Bounds for sine and cosine via eigenvalue estimation

©2014 Jo ma K. Me ikoski e al., licensee De G uy e Open. This wo k is licensed unde he C ea i e Commons A ibu ion-NonComme cial- NoDe i s 3.0 License. DOI 10.2478/spma-2014-0003 |Spec. Ma ices 2014; 2:19–29 Resea ch A icle Open Access Pen i Haukkanen, Mika Ma ila, Jo ma K. Me ikoski*, and Alexande Ko ačec Bounds o sine and cosine ia eigen alue es ima ion Abs ac : De ine n×n idiagonal ma ices Tand Sas ollows: All en ies o he main diagonal o Ta e ze o and hose o he i s supe - and subdiagonal a e one. The en ies o he main diagonal o Sa e wo excep he (n,n)en y one, and hose o he i s supe - and subdiagonal a e minus one. Then, deno ing by λ(·) he la ges eigen alue, λ(T) = 2 cos π n+ 1,λ(S−1) = 1 4 cos2nπ 2n+1 . Using ce ain lowe bounds o he la ges eigen alue, we p o ide lowe bounds o hese exp essions and, u he , lowe bounds o sin xand cos xon ce ain in e als. Also uppe bounds can be ob ained in his way. Keywo ds: eigen alue bounds, igonome ic inequali ies MSC: 15A42, 26D05 || Pen i Haukkanen, Mika Ma ila: School o In o ma ion Sciences, FI-33014 Uni e si y o Tampe e, Finland, E-mail: pen i.haukkanen@u a. i; mika.ma ila@u a. i *Co esponding Au ho : Jo ma K. Me ikoski: School o In o ma ion Sciences, FI-33014 Uni e si y o Tampe e, Finland, E-mail: jo ma.me ikosk[email p o ec ed] Alexande Ko ačec: Depa men o Ma hema ics, Uni e si y o Coimb a EC San a C uz, 3001-501 Coimb a, Po ugal, E-mail: [email p o ec ed] 1In oduc ion Gi en n≥2, le idiag (a,b)deno e he symme ic idiagonal n×nma ix wi h diagonal aand i s supe - and subdiagonal b. De ine T= ( ij) = idiag (0,1). Also de ine S= (sij) = idiag (2,−1) −F, whe e he en ies o Fa e ze o excep he (n,n)en y one. Le λ(·)and µ(·)deno e he la ges and espec i ely smalles eigen alue. Then λ(T) = 2 cos π n+ 1 (1) and µ(S) = 4 cos2nπ 2n+ 1, due o Ru he o d [14, p. 230] (see also [2, 17]). Then λ(S−1) = 1 4 cos2nπ 2n+1 .(2) The e a e se e al eigen alue bounds in he li e a u e. Using hem, can we ind easonably good bounds o he igh -hand sides o (1) and (2)? Many eigen alue bounds a e oo ough o his pu pose, bu he ollow- ing bounds ha e some in e es . 20 |Pen i Haukkanen, Mika Ma ila, Jo ma K. Me ikoski, and Alexande Ko ačec Le Abe a complex He mi ian n×nma ix and le 0=x∈Cn. Then (see, e.g., [5, Theo em 4.2.2]) λ(A)≥x*Ax x*x(3) wi h equali y i and only i xis an eigen ec o o Aco esponding o λ(A). In pa icula , choosing x= (1 . . . 1)T=e, we ob ain λ(A)≥su A n,(4) whe e su deno es he sum o en ies. Equali y holds i and only i eis an eigen ec o co esponding o λ(A). I Ais (en ywise) nonnega i e, hen his bound is o en a he good. The explana ion is ha he e is a nonneg- a i e eigen ec o zco esponding o λ(A). Since eis posi i e, he di ec ions o eand zcanno be comple ely di e en . Each ow o Ais in e“wi h equal weigh ”, bu be e “weigh s” may be he ow sums o A; deno e hem by 1,..., n. So assume A=Oand subs i u e x= ( 1. . . n)T=Ae in (3). Then λ(A)≥su A3 su A2.