Recent content by golanor

Yes, I think it is Wightman QFT.

Thanks for the reference. I skimmed over it, seems interesting, but it is definitely not what I was talking about. Not at all what I meant.I tried to look up what I was talking about - apparently the idea was to "smear" the annihilation/creation operators, which modifies the CCR. Another place...

Hi, I remember seeing a few months ago, at a lecture about statistical signal processing, something which looked similar to commutation relations, only with a gaussian, instead of a delta function. Basically, it looked like this: $$\left[\phi(x),\phi(y)\right] = ie^{-\alpha(x-y)^2}$$ This...

Sorry, guess I was kind of lazy with copying it. a) \int_{200}^{300} {\frac{300 - T}{300}}{dT} = 100 - 250/3 = 16.67 b) \int_{300}^{200} {\frac{T - 200}{T}}{dT} = -100 + 200 * \log(3/2) = 19

Homework Statement We have two different set ups: a) A 200K brick in a 300K environment b) A 300K brick in a 200K environment From which one of these can you extract more work? Assume equal mass and heat capacity. Homework Equations Efficiency - η=\frac{T_H - T_L}{T_H} and: W = ΔQ => W≤ Q_H *...

The remainder must approach 0 for every c between x and 100. If i can find even 1 such number for which the remainder does not approach 0, it means that the series is not a correct estimation for the function.

It's very confusing. If the Lagrange remainder doesn't approach 0 it means you cannot estimate the function correctly, but in this case the calculation of the series does give accurate estimations. It's very interesting. Does anyone know the answer to this?

The point I missed was the relation between the cylinder's acceleration and the weight's acceleration.

yes, but since c is between x and 100, if x is bigger than 50 or smaller than 200 it won't diverge for any x and for any c. when i pick x=1, for example, for 1<c<99 the limit diverges. you cannot talk about a series in that case, because the remainder does not necessarily converge to 0.

That's why it is undetermined and not infinity. x=50 is the smallest x for which the limit will always be 0. anything lower than 50, and the nominator could be larger than the denominator => the limit will diverge.

That's the thing. \lim_{n\to \infty } \frac{(x-100)^n}{c^{n+1}*(n+1)} goes to 0 only for x≥50. x≤c≤100 for x<50 it cannot be determined.

I just realized something - can we even talk about a taylor series in this case? The Lagrangian remainder doesn't approach 0 as x approaches 0 (or even 1). The interval in which the remainder approaches 0 is (50,200), and not (0,200).

They have the same density. I=\rho \left(\int _0^h\int _0^{2\pi }\int _0^Rr^3drd\phi dz+\int _0^h\int _0^{2\pi }\int _0^rr^3drd\phi dz\right)=\frac{M}{\pi *h\left(R^2-r^2\right)}h*\frac{\pi }{2}\left(R^4-r^4\right)=\frac{M}{2}\left(R^2+r^2\right)

Well, i forgot the 1/2 there. it's basically the MI of the large cylinder minus the one we took out. They are co-centric so there is no distance to add. if i look at extreme cases - r<<R it's as if the cylinder is complete, r=R - there is no mass. = the equation makes sense...

I understand..just got confused by something we proved in class. So, the linear acceleration should be: \sum F=T-f=M*a_1 and the torque around the center of mass: \sum \tau =R(T+f)=\text{I$\alpha $} Around the point of contact: \sum \tau =2R*T=\text{I$\alpha $} And in this case, the moment...