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: \prod (x^2) = \prod (\frac{x}{\sqrt{2}})
Recall that the rectangular function is given by:
\prod (x) = \begin{cases} 1 & if |x| < 1/2 \\ 0 & if |x| >1/2 \end{cases}
if x -> x/T then
as a general case:
\prod (\frac{x}{T}) = \begin{cases} 1 & if |x| < T/2 \\ 0 & if |x| >T/2 \end{cases}

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

x \rightarrow x^2
trying to solve this ineqality
\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}
Therefore:
\prod (x^2) = \prod (\frac{x}{\sqrt{2}})

 

[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: \prod (x^2) = \prod (\frac{x}{\sqrt{2}})
Recall that the rectangular function is given by:
\prod (x) = \begin{cases} 1 & if |x| < 1/2 \\ 0 & if |x| >1/2 \end{cases}
if x -> x/T then
as a general case:
\prod (\frac{x}{T}) = \begin{cases} 1 & if |x| < T/2 \\ 0 & if |x| >T/2 \end{cases}

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

x \rightarrow x^2
trying to solve this ineqality
\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}
Therefore:
\prod (x^2) = \prod (\frac{x}{\sqrt{2}})

[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
|\frac{x}{\sqrt{2}}|&lt; \frac{1}{2}

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
-1/2 &lt; x/\sqrt{2} &lt; 1/2

 

[radiator]

Thanks Mark, Really appreciate it :)