Without the intention to insult: "some people" are those who provide content.

"it is in the plan ..." simply says: we ignored you so far, there was s th more
important. For us.

It is demanded for ages.

Just do it.

Very nice!

I remember, that Madan et al did something towards Bessel for that model by
paper + pencil (but practical in Finance it is used without slow Bessel functions)

Very nice!

I remember, that Madan et al did something towards Bessel for that model by
paper + pencil (but practical in Finance it is used without slow Bessel functions)

@Markiyan Hirnyk 

http://www.math.nyu.edu/research/carrp/papers/pdf/VGEFRpub.pdf

http://maplesoft.com/documentation_center/

@Markiyan

Ah, I guess I understand what you mean:

Sum(f(n)) = Sum(f(3k*+0)) + Sum(f(3k*+1)) + Sum(f(3k*+2)).

That splitting idea for the summation is done by hand, yes. For the special case.

At the 'positive side': Math questions would become boring if computers could
answer all of them :-)

I am not aware of a systematic approach (though I vaguely remember floor et al and
the relations to harmonic sums or similar, but can not find it).


PS: where - as minor critics to your post using gfun - that gives a solution, but
not a proof (since gfun only verifies the formula on the finite input, but will
not proof the formula in general).

@Markiyan

Ah, I guess I understand what you mean:

Sum(f(n)) = Sum(f(3k*+0)) + Sum(f(3k*+1)) + Sum(f(3k*+2)).

That splitting idea for the summation is done by hand, yes. For the special case.

At the 'positive side': Math questions would become boring if computers could
answer all of them :-)

I am not aware of a systematic approach (though I vaguely remember floor et al and
the relations to harmonic sums or similar, but can not find it).


PS: where - as minor critics to your post using gfun - that gives a solution, but
not a proof (since gfun only verifies the formula on the finite input, but will
not proof the formula in general).

Here is a 'maple-ish' proof for Preben's "modulo 3", more detailed appended.


As just seen one has 2^(2*n+2) = 2^(3*k+p) = 1, 2 or 4 mod 7 for p=0, 1 or 2.

Switching to frac (by floor(x) = x - frac(x)) the toublesome expressions now
simply become 1/7, 2/7 or 4/7. Splitting summation index modulo 3 with those
p one can find the value.

One gets the asserted 27/26 since the series without floor is found as well.

hirnyk_nachtmusi.mws

Here is a 'maple-ish' proof for Preben's "modulo 3", more detailed appended.


As just seen one has 2^(2*n+2) = 2^(3*k+p) = 1, 2 or 4 mod 7 for p=0, 1 or 2.

Switching to frac (by floor(x) = x - frac(x)) the toublesome expressions now
simply become 1/7, 2/7 or 4/7. Splitting summation index modulo 3 with those
p one can find the value.

One gets the asserted 27/26 since the series without floor is found as well.

hirnyk_nachtmusi.mws

@Markiyan Hirnyk 

Yes, it was just a sketch, what his formula S1,S2,S3 mean.

I you last post you have m0=1, here are the other 2 cases

m0:=2; # or m0:=0
2^(3*k + m0)*4;  simplify(%); % mod 7 assuming k::posint;

@Markiyan Hirnyk 

Yes, it was just a sketch, what his formula S1,S2,S3 mean.

I you last post you have m0=1, here are the other 2 cases

m0:=2; # or m0:=0
2^(3*k + m0)*4;  simplify(%); % mod 7 assuming k::posint;

@Markiyan Hirnyk 

I think he says:
that floor (or frac) depends on the reminder mod 3 and he adds just that 3 cases.

[seq(2^k*4 mod 7, k=1 .. 21)];
   [1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4]

@Markiyan Hirnyk 

I think he says:
that floor (or frac) depends on the reminder mod 3 and he adds just that 3 cases.

[seq(2^k*4 mod 7, k=1 .. 21)];
   [1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4, 1, 2, 4]

@Kitonum:

Do you see any rational solutions?

@Kitonum:

Do you see any rational solutions?