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seen Nov 3 at 18:48

Nov
14
comment Is this language and its complement is Context free Language?
@BrianM.Scott: I know that - my first answer was quite messy and everything is now... yeah. Thanks for the remark.
Nov
14
comment LaTeX help - multiple answers using set bracket / cases
@Vafa: You are absolutely right and I suggest this to the OP, too. However we could argue that the OP wants to know about MathJaX ;)
Nov
14
comment LaTeX help - multiple answers using set bracket / cases
@Dexter: You're welcome. I have added a link to a good example in my answer. Please have a look.
Nov
13
comment Is there a subset of a non regular language that is regular
@user34790: Every finite language is regular because you can create a enumerate all elements extensionally. Furthermore there exists a regular expression of the form "word1 | word2 | ... | wordN" which implies the regularity of all finite languages.
Nov
13
comment Is there a subset of a non regular language that is regular
I can provide a cheat sheet from my "Theoretical Computer Science" course (in German). Just have a look at the top picture with all the boxes. From top to bottom (on thr right hand side) the descriptions are: all languages $\Sigma^*$, semi-decidable (Type 0), decidable, LOOP-decidable, NP, P, context free (Type 2), regular (Type 3), finite. Hope this might help :)
Nov
13
comment Is this language and its complement is Context free Language?
@Sudhir: Please have a look at my updated answer, I did a mistake before!
Nov
12
comment $\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}$
I did replace it with $\sqrt{x^2+2x+1}=\sqrt{(x+1)^2}=(x+1)$ thinking of sufficiently large $x$.
Nov
12
comment $\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}$
@icurays1: I think I need a break... fixed this one, too.
Nov
12
comment $\lim\limits_{x\to\infty}\frac{(x+7)^2\sqrt{x+2}}{7x^2\sqrt{x}-2x\sqrt{x}}$
@icurays1: That was just a typo, thanks for reminding me of that!
Nov
11
comment Show that a matched set of nodes forms a matroid
It will take some time to fully understand what you have written, i will come back in a few minutes if I have any questions - meanwhile you can have a look whether I made a mistake during the translation (you're from Berlin and probably know German ;) ) of the problem and confused everyone: i.imgur.com/sO8IU.png
Nov
11
comment Show that a matched set of nodes forms a matroid
I was now able to understand (1) and (2) but I have problems with the proof of (3). I was thinking about defining a basis $\mathcal{B}\in\mathcal{I}$ which contains all nodes of a perfect matching (which would be the entire set $V$), however this sounds incorrect because I think something would be missing here. Do you have any hint what I did misunderstand?
Nov
11
comment Show that a matched set of nodes forms a matroid
@joriki: I don't think that this is exactly what makes it difficult for me. We never had any exmaples in class because our prof just shows us some definitions without even explaining how to work with such structures. In the end this concept of a matroid is still a mystery to me and I don't know whether my thoughts on what to do suffice and furthermore I don't know how to proove this while having such abstract definitions without any imagination how my sets might look like.
Nov
10
comment Extend functions such that they are continuous
@martini: Oh again a typo... but nevertheless thanks.
Nov
10
comment Extend functions such that they are continuous
Oh typo. Fixed it now - may I ask you whether you can confirm my second result $b=4$ which I did determine the same way as $a$?
Nov
9
comment Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
@BrianM.Scott: Did you forgot to add a $1/2$ to the partial sum at the end? Nevertheless this looks more familiar and comprehensible and I will try working on that later on. - Thanks!
Nov
9
comment Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
@BrianM.Scott: We haven't even defined derivatives and integrals yet, currently attending "Calc. I" if you could say so. (Everything is different in Germany...) - However I still don't see how to determine the $1/4$ WolframAlpha has computed. By now I only know how to show, that my series converges.
Nov
9
comment Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
@BrianM.Scott: Looking up in my textbook the $p$-series is prooved (in a part we did not work through yet) with approximations $1/n^k\leq 1/n^2\leq 2/(n(n+1))$ for $k>1,k\in\mathbb{N}$ and the last term converges because the telescopic series converges. You claim, that my series I have to work with does not converge, however WolframAlpha claims that it converges to $1/4$ - am I missing something?
Nov
9
comment Convergence of semi-telescopic series $\sum\limits_{k=1}^\infty\frac{1}{k(k+1)(k+2)}$
We haven't discussed integrals yet in the context of limits and sequences and series so your hint does not help that much - so any easier way to show that? Furthermore I understand that I can show that the series will converge by your mentioned approximation, however I still don't know how to get the limit in the second step.
Nov
5
comment $\lim\limits_{n\to+\infty}\frac{2^{n^3}}{n!5^{n^2}-n^n}$ without special means
Oh I did some stupid mistakes - I have been able to get the desired result!
Nov
5
comment $\lim\limits_{n\to+\infty}\frac{2^{n^3}}{n!5^{n^2}-n^n}$ without special means
My results are always converging to 0 like $1/n!$ or $1/(5^{n^2}-1)$. Do you know what I might have done wrong?