Ramanujan-Type Series Related to Clausen's Product

0.19 MB·English

by  John M. Campbell

Ramanujan-Type Series Related to Clausen’s
Product
John M. Campbell
York University
Toronto, ON
Canada
1
1 [email protected]
0
2
Abstract

n
a Infinite series are evaluated through the manipulation of a series for cos(2tsin 1x)
J −
resulting from Clausen’s Product. Hypergeometric series equal to an expression in-
7
volving 1 are determined. Techniques to evaluate generalized hypergeometric series
π
]
are discussed through perspectives of experimental mathematics.
T
N
.
h 1 Introduction
t
a
m
Clausen’s product is discussed in the fascinating book by Borwein, Bailey, and Girgensohn,
[
Experimentation in Mathematics: Computational Paths to Discovery [4]. New Ramanujan-

3
type series derived from Clausen’s product such as the following are given in this paper:
v
1 ∞n=2(41)3n−1((2n)!(n(2+n1−)(32)n!(+4n1)+)23()n! 2)!n!; the following section (Ramanujan-Type Series) ex-
4 pPlains the derivation of this ser−ies and a related series. We discuss the evaluation of hyperge-
9
ometric series through integral substitution. For example, we shall soon consider an infinite
1
. array of hypergeometric series of the form F (1, 2k, 2n 2k; 3,2;1). Furthermore, we discuss
0 3 2 2 n −n 2
1 methods to incorporate the Riemann-zeta function into hypergeometric-like series, and dis-
0
cuss related new hypergeometric generalizations, evaluating hypergeometric series such as
1
: F (1,1, 9, 5, 11,3;2, 37, 41, 45, 49;1).
v 6 5 4 2 4 16 16 16 16
i Weareinterestedintheevaluationofinfiniteseries. Therehasbeenmuchworkdiscussing
X
hypergeometric identities of theform F (a ,a ,...,a ;b ,b ,...,b ;x) forconstant x such that
r p q 1 2 p 1 2 q
a a ,a ,...,a and/or b ,b ,...,b contain a variable/variables. There does not seem to be as
1 2 p 1 2 q
much work on hypergeometric series of the form F (a ,a ,...,a ;b ,b ,...,b ;x) for variable
p q 1 2 p 1 2 q
x, which can be manipulated through integration with respect to x (and by replacing x by
f(x) and integrating). For example, only series of the form F (a ,a ,...,a ;b ,b ,...,b ;x)
p q 1 2 p 1 2 q
for constant x are discussed in [7] (which is cited in [10]).
“Common”, modern computational software is usually unable to directly evaluate hy-
pergeometric series of the form F (a ,a ,...,a ;b ,b ,...,b ;x) for variable x; we thus con-
p q 1 2 p 1 2 q
sider the evaluation of such series to be important. Consider infinite series of the form
g(n)(1 xa)bnxdn, where g(n) represents a combination of a finite number of elementary
−
Pfunctions (and a, b, and d are positive integers). Integration of the expression equal to such
series isoften very difficult, yielding complicated results, some of which we shall here present.
1
The evaluation of a series such as ∞ 4k is an example of a tangible ap-
k=0 (2k)(2k+1)(k+1)2
k
proach to exploring series representatioPns of important constants through integral substitu-
tion. Although the theorems below have somewhat more cumbersome derivations, briefly
describing our evaluation of the aforementioned series will hopefully serve as a straightfor-
ward introduction to this paper. The following well-known and important series (for x2 1)
≤
is given in [8]: (sin−1x)2 = ∞k=0 (42kk(+k!1))2!x(k2k++12). Divide both sides of this equation by x and
integrate: P
(sin 1x)2 Li (e2isin−1x) 1
− dx = 3 +isin 1xLi (e2isin−1x)+ i(sin 1x)3+(sin 1x)2ln(1 e2isin−1x)
− 2 − −
Z x 2 3 −
Give the above integral an upper limit of one and a lower limit of zero, resulting in an
integral evaluated in [6] (described as “an interesting definite integral”). It is now clear that
∞ 4k = π2 ln(2) 7ζ(3). The following related series is elegantly proven in [5]:
k=0 (2k)(2k+1)(k+1)2 2 − 4
k
P∞ 4n = π2ln(2) 7ζ(3).
n=1 n3(2n) − 2
n
P Let us continue by providing the infinite array of F series we have in mind, which is
3 2
based upon Clausen’s product. Consider:
k n k 3 cscxsin(( 1+2(1 k))sin 1(sinx))
( F ( , − ; ;sin2x))2 = ( − − n − )2
2 1 n n 2 1+2(1 k)
− − n
Theorem 1.
