Algorithmic Manipulation of Fibonacci Identities

Transcription

1 Algorthmc Manpulaton of Fbonacc Identtes 1 Stanley Rabnowtz 1 Vne Brook Road Westford, MA USA 1 Introducton Methods for manpulatng trgonometrc expressons, such as changng sums to products, changng products to sums, expandng functons of multple angles, etc, are wellknown [1] In fact, the process of verfyng trgonometrc denttes s algorthmc (see [] or [5] Roughly speakng, all trgonometrc denttes can be derved from the basc dentty sn x + cos x =1 Methods for manpulatng expressons nvolvng Fbonacc and Lucas numbers are also known; however, they do not seem to be systematcally collected n one place The fact that verfyng Fbonacc denttes s algorthmc seems to be less well-known, especally consderng the strong relatonshp there s between Fbonacc numbers and trgonometrc functons (see equaton (7 below For example, the problem of establshng the dentty F 14r (F 7 n+4r + F 7 n 7F 10r (F 6 n+4rf n + F n+4r F 6 n+1f 6r (F 5 n+4rf n + F n+4rf 5 n 35F r (F 4 n+4rf 3 n + F 3 n+4rf 4 n=f 7 4rF 7n+14 (1 s stll consdered hard as wtnessed by the fact that ths problem appeared n the advanced problem secton of The Fbonacc Quarterly [10] Yet ths dentty (and all such denttes can be proven usng a straghforward algorthm that we shall descrbe below Roughly speakng, all Fbonacc denttes can be derved from the basc dentty L n 5Fn =4( 1 n The Basc Algorthms The key dea to algorthmcally provng denttes nvolvng polynomals n F an+b and L an+b s to frst reduce them to polynomals n F n and L n Todothat, we need reducton formulas for F m+n and L m+n These are well-known In fact, all the formulas presented n ths paper can be found n the lterature (see, for example, [9] Most of these formulas were known to Lucas n the 19th century [6] ALGORITHM FbEvaluate TO NUMERICALLY EVALUATE F k AND L k : Gven an nteger constant k, toevaluate F k or L k numercally, apply the followng algorthm: STEP 1: [Make subscrpt postve] If k<0, apply algorthm FbNegate (gven below STEP : [Recurse] If k>1, apply the recurson: F n = F n 1 + F n L n = L n 1 + L n Reprnted from Applcatons of Fbonacc Numbers, volume 6 (edted by G E Bergum, et al, Kluwer Academc Publshers, Dordrecht, The Netherlands: 1996, pp (

2 Repeat step untl k < Step reduces the subscrpt by 1, so the recurson must eventually termnate STEP 3: [Intal values] If k s 0 or 1, then use the followng ntal values: F 0 =0, F 1 =1; L 0 =, L 1 =1 (3 Reprnted from Applcatons of Fbonacc Numbers, volume 6, (edted by G E Bergum, et al, pp Corrected 6/98 NOTE: Whle ths may not be the fastest way to evaluate F k and L k,tsnevertheless an effectve algorthm It s our purpose here to present effectve methods to show that the manpulatons descrbed are all algorthmc and t s not our purpose to fnd the most effcent algorthms possble Faster algorthms can be obtaned by usng the double argument formulas gven below n dsplay (1 For a descrpton and analyss of fast methods of numercally evaluatng F k and L k, see chapter 7 of [7] ALGORITHM FbNegate TO REMOVE NEGATIVE SUBSCRIPTS: F n =( 1 n+1 F n L n =( 1 n L n (4 ALGORITHM FbReduce TO REMOVE SUMS IN SUBSCRIPTS: F m+n = F ml n + L m F n L m+n = 5F mf n + L m L n These are also called the addton formulas 3 The Fundamental Identty Connectng F and L The Fbonacc and Lucas numbers are connected by the well-known dentty: (5 L n 5F n =4( 1 n (6 In the same way that all trgonometrc denttes are consequences of the fundamental dentty sn x + cos x =1,all Fbonacc/Lucas denttes are consequences of the fundamental dentty (6 That s, equaton (6 s the unque dentty connectng F n and L n The proof s a consequence of the fact that F n = 5 n+1 sn( n ln L n = n cos( n ln or F n = ( n n snh 5 ln ( n L n = n cosh ln (7

