On physical property tensors invariant under line groups

ON PHYSICAL PROPERTY TENSORS INVARIANT UNDER LINE GROUPS Daniel B. Litvin Department of Physics, Eberly College of Science, The Pennsylvania State University, Penn State Berks, P.O. Box 7009, Reading, PA 19610-6009, U.S.A., [email protected] Supporting information Table 5: Rank two tensors invariant under line point groups For each line point group family, the form of the twelve rank two physical property tensors listed in Equation (9) are tabulated. (Subsets of these tabulations can be found in the work of Milošević (1995), Dmitriev (2003), and Sirotin & Shaskolskaya (1982).)On the left-hand-side is the symbol of the line point group family followed by the form of the twelve tensors invariant under a representative line point group of the line point group types of index n = 1, 2, and n = ∞ of that family. For each line point group, the twelve tensor forms are given in an array corresponding, respectively, to the array of tensor symbols in Equation (9). Families of line groups G1' are not explicitly listed as for these groups, the form of tensors types not containing a are the same as for corresponding groups G, and for tensor types containing a, the form has only zero entries. Appendix A: We first show that for a tensor Vm and a group Cn that the components which satisfy the 3m conditions m-products = Cn (m-products ) (A1) s = 1 ,2, …, 3m , also satisfy, the 3m (n - 1) conditions m-products = (Cn)j(m-products ) (A2) s = 1 ,2, …, 3m and j = 2, 3, …, n: Substitute equation (A1) into the right-hand-side of equation (A1): m-products = Cn (Cn (m-products )) =(Cn)2(m-products ) Repeat an additional j-2 times to obtain equation (A2). To show that for a tensor Vm and a group Cn that the components which satisfy the 3m conditions n n(m-products) = ∑ (C ) (m − product j n j=1 s ) (A3) satisfy each of the equations (A1) and (A2), we operate on both sides of (A3) with (Cn)t where "t" is an integer: (Cn)t n(m-products) = (Cn)t n ∑ (C ) (m − product j n j=1 n(Cn)t (m-products) = n ∑ (C ) j=1 j+ t n s ) (m − product s ) (Cn)j+t can be replaced with (Cn)(j+t)mod(n) , and as j runs from 1 to n, (j+t)mod(n) also runs from 1 to n. Therefore: n(Cn)t (m-products) = n ∑ (C ) (m − product j j=1 n and finally: (m-products) = (Cn)t (m-products) Taking t = 1, 2, …, n we have equations (A1) and (A2). s ) = n(m-products) Appendix B: Form of V2 invariant under Cn For n = 1 ,we have invariant under C1 , the general form of the tensor is  xx xy xz    V2 =  yx yy yz  . For n > 2, applying equation (5) to each of the components, one  zx zy zz    has using equations (6a,b): 2π 2π )j − y sin( )j )z; n n (B1) 2π 2π )j − y sin( )j ); n n (B2) 2π 2π )j + y cos( )j )z; n n (B3) 2π 2π )j + y cos( )j ); n n (B4) n nxz = ∑ ( x cos( j=1 n nzx = ∑ z( x cos( j=1 n nyz = ∑ ( x sin( j=1 n nzy = ∑ z( x sin( j=1 nzz = nzz; (B5) 2π 2π )j − y sin( )j )2; n n (B6) 2π 2π )j + y cos( )j )2; n n (B7) 2π 2π 2π 2π )j − y sin( )j )( x sin( )j + y cos( )j ); n n n n (B8) 2π 2π 2π 2π )j + y cos( )j )( x cos( )j − y sin( )j ); n n n n (B9) n nxx = ∑ ( x cos( j=1 n nyy = ∑ ( x sin( j=1 n nxy = ∑ ( x cos( j=1 n nyx = ∑ j=1 ( x sin( n Since ∑ sinaj = j=1 sin n +1 na a sin 2 2 and a sin 2 n ∑ cosaj = j=1 n Ryshik, 2007), we have for n>2 that cos ∑ cos j=1 n +1 na a sin 2 2 (Gradshteyn and a sin 2 n 2π j =0 and n ∑ sin j=1 2π j =0 and from n equations (B1-B4) we have: xz = zx = yz = zy = 0. (B10) From equation (B5) we have the trivial zz = zz, i.e. no condition on component zz. The remaining four conditions can be rewritten as: n nxx = xx ∑ cos2 j=1 n nyy = yy ∑ cos2 j=1 n 2π 2π j - (xy + yx) j + yy ∑ sin2 n n j=1 n ∑ cos j=1 n 2π 2π j + (xy + yx) j + xx ∑ sin2 n n j=1 n 2π 2π jsin j n n ∑ cos j=1 2π 2π jsin j n n (B11) n nxy = xy ∑ cos2 j=1 n nyx = yx ∑ cos2 j=1 n 2π 2π j + (xx - yy) j - yx ∑ sin2 n n j=1 n 2π 2π j - xy ∑ sin2 j + (xx - yy) n n j=1 n ∑ cos j=1 n ∑ cos j=1 2π 2π jsin j n n 2π 2π jsin j n n The values of the summations in these conditions are for n=2 2 ∑ cos2 πj = 2, j=1 2 2 ∑ sin2 πj = 0, and ∑ cos πjsin πj = 0 j=1 j=1 and consequently equations (B11) give no conditions on the components xx, yy, xy, and yx. For n > 2, using (Gradshteyn and Ryshik, 2007) n ∑ cos2 kx= k =1 n n cos(n + 1)x sinnx , + 2 2 sin x ∑ sin k =1 2 kx= n cos(n + 1)x sinnx , and − 2 2 sin x 1 n 1 sin(n + 1)x sinnx , we have: = sin 2kx ∑ 2k 1 2 sin x 1= = sinkx ∑ coskx k n n 2π cos j = ∑ n j=1 2 n 2π n sin j = and ∑ n 2 j=1 2 n ∑ cos j=1 2π 2π jsin j = 0. Using these summations n n and equations (B11) we have no conditions on components xx and xy and the condition yx = -xy. Using these summations and conditions we obtain that the form of the physical  xx xy 0    