## On useful study habits

I’ve been teaching MAM1000W for around 9 years now, and I am learning all the time. I learn both about new ways to think about old subjects (and how to try and best explain them), and I learn about the way students study, about what works and what doesn’t, and what are some of the habits of students who succeed. Not all of these ideas will be perfect for everyone, but I hope that they will help.

Passive versus active learning

Trying to teach as clearly as possible is a double-edged sword. Of course I want students to come away feeling like they have understood the subject, but if they come away with too much confidence, then they won’t do the one thing which they have to do to actually understand it…and that is practice, but practice of a very particular kind. There is a balance that we should all be thinking about when trying to improve on something (be it sports, music, languages, or maths), and that is finding the right questions to practice on which are hard enough to make us have to sweat a little, but not so hard so as to make us give up completely.…

## A challenging limit

This post comes mostly from the youtube video by BlackPenRedPen found here: https://www.youtube.com/watch?v=89d5f8WUf1Y&t=3s

This in turn comes from Brilliant.com – details and links can be found in the original video

In this post we will have a look at a complicated-looking limit that has an interesting solution. Here it is:

$\lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}}$

This looks pretty daunting – but we will break the solution down into sections:

• taking the logarithms and rearranging
• recognising something familiar
• finding the numerical value

Step 1: Taking the Logarithm

The first step here is to take the logarithm, a generally useful trick when applying limits. First we assign the variable L to the limit (so that we can solve for it in the end). Now lets do some algebra:

$L = \lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}}$

$\ln(L) = \ln(\lim_{n \rightarrow \infty} ( \frac{n!}{n^n})^{\frac{1}{n}})$

Noting that the natural logarithm $\ln$ is a continuous function and therefore we can take the limit outside of the function:

$\ln(L) = \lim_{n \rightarrow \infty} \ln( (\frac{n!}{n^n})^{\frac{1}{n}})$

Next we can use the logarithm laws to bring down the exponent:

$\ln(L) = \lim_{n \rightarrow \infty} \frac{1}{n} \ln(\frac{n!}{n^n})$

Alright, now we have taken the logarithm, step 1 is complete.…

## The definite integral

I realise now, in all the excitement of the FTC that I hadn’t written a post about the definite integral…that’s shocking! ok, here we go…the plan for this post:

• Look at our Riemann sums and think about taking a limit of them
• Define the definite integral
• Look at a couple of theorems about the definite integral
• Do an example
• Look at properties of definite integrals

That’s quite a lot, but we are more or less going to follow along with Stewart. Stewart just has a slightly different style to mine, so I recommend reading his for more detail, and mine for potentially a bit more intuition.

So, let’s begin…

We have seen in previous lectures/sections/semesters/lives that we can approximate the area under a curve by splitting it up into rectangular regions. Here are examples of splitting up one function into rectangles (and, in the last way trapezoids, but you don’t have to worry about this).…

## The Fundamental Theory of Calculus part 2 (part ii)

OK, get ready for some Calculus-Fu!

We have now said that rather than taking pesky limits of Riemann sums to calculate areas under curves (ie. definite integrals), all we need is to find an antiderivative of the function that we are looking at.

As a reminder, to calculate the definite integral of a continuous function, we have:

$\int_a^b f(x)dx=F(b)-F(a)$

where $F$ is any antiderivative of $f$

Remember that to calculate the area under the curve of $f(x)=x^4$ from, let’s say 2 to 5, we had to write:

$\int_2^5 x^4 dx=\lim_{n\rightarrow \infty}\sum_{i=1}^n f(x_i)\Delta x=\lim_{n\rightarrow \infty} f\left(2+\frac{3i}{n}\right)\frac{3}{n}=\lim_{n\rightarrow\infty}\frac{3}{n}\left(2+\frac{3i}{n}\right)^4$

And at that point we had barely even started because we still had to actually evaluate this sum, which is a hell of a calculation…then we have to calculate the limit. What a pain.

Now, we are told that all we have to do is to find any antiderivative of $f(x)=x^4$ and we are basically done.

Can we find a function which, when we take its derivative gives us $x^4$?…

## The Fundamental Theory of Calculus part 2 (part i)

OK, now we come onto the part of the FTC that you are going to use most. We are finally going to show the direct link between the definite integral and the antiderivative. I know that you’ve been holding your breaths until this moment. Get ready to breath a sign of relief:

The Fundamental Theorem of Calculus, Part 2 (also known as the Evaluation Theorem)

If $f$ is continuous on $[a,b]$ then

$\int_a^b f(x) dx=F(b)-F(a)$

where $F$ is any antiderivative of $f$. Ie any function such that $F'=f$.

