Question

Upper and lower bounds on the generalized arithmetic derivative:

Answer 1

\[\Large \binom{\Omega(n) }{k} n^{1- \frac{k}{\Omega(n)}} \le n^{[k]} \le \binom{\Omega(n)}{k} \frac{n}{LD(n)^k}\] where Omega(n) is the number of prime factors in n and LD(n) is the Least Divisor (smallst prime) in n.

Answer 2

is that headphones ??

Answer 3

Haaha

Answer 4

Setting the left and right equal also gives us something that's obviously true: \[\Large n \ge LD(n) ^{\Omega(n)}\] All this says is n is greater than the smallest divisor the same amount of times, it's like saying: \[\Large 12 \ge 2^3\]

Answer 5

Ok I'm gonna go downstairs to grab a snack, if anyone wants to see a proof of this say so before I get back in a minute. =P

Answer 6

Actually I believe that upper bound there is better than the wikipedia article for the regular old arithmetic derivative.

Answer 7

hmm not coffee ttodday ??

Answer 8

Nahhh I think I drank way too much coffee this week lol

Answer 9

@Kainui where's my coffee and my snack? How dare you not invite me to the party!

Answer 10

Ω so what is it ??

Answer 11

Commenting here so I can get notified :P
@Kainui ughgh! You're killing me with the suspense :P

Answer 12

Woah woah party? Pretty sure there's no party happening lol. Like I'm just sitting at a desk working out some math on a dreary rainy day here haha, nothing exciting at all XD

Answer 13

:p That's what it sounds like. In fact, I'm having one tomorrow. I wish I could party at college. Those Professors are pretty stricked. No coffee allowed. #1 rule. Still bring it in anyways. #2 no cellphones. That sucks. #3 Idk, lol.