Higher order analogue of Yoshio Matsuoka's formula on Basel Problem

Shigeki Matsumoto, Konan University, Japan

Yoshio Matsuoka's formula

Basel problem : What is the value (∑_(k=1)^∞)(1/(k^2)) ?

The Basel problem asks for the precise summation of the reciprocals of the squares of the natural numbers.

Mathematica knows the value.

(∑_(k=1)^∞)(1/(k^2))

(∑_(k=1)^∞)(1/(k^2))==((π^2)/6)

Yoshio Matsuoka showed the following integral formula (1961) and solved Basel Problem.

((2(∫_0^(π/2))(x^2)(cos^(2n))xdx)/((∫_0^(π/2))(cos^(2n))xdx))=((π^2)/6)-(∑_(k=1)^n)(1/(k^2))

(((∫_0^(π/2))(x^2)((Cos[x])^(2n))\[DifferentialD]x)/((∫_0^(π/2))((Cos[x])^(2n))\[DifferentialD]x))//Timing

Simplify[(((∫_0^(π/2))(x^2)((Cos[x])^(2n))\[DifferentialD]x)/((∫_0^(π/2))((Cos[x])^(2n))\[DifferentialD]x))-(((π^2)/6)-(∑_(k=1)^n)(1/(k^2))),Assumptions->n∈Integers&&n≥1]

The above tends to zero if n -> ∞.

This is the starting point of my argument.

We explore higher order analogue of the above using Mathematica.

"the fourth power" case

I think it is natural to ask what the following integral("the fourth power of x") is.

(((∫_0^(π/2))(x^4)(cos^(2n))xdx)/((∫_0^(π/2))(cos^(2n))xdx))

(((∫_0^(π/2))(x^4)((Cos[x])^(2n))\[DifferentialD]x)/((∫_0^(π/2))((Cos[x])^(2n))\[DifferentialD]x))//Timing

For small n, Mathematica gives the folllowing result.

Table[{n,(((∫_0^(π/2))(x^4)((Cos[x])^(2n))\[DifferentialD]x)/((∫_0^(π/2))((Cos[x])^(2n))\[DifferentialD]x))},{n,1,5}]//Apart//TableForm

Q :What is the closed form for the above?

A: We find rules of these coefficients of polynomials of π.

Table[(1/4)(∑_(k=1)^n)(1/(k^2)),{n,1,5}]

Table[(3/2)(∑_(b=1)^n)(∑_(a=1)^b)(1/((a^2)(b^2))),{n,1,5}]

Table[(3/4)((((∑_(a=1)^n)(1/(a^2))))^2)+(3/4)(∑_(k=1)^n)(1/(k^4)),{n,1,5}]

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Remark: WolframAlpha can not find the rule of the sequence {1,(5/4),(49/36),(205/144),(5269/3600)}.

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The following equation is expected to be correct.

(((∫_0^(π/2)) (x^4) (cos^(2n))xdx)/((∫_0^(π/2))(cos^(2n))xdx))=((π^4)/80)-((π^2)/4)(∑_(a=1)^n)(1/(a^2))+(3/2)(∑_(b=1)^n)(∑_(a=1)^b)(1/((a^2)(b^2)))

↑multiple zeta sum

I will expalain the coefficients   {((π^4)/80),-((π^2)/4),(3/2)}.

If the above is correct for every positive integer n, we can have

Solve[0==((π^4)/80)-((π^2)/4)(∑_(a=1)^∞)(1/(a^2))+(3/2)sol]

(∑_(b=1)^∞)(∑_(a=1)^b)(1/((a^2)(b^2)))==((7 (π^4))/360)

↑multiple zeta value

Mathematica knows the value!

(∑_(b=1)^∞)(∑_(a=1)^b)(1/((a^2)(b^2)))

Evaluating (∑_(n=1)^∞)(1/(n^4))

Solve[0==((π^4)/80)-((π^2)/4)(∑_(a=1)^∞)(1/(a^2))+(3/4)((((∑_(a=1)^∞)(1/(a^2))))^2)+(3/4)sol]

(∑_(n=1)^∞)(1/(n^4))==((π^4)/90)

"the sixth power" case

We are also interested in the following integral("the sixth power of x").

