Recent content by sat

Seems quite reasonable. Thanks. I'll look into this tomorrow.

It's fairly well known that isometries in Euclidean space are composed of only translations, reflections and rotations. However, I'm finding it difficult to locate a proof of that. As usual, it's "intuitively obvious" but formally I'm not sure where to start. Does anyone know of a good...

We do seem to have a group, led by Wilson SIbbet, interested in ultrafast pulse lasers. Much of the department's research is in optics of one sort or another.

http://www.st-andrews.ac.uk/~ulf/invisibility.html For a theorist's approach.

Thanks Thanks for that. I do agree with what you've said though I thought perhaps there might be some way of manipulating it so that in this special case you could show it. b^{2}=b^{3} certainly sounds like a way forward.

Would it be possible to infer that b^5 = e (where e is the group's identity element) from b^{5} a = ab^{5} given that a^{2}=e? (Basically we are given b^{2}a=ab^{3} and a^{2}=e and asked to show that b^{5}=e, though I've managed to infer the "equation" above and I can't quite see how...

Thanks. That seems to solve the problem of making it work for x_{1}x_{2}\ldots x_{n}\neq 1

Well, 7 pi/6 is "the same as" pi/6 but in the 3rd quadrant. If you think of a cosine graph, then when you get to pi, the graph is -1. pi is equal to 6pi/6, so 7pi/6 is just that plus pi/6. So you evaluate cos pi/6 but make it negative since the function is negative here. That's not a very...

Let x_{1},x_{2},\ldots,x_{n} be real and positive. Show that g\equiv \sqrt[n]{x_{1}x_{2}\ldots x_{n}} < \frac{x_{1}+x_{2}+\ldots+x_{n}}{n} \equiv a except when x_{1}=x_{2}=\ldots = x_{n} in which case a=g. We are given the results (from previous exercises) that \forall x>0 ...

Thanks. I'll look at that.

I just did: du/dx = (du/dr)(dr/dx) + (du/dtheta)(dtheta/dx) though I think that's probably not quite right when I think about the dependence of the variables on each other... Though the "units" look OK at first sight, I'm not sure that I can do that.

Take x=r\cos\theta and y=r\sin\theta If f(z)=u(r,\theta) + iv(r,\theta), is analytic with u and v real, show that the derivative is given by f'(z) = \left( \cos\theta \frac{\partial u}{\partial r}- \sin\theta\frac{1}{r}\frac{\partial u}{\partial\theta}\right) + i\left(...

I am assuming that by the Lagrange error you mean the expression \frac{f^{(n+1)}(c)}{(n+1)!}(x-x_{0})^{n+1} This is almost the same as the expression for the (n+1)th term, but (as you seem to have noticed) the derivative of f is evaluated at some c between x and x_0. Basically the c can...