APM346-2012 > Home Assignment X

Problem 4

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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:

--- Quote from: Aida Razi on October 15, 2012, 01:58:43 AM ---Solution of part a is attached,

--- End quote ---

Where k =1 on the solution.

Aida Razi:
Solution of part b is attached,

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