Jonathan Shock – Page 18

About Jonathan Shock

I'm a lecturer at the University of Cape Town in the department of Mathematics and Applied Mathematics. I teach mathematics both at undergraduate and at honours levels and my research interests lie in the intersection of applied mathematics and many other areas of science, from biology and neuroscience to fundamental particle physics and psychology.

A little medical statistics

Originally written by John Webb

tw: fictionalised statistics of disease rates.

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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 Jonathan Shock| November 2nd, 2015|English, Level: Simple|0 Comments

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?

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

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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.

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Originally by John Webb

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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

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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

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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.

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

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

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UCT MAM1000 lecture notes part 50 – linear algebra part iii

Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate

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 Jonathan Shock| October 4th, 2015|Courses, English, First year, Level: intermediate, MAM1000, Uncategorized, Undergraduate|1 Comment

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