The Number of Ways to Write n as a Sum of ` Regular Figurate Numbers

Transcription

1 Syracuse Unversty SURFACE Syracuse Unversty Honors Program Capstone Projects Syracuse Unversty Honors Program Capstone Projects Sprng The Number of Ways to Wrte n as a Sum of ` Regular Fgurate Numbers Seth Jacob Rothschld Follow ths and addtonal works at: Part of the Appled Mathematcs Commons, and the Mathematcs Commons Recommended Ctaton Rothschld, Seth Jacob, "The Number of Ways to Wrte n as a Sum of ` Regular Fgurate Numbers" (01) Syracuse Unversty Honors Program Capstone Projects 19 Ths Honors Capstone Project s brought to you for free and open access by the Syracuse Unversty Honors Program Capstone Projects at SURFACE It has been accepted for ncluson n Syracuse Unversty Honors Program Capstone Projects by an authorzed admnstrator of SURFACE For more nformaton, please contact surface@syredu

2 The Number of Ways to Wrte n as a Sum of l Regular Fgurate Numbers A Capstone Project Submtted n Partal Fulfllment of the Requrements of the Renée Crown Unversty Honors Program at Syracuse Unversty Seth Jacob Rothschld Canddate for a BS Degree n Mathematcs and Renée Crown Unversty Honors May 01 Honors Capstone Project n Mathematcs Capstone Project Advsor: Capstone Project Reader: Honors Drector: Professor Jeffrey Meyer Professor Clauda Mller Stephen Kuussto, Drector Date: Aprl 5, 01 1

3 Abstract The elementary approaches to fndng the number of nteger solutons (x, y) to the equaton x + y = n wth n N are well known We frst examne a generalzaton of ths problem by fndng the ways a natural number n can be wrtten as a sum of two k-sded regular fgurate numbers The technque s then adapted to fnd the number of ways n can be wrtten as a sum of l regular fgurate numbers

4 Contents 1 Introducton 11 Fgurate numbers 1 Bg O notaton 6 13 The Gauss Crcle Problem 7 Sums of Two Fgurate Numbers 10 1 Sums of two trangular numbers 10 Sums of two pentagonal numbers 1 3 Sums of two regular fgurate numbers 1 3 Sums of Multple Fgurate Numbers Sums of three regular fgurate numbers 16 3 Sums of l regular fgurate numbers 18 Summary of Capstone Project 3

5 1 Introducton 11 Fgurate numbers Regular fgurate numbers have been an object of study snce Fermat Specal cases of fgurate numbers, trangular and square numbers were studed far before that [] Defnton 111 Let n N and k N wth k Then the n th k-sded regular fgurate number f k (n) s defned recursvely by f k (1) = 1 and f k (n) = f k (n 1) + (k )n (k 3) Consder when k = 3 From the defnton, f 3 (n) = f 3 (n 1) + n + 0 Snce f 3 (1) = 1 and f 3 () = 1 +, f 3 (3) = Inductvely, we see that n 1 f 3 (n) = n + = =1 n =1 Snce we can arrange f 3 (n) dots nto a trangle wth n dots on each sde (see Fgure 1), f 3 (n) are also known as the trangular numbers Smlarly, f (n) dots can be arranged nto squares (see Fgure ) and f 5 (n) dots can be arranged nto pentagons (see Fgure 3) In fact, f k (n) dots wll always gve the k sded regular polygon wth sdes of length n The formula gves f k (1) = 1 and f k () = k Then, f k (n) s f k (n 1) dots wth k sdes of length n added to t and k 3 subtracted from t for the overlap at corners For a specfc example, examne Fgure 3 If we want to get f 5 (n), we start wth wth f 5 (n 1) and add 3 more sdes of length n, and subtract for the overlap at the bottom two corners Fgure 1: The frst fve f 3 (n), each arranged as a trangle The numbers f 3 (n) are also known as trangular numbers Note that f 3 (n) = f 3 (n 1) + n Theorem 11 Let n N and k N wth k Then f k (n) = n [(k )n + ( k)]

