Jackson Walters
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 Jul 10 awarded Yearling Sep 30 awarded Explainer Jul 10 awarded Yearling Jul 2 awarded Curious Jul 11 answered Infinity and structures Jul 11 answered Finding the value of $\sqrt{1+2\sqrt{2+3\sqrt{3+4\sqrt{4+5\sqrt{5+\dots}}}}}$ Jul 10 awarded Yearling Jun 25 comment Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$ If $n=1,2,3,6$, $s_{\pm}=\sqrt{2n\pm 1}$, and $\eta_{\pm} \in \mathbb{Z}[s_{\pm}]$ are fundamental units, respectively, then the integral $$I_{n}=\frac{\pi^{2}}{12}\left(n-s_{+}s_{-}\right)+\log(\eta_{+})\log(\eta_{-})‌​+\log(2)\log(s_{+}+s_{-}).$$ Numerically only works for those $n$. Jun 25 comment Evaluate $\int_0^1 \frac{\log \left( 1+x^{2+\sqrt{3}}\right)}{1+x}\mathrm dx$ Looks like if $n$ is such that $2n-1$ and $2n+1$ are twin primes, $\int_{0}^{1} \frac{\log(1+x^{2n+\sqrt{4n^{2}-1}})}{1+x}dx$ should be $$\frac{\pi^{2}}{12}\left(n-\sqrt{4n^{2}-1}\right)+\log(a+b\sqrt{2n-1})\log(c+d \sqrt{2n+1})+\log(2)\log(\sqrt{2n-1}+\sqrt{2n+1})$$ for some $a,b,c,d$. Jun 17 comment Does there exist rational $a,b,c$, such that $\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$ Yes, $\Delta_{K}$ is the discriminant of the number field $K$. I think the question is really driving at the linear independence of radicals for the case $n=3$, from which it would follow that $(1,2,4)$ is the unique solution (see this question: math.stackexchange.com/questions/158722/…). To actually prove the linear independence you need to circle back to field theory and Galois theory, which I am only just beginning to learn about. Jun 17 answered Does there exist rational $a,b,c$, such that $\sqrt[3]{1}+\sqrt[3]{2}+\sqrt[3]{4}=\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}$ Jun 10 comment If $L$ commutes with all isomorphisms of $\mathbb{F}_2^n$, is $L=\lambda I$ for some $\lambda$? You're looking for the center of the group $GL(n,\mathbb{F}_{2})$, which is indeed scalar multiplies of the identity: groupprops.subwiki.org/wiki/… Jun 9 comment Computing limit of $(1+1/n^2)^n$ @PJMiller He raised both sides to the $n$, so $\exp(\frac{1}{n^{2}})^{n}=\exp(\frac{1}{n})$. Jun 7 revised Closed form for the series $\sum\limits_{n=0}^\infty \frac{\exp(\cos(n))}{n!}$ deleted 2 characters in body Jun 7 answered Closed form for the series $\sum\limits_{n=0}^\infty \frac{\exp(\cos(n))}{n!}$ May 9 awarded Caucus Apr 21 revised Trying to extend Vandermonde's formula to the case where m and n are not integers TeXified everything. Apr 21 suggested approved edit on Trying to extend Vandermonde's formula to the case where m and n are not integers Apr 11 accepted $x \in \left[L,L\right] \Rightarrow tr(ad \, x)=0$ Apr 3 comment $x \in \left[L,L\right] \Rightarrow tr(ad \, x)=0$ Why do we get to assume the Lie bracket is a commutator? Surely there are more general brackets.