[math-fun] More power series Q's
We have sin(3*x)=3*sin(x)-4*sin(x)^3 and sinh(3*x)=3*sinh(x)+4*sinh(x)^3 These give us recursive approximations to sin(x), sinh(x), thanks to RWG/Hakmem. So we have f(3*x)=3*f(x)+-4*f(x)^3 What are the functions f(3*x)=3*f(x)+b*f(x)^3, for some real b ? They seem to live in the land between sin(x) and sinh(x). Actually, I'm more interested in f(x), s.t. f(2*x) = 2*f(x)+b*f(x)^3, for some real (probably integer?) constant b, because I want a representation in which binary shifting is easy(er). If f(x) = x + ..., then we have f(x) = 8 17 7 15 6 13 5 11 4 9 3 7 3035153449 b x 32516717 b x 151 b x 163 b x 37 b x b x --------------------- + ----------------- + ---------- + ---------- + -------- + ----- 137019759699497875200 29401699409082000 3204726525 97389600 771120 945 2 5 3 b x b x + ----- + ---- + x 60 6 How to easily compute these coefficients? If b=4, the coefficient of x^3 becomes 2/3, which matches that in sinh(2x)/2, although f(x) doesn't converge quite as fast. If b=1, f(x) converges somewhat faster, but how fast? Does it converge everywhere? What is its radius of convergence? If f(2*x)=2*x+b*x^3, what is the/a differential equation for f(x)?
If f(2*x)=2*f(x)+b*f(x)^3, then we can substitute x=2^(y-1): f(2*2^(y-1))=2*f(2^(y-1))+b*f(2^(y-1))^3, so f(2^y) = 2*f(2^(y-1)) + b*f(2^(y-1))^3, so if g(y)=f(2^y) then g(y) = 2*g(y-1) + b*g(y-1)^3 So how do we solve this non-linear recurrence relation for g(y)?
On Mar 28, 2017, at 9:29 AM, Henry Baker <hbaker1@pipeline.com> wrote:
If f(2*x)=2*f(x)+b*f(x)^3, then we can substitute x=2^(y-1):
f(2*2^(y-1))=2*f(2^(y-1))+b*f(2^(y-1))^3, so
f(2^y) = 2*f(2^(y-1)) + b*f(2^(y-1))^3, so if g(y)=f(2^y) then
g(y) = 2*g(y-1) + b*g(y-1)^3
So how do we solve this non-linear recurrence relation for g(y)?
Failed attempt: --------------- Set g(y) = Sum_{n=0...oo} c_n y^n. The last equation above gives (*) Sum_{n=0...oo} c_n y^n = 2 * Sum_{n=0...oo} c_n (y-1)^n + b * Sum_{n=0...oo} c_n (y-1)^3n or setting z = y-1: Sum_{n=0...oo} c_n (z+1)^n = 2 * Sum_{n=0...oo} c_n z^n + b * Sum_{n=0...oo} c_n z^3n Now setting constants equal gives: Sum_{n=0...oo} c_n = 2 c_0 + b c_0 or c_0 = (1/(b+1)) Sum_{n=1...oo} c_n assuming b <> -1 (not that helpful!) Now setting z^1 powers equal gives: (Sum_{n=0...oo} n c_n) z = 2 c_1 z or c_1 = Sum_{n=2...oo} n c_n Hmm, this doesn't look easily solvable for the coefficients. —Dan
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Dan Asimov -
Henry Baker