Apply the same bound to the other region.
authorTaylor R Campbell <campbell@mumble.net>
Thu, 25 Oct 2018 05:33:57 +0000 (05:33 +0000)
committerTaylor R Campbell <campbell@mumble.net>
Thu, 25 Oct 2018 05:33:57 +0000 (05:33 +0000)
src/runtime/arith.scm

index db95eb48667544e8127e06e821c334af9c1f8827..e1afa94c13ed2e75f24d2660fbd0d5db73f95a20 100644 (file)
@@ -2304,7 +2304,9 @@ USA.
          ;;     = -log(e^{-t} - 1)
          ;;       * (1 + d0 + (1 + d0) log(1 + d1)/log(e^{-t} - 1))
          ;;
-         ;; If t <= -log(1 + e), then log(e^{-t} - 1) >= 1, so
+         ;; If t <= -log(1 + e), then log(e^{-t} - 1) >= 1; similarly,
+         ;; if t >= -log(1 + 1/e), then log(e^{-t} - 1) <= -1.  Hence,
+         ;; in both cases, |log(e^{-t} - 1)| >= 1, so that
          ;;
          ;;     |d0 + (1 + d0) log(1 + d1)/log(e^{-t} - 1)|
          ;;     <= |d0| + |(1 + d0) log(1 + d1)/log(e^{-t} - 1)|