Pell's equation and series expansions for irrational numbers

0.14 MB·

by  Chuanan Wei

PELL’S EQUATION AND SERIES EXPANSIONS
FOR IRRATIONAL NUMBERS
6
1 1,2CHUANAN WEI
0
2 1Department of Mathematics
Shanghai Normal University, Shanghai 200234, China
n
a 2Department of Information Technology
J Hainan Medical College, Haikou 571199, China

1
2

Abstract. SolutionsofPell’sequationandhypergeometricseries
]
T identities are used to study series expansions for √p where p are
N arbitrary prime numbers. Numerous fast convergent series expan-
. sions for this family of irrationalnumbers are established.
h
t
a
m
[

1 1. Introduction
v
1
9 Pell’s equation(also called the Pell-Fermat equation) is any Diophantine equation
4 of the form
5
0 x2 py2 =1, (1)
. −
1
where p is a given positive nonsquare integer and integer solutions are sought for
0
x and y. Joseph Louis Lagrange proved that Pell’s equation has infinitely many
6
1 distinct integer solutions. Furthermore, there holds the following relation.
:
v Lemma 1. Let s be a positive integer. If (x1,y1) is the integer solution to (1),
i
X then (xs,ys) with
ar x = ⌊2s⌋ s pky2kxs−2k,
s k=0 2k 1 1
(ys = P⌊k=s−201(cid:0)⌋ 1(cid:1)+s2k pky11+2kxs1−1−2k
is also the integer solution ofP(1). (cid:0) (cid:1)
Proof. Because that (x1,y1) is the integer solution to (1), we obtain
(x1+√py1)s(x1 √py1)s =1.
−
In terms of the binomial theorem
t
t
(u+v)t = ukvt−k,
k
k=0(cid:18) (cid:19)
X
2010 Mathematics Subject Classification: Primary65B10andSecondary33C20.
Key words and phrases. Pell’s equation; Hypergeometric series; Series expan-
sions for irrationalnumbers.
Email address: [email protected].
2 Chuanan Wei
we get the following two expansions
s
s
(x1+√py1)s = k (√py1)kx1s−k
k=0(cid:18) (cid:19)
X
⌊s2⌋ s
= pky2kxs−2k
2k 1 1
k=0(cid:18) (cid:19)
X
⌊s−21⌋ s
+√p pky1+2kxs−1−2k,
1+2k 1 1
k=0 (cid:18) (cid:19)
X
s
s
(x1−√py1)s = k (−√py1)kx1s−k
k=0(cid:18) (cid:19)
X
⌊s2⌋ s
= pky2kxs−2k
2k 1 1
k=0(cid:18) (cid:19)
X
⌊s−21⌋ s
√p pky1+2kxs−1−2k.
− 1+2k 1 1
k=0 (cid:18) (cid:19)
X
So we gain
⌊2s⌋ s 2 ⌊s−21⌋ s 2
pky2kxs−2k p pky1+2kxs−1−2k =1.
(k=0(cid:18)2k(cid:19) 1 1 ) − ( k=0 (cid:18)1+2k(cid:19) 1 1 )
X X
This completes the proof of Lemma 1. (cid:3)
The circumference ratio π = 3.1415926535 is one of the most important irra-
···
tional numbers. For centuries, the study of π-formulas attracted many persons.
Recently, Chu [4], Liu [12, 13] and Wei and Gong [14] gave many π-formulas in
terms of the hypergeometricmethod. Different methods and results canbe seen in
the papers [5, 7, 8, 10, 11, 15]. For historical notes and introductory informations
on this kind of series, the readers may refer to four surveys [2, 3, 9, 16].
Itiswellknownthat√pareirrationalnumberswhenparearbitraryprimenumbers.