3 so that the result s equvalent to the result for trgonometrc polynomals, whose proof can be found n [] The fundamental dentty allows us to take any polynomal n F n and L n and remove all powers of L n (of degree or hgher by usng the followng algorthm ALGORITHM RemovePowersOfL : Use the dentty: L n =5Fn +4( 1 n (8 Contnue applyng ths substtuton untl no L n term has an exponent larger than 1 Smlarly, we could remove powers of F n nstead ALGORITHM RemovePowersOfF : Use the dentty: Fn = L n 4( 1 n (9 5 Contnue applyng ths substtuton untl no F n term has an exponent larger than 1 4 The Smplfcaton Algorthm Let us be gven a polynomal functon of elements of the form F x and L x, where the subscrpts of F and L are of the form a 1 n 1 + a n + + a k n k + b where b and the a are nteger constants and the n are varables To put ths expresson n canoncal form, we apply the followng algorthm: ALGORITHM FbSmplfy TO TRANSFORM AN EXPRESSION TO CANON- ICAL FORM STEP 1: [Remove sums n subscrpts] Apply algorthm FbReduce to remove any sums (or dfferences n subscrpts STEP : [Make multplers postve] All subscrpts are now of the form cx where c s an nteger For any term n whch the multpler c s negatve, apply algorthm FbNegate After performng ths step, you must take care to replace any expressons of the form ( 1 an+b where a and b are constants by a properly evaluated [( 1 a ] n ( 1 b,toavod wndng up wth expressons lke ( 1 n n your fnal canoncal form STEP 3: [Remove multplers] All subscrpts are now of the form cx where c s a postve nteger For any term n whch the multpler c s not 1, apply algorthm FbReduce successvely untl all subscrpts are varables STEP 4: [Evaluate constants] If any term nvolves an expresson of the form Fc k or L k c where c s an nteger constant, use algorthm FbEvaluate to replace F c and L c by ther numercal equvalents STEP 5: [Remove Fbonacc Powers] If any term nvolves an expresson of the form Fn k where k > 1 and n s a varable, apply algorthm RemovePowersOfF to leave only lnear terms n F n PROVING IDENTITIES To prove that an expresson s dentcally 0, apply algorthm FbSmplfy The expresson s dentcally 0 f and only f algorthm FbSmplfy transforms t to 0 3

4 Ths s because algorthm FbSmplfy leaves us wth a polynomal n varables of the form F n and L n wth the degree of F n beng 0 or 1 Convertng to trgonometrc form usng formula (7, we get a polynomal n varables of the form sn cx and cos cx wth the sne terms havng degree 0 or 1 Such polynomals cannot be dentcally 0 snce t s known that all trgonometrc denttes are consequences of the fundamental dentty sn θ+cos θ =1(see, for example, [5] and so some sne term would have to have degree or more Example Let us see how to prove F n+w F n+w L w F n+w F n w F n w F n w =(L 3w L w F n wth w odd, an dentty that comes from [11] Algorthmc proof Snce we are gven that w s odd, we make the substtuton w =m+1 before we begn We must show that F n+4m+ F n+m+1 L m+1 F n+m+1 F n m 1 F n m 1 F n 4m Fn(L 6m+3 L m+1 sdentcally 0 Frst we apply algorthm FbReduce (step 1 to get: (Fn(80F m +16L m 80F 6m 3L 6m ((F n ( 5F 4m +3L 4m +(3F 4m L 4m L n (5F m F n F n L m F m L n + L m L n (5F m + L m (5F m F n F n L m F m L n + L m L n (5F m F n + F n L m + F m L n + L m L n +(5F m F n + F n L m + F m L n + L m L n (F n (5F 4m +3L 4m +(3F 4m + L 4m L n /16 Then we apply algorthm FbNegate (step to get: (80F m Fn + 15FmF 3 n + 50F m F 4m Fn 80F 6m Fn +16FnL m +75FmF nl m +10F 4m FnL m +15F m FnL m + FnL 3 m+30f m FnL 4m +6FnL m L 4m 3FnL 6m 5FmL 3 n+6f m F 4m L n 11FmL m L n+ 6F 4m L m L n 7F m L ml n L 3 ml n +F m L 4m L n +L m L 4m L n/16 Applyng algorthm FbReduce agan (step 3 to get rd of scalar multples n subscrpts yelds: (80FmF n 15FmF 6 n + 160F m FnL m 50FmF 5 nl m +16FnL m + 5FmF 4 nl m + 100FmF 3 nl 3 m +5FmF nl 4 m 10F m FnL 5 m FnL 6 m/3 Havng evaluated constants lke F 1 all along, we proceed to step 5 and apply algorthm RemovePowersOfF Upon expandng and gatherng lke terms together, we fnd that the result s 0 Thus our example s n fact an dentty 5 Other Algorthms Sometmes we want to transform an expresson from one form to another, rather than put t n canoncal form The followng algorthms can be used for such purposes ALGORITHM ConvertToF TO REMOVE LUCAS NUMBERS: Use the dentty: L n = F n 1 + F n+1 (10 ALGORITHM ConvertToL TO REMOVE FIBONACCI NUMBERS: Use the dentty: F n = L n 1 + L n+1 (11 5 4

