Problem 12099
(American Mathematical Monthly, Vol.126, March 2019) Proposed by M. Bataille (France).
Letm and n be integers with 0 ≤ m ≤ n − 1. Evaluate
n−1
X
k=0,k6=m
cot2 (m − k)π n
.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let j = m − k then
n−1
X
k=0,k6=m
cot2 (m − k)π n
=
−1
X
j=m−n+1
cot2 jπ n
+
m
X
j=1
cot2 jπ n
=
n−1
X
j=m+1
cot2 (j − n)π n
+
m
X
j=1
cot2 jπ n
=
n−1
X
j=1
cot2 jπ n
which means that the given sum does not depend on m.
Moreover, by De Moivre’s formula,
sin(nt) = Im ((cos(t) + i sin(t))n) =
⌊n/2⌋
X
k=0
(−1)k
n 2k + 1
cosn−(2k+1)(t) sin2k+1(t).
Hence, by letting t = jπ/n with 1 ≤ j ≤ n − 1, we have that
⌊n/2⌋
X
k=0
(−1)k
n 2k + 1
cotn−(2k+1) jπ n
= 0,
which implies that the polynomial
P (z) :=
⌊n/2⌋
X
k=0
(−1)k
n 2k + 1
zn−(2k+1)= nzn−1−n 3
zn−3+ . . .
has degree n − 1 and it has n − 1 distinct solutions given by zj := cot jπn for 1 ≤ j ≤ n − 1.
Therefore, by Vieta’s formulas,
n−1
X
k=0,k6=m
cot2 (m − k)π n
=
n−1
X
j=1
z2j =
n−1
X
j=1
zj
2
−2
X
1≤j<k≤n−1
zjzk
= 0 − 2− n3
n =(n − 1)(n − 2)
3 .
