;;
;; 1. No inputs. Empty sum is zero, so yield log(0) = -inf.
;; 2. One input. Computation is exact. Preserve it.
- ;; 3. Maximum is infinite:
- ;; - if +inf, sum overflows, so +inf.
- ;; - if -inf, all inputs are -inf, so -inf.
- ;; 4. NaN among the inputs: invalid operation; result is NaN.
+ ;; 3. Maximum +inf. Sum overflows, so yield it (unless NaN).
;;
;; Most likely all the inputs are finite, so prioritize that case by
- ;; checking for an infinity first -- if there is a NaN, the usual
+ ;; checking for +inf first -- if there is a NaN, the usual
;; computation will propagate it.
;;
;; Overflow is not possible because everything is normalized to be