Average Error: 14.6 → 0.3
Time: 9.4s
Precision: 64
\[\tan^{-1} \left(N + 1\right) - \tan^{-1} N\]
↓
\[\tan^{-1}_* \frac{1}{1 + \left(N + 1\right) \cdot N}\]
\tan^{-1} \left(N + 1\right) - \tan^{-1} N
↓
\tan^{-1}_* \frac{1}{1 + \left(N + 1\right) \cdot N}
double f(double N) {
        double r103486 = N;
        double r103487 = 1.0;
        double r103488 = r103486 + r103487;
        double r103489 = atan(r103488);
        double r103490 = atan(r103486);
        double r103491 = r103489 - r103490;
        return r103491;
}
↓
double f(double N) {
        double r103492 = 1.0;
        double r103493 = 1.0;
        double r103494 = N;
        double r103495 = r103494 + r103492;
        double r103496 = r103495 * r103494;
        double r103497 = r103493 + r103496;
        double r103498 = atan2(r103492, r103497);
        return r103498;
}

Error

Bits error versus N

Try it out

Your Program's Arguments

Results

Enter valid numbers for all inputs

Target

Original14.6
Target0.3
Herbie0.3
\[\tan^{-1} \left(\frac{1}{1 + N \cdot \left(N + 1\right)}\right)\]

Derivation

  1. Initial program 14.6

    \[\tan^{-1} \left(N + 1\right) - \tan^{-1} N\]
  2. Using strategy rm

  3. Applied diff-atan13.5

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

  4. Simplified0.3

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

  5. Final simplification0.3

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

Reproduce

herbie shell --seed 2019212 
(FPCore (N)
  :name "2atan (example 3.5)"
  :precision binary64

  :herbie-target
  (atan (/ 1 (+ 1 (* N (+ N 1)))))

  (- (atan (+ N 1)) (atan N)))