6 Fgure : The frst fve f (n), each as a square wth sdes of length n By defnton, f (n) = f (n 1) + n 1 Fgure 3: The frst four pentagonal numbers The nth pentagonal number has 5 sdes of length n Notce that f 5 (n) = f 5 (n 1) + 3n Proof Defne n[(k )n + ( k)] g k (n) = We show that g k (n) = f k (n) by showng that they satsfy the same ntal condtons and the same recurrence relaton Frst, notce that g k (1) = 1 [(k ) 1 + k] = = 1 = f k(1) Now we need to show that g k (n) = g k (n 1) + (k )n (k 3) To check ths, note that g k (n 1) + (k )n (k 3) = (n 1)[(k )(n 1) + ( k)] = 1 (n (k ) + n(k + k k)) n[(k )n + ( k)] = = g k (n) + (k )n (k 3) 5

7 Therefore g k (n) = f k (n) and thus f k (n) = n [(k )n + ( k)] When k =, the formula gves f (n) = n[( )n + ] = n ; thus the n th square number s n squared For k = 3 and k = 5 the formulas are f 3 (k) = n + n and f 5 (k) = 3n n respectvely 1 Bg O notaton It turns out to be mpossble to fnd a general explct formula for certan functons n whch we are nterested, so we utlze bg O notaton to descrbe the growth of those functons n general Defnton 11 Let f be a functon over the real numbers We say that f(x) = O(g(x)) for some functon on the real numbers g f there exst constants c and x 0 so that f(x) c g(x) for all x x 0 For example, let x be a real number such that x 1 Snce x 1, 1 x + x + 1 x ( ) 1 x = + = c x, where c = + 1 Therefore, x + 1 = O( x) However, keep n mnd that t s equally vald to say x + 1 = O(x) We may do that because the bg O s dependent on an nequalty In general, we seek the smallest possble g(x) to gve the best approxmaton 6

8 Defnton 1 Let f, g, and h be functons on the real numbers We say f(x) = h(x) + O(g(x)) f there exsts c such that f(x) h(x) c g(x) In ths case, we thnk of g(x) as an error term as long as g(x) lm x h(x) = 0 13 The Gauss Crcle Problem Defnton 131 Let n N {0} Let r (n) be the number of ways to wrte n as a sum of two squared nonnegatve ntegers Notce that f an nteger n can be wrtten as a sum of two squares, then the crcle x + y = n wll pass through the pont (h, k) where h + k = n In ths way, we gan a geometrc representaton of numbers that can be descrbed as a sum of two squares (see Fgure ) Fgure : The crcles x + y = n wth n 9 Note that r (n) s exactly the number of lattce ponts on a crcle centered at the orgn wth radus n Defnton 13 Let n N {0} Then defne p (n) by p (n) = n r () =0 7

9 Fgure 5: The frst 1 values of r (n) Notce that there s no dscernble pattern Fgure 6: The frst 1 values of p (n) We create a patten out of the r (n) by addng them together We study p (n) because t s a more predctable functon than r (n) Whle r (n) can vary wldly from one postve nteger to the next, p (n) has a notceable ncreasng trend (see Fgures 5 and 6) If we set lattce ponts n R at every (h, k) where h, k Z 0, then p (n) s the number of frst quadrant lattce ponts nsde a crcle of radus n We dentfy each lattce pont wth a square of area one, whose bottom left corner on a lattce pont nsde the crcle That s, the square for the lattce pont (h, k) has ts four corners at the ponts (h +, k + j) where, j {0, 1} From ths pont forward, we treat p (n) as the area created by the lattce ponts of p (n) by that method, (see Fgure 7) Lemma 133 Let n 0 Then πn p (n) π( n + ) Proof The maxmum lnear dstance between two ponts n a square wth sde 1 s Snce we are only workng wth the frst quadrant, p (n) s strctly greater than the area of the crcle of radus n and strctly less than the area of the crcle of radus n + Theorem 13 Let n 0 Then p (n) = πn + O( n) Proof From Lemma 133, we have that πn p (n) π( n + ) In the 8

