2atan (example 3.5)

Average Error: 15.4 → 0.4

Time: 7.1s

Precision: binary64

Cost: 6912

Math

\[\tan^{-1} \left(N + 1\right) - \tan^{-1} N \]

↓

\[\tan^{-1}_* \frac{1}{1 + N \cdot \left(1 + N\right)} \]

FPCore

(FPCore (N) :precision binary64 (- (atan (+ N 1.0)) (atan N)))

↓

(FPCore (N) :precision binary64 (atan2 1.0 (+ 1.0 (* N (+ 1.0 N)))))

C

double code(double N) {
	return atan((N + 1.0)) - atan(N);
}

↓

double code(double N) {
	return atan2(1.0, (1.0 + (N * (1.0 + N))));
}

Fortran

real(8) function code(n)
    real(8), intent (in) :: n
    code = atan((n + 1.0d0)) - atan(n)
end function

↓

real(8) function code(n)
    real(8), intent (in) :: n
    code = atan2(1.0d0, (1.0d0 + (n * (1.0d0 + n))))
end function

Java

public static double code(double N) {
	return Math.atan((N + 1.0)) - Math.atan(N);
}

↓

public static double code(double N) {
	return Math.atan2(1.0, (1.0 + (N * (1.0 + N))));
}

Python

def code(N):
	return math.atan((N + 1.0)) - math.atan(N)

↓

def code(N):
	return math.atan2(1.0, (1.0 + (N * (1.0 + N))))

Julia

function code(N)
	return Float64(atan(Float64(N + 1.0)) - atan(N))
end

↓

function code(N)
	return atan(1.0, Float64(1.0 + Float64(N * Float64(1.0 + N))))
end

MATLAB

function tmp = code(N)
	tmp = atan((N + 1.0)) - atan(N);
end

↓

function tmp = code(N)
	tmp = atan2(1.0, (1.0 + (N * (1.0 + N))));
end

Wolfram

code[N_] := N[(N[ArcTan[N[(N + 1.0), $MachinePrecision]], $MachinePrecision] - N[ArcTan[N], $MachinePrecision]), $MachinePrecision]

↓

code[N_] := N[ArcTan[1.0 / N[(1.0 + N[(N * N[(1.0 + N), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]

Target

Original	15.4
Target	0.4
Herbie	0.4

\[\tan^{-1} \left(\frac{1}{1 + N \cdot \left(N + 1\right)}\right) \]

Derivation

1. Initial program 15.4

   \[\tan^{-1} \left(N + 1\right) - \tan^{-1} N \]

2. Applied egg-rr 14.3

   \[\leadsto \color{blue}{\tan^{-1}_* \frac{N + \left(1 - N\right)}{1 + N \cdot \left(N + 1\right)}} \]

3. Taylor expanded in N around 0 0.4

   \[\leadsto \color{blue}{\tan^{-1}_* \frac{1}{1 + N \cdot \left(N + 1\right)}} \]

4. Final simplification 0.4

   \[\leadsto \tan^{-1}_* \frac{1}{1 + N \cdot \left(1 + N\right)} \]

Alternatives

Alternative 1
Error	1.6
Cost	6784

\[\tan^{-1}_* \frac{1}{1 + N \cdot N} \]

Alternative 2
Error	30.6
Cost	6656

\[\tan^{-1}_* \frac{1}{1 + N} \]

Reproduce

herbie shell --seed 2022291 
(FPCore (N)
  :name "2atan (example 3.5)"
  :precision binary64

  :herbie-target
  (atan (/ 1.0 (+ 1.0 (* N (+ N 1.0)))))

  (- (atan (+ N 1.0)) (atan N)))