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...
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...
Well, I get that. Just so I'm making myself clear, I actually thought about the circularity for some time and I tried to resolve it by adhering strictly on what the principle says, and that (i) and (ii) should be true; (ii) which states that the implication itself must be true and not just...
Actually, I've found out that to 'parametrize' the variables into x=t, y=t is a more comforting method to do it. At least intuitively, I see it as tracing the path of integration when we set the x and y variables into that parametric equation.
Edit: Yes, I didn't see it, but I was using x as...
Suppose I have a vector V and I want to compute for the line integral from point (1,1,0) to point (2,2,0) and I take the path of the least distance (one that traces the identity function).
The line integral is of the form:
\int _a ^b \vec{V} \cdot d\vec{l}
Where:
x=y, \ d\vec{l}...