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

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