Is f Differentiable at (0,0)?

In summary, it is shown that all directional derivatives of $f$ in $(0,0)$ exist, but $f$ is not differentiable at $(0,0)$ since the limit of the difference quotient along the path $h_1 = h_2$ does not converge to $0$.
  • #1
mathmari
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Hey! :giggle:

We consider the function\begin{align*}f:\mathbb{R}^2 &\rightarrow\mathbb{R} \\ (x,y)&\mapsto \begin{cases}\frac{x^3}{x^2+y^2} & \text{ if } (x,y)\neq(0,0) \\ 0 & \text{ if } (x,y)=(0,0) \end{cases}\end{align*}
(a) Show that all directional derivatives of $f$ in $(0,0)$ exist.
(b) Is $f$ differentiable in $(0,0)$ ? For (a) Ihave donethe following:
$$\partial_vf(0,0)=\lim_{h\rightarrow 0}\frac{f(hv_1, hv_2)-f(0,0)}{h}=\lim_{h\rightarrow 0}\frac{v_1^3}{v_1^2+v_2^2}=\frac{v_1^3}{v_1^2+v_2^2}\in \mathbb{R}$$ The limit exists for all $(v_1,v_2)\neq (0,0)$ and so all directional derivatives of $f$ in $(0,0)$ exist.

Is that correct? For (b) I thought to find a sequence $a_n$ that converges to $(0,0)$ such that $f(a_n)$ doesn't converge to $0$, but I haven't found yet such an examle. Can we maybe not find such an example because $f$ is differentiable in $(0,0)$ ?:unsure:
 
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  • #2
mathmari said:
For (a) Ihave donethe following:
$$\partial_vf(0,0)=\lim_{h\rightarrow 0}\frac{f(hv_1, hv_2)-f(0,0)}{h}=\lim_{h\rightarrow 0}\frac{v_1^3}{v_1^2+v_2^2}=\frac{v_1^3}{v_1^2+v_2^2}\in \mathbb{R}$$ The limit exists for all $(v_1,v_2)\neq (0,0)$ and so all directional derivatives of $f$ in $(0,0)$ exist.

Hey mathmari!

Looks correct to me. (Nod)

mathmari said:
For (b) I thought to find a sequence $a_n$ that converges to $(0,0)$ such that $f(a_n)$ doesn't converge to $0$, but I haven't found yet such an examle. Can we maybe not find such an example because $f$ is differentiable in $(0,0)$ ?
How is it relevant whether $f(a_n)$ converges to $0$ or not? :unsure:
We're not checking continuity are we?
Or are you trying to prove that $f$ is not continuous in $(0,0)$, and therefore it is not differentiable either?
Either way, $f$ is continuous in $(0,0)$, so that won't work. (Shake)

Shouldn't we check instead whether $\frac{\|f(a_n)-f(0,0)-\nabla f(0,0)\cdot a_n\|}{\|a_n\|}$ converges to $0$ or not? :unsure:

Note that since the directional derivatives exist, the partial derivatives exist as well, and $\nabla f(0,0)$ is well defined. 🧐
 
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  • #3
mathmari said:
Hey! :giggle:

We consider the function\begin{align*}f:\mathbb{R}^2 &\rightarrow\mathbb{R} \\ (x,y)&\mapsto \begin{cases}\frac{x^3}{x^2+y^2} & \text{ if } (x,y)\neq(0,0) \\ 0 & \text{ if } (x,y)=(0,0) \end{cases}\end{align*}
(a) Show that all directional derivatives of $f$ in $(0,0)$ exist.
(b) Is $f$ differentiable in $(0,0)$ ?For (a) Ihave donethe following:
$$\partial_vf(0,0)=\lim_{h\rightarrow 0}\frac{f(hv_1, hv_2)-f(0,0)}{h}=\lim_{h\rightarrow 0}\frac{v_1^3}{v_1^2+v_2^2}=\frac{v_1^3}{v_1^2+v_2^2}\in \mathbb{R}$$ The limit exists for all $(v_1,v_2)\neq (0,0)$ and so all directional derivatives of $f$ in $(0,0)$ exist.

Is that correct?[/tex]
Except that you jumped over the most important point- showing that the limit does exist!

I would have done this: every line through (0, 0) (except the vertical line) can be written y= mx for some constant m. Then [tex]\frac{x^3}{x^2+ y^2}= \frac{x^3}{x^3+ m^3x^2}= \frac{x^3}{x^2(m^2+ 1}= \frac{1}{m^2+ 1} x[/tex].
mathmari said:
Hey! :giggle:

We consider the function\begin{align*}f:\mathbb{R}^2 &\rightarrow\mathbb{R} \\ (x,y)&\mapsto \begin{cases}\frac{x^3}{x^2+y^2} & \text{ if } (x,y)\neq(0,0) \\ 0 & \text{ if } (x,y)=(0,0) \end{cases}\end{align*}
(a) Show that all directional derivatives of $f$ in $(0,0)$ exist.
(b) Is $f$ differentiable in $(0,0)$ ?For (a) Ihave donethe following:
$$\partial_vf(0,0)=\lim_{h\rightarrow 0}\frac{f(hv_1, hv_2)-f(0,0)}{h}=\lim_{h\rightarrow 0}\frac{v_1^3}{v_1^2+v_2^2}=\frac{v_1^3}{v_1^2+v_2^2}\in \mathbb{R}$$ The limit exists for all $(v_1,v_2)\neq (0,0)$ and so all directional derivatives of $f$ in $(0,0)$ exist.

Is that correct?
It's good except that you have skipped over the important part- showing that the limit does exist. You just stated that it did.

