MaplePrimes - answers and comments on Question, Eine kleine Nachtmusik

27/26

Preben Alsholm — Fri, 16 Nov 2012 20:08:05 Z

It seems that the answer is 27/26 based on

restart;
Digits:=100:
evalf(Sum(3^(-n)*floor((1/7)*2^(n+2)), n = 1 .. infinity));
identify(%);

Proof:

Based on the familiar formula for (x^n-1)/(x-1) with x=8.

We have
S1:=floor(2^(3*k)/7)=(2^(3*k)-1)/7;
S2:=floor(2^(3*k+1)/7)=2*(2^(3*k)-1)/7;
S3:=floor(2^(3*k+2)/7)=4*(2^(3*k)-1)/7;
Sum(3^(-n)*floor((1/7)*2^(n+2)), n = 1 .. infinity)=add(Sum(3^(-3*k+2-p)*floor((1/7)*2^(3*k+p)), k = 1 .. infinity),p=0..2);
subs(S1,S2,S3,%);
value(rhs(%));

Another proof

Markiyan Hirnyk — Fri, 16 Nov 2012 22:23:17 Z

Download another_proof.mw

p=17 or even larger

Axel Vogt — Sat, 17 Nov 2012 23:54:18 Z

I was playing a bit more with that nice task, Sum(3^(-n)*floor(2^(n+2)/p), n = 1 .. infinity)
and for p=17 that does not evaluate numerically, only after restrichting to 100 terms or so.

The usage of gfun as shown by Markiyan seems to run into troubles as well for increasing p.

Let me consider the problem with 'frac', which is more or less equivalent (one can use gfun
for that as well).

By looking at examples one sees that frac(2^(n+2)/p) has only finitely many values and those
are periodic, in the case p=7 the period was 3.

Now remember Fermat's little theorem: a^(p-1) = 1 modulo p for p a prime number.

Hence 2^(n+2) - 2^(n+2+(q-1)) = 2^(n+2)*(1 - 2^(q-1)) = 0  modulo p. Which means, that all
the rationals frac(2^(n+2)/p) are periodic of period p-1 (there is no zero, since 2 is a unit
modulo p, it is a field).

For p=7 we had a shorter period (p-1)/2, but that is not true in general (p=11).

But that already allows to write down the recursion without using gfun: u(k+p-1)=u(k) and
the (p-1) initial values.

For that rsolve replies in terms, which are difficult to be used in the infinite series
and for p=23 I ran out of memory (and patients).


Thus returned to the idea of splitting the summation.

Write 2+n = (p-1)*k + j. Then j = 2+n modulo (p-1) and k = (n+2 - j)/(p-1).

One can convince oneself, that therefore the following is a re-arragement of the series:

Sum( Sum(3^(-(p-1)*k-j) * L[j],k = 0 .. infinity),j = 1 ..  p-1)

The inner series evaluates to 3^(-j+ p-1)* L[j]/(3^(p-1)-1) and adding p-1 such terms
is quick and without problems, even for larger p. Hence we have

  Sum(3^(-n)*frac(2^(n+2)/p), n = 1 .. infinity) = 
  Sum( 3^(-j+ p-1)* L[j]/(3^(p-1)-1), j = 1 ..  p-1)

One only has to provide L:=[seq(frac(2^(n+2)/p), n=1 .. (p-1))].

Probably true

Markiyan Hirnyk — Fri, 16 Nov 2012 20:22:05 Z

How to ground 27/26 in a more reliable way?

PS. I have in mind explicitly. Compare with http://www.mapleprimes.com/questions/127391-Summa-Technologiae

See the proof added above

Preben Alsholm — Fri, 16 Nov 2012 20:59:52 Z

@Markiyan Hirnyk Please see the proof added above.

Don't understand it

Markiyan Hirnyk — Fri, 16 Nov 2012 21:19:31 Z

@Preben Alsholm Can you kindly explain that in details? I don't understand why floor(2^(3*k)/7)=(2^(3*k)-1)/7 .

I don't understand why Sum(3^(-n)*floor((1/7)*2^(n+2)), n = 1 .. infinity)=add(Sum(3^(-3*k+2-p)*floor((1/7)*2^(3*k+p)), k = 1 .. infinity),p=0..2).

PS. I don't see how to prove a similar thing with 17 instead of 7 in the denominator.