On certain finite and infinite sums of inverse tangents
aa r X i v : . [ m a t h . G M ] O c t ON CERTAIN FINITE AND INFINITE SUMS OF INVERSE TANGENTS
MARTIN NICHOLSON
Abstract.
An identity is proved connecting two finite sums of inverse tangents. This identity is dis-cretized version of Jacobi’s imaginary transformation for the modular angle from the theory of ellipticfunctions. Some other related identities are discussed. Introduction
Sums of inverse tangents have attracted a lot of attention. For example, the following sums of inversetangents can be calculated in closed form: ∞ X n =0 arctan 2(2 n + 1) = π , (1.1) ∞ X n =0 ( − n +1 arctan 1 F n = arctan √ − , (1.2) ∞ X n =0 arctan sinh x cosh nx = 3 π − arctan e x . (1.3)(1.1) is a classic sum evaluated first by Glaisher in [1]. The sum (1.2), where F n is n -th Fibonacci number,was calculated by Hoggatt and Ruegels [6]. The sum (1.3) was noted in [8]. See [2] for further referencesand a brief summary of research in this direction.All summations of the type (1.1) and (1.2) seem to be based on two methods: the telescopic principle,and the method of zeroes, as was noted in [7].Even earlier, in his studies of elliptic functions, Jacobi proved identity of which he wrote in his treatiseon elliptic functions “one is obliged to rank among the most elegant formulas” [9],[10]:
14 arcsin k = arctan q / − arctan q / + arctan q / − . . . (1.4)Here q = e − πK ′ /K , K is the complete elliptic integral of the first kind with modulus kK = K ( k ) = Z π/ dϕ p − k sin ϕ ,K ′ = K ( k ′ ) with k ′ = √ − k being the complementary modulus. The quantity arcsin k is called modularangle. Together with the obvious relation arcsin k + arcsin k ′ = π , this implies ∞ X n =1 χ ( n ) arctan e − αn + ∞ X n =1 χ ( n ) arctan e − βn = π , αβ = π , (1.5)where χ ( n ) = sin πn is Dirichlet character modulo ([3], ch.14, entry 15). (1.5) is Jacobi’s imaginarytransformation for the modular angle.Another arctan series related to elliptic functions was found in an unpublished manuscript by B. Cais[4]: ∞ X n =1 (cid:16) n (cid:17) arctan √
31 + 2 e αn + ∞ X n =1 (cid:16) n (cid:17) arctan √
31 + 2 e βn = π , αβ = 4 π , (1.6)where (cid:16) j (cid:17) = √ sin πj is Legendre symbol modulo .The focus of this paper will be two reciprocal identities for finite sums of inverse tangents, Theorems1 and 3 below and another transformation formula with two continuous parameters, Theorem 5. We givetwo proofs of Theorem 1 in sections 2, 3. Theorems 3 and 5 are proved in sections 4 and 5, respectively.In section 6 we mention another transformation formula for sum of two finite reciprocal sums related tosolution of Dirichlet problem on a finite rectangular grid. Theorem 1.
Let n, m ∈ N and αβ = 1 , α > . Then X | j |≤ n ( − n + j arctan (cid:18)q α cos πj n +1 − α cos πj n +1 (cid:19) m +1 + X | k |≤ m ( − m + k arctan (cid:18)q β cos πk m +1 − β cos πk m +1 (cid:19) n +1 = π . (1.7)Note that when n = m and α = 1 both sums in (1.7) are equal and we get a closed form summation: Corollary 2.
For n ∈ N : X | j |≤ n ( − n + j arctan (cid:18)q πj n +1 − cos πj n +1 (cid:19) n +1 = π . (1.8)It is instructive to write (1.7) in another form by shifting the summation variable and simple rearrangementof terms n X j =1 χ ( j ) arctan (cid:18)q α sin πj n +2 − α sin πj n +2 (cid:19) m +1 + m X k =1 χ ( k ) arctan (cid:18)q β sin πj m +2 − β sin πj m +2 (cid:19) n +1 = π −
12 ( − n arctan (cid:16)p α − α (cid:17) m +1 −
12 ( − m arctan (cid:16)p β − β (cid:17) n +1 . From this form of (1.7), it is evident that letting n = m → ∞ and redefining α and β , one recovers (1.5).Thus, (1.7) is discretized version of (1.5). Our proof is completely elementary and provides an elementaryproof of the modular relation (1.5). Theorem 3.
