Bounds for sine and cosine via eigenvalue estimation

©2014 Jorma K. Merikoski et al., licensee De Gruyter Open. This work is licensed under the Creative Commons Attribution-NonCommercialNoDerivs 3.0 License. DOI 10.2478/spma-2014-0003 |Spec. Matrices 2014; 2:19–29 Research Article Open Access Pentti Haukkanen, Mika Mattila, Jorma K. Merikoski*, and Alexander Kovačec Bounds for sine and cosine via eigenvalue estimation Abstract: Define n×ntridiagonal matrices Tand Sas follows: All entries of the main diagonal of Tare zero and those of the first superand subdiagonal are one. The entries of the main diagonal of Sare two except the (n,n)entry one, and those of the first superand subdiagonal are minus one. Then, denoting by λ(·)the largest eigenvalue, λ(T) = 2 cos π n+ 1,λ(S−1) = 1 4 cos2nπ 2n+1 . Using certain lower bounds for the largest eigenvalue, we provide lower bounds for these expressions and, further, lower bounds for sin xand cos xon certain intervals. Also upper bounds can be obtained in this way. Keywords: eigenvalue bounds, trigonometric inequalities MSC: 15A42, 26D05 || Pentti Haukkanen, Mika Mattila: School of Information Sciences, FI-33014 University of Tampere, Finland, E-mail: [email protected]; [email protected] *Corresponding Author: Jorma K. Merikoski: School of Information Sciences, FI-33014 University of Tampere, Finland, E-mail: jorma.merikosk[email protected] Alexander Kovačec: Department of Mathematics, University of Coimbra EC Santa Cruz, 3001-501 Coimbra, Portugal, E-mail: [email protected] 1Introduction Given n≥2, let tridiag (a,b)denote the symmetric tridiagonal n×nmatrix with diagonal aand first superand subdiagonal b. Define T= (tij) = tridiag (0,1). Also define S= (sij) = tridiag (2,−1) −F, where the entries of Fare zero except the (n,n)entry one. Let λ(·)and µ(·)denote the largest and respectively smallest eigenvalue. Then λ(T) = 2 cos π n+ 1 (1) and µ(S) = 4 cos2nπ 2n+ 1, due to Rutherford [14, p. 230] (see also [2, 17]). Then λ(S−1) = 1 4 cos2nπ 2n+1 .(2) There are several eigenvalue bounds in the literature. Using them, can we find reasonably good bounds for the right-hand sides of (1) and (2)? Many eigenvalue bounds are too rough for this purpose, but the following bounds have some interest. 20 |Pentti Haukkanen, Mika Mattila, Jorma K. Merikoski, and Alexander Kovačec Let Abe a complex Hermitian n×nmatrix and let 0=x∈Cn. Then (see, e.g., [5, Theorem 4.2.2]) λ(A)≥x*Ax x*x(3) with equality if and only if xis an eigenvector of Acorresponding to λ(A). In particular, choosing x= (1 . . . 1)T=e, we obtain λ(A)≥su A n,(4) where su denotes the sum of entries. Equality holds if and only if eis an eigenvector corresponding to λ(A). If Ais (entrywise) nonnegative, then this bound is often rather good. The explanation is that there is a nonnegative eigenvector zcorresponding to λ(A). Since eis positive, the directions of eand zcannot be completely different. Each row of Ais in e“with equal weight”, but better “weights” may be the row sums of A; denote them by r1,...,rn. So assume A=Oand substitute x= (r1. . . rn)T=Ae in (3). Then λ(A)≥su A3 su A2.(5) Equality holds if and only if Ae is an eigenvector of Acorresponding to λ(A). Usually (5) is better than (4) but not always [8]. For further discussion on this topic, see [6]. We will in Sections 2 and 3 underestimate λ(T)and λ(S−1), respectively. In studying λ(T), we apply (5) because it is better than (4) and easy to compute. In studying λ(S−1), we apply (4) because (5) is rather complicated. Using these lower bounds, we will obtain also lower bounds for sin xand cos xon certain intervals. We will in Section 4 improve the lower bound for λ(T)by a suitable shifting. To see how good our bounds are, we will compare them with certain other bounds in Section 5. Finally, we will outline some further developments in Section 6, and draw conclusions and make remarks in Section 7. 