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copied!<p>The derivative of <code>ArcTan(X, Sqrt(1-X*X))</code> with respect to X is <code>1/Sqrt(1-X*X)</code>. This goes to infinity as |X| goes to 1. Therefore, when X is near 1 or -1, any error in evaluation has a huge effect on the result. Thus, it is critical that evaluation minimize error in these cases.</p> <p>When X is near 1, evaluation of <code>1-X</code> has no error (in IEEE 754 or any good floating-point system, because the scale of the result is such that its least significant bit is at least as low as the least significant bit in 1 or X, so the exact mathematical result has no bits outside the available significand bits). Since <code>1-X</code> is exact, consider the effect of the error in <code>1+X</code> by considering the derivative of <code>ArcTan(X, Sqrt((1-X)*(1+X+e))</code> with respect to e, where e is the error introduced in the <code>1+X</code> operation. The derivative is, when X is near 1 and e is small, approximately -1/10. (Taking the derivative with Maple and substituting 1 for x yields <code>-1/(sqrt(4+2e)*(5+2e)</code>. Then substituting 0 for e yields -1/10.) Thus, the error in <code>1+X</code> is not critical.</p> <p>Therefore, evaluating the expression as <code>ArcTan(X, Sqrt((1-X)*(1+X))</code> is a good way to evaluate it.</p> <p>The situation is symmetric for X near -1. (<code>1+X</code> has no error, and <code>1-X</code> is not critical.)</p> <p>Conversely, if we consider the error in <code>X*X</code>, the derivative of <code>ArcTan(X, Sqrt(1-X*X+e))</code> with respect to e is, when X is near 1, approximately -1/(2*sqrt(e)*(1+e)), so it is large when e is small. So a small error in evaluating <code>X*X</code> will cause a large error in the result, when X is near 1.</p> <hr> <p>Ask Pascal Cuoq points out, when evaluating a function f(x), we are generally interested in minimizing the relative error in the final result. And, as I have pointed out, the errors that occur during calculation are generally relative errors in intermediate results due to floating-point rounding. I was able to ignore this in the above because I was considering the function when X is near 1, so both the intermediate values under consideration (1+X and X*X) and the final value had magnitudes near 1, so dividing the values by those magnitudes would not change anything significantly.</p> <p>However, for completeness, I examined the situation more closely. In Maple, I wrote <code>g := arctan(x, sqrt((1-x*x*(1+e0))*(1+e1))*(1+e2))</code>, thus allowing for relative errors e0, e1, and e2 in the calculations of <code>x*x</code>, <code>1-x*x</code>, and the <code>sqrt</code>, respectively, and I wrote <code>h:= arctan(x, sqrt((1-x)*(1+x)*(1+e0))*(1+e2))</code> for the alternative. Note that e0 in this case combines the three errors in <code>1-x</code>, <code>1+x</code>, and the multiplication of them; the full error term could be <code>(1+ea)*(1+eb)*(1+ec)</code>, but this is effectively <code>1+e0</code> with a larger possible range for e0.</p> <p>Then I examined the derivatives of these functions with respect to (one at a time) e0, e1, and e2 divided by abs(f(x)), where <code>f</code> was the ideal function, <code>arctan(x, sqrt(1-x*x))</code>. E.g., in Maple, I examined <code>diff(g, e0) / abs(f(x))</code>. I did not perform a full analytic evaluation of these; I examined the values for some values of x near 0 and near 1 and for values of e0, e1, and e2 at one of their limits, -2<sup>-54</sup>.</p> <p>For x near 0, the values were all of magnitude about 1 or less. That is, any relative error in calculation resulted in a similar relative error in the result, or less.</p> <p>For x near 1, the values with the derivatives of e1 and e2 were tiny, about 10<sup>-8</sup> or less. However, the values with the derivatives of e0 were hugely different for the two methods. For the <code>1-x*x</code> method, the value was about 2•10<sup>7</sup> (using x = 1-2<sup>-53</sup>). For the <code>(1-x)*(1+x)</code> method, the value was about 5•10<sup>-9</sup>.</p> <p>In summary, the two methods do not differ much near x = 0, but the latter method is significantly better near x = 1.</p>

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