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Rectangular function & Inequalities !

  1. Oct 24, 2013 #1
    Note: I think I solved this while writing this topic, did not want to scrap it! if you think its wrong let me know!

    I am trying to manipulate the rectangular function with different arguments and came across a confusing one
    Trying to show: [tex] \prod (x^2) = \prod (\frac{x}{\sqrt{2}}) [/tex]
    Recall that the rectangular function is given by:
    [tex] \prod (x) = \begin{cases} 1 & if |x| < 1/2 \\ 0 & if |x| >1/2 \end{cases}[/tex]
    if x -> x/T then
    as a general case:
    [tex] \prod (\frac{x}{T}) = \begin{cases} 1 & if |x| < T/2 \\ 0 & if |x| >T/2 \end{cases}[/tex]

    this still gives the the rect function a width of T by solving
    [tex] -T/2 < x < T/2 [/tex]
    How about an argument such as

    [tex] x \rightarrow x^2 [/tex]
    trying to solve this ineqality
    [tex] \begin{eqnarray} |x^2| & < & 1/2 \ if x \in R \\ x^2-1/2 & < & 0 \\ -\frac{1}{\sqrt{2}} < x & < & \frac{1}{\sqrt{2}} \\ -\frac{\sqrt{2}}{2} < x & < & \frac{\sqrt{2}}{2} \\ -\frac{1}{2} < \frac{x}{\sqrt{2}} & < & \frac{1}{2} \\ | \frac{x}{\sqrt{2}} |< | \frac{1}{2} | \end{eqnarray} [/tex]
    Therefore:
    [tex] \prod (x^2) = \prod (\frac{x}{\sqrt{2}}) [/tex]
     
  2. jcsd
  3. Oct 24, 2013 #2

    Mark44

    Staff: Mentor

    [STRIKE]They're not the same.[/STRIKE]
    For ∏(x2), ∏(x2) = 1 if x2 < 1/2
    The inequality on the right is equivalent to -1/√2 < x < 1/√2.

    So ∏(x2) = 1 for x ##\in## (-1/√2, 1/√2).

    For ∏(x/√2), ∏(x/√2) = 1 if x/√2 < 1/2.

    Edit: I didn't notice that the ∏ function used absolute values.

    [STRIKE]This inequality is equivalent to x < 1/√2.
    Here ∏(x/√2) = 1 if x ##\in## (-∞, 1/√2).[/STRIKE]
     
    Last edited: Oct 24, 2013
  4. Oct 24, 2013 #3
    why don't we put the argument x/√2 in an absolute value and have it
    [tex] |\frac{x}{\sqrt{2}}|< \frac{1}{2} [/tex]

    I am kind of confused now! how do we prove them to be the same?
     
  5. Oct 24, 2013 #4

    Mark44

    Staff: Mentor

    Sorry, I missed the absolute values in your function definition. What you have is fine.
     
  6. Oct 24, 2013 #5
    so that would justify them to be equivalent ? since the absolute value would mean that
    [tex] -1/2 < x/\sqrt{2} < 1/2 [/tex]
     
  7. Oct 24, 2013 #6
    Thanks Mark, Really appreciate it :)
     
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