Advanced Higher Maths 2011-2012 : Discussion and Help

At school I would wildly have disagreed, as putting in more stats, would involve taking out more maths

I admit, the above makes me think this may be in part because I like the idealised 'maths for, and only for, it's own sake' idea..

$Maths \Leftrightarrow Pure \ Maths$?

I'd be interested in hearing such a speech.
I'll be the first to say that I'm completely ignorant of anything in the field of statistics, not that I'm disinterested by it, I've just never been given a path into it and I wouldn't know where to start by myself.

Let us hear it!

I love Stats but there still is sometimes when I'm like, when will we use this/what is this for.

It's getting late, so I'll skimp on the eloquence if that's alright with you folks. But, basically, an education in statistics will allow you to look critically at the empirical arguments people make in attempts to influence you. The statistics needed to make it much more difficult to pull the wool over your eyes really is pretty basic, but you're pretty much given no serious exposure to it at all, at any point in the curriculum. Given the rife misuse of statistics in the media, I think the ability to intelligently filter this and inform yourself is vital.

Part of it comes from taught knowledge. Fundamentals like statistical significance, and the existence, suitabilities and strengths (in the sense that e.g. a parametric test is weaker than a parameterless one) of different tests for such. Confidence intervals, correlation without causation, variance and covariance. Independence, distributions, the Central Limit Theorem. Bayesian analysis, and a handful of regression methods. All of these are things you cover, if not in Advanced Higher then certainly in the first few weeks of a university course. This isn't rocket surgery!

Some of it is from increased exposure to things which you might not be likely to encounter in day-to-day life. At least, until you're made aware of them and then you see them everywhere Read the article on Simpson's paradox. Do it now. It will take you maybe five minutes to read, another five to grasp. Again, this is Statistics 101 but you will see reports written which manage to forget it and draw false conclusions (cf the Berkeley example in that link) all the freaking time. There's a whole bunch of stuff like this which an introductory course would cover, which really makes you think, and which is just ignored.

But I think some of the real value is less easily definable, is easily written off as 'soft' skills, and really comes just from the experiences you're put through. You have critical thinking, but with a set of tools that let you formalize it. Maybe I'm straying a litle towards the scientific method, but some of it comes from the way you have to ask 'why?' and understand that you don't just apply these tools blindly. Experience means you can rattle off a list of reasons that correlation might not imply causation, and see how to control for these. Similarly, you can look at a survey presented in the news and criticize its method and highlight sources of bias. And so on.

Now, I really am talking about the fundamentals here, nothing fancy. But, as you break out the gradually bigger guns of statistics, there is much more that you can analyse and so much more cool stuff that you can do! When I graduate, I've got an idea for a text-input system for people with limited motor control using a Markov model for prediction (the prediction isn't a new idea, the interface is). My dissertation project involves using a bunch of statistical algorithms to take a moving camera and figure out its location and orientation and a low-resolution map of the room. My group project last year involved doing as much analysis as we could of a corpus of financial data to see if there was a way to make money out of it (there wasn't, sadly!).

Finally, this is the information age. People have been talking about this for years, but I think the sun's really only rising on it now. Every business - every business - is looking at their Big Data and wondering what they can do with it. There are engineering problems here (how to store it, how to query it, how to get it where it needs to be) but it is statistics which will tell you what data you need, what data you can throw away, and how to learn from what you've got. If you've got a solid grasp of statistics, and ideally the ability to throw together some code, you are a both incredibly valuable and in short supply. The demand is there now.



To be fair, it is true that many students will never need to e.g. solve a differential equation themselves. That said, I found an effective way of staying such complaints in short order was to respond with a sizeable list of applications of each topic so declared until the questions became more "why is this useful?" Moving from "I don't need this, therefore it is useless" to "this is probably useful for something, even if I don't need to do that something itself" means that you can then inspire interest by going "look at this really awesome thing; the tool you've just learnt is the one that lets you do this thing - you could, at least in principle, do this yourself!"

I really can't think of much in the syllabus that isn't particularly practical.



Anything in particular in mind? I would say 'statistics' is probably the tool I use most often, although since doing so usually implies employing at least some mathematical nous as well, perhaps I'm somewhat biased.

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And, for you... Why does this formula for Spearman's rho have a six on the top?



(Hint: this is actually pretty simple, although ukd will doubtless swoop in with some much deeper insight than I have in mind )

No "deep insight" I'm afraid. If there are no ties it comes directly from xi and yi being consecutive. In general the coefficient comes from Faulhaber's formula and Bernoulli numbers.

PRSOM on the speech by the way

^ that's true.

You two come in whenever there's a maths query, solve it 15 different ways and then leave everyone wondering how anyone can ever know all that stuff.

And that's why we love you for it.

Edit: crap, ukd, I just remembered that I didn't reply to your last PM and I deleted all of them. Can you send me the question again so I can actually have a stab at it? (Some calculus if it's of interest to anyone.)

You and ukd have become gods around here...

Quick question. If you had a complex number $z=2+2\sqrt{3}i$ then putting it in polar form would yield $4(cos(\frac{\pi}{3})+isin(\frac{\pi}{3}))$.

If we were then given $z^2=2+2\sqrt{3}i$ then it's just as simple as taking the square root of $2+2\sqrt{3}i$ or, in polar form, taking the square root of r and halving the values for theta by de Moivre's theorem, yes? So in that case it'd be $2(cos(\frac{\pi}{6})+isin(\frac{\pi}{6}))$, yes?

I hope that makes sense. I was screwed with tiredness and a headache when I first looked at this so it made no sense to me!

There would be two roots.

So, yes one of the roots would be what you said then the other one would be

z= 2(cos(pi/6 + 2pi) + isin(pi/6 + 2pi) ) then simplify.

So if it was z^3 = .... you would anticipate 3 roots, z^4 - 4 roots and so on.

Edit: I've been crazy revising complex numbers at the moment :P

There would be two roots.

So, yes one of the roots would be what you said then the other one would be

z= 2(cos(pi/6 + 2pi) + isin(pi/6 + 2pi) ) then simplify.

You're almost right.
Should be pi, not 2pi.

Quick question. If you had a complex number $z=2+2\sqrt{3}i$ then putting it in polar form would yield $4(cos(\frac{\pi}{3})+isin(\frac{\pi}{3}))$.

If we were then given $z^2=2+2\sqrt{3}i$ then it's just as simple as taking the square root of $2+2\sqrt{3}i$ or, in polar form, taking the square root of r and halving the values for theta by de Moivre's theorem, yes? So in that case it'd be $2(cos(\frac{\pi}{6})+isin(\frac{\pi}{6}))$, yes?

I hope that makes sense. I was screwed with tiredness and a headache when I first looked at this so it made no sense to me!

Note that for this question you don't *have* to transform it to polar form. If you can see sufficiently far ahead then you can do:

${z^2} = {\left( {a + bi} \right)^2} = {a^2} - {b^2} + 2abi = 2 + 2\sqrt 3 i$

Hence:



So by inspection $a = \pm \sqrt 3$ and $b = \pm 1$ and the answer is $z = \pm \left( {\sqrt 3 + i} \right)$