Author Topic: Problem 4  (Read 5880 times)

Calvin Arnott

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Problem 4
« on: October 13, 2012, 10:02:10 PM »
In problem 4, we have the functions:

[ ÆŸ(x) = 0  \{x < 0\}, \space ÆŸ(x) = 1  \{x > 0\}]            [ f(x) = 1   \{|x| > 1\},    f(x) = 0   \{|x| < 1\}]

and an idea to use the functional relation [f(x) = ÆŸ(x+1) - ÆŸ(x-1)]

But this identity does not hold. For instance, [f(0) = ÆŸ(1) - ÆŸ(-1) = 1 - 0 = 1 \ne 0]

An identity which I found to work instead for the integral is: [f(x) = ÆŸ(x - 1) + ÆŸ(-x - 1) ]

Have I made an error?
« Last Edit: October 13, 2012, 10:14:47 PM by Calvin Arnott »

Victor Ivrii

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Re: Problem 4
« Reply #1 on: October 13, 2012, 10:17:19 PM »
No, you are right. I meant $f(x)=1$ as $|x|<1$ and $f(x)=0$ as $|x|>1$ but typed opposite. Basically your $f$ complements my intended $f$ to $1$

Aida Razi

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Re: Problem 4
« Reply #2 on: October 15, 2012, 01:58:43 AM »
Solution of part a is attached,

Aida Razi

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Re: Problem 4
« Reply #3 on: October 15, 2012, 02:00:33 AM »
Solution of part a is attached,

Where k =1 on the solution.

Aida Razi

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Re: Problem 4
« Reply #4 on: October 15, 2012, 02:13:58 AM »
Solution of part b is attached,

Peishan Wang

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Re: Problem 4
« Reply #5 on: October 15, 2012, 06:24:49 AM »
I'm not sure if I did it wrong but I got a different answer to part(a): 1/2 + 1/2*erf(x/√2t), where √2t stands for "square root of 2t".
« Last Edit: October 16, 2012, 11:40:30 AM by Peishan Wang »

Peishan Wang

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Re: Problem 4
« Reply #6 on: October 15, 2012, 06:53:33 AM »
And I got a different answer for part (b) as well, which is

1/2*erf⁡((1-x)/√2t) + 1/2*erf⁡((1+x)/√2t), where √2t stands for "square root of 2t".
« Last Edit: October 16, 2012, 11:40:57 AM by Peishan Wang »

Victor Ivrii

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Re: Problem 4
« Reply #7 on: October 15, 2012, 07:10:24 AM »
OK guys, you know how to do, the rest I am not checking

Aida Razi

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Re: Problem 4
« Reply #8 on: October 16, 2012, 12:09:26 PM »
For part b, you are right.

And I got a different answer for part (b) as well, which is

1/2*erf⁡((1-x)/√2t) + 1/2*erf⁡((1+x)/√2t), where √2t stands for "square root of 2t".