Several ones of this family of irrational numbers are closely related to π. For
example, we have the following relations:
π √2
sin = , (2)
4 2
π √3
sin = , (3)
3 2
π √5 1
sin = − , (4)
10 4
√7+1 ∞ ( 1)(k2) 3 k
π = − , (5)
6√3 2k+1 4+√7
k=0 (cid:18) (cid:19)
X
π ∞ 1 √13 3 109√13 393
= − − , (6)
3√3 (649+180√13)k 6k+1 − 6k+5
k=0 (cid:26) (cid:27)
X
where (2)-(4) are proverbialand (5), (6) can be seen in Wei [15].
For a complex number x, define the shifted factorial to be
(x)0 =1 and (x)n =x(x+1) (x+n 1) when n N.
··· − ∈
Series expansions for irrational numbers 3
Following Bailey [1], define the hypergeometric series by
1+rFs a0, a1, ··· , ar z = ∞ (a0)k(a1)k···(ar)kzk,
(cid:20) b1, ··· , bs (cid:12) (cid:21) kX=0 (1)k(b1)k···(bs)k
(cid:12)
where ai i≥0 and bj j≥1 are comple(cid:12)x parameters such that no zero factors ap-
{ } { }
pear in the denominators of the summand on the right hand side. Then three
hypergeometric series identities can be stated as follows:
a 1
1F0 x = with x <1, (7)
(1 x)a | |
(cid:20)− (cid:12) (cid:21) −
1a(cid:12),1 + 1a 4x 4x
2F1 2 (cid:12)12+a2 (1+x)2 =(1+x)a with (1+x)2 <1, (8)
(cid:20) (cid:12) (cid:21) (cid:12) (cid:12)
3F2 31a1,31++1a31,a1,(cid:12)(cid:12)+23 +1a31a 4(12+7xx)3 =(1+x)a(cid:12)(cid:12)(cid:12) with (cid:12)(cid:12)(cid:12)4(12+7xx)3 <1, (9)
(cid:20) 2 2 2 (cid:12) (cid:21) (cid:12)(cid:12) (cid:12)(cid:12)
where (7) is a well-knownide(cid:12)ntity and (8), (9) can be found i(cid:12)n Gessel a(cid:12)nd Stanton
(cid:12) (cid:12) (cid:12)
[6].
Inspiredby the worksjustmentioned, we shallsystematicallyexploreseriesexpan-
sions for √p with p being prime numbers by means of Lemma 1 and (7)-(9). The
structure of the paper is arranged as follows. We shall establish six theorems in
Section 2. Then they and Mathematica program are utilized to produce concrete
series expansions for √p in Sections 3-8.
2. Six Theorems
Theorem2. Letpbeapositivenonsquareintegerandm,nbebothpositiveintegers
satisfying n2 pm2 =1. Then
−
∞ k
mp (1/2) 1
k
√p= .
n k! pm2+1
k=0 (cid:18) (cid:19)
X
Proof. The case a=1/2 of (7) reads as
∞
1 (1/2)
= kxk with x <1. (10)
√1 x k! | |
− Xk=0
Setting x=1/(pm2+1) in (10), we achieve
pm2+1 ∞ (1/2) 1 k
k
= .
s pm2 k! pm2+1
k=0 (cid:18) (cid:19)
X
When pm2+1=n2, the last equation becomes
∞ k
n 1 (1/2) 1
k
= .
m√p k! pm2+1
k=0 (cid:18) (cid:19)
X
Multiplying both sides by mp/n, we attain Theorem 2 to finish the proof. (cid:3)
Theorem3. Letpbeapositivenonsquareintegerandm,nbebothpositiveintegers
provided that n2 pm2 =1. Then
−
∞ k
n (1/2) 1
k
√p= .
m k! − pm2
k=0 (cid:18) (cid:19)
X
4 Chuanan Wei
Proof. Taking x= 1/pm2 in (10), we obtain
−
pm2 ∞ (1/2) 1 k
k
= .
spm2+1 k! − pm2
k=0 (cid:18) (cid:19)
X
When pm2+1=n2, the last equation creates
∞ k
m (1/2) 1
k
√p= .
