Consider the half-iterate of exp(x) : \( Exp^{[\frac {1}{2}]}(x). \)
In particular we consider the real-analytic half-iterate of exp(x) such that for all real x : \( \frac{d Exp^{[\frac {1}{2}]}(x)}{dx} > 0 \) and also \( \frac{d^2 Exp^{[\frac {1}{2}]}(x)}{d^2 x} > 0 \).
So far so good. But then I get confused ...
\( \frac{d Exp^{[\frac {1}{2}]}(x=-oo)}{dx} = 0 \)
and for some 100 > y > -oo :
\( \frac{d Exp^{[\frac {1}{2}]}(x=y)}{dx} = 1 \)
SO for C in the neighbourhood of y we get that \( Exp^{[\frac {1}{2}]}(x=C). \) is approximated by the linear function
f1(x) = A + (1) x.
(A = \( Exp^{[\frac {1}{2}]}(y). \) and "x" follows from \( \frac{d Exp^{[\frac {1}{2}]}(x=y)}{dx} = 1 \) )
Now clearly \( Exp^{[\frac {1}{2}]}(x=A) = e^y \).
By analogue let \( Exp^{[\frac {1}{2}]}(x=A) = e^y \).
be approximated by the linear function f2(x) = A_2 + B x.
Now my idea was that since exp is its own derivative and composition of linear functions is simple we get :
A_2 = exp(y) and (1) B = exp(y).
HOWEVER (!!!) this implies that we have the derivative of exp(y) at both \( Exp^{[\frac {1}{2}]}(y) \) and \( Exp^{[\frac {1}{2}]}(x=A) = exp(y) \) !?
This violates the initial condition (above) that for all real x : \( \frac{d Exp^{[\frac {1}{2}]}(x)}{dx} > 0 \) and also \( \frac{d^2 Exp^{[\frac {1}{2}]}(x)}{d^2 x} > 0 \).
So this confuses me.
I had the idea this composition structure is only valid for derivatives above exp(1) but Im unable to show and understand this completely ...
I made pictures to help understand it but to my amazement that did not solve my confusion. ( pictures usually help for me )
Maybe you guys here can explain this.
regards
tommy1729
In particular we consider the real-analytic half-iterate of exp(x) such that for all real x : \( \frac{d Exp^{[\frac {1}{2}]}(x)}{dx} > 0 \) and also \( \frac{d^2 Exp^{[\frac {1}{2}]}(x)}{d^2 x} > 0 \).
So far so good. But then I get confused ...
\( \frac{d Exp^{[\frac {1}{2}]}(x=-oo)}{dx} = 0 \)
and for some 100 > y > -oo :
\( \frac{d Exp^{[\frac {1}{2}]}(x=y)}{dx} = 1 \)
SO for C in the neighbourhood of y we get that \( Exp^{[\frac {1}{2}]}(x=C). \) is approximated by the linear function
f1(x) = A + (1) x.
(A = \( Exp^{[\frac {1}{2}]}(y). \) and "x" follows from \( \frac{d Exp^{[\frac {1}{2}]}(x=y)}{dx} = 1 \) )
Now clearly \( Exp^{[\frac {1}{2}]}(x=A) = e^y \).
By analogue let \( Exp^{[\frac {1}{2}]}(x=A) = e^y \).
be approximated by the linear function f2(x) = A_2 + B x.
Now my idea was that since exp is its own derivative and composition of linear functions is simple we get :
A_2 = exp(y) and (1) B = exp(y).
HOWEVER (!!!) this implies that we have the derivative of exp(y) at both \( Exp^{[\frac {1}{2}]}(y) \) and \( Exp^{[\frac {1}{2}]}(x=A) = exp(y) \) !?
This violates the initial condition (above) that for all real x : \( \frac{d Exp^{[\frac {1}{2}]}(x)}{dx} > 0 \) and also \( \frac{d^2 Exp^{[\frac {1}{2}]}(x)}{d^2 x} > 0 \).
So this confuses me.
I had the idea this composition structure is only valid for derivatives above exp(1) but Im unable to show and understand this completely ...
I made pictures to help understand it but to my amazement that did not solve my confusion. ( pictures usually help for me )
Maybe you guys here can explain this.
regards
tommy1729