For any monic polynomial f ∈ Q[X] of even degree, prove that there exists

3.3.8. For any monic polynomial f ∈ Q[X] of even degree, prove that there exists

a polynomial g ∈ Q[X] such that lim |x| → ∞ (

f (x) − g(x)) = 0 . Gabriel Dospinescu

Solution. If f has degree 2n, define g (x) = x n +a 1 x n −1 + ··· + a 0 . Next, we compute the coefficients a j inductively on j such that f (x) − g 2 (x) has degree at most n − 1. This is always possible, since we obtain a linear equation in a j that

3.3 Qualitative Results 129 ensures that a j is rational. To conclude, it is enough to observe that

f (x) − g 2 (x)

f (x) − g(x) =

f (x) + g(x)

be distinct positive num- bers with b 1 being the greatest of them, and let α ∈R\N . If there exists an infinite

3.3.9. Let a 1 ,...,a p

be real numbers and let b 1 ,...,b p

set of real numbers A ⊂ (−b 1 , ∞) such that

∑ a i (x + b i ) α =0 for all x ∈ A,

i =1

prove that a 1 = ··· = a p =0 . Marian Tetiva, Problem 122, GMA 4(2002)

Solution. The infinite set A has an accumulation point x 0 , which lies in [−b 1 , ∞)∪ {∞}. We then can find a sequence (x n ) n ≥1 of elements from A (which can be as-

sumed to be distinct and all different from x 0 ) which has the limit x 0 . There are three possible cases.

(i) x 0 = ∞. Since

n →∞ x n = ∞, we get (passing to the limit for n → ∞ in the above equality) ∑ i =1 a i = 0. Using this and again the assumed equality we have

x n ∑ a i 1 + i −1 = 0,

i =1

for all n ≥ 1. Passing to the limit here, and using

x lim x

−1 = α b

(for some real number b), we get α p ∑ i =1 a i b i p = 0; thus ∑ i =1 a i b i = 0 (because α Now one can write

= 0, ∀n ∈ N ∗ ,

i =1

and in this new relation one can pass to the limit for n → ∞, to obtain ∑ p

i =1 a i b 2 i =0 (after division by α ( α

lim x 2 1 − 1)

130 3 Limits of Functions And so on. So clearly, we will obtain p ∑ i =1 a i b k i = 0, for all natural numbers k,

hence the conclusion a

1 = ··· = a p = 0 is plain. (Indeed, the equations ∑ i =1 b k i a i =0 for k = 0, 1, . . . , p − 1 form a homogeneous system with nonzero determinant, the

Vandermonde determinant of the distinct numbers b 1 ,...,b p .) (ii) Let now x 0 ∈ (−b 1 , ∞). From the given conditions we infer

∑ a i (x n +b i ) α = 0, ∀n ∈ N ∗ ,

i =1

which yields (by letting n tend to infinity)

∑ a i (x 0 +b i ) α = 0.

i Then we have =1

∑ a i − (x 0 +b i i = 0,

and passing to the limit, we get α p ∑ i =1 a i (x 0 +b i ) α−1 = 0, due to the limit (x + b) α

But α

i =1 a i (x 0 +b i ) α−1 = 0, and so we get

(x n +b i ) α − (x 0 +b i ) α

−x a ) − α (x 0 +b i ) α−1 (x n 0

Together with the limit (x + b) α − (x 0 + b) α − α (x 0 + b) α−1 (x − x 0 ) α ( α − 1)

2 0 + b) (x − x α−2 0 ) this leads us to

2 ∑ a i (x 0 +b i ) i = 0, =1

hence to p ∑ a i (x 0 +b i ) i α−2 =1 = 0 (as long as we know that α α We will get, continuing this process, p ∑ i =1 a i (x 0 +b i ) α−k = 0, for all natural num- bers k. Considering the first p (for k = 0, 1, . . . , p − 1) equations of this type as a homogeneous system with unknowns a 1 ,...,a p , we see that its determinant is

x 0 +b 1 ...

(x +b ) α

0 1 ··· (x 0 +b p )

x 0 +b p

(x 0 +b 1 ) p −1 ...

1 (x 0 +b p ) p −1

3.3 Qualitative Results 131 because 1 /(x 0 +b 1 ), . . . , 1/(x 0 +b p ) are distinct; consequently the system has the

sole solution a 1 = ··· = a p = 0. (iii) The case x 0 = −b 1 remains, but this does not differ very much from the previ-

ous one. Indeed, we have now x 0 +b 1 = 0, and by the same procedure as above we can obtain

∑ a i (x 0 +b i ) α−k = 0, ∀k ∈ N.

i =2

Then, with the same reasoning, we get a 2 = ··· = a p = 0, which, together with the given condition, leads to a 1 (x + b 1 ) α = 0 for all x ∈ A and thus to a 1 = 0. The proof is now complete. ⊓ ⊔

We point out that the above statement does not remain true if α is a natural number. Indeed, in such a case we can choose nonzero numbers a

1 ,...,a p such that ∑ i =1 a i b k i = 0 for k = 0, 1, . . . , p − 2 (b 1 ,...,b p being given), and we obtain, with Newton’s binomial, p ∑ i =1 a i (x + b i ) α = 0, for all positive integers α ≤ p − 2.

be an increasing sequence of positive numbers. Prove that the series ∞ ∑ n =1 arccos 2 a n /a n +1 converges if and only if the sequence (a n ) n ≥1 is bounded.

3.3.10. Let (a n ) n ≥1

Solution. If a n /a n +1 does not converge to 1, then the series diverges. Let us now assume that a n /a n +1 →1 as n→∞. The basic idea in the proof is that

arccos x

lim

x → 1 −0 2(1 − x) Hence, the series ∞ ∑ n =1 arccos 2 a n /a n +1 converges simultaneously with the series

∑ ∞ n =1 (1 − a n /a n +1 ), and the latter converges simultaneously with the series

n → ∞ ln a .⊓ n ⊔ This shows that the given series converges if and only if the sequence (a n ) n ≥1 is