Dork Scratchings

If the sky were covered in moons it'd be almost a bright as day! Walking to the pub through the Yorkshire Dales on a particularly brilliant moonlit evening, when everything was clearly visible, I wondered just how much less light there was than during daytime?  It turns out to be quite easy to estimate an upper bound - all you need to do is measure the angle subtended by the moon! Let $A_e$ and $A_m$ be the cross sectional areas of the Earth and moon, $d$ be the distance to the moon, and $r$ be the distance to the Sun.  Now, suppose the Sun releases some energy $E_s$ then the amounts $E_{se}$ and $E_{sm}$ which land on the Earth and the moon are given by: $$ \begin{align} E_{se} &= \frac{A_e E_s}{4\pi r^2}\\ E_{sm} &\approx \frac{A_m E_s}{4\pi r^2} \end{align} $$ On a full moon, let's assume for the sake of calculating an upper bound that the moon reflects all of $E_{sm}$ equally in all hemispheric directions.  Then the energy reflected to the Earth is giv

Theorem of the week This week's theorem of the week is the fundamental theorem of algebra , and the picture is the proof! Theorem Every non degree-zero polynomial $p(z) = a_nz^n + ...+ a_1z +  a_0$ has a root in $\mathbb{C}$. Picture proof To see how the picture proves this, write $z$ as $Re^{i\theta}$, then for all $k$ $$ z^k = R^ke^{k i \theta} $$ So for sufficiently large $R$ the $a_nz^n$ term dwarfs all the others and so the image of $\{z\in\mathbb{C}: \lvert z \rvert = R\}$ must go around the origin $n$ times, like the rubber band in the photo.  But when $R= 0$ the image is just $\{a_0\}$ which goes around the origin zero times.  So, for some $0 < r < R$ the image of $\{z\in\mathbb{C}: \lvert z \rvert = r\}$ must cross the origin.  QED. Less handwavy proof In order to obtain a contradiction assume $p(z)$ has no zeros.  Then $\frac{z^{n-1}}{p(z)}$ is everywhere differentiable, which in turn means that its closed loop integrals are zero$^{(\dagger)}$.