10 Fgure 7: The sold black lne gves a crcle of radus 18 centered at (0,0), and the dotted black lne gves a crcle of radus 18 + Each blue square has s assgned to the lattce pont at t s bottom left corner The total area of the blue squares s p (18) = 0 rght hand nequalty, we have that p (n) π( n + ) = πn + π n + π whch gves us p (n) πn π n + π (1) The left hand nequalty gves p (n) πn whch mples that p (n) πn 0 π n + π From ths, and from statement (1) we know that p (n) πn π n π, 9

11 whch means that p (n) = πn + O( n) The Gauss Crcle Problem s the quest for the smallest possble g(x) so that p (n) = πn + O(g(x)) Sums of Two Fgurate Numbers The problem of determnng p (n) was frst proposed by Gauss, who also found ths relatonshp to the area of the crcle We generalze hs soluton from countng the sums of two squares, f (m), to countng the sums of two regular fgurate numbers, f k (m) Defnton 05 Fx n 0 and k 3 Then r k (n) s the number of ways n can be wrtten as a sum of two k-sded regular fgurate numbers In other words r k (n) counts nonnegatve nteger solutons (x, y) of n = x[(k )x + k] + y[(k )y + k] () Defnton 06 Fx n 0 and k 3 Then defne p k (n) by p k (n) = n r k () =0 We agan attach the bottom left corner of a square wth area 1 to each lattce pont n the frst quadrant We thnk of p k (n) as the sum of the areas of all of the squares attached to frst quadrant lattce ponts nsde the crcle defned by equaton () 1 Sums of two trangular numbers We set k = 3 n equaton () to get x(x + 1) and complete the square so that + y(y + 1) = n ( x + 1 ( + y + ) 1 ) = n

12 Fgure 8: The sold black lne gves a crcle of radus centered at ( 1, 1 ) Each blue square s assgned to a lattce pont ncluded n p 3 (18), so ther sum s p 3 (18) = 30 The dotted black lne quarter crcles enclose areas that are upper and lower bounds for p 3 (18) Notce that ths s a crcle wth radus n + 1 and center ( 1, 1 ) However, the radus s dfferent from that of the Gauss Crcle Problem, and the center of ths crcle s not at the orgn Lemma 11 Let n be gven Then ( n ) π + 1/ 1/ p 3 (n) π( n + 1/ + ) Proof As n the proof of Lemma 133, ths s geometrcally clear We bound the crcle from p 3 (n) above and below by crcles centered at (0, 0) We compensate for the center beng n the thrd quadrant by decreasng the radus of the lower 1 bound by We ncrease the radus of the upper bound by for the same reason as n the proof of Lemma 133 (see Fgure 8) Theorem 1 Let n Then p 3 (n) = πn + O( n) 11

13 Proof From Lemma 11, we have p 3 (n) ( n ) π + 1/ + = π(n + 1/) + π n + 1/ + π so that p 3 (n) πn π(1/) + π n 1/ + π (3) Also from Lemma 11 we get so that p 3 (n) ( n ) π + 1/ 1/ = π(n + 1/) π 1/ n + 1/ + π/ () p 3 (n) πn From (3) and () we have that Thus π(1/) π 1/ n + 1/ + π/ π(1/) + π n 1/ + π p 3 (n) πn π(1/) + π n 1/ + π p 3 (n) = πn + O ( n ) Sums of two pentagonal numbers By Theorem 11, a pentagonal number s a regular fgurate number of the form x(3x 1) where x N {0} We set k = 5 n equaton () and complete the square to get ( x 1 ( + y 6) 1 ) = n Lemma 1 Let n 0 Then ( π n ) ( (n p 5 (n) π ) + + 1/) 18 1