n k! − pm2
k=0 (cid:18) (cid:19)
X
Multiplying both sides by n/m, we get Theorem 3 to complete the proof. (cid:3)
Theorem4. Letpbeapositivenonsquareintegerandm,nbebothpositiveintegers
satisfying n2 pm2 =1. Then
−
mp ∞ (1/4) (3/4) 4pm2 k
k k
√p= .
n k!(3/2) (pm2+1)2
k=0 k (cid:26) (cid:27)
X
Proof. The case a=1/2 of (8) gives
∞ k
(1/4) (3/4) 4x 4x
√1+x= k k with <1. (11)
k!(3/2) (1+x)2 (1+x)2
k=0 k (cid:26) (cid:27) (cid:12) (cid:12)
X (cid:12) (cid:12)
Fixing x=1/pm2 in (11), we gain (cid:12) (cid:12)
(cid:12) (cid:12)
pm2+1 ∞ (1/4) (3/4) 4pm2 k
k k
= .
s pm2 k=0 k!(3/2)k (cid:26)(pm2+1)2(cid:27)
X
When pm2+1=n2, the last equation produces
n 1 ∞ (1/4) (3/4) 4pm2 k
k k
= .
m√p k=0 k!(3/2)k (cid:26)(pm2+1)2(cid:27)
X
Multiplying both sides by mp/n, we achieve Theorem 4 to finish the proof. (cid:3)
Theorem5. Letpbeapositivenonsquareintegerandm,nbebothpositiveintegers
provided that n2 pm2 =1 and p2m4 >4pm2+4. Then
−
n ∞ (1/4) (3/4) 4(pm2+1) k
k k
√p= .
m k!(3/2) − p2m4
k=0 k (cid:26) (cid:27)
X
Proof. Setting x= 1/(pm2+1) in (11), we attain
−
pm2 ∞ (1/4) (3/4) 4(pm2+1) k
k k
= .
spm2+1 k=0 k!(3/2)k (cid:26)− p2m4 (cid:27)
X
When pm2+1=n2, the last equation becomes
m ∞ (1/4) (3/4) 4(pm2+1) k
k k
√p= .
n k!(3/2) − p2m4
k=0 k (cid:26) (cid:27)
X
Multiplying both sides by n/m, we obtain Theorem 5 to complete the proof. (cid:3)
Theorem6. Letpbeapositivenonsquareintegerandm,nbebothpositiveintegers
satisfying n2 pm2 =1 and 4(pm2+1)3 >27p2m4. Then
−
mp ∞ (1/2) (1/6) (5/6) 27p2m4 k
k k k
√p= .
n k!(3/4) (5/4) 4(pm2+1)3
k=0 k k (cid:26) (cid:27)
X
Series expansions for irrational numbers 5
Proof. The case a=1/2 of (9) offers
∞ k
(1/2) (1/6) (5/6) 27x 27x
√1+x= k k k with <1. (12)
k!(3/4) (5/4) 4(1+x)3 4(1+x)3
k=0 k k (cid:26) (cid:27) (cid:12) (cid:12)
X (cid:12) (cid:12)
(cid:12) (cid:12)
Taking x=1/pm2 in (12), we get (cid:12) (cid:12)
pm2+1 ∞ (1/2) (1/6) (5/6) 27p2m4 k
k k k
= .
s pm2 k=0 k!(3/4)k(5/4)k (cid:26)4(pm2+1)3(cid:27)
X
When pm2+1=n2, the last equation creates
n 1 ∞ (1/2) (1/6) (5/6) 27p2m4 k
k k k
= .
m√p k=0 k!(3/4)k(5/4)k (cid:26)4(pm2+1)3(cid:27)
X
Multiplying both sides by mp/n, we gain Theorem 6 to finish the proof. (cid:3)
Theorem7. Letpbeapositivenonsquareintegerandm,nbebothpositiveintegers
provided that n2 pm2 =1 and 4p3m6 >27(pm2+1)2. Then
−
n ∞ (1/2) (1/6) (5/6) 27(pm2+1)2 k
k k k
√p= .
