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Messages - Ian Kivlichan

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Home Assignment 6 / Re: Problem 4
« on: November 07, 2012, 04:32:43 PM »
Is there a way to go from $\int_{-\infty}^{\infty}{\frac{\sin^2(x)}{x^2}dx}$ to $\int_{-\infty}^{\infty}{\frac{\sin(x)}{x}dx}$?

APM346 Misc / Re: Should be the night class offered at Fall or at Spring?
« on: November 03, 2012, 04:51:21 AM »
As is, students can only take this course in Fall semester -- for that reason alone it would be better to offer a section in Spring semester.

But yes, constant holidays on Mondays in Fall mean that Monday night lectures shouldn't really be offered... the Faculty Registrar probably doesn't care though...

APM346 Misc / Re: Are the grades going to be bellcurved?
« on: November 03, 2012, 04:44:13 AM »
So, the average is pretty low. The question is: Are the grades going to be bellcurved?

Levon: Bell curving is absolutely forbidden by the ArtSci Faculty (
Calibration of test scores should be done fairly and equitably, and bear a justifiable relation to academic performance. Policy explicitly forbids manipulating marks to fit into a “normal curve” or any other prior expectation – in the language of the Policy: “academic assessment must not be predetermined by any system of quotas that specifies the number or percentage of grades allowable at any grade level. ”

As far as I know, linear adjustments are most commonly used to change course averages (via new_mark = old_mark * multiplier + shift), but this depends entirely on whether Prof. Ivrii thinks the marks are fair.

Usually, course coordinators are pressured to maintain a C or C+ average (the lowest course average on my transcript is C, and the lowest I've heard of is a C-), so since the TT1 average is 57.5% (D+) it's likely that the final average will be higher.

Home Assignment 5 / Re: Problem 4
« on: October 31, 2012, 09:53:35 PM »
Hopeful solution for 4.b) attached!

Home Assignment 5 / Re: Problem 4
« on: October 31, 2012, 09:44:27 PM »
Additional solution for 4.a) (essentially the same as Aida's post here , but showing more of the sketch, as well as with details on the odd continuation used for sin Fourier series).

Home Assignment 5 / Re: Problem 6
« on: October 31, 2012, 09:38:35 PM »
Hopeful solution to 6.e) attached!

Home Assignment 5 / Re: Problem 5
« on: October 31, 2012, 09:32:43 PM »
Hopeful solution for 5.a)!

Home Assignment 5 / Problem 4
« on: October 31, 2012, 09:32:02 PM »
Hopeful solutions for 4.c)! :)

edit: Note that sketch is for m=1.

Home Assignment 5 / Re: Problem 1
« on: October 31, 2012, 02:27:03 AM »
Jinchao: you can integrate (e^x)sin(x) by parts. Set u = e^x, dv = sinx dx, and go through. You'll have to integrate by parts a second time, but you'll end up with (e^x)sinx integrals on both sides. Hope that helps! :)

APM346 Misc / Dropping two worst assignments
« on: October 24, 2012, 03:40:08 PM »
Hi all,

I noticed on that it says

  • Two worst assignments (included those you did not submit and got 0) will be dropped;
  • So, if there will be 10 assignments, only 8 will be counted and each would contribute 2.5 to the Term Mark;

To be clear, does this mean that no matter how many assignments there are, 2 will be dropped? (Ex. if we had 8 assignments, only 6 would be counted, and each would contribute 3.3% to the Term Mark.)

Misc Math / Lecture 12 Eqn 9 Question
« on: October 24, 2012, 01:59:26 AM »
Hi all,

Just to clarify - should equation 9 in the notes for lecture 12 ( read

$\lambda_n = - n^2 \pi^2 / l^2$ ?



Term Test 1 / Re: TT1 = Problem 3
« on: October 17, 2012, 01:11:39 AM »
PS. Ian, your posts are virtually useless for a class: too poor handwriting makes it almost impossible to read for anyone who does not know solution. Could you repost?

Sorry!!  :(

I have tried my best to re-write it nicely (edited original post).

Term Test 1 / Re: TT1 = Problem 2
« on: October 16, 2012, 09:53:02 PM »
Ian, while explanation is basically correct I would like to see more convincing arguments. In particular: where solution will be defined uniquely?
With the given conditions, the I think solution is defined for -inf < x < -t, . The given conditions on u and u_t restrict it there, as any wave starting early in time would have to pass through (x,t)=(x,0). 0 < x < -t, however, does not have the required conditions for uniqueness.

Term Test 1 / Re: TT1 = Problem 1
« on: October 16, 2012, 09:24:06 PM »
Subqueston (d):

$ \frac{dt}{1} = \frac{dx}{x^2} $
$ t = -x^{-1}+c $
so the general solution is $ u(t,x)=f(t+x^{-1})$.
Since $y=x^{-1}$, so when $x>0$ we have $y>0$ as well.
We need $t+x^{-1}>0$
Since $x>0$,
therefore $tx+1>0$
so the domain be defined is $\{(t,x) | tx>-1 \}.
I think Jinchao has the most correct solution.

Qitan, is it possible to only have the one discontinuity in your solution - won't your characteristic curves be "blocked" by the discontinuity at tx=-1, and not able to go any further?

Term Test 1 / Re: TT1 = Problem 5
« on: October 16, 2012, 08:25:25 PM »
Solution is attached,

Aida, I think your solution is not correct - the integral from -1 to 0 will only be 0 if x < y in H(x-y), i.e. x < -1.

Also, consider extreme cases - for C(-100), H(-100-y) can never be greater than 0, so a solution with C(x) = 1 cannot be right (unless I've totally misunderstood something).

I believe your solution is not correct; H(x) and H(x-y) has the same value for these two different domain because it is a constant function.

Shouldn't they be different though? For H(x), it's 1 for x>0 and 0 for x<=0, but H(x-y) is 1 for x-y>0, or x>y, and 0 for x<=y.

In any case, how can C(-100) be nonzero? x-y=-100-y>0 is impossible for -1<=y<=1, which is the only area where Q(y) is nonzero.

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