14 3 Fgure 9: The sold black lne gves a crcle of radus centered at ( 1 6, 6) 1 The total area of the blue squares s p5 (17) = 15 The dotted black lne quarter crcles enclose areas whch are upper and lower bounds bounds for p 5 (17) Proof The bounds are geometrcally clear The center of ths crcle s at the pont ( 1 6, 6) 1 As n the Gauss Crcle Problem, we use the area of the quarter crcle wth ths radus as a lower bound for p 5 (n) We can do ths because the area of the quarter crcle wth ths radus centered at the orgn s less than the crcle wth the same radus centered at ( 1 6, 1 6) (see Fgure 9) For the upper bound we ncrease the radus by 1 +, whch s 3, where 1 the s an overcompensaton for movng the center a dstance of 1 18 nto the frst quadrant Theorem Let n 1 Then p 5 (n) = πn 6 + O( n) Proof We begn n the same way, wth the rght hand nequalty n of Lemma 1; ( (n p 5 (n) π [ (n = π ) + 3 ) + 3 ) n ] 13

15 whch gves us ( p 5 (n) πn 6 π ) 1 n (5) We rearrange the left nequalty from Lemma 1 to see that p 5 (n) π ( n ) 18 and subtract the leadng term from both sdes to get that p 5 (n) πn 6 π 7 ( π ) 1 n (6) Fnally, we combne (5) and (6) and see that ( p 5 (n) π π ) 1 n or p 5 (n) = πn 6 + O ( n ) 3 Sums of two regular fgurate numbers We use the same methodology as n the proof of Lemma 1 to fnd upper and lower bounds for p k (n) Lemma 31 Let n 1 and k Then [ π n k + ( ) ] k p k (n) π k n k + ( k k ) + 3 Proof We begn by completng the square of equaton () to get ( x + k ) ( + y + k ) = n ( ) k k k k + k Snce k the crcle s centered n the frst quadrant and the lower bound can be the area of the quarter crcle wth the gven radus 1

16 To ( fnd an upper bound on the center we examne the behavor of the center k k, k ) Frst we see that the expresson k s strctly k k monotoncally ncreasng n k snce k k < 3 k (k + 1) = k (k + 1) Then we take the lmt as k goes to nfnty to get lm k k k = 1 Therefore, our upper bound must be adjusted not just by as n the Gauss Crcle Problem, but by 1 + Theorem 3 Let n 1 and k 3 Then p k (n) = πn k + O( n) Proof Theorem 1 proves ths when k = 3 If k > 3 we frst manpulate the rght nequalty of Lemma 31 to get that p k (n) π ( ) n k k k = π n ( ) k k ( ) n k k k k Thus p k (n) = πn k + π ( ) k + 3π k πn k π ( ) k + π k The left nequalty of Lemma 31 gves [ p k (n) π n k + ( ) n k k + + 9π k 8 ( ) n k k + + 9π k 8 ( ) ] k, k so that p k (n) πn k π ( ) k k Note that π p k (n) ( ) k s a postve number Therefore k πn ( k k π k ) π 15 n k + ( ) k 9π k 8

17 Combnng the results from above, we get p k(n) πn k π ( ) k + π k n k + ( ) k + 9π k 8 Thus, p k (n) = πn k + O ( n ) Corollary 33 Let k Then r k (n) = 0 for nfntely many n Proof Frst, notce that r k (n) s always a non-negatve nteger average value of r k (n) to be r k (n) where Defne the By Theorem 3, r k (n) = r k (n) = p k(n) n π ( ) 1 k + O n Notce that snce k, the leadng term s less than 1 As n goes to nfnty, the error goes to 0 Therefore, lm r k(n) = n therefore nfntely many r k (n) must be 0 π k < 1 3 Sums of Multple Fgurate Numbers 31 Sums of three regular fgurate numbers Defnton 311 Let n 0 and k 3 be gven Then r k,3 (n) s the number of ways n can be wrtten as a sum of three k-sded regular fgurate numbers In other words t counts postve nteger solutons (x 1, x, x 3 ) of n = x 1[(k )x 1 + ( k)] + x [(k )x + ( k)] Defnton 31 Let n 0 and k > Defne p k,3 (n) by + x 3[(k )x 3 + ( k)] (7) p k,3 (n) = n r k,3 () =0 16

