Problem 12077
(American Mathematical Monthly, Vol.125, December 2018) Proposed by M. A. Alekseyev (USA).
Let f(x) be a monic polynomial of degree n with distinct zeros a1, . . . , an. Prove
n
X
k=1
ank−1 f′(ak) = 1.
Solution proposed by Roberto Tauraso, Dipartimento di Matematica, Universit`a di Roma “Tor Vergata”, via della Ricerca Scientifica, 00133 Roma, Italy.
Solution. We will show the following more general result:
n
X
k=1
amk f′(ak) =
0 if 0 ≤ m < n − 1 1 if m = n − 1, Pn
j=1aj if m = n.
We first consider the polynomial
F(x) := xn−1−
n
X
k=1
ank−1
f′(ak)· f(x) x− ak
.
Note that the degree of F is n − 1 and that for 1 ≤ j ≤ n, f(x)
x− ak
x=aj
=
(0 if j 6= k, f′(ak) if j = k,
which implies that F (ak) = 0 for k = 1, 2, . . . , n. Since F has n different roots, we may conclude that F is identically zero. Therefore
xn−1 =
n
X
k=1
ank−1
f′(ak)· f(x) x− ak
.
For m = n − 1,
1 = [xn−1]xn−1 =
n
X
k=1
ank−1
f′(ak)· [xn−1] f(x) x− ak
=
n
X
k=1
ank−1 f′(ak). In similar way, we show the case m = n,
0 = [xn−2]xn−1=
n
X
k=1
ank−1
f′(ak)· [xn−2] f(x) x− ak
=
n
X
k=1
ank−1 f′(ak)·
ak−
n
X
j=1
aj
=
n
X
k=1
ank f′(ak)−
n
X
j=1
aj.
Finally for 0 ≤ m < n, let R > max1≤j≤n|aj|. Then, by Cauchy’s theorem, 1
2πi I
|z|=R
zm f(z)dz=
n
X
k=1
Res zm f(z); ak
=
n
X
k=1
amk f′(ak)
where we used the fact that a1, . . . , an are simple poles. Since m < n − 1, the limit of the left-hand
side as R → +∞ is zero and we are done.
