@Yong Once again many thanks! I went along and checked as you suggested, and I saw that for n>4, 2n-2 and 2n-3 have odd degrees (along with 4 and 5), thus there are two at the very least two trails needed to cover all the vertexes, thus there is no eulerian trail for n>4, only for n=4. Yet I feel very unsure about it, as if I miss something there.. is it so or am I just paranoid?
@Charlie The "spinning bit" represents an artistic depiction of quantum coherence, wherein a qubit attains the classically forbidden superposition of both logical value zero and logical value one, concurrently.
@user126885: Sorry I disappeared earlier. My connection screwed up. But I saw you had asked the question separately so I recommend you pursue a solution there. And you can always use this theorem to prove the limit of x^n is equal to that of x around 0. math.lsa.umich.edu/courses/185F08/composition.pdf
@fpqc I am not taking this conversation seriously since you're deleting half of it... I just argued hueristically. I figured $R^\ast$ has cardinality smaller than or equal to $K^\ast$, while $K^\ast$ is easily to characterize.
@KarlKronenfeld I think that once they get past victory road it will be 1-3 days. Grinding will be very fast when they're at E4. But I can't estimate victory road. The hard part here is puzzles, not battles.
Show that this function, defined in $(1,+\infty)$, is continuous on every compact subset of that ray, and conclude thus it must be continuous over that said say.
@Pedro Okay, so the problem is to prove that $\zeta'(s)$ is continuous on every compact subset of $(1, \infty)$ and thus is continuous on that interval whereas $\zeta'$ is the $\Re[s] > 1$ counterpart, i.e., the p-series.
@Mike Already knows the solution I gave. So I'll leave now. Hint Every compact $K\subseteq (1,+\infty)$ is bounded away from $1$. Consider $$\sum_{n\geqslant 1}\frac{\log n}{n^{1+\varepsilon}}$$ and invoke your friend Weiertrass.
@PedroTamaroff The only thing you are going to get is either "Lindemann-Weiestrass" or "Weierstrass–Casorati theorem" by saying "Weierstrass" to me. =D.
@Mike $L$ be the galois closure of the extension $\Bbb Q(\alpha)$ over $\Bbb Q$. Prove that for any prime $p$ s.t. $p$ divides $|\text{Gal}(L/\Bbb Q)|$ there exists a subfield $F$ of $L$ with degree $p$ and $L = F(\alpha)$
I think you first break down a list: text, number, text, number,.. then you need something special to handle text-only strings, which applies to all text units
@Alex That formula holds generally for linear subspaces, $\dim A + \dim B = \dim (A+B) + \dim (A\cap B)$. That the subspaces here are the ranges of two linear operators is irrelevant.
@Mike $L$ be the galois closure of the extension $\Bbb Q(\alpha)$ over $\Bbb Q$. Prove that for any prime $p$ s.t. $p$ divides $|\text{Gal}(L/\Bbb Q)|$ there exists a subfield $F$ of $L$ with degree $p$ and $L = F(\alpha)$
