## MATHS BITE: Brouwer Fixed Point Theorem

The Brouwer Fixed Point Theorem is a result in topology that has proven to be extremely useful in mathematics.

### 2D

Suppose you take two sheets of paper with one lying directly above the other. If you crumple the top sheet and place it on top of the other sheet, then Brouwer’s theorem states that there is at least one point that has not moved, i.e. there is at least one point on the top sheet that is directly above the corresponding point on the bottom sheet.

### 3D

Take a cup of coffee, and mix it around. The theorem states that after mixing there must be some point in the coffee, which is in the exact spot that it was before the mixing. Furthermore, if you try to move that point out of its original position then you will inevitably move another point into its original position.

### Formally

A continuous function from an n-ball into an n-ball must have a fixed point. Note that continuity of the function is essential.

## MATHS BITE: Steinmetz Solid

Named after mathematician Charles Proteus Steinmetz, a Steinmetz solid is the solid body generated by the intersection of two or three cylinders with the same radii at right angles.

These solids are named after Steinmetz as he managed to determine the volume of the intersection, though these solids were known long before he studied them.

If two/three cylinders intersect then the intersection is called a bicylinder/tricylinder.

### Bicylinder

Surface Area: 16r2

Volume: , where r is the radius of both cylinders.

The volume of the cube minus the volume of the eight pyramids is the volume of the bicylinder: the volume of 8 pyramids is .

Then the bicylinder volume is:

Surface Area:

Volume:

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## MATHS BITE: Stars and Bars

A common problem in combinatorics is when we are asked to count the number of ways to group identical objects, such as placing indistinguishable balls into labelled urns. Popularised by William Feller in his book on probability, the Stars and Bars methods aims to solve such problems.

### Theorem

The number of ways to put n indistinguishable balls into k labelled urns is

Proof using Stars and Bars:

Represent n balls by n adjacent stars and consider inserting k – 1 bars between the starts to separate them into k groups.

For example, for n = 12 and k = 5 the following is a possible arrangement:

**|****||***|***

Here, urns 1, 2, 3, 4 and 5 are of size 2, 4, 0, 3 and 3 respectively.

There are a total of n + k – 1 positions, of which n are stars and k – 1  are bars. So, the number of ways to place n indistinguishable balls into k labeled urns in the same as the number of ways of choosing n positions (or k – 1 positions) amount n + k – 1 spaces, with all the remaining positions taken as bars (stars), i.e.

ways.

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## MATHS BITE: Sierpinski Number

A Sierpinski number is an odd natural number k such that  is not prime for all natural numbers n. In 1960, Sierpinski proved that there are infinitely many odd integers k with this property, but failed to give an example. Numbers in such a set with odd k and k < 2^n are called Proth numbers.

78557, 271129, 271577, 322523, 327739, 482719, 575041, 603713, 903983, 934909,….

-List of some known Sierpinski Numbers

### Sierpinski Problem

The Sierpinski problem asks what the smallest Sierpinski number is. In 1967, Sierpinski and Selfridge conjectured that 78557 is the smallest Sierpinski number. To show this is true, we need to show that all the odd numbers smaller than 78557 are not Sierpinski numbers, i.e. for every odd k below 78557 there is a positive interger n such that  is prime. There are only five numbers which have not been eliminated:

k = 21181, 22699, 24737, 55459, and 67607

### Numberphile Video

Find out more here. M x

## MATHS BITE: Wythoff’s Game

Wythoff’s game is a two-player mathematical strategy game. It is played with two piles of counters and goes as follows:

Players take turns to remove counters from one, or both piles (if they are removing stones from both piles, they must remove the same number of stones from each). The game ends when one person removes the last counter, or counters, thus winning.

It is name Wythoff’s game, as it was Dutch mathematician W. A. Wythoff that published a mathematical analysis of the game in 1907, although it was previously know in China as tsyan-shidzi (“choosing stones”).

### Optimal Strategy?

Any position of the game can be described by a pair of integers (x,y) where x≤y. This describes the size of both piles in the position.

The strategy of the game involves cold and hot positions:

• A cold position is one where the player whose turn it is to move will lose, even with his best move;
• A hot position is one where the player whose turn it is to move will win with his best move.

Thus, the optimal strategy for a person in a hot position is to move to a cold position.

We can classify the positions as hot or cold recursively, using 3 simple rules:

1. (0,0) is a cold position;
2. Any position from which a cold position can be reached in a single move is a hot position;
3. If every move leads to a hot position then the position is cold.

For example (taken from Wikipedia):

“all positions of the form (0, y) and (y, y) with y > 0 are hot, by rule 2. However, the position (1,2) is cold, because the only positions that can be reached from it, (0,1), (0,2), and (1,1), are all hot. The cold positions (x, y) with the smallest values of x and y are (0, 0), (1, 2), (3, 5), (4, 7),(6,10) and (8, 13).”

### Connection with the Golden Ratio

Wythoff discovered that the cold positions can be determined using the Golden Ratio. Namely, if n is any natural number, then

is a cold position (where we use the floor function and φ is the Golden Ratio).

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## MATHS BITE: Kaprekar Numbers

Consider an n-digit number k. Square it, and then add the right n digits to the left n or n-1 digits (by convention, the second part may start with the digit 0, but must be nonzero). If the result is k then it is called a Kaprekar number. They are named after D. R. Kaprekar, a recreational mathematician from India.

We can extend the definition to any base b:

Let  X  be a non-negative integer and  n a positive integer.  X  is an n-Kaprekar number for base  b  if there exist non-negative integer A, and positive integer B  satisfying:

X2 = Abn + B, where 0 < B < bn
X = A + B

-Wikipedia

### Examples in Base 10

• 297: 2972 = 88209, which can be split into 88 and 209, and 88 + 209 = 297.

