1. Introduction

The explicit formula of Riemann zeta function over \( 2n (n∈N) \) is originally given by:

\( ζ(2n)=\frac{|{B_{2n}}|{(2π)^{2n}}}{2(2n)!} \)

Where \( {B_{n}} \) denotes as the nth Bernoulli number.

Also, the values of \( {B_{n}} \) can be evaluated recursively by this following relationship, and it is the only simplest way to calculate their exact values:

\( {B_{n}}=-\frac{1}{n+1}\sum _{k=0}^{n-1}( \begin{array}{c} n+1 \\ k \end{array} ){B_{k}} \)

Which is just being rewritten from the identity that \( \sum _{k=0}^{n}( \begin{array}{c} n+1 \\ k \end{array} ){B_{k}}=0 . \)

However, when we are evaluating \( ζ(2n) \) based on Bernoulli numbers, then we have to find the Bernoulli numbers at corresponding term first, then plug them in into the explicit formula. It will be very complex especially when \( n \) becomes larger.

However, there is a way to eliminate the Bernoulli numbers in the formula when we are calculating \( ζ(2n) \) , which can simplify the process at some extend.

That is, the recursive formula of Riemann zeta function \( ζ(n) \) at even natural numbers can be expressed by the following equation:

\( ζ(2n)={(-1)^{n-1}}(\frac{n{π^{2n}}}{(2n+1)!}+\sum _{k=1}^{n-1}{(-1)^{k}}\frac{{π^{2(n-k)}}ζ(2k)}{(2(n-k)+1)!}) for n≥2 \)

From using this recursive formula:

\( ζ(2)=\frac{{π^{2}}}{3!}=\frac{{π^{2}}}{6} \)

\( ζ(4)=-\frac{{2π^{4}}}{5!}+\frac{{π^{2}}}{3!}ζ(2)=\frac{{π^{4}}}{90} \)

\( ζ(6)=\frac{{3π^{6}}}{7!}-\frac{{π^{4}}}{5!}ζ(2)+\frac{{π^{2}}}{3!}ζ(4)=\frac{{π^{6}}}{945} \)

\( ζ(8)=-\frac{{4π^{8}}}{9!}-\frac{{π^{6}}}{7!}ζ(2)+\frac{{π^{4}}}{5!}ζ(4)-\frac{{π^{2}}}{3!}ζ(6)=\frac{{π^{8}}}{9450} \)

…

This formula is derived from the principles of elementary symmetric polynomials and the associations between them and the values of zeta function. Also, it is equivalent to the original explicit formula of zeta function, that is, \( ζ(2n)=\frac{|{B_{2n}}|{(2π)^{2n}}}{2(2n)!} \) .

2. Elementary symmetric polynomials & Newton’s identities

For variables \( {x_{1}},…,{x_{k}} (k \gt 1), \) let \( {p_{k}}({x_{1}},…,{x_{n}}) \) be the power sum \( \sum _{i=1}^{n}x_{i}^{k} \) and \( {e_{k}}({x_{1}},…,{x_{n}}) \) be the elementary symmetric polynomial such that \( {e_{k}}({x_{1}},…,{x_{n}})=\sum _{1≤{i_{1}} \lt {i_{2}} \lt … \lt {i_{k}}}{x_{{i_{1}}}}\cdot {x_{{i_{2}}}}…{x_{{i_{k}}}} \) .

For example, we know that \( {(a+b)^{2}}={a^{2}}+2ab+{b^{2}} \) , replace \( a+b \) by \( {p_{1}} \) ; \( {a^{2}}+{b^{2}} \) by \( {p_{2}};ab by {e_{2}} and we obtain p_{1}^{2}=2{e_{2}}+{p_{2}}⇒{e_{2}}=\frac{1}{2}(p_{1}^{2}-{p_{2}}) . \) This is important because it generalizes the following summation identity:

\( \sum _{1≤i \lt j}ij=\frac{1}{2}[{(\sum _{n=1}^{∞}i)^{2}}-\sum _{n=1}^{∞}{j^{2}}] \)

Similarity for \( {e_{3}}, \) expanding \( {(a+b+c)^{3}} \) and do some elementary algebra:

