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A function which doesn't equal its Taylor series, part 2

A more sophisticated example than the last one is the following. Define f(x) by


Now, f(x) is continuous at x=0, since as tex2html_wrap_inline451 , tex2html_wrap_inline453 , so tex2html_wrap_inline455 . f also has a derivative at x=0:


which you can show goes to zero by using L'Hôpital's rule on tex2html_wrap_inline463 . In fact, using mathematical induction and a considerable amount of work, you can show that all of f's derivatives exist at 0 and are equal to zero.

This says that f has a very nice Taylor expansion around the origin. Since tex2html_wrap_inline469 for all n, the Taylor series around the origin is simply a sum of zeroes, so it is identically zero. However, this equals f(x) only at the point x=0! Here's a picture of f. Notice that f becomes very flat at the origin. However, it only equals 0 when x=0, since e raised to any finite power is positive.


Tom Vogel
Mon May 5 12:53:33 CDT 1997