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+--- Thread: " tommy integreable " (?) (/showthread.php?tid=861)
" tommy integreable " (?) - tommy1729 - 05/04/2014
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