property tensor V invariant under C2 is V =  yx yy 0  and that the form invariant  0 0 zz    2 2  xx xy 0    under Cn , for n>2, is V =  − xy xx 0  . These results agree with form of the  0 0 zz   2 physical property tensor V2 invariant under Cn for n=2,3,4, and 6 given by Sirotin & Shaskolskaya (1982). Appendix C: Form of V2 invariant under Cθ  xx xy xz    The tensor V in terms of the m-products of its components is  yx yy yz  .  zx zy zz    2 Applying equation (9) to each of the m-products: 2πxx = ∫ θ = 2π θ =0 (x cosθ − y sinθ)2 dθ = θ 2π = θ 2π 2 = xx ∫ cos θdθ + yy ∫ θ 0= θ 0 = 2πyy = ∫ θ = 2π θ =0 = yy ∫ cos θdθ + xx ∫ = θ 0= θ 0 θ = 2π θ =0 2πxy = ∫ θ = 2π θ =0 ∫ θ = 2π θ =0 ∫ θ = 2π ∫ θ = 2π ∫ θ = 2π θ =0 cosθ sinθdθ (C2) (x cosθ − y sinθ)(x sinθ + y cosθ)dθ cos θdθ − yx ∫ = θ 2π sin2 θdθ + (xx − yy)∫ θ 0 = cosθ sinθdθ (C4) cosθ sinθdθ (C5) (x sinθ + y cosθ)(x cosθ − y sinθ)dθ cos θdθ − xy ∫ = θ 0= θ 0 2πxz = = θ 0 (C3) = θ 2π = θ 2π 2 = yx ∫ (C1) dθ θ 0= θ 0 = 2πyx = = θ 2π sin2 θdθ + (xy + yx)∫ = θ 2π = θ 2π 2 = xy ∫ cosθ sinθdθ θ 0 = (x sinθ + y cosθ)2 dθ = θ 2π = θ 2π 2 2πzz = zz ∫ = θ 2π sin2 θdθ − (xy + yx)∫ = θ 2π sin2 θdθ + (xx − yy)∫ = θ 0 = θ 2π = θ 2π (x cosθ − y sinθ)zdθ = xz ∫ cosθdθ − yz ∫ = θ 0= θ 0 sinθdθ (C6) 2πzx = 2πyz = (C8) θ =0 θ =0 = θ 2π = θ 2π z(x cosθ − y sinθ)dθ = zx ∫ cosθdθ − zy ∫ = θ 0= θ 0 = θ 2π = θ 2π (x sinθ + y cosθ)zdθ = yz ∫ cosθdθ + xz ∫ = θ 0= θ 0 sinθdθ sinθdθ (C7) ∫ 2πzy = Using θ = 2π θ =0 ∫ θ = 2π θ =0 = θ 2π = θ 2π z(x sinθ + y cosθ)dθ = zy ∫ cosθdθ + zx ∫ = θ 0= θ 0 cos2 θdθ = π , ∫ θ = 2π θ =0 sinθ2dθ = π , and ∫ θ = 2π θ =0 sinθdθ (C9) cosθ sinθdθ = 0 , we have xx = yy from Equations (C1,C2), no condition on zz from Equation (C3), yx = -xy from Equations (C4, C5), and xz = zx = yz =zy =0 from equations (C6 – C9). Therefore, the form of  xx xy 0    tensor V invariant under C∞ is  − xy xx 0  .  0 0 zz   2 Appendix D: Proof of the Theorem: The form of the physical property tensor Vm invariant under a group Cn , with n > m, is independent of n. The proof is by showing that when n > m all conditions on the components of the tensor, which determine the form of the tensor, are independent of n. This is done , as n explained in the text, by showing that for s<m, 0<b<s, and n>m ∑ cos j=1 s −b 2π 2π jsinb j = n n 0 or ∝ n. The proof is divided into eight parts depending on the