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

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Discussion Overview

The discussion revolves around finding a function f such that f(xy)≠f(yx) specifically at the point (0,0). Participants explore the implications of mixed partial derivatives and the conditions under which such a function might exist, focusing on theoretical aspects of calculus and differentiability.

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • Some participants clarify the notation f(xy) and question whether it refers to the product of xy versus yx.
  • One participant asserts that any regular function of x and y will not satisfy the condition, as continuously differentiable functions will have equal mixed partial derivatives at (0,0).
  • Another participant suggests that a function must be defined differently based on conditions in its domain to achieve the desired inequality in mixed partials.
  • Some participants express difficulty in finding a suitable function, noting that functions they consider either tend to zero or result in f(xy) being equivalent to f(yx).
  • One participant proposes using functions with different powers of the variables but notes that exceeding a power of 1 leads to zero after evaluating mixed partials.
  • Links to external resources are provided for further reading on the topic.

Areas of Agreement / Disagreement

Participants do not reach a consensus on a specific function that meets the criteria. There are multiple competing views on the nature of functions that could potentially satisfy the condition, and the discussion remains unresolved.

Contextual Notes

Participants highlight limitations related to the definitions of derivatives and the behavior of functions near (0,0), indicating that assumptions about continuity and differentiability play a significant role in the discussion.

hivesaeed4
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Mixed Partials:
Find a function f such that f(xy)≠f(yx) at (0,0).
Help?
 
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hivesaeed4 said:
Mixed Partials:
Find a function f such that f(xy)≠f(yx) at (0,0).
Help?

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?
 
He means
\frac{\partial^2 f(x,y)}{\partial x \partial y} \ne \frac{\partial^2 f(x,y)}{\partial y \partial x}
 
Hi hivesaeed4! :smile:

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:
$$f_x(0,y) = \lim_{h \to 0} \frac{f(h,y)-f(0,y)}{h}$$
$$f_{xy}(0,0) = \lim_{h \to 0} \frac{f_x(0,h)-f_x(0,0)}{h}$$
 
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?
 
hivesaeed4 said:
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?



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

DonAntonio
 
Last edited by a moderator:
Thanks.
 

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