Using integration to calculate the volume of a solid with a known cross-sectional area.

Hi there again, I have not written a post in while, here goes my second post.

I would like us to discuss one of the important applications of integration. We have seen how integration can be used to solve the area problem, in this post we are going to see how we can use a similar idea to solve the volume problem. I suggest that we start by looking at the solids whose volume we know very well. You should be able to calculate the volumes of the cylinders below (yes,  they are all cylinders.)


circular cylinder                                 rectangular cylinder                triangular cylinder

Cylinders are nice, we only need to multiply the cross-sectional area by the height/length to find the volume. This is because they have two identical flat ends and the same cross-section from one end to the other. Unfortunately, not all the solid figures that we come across everyday are cylinders. The figures below are not cylinders.…

An integral expression for n!

I gave a challenge question at the end of class a week or so ago. Here I will give the solution and show that it gives us something rather strange and surprisingly useful.

I wrote down the following, and asked you to prove it:


\int_0^\infty e^{-t} t^N dt=N!


For N\ge 0, N\in \mathbb{Z}. Now, N! can be thought of as the number of different orderings of pulling N objects out of a bag (without replacement) when they are all different. If you have N things in a bag, then there are N possible things that you can pull out first. There are then N-1 ways of pulling out the next object, N-2 ways of pulling out the next, etc. and finally, when you’ve pulled out N-1 objects there’s only a single possibility of pulling out the last. So:




And the number of ways of pulling no objects out of a bag is 1, because you just don’t pull anything out.…

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

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

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

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.



-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

How clear is this post?
By | 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 | 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.




Every fifth Fibonacci number is divisible by 5.




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.




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

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:




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:




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

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 published in frontiers in human neuroscience.

How clear is this post?
By | October 14th, 2015|English, Level: intermediate, News|0 Comments