\( ⇒{(a+b+c)^{3}}={a^{3}}+{b^{3}}+{c^{3}}+6abc+{3a^{2}}b+{3a^{2}}c+{3b^{2}}a+{3b^{2}}c+{3c^{2}}a+{3c^{2}}b \)

\( =({a^{3}}+{b^{3}}+{c^{3}})+6abc+{3a^{2}}(b+c+a-a)+{3b^{2}}(a+c+b-b)+{3c^{2}}(a+b+c-c) \)

\( =({a^{3}}+{b^{3}}+{c^{3}})+6abc+({3a^{2}}+{3b^{2}}+{3c^{2}})(a+b+c)-({3a^{3}}+{3b^{3}}+{3c^{3}}) \)

\( =-2({a^{3}}+{b^{3}}+{c^{3}})+6abc+3({a^{2}}+{b^{2}}+{c^{2}})(a+b+c) \)

\( ⇒ p_{1}^{3}=-2{p_{3}}+6{e_{3}}+3{p_{2}}{p_{1}}⇒{e_{3}}=\frac{1}{6}(p_{1}^{3}-3{p_{1}}{p_{2}}+2{p_{3}}) \)

We can keep deriving, but it becomes complicated when \( k \) goes larger since it involves a lot of substitution and the coefficients are irregular.

However, there is a way to make it simpler:

Define \( E(t) \) to be the generating function of \( {e_{k}}, \) that is,

\( E(t)=\sum _{k=1}^{∞}{e_{k}}{t^{k}} \)

It can be observed that by the definition of \( {e_{k}} \) ,

\( E(t)=\prod _{i=1}^{n}(1+{x_{i}}t) \)

Taking logarithms from both sides:

( \( Note that ln{X} \) is an alternative notation for \( {log_{e}}{X} \) )

\( ln{E(t)}=ln{\prod _{i=1}^{n}(1+{x_{i}}t)=\sum _{i=1}^{n}ln{(1+{x_{i}}t)}} \)

\( ∵ln{(1+{x_{i}}t)=}\sum _{k=1}^{∞}\frac{{(-1)^{k-1}}{({x_{i}}t)^{k}}}{k} (Taylor \prime s expansion for ln{(1+X)}) \)

\( ∴ln{E(t)=\sum _{k=1}^{∞}\frac{{(-1)^{k-1}}{p_{k}}{t^{k}}}{k}} \)

And taking derivatives respect to \( t \) from both sides of the equation:

\( \frac{E \prime (t)}{E(t)}=\sum _{k=1}^{∞}{(-1)^{k-1}}{p_{k}}{t^{k-1}} \)

\( ⇒E \prime (t)=E(t)\sum _{k=1}^{∞}{(-1)^{k-1}}{p_{k}}{t^{k-1}} \)

\( =(1+{e_{1}}t+{e_{2}}{t^{2}}+…)({p_{1}}-{p_{2}}t+{{p_{3}}t^{2}}-…) \)

By comparing and equating the coefficient of \( {t^{k-1}} \) and therefore obtain:

\( k{e_{k}}=\sum _{i=1}^{k}{(-1)^{k-1}}{{e_{k-i}}p_{i}} \)

Which is so called Newton’s Identities for elementary symmetric polynomials. From using this principle, we may express \( {e_{k}} \) in terms of \( {p_{1}}, {p_{2}},…,{p_{k}} \) (Note that \( {e_{1}}={p_{1}}) \) :

\( {e_{1}}={p_{1}} \)

\( {e_{2}}=\frac{1}{2}(p_{1}^{2}-{p_{2}}) \)

\( {e_{3}}=\frac{1}{6}(p_{1}^{3}-3{p_{1}}{p_{2}}+2{p_{3}}) \)

\( {e_{4}}=\frac{1}{24}(p_{1}^{4}-6p_{1}^{2}{p_{2}}+8{p_{1}}{p_{3}}+3p_{2}^{2}-6{p_{4}}) \)

…

This property will further be using to derive the recursive formula of \( ζ(2k) \) essentially.

