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What happens when $f(t)=rect(\frac{x-10}{2})rect(\frac{x}{2}) $ ? Which takes priority?

The first function occurs between $9<x<11$ while the second rect occurs when $-1<x<1$. What exactly happens in this function?

EDIT:

Rect is defined as:

$$-\frac12<x<\frac12: 1 $$ $$else: 0$$

RonaldB
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  • What is "rect"? – kimchi lover Aug 13 '17 at 18:39
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    It is impossible for both terms in the multiplication to be non-zero as there is no such $x$ which satisfies both inequalities. Therefore $f$ is $0$. I think you might have some sort of confusion as you talk about 'priority' when none exists and this term makes little sense/is irrelevant. – Shuri2060 Aug 13 '17 at 18:42

1 Answers1

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Just multiply point by point. You will see that $f(x)=0$.

Math Lover
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  • Could you elaborate on multiplying point by point? – RonaldB Aug 13 '17 at 18:41
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    First, I believe it is $f(x)$ instead of $f(t)$. By point-by-point multiplication, I mean that for any $x$ evaluate rect((x-10)/2) and rect(x/2) and multiply them. For $-1 < x <1$, the first rect() is zero, and for other values of $x$ the second rect() is zero. – Math Lover Aug 13 '17 at 18:43