Euler calculates like we breath.
François Arago
https://en.wikiquote.org/wiki/Leonhard_Euler
In the textbook “Klassische elementare Analysis 1987. Max Koecher” the following elegant recurrence relation is proved for
ζ(2n)
on p. 157:
n+
1
2
ζ(2n)=
n-1
∑
k=1
ζ(2k)ζ(2n-2k)
for
n>1
https://en.wikipedia.org/wiki/Particular_values_of_the_Riemann_zeta_function
The main purpose of this notebook is to proof the above recurrence relation inspired by [2]. But it was already known to Euler 1739, see E130 §23 and the Appendix.
We start with the partial fraction expansion of Cot:
​https://en.wikipedia.org/wiki/Trigonometric_functions
Out[]=
(1)
∞
∑
k=1
x
x+k
+
x
x-k
πxcot(πx)-1
The partial fraction can be rewritten as:
Out[]=
(2)
x
x+k
+
x
x-k

2
2
x
2
x
-
2
k
So (1) turns into the definition of the function
f(x)
:
​This equation was known to Euler by 1739, see the Appendix.
Out[]=
(3) 1-
∞
∑
k=1
2
2
x
2
k
-
2
x
πxcot(πx)f(x)
And here is a ComplexPlot of
f(x)
with the zero’s and poles on the real axis:
Out[]=
And Plot of
f(x)
for real
x
:
Out[]=
(2) can be written as an infinite sum:
Out[]=
(4)
∞
∑
n=1
2
2n
x
2n
k

2
2
x
2
k
-
2
x
and insert (4) into (3) to get:
Out[]=
(5) 1-
∞
∑
k=1
∞
∑
n=1
2
2n
x
2n
k
πxcot(πx)
The sums can be swapped, because
2n
x
2n
k
is bounded (
k,nϵ,x<1
)
Out[]=
(6) 1-
∞
∑
n=1
∞
∑
k=1
2
2n
x
2n
k
πxcot(πx)
The term
2n
x
can be extracted from the second sum and 2 from both sums:
Out[]=
(7) 1-2
∞
∑
n=1
2n
x
∞
∑
k=1
1
2n
k
πxcot(πx)
The second sum is:
Out[]=
(8)
∞
∑
k=1
1
2n
k
ζ(2n)
And (7) can now be written as:
Out[]=
(9) 1-2
∞
∑
n=1
2n
x
ζ(2n)πxcot(πx)
The cot has the following derivative:
Out[]=
(10) -
∂cot(x)
∂x

2
cot
(x)+1
Proof of identity (10):
In[]:=
FullSimplify[Activate[eqn10]]
Out[]=
True
f(x)
from (3) has this property:
Out[]=
(11) x
′
f
(x)f(x)-
2
f(x)
-
2
π
2
x
Proof of identity (11):
In[]:=
Simplify[ReleaseHold@eqn11/.f->Function[{x},πxCot[πx]]]
Out[]=
True
Now we insert the l.h.s. of (9) into (11) and extract the coefficients:
Out[]=
6ζ(2)
2
π
2
2
ζ(2)
5ζ(4)
4ζ(2)ζ(4)7ζ(6)
2
2
ζ(4)
+4ζ(2)ζ(6)9ζ(8)
4ζ(4)ζ(6)+4ζ(2)ζ(8)11ζ(10)
2
2
ζ(6)
+4ζ(4)ζ(8)+4ζ(2)ζ(10)13ζ(12)
30ζ(14)8(ζ(6)ζ(8)+ζ(4)ζ(10)+ζ(2)ζ(12))
It can be solved for
ζ(2n)
:
Out[]=
ζ(2)
2
π
6
ζ(4)
2
2
ζ(2)
5
ζ(6)
4
7
ζ(2)ζ(4)
ζ(8)
2
9
(
2
ζ(4)
+2ζ(2)ζ(6))
ζ(10)
4
11
(ζ(4)ζ(6)+ζ(2)ζ(8))
ζ(12)
2
13
(
2
ζ(6)
+2ζ(4)ζ(8)+2ζ(2)ζ(10))
ζ(14)
4
15
(ζ(6)ζ(8)+ζ(4)ζ(10)+ζ(2)ζ(12))
The building law is:
​This recurrence relation was also known to Euler! See the Appendix.
In[]:=
recurrencerelation[m_Integer?EvenQ]:=ζ[m]->
2
m+1
Sum[ζ[k]·ζ[m-k],{k,2,m-2,2}]
Here Centerdot, . is used for non commutative multiplication, so you can see better, how the calculation is performed.
Example:
Out[]=
ζ(10)
2
11
(ζ(2)·ζ(8)+ζ(4)·ζ(6)+ζ(6)·ζ(4)+ζ(8)·ζ(2))
Allow commutation with Times:
In[]:=
%/.CenterDot->Times
Out[]=
ζ(10)
2
11
(2ζ(4)ζ(6)+2ζ(2)ζ(8))
In[]:=
Simplify[%]
Out[]=
ζ(10)
4
11
(ζ(4)ζ(6)+ζ(2)ζ(8))
Now the explicit recurrence relation is (*):
In[]:=
zeta[2]:=
ζ
2
(*=
2
π
/6*)​​zeta[m_Integer?EvenQ]:=zeta[m]=
2
m+1
Sum[zeta[k]zeta[m-k],{k,2,m-2,2}]
Example:
In[]:=
Map[zeta,Range[2,14,2]]
Out[]=

