Finding $Aeven(x)$ and $Aodd(x)$ for a Fast Fourier Transform (FFT) problem?

Question

In this article about FFT the author used FFT on this polynomial.

$A(x) = 3+2x+3x^2+4x^3$

Using $ A(x)= Aeven(x^2) + xAodd(x^2)$ the author determined the following for $Aeven(x)$ and $Aodd(x)$

$ Aeven(x) = 3+3x$

$ Aodd(x) = 2+4x $

How were $Aeven(x)$ and $Aodd(x)$ determined? Shouldn't $Aeven(x) = 3+3x^2$ and $Aodd(x) = 2x+4x^3$ ?

Nathaniel · Accepted Answer · 2021-09-17 16:48:04Z

2

You have $Aeven(x^2) = 3 + 3x^2$, so that means that $Aeven(x) = 3 + 3x$.

For $Aodd$, you have $xAodd(x^2) = 2x + 4x^3$, so $Aodd(x^2) = 2 + 4x^2$, hence $Aodd(x) = 2 + 4x$.

answered Sep 17, 2021 at 16:48

Nathaniel

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$\begingroup$ So for $Aodd(x)$ did you simply replace $x^2$ with $x$ instead of doing a $\sqrt{}$ of $Aodd(x)$ ? $\endgroup$
– calico jack
Commented Sep 17, 2021 at 18:03
$\begingroup$ Why would I calculate the square root of $Aodd(x)$?? The equation is $Aodd(x^2) = 2 + 4x^2$, not $Aodd(x)^2 = 2 + 4x^2$. Just suppose $X=x^2$ and replace each $x^2$ by $X$. $\endgroup$
– Nathaniel
Commented Sep 17, 2021 at 20:08
$\begingroup$ Sorry. I meant $ Aodd(x^2) $. But if we are merely replacing $X^2$ with $X$, then if $Aodd(x^2) = 2x$ , what would $Aodd(x)$ be? @Nathaniel $\endgroup$
– calico jack
Commented Sep 17, 2021 at 20:15
$\begingroup$ I sincerely (without trying to offend you) suggest that you try to review basic math and function manipulation before trying to understand FFT which is quite complicated. $\endgroup$
– Nathaniel
Commented Sep 17, 2021 at 20:28
$\begingroup$ If $Aodd(x^2) = 2x$ then $Aodd(x) = 2 $. That is the answer . See reference at youtu.be/bR1PD-Hn6t8?t=600 $\endgroup$
– calico jack
Commented Sep 17, 2021 at 20:58

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