(5) Equali y holds i and only i Ae is an eigen ec o o Aco esponding o λ(A). Usually (5) is be e han (4) bu no always [8]. Fo u he discussion on his opic, see [6]. We will in Sec ions 2 and 3 unde es ima e λ(T)and λ(S−1), espec i ely. In s udying λ(T), we apply (5) because i is be e han (4) and easy o compu e. In s udying λ(S−1), we apply (4) because (5) is a he com- plica ed. Using hese lowe bounds, we will ob ain also lowe bounds o sin xand cos xon ce ain in e als. We will in Sec ion 4 imp o e he lowe bound o λ(T)by a sui able shi ing. To see how good ou bounds a e, we will compa e hem wi h ce ain o he bounds in Sec ion 5. Finally, we will ou line some u he de elop- men s in Sec ion 6, and d aw conclusions and make ema ks in Sec ion 7. 2Unde es ima ing λ(T) Assume n≥3. Since Tis he adjacency ma ix o he linea g aph 1−2− · · · − n, he (i,j)en y o Tkcoun s he pa hs om i o jo leng h k. So he main diagonal o T2is (1,2,. . . ,2,1), he second supe - and subdi- agonal is (1,. . . ,1), and he emaining en ies a e ze o. Mo eo e , he i s supe - and subdiagonal o T3is (2,3,. . . ,3,2), he hi d supe - and subdiagonal is (1,. . . ,1), and he emaining en ies a e ze o. Hence su T2= 2 + (n−2) ·2 + 2(n−2) = 4n−6, su T3= 2[2 ·2 + (n−3) ·3 + n−3] = 8n−16. Since Te is no an eigen ec o co esponding o λ(T), we he e o e ha e by (1) and (5) cos π n+ 1 >2n−4 2n−3,(6) which i ially holds also o n= 2. Thus (6) is alid o all in ege s n≥2. We show ha in ac cos π x+ 1 >2x−4 2x−3(7) o all eal numbe s x>3 2.(8) Because lim x→∞ 2x−4 2x−3 cos π x+1 = 1,(9) Bounds o sine and cosine ia eigen alue es ima ion |21 he bound (7) is good when xis la ge. Since cos x= cos π π−x x+ 1,2π−x x−4 2π−x x−3=2π−6x 2π−5x, he claim (7) is equi alen o ha in he ollowing Theo em 1. I 0<x<2 5π,(10) hen cos x>2π−6x 2π−5x.(11) P oo . Assume (10). Since 2π−6x 2π−5x= 1 −x 2π−5xand cos x>1−x2 2, he claim ollows i x2 2≤x 2π−5x, i.e., 5x2−2πx + 2 ≥0. This holds, because he disc iminan D= 4π2−40 <0. Co olla y 1. I π 10 <x<π 2,(12) hen sin x>2π−12x π−10x.(13) P oo . Assume (12); hen π 2−xsa is ies (10). Apply (11) o i . By (9), he bound (11) is good when π−x xis la ge, i.e., x≈0, and (13) is good when x≈π 2. 3Unde es ima ing λ(S−1) Since Scon ains nega i e en ies, i is no easonable o apply (4) in unde es ima ing λ(S). Indeed, he bound so ob ained appea s o be e y poo . Bu S−1= (min (i,j)) is posi i e; so le us y (4) o unde es ima e λ(S−1). Fo k= 1,. . . ,n, deno e by Ek he k×kma ix wi h all en ies one. Fo k= 1,. . . ,n−1, de ine he n×n ma ix Fkby Fk= O O O Ek!. Then S−1=En+Fn−1+· · · +F1, and so su S−1= su En+ su Fn−1+· · · + su F1=n2+ (n−1)2+· · · + 12=1 6(2n3+ 3n2+n). 