03/01/2016, 10:38 PM
In principle a strategy is simple.
Show D^n ln(f(exp(x))) is close to D^n f(x) for all n > -1.
Write f(x) in the product form prod_i (1 + x^i / a_i) and use log(a b) = log a + log b.
So we get ( ignoring radius , continuation , summability ,...)
D^n F(x) ~~ D^n Sum_i ln( 1 + exp( i x) / a_i).
Use Taylor for ln(1 + x^t) and ln(x^3) = 3 ln(x) etc.
Should work.
In principle ... In theory ...
However in practice this is near impossible !?
See here at Gottfried's paper
http://go.helms-net.de/math/musings/drea...quence.pdf
Also posted on MSE by our friend Mick
http://math.stackexchange.com/questions/...of-lnfexpx
Regards
Tommy1729
Show D^n ln(f(exp(x))) is close to D^n f(x) for all n > -1.
Write f(x) in the product form prod_i (1 + x^i / a_i) and use log(a b) = log a + log b.
So we get ( ignoring radius , continuation , summability ,...)
D^n F(x) ~~ D^n Sum_i ln( 1 + exp( i x) / a_i).
Use Taylor for ln(1 + x^t) and ln(x^3) = 3 ln(x) etc.
Should work.
In principle ... In theory ...
However in practice this is near impossible !?
See here at Gottfried's paper
http://go.helms-net.de/math/musings/drea...quence.pdf
Also posted on MSE by our friend Mick
http://math.stackexchange.com/questions/...of-lnfexpx
Regards
Tommy1729