Jonathan Shock – Page 24

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.

How smart do you think your classmates think your classmates think your classmates think your classmates are

Here are the results of the numbers game we played in class. You were asked to guess a number between 0 and 100. The winner would be the person who guessed closest to two thirds of the average of everyone in the class. Here is a histogram of the guesses.

There was a single guess of 20, which was the eventual winner. There was also one of 20.333 but that was a smidgin off…

Congratulations to class rep Nic, who clearly had the best handle on the class!

See the history of this game here.

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By Jonathan Shock| August 9th, 2015|Uncategorized|2 Comments

A plea for LaTeX Formatting help

If anybody out there knows how to get the typesetting for LaTeX within WordPress to look more sightly, I would be very grateful for any help.

In particular, single symbols get strangely raised in the text. For instance $n$ is supposed to be inline, but clearly isn’t…

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By Jonathan Shock| August 9th, 2015|Uncategorized|0 Comments

UCT MAM1000 lecture notes part 18 – Friday 14th August

How to choose $r$ objects from $n$ objects

Let’s say that we have 6 objects ( ${\{a,b,c,d,e,f\}}$ ) and we want to know the number of ways of picking 4 of them (where the order doesn’t matter, so ${\{a,b,c,d\}}$ is no different from ${\{a,b,d,c\}}$ ). We can imagine writing down all of the possible permutations of the 6 objects (of which there are ${6!}$ as we know) and then just taking the first four items. For instance, these are the first few permutations and the items in brackets are the ones that we pick as our 4:

$\begin{array}{c} \text{(abcd)ef} \\ \text{(abcd)fe} \\ \text{(abce)df} \\ \text{(abce)fd} \\ \text{(abcf)de} \\ \text{(abcf)ed} \\ \text{(abdc)ef} \\ \text{(abdc)fe} \\ \text{(abde)cf} \\ \text{(abde)fc} \\ ... \\ \end{array}$

If we pick the items in the brackets then clearly we are going to get many repetitions of the same thing. For instance, the first two items are the same (both (abcd)). Also many of the items in the brackets are the same, up to a reordering, and we said that we didn’t care about the order, so (abcd) is the same as (abdc).…

By Jonathan Shock| August 9th, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|1 Comment

UCT MAM1000 lecture notes part 17 – Thursday 13th August

Here we’re going to continue looking at how many different ways there are to order collections of objects, but with certain constraints.

We saw previously that if you have $n$ (distinguishable) objects, you can arrange them in $n!$ ways ( $n$ positions for the first object $(n-1)$ for the second, etc, with a single position for the last object).

Now we can ask what is the way of arranging $r$ of $n$ objects. That is, if I give you a deck of cards and ask you to take any five cards and put them in some order, how many possibilities are there in total?

Well, clearly the first card is going to be one of 52 different possibilities (in a 52 card deck). The second card is going to be one of 51 possibilities, etc. So the answer is going to be:

$52\times 51\times 50\times 49\times 48$

can we write this in terms of factorials? Let’s multiply and divide this number by $47!$ :

$\frac{52\times 51\times 50\times 49\times 48\times 47!}{47!}$

But the top is just $52!$ , so this is given by:

$\frac{52!}{47!}$

Where did the $47$ come from?…

By Jonathan Shock| August 9th, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|3 Comments

UCT MAM1000 Lecture notes part 15 – Tuesday 11th August

This section is going to be a pretty diverse one, but is going to give you a very powerful set of tools for solving diverse problems in the real world.

You will be able to calculate the probability of two people in a given sized room having the same birthday, you will be able to calculate the combinatorics of all sorts of permutations and combinations, and you’ll learn about how to approximate many functions by polynomial expressions – this is an incredibly powerful technique and is used throughout the sciences.

Combinatorics

Combinatorics is the study of the different ways to rearrange objects given certain constraints. A question might be:

If a teacher is randomly picking two different students out per class for a year to get them to explain an answer on the board , what is the likelihood that you will be picked exactly five times?

How about the number of times Jodhi is likely to be called in a given game of Thunee?…

By Jonathan Shock| August 8th, 2015|Courses, First year, MAM1000, Uncategorized, Undergraduate|5 Comments

Infinite sums, continued fractions and Srinivasa Ramanujan

I mentioned Srinivasa Ramanujan today in class. He was really a truly exceptional mathematician, even amongst the greats. I haven’t had a chance to watch this documentary, but browsing through it, it looks pretty good (though quite deep in terms of the mathematics):

There’s also a very nice radio podcast from the BBC which can be found here.

