?

Average Accuracy: 75.2% → 99.4%
Time: 6.0s
Precision: binary64
Cost: 6912

?

\[\tan^{-1} \left(N + 1\right) - \tan^{-1} N \]
\[\tan^{-1}_* \frac{1}{1 + \left(N + N \cdot N\right)} \]
(FPCore (N) :precision binary64 (- (atan (+ N 1.0)) (atan N)))
(FPCore (N) :precision binary64 (atan2 1.0 (+ 1.0 (+ N (* N N)))))
double code(double N) {
	return atan((N + 1.0)) - atan(N);
}
double code(double N) {
	return atan2(1.0, (1.0 + (N + (N * N))));
}
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 + (n * n))))
end function
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 + (N * N))));
}
def code(N):
	return math.atan((N + 1.0)) - math.atan(N)
def code(N):
	return math.atan2(1.0, (1.0 + (N + (N * N))))
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(N * N))))
end
function tmp = code(N)
	tmp = atan((N + 1.0)) - atan(N);
end
function tmp = code(N)
	tmp = atan2(1.0, (1.0 + (N + (N * N))));
end
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[(N * N), $MachinePrecision]), $MachinePrecision]), $MachinePrecision]], $MachinePrecision]
\tan^{-1} \left(N + 1\right) - \tan^{-1} N
\tan^{-1}_* \frac{1}{1 + \left(N + N \cdot N\right)}

Error?

Try it out?

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original75.2%
Target99.4%
Herbie99.4%
\[\tan^{-1} \left(\frac{1}{1 + N \cdot \left(N + 1\right)}\right) \]

Derivation?

  1. Initial program 79.5%

    \[\tan^{-1} \left(N + 1\right) - \tan^{-1} N \]
  2. Applied egg-rr82.0%

    \[\leadsto \color{blue}{\tan^{-1}_* \frac{N + \left(1 - N\right)}{\mathsf{fma}\left(N, N + 1, 1\right)}} \]
  3. Simplified99.5%

    \[\leadsto \color{blue}{\tan^{-1}_* \frac{1}{\mathsf{fma}\left(N, 1 + N, 1\right)}} \]
  4. Applied egg-rr99.5%

    \[\leadsto \tan^{-1}_* \frac{1}{\color{blue}{\left(N + N \cdot N\right) + 1}} \]
  5. Final simplification99.5%

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

Alternatives

Alternative 1
Accuracy98.5%
Cost7049
\[\begin{array}{l} \mathbf{if}\;N \leq -1 \lor \neg \left(N \leq 1\right):\\ \;\;\;\;\tan^{-1}_* \frac{1}{N + N \cdot N}\\ \mathbf{else}:\\ \;\;\;\;\tan^{-1}_* \frac{1}{1 + N}\\ \end{array} \]
Alternative 2
Accuracy97.5%
Cost6921
\[\begin{array}{l} \mathbf{if}\;N \leq -1 \lor \neg \left(N \leq 1\right):\\ \;\;\;\;\tan^{-1}_* \frac{1}{N \cdot N}\\ \mathbf{else}:\\ \;\;\;\;\tan^{-1}_* \frac{1}{1}\\ \end{array} \]
Alternative 3
Accuracy97.9%
Cost6921
\[\begin{array}{l} \mathbf{if}\;N \leq -0.62 \lor \neg \left(N \leq 1.6\right):\\ \;\;\;\;\tan^{-1}_* \frac{1}{N \cdot N}\\ \mathbf{else}:\\ \;\;\;\;\tan^{-1}_* \frac{1}{1 + N}\\ \end{array} \]
Alternative 4
Accuracy50.7%
Cost6528
\[\tan^{-1}_* \frac{1}{1} \]

Error

Reproduce?

herbie shell --seed 2023157 -o generate:proofs
(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)))