# Find a function f such that f(xy)≠f(yx) at (0,0).

1. May 5, 2012

### hivesaeed4

Mixed Partials:
Find a function f such that f(xy)≠f(yx) at (0,0).
Help?

2. May 6, 2012

### brydustin

What do you mean by f(xy)? You didn't use any commas and surely we're not supposed to assume you mean the product of xy vs yx?

3. May 6, 2012

### D H

Staff Emeritus
He means
$$\frac{\partial^2 f(x,y)}{\partial x \partial y} \ne \frac{\partial^2 f(x,y)}{\partial y \partial x}$$

4. May 6, 2012

### I like Serena

Hi hivesaeed4!

Obviously any regular function of x and y won't work, since any function that is continuously differentiable two times wil have fxy(0,0)=fyx(0,0).

Did you try anything?
What kind of functions would you think of?
Can you give an example?
If you show some work, some stuff that you tried, we can help you further.

To help you on the way, you will need to define a function that is different based on conditions on its domain.
This also means you have to go back to the actual definition of the derivative to see how it works out.

To get to for instance fxy(0,0), you need:
$$f_x(0,y) = \lim_{h \to 0} \frac{f(h,y)-f(0,y)}{h}$$
$$f_{xy}(0,0) = \lim_{h \to 0} \frac{f_x(0,h)-f_x(0,0)}{h}$$

5. May 6, 2012

### hivesaeed4

Every function that I think of either gor=es to zero and/or f(xy) becomes equivalent to f(yx). I was thinking of using a function which had different powers of the two variables but then if the powers exceed 1 then always after the mixed partials are evaluated the function goes to zero cause one of the variables was left behind.
Any pointers in the right direction?

6. May 6, 2012

### DonAntonio

Check this: http://www.math.uconn.edu/~leibowitz/math2110f09/mixedpartials.pdf [Broken]

DonAntonio

Last edited by a moderator: May 6, 2017
7. May 6, 2012

Thanks.

8. May 7, 2012