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 " tommy integreable " (?) tommy1729 Ultimate Fellow Posts: 1,421 Threads: 346 Joined: Feb 2009 05/04/2014, 03:16 PM Consider a C^oo function that is nowhere analytic. Now any Taylor series has radius 0. And a Taylor series expanded at (any) point A is unrelated to a Taylor series expanded at a(any) different point B. In other words a (divergent) Taylor series for a C^oo function does NOT uniquely define that C^oo function. ( If this confused you remember that you could add a (C^oo) function that is flat for x<0 and increasing for x>0. If you expand your taylors at -1 and +1 ... Also f(x+v) does not work when expaned at x because the radius = 0 ) Thats intresting. But are there properties between analytic and C^oo ? If for all x,y in the domain where f is C^oo , and x,y are connected by a C^oo path : the taylor for f(x) corresponds to the Taylor for f(y) in the sense that : integral from 0 to x of f(z) dz = F(x). the Taylor of F ' (x) = the Taylor of f(x). And the same is true for any y ( x replaced by y ). Maybe this is Always the case ?? If it is not Always the case , I call this " tommy integreable ". Why, you might wonder. Well, I wanted to integrate sexp^[1/2](x) based on my 2sinh method. So that is where this is all coming from. Basicly trying to integrate C^oo functions. Any other ideas are welcome. regards tommy1729 « Next Oldest | Next Newest »

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