Continuity of the derivative Df

quasar987
Science Advisor
Homework Helper
Gold Member
Messages
4,796
Reaction score
32

Homework Statement


I'm reading this at the moment: "Let f:R^n-->R^n be of class C^1 (that is, assume Df exists and is continuous)"

What does it mean?? If it means that for all x in R^n, the linear map Df(x):R^n-->R^n is continuous, then it's a triviality since all linear maps from R^n to R^m are continuous. So I am skeptical that this is what it means!

What other option is there? That map that send x in R^n to the point Df(x) in the space of linear map is a continuous map? I highly doubt that!

So what does it mean??
 
Physics news on Phys.org
C^1 usually means continuous partials.
 
Thanks for the tip!
 
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...
Back
Top