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    <p>Well, <code>exp(log_b)</code> is so much smaller than <code>exp(log_a)</code> that you can completely ignore that term and still get the correct answer with respect to double-precision:</p> <pre><code>exp(log_a) = 2.845550077506*10^-714 exp(log_b) = 4.05118588390*10^-1663 </code></pre> <p>If you are actually trying to compute <code>(exp(log_a) + exp(log_b)) / 2</code>, the answer would underflow to zero anyways. So it wouldn't really matter unless you're trying to take another logarithm at the end.</p> <p>If you're trying compute:</p> <pre><code>log((exp(log_a) + exp(log_b)) / 2) </code></pre> <p>Your best bet is to examine the difference between <code>log_a</code> and <code>log_b</code>. If the difference is large, then simply take the final value as equal to the larger term - log(2) since the smaller term will be small enough to completely vanish.</p> <p>EDIT:</p> <p>So your final algorithm could look like this:</p> <ol> <li>Check the magnitudes. If <code>abs(log_a - log_b) &gt; 800</code>. Return <code>max(log_a,log(b)) - log(2)</code>.</li> <li>Check either magnitude (they will be close together at this point.). If it is much larger or smaller than 1, add/subtract a constant from both <code>log_a</code> and <code>log_b</code>.</li> <li>Perform the calculation.</li> <li>If the values were scaled in step 2. Scale the result back.</li> </ol> <p>EDIT 2:</p> <p>Here's an even better solution:</p> <pre><code>if (log_a &gt; log_b) return log_a + log(1 + exp(log_b - log_a)) - log(2) else return log_b + log(1 + exp(log_a - log_b)) - log(2) </code></pre> <p>This will work if <code>log_a</code> and <code>log_b</code> are not too large or are negative.</p>
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    1. COThis is a good answer, and does comfort me somewhat. However, in general the values could be somewhat closer together in other runs of the algorithm (see the clarification in my question). I think a combination of this and my previous idea might work, though. In particular: add `c` to all the values such that largest ones become computable; compute `log(mean(exp(x+c)))-c`; the values that still underflow can be assumed to be too small to affect the calculation. Let me ponder a little more, but I think this will work out. Thanks!
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    2. COIf the values are close enough to where it matters, but large/small enough to where `exp()` will over/underflow, then yes, you will need to scale the values up/down so that they won't overflow.
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