Solution to Exercise 8.4-3 © 1999 by Karl Hahn

The problem was to construct a Taylor series for

           _
   f(x) = √x

Step 0: Turn the problem into a Maclaurin series problem. The recipe we discussed in the text suggested that you substitute variables to turn the problem into one of making a Maclaurin series. That substitution would be  u = x - 1  or equivalently  u + 1 = x.  So now you would be finding a Maclaurin series for

             _____
   g(u)  =  √u + 1

Step 1: Find the derivatives of the function. You ought to be well practiced at taking derivatives by now.

                1
   g'(u)  =         
             2√u + 1

                 -1
   g"(u)  =             
             4(u + 1)3/2

   g(3)(u)  =

       3
             
  8(u + 1)5/2

   g(4)(u)  =

      -15
              
  16(u + 1)7/2

g(n)(u) =

(-1)n-1(2n - 1)!! 2n(u + 1)n-(1/2)

where  (2n - 1)!!  indicates the product of only the odd integers from 1 to  2n - 1. But you don't need to know that just to come up with the first four terms of the Maclaurin series (you would, though, if the problem had asked for the entire Maclaurin series. Finding the pattern would have constituted step 2. So now we skip to step 3).

Step 3: Evaluate the derivatives at zero. Just plug zero in for u into g(u) and its derivatives.

   g(0)  =  1       =  A0

             1
   g'(0)  =         =
             2

  A1

               1
   g"(0)  =  -      =
               4

  A2

   g(3)(0)  =

  3
        =
  8

  A3

   g(4)(0)  =  -

 15
     =
 16

  A4

g(n)(0) =

(-1)n-1(2n - 1)!! 2n

Step 4: Put it into the Maclaurin formula. For the first five terms (I'm giving you an extra one here) will be:

g(u) ≈ A0 + A1u -

A2 2!

u2 +

A3 3!

u3 -

A4 4!

u4

Putting in the numbers from the last step (working out the factorials and cancelling common factors) you get:

g(u) ≈ 1 +

1 u - 2

1 8

u2 +

1 16

u3 -

5 128

u4

which is a truncated Maclaurin series. If you put  u = 1/2  into the above, it should yield an approximation for square root of 1.5. What it does yield is  g(1/2) ≈ 1.2241211.  The actual value for the square root of 1.5 is 1.2247449, to eight figures.

The series given above is already more than the problem asks for. But if you wanted to write a general expression in sigma form for this entire series, you would have

∞ g(u) = 1 + ∑ k=1

(-1)k-1(2k - 1)!! uk 2k k!

Again the  (2k - 1)!!,  indicates the product of the odd integers from 1 to 2k - 1. Observe that the denominator,  2^k k!,  is, when you work it out, the product of the even integers from 2 to 2k.

Recall that g'(u) has a discontinuity at  u = -1,  so the Maclaurin series above cannot have a radius of convergence greater than 1. That means you can't expect this series to work outside the domain of  -1 < u < 1.

Step 5: Substitute back. The problem asked for a Taylor series taken around  x = 1.  What we have done so far is to substitute variables and then find a Maclaurin series. So now we substitute back  u = x - 1.  That gives

f(x) ≈ 1 +

1 (x-1) - 2

1 8

(x-1)2 +

1 16

(x-1)3 -

5 128

(x-1)4

which is the Taylor series.

If you are interested in the general sigma notation for the entire series, I'll let you translate from the sigma notation given for the equivalent Maclaurin series into the Taylor series yourself. It's pretty easy to do.

By adapting the maximum possible domain in which the Maclaurin series converges you find that the Taylor series cannot converge whenever x outside the range  0 < x < 2.

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