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I would like to show that the function:

$$f:(x,y)\mapsto \frac{x\sin(y)-y\sin(x)}{x^2+y^2}$$

is a $C^1$-function.

$$ \frac{\partial f}{\partial x}(x,y)=\frac{\sin(y)-y\cos(x)}{x^2+y^2}+\frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}$$

$$ \frac{\partial f}{\partial y}(x,y)=-\frac{\partial f}{\partial y}(y,x)=... $$

So I just have to show that:

$$ \frac{\partial f}{\partial x}(x,y)\rightarrow_{(0,0)}0$$

When $y\geq0$ :

$$ -\frac{y^3}{6(x^2+y^2)}+\frac{x^2y}{x^2+y^2}-\frac{x^4y}{4!(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}$$

When $y<0$ :

$$ -\frac{y^3}{6(x^2+y^2)}+\frac{y^5}{5!(x^2+y^2)}+\frac{x^2y}{2(x^2+y^2)} \leq \frac{\sin(y)-y\cos(x)}{x^2+y^2} \leq \frac{yx^2}{2(x^2+y^2)}-\frac{yx^4}{4!(x^2+y^2)}$$

So $$ \frac{\sin(y)-y\cos(x)}{x^2+y^2}\rightarrow_{(0,0)}0 $$

How can I directly find an upper bound of $$ \left| \frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2} \right|$$ that tends to 0 ?

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2 Answers

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Because of the Taylor expansion $\sin(x) = x - {x^3 /6} + ...$, if $x$ and $y$ are small enough you can write $\sin(x) = x + E(x)$ and $\sin(y) = y + E(y)$, where $|E(x)| < |x|^3$ and $|E(y)| < |y|^3$. So you have $${2x(y\sin(x) - x\sin(y)) \over (x^2 + y^2)^2} = {2x(yx + yE(x) - xy - xE(y)) \over (x^2 + y^2)^2}$$ $$= {2x(yE(x) - xE(y)) \over (x^2 + y^2)^2}$$ Taking absolute values and bounding, this is at most $$= 2|x|{|yE(x)| + |xE(y)| \over (x^2 + y^2)^2}$$ Inserting $|E(x)| < |x|^3$ and $|E(y)| < |y|^3$ this is bounded by $$= 2|x|{|x^3y| + |xy^3| \over (x^2 + y^2)^2}$$ $$= 2|x|{|xy|(x^2 + y^2) \over (x^2 + y^2)^2}$$ $$= 2|x| {|xy| \over x^2 + y^2}$$ I think you can take it from there...

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So we can write the Taylor expansion of $f(x,y)$ with respect to each variable $x$ and $y$ at the same time to find the limit in $(0,0)$ ? So: $$ \frac{2x(y\sin(x)-x\sin(y))}{(x^2+y^2)^2}=\frac{2x(x(y-y^3/6+o(y^4))-y(x-x^3/6+o‌​(x^4)))}{(x^2+y^2)^2}...\rightarrow 0$$ ? –  Chon Mar 19 '12 at 18:34
    
@Chon Yes, exactly. –  Zarrax Mar 19 '12 at 21:20
    
Thank you for your answer! –  Chon Mar 20 '12 at 21:52
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A partial answer to at least show that $f$ is differentiable at $(0,0)$. Then it remains to show that its derivative is continuous. I didn't check if that also follows from this bound directly.

Combine $\tan(x)\geq x$ and $\sin(x) \leq x$ for $x \in [0,\pi/2)$ to get

$$ \cos(x) \leq \frac{\sin(x)}{x} \leq 1 $$

for $x \in (-\pi/2,\pi,2)$. Then for $x,y \in (-\pi/2,\pi/2)$

$$ -\frac{y^2}{2} \leq \cos(y)-1 \leq \frac{\sin(y)}{y} - \frac{\sin(x)}{x} \leq 1 - \cos(x) \leq \frac{x^2}{2}. $$

Taking the absolute value:

$$ \left| \frac{\sin(y)}{y} - \frac{\sin(x)}{x} \right| \leq \frac{\max(x^2,y^2)}{2}\leq \frac{x^2+y^2}{2}. $$

This results in the following estimate of $f$:

$$ \left|\frac{x\sin(y) - y\sin(x)}{x^2+y^2}\right| \leq \left| \frac{xy}{x^2+y^2}\right| \cdot \left| \frac{\sin(y)}{y} - \frac{\sin(x)}{x}\right| \leq \frac{x^2+y^2}{4} $$

This is sufficiently sharp to conclude that $f$ is differentiable in $(0,0)$.

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Thank you for your answer WimC, even if I already know that $f$ is continuous in $(0,0)$ (it was the subject of my previous post). –  Chon Mar 18 '12 at 21:19
    
@Chon I know, but this shows more than mere continuity. It shows differentiability for one. It remains to show that the derivative is continuous. –  WimC Mar 19 '12 at 5:42
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