Box 8.4-2: Solution to Coached ExerciseKCT logo

© 1999 by Karl Hahn

Step 1: Finding the first three derivatives of 

              1
   f(x)  =       
            x + 1
You can see that this is a composite of  g(x) = x + 1  with  h(g) = 1/g = g-1.  Following the chain rule you find that  g'(x) = 1,  and  h'(g) = -1/g2 = -g-2.  So 
                                 -1
   f'(x)  =  h'(g(x))g(x)  =          
                              (x + 1)2
You should be able to continue to apply the chain rule to arrive at 
                 2
   f"(x)  =          
             (x + 1)3
and 
                  -6
   f(3)(x)  =          
               (x + 1)4
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Step 2: As you multiply successive negative numbers, you get (-1),   (-1)×(-2),   (-1)×(-2)×(-3),   (-1)×(-2)×(-3)×(-4),  and so on. If you factor out all the (-1)'s, you get, for the nth number in this series, (-1)n×n!. Hence to find the nth derivative of f(x), you will need this as a coefficient. And since the power in the denominator increases by 1 with each derivative, you will be taking the nth power in the denominator. So 

                (-1)n n!
   f(n)(x)  =            
               (x + 1)n+1
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Step 3: You have 

              1
   f(x)  =       
            x + 1
Put in zero for x and the denominator is 1. So  f(0) = 1 = A0. 

And you have 

                -1
   f'(x)  =          
             (x + 1)2
Put in zero for x and the denominator is still 1, and 1 raised to any power is still 1. So  f'(0) = -1 = A1. 

And you have 

                 2
   f"(x)  =          
             (x + 1)3
so  f"(0) = 2 = A2. 

And you have 

                  -6
   f(3)(x)  =          
               (x + 1)4
so  f(3)(0) = -6 = A3. 

And in general you have 

                (-1)n n!
   f(n)(x)  =            
               (x + 1)n+1
so f(n)(0) = (-1)n n! = An.

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Step 4: From the previous step you have that 

   Ak  =  (-1)k k!
So plug that into 
             ¥       1
   f(x)  =   å   Ak    xk
            k=1     k!
which is the Maclaurin formula. Substituting gives you 
             ¥      (-1)k
   f(x)  =   å   k!       xk
            k=0       k!
Now cancel the k!'s. 
             ¥   (-1)k
   f(x)  =   å         xk
            k=0    k!
Which is the Maclaurin series for 
              1
   f(x)  =       
            x + 1
If you wanted to write the series without the sigma notation it would be 
   f(x)  =   lim    1  -  x  +  x2  -  x3  +  x4  -  ...  ±  xn
            n  > ¥

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