MATH2111 Higher Several Variable Calculus (5 Viewers)

InteGrand · Mar 6, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Bit confused. How is it possible for }\lim_{x\to 0}\lim_{y\to 0}f(x,y)\\ \text{and the reverse order of limiting to not exist}$

$\text{but then }\lim_{(x,y)\to (0,0)}f(x,y)\text{ exists}\\ \text{for }f(x,y)=(x+y)\sin \frac1x \sin \frac1y$

$\noindent That example is showing that a multivariable limit can exist even though the corresponding iterated limits may not exist. If we hold $x$ fixed and non-zero and limit $y$ to $0$, the limit won't exist because $f(x, y) = x \sin \frac1x \sin \frac1y + y\sin \frac1x \sin \frac1y$. The first term here doesn't have a limit as $y\to 0$, but the second one does (has limit $0$, as you can show by the squeeze law), so overall the limit doesn't exist.$

$\noindent Of course, $\lim_{(x,y)\to (0,0)}f(x,y)$ exists and equals $0$, basically because $\left| f(x,y)\right| \leq |x+y|$ for all $x\neq 0$ and $y\neq 0$.$

leehuan · Mar 6, 2017

Re: Several Variable Calculus

InteGrand said:
$\noindent That example is showing that a multivariable limit can exist even though the corresponding iterated limits may not exist. If we hold $x$ fixed and non-zero and limit $y$ to $0$, the limit won't exist because $f(x, y) = x \sin \frac1x \sin \frac1y + y\sin \frac1x \sin \frac1y$. The first term here doesn't have a limit as $y\to 0$, but the second one does (has limit $0$, as you can show by the squeeze law), so overall the limit doesn't exist.$

$\noindent Of course, $\lim_{(x,y)\to (0,0)}f(x,y)$ exists and equals $0$, basically because $\left| f(x,y)\right| \leq |x+y|$ for all $x\neq 0$ and $y\neq 0$.$

Appears so counterintuitive though. I can't visualise what's going on here

Paradoxica · Mar 6, 2017

Re: Several Variable Calculus

leehuan said:
Appears so counterintuitive though. I can't visualise what's going on here

nobody can.....

QuantumRoulette · Mar 6, 2017

Re: Several Variable Calculus

Consider the two metrics $d(\mathbf{x}, \mathbf{y}) = ||\mathbf{x} - \mathbf{y}||$ and
$\delta(\mathbf{x}, \mathbf{y}) = \frac{d(\mathbf{x}, \mathbf{y})}{1 +d(\mathbf{x}, \mathbf{y})}$ (you may assume they are metrics).
i) Show that d and δ are not equivalent.

seanieg89 · Mar 7, 2017

Re: Several Variable Calculus

QuantumRoulette said:
Consider the two metrics $d(\mathbf{x}, \mathbf{y}) = ||\mathbf{x} − \mathbf{y}||$ and
$\delta(\mathbf{x}, \mathbf{y}) = \frac{d(\mathbf{x}, \mathbf{y})}{1 +d(\mathbf{x}, \mathbf{y})}$ (you may assume they are metrics).
i) Show that d and δ are not equivalent.

I assume d(x,y) is supposed to be ||x-y|| and the set is some normed vector space V with norm ||.|| (e.g. R^d). (Please specify more if this is not the intended setting.)

Then these two metrics are not (strongly) equivalent because V is bounded with the delta metric but unbounded with the d metric.

That V is bounded with the delta metric follows immediately from the definition of delta, which must always lie in [0,1). On the other hand d(tx,0)=|t|d(x,0) can be made arbitrarily large for nonzero x.