(((∫_0^(π/2))(x^6)(cos^(2n))xdx)/((∫_0^(π/2))(cos^(2n))xdx))

The following equalities are brought from the result of a numeric experiment by using Mathematica.

(((∫_0^(π/2)) (x^6) (cos^(2n))xdx)/((∫_0^(π/2))(cos^(2n))xdx))=((π^6)/448)-((3 (π^4))/32)(∑_(a=1)^n)(1/(a^2))+((15 (π^2))/8)(∑_(b=1)^n)(∑_(a=1)^b)(1/((a^2) (b^2)))-(45/4) (∑_(c=1)^n)(∑_(b=1)^c)(∑_(a=1)^b)(1/((a^2) (b^2) (c^2)))

I will expalain also the coefficients   {((π^6)/448),-((3 (π^4))/32),((15 (π^2))/8),-(45/4)}

Especially

(∑_(c=1)^∞)(∑_(b=1)^c)(∑_(a=1)^b)(1/((a^2) (b^2) (c^2)))=((31 (π^6))/15120).

Solve[((π^6)/448)-((3 (π^4))/32)(∑_(a=1)^∞)(1/(a^2))+((15 (π^2))/8)(∑_(b=1)^∞)(∑_(a=1)^b)(1/((a^2) (b^2)))-(45/4) sol==0]

Mathematica can not evaluate the following

(∑_(c=1)^∞)(∑_(b=1)^c)(∑_(a=1)^b)(1/((a^2) (b^2) (c^2)))

Mathematica evaluates partial sums very slowly.

N[(∑_(c=1)^1000)(∑_(b=1)^c)(∑_(a=1)^b)(1./((a^2)(b^2)(c^2)))]//Timing

N[(∑_(c=1)^1000)(∑_(b=1)^c)(∑_(a=1)^b)(1/((a^2)(b^2)(c^2)))]//Timing

((31 (π^6))/15120)//N

Here, I finally noticed an important following facts.

The sequence {(((∫_0^(π/2))(x^(2p))(cos^(2n)) xdx)/((∫_0^(π/2))(cos^(2n)) xdx))} satisfies the following difference equation:

(((∫_0^(π/2))(x^(2p))(cos^(2n)) xdx)/((∫_0^(π/2))(cos^(2n)) xdx))-(((∫_0^(π/2))(x^(2p))(cos^(2(n-1)))xdx)/((∫_0^(π/2))(cos^(2(n-1)))xdx))=-((2p (2p-1))/((2^2)(n^2)))(((∫_0^(π/2))(x^(2(p-1)))(cos^(2n)) xdx)/((∫_0^(π/2))(cos^(2n)) xdx))       （＊）

general case (the 2p-th power case).

Using the defference equation

(((∫_0^(π/2))(x^(2p))(cos^(2n)) xdx)/((∫_0^(π/2))(cos^(2n)) xdx))-(((∫_0^(π/2))(x^(2p))(cos^(2(n-1)))xdx)/((∫_0^(π/2))(cos^(2(n-1)))xdx))=-((2p (2p-1))/((2^2)(n^2)))(((∫_0^(π/2))(x^(2(p-1)))(cos^(2n)) xdx)/((∫_0^(π/2))(cos^(2n)) xdx))

we generate the red lists of coefficients.

p=3;Prepend[-((2p(2p-1))/(2^2)){((π^4)/80),-((π^2)/4),(3/2)},(((∫_0^(π/2)) (x^(2p))\[DifferentialD]x)/((∫_0^(π/2))\[DifferentialD]x))]

