Toronto Math Forum
APM3462012 => APM346 Math => Home Assignment X => Topic started 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(x1)]
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?

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 $1$

Solution of part a is attached,

Solution of part a is attached,
Where k =1 on the solution.

Solution of part b is attached,

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".

And I got a different answer for part (b) as well, which is
1/2*erfâ¡((1x)/âˆš2t) + 1/2*erfâ¡((1+x)/âˆš2t), where âˆš2t stands for "square root of 2t".

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

For part b, you are right.
And I got a different answer for part (b) as well, which is
1/2*erfâ¡((1x)/âˆš2t) + 1/2*erfâ¡((1+x)/âˆš2t), where âˆš2t stands for "square root of 2t".