22 |Pen i Haukkanen, Mika Ma ila, Jo ma K. Me ikoski, and Alexande Ko ačec Since eis no an eigen ec o o S−1co esponding o λ(S−1), we he e o e ge by (2) and (4) 1 4 cos2nπ 2n+1 >2n2+ 3n+ 1 6, which simpli ies in o cos π 2n+ 1 >2n2+ 3n−2 2n2+ 3n+ 1 . We show ha in ac cos π 2x+ 1 >2x2+ 3x−2 2x2+ 3x+ 1 (14) o all eal numbe s xsa is ying x<−1∨−1 2<x<1 2∨x>1. Because lim x→±∞ 2x2+3x−2 2x2+3x+1 cos π 2x+1 = 1, he bound (14) is good when |x|is la ge. Since cos x= cos π 2π−x 2x+ 1,2(π−x 2x)2+ 3π−x 2x−2 2(π−x 2x)2+ 3π−x 2x+ 1 =π2+πx −6x2 π2+πx , he claim (14) is equi alen o ha in he ollowing Theo em 2. I −π<x<0∨0<x<π 3∨x>π 2,(15) hen cos x>π2+πx −6x2 π2+πx .(16) P oo . We di ide he p oo in h ee cases. Case 1.−π<x<0∨0<x<12 π−π. Then 6x2 π2+πx −x2 2=x212 −π2−πx 2(π2+πx)>0, and so cos x>1−x2 2>1−6x2 π2+πx =π2+πx −6x2 π2+πx . Case 2.12 π−π≤x<π 3. W i e he claim (16) as cos x>(π−2x)(3x+π) π(x+π), equi alen ly d(x) = π(x+π) cos x−(π−2x)(3x+π)>0.(17) Deno e x=π 3− , hen 0< ≤4π 3−3 π= 0.369.(18) (This and co esponding equali y signs la e deno e equali y in he p ecision o he numbe o digi s shown.) Since cos x= cos ( π 3− ) = 1 2cos +√3 2sin >1 21− 2 2+√3 2 − 3 6=c( ), Bounds o sine and cosine ia eigen alue es ima ion |23 we ha e d(π 3− )>ππ 3− +πc( )−[π−2(π 3− )][3(π 3− ) + π] = π 4√3 4+π 4−π2 3√3 3+6−π2 3−π√3 2 2+2π2 √3−7π 2 =g( ). Because he exac coe icien s o g( )a e qui e in ol ed, we unde es ima e g( )>0.4 4−1.2 3−0.02 2+ 0.4 =h( ). The ze os o h( )a e 1=−0.5387, 2= 0, 3= 0.6405, 4= 2.898. Since h(0.5) = 0.07 >0, we ha e h( )>0 o all sa is ying 0< < 3, in pa icula , unde (18). Then also g( )>0, and (17) ollows. Case 3.x>π 2. Deno e x=π 2+ , hen >0. Because cos x= cos ( π 2+ ) = −sin >− , we ha e d(π 2+ )>π(π 2+ +π)(− )−[π−2(π 2+ )][3(π 2+ ) + π] = [(6 −π) +π(5 −3 2π)] >0. The p oo is comple e. Co olla y 2. I x<0∨π 6<x<π 2∨π 2<x<3π 2,(19) hen sin x>10πx −12x2 3π2−2πx .(20) P oo . Assume (19); hen π 2−xsa is ies (15). Apply (16) o i . 4Imp o ing (6) Fo all eal numbe s , we ha e λ(T) = λ(T+ I)− . Since (T+ I)eis no an eigen ec o o T+ I, we ha e by (5) λ(T)>su (T+ I)3 su (T+ I)2− = ( ).(21) To imp o e (6), we y o ind = 0maximizing he igh -hand side o (21). Assuming n≥3, we ha e ( ) = n 3+ 6(n−1) 2+ 6(2n−3) + 8(n−2) n 2+ 4(n−1) + 2(2n−3) − =2(n−1) 2+ 4(2n−3) + 8(n−2) n 2+ 4(n−1) + 2(2n−3) . I is s aigh o wa d o show ha ′( ) = 0 i and only i (n−2) 2+ 2(n−3) −2 = 0 and ha 0is he posi i e oo o his equa ion. Thus 0=3−n+√n2−4n+ 5 n−2, 24 |Pen i Haukkanen, Mika Ma ila, Jo ma K. Me ikoski, and Alexande Ko ačec which, howe e , is oo complica ed. The e o e we eplace 5 wi h 4 he e and ake =3−n+√n2−4n+ 4 n−2=3−n+n−2 n−2=1 n−2. Subs i u ing in (21), we ge cos π n+ 1 >4n3−20n2+ 35n−21 4n3−18n2+ 29n−16.(22) The co esponding equali y holds o n= 2. Ex ending (6) o (7) wo ks unde (8), bu his condi ion does no allow ex ending (22) o cos π x+ 1 >4x3−20x2+ 35x−21 4x3−18x2+ 29x−16.