m k!(3/4) (5/4) − 4p3m6
k=0 k k (cid:26) (cid:27)
X
Proof. Fixing x= 1/(pm2+1) in (12), we achieve
−
pm2 ∞ (1/2) (1/6) (5/6) 27(pm2+1)2 k
k k k
= .
spm2+1 k=0 k!(3/4)k(5/4)k (cid:26)− 4p3m6 (cid:27)
X
When pm2+1=n2, the last equation produces
m ∞ (1/2) (1/6) (5/6) 27(pm2+1)2 k
k k k
√p= .
n k!(3/4) (5/4) − 4p3m6
k=0 k k (cid:26) (cid:27)
X
Multiplying both sides by n/m, we attain Theorem 7 to complete the proof. (cid:3)
3. Series expansions for √2
Setting p=2 in (1), we obtain
x2 2y2 =1. (13)
−
It is easy to know that x1 =3,y1 =2 is the solution to (13). So
x = ⌊s2⌋ s 23k3s−2k,
s k=0 2k
(ys = Pk⌊=s−201(cid:0)⌋ 1(cid:1)+s2k 21+3k3s−1−2k
P (cid:0) (cid:1)
is also the solution of (13) thanks to Lemma 1. Now we choose x4 =577,y4=408
and x7 =114243,y7=80782 to give 12 series expansions for √2.
6 Chuanan Wei
Substituting p=2, n= x4 =577 and m =y4 = 408 into Theorems 2-7, we get the
following six series expansions for √2:
∞ k
816 (1/2) 1
√2= k ,
577 k! 332929
k=0 (cid:18) (cid:19)
X
∞ k
577 (1/2) 1
√2= k ,
408 k! − 332928
k=0 (cid:18) (cid:19)
X
∞ k
816 (1/4) (3/4) 1331712
√2= k k ,
577 k!(3/2) 110841719041
k=0 k (cid:18) (cid:19)
X
∞ k
577 (1/4) (3/4) 332929
√2= k k ,
408 k!(3/2) − 27710263296
k=0 k (cid:18) (cid:19)
X
∞ k
816 (1/2) (1/6) (5/6) 748177108992
√2= k k k ,
577 k!(3/4) (5/4) 36902422678601089
k=0 k k (cid:18) (cid:19)
X
∞ k
577 (1/2) (1/6) (5/6) 110841719041
√2= k k k .
408 k!(3/4) (5/4) − 5466976319176704
k=0 k k (cid:18) (cid:19)
X
Substituting p=2, n = x7 = 114243 and m = y7 = 80782 into Theorems 2-7, we
gain the following six series expansions for √2:
∞ k
161564 (1/2) 1
√2= k ,
114243 k! 13051463049
k=0 (cid:18) (cid:19)
X
∞ k
114243 (1/2) 1
√2= k ,
80782 k! − 13051463048
k=0 (cid:18) (cid:19)
X
∞ k
161564 (1/4) (3/4) 52205852192
√2= k k ,
114243 k!(3/2) 170340687719412376401
k=0 k (cid:18) (cid:19)
X
∞ k
114243 (1/4) (3/4) 13051463049
√2= k k ,
80782 k!(3/2) − 42585171923327362576
k=0 k (cid:18) (cid:19)
X
∞ k
161564 (1/2) (1/6) (5/6) 42585171923327362576
√2= k k k ,
114243 k!(3/4) (5/4) 82340562648561433725590040987
k=0 k k (cid:18) (cid:19)
X
∞ k
114243 (1/2) (1/6) (5/6) 4599198568424134162827
√2= k k k .
80782 k!(3/4) (5/4) − 8892780764000546589887393466368
k=0 k k (cid:18) (cid:19)
X
Numerous different series expansions for √2 can be derived in the same way. Due
to limit of space,the correspondingresults will not be displayedin the paper. The
discuss is also adapt to series expansions for √p with p>2.
