# Toronto Math Forum

## APM346-2012 => APM346 Math => Home Assignment X => Topic started by: Calvin Arnott on October 13, 2012, 10:02:10 PM

Title: Problem 4
Post by: Calvin Arnott 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) ]

Title: Re: Problem 4
Post by: Victor Ivrii 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 $1$
Title: Re: Problem 4
Post by: Aida Razi on October 15, 2012, 01:58:43 AM
Solution of part a is attached,
Title: Re: Problem 4
Post by: Aida Razi on October 15, 2012, 02:00:33 AM
Solution of part a is attached,

Where k =1 on the solution.
Title: Re: Problem 4
Post by: Aida Razi on October 15, 2012, 02:13:58 AM
Solution of part b is attached,
Title: Re: Problem 4
Post by: Peishan Wang 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/âˆš2t), where âˆš2t stands for "square root of 2t".
Title: Re: Problem 4
Post by: Peishan Wang on October 15, 2012, 06:53:33 AM
And I got a different answer for part (b) as well, which is

1/2*erfâ¡((1-x)/âˆš2t) + 1/2*erfâ¡((1+x)/âˆš2t), where âˆš2t stands for "square root of 2t".
Title: Re: Problem 4
Post by: Victor Ivrii on October 15, 2012, 07:10:24 AM
OK guys, you know how to do, the rest I am not checking
Title: Re: Problem 4
Post by: Aida Razi 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*erfâ¡((1-x)/âˆš2t) + 1/2*erfâ¡((1+x)/âˆš2t), where âˆš2t stands for "square root of 2t".