All entries for November 2011

November 27, 2011

P4: Giving Presentations and Whatnot

Tutor was Bev Walshe.

Attended workshop on 24th; was good. Material was presented well. Would recommend.

Points:
Calm down
This point is pretty general but I’ll try whatever works. My heart starts racing easily enough, even when I’m alone. On the plus side, this means there’s practically no difference in talking in front of some people or no people: I tend to tune them out once I get going. Points would be breathing (deeply or not) and drinking water to have an excuse to not talk for a while.

Watch people
This comes in tune with “listen to people talk” from A4. I’ve consciously observed some of how body language affects appearance: time to watch some more.


November 16, 2011

Combinatorics – a misread question (graph theory)

While doing combinatorics questions, we come across interesting results by doing the wrong question, whether as a stepping stone (we hope!) to the correct answer, or merely completely misreading the question. Below is one that came up last week:

Find the number of labelled (n-2)-regular graphs with n vertices.
(a k-regular graph is one in which all vertices have degree k)

Clearly for n odd or less than 2 there are no such graphs. The case for even n greater than or equal to 2 runs as follows:

Note that if a vertex has degree (n-2), it is connected to every vertex apart from one other, which is necessarily connected to every vertex bar the first. Thus, we can ‘pair off’ vertices in this manner, obtaining {n \choose 2} {n-2 \choose 2} {n-4 \choose 2}  \cdots = \frac{n!}{(n-2)!2!} \frac{(n-2)!}{(n-4)!2!} \frac{(n-4)!}{(n-6)!2!} \cdots = \frac{n!}{2^{\frac{n}{2}}}.

Edit
We then (thanks Nick!) note that order of pairs doesn’t matter, so we need to divide by the number of ways of choosing the \frac{n}{2} pairs—\frac{n}{2}! – giving us \frac{n!}{2^{\frac{n}{2}}\frac{n}{2}!}

This gives the same result as a different way of thinking about it—(n-1)(n-3)(n-5) \cdots = \sum_{k=1}^{\frac{n}{2}} (n-2k+1)


A4: Last entry: time is up.

Follow-up to A4: Memory and Confusion from Midgley, Christopher - Pointless twaddle and meaningless diatribes

Tutor was Ceri Marriott.

The listening part went well – I can now pay attention and get something out of listening to people talk. Focusing I can even remember them standing and saying it, which is rather nice. It was surprising that something so simple took so long.

The colour/sound part of memory was abandoned – I simply couldn’t get it to work. I’ve moved on to thinking about position but it’s likely that I remember best based on something completely different I’m not yet aware of. I can remember whether we talked about something close to something else, or if there’s a vague link between them (say, the IVT and MVT are both frequently used in Analysis, so that’s a ‘link’ of sorts), but not in subjects where we do incredibly similar things for a long time (e.g. Math Stats); this positioning does help me to remember the material itself.

I find it interesting that the task I thought would be harder turned out to be the only one I actually accomplished :)


November 09, 2011

Emptiness

Consider the definition of a graph – does this preclude the idea of a graph with 0 nodes? No, as both the set of vertices and the set of edges may be empty. This graph exists mostly to make you consider an extra case for every problem you do :)

How many automorphisms are there on the empty graph (how many isomorphisms from the empty graph to the empty graph)? This is equivalent to asking how many bijections there are from the empty set to itself. For a function to be well-defined, we only require that every element in the domain is mapped somewhere in the codomain – if the domain is the empty set, this condition is trivially true (if vacuous). It turns out the answer is one – the empty function.

I wondered about this a while in first year – how would you go about drawing a function if the first set in question was empty? – and I’m glad to see it’s actually something that’s been considered and given an answer.

Edit on 10/11 in reply to Nick:
Well, the definition of a matrix doesn’t preclude the existence of a 0×0 matrix, and I’d suppose that different matrices differ in at least one element, so I agree that that’s the only one.

The question on determinant is much more interesting due to how many different ways there are to approach it. Primarily, I want to check it doesn’t run opposite to my intuition in other areas.

Going back to the definition, we get that \det(A)=\sum_{\phi \in S_n} sign(\phi) \alpha_{1 \phi (1)}\alpha_{2 \phi (2)} \cdots \alpha_{n \phi (n)}. S_0 would be the symmetric group on 0, having 0! = 1 elements (the empty function). So the determinant would be the sum of one number that was the product of no numbers. I’m of the opinion that the sum of no numbers is 1 (the multiplicative identity), so that would make the determinant one in this case.

Considered geometrically, the determinant represent area or volume in two or three dimensions. Extending backwards, we can get that the determinant of a 1×1 matrix acting on a line gives the length of the line under the transformation, and hence that a 0×0 matrix should act on a point. However, as the point is the entire space, this tells us nothing of what the determinant should be.

Considering that every matrix represents a linear transformation, and also that the empty function is bijective, we obtain that ( ) is nonsingular and hence the determinant isn’t 0, which is fine.

Going back to the definition I learnt in high school, where \begin{vmatrix} a&b\\c&d \end{vmatrix}=ad-bc and determinants of matrices of dimension greater than two are defined using minors, cofactors and the above definition, let us try going backwards – can we see what the determinants of 1×1 and 0×0 matrices /should/ be based only on this? Expanding the 2×2 matrix above by the first row, we find that the determinant of (d) should be d and that the determinant of (c) should be c, which fits nicely with the actually definition in addition to the geometric one above. Expanding the 1×1 matrix by the first row, we require the determinant of any 0×0 matrix to be 1 (so that the determinant of the whole thing can be the sole entry), which also fits with our definition above.

