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Jan
2
comment Derivative of the function $(x)!$.
the $n^{\rm{th}}$ derivative of $x^n$ is $n!$, which is a natural number, it's not a function in $x$
Dec
29
comment Any transformation that commutes with a transformation commuting with $S$ must be a polynomial in $S$
That makes sense.. Had troubles using the fact S is diagonalizable, but this makes it doable for me.
Dec
29
comment Any transformation that commutes with a transformation commuting with $S$ must be a polynomial in $S$
@orangeskid love to see a proof of the general case. As I'm still shakey on this specific case don't think I'll be able to prove it myself
Dec
29
comment Any transformation that commutes with a transformation commuting with $S$ must be a polynomial in $S$
@Omnomnomnom didn't know that. Thanks. English is not my native language
Dec
21
comment All the ternary n-words with an even sum of digits and a zero.
About your addition: This is exactly what I did originally, but what about even strings of lenght n-1 with zeroes? you can add 1 to then, so I also need $f(n) $ in the definition of f-bar, which puts me in a loop
Dec
21
comment All the ternary n-words with an even sum of digits and a zero.
got it specifically for n=1 with starting condition f(0)=1 because of the 2. And yeah.. f-bar is wrong. always confused by these...
Dec
5
comment Finding suitable basis for a free abelian finitely generated group.
It looks like I am taking the exact same course exactly a year later, as the question fits the one appearing on my homework, no one explained anything about Smith normal form to us, and I have no idea what do to.. (also your name fits the area :P). Let's hope the single answer here will help me get on my way...
Dec
5
comment Finding the function of the power series $\sum\limits _{n=1}^{\infty}\frac{x^{2n+1}}{n}$
That's a lot simpler than my way...
Dec
5
comment Finding the function of the power series $\sum\limits _{n=1}^{\infty}\frac{x^{2n+1}}{n}$
@r9m yep, you are right...
Nov
28
comment Proving differentiability on every interval equals proving differentiability on all of $\mathbb{R}$?
Well, that should have been pretty obvious to me looking back at it.. Thanks for clearing it all up anyway
Nov
28
comment Proving differentiability on every interval equals proving differentiability on all of $\mathbb{R}$?
@JohnMa don't really think I am.. The theorem in Wikipedia doesn't mention anything on a bound for any closed interval. Would it have any affect on all of $\mathbb{R}$ to have such a bound? en.wikipedia.org/wiki/Uniform_convergence#To_differentiability
Nov
24
comment If $\lim\limits _{x\to\infty}\left(\log f\right)'\left(x\right)$ is negative, then $\int_{0}^{\infty}f\left(x\right)dx $ converges?
@Did I didn't really understand your second comment though, but I did take the first one somewhere, and I'd love to know if my answer is correct :P
Nov
24
comment If $\lim\limits _{x\to\infty}\left(\log f\right)'\left(x\right)$ is negative, then $\int_{0}^{\infty}f\left(x\right)dx $ converges?
@Did Thinking about it so far, still not sure.. Definitely feels like I'm missing something simple though..
Nov
19
comment Powers of linear transformation and minimal polynomial
@MarcvanLeeuwen well, I give my reasoning in the question. Setting $T^4$ in the minimal polynomial of $T$ gives you zero, so the minimal polynomial of $T^4$ must divide it...
Nov
19
comment Powers of linear transformation and minimal polynomial
Well took me a while, but I understand everything. Thank you
Nov
19
comment Powers of linear transformation and minimal polynomial
Also, I'm not quite sure about the relation $T^14=T^13$. This is obviously true if you look only on the generalized eigenspace of $0$, but what makes it hold true for all of $V$?
Nov
19
comment Powers of linear transformation and minimal polynomial
Unfortunately I did not understand the answer. How from $(T^4)^5-(T^4)^4=0$ did you get to the minimal polynomial?
Nov
10
comment Union of conjugates of a subgroup of a finitely generated group.
@Derek Holt Oh I see. That's interesting. Though guess I'll need to come back to it aftrr I've studied the subject for a bit longer..
Nov
6
comment If $H<G$ is of finite index, and for some $x\in G$, $xHx^{-1}\subset H$, prove that $xHx^{-1}=H$
I see.. Thanks!
Nov
6
comment If $H<G$ is of finite index, and for some $x\in G$, $xHx^{-1}\subset H$, prove that $xHx^{-1}=H$
Why does the second statement, about $x^m$ being in $H$ for some $m$, follow from the finite index of $H$?