2Underestimating λ(T) Assume n≥3. Since Tis the adjacency matrix of the linear graph 1−2− · · · − n, the (i,j)entry of Tkcounts the paths from ito jof length k. So the main diagonal of T2is (1,2,. . . ,2,1), the second superand subdiagonal is (1,. . . ,1), and the remaining entries are zero. Moreover, the first superand subdiagonal of T3is (2,3,. . . ,3,2), the third superand subdiagonal is (1,. . . ,1), and the remaining entries are zero. 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 not an eigenvector corresponding to λ(T), we therefore have by (1) and (5) cos π n+ 1 >2n−4 2n−3,(6) which trivially holds also for n= 2. Thus (6) is valid for all integers n≥2. We show that in fact cos π x+ 1 >2x−4 2x−3(7) for all real numbers x>3 2.(8) Because lim x→∞ 2x−4 2x−3 cos π x+1 = 1,(9) Bounds for sine and cosine via eigenvalue estimation |21 the bound (7) is good when xis large. Since cos x= cos π π−x x+ 1,2π−x x−4 2π−x x−3=2π−6x 2π−5x, the claim (7) is equivalent to that in the following Theorem 1. If 0<x<2 5π,(10) then cos x>2π−6x 2π−5x.(11) Proof. Assume (10). Since 2π−6x 2π−5x= 1 −x 2π−5xand cos x>1−x2 2, the claim follows if x2 2≤x 2π−5x, i.e., 5x2−2πx + 2 ≥0. This holds, because the discriminant D= 4π2−40 <0. Corollary 1. If π 10 <x<π 2,(12) then sin x>2π−12x π−10x.(13) Proof. Assume (12); then π 2−xsatisfies (10). Apply (11) to it. By (9), the bound (11) is good when π−x xis large, i.e., x≈0, and (13) is good when x≈π 2. 3Underestimating λ(S−1) Since Scontains negative entries, it is not reasonable to apply (4) in underestimating λ(S). Indeed, the bound so obtained appears to be very poor. But S−1= (min (i,j)) is positive; so let us try (4) to underestimate λ(S−1). For k= 1,. . . ,n, denote by Ekthe k×kmatrix with all entries one. For k= 1,. . . ,n−1, define the n×n matrix 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 |Pentti Haukkanen, Mika Mattila, Jorma K. Merikoski, and Alexander Kovačec Since eis not an eigenvector of S−1corresponding to λ(S−1), we therefore get by (2) and (4) 1 4 cos2nπ 2n+1 >2n2+ 3n+ 1 6, which simplifies into cos π 2n+ 1 >2n2+ 3n−2 2n2+ 3n+ 1 . We show that in fact cos π 2x+ 1 >2x2+ 3x−2 2x2+ 3x+ 1 (14) for all real numbers xsatisfying x<−1∨−1 2<x<1 2∨x>1. Because lim x→±∞ 2x2+3x−2 2x2+3x+1 cos π 2x+1 = 1, the bound (14) is good when |x|is large. 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 , the claim (14) is equivalent to that in the following Theorem 2. If −π<x<0∨0<x<π 3∨x>π 2,(15) then cos x>π2+πx −6x2 π2+πx .(16) Proof. We divide the proof in three 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. Write the claim (16) as cos x>(π−2x)(3x+π) π(x+π), equivalently d(x) = π(x+π) cos x−(π−2x)(3x+π)>0.(17) Denote x=π 3−t, then 0<t≤4π 3−3 π= 0.369.(18) (This and corresponding equality signs later denote equality in the precision of the number of digits shown.) Since cos x= cos ( π 3−t) = 1 2cos t+√3 2sin t>1 21−t2 2+√3 2t−t3 6=c(t), Bounds for sine and cosine via eigenvalue estimation |23 we have d(π 3−t)>ππ 3−t+πc(t)−[π−2(π 3−t)][3(π 3−t) + π] = π 4√3t4+π 4−π2 3√3t3+6−π2 3−π√3 2t2+2π2 √3−7π 2t=g(t). Because the exact coefficients of g(t)are quite involved, we underestimate g(t)>0.4t4−1.2t3−0.02t2+ 0.4t=h(t). The zeros of h(t)are t1=−0.5387,t2= 0,t3= 0.6405,t4= 2.898. Since h(0.5) = 0.07 >0, we have h(t)>0for all tsatisfying 0<t<t3, in particular, under (18). Then also g(t)>0, and (17) follows. Case 3.x>π 2. Denote x=π 2+t, then t>0. Because cos x= cos ( π 2+t) = −sin t>−t, we have d(π 2+t)>π(π 2+t+π)(−t)−[π−2(π 2+t)][3(π 2+t) + π] = t[(6 −π)t+π(5 −3 2π)] >0. The proof is complete. Corollary 2. If x<0∨π 6<x<π 2∨π 2<x<3π 2,(19) then sin x>10πx −12x2 3π2−2πx .