But now the geometric line of thinking has lead me into vector spaces! To start, there is no vector space consisting of no vectors (fails presence of an identity). Consider the vector space consisting only of the zero vector. It is a vector space, but we cannot find a basis for it – the zero vector is not linearly independent. However, the vector space is at most of dimension 1 (it has one vector in it), but its basis of length one is linearly depedent, so removing the linearly dependent vectors, we obtain a basis of zero vectors. So this way (by a rather weak argument), the vector space consisting only of the zero vector has dimension zero, as we’d expect.


November 08, 2011

An Introduction to Skills Development Part IV – Afterthoughts

Follow-up to An Introduction to Skills Development Part III – I'm Actually Finding it Difficult to Tell from Midgley, Christopher - Pointless twaddle and meaningless diatribes

Tutor was Samena Rashid.

According to my booklet thing today is the date for the final blog entry! Seems slightly early but doesn’t really matter.

First improvement: I found later on that I should have some way of checking progress that definitely only checks that. When I’m trying to change something about how I think when I’m not thinking about it this is rather vague, and would have been helped had I thought about how much I thought about things before responding before I started this thing, because now I’ve no benchmarks so I can’t really evaluate whether I’ve done better or not on that point.

What went well? I achieved perhaps half what I tried. Trying P12 has lead me to find links, which is always nice. Trying T23 has probably not done much – it may have increased my speed of thinking about a problem, but as I never timed myself or anything before starting I won’t be able to tell. A13 has lead me to puzzle a few ways where my answers disagree with other people’s (example: probability: a different interpretation of “no preference for direction”). Additionally answering the wrong question (or a slight variation) is an interesting aside.

I still don’t think I’ve finished with the plan – it seems a fairly long term thing, and I’ll probably revert to existing behaviours if I stop actively thinking about it. I also may have tried to do too much – while I’ve started on the path, I shouldn’t have expected to be anywhere near the end.


November 06, 2011

A4: Memory and Confusion

Follow-up to A4: Memory and Sense from Midgley, Christopher - Pointless twaddle and meaningless diatribes

Tutor was Ceri Marriott.

Try to remember things based on what was said, as opposed to what was written
I can now listen to people talk (hooray!). I’m still not getting much out of it, and I still definitely prefer images and words, but I can now at least get something out of a speech. Still having difficulty holding things in active memory, but I now find I can remember some of the lecture from looking at my notes.

Take advantage of synaesthesia of memory by associating words with colours/sounds/whatever – give yourself more mental hooks to connect ideas with
This one proceeds in a confusing manner – thinking back to the workshop, I can remember the colour I highlighted the words in, and the position of that on the page in addition to some of the words, but not the ones I highlighted. I’ve also managed to remember a proof fairly well, and also that I associated it with the colours blue and green, but not /why/ I did that or any other information I was trying to convey to myself. This and the above makes me think I should perhaps have focused on /position/ instead of anything else – I think that could be what I currently remember best according to? It’s rather difficult to tell.


An Introduction to Skills Development Part III – I'm Actually Finding it Difficult to Tell

Follow-up to An Introduction to Skills Development Part II – Developing Skills? from Midgley, Christopher - Pointless twaddle and meaningless diatribes

Tutor was Samena Rashid.

T23: I prefer to evaluate the soundness of my ideas before sharing them.
It occurs to me that this one is rather difficult to tell whether I’ve been successful at it, or at the closely related I have better ideas in the first place. If I go a way in and find the idea sound, am I going further than before, or is it just that I’m having better ideas earlier than before? This one’s going well, but I probably should have found some way of benchmarking how much I thought about problems before speaking before – that’s not really something I thought about.

P12, slightly modified: When I hear about a new idea or technique, I immediately start working out how to apply it to other situations/problems – the benefits of said idea/technique.
Example, as requested, is the balls in boxes problem I wrote about down there V, on the 30th. We encountered Stirling numbers and Bell numbers in lecture and I got the inkling that they were relevant, but I didn’t actually get to a solution until I searched the partition numbers so I could find what the sequence was, and found the OEIS link.
MathStats presents new ways of finding out the value of integrals, even if I take down nothing about the probabilistic part of the course :)
I’m still heavily modularising my knowledge, though – I occasionally spot links, but only after the fact. I also don’t query what I know from other subjects while trying to solve a problem (most recently, Jacobian determinants in variable changes in Stats).

A13: I like the challenge of trying out different ways of doing things.
I only sit and think of other possibilities if I’m of the opinion that my current way is rather more difficult than other routes. I still tend to stick to one thing when not actively thinking about it, although I do follow other people’s routes I didn’t try if I like the idea.
Example: Manually finding a matrix to convert the transpose of a Jordan matrix to Jordan form as opposed to proving the Jordan form of the transpose is the Jordan matrix by generalized eigenspaces. The second method is easier and more elegant, but I didn’t consider it, but having the matrix itself is nice.


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  • Nice proof! Does this mean you're going to specialize in analysis and differential equations next ye… by Nick on this entry
  • Hi Chris, It was most interesting to read your various reflections – thank you for sharing them. I'm… by Ceri Marriott on this entry
  • Feel free. Chris by Christopher Midgley on this entry
  • Hi Chris This is an honest final entry for the WSPA. Im glad that you have found the WSPA journey wo… by Samena Rashid on this entry
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