319 reputation
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age 19
visits member for 1 year, 11 months
seen Jul 14 at 4:38

Dec
8
asked Proving that $(1+\frac{x}{n})^n\to e^x$?
Dec
8
comment Partial limits of sequences
@yoyo for a private case or the general case, how would you prove this is indeed the number of partial limits? (sorry for coming back to such an old question, but I got it as related when about to ask a similar question)
Dec
5
comment $\mathbb{R}\subset\mathbb{F}\subset\mathbb{C}$ implies $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$?
hmm.. Interesting. But I'm not sure how to conclude $i\in\mathbb{F}$ from that. mind clarifying?
Dec
5
accepted $\mathbb{R}\subset\mathbb{F}\subset\mathbb{C}$ implies $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$?
Dec
5
comment $\mathbb{R}\subset\mathbb{F}\subset\mathbb{C}$ implies $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$?
Oh yeah... You reminded me that we did prove that fields are vector spaces above subfields of themselves. This really does make it a lot simpler. Thank you!
Dec
5
asked $\mathbb{R}\subset\mathbb{F}\subset\mathbb{C}$ implies $\mathbb{F}=\mathbb{R}$ or $\mathbb{F}=\mathbb{C}$?
Dec
1
awarded  Enthusiast
Nov
28
comment $a_{n+1}=\frac{1}{4}+a_n^2$ is converging and have a limit
Thank you! That's a nice trick i didn't know... Too bad classes here doesn't bother to teach you how to solve the exercises you get...
Nov
28
accepted $a_{n+1}=\frac{1}{4}+a_n^2$ is converging and have a limit
Nov
28
asked $a_{n+1}=\frac{1}{4}+a_n^2$ is converging and have a limit
Nov
25
revised Proving that if $a_n\in\mathbb{Z}$ for all $n$ then it's limit is also in $\mathbb{Z}$?
added 224 characters in body
Nov
25
asked Proving that if $a_n\in\mathbb{Z}$ for all $n$ then it's limit is also in $\mathbb{Z}$?
Nov
21
comment Proving $0$ is the limit of $\frac{n}{2^n-1}$ from definition alone
Well, the second way is pretty similar to Pambos's idea, but I really don't feel comfortable with logs. Is it something you expect I'd be introduced to or should I make an effort learn it on my own?
Nov
21
accepted Proving $0$ is the limit of $\frac{n}{2^n-1}$ from definition alone
Nov
21
comment Proving a vector space over itself have no subspaces
In the end, after actually understanding how to use the dimensions, I found the other way simpler. But thanks for the help!
Nov
21
accepted Proving a vector space over itself have no subspaces
Nov
20
comment Proving $0$ is the limit of $\frac{n}{2^n-1}$ from definition alone
Both answers say to use logs, but as of now we never had to use stuff that wasn't thought (or at least shown) in the course apart from basic algebra. Do you think logs enter this definition? Since I most definitly remeber nothing about them from high school...
Nov
20
asked Proving $0$ is the limit of $\frac{n}{2^n-1}$ from definition alone
Nov
19
comment Proving a vector space over itself have no subspaces
Thanks for the edits! English is not my mother language so translating is sometimes difficult for me.
Nov
19
comment Proving a vector space over itself have no subspaces
Ok, that did help me. I now know that if $1_F\in U$ it can't be a vector subspace. But I'm not sure what happens if $1_F\notin U$, since the axioms for vector spaces don't require a vector identity element, right?