This is motivated by a calculation using Fourier transforms of filters / mollifiers. Since it is such a classical question, I suspect there is a classical answer out there and would appreciate help satisfying my curiosity.
Dear Bill,
2 examples: 1/(1+x^2) --> coefficients are alternating
1/(1-x^2) --> c.'s are not alternating
Best regards,
Friedhelm
Dear Bill,
Generalizing F. Schieweck example you can see that if the function can be written in the form
f(x)=1/(M-g(x)) ,
where g(x) is a positive function less than M>0 then its power series is alternating.
Otherwise, in terms of Taylor series, if the sequence of n- th derivatives f(a),f'(a),f''(a),...
is alternating at some point a, then the Taylor series will be alternating, if it is convergent.
Dear Bill,
another example is: cos(x) has alternating coefficients.
If f(x) is an even function, that is, f(x) = f(- x), then f ' (x ) , f ''' (x ), fv (x) etc..will be odd functions and will be zeros at x=0. So power series or McLaurin's series will have only even degree terms, that is,
f(x) = a0 + a2 x2 + a4 x4 + ............ For alternating sign f 2m (x) must have alternate signs at x = 0. For example f(x) = cos x .
Dear Bill:
If I understand it correctly, you are expanding a function f in Taylor series at a=0, which is natural, as the function is even. As such, all odd-order derivatives of f at 0 must be zero. Thus, we deal only with f^(2k)(0) and this sequence should alternate. That's it. It seems not to exist any reasonable classification of functions with this property.
All the best,
Roman
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