MATH2111 Higher Several Variable Calculus (1 Viewer)

leehuan · May 29, 2017

Re: Several Variable Calculus

My lecture slides. I haven't used a maths textbook before at uni. I'm just going to have to take it up with the lecturer

leehuan · Jun 4, 2017

Re: Several Variable Calculus

$\nabla \cdot \textbf{F}\text{ represents the divergence of the vector field}$

$\text{What is this supposed to represent? }\textbf{F} \cdot \nabla$

InteGrand · Jun 4, 2017

Re: Several Variable Calculus

leehuan said:
$\nabla \cdot \textbf{F}\text{ represents the divergence of the vector field}$

$\text{What is this supposed to represent? }\textbf{F} \cdot \nabla$

$\noindent It's an operator that acts on scalar functions. Let $\textbf{F} = \left(F_{x}, F_{y}, F_{z}\right)$, then as $\nabla = \left(\partial_{x}, \partial_{y}, \partial_{z}\right)$, we have $\textbf{F} \cdot \nabla = F_{x}\partial_{x}+F_{y}\partial_{y}+F_{z}\partial_{z}$. This acts on a scalar function $\phi$ as follows:$

$\noindent \left(\textbf{F} \cdot \nabla\right)\phi \equiv \left(F_{x}\partial_{x}+F_{y}\partial_{y}+F_{z} \partial_{z} \right)\phi \equiv F_{x}\partial_{x}\left(\phi\right) + F_{y}\partial_{y}\left(\phi\right) + F_{z}\partial_{z}\left(\phi\right) = \mathbf{F}\cdot \left(\nabla \phi\right),$

$\noindent where $\nabla \phi = \left(\partial_{x}\left(\phi\right),\partial_{y} \left(\phi\right),\partial_{z}\left(\phi \right) \right)$ is of course the gradient of $\phi$, i.e. $\mathrm{grad}(\phi)$.$

leehuan · Jun 5, 2017

Re: Several Variable Calculus

Can someone compute the directional derivative at (0,0) of this function?

$f(x,y) = \begin{cases}\frac{xy^2}{x^2+y^4} & \text{if }x\neq 0\\ 0&\text{otherwise}\end{cases}$

$\text{In some cases I end up with }\frac{\partial f}{\partial \textbf{u}} = 0\text{ but for most cases I get }\frac{{u_2}^2}{u_1}\\ \text{yet the answers say it's 0 in every case}$

(Assume u is a unit vector already)

seanieg89 · Jun 5, 2017

Re: Several Variable Calculus

The partial derivatives are all zero, but the directional derivatives are as you say for u1 nonzero, and zero if u1=0.

This is one of those funny functions that you don't see the full weirdness of by just looking at 2d slices of its graph. Each partial derivative exists at the origin, but the function f isn't even continuous at the origin!

leehuan · Jun 11, 2017

seanieg89 · Jun 11, 2017

leehuan said:
$\text{Trying to prove that the mixed partial theorem doesn't hold for this function}\\ f(x,y)=\begin{cases}0&\text{ if }(x,y)=(0,0)\\ \frac{xy(x^2-y^2)}{x^2+y^2}&\text{otherwise}\end{cases}$

$\text{But I thought that }f, f_x\text{ and }f_y\text{ are continuous here?}$

They are, but so what? These are not the hypotheses of Clairaut's theorem.

leehuan · Jun 11, 2017

seanieg89 said:
They are, but so what? These are not the hypotheses of Clairaut's theorem.

Maybe it was just my unwillingness to compute any second order partial derivatives because the first order partial derivatives were thoroughly untidy.

Is there any easy way of contradicting the criteria of Clairaut's theorem then?

glittergal96 · Jun 11, 2017

Sorry about alt, am on phone. Showing that typical Clairaut hypotheses don't hold doesn't prove that mixed partials differ because its not iff. Just compute mixed partials at 0, it really shouldn't take long.

Sent from my GT-I9506 using Tapatalk

leehuan · Jun 11, 2017

Oh, that was done in a previous part. I understand your point there...

...albeit this was the question. It's the last part that's of interest.

InteGrand · Jun 11, 2017

leehuan said:
Oh, that was done in a previous part. I understand your point there...

