OK, we know if a > b, then b^a > a^b when a, b > e. So we can discard half the range, easily--- 39^41> 41^39, 38^42 > 42^38, etc.

We can also calculate the derivative of f = x^(80-x) as:

ln f = (80-x) ln x

d/dx ln f = d/dx 80 ln x - d/dx x ln x

1/x * df/dx = 80/x - (1 + ln x )

df/dx = 80 - x - x ln x

x( 1 + ln x) = 80 is a little hard to approximate. If we know ln 10 ~= 3 and ln 2 ~= 0.7, then we can quickly see that x=40 is too large and x=10 is too small, but x = 20 is about right, maybe a bit too large: 20( 1 + 3 + 0.7 ) = 20 * 4.7

So how can we compare 19^61 with 20^60?

The first few powers of 19 are

361

6859

130321

2476009

Not a lot of help there.

Playing around with the binomial expansion didn't seem to lead to any good result. If we want to compare 19 and (1+1/19)^60, the first few terms are

1 + 60/19 + 30 * 59/(19^2) + 20 * 59* 29/(19^3) + 15 * 59 * 29 * 19/(19^4)

which does not seem to be converging quickly enough.

So I'm missing the clever trick here, whatever it is.

rycharde (59)· 5 months agoThat's very good, thanks.

I also tried a binomial expansion and it just doesn't converge quickly enough to be useful.

The "insight" was earlier: that e^3 = 20 ;-) to less than 1% error.

I chose the numbers carefully, but in general, I think one would need to use logs for the final check.

I just find it an interesting counter-intuitive function :-)