Rectangular function & Inequalities

[radiator]

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]

 

[Mark44]

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]

[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]

 

[radiator]

For ∏(x/√2), ∏(x/√2) = 1 if x/√2 < 1/2.
This inequality is equivalent to x < 1/√2.

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?

 

[Mark44]

Sorry, I missed the absolute values in your function definition. What you have is fine.

 

[radiator]

so that would justify them to be equivalent ? since the absolute value would mean that
[tex] -1/2 < x/\sqrt{2} < 1/2 [/tex]

 

[radiator]

Thanks Mark, Really appreciate it :)