3. Associations with Riemann zeta functions

According to Euler’s product formula, for any polynomial function \( P(x) \) , it can be written as

\( P(x)={a_{n}}\prod _{i=1}^{n}(x-{x_{i}}) \)

where \( {x_{1}},{x_{2}},…,{x_{i}} \) are the roots of \( P(x) \) and \( {a_{n}} \) is the leading coefficient of \( P(x) \) .

Let \( P(x)=sin{x} \) , then

\( sin{x}=x\prod _{n=1}^{∞}(1-\frac{{x^{2}}}{{(nπ)^{2}}}) \)

\( ⟹\frac{sin{x}}{x}=\prod _{n=1}^{∞}(1-\frac{{x^{2}}}{{(nπ)^{2}}}) \)

And associate it with Taylor’s expansion of \( \frac{sin{x}}{x} \) :

\( \prod _{n=1}^{∞}(1-\frac{{x^{2}}}{{(nπ)^{2}}})=\sum _{n=1}^{∞}\frac{{(-1)^{n-1}}{x^{2n-2}}}{(2n-1)!} \)

Equating the coefficients of \( {x^{2}},{x^{4}},…,{x^{2n}}: \)

\( {x^{2}}:-\frac{1}{3!}=-\frac{1}{{π^{2}}}\sum _{i=1}^{∞}\frac{1}{{i^{2}}} \)

\( {x^{4}}:\frac{1}{5!}=\frac{1}{{π^{4}}}\sum _{1≤i \lt j}\frac{1}{{i^{2}}{j^{2}}} \)

\( ⋮ \)

\( {x^{2n}}:\frac{{(-1)^{n}}}{(2n+1)!}=\frac{{(-1)^{n}}}{{π^{2n}}}\sum _{1≤{i_{1}} \lt {i_{2}} \lt … \lt {i_{n}}}\frac{1}{i_{1}^{2}i_{2}^{2}…i_{n}^{2}}(n \gt 2) \)

\( {⇒x^{2n}}:\frac{1}{(2n+1)!}=\frac{1}{{π^{2n}}}\sum _{1≤{i_{1}} \lt {i_{2}} \lt … \lt {i_{n}}}\frac{1}{i_{1}^{2}i_{2}^{2}…i_{n}^{2}}(n \gt 2) \)

The summation part

\( \sum _{1≤{i_{1}} \lt {i_{2}} \lt … \lt {i_{n}}}\frac{1}{i_{1}^{2}i_{2}^{2}…i_{n}^{2}}(n \gt 2) \)

can be considered as an elementary symmetric polynomial \( {e_{n}}({x_{1}},…,{x_{k}}) \) where \( {x_{1}},{x_{2}},…,{x_{k}}=\frac{1}{i_{1}^{2}},\frac{1}{i_{2}^{2}},…,\frac{1}{i_{n}^{2}}(n=k) \)

\( ∴{e_{n}}=\sum _{1≤{i_{1}} \lt {i_{2}} \lt … \lt {i_{n}}}\frac{1}{i_{1}^{2}i_{2}^{2}…i_{n}^{2}}=\frac{{π^{2n}}}{(2n+1)!} \)

Therefore, the power sum \( {p_{n}}({x_{1}},…,{x_{k}}) \) has

\( {p_{n}}=\sum _{i=1}^{∞}\frac{1}{{i^{2n}}}=ζ(2n) \)

Associate with Newton’s identities which has been proven in 1.1:

\( n{e_{n}}=\sum _{k=1}^{n}{(-1)^{k-1}}{{e_{n-k}}p_{n}} \)

\( ⇒\frac{{nπ^{2n}}}{(2n+1)!}=\sum _{k=1}^{n}{(-1)^{k-1}}\frac{{π^{2(n-k)}}}{(2(n-k)+1)!}ζ(2k) \)

\( ⇒\frac{{nπ^{2n}}}{(2n+1)!}=\sum _{k=1}^{n-1}{(-1)^{k-1}}\frac{{π^{2(n-k)}}}{(2(n-k)+1)!}ζ(2k)+{(-1)^{n-1}}ζ(2n) \)

\( ⇒{(-1)^{n-1}}ζ(2n)=\frac{{nπ^{2n}}}{(2n+1)!}+\sum _{k=1}^{n-1}{(-1)^{k}}\frac{{π^{2(n-k)}}}{(2(n-k)+1)!}ζ(2k) \)