I would have done this: any line (except a vertical line) through the origin can be written y= mx for some constant m. Then [tex]f(x,y)= f(x,mx)= \frac{x^3}{x^2+ m^2x^2}= \frac{x^3}{x^2(m^2+ 1)}= \frac{1}{m^2+ 1}x[/tex]. So the difference quotient, along that line is [tex]\frac{\frac{1}{m^2+ 1}h}{h}= \frac{1}{1+ m^2}[/tex]. The directional derivative is the limit of that as h goes to 0, [tex]\frac{1}{m^2+ 1}[/tex] which certainly exists!

(Along the vertical line through the origin, x= 0 so the function is [tex]\frac{0^3}{0^2+ y^2}= 0[/tex], a constant so the derivative is 0.)

For (b) I thought to find a sequence $a_n$ that converges to $(0,0)$ such that $f(a_n)$ doesn't converge to $0$, but I haven't found yet such an examle. Can we maybe not find such an example because $f$ is differentiable in $(0,0)$ ?:unsure:
For (b) I thought to find a sequence $a_n$ that converges to $(0,0)$ such that $f(a_n)$ doesn't converge to $0$, but I haven't found yet such an examle. Can we maybe not find such an example because $f$ is differentiable in $(0,0)$ ?:unsure:
 
  • #4
Klaas van Aarsen said:
Shouldn't we check instead whether $\frac{\|f(a_n)-f(0,0)-\nabla f(0,0)\cdot a_n\|}{\|a_n\|}$ converges to $0$ or not? :unsure:

Note that since the directional derivatives exist, the partial derivatives exist as well, and $\nabla f(0,0)$ is well defined. 🧐

We have that (ifI have doneall calculations correctly) $$\frac{\|f\left ((0,0)+(h_1, h_2)\right )-f(0,0)-\nabla f(0,0)\cdot (h_1,h_2)\|}{\|h\|}=\frac{\|f\left (h_1, h_2\right )-0-(1,0)\cdot (h_1,h_2)\|}{\sqrt{h_1^2+h_2^2}}=\frac{\|f\left (h_1, h_2\right )-h_1\|}{\sqrt{h_1^2+h_2^2}}=\frac{|h_1|\cdot h_2^2}{\sqrt{h_1^2+h_2^2}^3}$$ How do we calculate the last limit? :unsure:
 
  • #5

So can we not just say, at the way I did it, that for every $(v_1, v_2)\neq (0,0)$ the limit is a real number and so it exists? :unsure:
 
  • #6
mathmari said:
$$\frac{\|f\left ((0,0)+(h_1, h_2)\right )-f(0,0)-\nabla f(0,0)\cdot (h_1,h_2)\|}{\|h\|}=\frac{\|f\left (h_1, h_2\right )-0-(1,0)\cdot (h_1,h_2)\|}{\sqrt{h_1^2+h_2^2}}=\frac{\|f\left (h_1, h_2\right )-h_1\|}{\sqrt{h_1^2+h_2^2}}=\frac{|h_1|\cdot h_2^2}{\sqrt{h_1^2+h_2^2}^3}$$ How do we calculate the last limit?
Can we find a path $(h_1(t), h_2(t))$ such that the limit is not $0$? 🤔
 
  • #7
Klaas van Aarsen said:
Can we find a path $(h_1(t), h_2(t))$ such that the limit is not $0$? 🤔

For $h_1=h_2$ we get $$\frac{|h_1|\cdot h_1^2}{\sqrt{h_1^2+h_1^2}^3}=\frac{|h_1|^3}{\sqrt{2h_1^2}^3}=\frac{|h_1|^3}{\sqrt{2}\cdot |h_1|^3}=\frac{1}{\sqrt{2}}\rightarrow \frac{1}{\sqrt{2}}\neq 0$$ Since this doesn't converge to $0$, it follows that $f$ is not differentiable.

Is that correct? Or can we not just take $h_1=h_2$ ? :unsure:
 
  • #8
mathmari said:
For $h_1=h_2$ we get $$\frac{|h_1|\cdot h_1^2}{\sqrt{h_1^2+h_1^2}^3}=\frac{|h_1|^3}{\sqrt{2h_1^2}^3}=\frac{|h_1|^3}{\sqrt{2}\cdot |h_1|^3}=\frac{1}{\sqrt{2}}\rightarrow \frac{1}{\sqrt{2}}\neq 0$$ Since this doesn't converge to $0$, it follows that $f$ is not differentiable.
All correct. (Sun)
 
  • #9
Klaas van Aarsen said:
All correct. (Sun)

Great! Thank you very much! (Sun)
 

Related to Is f Differentiable at (0,0)?

1. What does it mean for a function to be differentiable?

Differentiability of a function refers to the existence of a derivative at a given point. In other words, a function is differentiable at a point if it has a defined slope or rate of change at that point.

2. How do you determine if a function is differentiable at a specific point?

A function is differentiable at a point if the limit of its derivative as the input approaches that point exists and is finite. This can be determined using the definition of the derivative or by using the rules of differentiation.

3. What is the significance of a function being differentiable in (0,0)?

A function being differentiable in (0,0) means that it has a well-defined slope at that point, which is often referred to as the instantaneous rate of change. This is important in many applications, such as optimization and curve fitting.

4. Can a function be differentiable at a point but not continuous?

Yes, it is possible for a function to be differentiable at a point but not continuous. This can occur if the function has a sharp corner or vertical tangent at that point, which violates the definition of continuity.

5. What are the conditions for a function to be differentiable in (0,0)?

In order for a function to be differentiable in (0,0), it must first be defined and continuous at that point. Additionally, the left and right derivatives must exist and be equal at that point.

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