Recent content by projektMayhem

There is no evidence of the existence or non-existence of a god, nor is it wise to expect such. You are wasting time debating with religious people - the very nature of faith is the rejection of reason and hence no argument you bring can be successful, by definition. In fact, you will have the...

I'm sorry but I have to take issue with this. The greed and corruption inherent in big business has poisoned our political system. Until we limit politicians to only accepting donations from living breathing citizens (plus a set amount of tax payer dollars) the corporations will be able to exert...

Say what you will about the man's politics... the fact is that he is one of the few who actually speaks his mind - even if his ideas lack practicality. Moreover, he would appear to be one of the few who respects constitutional boundaries. I'll agree he's not presidential material, but congress...

Nevermind, problem resolved.

I am trying to iterate a solution to a non-linear differential equation - but I am unable to evaluate a statement within a Do loop Code: Do[q = Cos[Evaluate[x'[i]/.up]]; If[ q <= 1, Print[q], Print[i], Print[i^2]], {i, 0.1, 1, .1}] x'[i] is a solution to NDSolve The output is i^2, which is...

I don't think it's so far fetched. What do you suppose would happen if they buy gold instead of our bonds?

I'm sorry, but I must point out the one critical point people seem to have missed here: we are not only running MASSIVE budget deficits, we are also running incredibly large trade deficits - to china in particular. IMHO, yes, an economic collapse is imminent. We are, in fact, one of the...

Pfft, yeah what he said. sorry for the bogus post.

That won't be enough. The theorem relates (in this particular case) the time rate of change of the energy density to to the integral you have given. More generally, you'd also have to integrate over the poynting vector itself. In this case, however, the poynting vector is zero because of the...

so, tell me if this is absurd: for t1 < t < t2: |X(t)> = (cos x)|E1>exp(-i E1 (t1 - t) / h) + (sin x)|E2>exp(-i E2 (t1 - t) / h) then: <Q1|X(t2)> = exp(-iE1 (t2 - t1)/h) [ (cos x)^2 + (sin x)^2 exp(-i(E2 - E1)(t2 - t1)/h) so, P(Q1) = (cos x)^4 + (sin x)^4 + 2(cos x)^2 (sin x)^2 cos[(E2 -...

i think the answer to my last question was no.

belliot, thanks for the response. for case 1, is the probability unity?

A two state system is described by linear superpositions of two stationary states: |E1> and |E2>, with corresponding energies E1 and E2. An observable Q has the eigenstates |Q1> and |Q2>, corresponding to eigenvalues Q1 and Q2 (both real). All states are normalized. Given: |Q1> = (cos x)|E1> +...

What good will that do?

Ivan, to compare our current situation to our post WW2 situation is fallacious. Yes, during WW2 we racked up a huge deficit and subsequently entered into a period of economic growth. However, there is nothing in common with post ww2. 1) We don't have a large manufacturing base and are primarily...