38,340 reputation
177146
bio website
location Dubuque, IA
age
visits member for 3 years, 7 months
seen Dec 20 '13 at 3:42

As of August 2011 I am an assistant professor of mathematics at Loras College in Dubuque, Iowa, USA.


Apr
7
awarded  Nice Answer
Mar
31
awarded  Enlightened
Mar
31
awarded  Nice Answer
Mar
26
awarded  Good Answer
Mar
8
awarded  Nice Answer
Feb
24
awarded  functions
Jan
22
awarded  Enlightened
Jan
22
awarded  Nice Answer
Dec
4
awarded  Good Answer
Nov
30
comment How to prove the closed form $\left(\frac {1}{1-x}\right)^2 = \sum_{n=0}^{\infty}(n+1)x^n$
math.stackexchange.com/q/30732
Nov
30
comment Dimension of diagonal matrices
What was the result of your counting? Can you answer this when $n=1$, when $n=2$, or when $n=3$?
Nov
30
answered when convergence in measure implies convergence almost surely
Nov
30
comment Can't argue with success? Looking for “bad math” that “gets away with it”
Depending on your perspective, often such use is not simply (shockingly) wrong, just lacking in proper justification and rigor. If it works each time, there is something to it.
Nov
30
comment when convergence in measure implies convergence almost surely
Do you have a question?
Nov
30
comment Seeking for construction s.t. every intersection contains at least 3 lines
Interesting problem. I am curious, what led you to it? Although tagged (euclidean-geometry), it might help clarify the problem if you state explicitly that you are talking about lines in the Euclidean plane (if that is the case).
Nov
30
comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0 $ implies $f(x)$ has a limit in $0$?
(continued) @user83081: And if they don't, then all the terms are $0$.
Nov
30
comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0 $ implies $f(x)$ has a limit in $0$?
@user83081: (See also my comment on Ewan's answer.) To show that a given function $f$ satisfies the hypothesis means showing that if you start with arbitrary $a\in\mathbb R$, the sequence $f(a), f(a/2),f(a/3),\ldots$ converges to $0$. Pick any $a$ you want to, and it may or may not be the case that there exists some $n$ and some $k$ such that $1/e^k = a/n$ (again, there is no reason to assume $k=n$). Let your argument depend on whether or not such $n$ and $k$ exist. If they do, show that it happens only once, so all further terms in the sequence vanish.
Nov
30
comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0 $ implies $f(x)$ has a limit in $0$?
@user83081: In order to show that $f(a/n)$ converges to $0$ for each $a$, you must start with an arbitrary $a$, then consider the sequence $f(a), f(a/2), f(a/3),\ldots$. Ewan has given an example where you can give explicit reason why this sequence converges to $0$. You can break up your argument into $2$ cases: either $a$ is a rational multiple of $\dfrac{1}{\sqrt p}$ for some prime $p$, or it is not. The "$x$" appearing in the definition of the function is just for purposes of defining the function. Your question does not really make sense.
Nov
30
comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0 $ implies $f(x)$ has a limit in $0$?
@ParamanandSingh: Yes.
Nov
30
comment Is it true that "$\lim_{n \rightarrow \infty} f(\frac{a}{n}) = 0 $ implies $f(x)$ has a limit in $0$?
@user83081: I wonder if you might be overloading $n$ there. Taken literally, of course that would be the case when $a=n/e^n$. But then $f(a/k)=0$ when $k>n$, so there is no problem. More to the point, for every $a$, there is at most one $k$ such that there exists $n$ such that $a/n=1/e^k$, and therefore this $n$ is also unique, and thus $f(a/m)=0$ when $m>n$. This uses the fact that no power of $e$ is an integer.