4. Series expansions for √3
Taking p=3 in (1), we have
x2 3y2 =1. (14)
−
It is not difficult to see that x1 =2,y1 =1 is the solution to (14). Thus
x = ⌊2s⌋ s 3k2s−2k,
s k=0 2k
(ys = Pk⌊=s−201(cid:0)⌋ 1(cid:1)+s2k 3k2s−1−2k
P (cid:0) (cid:1)
Series expansions for irrational numbers 7
isalsothesolutionof (14)accordingtoLemma1. Nowweselectx5 =362,y5=209
and x9 =70226,y9=40545 to create 12 series expansions for √3.
Substituting p=3, n= x5 =362 and m =y5 =209 into Theorems 2-7, we achieve
the following six series expansions for √3:
∞ k
627 (1/2) 1
√3= k ,
362 k! 131044
k=0 (cid:18) (cid:19)
X
∞ k
362 (1/2) 1
√3= k ,
209 k! − 131043
k=0 (cid:18) (cid:19)
X
∞ k
627 (1/4) (3/4) 131043
√3= k k ,
362 k!(3/2) 4293132484
k=0 k (cid:18) (cid:19)
X
∞ k
362 (1/4) (3/4) 524176
√3= k k ,
209 k!(3/2) − 17172267849
k=0 k (cid:18) (cid:19)
X
∞ k
627 (1/2) (1/6) (5/6) 463651231923
√3= k k k ,
362 k!(3/4) (5/4) 9001428051732736
k=0 k k (cid:18) (cid:19)
X
∞ k
362 (1/2) (1/6) (5/6) 4293132484
√3= k k k .
209 k!(3/4) (5/4) − 83344647990241
k=0 k k (cid:18) (cid:19)
X
Substituting p=3, n = x9 = 70226 and m = y9 = 40545 into Theorems 2-7, we
attain the following six series expansions for √3:
∞ k
121635 (1/2) 1
√3= k ,
70226 k! 4931691076
k=0 (cid:18) (cid:19)
X
∞ k
70226 (1/2) 1
√3= k ,
40545 k! − 4931691075
k=0 (cid:18) (cid:19)
X
∞ k
121635 (1/4) (3/4) 4931691075
√3= k k ,
70226 k!(3/2) 6080394217274509444
k=0 k (cid:18) (cid:19)
X
∞ k
70226 (1/4) (3/4) 19726764304
√3= k k ,
40545 k!(3/2) − 24321576859234655625
k=0 k (cid:18) (cid:19)
X
∞ k
121635 (1/2) (1/6) (5/6) 656682575199335701875
√3= k k k ,
70226 k!(3/4) (5/4) 479786014398315252276040347904
k=0 k k (cid:18) (cid:19)
X
∞ k
70226 (1/2) (1/6) (5/6) 6080394217274509444
√3= k k k .
40545 k!(3/4) (5/4) − 4442463093578299350981890625
k=0 k k (cid:18) (cid:19)
X
5. Series expansions for √5
Fixing p=5 in (1), we obtain
x2 5y2 =1. (15)
−
It is ordinary to find that x1 =9,y1 =4 is the solution to (15). Therefore
x = ⌊s2⌋ s 5k42k9s−2k,
s k=0 2k
(ys = Pk⌊=s−201(cid:0)⌋ 1(cid:1)+s2k 5k41+2k9s−1−2k
P (cid:0) (cid:1)
8 Chuanan Wei
is also the solution of (15) in accordance with Lemma 1. Now we choose x3 =
2889,y3 = 1292 and x4 = 51841,y4 = 23184 to produce 12 series expansions for
√5.
