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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) −

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