talk.tryitonline.net

For general discussion and feature requests regarding tryitonl...
Dec 22, 2020 06:06
Thanks!
Dec 22, 2020 06:01
Oh wow
Dec 22, 2020 05:52
@Bubbler Could you see this?
Dec 22, 2020 05:52
Dec 22, 2020 05:48
(Actually I have tried it once)
Dec 22, 2020 05:48
wait I'll do that right away
Dec 22, 2020 05:48
okay
Dec 22, 2020 05:25
I just want to do "SELECT * FROM topics LIMIT 10"
Dec 22, 2020 05:25
How do I use "Input"?
Dec 22, 2020 05:24
Hello I have a question about using TIO SQLite.
 

 Mathematics

Associated with Math.SE; for both general discussion & math qu...
Jun 24, 2018 09:58
Ill leave for dinner. Plz tag me when you reply!
Jun 24, 2018 09:37
Hello? Nobody?
Jun 24, 2018 09:34
I'm studying with Evan's PDE (which is famous) in chapter calculus of variations. But it is too boring. So is there any good stuff to read about calculus of variations?
Jun 24, 2018 09:34
Hey everyone Hi.
Jun 8, 2017 16:38
math.stackexchange.com/q/2314906/348589 This question looks more HW but I got +1 lol
Jun 8, 2017 16:31
@TedShifrin Thanks for your suggestion. Is it not allowed or not suggested to ask true or false questions? It could be a big deal to solve something because it can give someone whether this direction is right.
Jun 8, 2017 16:29
@TedShifrin Definitely its not related to my HW. It comes from my head and trial and error.
Jun 8, 2017 16:27
@TedShifrin Ohh..
Jun 8, 2017 16:25
But it is well answered and nothing missed.
Jun 8, 2017 16:25
Saying : "This question is missing context or other details"
Jun 8, 2017 16:24
That is put on hold.
Jun 8, 2017 16:24
math.stackexchange.com/q/2313791/348589 This question is what I asked.
Jun 8, 2017 16:24
I have a question about MSE
Jun 8, 2017 16:24
Hi there
Mar 29, 2017 14:47
Check this for me.
Mar 29, 2017 14:47
0
Q: Show the following two growth orders are same.

kayak Let $m(n)=\min\{m:2^{-m}m^{-3/2}\leq n^{-1}\}$. Then $$ 2^{m(n)}\sim\frac{n}{\log_{2}n}. $$ This problem is a piece of solving : An unfair "fair game."

Mar 29, 2017 14:33
I have short
Mar 29, 2017 14:27
Plz put bounty on this.
Mar 29, 2017 14:26
4
Q: An unfair "fair game."

User112358This is problem 2.2.8 from Durrett's Probability Theory and Examples 4th edition, I am using the version of this book that can be found on his website. Let $p_k=\frac{1}{2^k k (k+1)}, \ k=1,2,\dots$ and $p_0=1-\sum_{k\geq 1}p_k.$ $$ \sum\limits_{k=1}^\infty 2^k p_k = \left(1-\frac{1}{2}\rig...

Mar 29, 2017 14:26
Guys
Feb 26, 2017 13:38
Hey @DHMO Long time no see
Jan 4, 2017 11:40
@DHMO Hi!
Jan 4, 2017 01:35
Please do not tell anyone where I'm from. @DHMO
Jan 3, 2017 13:49
@DHMO
Jan 3, 2017 13:49
Oh god.
Jan 3, 2017 04:30
@user21820 It says it is because site association bonus
Jan 3, 2017 04:29
@user21820 I don't trust them.
Jan 3, 2017 04:29
@JessyCat because I want more
 
Mar 29, 2017 15:21
It seems clear but I want to see the details.
Mar 29, 2017 15:20
how did you get the first line of your answer?
Mar 29, 2017 15:20
Then
Mar 29, 2017 15:19
(lol)
Mar 29, 2017 15:19
Please say you are sure.
Mar 29, 2017 15:19
Ahhh
Mar 29, 2017 15:15
What do you mean?
Mar 29, 2017 15:15
Seems like a bug of sorts?
Mar 29, 2017 15:15
Hi.
Mar 29, 2017 15:15
From your final line, $$ \lim_{n\rightarrow\infty}\frac{2^{m(n)}\log n}{n}=\lim_{n\rightarrow\infty}\frac{m+\frac{3}{2}\log m}{m^{3/2}}. $$ Then it converges to 0. So this is not what I want to get.
Mar 29, 2017 15:15
I got to your final line. But how does that imply the answer of my question?
Mar 29, 2017 15:15
Sorry but could you check my question again? I edited it just before you answer.