the higher derivatives indefinitely. Since we have found the pattern, we have completed step 2 of this one, so we proceed to step 3.

Step 3: Evaluate the function and its derivatives at zero. Since  e0 = 1,  we get 

   f(0)      =  1  =  A0

   f'(0)     =  0  =  A1

   f"(0)     =  1  =  A2

   f(3)(0)    =  0  =  A3
And you should be able to see why the pattern of zero on the odd derivatives and 1 on the even derivatives continues indefinitely.

Step 4:Apply the Maclaurin Formula. The problem asks only for the first four nonzero terms. That would be

   ex + e-x
             ≈
       2

         1
  1  +    
        2!


 x2  +


   1
    
  4!


 x4  +


   1
    
  6!


 x6
The sigma notation for the entire series would be

   ex + e-x
             =
       2

   ∞
   ∑
  k=0

    1
       
  (2k)!


 x2k
Do you expect that the hyperbolic cosine is an even function, an odd function, or neither?

The hyperbolic cosine function is abbreviated by the notation,  cosh(x).  Would it surprise you that there is also a hyperbolic sine function as well? Its notation is  sinh(x).  

               ex + e-x
   cosh(x)  =          
                   2

               ex - e-x
   sinh(x)  =          
                   2
Observe that the  cosh(x)  function is the derivative of the  sinh(x)  function and vice versa. And not surprisingly, the Maclaurin series for  sinh(x)  is

                      1
   sinh(x)  =  x  +    
                     3!


 x3  +


   1
    
  5!


 x5  +


   1
    
  7!


 x7  +  ...
which you can gather simply by taking the term-by-term derivative of the Maclaurin series for  cosh(x).  In sigma notation the Maclaurin series for  sinh(x)  is

                ∞
   sinh(x)  =   ∑
               k=0

      1
           
  (2k + 1)!


 x2k+1
Do you expect that hyperbolic sine is an even function, an odd function, or neither?

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