Recent content by crazyjimbo

When I applied for a PhD this year they told me the deadline for accepting or rejecting the offer was the 31st of March. Apparently this was set by the ESPRC/STFC. Good luck with your applications

You need to be clear enough about what comes from the book and what is your take on it that you can't be accused of plagiarism. If you are worried that your wording is still too close to the authors, maybe say something like: To paraphrase T. Author (2009), this idea can be explained by...

You need to make it very clear that you are quoting the text verbatim. If it is a sizeable passage it's probably best to use a block quote (http://www.englishdiscourse.org/block.quotes.htm ) with the citation either at the end of the quote or in the text leading up to it.

H is not a subgroup since (14)(123)(456) = (123456) which is not in H. Are you sure you've written down the correct group?

If \phi is an isomorphism between groups then \phi(g^n) = \phi(g)^n so \phi must preserve orders. If you can demonstrate there is an element in one group with an order which doesn't appear in the other group then that means there can be no isomorphism.

If you are having trouble getting LaTeX installed and set up (it can be a pain on Windows) then there are sites like http://www.scribtex.com" which let you use LaTeX completely online. There are plenty of tutorials around too and once you get started it's not too bad, honest!

If you remove (a,b) then since you have four elements you can pick an element c which isn't equal to a or b. Try going from there. This is the condition which doesn't hold with only two elements.

If you pick one ordered pair and remove it from that relation you have listed, can you find an example of ordered pairs such that (a,b) and (b,c) is in the relation but (a,c) isn't? I.e. the relation isn't transitive. At first pick an actual pair. You should then see that the case is the same...

What about {(1,1), (1,2), (2,2)}? EDIT: Sorry, I'm just being nit-picky. The general case and idea is sound. It works for 3 elements and above.

I should point out that I hated the statistics I did at university but I was really helped by a book called 'A Basic Course in Statistics' by G M Clarke and D Cooke. I ended up learning most of my course directly from the book rather than the lectures so I can recommend it for self-study. I...

I'm not sure where linear independence comes into it but if X1, ..., X6 are the components of your vector then like christianjb says you will only be able to fix a 3D subspace, where all vectors in it satisfy your equation. If you just need one specific solution, try fixing X1, X2 and X3...

I'm sure you're familiar with the definitions but I'll repeat them for clarity. If we're working on the real line then a test function is an infinitely differentiable function with compact support. The space of all such test functions is denoted D(\mathbb{R}). A distribution is then a...

My understanding has always been that the inner product of a 1-form and a vector is given by \langle \omega, X \rangle = \omega(X) For the inner product of two vectors, we first have to convert the first vector to a 1-form. If X is a vector then the corresponding 1-form or covector is denoted...