Example 3.  Consider the function  .  
3 (a).  Find the terms up to    in the Maclaurin series for  f[x].
Solution 3 (a).

3 (a).  Find the terms up to    in the Maclaurin series for  f[x].

Remark.  If you just find the "Series" it will include a "Big "  term, which cannot be used in either evaluations or graphing, we eliminate it with the command "Normal."  The "Big   term lets us know the power of x in the "remainder.

Aside.  Of course the "full" Maclaurin series has an infinite number of terms.  Mathematica is capable of finding sums of infinite series.  

The "symbolic" command is:

Use Mathematica to sum the infinite series  .
Check it out for  .

Check it out for  .

What is happening to the terms in the last series ?

The n-th term does not go to zero, therefore     diverges.

Do we really need to understand the "interval of convergence" ?  What is it ?

The "interval of convergence" is [-1,1).  Look at what happens in this interval ?  

Aside.  The logarithm of negative numbers is allowed in the study of complex analysis.  For example

This leads to one of the most curious equations in mathematics  ,  because it involves many of our basic numbers  .  You are welcome to experiment with complex numbers in this course.  For example, Newton's method will converge to those "complex roots" if you give it an initial starting value that is complex and reasonably close to the desired root.

Background for part 3 (b).  Use the fact that the series is "alternating" to investigate the error for the Maclaurin polynomial of degree n = 10 over the interval  [-0.5, 0.5].  

Now look closely at the "error" when the series is used to approximate the function.  It becomes infinite near x = 1.  For this reason we work on smaller intervals, in this case we chose  [-0.5, 0.5].  How close are were the two curves in part (a) ?

Caveat. The series is not alternating and we cannot use the magnitude of the "next term" in the series to determine an error bound.

The error bound for the entire interval  [-0.5, 0.5] is

However, the estimate does not work on the interval    where the series is not alternating.

However, the estimate will work on the interval   where the series is alternating.

(c) John H. Mathews 2004