value and parity of the exponents of the trigonometric functions: We use the following formulae each specialized from general trigonometric summations given in Gradshteyn & Ryshik (2007). The A, B, C, and D's are real numbers, and a, b, and N are integers. b even: sinb coss-b 2π j = Boo + n b odd: sinb coss-b b/2 −1 2π j = Aoo + n 2π j= n 2π j= n ∑A k =0 (s −b)/2 −1 ∑ k =0 (b −1)/2 ∑ k =0 (s −b −1)/2 ∑ k =0 k cos(b − 2k) Bk cos(s − b − 2k) Ck sin(b − 2k) 2π j n Dk cos(s − b − 2k) N sin[Nπ + ( )π]sinNπ 2π n sinN j = ∑ N n j=1 sin( )π n n 2π j n 2π j n 2π j n N cos[Nπ + ( )π]sinNπ 2π n cosN j = ∑ N n j=1 sin( )π n n Proof: 1) b=0 s even n ∑ coss j=1 n s/2 −1 2π 2π j = nBoo + ∑ ∑ Bk cos(s − 2k) j n n =j 1= k 0 s/2 −1 n 2π j) n k 0=j 1 = s − 2k cos[(s − 2k)π + ( )π]sin(s − 2k)π s/2 −1 n = nBoo + ∑ Bk s − 2k k =0 sin( )π n = nBoo + ∑ Bk (∑ cos(s − 2k) and since the cosine function is periodic with cos(θ+2π) = cosθ, we have: cos[( s/2 −1 = nBoo + ∑B k =0 k s − 2k )π]sin(s − 2k)π n s − 2k sin( )π n s − 2k π < π and the denominator in the second term is not zero. n The numerator is zero since s-2k is even and then sin(s-2k)π=0. Therefore: Since n>m>s>s-2k, n ∑ cos s j=1 2π j ∝n . n 2) b=0 s odd n ∑ coss j=1 (s −1)/2 n 2π 2π j = ∑ ∑ Dk cos(s − 2k) j n =j 1=k 0 n (s −1)/2 n 2π j) n = k 0=j 1 s − 2k cos[(s − 2k)π + ( )π]sin(s − 2k)π (s −1)/2 n = ∑ Dk s − 2k k =0 sin( )π n = ∑ Dk ( ∑ co s(s − 2k) s-2k= s, s-2,s-4,….,1 are all odd, (s-2k)π is π+a multiple of 2π. since cos is periodic in 2π and cos(θ+π) = -cosθ we have: cos( (s −1)/2 ∑ =- Dk k =0 s − 2k )π sin(s − 2k)π n s − 2k sin( )π n Since the denominator is never zero and sin(s-2k)π=0 : n ∑ cos j=1 s 2π j =0 n 3) b=s s even n s/2 −1 2π 2π j = nAoo using the analogous argument j = nAoo + ∑ ∑ A k cos(s − 2k) n n =j 1= j=1 k 0 as in the case of b=0 and s even, we have: n ∑ sins n ∑ sin s j=1 2π j ∝n n 4) b=s s odd n 2π sin j ∑ n j=1 s (s −1)/2 n = ∑ ∑ Ck sin(s − 2k) (s −1)/2 n k 0 =j 1= 2π j n 2π j) n = k 0=j 1 s − 2k sin[(s − 2k)π + ( )π]sin(s − 2k)π (s −1)/2 n = ∑ Ck s − 2k k =0 sin( )π n = ∑ Ck ( ∑ sin(s − 2k) s-2k = s,s-2,s-4,….,1 all odd, (s-2k)π is π+a multiple of 2π, since sin is periodic in 2π and sin(θ+π) = -sinθ we have: sin( (s −1)/2 =- ∑ k =0 Ck s − 2k )π sin(s − 2k)π n s − 2k sin( )π n Since denominator is never zero and sin(s-2k)π=0 we have: n ∑ sin 2π j =0 n s j=1 