Note that these two metrics ARE topologically equivalent though, in the sense that convergence in one metric implies convergence in the other. This follows from from the map x->x/(1+x) being a homeomorphism from [0,inf) to [0,1).

seanieg89 · Mar 7, 2017

Re: Several Variable Calculus

leehuan said:
Appears so counterintuitive though. I can't visualise what's going on here

Might not be easy to visualise the graph of the function on all of R^2, but you should certainly be able to visualise what it looks like on the slices x=const. or y=const which is all that matters for seeing/proving the nonexistence of the iterated limit. It is an oscillatory expression that oscillates faster as you approach axes. One of the two summands becomes irrelevant as you get close to the axes, so the other one dominates. This thing behaves like (const).sin(1/x), which of course does not converge unless that const is zero.

The boundedness of sine makes it clear that f(x,y) tends to zero as (x,y) tends to zero though.

Long story short: don't be too hasty to form intuitions in analysis, lots of things can behave weirdly...you kind of have to slowly build up a list of things that ARE true (via proof!) rather than assuming innocuous statements are true and ruling these things out as you come across pathological counterexamples.

And when you are looking at functions like this, try to isolate the terms that actually matter for the property you are trying to prove. A large part of analysis is just approximating ugly things by nice things, throwing away small sets on which a function behaves badly, etc etc.

leehuan · Mar 8, 2017

Re: Several Variable Calculus

$\text{Prove that }\lim_{(x,y)\to (0,a)}\frac{\sin (xy)}{x}=a$

$\text{This result may be assumed: }\left| \frac{\sin a}{a}-1\right|=|a|^2, \, a\in \mathbb{R}$

InteGrand · Mar 8, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Prove that }\lim_{(x,y)\to (0,a)}\frac{\sin (xy)}{x}=a$

$\text{This result may be assumed: }\left| \frac{\sin a}{a}-1\right|=|a|^2, \, a\in \mathbb{R}$

$\noindent Here's the gist of a way to do it. For $x\neq 0$ and $y\neq 0$, we have$

$\begin{align*}\left|\frac{\sin (xy)}{x} - a \right| &= \left|y\right|\left|\left(\frac{\sin (xy)}{xy} -1\right) + \left(1 - \frac{a}{y}\right) \right| \\ &\leq \left|y \right| \left(\left|\frac{\sin (xy)}{xy} -1 \right| + \left|1 - \frac{a}{y} \right| \right) \quad (\text{triangle inequality}) \\ &\leq \left|y\right| \left|xy\right|^{2} + \left|y-a\right| \quad (\text{using the given assumable result}). \end{align*}$

$\noindent You should be able to finish from here.$

seanieg89 · Mar 9, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Prove that }\lim_{(x,y)\to (0,a)}\frac{\sin (xy)}{x}=a$

$\text{This result may be assumed: }\left| \frac{\sin a}{a}-1\right|=|a|^2, \, a\in \mathbb{R}$

|sin(x)/x-1|=|x|^2 ?

Surely you mean something like:

|sin(x)/x-1| = O(|x|^2).

leehuan · Mar 11, 2017

Re: Several Variable Calculus

seanieg89 said:
|sin(x)/x-1|=|x|^2 ?

Surely you mean something like:

|sin(x)/x-1| = O(|x|^2).

There was a typo; it was meant to be a less than or equals to. I think InteGrand just ignored or figured I had typo'd and just used the correct inequality.
________________________________________________________________

$\text{Prove that }\mathbb{Q}\text{ is not open nor closed}$

$\text{So I proved it was not open by showing that for }0\in \mathbb{Q}\\ \forall r > 0,\, \frac{r}{\sqrt2}\text{ fails for }r\neq \alpha\sqrt{2}, \, \frac{r}{\sqrt3}\text{ fails for }r=\alpha \sqrt2\\ \text{where }\alpha \in \mathbb{Q}$

As in, for every ball around 0 there's always an irrational number in it, therefore 0 is not an interior point so it cannot be open. (I hope I did not screw this up.)
____________________________

$\text{But now I'm a bit stuck on proving it's not closed, i.e. }\mathbb{R} - \mathbb{Q}\text{ is not open}$

$\text{Progress: Take }\sqrt{2},\text{ then }\forall r > 0\\ \text{If }y\in B(\sqrt{2},r)\\ \text{Then }|y-\sqrt{2}|<r \implies \sqrt{2}-r < y < \sqrt{2}+r$

I wanted to use the floor/ceiling functions but then I realised that'd be problematic if r = 0.1, so what might be a good choice for a rational number that is in every ball B(sqrt2, r)?

