5

In $\mathbb R$ for a derivative to exist (or a limit generally) it is necessary that the limit be the same in both directions (from below and above) and this is the same in $\mathbb C$ where for a function to be differentiable at a point the difference quotient limit must be the same independent of the way $h$ approach the point, which is where the Cauchy-Riemann equations come from. Why is this only the case in $\mathbb R$ and $\mathbb C$ and not in $\mathbb R^2$? In $\mathbb R^2$ we just make a matrix/vector of partial derivatives and say the derivative is just a linear map. What is the fundamental difference that requires such different ways of defining differentiability?

user85798
  • 1
  • In higher dimensions you have more freedom to approach a point as a limit. So different notions of differentiability exist. For example functions which have partial derivative but no total derivative show that limit (defining the derivative) exists only for particular sequences not all of them. – DBS Feb 04 '16 at 13:34

3 Answers3

12

A function ${\bf f}:\>{\mathbb R}^2\to{\mathbb R}^2$ is differentiable at $p$ if there is a linear map $A:\>{\mathbb R}^2\to{\mathbb R}^2$ such that $${\bf f}({\bf p}+{\bf h})-{\bf f}({\bf p})=A{\bf h}+o(|{\bf h}|)\qquad({\bf h}\to{\bf 0})\ .\tag{1}$$ This $A$ is then called the derivative of ${\bf f}$ at ${\bf p}$, and is denoted by $d{\bf f}({\bf p})$, or similar. If $d{\bf f}({\bf p})$ exists then the four partial derivatives $f_{i.k}:={\partial f_i\over\partial x_k}({\bf p})$ exist, and the matrix of $d{\bf f}({\bf p})$ with respect to the standard basis in ${\mathbb R}^2$ is the matrix $[f_{i.k}]$.

Let such an ${\bf f}=(f_1,f_2)$ be given. The complexified version $f:=f_1+i f_2$ of this ${\bf f}$ is a differentiable function of the complex variable $z:=x_1+i x_2$ at the "complex point" $p$ iff the matrix of $d{\bf f}({\bf p})$ has the special form $$\left[\matrix{a&-b \cr b& a\cr}\right]\ .$$ The complex number $c:=a+ib$ is then the derivative of $f$ at $p$, and $(1)$ assumes the form $$f(p+h)-f(p)=c\>h+o(|h|)\qquad(h\to0\in{\mathbb C})\ .$$ In short: Complex differentiability is a special case of ${\mathbb R}^2$-differentiability, expressed in terms of other fonts.

Christian Blatter
  • 226,825
3

It is not sufficient to make a matrix/vector of partial derivatives to say that a function is differentiable in a point. You need to prove that the limit of the difference quotient is indipendent of the way $h$ approaches the point whatever is the set where the function is defined.

trying
  • 4,756
  • 1
  • 13
  • 23
1

The fundamental difference lies in the fact that in R^2 or any other higher dimension, there is no notion of divisibility by a vector. So the basic definition of differentiability at a point by difference quotient limit cannot be extended due to the non-existence of division by vectors. However, in the case of complex numbers, divisibility is well defined. Hence we were able to extend the definition of differentiability in R to C.

Mohammad Areeb
  • 11