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Mar
18
comment least common multiple $\lim\sqrt[n]{[1,2,\dotsc,n]}=e$
It was a pleasure to read this answer. It is truly a textbook quality work.
Mar
12
answered Two simple questions on quotient rings.
Mar
4
comment Infinitely many prime divisors of $f(a)$
@Greg Thanks, got it! That is a pretty slick proof.
Mar
4
comment Infinitely many prime divisors of $f(a)$
@Greg This sounds like a fun approach. So if $S=\{p_1, …, p_m\}$ is a finite set of primes, the number of integers in the range $\{1, 2, …, N\}$ all of whose prime factors come from $S$ would be at most $(\log_{2}N+1)^{m}$. But perhaps this is too weak of a bound. More interestingly, how do you bound the number of values of $f(x)$ that lie in the range $\{1, 2, …, N\}$? Intuitively, this should be a big proportion when $N$ is large with respect to the degree of $f$, but I am not sure how to make it precise.
Feb
14
accepted Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin
Feb
14
comment Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin
This is beautiful! Thanks so much Daniel
Feb
14
comment Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin
@DanielFischer: Apparently, this rational function works. But I do wanna hear the counter-example you have in mind as well :)
Feb
14
comment Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin
@Daniel: Right. I cannot come up with simple counter-examples. Is the counter-example you have mind a rational function of $x$ and $y$?
Feb
14
comment Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin
@Daniel Would something like $\frac{x^2+y^2}{x}$ work? It is not continuous at $(0, 0)$ but the limit exists along every straight line.
Feb
14
comment Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin
@DanielFischer: Thanks! That answers the question, namely that such a function $f$ does not exist (because in some sense $\mathbb{R^2}$ is too nice). I would be happy to accept your answer if you could say a few words about the proof (for $X=\mathbb{R}^2$ or more generally for $X$ locally path connected + first countable).
Feb
14
revised Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin
deleted 4 characters in body
Feb
14
comment Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin
@David: Ah good point. I have edited the question accordingly.
Feb
14
asked Example of a function that is not continuous at $(0, 0)$ but continuous when restricted to any curve approaching the origin
Feb
5
awarded  Notable Question
Feb
2
comment Liftings of curves $u\cdot v$ and $v\cdot u$ with respect to the sine covering map.
@ts375_zk26: Yes, $u\cdot v$ means the curve obtained by first tracing $u$ and then $v$.
Jan
26
comment Show that $\sum_{n=0}^\infty |a_n|<\infty.$
@RRL: I know $t_N$ is well-defined because the series it represents converges. Writing $\sum_{n=1}^{\infty} |a_n|$ inside $t_{N}$ is just algebraic manipulation. I am not assuming that $\sum_{n=1}^{\infty} |a_n|$ is finite there. I think there ought to be a better write-up of this approach that doesn't make it seem circular.
Jan
26
revised Show that $\sum_{n=0}^\infty |a_n|<\infty.$
corrected a mistake
Jan
26
comment Show that $\sum_{n=0}^\infty |a_n|<\infty.$
@RRL: I say $t_N$ is well-defined because we have proved it to be! I don't think there is a circular logic here, but perhaps my presentation is poor. Also, thanks for the catch. It should be $\sum (a_n - |a_n|) \leq \sum 2a_n$ because if $a_n$ is positive, then $a_n-|a_n|=0$, and if $a_n$ is negative, then $a_n-|a_n|=2a_n$. Thus, $\sum (a_n-|a_n|)\leq \sum 2a_n$. I will edit the answer accordingly.
Jan
26
comment Show that $\sum_{n=0}^\infty |a_n|<\infty.$
@EpsilonDelta I have now edited my answer. Thanks for the heads-up, and I hope the argument is simpler now.
Jan
26
revised Show that $\sum_{n=0}^\infty |a_n|<\infty.$
the argument is now simpler