Where did that substitution come from?

If you want to understand maths, you really have to do it. I recommend going through these examples and using the substitutions given here as hints. Get a blank piece of paper, put your notes away and try to do these examples and see if you get the same answers as in class. If you don’t, write in the comments, and we can see where things may have gone astray.

 

I’ve been teaching integration by substitution, including by trig substitutions over the last few days, and a frequent question which a newbie substituter will ask is “how did you know to make that substitution?”. It’s a very reasonable question, and one that takes practice to build the correct intuition, but I’ll do my best to give some motivation now as to why we made some of the substitutions we made. We won’t solve the integrals, but we will motivate here why we make particular choices for substitutions.…

By | July 22nd, 2016|Courses, English, First year, MAM1000, Undergraduate|2 Comments

The confusion about discontinuity

Whilst reading Mícheál Ó Searcóid’s book Metric Spaces, I found out about a nuance in the definition of continuity that I was not previously aware of, and something which may be taught incorrectly at high schools. M. Searcóid states that a function such as tan(x) is continuous (read the page here). The definition of continuity at a point is based on the fact that the function has a value at that point (if a function is continuous at x = a, then f(a) has a value in the expression |f(x)-f(a)|<ϵ). However, following M. Searcóid’s line of thought about continuous functions, it does not make sense to consider points at which a function is not defined. If we were asked to prove that the function is ‘discontinuous’ at a point, we would need to show that the condition for continuity at that point is false. And the negation* of the condition for continuity would not make sense at a point where the function value is not defined.…

By | July 21st, 2016|Courses, Level: intermediate|2 Comments

Square roots: in your head

I came across the following YouTube video a while back which uses a strange trick to accurately approximate square roots. I suggest watching at least the first minute where the presenter explains how it’s done:

Let’s do an example. Approximate to 2 decimal places: \sqrt{40}

First, YouTube tells us to find the nearest perfect square that’s less than 40, that’s 36, and take the root, giving us 6. So our answer is, obviously, 6 point something. That something is a fraction, where the numerator is the difference between 40 and 36, and the denominator is 2 times 6. So we have:

 \sqrt{40} \approx \sqrt{36} +\frac{4}{12} = 6+ \frac{1}{3} \approx 6.33

So what’s the actual answer to 2 decimals? It’s 6.32. That’s what I like to call: pretty darn close. Naturally, there are a few catches to this technique: you need to know your perfect squares, you need to know your fractions, and things tend to get hard with larger numbers.

But still, I thought this was surprisingly effective for such a simple piece of arcane trickery.…

By | June 23rd, 2016|Uncategorized|1 Comment

Type Theory, Logic, and Programming

Hello Internet. In this blog post we are going to learn some type theory, which is a field that lies in the intersection of mathematics and computer science. We are eventually going to learn the variant called Martin-Loef Type Theory, or MLTT for short. To motivate it, we are going to try answering the question, “What is a valid program?”

Not every program makes sense, for example what would a program consisting of the single expression x + 2 mean? If we don’t know what x is, then we don’t know what the meaning of that expression is. More importantly, a program can combine expressions that are incompatible with each other. For example, if + is the usual operation that adds two numbers then an expression like "abc" + 5 can’t be given a meaning. To avoid such meaningless programs, we should talk about the different kinds of data and the valid operations on these pieces of data.…

By | June 19th, 2016|Uncategorized|2 Comments

Down the rabbit hole

The following has been a rather interesting journey – from a test question which seemed fine, to a subtlety which seemed easy, to a discussion with a number of different mathematicians about the nature of distributions, measure theory and regularisation. I will try and make it as clear as possible in the post below. Note, as mentioned in the comments, we have actually only found the solution to this problem for a constrained range of x, and not x \in \mathbb{R}. I didn’t want to complicate things any more than necessary here for first year students, but the comments are very important too.

 

In a recent class test, there was a question, written by me, which was not quite the question that I wanted to ask. It turns out that it does have an answer, but it’s not an answer that can yet be found by the means at the class’s disposal.…

On Convergent Sequences and Prime Numbers

Ever since Euclid first proved that there are infinitely many prime numbers, mathematicians have found ever more creative ways to prove the same result, and also various stronger theorems that imply it. Dirichlet’s Theorem, for example, states that ifm and n are relatively prime integers, then there are infinitely many prime numbers of the form mk + n for some integer k. It is also known that the sum of the reciprocals of the prime numbers diverges, that the sum

\displaystyle \sum_{\substack{p \leq n \\ p \text{ prime}}} \frac{1}{p} \sim \log(\log(n))

and that the number of prime numbers less than n is asymptotically equal to \displaystyle \frac{n}{\log(n)}. In this blog post, we will continue this proud tradition by proving that there are infinitely many prime numbers which have your phone number somewhere in their digits, and which simultaneously have a prime number of digits.