18 For each lattce pont (h 1, h, h 3 ) wth h 1, h, h 3 N {0} we attach a cube of volume 1 so that all vertces of the cube are of the form (h 1 + j 1, h + j, h 3 + j 3 ) where j {0, 1} We thnk of p k,3 (n) as the sum of the volume of each of the cubes attached to lattce ponts nsde the sphere defned by equaton (7) Lemma 313 Let n 0 be gven Then πn 3 6 r,3 (n) π ( n + 3 ) 3 6 Proof The lower bound s the volume of the sphere n the frst octant, π ( n ) 3 The upper bound s adjusted by the maxmum dstance from one pont of our lattce-countng-cubes to another, = 3 Theorem 31 Let n 0 be gven Then p,3 (n) = π 6 n 3 + O(n) Proof We manpulate the rght sde of the Lemma 313 nequalty to get p,3 (n) π ( n + 3 ) 3 6 = π (n 3 ) 3 + 3n n so that p,3 (n) π 6 n 3 π ( 3n ) n (8) 6 We rearrange the left nequalty from Lemma 313 to get p,3 (n) π 6 n 3 0 π 6 We put (8) and (9) together to get p,3 (n) π 6 n 3 π 6 ( 3n ) n (9) ( 3n ) n Thus, p,3 (n) = π 6 n 3 + O(n) We use the same methodology as n Subsecton 3 to fnd upper and lower bounds for p k,3 (n) 17

19 Lemma 315 Let n 1 and k Then [ π 6 n k + 3 ( ) ] 3 k k p k,3 (n) π 6 n k + 3 ( k k ) Proof The proof s analogous to that of Lemma 31 Frst we complete the square n equaton (7) to get that ( x 1 + k ) ( + x + k ) ( + x 3 + k ) = n ( ) k k k k k + 3 k The, the bounds follow geometrcally as before Theorem 316 Let n 1 and k Then p k,3 (n) = π 6 ( ) 3 n + O(n) k We omt the proof n favor of the proof for Theorem 3, whch s a generalzaton 3 Sums of l regular fgurate numbers Defnton 31 Let n 0 and k 3 be gven Then r k,l (n) s the number of ways n can be wrtten as a sum of l, k-sded regular fgurate numbers, the postve nteger solutons (x 1,, x l ) of n = x 1[(k )x 1 + ( k)] Defnton 3 Let n 0 and k 3 Defne p k,l (n) by + + x l[(k )x l + ( k)] (10) p k,l (n) = n r k,l () =0 For each lattce pont (h 1,, h l ) wth h 1,, h l N {0} we attach a hypercube of volume 1 so that all vertces of the cube are of the form (h 1 +j 1,, h l + j l ) where j {0, 1} In the same way as before, we can thnk p k,l (n) as the sum of the volumes of each of the hypercubes attached to lattce ponts nsde the hypersphere defned by equaton (10) The maxmum dstance between two ponts n a unt hypercube s l 18

20 Lemma 33 Let n 1, k and l be gven If l s even, and If l s odd, and [ π l ( ) l l! p k,l (n) π l ( l l p k,l (n) ( l 1 ) l 1! π l! n k + l )! ( l 1 ) l 1! π l! ( ) ] l k k n k + l ( k k [ ( ) ] l n k k + l k n k + l ( k k p k,l (n) ) + 3 l p k,l (n) ) + 3 l l l Proof We begn by completng the square n equaton (10) to see that ( x 1 + k ) ( + + x l + k ) = n ( ) k k k k + l k From [3], the volume of the l-dmensonal hypersphere wth radus r s π l ( l )! r l, ( ) f l s even, and l l 1 l 1! π r l, f l s odd Snce we are restrcted to postve l! solutons, we dvde the volume by l We use the radus r lower = ( ) n k k + l k for the lower bound For the upper bound, we use the radus ( ) n k l r upper = k + l + k + l Indeed, as n the proof of Lemma 31 as k goes to nfnty, the center s ( 1,, 1 ) We take nto account ths dstance of l from the orgn, and add an addtonal l for the maxmum dstance between two corners of one of volumecountng hypercube Theorem 3 Let n 1, k, and l be gven Then, f l s even, p k,l (n) = π l ( l l ( ) l n )! k 19 + O (n l 1 )

