October 2015 – Mathemafrica

The Binary Leap Year

Originally written by John Webb

The year is 365.2422 days long. To make this inconvenient number fit into calendars with whole numbers of days in a year, a system of Leap Years is needed.

Julius Caesar decreed a calendar with years of 365 days, with an extra day every four years. That was equivalent to using $365 +\frac 14 = 365.25$ as the number of days in a year. This number was a little too large, so that more Leap Years were added than needed.

Over the years, the error built up to ten days, until in 1582 Pope Gregory XIII decreed that there should be a Leap Year in every year divisible by 4, except in years that were divisible by 100 but not by 400.

That amounted to using $365 + \frac 14 - \frac 3{400} =365.2425$ as the length of the year, which is a very good approximation to the true year of 365.2422 days. Today’s Western Calendar follows Pope Gregory’s rule.…

By Jonathan Shock| October 31st, 2015|Uncategorized|0 Comments

DAAD PhD scholarships – Applications for 2016 are now open

DAAD, the German Academic Exchange Service, in association with AIMS, provides five in-region scholarships for a PhD programme in Mathematical Sciences.

These scholarships are available to students wishing to commence a PhD programme at a South African University. Funds awarded by the scholarship must be used towards covering the costs of tuition fees, living expenses, medical insurance, books, stationery, research literature expenses, study permit and re-search.

Eligibility: African nationals from Sub Saharan countries (excluding South African nationals), and not older than 36 years of age on the date of nomination may apply. Women are particularly encouraged to submit an application.

For more information and to apply please visit the website.

How clear is this post?

…

By Jonathan Shock| October 24th, 2015|Advertising|0 Comments

Permalink
Gallery
Fibonacci and the Golden Ratio

English, Level: intermediate

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

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

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

————

Originally by John Webb

How clear is this post?

…

By Jonathan Shock| October 20th, 2015|English, Level: intermediate|2 Comments

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 Jonathan Shock| October 19th, 2015|English, Level: intermediate, Uncategorized|2 Comments

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 Jonathan Shock| October 18th, 2015|English, Level: intermediate|3 Comments

Permalink
Gallery
How Pringles are made (or alternatively hyperbolic paraboloids)

English, Level: intermediate

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:

How about at $z=1$ ?…

By Jonathan Shock| October 14th, 2015|English, Level: intermediate|0 Comments

Permalink
Gallery
The experience of mathematical beauty and its neural correlates

English, Level: intermediate, News

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.

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

How clear is this post?

…

By Jonathan Shock| October 14th, 2015|English, Level: intermediate, News|0 Comments

More on misunderstanding maths

The difference between mathematicians and other people is not that they are specially clever, it is that they have learned to suffer the journey from misunderstanding to understanding. When encountering new mathematics, mathematicians accept that there will be a time that they are mystified, uncomprehending, in a foreign country where a language is spoken they barely understand. They also know that with persistence, understanding will come.

Of course they are better at this journey than non-mathematicians. How could it be otherwise? But this is not a matter of gift (though gift helps), it is that they have learned how to learn mathematics, how not to get stuck in misunderstanding, how to recognise misunderstanding. A few simple tricks are part of this, like writing out your own examples and trying to explain the new ideas to a friend. Other methods are elaborate, such as making complicated diagrams that relate old ideas to new, or that relate unsolved problems to each other in detail.…

By Henri Laurie| October 14th, 2015|Uncategorized|0 Comments

Is misunderstanding important in mathematics?

Yes. It’s at the heart of the experience of learning and doing maths.

But it’s also the great unspoken, almost like a taboo. I’ve even heard the following: “When a mathematician says he* doesn’t understand something, it’s because he thinks it’s wrong”.

We are not supposed to admit that we don’t understand. It almost seems that if you misunderstand, you don’t exist, mathematically speaking, you’re not there, not in the place where mathematics happens, that austere realm where IF YOU CAN BUT SEE, beautiful objects exist in timeless purity.

“That is a pack of lies!” Ah, where is Zorba the Greek when we most need him, to protect us against Plato the Greek?

We do misunderstand. It’s an essential stage. When we first encounter a new idea or a difficult problem or even when we do a calculation that should be routine but seems to go wrong, we always reach understanding through misunderstanding.…

By Henri Laurie| October 10th, 2015|Uncategorized|0 Comments

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.

How clear is this post?

…

By Jonathan Shock| October 5th, 2015|English, Level: intermediate|1 Comment

1 2 Next