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f ' (x) = f(exp(x)) ?
#2
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.
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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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