Proving Continuous Extension of $f(x,y)$ Function

In summary, we can extend the function $f(x,y)=(x^2+y^2)\arctan\dfrac{1}{|xy|}$ to a continuous function by finding the limit along the line $x=k$ and using local polar coordinates. The limit is equal to $k^2\frac{\pi}{2}$, which allows us to extend $f$ to a continuous function with $f(x,0)=x^2\frac\pi 2$, $f(0,y)=y^2\frac\pi 2$, and $f(0,0)=0$.
  • #1
laura1231
28
0
Can I extend the function $f(x,y)=(x^2+y^2)\arctan\dfrac{1}{|xy|}$ to a continuous function?
If I consider the restriction of $f$ along the line $x=k$ i find $\lim_{(x,y)\rightarrow(k,0)}(x^2+y^2)\arctan\dfrac{1}{|xy|}=k^2\dfrac{\pi}{2}$
how can i prove that?
 
Physics news on Phys.org
  • #2
laura123 said:
Can I extend the function $f(x,y)=(x^2+y^2)\arctan\dfrac{1}{|xy|}$ to a continuous function?
If I consider the restriction of $f$ along the line $x=k$ i find $\lim_{(x,y)\rightarrow(k,0)}(x^2+y^2)\arctan\dfrac{1}{|xy|}=k^2\dfrac{\pi}{2}$
how can i prove that?

Hey laura123!

To extend $f$ to a continuous function on $\mathbb R^2$ we need that for every $x_0$ the limit $\displaystyle\lim_{(x,y)\to(x_0,0)} f(x,y)$ exists. And we need the same thing for $y$ values, although that follows by symmetry.

We can try to find such a limit by switching to local polar coordinates.
That is:
$$\lim_{(x,y)\to(x_0,0)} f(x,y)
= \lim_{r\to 0} f(x_0+r\cos\phi, r\sin\phi)
$$
Can we find that limit? (Wondering)
 
  • #3
$$\displaystyle\lim_{(x,y)\rightarrow(x_0,0)}(x^2+y^2)\arctan\dfrac{1}{|xy|}=$$
$$=\displaystyle\lim_{r\rightarrow0}[(x_0+r\cos\phi)^2+(r\sin\phi)^2]\arctan\dfrac{1}{|(x_0+r\cos\phi)r\sin\phi|}=$$
$$=\displaystyle\lim_{r\rightarrow0}(x_0^2+r^2\cos^2\phi+2rx_0\cos\phi+r^2\sin^2\phi)\arctan\dfrac{1}{|x_0r\sin\phi+r^2\cos\phi \sin\phi|}=$$
$$=\displaystyle\lim_{r\rightarrow0}(x_0^2+2rx_0\cos\phi+r^2)\arctan\dfrac{1}{|r(x_0\sin\phi+r\cos\phi \sin\phi)|}=x_0^2\frac{\pi}{2},\ \ \forall\phi\in]0;2\pi],\phi\neq\pi$$
 
  • #4
Yep. (Nod)
So we can indeed extend f to a continuous function with $f(x,0)=x^2\frac\pi 2$, $f(0,y)=y^2\frac\pi 2$, and $f(0,0)=0$.
 

FAQ: Proving Continuous Extension of $f(x,y)$ Function

What is the definition of continuous extension?

Continuous extension is a mathematical concept that refers to the ability to extend a function beyond its original domain while maintaining continuity. In other words, the extended function should have no sudden jumps or breaks in its graph.

Why is proving continuous extension important?

Proving continuous extension is important because it allows us to extend the applicability of a function to a larger domain, making it more useful and applicable in various mathematical and real-world situations. It also helps us to better understand the behavior of a function and its limits.

What are the common methods used to prove continuous extension?

The most commonly used methods to prove continuous extension are the ε-δ definition of continuity and the intermediate value theorem. Other methods include the use of limits and piecewise functions.

Under what conditions can a function be extended continuously?

A function can be extended continuously if it satisfies the existence of limits and continuity of the limits criteria. This means that the limit of the function as it approaches a point must exist, and that the value of the limit must be equal to the value of the function at that point.

What are some common challenges in proving continuous extension?

Some common challenges in proving continuous extension include dealing with discontinuous functions, finding the right domain of the extension, and handling complex mathematical expressions. It is also important to pay attention to the epsilon and delta values used in the proof and ensure they are chosen carefully to meet the definition of continuity.

Back
Top