5 5 THE DOUBLE ARGUMENT FORMULAS Lettng m = n n formula (5 gves us the followng formulas F n = F n L n L n = 5F n + L n These correspond to the double angle formulas n trgonometry TO REMOVE SCALAR MULTIPLES OF ARGUMENTS IN SUBSCRIPTS: Repeatedly apply the reducton formulas successvely In ths manner, we obtan the denttes: F 3n = 3F nl n +5Fn 3 4 L 3n = L3 n +15FnL n 4 F 4n = F nl n (5Fn + L n L 4n = 5F (13 n 4 +30FnL n + L 4 n 8 F 5n = 5F n(5fn 4 +10FnL n + L 4 n 16 SHORTCUT: Apply the followng recurrences: L 5n = 15F nl 4 n +50FnL 3 n + L 5 n 16 (1 F kn = F (k 1nL n + L (k 1n F n L kn = 5F (k 1nL n + L (k 1n F n Even more straghtforward s to use a drect formula: (14 Example: F kn = 1 k 1 L kn = 1 k 1 k 1 k/ ( k +1 ( k 5 Fn L k n 5 Fn +1 L k 1 n (15 56F 9n = 65F 9 n F 7 nl n F 5 nl 4 n + 40F 3 nl 6 n +9F n L 8 n 56L 9n = 565F 8 nl n F 6 nl 3 n F 4 nl 5 n + 180F nl 7 n + L 9 n

6 Because of the fundamental dentty, these can also be put n other forms, such as the followng, whch converts drectly to our canoncal form 6 F kn = F n L kn = k/ k 1 k k ( k 1 ( k ( 1 (n+1 L k 1 n ( 1 (n+1 L k n (16 In general, F kn cannot be replaced by sums of powers of F n alone, so no formula of ths type s gven Also of nterest are the followng compact formulas F kn = L kn = k ( k F nfn 1 k F k ( k F nfn 1 k L (17 ALGORITHM FbExpand TO TURN PRODUCTS INTO SUMS: F m F n = L m+n ( 1 n L m n 5 L m L n = L m+n +( 1 n L m n F m L n = F m+n +( 1 n F m n (18 To remove products of more than two terms, the above expanson formulas can be used repeatedly, expandng out the results after each step Removng powers s the same as removng products For example, L n = L n L n = L n +( 1 n L 0 Smlarly for F nsowehave F n = L n ( 1 n 5 L n = L n +( 1 n (19