π2(cscxsin((−1+2(1− nk))sin−1(sinx)))2dx =
Z 1+2(1 k)
0 − − n
4ikπ
1 2ikπ n e n n 2ikπ 2kπ 1 k k
ie n n( + 2e n πcot( ) ψ0( )+ψ0(1 )
−4( 2k +n) 2k n 2k n − n − 2 − n − n
− − −
1 k k
4ikπ 4ikπ
+e n ψ0( + ) e n ψ0( ))
−2 n − n
=
1 2k 2n 2k 3 π
F ( , , − ; ,2;1)
3 2
2 n n 2 2
Proof. Given Clausen’s product, the above theorem holds.
Perhaps it is probable that the above theorem has been considered before.
Integrating ( F (2 k, k; 5;sin2x))2 proves to be much more cumbersome, involving
2 1 − n n 2
multifarious hypergeometric series.
Now, let us briefly discuss some more simple results.
2
Clausen’s product can be used to obtain the following relationships [4]:
∞ (t)n(−t)n(2x)2n = cos(2tsin 1x) (1)
−
(2n)!
Xn=0
1 ∞ (t)n(−t)n(4sin2x)n = sin2(tx)
−2 (2n)!
Xn=1
A generalization of (1) which does not involve the Pochhammer symbol is given in [9].
Consider integrating (1) for cos(sin 1(√x)):
−
∞ (12)n(−12)n 4nxn+1 = 2(1 x)32 + 2 (2)
(2n)!(n+1) −3 − 3
Xn=0
Consider integrating (2):
∞ (21)n(−12)n 4nxn+2 = 2( 2(1 x)52 x) 4 (3)
(2n)!(n+1)(n+2) −3 −5 − − − 15
Xn=0
(2) can be transformed as follows:
∞ (12)n(−12)n 4nx2n+3 = 2(√1 x2(5x x3) x+ 3 sin 1x) (4)
−
(2n)!(n+1)(2n+3) −3 − 8 − 4 − 8
Xn=0
Using such techniques, it is not difficult to evaluate series such as ∞ (16)n(−16)n or
n=0 (2n)!(2n+3)
∞ (18)n(−81)n . Simple variations of series such as (2) often lead to outlaPndish results.
n=0 (2n)!(n+2)
P Consider multiplying both sides of (4) by x and integrating:
∞ (21)n(−12)n 4nx2n+5 = 1 (32x3 18x2sin 1(x)+√1 x2(4x2 16x2 3)x+3sin 1x)
− −
(2n)!(n+1)(2n+3)(2n+5) 144 − − − −
Xn=0
(5)
The Ramanujan-type series which follow are proven using the following very important
definite integral:
π2 sin2ntdt = π(21)n (6)
Z 2 n!
0
Consider substituting (6) into (1) and variations of (1), yielding fairly simple series,
the partial sums of many of which are known. For example consider the partial sums of
∞n=04n(312)nn(−!(231n2))!n(21)n or ∞n=2(41)2n−1(n2n(nn−−+131))((2nn+n+21)).
P Relationships such asP(5) can be manipulated using (6), resulting in series which are
perhaps new; even if such series are not outright new, some computational software is unable
3
to evaluate such series, and they do not appear in authoritative mathematical literature on
the topic.
The following integral is worthy of mention:
π π23Γ(n+1)
xsinnxdx = 2 (7)
Z 2Γ(1+ n)
0 2
Now, consider using relationships such as the following to manipulate series such as (2):
π 2 2 π
x( (1 sinx)32)dx = (20 16√2+3π) (8)
Z 3 − 3 − 9 −
0
Let us begin by acknowledging that it seems to us that the following relationship is highly
important, at least in terms of the evaluation of hypergeommetric series:
∞
(((1 xa)(bn)+c)(n+g)dfn = f(1 xa)b+cΦ(f(1 xa)b, d,1+g)
− − − −
Xn=1
So, the above series can be transformed into a hypergeometric series. If Re(a) > 0 and
Re(c+bn) > 1, then:
−
1 Γ(1+ 1)Γ(1+c+bn)
(((1 xa)(bn)+c)(n+g)dfndx = fn(g +n)d a (9)
Z − Γ(1+ 1 +c+bn)
0 a
The above relationships lead to an extremely important question, which is author is
presently unable to answer: Is it possible to determine a meaningful, general evalu-
ation of the following integral?
1
f(1 xa)b+cΦ(f(1 xa)b, d,1+g)dx
Z − − −
0
The above integral leads to very beautiful evaluations of hypergeometric series.
