"Modular" complex arithmetic

[David Makin]

^fixed. But in any case, do you see my point as to why it's wrong?

Not really as I haven't studied what's being done in this thread in detail.

In case it's relevant I think UF's mod function maybe slightly different from the norm - in UF:

  -1.6%1.0  ->   -0.6
  1.6%1.0   ->   0.6

[jehovajah]

just a quick question: are you referring to mod or modulo or some combination of the two?

[jehovajah]

The hangups we "mathematicians" have are due to us being taught to be hypercritical,usually by some sadist who delighted in pointing out our mistakes! Any way having got that off my chest i think playfulness means lots of wrong turns down the road to interesting stuff and your colour algorithm is veeeery interesting!

[kram1032]

yup, the simplification seemingly doesn't even hold for the -1.6 example...

the sum-free formula returns 0.4, so it seems correct, the simplified formula some complex number and the log and exp free one actually even totally simplifys to -z, which makes it totally useless, of course...



if that formula indeed is the correct one, the cool thing about it would be, that a complex a also is defined that way.

[Timeroot]

Hmm, yeah, I tried the exp/log formula, it doesn't seem to work for me...   

I actually got that fourier series from a math book my dad gave me, but now I see if you type "-1.6 mod 1" into W|A you get the same thing under series representations.

When you say "the simplified formula", what do you mean? You'll see that both of the closed form formulas I gave give a final result of 0.4...

[kram1032]

EDIT!
Since LaTex apparently is (or at some point was) broken, all the results are lost...
Here is the final Complex Mod Function with z and a being complex numbers:

z mod a = CMod(z,a)=π+ i (e^{2 i π z/a} Log(1-e^{-2 i π z/a})-e^{-2 i π z/a} Log(1-e^{2 i π z/a}))
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the simplified formula was meant to be the one where I converted log()-log() to log(/).
That one somehow does not return correct results.

but if we stick to log()-log(), everything seems to work nicely.
Now I'd like to see an Mset with that formula rather than the symmetric approach.
Reading about mod, I found out that both variants are kind of valid. One, however, is called "mathematical" and the other one "symmetrical"

The symmetrical one is mod(|z|,a)*sign(z) and doesn't really allow for a complex a.
The mathematical one is the more complicated one...
, which has the big advantage to even work in case of a complex a. No idea if that would make any sense, though