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 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 $b<. Thank you Jun 29 comment Proving that$\ln ^3|x|=x$has exactly 3 real solutions @ClementC. I actually tried that, from the second derivative I found out the minimum and maximum of$f'$Now I could go to a third derivative as it is already pretty simple, but it's so far I have no idea how to use it in my original problem Jun 17 comment For any closed set$A$of$\mathbb R$, does there exists a function$f:\mathbb R \to \mathbb R$such that,$f$is discontinuous exactly on$A$? That what happens when I don't think... Removed Jun 17 comment For any closed set$A$of$\mathbb R$, does there exists a function$f:\mathbb R \to \mathbb R$such that,$f$is discontinuous exactly on$A$? What about$f(x)=0$for$x\in\mathbb{R}\setminus A$and$f(x)=1$if$x\in\mathbb{Q}\cap{A}$and$f(x)=2$if$x\in\mathbb{I}\cap{A}$. It will be continuous at each point$x_0$outside of$A$as it will be$0$at an environment of$x_0$(As$\mathbb{R}\setminus A$is open), and obviously not continuous at$A$. Jun 11 comment Confused with$x$and$a$of Taylor Series @user247433$f(x) $would stay the same, the Taylor polynomial is equal to the function (as long as it converges to it). Yves gave a good answer by showing you the power series of$sin$at a general$a$. Notice that (alao as Yves stated) the power series converges for any$a\in \mathbb{R} $thus$sin(x) =T_a(x) $for any$a$and any$x$Jun 11 comment Confused with$x$and$a$of Taylor Series You can take any origin you want as long as the the power series around it converges to the function. People usually expand around 0 as it's the easiest to evaluate. If you want to expand around 1, for example, you need to be able to calculate$sin(1)$and$cos(1)$Jun 9 comment Using second derivative to find a bound for the first derivative @TedShifrin That's a fair point, changed it. About the maximum, the hint I was given was to use the minimum, when I wasn't able to use that I tried thinking of the maximum, but even if I do, everything else is pretty much the same, and in the given expansion I'm not sure how to prove it's smaller than$\frac{1}{2}$Jun 9 comment Where did I go wrong in this limit? I think the first answer here math.stackexchange.com/questions/46065/… is relevant Jun 9 comment Evaluate$\lim \limits_{n\to \infty }\sin^2 (\pi \sqrt{(n!)^2-(n!)})$Seems like${\pi}\over{2}$to me as well Jun 7 comment Proving the inequality$2^{-1+\frac{1}{n}}\left(x+1\right)\leq\left(x^{n}+1\right)^{\frac{1}{n}}$for$n\in\mathbb{N} $and$x>0\$ Well that does it...