the no-normie zone - 2017-10-14

Anonymous

07:28

@BalarkaSen So, I got confused again. Why is continuity of partial derivatives necessary for the Jacobian to exist?

Anonymous

in The h Bar, Sep 17 at 8:22, by Balarka Sen

Jut having the partials is not enough, but partials being continuous is enough to guarantee that the Jacobian is the derivative.

Anonymous

Isn't the Jacobian just the matrix consisting of partial at a point $\mathbf{a}$ ?

Anonymous

08:37

From what I understood:

Anonymous

1) The existence of the Jacobian matrix does not guarantee differentiability. It is a necessary but not sufficient condition.

Anonymous

2) Existence of Jacobian matrix and continuity of the partial derivatives is a sufficient but not necessary condition for differentiability.

Anonymous

That just says the partial derivatives exist. That doesn't even give continuity. — zhw. Sep 25 at 17:44

Anonymous

Sure, but then you are assuming something stronger than differentiability. — zhw. Sep 25 at 17:48

Anonymous

@BalarkaSen My question now is: What is the necessary and sufficient condition for the function $f(x,y)$ to be differentiable at $(a,b)$ ?

Anonymous

08:52

According to what I understood the necessary and sufficient is

Anonymous

a) $f(a+h,b+k) = f(a,b) + T(h,k) + o[(h^2+k^2)^{1/2}]$ where $T:\Bbb R^2 \to \Bbb R$

Anonymous

b) $f(a+h,b+k)=f(a,b)+Ah+Bk+h\phi(h,k)+k\psi(h,k)$

Anonymous

where $\phi,\psi \to 0$ as $(h,k)\to (0,0)$

Anonymous

c)$\lim_{(h,k)\to(0,0)}\frac{f(a+h,b+k)-f(a,b)-\mathbf{A}(h,k)}{\sqrt{h^2+k^2}}=‌0$ where $\mathbf{A}:\Bbb R^2\to \Bbb R$. $\mathbf{A}$ is the linear map which is the tangent plane at $(a,b)$

Anonymous

DOUBTS: We now need to show that (a),(b) and (c) are identical. Also, we need to explain why (a),(b) and (c) are the necessary and sufficient conditions. Moreover we need to explain why existence of Jacobian and continuity of partial derivatives is a condition stronger than differentiability as zhw says.

Anonymous

09:29

I think I can understand why a and c are equivalent conditions. Now need to show b is equivalent to them.

Anonymous

09:46

I think zhw is correct in saying that partial derivatives need not be continuous for differentiability at a point. Take $$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) & \text{ if $(x,y) \ne (0,0)$}\\0 & \text{ if $(x,y) = (0,0)$}.\end{cases}$$ for example.

Balarka Sen

10:50

@Blue Necessary? No. Just sufficient.

You quoted me as saying "enough". That means sufficient, not necessary

Anonymous

@BalarkaSen Yeah, that's what I wrote ^ :P

Anonymous

2 hours ago, by Blue

DOUBTS: We now need to show that (a),(b) and (c) are identical. Also, we need to explain why (a),(b) and (c) are the necessary and sufficient conditions. Moreover we need to explain why existence of Jacobian and continuity of partial derivatives is a condition stronger than differentiability as zhw says.

Anonymous

See these though ^

Balarka Sen

This is for differentiability of a function $f : \Bbb R^2 \to \Bbb R$?

Anonymous

@BalarkaSen

Anonymous

10:53

@BalarkaSen Yes

Anonymous

The necessary and sufficient conditions

Anonymous

I just need to prove that (a),(b) and (c) are identical

Anonymous

And also that continuity of partials is a stronger condition that differentiability

Balarka Sen

(a) and (b) is equivalent because $T(h, k) = T(h(1, 0) + k(0, 1)) = hT(1, 0) + kT(0, 1)$

Write $A = T(1, 0)$ and $B = T(0, 1)$

Anonymous

Aha. Brilliant. How do you get the $\psi$ and $\phi$ part though ?

Balarka Sen

10:57

So you want to prove $h\phi(h, k) + k\psi(h, k)$ is of the form $o(\sqrt{h^2+k^2})$.

Anonymous

@BalarkaSen Yup!

Balarka Sen

Look at $(h \phi(h, k) + k\psi(h, k))/\sqrt{h^2 + k^2}$

The first term is $\displaystyle \frac{h}{\sqrt{h^2+k^2}} \phi(h, k)$

The factor there is bounded by $1$, because $h < \sqrt{h^2+k^2}$.

Anonymous

I agree. Then?

Balarka Sen

Then this is bounded below by $0$ and above by $|\phi(h, k)|$ in absolute values. Take limit as $(h, k) \to 0$ and squeeze theorem it.