(23) Fo example, i x= 3.5, hen he le -hand side is 0.7660, less han he igh -hand side 0.7671. To ind a condi ion o (23), we apply ideas o Lague e de eloped la e in an exchange o le e s be ween Feke e and Pólya, see [10, p. 69] and [4, p. 12]. The ollowing heo em holds ac ually o Lau en se ies, bu powe se ies a e enough o us. Theo em 3. Gi en eal numbe s α0,α1,. . . , no all ze o, conside he se ies ϕ(x) = α0+α1x+α2x2+· · · wi h con e gence adius R>0. Le 0< <R, deno e by ϕ he es ic ion ϕ|]0, [, and le kbe a nonnega i e in ege . The numbe o sign changes o he sequence (β(k) 0,β(k) 1,β(k) 2,. . . ), de ined by ϕ( x) (1 −x)k=β(k) 0+β(k) 1x+β(k) 2x2+· · · , is an uppe bound o he numbe o ze os o ϕ . We do no use he ull o ce o his heo em. I is enough ha we can conclude: I β(k) 0,β(k) 1,β(k) 2,. . . ≥0(no all ze o) o some k, hen ϕ(x)>0 o all xsa is ying 0<x< . Theo em 4. I x>π 0.63 −1 = 3.98666 . . . ,(24) hen (23) holds. P oo . Subs i u ing x7→ π x+ 1, he claim (23) eads cos x+80x3−87πx2+ 32π2x−4π3 −67x3+ 77πx2−30π2x+ 4π3= cos x+p(x) q(x)>0(25) o all xsa is ying 0<x<0.63.(26) Since he disc iminan o q′(x) = −201x2+ 154πx −30π2 is 1542−4·201 ·30 = −404 <0, we ha e q′(x)<0 o all x. Assume (26). Since q(x)>q(0.63) = 16.75 >0, an equi alen claim o (25) is q(x) cos x+p(x)>0. Bounds o sine and cosine ia eigen alue es ima ion |25 We p o e a s onge claim (x) = q(x)1−x2 2! +x4 4! −x6 6! +p(x)>0. Le us apply Theo em 3 o ϕ= , = 0.63. We ind he β(0) i’s om ϕ( x) = α0+ α1x+ 2α2x2+· · · =β(0) 0+β(0) 1x+β(0) 2x2+· · · , so β(0) i=αi i,i= 0,1,2,. . . . We cons uc he β(k) i’s ecu si ely. Since ( x) (1 −x)k+1 =1 1−x ( x) (1 −x)k= (1 + x+x2+· · · )(β(k) 0+β(k) 1x+β(k) 2x2+· · · ) = β(k) 0+ (β(k) 0+β(k) 1)x+ (β(k) 0+β(k) 1+β(k) 2)x2+· · · =β(k+1) 0+β(k+1) 1x+β(k+1) 2x2+· · · , we ge β(k+1) i=β(k) 0+· · · +β(k) i,i,k= 0,1,2,. . . .(27) Now a simple compu a ion yields (0.63x) = 0.00145481x9−0.00833744x8−0.0937648x7+ 0.619422x6+ 2.10029x5−18.2393x4+ 40.2686x3−37.0818x2+ 12.4357x.(28) The e o e β(0) 0=β(0) 10 =β(0) 11 =· · · = 0, which implies by (27) ha β(1) 0= 0 and β(1) 9=β(1) 10 =· · · = 0.00145481 −0.00833744 −0.0937648 + 0.619422 + 2.10029 −18.2393 + 40.2686 −37.0818 + 12.4357 = 0.0022 >0. Hence, by (27), β(k) 0= 0 and β(k) 9,β(k) 10 ,. . . >0 o all k≥1. I emains o show ha β(k) 1,. . . ,β(k) 8≥0 o some k. Le Lbe he 8×8lowe iangula ma ix wi h diagonal and lowe iangle one, and deno e bk= (β(k) 1. . . β(k) 8)T. We ind b0 om (28) and ob ain b3=L3b0= (12.4 0.225 3.64 4.43 4.71 5.09 5.48 5.88)T. Now he p oo is comple e. As in he p oo o (11) and (13), we can ind lowe bounds o sin xand cos x, bu hey a e qui e complica ed. Shi ing does no imp o e (4), because su (A+ I) n− =su A n o all . The e o e we canno apply his ick o (14). 5Compa isons We compa e ou bounds o sin xwi h ce ain o he bounds. Because ou bounds wo k well nea o π 2, we choose o compa ison only such bounds ha a e de ined he e. Mos o hem a e imp o emen s o Jo dan’s inequali y sin x>2 πx,0<x<π 2.