ζ
2
,
2
2
ζ
2
5
,
8
3
ζ
2
35
,
24
4
ζ
2
175
,
32
5
ζ
2
385
,
44224
6
ζ
2
875875
,
768
7
ζ
2
25025

In[]:=
%/.
ζ
2
->
2
π
/6
Out[]=

2
π
6
,
4
π
90
,
6
π
945
,
8
π
9450
,
10
π
93555
,
691
12
π
638512875
,
2
14
π
18243225

Compare with built in Zeta:
In[]:=
Map[Zeta,Range[2,14,2]]
Out[]=

2
π
6
,
4
π
90
,
6
π
945
,
8
π
9450
,
10
π
93555
,
691
12
π
638512875
,
2
14
π
18243225


Appendix: Euler E130 and the Cot series

Leonhard Euler, Princeps Mathematicorum.
Salutation in a Letter from 1745 by
Johann Bernoulli (1667-1748).
http://eulerarchive.maa.org/correspondence/correspondents/JBernoulli.html​
​http://eulerarchive.maa.org/correspondence/letters/OO0226.pdf
S.P.D. salutem plurimam dicit.
https://scholarlycommons.pacific.edu/euler-works/130/​
E130 De seriebus quibusdam considerationes (Considerations on certain series). Written 1739. Published 1750.
​Euler factors a power series into an infinite product, applying it to
f'/f
, and evaluates a number of series using what look like trigonometric substitutions. He also applies this to the differential equation
zs'–s–sz+
2
s
=0
by recursion. Then he attempts to evaluate
ζ(2k+1)
.
There is an english translation:
​https://download.uni-mainz.de/mathematik/Algebraische%20Geometrie/Euler-Kreis%20Mainz/E130RevisedVersion1.pdf
In §9 Euler derives:
Euler writes the right side differently:
But it is the same:
The cot series appears with:
In §17 Euler derives:
and:
and:
and:
and:
In §22 Euler derives (3) from above:
but writes the r.h.s. as:
In §23 Euler writes down the recurrence relation.
​https://scholarlycommons.pacific.edu/euler-works/130/
euler[1] is A, euler[2] is B etc.:
It is the same as above (*):
In § 24 Euler writes:
​https://scholarlycommons.pacific.edu/euler-works/130/
It’s the same:
https://mathoverflow.net/questions/306898/a-recurrence-relation-for-zeta2n-reference-request​
[1] G. T. Williams, “A new method of evaluating ζ(2n)“, Amer. Math. Soc, 60, (1953) 19--25.
[2] Max Koecher, Klassische elementare Analysis (1987) p. 157.
​[0] Euler E130 §23 (Written 1739, published 1750)

CITE THIS NOTEBOOK

An elegant recurrence relation for Zeta(2n), Euler E130​
by Oliver Seipel​
Wolfram Community, STAFF PICKS, January 3, 2024
​https://community.wolfram.com/groups/-/m/t/3094943