Global maximum of ?(x)-x

[Gerald McGarvey]

Based on a laborious semi-manual process (not recommended) of
narrowing down the maximum, I believe the maximum of ?(x)-x
occurs at a constant c whose continued fraction begins
[0; 1, 3, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 
1, 4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 
4, 1, 4, 1, 4, 1, 5, 1, 4, 1, 4, 1, 4, 1, 4, 1, ...
I hope I didn't miss a larger ?(x)-x along the way.
Here's a conjecture: the maximum occurs at a constant c whose continued 
fraction
starts as [0; 1, 3, 1, 4, 1, 4, 1, 5, then has the repeating sequence 1, 4, 
1, 4, 1, 4, 1, 4, 1, 4, 1, 5
so c = 1/(1+1/(3+1/(1+1/(4+1/(1+1/(4+1/(1+1/(5+1/((17325 + 
3*sqrt(59115595))/33461)))))))))
or (2343*sqrt(59115595) + 18014599)/(2955*sqrt(59115595) + 22720034)
or
.79289414860601842644318261515336048837629147616654626681193944753721958311485553800434172526654160238532789219294579...
If this is correct, the value of ?(c) appears to be 2008938429 / 2147483647.
Of course, it could very well be incorrect.

Has it be proven that ?(x) + ?(1-x) = 1 for all x?

- Gerry


At 12:15 AM 6/11/2006, Joseph Biberstine wrote:
>         Indeed.  I have been using the following Mathematica function for 
> rough work:
>
>mqf[x_] := If[x < 0, -mqf[-x], (cf = ContinuedFraction[(*some bug causes 
>certain decimals to be misinterpreted*)Rationalize[x]]; cf[[1]] + 
>Sum[(-1)^(k)/2^(Sum[cf[[i]], {i, 2, k}] - 1), {k, 2, Length[cf]}])];
>
>         (It is interesting to note there is something wrong with Mma 5.1 
> or my installation specifically that, without the Rationalize comment 
> noted in the comment above, even rational numbers can be completely 
> bugged.  For example, without the Rationalize command above, I get 
> mqf[0.4]=1/2 which is wrong.  mqf[2/5] yields the correct value of 
> 3/8.  Thus why I say "rough" work above.)
>         This implementation is intended for rational x only (mqf is for 
> mq finite cf).  Generalizing to a function approximating reals with 
> non-terminating c.f.'s is trivial.  Finally, recall that Mma one-indexes 
> its lists.
>         My implementation similarly yields the correct special values and 
> graphs appropriately.  I was going to attach a pair of images to my last 
> message but worried the list might bounce for anti-solicitation reasons, 
> but I think I'll try this time.
>         I'm sure that the function is fractal (beautifully I might 
> add).  This is evident from the plot.
>
>         I'm glad you also noticed how well dyadics approximate many 
> interesting constants under application of Minkowski's terrific 
> function.  This of course stems from that large terms in an input's c.f. 
> expansion (such as 292 early in Pi and so 1/Pi) results in a drastic 
> reduction of all following summands.  For further reference, see my OEIS 
> submissions under 
> http://www.research.att.com/~njas/sequences/?q=expansion+minkowski+biberstine&sort=2 
> , especially http://www.research.att.com/~njas/sequences/A119929 ! Also, 
> a notice to all: I recently corrected/modified several of these entries 
> and Neil hasn't had time to apply the changes yet, so I suggest these be 
> taken with a grain of salt for a few weeks (specifically,
>A119928 and A119929 which I warn you are entirely incorrect due to a code 
>artifact (namely I had been using a conditional on Element[x,Rationals] in 
>my function for unification purposes and, of course, Khinchin does not 
>(yet) have a Boolean evaluation in Mma).
>
>Gerald McGarvey wrote:
> >
> > I defined the following function in PARI/GP for Minkowski's question
> > mark function,
> > using formula (1) on
> > http://mathworld.wolfram.com/MinkowskisQuestionMarkFunction.html
> > then tested it using the special values shown on the web page, it checks
> > out ok for
> > the special values.
> >
> > 
> mq(x)=cf=contfrac(x);return(sum(k=1,length(cf)-1,(-1.)^(k-1)/2^(sum(m=2,k+1,cf[m])-1))) 
>
> >
> >
> > contfrac(x) creates an array, the rest uses the array.
> > Something similar could be done in Mathematica or Maple.
> >
> > then defined f(x) = mq(x)-x for ?(x)-x
> >
> > I used a crude approach (using print in a for loop) to find larger
> > values for
> > ?(x)-x near the value .79285714 and found (using a precision of \p 115)
> >
> > f(.79289)    = .142588184488601982593536376953125 (not sure if exact)
> >
> > f(.79285714) = .1425676646875 (not sure if exact)
> >
> > so I think f(.79285714) is near a local maximum, but that the function
> > then decreases before going to the global maximum, wherever that is.
> > The problem is that f(x) is not strictly increasing.
> > Could it be fractal in nature? If so, finding the global maximum could
> > be very difficult, but maybe there is an easy way. It could require
> > a bit of programming that iteratively creates arrays of values and finds
> > the maximum of the array to narrow the search. How fine-grained would
> > each step need to be, how erratic is ?(x)-x ?
> >
> > Using the mq function I get the following puzzling values.
> > Are these values accurate or due to a calculation error?
> >
> > ?(1/Pi) =
> > 
> 0.248046904802322387695312500000000000000000000000000000000000000000000000000000000000000000000002403016875... 
>
> >
> >
> > ?(1/Pi^3) =
> > 
> 0.000000000931322574615478515624999999999999999999999999998884213294804113213548547849861136557254326556794... 
>
> >
> >
> > - Gerry
> >
> > At 07:10 PM 6/10/2006, Joseph Biberstine wrote:
> >>         Not clever enough to figure this out a reliable computation of
> >> this myself.  Please help find enough accurate terms for OEIS and I
> >> will of course credit you.
> >>
> >>         Consider f(x) := ?(x)-x where ?(x) is Minkowski's question
> >> mark function.
> >>         Recently I've wondered at what value xmax f will obtain its
> >> global maximum (mod 1).  Specifically I'm curious how the c.f.
> >> expansion for xmax grows (based on the definition of f it should make
> >> for a very interesting sequence!)
> >>         To the few places I've been able to calculate, neither c.f.
> >> nor decimal expansions for xmax or f(xmax) are in OEIS.  Note the c.f.
> >> expansion looks about as we'd expect so far (that is, when
> >> zero-indexed, odd entries are 1 and even entries slowly grow).
> >>
> >> xmax =~ {0; 1, 3, 1, 4, 1, 4, ?1, ...} =~ 0.79285714...
> >>
> >> - Joseph
> >
>
>
>
>