Questions related to the above question such as the following arise: “Is it possible to
π
meaningfully evaluate 2 sin 1sin2xdx?”
−
0
Given that ∞ (1R−xa)bn = (1−xa)2b , is it possible to determine a meaningful,
n=2 cn −c( c+(1 xa)b)
general evaluatiPon of 1 (1−xa)2b −dx?−“General” evaluations of this integral for only a
0 −c( c+(1 xa)b)
single variable are verRy comp−licat−ed: consider Theorem 17.
Given that ∞n=2 (1−nxca)bn = −(1 − xa)b + Lic((1 − xa)b), is it possible to determine a
meaningful, genPeral evaluation of 1 (1 xa)b + Li ((1 xa)b)dx? These questions seem
0 − − c −
to us to be highly important. InR[13] it is stated that “No general algorithm is known
for integration of polylogarithms of functions.” Such general series can be used to evaluate
generalized hypergeometric series F for relatively arbitrary p and q.
p q
Moregenerally, comparedto(9),weareinterestedinintegralsoftheform p(f(x))ng(x)dx
q
which are equal to a finite number of combinations of elementary functions,Rwhere f(x) and
g(x) represent a finite number of combinations of elementary functions such that n Z+;
∈
4
such integrals can be used to establish new hypergeometric series. Let h(x) = (f(x))ng(x).
Consider the following straightforward two steps:
1) If possible, evaluate ∞ h(x)j(n), where j(n) represents a finite combination of
n=1
elementary functions (e.g. j(Pn) = n45+1)
n+2
p
2) Integrate the above sum (i.e. using h(x)dx) and its equivalent expression
q
Let the above simple two-step proceduRre be called the HJ-algorithm. In this paper, we
present some general tendencies of the HJ-algorithm, through perspectives of experimental
mathematics. As stated in the excellent Mathematics by Experiment: Plausible Reasoning
in the 21st Century [3],
Thecomputer provides themathematicianwitha“laboratory”inwhich heorshe
canperformexperiments: analyzing examples, testing outnewideas, orsearching
for patterns.
Most of the theorems in this paper are not given rigorous analyses; indeed, most of the
series are presented for their own sake. Although the ideas (i.e. theorems) being “tested”
in this paper are new in the sense that they may not have been previously published or
widely recognized, the HJ-algorithm is so simple that it can hardly be described as a “new
idea”. Finally, although we present some seemingly unpredictable tendencies of the HJ-
algorithm, we presently do not claim to have considerable knowledge of general patterns
in the hypergeometric series. Indeed, some modern computational software is unable to
evaluate extremely important integrals such as 1 (1−xa)b dx.
0 c+(1 xa)b
Many oftheevaluations ofthehypergeometriRc se−ries−oftheformswe analyze inthispaper
involve special functions such as the polylogarithm function; for the time being, we mostly
discuss hypergeometric series which have a “nice” (although often long and intricate) form
consisting of a finite number of combinations of elementary functions; evaluating an infinite
series (i.e. a hypergeometric series) using another infinite series (i.e. the polylogarithm
function) seems somwhat redundant. Having said that, we present a way to incorporate the
Riemann-zeta function into certain forms of hypergeometric series.
We will now present a variety of infinite series. Theorem 2 is particularly worthy of
note, because it is most likely to be original, and does not seem to be equal to a single
hypergeometric function.
2 Ramanujan-Type Series
Theorem 2.
∞ (1)3n 1 (2n−3)!(4n+3)! = 553 42ln(22−+√√22)−34√8 (10)
−
4 ((2n)!(n+1)(2n+1))2(n 2)!n! − 96 − 9π
Xn=2 −
Proof. Theorem 2 spawns from (2). Determine (1 sin4x)32dx and substitute (6) into
−
(2). R
5
Theorem 3.
5 1 3 1 1141 103ln(2−√2)
F ( , , ; ,1;1) = 2+√2 (11)
3 2
−2 4 4 2 960√2π − 256π
Proof. Use the same technique as in Theorem 2, using (3).
