3 ms·
It's accurate to about a 100 digits, admittedly not quite as good as the expression in the OP, but it's the same principle. It's taking a fairly standard approx
by messe 8y ago
It's accurate to about a 100 digits, admittedly not quite as good as the expression in the OP, but it's the same principle. It's taking a fairly standard approximation, in the OP's case a Riemann sum, and inserting concrete but aesthetically pleasing numbers and acting as if something is incredible about it.
- saagarjha 8y agoYes, I get that. It seems like Wolfram|Alpha thinks 10^100 is too large of a number for it to have to care about, and is rounding it to infinity internally somewhere, hoping it doesn't change the final result. But, of course, it does matter in this case.
- kidme5 8y agoYou asked WA for the result with 100 digits of precision. It probably is 1 to that precision.
- saagarjha 8y agoI don't think it is. Try N[((1+10^-n)^(10^n))/e, n] for smaller values of n. At first it approaches e faster than the precision can "catch up", and up until around n=50 it will result in 1. But at 10^60 and above it will return 1.0000000…
- btilly 8y agoThere is no need to guess. First of all: (1 + 1/n)^n = e^(log(1 + 1/n) * n) Now from the Taylor series approximation: log(1 + t) = t - t^2/2 + t^3/3 - t^4/4 + ... Substitute that in and we get: e^(log(1 + 1/n) * n) = e^((1/n - 1/(2 n^2) + 1/(3 n^3) - ...) n) = e^(1 - 1/(2n) + 1/(3n^2) - ...) And now you can apply the tangent line approximation to find that this is: e - e/(2n) + O(1/n^2) So if n = 10^N, then the first error should be around the N'th digit.
- NegativeLatency 8y agoIt will say they're the same out to more than 100 digits: http://www.wolframalpha.com/input/?i=N%5B((1%2B10%5E-100)%5E(10%5E100))%2Fe,+10000000000000000000000000000000000000000000%5D http://www.wolframalpha.com/input/?i=N%5B((1%2B10%5E-100)%5E...