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Apr
15
comment Maximizing profit (dynamic programming)
More precisely: how many of questions up to 18 did you solve? Why attempt 19?
Apr
15
comment Maximizing profit (dynamic programming)
Guess you need to first read about dynamic programming before solving exercises. Did you manage to solve all (or most) of questions 1 to 18, before attempting question 19?
Apr
14
awarded  Nice Answer
Mar
22
comment Is this 5th root in the set of natural numbers?
What have you tried? What leads you to the belief that no such $x$ exists? If you include that in your question you are more likely to get an answer.
Mar
8
comment I don't get a 'What if' math joke
@bcrist: Ah I see... I was wrong, thanks for that!
Mar
7
comment I don't get a 'What if' math joke
@bcrist: Ah good point; perhaps one should say that though the What-If series contains references to the XKCD comics, there is none in the other direction. (Sort of how I can make references to a TV show, but a TV show will never make references to me.) At any rate, the subject of this question definitely doesn't count as "an xkcd", whatever that means. Nor is it even a joke, IMO.
Mar
7
answered I don't get a 'What if' math joke
Mar
7
comment I don't get a 'What if' math joke
@Ant: what-if.xkcd.com has nothing to do with the xkcd comic either, besides being hosted on the same domain. :-)
Mar
7
comment I don't get a 'What if' math joke
This has nothing to do with xkcd, besides being by the same author.
Mar
3
awarded  Great Answer
Feb
6
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Feb
2
awarded  Good Answer
Feb
2
awarded  Enlightened
Feb
2
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Jan
23
awarded  Good Answer
Jan
23
comment Curious Binomial Coefficient Identity
Just for completeness, on deriving $B(x)$: note that $\sum_{n=0}^{\infty}\binom{n}{k}x^n = \sum_{n=k}^{\infty}\binom{n}{k}x^n = x^k\sum_{n=0}^{\infty}\binom{n+k}{k}x^n =x^k\sum_{n=0}^{\infty}(-1)^n\binom{-k-1}{n}x^n=x^k(1-x)^{-k-1}$ as $$\binom{-k-1}{n} = \frac{(-k-1)(-k-2)\cdots(-k-n)}{n!}=(-1)^n\frac{(n+k)\cdots(k+1)}{n!} =(-1)^n\binom{n+k}{k}.$$
Jan
22
answered Curious Binomial Coefficient Identity
Jan
22
comment Curious Binomial Coefficient Identity
@anorton: Quite clearly from context, $a_n = \binom{n}{k}$ (for some/any fixed $k$).
Jan
3
awarded  Nice Answer
Dec
8
awarded  Caucus