Solution of problem U583

Let k ≥ 1 be a fixed integer and let

P_{n}(x) = x^{n}(x^{k} − x^{k−1} − … − x − 1) − 1

Prove that each polynomial P_{n}(x) has a single positive root, r_{n} and the sequence r_{1}, r_{2}, …, r_{n} is decreasing

Proof

Q_{k}(x) = x^{k} − x^{k−1} − … − x − 1 = x^{k} − \frac{1 − x^{k}}{1 − x} = \frac{x^{k+1} − 2x^{k} + 1}{x − 1}, x ≠ 1

Q_{k}(0) = −1 and let’s assume Q_{k}(1) = 1 − k.

Let f_{k}(x) = x^{k+1} − 2x^{k} + 1, f_{k}(0) = 1, f_{k}(1) = 0, lim_{x→∞} f_{k}(x) = ∞

(x^{k+1} − 2x^{k} + 1)′ = (k + 1)x^{k} − 2kx^{k−1} ≥ 0 ⟺ x ≥ (2k)/(k + 1) ≥ 1

It follows that f_{k}(x) is positive for 0 ≤ x ≤ 1, negative for 1 < x < \overline{x} where \overline{x} which is the second solution besides x = 1 of the equation x^{k+1} − 2x^{k} + 1 = 0 and positive for x > \overline{x} (\overline{x} > x_{m} ≐ (2k)/(k + 1))

This means that Q_{k}(x) ≤ 0 for 0 ≤ x ≤ \overline{x} and Q_{k}(x) > 0 for x > \overline{x}. We want to show that Q_{k}(x) increases for x ≥ \overline{x} and it suffices to show that it increases for x ≥ (2k)/(k + 1).

Q′_{k}(x) = \frac{x^{k+2} − 4x^{k−1} + 4x^{k} + x − 2}{(x − 1)^{2}} \underbrace{≥}_{AGM} \frac{4x^{k+1} − 4x^{k−1} + x − 2}{(x − 1)^{2}} ≥ 0

if and only if 4x^{k+1} − 4x^{k−1} + x − 2 = 4x^{k−1}(x^{2} − 1) + x − 2 ≥ 0. Moreover

4x^{k−1}(x^{2} − 1) + x − 2 ≥ 4x_{m}^{k−1}(x_{m}^{2} − 1) + x_{m} − 2 =

= 4 (\frac{2k}{k + 1})^{k−1} (\frac{4k^{2}}{(k + 1)^{2}} − 1) + \frac{2k}{k + 1} − 2 ≥ 4 (\frac{4k^{2}}{(k + 1)^{2}} − 1) − \frac{2}{k + 1} =

≥ \frac{12k^{2} − 10k − 6}{(k + 1)^{2}} > 0 k ≥ 2

Thus if k ≥ 2 Q_{k}(x) is positive and increases for x ≥ (2k)/(k+1) and a fortiori for x > \overline{x}. It follows that also the function P_{n}(x) = x^{n}Q_{k}(x) is positive and increases and then the equation P_{n}(x) = 1 has the unique solution r_{n}. Since (2k)/(k+1) > 1 for k ≥ 2, P_{n+1}(x) > P_{n}(x) and then r_{n+1} < r_{n} so concluding the proof for k ≥ 2.

The case k = 1 is studied apart. P_{n}(x) = x^{n+1} − x^{n} − 1 decreases for 0 ≤ x ≤ n/(n + 1) and increases for x > n/(n + 1) thus there exists the point \overline{x} such that P_{n}(\overline{x}) = 0 with P_{n}(x) ≥ 0 if x ≥ \overline{x} monotonically increasing. The equation P_{n}(x) = 0 has a unique solution r_{n} > \overline{x} and r_{n+1} < r_{n} since x^{n+1}(x − 1) ≥ x^{n}(x − 1).

The proof is complete