Recent content by FatPhysicsBoy

Ooooooooooooooooooooooooooooooooooooooooooooooohhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhhh! :D Oh! :D Oh my lord okay I understand :D That is an insanely good explanation thanks so much! :D I guess because it is a show that question too it is clear from the result that one should expand in...

Hi Bruce, so if we look at the terms either side of the first equals sign I see that as the definition of 'expanding ##f(x)## around the value ##x=x_0##. So using that 'definition', if we wanted to say 'what does ##f(x)## look like around ##x=z##?' we'd substitute ##x_0 = z## right? Where I've...

See, reading this again in the context of what I posted earlier just reaffirms the thing I'm confused about. If I read your first sentence I can see why we have the ##x_0##'s and ##\delta a##'s where they are on the rhs of 3) but then if I try to think of a 'variable transformation' which...

Thanks for the replies! :) Why are we looking at what's happening to ##f(x_0-\delta a)## around ##x_0##. It's just confusing because if we had ##f(x)## and we wanted to expand about a point ##x=x_0##.. I guess I'm confused because in 1) ##x_0## doesn't appear on the left hand side, but all...

Anyone? Parts 1) and 2) can pretty much be ignored they just provide context for the problem.. I think ultimately it's just a complex number/conjugation question which I haven't understood properly.

Seriously, thanks for the help! I've got another question I've just posted over in advanced physics on the taylor expansion if you fancy it! :)

Homework Statement I have quite a straightforward question on the taylor expansion however I will try to provide as much context to the problem as possible: ##T(a)## is unitary such that ##T(-a) = T(a)^{-1} = T(a)^{\dagger}## and operates on states in the position basis as ##T(a)|x\rangle =...

Apologies that's what I meant, that the 'area' is then just 1 scaled by the 'area' of ##g(u)##. Thanks for all your help guys!

Is this because, ##\delta(x-u)## is a delta function centered on u? Is it sort of like an orthonormal function where if we turn the integral into a sum along ##x## then the only place where ##\delta(x-u)dx## is non-zero is at ##x=u##? So if we're essentially summing over ##\delta(x-u)g(u)## then...

##\int_\infty^\infty \delta(u-u_0)f(u-a)du = f(u_0 - a)## I still don't understand? So ##\int_\infty^\infty \delta(x-u)f(x-a)dx = f(-u-a) = f(-x-2a)## I think I'm seriously missing something :S

I don't follow? Why do I do this and also I don't understand, so I have ##\int^{\infty}_{\infty}\delta(x-u)g(u)du## but ##x=u## means I have ##\delta(0)##?

Homework Statement Having trouble understanding dirac deltas, I understand what they look like and how you can express one (i.e. from the limiting case of a gaussian) but for the life of me I can't figure out why the results of some integrals featuring dirac deltas equate to what they do...

Homework Statement Given two arbitrary vectors |\phi_{1}\rangle and |\phi_{2}\rangle belonging to the inner product space \mathcal{H}, the Cauchy-Schwartz inequality states that: |\langle\phi_{1}|\phi_{2}\rangle|^{2} \leq \langle\phi_{1}|\phi_{1}\rangle \langle\phi_{2}|\phi_{2}\rangle...

Ultimately all the power is in my hands at the moment since everything we are discussing is in-fact being done in simulation. We have the full spherical detector geometry and can run events as and when and where we like through monte carlo. I think this method has broken down for the reasons you...

Thank you for your help Stephen, I will compute the variances using the method you suggested. I understand, this is supposed to be the first stage though. See how far we can go using chi-square and possibly replace this method with a likelihood method or some other determining method. The aim...