Problem 4

[Calvin Arnott]

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?

[Victor Ivrii]

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]

Solution of part a is attached,

[Aida Razi]

Solution of part a is attached,

Where k =1 on the solution.

[Aida Razi]

Solution of part b is attached,

[Peishan Wang]

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".

[Peishan Wang]

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".

[Victor Ivrii]

OK guys, you know how to do, the rest I am not checking

[Aida Razi]

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".