Episode 2

OOMC ep 2 Unseen MAT

In episode 2, we look at some unseen MAT questions with help from David (first-year student), before returning to the square problem from episode 0 and the cube problem from episode 1.


Further Reading

MAT questions

We looked a bit at the function $\frac{x}{x^2+4}$. David made an interesting observation, which is that when $x$ is very large, this function is approximately zero. If you haven’t seen that sort of argument before, it might be worth taking a moment to think about what happens to the numerator and denominator of the function when $x$ is a very large number. Can you extend this idea to think about what would happen to the function $x^2+\frac{3}{x^2}+1$? What about $\frac{\sin(x)}{x^2}$? This is related to the idea of finding the “limit of the function as $x$ tends to infinity”.

David also challenged the Y13s to work out the derivative of $\cos\left(\frac{240 x}{x^2+4}\right)$. Enjoy! If you haven’t met the quotient rule and the chain rule yet, ignore this challenge (those links go to Wolfram MathWorld. Note that the quotient rule has a very short article, and the chain rule has a rather longer article, only the first sentence of which is relevant here).

Return of the squares

I really struggled to work out how big the square root of 10 is at the start of this segment. Here’s a good way I could have worked out an approximation for the square root of 10 in my head.

10=9+1 so the square root of 10 is a bit more than three. In fact, if we take out a factor of three, then the square root of 10 is 3 times the square root of (1+1/9). Now the square root of (1+1/9) is about 1+1/18, so the square root of 10 is about 3+1/6=3.17 or thereabouts, which would have been good enough for what I was trying to do.

Now hang on a minute, you might be thinking, how do you know that the square root of (1+1/9) is about 1+1/18 ? That’s an approximation that we get from the binomial expansion of $(1+x)^{1/2}$. You might have already seen how to expand $(1+x)^n$ when $n$ is a whole number, in which case you get a nice expansion of $n+1$ terms. When $n$ isn’t a whole number, you get infinitely many terms, which is a bit harder to imagine! The terms get smaller and smaller (provided that $-1<x<1$). It’s a bit like the ideas we explored last week with a series of infinitely many things to add up. All that I’m going to say here is that $(1+x)^n$ is approximately $1+nx$ when $-1<x<1$.

I’ve noticed lots of Desmos fans in chat, so here’s a Desmos investigation for you to try.

Open desmos.com and draw a graph of $(1+x)^{1/2}$

Add a graph of $1+x/2$. Zoom in and have a look at the two graphs between $-1$ and $1$.

Add a graph of $1+x/2-x^2/8$. Is that better or worse?

Extension for people who like Desmos; draw graphs of $(1+x)^n$ and $1+nx$ and $1+nx+n(n-1)x^2/2$. Add a slider for $n$. Add extra Desmos features that I don't know about.

Extension; plot $\frac{x}{x^2+4}$ and $\frac{1}{x} - \frac{4}{x^3}$. What’s going on with this one?


If you want to get in touch with us about any of the mathematics in the video or the further reading, feel free to email us on oomc [at] maths.ox.ac.uk.

Please contact us with feedback and comments about this page. Last updated on 29 Apr 2022 12:07.