Substituting p=5, n = x3 = 2889 and m = y3 = 1292 into Theorems 2-7, we get
the following six series expansions for √5:
∞ k
6460 (1/2) 1
√5= k ,
2889 k! 8346321
k=0 (cid:18) (cid:19)
X
∞ k
2889 (1/2) 1
√5= k ,
1292 k! − 8346320
k=0 (cid:18) (cid:19)
X
∞ k
6460 (1/4) (3/4) 33385280
√5= k k ,
2889 k!(3/2) 69661074235041
k=0 k (cid:18) (cid:19)
X
∞ k
2889 (1/4) (3/4) 8346321
√5= k k ,
1292 k!(3/2) − 17415264385600
k=0 k (cid:18) (cid:19)
X
∞ k
6460 (1/2) (1/6) (5/6) 17415264385600
√5= k k k ,
2889 k!(3/4) (5/4) 21533840250758579043
k=0 k k (cid:18) (cid:19)
X
∞ k
2889 (1/2) (1/6) (5/6) 1880849004346107
√5= k k k .
1292 k!(3/4) (5/4) − 2325653911149135872000
k=0 k k (cid:18) (cid:19)
X
Substitutingp=5,n=x4 =51841andm=y4 =23184intoTheorems2-7,wegain
the following six series expansions for √5:
∞ k
115920 (1/2) 1
√5= k ,
51841 k! 2687489281
k=0 (cid:18) (cid:19)
X
∞ k
51841 (1/2) 1
√5= k ,
23184 k! − 2687489280
k=0 (cid:18) (cid:19)
X
∞ k
115920 (1/4) (3/4) 10749957120
√5= k k ,
51841 k!(3/2) 7222598635489896961
k=0 k (cid:18) (cid:19)
X
∞ k
51841 (1/4) (3/4) 2687489281
√5= k k ,
23184 k!(3/2) − 1805649657528729600
k=0 k (cid:18) (cid:19)
X
∞ k
115920 (1/2) (1/6) (5/6) 48752540753275699200
√5= k k k ,
51841 k!(3/4) (5/4) 19410656413844324266481975041
k=0 k k (cid:18) (cid:19)
X
∞ k
51841 (1/2) (1/6) (5/6) 7222598635489896961
√5= k k k .
23184 k!(3/4) (5/4) − 2875652798840967165640704000
k=0 k k (cid:18) (cid:19)
X
6. Series expansions for √7
Setting p=7 in (1), we have
x2 7y2 =1. (16)
−
It is easy to know that x1 =8,y1 =3 is the solution to (16). So
x = ⌊s2⌋ s 7k32k8s−2k,
s k=0 2k
(ys = Pk⌊=s−201(cid:0)⌋ 1(cid:1)+s2k 7k31+2k8s−1−2k
P (cid:0) (cid:1)
Series expansions for irrational numbers 9
is also the solution of (16) thanks to Lemma 1. Now we select x3 =2024,y3=765
and x4 =32257,y4=12192 to give 12 series expansions for √7.
Substituting p=7, n=x3 =2024andm=y3 =765into Theorems2-7, we achieve
the following six series expansions for √7:
∞ k
5355 (1/2) 1
√7= k ,
2024 k! 4096576
k=0 (cid:18) (cid:19)
X
∞ k
2024 (1/2) 1
√7= k ,
765 k! − 4096575
k=0 (cid:18) (cid:19)
X
∞ k
5355 (1/4) (3/4) 4096575
√7= k k ,
2024 k!(3/2) 4195483730944
k=0 k (cid:18) (cid:19)
X
∞ k
2024 (1/4) (3/4) 16386304
√7= k k ,
765 k!(3/2) − 16781926730625
k=0 k (cid:18) (cid:19)
X
∞ k
5355 (1/2) (1/6) (5/6) 453112021726875
√7= k k k ,
2024 k!(3/4) (5/4) 274993887369210363904
k=0 k k (cid:18) (cid:19)
X
∞ k
2024 (1/2) (1/6) (5/6) 4195483730944
√7= k k k .