21 and, f l s odd, p k,l (n) = ( l 1 ) l 1!π l l! ( ) l n k + O (n l 1 ) Proof We consder the even and odd cases separately for clarty Let l be even Defne a = n ( ) k k + l, k and b = 3 l c = π l ( ) l l! We manpulate the rght nequalty from the even case of Lemma 33 ( ) l p k,l (n) c n k k + l + 3 l k We can expand a l a l = ca l + c l =1 ( ) l a l b wth the bnomal theorem [ ( ) n = + k so that for n suffcently large, l =1 ( ) ( ) l n l ( ) ] 1 k l k k a l [ ( ) n l l + k =1 [ ( ) n l = + ( l 1) k ( ) ( ) l n l 1 ( ) ] 1 k l k k ( ) n l 1 ( ) ] 1 k l k k From there, the trangle nequalty gves ) l l 1 + l a l Therefore, ( n p k,l (n) c k [ ( n k c ) l l l 1 ( n k ) l 1 ( ) l 1 ( ) n k + k k ( ) k k l =1 ( ) ] l a l b (11) 0

22 We rearrange the left nequalty from the even case of Lemma 33 to get so that ( ) l n p k,l (n) c 0 k [ c l l 1 p k,l (n) ca l We put (11) and (1) together to get p k,l(n) π l ( ) l n l ( l )! k [ c c ( n k ) l ( ) l 1 ( ) n k + k k l l 1 So, when l s even, we conclude that Now let l be odd Set p k,l (n) = π l ( l l a = l =1 ( ) l 1 ( ) n k + k k ( ) l n )! k + O (n l 1 n ( ) k k + l, k ) ( ) ] l a l b (1) l =1 ( ) ] l a l b b = 3 l as before, and set ( l 1 ) l 1!π c = l l! We manpulate the rght nequalty from the odd case of Lemma 33 p k,l (n) c n k + l = ca l + c l =1 ( k k ( ) l a l b ) + 3 l l 1

23 ( n p k,l (n) c k c ) l [ l l 1 ( ) l 1 ( ) n k + k k l =1 ( ) ] l a l b (13) We rearrange the left nequalty from the odd case of Lemma 33 to get so that ( ) l n p k,l (n) c 0 k [ c l l 1 p k,l (n) ca l We put (13) and (1) together to get ( l 1 ) l 1 ( ) l p!π n k,l(n) l l! k [ c c ( n k ) l ( ) l 1 ( ) n k + k k l l 1 So, when l s odd, we conclude that l =1 ( ) l 1 ( ) n k + k k ( ) ] l a l b (1) l =1 ( ) ] l a l b p k,l (n) = ( l 1 ) l 1!π l l! ( ) l n k + O (n l 1 )

24 References [1] George E Andrews, Number Theory Phladelpha: Saunders, 1971 [] W Duke, Some old and new results about quadratc forms, Notces Amer Math Soc, 1997, [3] AE Lawrence, The volume of an n-dmensonal hypersphere Web coael/hyperspherepdf [] A Wel, Number Theory: An approach through hstory, Brkhäuser, 198 3

25 Summary of Capstone Project The problem of fndng the number of nteger solutons (x, y) to the equaton x + y = n wth n as a postve nteger s more than 00 years old [1] Ths capstone examnes some generalzatons of ths problem We examne elementary solutons to the sum of two fgurate numbers, an expresson motvated by a polygon wth k sdes of the same length From there we adapt the technque to fnd the number of ways n can be wrtten as a sum of l regular fgurate numbers, nstead of only two Vocabulary We defne regular fgurate numbers by a formula, and then gve examples of how the formula s related to a polygon wth k sdes The proof that they are the same s n the paper Defnton The equaton of the n th fgurate number wth k sdes s defned by f k (n) = n [(k )n + ( k)] As an example, consder when k =, the four sded fgurate numbers whch we call f (n) We would lke f (n) to form a polygon wth four sdes of equal length Computng gves f (n) = n[( )n + ] = n(n) 0 = n You mght recognze f (n) as n squared but t s mportant to see (check Fgure 10) that these are lterally squares For each n, f (n) gves a square wth sdes of length n Fgure 10: The frst fve four sded fgurate numbers Notce that f (n) dots can be arranged n a square, wth each of the four sdes havng a length of n Ths s the ratonale for callng f (n) square numbers

