AIMS-Senegal – picture story of a maths communication adventure

Dear all,

this is a picture story, so be prepared to see many pictures! And it is an adventure too, since we (I will explain later who is “we”) tried something quite new: an interactive highly technical mathematical exhibition in Senegal, including a road show with high school students and Master maths students, talks, conferences, workshops and discussions, and games! And: all in three days!

Let’s start:

Day 0:

The day before the opening of the exhibition and the start of the roadshow. And my first day in Senegal. I arrive around 1:30 am in the morning and spend the night in Dakar. At 11:00 I am picked up by a car from AIMS-Senegal, who is organising the event together with the Next Einstein Initiative and IMAGINARY. In the car, I meet two other international participants: Niane Kode from Senegal, who works at AIMS-Cameroon, and Marcos Cherinda from Mocambique, an ethno-matematician, who also attended the AIMS-IMAGINARY workshop in 2014.…

By | November 16th, 2015|English, Event, Level: Simple, Uncategorized|6 Comments
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A little medical statistics

Originally written by John Webb

tw: fictionalised statistics of disease rates.

———

Today there are many tests that are widely used to detect life-threatening diseases early. How effective are they? Should they be believed?

At a routine checkup, your doctor tells you that there is a simple and inexpensive blood test that can detect a rare but particularly
nasty form of cancer. You agree to have the test done, and the doctor takes a blood sample and sends it off to the pathology laboratory.

Two days later the doctor calls to tell you that the test has come up positive. The good news is that the cancer can be cured since it
has been caught at an early stage. The bad news is that the treatment, though effective, is very expensive and has a number of unpleasant side-effects.

Before agreeing to treatment you need to do a little bit of basic arithmetic.…

By | November 2nd, 2015|English, Level: Simple|0 Comments
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Fibonacci and the Golden Ratio

The Fibonacci numbers 1, \ 1, \ 2,\ 3, \ 5,\ 8, 13,\ 21,\ 34, \ 55,\ 89,\ 144, \ \cdots are defined by the recurrence relation

 

F_1 = F_2 = 1, \ F_{n+2} = F_{n+1} + F_n \ \textrm{ for } n \ge 1.

 

Consider now the ratios \dfrac{F_{n+1}}{F_n}:

 

\dfrac 11, \ \dfrac 21, \ \dfrac 32, \ \dfrac 53, \ \dfrac 85, \ \dfrac {13}8, \dfrac {21}{13}, \ \dfrac {21}{13}, \ \dfrac {34}{21}, \dfrac {55}{34}, \dfrac {89}{55}, \ \dfrac {144}{89}, \ \cdots \ .

 

The sequence of fractions appears to rise and fall alternately. This observation can be confirmed by reference to the equation

 

F_{n+1}^2 - F_nF_{n+2} = (-1)^n

 

which, when divided by F_{n+1}F_n gives

 

\dfrac{F_{n+1}}{F_n} - \dfrac {F_{n+2}}{F_{n+1}} = \dfrac {(-1)^n}{F_{n+1}F_n } \ ,

 

showing that the difference between successive terms in the sequence of ratios is alternately positive and negative.

Calculating the first few terms suggest that successive ratios converge to 1.618 … . This may be established by using Binet’s formula (see the previous post):

 

F_n = \dfrac 1{\sqrt 5}\Big(\Big(\dfrac{1+\sqrt 5}2 \Big)^n - \Big(\dfrac{1-\sqrt 5}2 \Big)^n\Big) \

 

from which we have

 

\dfrac {F_{n+1}}{F_n} = \dfrac {\Big(\dfrac{1+\sqrt 5}2 \Big)^{n+1} - \Big(\dfrac{1-\sqrt 5}2 \Big)^{n+1}} {\Big(\dfrac{1+\sqrt 5}2 \Big)^n - \Big(\dfrac{1-\sqrt 5}2 \Big)^n}

 

Now look at the second terms in the numerator and denominator.

Since

 

-1< \dfrac{1-\sqrt 5}2 < 1,

 

these terms tend to zero as n tends to infinity, so are insignificant in comparison with the first terms. It follows that

 

\frac {F_{n+1}}{F_n} \rightarrow \frac{1+\sqrt 5}2 \ \ - \ \ \textrm{the Golden Ratio}.

UntitledCheck out the link between the golden spiral and the Fibonacci sequence here.

————

Originally by John Webb

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By | October 20th, 2015|English, Level: intermediate|2 Comments
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Fibonacci and Binet

The Fibonacci numbers 1, \ 1, \ 2,\ 3, \ 5,\ 8, 13,\ 21,\ 34, \ 55,\ 89,\ 144, \ \cdots

are defined by a recurrence relation

 

F_1 = F_2 = 1, \ F_{n+2} = F_{n+1} + F_n \ \textrm{ for } n \ge 1.

 

This definition allows one to find any Fibonacci number, but for large values of n it would be time-consuming to compute F_n this way. Worse, any error of calculating a term would be repeated and magnified in every subsequent term. So a natural question to ask is whether there is a simple formula, in terms of n, which enables one to calculate F_n without having to find all the earlier Fibonacci numbers.

In 1843 the French mathematician Jacques Philippe Marie Binet announced that he had found a Fibonacci Formula. Although it was later revealed that other mathematicians, including Leonhard Euler and Abraham de Moivre, had worked out the same formula a hundred years earlier, the formula is today still labelled Binet’s Formula.

Deriving the Binet’s formula starts with a simple little quadratic equation x^2 = x + 1.

