400 Algebra
IfP′(x^1 + 2 x^2 )=0, then this identity gives0 =
1
x 1 +x 2
2 −x^3+
1
x 1 +x 2
2 −x^4+···+
1
x 1 +x 2
2 −xn< 0 + 0 +···+ 0 = 0 ,
a contradiction. Similarly, ifP′(xn−^12 +xn)=0, then
0 =
1
xn− 1 +xn
2 −x^1+
1
xn− 1 +xn
2 −x^2+···+
1
xn− 1 +xn
2 −xn−^2> 0 + 0 +···+ 0 = 0 ,
another contradiction. The conclusion follows.
(T. Andreescu)
173.The equationP(x)=0 is equivalent to the equationf(x)=1, wheref(x)=
a 1
x+
a 2
x^2 +···+an
xn. Sincefis strictly decreasing on(^0 ,∞), limx→^0 +f(x)=∞and
limx→∞f(x)=0, the equation has a unique solution.
Remark.A more general principle is true, namely that if the terms of the polynomial
are written in decreasing order of their powers, then the number of sign changes of the
coefficients is the maximum possible number of positive zeros; the actual number of
positive zeros may differ from this by an even number.
174.Assume to the contrary that there iszwith|z|≥2 such thatP(z)=0. Then by the
triangle inequality,
0 =
∣∣
∣∣P(z)
z^7∣∣
∣∣=
∣∣
∣∣ 1 +^7
z^3+
4
z^6+
1
z^7∣∣
∣∣≥ 1 −^7
|z|^3−
4
|z|^6−
1
|z|^7≥ 1 −7
8
−
4
64
−
1
128
=
7
128
> 0 ,
a contradiction. Hence our initial assumption was false, and therefore all the zeros of
P(z)lie inside the disk of radius 2 centered at the origin.
175.Letz=r(cost+isint), sint =0. Using the de Moivre formula, the equality
zn+az+ 1 =0 translates to
rncosnt+arcost+ 1 = 0 ,
rnsinnt+arsint= 0.View this as a system in the unknownsrnandar. Solving the system gives
rn=∣
∣∣
∣
−1 cost
0 sint∣
∣∣
∣
∣∣
∣
∣
cosntcost
sinntsint∣∣
∣
∣
=
sint
sin(n− 1 )t