(06/18/2014, 12:59 PM)sheldonison Wrote:(06/18/2014, 12:23 PM)tommy1729 Wrote: Then how do you explain the violation of the chain law ?
regards
tommy1729
The flaw in your proof is that k is an integer, but you state k as a real number. "sexp ' (w+k) = exp^[k] ' (sexp(w)) * sexp ' (w) = 0". For k as a fraction, the chain rule does not apply.
What ? I never saw any exceptions to the chain rule for analytic functions ?
sexp(w-1) chain law works. sexp(w+1) chain law works.
In fact the idea that the derivative of sexp relates to a product is the result of the chain law.
So you say :
1) sexp(w+k) is analytic in w,k,w+k and sexp(w) is analytic in w.
2) sexp(w+k) = exp^[k](sexp(w))
3) exp^[k] is also analytic.
4) Yet the derivative of exp^[k](sexp(w)) IS NOT exp^[k] ' (sexp(w)) * sexp ' (w) despite that all functions involved are analytic and the conditions for the chain law are fullfilled ?
Even for the logarithm the chain rule applies even though it has a singularity :
D ln(f(z)) = f ' (z)/f(z) ... = chain law !
The chain law applies to all analytic functions and even many more.
There are many proofs of the chain law and I never read " an exception " ??
Could I be so so wrong in Calc 101 ??
I can imagine going to 0 superexponentially fast but reaching it seems impossible.
Maybe the functional equation is no longer satisfied in the regions or variants of sexp that you use ? That could explain it.
I would like to note that if f ' (h) as h approaches 0 could have as limit 0 BUT that is not the derivative f ' (0).
So in other words , do not confuse the derivative and infinitesimal with the ball from analysis.
As a concrete example : a very mild singularity that is very flat.
There is something known as the Cauchy-Riemann equations.
Maybe Im making a fool of myself if im terribly wrong but calculus 101 and analysis 101 make a strong case imho.
In math one cannot simply say : Ok you have proved this but my numeric example is counterproof.
The proofs of the chain rule are the most beautiful short proofs.
regards
tommy1729