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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 $H$?
Oct
28
comment Show that if $E\subseteq F$ is a subfield and $f,g\in E[x]$ then $\gcd(f,g)$ (relative to $F$) is in $E$
@Wojowu Tried to think about it, but not sure how to use it... I mean I can just theoretically talk about the reminder at the end, so I assume I need to show that the algorithm gives the same result no matter the field, but I'm not actually sure how to show it...
Oct
23
comment Integrating $\frac{\ln{ax}}{x\ln{bx}}$
@JackD'Aurizio well... that was stupid of me...
Jul
23
comment Fundamental Theorem of Calculus and the left endpoint of the interval.
Well, that makes sense. Thank you
Jul
23
comment Fundamental Theorem of Calculus and the left endpoint of the interval.
He does mention that the value of $b$ can be negative, but that is mentioned as notation for the integral itself, and here in the definitions we have actual intervals where it makes no sense to have $b<a$ (or at least it wasn't defined)
Jun
30
comment A negative third derivative implies a positive first derivative at a point.
@Tryss Well I haven't learned about integration yet, so I don't know what the last part tells me :P
Jun
30
comment A negative third derivative implies a positive first derivative at a point.
@MattSamuel I wrote the last derivative using Lagrange's remainder, which is why it's $c$ instead of $x$. Added clarification. Is it wrong?
Jun
30
comment A negative third derivative implies a positive first derivative at a point.
@HagenvonEitzen Why must $f'$ average around 0?
Jun
29
comment Calculating $\lim_{x\to\infty} (\sin\frac{1}{x}+\cos\frac{1}{x})^x$ without l'Hopital
@anomaly I agree, but I'm preparing for a test, and the instructions say "without l'Hopital's rule"
Jun
29
comment Proving that $\ln ^3|x|=x$ has exactly 3 real solutions
@NikolajK Correct me if I'm wrong, but I am supposed to accept an answer once I understood the solution to my question, am I not? I really wouldn't mind leaving questions open for longer but it does seem like I'm supposed to accept it when I solved it to prevent people putting in effort for nothing instead of answering still open questions.
Jun
29
comment Proving that $\ln ^3|x|=x$ has exactly 3 real solutions
That is exactly what I got to at the end, but thanks for the confirmation :)
Jun
29
comment Proving that $\ln ^3|x|=x$ has exactly 3 real solutions
Actually just the fact that the function must be negative on $(0,1)$ is the obvious thing I was missing ><. Thank you