## Prove that for every positive integer n, 9^n – 8n -1 is divisible by 64.

Prove that for every positive integer $n$, $9^n-8n-1$ is divisible by 64.

This question screams proof by induction, so we start with the base case, which in this case is $n=1$:

$9^1-8-1$ which is indeed divisible by 64.

Now, let’s assume that it holds true for some positive integer $n=k$. ie:

$9^k-8k-1=64p$ for $p\in\mathbb{Z}$.

Now let’s see how we can use this to prove that the statement holds true for $n=k+1$. For $n=k+1$ we have:

$9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(9^k-8k-1)+64k$

where we have manipulated the expression to contain the left hand side of the inductive hypothesis. Thereby, plugging in the inductive hypothesis, we get:

$9^(k+1)-8(k+1)-1=9(9^k)-8k-8-1=9(64p)+64k=64(9p+k)$

but clearly $9p+k$ is an integer, so this is divisible by 64 and thus the statement holds true for $n=k+1$, thus it holds true for all positive integers $k$

## A tricky complex numbers problem

The question is as follows:

If $\frac{\pi}{6}\in arg(z+a)$ and $\frac{2\pi}{3}\in arg(z-a)$ and $a\in \mathbb{R}$, find $z$.

So, let’s think about the information given and what we are trying to find. We want to find the complex number $z$ which satisfies this slightly strange set of constraints, and the constraints are given in terms of $z$ and $a$. So, by the looks of things, the answer will depend on $a$ and so the final expression should be a function of $a$.

Now let’s explore the constraints. In fact, let’s simply take $z+a$ and $z-a$ as two complex numbers, but importantly two numbers which differ only by a real number $2a$, so wherever they lie in the complex plane, they have the same imaginary part and differ only by an real part.

Now, the constraints are about the arguments of the two complex numbers. It doesn’t tell us anything about the magnitude of the numbers, so all the information tells us is the direction are in relation to the origin.…

## Using polynomials to solve differential equations

One of the aims of MAM1000W isn’t just to teach you individual mathematical topics, but over time to allow you to see the links between these subjects. Sometimes we do this explicitly, and sometimes you should notice the connections yourself simply by seeing one topic pop up in the middle of another. As I’ve written before, so much of it is about noticing patterns.

Today in class I gave a differential equation which wouldn’t be solvable by any of the methods we have looked at.

$y''(x)+\cos(x)y'(x)+e^xy(x)=x^2$

This is second order linear but its coefficients are not constant. We don’t have any way in to solve this. We actually wanted to solve this with the initial conditions $y(0)=1$ and $y'(0)=-1$.

Actually, that’s a lie. We didn’t want to solve it, but we wanted to get an approximation for the solution close to $x=0$. This is like saying: OK, so we have a differential equation for population dynamics, or climate change, or the heating of an object, and we don’t worry too much about the very far future, but we want to know what it’s going to do in the short term.…

## The Diagnostic Mathematics Information for Student Retention and Success (DMISRS) Project

Presentation by Robert Prince, UCT at the Teaching and Learning of Mathematics Communities of Practice meeting at UJ, 29 – 30 August 2018

The Diagnostic Mathematics Information for Student Retention and Success (DMISRS) Project

The problem: Only 27% of students entering full-time university in 2006 graduated in minimum time.

40% leave higher education.

41% of engineering and 48% of science 2006 entrants graduated in 5 years.

Educational diagnostic testing is assessment before instruction begins.

DMISRS – a collaboration by mathematicians to improve graduation rates.

• Make use of NBT data to inform students and lecturers about what areas of weakness and strength are.
• Share practices, leverage best practices.
• Extend the reach of academics beyond a physical classroom.
• Create supportive environments for maths learning [maybe using positive psychology]. What kinds of things will make our classrooms more welcoming to students.

Objectives

1. Get more institutions on board to collect diagnostic information in the same style/language.

## Future Planning of the USAf Teaching and Learning of Mathematics Community of Practice

Professor Rajendran Govender from the University of the Western Cape presented the objectives and future plans of the Universities South Africa (USAf) Teaching and Learning of Mathematics Community of Practice (TLM CoP) at the 2-day meeting at the University of Johannesburg, 29 – 30 August 2018

• In accordance with the principles guiding a community of practice, the TLM CoP provides an opportunity for academics and relevant other university staff members to collaborate, network and share knowledge on issues of common interest or concern.
• The objectives of the TLM CoP are to promote and strengthen the teaching and learning of Mathematics in public universities in South Africa by:

1.Developing and recommending strategies for the sector to ensure improved access and success in the teaching and learning of Mathematics, thus contributing directly to the transformation needs of South Africa. (session dedicated to this at the next meeting)

2.Providing a shared, common platform from which successful initiatives may be disseminated.…

## Radically transforming mathematics learning experiences: Lessons from the Carnegie Math Pathways

Siyaphumelela Conference 2017, The Wanders Club, Johannesburg

Andre Freedman, Capital Community College

Bernadine Chuck Fong, Carnegie Math Pathways

Workshop goals:

• Learn about the design, goals, implementations of Carnegie Math Pathways
• Experience Pathways lessons
• Engage in design tasks to improve student success in maths and college
• Engage in conversations about professional learning to address issues and concerns that are specific to local contexts

Faculty had to learn new ways to teaching maths, there had to be ‘buy in’ for it to be successful.