7 7 Contnung n ths way, we fnd: F 3 n = F 3n 3( 1 n F n 5 L 3 n = L 3n +3( 1 n L n F 4 n = L 4n 4( 1 n L n +6 5 L 4 n = L 4n +4( 1 n L n +6 F 5 n = F 5n 5( 1 n F 3n +10F n 5 L 5 n = L 5n +5( 1 n L 3n +10L n F 6 n = L 6n 6( 1 n L 4n +15L n 0( 1 n 15 L 6 n = L 6n +6( 1 n L 4n +15L n + 0( 1 n (0 To remove hgher powers, the expanson formulas (18 can be used repeatedly, expandng out the results after each step CHANGE OF BASIS (Shft Formulas ALGORITHM FbShft TO TRANSFORM AN EXPRESSION INVOLV- ING F n, L n INTO ONE INVOLVING F n+a, L n+b : ( Fn L n ( Lb F = a L a L b 5F a F b 5F b L a ( Fn+a L n+b (1 Proof Solve the lnear equatons F n+a = 1 (F al n + L a F n L n+b = 1 (5F bf n + L b L n for F n and L n In a smlar manner, we fnd ( ( Fn Fb F = a L n F a L b L a F b L b L a ( Fn L n = ( Fn+a F n+b ( ( ( Lb L a Ln+a (3 5(L a F b F a L b 5F b 5F a L n+b To change from an arbtrary bass to another, apply algorthm FbReduce to transform the gven expresson to the bass (F n,l n Then use one of the above shft formulas

8 ALGORITHM ConvertToAlphaBeta TO EXPRESS F n AND L n IN TERMS OF α AND β: Use the well-known Bnet forms: F n = αn β n 5 (4 L n = α n + β n to express F n and L n n terms of α and β, the roots of the characterstc equaton x = x+1 ALGORITHM RemoveAlphaBeta TO REMOVE α s AND β s: Sometmes a formula nvolves the quanttes α and β Totransform such expressons nto canoncal form, apply the denttes: α n = L n + F n 5 β n = L n F n 5 Expand out the resultng polynomal and express t n the form x + y 5 If y s not dentcally 0, then the resultng expresson cannot be expressed n terms of Fbonacc and Lucas numbers alone The quanttes α and β can be rentroduced nto the resultng expresson by means of the formula 5=α β (6 8 (5 and ether α or β may be removed by use of one of the denttes α + β =1 or αβ = 1 (7 whchever s preferred TO REMOVE POWERS OF α AND β: The followng denttes are frequently useful: α n = αf n + F n 1 β n = βf n + F n 1 (8 THE SUBTRACTION FORMULAS: Combnng the reducton and negaton formulas gves us the followng: F m n =( 1 n FmL n F n L m L m n =( 1 n LmL n 5F m F n

9 9 6 Arbtrary Startng Condtons Let {H n } be asequence that satsfes H n+ = H n+1 + H n (9 wth ntal condtons H 0 = a and H 1 = b We can formally manpulate ths sequence by frst convertng to Fbonacc numbers and then convertng back ALGORITHM ConvertHToF TO EXPRESS H n IN TERMS OF F n AND F n 1 : Use the dentty: H n = af n 1 + bf n (30 To prove an dentty nvolvng H n, use equaton (30 to convert H s to F s, and then apply algorthm FbSmplfy After applyng algorthms ConvertHToF and FbSmplfy to obtan a canoncal form, use the followng algorthm to transform the result (expressed wth F s and L s back to an expresson nvolvng H s ALGORITHM ConvertToH TO EXPRESS F n and L n IN TERMS OF H n AND H n+1 : where e = b a ab F n = 1 e (bh n ah n+1 L n = 1 e [(b ah n (a bh n+1 ] ALGORITHM HExpand TO EXPRESS H n+m IN TERMS OF H n AND H n+1 : (31 A more symmetrcal form s: H n+m = F m 1 H n + F m H n+1 (3 H n+m = 1 (F mg n + L m H n where G n = H n 1 + H n+1

