Recent content by cqfd

Oh, ok. So there's just a constant offset between the two solutions that we do not care about. :)

I am encountering a paradox when calculating the integral ##\int sin(x)\cos(x)\,dx## with integration by parts: Defining ##u = sin(x), v' = cos(x)##: ##\int sin(x)cos(x) dx = sin^2(x) - \int cos(x) sin(x) dx## ##\Leftrightarrow \int sin(x) * cos(x) dx = +1/2*sin^2(x)##. On the other hand...

I'm always a little uncertain about such notational things, so just before i start, what you mean by your concrete notation, is that the experiment is performed 2 times, each times taking a sample of size N. In that case, the distribution of the position is given by the product of the total...

Sorry, i first thought i wanted to add something, but it ended up being wrong. :/ So this post can be deleted.

Oh, ok i see what you mean. I should have stated it like in post #9 from the beginning. ^^ Now I've thought about the RMS part some more, but i can't get my head around it. Has it something to do with this property? Because then we would have ##RMS(X_1+X_2+...+X_N) =...

But you're saying basically the same thing, and i don't understand how your way should be a valid proof, but not mine? And shouldn't the expected value for each random variable that is corresponding to a draw from the sample ##X_i## be already ##x_0##? I.e. ##\forall i \in [1,N], E[X_i] =...

I've given the problem a second try, from the beginning, thinking about what you said. Now this is what i came up with, i tried to order all the different terms, and hope you correct me if there's something wrong: We define a random variable, X, the position of the object. We know the pdf of...

Still can't figure this one out... :/

Thanks for your help already. This all makes sense to me intuitively, I mean it is clear to me that when we make the measurements many times and take the mean value, that this value is going to converge to x0, but I just can't figure out how to write this down formally. And I still can't figure...

Hmm, this makes sense. In fact his was what I thought of in the first place. So I get the mean like this: <x>N = 1/N*sum(xi) I'm sorry but I don't understand what this means. All I've calculated so far are the means of the individual samples <x>N,j with the now hopefully correct formula. But...

Hi everyone, I'm new here and this is my first post in this forum. ^^ Homework Statement Suppose that you observe a fluorescent object whose true location is x0. Individual photons come from this object with apparent locations xi in an approximately Gaussian distribution about x0...