5) bǂ0 or s n ∑ cos s-b even, b even 2π 2π jsinb j n n s −b j=1 (s −b)/2 −1 n ∑ = ∑ (Boo + j=1 k =0 Bk cos(s − b − 2k) 2π j ) (Aoo + n b/2 −1 ∑A t =0 k cos(b − 2t) 2π j) n = nBooAoo n + ∑ Boo b/2 −1 ∑A =j 1=t 0 n + ∑ A oo (s −b)/2 −1 ∑ =j 1= k 0 (s −b)/2 −1 n +∑ k ∑ =j 1= k 0 cos(b − 2t) 2π j n Bk cos(s − b − 2k) Bk cos(s − b − 2k) 2π j n 2π b/2−1 2π j ∑ A k cos(b − 2t) j n=t 0 n The second term of the four terms: b/2 −1 n ∑ ∑A Boo =j 1=t 0 k cos(b − 2t) b/2 −1 2π j n n 2π j) n =t 0=j 1 b − 2t cos[(b − 2t)π + ( )π]sin(b − 2t)π b/2 −1 n = Boo ∑ A k b − 2t t =0 sin( )π n = Boo ∑ A k ( ∑ cos(b − 2t) b-2t is even , (b-2t)π is then a multiple of 2π and cosine is periodic in 2π, so b − 2t cos( )π sin(b − 2t)π b/2 −1 n = Boo ∑ A k b − 2t t =0 sin( )π n the denominator is never zero and sin(b-2t)π=0 , so: b/2 −1 n ∑ Boo ∑A =j 1=t 0 k cos(b − 2t) 2π j= 0 n The third term is: (s −b)/2 −1 n ∑ A oo ∑ k 0 =j 1= Bk cos(s − b − 2k) 2π j n (s −b)/2 −1 n 2π j) n = k 0=j 1 s − b − 2k cos[(s − b − 2k)π + ( )π]sin(s − b − 2k)π (s −b)/2 −1 n = A oo ∑ Bk s − b − 2k k =0 sin( )π n ∑ = A oo Bk ( ∑ cos(s − b − 2k) s-b-2k is even, (s-b-2k)π is then a multiple of 2π and cosine is periodic in 2π, so s − b − 2k cos( )π sin(s − b − 2k)π (s −b)/2 −1 n = A oo ∑ Bk s − b − 2k k =0 sin( )π n The denominator is never zero, sin(b-2k)π=0 so (s −b)/2 −1 n ∑ ∑ A oo =j 1= k 0 Bk cos(s − b − 2k) 2π j =0 n The fourth term: n (s −b)/2 −1 ∑ ∑ =j 1= k 0 Bk cos(s − b − 2k) (s −b)/2 −1 ∑ = b/2 −1 Bk n ∑A ∑ cos(s − b − 2k) k = k 0 =t 0=j 1 (s −b)/2 −1 =½ ∑ b/2 −1 n ∑A ∑ Bk k k 0 =t 0=j 1 = (s −b)/2 −1 =½ ∑ b/2 −1 Bk 2π b/2−1 2π j ∑ A k cos(b − 2t) j n=t 0 n [cos(s − 2b − 2k + 2t) n ∑ A (∑ cos(s − 2b − 2k + 2t) k = k 0 =t 0=j 1 (s −b)/2 −1 +½ 2π 2π jcos(b − 2t) j n n ∑ b/2 −1 Bk ∑ n 2π j) n A k ( ∑ cos(s − 2k − 2t) = k 0 =t 0=j 1 2π 2π j + cos(s − 2k − 2t) j] n n 2π j) n In the first half of this: n ∑ cos(s − 2b − 2k + 2t) j=1 n ∑ cos(s − 2b − 2k + 2t) j=1 2π j = n if s=2b-2k+2t=0. If s-2b-2k+2t ≠ 0 then n 2π j= n s − 2b − 2k + 2t )π]sin(s − 2b − 2k + 2t)π n s − 2b − 2k + 2t sin( )π n cos[(s − 2b − 2k + 2t)π + ( = s-2b-2k+2t is even so cos( = s − 2b − 2k + 2t )π sin(s − 2b − 2k + 2t)π n s − 2b − 2k + 2t sin( )π n and since the denominator not zero, sin(s-2b-2k+2t)π=0 we have: n ∑ cos(s − 2b − 2k + 2t) j=1 2π j =0 . n In the second half of the fourth term: n ∑ j=1 n ∑ j=1 2π