Talk:Multiplicative function

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Somebody wrote:

the function d(n) - the number of positive divisors of n, σ(n) - the sum of all the positive divisors of n, σk(n) - the sum of the k-th powers of all the positive divisors of n (where k may be any complex number)

How ks can look like? Any example.? Aren't ks just simple natural numbers here?
XJam [2002.04.16]] 2 Tuesday (0)

Normally, the k is taken to be a natural number, that's correct, but the definition works fine for complex numbers, and yields a multiplicative function. Here's how to work with complex exponents, for instance k=i: 5i = eln(5)i = cos(ln(5)) + i sin(ln(5)) (using Eulers formula in complex analysis). AxelBoldt

Yes Axel that is quite OK. But I really don't see any necessity to extend ks over the set of natural numbers. We just have examples for σk(n) where k goes from 0, 1, 2 and so on. In fact special cases with k=0 and k=1 are functions d(n) ≡ σ0(n) and σ(n) ≡ σ1(n). It is for example with n=144, d(144)= 15, σ(144)=403. For k=2 and for n=24·32=144 we have:
σ2(144) = σ2(24)σ2(32) = (12+22+42+82+162)(12+32+92) = 341 · 91 = 31031.
Because of the multiplicativity of σk(n) it is easier to calculate product of powers than adding all the squares of all the positive divisors of n. Futher on we have:
σ3(144) = σ3(24)σ3(32) = (13+23+43+83+163)(13+33+93) = 4681 &middot 757 = 3543517,
σ4(144) = ... = 464378915,
σ5(144) = ... = 64178802493,
σ6(144) = ... = 9070067614091,
σ7(144) = ... = 1294620020196997,
σ8(144) = ... = 185637589303481315,
σ9(144) = ... = 26676789058694821933,
σ10(144) = ... = 3837572548050547502651,
σ11(144) = ... = 552334249790915518944277 and all the way to "funny" ∞.
We don't have function σ5i(n) or do we? k is just counter here for this arithmetical function. And finally according to extension into the complex the value for σ2i(144) is (210i-1)(36i-1)/(22i-1)(32i-1). But is this a mathematical reality in this particular case? What in fact ki-th powers of all the positive divisors of n are doing in arithmetical function? I guess nothing. If there are any mathematical meanings, then we should put these facts in the article, don't you agree. --XJam [2002.04.17]] 3 Wednesday (0)

σ5i(n) is just as real a multiplicative arithmetical function as σ11(n), σπ(n) or σ-2(n) for that matter. Personally, I don't know what they are being used for, if at all. AxelBoldt

Yes, strange. Mathematicians can 'produce' something and they can't use it anywhere. Just what Hardy dreamed about and also worked and lived on about. But perhaps we should somehow consider these cases too. Let us just not forget on negative numbers and on complex ones. There were times when people din't know what to do with them and nowadays we can find them in physical reality. And values of σξ(n) become into the set of ξ and are no more part of the set of natural numbers as they are when ξ is natural number. --XJam [2002.04.18]] 4 Thur's day (0)

Convolution equations are not easier without an argument by no means. We have to be more carefull to read them - because e (which is defined elsewhere) can mean many things in math. And accidentally e also means more famous base of natural logarithm. So I guess we have to write e(n) to distinguish anyway. If I would write in this way at math exam, I wouldn't finish my mechanical engineering after all. I can recall all those long terms in Taylor's formulas for several variables. But every eyes have their own whitewasher... --XJam [2002.04.18]] 4 Thur's day (1st ed)


Good point, I'll call it ε. The arguments were all in the wrong places: it's not f(n)*g(n), but (f*g)(n). AxelBoldt

Yes - I was a little bit confused with (f*g)(n) too and I believe other Wikipedians also. Axel can you please clear one more another thing. Why is multiplicative function more general term than arithmetic one? Or is this just your opinion? I prefer arithmetical one because it tells me more. (And futhermore I am constantly confuseing in English natural number (N) and integer (Z).) --XJam [2002.04.18]] 4 Thur's day (2nd ed)

Contents

[edit] Classic totient identity

The part on proving the totient identity seems to be flawed... what if n is not a power of a prime? I'm not sure how to prove this identity... could somebody fix this? There might be a proof on the totient function page. CecilBlade 12:23, 6 December 2006 (UTC)

[edit] Proving convolution identities deleted

I don't think the section on proving convolution identities belongs in Wikipedia. Deleted. -Zahlentheorie 21:46, 18 May 2007 (UTC)

[edit] Proving convolution identities re-instated

After careful study of the argument concerning deletion of the proof of totient identities page, Wikipedia:Articles_for_deletion/Totient_function/Proofs, I decided to re-instate the examples and add some more material concerning Dirichlet series, the goal being that we ultimately have a set of examples for the article on Perron's formula. -Zahlentheorie (talk) 20:12, 8 January 2008 (UTC)

[edit] Fixing definition of (incomplete) multiplicativity

a,b must be coprime as explained on Planet Math. Otherwise this allows things like f(x) = gcd(x,2), so f(4) = f(2)*f(2) = 2 * 2 = 4, but clearly f(4) = 2. Arwest 00:20, 25 June 2007 (UTC)

The article reads strangely now: it says that multiplicative means f(ab)=f(a)f(b) for coprime a, b except outside number theory where it means the same but for coprime a, b. If the article was originally correct it should perhaps says "Outside number theory 'multiplicative' is used to mean 'completely multiplicative'."; if that's not really used then the sentence should be deleted. As it stands it's just confusing.
CRGreathouse (t | c) 14:17, 16 July 2007 (UTC)
Yes. Spoted that. Changed. Rich Farmbrough, 18:02 4 March 2008 (GMT).

[edit] Comment on old discussion

The first remarks on this page are six years old but XJam is mistaken; nonreal values of the subscript k are actually used. Ramanujan's formula

\sum_{n=1}^{\infty} \frac{\sigma_a(n)\sigma_b(n)}{n^s}=\frac{\zeta(s)\zeta(s-a)\zeta(s-b)\zeta(s-a-b)}{\zeta(2s-a-b)}

is used in Ingham's proof that the zeta function does not vanish on the line real part = 1. It's a proof by contradiction which starts off by assuming a is a zero with real part 1 (and iirc b is its complex conjugate). The proof is in Titchmarsh, but I don't have it in front of me right now to give a page reference. Virginia-American (talk) 01:58, 1 April 2008 (UTC)

Here is the reference

Titchmarsh, Theory of the Riemann Zeta Function, Oxford, 1986, 0-19-853369-1, § 3.4 pp. 48–49.

What he actually does is set a and b to the imaginary values ai and −ai

\sum_{n=1}^{\infty} \frac{|\sigma_{ai}(n)|^2}{n^s} =\frac{\zeta^2(s)\zeta(s-ai)\zeta(s+ai)}{\zeta(2s)}

Then assuming that 1 ± ai is a zero he derives a contradiction.

Virginia-American (talk) 16:51, 1 April 2008 (UTC)