 

Multiplying the equation by successive powers of x, and simplifying at each step, gives

 

x^3 = x(x^2) = x(x+1) = x^2 + x = (x+1) + x = 2x + 1

x^4 = x(x^3) = x(2x+1) = 2x^2 + x = 2(x+1) + x = 3x+2

x^5 = x(x^4) = x(3x+2) = 3x^2 + 2x = 3(x+1) + 2x = 5x + 3

x^6 = x(x^5) = x(5x+3) = 5x^2 + 3x = 5(x+1) + 3x = 8x + 5

 

and so on.…

By | October 19th, 2015|English, Level: intermediate, Uncategorized|2 Comments
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Fibonacci and induction

The Fibonacci numbers F_n are defined by: F_1 = F_2 = 1, F_{n+2} = F_{n+1} + F_n \textrm{ for } n\ge 1.

 

The numbers are 1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, 144, 233, 377, 610, …

 

The Fibonacci numbers have many interesting properties, and the proofs of these properties provide excellent examples of Proof by Mathematical Induction. Here are two examples. The first is quite easy, while the second is more challenging.

 

Theorem

 

Every fifth Fibonacci number is divisible by 5.

 

Proof

 

We note first that F_5 = 5 is certainly divisible by 5, as are F_{10} = 55 and F_{15} = 610. How can we be sure that the pattern continues?

 

We shall show that: IF the statement “F_n is divisible by 5″ is true for some natural number m, THEN the statement is also true for the natural number m+5.

 

Now

 

F_{m+5} = F_{m+4} + F_{m+3}

= (F_{m+3} + F_{m+2}) + F_{m+3}

= 2F_{m+3} + F_{m+2}

= 2(F_{m+2} + F_{m+1}) + (F_{m+1} + F_m)

= 2F_{m+2} + 3F_{m+1} + F_m

= 2(F_{m+1} + F_m) +3F_{m+1} + F_m

= 5F_{m+1} + 3F_m.

 

Since we know that F_m is divisible by 5, it is now clear that F_{m+5} is also divisible by 5.…

By | October 18th, 2015|English, Level: intermediate|3 Comments
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How Pringles are made (or alternatively hyperbolic paraboloids)

We are currently looking at functions of 2 variables and their graphs. Today we looked at cross-sections through a couple of different surfaces to try and figure out what they looked like in three dimensions. We did this by looking at slices in different directions and then worked out how they all fitted together. In the following animation I have taken horizontal slices of the function:

 

f(x,y)=x^2-y^2

 

Remember the graph of this is the set of points \{(x,y,z)\in{\mathbb R}^3|z=f(x,y) and (x,y)\in D\}

 

We can take horizontal slices of the surface by fixing the z-value and seeing how x and y are constrained. For instance, let’s fix the z-value to 0. Then we have:

 

0=x^2-y^2

 

This actually gives us two functions in the xy-plane: y=\pm x. This of course is just given by two lines of gradient +1 and -1 which pass through the origin in the xy-plane. In three dimensions then, if we slice through our surface at z=0 we should find these two lines which look like:

 

p0

 

How about at z=1?…

By | October 14th, 2015|English, Level: intermediate|0 Comments
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The experience of mathematical beauty and its neural correlates

Check out a mathematics and neuroscience paper here, whose authors include Fields and Abel medalist Michael Atiyah.

Untitled 2

http://journal.frontiersin.org/article/10.3389/fnhum.2014.00068/full published in frontiers in human neuroscience.

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By | October 14th, 2015|English, Level: intermediate, News|0 Comments
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Pi and the Mandelbrot set

For those studying calculus, can you see what Pi has to do with the Mandelbrot set? I’ll give you a hint, it has to do with certain integrals which you have almost certainly studied in class. We will have a post explaining exactly why this number pops up in this particular place, soon!

If you don’t know about the Mandelbrot set, take a look here, then have a look at the video above.

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By | October 5th, 2015|English, Level: intermediate|1 Comment
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UCT MAM1000 lecture notes part 50 – linear algebra part iii

Gauss reduction

So far we have seen that we have a way to translate a system of linear equations into a matrix. We can manipulate the matrix in ways which correspond to operations on the equations which keep the important information in the system of equations the same (ie. the solution of the equations before and after the operations is the same). We have seen a couple of examples of when we can read off the solution from the matrix having performed the operations. So far the order with which we perform the operations feels a bit arbitrary, although we know that we would like to get the matrix into reduced row echelon form. There is however a very systematic way of going about this, and the term for the process is called Gauss Reduction.

Here is a detailed view of what Gauss Reduction will give us:

Gauss Reduction:

To solve a system of linear equations:

 

1) First find the augmented coefficient matrix of the system of equations.…

By | October 4th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment
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UCT MAM1000 lecture notes part 49 – linear algebra part ii

Matrices

Solving a system of linear equations is not technically difficult: just eliminate the variables in a systematic fashion. When there are only two or three variables, this is easy to manage. But for a bigger system, things can quickly get confusing. We need to develop a systematic method.

The first thing to notice is that the names of the variables don’t matter. Consider, for example, the two systems

 

\begin{array}{cc} x + y &=3\\ 2x-y &= 4 \end{array}

 

and

 

\begin{array}{cc} u + v &=3\\ 2u-v &= 4 \end{array}

 

It’s clear that if we ignore the names of the variables — x and y versus u and v — these two systems are the same. The reason we can tell that they’re the same is because the {\em coefficients} of the variables are the same and the numbers on the right hand side are the same. These are really the only things about a system of linear equations that matter, and so what we can do is strip the system down to its bare bones and rewrite it like this:

 

\left( \begin{array}{cc|c} 1&1&3\\ 2&-1&4 \end{array} \right)

 

This is an augmented coefficient matrix (in general, a rectangular array of numbers, like the above, is called a matrix; a matrix with an additional vertical line, which plays the same role as the equals signs in the original equations, is augmented).…

By | October 3rd, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment
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A project initiated at the AIMS-IMAGINARY workshop, November 2014, Cape Town | All Rights Reserved
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