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copied!<p>How accurate and how fast does this need to be? You can do something pretty good with a piecewise Chebyshev approximation. You control the accuracy using the order of the Chebyshev approximation (higher order = slower but more accurate). I would also suggest doing argument reduction by decomposing your double into its mantissa (between 1 and 2) and its exponent (a power of two whose logarithm is simply the exponent times <code>log(2)</code>, which you can precompute.)</p> <p>I don't think you can come up with anything very accurate on [0.1, 2] without either doing more arithmetic whenever you want a log or using a huge table and incurring all the caching and unpredictable memory access issues that entails. But do consider doing a piecewise Chebyshev approximation if you have enough time. (If you want me to show you code that uses a Chebyshev approximation, let me know with a comment and I'll update this post.)</p> <p><strong>EDIT</strong>: Code for logarithm using Chebyshev approximation. Accurate within 1e-5.</p> <pre><code>double mylog(double x) { static double logtwo = log(2); static double tbls[4][5] = {{0,0,0,0,0}, {-.9525934480e-2,-.87402539075,-1.135790603,1.1519051721,-1.7063112037}, {.53892330786e-3,-1.0117355213,-.4085197450,-.6242237228,0}, {.60393290013e-6,-1.0001523639,-.4940510719,-.4058961978,0}}; if (x>1) return -mylog(1/x); int expo,e2; x = 1-frexp(x, &expo); double y = frexp(x, &e2); if (e2 < -3) e2 = -3; double *tbl = tbls[-e2]; return expo*logtwo + tbl[0]+x*(tbl[1]+x*(tbl[2]+x*(tbl[3]+x*tbl[4]))); } </code></pre> <p>I calculated the Chebyshev approximations using Maple and expanded them as traditional polynomials for speed.</p> <p>If you want good accuracy very close to 1, you can change <code>if (e2 < -3) e2 = -3</code> and add the row <code>{0,-1,-.5,-1/3.,-.25}</code> to the end of <code>tbls</code> corresponding to Taylor's approximation. If you want it faster, compute a better cubic approximation to log(x) between 1/2 and 3/4 and store it in the first row of <code>tbls</code>.</p>

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