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Messages - Victor Ivrii

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APM346 Announcements / Re: Term Test 2
« on: November 09, 2012, 11:02:07 AM »
OK, so how it will be organized:

I will be in my office from 2 pm - HU1008, so anyone can come here from 2:00 to 3:30 (there is room for 3 students only, but only 3 wanted it earlier than  4:00, right)

From 4:00 to 4:30 come to HU1018. Don't be late!

At  8:10 come to HU1018 (I will come at  8:10--8:20 from Galbraith)--so it will be 8:20--10:20

Home Assignment 6 / Re: Problem 4
« on: November 08, 2012, 04:32:56 PM »
Fanxun, I decided that you deserve a karma, for raising a good question. In Lecture 15 we proved Theorem 2:
Theorem 2. Let $f$ be a piecewise continuously differentiable function. Then the Fourier series
\frac{a_0}{2}+\sum_{n=1}^\infty \Bigl(a_n\cos \bigl(\frac{\pi n x}{l}\bigr) + a_n\cos \bigl(\frac{\pi n x}{l}\bigr)\Bigr)
converges to

(b) $\frac{1}{2}\bigl(f(x+0)+f(x-0)\bigr)$  if $x$ is internal point and $f$ is discontinuous at $x$.

Exactly the same statement holds for Integral Fourier
\int_0^\infty \Bigl(A(\omega) \cos (\omega x) + B(\omega)\sin (\omega x)\Bigr)\,d\omega
where $A(\omega)$ and $B(\omega)$ are $\cos$- and $\sin$-Fourier transforms.

None of them however holds for Fourier series or Fourier Integral in the complex form:
\sum_{n=-\infty}^\infty c\_n e^{i\frac{\pi n x}{l}},\label{eq-3}\\
\int_{-\infty}^\infty C(\omega)e^{i\omega x}\,d\omega.\label{eq-4}

Why and what remedy do we have? If we consider definition of the partial sum of (\ref{eq-1}) and then rewrite in the complex form  and similar deal with (\ref{eq-4}) we see that in fact we should look at
\lim_{N\to \infty} \sum_{n=-N}^N c_n e^{i\frac{\pi n x}{l}},\label{eq-5}\\
\lim_{N\to \infty} \int_{-N}^N C(\omega)e^{i\omega x}\,d\omega\label{eq-6}.

Meanwhile convergence in (\ref{eq-3}) and (\ref{eq-4}) means more than this:
\lim_{M,N\to \infty} \sum_{n=-M}^N c_n e^{i\frac{\pi n x}{l}},\label{eq-7}\\
\lim_{M,N\to \infty} \int_{-M}^N C(\omega)e^{i\omega x}\,d\omega\label{eq-8}
where $M,N$ tend to $\infty$ independently. So the remedy is simple: understand convergence as in (\ref{eq-5}), (\ref{eq-6}) rather than as in (\ref{eq-7}), (\ref{eq-8}) . For integrals such limit is called essential value of integral and is denoted by
\operatorname{ess}-\int_{-\infty}^\infty G(\omega)\,d\omega
\operatorname{vrai}-\int_{-\infty}^\infty G(\omega)\,d\omega

BTW similarly is defined \begin{equation*}
\operatorname{ess}-\int_{a}^b G(\omega)\,d\omega:= \lim_{\epsilon\to +0} \Bigl(\int_a^{c-\epsilon}G(\omega)\,d\omega+
if there is a singularity at $c\in (a,b)$.

This is more general than improper integrals studied in the end of Calculus I. Those who took Complex Variables encountered such notion.

Home Assignment 6 / Re: Bonus Web Problem--4
« on: November 08, 2012, 02:35:16 AM »
Nice! $\renewcommand{\Re}{\operatorname{Re}}$

BTW one can consider for $\lambda \in \mathbb{C}\setminus \{-1,-2,-3,\ldots\}$  function $f_\lambda(x):= \frac{1 }{\Gamma (\lambda+1)}x_+^\lambda$ where $\Gamma$ is Euler's gamma-function (both $\Gamma(\lambda+1)$ and $f_\lambda(x)$ are first defined for $\Re \lambda >-1$ and then analytically continued to $\mathbb{C}$ with the poles ($f_\lambda$ is here a distribution rather than an ordinary function) and one can consider "fractional integration" $I_\lambda u:=f_{\lambda-1} * u$.