Use similar techniques to prove the following very similar equation:
5 1 3 3 1067 383tanh−1( 1 )
F ( , , ; ,2;1) = + √2
3 2
−2 4 4 2 6720√2π 128π
Considerhowzeta-typesumscanbedeterminedusing(6). InthegreatbookComputation
Techniques for the Summation of Series [11], Anthony Sofo provides the following series,
which results from integrating 1 ∞ (2x)2n repeatedly:
2 n=1 n2(2n)
n
P
∞ (2x)2n+3 x(2x2 +3)(sin−1x)2 2√1 x2(11x2 +4)sin−1x 85x3 8x
= + −
n2 2n 4(2n+3)(2n+2)(2n+1) 3 9 − 27 − 9
Xn=1 n
(cid:0) (cid:1)
We leave it as an exercise to demonstrate how (6) and the above series can be used to
prove the equation ∞n=1 n2(n+11)2(n+2) = 2π28−19, the partial sums of the series of which are
known. P
The above series may be simplified as follows:
7 x5 x(2x2 +3)(sin 1x)2 2√1 x2(11x2 +4)sin 1x 85x3 8x
F (1,1,1;2, ;x2) = − + − −
3 2
2 15 3 9 − 27 − 9
It is worthy of note that the following infinite series, which is reminiscent of the series
under discussion, is indicated in [9]:
∞ (−1)nθ2n−1(−m)2n−1 = sin(mtan−1θ)(1+θ2)m2
(2n 1)!
Xn=1 −
(2j) ′
Considerincorporatingalternatingharmonicnumbersintosimilarsuchseries: ∞j=0 4j(2jj+1)H2n+1 =
2C + 3πln2. Consider using the integrals 1 x2n+1dx = H′ ln2 and 1Psin−1xdx =
− 2 0 x+1 2n+1 − 0 x+1
2C +πln2 to establish the above equation.R R
−
It can be established that:
1 4n
x2n(1 x2)ndx =
Z − (2n+1) 4n+1
0 2n
(cid:0) (cid:1)
Integrals of such forms are very important identities. We are presently interested in
series involving the expression x2n(1 x2)n. In the excellent Some New Formulas for π [1],
−
6
integralsoftheform xpn(1 x)(m p)ndxareanalyzed. In[2], theequation 1xm(1 x)ndx =
− − 0 −
Γ(m+1)Γ(n+1) is given,Rand is described as “...an extremely well-known formRula for the beta-
Γ(m+n+2
function... .” Let us consider the integral 1x2n(1 x2)ndx and similar integrals. We intend
0 −
to establish new infinite series, and classifiRcationss of integrals (which are similar to the beta
function).
Theorem 4.
3 7 9 1 15
F (1,1,1, ; ,2, ; ) = ( 64+π(8+π)+(cos 12)2+8√3cosh 1(7)+cosh 1(26)ln(2 √3))
4 3 − − −
2 4 4 4 8 − −
Proof.
∞ x2n(1 x2)n
− = Li (x2 x4)
n2 2 −
Xn=1
Integrate both sides of the above equation.
Example 1.
3 7 9 1 15
F ( ,2,2; , ; ) = (6 √3ln(2+√3))
3 2
2 4 4 4 36 −
Example 2.
∞ 4n(n+1)2 36+45π+10√3cosh−12
= −
(2n+1) 4n+1 144
Xn=1 2n
(cid:0) (cid:1)
It can be established that:
1 (2n 1)!!(4n+1)!!
x4n+1(1 x4)2n+12dx = π −
Z − 23n+3(3n+1)!
0
The above equation can be proven using the gamma function of half-integer expressions.
It is not difficult to consider an infinitude of integrals of the form 1xa(1 xb)c equal to an
0 −
expression involving the gamma function of half-integer expressionRs.
Consider:
∞ 1 2n 2n−3 (2n 1) (Γ(1))2 9
( )n n n 2 − = 4
Xn=2 32 (cid:0) (cid:1)(cid:0) − n(cid:1) 8π32 − 32
This series is determined through the complete elliptic integral of the first kind, being
based upon F (1, 1;2; 1). In Theory and Problems of Complex Variables with an Intro-
2 1 2 2 −
duction to Conformal Mapping and its Applications [12] it is indicated that:
π2 1 (Γ(1))2
dx = 4
Z0 1 sin2x 4√π
q − 2
7
Use the binomial theorem as follows:
1 = ∞ xn(−21)n(2n−1)
√1 x − n!
− Xn=0
Substitute (6):
1 = ∞ sin2nx(−12)n(2n−1)
1 sin2x Xn=0 2nn!
q − 2
3 Computational Perspectives
Let us discuss some results, some of which are possibly original (given that computational
software is unable to evaluate them), from perspectives of experimental mathematics.
Example 3.
1 2 4 3 3 3√3coth 1√3
−
F ( , , ; , ;1) =
3 2
2 3 3 2 2 2
Proof. Consider:
1 2 3 3sin(sin−1z)
F ( , ; ;z2)2 = ( 3 )2
2 1
3 3 2 z
Integrate both sides of the above equation
Example 4.
