Dork Scratchings

Theorem of the week: The Hairy Ball Theorem This says you can't comb a hairy ball without introducing discontinuities such as partings or whorls, unless there's a bald spot.  (There's a more mathematical statement below under the heading "theorem".)  The proof is from An Extremely Short Proof of the Hairy Ball Theorem, by P McGrath , but I've put it into my own words, completely removed all maths notation, and added pictures to make it as accessible as possible.  In addition to being extremely short, it's extremely elegant, and somewhat reminiscent of the Ham Sandwich Theorem . Theorem It is not possible to impose a continuous vector field onto a sphere, such that the vectors are all tangential to the surface, unless the field is zero somewhere Proof Let's assume the sphere does have a continuous, tangential, everywhere non-zero vector field, and attempt to derive a contradiction. Draw a small circle around a point p.  Do one lap around...

Yay! Clickbait for nerds! What's this?  Read on to find out!

I've been wondering: is it possible to visualize the rules of calculus? 1. Integration by parts To simplify things, in the above visualization $v(x_1) = 0$, so that $\int^{x'}_{x_1}\frac{dv}{dx}dx = v(x')$.  This means that $v(x_2)$ is simply the area of the cross-section facing us on the LHS, and the volume  on the LHS is $u(x_2)v(x_2)$. The RHS shows the same volume split into two parts.  The rectangle embedded in the first is $u\frac{dv}{dx}$, and so its volume is $\int^{x_2}_{x_1}u\frac{dv}{dx}dx$.  The cross section of the 2nd part is $v(x)$, so its volume is $\int^{u_2}_{u_1}vdu$, which becomes $\int^{x_2}_{x_1}v\frac{du}{dx}dx$ when rewritten as an integral over $x$. So the picture is visual proof that the equation in pink holds, provided that $v(x_1) = 0$.  To prove it in general we just need to check that when we replace $v$ with $v+v_1$ in the equation, both sides change by the same amount. 2. The chain rule This picture demonstr...

Connecting Geometry and Algebra Matrix determinants have the following strange definition that seems to have been pulled out of thin air: $$ det(A) = \sum_{\sigma}{sign(\sigma)a_{1\sigma(1)}...a_{n\sigma(n)}} $$ where $A = (a_{ij})$ is a real $n\times n$ matrix $\sigma$ ranges over all permutations of $\{1,...,n\}$ $sign(\sigma)$ is $+1$ if $\sigma$ is a product of an even number of transpositions and $-1$ otherwise$^\dagger$ However, in the geometric world the definition is far more intuitive: The determinant of A is the volume of the parallelepiped formed by its columns, multiplied by minus one if these have the opposite handedness to the unit vectors. Why are these two definitions the same?  To begin to answer this we need to first define elementary matrices and then show that every square matrix can be written as a product of these. Definition The elementary matrices are $E_{i, j}$ for $1 \le i,j \le n$ $E_{i,\lambda}$ for every real $\lambda$ and $...

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 ...

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...

A few Christmases ago I was playing with a toy xylophone meant for my nephew when I noticed something odd.  I expected going up an octave to halve the bar length, but it didn't.  A quick measurement confirmed my suspicion: to go up one octave you divide by $\sqrt{2}$ instead of by $2$.  I remembered this earlier this week while at my son's school concert, and decided to see if I could work out why! Strings First a bit of background.  Why did I expect halving the length to result in a note one octave higher (i.e. double the frequency)?  The answer is because I'd learned that this was the case with stringed instruments.  Assume that the tension is a constant $T$, and let $z(y,t)$ be the vertical displacement of the string at position $y$ and time $t$.  Then the upward force on a small element of  size $\delta y$ is approximately $$ T \left(\left. \frac{\partial z}{\partial y}\right|_{y+\delta y} - \left.\frac{\partial z}{\partial y}\right|_y...

Imagine you were a hamster in a hamster ball living on a hilly surface.  Base camp is surrounded on all sides by high summits, but you have a powerful catapult there that can fire you to the top of any of them.  Once fired you can influence your trajectory, but it's hard work and you don't have much energy in your little legs.  Suppose you know where you want to end up beyond the hills.  What's the best strategy for getting there?

How much can you make the top Jenga brick overhang the base by stacking them together? Surprisingly, you can go as far as you want. Suppose your bricks are length $l$ and you have one brick (not including the base).  Obviously you can overhang by $\frac{l}{2}$ without the centre of mass being unsupported.  What if you have two?  Now you have two conditions The centre of mass of the top brick is supported The centre of mass of the top 2 bricks are supported A quick calculation gives us that if the top brick is displaced (relative to the one below) by $\frac{l}{2}$ then the one below could be displaced by at most $\frac{l}{4}$ (relative to the one below it). Now suppose you have $n$ bricks (not including the base), then  you have $n$ conditions.  Let's guess the answer based on the result for $n=2$ and let's set $d_k = \frac{l}{2}\frac{1}{k}$ where $d_k$ is the displacement relative to the brick below and $k$ is the brick number starting at the top....

When the 2nd law of thermodynamics was discovered a lot of theorists were understandably suspicious.  It said that entropy always increases with time, or $\frac{dS}{dt} > 0$, but this seems to go against time reversal symmetry, which is a feature of all physical laws. (Okay... in QM you also have to reverse charge and parity in the system too....)  Amongst those who challenged the new theory was James Clerk Maxwell (he who shed light on light) and he did so with a thought experiment.  Imagine you have a box full of air at a given temperature, and it is divided into two parts.  Between the two is a little door which can be opened and shut by a daemon.  This daemon watches the air molecules speeding towards the door and if the molecule is faster than average and on the LHS he opens the door briefly to let it pass through to the right hand side.  Conversely if the molecule is slower than average and on the RHS the daemon will open the door to let is...

A proof that the net force due to pressure on a (fully or partially) submerged object is equal to the weight of the water displaced. The facts/approximations used were pressure is isotropic density is constant gravitational force is constant And the method of proof was to apply the Divergence Theorem. POSTSCRIPT I was just wondering why pressure is isotropic when I found this post and  I realised that the scratchings above actually prove it!  Imagine a blob of water submerged in water: the above proof shows that the net force on it is zero, but only if pressure is isotropic !

I was never really that interested in economics until after 2008 when suddenly technical terms like structural deficit started appearing in the news.  Oddly, with anything relating to macro economics, journalists have to pretend they were born understanding all the concepts perfectly and they're not going to patronize you by explaining them.  This is in stark contrast to anything related science where they have to pretend to understand even less than they do!  Somehow this attitude has leaked out to the wider world, so that friends and colleagues down the pub - or politicians on Question time - will b******t eternally about the effect of interest rate rises, but happily or even boastfully admit to knowing nothing about how the rest of the universe works.  I, on the other hand, knew I didn't know anything about economics, but thought it was less important than all the other stuff I didn't know.  But when economics stories started to become the main content of th...

In More Mathematical Puzzles and Diversions , Martin Gardner makes a passing reference to the Ham Sandwich Theorem.  It goes like this Any 3 shapes in 3 dimensional space can be simultaneously bisected by a single plane So imagine you have two roughly cut pieces of bread and a slice of ham, then you can always cut the sandwich in half such that each half has exactly half of each piece of bread and half of the ham, no matter how roughly strewn the pieces are. According to Gardner the generalized version has been proved by Tukey and Stone: any n shapes in $\mathbb{R}^n$ can be simultaneously halved by a single $n-1$ dimensional hyperplane.  But I thought I'd have a go at proving it myself in the 3D case, just for kicks. First observe that there are at least enough degrees of freedom to make it not impossible .  A plane (other than one going through the origin) can be described by the equation $k_xx+k_yy+k_zz = 1$ for some $\boldsymbol{k} \neq \boldsymbol{0}$ so the...

The last time I visited King's College Chapel Cambridge I noticed a small poster which talked about the construction.  It turns out that just before they built it they discovered an innovation which allowed them to use much less masonry.  The poster said that if you build an arch such that it is possible to draw a catenary between the inside and outside wall, then the arch will be stable.  This discovery meant that they could make the walls of King's College Chapel as thin as they wanted, subject to their ability to measure accurately. This got me thinking.  Why should that be?  Then I realised that it all has to do with the name: catenary .  A catenary is a hyperbolic cosine like $cosh(x)$.  (Obviously if you stretch or translate along the x or y axes it is still a catenary.)  The name comes from catena or "chain", because it is the shape a chain makes if held in two places at the same height. Now imagine a chain being held in that manne...

I remember once watching a wildlife documentary which had some amazing footage of a raptor catching a small bird mid air.  I can't remember the name of the program now, and I've completely failed to find a reusable picture online - it must be quite a rare thing to capture on film.  (The picture above shows a bird catching an insect, which is less impressive but illustrates the same thing.) I was impressed at how the bird of prey changed from a diving posture to one in which its wings were spread and its claws were forward, at exactly the right time.  There were no land marks of any kind and it was impossible for the hunter to know how fast it was moving how fast its prey was moving how large its prey was  Without these key bits of information, how is it possible to judge when to put the on (air) brakes and open the claws?  I decided to try and work it out: If the size of the prey is $h$, the distance to the prey is $r(t)$, and the angle subtended by ...