Let n and m be positive odd numbers and αβ = 1 . Then n/ X j =1 (cid:18) j (cid:19) arctan √
31 + 2 α + tan πj n α − tan πj n ! m + m/ X k =1 (cid:18) k (cid:19) arctan √
31 + 2 β + tan πk m β − tan πk m ! n = − π , (1.9) where (cid:16) j (cid:17) is Legendre symbol modulo . If n = m and α = 1 the two sums in Theorem 3 are equal and we get closed form summation Corollary 4.
For an odd positive integer n : n/ X j =1 (cid:18) j (cid:19) arctan √
31 + 2 cot n (cid:18) π − πj n (cid:19) = − π . (1.10)As an illustration of (1.10) note the case n = 3 : arctan √
31 + 2 cot π − arctan √
31 + 2 cot π − arctan √
31 + 2 cot π = π . Although (1.9) has a structure similar to (1.6) it is not clear if (1.6) can be derived from (1.9) asa limiting case. However, by combining the limiting case of Theorem 3 with (1.6) one can find thetransformation formula for another infinite arctan series: ∞ X n =0 (cid:18) n − (cid:19) arctan √ − e α (2 n +1) + ∞ X n =0 (cid:18) n − (cid:19) arctan √ − e β (2 n +1) = 2 π , αβ = π . (1.11)More generally one has as a consequence of imaginary transform for theta functions [10] (see Theorem 5below for definition of s ( x ) ) ∞ X j = −∞ s ( j ) arctan sin 2 θe α ( π | j | + ϕs ( j )) − cos 2 θ + ∞ X k = −∞ s ( k ) arctan sin 2 ϕe β ( π | k | + θs ( k )) − cos 2 ϕ = 2 π (cid:16) π − θ (cid:17) (cid:16) π − ϕ (cid:17) , which suggests the following generalization of Theorem 3 with two additional continuous parameters. Theorem 5.
Let n and m be positive odd integers, αβ = 1 ( α > , and θ, ϕ ∈ (0 , π/ . Define function s ( x ) = ( , x ≥ − , x < , which differs from the sgn function only at where it takes value instead ofvalue . Then X | j |≤ n − s ( j ) arctan sin 2 θ α + tan ϕ + πjn α − tan ϕ + πjn ! ms ( j ) − cos 2 θ + X | k |≤ m − s ( k ) arctan sin 2 ϕ β + tan θ + πkm β − tan θ + πkm ! ns ( k ) − cos 2 ϕ = π − θ − ϕ. (1.12)To see that Theorem 3 is a particular case of Theorem 5 let θ = ϕ = π/ in 1.12. The first sum in 1.12becomes n − X j =0 arctan √
31 + 2 α + tan π (1+3 j )3 n α − tan π (1+3 j )3 n ! m − n − X j =1 arctan √
31 + 2 α − tan π (1 − j )3 n α + tan π (1 − j )3 n ! m . After shifting the summation index of the second sum in this expression, one obtains using the definitionof Legendre symbol n − X j =0 arctan √
31 + 2 α + tan π (1+3 j )3 n α − tan π (1+3 j )3 n ! m − n − X j =0 arctan √
31 + 2 α + tan π (2+3 j )3 n α − tan π (2+3 j )3 n ! m = n − X j =1 (cid:18) j (cid:19) arctan √
31 + 2 α + tan πj n α − tan πj n ! m . When m = n , ϕ = θ , α = β = 1 in Theorem 5 one gets the summation formula Corollary 6.
Let n be a positive odd integer, θ ∈ (0 , π/ , and the function s ( x ) defined as in Theorem5. Then X | j |≤ n − s ( j ) arctan sin 2 θ n tan (cid:16) π + θ + πjn (cid:17)o ns ( j ) − cos 2 θ = π − θ. (1.13)As an illustration of the corollary above note the case n = 3arctan sin 2 θ cos 2 θ + cot (cid:0) π + θ (cid:1) − arctan sin 2 θ cos 2 θ + cot (cid:0) π − θ (cid:1) + arctan sin 2 θ cos 2 θ + cot (cid:0) θ − π (cid:1) = θ − π . (1.14)Technically, one could generalize Theorem 1 too, but the resulting identity is not nice. We give it herefor illustration purposes only without proof. Assuming θ, ϕ ∈ (0 , π/ , α = 1 /β > , n, m ∈ N one has Re " X | j |≤ n ( − n + j arctan ((cid:18)q α cos θ + πj n +1 − α cos θ + πj n +1 (cid:19) m +1 e iϕ ) + X | k |≤ m ( − m + k arctan ((cid:18)q β cos ϕ + πk m +1 − β cos ϕ + πk m +1 (cid:19) n +1 e iθ ) = π . First proof of Theorem 1
We break the proof into a series of lemmas.