10 CHANGE OF BASIS ALGORITHM HShft TO TRANSFORM AN EXPRESSION INVOLVING H n, H n+1 INTO ONE INVOLVING H n+c, H n+d : ( Hn = H n+1 1 F c 1 F d F c F d 1 ( Fd F c F d 1 F c 1 ( Hn+c H n+d 10 (33 To convert expressons nvolvng arbtrary subscrpts of H to expressons nvolvng H n+c and H n+d, apply algorthm ConvertHToF to put everythng n terms of F Then apply algorthm FbReduce to get everythng n terms of F n and L n Use algorthm ConvertToH to get the expresson n terms of H n and H n+1 ; and then perform an HShft to transform to the desred bass 7 Generalzed Fbonacc Numbers We may consder the sequences {u n } and {v n } defned by the recurrences: u 0 =0, u 1 =1, u n+ = Pu n+1 Qu n v 0 =, v 1 = P, v n+ = Pv n+1 Qv n (34 where P and Q are gven constants (usually ntegers wth Q 0 Let r 1 and r be the roots of the characterstc equaton so that and let so that x = Px Q (35 r 1 + r = P and r 1 r = Q, (36 r 1 = P + D D = P 4Q (37 and r = P D (38 We wll also assume that P and Q are chosen so that D 0 Ths assures that the roots are dstnct The Bnet form for u n and v n s u n = rn 1 r n r 1 r, v n = r n 1 + r n (39 Powers of r 1 and r can be removed from an expresson by means of the formulas: r1 n = v n + u n D r n = v n u n D (40

11 If a term of the form D s present, r 1 and r can be rentroduced, f desred, by the formula r 1 r = D (41 Also of nterest are the denttes: 11 r1 = Pr 1 Q r = Pr Q (4 and r n 1 = r 1 u n Qu n 1 r n = r u n Qu n 1 For these sequences, we have the followng algorthms (from [6]: ALGORITHM LucasNegate TO REMOVE NEGATIVE SUBSCRIPTS: u n = u n Q n v n = v (43 n Q n ALGORITHM LucasReduce TO REMOVE SUMS IN SUBSCRIPTS: u m+n = u mv n + u n v m v m+n = v mv n + Du m u n We also have the subtracton formulas : u m n = u mv n u n v m Q n v m n = v mv n Du m u n Q n (44 (45 THE FUNDAMENTAL IDENTITY: The fundamental dentty that connects u n and v n s: vn Du n =4Q n (46 Ths can be used to gve us the followng three algorthms ALGORITHM RemovePowersOfV : Apply the substtuton vn = Du n +4Q n (47

12 1 repeatedly untl the exponent of all v n terms s less than ALGORITHM RemovePowersOfU : Apply the substtuton u n = v n 4Q n D repeatedly untl the exponent of all u n terms s less than (48 ALGORITHM RemovePowersOfQ : Apply the substtuton Q n = v n Du n (49 4 ALGORITHM LucasExpand TO TURN PRODUCTS INTO SUMS: u m u n = v m+n Q n v m n D v m v n = v m+n + Q n v m n u m v n = u m+n + Q n u m n (50 Ths algorthm can be repeated to turn products of any number of u and v terms nto smple sums of such terms ALGORITHM ConvertToU TO REMOVE ALL v s: Use the dentty: v n = u n+1 Qu n 1 (51 ALGORITHM ConvertToV TO REMOVE ALL u s: Use the dentty: u n = v n+1 Qv n 1 (5 D RELATIONSHIP TO TRIGONOMETRIC EXPRESSIONS: To convert to trgonometrc form, use the denttes: ( n u n =Q n/ sn log r 1 / D r ( n v n =Q n/ cos log r 1 r (53

13 13 TO REMOVE SCALAR MULTIPLES IN SUBSCRIPTS: Repeatedly apply algorthm LucasReduce or use the denttes: u kn = u n v kn = k/ k 1 k k ( k 1 ( k ( 1 Q n v k 1 n ( 1 Q n v k n (54 These formulas convert drectly to our canoncal form In partcular, the double argument formulas are: u n = u n v n v n = v n Q n (55 Other formulas of nterest: u kn = 1 k 1 v kn = 1 k 1 k 1 k/ ( k +1 ( k D u n vn k D u +1 n vn k 1 u kn+s = v kn+s = k ( k u n( Qu n 1 k u +s k ( k u n( Qu n 1 k v +s TO REMOVE POWERS: Repeatedly apply algorthm LucasExpand or use the denttes: u k n = 1 v k n = 1 1 D k/ k k ( { k ( 1 Q n u(k n, f k s odd, v (k n, f k s even; ( k Q n v (k n (56 Any negatve subscrpts ntroduced can be removed by usng algorthm LucasNegate, f desred

