Multiple zeta function (Values) Mzv

Multiple zeta function (Values) Mzv

Abstract

I just think that multiple zeta function is interesting topic and I just saw the Don Zagier's Video and other videos discussing it. I want to build my note for this interesting series. Hope I can understand motives, periods one day.

Definition:

$$\zeta(s_1, s_2,..s_m)=\sum_{n_1>n_2>...>n_m \geq 1}\frac{1}{n_1^{s_1}n_2^{s_2}..n_m^{s_m}}$$ To be convergent $$s_1, s_2 , s_{m-1}\geq 2 , s_m \geq 1$$ This is famous generalization for riemann zeta function: $$\zeta(s)=\sum_{n=1}^{\infty} \frac{1}{n^s}$$ If you are interested in this. You can see my another blog Note for Riemann Hypothesis The famous number traced back to Euler is: $$\zeta(2)=\sum_{n=1}^{\infty} \frac{1}{n^2}=\frac{\pi^2}{6}$$ And Euler also proof that $k$ is positive integer: $$\zeta(2k)=\sum_{n=1}^{\infty} \frac{1}{n^{2k}}=\text{Rational Numbers }\pi^{2k}$$ I will give a proof in this blog. But up to now, no one can give a closed form for zeta function with odd integers which led to the following conjecture:

Conjecture 1

$$\zeta(3), \zeta(5),,,,\pi^{2k} \text{ are algebraic independent}$$

Dimension

There are many relations for mzv. We can define dimension for weight k is a vector space spanned by $\mathbb{Q}$ coefficient: $$\mathbb{Q} \zeta(s_1, s_2,..s_m) : s_1+s_2+...s_m=k$$ . So what is the dimension for this vector space?

Weight=3 --- dimension=1

It seems like we have two dimensional space since we have $$\zeta(3), \zeta(2,1)$$ But actually they are the same: $$\zeta(3)= \zeta(2,1)$$ . proof: Write $$\sum_{mn}\frac{1}{mn(m+n)}=\sum_{mn}\frac{1}{m(m+n)^2}+\sum_{mn}\frac{1}{n(m+n)^2}$$ We can see (after some mediation) $$\sum_{mn}\frac{1}{m(m+n)^2}=\sum_{n>m=1}\frac{1}{m(n)^2}=\zeta(2,1)$$ So we have: $$\sum_{mn}\frac{1}{mn(m+n)}=2\zeta(2,1)$$ On the other hand: $$\sum_{mn}\frac{1}{m(m+n)^2}=\sum_{m}\sum_{n}\frac{1}{m}(\frac{1}{m(m+n)})$$ Use $$\sum_{m}\frac{1}{m(m+n)}=\frac{1}{n}(1+\frac{1}{2}+....\frac{1}{n})=1+\frac{1}{4}(1+\frac{1}{2})+\frac{1}{9}(1+\frac{1}{2}+\frac{1}{3})+...$$ $$=\zeta(2,1)+\zeta(3)$$ So we have $$\zeta(3)= \zeta(2,1)$$ .

Nielsen Reflexion Formula

$$\zeta(a)\zeta(b)=\zeta(a,b)+\zeta(b,a)+\zeta(a+b)$$ proof: $$\zeta(a)\zeta(b)=\sum_{n,m}\frac{1}{n^a m^b}=\sum_{n>m>=1}\frac{1}{n^a m^b}+\sum_{m>n>=1}\frac{1}{n^a m^b}+\sum_{n=m>=1}\frac{1}{n^a n^b}=\zeta(a,b)+\zeta(b,a)+\zeta(a+b)$$

Continuing: