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## Runge's Phenomenon

This is a post I wrote for round 2 of The Aperiodical's Big Internet Math-Off 2018. As I went out in round 1 of the Big Math-Off, you got to read about the real projective plane instead of this.

Polynomials are very nice functions: they're easy to integrate and differentiate, it's quick to calculate their value at points, and they're generally friendly to deal with. Because of this, it can often be useful to find a polynomial that closely approximates a more complicated function.

Imagine a function defined for \(x\) between -1 and 1. Pick \(n-1\) points that lie on the function. There is a unique degree \(n\) polynomial (a polynomial whose highest power of \(x\) is \(x^n\)) that passes through these points. This polynomial is called an

*interpolating polynomial*, and it sounds like it ought to be a pretty good approximation of the function.So let's try taking points on a function at equally spaces values of \(x\), and try to approximate the function:

$$f(x)=\frac1{1+25x^2}$$
I'm sure you'll agree that these approximations are pretty terrible, and they get worse as more points are added. The high error towards 1 and -1 is called Runge's phenomenon, and was discovered in 1901 by Carl David Tolmé Runge.

All hope of finding a good polynomial approximation is not lost, however: by choosing the points more carefully, it's possible to avoid Runge's phenomenon. Chebyshev points (named after Pafnuty Chebyshev) are defined by taking the \(x\) co-ordinate of equally spaced points on a circle.

The following GIF shows interpolating polynomials of the same function as before using Chebyshev points.

Nice, we've found a polynomial that closely approximates the function... But I guess you're now wondering how well the Chebyshev interpolation will approximate other functions. To find out, let's try it out on the votes over time of my first round Big Internet Math-Off match.

The graphs below show the results of the match over time interpolated using 16 uniform points (left) and 16 Chebyshev points (right). You can see that the uniform interpolation is all over the place, but the Chebyshev interpolation is very close the the actual results.

But maybe you still want to see how good Chebyshev interpolation is for a function of your choice... To help you find out, I've written @RungeBot, a Twitter bot that can compare interpolations with equispaced and Chebyshev points. Just tweet it a function, and it'll show you how bad Runge's phenomenon is for that function, and how much better Chebysheb points are.

For example, if you were to tweet @RungeBot f(x)=abs(x), then RungeBot would reply: Here's your function interpolated using 17 equally spaced points (blue) and 17 Chebyshev points (red). For your function, Runge's phenomenon is terrible.

A list of constants and functions that RungeBot understands can be found here.

### Similar posts

Mathsteroids | Christmas (2018) is over | Christmas cross stitch | MENACE in fiction |

### Comments

Comments in green were written by me. Comments in blue were not written by me.

**2018-07-07**

## World Cup stickers 2018, pt. 3

So you've calculated how much you should expect the World Cup sticker book to cost
and recorded how much it actually cost. You might be wondering what else you can do with your sticker book.
If so, look no further: this post contains 5 mathematical things involvolving your sticker book and stickers.

### Test the birthday paradox

In a group of 23 people, there is a more than 50% chance that two of them will share a birthday. This is often called the birthday paradox, as the number 23 is surprisingly small.

Back in 2014 when Alex Bellos suggested testing the birthday paradox on World Cup squads, as there are 23 players in a World Cup squad. I recently discovered that even further back in 2012, James Grime made a video about the birthday paradox in football games, using the players on both teams plus the referee to make 23 people.

In this year's sticker book, each player's date of birth is given above their name, so you can use your sticker book to test it out yourself.

### Kaliningrad

One of the cities in which games are taking place in this World Cup is Kaliningrad. Before 1945, Kaliningrad was called Königsberg. In Königsburg, there were seven bridges connecting four islands. The arrangement of these bridges is shown below.

The people of Königsburg would try to walk around the city in a route that crossed each bridge exactly one. If you've not tried this puzzle before, try to find such a route now before reading on...

In 1736, mathematician Leonhard Euler proved that it is in fact impossible to find such a route. He realised that for such a route to exist, you need to be able to pair up the bridges on each island so that you can enter the island on one of each pair and leave on the other. The islands in Königsburg all have an odd number of bridges, so there cannot be a route crossing each bridge only once.

In Kaliningrad, however, there are eight bridges: two of the original bridges were destroyed during World War II, and three more have been built. Because of this, it's now possible to walk around the city crossing each bridge exactly once.

I wrote more about this puzzle, and using similar ideas to find the shortest possible route to complete a level of Pac-Man, in this blog post.

### Sorting algorithms

If you didn't convince many of your friends to join you in collecting stickers, you'll have lots of swaps. You can use these to practice performing your favourite sorting algorithms.

#### Bubble sort

In the bubble sort, you work from left to right comparing pairs of stickers. If the stickers are in the wrong order, you swap them. After a few passes along the line of stickers, they will be in order.

In the insertion sort, you take the next sticker in the line and insert it into its correct position in the list.

In the quick sort, you pick the middle sticker of the group and put the other stickers on the correct side of it. You then repeat the process with the smaller groups of stickers you have just formed.

### The football

Sticker 007 shows the official tournament ball. If you look closely (click to enlarge), you can see that the ball is made of a mixture of pentagons and hexagons. The ball is not made of only hexagons, as road signs in the UK show.

Stand up mathematician Matt Parker started a petition to get the symbol on the signs changed, but the idea was rejected.

If you have a swap of sticker 007, why not stick it to a letter to your MP about the incorrect signs as an example of what an actual football looks like.

### Psychic pets

Speaking of Matt Parker, during this World Cup, he's looking for psychic pets that are able to predict World Cup results. Why not use your swaps to label two pieces of food that your pet can choose between to predict the results of the remaining matches?

### Similar posts

World Cup stickers 2018, pt. 2 | World Cup stickers 2018 | World Cup stickers | Euro 2016 stickers |

### Comments

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**2018-07-06**

## Mathsteroids

This is a post I wrote for round 1 of The Aperiodical's Big Internet Math-Off 2018, where Mathsteroids lost to MENACE.

A map projection is a way of representing the surface of a sphere, such as the Earth, on a flat surface. There is no way to represent all the features of a sphere on a flat surface, so if you want a map that shows a certain feature of the world, then you map will have to lose some other feature of the world in return.

To show you what different map projections do to a sphere, I have created a version of the game

*Asteroids*on a sphere. I call it*Mathsteroids*. You can play it here, or follow the links below to play on specific projections.### Mercator projection

The most commonly used map projection is the Mercator projection, invented by Gerardus Mercator in 1569. The Mercator projection preserves angles: if two straight lines meet at an angle \(\theta\) on a sphere, then they will also meet at an angle \(\theta\) on the map. Keeping the angles the same, however, will cause the straight lines to no longer appear straight on the map, and the size of the same object in two different place to be very different.

The angle preserving property means that lines on a constant bearing (eg 030° from North) will appear straight on the map. These constant bearing lines are not actually straight lines on the sphere, but when your ship is already being buffeted about by the wind, the waves, and the whims of drunken sailors, a reasonably straight line is the best you can expect.

The picture below shows what three ships travelling in straight lines on the sphere look like on a Mercator projection.

To fully experience the Mercator projection, you can play

*Mathsteroids*in Mercator projection mode here. Try flying North to see your spaceship become huge and distorted.### Gall–Peters projection

The Gall–Peters projection (named after James Gall and Arno Peters) is an area-preserving projection: objects on the map will have the same area as objects on the sphere, although the shape of the object on the projection can be very different to its shape on the sphere.

The picture below shows what three spaceships travelling in straight lines on a sphere look like on the Gall–Peters projection.

You can play

*Mathsteroids*in Gall–Peters projection mode here. I find this one much harder to play than the Mercator projection, as the direction you're travelling changes in a strange way as you move.### Azimuthal projection

An azimuthal projection makes a map on which the directions from the centre point to other points on the map are the same as the directions on the sphere. A map like this could be useful if, for example, you're a radio operator and need to quickly see which direction you should point your aerial to communicate with other points on the map.

The azimuthal projection I've used in

*Mathsteroids*also preserves distances: the distance from the centre to the another points on the map is proportional to the actual distance on the sphere. This projection is used as the emblem of the UN.The picture below shows three straight lines on this projection. You can play

*Mathsteroids*in azimuthal mode here.A retroazimuthal projection makes a map on which the directions to the centre point from other points on the map are the same as the directions on the sphere. If you're thinking that this is the same as the azimuthal projection, then you're too used to doing geometry on flat surfaces: on a sphere, the sum of the angles in a triangle depends on the size of the triangle, so the directions from A to B and from B to A aren't as closely related as you would expect.

The Craig retroazimuthal projection was invented by James Ireland Craig in 1909. He used Mecca as his centre point to make a map that shows muslims across the world which direction they should face to pray.

The picture below shows what three spaceships travelling in a straight lines on a sphere looks like on this projection.

You can play

*Mathsteroids*in Craig retroazimuthal mode here to explore the projection yourself. This is perhaps the hardest of all to play, as (a) two different parts of the sphere overlap on the map, and (b) the map is actually infinitely tall, so quite a bit of it is off the edge of the visible game area.### Stereographic projection

The final projection I'd like to show you is the stereographic projection.

Imagine that a sphere is sitting on a 2D plane. Take a point on the sphere. Imagine a straight line through this point and the point at the top of the sphere. The point where this line meets the 2D plane is stereographic projection of the point on the sphere.

This projection (backwards) can be used to represent the every complex number as a point on a sphere: this is called the Riemann sphere.

To make

*Mathseteroids*playable after this projection, I split the sphere into 2 hemisphere and projected each seperately to give two circles. You can play*Mathsteroids*in stereographic projection mode here. Three spaceships travelling in straight lines on this projection are shown below.... and if you still don't like map projections, you can still enjoy playing ~~head over to The Aperiodical and vote~~ (voting now closed).

*Mathsteroids*on an old fashioned torus. Or on a Klein bottle or the real projective plane. Don't forget to take a short break from playing to### Similar posts

Video game surfaces | Optimal Pac-Man | MENACE in fiction | Runge's Phenomenon |

### Comments

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**2018-06-16**

## World Cup stickers 2018, pt. 2

This year, like every World Cup year, I've been collecting stickers to fill the official Panini World Cup sticker album.
Back in March, I calculated that I should expect it to cost £268.99 to fill this year's album (if I order the last 50 stickers).
As of 6pm yesterday, I need 47 stickers to complete the album (and have placed an order on the Panini website for these).

### So... How much did it cost?

In total, I have bought 1781 stickers (including the 47 I ordered) at a cost of £275.93. The plot below shows
the money spent against the number of stickers stuck in, compared with the what I predicted in March.

To create this plot, I've been keeping track of exactly which stickers were in each pack I bought. Using this data, we can
look for a few more things. If you want to play with the data yourself, there's a link at the bottom to download it.

### Swaps

The bar chart below shows the number of copies of each sticker I got (excluding the 47 that I ordered). Unsurprisingly, it looks a lot like
random noise.

The sticker I got most copies of was sticker 545, showing Panana player Armando Cooper.

I got swaps of 513 different stickers, meaning I'm only 169 stickers short of filling a second album.

### First pack of all swaps

Everyone who has every done a sticker book will remember the awful feeling you get when you first get a pack of all swaps.
For me, the first time this happened was the 50th pack. The plot below shows when the first pack of all swaps occurred in 500,000 simulations.

Looks like I was really quite unlucky to get a pack of all swaps so soon.

### Duplicates in a pack

In all the 345 packs that I bought, there wasn't a single pack that contained two copies of the same sticker.
In fact, I don't remember

*ever*getting two of the same sticker in a pack. For a while I've been wondering if this is because Panini ensure that packs don't contain duplicates, or if it's simply very unlikely that they do.If it was down to unlikeliness, the probability of having no duplicates in one pack would be:

\begin{align}
\mathbb{P}(\text{no duplicates in a pack}) &= 1 \times\frac{681}{682}\times\frac{680}{682}\times\frac{679}{682}\times\frac{678}{682}\\
&= 0.985
\end{align}
and the probability of none of my 345 containing a duplicate would be:

\begin{align}
\mathbb{P}(\text{no duplicates in 345 packs})
&= 0.985^{345}\\
&= 0.00628
\end{align}
This is very very small, so it's safe to conclude that Panini do indeed ensure that packs do not contain duplicates.

### The data

If you'd like to have a play with the data yourself, you can download it here. Let me know if
you do anything with it...

### Similar posts

World Cup stickers 2018, pt. 3 | World Cup stickers 2018 | World Cup stickers | Euro 2016 stickers |

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**2018-05-02**

## A bad Puzzle for Today

Every morning just before 7am, one of the Today Programme's presenters reads out a puzzle. Yesterday, it was this puzzle:

In a given month, the probability of a certain daily paper either running a story about inappropriate behaviour at a party conference or running one about somebody's pet being retrieved from a domestic appliance is exactly half the probability of the same paper containing a photo of a Tory MP jogging. The probability of no such photo appearing is the same as that of there being a story about inappropriate behaviour at a party conference. The probability of the paper running a story about somebody's pet being retrieved from a domestic appliance is a quarter that of its containing a photo of a Tory MP jogging. What are the probabilities the paper will (a) run the conference story, (b) run the pet story, (c) contain the jogging photo?

