Defining the continuity and differentiability of multi variate functio

trap101
Messages
339
Reaction score
0
Let f: R2-->R be defined by f(x,y) = xy2/(x2+y2 if (x,y) ≠ 0, f(0,0) = 0

a) is f continuous on R2?
b) is f differentiable on R2?
c) Show that all the dirctional derivatives of f at (0.0 exist and compute them

Attempt:

a) I had an idea to show that multivariate functions are continuous, but now I think it's faulty. I did this:

xy2/(x2+y2 ≤ |xy2|/|(x2+y2| < |xy2|

now taking the lim (x,y)-->(0,0) of |xy2| is 0 as such the function is also continuous.

But looking at a similar example from my notes where the only difference in the function was:


xy2/(x2+y6

We had shown that this was not continuous. So now I'm not so sure if the method I used was valid.

b) I'm a little stumped on showing differentiability in this setting, it doesn't seem to involve using the definition directly because of all the possible paths. So what other option may I have?

This is all in terms of discussing the trouble spot. i.e approaching (0,0)
 
Physics news on Phys.org
Find the limit when ##x_n = y_n = \frac 1 n ##, and when ##x_n = y^2_n = \frac 1 n ##.
 
I don't follow. By doing that i get different solutions taking n-->0.
 
trap101 said:
Let f: R2-->R be defined by f(x,y) = xy2/(x2+y2 if (x,y) ≠ 0, f(0,0) = 0

a) is f continuous on R2?
b) is f differentiable on R2?
c) Show that all the dirctional derivatives of f at (0.0 exist and compute them

Attempt:

a) I had an idea to show that multivariate functions are continuous, but now I think it's faulty. I did this:

xy2/(x2+y2 ≤ |xy2|/|(x2+y2| < |xy2|

now taking the lim (x,y)-->(0,0) of |xy2| is 0 as such the function is also continuous.

But looking at a similar example from my notes where the only difference in the function was:


xy2/(x2+y6

We had shown that this was not continuous. So now I'm not so sure if the method I used was valid.

b) I'm a little stumped on showing differentiability in this setting, it doesn't seem to involve using the definition directly because of all the possible paths. So what other option may I have?

This is all in terms of discussing the trouble spot. i.e approaching (0,0)

So you have :

##f(x,y) = \frac{xy^2}{x^2 + y^2}## if ##(x,y) ≠ (0,0)##
##f(x,y) = 0## if ##(x,y) = (0,0)##

For part a) where you're asked if ##f## is continuous, this should ring bells about the ##ε - δ## definition if you want to show it rigorously. Notice you can already tell that the function is continuous everywhere ( since it is a quotient of polynomials and polynomials are continuous ) except at (0,0). So using the definition with the information we're given :

##\forall ε > 0, \exists δ > 0 \space | \space 0 < |(x,y) - (0,0)| < δ \Rightarrow |\frac{xy^2}{x^2 + y^2} - 0| < ε##

So now you want to massage the expression into a suitable form :

##|f(x,y) - L| = \frac{|x|y^2}{x^2 + y^2}##

Now, using the fact that ##|x| < δ## and ##|y| < δ## you should be able to continue.
 
trap101 said:
I don't follow. By doing that i get different solutions taking n-->0.

n was supposed to be integer, going to infinity. Does that make (x, y) go to zero?
 
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