Recent content by Marioeden

Oh, if the second one is partial as well then substitute it into the first one. So assuming k2 is non-zero, you have dC/dx = [k1*k3/k2]*C The solution to this is the standard exponential as in the ODE case, only your constant is now a function of time. Then just plug this back into the...

I'm assuming for your PDE you mean both as partial derivatives and for the ODE you mean total derivative? In which case (writing D for total derivative and d for partial), the ODE gives you: DC/Dt = dC/dt + [dC/dx][dx/dt] Further, you know dC/dx in terms of dC/dt from the PDE, and you can...

Exactly, the only difference is the operation i.e. sigma means to sum and U means to take the union. So Sigma over n means summing over all n in the given range, and U over n means taking the union of the sets over the given range.

Well, your Fourier series is periodic. If you look at the sketch posted above you can see that the Fourier series replicates the function in the period given and then just repeats itself as a periodic function outside that interval (in the same way the simple trig functions do). Moreover, if...

I think (and I may be wrong) that if they're using standard convention, then := means "define". In other words, that statement means "Define A as the union of the sets A(n) over all n" And I believe that the up arrow indeed means the limit.

The best way to derive these results is to use summation convention. Everything drops out quite nicely. The vector identity you've stated is also derived in the same way but you'll notice the subtleties of the fact that you're dealing with an operator when doing the derivation.

Here's a simple example: Suppose you have a function f(t) defined on the range 0<t<a Then two possible Fourier series for f(t) are the Fourier sine series and the Fourier cosine series. You can obtain the sine series by defining f(t) to be an odd function i.e. define it on -a<t<a with f(-t) =...

Oh, I'm not doing this from anywhere near a mathematically rigorous point of view (far to many subtleties as you've mentioned), wouldn't know where to begin with constructing anything sensible. I'm simply stating the best you can do with the information you've been given (and even then there...

Ah yes, sorry I missed the special case where b is zero.

So I'm intending to teach myself some Particle Physics and Standard Model type stuff, I was wondering if someone who's already covered this could give me some advice. I did some Group Theory a few years back and looking over content pages of lecture notes I occasionally spot references to...

The weird thing about maths is that once you get to higher level stuff, the first time you see it you have no idea what's going on. Then after a while you get your head around what the definitions actually mean, but still can't seem to answer the questions. After a long time spent of having...

It's been a while since I did this, but as far as I remember you don't necessarily have to derive the operator A, you just need to find one which works. I think I always used A = 4[(d/dx)^3] - 3 (u[d/dx] + [d/dx]u), or something along those lines

Hmmm... let's see what we can do. Suppose you have a dot product a.x = A where a and x are vectors and A is some number. Now, given a and A, what can we say about x? Well a.x = |a||x|cos(theta), where theta is the angle between a and x. So |a||x|cos(theta) = A For a non-zero, we can say...

I think what you're referring to as nabla is what I call grad. Just do the vector taylor expansion as I mentioned, this was the box standard thing to do back in electrodynamics exams. Oh, and use summation convention to make life easier.

Use the vector form of Taylor Expansion i.e. f(x+h) = f(x) + (h.grad)f(x) + [(h.grad)^2]f(x) + ... where x and h are vectors, grad is the usual gradient operator and "." indicates the dot product.