Recent content by mathsciguy

I might be necroing, but this question is also relevant for me. I am also reading the same example from Griffiths. If the uncharged metal sphere has a zero potential, then why does it imply that the xy plane has zero potential? You are not given any information about the charge density right?

Is my question too cluttered and messy? I'll revise my question if it is. Because I'd really appreciate it if someone could give their insight on this.

I was reading about Cauchy-Goursat theorem and one step in the proof stumped me. It's the easier one, that is, Cauchy's proof that requires the complex valued function f be analytic in R, and f' to be continuous throughout the region R interior to and on some simple closed contour C. So that the...

Thanks, I think that kind of proof is something I would think of if I was working with matrices.

Hm, well, I also thought so. But see, I was trying to figure out how to show that an operator like L = i(d^3/dx^3) is hermitian for functions f(x) in the interval -(infinity)< x < (infinity) where as f approaches infinity, it also approaches zero. I tried using the straightforward method where I...

Thanks, that's pretty straightforward huh. Anyway, I wonder if you guys can answer a barely tangential question? Let's say I have a complex valued function f(x) of a real variable. If the limit f(x) as x-> infinity is zero, are the derivatives of f(x) as x-> infinity also zero?

I have a question about complex valued functions, say f(z) where z=x+iy is a complex variable. Can every such complex valued function be represented by: f(z)=u(x,y)+iv(x,y)? Also, is the limit of the conjugate such a function equal to the conjugate of the limit of the function? Something like...

That's cool, I get it now. Then that means the roots are either purely imaginary or complex (but not purely real) right? Then why is it required that the discriminant be less than zero? Is it because it will make sure that part of the solution will have an imaginary part?

Hello, I was looking at Riley's solution manual for this specific problem. Along the way, he ended up with a quadratic inequality: If you looked at the image, he said it is given that λ is real, but he asserted that λ has no real roots because of the inequality. Doesn't that mean λ is...

I have studied multivariable calculus of course, along with some basic calculus of vector fields, and I am reviewing some ODEs too (I am forgetting the theory, for some reason, I don't like the idea that I am forgetting what little I know about math theory). Well, coincidentally, I am reviewing...

^ I can understand these, at least mechanically, but well, I guess, what I am desperately trying to find is a general method that applies to all cases; something in which the usual basic methods is actually a special case of it. What you outlined above is intuitive, but it doesn't seem to fit...

Oh this, I did some searching. Apparently, what I said above about straight forward integrating over the surface S isn't exactly correct. You parametrize the surface (to vector valued function), I think, usually using spherical/cylindrical coordinates then transform it to rectangular components...

Yes, exactly. But the problem here is, he didn't actually tell that the surface integral is a double integral (at least from what I currently know) that can be solved by actually integrating over the surface. In the textbook, after constructing something that's starting to look like a Riemann...

I am saying all those given that my textbook only offers an introduction to calculus of vector fields, so I know asking for a more general approach is a bit unreasonable. Anyway here's an example of one: For a function of three variables G(x,y,z), the surface integral of G over the surface S...

I have been restudying vector calculus, especially on topics pertaining to line integrals, surface integrals (and the accompanying vector forms). One problem I have encountered from the book I have been using is that it seems there are some theorems and results that are only restricted to...