Exponential Sums and the Multisection Formula. Brian David Sittinger

Transcription

1 Exponential Sums and the Multisection Formula Brian David Sittinger 27 January 2010

2 Outline: Exponential Sums The Multisection Formula Examples and Related Results Parting Shots 1

3 Exponential Sums I ll refer to any series of the form n=0 a n x n n! as an exponential sum = a 0 + a 1 x + a 2x 2 + 2! In particular, consider special subseries of e x = 1 + x + x2 2! + x3 3! + 2

4 Motivating Problem (Stewart, Calculus, 3rd Edition, p680, #15) Let u = 1 + x3 3! + x6 6! + x9 9! + v = x + x4 4! + x7 7! + x10 10! + w = x2 2! + x5 5! + x8 8! + x11 11! + Show that u 3 + v 3 + w 3 3uvw = 1 3

5 Solution (sketch): Note that these series are derivatives of one another: u = w, v = u, and w = v Letting F = u 3 + v 3 + w 3 3uvw, we see that df/dx = 0 Hence F is constant Using u(0) = 1, v(0) = 0, and w(0) = 0, we see that this constant equals 1 4

6 More interesting question: Can we find closed forms for u, v, and w? Feasibility: We know that cosh x = ex + e x 2 = x 2k (2k)! sinh x = ex e x 2 = x 2k+1 (2k + 1)! 5

7 A solution via Differential Equations u satisfies the linear differential equation u = u This DE has characteristic equation r 3 1 = 0, with solutions r = 1, 1 ± 3 2 Note: These are all cube roots of unity! Let ω = Then, r = 1, ω, ω 2 2 General solution: u = C 1 e x +C 2 e ωx +C 3 e ω2x 6

8 DE solution, continued Find C 1, C 2, and C 3 through the initial conditions u(0) = 1, u (0) = 0, and u (0) = 0 Solving, C 1 = C 2 = C 3 = 1 3 Therefore, u = ex + e ωx + e ω2 x 3 7

9 Conclusion: u = x 3k (3k)! = ex + e ωx + e ω2x 3 v = u = x 3k+1 (3k + 1)! = ex + ω 2 e ωx + ωe ω2x 3 w = u = x 3k+2 (3k + 2)! = ex + ωe ωx + ω 2 e ω2x 3 8

10 Observations: Note the similarities between these series closed forms and those for hyperbolic sine and cosine This would have been lost if we had used real numbers instead of complex numbers For example, u = ex + 2e x/2 cos( 3x/2) 3 Moral: Complex numbers often simplify matters! 9

11 A generalisation: Let ω denote a primitive n-th root of unity (such as e 2πi/n ), and fix l N where 0 l < n x nk+l (nk + l)! = ex + ω l e ωx + + ω l(n 1) e ωn 1x n In particular, x nk (nk)! = ex + e ωx + e ω2x + + e ωn 1x n Remark: Note that the above series sum over l to e x (Compare to e x = cosh x + sinh x) 10

12 Alternate proof (sketch) We didn t need to use DE s to solve the previous problem Since ω satisfies x n 1 = (x 1)(x n 1 +x n 2 ++x+1) = 0, we see that ω n 1 + ω n ω + 1 = 0 In fact, for any s N, (ω s ) n 1 +(ω s ) n 2 ++(ω s )+1 = 0 if s n n if s n Using this last fact, the proof is a straightforward computation 11

13 What about other series, besides that of e x? Using (ω s ) n 1 + (ω s ) n (ω s ) + 1 = 0 whenever s is not a multiple of n, we may easily deduce the following theorem Multisection Formula: Let ω = e 2πi/n and l N such that 0 l < n If f(x) = a k x k, then j=0 a nj+l x nj+l = 1 n n 1 s=0 ω ls f(ω s x) Remark: Note that the above series sum over l to f(x) (and hence its name) 12

14 First published by Thomas Simpson (1759): The invention of a general method for determining the sum of every second, third, fourth, or fifth, etc terms of a series taken in order the sum of the whole series being known, Philosophical Transactions of the Royal Society of London, Vol 50, pp

15 Special case m = 2: If f(x) = a k x k, then j=0 a 2j x 2j = 1 [f(x) + f( x)] 2 j=0 a 2j+1 x 2j+1 = 1 [f(x) f( x)] 2 This is essentially the classic result that a function may be written in terms of an even function and an odd function! 14

16 An example (1 + x) n = n ( n k ) x k with x = 1 yields n ( n k ) = 2 n This, with letting x = 1, yields ) ( n) ( n) = 2 n ) ( n) ( n) = 2 n 1 ( n ( n What is ( n k k 0(3) ) ( n) ( n) ( n) = + + +?

17 An example (continued) Applying the Multisection Formula to (1 + x) n n = ( n) x k with 0 k < m and k x = 1: ( n k k 0(3) ) = s=0 (1 + ω s ) n Since ω 2 = ω, applying De Moivre s Theorem yields k 0(3) ( n k ) = 1 3 [ 2 n + 2 cos ( nπ 3 )] 16

18 An example (concluded) In general, we get the following result (C Ramus, 1834) ( n k k r(m) ) = 1 m m 1 j=0 ( 2 cos ( )) jπ n j(n 2r)π cos m m 17

19 Related problems Consider alternating exponential sums of the form ( 1) k x nk+l (nk + l)! When n = 1, this is e x When n = 2, these are cos x (l = 0) and sin x (l = 1) What about n > 2? 18

20 Alternating exponential sums (continued) Let ω denote a primitive (2n)-th root of unity, and fix l N where 0 l < n ( 1) k x nk+l (nk + l)! = 1 n n 1 s=0 ω l(2s+1) e ω2s+1x In particular, ( 1) k x nk (nk)! = eωx + e ω3x + + e ω2n 1 x n Let s l (x) = ( 1) k x nk+l (nk + l)! Then, e ωx = n 1 l=0 ω l s l (x) In the case when n = 2, this reduces to Euler s identity e ix = cos x + i sin x 19

21 Parting shots Generalise the Multisection Formula to polynomials/series in more than one variable Develop an analogue of the Multisection Formula for j=0 ( 1) j a nj+l x nj+l What happens if we use other groups to act on a variable instead of the n-th roots of unity (ie, the cyclic group of order n) in the Multisection Formula? 20