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/sci/ - Science & Math

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2172371 No.2172371 [Reply] [Original]

I pose a question and a challenge to you.

There are a set of functions which have a certain periodicity. Let me define a variable n which is contained on the set of integers.

The function which has a periodicity of 1 is 1^n. Basically after 1 alteration of n you get back to the same answer.

The function which has a periodcity of 2 is (-1)^n. After altering n twice you get back to the same answer (i.e. (-1)^1=-1, (-1)^2=1, (-1)^3=1).

The function which a periodicity of 4 is (i)^n (i being the complex number). After altering n 4 times you get back to the same answer (i.e (i)^1=i, (i)^2=-1, (i)^3=-i, (i)^4=1, (i)^5=1)

What is the function which has a periodicity of 6?

>> No.2172378

inb4 do your own homework. This isn't my homework, just a problem I found interesting. I already found the function in question, but if you want to find one for other periodicities, be my guest. I only know the function with periodicity of 6 so far.

>> No.2172390

Can nobody answer? Have I outsmarted /sci/?

>> No.2172393

I'm thinking about sin or cos...

>> No.2172397

The function(s) with periodicity k are just the kths roots of unity.

>> No.2172402

>>2172393
eh.. kind of but not really.

>>2172397
Care to expand on that?

>> No.2172404

>>2172378

e^(2pi)/6

>> No.2172407

>>2172404
Thats just a number. But you're on the right track. Specifically I was looking for functions of the form (a)^b like the ones listed in the original problem.

>> No.2172410

>>2172404
this is analogous to 1/6 rotation about the origin. in general this is true for any periodicity you want.
formal would be

f(n)=n*e^(2*pi*i/6)

replace 6 with k for the general case.
there might also be others.
three fold rotoinversion perhaps.

>> No.2172412

>>2172407
f(x) = i
That's a function.

>> No.2172413
File: 366 B, 46x22, 1.png [View same] [iqdb] [saucenao] [google]
2172413

>>2172402
<-- this thing for each k from 0 to n-1.
If you only want new functions, then choose k which make them primivite roots, i.e. you need k and n to be coprime.

>> No.2172418

>>2172413
>>2172410
ok but can you now relate that to a form that is (a)^b?

>> No.2172419

>>2172418
And I don't me exponentials.

>> No.2172429

>>2172418
Sigh.
f(m) = (that pic)^m.
The function has period n for any integer k.
If you only want functions with exactly that period, choose any k coprime to n.
This is EVERY complex function with that period, too.

captcha: ficklece PROOF