Talk:Dedekind sum

[Untitled]

Sorry about the big mod below the sum but I don't know how to make it smaller. Any suggestions?

The 'see also' to Dedekind cut is to a totally unrelated page. Charles Matthews 10:31, 11 Feb 2005 (UTC)

cotangent formula

Could you please check if the formula you added under Alternative forms on Dedekind sum for s(b,c) involving cots is correct for the saw-tooth function as now (and previously) defined on the page. I've corrected the other alternative form of s(b,c) myself. Ncik 18:42, 25 November 2005 (UTC)[reply]

This formula is cited in the Beck and Robbins PDF. Do you believe its wrong? Just right now, I can't think of any quick and obvious way to verify this. Anyway, we should have this conversation on Talk:Dedekind sum linas 20:15, 25 November 2005 (UTC)[reply]

Also, not sure what you mean by "I corrected the other formula"; I am a bit lazy in trying to decipher the changelog. What was the problem, and how was it corrected? linas 20:18, 25 November 2005 (UTC)[reply]

Well, the formula stayed actually the same, just the following comment was changed: omega != 1, of course. Ncik 21:00, 2 December 2005 (UTC)[reply]

I corrected the "Aternate forms" section--it was lacking the assumptions of coprimeness, for both formulas. mattbeck 29 November 2005

Sorry, wasn't quite alert as it seems. Ncik 21:00, 2 December 2005 (UTC)[reply]

"Another useful formula"

I don't think the "another useful formula" (at the end of "simple formulae") is correct--for once, the second factor in the sum clearly depends on which residue class we're summing over (which it shouldn't). If we change the summation condition to n running from 1 to c-1, then the formula is correct. But beyond that, I don't quite see its usefulness: one can distribute to separate n/c and x, and then a quick application of the identity just before this one shows that the x plays no role. I suggest taking the "another useful formula" out. (But not the "Furthermore..."!) mattbeck 29 November 2005

Hmm. Please make the correction. I am not studying this stuff at this time, so I cannot just tell be looking at it whether its correct or not. However, several remarks:

If I added this formula (I don't remember), I would have copied it from a reputable source, e.g. one of Tom Apostol's books, and so don't be too quick to assume there's an error. (There may be one, but don't assume it unless you can prove it).

One of the many roles a WP math article can take is as a reference, which can include formulas that, while "obvious" to the specialist, may not be obvious to the casual reader. Think of a table of integrals: A lot of the integrals listed in a table, or even many of the formulas in Abramowitz and Stegun are "obvious" when you've been doing it for a while. However, "another useful formula" doesn't seem at all to be immediately obvious, so keeping it seems like a good idea to me. linas 00:04, 1 December 2005 (UTC)[reply]

I outlined a proof that the "another useful formula" was wrong: as stated, the formula violated the fact that s(a+b,b) = s(a,b). I took it out. By the way, I couldn't agree more to your second comment, and it is for these reasons that I despise the word "obvious". mattbeck 1 December 2005

OK, this time, I'm putting my brain in gear. The incorrect formula(s) were added by User:Ncik on 21 November 2005. We need to ask him to explain himself.

In the meanwhile, I'm finding that

s(b,c)-p{\frac {x}{2}}=\sum _{n\mod c}\left(\left({\frac {bn}{c}}\right)\right)\left({\frac {n}{c}}+x\right),\qquad \forall x\in \mathbb {R} .

for some integer p that depends on b,c ~~in a way that I can't quite make ou~~ so that p=gcd(b,c). So in some way it is kind-of an interesting formula, ~~esp. If I can ping down p~~. linas 18:19, 1 December 2005 (UTC) -- it seems to be gcd at least for positve b,c -- linas 18:33, 1 December 2005 (UTC)[reply]

Hmmm... I think there are similar problems as before. You'll get different answers for the sum on the right, depending on how you let n vary (e.g., you get two different answers as n ranges from 0 to c-1 or from c to 2c-1).

But even if we agree on letting n range from 0 to c-1, I don't see your formula... then

\sum _{n=0}^{c-1}\left(\left({\frac {bn}{c}}\right)\right)\left({\frac {n}{c}}+x\right)=\sum _{n=0}^{c-1}\left(\left({\frac {bn}{c}}\right)\right)\left({\frac {n}{c}}-{\frac {1}{2}}\right)+\sum _{n=0}^{c-1}\left(\left({\frac {bn}{c}}\right)\right)\left(x+{\frac {1}{2}}\right)=s(b,c)+\left(x+{\frac {1}{2}}\right)\sum _{n=0}^{c-1}\left(\left({\frac {bn}{c}}\right)\right),

and this last sum on the right is zero, independent of the relationship of b and c... mattbeck 1 December 2005

Argh. We are both confused and going in circles. First, I mislead you and myself, since I use a definition of ((x)) that differs from the article (I take it to be -1/2 for integers, and not 0. I do this for numerical exploration, and I am now realizing that this is actually a subtle point. Among other things, it leads to that curious gcd identity).

Second, using the article definition of ((x)), you have just proved that the original formula was correct, i.e. that the result does not depend on x.

Third, I am not sure why you are insisting that $\sum _{n=0}^{c-1}$ is somehow more valid than the sum $\sum _{nmodc}$ , it seems that both give the same answers. That's kind of the point of having a sawtooth, to get rid of the integer part ... linas 21:14, 1 December 2005 (UTC)[reply]

I see where the confusion was and I'm glad that's clear now. (The literature is also confusing regarding this notation--and I'm one of the culprits who used different notations in different papers...) However, the summation condition is still an issue. I'm not claiming that one is more valid than the other, but that the sum over n mod c could give a wrong identity--at least I don't see why it should be correct. The second factor in the sum has no sawtooth function around the n, so that factor is not independent on which residue system we're summing over. (It could very well happen that the sum over n=0..c-1 gives the same result as a sum over n=c..2c-1, but those are not the only complete residue systems we may choose...) mattbeck 1 December 2005

Sorry, guys. I forgot the condition (b,c)=1. I reinstated the formula, but if you think it's not worth being mention because it indeed is just a simple corollary of the formula above, feel free to remove it. Ncik 21:06, 2 December 2005 (UTC)[reply]

As stated, the "formula" is wrong, no matter whether b and c are relatively prime or not. Plug in b=1, c=4, x=0. Take the sum on the right-hand side over n=0,1,2,3, and then over n=0,1,6,7. Both are complete residue systems mod 4, but they give different sums. We can fix things by demanding that n runs from 0 to c-1, but then the "formula" is a simple corollary of the formula right before, as shown above. mattbeck 2 December 2005