(29) 26 |Pen i Haukkanen, Mika Ma ila, Jo ma K. Me ikoski, and Alexande Ko ačec Kobe ’s inequali y cos x>1−2 πx,0<x<π 2, is equi alen o his (simply subs i u e x7→ π 2−xin one o hem o ge he o he ), and so b ings no hing new o us. The e is an ex ensi e li e a u e on e ining and ex ending hese inequali ies. Qi, Niu and Guo [11] su eyed his opic conce ning (29). We compa e ou bounds (13) and (20) wi h each o he and wi h he ollowing bounds: sin x>π2x−x3 π2+x2,0<x<π,(Redhe e [12, 13], Williams [18]); (30) sin x>3 πx−4 π3x3,0<x<π 2,(Caccia [1]); (31) sin x>x+2(2 −π) π2x2,0<x<π 2,(Sándo [15]); (32) sin x>(√2−1)2√2 πx+ 1,π 4<x<π 2,(Sándo [16]); (33) sin x>x+12 −4π π2x2+4π−16 π3x3,0<x<π 2,(Özban [19]); (34) sin x>9π 80 +2 πx−1 2πx3+1 5π3x5,0<x<π 2,(Kuo [7]).(35) In s udying (13), we es ic o π 10 <x<π 2, and in s udying (20) o π 6<x<π 2. In compa ing hem wi h (33), we es ic o π 4<x<π 2. We lis he condi ions unde which he i s -men ioned bound is be e han he second. (13) s. (20): 3π 10 <x<π 3. (13) s. (30): x>0.8622. (20) s. (30): Always. (13) s. (31): 0.8579 <x<1.1181. (20) s. (31): Always. (13) s. (32): x>0.7449. (20) s. (32): Always. (13) s. (33): x>0.8505. (20) s. (33): x>0.8085. (13) s. (34): 0.9205 <x<1.0482. (20) s. (34): x<1.0526. (13) s. (35): Ne e . (20) s. (35): x<0.6815 o x>1.4798. 6Fu he de elopmen s We ex end (11). Le b>a>0. We de e mine d(≤1/a)so ha cos x>1−bx 1−ax (36) o all xsa is ying 0<x<d.(37) Bounds o sine and cosine ia eigen alue es ima ion |27 As in he p oo o Theo em 1, we can see ha (36) holds i x2 2≤(b−a)x 1−ax . Unde (37), his is equi alen o p(x) = ax2−x+ 2(b−a)≥0.(38) The disc iminan D= 1 −8a(b−a). Case 1.D≤0, i.e., b≥a+1 8a. Then (38) holds o all x. Gi en a>0, he choice b=a+1 8a is clea ly op imal. So we ha e p o ed ha cos x>1−(a+1 8a)x 1−ax , assuming (37) wi h d= 1/a. In pa icula , ake a=5 2π; hen cos x>1−(5 2π+π 20 )x 1−5 2πx=2π−(5 + π2 10 )x 2π−5x o all xsa is ying 0<x<2π 5. This imp o es (11) sligh ly. Case 2.D>0. Since bo h ze os o p(x)a e posi i e, xmus be less han o equal o he smalle ze o. We ha e now p o ed he ollowing Theo em 5. Le b>a>0. I D= 1 −8a(b−a)≤0, hen cos x>1−bx 1−ax (39) o all xsa is ying 0<x<1 a. I D>0, hen (39) holds o all xsa is ying 0<x≤1−p1−8a(b−a) 2a. The e e eesugges ed ha pe haps,byconside ingce ainma iceswi hcomplexen ies,hype bolic e sions o ou bounds can be ound. We lea e he ques ion conce ning such ma ices open (see Rema k 8) bu s udy wha happens in an a emp o ind he hype bolic e sion o (39) by using powe se ies. Le b>a>0. We y o ind a easonable condi ion conce ning x(>0) so ha cosh x>1 + bx 1 + ax . Applying he inequali y cosh x>1 + 1 2x2and p oceeding as abo e, we ob ain a su icien condi ion p(x) = ax2+x−2(b−a)≥0.(40) Since p(x)has bo h posi i e and nega i e ze o, xmus be g ea e han o equal o he posi i e ze o. Thus we ha e p o ed he ollowing