Manuscript I-B-VP-0@1 · I-B-VP-0-v1.tex · lemma · 8a9d0682

Lemma: Finite character identity

An independent exact quotient check · lines 976–1014 · proof included or linked

I-B-VP-0@1#lem:finiteidentity, I-B-VP-0@1#eq:finiteidentity

For $p = 0, \dots, N$ ,

k = 1 \sum^{N - 1} sin (\frac{2 π p k}{N}) cot (\frac{π k}{N}) csc^{2} (\frac{π k}{N}) = \frac{2}{3} p (N - p) (N - 2 p) .

Proof. Call the left side $S_{p}$ , and put $x_{k} = π k / N$ . Then $S_{0} = S_{N} = 0$ , while

S_{p + 1} - 2 S_{p} + S_{p - 1} = - 4 k = 1 \sum^{N - 1} sin (2 p x_{k}) cot x_{k} .

For $1 \leq p \leq N - 1$ ,

sin (2 p x) cot x = 1 + cos (2 p x) + 2 j = 1 \sum^{p - 1} cos (2 j x) .

Summing over $k$ , and using the finite root-of-unity sums, gives

k = 1 \sum^{N - 1} sin (2 p x_{k}) cot x_{k} = N - 2 p .

Thus the second difference of $S_{p}$ is $8 p - 4 N$ . The polynomial

f_{p} = \frac{2}{3} p (N - p) (N - 2 p)

has the same second differences and endpoint values. Their difference has zero second difference and zero values at $0, N$ , so it vanishes identically.

Displayed formulas in this passage

I-B-VP-0@1#eq:finiteidentity · line 979
$k = 1 \sum^{N - 1} sin (\frac{2 π p k}{N}) cot (\frac{π k}{N}) csc^{2} (\frac{π k}{N}) = \frac{2}{3} p (N - p) (N - 2 p) .$
unlabeled · line 992
$S_{p + 1} - 2 S_{p} + S_{p - 1} = - 4 k = 1 \sum^{N - 1} sin (2 p x_{k}) cot x_{k} .$
unlabeled · line 997
$sin (2 p x) cot x = 1 + cos (2 p x) + 2 j = 1 \sum^{p - 1} cos (2 j x) .$
unlabeled · line 1003
$k = 1 \sum^{N - 1} sin (2 p x_{k}) cot x_{k} = N - 2 p .$
unlabeled · line 1008
$f_{p} = \frac{2}{3} p (N - p) (N - 2 p)$

Lemma: Finite character identity

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