Recent content by jacophile

Care to elaborate? My question is about entanglement so yep, I'm sure the answer is too...

Hi, can someone help me understand the firewall idea? I have a heuristic understanding that the firewall is created due to breaking entanglement between particles (virtual particles?) inside the EH and old Hawking radiation that emanated from the BH previously. I don't understand how the...

Um, so the answer to the original question is neither. You get y2.

Sorry, I guess I have the notation confused, I thought the fx was the derivative wrt x.

Huh? Isn't the answer y^{2}

Oh, and by the way, the CDF is the integral of the PDF not the derivative...

Well the integral of the PDF is the area enclosed between it and the x-axis. So if you accept that the a delta function centered on x has finite area 1 f(x) then why don't you accept that the integral increases by a finite step of f(x)? Also, just to restate mathman's earlier comment, the...

Although the delta function has infinite value, the area under it is 1, yeh? It is defined as having width \epsilon and height \frac{1}{\epsilon} in the limit that \epsilon aproaches 0. When you multiply it by f(x), given that the delta function is centred on x, the result is a delta function...

Hmm, I guess my post was too confusing. Any way I think I worked it out.. Just to re-state the problem (hopefully more clearly), What is the distribution of the sales demand for a stock item during the replenishment lead time (DLT), given that the daily demand D and the lead time N are...

Hi, I am trying (in vain) to understand a common result in logistics due to Hadley and Whitin. It is about the variance of the reorder point process in an inventory management system. It is assumed that the demand X during lead time has a normal distribution with mean E(X) and variance...

Yes, sorry, that is what I meant in my last post. I got e^{\mu + \sigma^2/2} as well. It was just a typo in my previous posts.

Well its straight forward to get it into the form \int_{0}^{\pi}\frac{Adx}{(sin(x+\alpha)-B)^\frac{3}{2}} where A, B and alpha are functions of R and the K's you could then try a substitution like y=sin(x+\alpha)+A\;\;\;\;;dx=\frac{dy}{cos(arcsin(y-A)))} to give...

yeh, sorry, that is the result I got, neglected to fix the typo...

Thanks! Very much appreciated: you assume correctly! Thanks for you help.

Thanks, you are right, I have fixed the typos. Not sure I understand your suggestion though... Do you mean combine the two exponents into one and re-factorise the resultant polynomial? The reason I am trying to understand this is that...