Recent content by Volt

Can anyone manage to prove the result in my original post? I get very close but end up with a B in the numerator and I can't figure out where the mistake is. (it comes from factoring the quadratic, with B in the denominator).

Sorry, I'm not familiar with this formula, where does it come from? This is probably a really dumb question, but how do I divide an infinite series by another infinite series? I'm not following how you are doing the division in your examples (for example, how you end up with a/z^2 as the first...

I don't understand your point that it doesn't "work well" with a Laurent series. As far as I know Laurent series are supposed to be used in these cases where a function fails to be analytic at a point, and when you find the Laurent series for the function you'll see what type the singularity is...

I received clarification that for this question A > B > 0. After getting that information I *almost* managed to get the result that they give, except there was a B in the numerator (i.e. (B*pi) / sqrt(A^2 - B^2)). Also, I don't know what you're confused about for the integral I got in the...

Are you sure it's infinite? If you substitute z = 0, then the function is of the form 0/0 where it's not immediately clear what's going on.

Homework Statement Use residue theory to establish the result: \int^{\pi}_{0}\frac{dx}{A + Bcosx} = \frac{\pi}{\sqrt{A^2 - B^2}} The Attempt at a Solution So I've gotten to the point that the above integral = \frac{1}{2} \oint^{2\pi}_{0} \frac{-2i}{Bz^{2} + 2Az + B} dz...

Homework Statement Find the Laurent series expansion of f(z) = (e^z - 1) / (sinz)^3 at z = 0.The Attempt at a Solution Ok, so I'm confused on a number of fronts here. For e^z - 1, I assume you just use the standard power series expansion of e^z and then tack on a -1 at the end, which would...

Sorry, I'll try to be more clear. What I mean is that for the gradient to exist, the function only has to have partial derivatives for x and y (for a function of two variables). This means that it has to at least be continuous in the x and y directions at any point where the gradient exists. But...

First of all, I don't have a concrete example for this, but I hope it's not too hard to understand what I'm trying to get at. For a multivariable function of, say, 2 variables x and y, the gradient at a point only depends on the existence of partial x and partial y, right? In other words, if...

Homework Statement Verify the following limit by using delta-epsilon arguments Homework Equations lim (x, y) -> (1, -1) of xy^2 = 1 The Attempt at a Solution Right, so I'm having some trouble with these delta-epsilon proofs for multivariable limits. Some of them are easier than others...