8:52 PM

The Taylor series approximation of e^x converges rapidly for small positive values of x, but gives incorrect results for negative values, because computing the series requires subtracting large numbers from each other to get a small number, and on a computer this will introduce rounding errors.

The solution is to rewrite e^x as 1/e^−x when x<0. This can be computed without rounding errors reducing the accuracy of the final result. This solution is straightforward, but it was not obviously necessary in the original computation. The lesson, again, is to watch out for rounding errors.

The Taylor series approximation of e^x is accurate for larger positive values of x, but the number of terms that must be computed in order for the approximation to converge increases as x increases. This is a problem, because each term of the Taylor series has xⁿ in the numerator. If the number of required terms gets too large, xⁿ will become too large for the computer to represent and the computation will fail. My calculator gives e⁷⁹ as 2.038281066651×10³⁴. The Taylor series approximation returns infinity, because it requires more than 163 terms and 79¹⁶³ is greater than the computer can represent.

Once again, the solution is to rewrite the original function. x can be broken up into an integer part and a fractional part, and then e^x=e^{integer part}×e^{fractional part}. e^{fractional part} can be calculated as a Taylor series expansion. If e is stored as a constant, then e^{integer part} can be directly multiplied out, and this can be computed when a strict Taylor series approximation of e^x fails.

Taylor series can also be used to approximate sin(x) and cos(x). Like with e^x, the series require fewer terms to converge near 0, so the identities sin(x)=sin(x+2πn) and cos(x)=cos(x+2πn) for all positive and negative integer values of n should be used so the Taylor series is only computed for −π<x<π.

Finally, Taylor series can be used to approximate ln(x). The catch here is that e^x, sin(x), and cos(x) are all computed based the difference between x and 0, while ln(x) must be computed based on the difference between x and 1. This makes the math a little messier but isn't otherwise a big deal. However, the Taylor series for ln(x) does not converge for x>2. Since ln(x) = −ln(1/x), to compute ln(x) for x≥2, compute −ln(1/x) instead.

0 comments