Recent content by Theraven1982

I was afraid of that ;). I'll find my linear algebra book ;). Is there anyone who has experience in solving these equations in Matlab? There are probably functions for this purpose ; again, any kick in the right direction is welcome ;)

Homework Statement I want to find certain coefficients \alpha_{uv} by minimizing a error function. In the end, I want to make a function of this, so iteration is not a problem. I want to estimate these coefficients to find the best estimate for f(x,y). Homework Equations The error...

I'm sorry for the (very) late reply, but thanks a lot! I now fully understand. Sometimes I need to ripen new information ;). Thanks again all!

Thanks. But why is this conducting channel needed? This inversion layer is above the depletion region which fills up with charge; when the device is clocked (for measuring how much charge is in the well), the charges move from well to well. I don't see the need for this conducting channel.

Thanks a lot for your answer. It is not necessary that I know exactly how it is derived, but just wanted to know where it came from. I've seen the name 'Fowler-Nordheim' before in a paper, and now I at least understand where it comes from. Any takers for the other questions? ;)

[SOLVED] inversion layer in a MOS Hello, I have some questions about MOS devices, and CCD In a MOS or MIS after applying a voltage on the metal, the valenceband and the conduction band bend downwards (in the usual band diagram) when a positive voltage is applied. (p-type semiconductor)...

The philoshophy is confusing me. If i understand correctly, it comes down to the following: when a pion decays (which decays to an electron and positron), we can make a measurement of the spin of both resulting particles. If I measure the spin in the z-direction of the electron, I instantly...

I finally found the right solution (well, my teacher did.. when I got to the Riemann zeta function, I always thought: This can't be correct... but it was). But it appears the above mentioned answer was correct as well; or at least it gets very close to the result my teacher gave me. Maybe the...

Hmm ... that's the same answer I get now... I don't know why the book says 5x10^5 photons per litre :\. I tried everything; maybe I should go with this answer instead. Thanks for all the help! Much appreciated! (You all do a great job, helping untangle the web of science ;) )

You're a lot closer than me; the answer should be 5*10^5 photons per litre ... I'll look again at your calculation, maybe I can find the reason why there's a difference. But a big thanks! edit: energy density is (according to my book): pi^2 *k^2*T^4/15c^3*hbar^3 There's an extra pi ;)...

Homework Statement I have a fairly simple question, but for some reason I can't find the answer. It's about Planck's law, and cosmic background radiation. We know Planck's law, and we know the temperature of the universe (approx 2.7 K). Furthermore, we know the mean number of photons in a...

Off course... now that I see it, it's all very simple. Guess sometimes my mind gets confused after too much maths ;). Thank you, W.

Hello, it's me again ;) Problem: ------- Define f_{n}(x)=n^{\alpha}, |x|\leq 1/n, f_{n}=0 elsewhere Give all \alpha \in \Re for which \lim_{n \to \infty} \int_{\Re}f_{n}(x)dx=+\infty ------- Can i change this last integral to: \lim_{x \to 0} \int_{0}^{\infty}...

So you just say: if n goes to infinity, the formula goes to zero, and therefore the integral goes to zero? edit Thanks to all who replied; really appreciated. W.

Yes, this is just the way of filling in the integral; but i don't know what changes when you first take the limit. You could equally well take the n^2 outside of the integral: \int_{1}^{\infty} \frac{1}{n^2}\frac{1}{1+(x/n)^2} from which it can also be seen. But maybe i should be...