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 Confused about d exp^[1/2](x) / dx tommy1729 Ultimate Fellow Posts: 1,493 Threads: 356 Joined: Feb 2009 04/21/2014, 10:07 PM (This post was last modified: 04/21/2014, 10:09 PM by tommy1729.) 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 « Next Oldest | Next Newest »

 Messages In This Thread Confused about d exp^[1/2](x) / dx - by tommy1729 - 04/21/2014, 10:07 PM RE: Confused about d exp^[1/2](x) / dx - by sheldonison - 04/22/2014, 08:19 PM RE: Confused about d exp^[1/2](x) / dx - by tommy1729 - 04/23/2014, 08:44 PM

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