(20) Proof. Assume (19); then π 2−xsatisfies (15). Apply (16) to it. 4Improving (6) For all real numbers t, we have λ(T) = λ(T+tI)−t. Since (T+tI)eis not an eigenvector of T+tI, we have by (5) λ(T)>su (T+tI)3 su (T+tI)2−t=f(t).(21) To improve (6), we try to find t=t0maximizing the right-hand side of (21). Assuming n≥3, we have f(t) = nt3+ 6(n−1)t2+ 6(2n−3)t+ 8(n−2) nt2+ 4(n−1)t+ 2(2n−3) −t=2(n−1)t2+ 4(2n−3)t+ 8(n−2) nt2+ 4(n−1)t+ 2(2n−3) . It is straightforward to show that f′(t) = 0 if and only if (n−2)t2+ 2(n−3)t−2 = 0 and that t0is the positive root of this equation. Thus t0=3−n+√n2−4n+ 5 n−2, 24 |Pentti Haukkanen, Mika Mattila, Jorma K. Merikoski, and Alexander Kovačec which, however, is too complicated. Therefore we replace 5 with 4 there and take t=3−n+√n2−4n+ 4 n−2=3−n+n−2 n−2=1 n−2. Substituting in (21), we get cos π n+ 1 >4n3−20n2+ 35n−21 4n3−18n2+ 29n−16.(22) The corresponding equality holds for n= 2. Extending (6) to (7) works under (8), but this condition does not allow extending (22) to cos π x+ 1 >4x3−20x2+ 35x−21 4x3−18x2+ 29x−16.(23) For example, if x= 3.5, then the left-hand side is 0.7660, less than the right-hand side 0.7671. To find a condition for (23), we apply ideas of Laguerre developed later in an exchange of letters between Fekete and Pólya, see [10, p. 69] and [4, p. 12]. The following theorem holds actually for Laurent series, but power series are enough to us. Theorem 3. Given real numbers α0,α1,. . . , not all zero, consider the series ϕ(x) = α0+α1x+α2x2+· · · with convergence radius R>0. Let 0<r<R, denote by ϕrthe restriction ϕ|]0,r[, and let kbe a nonnegative integer. The number of sign changes of the sequence (β(k) 0,β(k) 1,β(k) 2,. . . ), defined by ϕ(rx) (1 −x)k=β(k) 0+β(k) 1x+β(k) 2x2+· · · , is an upper bound for the number of zeros of ϕr. We do not use the full force of this theorem. It is enough that we can conclude: If β(k) 0,β(k) 1,β(k) 2,. . . ≥0(not all zero) for some k, then ϕ(x)>0for all xsatisfying 0<x<r. Theorem 4. If x>π 0.63 −1 = 3.98666 . . . ,(24) then (23) holds. Proof. Substituting x7→ π x+ 1, the claim (23) reads cos x+80x3−87πx2+ 32π2x−4π3 −67x3+ 77πx2−30π2x+ 4π3= cos x+p(x) q(x)>0(25) for all xsatisfying 0<x<0.63.(26) Since the discriminant of q′(x) = −201x2+ 154πx −30π2 is 1542−4·201 ·30 = −404 <0, we have q′(x)<0for all x. Assume (26). Since q(x)>q(0.63) = 16.75 >0, an equivalent claim to (25) is q(x) cos x+p(x)>0. Bounds for sine and cosine via eigenvalue estimation |25 We prove a stronger claim f(x) = q(x)1−x2 2! +x4 4! −x6 6! +p(x)>0. Let us apply Theorem 3 to ϕ=f,r= 0.63. We find the β(0) i’s from ϕ(rx) = α0+rα1x+r2α2x2+· · · =β(0) 0+β(0) 1x+β(0) 2x2+· · · , so β(0) i=αiri,i= 0,1,2,. . . . We construct the β(k) i’s recursively. Since f(rx) (1 −x)k+1 =1 1−x f(rx) (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 get β(k+1) i=β(k) 0+· · · +β(k) i,i,k= 0,1,2,. . . .(27) Now a simple computation yields f(0.63x) = 0.00145481x9−0.00833744x8−0.0937648x7+ 0.619422x6+ 2.10029x5−18.2393x4+ 40.2686x3−37.0818x2+ 12.4357x.