(Sorry, I think I worded my question horribly)

InteGrand · Mar 11, 2017

Re: Several Variable Calculus

leehuan said:
There was a typo; it was meant to be a less than or equals to. I think InteGrand just ignored or figured I had typo'd and just used the correct inequality.
________________________________________________________________

$\text{Prove that }\mathbb{Q}\text{ is not open nor closed}$

$\text{So I proved it was not open by showing that for }0\in \mathbb{Q}\\ \forall r > 0,\, \frac{r}{\sqrt2}\text{ fails for }r\neq \alpha\sqrt{2}, \, \frac{r}{\sqrt3}\text{ fails for }r=\alpha \sqrt2\\ \text{where }\alpha \in \mathbb{Q}$

As in, for every ball around 0 there's always an irrational number in it, therefore 0 is not an interior point so it cannot be open. (I hope I did not screw this up.)
____________________________

$\text{But now I'm a bit stuck on proving it's not closed, i.e. }\mathbb{R} - \mathbb{Q}\text{ is not open}$

$\text{Progress: Take }\sqrt{2},\text{ then }\forall r > 0\\ \text{If }y\in B(\sqrt{2},r)\\ \text{Then }|y-\sqrt{2}|<r \implies \sqrt{2}-r < y < \sqrt{2}+r$

I wanted to use the floor/ceiling functions but then I realised that'd be problematic if r = 0.1, so what might be a good choice for a rational number that is in every ball B(sqrt2, r)?

(Sorry, I think I worded my question horribly)

Some rational points arbitrarily close to sqrt(2) are points where we truncate the decimal expansion of sqrt(2) arbitrarily far. I.e.

$\lfloor 10^{n}\sqrt{2}\rfloor 10^{-n}$

for positive integers n.

leehuan · Mar 11, 2017

Re: Several Variable Calculus

InteGrand said:
Some rational points arbitrarily close to sqrt(2) are points where we truncate the decimal expansion of sqrt(2) arbitrarily far. I.e.

$\lfloor 10^{n}\sqrt{2}\rfloor 10^{-n}$

for positive integers n.

That is one beautiful trick.
_____________________________________

$\text{RTP: }\Omega= \{(x,y)\in \mathbb{R}^2: \, x^2-y^2<1 \}\text{ is open.}$

I definitely don't want to use the Lagrange multiplier theorem to find the closest point (x,y) to work off, so any pointers on what radius I should choose my ball to be? (to prove an arbitrary point (x,y) is interior)

InteGrand · Mar 11, 2017

Re: Several Variable Calculus

leehuan said:
That is one beautiful trick.
_____________________________________

$\text{RTP: }\Omega= \{(x,y)\in \mathbb{R}^2: \, x^2-y^2<1 \}\text{ is open.}$

I definitely don't want to use the Lagrange multiplier theorem to find the closest point (x,y) to work off, so any pointers on what radius I should choose my ball to be? (to prove an arbitrary point (x,y) is interior)

That set Omega is an open ball already (using the Euclidean metric), and an open ball is an open set (you should prove this as an exercise if you haven't before).

MATH2111 · Mar 11, 2017

Re: Several Variable Calculus

Mate, stop cheating on these assignment questions. Do it yourself ffs.

leehuan · Mar 11, 2017

Re: Several Variable Calculus

MATH2111 said:
Mate, stop cheating on these assignment questions. Do it yourself ffs.

...?

These are homework questions I'm stuck on. I'd be too worried about getting a 0 mark for the course if these were assignment questions.

leehuan · Mar 11, 2017

Re: Several Variable Calculus

InteGrand said:
That set Omega is an open ball already (using the Euclidean metric), and an open ball is an open set (you should prove this as an exercise if you haven't before).