Lemma 7.
The following identity holds for α > , n, m ∈ N and j ∈ Z (cid:18)q α cos πj n +1 − α cos πj n +1 (cid:19) m +1 = π − arctan (sinh(2 m + 1) α j ) where α j is the positive solution of sinh α j = α cos πj n +1 .Proof. By denoting s = 2 m + 1 for brevity we obtain (cid:18)q α cos πj n +1 − α cos πj n +1 (cid:19) m +1 = 2 arctan (cosh α j − sinh α j ) s = 2 arctan e − sα j = π − (cid:0) arctan e sα j − arctan e − sα j (cid:1) = π − arctan e sα j − e − sα j . Since e x − e − x x the proof is complete. (cid:3) Lemma 8.
For n, m ∈ N , j ∈ Z , and α j as was defined in the previous lemma, one has π − arctan (sinh(2 m + 1) α j ) = ( − m X | k |≤ m arctan cos πk m +1 α cos πj n +1 , Proof.
Using properties of complex numbers we write π − arctan (sinh(2 m + 1) α j ) = arg ( i ) + arg (1 − i sinh(2 m + 1) α j )= arg (sinh(2 m + 1) α j + i )= ( − m arg (cid:16) sinh(2 m + 1) α j + sinh πi (2 m +1)2 (cid:17) . This expression can be factorised according to the formula sinh(2 m + 1) a + sinh(2 m + 1) b = 2 m Y | k |≤ m (cid:18) sinh a + sinh (cid:18) b + 2 πik m + 1 (cid:19)(cid:19) . Its validity is easy to check by standard methods: both sides are polynomials in sinh a of order m + 1 with leading coefficient m and zeroes − sinh (cid:16) b + πik m +1 (cid:17) , | k | ≤ m .Thus π − arctan (sinh(2 m + 1) α j ) = ( − m arg m Y | k |≤ m (cid:18) sinh α j + sinh (cid:18) πi πik m + 1 (cid:19)(cid:19) = ( − m arg Y | k |≤ m (cid:18) α cos πj n + 1 + i cos 2 πk m + 1 (cid:19) = ( − m X | k |≤ m arctan cos πk m +1 α cos πj n +1 , as required. (cid:3) Lemma 9.
For n, m ∈ N , j ∈ Z , one has X | k |≤ m arctan cos πk m +1 α cos πj n +1 = X | k |≤ m ( − k arctan cos πk m +1 α cos πj n +1 . Proof.
Let f be an odd function. Then X | k |≤ m ( − k f (cid:18) cos πk m + 1 (cid:19) = X | k |≤ m f (cid:18) cos (cid:18) πk m + 1 − πk (cid:19)(cid:19) = X | k |≤ m f (cid:18) cos 2 πkm m + 1 (cid:19) = X | k |≤ m f (cid:18) cos 2 πk m + 1 (cid:19) . The last equality is explained as follows. First, note that cos has period π . The sum P | k |≤ m is overresidue class mod m + 1 . When m > , the numbers m and m + 1 are coprime. Hence, when k runsover residue class mod m + 1 , the set of numbers km runs over residue class mod m + 1 .To complete the proof of the lemma set f ( x ) = arctan xα cos πj n +1 . (cid:3) Lemma 10. X | j |≤ n ( − j = ( − n , n ∈ N . Proof.
The sum is trivial when n = 0 . Let’s assume that n > . Then X | j |≤ n ( − j = ( − n − ( − n +1 − ( −
1) = ( − n . (cid:3) Now, we are in a position to prove Theorem 1. According to lemmas 7-10 we have that the LHS ofequation (1.7) equals X | j |≤ n ( − n + j X | k |≤ m ( − m + k arctan cos πk m +1 α cos πj n +1 + X | k |≤ m ( − m + k X | j |≤ n ( − n + j arctan cos πj n +1 β cos πk m +1 = 12 ( − n + m X | j |≤ n X | k |≤ m ( − j + k π (cid:18) α cos πj n + 1 cos πk m + 1 (cid:19) = π − n + m X | j |≤ n ( − j X | k |≤ m ( − k = π . Second proof of Theorem 1
Lemma 11.
