How "messy" can a multivariable function be? f(x,y)=(2xy)/(X^2+y^2) f(0,0)=0

Sophie Marks

Sophie Marks

Answered question

2022-11-16

How "messy" can a multivariable function be?
f ( x , y ) = 2 x y x 2 + y 2 f ( 0 , 0 ) = 0

Answer & Explanation

Liehm1mm

Liehm1mm

Beginner2022-11-17Added 13 answers

Here's a mind bendy function:
f ( x , y ) = { 1 ( x , y ) R 2 Q 2 0 ( x , y ) R 2 Q 2
Try to think about what your function does in any ball around the origin.
Rigoberto Drake

Rigoberto Drake

Beginner2022-11-18Added 2 answers

In fact, the situation is quite bad. Consider for example the function
f ( x , y ) = x x 2 + y 2 .
This is a continuous function on R { ( 0 , 0 ) } and it is constant on any ray emanating from the origin because for all t > 0 and ( a , b ) ( 0 , 0 ) we have
f ( t ( a , b ) ) = t a t 2 a 2 + t 2 b 2 = a a 2 + b 2 = f ( a , b ) .
Hence, if you approach the origin via all possible rays, you'll get any possible limit in [ 1 , 1 ]. This can be seen more geometrically by noting that in polar coordinates, f has the form
f ( r , θ ) = cos θ .
In fact, the function you presented in the post suffers from the same problem because in polar coordinates it is given as f ( r , θ ) = sin ( 2 θ ).

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