Answer fractional powers of exact numbers past a double's range (closes #16)

A fractional power turned its base into a double first and refused a
base outside the normal range of a double, so (2^1200)^0.5, (2^1024)^0.5,
(2^-1200)^0.5 and 1e400^0.5 were refused although each answer is an
ordinary double. Such a base is now brought into that range by square
roots taken from its exact value in a big.Float, at most three under the
4096-bit limit, with the exponent doubled for each, and only a result
that is not a normal double is refused. The tests give these powers and
the edges of the range their values, and the bounded-work test covers a
base that needs three roots. The README's refusal sentence and example
are updated.

Model: opus-5-5

README.md

@@ -438,8 +438,11 @@ container.
   `4.9406564584124654e-324`. Refused as too large or too small: any
   number whose numerator or denominator reaches 4096 bits, wherever it
   appears, as `go/constant` rounds a fraction that grows that large
-  (`2^4095`, `1e-1300 + 1`); and a power computed as a double whose base
-  or result is outside the normal range of a double (`1e-400^0.5`).
+  (`2^4095`, `1e-1300 + 1`); and a power computed as a double whose
+  result is outside the normal range of a double (`2^1500.5`). The base
+  of such a power may be outside that range: square roots taken from its
+  exact value bring it inside first, so `(2^1200)^0.5` is
+  `4.149515568880993e+180` and `1e-400^0.5` is `1e-200`.
 - **Failure is an exit.** If the chat client exits or the connection to
   it drops, the bot exits with an error and the container's restart
   policy starts both again. `SIGTERM` stops the bot, which stops the