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...
Homework Statement
How does a picoammeter work? I know it can measure small currents, but how is it different from the design of an ammeter and why can it measure small currents?
Homework Equations
Q = CV = CIR
I = Q/(RC)
The Attempt at a Solution
For example, the circuit...
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...
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...
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?