Wait, I can't see it - Can't tell how a region cut off by the rectangular hyperbola is a ball

InteGrand · Mar 11, 2017

Re: Several Variable Calculus

leehuan said:
Wait, I can't see it - Can't tell how a region cut off by the rectangular hyperbola is a ball

Sorry misread it (saw + instead of -).

One way is to recall that the preimage of an open set under a continuous map is open.

leehuan · Mar 15, 2017

Re: Several Variable Calculus

$\text{Suppose that }f:\mathbb{R}^n\to \mathbb{R}\text{ is such that the sets}\\ \{\textbf{x}\in \mathbb{R}^n:f(x)>d \}\\ \{\textbf{x}\in \mathbb{R}^n:f(x)<d \}\\ \text{are open for every }d\in \mathbb{R}$

$\text{Prove that }f\text{ is continuous}$

InteGrand · Mar 15, 2017

Re: Several Variable Calculus

leehuan said:
$\text{Suppose that }f:\mathbb{R}^n\to \mathbb{R}\text{ is such that the sets}\\ \{\textbf{x}\in \mathbb{R}^n:f(x)>d \}\\ \{\textbf{x}\in \mathbb{R}^n:f(x)<d \}\\ \text{are open for every }d\in \mathbb{R}$

$\text{Prove that }f\text{ is continuous}$

$\noindent Let $\mathbf{a} \in \mathbb{R}^{n}$ be fixed and arbitrary. By definition of continuity, RTP:$

$for all $\varepsilon > 0$ there exists an open ball $B\left(\mathbf{a},\delta\right)$ (i.e. some radius $\delta > 0$) such that $\underbrace{|f\left(\mathbf{x}\right) - f\left(\mathbf{a}\right)| < \varepsilon}_{\iff f\left(\mathbf{a}\right) - \varepsilon < f\left(\mathbf{x}\right) < f\left(\mathbf{a}\right) + \varepsilon}$ whenever $\mathbf{x} \in B\left(\mathbf{a},\delta\right)$.$

$\noindent So fix $\varepsilon > 0$. Let $S_{1} = \left\{ \mathbf{x}\in \mathbb{R}^{n} : f\left(\mathbf{x}\right) < f\left(\mathbf{a}\right) + \varepsilon \right\}$ and $S_{2} = \left\{ \mathbf{x}\in \mathbb{R}^{n} : f\left(\mathbf{x}\right) > f\left(\mathbf{a}\right) - \varepsilon \right\}$ (note that a vector $\mathbf{x}$ is in both $S_{1}$ and $S_{2}$, i.e. in $S_{1}\cap S_{2}$, iff $\left|f\left(\mathbf{x}\right) - f\left(\mathbf{a}\right)\right| < \varepsilon$). Observe that $\mathbf{a}$ belongs to both $S_{1}$ and $S_{2}$ (because $\varepsilon > 0$). So for $i = 1,2$, $\mathbf{a} \in S_{i}$ and $S_{i}$ being open (by assumptions of the question) implies that there exists a $\delta_{i}$ such that$

$$\mathbf{x}\in S_{i}$ whenever $x \in B\left(\mathbf{a},\delta_{i}\right)$.$

$\noindent Set $\delta = \min \left\{\delta_{1}, \delta_{2}\right\}$, then we get that $\mathbf{x}$ belongs to both $S_{1}$ and $S_{2}$ (i.e. to $S_{1}\cap S_{2}$) whenever $\mathbf{x} \in B\left(\mathbf{a}, \delta\right)$. This is exactly what we had to show, and the proof is complete.$

leehuan · Mar 19, 2017

Re: Several Variable Calculus

Just a brief sketch-out please

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

$\text{Why is it that }\frac{\partial^2 f}{\partial x \partial y}(0,0)\text{ DNE?}$

I know that f is not continuous at 0 but I'm not sure if that helps since we're talking about partial derivatives here.

MATH2111 Higher Several Variable Calculus (5 Viewers)

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