Apologies if I am seem pedantic I programmed this in C++, and got essentially the same values for c(c(x)) plotted by rwg. For any x < 2, the evaluation of c(c(x)) ultimately ends up computing a^√2 for complex a. Unfortunate, for complex a, a^√2 = exp(√2 (log(a) + 2πi k)) has an infinitude of values (for integer k). For C++, and presumably Mma, log(a) returns the principle value with -π < Im(log(a)) <= π. This happily coincides with real log(a) for positive real a. Regarding the c(c(x)) calculation: If x >= 2, we can restrict calculations to the real realm, take the principle (real) value of a^√2, and end up with c(c(x)) = x^2 -2. If x < 2, however, a^√2 must be evaluated for non-positive-real x, and it is not clear that the principle value is appropriate in any particular calculation. I submit that, when c(c(x)) differs from x^2 - 2, we might restore c(c(x)) = x^2 -2 by using the appropriate (non-principle) value of a^√2 in the appropriate calculations, though it is unclear which value of a^√2 (if any) works for any particular evaluation of c(x).
-----Original Message----- From: math-fun [mailto:math-fun-bounces@mailman.xmission.com] On Behalf Of Bill Gosper Sent: Saturday, March 25, 2017 11:35 PM To: Wolfram Technical Support Cc: math-fun@mailman.xmission.com Subject: [math-fun] Brutal FullSimplify challenge II
In[540]:= $Version
Out[540]= "11.0.1 for Mac OS X x86 (64-bit) (September 21, 2016)"
cc[x_] := Nest[((# + √(#^2 - 4))/2)^√2 + ((# - √(#^2 - 4))/2)^√2 &, x, 2]
In[653]:=FullSimplify@cc@-Sqrt[2 (1 - Sin[√2 π])] == 2 Sin[√2 π]
Out[653]= 2^(-1 - Sqrt[ 2]) (((-Sqrt[1 - Sin[Sqrt[2] \[Pi]]] - I Sqrt[1 + Sin[Sqrt[2] \[Pi]]])^Sqrt[ 2] + (-Sqrt[1 - Sin[Sqrt[2] \[Pi]]] + I Sqrt[1 + Sin[Sqrt[2] \[Pi]]])^Sqrt[ 2] - \[Sqrt](-2^( 1 + Sqrt[2]) + (-Sqrt[1 - Sin[Sqrt[2] \[Pi]]] - I Sqrt[1 + Sin[Sqrt[2] \[Pi]]])^( 2 Sqrt[2]) + (-Sqrt[1 - Sin[Sqrt[2] \[Pi]]] + I Sqrt[1 + Sin[Sqrt[2] \[Pi]]])^(2 Sqrt[2])))^Sqrt[ 2] + ((-Sqrt[1 - Sin[Sqrt[2] \[Pi]]] - I Sqrt[1 + Sin[Sqrt[2] \[Pi]]])^Sqrt[ 2] + (-Sqrt[1 - Sin[Sqrt[2] \[Pi]]] + I Sqrt[1 + Sin[Sqrt[2] \[Pi]]])^Sqrt[ 2] + \[Sqrt](-2^( 1 + Sqrt[2]) + (-Sqrt[1 - Sin[Sqrt[2] \[Pi]]] - I Sqrt[1 + Sin[Sqrt[2] \[Pi]]])^( 2 Sqrt[2]) + (-Sqrt[1 - Sin[Sqrt[2] \[Pi]]] + I Sqrt[1 + Sin[Sqrt[2] \[Pi]]])^(2 Sqrt[2])))^Sqrt[2]) == 2 Sin[Sqrt[2] \[Pi]]
In[654]:= Chop@N@# & /@ %
Out[654]= True --Bill Gosper _______________________________________________ math-fun mailing list math-fun@mailman.xmission.com https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun