MATH2111 Higher Several Variable Calculus (5 Viewers)

InteGrand · May 26, 2017

Re: Several Variable Calculus

$\noindent Let $\alpha < 1$ be fixed, and $f_{n}:[0,1]\to \mathbb{R}$ is given by $f_{n}(x) = n^{\alpha}x^{n}(1-x)$. That this converges to the zero function on $[0,1]$ uniformly means that for each $\varepsilon > 0$, there exists $N$ such that \textbf{for all} $x \in [0,1]$, we have $|f_{n}(x) - 0| < \varepsilon$ if $n > N$. So this is what you'd be aiming to show.$

$\noindent The key difference with pointwise convergence is that with uniform convergence, the $N$ \textbf{can't depend on} $x$. I.e. the same $N$ must work for all $x$, given an $\varepsilon$. (So the $N$ still can depend on $\varepsilon$, but can't depend on $x$.) For pointwise convergence, it is OK to have a different $N$ for different $x$ (i.e. the $N$ is allowed to depend on $x$.)$

leehuan · May 26, 2017

Re: Several Variable Calculus

To be honest, all I could really write out was the definition you provided. I realised I had to work with

$n> N \implies n^\alpha x^n(1-x) < \epsilon$

I'm not sure what should delete to deduce my N.

InteGrand · May 26, 2017

Re: Several Variable Calculus

leehuan said:
To be honest, all I could really write out was the definition you provided. I realised I had to work with

$n> N \implies n^\alpha x^n(1-x) < \epsilon$

I'm not sure what should delete to deduce my N.

$\noindent One way you can do it is to show via calculus that $n^{\alpha}x^{n} (1-x)$ has maximum value when $x = \frac{n}{n+1}$ for $x\in [0,1]$, and thus find this max. value, and show it goes to $0$ as $n\to \infty$.$

leehuan · May 26, 2017

Re: Several Variable Calculus

$f:\mathbb{R}\to \mathbb{R},\, f(x) = 1-6x^2+4x^3\text{ for }0\le x \le 1$

$f(x+2)=f(x)\text{ and }f(-x)=f(x) \, \forall x\in \mathbb{R}$

Required: The Fourier series Sf(x).

Hints: I found that f'(0) = f'(1) = 0 and f'''(x) = 24

$\\\text{Deduced: }a_0[f] = -2, \, a_k[f]=\frac{96}{k^4\pi^4}\\ b_[f]=0\\ \text{for }k \ge 1$

$\text{So in the formula }Sf(x) = \frac{a_0[f]}{2}+ \sum_{k=1}^n \left(a_k [f]\cos \left(k\frac{\pi x}L\right)+b_k[f]\sin \left(k\frac{\pi x}L\right)\right)\\ \text{I arrive at }Sf(x) = -1+\frac{96}{\pi^4}\sum_{k=1}^\infty \frac{\cos k\pi x}{k^4}$

$\text{Given answer: }Sf(x) = \frac{\pi^4}{96}\left(\cos \pi x + \frac{\cos 3\pi x}{3^4} + \frac{\cos 5\pi x}{5^4}+\dots\right)$

Have I made a mistake? If not then how do I reunite the solutions (by somehow binning -1, cos(2pi x), cos(4pi x), ...?)

InteGrand · May 26, 2017

Re: Several Variable Calculus

leehuan said:
$f:\mathbb{R}\to \mathbb{R},\, f(x) = 1-6x^2+4x^3\text{ for }0\le x \le 1$

$f(x+2)=f(x)\text{ and }f(-x)=f(x) \, \forall x\in \mathbb{R}$

Required: The Fourier series Sf(x).

Hints: I found that f'(0) = f'(1) = 0 and f'''(x) = 24

$\\\text{Deduced: }a_0[f] = -2, \, a_k[f]=\frac{96}{k^4\pi^4}\\ b_[f]=0\\ \text{for }k \ge 1$

$\text{So in the formula }Sf(x) = \frac{a_0[f]}{2}+ \sum_{k=1}^n \left(a_k [f]\cos \left(k\frac{\pi x}L\right)+b_k[f]\sin \left(k\frac{\pi x}L\right)\right)\\ \text{I arrive at }Sf(x) = -1+\frac{96}{\pi^4}\sum_{k=1}^\infty \frac{\cos k\pi x}{k^4}$

$\text{Given answer: }Sf(x) = \frac{\pi^4}{96}\left(\cos \pi x + \frac{\cos 3\pi x}{3^4} + \frac{\cos 5\pi x}{5^4}+\dots\right)$

Have I made a mistake? If not then how do I reunite the solutions (by somehow binning -1, cos(2pi x), cos(4pi x), ...?)

$$\noindent One quick test is to just check the constant term first (since your answer and the given answer have different constant terms; your one is $-1$ and the given answer's is 0).$

$\noindent The constant term in a Fourier series is the \emph{average value} of the function over a period. In other words, it is$

$\frac{1}{2L}\int_{-L}^{L} f(x)\, \mathrm{d}x,$

$\noindent where $L$ is the \textsl{half-period} of the function. In our case $L = 1$. Since the function in our case is even, this becomes$

