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