Fast Exp calculation: possible to improve accuracy without losing too much performance?

Question

I am trying out the fast Exp(x) function that previously was described in this answer to an SO question on improving calculation speed in C#:

public static double Exp(double x)
{
  var tmp = (long)(1512775 * x + 1072632447);
  return BitConverter.Int64BitsToDouble(tmp << 32);
}

The expression is using some IEEE floating point "tricks" and is primarily intended for use in neural sets. The function is approximately 5 times faster than the regular Math.Exp(x) function.

Unfortunately, the numeric accuracy is only -4% -- +2% relative to the regular Math.Exp(x) function, ideally I would like to have accuracy within at least the sub-percent range.

I have plotted the quotient between the approximate and the regular Exp functions, and as can be seen in the graph the relative difference appears to be repeated with practically constant frequency.

Is it possible to take advantage of this regularity to improve the accuracy of the "fast exp" function further without substantially reducing the calculation speed, or would the computational overhead of an accuracy improvement outweigh the computational gain of the original expression?

(As a side note, I have also tried one of the alternative approaches proposed in the same SO question, but this approach does not seem to be computationally efficient in C#, at least not for the general case.)

UPDATE MAY 14

Upon request from @Adriano, I have now performed a very simple benchmark. I have performed 10 million computations using each of the alternative exp functions for floating point values in the range [-100, 100]. Since the range of values I am interested in spans from -20 to 0 I have also explicitly listed the function value at x = -5. Here are the results:

      Math.Exp: 62.525 ms, exp(-5) = 0.00673794699908547
Empty function: 13.769 ms
     ExpNeural: 14.867 ms, exp(-5) = 0.00675211846828461
    ExpSeries8: 15.121 ms, exp(-5) = 0.00641270968867667
   ExpSeries16: 32.046 ms, exp(-5) = 0.00673666189488182
          exp1: 15.062 ms, exp(-5) = -12.3333325982094
          exp2: 15.090 ms, exp(-5) = 13.708332516253
          exp3: 16.251 ms, exp(-5) = -12.3333325982094
          exp4: 17.924 ms, exp(-5) = 728.368055056781
          exp5: 20.972 ms, exp(-5) = -6.13293614238501
          exp6: 24.212 ms, exp(-5) = 3.55518353166184
          exp7: 29.092 ms, exp(-5) = -1.8271053775984
      exp7 +/-: 38.482 ms, exp(-5) = 0.00695945286970704

ExpNeural is equivalent to the Exp function specified in the beginning of this text. ExpSeries8 is the formulation that I originally claimed was not very efficient on .NET; when implementing it exactly like Neil it was actually very fast. ExpSeries16 is the analogous formula but with 16 multiplications instead of 8. exp1 through exp7 are the different functions from Adriano's answer below. The final variant of exp7 is a variant where the sign of x is checked; if negative the function returns 1/exp(-x) instead.

Unfortunately, neither of the expN functions listed by Adriano are sufficient in the broader negative value range I am considering. The series expansion approach by Neil Coffey seems to be more suitable in "my" value range, although it is too rapidly diverging with larger negative x, especially when using "only" 8 multiplications.

Answer 1

Taylor series approximations (such as the expX() functions in Adriano's answer) are most accurate near zero and can have huge errors at -20 or even -5. If the input has a known range, such as -20 to 0 like the original question, you can use a small look up table and one additional multiply to greatly improve accuracy.

The trick is to recognize that exp() can be separated into integer and fractional parts. For example:

exp(-2.345) = exp(-2.0) * exp(-0.345)

The fractional part will always be between -1 and 1, so a Taylor series approximation will be pretty accurate. The integer part has only 21 possible values for exp(-20) to exp(0), so these can be stored in a small look up table.