$\begin{align*}\frac{1}{2L}\times 2 \int_{0}^{L}f(x) \, \mathrm{d} x &= \int_{0}^{1}\left(1 - 6x^{2} + 4x^{3}\right)\, \mathrm{d}x \quad (\text{as } L = 1 \text{ and }f(x) = 1-6x^2+4x^3\text{ for }0\le x \le 1) \\ &= 1 - \frac{6}{3} + \frac{4}{4} \\ &= 1 - 2 + 1 = 0.\end{align*}$

$\noindent So the constant term is $0$, so the given answer can be correct, but the one you got cannot be.$

leehuan · May 26, 2017

Re: Several Variable Calculus

InteGrand said:
$$\noindent One quick test is to just check the constant term first (since your answer and the given answer have different constant terms; your one is $-1$ and the given answer's is 0).$

$\noindent The constant term in a Fourier series is the \emph{average value} of the function over a period. In other words, it is$

$\frac{1}{2L}\int_{-L}^{L} f(x)\, \mathrm{d}x,$

$\noindent where $L$ is the \textsl{half-period} of the function. In our case $L = 1$. Since the function in our case is even, this becomes$

$\begin{align*}\frac{1}{2L}\times 2 \int_{0}^{L}f(x) \, \mathrm{d} x &= \int_{0}^{1}\left(1 - 6x^{2} + 4x^{3}\right)\, \mathrm{d}x \quad (\text{as } L = 1 \text{ and }f(x) = 1-6x^2+4x^3\text{ for }0\le x \le 1) \\ &= 1 - \frac{6}{3} + \frac{4}{4} \\ &= 1 - 2 + 1 = 0.\end{align*}$

$\noindent So the constant term is $0$, so the given answer can be correct, but the one you got cannot be.$

Hmm, not sure if I'm missing something but I think there's an x-cubed there. I don't think the original function is even?

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=integrate+1/2(1-6x^2+4x^3)+from+-1+to+1

Although when I split up the x^3 and the 1-6x^2 the latter is even.

InteGrand · May 26, 2017

Re: Several Variable Calculus

leehuan said:
Hmm, not sure if I'm missing something but I think there's an x-cubed there. I don't think the original function is even?

http://www.wolframalpha.com/input/?t=crmtb01&f=ob&i=integrate+1/2(1-6x^2+4x^3)+from+-1+to+1

Although when I split up the x^3 and the 1-6x^2 the latter is even.

The function has been defined on [0, 1] by that formula, and then we are told f(-x) = f(x) for all x and f(x) = f(x+2) for all x, so this means the function is an even periodic function with period 2, and on [0, 1] it looks like 1 - 6x^2 + 4x^3.

In other words, its graph is as follows: sketch 1 - 6x^2 + 4x^3 on [0, 1], then reflect this section about the y-axis so it's even on [-1, 1], and finally imagine "stamping" this picture on [-1, 1] across the whole graph with spacing 2, so it becomes a repeating picture with period 2.

InteGrand · May 26, 2017

Re: Several Variable Calculus

In other words, take this graph ( http://www.wolframalpha.com/input/?i=graph+(1-6|x|^2+4|x|^3)+from+-1+to+1 ) and imagine repeating that same picture on [1, 3], and [3, 5] and [5, 7] and ..., and also on the negative side (so on [-3, -1], [-5, -3], ...). So it becomes a periodic even function with period 2, and looks like a sort of wave, and each cycle of the graph looks like the graph in the link.

leehuan · May 26, 2017

Re: Several Variable Calculus

Oh my bad, I see now. That was foolish of me.

leehuan · May 27, 2017

Re: Several Variable Calculus

$Sf(x) = \frac12 + \frac2\pi \sum_{k=1}^\infty \frac{\sin [(2k-1)x]}{2k-1}$

How would I show that this series does not converge uniformly? I don't think I can use the Weierstrass M-test for divergence

$\text{Or if I use the definition, I really just don't know how to negate it}\\ \forall \epsilon > 0 \quad \exists N > 0:\quad |S_n(x) - S(x)| < \epsilon\, \,\quad \forall n\ge N, x\in \mathbb{R}$

Exists eps > 0, such that for all N > 0, |S_n(x) - S(x)| < eps then what?

InteGrand · May 27, 2017

Re: Several Variable Calculus

leehuan said:
$Sf(x) = \frac12 + \frac2\pi \sum_{k=1}^\infty \frac{\sin [(2k-1)x]}{2k-1}$

How would I show that this series does not converge uniformly? I don't think I can use the Weierstrass M-test for divergence

$\text{Or if I use the definition, I really just don't know how to negate it}\\ \forall \epsilon > 0 \quad \exists N > 0:\quad |S_n(x) - S(x)| < \epsilon\, \,\quad \forall n\ge N, x\in \mathbb{R}$

Exists eps > 0, such that for all N > 0, |S_n(x) - S(x)| < eps then what?

Do you know where this series comes from? If so, you can use that to help prove it.

leehuan · May 27, 2017

Re: Several Variable Calculus

InteGrand said:
Do you know where this series comes from? If so, you can use that to help prove it.