• 999: 9992 = 998001, which can be split into 998 and 001, and 998 + 001 = 999.
• In particular, 9, 99, 999… are all Kaprekar numbers.
• More generally, for any base b, there exist infinitely many Kaprekar numbers, including all numbers of the form bn − 1.
• 100: 100 is NOT a Kaprekar number as, although 1002 = 10000 and 100 + 00 = 100, the second part here is zero.

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(P.S. Another post on Kaprekar is coming soon!)

## MATHS BITE: Even or Odd?

What are even and odd functions? These two terms get used very frequently in order to simplify problems such as integration or when finding Fourier coefficients, for example.

### Even Function

An even function is such that for all x in the domain

f(x) = f(-x)

An even function is symmetric with respect to the y-axis.

Examples:

• Any polynomial p, where n is even for all xn. For example f(x) = x2+2x4 – 6. Note that (x+1)2 is not even.
• f(x) = cos(x)

### Odd Function

An odd function is such that for all x in the domain

f(x) = -f(-x)

An odd function is symmetric with respect to the origin.

Examples:

• x, x3, x5,… etc. Note that unlike even functions, an expression such as x3 + 1 is not odd.
• f(x) = sin(x)

### Odd and/or Even?

A function can be neither odd nor even, for example x− x + 1:

Additionally, the only function that is even and odd is f(x) = 0.

### Properties:

• Adding two even (odd) functions will give an even (odd) function.
• Adding an even and an odd function will give a function that is neither odd nor even.
• Both the product of two even functions and the product of two odd functions is even.
• The product of an even function and an odd function is odd.

A heptadecagon (or a 17-gon) is a seventeen sided polygon.

In 1796, Gauss proved, at the age of 19 (let that sink in…) that the heptadecagon is constructible with a compass and a straightedge, such as a ruler. His proof of the constructibility of an n-gon relies on two things:

• the fact that “constructibility is equivalent to expressibility of the trigonometric functions of the common angle in terms of arithmetic operations and square root extractions“;
• the odd prime factors of n are distinct Fermat primes.

Constructing the regular heptadecagon involves finding the expression for the cosine of   in terms of square roots, which Gauss gave in his book Disquistiones Arithmeticae:

An explicit construction was given by Herbert Willian Richmond in 1893.

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## MATHS BITE: Ford Circles

A ford circle is a circle with centre , and radius  where p and q are coprime integers.

Notice that each Ford Circle is tangent to to the horizontal axis and any two Ford circles are either tangent or disjoint. The latter statement can be proven by finding the squared distance d^2 between the centres of the circles with (p,q) and (p’,q’) as the pairs of coprime integers.

Let s be the sum of the radii:

Then

However, we have that  and so , thus the distance between circles is greater or equal to the sum of the radii of the circles. There is equality iff

In this case, the circles are tangent to one another.

#### Total area of Ford Circles

(taken from Wikipedia)

As no two ford circles intersect, it follows immediately that the total area of the Ford circles:

is less than 1.

From the definition, the area is

Simplifying this expression gives us

noting that the last equality is given by considering the Dirichlet generating function for Euler’s totient function φ(q).

Given that ζ(4) = π^4/90, we get

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## MATHS BITE: Shoelace Theorem

The Shoelace theorem is a useful formula for finding the area of a polygon when we know the coordinates of its vertices. The formula was described by Meister in 1769, and then by Gauss in 1795.

### Formula

Let’s suppose that a polygon P has vertices (a1, b1), (a2, b2), …, (an, bn), in clockwise order. Then the area of P is given by

$$\dfrac{1}{2} |(a_1b_2 + a_2b_3 + \cdots + a_nb_1) - (b_1a_2 + b_2a_3 + \cdots + b_na_1)|$$

The name of this theorem comes from the fact that if you were to list the coordinates in a column and mark the pairs to be multiplied, then the image looks like laced-up shoes.

### Proof

(Note: this proof is taken from artofproblemsolving.)

Let $\Omega$ be the set of points that belong to the polygon. Then

$$A=\int_{\Omega}\alpha,$$

where $\alpha=dx\wedge dy$.

Note that the volume form $\alpha$ is an exact form since $d\omega=\alpha$, where

$$\omega=\frac{x\,dy}{2}-\frac{y\,dx}{2}.\label{omega}$$

Substitute this in to give us

$$\int_{\Omega}\alpha=\int_{\Omega}d\omega.$$

and then use Stokes’ theorem (a key theorem in vector calculus) to obtain

$$\int_{\Omega}d\omega=\int_{\partial\Omega}\omega.$$

where

$\partial \Omega=\bigcup A(i)$

and $A(i)$ is the line segment from $(x_i,y_i)$ to $(x_{i+1},y_{i+1})$, i.e.  is the boundary of the polygon.

Next we substitute for $\omega$:

$$\sum_{i=1}^n\int_{A(i)}\omega=\frac{1}{2}\sum_{i=1}^n\int_{A(i)}{x\,dy}-{y\,dx}.$$

Parameterising this expression gives us

$$\frac{1}{2}\sum_{i=1}^n\int_0^1{(x_i+(x_{i+1}-x_i)t)(y_{i+1}-y_i)}-{(y_i+(y_{i+1}-y_i)t)(x_{i+1}-x_i)\,dt}.$$

Then, by integrating this we obtain

$$\frac{1}{2}\sum_{i=1}^n\frac{1}{2}[(x_i+x_{i+1})(y_{i+1}-y_i)- (y_{i}+y_{i+1})(x_{i+1}-x_i)].$$

This then yields, after further manipulation, the shoelace formula:

$$\frac{1}{2}\sum_{i=1}^n(x_iy_{i+1}-x_{i+1}y_i).$$

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