## 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 $k$

 How clear is this post?

## 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.…

## 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.…

## Fundamental theorem of calculus example

We did an example today in class which I wanted to go through again here. The question was to calculate

$\frac{d}{dx}\int_a^{x^4}\sec t dt$

We spot the pattern immediately that it’s an FTC part 1 type question, but it’s not quite there yet. In the FTC part 1, the upper limit of the integral is just $x$, and not $x^4$. A question that we would be able to answer is:

$\frac{d}{dx}\int_a^{x}\sec t dt$

This would just be $\sec x$. Or, of course, we can show that in exactly the same way:

$\frac{d}{du}\int_a^{u}\sec t dt=\sec u$

That’s just changing the names of the variables, which is fine, right? But that’s not quite the question. So, how can we convert from $x^4$ to $u$? Well, how about a substitution? How about letting $x^4=u$ and seeing what happens. This is actually just a chain rule. It’s like if I asked you to calculate:

$\frac{d}{dx} g(x^4)$.

You would just say: Let $x^4=u$ and then we have:

$\frac{d}{dx} g(x^4)=\frac{du}{dx}\frac{d}{du}g(u)=4x^3 g'(u)$.…

## Advice for MAM1000W students from former MAM1000W students – part 5

While I resisted Mam1000W every single day, I even complained about how it isn’t useful to myself. Little did I know when it all finally clicked towards the end that even though I wasn’t going to be using math in my life directly, the methodology of thinking and applying helps me to this day.

Surviving Mam1000W isn’t really a miraculous thing. While everyone tends to make it seem like it’s impossible, it is challenging (Not hard) and I said that because I have seen first-hand that practice makes it better each time. Getting to know the principles by actually doing the tuts which is the most important element of the course in my opinion will make sure that even though you feel like you aren’t learning anything when the time comes (usually 2nd semester) it will all click on how you actually are linking the information together.

Another important aspect is playing the numbers game.

## Missing lectures after writing a challenging test – thoughts from a recent MAM1000W student

The following is by one of the current undergraduate tutors for MAM1000W, Nthabiseng Machethe, who has been providing me with extremely useful feedback and her thoughts on the course from the perspective of a recent student of it. She wrote this to me after a lot of students were disappointed with their marks from the last test.

——

This is based on my perspective as a student. I always plan to attend lectures, however as the work load increases and exhaustion kicks in, it is difficult to keep up with the plan.

It is easy to think of the things one may want (like excelling in MAM1000W) but realistically, it is hard to achieve them. In most instances, you find that students are studying a certain concept with a short term vision (passing a test), which can give one instant gratification but may not sustain in the long run (exam). Hence, one tends to quarrel about the time spent studying for the test not equating to the marks.

## 2017 2/3rds numbers game

This is the fourth year that I’ve played the 2/3rds numbers game with my first year maths class. I’m always interested to see how, knowing previous results will affect this year’s results. Of course I am sure that a great deal depends on exactly how I explain the game, and so I imagine that this is the largest confounding factor in this ‘study’.
If you don’t know about the 2/3rds numbers game, take a look at the post here.

Here are the histograms from the last three years:

This year I told the class the mean results from the previous years to see if it would make a difference (as it seemed to last year). This year, the results are somewhat lower:

The winner was thus the person who got closest to 2/3 of 24.4=16.3. This year one person guessed 16, and one person guessed 16.2. Because everyone was asked to write down an integer, unfortunately I can’t claim that 16.2 is the winner, but they will get a second prize.…

## Some more volume visualisations

Here is an animation which may help you imaging a shape which has a circular base, with parallel slices perpendicular to the base being equilateral triangles:

The same thing, where the slices are squares.

And here is the region in the (x,y) plane between $y=\sqrt{x}$, the x-axis and the line x=1. rotated about the y-axis. Here a thin shell is drawn in the volume, then pulled out. Then it is replaced, then the volume is filled with shells, and each of them is pulled out of the volume vertically. This is to give you an idea about how to visualise the method of cylindrical shells.

 How clear is this post?
Gallery

## Guidelines for visualising and calculating volumes of revolution

I have seen some people try to blindly use the formulae for volumes of revolution by cylindrical cross-sections and by cylindrical shells, and I thought that I would write a guide as to how I would recommend tackling such problems, as generally just using the formulae will lead you down blind alleys.

I’ve created an example, with an animation, which I hope will help to master this technique.

So, here is a relatively fool-proof strategy:

1. Draw the region which you are going to have to rotate around some axis. This will generally be a matter of:
• Drawing the curves that you have been given
• Finding where they intersect
2. Draw the line about which you are supposed to rotate the region
3. Draw the reflection of the region about the line of rotation: This gives you a slice through the volume that will be formed
4. Now you have to decide which method to use:
• Take a slice through the volume perpendicular to the axis of rotation.