What is the Theorem for Differentiability in Advanced Calculus?

nateHI
Messages
145
Reaction score
4

Homework Statement


This isn't a standard homework problem. We were asked to do research and to find a theorem of the form:
If something about the partial derivatives of u and v is true then the implication is ##D(u,v)## at ##(x_0,y_0)## exists from ##R^2## to ##R^2##

Homework Equations

The Attempt at a Solution


I've done a lot of reading on the the difference in differentiability between ##R^2## and ##\mathbb{C}## but haven't been lucky enough to stumble upon an exact theorem. Anyway, a push in the right direction would save me a lot of time.
 
Last edited by a moderator:
Physics news on Phys.org
A little hint: In ℝ2 , you may calculate the partial derivatives \frac{\partial u}{\partial x}, \frac{\partial u}{\partial y}, \frac{\partial v}{\partial x}, \frac{\partial v}{\partial y} and you do not expect any obvious relation between them. On the other hand, in ℂ, you expect the derivative of f(z) to be just a function of z and nothing else.
So, if you put z = x + iy and f(z) = u(z) + iv(z), this implies a relation between u and v...
 
I found it thanks! It's kind of a long theorem but if you're interested to know what it is let me know and I'll type. It doesn't have a distinct name that I can just reference for you.
 
Wouldn't they be called the Cauchy-Riemann equations ?
 
BvU said:
Wouldn't they be called the Cauchy-Riemann equations ?
That's what I thought at first and I suppose the class will get to Cauchy-Riemann eventually. But the instructor stressed the fact that we were not working in the complex numbers for this problem. The Theorem he was looking for is from Advanced calculus. He probably wants to demonstrate the advantages of ##\mathbb{C}##.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

Similar threads

Replies
9
Views
2K
Replies
16
Views
2K
Replies
3
Views
2K
Replies
10
Views
5K
Replies
2
Views
1K
Replies
21
Views
2K
Back
Top