Apologies Fred--I just found this under a massive spamslide: On 4/3/08, Bill Gosper <rwmgosper@yahoo.com> wrote:
rwg>It's pretty clear that any product or sum over a period of a rational function of trigs comes out in closed form. You can interconvert such products and sums via trig, calculus, and generating function hacks. [...] So we get sums/prods over periods of some algebraic functions of trigs as well as rational.
I stand amazed. Is there some automatic algorithm lurking under there? I'd love to find it--it would make a nifty enhancement to the definite summation capabilities of computer algebra systems. In principal, you could convert all the trigs to complex exponentials, algebraically factor everything down to terms linear in cis(2 pi k/n), and then repeatedly use n 2 %i %pi k /===\ ---------- | | n n n (*) | | (a - b %e ) = a - b | | k = 1 and its log derivatives. But this will produce gruesomely unsimplified results. So I have a small collection of generalizations of (*). I can produce algebraic summands and prodands with nonlocal derangement (http://gosper.org/fst.dvi again), but not on cue, at least not without a lot more practice. But it can sure make for screwy summands, e.g.: inf ==== 2 \ 1 9 pi > -- cos(-----------------------) = - -----. / 2 2 2 3 ==== n sqrt(n pi - 9) + n pi 12 e n = 1
(Reminder: T_n(x) = cos(n acos x) = cosh(n acosh x) = 2 x T_n-1 - T_n-2, T_0(x)=1, T_1(x)=x. T_n(T_m(x)) = T_nm(x), so e.g. T_(n/2)(x) := T_n(sqrt((x+1)/2)) = sqrt((T_n(x)+1)/2) .)
This last line does seem to work --- e.g. T_2(x) = 2x^2 - 1, T_4(x) = 8x^4 - 8x^2 + 1, 2x^2 - 1 = 2((x+1)^2) - 4(x+1) + 1 = sqrt(4x^4 - 4x^2 + 1)
The very last line is even more interesting: Using T_(1/2)(x):=sqrt((x+1)/2) = inverse of T_2, e.g., (c187) chebyshev_t[3](sqrt((x+1)/2)) = sqrt((chebyshev_t[3](x)+1)/2) 3/2 3 2 (x + 1) 3 sqrt(x + 1) sqrt(4 x - 3 x + 1) (d187) ------------ - ------------- = -------------------- sqrt(2) sqrt(2) sqrt(2) so T_2n(x)+1 is a square, and T_(2n+1)(x)+1 is x+1 times a square. Even the function 2 1/3 (sqrt(y - 1) + y) 1 T (y) := --------------------- + ----------------------- 1/3 2 2 1/3 2 (sqrt(y - 1) + y) commutes with the Chebs.
--- but I still don't quite understand why. Something to do with underlying exponential functions? --WFL
Sort of. The acosh cancels the cosh, so the m multiplies the n. --rwg __________________________________________________ Do You Yahoo!? Tired of spam? Yahoo! Mail has the best spam protection around http://mail.yahoo.com