Dealing with large numbers
can be confusing
, once the numbers get so big you need to use exponents to express their exponents. I recently dealt with a number about the size of 102.1 x 10343
, call it N
, a number whose exponent
has 343 digits. I wanted to see what it was when raised to its own power.
Firing up hypercalc ("just try and MAKE me overflow!") I was surprised to see that my number (N) when raised to its own power, gave the result 10N.
(That's 10102.1 x10343.)
But in fact hypercalc is correct, and, to normal levels of accuracy, NN = 10N when N is very large, as in the present case.
To raise to the power of N, we multiply a number by itself N times. If the number is a power of 10, with exponent exp, then we add exp digits for each such multiplication. So 10N will have N digits, since 10 = 101 and 1 times N is N.
N itself has an exponent of log10(N), and so NN has log10(N) * N digits.
So NN has log10(N) times as many digits as 10N.
So the exponent of NN (putting NN into the form 10exp) has log10(log10(N)) times as many digits as N.
Which means it has around 343 times as many digits as N (because N = 102.1 x 10343.)
So the second exponent has less than 3 times as many digits as the second exponent of 10N (because log10(343) < 3.)
So the third exponent is less than log10(3) out.
Which means that increasing the third exponent of 10N by one would give a value significantly larger than NN, since log10(3) < 1.
So, given N large enough to force us to use these 'third exponents', 10N is pretty much "equal" to NN, which seems surprising, shocking even, especially considering how large N is in the first place.
Now I'm not saying that NN and 10N are converging - the difference is plainly growing.
What's happening is that the difference is growing more slowly than the numbers themselves are, with the result that at some point, our notation gives out, and to record the difference, we'd have to include more digits than it will allow.
If you're having problems believing this (and I did myself) here's a different proof, showing the same thing in more detail, and more understandably, from hypercalc author mrob27 (whose large number pages at http://home.earthlink.net/~mrob/pub/math/largenum.htm will be of undoubted interest to anyone still reading. Go there.)
Your explanation of how the exponents work is fine (based on an intuitive
notion of "number of digits") but I would prefer something more like this:
Given R2 = 10 ^ (2 x 10^343), compare 10^R2 to R2^R2.
We need to use the fact that (a^b)^c = a^(b*c), and also (a^b)*(a^c) =
10^R2 = 10 ^ [10 ^ (2 x 10^343)]
R2^R2 = [10 ^ (2 x 10^343)] ^ [10 ^ (2 x 10^343)]
= 10 ^ [(2 x 10^343) x 10 ^ (2 x 10^343)]
= 10 ^ [2 x 10^(343 + 2 x 10^343)]
= 10 ^ [10 ^ (log(2) + 343 + 2 x 10^343)]
Now notice that log(2) + 343 + 2*10^343 is only a little bit more than
2*10^343. In fact, when the addition is performed, the log(2) and the
343 will get completely lost in the roundoff error unless at least 342
digits of precision are used in the calculation. So the most accurate
value we can calculate for R2^R2 is 10^[10^(2x10^343)], which is
This is a particular instance of the general rule that as the exponent gets larger and larger, the value of the mantissa becomes increasingly irrelevant for the size of the result.