f(x,y) = 2xy/(x^2+y^2)^n when (x,y) is not equal to (0,0)
= 0 when (x,y) when (x,y) is equal to (0,0)
Prove that f is cont. at (0,0) if and only if n=1/2 (n>0)......Thank you very much for answering
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I have shown that n can not be greater than 1...that's all...but I do not know the trick to show n=1/2 only...@moorish – Abhisek Mukherjee Dec 06 '15 at 06:43
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Change to polar coordinates, so $x=r\cos\theta$, $y=r\sin\theta$. Our expression becomes $$\frac{r^2\cos\theta\sin\theta}{r^{2n}}.$$ If $n\lt 1$, this has limit $0$ as $r\to 0$.
André Nicolas
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The only if claimed is incorrect.
Note the following:
- $\forall (x,y)\in \mathbb{R}^2, ~2|xy| \le x^2 + y^2$
- $\forall \alpha > 0, 0<a<b \implies 0<a^\alpha<b^\alpha$
For $n<1$, Let $\delta(\epsilon) = \epsilon^{1/{2 - 2n}}$ $$\left|\frac{2xy}{(x^2 + y^2)^n} - 0\right| \le (x^2 + y^2)^{1-n} \le \delta^{2 - 2n} = \epsilon$$
demonstrates the continuity of $f_n(x,y)$.
stochasticboy321
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I also think that only if condition is not true...thank you for confirming sir – Abhisek Mukherjee Dec 06 '15 at 07:04