Yep

$f(x) = \begin{cases}0& -\pi \le x \le 0\\ 1& 0 < x < \pi\end{cases}$

If we ensure that f(x + 2pi) = f(x) I can visually see why it fails to work, but I still can't figure out where to get started

InteGrand · May 27, 2017

Re: Several Variable Calculus

leehuan said:
Yep

$f(x) = \begin{cases}0& -\pi \le x \le 0\\ 1& 0 < x < \pi\end{cases}$

If we ensure that f(x + 2pi) = f(x) I can visually see why it fails to work, but I still can't figure out where to get started

Correct! This is the Fourier series of a discontinuous function (a square wave) on [-pi, pi) say. What do you know about Fourier series of functions with jump discontinuities? (If you haven't seen it before, you may want to read about this: https://en.wikipedia.org/wiki/Gibbs_phenomenon .)

Also recall that if a sequence of continuous functions (like partial sums of a Fourier series) converges to a discontinuous function, the convergence cannot be uniform.

leehuan · May 27, 2017

Re: Several Variable Calculus

Ohhhhhhh I see now, completely forgot about that!

leehuan · May 27, 2017

Re: Several Variable Calculus

I don't want this question done just yet, I just want to know why the hint works

$\text{Show that for any planar region }\Omega\\ \text{area}(\Omega) = \frac{1}{2}\oint_{\partial \Omega}(x\, dy - y \, dx)$

$\text{Hint: Apply Green's Theorem with }\boxed{P=-y,\quad Q=x}$

I only partially get it. Why does this choice of P and Q ensure that we cover all types of planar regions?

$\text{Green's theorem as we were taught (reference): }\\ \iint_{\Omega}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy = \oint_{\partial \Omega}(P\, dx + Q\, dy)\\ \text{(subject to relevant conditions)}$

seanieg89 · May 28, 2017

Re: Several Variable Calculus

leehuan said:
I don't want this question done just yet, I just want to know why the hint works

$\text{Show that for any planar region }\Omega\\ \text{area}(\Omega) = \frac{1}{2}\oint_{\partial \Omega}(x\, dy - y \, dx)$

$\text{Hint: Apply Green's Theorem with }\boxed{P=-y,\quad Q=x}$

I only partially get it. Why does this choice of P and Q ensure that we cover all types of planar regions?

$\text{Green's theorem as we were taught (reference): }\\ \iint_{\Omega}\left(\frac{\partial Q}{\partial x}-\frac{\partial P}{\partial y}\right)dx\,dy = \oint_{\partial \Omega}(P\, dx + Q\, dy)\\ \text{(subject to relevant conditions)}$

Are you sure you are writing the question correctly? Is there an established convention in your book/notes for what "planar region" means?

My guess is that it means things like the regions bounded between two lines x=a,b and g(x) < y < h(x).

Otherwise your notion of planar region has to be nice enough for "area" to make sense, so more information is required about this niceness.

leehuan · May 28, 2017

Re: Several Variable Calculus

seanieg89 said:
Are you sure you are writing the question correctly? Is there an established convention in your book/notes for what "planar region" means?

My guess is that it means things like the regions bounded between two lines x=a,b and g(x) < y < h(x).

Otherwise your notion of planar region has to be nice enough for "area" to make sense, so more information is required about this niceness.

Couldn't find anything else though
___________

Not sure at all if this helps:

$\text{Q31: Use that result to find the area of the planar region enclosed by the astroid}\\ x^{\frac23}+y^{\frac23}=a^{\frac23}\text{ for }a>0$

seanieg89 · May 28, 2017

Re: Several Variable Calculus

Well, what kind of regions was Green's theorem stated/proved for? For any domain A that is nice enough that Greens theorem works, you will get that the RHS boundary integral is equal to the integral of the indicator function 1_A of A. This can be taken as the definition of area for sets A such that 1_A is integrable.

(I presume you are working with the Riemann/Darboux integeral at this stage, so such sets A are said to be Jordan measurable.)

If the set isn't nice enough though it is impossible to assign a meaning to either the LHS or RHS, so the result as stated will only apply to sufficiently nice planar regions such that these things work out.

leehuan · May 28, 2017

Re: Several Variable Calculus

For our version, P and Q (as used above) have to be C¹ on $\Omega \cup \partial \Omega$

That was really the only condition given.

seanieg89 · May 28, 2017

Re: Several Variable Calculus

leehuan said:
For our version, P and Q (as used above) have to be C¹ on $\Omega \cup \partial \Omega$

That was really the only condition given.

What book/notes are these? Anyway, yeah as I said, you need some kind of niceness condition on the domain for these things to be well-defined. You should be able to see what this condition is by going through the proof of Green's theorem, unless it is presented pretty sloppily/nonrigorously.

I don't know what you mean by "why this choice of P and Q covers all planar domains". P and Q have nothing to do with the domain, they are just C^1 functions. The class of domains for which the stated result holds is just the class of domains for which the version of Green's theorem you are using is valid and for which area makes sense.

MATH2111 Higher Several Variable Calculus (5 Viewers)

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