The particular puzzle I was talking about today with the house numbers can be found here: https://www.math.auckland.ac.nz/~butcher/miniature/miniature2.pdf.

I wrote up a strange sum on the board the other day and didn’t discuss it any more. This is another example of the really weird properties of infinities and some discussions about it can be found here:

Part 1:

Part 2:

This particular mind-boggling sum is actually used within superstring theory when we find that the number of dimensions that superstring theory is consistent in is 10.

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By Jonathan Shock| August 7th, 2015|Uncategorized|0 Comments

The mean value theorem for integrals – proof

I left out one line at the beginning, which was pointed out in class.

We start with some function, $f(t)$ which is continuous on $[a,b]$ . We define:

$F(x)=\int_a^x f(t)dt$

By the fundamental theorem of calculus, this is continuous of $[a,b]$ and differentiable on $(a,b)$ .

Now, we use the mean value theorem for derivatives (somewhere in an interval, a function has a gradient which is equal to its average gradient) which say that if $g$ is differentiable on $(a,b)$ and continuous on $[a,b]$ then there exists some $c$ between $a$ and $b$ such that $g'(c)=\frac{g(b)-g(a)}{b-a}$ .

Now simply using our function $F(x)$ in the mean value theorem for derivatives, we have that:

$F'(c)=\frac{F(b)-F(a)}{b-a}$

But we know that $F'(c)=f(c)$ . We also have defined $F(x)$ above, so we can plug in $a$ and $b$ to get:

$f(c)=\frac{\int_a^b f(t) dt-\int_a^a f(t) dt}{b-a}$

The second term on the right hand side is zero as the two limits are the same, so we have:

$f(c)=\frac{\int_a^b f(t) dt}{b-a}$

But we know that $\int_a^b f(t) dt=\int_a^b f(x) dx$ as $t$ and $x$ are just dummy variables, and so we have proved that:

$f(c)=\frac{\int_a^b f(x) dx}{b-a}$

This finishes the proof.…

By Jonathan Shock| August 7th, 2015|English, Level: intermediate|0 Comments

On being comfortable with being uncomfortable

This sounds like some kind of zen koan, but it’s actually an integral part of learning anything within the sciences. It’s ok not to know, and it’s ok not to understand.

The process of science is the gradual chipping away at those things which we don’t understand until we uncover the truth, and it’s very important to be comfortable with the state of mind that you find yourself in most of the time: That of not knowing.

This is the third year that I’ve taught at UCT (having never taught a full lecture course before arriving here just over two years ago), and the challenges of teaching university level mathematics to a diverse and large audience are very different from what I had expected.

As I stand and teach, the act is, hopefully, a reactive one. I like to get feedback from the class as to whether they follow what I’m saying, and when I get questions I’m extremely happy.…

By Jonathan Shock| August 6th, 2015|Uncategorized|3 Comments

Translation competition: Galileoscope give away!

With many thanks to Kechil Kirkham of Fine Music Radio’s Looking Up series on astronomy and cosmology (twitter feed here), I have a beautiful Galileoscope to give away.

The Galileoscope is a small telescope with the power to resolve Jupiter and the four Galilean moons, the finer features of our moon as well as Saturn’s rings and more besides.

We are offering this scope for the first person who sends in a translation of any of these blogposts:

Ishango, The Cradle of Mathematics
A Mathematics Problem from the SBITC
Mathematics or dreams, which is more real?

In any of the official languages of South Africa other than English and Afrikaans. If you send in a translation and are not the first, then we will ask your permission to post your translation, but you will be under no obligation. The winning translation will be posted on the blog.

We want to be quite open that we are running this competition to try and get Mathemafrica running in other languages.…

By Jonathan Shock| August 6th, 2015|Uncategorized|0 Comments

What to compare when you have to use the comparison theorem

I have been asked by a lot of students how to tell what to do when faced with a comparison theory question and they can’t pick the comparison function.

The comparison theorem is used to tell whether a given integral is divergent or convergent if you can’t actually solve the integral itself.

Let’s say we have an integral which is an improper integral of the first kind. eg. the limit of integration goes up to $\infty$ . If the function has no divergences at any point, then we know that the only behaviour we have to be concerned about is how the function goes as $x\rightarrow \infty$ . In general we are concerned as to whether the function falls off fast enough in this limit. Clearly if the function goes to a constant, then it’s not going to converge, and if it falls off any slower than $\frac{1}{x}$ then it’s also not going to converge.…

By Jonathan Shock| August 5th, 2015|Uncategorized|1 Comment

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