Same for the second term.

Anonymous

Oooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌ooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooooo‌oooooooooooooooooooooooooooooooooooooooooooooooooooooooooo :'D

Anonymous

11:03

That was easy

Anonymous

I was overthinking

Anonymous

$\phi,\psi \to 0$ anyway

Balarka Sen

Right.

Anonymous

as $h,k\to 0$

Anonymous

Yay :)

Balarka Sen

11:04

Also yes there are a lot of functions that are not $C^1$ but the Jacobian exists.

Anonymous

One thing. Is continuity a necessary condition for all directional derivatives to exist?

Balarka Sen

No...

The directional derivatives might exist but not be continuous

Anonymous

@BalarkaSen I'm not talking about continuity of directional derivative

Balarka Sen

Read my sentence.

You can also just take the one dimensional version of what you wrote, $f(x) = x^2 \sin(1/x)$ for $x \neq 0$ and $0$ if $x = 0$.

All the directional derivatives exist at $x = 0$ but is not continuous :P

Anonymous

@BalarkaSen But $\lim_{x\to 0} x^2\sin(1/x)$ is 0

Anonymous

11:08

How is it not continuous ?

Balarka Sen

... we are thinking about the derivative not being continuous.

$C^1$ means continuous derivatives.

Anonymous

2 mins ago, by Blue

@BalarkaSen I'm not talking about continuity of directional derivative

Anonymous

I was talking about continuity of the function itself

Anonymous

Not the derivative

Balarka Sen

What? Pay attention to what you are writing, man.

"I think zhw is correct in saying that partial derivatives need not be continuous for differentiability at a point."

>Partial derivatives need not be continuous

Anonymous

11:10

4 mins ago, by Blue

One thing. Is continuity a necessary condition for all directional derivatives to exist?

Balarka Sen

Yes, sure you are talking about continuity of the derivatives

Anonymous

I was asking a different question :P

Balarka Sen

Then you should explicitly write that. I can't keep track of a hundred questions at the same time.

Anonymous

@BalarkaSen Sorry :P

Anonymous

Let's do it one at a time

Balarka Sen

11:11

@Blue No, it is not. You already know lots of examples.

Anonymous

@BalarkaSen Even I think it is not. But I can't think of an example right now

Balarka Sen

Think about $f(x, y) = 1/x$ if $y = x^2 \neq 0$ and $f(0, 0) = 0$, the infamous example.

There are also standard examples. Looking through Ted's book tells me $f(x, y) = |x|y/\sqrt{x^2+y^2}$ and $f(0, 0) = 0$

Anonymous

@BalarkaSen Ah. We had done that example earlier I think. I should bookmark conversations from now on

Anonymous

Also found this math.jhu.edu/~js/Math202/example2.pdf

Balarka Sen

Sure. There are many examples.

Anonymous

11:15

Okay, so this is clear now

Anonymous

Now the next question is:

Anonymous

Is continuity of function at a point necessary for differentiability at that point ?

Balarka Sen

Yes, if a function is differentiable at $a$, it is automatically continuous at $a$

Anonymous

@BalarkaSen I suppose that is because the derivative(tangent plane) acts as a linear approximation about $a$

Balarka Sen

Yes, kind of. You can try to prove that using the definition of differentiability, it's not hard.

Anonymous

11:20

Alright. I'll try that as an exercise!

Anonymous

Now to prove zhw's claim =P

Balarka Sen

OK

Anonymous

I think it feels intuitive now

Balarka Sen

Yeah it's believable

Anonymous

But perhaps I should prove it rigorously that continuous partials is stronger than differentiability

Balarka Sen

11:22

You just need to produce an example where it is differentiable but does not have continuous partials. $f(x) = x^2 \sin(1/x)$ does the trick

Anonymous

$f(x,y)=\begin{cases}(x^2+y^2)\sin\left(\frac{1}{\sqrt{x^2+y^2}}\right) & \text{ if $(x,y) \ne (0,0)$}\\0 & \text{ if $(x,y) = (0,0)$}.\end{cases}$ doesn't have continuous partials but it perfectly satisfies (a),(b) or (c)

Anonymous

@BalarkaSen Yep, that's an example :)

Balarka Sen

Why work with multiple variables when you have one dimensional counterexamples?

It just complicates calculations

Anonymous

@BalarkaSen I agree

Anonymous

I shouldn't

Anonymous

11:27

Thanks a lot. I should study some other subject now. See you :)

Balarka Sen

OK, see you

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♦ $Mathematics$ the no-normie zone

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♦ the no-normie zone

♦ $Mathematics$ the no-normie zone