Problem 12171
(American Mathematical Monthly, Vol.127, March 2020) Proposed by U. Abel and V. Kushnirevych (Germany).
LetTn be the n-th Chebyshev polynomial, defined by Tn(cos(θ)) = cos(nθ). Prove Tn′(1/z)
Tn(1/z) = nz
√1 − z2 + O(z2n+1) asz → 0.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. Let θ = it, then cos(it) = cosh(t) and we have that Tn(cosh(t)) = cosh(nt). It follows that Tn′(cosh(t))
Tn(cosh(t)) = n tanh(nt) sinh(t) . In order to show the power series identity
Tn′(1/z)
Tn(1/z)− nz
√1 − z2 = O(z2n+1), it suffices to show that
lim
z→0+
1 z2n+1
Tn′(1/z)
Tn(1/z)− nz
√1 − z2
exists and it is finite. As z → 0+, t := arccosh(1/z) → +∞, √1−zz 2 = sinh(t) and
lim
z→0+
1 z2n+1
Tn′(1/z)
Tn(1/z) − nz
√1 − z2
= lim
t→+∞
n cosh2n+1(t)(tanh(nt) − 1) sinh(t)
= 1 22n lim
t→+∞
n(et+ e−t)2n+1(−2e−nt) (et− e−t)(ent+ e−nt)
= − n 22n−1 lim
t→+∞
(1 + e−2t)2n (1 − e−2t)(1 + e−2nt)
= − n 22n−1
and the proof is complete.
