Recent content by twotaileddemon

I tried doing something like indexfinal=99; span = 0:1/indexfinal:1; row(2,:) = span; column(:,2) = span; for n=2:indexfinal+1; for m=2:indexfinal+1; row(n,:) = span*span(n); column(:,m) = span*span(m); table = row * column; end end But it still doesn't give me the desired...

Hi, I'm new to MATLAB and I have a problem: I am trying to create a multiplication matrix of sorts. The first row will have numbers from 0 to 1 in steps of .0101 The first column will have the same, from 0 to 1 in steps of .0101 Then to generate the rest I will multiple each entry in the...

Hilbert's Satz 90 states that, “for a finite cyclic field extension K/F with Galois group (S): An element b є K has norm 1 (with respect to the extension K/F) iff b = S(c)/c for some c є K*. Can someone please explain this to me in simpler terms? I'm not sure what he means by norm 1 or S(c)/c.

that's really quite interesting and good to know - thank you! I checked my work over, and I don't -think- I'm missing a factor of 1/2. When I converted from sin[(at)^(1/2)] to sin u, I made the substitution u = (at)^(1/2), and so du = (a/2u)dt and dt = (2u/a)du. I brought this 2 out front, and...

I did bring that out front, but it ended up canceling with the 2 I had after making the u substitution dt = 2u/a du (at the very beginning)

oh ok! so after combining the two integrals with (ia/2s) in that way, I ended up getting just a pi1/2 for that. the other two with v*(exp(-v2) ended up canceling out, and so I was left with 1/(as)1/2*exp(-a/4s)*(ia/2s)*pi1/2 = (a*pi)1/2 * exp(-a/4s) / s3/2 It looks weird because I...

hmm ok. any other limits would fail though, right?

isn't this some kind of error function though? from what I understand it's impossible to integrate exp(-x^2)?

I have some work, but I cannot understand how to do the latex for it (i did read your signature, but I easily get lost in all the latex text) But basically, you distribute the integrals, then for the one that has -(ia/2s)*exp(-v^2), change the limits of the integral from -alpha to -infinity...

I don't even know, to be honest. I've been working on this problem for a long, long time today and it's really confusing me =|. I got to a point where I had to integrate exp(-x2) which is apparently not integrable, and wolfram suggested to use the error function and gave me some weird...

\int\left[\sqrt{\frac{a}{s}}v + \frac{ia}{2s}\right]e^{-v^2}dv=\sqrt{\frac{a}{s}}\int ve^{-v^2}dv +\frac{ia}{2s}\int e^{-v^2}dv --- \int\left[\sqrt{\frac{a}{s}}v - \frac{ia}{2s}\right]e^{-v^2}dv=\sqrt{\frac{a}{s}}\int ve^{-v^2}dv -\frac{ia}{2s}\int e^{-v^2}dv then I can combine the two...

so if the limits of the first integral were -i(1/2)(a/s)1/2 to infinity and the second were +i(1/2)(a/s)1/2 to infinity, I can just make it 0 to infinity in both cases by ignoring the imaginary parts? I do see how there are no poles, as we covered this a while ago. I never knew this could...

I did separate into two integrals, using different substitutions for v, namely the one with the +(ia/2s) in the equation and the other with -(ia/2s) in the equation. but then the limits of integration are different for each integral and you can't combine and cancel. hence.. I was thinking of...

I do know how to integrate that, but how can you make a substitution twice using the same variable? you set v equal to two different things in two different equations?

a big mess! the limits of integration are from -(i/2)(a/s)1/2 to infinity now, and the equation under the integral is (a/s)1/2 * [(a/s)1/2v + (ia/2s)]exp(-v2) dv note: there was a u term before the whole exponential term, so I accounted for that. and then made a substitution for du as well