268,357 reputation
25268549
bio website
location Cleveland Heights, OH
age 66
visits member for 3 years, 7 months
seen 16 mins ago

Professor emeritus at Cleveland State University. I’m a set-theoretic and general topologist with an interest in combinatorics. I’m also interested in linguistics, especially historical linguistics.


8h
comment What's the term for antisymmetry where equal elements are not in the relation?
@limp_chimp: You're welcome.
9h
answered Difference of two graphs
10h
answered What's the term for antisymmetry where equal elements are not in the relation?
12h
answered To show that $X = (0,1]$ is complete .
12h
answered Proof Verification (Set Theory)
13h
answered Isolate points of a metric space with some properties?
13h
comment 2003 Putnam A-1 Help needed about sequences
@Amad27: You’re welcome! Take a closer look: those inequalities are all $\le$, not $<$, just as in the statement of the problem. In fact, that solution is the one that I was hinting at in my answer and in the example in the long comment.
13h
comment Issue with Spivak's Solution
@Amad27: We have the $y_n$’s converging to $u$, and $f$ is continuous, so the $f(y_n)$’s have to converge to $f(u)$. But all of the $f(y_n)$’s are $0$, so they converge to $0$, and $f(u)$ must be $0$. But then by definition $u\in A$.
13h
comment 2003 Putnam A-1 Help needed about sequences
... that this idea generalizes. A lot.
13h
comment 2003 Putnam A-1 Help needed about sequences
@Amad27: As an example, suppose that you want $3$ terms that meet the conditions and sum to $7$. Clearly $2+2+3$ works. Suppose that we try to start with $1$: the biggest the other two terms can be is $2$, and $1+2+2$ is only $5$, so that won’t work. And if we try to start with $3$, the smallest we can manage is $3+3+3=9$, so that won’t work, either. Thus, for $n=7$ and $k=3$ the only possible value of $a_1$ is $2$. Moreover, the other two numbers have to be $2+2$, $2+3$, or $3+3$, and only one of those adds up to the missing $5$, so $2+2+3$ is the only solution with $k=3$. It turns out ...
14h
comment Combinatorial Proof of Identity b_n
The upper limit of $\infty$ doesn’t matter: $\binom{n-1}k$ is non-zero only for $0\le k\le n-1$ anyway, so in effect the summation is from $1$ through $n-1$.
14h
answered 2003 Putnam A-1 Help needed about sequences
14h
comment 2003 Putnam A-1 Help needed about sequences
And $n=3$ has $3$ solutions: $1+1+1$, $1+2$, and $3$.
14h
comment Give me an idea
I believe that the question is as follows. You have an unlimited supply of cards, on each of which is a positive integer. Show that any set of $2015$ cards whose numbers sum to $4028$ can be divided into two sets, each of which sums to $2014$. Is this still true if $2015$ is replaced by $2014$? If fits the wording, the stated result is true, and the answer to the last question is no.
14h
comment Prove that $\sum_{n≥0} a_k(n)x^n = \frac{1-x}{1- 2x + x^{k+1}}$
@Karen: You’re welcome!
14h
comment Induction proof concerning Pell numbers
@Kamil: It can indeed; you're welcome.
22h
answered Prove that $\sum_{n≥0} a_k(n)x^n = \frac{1-x}{1- 2x + x^{k+1}}$
1d
comment Sum $(1-x)^n$ $\sum_{r=1}^n$ $r$ $n\choose r$ $(\frac{x}{1-x})^r$
@learnmore: I really am sorry. Good luck with the interview! Oh, I just had a thought: it’s possible that someone in chat might have some suggestions, if you’re not too uncomfortable asking there. (I’m going to make the moderators happy and delete my earlier comments in this thread.)
1d
comment Prove that the following statement is false
@Orland: You’re welcome.
1d
comment Prove that the following statement is false
@Orland: That’s right. And there’s the counterexample that you need to disprove the assertion.