On Tue, Apr 9, 2019 at 1:09 AM Albert Chan wrote:
> The most straightforward approach is
> to build Taylor series of cin(x) one term at a time.
>
> function maclaurin_of_cin(k)
> for n = #c + 1, k do
> a = {}
> local e, h = f(c), f(d)
> s, a = -s/(2*n)/(2*n+1)
> local t = (s-h-e)/3
> assert(math.abs(t) < 0.056)
> c[n], d[n] = t, e + 2*t
> end
> return c[k]
> end
Say, I wanted a function, din(x), such that din(din(din(din(x)))) = sin(x)
I don't use smart methods like Newton's (mentioned by Dirk).
My implementation is dumb and straightforward.
I solve trivially constructed system of equations to find Tailor coefficients of din().
(I simply substitute one Tailor series into another.)
If n first coefficients are already known, the formula for the (n+1)-th one is not very complex.
local maclaurin_of_din
do
local c, d, s, a = {[0] = 1}, {[0] = 1}, 1
local function g(k, m, o)
if k == 0 or m < 0 then
return m == 0 and 1 or 0
else
local i = k..";"..m
local r = a[i]
if not r then
r = 0
for j = 0, #o do
r = r + o[j] * g(k-1, m-j, o)
end
a[i] = r
end
return r
end
end
local function f(o)
local r = 0
a = {}
for j = 0, #o do
r = r + o[j] * g(2*j+1, #o+1-j, o)
end
a = nil
return r
end
function maclaurin_of_din(k)
for n = #c + 1, k do
local e, h = f(c), f(d)
s = -s/(2*n)/(2*n+1)
local t = (s-h-2*e)/4
assert(t*t < 1)
c[n], d[n] = t, e + 2*t
end
return c[k]
end
end
local function din(x)
local r, p, s, n, R = x, x, x*x, 0
repeat
R, n, p = r, n+1, p*s
r = r + maclaurin_of_din(n) * p
until r == R
return r
end
local x = 0.7
print("x = "..x)
print("din(x) = "..din(x))
print("din(din(x)) = "..din(din(x)))
print("din(din(din(x))) = "..din(din(din(x))))
print("din(din(din(din(x)))) = "..din(din(din(din(x)))))
print("sin(x) = "..math.sin(x))
