Young friend Rohan demanded the sin Fourier coefficients of Warut's non-rational-function: Out[439]= -8 Mod[k, 2]/(k^3 Pi^3) . So (-1)^Floor[t]*Mod[Abs[t], 1]*(Mod[Abs[t], 1] - 1) == Sum[% Sin[k*\[Pi]*t], {k, \[Infinity]}] Out[456]= (-1)^Floor[t] (-1 + Mod[Abs[t], 1]) Mod[Abs[t], 1] == ( 2 I (PolyLog[3, -E^(-I \[Pi] t)] - PolyLog[3, E^(-I \[Pi] t)] - PolyLog[3, -E^(I \[Pi] t)] + PolyLog[3, E^(I \[Pi] t)]))/\[Pi]^3 a sum of four trilogs, now defined for complex t due to the loss of (-1)^Floor[t]. The even Fourier terms vanish, so manually bisecting the series, In[443]:= Assuming[k \[Element] Integers, Simplify[%439 /. k -> 2 k - 1]] Out[443]= -8 Sin[(-1 + 2 k) Pi t]/((-1 + 2 k)^3 Pi^3) gives yet another expression: In[444]:= Sum[%, {k, \[Infinity]}] Out[444]= (I E^(-I \[Pi] t) (-LerchPhi[E^(-2 I \[Pi] t), 3, 1/2] + E^(2 I \[Pi] t) LerchPhi[E^(2 I \[Pi] t), 3, 1/2]))/(2 \[Pi]^3) Trying a Gaussian rational, In[505]:= Rationalize[N[%444 /. t -> 3/5 + 4 I/5, 33]] Out[505]= -22/25 + 4 I/25 so we seem to have a non-rational-function preserving Gaussian rationals. With pi^3 in its denominator. But even with t declared real, Mathematica is unable to relate the three equivalent expressions. On 2016-02-10 05:03, rwg wrote:
(Sniff) but these are all _piece_wise rational. How about not even that? gosper.org/baz.png The continuous function defined on the triadic rationals to satisfy In[88]:= Clear@baz; baz[0] = 0; baz[1] = 1; baz[x_] := baz[3 x]/2 /; x <= 1/3; baz[x_] := 1/2 + baz[3 x - 2]/2 /; x >= 2/3; baz[x_] := 1/2 + baz[6 x - 2]/2 /; 1/3 <= x <= 1/2; baz[x_] := 1/2 + baz[4 - 6 x]/2 /; 1/2 <= x <= 2/3
In[89]:= baz /@ Range[0, 1, 1/9]
Out[89]= {0, 1/4, 1/4, 1/2, 3/4, 3/4, 1/2, 3/4, 3/4, 1}
I.e., baz maps the triadic rationals onto dyadic rationals. That it maps rationals to rationals follows from the finite state preservation of periodicity. E.g., inactivating baz: Clear@ibaz; ibaz@0 = 0; ibaz@1 = 1; ibaz[x_] := Inactive[ibaz][3 x]/2 /; x <= 1/3; ibaz[x_] := 1/2 + Inactive[ibaz][3 x - 2]/2 /; x >= 2/3; ibaz[x_] := 1/2 + Inactive[ibaz][6 x - 2]/2 /; 1/3 <= x <= 1/2; ibaz[x_] := 1/2 + Inactive[ibaz][4 - 6 x]/2 /; 1/2 <= x <= 2/3
Then In[472]:= Inactive[ibaz][1/4]
Out[472]= Inactive[ibaz][1/4]
In[473]:= Activate@%
Out[473]= 1/2 Inactive[ibaz][3/4]
In[474]:= Activate@%
Out[474]= 1/2 (1/2 + 1/2 Inactive[ibaz][1/4])
In[475]:= Solve[% == %%%, %%%]
Out[475]= {{Inactive[ibaz][1/4] -> 1/3}}
a nondyadic rational. But, ironically, triadic. Less trivially,
Out[487]= Inactive[ibaz][7/22]
In[488]:= Activate@%
Out[488]= 1/2 Inactive[ibaz][21/22]
In[489]:= Activate@%
Out[489]= 1/2 (1/2 + 1/2 Inactive[ibaz][19/22])
In[490]:= Activate@%
Out[490]= 1/2 (1/2 + 1/2 (1/2 + 1/2 Inactive[ibaz][13/22]))
In[491]:= Activate@%
Out[491]= 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 Inactive[ibaz][5/11])))
In[492]:= Activate@%
Out[492]= 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 Inactive[ibaz][8/11]))))
In[493]:= Activate@%
Out[493]= 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 Inactive[ibaz][2/11])))))
In[494]:= Activate@%
Out[494]= 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/4 Inactive[ibaz][6/11])))))
In[495]:= Activate@%
Out[495]= 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/2 (1/2 + 1/4 (1/2 + 1/2 Inactive[ibaz][8/11]))))))
In[496]:= Solve[% == %%%%, Inactive[ibaz][8/11]]
Out[496]= {{Inactive[ibaz][8/11] -> 5/7}}
In[497]:= %%%%%%%%%% == %% /. %
Out[497]= {Inactive[ibaz][7/22] == 55/112} a reasonably generic rational. --rwg
On 2016-02-09 21:50, Dan Asimov wrote: Warut, nice formula.
For any n in Z+, here's an (2n-1)-times differentiable example that is not a rational function:
H(x) = sgn(x) x^(2n).
Which suggests a question:
Question: ---------
Does there exist a real analytic function
f: R --> R
that takes rationals to rationals, but is not a rational function?
--Dan
On Feb 9, 2016, at 8:45 PM, Warut Roonguthai <warut822@gmail.com> wrote:
Here is my example of a continuous function which maps rationals to rationals, but is not a rational function:
F(x) = (-1)^floor(x) * frac(|x|) * (frac(|x|) - 1),
where frac(x) is the fractional part of x.
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