How to radically transform outcomes for all mathematics students?

• Completion
• persistence
• quality of learning (e.g. students can explain what a function is years after taking a maths course)
• identities of learning (students see themselves as someone who can do maths).

Make maths a Gateway not a Gatekeeper.

How?

• Acceleration: Rather not have 3 developmental courses (pre-algebra, algebra1, 2) before taking the required course
• Problem-centered curriculum
• Student-focused, collaborative pedagogy
• ‘Productive Persistence’ interventions / practices to give students belief that they can succeed
• Language and literacy supports
• Train faculty so they can feel comfortable about the new approach
• Use networking to support staff to get running and sustain change
• Keep cohort together for 2 semesters in classes of 30 – 40, or as one institution did teach the 2 semesters in one term (quarter of a year) with about 5 hours a day and one other course -success rate was very high at 78%.

## How Behavior Spreads: The Science of Complex Contagions, by Damon Centola, a review

NB. This book was sent to me as a review copy.

The idea of this book is relatively simple, but the consequences are huge, and in fact some of the ideas are far more subtle and complex than they may first appear.

Essentially this book is based on a series of experiments which Damon Centola has run, which are all related to changes in behaviour which can be tracked, and made to occur, through a social network (in the broadest sense of the word). This is the study diffusion in a network.

The fundamentals of the research lie on two distinctions: One in the complexity of a contagion/behaviour, meaning how many connections with others who have the contagion/behaviour do you need until you adopt it, and the other in the topology of the social network, meaning loosely, how much like a street where each person only talks to their neighbours, versus a small world-network where there are a lot of disparate connections does the network look like.…

## Why did we choose that range for theta when doing trig substitutions?

Remember when we are doing a trig substitution, for instance for an integral with:

$\sqrt{a^2-x^2}$

We said that we should choose $x=a\sin\theta$, which seemed reasonable, but we also said that $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$. Where did this last bit come from?

Well, we want a couple of things to hold true. The first is that any substitution that we make, we have to be able to undo. That is, we will substitute $x$ for a function of $\theta$ but in the end we need to convert back to $x$ and so to do that we have to be able to write the inverse function of, in this case $x=a\sin\theta$. The $\sin$ function is itself not invertible because it’s not one to one, so we have to choose a range over which it is one to one. We could choose $-\frac{\pi}{2}\le\theta\le\frac{\pi}{2}$ or we could choose $\frac{\pi}{2}\le\theta\le\frac{3\pi}{2}$ (amongst an infinite set of possibilities). That would also be invertible. However, remember that we are going to end up with a term of the form:

$\sqrt{1-\sin^2\theta}=\sqrt{\cos^2\theta}$

So if we want this to simplify, we had better choose our range of $\theta$ such that $\cos\theta$ is positive, so that we can write $\sqrt{\cos^2\theta}=\cos\theta$.…

## Integrals with sec and tan when the power of tan is odd

We went through an example in class today which was

$\int tan^6\theta \sec^4\theta d\theta$

In this case we took out two powers of sec and then converted all the other $\sec$ into $latex\ tan$, which left a function of tan times $sec^2\theta d\theta$. We wanted to do this because the derivative of $\tan$ is $\sec^2$ and so we can do a simple substitution. If we have an odd power of $\tan$, we can employ a different trick. Let’s look at:

$I=\int \tan^5\theta\sec^7\theta d\theta$.

Here, sec is an odd power and so we can’t employ the same trick as before. Now we want to convert everything to a function of $\sec$ and have only a factor which is the derivative of $\sec$ left over. The derivative of $\sec$ is $\sec\tan$, so let’s try and take this out:

$I=\int \tan^5\theta\sec^7\theta d\theta=\int \tan^4\theta\sec^6\theta (\sec\theta\tan\theta)d\theta$.

Now convert the $\tan$ into $\sec$ by $\tan^2\theta=\sec^2\theta-1$:

$I=\int (\sec^2\theta-1)^2\sec^6\theta (\sec\theta\tan\theta)d\theta=\int (\sec^{10}\theta-2\sec^8\theta+\sec^6\theta) (\sec\theta\tan\theta)d\theta$

where here we have just expanded out the bracket and multiplied everything out.…