We have the partial fractions expansion for arbitrary non-negative integer m : m + 1cosh (cid:0) (2 m + 1) sinh − z (cid:1) √ z + 1 = X | k |≤ m ( − m − k cos πk m +1 z + cos πk m +1 . (3.1) Proof. cosh (cid:0) (2 m + 1) sinh − z (cid:1) √ z + 1 is a polynomial in z of order m + 2 with roots z s = i sin π (2 s + 1)2(2 m + 1) , s = − m − , ..., m. Let us denote the LHS of 3.1 by f ( z ) . Residues of f ( z ) at the points z s are res f ( z ) (cid:12)(cid:12) z s = ( − s i (1 + δ s,m + δ s, − m − ) , s = − m − , ..., m, where δ s,r = ( , s = r , s = r is the Kronecker delta. It is easy to see this for s = − m, ..., m − . The points z m = i and z − m − = − i are more tricky, in which case we write f ( z ) = 2 m + 1 z + 1 · √ z + 1cosh (cid:0) (2 m + 1) sinh − z (cid:1) , where the second multiplier does not have singularities at z = ± i .Now we can write the partial fractions expansion m + 1cosh (cid:0) (2 m + 1) sinh − z (cid:1) √ z + 1 = 1 i m − X k = − m ( − s z − i sin π (2 s +1)2(2 m +1) + ( − m i (cid:18) z − i − z + i (cid:19) . It is quite easy to bring this to the form stated in the lemma. (cid:3)
Lemma 12.
For arbitrary non-negative integers n and m we have the transformation formula X | j |≤ n ( − n + j m + 1cosh (cid:16) (2 m + 1) sinh − n z cos πj n +1 o(cid:17) cos πj n +1 q z cos πj n +1 = 1 z X | k |≤ m ( − m + k n + 1cosh (cid:16) (2 n + 1) sinh − n z − cos πk m +1 o(cid:17) cos πk m +1 q z − cos πk m +1 . (3.2) Proof.
Multiply 3.1 by z and replace z with z cos πj n +1 . Then summing over j one gets X | j |≤ n ( − n + j m + 1cosh (cid:16) (2 m + 1) sinh − n z cos πj n +1 o(cid:17) z cos πj n +1 q z cos πj n +1 = X | j |≤ n X | k |≤ m ( − n + m − j − k cos πj n +1 cos πk m +1 z cos πj n +1 + z − cos πk m +1 . The RHS of this expression is symmetric under the transformation z → z − , n ↔ m . As a result the LHSis also symmetric under this transformation, which implies 3.2. (cid:3) Integrating both sides of 3.2 wrt α using the elementary formulas R dx √ x +1 = sinh − x = ln( √ x + x ) , R dy cosh y = − e − y one obtains X | j |≤ n ( − n + j arctan (cid:18)q α cos πj n +1 − α cos πj n +1 (cid:19) m +1 = X | k |≤ m ( − m + k " π − arctan (cid:18)q β cos πk m +1 − β cos πk m +1 (cid:19) n +1 . To complete the proof note that P | k |≤ m ( − m + k =1.4. Proof of Theorem 3
Despite the fact that Theorem 3 is a particular case of Theorem 5 it is instructive to give an independentproof. Again, as we did in the previous sections, it is convenient to break the proof into several parts.
Lemma 13.
We have the partial fractions expansion for arbitrary positive integer m : sinh (cid:0) m tanh − z (cid:1) sinh (cid:0) m tanh − z (cid:1) − z = 1 m √ m/ X k =1 (cid:18) k (cid:19) tan πk m z + tan πk m . (4.1) Proof.
Since sinh t sinh 3 t =
12 cosh 2 t +1 , tanh − z = ln z − z , and (cid:0) m tanh − z (cid:1) = (cid:18) z − z (cid:19) m + (cid:18) − z z (cid:19) m , the LHS of (4.1) is a rational function of z of the form f ( z ) = (1 − z ) m − P m ( z ) , where P m ( z ) is polynomialof degree exactly m . This rational function has poles at z k = i tan πk m , k = 3 r − or r − with r = 1 , , , ..., m . Rezidues of f ( z ) at z k are ( − k im sin πk im √ (cid:18) k (cid:19) . Hence, taking into account that (cid:0) k (cid:1) = 0 when k ≡ (cid:0) m tanh − z (cid:1) sinh (cid:0) m tanh − z (cid:1) − z = 12 im √ m X k =1 (cid:18) k (cid:19) z − i tan πk m . Due to (cid:0) m − k (cid:1) = − (cid:0) k (cid:1) and tan π (3 m − k )3 m = − tan πk m this is equivalent to (4.1). (cid:3) Lemma 14.
