# Continuous extension

1. Aug 11, 2010

### Alexx1

How can the function

f: ℝ² → ℝ : (x,y) |--> $${{x^2+y^2-(x^3y^3)}\over{x^2+y^2}}$$ if (x,y) ≠ (0,0)

be defined in the origin so that we get a continuous function?

When I take 'x=y' (so (y,y)) and 'y=x' (so (x,x)) I get:

$${{2-y^4}\over{2}}$$

and

$${{2-x^4}\over{2}}$$

So for the first one I get '1' when y=0
and for the second one I get '1' when x=0

So can I say that if (x,y)=0 --> f=1 ??

2. Aug 11, 2010

### Petr Mugver

Yes, you can extend your function to a continuous one if you define f(0,0)=1.

3. Aug 11, 2010

### Alexx1

Thx!

4. Aug 11, 2010

### arildno

In order to make this general, you should switch to polar coordinates.

Then, we can transforme the fraction to:
$$\frac{r^{2}-r^{6}\sin^{3}\theta\cos^{3}\theta}{r^{2}}=1-r^{4}\sin^{3}\theta\cos^{3}\theta, r\neq{0}$$
clearly, this is readily extendable for r=0.

5. Aug 11, 2010

### Alexx1

Thank you very much!
And how about this one?

https://www.physicsforums.com/showthread.php?t=421612

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