————-

This means that, very excitingly, now to calculate the area under the curve of a continuous function we no longer have to do any ghastly Riemann sums. We just have to find an antiderivative!

OK, let’s prove this one straight away.

We’ll define:

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

and we know from the FTC part 1 how to take derivatives of this. It’s just $g'(x)=f(x)$. This says that $g$ is an antiderivative of $f$.…

## The Fundamental Theorem of Calculus part 1 (part iii)

So, we are now ready to prove the FTC part 1. We’re going to follow the proof in Stewart and add in some discussion as we go along to motivate what we are doing. What we are going to prove is that:

$\frac{d}{dx} \int_a^x f(t) dt=f(x)$

for $x\in [a,b]$ when $f$ is continuous on $[a,b]$.

Proof:

we define $g(x)=\int_a^x f(t)dt$ and we want to find the derivative of $g$. We will do this by using the fundamental definition of the derivative, so let’s look at calculating this function at $x$ and $x+h$ – ie. how much does it change when we change $x$ by a little bit?

$g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt$

But remember that the definite integral is just the area, so this difference is the area between a and x+h minus the area between a and x. Which is just the area between x and x+h. Using the properties of integrals, we can write this formally as:

$g(x+h)-g(x)=\int_a^{x+h}f(t) dt-\int_a^x f(t) dt=\left(\int_a^{x}f(t)+\int_x^{x+h}f(t)\right)-\int_a^{x}f(t)=\int_x^{x+h}f(t)dt$

and we can write, for $h\ne 0$:

$\frac{g(x+h)-g(x)}{h}=\frac{1}{h}\int_x^{x+h}f(t)dt$

Restated, we can think of this as the area between x and x+h divided by h.…

## The Fundamental Theorem of Calculus part 1 (part ii)

OK, so up to now we can’t actually use the FTC (Fundamental Theorem of Calculus) to calculate any areas. That will come from the FTC part 2.

For now, let’s take some examples and see what the FTC is saying. I’ll restate it here:

The Fundamental Theorem of Calculus, part 1

If $f$ is continuous on $[a,b]$ then the function $g$ defined by:

$g(x)=\int_a^x f(t) dt$,     for $a\le x\le b$

is continuous on $[a,b]$ and differentiable on $(a,b)$, and $g'(x)=f(x)$.

——

Let’s look at some examples. We’re going to take an example that we can calculate using a Riemann sum. Let’s choose $f(x)=x^2$.

If we integrate this from $0$ to some point $x$ – ie. calculate the area under the curve, we get:

$\int_0^x t^2 dt=\frac{x^3}{3}$.

Make sure that you can indeed get this by calculating the Riemann sum.

So, what does the FTC part 1 tell us? It says that if we take the derivative of this area, with respect to the upper limit, $x$, then we get back $f(x)$.…

## The Fundamental Theorem of Calculus, part 1 (part i)

We’ve seen some intriguing things in this course so far, and we’ve developed some clever tricks, from how to find the gradient of just about any function we can throw at you, to proving statements to be true for an infinite number of cases.

To some extent, this is what we have looked at so far (at least in terms of calculus, and building up to calculus):

However, we’re about to see some magic. We’re about to see the most important thing yet on this course, and indeed one of the most important moments in all of mathematical history.

We are going to see…actually, we are going to prove, that there is a relationship between rates of change and the area under a graph. This doesn’t sound that amazing, but its consequences have essentially allowed for the development of much of modern mathematics over the last 350 years.

The link that we are going to prove will allow us to find the area under graphs of functions for which taking the Riemann sum would be really hard.…

## All you’ve ever wanted to know about absolute values (and weren’t afraid to ask)

I’ve been getting a lot of questions about absolute values, and so I thought I would try and clarify things here as much as possible. I’ll give some basic definitions and intuition, and then go through some examples, from easier to harder.

The absolute value function is just….a function. You give it a number, and it returns a number. In the same way that $f(x)=x^2$ is a function. You give it a number and it returns that number multiplied by itself. So the absolute value function, which we write as $f(x)=|x|$ takes a number and returns the same number if the number was positive, and the negative of the number if it was negative, thus returning always a positive number.

We can think of this as the function “how far away from the point 0 (the origin) on the real number line is x?”. It doesn’t care about what direction it is, only how far away it is.…