1 1 5 3 3 9 3
F ( , , ; , ;1) = √3coth 1√3
3 2 −
3 2 3 2 2 4 − 4
Example 5.
1 1 7 3 3 1
3F2( , , ; , ;1) = (12√2 3πi+12tanh−1( 1)34)
4 2 4 2 2 9 − −
Example 6.
1 1 15 3 3
F ( , , ; , ;1) =
3 2
8 2 8 2 2
16 2+√2 2iπ 8 4
p21 + 7 − 7itan−1((−1)18)+ 7i√2tan−1(1−√2)
4 4
i√2tan−1(1+√2) i√2tan−1(1 ( 1)18√2)
−7 − 7 − −
4
+ i√2tan−1(1+( 1)18√2)
7 −
8
Example 7.
1 1 11 3 3 9√3 √3πi 3√3tanh−1(3+2i√3)
F ( , , ; , ;1) = + 7
3 2
6 2 6 2 2 10 10 − 5
We consider integrals of the form 1(sin(asin−1x))2dx to be important.
0 x
Consider series for 1 using the aboRve latter technique presented in hypergeometric form:
π
Theorem 5.
1 2 4 3 3√3(3 ln4)
F ( , , ; ,2;1) = −
3 2
2 3 3 2 2π
Proof. Consider:
1 2 3 1
( F ( , ; ;sin2z))2 = (3csc(z)sin( sin 1(sinz)))2
2 1 −
3 3 2 3
Integrate both sides of the above equation.
Recall (6).
Theorem 6.
1 1 5 3 3√3(3+ln16)
F ( , , ; ,2;1) =
3 2
2 3 3 2 8π
Proof. Consider:
1 5 3 3 2
( F ( , ; ;sin2z))2 = ( csc(z)sin( sin 1(sinz)))2
2 1 −
6 6 2 2 3
Given the HJ algorithm, Theorem 6 holds.
Theorem 7.
8(√2+6tanh 1 2 √2)
1 1 7 3 − −
2+√2
F ( , , ; ,2;1) = q
3 2
2 4 4 2 9π
Proof.
1 7 3 4 3
( F ( , ; ;sin2z))2 = ( csc(z)sin( sin 1(sinz)))2
2 1 −
8 8 2 3 4
Given the HJ algorithm, Theorem 7 holds.
Theorem 8.
1 1 11 3 18 + 12√3coth−1√3
F ( , , ; ,2;1) = 25 5
3 2
2 6 6 2 π
9
Proof.
1 11 3 6 5
( F ( , ; ;sin2z))2 = ( csc(z)sin( sin 1(sinz)))2
2 1 −
12 12 2 5 6
Given the HJ algorithm, Theorem 8 holds.
The following very useful relationship is given in “Some New Formulas for π” [1]:
1 1
= x2n(1 x)2ndx
3n (3n+1) Z −
n 0
(cid:0) (cid:1)
Series of the form π = ∞ S(n) are discussed in [1].
n=0 (mn)an
pn
Consider another examPple of the above generalization:
Theorem 9.
1 8 2 3 5(5(5+3√5)+3(5+3√5)ln5 12(5+2√5)ln(5 √5)+6(5+√5)ln(5+√5))
F ( , , ; ,2;1) = − −
3 2
2 5 5 2
9π(1+√5) 2(5+√5)
q
Proof. Consider: ( F (1, 4; 3;sin2z))2 = (5 csc(z)sin(3 sin 1(sinz)))2.
2 1 5 5 2 3 5 −
The following integral is given in [8]: πxsinpxdx = π2 Γ(p+1) . We presently evaluate
0 2p+1(Γ(p+1))2
2
new hypergeometric series using such intRegrals through manipulations of the left-hand side
of the following equation discussed in [3]:
3 x
F (a,2b a 1,a 2b+2;b,a b+ ; ) =
3 2
− − − − 2 4
a a+1 a+2 3 27x
F ( , , ;b,a b+ ; − )(1 x) a
3 2 −
3 3 3 − 2 4(1 x)3 −
−
Replace x with 4sin2(x) and integrate. Let (n,x) = F ( n 1,n+2; 3;sin2x).
♥ 2 1 − − 2
4 Further Results
Theorem 10.
6 1 11 3
F ( , , ;1, ;1) =
3 2
−5 2 5 2
(5( 5(5+√5)(25 6ln( 1+√5))+ 5+√5( 25+6ln( 1+√5))
q − − q − −
+ 5 √5(1+√5)(ln(64) 6ln(3+√5))))/(204π√2)
q − −
10