Toronto Math Forum
APM346-2012 => APM346 Math => Home Assignment X => Topic started by: Calvin Arnott on October 13, 2012, 10:02:10 PM
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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?
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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 $1$
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Solution of part a is attached,
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Solution of part a is attached,
Where k =1 on the solution.
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Solution of part b is attached,
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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".
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And I got a different answer for part (b) as well, which is
1/2*erfâ¡((1-x)/√2t) + 1/2*erfâ¡((1+x)/√2t), where √2t stands for "square root of 2t".
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OK guys, you know how to do, the rest I am not checking
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For part b, you are right.
And I got a different answer for part (b) as well, which is
1/2*erfâ¡((1-x)/√2t) + 1/2*erfâ¡((1+x)/√2t), where √2t stands for "square root of 2t".