For arbitrary positive integers n and m we have the transformation formula m n/ X j =1 (cid:18) j (cid:19) sinh (cid:16) m tanh − (cid:16) z tan πj n (cid:17)(cid:17) sinh (cid:16) m tanh − (cid:16) z tan πj n (cid:17)(cid:17) tan πj n − z tan πj n − nz m/ X k =1 (cid:18) k (cid:19) sinh (cid:0) n tanh − (cid:0) z − tan πk m (cid:1)(cid:1) sinh (cid:0) n tanh − (cid:0) z − tan πk m (cid:1)(cid:1) tan πk m − z − tan πk m = 0 . (4.2) Proof.
In the previous lemma, replace z with z tan πj n , then multiply the resulting identity with zn (cid:18) j (cid:19) tan πj n , and sum wrt j from to n/ . It is easy to see the symmetry of the resulting double sum under thetransformation n → m , m → n , z → /z , from which the identity in the lemma follows. (cid:3) Lemma 15. √ Z ∞ s sinh t sinh 3 t dt = π − arctan tanh s √ √
31 + 2 e s . Proof.
The proof of this lemma is given in [4], but we reproduce it here for the sake of completeness.Since sinh t sinh 3 t = 12 cosh 2 t + 1 = 1cosh t
13 + tanh t , the integral can be easily calculated. The second equality follows from the elementary formula arctan x − arctan y = arctan x − y xy with x = √ , y = tanh s √ and the identity √ − tanh s √ tanh s = √
31 + 2 e s . (cid:3) Lemma 16.
For an odd positive integer n : n/ X j =1 (cid:18) j (cid:19) = 1 . Proof.
This is obvious for n = 1 . For arbitrary odd n its validity follows from the fact that the sum ofLegendre symbols mod for three consecutive integers is . (cid:3) The formula in Theorem 3 now follows easily from these lemmas. We integrate equation (4.2) wrt z from /α to ∞ using lemma 15. Then assuming that n and m are odd we complete the proof using lemma16.5. Proof of Theorem 5
With the help of the elementary formula arctan z = arg iz − iz one can recast the first sum in 1.12 inthe following form n − X j =0 arg (cid:18) α +tan ϕ + πjn α − tan ϕ + πjn (cid:19) m − e − iθ (cid:18) α +tan ϕ + πjn α − tan ϕ + πjn (cid:19) m − e iθ − n − X j =1 arg (cid:18) α − tan ϕ − πjn α +tan ϕ − πjn (cid:19) m − e − iθ (cid:18) α − tan ϕ − πjn α +tan ϕ − πjn (cid:19) m − e iθ . (5.1)In the second sum of 5.1, we make the change of the index of summation j → n − j , then rewrite bothsums as double sums using the fact that x m − Q mk =1 (cid:0) x − e πikm (cid:1) : n − X j =0 m X k =1 arg α +tan ϕ + πjn α − tan ϕ + πjn − e i πk − θm α +tan ϕ + πjn α − tan ϕ + πjn − e i πk + θm − n − X j = n +12 m X k =1 arg α − tan ϕ + πjn α +tan ϕ + πjn − e i πk − θm α − tan ϕ + πjn α +tan ϕ + πjn − e i πk + θm . (5.2)After simple algebraic manipulation of the summands, 5.2 becomes n − X j =0 m X k =1 arg ( e − i θm − iα sin πk − θm + tan ϕ + πjn cos πk − θm − iα sin πk + θm + tan ϕ + πjn cos πk + θm ) − n − X j = n +12 m X k =1 arg ( e − i θm iα sin πk − θm + tan ϕ + πjn cos πk − θm iα sin πk + θm + tan ϕ + πjn cos πk + θm ) . (5.3)The first sum in 5.3 can be simplified as − n + 12 θ + 12 n − X j =0 m X k =1 arctan ( α tan πk + θm tan ϕ + πjn ) − arctan ( α tan πk − θm tan ϕ + πjn )! , while the second as − n − θ − n − X j = n +12 m X k =1 arctan ( α tan πk + θm tan ϕ + πjn ) − arctan ( α tan πk − θm tan ϕ + πjn )! . Thus, 5.3 equals − θ + 12 n − X j =0 m X k =1 arctan ( α tan πk + θm tan ϕ + πjn ) − n − X j =0 m X