ShreevatsaR
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28,050
290/100 score
 May 27 answered Special representation of a number May 25 awarded Nice Answer 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 “What if” math joke: the derivative of $\ln(x)^e$ @bcrist: Ah I see... I was wrong, thanks for that! Mar 7 comment “What if” math joke: the derivative of $\ln(x)^e$ @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 “What if” math joke: the derivative of $\ln(x)^e$ Mar 7 comment “What if” math joke: the derivative of $\ln(x)^e$ @Ant: what-if.xkcd.com has nothing to do with the xkcd comic either, besides being hosted on the same domain. :-) Mar 7 comment “What if” math joke: the derivative of $\ln(x)^e$ This has nothing to do with xkcd, besides being by the same author. Mar 3 awarded Great Answer Feb 6 awarded Nice Answer Feb 2 awarded Good Answer Feb 2 awarded Enlightened Feb 2 awarded Nice Answer 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 $k$).