Hi All,
Anyone knows of a trig identity between the arc tangent of an angle and it's multiples, like:
arctan(2x) = function of arctan(x)
Thanks!
Namir
PS: I have found a polynomial approximation by Eurler to calculate arctan very accurately, but I require tens of thousands of terms!!!!! Ouch!
Used WolframAlpha with:
Taylor[arctan[2*tan[y]],y]
The result is:
2 y-2 y^3+4 y^5-(142 y^7)/15+(4612 y^9)/189-(312644 y^11)/4725+O(y^12)
But that's probably not what you are looking for: a simple formula.
HTH
Thomas
If you do it with the inverse order of the functions:
Taylor[tan[2*arctan[y]],y]
The result is:
2 y+2 y^3+2 y^5+2 y^7+2 y^9+2 y^11+O(y^12)
(converges when abs(y)<1)
This geometric series leads to the well known result:
Edited: 3 Mar 2013, 1:08 p.m.
I am looking for an equation that is similar to the one for tan(2x) but expressed in terms of arctan(2x) and arctan(x).
I have a good Pade Approximant for tan(x). Ultimately, I can can use the Pade approximant for tan(x) in an iterative process. That would still be much faster than calculating 3000 to 4000 terms in a slow converging polynomial.
Namir
Edited: 3 Mar 2013, 4:22 p.m.
Quote:
I am looking for an equation that is similar to the one for tan(2x) but expressed in terms of arctan(2x) and arctan(x).
This doesn't appear to be possible. From the identity #1 here I've obtained
and
None of which will serve your purpose, though.
Edited to correct a typo in the second equation
Edited: 3 Mar 2013, 7:07 p.m.
It may not be clear from my answer but the solution is:
arctan(2x) = 2 arctan(x) - 2 arctan3(x) + 4 arctan5(x) + ...
This is not the formula you are looking for but its Taylor expansion. So all you need to do is find a closed formula of this series. I didn't made any attempts to find it but just from looking at the coefficients of the higher orders I doubt that there is any.
It's not that you can expect that there is a closed formula in general. In the case of tan(x) it is related to the multiplication of complex numbers.
Sorry for the bad news.
Cheers
Thomas
I found the following useful identity:
arctan(x) = pi/2 - arctan(1/x)
This identity allows me to calculate the arctan for high values (just below 90 degrees) by calculating the arctan of 1/x.
I can calculate arctan with good accuracy for low x values using polynomial approximations or iteratively. Either way, I get good results for high x values!
Namir
Edited: 4 Mar 2013, 7:22 a.m.
I found an efficient algorithm to calculate arctan(x).
EPS = 1e-8
A1 = 1/SQRT(1+x*x)
B1 = 1
Do
A0 = A1
B0 = B1
A1 = (A0+B0)/2
B1 = SQRT(A1*B0)
Loop until |A1-A0|<=EPS and |B1-B0|<=EPS
arctan(x) = x/(SQRT(1+x*x)*A1)
Algorithm works as efficiently for small and large values of x. Number of iterations for the above loop range from 11 to 15.
Edited: 4 Mar 2013, 11:08 a.m.
Without coding and testing anything, I wonder if the difference is significant. The AGM converges pretty quickly which would mean A0 -> A1.
- Pauli
A quick check and they do converge different after all.
- Pauli
The CORDIC algorithm is another efficient way of computing arctangents with few iterations, but it requires a precomputed table of coefficients.
In another post you mentioned wanting the arctan of high values. If what you need is the angle given an x and y coordinate, i.e., the atan2(y,x) function, CORDIC is very good for that as well, since you don't have to do the y/x division, which can lose a lot of precision.
That's the same algorithm I found on the Wolfram web site. It works well even for large values of x. There is no need to use the trick:
arctan(x) = pi/2 - arctan(1/x)
Namir
Edited: 4 Mar 2013, 11:09 p.m.