j =n if s-2k-2t =0. If s-2k-2t ≠ 0 then: n s − 2k − 2t cos[(s − 2k − 2t)π + ( )π]sin(s − 2k − 2t)π 2π n cos(s − 2k − 2t) j= s − 2k − 2t n sin( )π n cos(s − 2k − 2t) s-2k-2t is even so s − 2k − 2t cos( )π sin(s − 2k − 2t)π n = s − 2k − 2t sin( )π n The denominator not zero, sin(s-2k-2t)π=0 and we have: n ∑ cos(s − 2k − 2t) j=1 2π j= 0 n Therefore, for bǂ0 or s, s-b even and b even: n ∑ cos s −b j=1 2π 2π jsinb j =0 or is proportional to n. n n 6) bǂ0 or s n ∑ cos j=1 s −b s-b even, b odd n 2π 2π jsinb j = ∑ (Boo + n n j=1 n = ∑ Boo (b −1)/2 j=1 ∑ k =0 ∑ t =0 Ck sin(b − 2k) n (s −b)/2 −1 j=1 t =0 +∑ (s −b)/2 −1 ∑ Bt cos(s − b − 2t) 2π (b −1)/2 2π j )( ∑ Ck sin(b − 2k) j) n n k =0 2π j n Bt cos(s − b − 2t) 2π (b −1)/2 2π j ∑ Ck sin(b − 2k) j n n k =0 In the first term (b −1)/2 n ∑ ∑ Boo j=1 k =0 Ck sin(b − 2k) (b −1)/2 n 2π 2π j) j = Boo ∑ Ck ( ∑ sin(b − 2k) n n = k 0=j 1 we have: n ∑ j=1 b − 2k sin[(b − 2k)π + ( )π]sin(b − 2k)π 2π n sin(b − 2k) j= b − 2k n sin( )π n Since b-2k is odd, (b-2k)π = π+ a multiple of 2π, and sin(θ+π) = -sinθ, we have: sin( =- b − 2k )π sin(b − 2k)π n b − 2k sin( )π n The denominator is not zero and sin(b-2k)π=0 , so: (b −1)/2 n ∑ ∑ Boo j=1 k =0 Ck sin(b − 2k) 2π j= 0 n In the second term n (s −b)/2 −1 j=1 t =0 ∑ ∑ Bt cos(s − b − 2t) (b −1)/2 (s −b)/2 −1 = ∑ t =0 ∑ Bt 2π (b −1)/2 2π j j ∑ Ck sin(b − 2k) n n k =0 n Ck ( ∑ cos(s − b − 2t) = k 0=j 1 2π 2π jsin(b − 2k) j) n n we have: n ∑ cos(s − b − 2t) j=1 2π 2π jsin(b − 2k) j n n n =½ ∑ sin(s − 2k − 2t) j=1 n 2π 2π j - ½ ∑ sin(s − 2b − 2t + 2k) j n n j=1 In the first part n ∑ j=1 sin(s − 2k − 2t) 2π j = 0 if s-2k-2t=0, if s-2k-2t≠0 then: n s − 2k − 2t )π]sin(s − 2k − 2t)π n s − 2k − 2t sin( )π n sin[(s − 2k − 2t)π + ( = Since s-2k-2t is odd, we have s − 2k − 2t )π sin(s − 2k − 2t)π n =s − 2k − 2t sin( )π n and since s-2k-2t≠0 : sin( n ∑ sin(s − 2k − 2t) j=1 2π j= 0 . n In the second part n ∑ sin(s − 2b − 2t + 2k) j=1 2π j = 0 if s-2b-2t+2k =0 n if s-2b-2t+2k≠0 then: s − 2b + 2k − 2t )π]sin(s − 2b + 2k − 2t)π n s − 2b + 2k − 2t sin( )π n sin[(s − 2b + 2k − 2t)π + ( = Since s-2b+2k-2t is odd sin( =- s − 2b + 2k − 2t )π sin(s − 2b + 2k − 2t)π n =0. s − 2b + 2k − 2t sin( )π n Therefore n (s −b)/2 −1 j=1 t =0 ∑ ∑ Bt cos(s − b − 2t) 2π (b −1)/2 2π j ∑ Ck sin(b − 2k) j =0 n n k =0 and n ∑ cos j=1 s −b 2π 2π jsinb j = 0 for bǂ0 or s, s-b even and b odd. n