Infinite arctangent sums involving Fibonacci and Lucas numbers Kunle Adegoke Department of Physics and Engineering Physics, Obafemi Awolowo University, Ile-Ife, 0005 Nigeria Saturday 3 rd July, 06, 6:43 Abstract Using a straightforward elementary approach, we derive numerous infinite arctangent summation formulas involving Fibonacci and Lucas numbers. While most of the results obtained are new, a couple of celebrated results appear as particular cases of the more general formulas derived here. Contents Introduction Preliminary result 4 3 Main Results 5 3. G F in identity (.3, that is, G 0 = 0, G =........ 5 3. G L in identity (.3, that is, G 0 =, G =........ 6 MSC 00: B39, Y60 adegoke00@gmail.com Keywords: Fibonacci numbers, Lucas numbers, Lehmer formula, arctangent sums, Infinite sums
4 Corollaries and special values 6 4. Results from Theorem 3..................... 6 4.. = F j, p = and k = 0 in identity (3......... 6 4.. = L j, p = and k = 0 in identity (3......... 7 4..3 = F j, k = j and p = 0 in identity (3......... 7 4..4 = F j and p = in identity (3............ 7 4..5 5 = L 4j, p = 0 and k = j in identity (3....... 8 4..6 5 = L 4j, p = 0 and k = j in identity (3...... 8 4..7 = L j / 5 and k = j in identity (3.......... 9 4..8 = L j / 5, p = 0 and k = j 0 in identity (3... 9 4. Results from Theorem 3..................... 9 4.. = F j and p = in identity (3........... 9 4.. = L j / 5 and k = j in identity (3......... 0 4..3 5 = L 4j and k = j in identity (3.......... 4.3 Results from Theorem 3.3.................... 4.3. = L 4j, k = 0 and p = in identity (3.3...... 4.3. = L j and p = in identity (3.3........... 4.3.3 = 5F j, p = and k = 0 in identity (3.3...... 4.4 Results from Theorem 3.4.................... 3 4.4. = L 4j and j = 0 = k in identity (3.4...... 3 4.4. = L j and p = in identity (3.4.......... 3 4.4.3 = L j and j = 0 = k in identity (3.4........ 4 4.4.4 = 5F j and j = 0 = k in identity (3.4...... 4 5 Conclusion 4 Introduction It is our goal, in this work, to derive infinite arctangent summation formulas involving Fibonacci and Lucas numbers. The results obtained will be found to be of a more general nature than one finds in earlier literature. Previously known results containing arctangent identities and/or infinite summation involving Fibonacci numbers can be found in references [,, 3, 4, 5] and references therein.
In deriving the results in this paper, the main identities employed are the trigonometric addition formula (y x tan = tan xy + x tan y, (. which holds for R such that either xy > 0 or xy < 0 and < xy, and the following identities which resolve products of Fibonacci and Lucas numbers F u v F u+v = Fu ( (u v Fv, L u v L u+v = L u + ( (u v L v, L u F v = F v+u + ( u F v u, F u L v = F v+u ( u F v u, L u L v = L u+v + ( u L v u, 5F u v F u+v = L u ( (u v L v. (.a (.b (.c (.d (.e (.f Also we shall make repeated use of the following identities connecting Fibonacci and Lucas numbers: F u = F u L u, L u ( u = 5Fu, 5Fu L u = 4( (u+, L u + ( u = L u. (.3a (.3b (.3c (.3d Identities (. and (.3 or their variations can be found in [6, 7, 8]. On notation, G i, i integers, denotes generalized Fibonacci numbers defined through the second order recurrence relation G i = G i + G i, where two boundary terms, usually G 0 and G, need to be specified. When G 0 = 0 and G =, we have the Fibonacci numbers, denoted F i, while when G 0 = and G =, we have the Lucas numbers, denoted L i. Throughout this paper, the principal value of the arctangent function is assumed. 3