26 Ths geometrc defnton s actually the ratonale behnd the term regular fgurate number For every k, f k (n) dots forms a polygon wth k sdes each of length n Settng k = 3 gves us the trangular numbers When lookng at the dagram, you should notce that each f k (n) s bult on the prevous fgurate number, f k (n 1) n a specfc way that s the same for all fgurate numbers For a more rgorous explanaton consult Subsecton 11 of the paper Fgure 11: The frst fve f 3 (n) n dots, each arranged as a trangle The n th trangular number can be represented as a trangle wth three sdes of length n The next tool we wll need to defne s called bg O notaton Ths s a way to talk about the relatve sze of a functon as compared wth an easer to understand functon Defnton Let f, g and h be functons on the real numbers We say f there exsts c such that f(x) = h(x) + O(g(x)) f(x) h(x) c g(x) We read ths f(x) s h(x) plus bg O of g(x) In other words, f the dstance between two functons f(x) and h(x) s less than a constant tmes g(x), we call f(x) equal to h(x) + O(g(x)) Ths notaton s useful n cases where t s mpossble to fnd an explct formula for a functon f(x) The number of ways to wrte a postve nteger n as a sum of some regular fgurate numbers s one such functon Defnton Let n be greater than or equal to 0, and k greater than or equal to 3 be gven Then r k,l (n) s the number of ways n can be wrtten as a sum of l, k-sded regular fgurate numbers It counts the number of ponts (x 1,, x l ) that satsfy n = f k (x 1 ) + + f k (x l ) or n = x 1[(k )x 1 + ( k)] + + x l[(k )x l + ( k)] 5

27 The number of ways to wrte n as a sum of l regular fgurate numbers, r k,l, s not easy predct In fact, the functon r k,l (n) cannot be wrtten as a polynomal As an example, consder when k = and l =, the functon r, (n) Ths s the number of ways a number n can be wrtten as a sum of two squares To fnd r, (n) we would calculate so that 1 = = ; = ; 3 a + b for all a and b; = + 0 = 0 + ; 5 = + 1 = 1 + ; 6 a + b for all a and b r, (1) = r, () = 1 r, (3) = 0 r, () = r, (5) = 1 r, (6) = 0 Snce there sn t an explct formula for r k,l (n), we add up all the r k,l () from = 0 to n We call ths sum p k,l (n) As t s the prmary object n ths paper, we provde a formal defnton here Defnton Let n 0 and k 3 Defne p k,l (n) by p k,l (n) = n r k,l () =0 It turns out that the functon p k,l (n) s closely related to the volume of a hypersphere n l dmensons Ths capstone captalzes on the nherent geometry of the problem to fnd a formula for p k,l (n) n bg O notaton The geometrc connecton to p k,l (n) s fascnatng, the nterested reader should consult the fgures n Secton 13 for more nformaton Results As early as 1638, Fermat stated that every number n can be wrtten as a sum of k, k-sded fgurate numbers That result was later proved by Cauchy [] The results from ths paper pont to that statement, but do not ad n provng t These are the two prmary results from the paper Theorem 3 Let n 1 and k 3 Then p k, (n) = πn k + O( n) Ths s the result for sums of two fgurate numbers, p k, (n) It says that the error term s O( n) regardless of whch two regular fgurate numbers are added together However, the more sdes each fgurate number has, the smaller the 6

28 leadng term So, as k ncreases, p k, (n) decreases The generalzaton of ths theorem s the fnal result from the paper Theorem 3 Let n 1, k, and l be gven Then, f l s even, and, f l s odd, p k,l (n) = π l ( ) l n ) + O (n l 1 l ( l )! k p k,l (n) = ( l 1 ) l 1!π l l! ( ) l n k + O (n l 1 ) In ths theorem the same pattern holds When l s constant, an ncrease n k makes p k,l (n) smaller Ths mples that a number n should be a sum of more k fgurate numbers than k 1 fgurate numbers Whle t can not prove Fermat s statement, ths theorem does gve an overvew of how arbtrary sums of fgurate numbers behave 7