However $I_\lambda$ as analytic continuation makes sense even for $\lambda=-n-1$: $f_{-1} = \delta$, $f_{-2}=\delta'$ etc $f_{-n-1}=\delta^{(n)}$.

Home Assignment 6 / Re: Problem 4
« on: November 08, 2012, 02:17:21 AM »
No votes. This course is really not very rigorous but APM351 is much less rigorous than any comparable math course and the reason is simple: the really rigorous and in the same time meaningful PDE course requires serious Real Analysis (Complex Analysis does not hurt either) which Math Specialist students have in the 4th year (so Graduate PDE class is rigorous). You attempts to bring more rigour here are well intended but not productive (sure, we should maintain some level of rigour but it is not very high).

Home Assignment 6 / Re: Problem 1--simplified
« on: November 08, 2012, 02:10:15 AM »
Calvin, first of all not "edited" but "converted" to png. Now why your png is rather invisible? The answer is simple: you generated pdf via TeX and therefore it has not a white but a transparent background.

Actually all graphics I generated via LaTeX with pgf package but I always put \pagestyle{empty}, then apply pdfcrop to the generated pdf,  and then convert by ImageMagick to png which inserts white background.

Home Assignment 6 / Re: Problem 4
« on: November 07, 2012, 10:14:17 PM »
Fanxun, thank you very much for the lecture what mathematics is :D  if you wanted more rigour you should take APM351Y.

I am not sure that everyone appreciates your suggestion about mark reduction and I am looking forward for a little flame war and myself issuing warnings and bans (so, my suggestion to everyone--don't start).

Home Assignment 6 / Re: Problem 2
« on: November 07, 2012, 10:07:01 PM »
Ziting, your jpgs are borked. Anyway, we have enough solutions.

Calvin, if you type your solutions in LaTeX or TeX, you could post just the source fixing some discrepancies on the fly

Home Assignment 6 / Re: Problem3
« on: November 07, 2012, 09:59:48 PM »
Again, simpler to use (as Calvin did) F.T. properties rather than calculate integral (especially because one really needs to prove that one can replace $x$ by $x+i \omega$ (thus going into complex plane)

Home Assignment 6 / Re: Problem 2
« on: November 07, 2012, 09:53:03 PM »
This integral can be computed using residues, but I think the whole point of the Fourier tranform is lost if one uses that approach...

Yes, residues work miracles here (see Calvin) but most have not taken Complex Analysis and there is a property for that :)

Calvin, it is $\sgn$ not $sgn$

See source

Home Assignment 6 / Re: Bonus Web Problem--4
« on: November 07, 2012, 09:51:29 PM »
Calvin: I would like to see the last property

Home Assignment 6 / Re: Problem 4
« on: November 07, 2012, 09:50:09 PM »
$k=0$ is trivial (and could be obtained by a limit transition).

Home Assignment 6 / Re: Problem 1--simplified
« on: November 07, 2012, 09:46:05 PM »
Calvin: 1) errors with the sign in denominator 2) Please don't upload pdfs as they are not displayed by Forum (so one needs to download them) 3) TeX advice: too long lines (not nice to read)

Everybody: I did not ask you to take all integrals. Instead after you found F.T. of $e^{-\alpha |x|}$ (say $G(\omega)$) you recalling properties of F.T understand that F.t. of $e^{-\alpha |x|}e^{\pm i\beta x}$ are  $G(\omega \mp \beta)$ and then F.T. of  $e^{-\alpha |x|}\cos(\beta x)$ and $e^{-\alpha |x|}\sin(\beta x)$ are $\frac{1}{2}[ G(\omega - \beta)+G(\omega + \beta)]$ and $\frac{1}{2i}[ G(\omega - \beta)-G(\omega + \beta)]$ respectively

Home Assignment 6 / Re: Problem 4
« on: November 07, 2012, 04:45:49 PM »
Is there a way to go from $\int_{-\infty}^{\infty}{\frac{\sin^2(x)}{x^2}dx}$ to $\int_{-\infty}^{\infty}{\frac{\sin(x)}{x}dx}$?

$$\int_0^{\infty} \frac{\sin^2(x)}{x^2}dx = -\int_0^{\infty} \sin^2(x) dx^{-1} $$
and integrate by parts.

This is not part of HW

APM346 Announcements / Re: TT1 graded
« on: November 07, 2012, 09:59:05 AM »
Who graded:

Problems 1, 5: Martin
Problems 3, 4: Daniel
Problems 2: Yannis

If you disagree with your mark speak first with TA who graded this problem.

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