What about -370 ? (Minus 370) Or -407 for that matter.
I thought about the negative numbers but the minus sign isn’t a digit is it?
So when you cube the digits and then sum them you aren’t cubing a minus sign each time correct?
So cubing the digits of -370 is still 3^3, 7^3, and 0^3 correct? Not -3^3, -7^3, and 0^3.
Mea culpa iterum.
I’m getting something closer to 1/2n - 11/24n[sup]2[/sup] + O(n[sup]-3[/sup]) - is the e needed there?
Short answer: yes.
Here are a few more terms:
e(1/2n - 11/24n² + 7/16n³ - 2447/5760n⁴ + …)
Let n = 100. Then e - (1 + 1/n)ⁿ = 0.01346799903751914…
while 1/2n - 11/24n² + 7/16n³ = 0.00495460416… (just shy of 1/200), while
e(1/2n - 11/24n² + 7/16n³) = 0.01346801047…
A good exercise is, use your preferred method (e.g., binomial theorem, Taylor series, some tricks with limits, whatever) to calculate just the order-1/n term.
Thanks. I think I see what I was missing (I saw something that looked like it was working without the e, without carrying the calculation far enough).
Also, perhaps think about exp(1 + Bx + Cx² + O(x³)) = e(1 + Bx + ((B² + 2C)/2) x² + …) [I don’t recall the name of this formula, at least in this form; please remind me!], so starting from (log(1+u))/u = 1 - u/2 + u²/3 - …, you get (1+u)[sup]1/u[/sup] = e(1 - u/2 + 11u²/24 - …); either way, the “e” should be there (consider the order-0 term!)
Thanks.
My mistake was to think that I could get the order 1/n term with just the first few terms of the binomial expansion.
(and maybe some algebra mistakes too)