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Mar
23
comment Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$
hmm I understand that, still not sure how to use it though :P
Mar
23
comment Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$
Well that's a really nice way of doing it. Didn't think about breaking the number into easier to manage numbers ><
Mar
23
comment Finding $\sup$ of $\{\sqrt{n} - \lfloor\sqrt{n}\rfloor | n\in\mathbb{N}\}$
We haven't actually learned limits yet so I'm not sure how to use this, but it seems to be going to 1? no idea how to prove it though.
Mar
16
comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$
Another question, if you don't mind. What was the sign that the original inequality wasn't strong enough for the induction? Is me getting stuck there it or was there something else I need to notice?
Mar
16
comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$
hmm, all clear now, though I'm still pretty sure I never would have figured it out myself. Thank you for the help!
Mar
16
comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$
Oh yeah I see it, I imagined the reversed order :P. Thank you very much!
Mar
16
comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$
I don't understand some of the steps actually, mainly the first (adding the square root of j to the denominator) and the last (going from the sum to n), but it looks pretty neat :)
Mar
16
comment Proving $2(\sqrt{n} - 1) < \sum\limits_{i=1}^n\frac{1}{\sqrt i}$
I'm looking at this and it seems pretty much like what I did, though obviously with \sqrt{n+1} instead of \sqrt{n}, but your induction step gave the elements \sqrt{k+1}-\sqrt{k} in the reverse order compared to mine. Even knowing that I still can't find what I did wrong...
Dec
23
comment limits calculus
This is about the same material where I'm at and I found the following to be a great source, examples are much better and more diversified than those given in class and you can really understand how they got to the solutions. math.ucdavis.edu/~kouba/CalcOneDIRECTORY/preciselimdirectory/…
Dec
23
comment Why does the definition of limits of a function have strict inequality?
Yeah about that part I know, every book and class did even explain it (rather then just giving it as is)
Dec
23
comment Why does the definition of limits of a function have strict inequality?
I actually just commented asking how exactly would you prove it, since I got into a bit of complications of getting rid of the cases of $|x-a|\leq \delta$ and $|f(x)-a|\leq \epsilon$ in the each direction respectively. This clarifies everything nicely.
Dec
23
comment finding a limit of a function by definition
Both solutions really helped me. From what you gave me I was able to realize what my delta needs to be, and the solution below allowed me to understand how I'm supposed to write a proof on these matter. Thank you both!
Dec
16
comment Proving that $m^p-m$ is divisible by $p$
A lot of concepts here I'm not familiar with...
Dec
9
comment Limit of $\left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n}$, is the following true?
lol yeah... that's right... So I guess it's wrong that $\lim \left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right)^{n} = (\lim \left(\frac{n^{2}+8n-1}{n^{2}-4n-5}\right))^n$ huh?... Oh after thinking about this for a while (using the answer given after I started typing as well) I think I see why it's wrong. Thanks!
Dec
8
comment Proving that $(1+\frac{x}{n})^n\to e^x$?
hmm that was simple enough. Thank you!
Dec
8
comment Proving that $(1+\frac{x}{n})^n\to e^x$?
Yeah I do. Though I mentioned it, my bad :p
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
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!
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...