14 14 CHANGE OF BASIS (Shft Formulas ALGORITHM LucasShft TO TRANSFORM AN EXPRESSION INVOLV- ING u n, v n INTO ONE INVOLVING u n+a, v n+b : ( ( ( un vb u = a un+a (57 v n v a v b Du a u b Du b v a v n+b PROVING IDENTITIES To effectvely prove an dentty nvolvng u s and v s, perform the followng algorthm whch s analogous to algorthm FbSmplfy ALGORITHM LucasSmplfy : Use algorthm LucasReduce to remove any sums n subscrpts Make all the multplers n subscrpts postve by usng algorthm LucasNegate Use LucasReduce agan to remove any constant multplers n the subscrpts Evaluate any numercal terms usng the recurrence (34 and ts ntal values (applyng algorthm LucasNegate frst for negatve subscrpts Fnally, use algorthm RemovePowersOfU and collect terms to arrve at a canoncal form Note that f the varables P, Q, and D all occur n the fnal expresson, you should replace D by ts equvalent value, P 4Q, from formula (37 The expresson s dentcally 0 f and only f ths canoncal form s 0 8 Arbtrary Second-Order Lnear Recurrences Let w n be an arbtrary second-order lnear recurrence wth constant coeffcents That s, w n = Pw n 1 Qw n, n (58 wth arbtrary ntal values w 0 and w 1, not both 0 Agan we requre that Q 0and P 4Q Snce u n and v n form a bass among all such sequences, w n can be expressed n terms of u n and v n Ths can be accomplshed by the followng formula: w n =(w 1 P w 0 u n + w 0 v n (59 Thus, to prove any dentty nvolvng w s, frst convert the w s to u s and v s and then apply algorthm LucasSmplfy An equvalent formula s w n =(w 1 Pw 0 u n + w 0 u n+1 = Qw 1 u n + w 0 u n+1 The reducton formula takes the form w n+m = Qw m 1 u n + w m u n+1 = Qw m u n 1 + w m+1 u n

15 9 Summatons We can perform ndefnte summatons of expressons nvolvng u n and v n any tme we can perform such summatons wth x n nstead, snce from the Bnet forms, these terms are actually exponentals wth bases r 1 and r ALGORITHM LucasSum : Frst, the expresson s converted to exponental form usng equaton (39 Then t s summed The result s converted back to u s and v s by usng equaton (40 Then r 1 and r are converted to expressons nvolvng P and Q usng equatons (38 and (36 The reader who has not kept abreast of developments n the area of symbolc algebra may be surprsed at the large class of functons that can now be automatcally summed Gosper s algorthm [3] can be used to sum complcated expressons nvolvng hypergeometrc seres Ths ncludes most of the common elementary functons See [4] pp 4 30 for an elementary exposton Example Let us see how to sum n k=1 L k cos kx algorthmcally Expandng L k by the Bnet form (4, and expandng cos kx by the formula cos x =(e x + e x /, turns the sum nto four terms each of whch s a smple geometrc progresson These are easly summed The result s turned back nto a nce form by usng formula (5 and the formula e x = cos x + sn x The result s the followng, whch we beleve to be new 15 n L k cos kx = k=1 1 [ 5+cosx 4cos x F n ( cos x 5 cos(n +1x cos x 3 ] + F n+1 (( cos x 3 cos nx +cos(n +x 4F n+ cos x cos nx (60 10 Dscoverng Identtes For the most part, the algorthms descrbed n ths paper can only be used to prove an dentty once the dentty s known or suspected It stll requres human ngenuty to dscover new and nterestng denttes There are, however, several ways these algorthms can be used to ad n dscoverng new denttes Algorthm LucasSum and ts Fbonacc counterpart, algorthm FbSum, are effectve algorthms for summng expressons Thus, they can be used, as n the prevous example, to dscover new results If the form of an dentty s suspected, algorthm FbSmplfy may frequently be used to advantage to dscover a new result For example, suppose that you suspect that F n+3 can be wrtten as a lnear combnaton of F n+, F n+1, and F ninthat case, you can use algorthm FbSmplfy to convert F n+3 af n+ bf n+1 cf n nto canoncal form The result s ( 1 n ( a + b +4c/5+L n(18 7a 3b c/10 + F n L n (8 3a b/

