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I am trying to understand the discrete Fast Fourier Transform. I get the idea of switching between coefficient and value representations to and then back but I am stuck in figuring out how the multiplication of 2 polynomials using their value representation works.

The screenshot below is from a Reducible video (at time 6:39) that explains the Fast Fourier transform. I see that we need 5 points to get $C(x)$, which is of degree $4$, but I don't see how multiplying say the first point of $A(x)$ and $B(x)$, that is, $(-2,1) \times (-2,9)$ gives $(-2,9)$

There's a similar answer here but the steps aren't provided just the final answer, which still baffles me.

Reducible screenshot

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If $P, Q, R \in \mathbb{R}[X], R = PQ$, then for any $x \in \mathbb{R}$, $R(x) = P(x)\times Q(x)$.

It means that if the value representation of $P$ is $[(x_1, P(x_1)); (x_2, P(x_2)); …, (x_n, P(x_n))]$ and the value representation of $Q$ is $[(x_1, Q(x_1)); (x_2, Q(x_2)); …, (x_n, Q(x_n))]$, then the value representation of $R$ is $[(x_1, P(x_1)Q(x_1)); (x_2, P(x_2)Q(x_2)); …, (x_n, P(x_n)Q(x_n))]$.

In your example, $(-2, 1)$ represents a couple $(x_i, P(x_i))$ and $(-2, 9)$ is $(x_i, Q(x_i))$, so it's normal that $(x_i, R(x_i))$ is $(-2, 1\times 9) = (-2, 9)$.

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  • $\begingroup$ Thank you. I didn't realize that I needed to just multiply the $y$ values, what is $R(x)$ in your example, and that the $x$ remains the same. $\endgroup$ Commented Apr 11, 2021 at 16:37

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