HP 17B Solver - another ARCTAN(Y/X) approximation

[Albert Chan]

(05-31-2021 04:42 PM)Albert Chan Wrote:  \(\displaystyle\arctan(x) = 2\arctan\left( {x \over \sqrt{2\sqrt{1+x^2}+x^2+2}} \right)\)

There is no need to use RC formulas to solve for h, for atan(x) = 2*atan(h)

Let h = tan(θ), |θ| ≤ pi/4

2*θ = atan(tan(2*θ))
2*atan(h) = atan(2*h/(1-h*h)) = atan(x)

2*h/(1-h*h) = x
x*h^2 + 2*h - x = 0

Remove quadratic solution of wrong sign (h and x sign should match), we have:

h = (√(1+x²)-1) / x = x / (√(1+x²)+1)

This h is same as above quoted formula, only simpler

\(\displaystyle\arctan(x) = 2\arctan\left( {x \over \sqrt{1+x^2}+1} \right)\)

Code:

from math import sqrt
def myatan(x, k=1):
    y = x*x
    if y > 1e-3: return myatan(x/(sqrt(1+y)+1), k+k)
    return k*(x - x*y*(1/3 - y*(1/5 - y*(1/7 - y*(1/9)))))

>>> myatan(1)
0.78539816339744839
>>> myatan(.5)
0.46364760900080615

>>> myatan(10) # |atan(x) = 2θ| ≤ pi/2
1.4711276743037345
>>> myatan(100)
1.5607966601082317

Update: Thomas Kleem beats me to it 2 years ago
https://www.hpmuseum.org/forum/thread-12...#pid113694