k =1 arctan ( α tan πk − θm tan ϕ + πjn ) . (5.4)We make the change of the summation variable k → m − k in the second sum of 5.4 − θ + 12 n − X j =0 m X k =1 arctan ( α tan πk + θm tan ϕ + πjn ) + 12 n − X j =0 m − X k =0 arctan ( α tan πk + θm tan ϕ + πjn ) . Because of the equivalences P n − j =0 = P nj =1 , P m − k =0 = P mk =1 we finally get the following symmetric formfor the first sum in 1.12 − θ + n X j =1 m X k =1 arctan ( α tan πk + θm tan ϕ + πjn ) . Similarly for the second sum in 1.12 − ϕ + n X j =1 m X k =1 arctan ( β tan ϕ + πjn tan πk + θm ) . Hence, because of the elementary formula arctan x + arctan x − = π sgn x the LHS of 1.12 equals − θ − ϕ + π n X j =1 sgn (cid:18) tan ϕ + πjn (cid:19) m X k =1 sgn (cid:18) tan πk + θm (cid:19) . When ϕ ∈ (0 , π/ , the expression tan ϕ + πjn takes negative values at n − points j = n +12 , ..., n − andpositive values at the rest n +12 points. This means n X j =1 sgn (cid:18) tan ϕ + πjn (cid:19) = 1 . Thus the LHS of 1.12 equals π − θ − ϕ , as required.6. Other reciprocal relations
In our previous paper [13], we have found many relations of the form P ( n, m ) = P ( m, n ) for finiteproducts of trigonometric functions. However, the identity in Theorem is of the type S ( n, m )+ S ( m, n ) = C , where C is independent of n and m . There is simple method to find other relations of this type. It isbased on the solution of Dirichlet problem on a finite rectangular grid [11]. For example m n X j =1 ( − j cot πj n sinh yα j sinh mα j sin πjxn + n m X k =1 ( − k cot πk m sinh xβ k sinh nβ k sin πkym = − xy, (6.1)where ≤ x ≤ n , ≤ y ≤ m are integers and cos πjn + cosh α j = 2 , cos πkm + cosh β k = 2 (1 ≤ j ≤ n, ≤ k ≤ m ) . (6.2)In particular, when x = y , n = m this gives the closed form summation n X j =1 ( − j cot πj n sinh xα j sinh nα j sin πjxn = − x n , sinh α j πj n . (6.3)Laplace operator on a finite rectangular grid is defined as ∆ f ( x, y, k ) = f ( x − , y ) + f ( x + 1 , y ) + f ( x, y −
1) + f ( x, y + 1) − f ( x, y ) . One can see that the RHS of (6.1) satisfies the discrete Laplace equation ∆ f ( x, y ) = 0 , (0 < ≤ x ≤ n, ≤ y ≤ m ) on a rectangular grid of size n × m . Also − xy = f ( x, y ) + f ( x, y ) , where f ( x, y ) and f ( x, y ) aresolutions of the Laplace equation with boundary conditions ( f (0 , y ) = f ( n, y ) = 0 , ≤ y ≤ m,f ( x,
0) = 0 , f ( x, m ) = xm, ≤ x ≤ n, (6.4) ( f ( x,
0) = f ( x, m ) = 0 , ≤ x ≤ n,f (0 , y ) = 0 , f ( n, y ) = ny, ≤ y ≤ m. (6.5)Partial solutions of Laplace equation corresponding to boundary conditions (6.4) and (6.5) are given by,respectively u (1) j ( x, y ) = sin πjxn sinh yα j , (1 ≤ j ≤ n ) .u (2) k ( x, y ) = sin πkym sinh xβ k , (1 ≤ k ≤ m ) . In fact this method is quite well known and there are many examples in electrodynamics and heatconduction problems in physics (e.g., [12]).One could generalize (6.1) to include one continuous parameter α by requiring that α j and β k be definedby sinh α j α sin πj n , sinh β k α sin πk m , (1 ≤ j ≤ n, ≤ k ≤ m ) instead of (6.2). However to obtain a closed form summation we would need α = 1 , so this does notgeneralize (6.3).0 References [1] J. W. L. Glaisher,
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