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### Messages - Aida Razi

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16
##### Home Assignment 6 / Re: Problem 2
« on: November 07, 2012, 09:30:35 PM »
Solution is attached!

17
##### Home Assignment 6 / Re: Problem 1--simplified
« on: November 07, 2012, 09:30:00 PM »
Solution is attached!

18
##### Misc Math / Fourier Transform
« on: November 05, 2012, 01:00:32 PM »
I was wondering (1/2 pi) in Fourier transform formula is an optional? (I mean, can we just ignore it?)

19
##### Home Assignment 5 / Re: Problem 2
« on: November 05, 2012, 01:53:33 AM »
Part (b) solution is attached!

20
##### Home Assignment 5 / Re: Problem 2
« on: November 04, 2012, 02:39:59 PM »
Part (a) solution is attached!

21
##### Home Assignment 5 / Re: Problem4
« on: November 01, 2012, 03:25:35 AM »
WTH you are bringing me from textbook the plots of the partial sums of the FS on the given interval rather than the complete sum on $(-\infty,\infty)$ which is due to continuation?

Professor Ivrii,

The interval is [0,Ï€]!

But F.s. converges everywhere!

Yes, I got it.
I am sorry,

22
##### Home Assignment 5 / Re: Problem4
« on: November 01, 2012, 02:09:44 AM »
WTH you are bringing me from textbook the plots of the partial sums of the FS on the given interval rather than the complete sum on $(-\infty,\infty)$ which is due to continuation?

Professor Ivrii,

The interval is [0,Ï€]!

23
##### Home Assignment 5 / Problem4
« on: October 31, 2012, 09:31:34 PM »
Part (a) solution is attached!

24
##### Home Assignment 5 / Re: Problem 2
« on: October 31, 2012, 09:30:42 PM »
Part (c) solution is attached!

25
##### Home Assignment 5 / Re: Problem 1
« on: October 31, 2012, 09:30:03 PM »
Part (a) solution is attached!

26
##### Home Assignment 4 / Re: Problem 1
« on: October 29, 2012, 09:46:48 PM »
Part (b) proof:

27
##### Home Assignment 4 / Re: Problem 1
« on: October 29, 2012, 06:57:50 PM »
Part (a) proof:

28
##### Term Test 1 / Re: TT1 = Problem 5
« on: October 16, 2012, 10:08:35 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.

Sorry Ian, You are right

29
##### Term Test 1 / Re: TT1 = Problem 1
« on: October 16, 2012, 08:27:12 PM »
My solution. Please check there might be mistakes. Sorry for quality of print and hand writing.

Djirar, I posted my solution 10 seconds after you

30
##### Term Test 1 / Re: TT1 = Problem 1
« on: October 16, 2012, 08:25:40 PM »