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--- math121a-f23:october_11_wednesday [2023/10/11 16:35]
pzhou
+++ math121a-f23:october_11_wednesday [2026/02/21 14:41] (current)
@@ Line 28: / Line 28: @@
 If we define hermitian inner product on $Fun(V_x,\C)$ as
-$$ \langle f(x), g(x) \rangle_x = (1/N) \sum_{x \in V_x} f(x) \overline{g(x)}, $$
+$$ \langle f, g\rangle_x = (1/N) \sum_{x \in V_x} f(x) \overline{g(x)}, $$
 and we define hermitian inner product on $Fun(V_p,\C)$ as
-$$ \langle F(p), G(p) \rangle_p =  \sum_{p \in V_p} F(p) \overline{G(p)}. $$
+$$ \langle F, G \rangle_p =  \sum_{p \in V_p} F(p) \overline{G(p)}. $$
 then we find that Fourier transformation is compatible with the two inner products, namely
 $$ \langle f, g \rangle_x = \langle F, G \rangle_p, \quad F = FT(f), G = FT(g). $$
 ==== Example 1: N = 2 ====
+A function $f(x)$ is determined by its values $f(0), f(1)$. Similarly for $F(p)$.
+We have relations
+$$ F(0) = (1/2) (f(0) + f(1)), \quad F(1) = (1/2) (f(0) - f(1)). $$
+So, we can reconstruct $f(x)$ from $F(p)$, by
+$$ f(0) = F(0) + F(1), \quad f(1) = F(0) - F(1). $$
+==== An important equality ====
+$1 + (-1) = 0. $
+and less obviously
+$1 + e^{2\pi i / 3} + e^{2\pi i (2/3)} = 0$
+more generally
+$$ \sum_{j=0}^{N-1} e^{2\pi i (j/N)} = 0 $$
+How to see this? You can say, this is the sum of all the $N$-th roots of unity, and we have
+$$ z^N - 1 = \prod_{j=0}^{N-1} ( z - e^{2\pi i (j/N)}). $$
+hence by looking at the coefficient of $z^{N-1}$, we see the sum of all the roots is 0.
+Or, draw these roots as vectors on the complex plane, they show up as evenly distributed on the unit circle, since the summands are invariant under rotation by $2\pi/N$, hence the result is invariant under such a rotation. And the only possible number is 0.
+==== Example 2: N = 3 ====
+try it yourself.

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