Recent content by Anthony

The equality is meant in the sense of distributions. For \varphi \in C^\infty_c(\mathbf{R}) the equality should be read as \int_0^\infty \left(\int_{-\infty}^\infty e^{\mathrm{i}\omega t}\varphi(\omega)\, \mathrm{d} \omega \right)\mathrm{d} t = \lim_{\epsilon\rightarrow 0}...

No problem - glad to be of service.

Alternatively, for arbitrary fixed vector \mathbf{a} \in \mathbf{R}^3 \mathbf{a} \cdot \oint \mathbf{r} \wedge \mathrm{d}\mathbf{r} = \oint \mathbf{a} \cdot (\mathbf{r} \wedge \mathrm{d}\mathbf{r}) = \oint (\mathbf{a} \wedge \mathbf{r}) \cdot \mathrm{d}\mathbf{r} = \iint \nabla \wedge (...

Your argument has the right idea. Let f\in C^\infty_c (\mathbf{R}) and choose \epsilon>0 sufficiently small so that \mathrm{supp}(f) \subset (-1/\epsilon, 1/\epsilon). Then f has the convergent Fourier series f(x) = \sum_{n\in \mathbf{Z}} c_n e^{\mathrm{i} \pi \epsilon n x} where c_n =...

You might like to prove the following: let X be compact and Y Hausdorff, then if f:X\rightarrow Y is bijective and continuous, then it is a homeomorphism.

That's not really true. It's straight forward to prove that as a subset of \mathbf{R}^{n^2}, the matrices in \mathrm{Mat}_n(\mathbf{R}) that are not diagonalisable form a set of Lebesgue measure zero. So in this sense, most matrices can be diagonalised.

Consider the vector field defined by: \mathbf{A}(\mathbf{x}) = \int_0^1 \mathbf{B}(\lambda \mathbf{x}) \wedge (\lambda\mathbf{x})\, \mathrm{d}\lambda. You might like to show that if \nabla\cdot\mathbf{B}=0, then \nabla \wedge \mathbf{A} = \mathbf{B}. Obviously this \mathbf{A} is not unique.

The book by Ablowitz and Fokas is very accessible and reaches a wide range of topics. :)

Apologies, I only read the first bit! That will teach me.

Well, using your notation we have: (\mathbf{x} - \mathbf{a})\cdot \mathbf{n} =0 \quad \textrm{means} \quad Ax+By+Cz = \mathbf{n}\cdot \mathbf{a} So your D is given by: D = - \mathbf{n}\cdot \mathbf{a} = -|\mathbf{a}| \cos \theta = |\mathbf{a}| \cos (\theta - \pi) where \theta is...

This can be seen as follows: if f(z)=u(x,y)+\mathrm{i} v(x,y) is complex analytic, then u and v are harmonic hence (real) analytic. Yes, given a real analytic function on an (a,b), it can be extended to a complex analytic function.

Excellent, well done! You might not need complex analysis to evaluate integrals of this form, but it's the standard treatment. I'm sure there's actually a book containing lots of integrals that are usually done via complex analysis, but explicitly calculates them using real analysis. I don't...

Well, you want to differentiate a distribution. You will be interchanging several limiting processes (the integral, the definition of the principal value and the differentiation itself) so you can't just differentiate under the integral sign and "expect" it to work. However, using...

Not easy enough, it would seem! You need Jordan's lemma type argument - your current version doesn't hold water, I'm afraid.

The equation of a plane with normal \mathbf{n} going through the point \mathbf{a} is given by (\mathbf{x} - \mathbf{a})\cdot \mathbf{n}=0. Can you see how that helps?