\( ⇒ζ(2n)={(-1)^{n-1}}(\frac{{nπ^{2n}}}{(2n+1)!}+\sum _{k=1}^{n-1}{(-1)^{k}}\frac{{π^{2(n-k)}}}{(2(n-k)+1)!}ζ(2k)),for n≥2 \)

Which obtain the recursive formula for \( ζ(2n) \) where \( n≥2 . \)

4. Associations with the explicit formula of Riemann zeta functions

For \( n∈{N^{*}} \) , the explicit formula of Riemann zeta function is given by:

\( ζ(2n)=\frac{|{B_{2n}}|{(2π)^{2n}}}{2(2n)!} \)

where \( {B_{n}} \) is the \( n \) th Bernoulli number.

Then according to the recursive formula derived before, the equation

\( \frac{|{B_{2n}}|{(2π)^{2n}}}{2(2n)!}={(-1)^{n-1}}(\frac{{nπ^{2n}}}{(2n+1)!}+\sum _{k=1}^{n-1}{(-1)^{k}}\frac{{π^{2(n-k)}}}{(2(n-k)+1)!}ζ(2k)) for n \gt 1 \)

must be satisfied.

Lemma:

\( xcot{x}=1-\sum _{n=1}^{∞}\frac{{2^{2n}}|{B_{2n}}|}{(2n)!}{x^{2n}} \)

Proof:

Recall that the generating function of Bernoulli numbers \( {B_{n}} \) is

\( \frac{x}{{e^{x}}-1}=\sum _{n=0}^{∞}\frac{{B_{n}}}{n!}{x^{n}} for |x| \lt 2π \)

\( ⟹\frac{x}{2}∙\frac{{e^{x}}+1}{{e^{x}}-1}=1+\sum _{n=2}^{∞}\frac{{B_{n}}}{n!}{x^{n}} \)

Since LHS is an even function and \( {B_{2n+1}}=0 \) for \( n=1,2,3… \)

\( ⟹\frac{x}{2}∙\frac{{e^{x}}+1}{{e^{x}}-1}=\sum _{n=0}^{∞}\frac{{B_{2n}}}{(2n)!}{x^{2n}} for |x| \lt 2π \)

\( ⟹\frac{x}{2}∙\frac{{e^{\frac{x}{2}}}+{e^{-\frac{x}{2}}}}{{e^{\frac{x}{2}}}-{e^{-\frac{x}{2}}}}=\sum _{n=0}^{∞}\frac{{B_{2n}}}{(2n)!}{x^{2n}} \)

\( ⟹\frac{x}{2}coth{\frac{x}{2}}=\sum _{n=0}^{∞}\frac{{B_{2n}}}{(2n)!}{x^{2n}} \)

\( ⟹xcoth{x}=\sum _{n=0}^{∞}\frac{{B_{2n}}{2^{2n}}}{(2n)!}{x^{2n}}, |x| \lt π \)

Since \( coth{ix}=-icot{x} \) , substitute \( ix→x \) to obtain:

\( ixcoth{ix}=\sum _{n=0}^{∞}\frac{{(-1)^{n}}{B_{2n}}{2^{2n}}}{(2n)!}{x^{2n}} \)

\( ⟹xcot{x}=\sum _{n=0}^{∞}\frac{{(-1)^{n}}{B_{2n}}{2^{2n}}}{(2n)!}{x^{2n}}, |x| \lt π \)

\( ∵{(-1)^{n+1}}{B_{2n}}=|{B_{2n}}| \)

\( ∴xcot{x}=-\sum _{n=0}^{∞}\frac{|{B_{2n}}|{2^{2n}}}{(2n)!}{x^{2n}} \)

\( =1-\sum _{n=1}^{∞}\frac{|{B_{2n}}|{2^{2n}}}{(2n)!}{x^{2n}}, |x| \lt π \)

As we successfully obtain the generating function of \( {B_{2k}} \) , we may use it to deduce the generating function of \( ζ(2k) \) :

