Calculating inverse trigonometric functions with formulas

The question is broad in scope, but here are some simple ideas (and code!) that might serve as a starting point for computing arctan. First, the good old Taylor series. For simplicity, we use a fixed number of terms; in practice, you might want to decide the number of terms to use dynamically based on the size of x, or introduce some kind of convergence criterion. With a fixed number of terms, we can evaluate efficiently using something akin to Horner’s scheme.

def arctan_taylor(x, terms=9):
    """
    Compute arctan for small x via Taylor polynomials.

    Uses a fixed number of terms. The default of 9 should give good results for
    abs(x) < 0.1. Results will become poorer as abs(x) increases, becoming
    unusable as abs(x) approaches 1.0 (the radius of convergence of the
    series).
    """
    # Uses Horner's method for evaluation.
    t = 0.0
    for n in range(2*terms-1, 0, -2):
        t = 1.0/n - x*x*t
    return x * t

The above code gives good results for small x (say smaller than 0.1 in absolute value), but the accuracy drops off as x becomes larger, and for abs(x) > 1.0, the series never converges, no matter how many terms (or how much extra precision) we throw at it. So we need a better way to compute for larger x. One solution is to use argument reduction, via the identity arctan(x) = 2 * arctan(x / (1 + sqrt(1 + x^2))). This gives the following code, which builds on arctan_taylor to give reasonable results for a wide range of x (but beware possible overflow and underflow when computing x*x).

import math

def arctan_taylor_with_reduction(x, terms=9, threshold=0.1):
    """
    Compute arctan via argument reduction and Taylor series.

    Applies reduction steps until x is below `threshold`,
    then uses Taylor series.
    """
    reductions = 0
    while abs(x) > threshold:
        x = x / (1 + math.sqrt(1 + x*x))
        reductions += 1

    return arctan_taylor(x, terms=terms) * 2**reductions

Alternatively, given an existing implementation for tan, you could simply find a solution y to the equation tan(y) = x using traditional root-finding methods. Since arctan is already naturally bounded to lie in the interval (-pi/2, pi/2), bisection search works well:

def arctan_from_tan(x, tolerance=1e-15):
    """
    Compute arctan as the inverse of tan, via bisection search. This assumes
    that you already have a high quality tan function.
    """
    low, high = -0.5 * math.pi, 0.5 * math.pi
    while high - low > tolerance:
        mid = 0.5 * (low + high)
        if math.tan(mid) < x:
            low = mid
        else:
            high = mid
    return 0.5 * (low + high)

Finally, just for fun, here’s a CORDIC-like implementation, which is really more appropriate for a low-level implementation than for Python. The idea here is that you precompute, once and for all, a table of arctan values for 1, 1/2, 1/4, etc., and then use those to compute general arctan values, essentially by computing successive approximations to the true angle. The remarkable part is that, after the precomputation step, the arctan computation involves only additions, subtractions, and multiplications by by powers of 2. (Of course, those multiplications aren’t any more efficient than any other multiplication at the level of Python, but closer to the hardware, this could potentially make a big difference.)

cordic_table_size = 60
cordic_table = [(2**-i, math.atan(2**-i))
                 for i in range(cordic_table_size)]

def arctan_cordic(y, x=1.0):
    """
    Compute arctan(y/x), assuming x positive, via CORDIC-like method.
    """
    r = 0.0
    for t, a in cordic_table:
        if y < 0:
            r, x, y = r - a, x - t*y, y + t*x
        else:
            r, x, y = r + a, x + t*y, y - t*x
    return r

Each of the above methods has its strengths and weaknesses, and all of the above code can be improved in a myriad of ways. I encourage you to experiment and explore.

To wrap it all up, here are the results of calling the above functions on a small number of not-very-carefully-chosen test values, comparing with the output of the standard library math.atan function:

test_values = [2.314, 0.0123, -0.56, 168.9]
for value in test_values:
    print("{:20.15g} {:20.15g} {:20.15g} {:20.15g}".format(
        math.atan(value),
        arctan_taylor_with_reduction(value),
        arctan_from_tan(value),
        arctan_cordic(value),
    ))

Output on my machine:

    1.16288340166519     1.16288340166519     1.16288340166519     1.16288340166519
     0.0122993797673      0.0122993797673   0.0122993797673002   0.0122993797672999
  -0.510488321916776   -0.510488321916776   -0.510488321916776   -0.510488321916776
    1.56487573286064     1.56487573286064     1.56487573286064     1.56487573286064

The simplest way to do any inverse function is to use binary search.

1. definitions

   let assume function

   x = g(y)
   

   And we want to code its inverse:

   y = f(x) = f(g(y))
   x = <x0,x1>
   y = <y0,y1>
   

2. bin search on floats

   You can do it on integer math accessing mantissa bits like in here:

   • Any Faster RMS Value Calculation in C?

   but if you do not know the exponent of the result prior to computation then you need to use floats for bin search too.

   so the idea behind binary search is to change mantissa of y from y1 to y0 bit by bit from MSB to LSB. Then call direct function g(y) and if the result cross x revert the last bit change.

   In case of using floats you can use variable that will hold approximate value of the mantissa bit targeted instead of integer bit access. That will eliminate unknown exponent problem. So at the beginning set y = y0 and actual bit to MSB value so b=(y1-y0)/2. After each iteration halve it and do as many iterations as you got mantissa bits n… This way you obtain result in n iterations within (y1-y0)/2^n accuracy.

   If your inverse function is not monotonic break it into monotonic intervals and handle each as separate binary search.

   The function increasing/decreasing just determine the crossing condition direction (use of < or >).

C++ acos example

so y = acos(x) is defined on x = <-1,+1> , y = <0,M_PI> and decreasing so:

double f64_acos(double x)
    {
    const int n=52;         // mantisa bits
    double y,y0,b;
    int i;
    // handle domain error
    if (x<-1.0) return 0;
    if (x>+1.0) return 0;
    // x = <-1,+1> , y = <0,M_PI> , decreasing
    for (y= 0.0,b=0.5*M_PI,i=0;i<n;i++,b*=0.5)  // y is min, b is half of max and halving each iteration
        {
        y0=y;                   // remember original y
        y+=b;                   // try set "bit"
        if (cos(y)<x) y=y0;     // if result cross x return to original y decreasing is <  and increasing is >
        }
    return y;
    }

I tested it like this:

double x0,x1,y;
for (x0=0.0;x0<M_PI;x0+=M_PI*0.01)  // cycle all angle range <0,M_PI>
    {
    y=cos(x0);              // direct function (from math.h)
    x1=f64_acos(y);         // my inverse function
    if (fabs(x1-x0)>1e-9)   // check result and output to log if error
     Form1->mm_log->Lines->Add(AnsiString().sprintf("acos(%8.3lf) = %8.3lf != %8.3lf",y,x0,x1));
    }

Without any difference found… so the implementation is working correctly. Of coarse binary search on 52 bit mantissa is usually slower then polynomial approximation … on the other hand the implementation is so simple …

[Notes]

If you do not want to take care of the monotonic intervals you can try

• approximation search

As you are dealing with goniometric functions you need to handle singularities to avoid NaN or division by zero etc …

If you’re interested here more bin search examples (mostly on integers)

• Power by squaring for negative exponents it contains