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Old 2008-11-24, 10:22   #3
tmorrow
 
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Jan 2004

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Kevin showed the way, here's an example I made up.

f(x) = sum{1,inf} x^n/[n(n+1)] ...(1)...

Using log(1-x) = - sum{1,inf} on |x|<r=1 and the fact that 1/n(n+1) = 1/n - 1/(n+1), the above power series can be expressed in terms of familiar functions:

f(x) = (1/x)(1-x)log(1-x) + 1 on |x|<1, x!=0 (x=0 works in the limit x->0) ...(2)...

Formal differentiation of the power series (1) gives:

f'(x) = sum{1,inf} x^(n-1)/(n+1) ...(3)...
which clearly diverges at x=1 by comparison with the harmonic series.

Routine differentiation of our elementary form (2) gives (after a tidy up)

f'(x) = -(1/x)[1+(1/x).log(1-x)] ...(4)...

From (4) it is clear that we are seeing the same difficulty with non differentiability at x=1. The equality between (3) and (4) can be easily established as well.

Using the above approach students can relate the problems of formal power series differentiation with that of routine differentiation of elementary functions and illustrates a power series can represent poorly behaved functions (a_n may look well behaved on the surface but looks are deceiving). I'm sure simpler function examples can be found if you look hard enough.
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