Since

\( ζ(2k)=\frac{|{B_{2k}}|{(2π)^{2k}}}{2(2k)!} \) (1)

and

\( xcot{x}=1-\sum _{k=1}^{∞}\frac{{2^{2k}}|{B_{2k}}|}{(2k)!}{x^{2k}} \) (2)

By substituting \( x=πx \) into (2):

\( ⟹πxcot{πx}=1-\sum _{k=1}^{∞}\frac{{2^{2k}}|{B_{2k}}|}{(2k)!}{(πx)^{2k}} \)

\( ⟹1-πxcot{πx}=\sum _{k=1}^{∞}\frac{{(2π)^{2k}}|{B_{2k}}|}{(2k)!}{x^{2k}} \)

\( ⟹\frac{1}{2}(1-πxcot{πx})=\sum _{k=1}^{∞}\frac{{(2π)^{2k}}|{B_{2k}}|}{2(2k)!}{x^{2k}} \)

And associate with (1):

\( ⟹\frac{1}{2}(1-πxcot{πx})=\sum _{k=1}^{∞}ζ(2k){x^{2k}} \)

Now consider another series which

\( xcot{x}=\sum _{k=1}^{∞}{a_{k}}{x^{2k}} \)

Where \( {a_{k}} \) is any real valued function

\( ⇒cos{x}=\frac{sin{x}}{x}\sum _{k=0}^{∞}{a_{k}}{x^{2k}} \)

\( ⇒\sum _{n=0}^{∞}\frac{{(-1)^{n}}{x^{2n}}}{(2n)!}=\sum _{n=0}^{∞}\frac{{(-1)^{n+1}}{x^{2n}}}{(2n+1)!}\sum _{k=0}^{∞}{a_{k}}{x^{2k}} \)

\( =\sum _{n=0}^{∞}(\sum _{k=0}^{n}{(-1)^{k}}\frac{{a_{n-k}}}{(2k+1)!}){x^{2n}} \)

By equating the coefficients of \( {x^{2n}} \) therefore obtain:

\( \frac{{(-1)^{n}}}{(2n)!}=\sum _{k=0}^{n}{(-1)^{k}}\frac{{a_{n-k}}}{(2k+1)!} \)

\( ⇒{a_{n}}=\frac{{(-1)^{n}}}{(2n)!}-\sum _{k=1}^{n}{(-1)^{k}}\frac{{a_{n-k}}}{(2k+1)!} \)

\( =\frac{{2n(-1)^{n}}}{(2n+1)!}-\sum _{k=1}^{n-1}{(-1)^{k}}\frac{{a_{n-k}}}{(2k+1)!} \)

From (3), it can be deduced that

\( {a_{n}}=-\frac{2ζ(2n)}{{π^{2n}}} \)

Therefore,

\( -\frac{2ζ(2n)}{{π^{2n}}}=\frac{{2n(-1)^{n}}}{(2n+1)!}-\sum _{k=1}^{n-1}{(-1)^{k}}\frac{2ζ(2(n-k))}{(2k+1)!{π^{2(n-k)}}} \)

\( ⇒ζ(2n)=\frac{{n{π^{2n}}(-1)^{n-1}}}{(2n+1)!}-\sum _{k=1}^{n-1}{(-1)^{k}}\frac{ζ(2(n-k))}{(2k+1)!} \)

\( ={(-1)^{n-1}}(\frac{{nπ^{2n}}}{(2n+1)!}+\sum _{k=1}^{n-1}{(-1)^{k}}\frac{{π^{2k}}}{(2(n-k)+1)!}ζ(2k)) \)

Which corresponds to the recursive formula of \( ζ(2n) \)

Therefore, it has been successfully proven the equivalent relationship between the recursive formula and the explicit formula for Riemann zeta function at even positive integers.

5. Conclusion

By associating the coefficients of the terms of Taylor’s expansion and Euler’s product of \( \frac{sin{x}}{x} \) with the essential properties of elementary symmetric polynomials (Newton’s identities), the recursive formula of Riemann zeta function at natural even numbers can be expressed by the previous values of it. This method of evaluating the values of zeta function at large even numbers, can be helpful in particular occasion than applying the explicit formula since it is difficult to calculate the values of large Bernoulli numbers.