06/30/2014, 01:27 AM
Related to mike's conjecture and some of my own about derivatives ,Im intrested in a fake alternating solution.
I use the following strategy :
S(x) = exp^[0.5](x)
For all x > 1 :
- a_n x^n + (a_n + b_n) x^(n+1) < S(x)
n ln(x) + ln( -a_n + (a_n + b_n) x ) < S(ln(x))
(x=exp(X) and b_n/a_n = c_n )
n X + ln(a_n) + ln( - 1 + (1 + c_n) exp(X) ) < S(X)
We could try the same with sexp too.
regards
tommy1729
I use the following strategy :
S(x) = exp^[0.5](x)
For all x > 1 :
- a_n x^n + (a_n + b_n) x^(n+1) < S(x)
n ln(x) + ln( -a_n + (a_n + b_n) x ) < S(ln(x))
(x=exp(X) and b_n/a_n = c_n )
n X + ln(a_n) + ln( - 1 + (1 + c_n) exp(X) ) < S(X)
We could try the same with sexp too.
regards
tommy1729