To do so, we will look at the convergence of two different sums: that of the reciprocals of the primes with a prime number of digits, and that of the reciprocals of the natural numbers which do not contain your phone number amongst their digits.…

By | May 22nd, 2016|Level: intermediate, Uncategorized|2 Comments

Nowhere Differentiable Functions

By: Jan Wuzyk

In this article I am going to show that nowhere differentiable functions do in fact exists and give a few examples, some of which are relatively modern. But first I’m going to try to answer a question that is, in my opinion, too rarely discussed in mathematics classes, ”Why do we care?”.

Why we care

To answer this question we have to look into the history of mathematics. In 1821 Augustin-Louis Cauchy published his seminal book, Cours d’Analyse, this is generally recognized as the first serious attempt to put calculus on a rigours footing[Com][1] , mainly through introducing rigorous definitions of limits,continuity and differentiability among others, and the definitions that go with them2. This was also time the integral was defined as an area instead of simply as the antiderivative.

It should be noted that Cauchy by no means closed the issue of rigour in analysis but he provided a starting point.…

By | May 17th, 2016|Uncategorized|0 Comments

The Newton-Raphson Method

How would you go about finding the value of \sqrt{3} if you didn’t have a square root button on your calculator? Well, the most obvious thing might be to try some values, based on your knowledge of the square root function. You are being asked to find that x for which:

 

x=\sqrt{3}

 

or, in other words, that x, which, when squared  gives 3. We have to be a little careful here because we know that there will actually be two numbers which satisfy this (one positive, one negative), and we are interested in the positive one only.

 

So, we try some values, but we don’t do it randomly, we can see that because 1^2=1 and 2^2=4 that whatever number squared gives 3 must be between 1 and 2. We can try something called a binary intersection. This just means taking the values which we know bound the right answer (ie. we know that 1<x<2), and trying the number in the middle.…

Welcome to Reproducing Kernel Hilbert Space

In a series of posts I hope to introduce Mathemafrica readers to some useful data analysis methods which rely on operations in a little back-water of Hilbert space, namely Reproducing Kernel Hilbert Space (or RKHS).

We’ll start with the “classic” example. Consider the data plotted in figure 1. Each data point has 3 “properties”: an x_1 coordinate, an x_2 coordinate and a colour (red or blue). Suppose we want to be able to separate all data points into two groups: red points and blue points. Furthermore, we want to be able to do this linearly, i.e. we want to be able to draw a line (or plane or hyperplane) such that all points on one side are blue, all points on the other are red. This is called linear classification.

Figure 1: A scatter of data with three properties: an x_1 coordinate, an x_2 coordinate and a colour.

Figure 1: A scatter of data with three properties: an x_1 coordinate, an x_2 coordinate and a colour.

Suppose for each data point we generate a representation of the data point \phi(x)=[x_1, x_2, x_1x_2] .…

By | April 30th, 2016|English, Level: intermediate, Uncategorized|1 Comment

Chain Rule.

Definition:

The chain rule is a method for differentiating a function of a function, or differentiating composite functions.

Consider the expression y = sin(x^2). We notice that this is not a normal sine function. It has an x^2 as argument for the sine function. Therefore, we can consider the x^2 in the sine function as a whole different function. This can be broken into two functions, f(x) and g(x).

If we consider f(x) =  sin(x); g(x) = x^2, we can write y = f(g(x)).

In order to differentiate a composite function, y = f(g(x)), i.e; to find y', we let

u = g(x)  and  y = f(u)

What this implies is that whatever g(x) is, u will be equal to that. Then, the process of differentiating is to find

  • \frac{du}{dx} (as u will be a function of x).
  • \frac{dy}{du}

Finally, we can write,

\frac{dy}{dx} = \frac{dy}{du} \cdot \frac{du}{dx}

Back to our example, y = sin(x^2); remember that f(x) = sin(x) and g(x) = x^2. We let

u =  x^2  and  y = f(u) = sin(u)

Therefore,  \frac{du}{dx} = 2x and  \frac{dy}{du} = cos(u)

This leaves us with  \frac{dy}{dx} = cos(u) \cdot 2x. We can simplify the equation by writing

\frac{dy}{dx} = 2xcos(x^2)

Note that the u in the cosine function is replaced with x^2.…

By | April 29th, 2016|Courses, First year, MAM1000, Undergraduate|0 Comments