ShreevatsaR
Reputation
25,272
153/100 score
 Jun10 comment Ramanujan's personification of small positive integers en.wikipedia.org/w/… — using things there are $n$ of, to denote the number $n$, was a standard practice in classical Indian mathematics. I'm not sure whether Ramanujan had encountered any classical Indian mathematics though, so I have a tiny suspicion that the author of The Man Who Knew Infinity just got things a bit mixed up. (Or maybe Ramanujan did arrive at the same convention by himself; we know he did that with a lot of mathematics!) Jun8 comment How does a row of zeros make a free variable in linear systems of equations? A row of zeroes (or, in general, the matrix's row rank being less than the number of columns) means that there are not enough equations to completely determine all the variables. In this case, you can pick any of the three variables as a free variable, and determine the other two in terms of it. Sorry if this isn't clear; maybe someone will post a more clear answer. Jun5 comment Is it true that $\binom{n}{m} \binom{k}{\ell} \le \binom{n+k}{m+\ell}$? Thanks, I get it now. In full detail (in addition to your explanation): $$\sum_{k\ge 0}\sum_{\ell=0}^k \frac{\binom{n}{m} \binom{k}{\ell}}{\binom{n+k}{m+\ell}}\frac{a^k s^k}{k!} = \sum_{k\ge 0}\frac{a^k s^k}{k!}\sum_{\ell=0}^k \frac{\binom{n}{m} \binom{k}{\ell}}{\binom{n+k}{m+\ell}} = \sum_{k\ge 0}\frac{a^k s^k}{k!} {n+1+k\over n+1}$$ $$= \sum_{k\ge 0}\frac{a^k s^k}{k!} + \frac1{n+1}\sum_{k\ge 0}k\frac{a^k s^k}{k!} = \exp(as)+\frac{as \exp(as)}{n+1}$$ Jun5 comment Is it true that $\binom{n}{m} \binom{k}{\ell} \le \binom{n+k}{m+\ell}$? How do you get the second equation from the first? Jun4 comment How do you evaluate $a^b$ where b is irrational using only basic operators. Pick a rational approximation $r$ to $b$, and evaluate $a^r$. For better results, start with better rational approximations. May29 comment The Probability of Catching Criminals @Anonymous: I'm not sure I understand the question. Could you clarify? (You could post it as a new question, too.) May29 comment The Probability of Catching Criminals @Anonymous: What is the difference you're thinking of between "number of pairs" and "possible number of pairs"? And yes, with $N$ people there are $\binom{N}{2}$ pairs, but over $T$ time units of buying, there are $T\binom{N}{2}$ opportunities for a pair of people to buy the same set of items at the same time. (For each pair of people, there are $T$ times when they could buy the same thing.) May27 comment Does anyone know why this inclusion exclusion calculation isn't working? @askyle: Sure, feel free. May27 comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? @DonAntonio: Similarly, $19=4!-5$ can be interpreted as using the integers $24$ and $5$, instead of two consecutive integers $4$ and $5$. Such interpretations seem perverse to me, but now I realize that's only because of my bias of trying to have the question make sense. (One can even literally interpret the wording in the title as saying that we just have to use at least two consecutive integers and then can use anything else at all—say $n=(n-3)+1+2$ where $1,2$ are consecutive—but I tend not to gravitate towards such "finding loopholes".) But I agree one's personal choice may be different. May27 comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? @DonAntonio: (…) I'd not prefer this interpretation as it invalidates the question; I'd prefer to interpret that it's the integers $1,2,3,4$ that are used. Similarly, there is also some leeway in interpreting the question based on the examples (the problem of induction). I chose to interpret the tightest question that fits the examples (and also match the remark at the end on triangular numbers), but yes you could interpret it differently: e.g. $16=(2/3)(4/5)(6+7+8+9)$, which uses consecutive integers $2,3,4,5,6,7,8,9$ can be interpreted as using fractions, and thus not fit the question. :-) May27 comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? @DonAntonio: Actually this discussion has been useful to me, thanks. Let me declare a bias I just realized: I prefer to interpret the examples in such a way that they fit the question. (It is possible to interpret them differently; that is your prerogative.) As the question is not specified unambiguously, nor is it specified how the examples map to the question (e.g. it doesn't say "in the example $10=1+2+3+4$, the consecutive integers used are $1,2,3,4$), there is some leeway in interpretation. E.g. you could say in $11=1-2+3\times4$ that the integers used are $1,-2,3,4$, not consecutive. (…) May27 comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? @DonAntonio When you divide two integers, you can get a fractional number. But you cannot use that fact to conclude that the example doesn't use integers. For example, in $10 = 1+2+3+4$, we have $3+4=7$, but it would be absurd to say that because the example "uses" $7$ which is not consecutive with $1$ and $2$, the example does not satisfy the constraints of the question. :-) May27 comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? @DonAntonio: I didn't mean to annoy you — yes I did read what you wrote, but I could ask you the same question. :-) To clarify: this discussion started when you said that "Some of the examples of the OP don't even use integers", and I pointed out that all the OP's examples could indeed be seen as using integers, true to the question title. Then again you said that the OP's examples for 16 and 17 use fractional numbers (rather than integers?), and I pointed out that that cannot be the way to interpret it, because the OP specifically mentions integers. Is something not clear? May27 comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? Rather, $(n+1)$, $(n+2)$ and $(n+3)$ are three consecutive integers. +1. May26 comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? @DonAntonio There is no such thing as a fractional integer, so the OP's examples for 16 and 17 must be interpreted as using the consecutive integers $2,3,4,\dots$ (with division), rather than as using arbitrary fractions like $2/3$. As for whether the consecutive integers must all be less than $n$ (as in all the examples) or not, we can await the OP's clarification, though there doesn't seem to be much doubt about which one is the less trivial question. :-) May26 comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? @DonAntonio: The title of the question mentions "integers". I interpret the examples as always using some set of consecutive positive integers in order (and in particular each integer exactly once), but the set of allowed operations includes negation/subtraction and division. May26 comment Does every integer $n > 2$ have an arithmetic expression involving at least two consecutive integers but excluding $n$ itself? I presume the OP means consecutive positive integers strictly smaller than $n$. A fact that you can use: Every number that is not a power of $2$ is a sum of consecutive positive integers (all smaller than itself). For instance, $11 = 5 + 6$, $12 = 3 + 4 + 5$, $13 = 6 + 7$, $17 = 8 + 9$, $18 = 5 + 6 + 7$, $19 = 9 + 10$, etc. (For proof, see here or here.) So it's enough to ask the (suitably modified) question for powers of $2$. May9 comment Handle $O(x^3)$ when finding limits $O(x^3)$ (as $x \to 0$) just means "some function $f$ such that for some constant $c$, we have $|f(x)| \le c|x^3|$ in a neighbourhood of $0$". That is, $O(x^3)$ means "something that is absolutely at most a constant multiple of $x^3$". That is all. May6 comment Proving a recurrence relationship for a sequence Really nice solution, congratulations. :-) The main trick seems to be making sure that the $(n-1)b_{n-2}$ term gets multiplied by $x^{n-1}/(n-1)!$ rather than say $x^n/n!$ or $x^{n-2}/(n-2)!$, so that the $(n-1)$ factor cancels cleanly and we don't have to solve too-complicated differential equations. May5 comment Expected number of loops Just for completeness: yes, $H_n = 1 + \frac12 + \dots + \frac1n$ (harmonic number), and the asymptotics $(\ln n)/2 + 0.98$ mentioned above does indeed give about $3.28$ for $n=100$.