1
$\begingroup$

I came across such problem:

$g:\mathbb{R}\rightarrow \mathbb{R}$ is a $C^1$-function with $g(\theta+\pi)=-g(\theta)$ for all $x$. Define a function $f: \mathbb{R}^2\rightarrow \mathbb{R}$ as $$f(x,y)=f(r\cos{\theta},r\sin{\theta})=r\cdot g(\theta). $$

Show that $f$ is differentiable everywhere in $\mathbb{R}^2$ except possibly at $(x,y)=(0,0)$ and also that all the directional derivatives of $f$ exist at $(0,0)$

Can anyone show me the differentiability here just by the definition that

$\displaystyle \lim_{||h||\rightarrow 0}{ \frac{||f(x_0+h)-f(x_0)-J\cdot h||}{||h||}}=0$

Since there is change of variable involved as well and I don't know what the use of the periodicity of $g$ here, I am confused even more about the question.

I'll really appreciate if someone can show me the computation process, because I really need a concrete example to understand this well. Thanks :)

$\endgroup$
  • $\begingroup$ Just rotate the function $\|x\|$ through the $z$-axis. $\endgroup$ – Pedro Tamaroff Apr 25 '13 at 1:31
  • $\begingroup$ What do you mean by this? Could please explain more explicitly? $\endgroup$ – Cancan Apr 25 '13 at 1:43
  • $\begingroup$ Look at the cone with vertex at the origin (or any point on the $z$ axis) and growing upwards (or downwards). It's like looking at the surface of revolution of $\|x\|$. $\endgroup$ – Pedro Tamaroff Apr 25 '13 at 1:56
1
$\begingroup$

You don't need a computation away from 0, because it is the product of differentiate functions. The function $r$ is not differentiable by itself at 0, so you have to try something else there. Along each line through the origin, $\theta$ is almst constant (it changes by $\pi$ at the origin), so taking the derivative is the same thing as finding the derivative of $mx$, where $m=g(\theta)$. This function is clearly differentiable.

An easy example is if $g=\sin(\theta)$. Then $f(x,y)=y$, and as I said above, the function is just a linear function $mx$ along each line through the origin.

$\endgroup$
  • $\begingroup$ That's why in the question it says "possibly". So, you just gave an example that is differentiable at origin but this doesn't prove the conclusion we want. So what I mean here is that I really expect an example that is not differentiable at origin. $\endgroup$ – Cancan Apr 25 '13 at 1:41
  • $\begingroup$ Try $g=\sin^3(\theta)$. Then the derivative along $\theta=\frac{\pi}{4}$ is not the product of the gradient and $(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$. So, for your explicit calculation, assume the derivative exists and take your limit in the vertical and horizontal directions to find $J$ explicitly, then go in the diagonal direction to show the limit is non zero. $\endgroup$ – Brian Rushton Apr 25 '13 at 1:53
  • $\begingroup$ Which gradient do you mean here? with respect to $x,y$ or $\theta , r$? and why do you set $h=(\frac{1}{\sqrt{2}},\frac{1}{\sqrt{2}})$ $\endgroup$ – Cancan Apr 25 '13 at 1:58
  • $\begingroup$ If the function is differentiable, then any directional derivative can be found by taking the product of a direction vector with the $x,y$ gradient. So the fact this doesn't hold for my example shows it is not differentiable. $\endgroup$ – Brian Rushton Apr 25 '13 at 2:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.