Interesting results obtained in this paper, for integers k, j 0 and p include tan r= r= r= F j L j L 4jr F4jr F j + F j tan F j L 4jr+j F 4jr+j = tan ( Lj, = tan ( L j ( tan F(j = tan Fj, F 4jr r+ F j ( tan L r = tan, 5 Fr L p tan r=, tan r= r= tan r= L j F j L 4jr F4jr F j + L j F(j F 4jr r+j F r+k = tan ( Fj, = tan ( L j ( = tan, F k ( tan F4j = tan Lj. F 4jr L j We also obtained the following special values tan L4r = π F r= 4r, tan L4r = π F r= 4r 4, 35 tan L4r = π L r= (4r, tan 3Lr = π L r= 4r 3, tan L r = π 5 F r= r 4, 5 tan = tan 5, L r= r 35 tan L4r 7 = L r= 8r 5, tan 5Fr = π L r= r, ( tan 5 7F 4r = tan ( 7, tan L4r+ = tan. L (4r F4r+ 3 r= Preliminary result Taking x = G mr+n m and y = G mr+n in the arctangent addition formula, identity (., gives r= ( tan (Gmr+n G mr+n m = tan G mr+n G mr+n m + 4 G mr+n m ( tan G mr+n. (.,
Summing each side of identity (. from r = p Z to r = N Z + and noting that the summation of the terms on the right hand side telescopes, we obtain N ( tan (Gmr+n G mr+n m = tan G mr+n G mr+n m + Now taking limit as N, we have Lemma. For R, n, m, p Z, m 0 holds G mp+n m ( tan (Gmr+n G mr+n m = tan G mr+n G mr+n m + 3 Main Results tan ( G mp+n m 3. G F in identity (.3, that is, G 0 = 0, G = G mn+n (.. (.3 Choosing m = 4j and n = k + j and using identities (.a and (.d we prove THEOREM 3.. For R, j, k, p Z and j 0 holds ( tan F j L 4jr+k F4jr+k F j + = tan F 4jp+k j, (3. while taking m = 4j and n = k + j and using identities (.a and (.c we prove THEOREM 3.. For R and j, k, p Z holds ( tan L j F 4jr r+k F4jr r+k F j + = tan F 4jp p+k j. (3.. 5
3. G L in identity (.3, that is, G 0 =, G = Choosing m = 4j and n = k + j and using identities (.b and (.f we prove THEOREM 3.3. For R, j, k, p Z and j 0 holds ( ( tan 5Fj F 4jr+k = tan L 8jr+4k L 4j + L 4jp+k j, (3.3 while taking m = 4j and n = k + j and using identities (.b and (.e we prove THEOREM 3.4. For R and j, k, p Z holds ( ( tan L j L 4jr r+k = tan L 8jr 4r+4k L 4j + 4 Corollaries and special values L 4jp p+k j+. (3.4 Different combinations of the parameters, j, k and p in the above theorems yield a variety of interesting particular cases. In this section we will consider some of the possible choices. 4. Results from Theorem 3. 4.. = F j, p = and k = 0 in identity (3. The choice = F j, p = and k = 0 in identity (3. gives tan r= F j L j L 4jr F 4jr F j + F j Thus, at j =, we obtain the special value r= = tan ( Lj. (4. tan L4r = π F4r 4. (4. 6
4.. = L j, p = and k = 0 in identity (3. The above choice gives tan r= L j F j L 4jr F 4jr F j + L j = tan ( Fj. (4.3 At j =, identity(4. is reproduced, while at j = we have the special value r= tan 9L8r = π F8r 4. (4.4 Note that identities (4. and (4.4 are special cases of identity (4.8 below, at j = and j =, respectively. 4..3 = F j, k = j and p = 0 in identity (3. This choice gives r= which, at j =, gives the special value F tan j L 4jr j = π F4jr j, (4.5 r= 4..4 = F j and p = in identity (3. This choice gives r= tan F j L 4jr+k F 4jr+k At k = 0 in identity (4.7 we have r= tan L4r = π F4r. (4.6 ( = tan Fj. (4.7 F j+k F tan j L 4jr = π F4jr 4. (4.8 7
Note that identities (4. and (4.4 are special cases of identity (4.8 at j = and j =, respectively. At k = j 0 in identity (4.7 we have r= yielding at j =, the special value F ( tan j L 4jr+j = tan, (4.9 F4jr+j L j r= ( tan L4r+ = tan F4r+ 3. (4.0 Finally, taking limit of identity (4.7 as j, we obtain F ( lim tan j L 4jr+k = tan. (4. j F4jr+k φ k r= 4..5 5 = L 4j, p = 0 and k = j in identity (3. Another interesting particular case of identity (3. is obtained by setting 5 = L 4j, p = 0 and k = j to obtain tan Fj 5L4j L 4jr j = π L (4jr j, (4. r= which at j = gives the special value r= 35 tan L4r = π L (4r. (4.3 4..6 5 = L 4j, p = 0 and k = j in identity (3. In this case Theorem 3. reduces to r= tan Fj 5L4j L 4jr L 8jr = tan ( 5 L4j F j. (4.4 8