(28) Therefore β(0) 0=β(0) 10 =β(0) 11 =· · · = 0, which implies by (27) that 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 ,. . . >0for all k≥1. It remains to show that β(k) 1,. . . ,β(k) 8≥0for some k. Let Lbe the 8×8lower triangular matrix with diagonal and lower triangle one, and denote bk= (β(k) 1. . . β(k) 8)T. We find b0from (28) and obtain b3=L3b0= (12.4 0.225 3.64 4.43 4.71 5.09 5.48 5.88)T. Now the proof is complete. As in the proof of (11) and (13), we can find lower bounds for sin xand cos x, but they are quite complicated. Shifting does not improve (4), because su (A+tI) n−t=su A n for all t. Therefore we cannot apply this trick to (14). 5Comparisons We compare our bounds for sin xwith certain other bounds. Because our bounds work well near to π 2, we choose for comparison only such bounds that are defined there. Most of them are improvements of Jordan’s inequality sin x>2 πx,0<x<π 2.(29) 26 |Pentti Haukkanen, Mika Mattila, Jorma K. Merikoski, and Alexander Kovačec Kober’s inequality cos x>1−2 πx,0<x<π 2, is equivalent to this (simply substitute x7→ π 2−xin one of them to get the other), and so brings nothing new to us. There is an extensive literature on refining and extending these inequalities. Qi, Niu and Guo [11] surveyed this topic concerning (29). We compare our bounds (13) and (20) with each other and with the following bounds: sin x>π2x−x3 π2+x2,0<x<π,(Redheffer [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,(Sándor [15]); (32) sin x>(√2−1)2√2 πx+ 1,π 4<x<π 2,(Sándor [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 studying (13), we restrict to π 10 <x<π 2, and in studying (20) to π 6<x<π 2. In comparing them with (33), we restrict to π 4<x<π 2. We list the conditions under which the first-mentioned bound is better than the second. (13) vs. (20): 3π 10 <x<π 3. (13) vs. (30): x>0.8622. (20) vs. (30): Always. (13) vs. (31): 0.8579 <x<1.1181. (20) vs. (31): Always. (13) vs. (32): x>0.7449. (20) vs. (32): Always. (13) vs. (33): x>0.8505. (20) vs. (33): x>0.8085. (13) vs. (34): 0.9205 <x<1.0482. (20) vs. (34): x<1.0526. (13) vs. (35): Never. (20) vs. (35): x<0.6815 or x>1.4798. 6Further developments We extend (11). Let b>a>0. We determine d(≤1/a)so that cos x>1−bx 1−ax (36) for all xsatisfying 0<x<d.(37) Bounds for sine and cosine via eigenvalue estimation |27 As in the proof of Theorem 1, we can see that (36) holds if x2 2≤(b−a)x 1−ax . Under (37), this is equivalent to p(x) = ax2−x+ 2(b−a)≥0.(38) The discriminant D= 1 −8a(b−a). Case 1.D≤0, i.e., b≥a+1 8a. Then (38) holds for all x. Given a>0, the choice b=a+1 8a is clearly optimal. So we have proved that cos x>1−(a+1 8a)x 1−ax , assuming (37) with d= 1/a. In particular, take a=5 2π; then cos x>1−(5 2π+π 20 )x 1−5 2πx=2π−(5 + π2 10 )x 2π−5x for all xsatisfying 0<x<2π 5. This improves (11) slightly. Case 2.D>0. Since both zeros of p(x)are positive, xmust be less than or equal to the smaller zero. We have now proved the following Theorem 5. Let b>a>0. If D= 1 −8a(b−a)≤0, then cos x>1−bx 1−ax (39) for all xsatisfying 0<x<1 a. If D>0, then (39) holds for all xsatisfying 0<x≤1−p1−8a(b−a) 2a. Therefereesuggestedthatperhaps,byconsideringcertainmatriceswithcomplexentries,hyperbolic versions of our bounds can be found. We leave the question concerning such matrices open (see Remark 8) but study what happens in an attempt to find the hyperbolic version of (39) by using power series. Let b>a>0. We try to find a reasonable condition concerning x(>0) so that cosh x>1 + bx 1 + ax . Applying the inequality cosh x>1 + 1 2x2and proceeding as above, we obtain a sufficient condition p(x) = ax2+x−2(b−a)≥0.(40) Since p(x)has both positive and negative zero, xmust be greater than or equal to the positive zero. Thus we have proved the following