I'm not the only one to notice that some of Radio 4's daily puzzles are not great.
I think this puzzle is a great example of a terrible puzzle. You can already see the first problem with it: it's long and words and very hard to follow on the radio.
But maybe this isn't so important, as you can
read it here after it's been read out.

Once you've done this, you can re-write the puzzle as follows:
there are three news stories (\(A\), \(B\) and \(C\)) that the newspaper might publish in a month. We are given the following information:

$$\mathbb{P}(A\text{ or }B)=\tfrac12\mathbb{P}(C)$$
$$1-\mathbb{P}(C)=\mathbb{P}(A)$$
$$\mathbb{P}(B)=\tfrac14\mathbb{P}(C)$$
To solve this puzzle, we need use the formula \(\mathbb{P}(A\text{ or }B)=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\).
These Venn diagrams justify this formula:

Using the information we were given in the question, we get:

\begin{align}
\tfrac12\mathbb{P}(C)&=\mathbb{P}(A\text{ or }B)\\
&=1-\mathbb{P}(C)+\tfrac14\mathbb{P}(C)-\mathbb{P}(A\text{ and }B)\\
\mathbb{P}(C)&=\tfrac45(1-\mathbb{P}(A\text{ and }B)).
\end{align}
At this point we have reached the second problem with this puzzle: there's no answer unless we make some extra assumptions, and the question doesn't make it clear what we can assume.
But let's give the puzzle the benefit of the doubt and try some assumptions.

### Assumption 1: The events are mutually exclusive

If we assume that the events \(A\) and \(B\) are mutually exclusive—or, in other words, only one of these two articles can be published,
perhaps due to a lack of space—then we can use the fact that

$$\mathbb{P}(A\text{ and }B)=0.$$
This means that
\(\mathbb{P}(C)=\tfrac45\),
\(\mathbb{P}(A)=\tfrac15\), and
\(\mathbb{P}(B)=\tfrac15\). There's a problem with this answer, though: the three probabilities add up to more than 1.

This wouldn't be a problem, except we assumed that only one of the articles \(A\) and \(B\) could be published.
The probabilities adding up to more than 1 means that either \(A\) and \(C\) are not mutually exclusive or \(A\) and \(B\) are not mutually exclusive,
so \(C\) could be published alongside \(A\) or \(B\). There seems to be nothing special about the three news stories to mean that only some combinations
could be published together, so at this point I figured that this assumption was wrong and moved on.

Today, however, the answer was posted, and
this answer was given (without an working out). So we have a third problem with this puzzle: the answer that was given is wrong, or at the very best
is based on questionable assumptions.

### Assumption 2: The events are independent

If we assume that the events are independent—so one article being published doesn't affect whether or not another is published—then
we may use the fact that

$$\mathbb{P}(A\text{ and }B)=\mathbb{P}(A)\mathbb{P}(B).$$
If we let \(c=\mathbb{P}(C)\), then we get:

\begin{align}
\tfrac12c&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A\text{ and }B)\\
&=\mathbb{P}(A)+\mathbb{P}(B)-\mathbb{P}(A)\mathbb{P}(B)\\
&=1-c+\tfrac14c-\tfrac14(1-c)c\\
\tfrac14c^2-\tfrac32c+1&=0.
\end{align}
You can use your favourite formula to solve this to find that \(c=3-\sqrt5\), and therefore
\(\mathbb{P}(A)=\sqrt5-2\) and
\(\mathbb{P}(B)=\tfrac34-\tfrac{\sqrt5}4\).

In this case, our assumption appears to be more reasonable—as over the course of a month the stories published by a paper probably don't have
much of an effect on each other—but we have the fourth, and probably biggest problem with the puzzle: the question and answer are not interesting or surprising, and
the method is a bit tedious.

### Similar posts

World Cup stickers 2018, pt. 3 | World Cup stickers 2018, pt. 2 | A 20,000-to-1 baby? | World Cup stickers 2018 |

### Comments

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**2018-05-03**

Matthew

**2018-05-03**

Stefan

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