Confused about d exp^[1/2](x) / dx tommy1729 Ultimate Fellow Posts: 1,480 Threads: 354 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 sheldonison Long Time Fellow Posts: 679 Threads: 24 Joined: Oct 2008 04/22/2014, 08:19 PM (This post was last modified: 04/23/2014, 05:07 PM by sheldonison.) (04/21/2014, 10:07 PM)tommy1729 Wrote: .... 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. ... f2(x) = A_2 + B x. ... 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 $\frac{d^2 Exp^{[\frac {1}{2}]}(x)}{d^2 x} > 0$. For Kneser's half iterate, $C \approx 0.241209414612060, \text{half}( C ) \approx 0.724688727528462$, where C is defined as the point with first derivative=1. The half iterate in the neighborhood of C $ f1(0.241209414612060+x)\approx 0.724688727528462 + x + 0.265039014265587x^2 + 0.0238149626989277x^3 -0.000607188663583442x^4$ The half iterate in the neighborhood of H, where we define H as the half iterate of C, $ f2(0.724688727528462 +x)\approx 1.27278754771200+ 1.27278754771200x+ 0.299055416840898x^2 + 0.0232971641989875x^3+ 0.000030326321314053x^4$ Notice that $\exp( C )\approx 1.27278754771200$, which matches both the value and the first derivative of f2, just as the OP has shown. I don't see any obvious contradictions in this data, concerning the second derivative, nor any reason to believe this shows the second derivative is less than zero at some points. I think the Op got his variables mixed up, and mistakenly believes sometimes that y=C, which is different than A, and other times that y=H=A? Here is a graph of the half iterate of x varying from -5 to +5, and the first and second derivatives. The half iterate is the red line, varying from -0.68423 to 14.143. The green line is the first derivative, and the second red line is the second derivative. It would appear from the graph that the second derivative is always greater than zero, and always increasing. Presumably, the third derivative could always be positive, but it is not always increasing since we know that the fourth derivative is sometimes less than zero, from f1. Also, note that H is clearly visible as the intersection of the half iterate and the first derivative of the half iterate.     The 3rd and 4th derivatives are not as well behaved as lower derivatives. Notice how small these 3rd and 4th derivatives of the half iterate of exp are, with the y axis varying from -0.0012, to +0.034. We know that the the half iterate has a singularity at the fixed point of exp, $L \approx 0.318\pm 1.337i$, so it should not be surprising that the higher derivatives misbehave but I have not graphed or investigated the derivatives for the half iterate of sexp before today. I think the branch singularity at L is fairly mild. The derivatives of the Kneser sexp(z) function, from which the half iterate was generated via sexp(slog(z)+1/2), are very well behaved, but there are no singularities in the complex plane for sexp(z), only at the real axis.     Update: The half iterate also has a singularity at -0.36237+iPi, which corresponds to sexp(-2.5). It may be that this more distant singularity has more impact on the lower derivatives than the singularity at L, since the singularity at L is locally quiet, with $\exp^{0.5}(L)=L$. See this relevant previous thread, which includes a complex plane plot of the half iterate of exp, and a discussion of the singularities. http://math.eretrandre.org/tetrationforu...4&pid=5401 tommy1729 Ultimate Fellow Posts: 1,480 Threads: 354 Joined: Feb 2009 04/23/2014, 08:44 PM (This post was last modified: 04/23/2014, 08:45 PM by tommy1729.) Thank you Sheldon for clarifying my confusion. Im sorry to have posted such a mistake. Guess I ate too many eggs for easter or so The data you provide seems correct. Also thank you for the link. regards tommy1729 « Next Oldest | Next Newest »