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f ' (x) = f(exp(x)) ?
#3
(04/17/2015, 02:15 AM)fivexthethird Wrote: I'm pretty sure that this function can't be analytic at the fix points of exp.
If it was, then if c is a fix point we could directly evaluate the derivatives of f at c, which are actually polynomials of c multiplied by f(c)







and in general the degree of the polynomial appears to be which obviously grows too fast to converge.

Sorry Guys but my 3rd derivative in the op was wrong as clearly shown here.

I guess these are named polynomials , not ?

Regards

Tommy1729
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Messages In This Thread
f ' (x) = f(exp(x)) ? - by tommy1729 - 04/15/2015, 12:18 PM
RE: f ' (x) = f(exp(x)) ? - by fivexthethird - 04/17/2015, 02:15 AM
RE: f ' (x) = f(exp(x)) ? - by tommy1729 - 04/17/2015, 08:32 AM
RE: f ' (x) = f(exp(x)) ? - by tommy1729 - 04/17/2015, 09:32 PM
RE: f ' (x) = f(exp(x)) ? - by tommy1729 - 04/17/2015, 09:40 PM



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