Well another month has passed and I'm about two thirds of the way through TMA05 for M208 on Group Theory and TMA03 for M337 Complex Analysis it's all coming together I really enjoyed the Calculus of Residues unit C1 on M337 it's amazing how productive the residue theorem is, all sorts of complicated integrals and series can be reduced to a simple contour integration and the calculation of a few residues for once I didn't feel I was having to scratch my hand vis a vis the 1st TMA question. Still the other two questions are a bit more stretching. Anyway Complex analysis is really clever and I hope to do it justice for the rest of the course having learnt some of the mechanics I hope to go back and look at it in more depth.

The M208 group theory was quite tricky in parts especially as it involves visualisation of symmetries of an object. Fortunately I found a way of systemising the symmetries as an adaption of the way Chemists treat molecules there are a few key points

1) The number of direct symmetries is identical to the number of indirect symmetries

Got my wrist slapped for pointing this out on the M208 forum rather unfairly I thought but heigh ho.

2) If there is one principal axis of rotation then the composite symmetries involve a rotation about the principal axis followed by a reflection in a plane orthogonal to the axis.

3) When finding normal subgroups as unions of conjugate classes you can immediately discount any subgroup which consists entirely of indirect symmetries as the composite of two indirect symmetries is a direct symmetry so that a proposed subgroup consisting entirely of indirect symmetries plus the identity will not satisfy closure so can't be a subgroup.

Would probably get my wrists slapped again for pointing these simple facts out on the M208 forum so I'll point them out here.

As well as that continuing with my distraction in trying to understand from a purely algebraic point of view the invariants of polynomials as a prelude to getting into Galois Theory. Those readers who want to amuse themselves might like to try to derive the Cardano formula for a cubic by carrying out the following steps

1) Reduce a cubic of the form

$$x^3 + bx^2 + cx +d = 0 $$ to the form

$$y^3 + py +q = 0$$

by making the transformation

$$y = x -\frac{b}{3} $$

You should be able to show that

$$p = \frac{-b^2}{3} + c $$

and

$$q = \frac{2b^3}{27} - \frac{bc}{3} + d $$

Then in the equation for y make the substitution

$$ y = z - \frac{p}{3z} $$

to transform the equation for y into a sextic for z

$$z^6 + qz^3 - \frac{p^3}{27} = 0 $$

This may appear to have made matters worse but if we substitute

$$ s = z^3$$ then the sextic becomes a quadratic in s

$$s^2 + qs - \frac{p^3}{27}=0 $$

with solutions

$$s1 = \frac{1}{2}(-q + \sqrt{q^2 + \frac{4p^3}{27}})$$

and

$$s2 = \frac{1}{2}(-q - \sqrt{q^2 + \frac{4p^3}{27}})$$

from which the original cubic can be solved.

This is relatively straightforward however as is well known applying such techniques to the quartic get a bit more complicated and Galois showed that such general formula could not be obtained for higher polynomials such as quntics. However whilst in general such general solutions are not available nevertheless a procedure was developed as an extension of the above techniques for deciding whether or not a quintic is solvable and if it is how can a solution be obtained. One such procedure was given by Watson and a summary of his studies which builds on the work of Cayley, P C Young and others is given here

http://www.math.carleton.ca/~williams/papers/pdf/244.pdf

Some of the expressions look horrendous but as it's 'just algebra' it would be a relatively straightforward exercise to verify these expressions. All of this makes no reference to Galois theory as it has developed and group theory is only hinted at so it should be an amusing exercise to try and verify the expressions and even maybe write a computer program to solve a quintic I'll let you know how I get on.

Chris,

ReplyDeleteHow are you finding the workload, studying the 60 points of Pure maths at the same time as the 30 points of complex analysis?

Not to bad actually but I did do a lot of preparatory work by reading Brannan. Also I don't tend to do all the exercises or get to bothered if I don't understand all the proofs. There is a three stage process

ReplyDelete1) Learn the mechanics sufficiently to do the TMA's reading through some of the proofs

2) Become proficient at the mechanics sufficient for the exam and reread the proofs

3) After the exam let it all sink in and come back to it maybe now the deep knowledge will sink in and some of the proofs won't be so complicated.

I would say I need to do about 2 hours every alternate day and 8 hours over the weekend.

Chris,

ReplyDeleteDon't know if you've seen this one?

http://www.wildd.freeserve.co.uk/

M337 exam paper solutions for 10yrs worth of OU exams.