06/16/2014, 10:57 PM
It seems sexp ' (T) = 0 implies sexp ' (T+1) = 0.
because sexp(T+1) = exp(sexp(T))
sexp ' (T+1) = exp ' (sexp(T)) * sexp ' (T) = 0
( chain rule used )
Notice T+1+theta(T+1) = T + 1 + theta(T).
Nice.
regards
tommy1729
because sexp(T+1) = exp(sexp(T))
sexp ' (T+1) = exp ' (sexp(T)) * sexp ' (T) = 0
( chain rule used )
Notice T+1+theta(T+1) = T + 1 + theta(T).
Nice.
regards
tommy1729