At j =, we have the special value r= 35 tan L4r 7 = L 8r 5. (4.5 4..7 = L j / 5 and k = j in identity (3. Setting = L j / 5 and k = j in identity (3. we have ( tan 5F4j = tan Lj, (4.6 L 4jr+j F 4jp 5 which at p = gives and at p = 0 yields r= ( tan 5F4j = tan L 4jr+j F j 5 r= (4.7 tan 5F4j = π L 4jr j. (4.8 4..8 = L j / 5, p = 0 and k = j 0 in identity (3. The above choice yields r= tan 5F4j L 4jr 4. Results from Theorem 3. 4.. = F j and p = in identity (3. The above choice gives tan r= F(j F 4jr r+k ( = tan Lj. (4.9 F j 5 ( = tan Fj. (4.0 F j+k At k = j in identity (4.0 we have the interesting formula 9
tan r= F(j F 4jr r+j ( = tan L j. (4. Note that identity (4., at j =, includes Lehmer s result (cited in [3, 5] as a particular case. Setting j = in identity (4.0 we obtain tan r= F r+k ( = tan F k. (4. Note again that identity (4. subsumes Lehmer s formula and the result of Melham (p = in identity(3.5 of [5], at k = and at k = 0 respectively. Finally, taking limit j in identity (4.0, we obtain lim j tan r= F(j F 4jr r+k 4.. = L j / 5 and k = j in identity (3. The above choice gives tan 5L j F 4jr r+j L 4jr r+j Setting p = in identity (4.4, we find 5L tan j F 4jr r+j L 4jr r+j r= while choosing j = leads to tan 5Fr+ L r+ which at p = 0 gives the special value ( = tan. (4.3 φ k ( = tan L 5 j. (4.4 F 4jp p ( = tan 5, (4.5 F j ( = tan 5, (4.6 F p 0
tan 5Fr = π L r. (4.7 r= 4..3 5 = L 4j and k = j in identity (3. The above substitutions give tan 5L4j L j F 4jr r+j L (4jr r+j ( = tan 5L4j. (4.8 5F 4jp p At p = 0 in identity (4.8 we have, for positive integers j, tan 5L4j L j F 4jr r j+ = π L (4jr r j+, (4.9 r= giving, at j =, the special value r= tan 5Fr = π L (r. (4.30 At p = in identity (4.8 we have, for positive integers j, r= tan 5L4j L j F 4jr r+6j 3 L (4jr r+6j 3 which gives, at j =, the special value r= = tan ( ( tan 5Fr+3 = tan L (r+3 5 4.3 Results from Theorem 3.3 4.3. = L 4j, k = 0 and p = in identity (3.3 The above choice gives, 5F4j F 8j 4 (4.3. (4.3
( 5 tan L4j F j F 4jr L 8jr r= which, at j =, gives r= ( = tan L4j, (4.33 L j ( tan 5 7F 4r = tan 7. (4.34 L (4r 4.3. = L j and p = in identity (3.3 Setting = L j and p = in identity (3.3 gives tan r= F4j F 4jr+k Taking limit as j in identity (4.35 gives lim j tan r= F4j F 4jr+k 4.3.3 = 5F j, p = and k = 0 in identity (3.3 r= ( = tan Lj. (4.35 L j+k ( = tan. (4.36 φ k Setting = 5F j, p = and k = 0 in identity (3.3 we obtain ( ( 5 5F tan j F 4jr = tan 5Fj, (4.37 L 4jr L j which gives the special value ( tan 5 5F 4r = tan 5, (4.38 L 4r at j =. r=
4.4 Results from Theorem 3.4 4.4. = L 4j and j = 0 = k in identity (3.4 With the above choice we obtain tan 3Lr L 4r which gives rise, at p =, to the special value r= 4.4. = L j and p = in identity (3.4 With the above choice we have r= tan L j 5 k = 0 in identity (4.4 gives r= ( 3 = tan, (4.39 L p tan 3Lr = π L 4r 3. (4.40 L 4jr r+k F 4jr r+k tan L j 5 which at j = gives the special value r= j = in identity (4.4 leads to tan L r+k 5 Fr+k r= which gives the special value r= ( = tan Lj. (4.4 L j+k L 4jr r = π F4jr r 4, (4.4 tan L r = π 5 Fr 4. (4.43 ( tan L r+ = tan 5 Fr+ 4 3 ( = tan, (4.44 L k+, (4.45
at k =. Taking limit j in identity (4.4, we obtain L ( lim tan j L 4jr r+k = tan. (4.46 j 5 F4jr r+k φ k r= 4.4.3 = L j and j = 0 = k in identity (3.4 This choice gives ( tan L r = tan, (4.47 5 Fr L p Note that identities (4.43 and (4.45 are special cases of (4.47 at p = and at p =. 4.4.4 = 5F j and j = 0 = k in identity (3.4 The above choice gives which at p = gives the special value 5 Conclusion ( 5 5 tan = tan, (4.48 L r L p r= 5 tan = tan 5. (4.49 L r Using a fairly straightforward technique, we have derived numerous infinite arctangent summation formulas involving Fibonacci and Lucas numbers. While most of the results obtained are new, a couple of celebrated results appear as particular cases of more general formulas derived in this paper. 4
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