765 k!(3/4) (5/4) − 2546237833204078125
k=0 k k (cid:18) (cid:19)
X
Substituting p=7, n = x4 = 32257 and m = y4 = 12192 into Theorems 2-7, we
attain the following six series expansions for √7:
∞ k
85344 (1/2) 1
√7= k ,
32257 k! 1040514049
k=0 (cid:18) (cid:19)
X
∞ k
32257 (1/2) 1
√7= k ,
12192 k! − 1040514048
k=0 (cid:18) (cid:19)
X
∞ k
85344 (1/4) (3/4) 4162056192
√7= k k ,
32257 k!(3/2) 1082669486166374401
k=0 k (cid:18) (cid:19)
X
∞ k
32257 (1/4) (3/4) 1040514049
√7= k k ,
12192 k!(3/2) − 270667371021336576
k=0 k (cid:18) (cid:19)
X
∞ k
85344 (1/2) (1/6) (5/6) 7308019017576087552
√7= k k k ,
32257 k!(3/4) (5/4) 1126532810779723715634459649
k=0 k k (cid:18) (cid:19)
X
∞ k
32257 (1/2) (1/6) (5/6) 1082669486166374401
√7= k k k .
12192 k!(3/4) (5/4) − 166893749263957816334352384
k=0 k k (cid:18) (cid:19)
X
7. Series expansions for √11
Taking p=11 in (1), we obtain
x2 11y2 =1. (17)
−
It is not difficult to see that x1 =10,y1 =3 is the solution to (17). Thus
x = ⌊s2⌋ s 11k32k10s−2k,
s k=0 2k
(ys = Pk⌊=s−201(cid:0)⌋ 1(cid:1)+s2k 11k31+2k10s−1−2k
P (cid:0) (cid:1)
10 Chuanan Wei
is also the solution of (17) according to Lemma 1. Now we choose x3 =3970,y3=
1197 and x4 =79201,y4=23880 to create 12 series expansions for √11.
Substituting p=11, n = x3 = 3970 and m = y3 = 1197 into Theorems 2-7, we get
the following six series expansions for √11:
∞ k
13167 (1/2) 1
√11= k ,
3970 k! 15760900
k=0 (cid:18) (cid:19)
X
∞ k
3970 (1/2) 1
√11= k ,
1197 k! − 15760899
k=0 (cid:18) (cid:19)
X
∞ k
13167 (1/4) (3/4) 15760899
√11= k k ,
3970 k!(3/2) 62101492202500
k=0 k (cid:18) (cid:19)
X
∞ k
3970 (1/4) (3/4) 63043600
√11= k k ,
1197 k!(3/2) − 248405937288201
k=0 k (cid:18) (cid:19)
X
∞ k
13167 (1/2) (1/6) (5/6) 6706960306781427
√11= k k k ,
3970 k!(3/4) (5/4) 15660406535270116000000
k=0 k k (cid:18) (cid:19)
X
∞ k
3970 (1/2) (1/6) (5/6) 62101492202500
√11= k k k .
1197 k!(3/4) (5/4) − 145003736614802587137
k=0 k k (cid:18) (cid:19)
X
Substituting p=11, n = x4 = 79201 and m = y4 = 23880 into Theorems 2-7, we
gain the following six series expansions for √11:
∞ k
262680 (1/2) 1
√11= k ,
79201 k! 6272798401
k=0 (cid:18) (cid:19)
X
∞ k
79201 (1/2) 1
√11= k ,
23880 k! − 6272798400
k=0 (cid:18) (cid:19)
X
∞ k
262680 (1/4) (3/4) 25091193600
√11= k k ,
79201 k!(3/2) 39347999779588156801
k=0 k (cid:18) (cid:19)
X
∞ k
79201 (1/4) (3/4) 6272798401
√11= k k ,
23880 k!(3/2) − 9836999941760640000
k=0 k (cid:18) (cid:19)
X
∞ k
262680 (1/2) (1/6) (5/6) 265598998427537280000
√11= k k k ,
79201 k!(3/4) (5/4) 246822070099948942419850075201
k=0 k k (cid:18) (cid:19)
X
∞ k
79201 (1/2) (1/6) (5/6) 39347999779588156801
√11= k k k .
23880 k!(3/4) (5/4) − 36566232589911843422208000000
k=0 k k (cid:18) (cid:19)
X
8. Series expansions for √13
Fixing p=13 in (1), we have
x2 13y2 =1. (18)
−