n 7) bǂ0 or s s-b odd, b odd n ∑ cos j=1 s −b 2π 2π jsinb j = n n ∑ k =0 Dk ∑ ( j=1 (b −1)/2 (s −b −1)/2 = (s −b −1)/2 n ∑ ∑ k =0 Dk cos(s − b − 2k) n Ct ( ∑ cos(s − b − 2k) =t 0=j 1 2π (b −1)/2 2π j) j )( ∑ Ct sin(b − 2t) n n t =0 2π 2π jsin(b − 2t) j) n n Here we have n ∑ cos(s − b − 2k) j=1 2π 2π jsin(b − 2t) j= n n n = ½ ∑ sin(s − 2k − 2t) j=1 n 2π 2π j - ½ ∑ sin(s − 2b − 2k + 2t) j n n j=1 In the first term n ∑ sin(s − 2k − 2t) j=1 2π j =0 if s-2k-2t = 0 and if s-2k-2t ≠0 : n s − 2k − 2t )π]sin(s − 2k − 2t)π n s − 2k − 2t sin( )π n sin[(s − 2k − 2t)π + ( = Since s-2k-2t is even s − 2k − 2t )π sin(s − 2k − 2t)π n = s − 2k − 2t sin( )π n and since s-2k-2t≠0: sin( n ∑ sin(s − 2k − 2t) j=1 2π j= 0 . n In the second term n ∑ sin(s − 2b − 2k + 2t) j=1 2π j = 0 if s-2b-2k+2t =0, and if s-2b-2k+2t≠0 n s − 2b − 2k + 2t )π]sin(s − 2b − 2k + 2t)π n s − 2b − 2k + 2t sin( )π n sin[(s − 2b − 2k + 2t)π + ( = Since s-2b-2k+2t is even sin( = s − 2b − 2k + 2t )π sin(s − 2b − 2k + 2t)π n =0. s − 2b − 2k + 2t sin( )π n Therefore: n ∑ cos s −b j=1 2π 2π jsinb j = 0 when bǂ0 or s, s-b odd and b odd. n n 8) bǂ0 or s n ∑ cos j=1 s −b s-b odd, b even 2π 2π jsinb j = n n ∑ j=1 k =0 ∑ (s −b −1)/2 ∑ Dk cos(s − b − 2k) Dk cos(s − b − 2k) ∑ j=1 = Aoo k =0 (s −b −1)/2 n + ∑ j=1 (s −b −1)/2 n = Aoo ∑ (s −b −1)/2 n ∑ k =0 n = k 0=j 1 (s −b −1)/2 ∑ + k =0 b/2 −1 Dk ∑ b/2 −1 ∑A t =0 t cos(b − 2t) 2π j n Dk cos(s − b − 2k) Dk ( ∑ cos(s − b − 2k) 2π j ( Aoo + n 2π j n b/2 −1 ∑A t =0 t cos(b − 2t) 2π j n 2π j) n n A t ( ∑ cos(s − b − 2k) =t 0=j 1 2π 2π jcos(b − 2t) j) n n In the first term s-b-2k≠0 because s-b is odd. Therefore: n ∑ j=1 cos(s − b − 2k) 2π j= n s − b − 2k )π]sin(s − b − 2k)π n s − b − 2k sin( )π n cos[(s − b − 2k)π + ( Since s-b-2k is odd s − b − 2k cos( )π sin(s − b − 2k)π n == 0. s − b − 2k sin( )π n In the second term 2π j) n n ∑ cos(s − b − 2k) j=1 2π 2π jcos(b − 2t) = n n n =½ ∑ cos(s − 2b − 2k + 2t) j=1 n 2π 2π j +½ ∑ cos(s − 2k − 2t) j n n j=1 In the first part s-2b-2k+2t ≠ 0 since s is odd. Therefore n ∑ cos(s − 2b − 2k + 2t) j=1 2π j= n s − 2b − 2k + 2t )π]sin(s − 2b − 2k + 2t)π n s − 2b − 2k + 2t sin( )π n cos[(s − 2b − 2k + 2t)π + ( = and since s-2b-2k+2t is odd cos( =- s − 2b − 2k + 2t )π sin(s − 2b − 2k + 2t)π n =0. s − 2b − 2k + 2t sin( )π n Therefore: n ∑ cos j=1 s −b 2π 2π jsinb j = 0 when bǂ0 or s, s-b odd and b even. n n