{((π^6)/448),-((3 (π^4))/32),((15 (π^2))/8),-(45/4)}

p=3;((Table[(((-1))^k) Binomial[2p+1,2k](2k)!(π^(2p-2k)),{k,0,p}])/((2p+1)(2^(2p))))

cc=Table[(1/((2p+1)(2^(2p))))Table[(((-1))^k) Binomial[2p+1,2k](2k)!(π^(2p-2k)),{k,0,p}],{p,1,6}];

cc//TableForm

Solve[
{
cc[[1]].{1,(Subscript[x, 1])}==0,
cc[[2]].{1,(Subscript[x, 1]),(Subscript[x, 2])}==0,
cc[[3]].{1,(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3])}==0,
cc[[4]].{1,(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3]),(Subscript[x, 4])}==0,
cc[[5]].{1,(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3]),(Subscript[x, 4]),(Subscript[x, 5])}==0,
cc[[6]].{1,(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3]),(Subscript[x, 4]),(Subscript[x, 5]),(Subscript[x, 6])}==0
},
{(Subscript[x, 1]),(Subscript[x, 2]),(Subscript[x, 3]),(Subscript[x, 4]),(Subscript[x, 5]),(Subscript[x, 6])}
]

We have the following values of multiple zeta sums.

(∑_(a=1)^∞)(1/(a^2))=((π^2)/6).

(∑_(b=1)^∞)(∑_(a=1)^b)(1/((a^2) (b^2)))=((7 (π^4))/360).

(∑_(c=1)^∞)(∑_(b=1)^c)(∑_(a=1)^b)(1/((a^2) (b^2) (c^2)))=((31 (π^6))/15120).

(∑_(d=1)^∞)(∑_(c=1)^d)(∑_(b=1)^c)(∑_(a=1)^b)(1/((a^2) (b^2) (c^2)(d^2)))=((127 (π^8))/604800).

(∑_(e=1)^∞)(∑_(d=1)^e)(∑_(c=1)^d)(∑_(b=1)^c)(∑_(a=1)^b)(1/((a^2) (b^2) (c^2)(d^2)(e^2)))=((73 (π^10))/3421440).

(∑_(f=1)^∞)(∑_(e=1)^f)(∑_(d=1)^e)(∑_(c=1)^d)(∑_(b=1)^c)(∑_(a=1)^b)(1/((a^2) (b^2) (c^2)(d^2)(e^2)(f^2)))=((1414477 (π^12))/653837184000).

Bernoulli numbers

Paul C Abbott pointed out a very important fact; a connection(relation) between the above summations (multiple zeta value) and Bernoulli numbers.

Paul >The general form of these nested sums appears to be

Table[{k,(((((-1))^(k-1))((2^(2k))-2)BernoulliB[2k])/((2k)!))(π^(2k)),(((2^(2k-1))-1)/(2^(2k-2)))(Zeta[2k])},{k,1,6}]

In other words, multiple zeta values appear as the coefficients in the Laurent expansion of cosec x.

Series[Csc[x],{x,0,12}]//Normal

Summary

Theorem(Paul C Abbott)

(lim_(n ->∞)) (MZ^((k)))[n]=(((((-1))^(k-1))((2^(2k))-2)BernoulliB[2k])/((2k)!))(π^(2k))             （＊＊）

e.g.     (MZ^((3)))[n]=(∑_(c=1)^n)(∑_(b=1)^c)(∑_(a=1)^b)(1/((a^2) (b^2) (c^2)))

Sketch of Proof. In the numerical computation for (((∫_0^(π/2)) (x^(2p)) (cos^(2n))xdx)/((∫_0^(π/2))(cos^(2n))xdx)) ( p=1,2,3,4,...) by using Mathematica we run into a difference equation (＊),which leads

(((∫_0^(π/2))(x^(2p))(cos^(2n))xdx)/((∫_0^(π/2))(cos^(2n))xdx))=(1/((2p+1)(2^(2p))))(∑_(k=0)^p)(((-1))^k) Binomial[2p+1,2k](2k)!(π^(2p-2k))(MZ^((k)))[n].

These integral formulas and the following (well known) linear relations of Bernoulli numbers reveals the identities (＊＊) of the theorem.

(∑_(k=1)^p)Binomial[2p+1,2k]((2^(2k))-2)BernoulliB[2k]==1

Reference

Yoshio Matsuoka, "An Elementary Proof of the Formula for Pi", The American Mathematical Monthly, vol. 68, no. 5, pp. 485-487, 1961.