16 Snce the orgnal expresson s an dentty f and only f ths canoncal form s dentcally 0, we must have the three equatons 9a + b +4c =16 7a +3b +c =18 3a + b =8 Solvng these equatons yelds a =,b =,and c = 1 Ths reveals the dentty F n+3 F n+ F n+1 + F n =0 (61 11 Implementaton of the Algorthms All the algorthms descrbed n ths paper have been mplemented n Mathematca TM They are avalable from the author by emal Smlar algorthms exst for thrd-order lnear recurrences [8] and hgher-order recurrences 1 Phlosophcal Implcatons Now that these effectve algorthms for provng Fbonacc denttes are known, the phlosophcal queston arses as to whether one should study and/or publsh newly dscovered denttes The answer s yes Although formula (1 can now be proven by computer, the computer proof gves no nsght nto how the formula came about What s the sgnfcance of the bnomal coeffcents appearng n the formula? Does the formula generalze? Can a smlar proof be used to dscover new results? Should one stop submttng pretty new denttes to problem columns? The answer s no There s always the challenge nvolved n fndng an elegant proof For example, an algorthmc proof of the trgonometrc dentty sn 17x + cos 17x =1would not recognze that t follows from the Pythagorean Theorem Instead, t would expand out by the multple angle formula and get a horrendous mess whch t would then attempt to show s dvsble by (sn x + cos x 1 Elegant new denttes wll always be welcome to journals such as The Fbonacc Quarterly Dscoverng new denttes s stll mportant Ingenous proofs are always apprecated What these algorthms mply, s that the next tme you need a complcated dentty as a lemma n the mddle of a research paper, you can safely state the dentty and say that Ths dentty s straghtforward to prove References [1] Davd E Dobbs, Provng Trg Identtes to Freshpersons, The MATYC Journal 14 (1980:39 4 [] Davd E Dobbs and Robert Hanks, AModern Course on the Theory of Equatons Passac, NJ: Polygonal Publshng House, 1980 [3] R Wllam Gosper, Jr, Decson Procedure for Indefnte Hypergeometrc Summaton, Proceedngs of the Natonal Academy of Scences of the Unted States of Amerca 75 (1978:

17 [4] Ronald L Graham, Donald E Knuth, and Oren Patashnk, Concrete Mathematcs Readng, MA: Addson-Wesley Publshng Company, 1989 [5] M S Klamkn, On Provng Trgonometrc Identtes, Mathematcs Magazne 56 (1983:15 0 [6] Edouard Lucas, Théore des Fonctons Numérques Smplement Pérodques, Amercan Journal of Mathematcs 1 (1878:184 40, 89 31; reprnted as The Theory of Smply Perodc Numercal Functons, Santa Clara, CA: The Fbonacc Assocaton, 1969 [7] Roman Maeder, The Mathematca Programmer Boston, MA: Academc Press, 1994 [8] Stanley Rabnowtz, Algorthmc Manpulaton of Thrd-Order Lnear Recurrences, The Fbonacc Quarterly 34 (1996: [9] S Vajda, Fbonacc & Lucas Numbers, and the Golden Secton West Sussex, England: Ells Horwood Lmted, 1989 [10] Gregory Wulczyn, Problem H-34, The Fbonacc Quarterly 19 (1981:93 [11] Gregory Wulczyn, Problem B-464, The Fbonacc Quarterly 0 (198:370 AMS Classfcaton Numbers: 11Y16, 11B39, 11B37 17