...albeit this was the question. It's the last part that's of interest.

It didn't work because the hypotheses of Clairaut's Theorem were not satisfied by that ƒ (I assume you know what typical hypotheses are for it).

leehuan · Jun 11, 2017

InteGrand said:
It didn't work because the hypotheses of Clariaut's Theorem were not satisfied by that ƒ (I assume you know what typical hypotheses are for it).

Goes back up to the above query I suppose. Is the only way of really proving it formally just to brute those second order derivatives?

InteGrand · Jun 11, 2017

leehuan said:
Goes back up to the above query I suppose. Is the only way of really proving it formally just to brute those second order derivatives?

You could probably also do it by computing the second partial derivatives at (x,y) ≠ (0, 0) and investigating their limit as (x, y) -> (0, 0), but there's no need to do that since you've already computed the mixed second-partials at the origin in an earlier part of the question (part b)).

leehuan · Jun 11, 2017

I suppose what's really bugging me is the fact that d) is actually asking to explain why b) is true. Looks like more of just a grind rather than testing some nice stuff but I suppose I'll just have to deal with it.

seanieg89 · Jun 11, 2017

It really doesn't take that long, especially since to prove discontinuity you can just show that approaching the origin via different axes gives a different result, so most terms in the mixed partials drop out from x or y being zero.

I would definitely go via this route rather than by restating the fact that the mixed partials do not match and using the contrapositive of Clairaut based on the wording of the question.

"Because the conclusion of Clairaut's theorem is not true" is NOT what I would consider a good answer to "Why does Clairaut's theorem not work here?".

leehuan · Jun 11, 2017

$f:[-\pi, \pi]\to \mathbb{R}\text{ satisfied }f(-x)=-f(x), \, f(\pi)=0, \, f^{\prime\prime\prime}(x) = -6$

$\text{Proven in b): The relevant Fourier series is}\\ Sf(x) = \sum_{k=1}^\infty \frac{12(-1)^k}{k^3}\sin(kx)$

c) Give a simple formula for f(x)

$\text{Proven in a): } f^{\prime\prime}(x)=-6x\\ \text{so I wanted to keep integrating up to get }f(x) = -x^3\\ \text{But then what? I still have to satisfy }f(\pm \pi) = 0\\ \text{Do I just define a piecewise function to get out of this issue?}$

InteGrand · Jun 11, 2017

leehuan said:
$f:[-\pi, \pi]\to \mathbb{R}\text{ satisfied }f(-x)=-f(x), \, f(\pi)=0, \, f^{\prime\prime\prime}(x) = -6$

$\text{Proven in b): The relevant Fourier series is}\\ Sf(x) = \sum_{k=1}^\infty \frac{12(-1)^k}{k^3}\sin(kx)$

c) Give a simple formula for f(x)

$\text{Proven in a): } f^{\prime\prime}(x)=-6x\\ \text{so I wanted to keep integrating up to get }f(x) = -x^3\\ \text{But then what? I still have to satisfy }f(\pm \pi) = 0\\ \text{Do I just define a piecewise function to get out of this issue?}$

It won't be -x^3. It'll be a different cubic. It will be an odd cubic that satisfies f(pi) = 0.

leehuan · Jun 11, 2017

InteGrand said:
It won't be -x^3. It'll be a different cubic. It will be an odd cubic that satisfies f(pi) = 0.

Oh, now that I think about it could it potentially be x(x^2-pi^2)?

InteGrand · Jun 11, 2017

leehuan said:
Oh, now that I think about it could it potentially be x(x^2-pi^2)?

Yeah but the third derivative was negative, so it'd be the negative of what you wrote there.

leehuan · Jun 19, 2017

(Stats is done for this sem)

$\\\text{Let }\textbf{f}:\mathbb{R}^n\to \mathbb{R}^n\text{ and }g:\mathbb{R}^n\to \mathbb{R}\text{ be differentiable, and define}\\ h = g\circ \textbf{f}$

$\text{If }\textbf{a}\in \mathbb{R}^n\text{ is a stationary point of }h\text{, what can you say about }\textbf{f}\text{ and }g?$

MATH2111 Higher Several Variable Calculus (1 Viewer)

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