Roots of a Type of Generalized Quasi-Fibonacci Polynomials
Roots of a Type of Generalized Quasi-Fibonacci Polynomials Daniel Thompson July 8, 2013 Abstract Let a be a nonnegative real number and define a quasi-Fibonacci polynomial sequence by F1a (x) = −a, F2a (x) = x − a, and Fna (x) = a (x) + xF a (x) for n ≥ 2. Let r a denote the maximum real Fn−1 n n−2 a } root of Fna . We prove for certain values of a that the sequence {r2n converges monotonically to βa = a2 + a from above and the sequence a {r2n+1 } converges monotonically to βa from below.
1
Introduction
Consider a Fibonacci type polynomial sequence defined recursively by G0 (x) = α, G1 (x) = x + β, and for n ≥ 2 Gn = xGn−1 + Gn−2 . Here α and β are integers with α 6= 0. The properties of these polynomials have been studied extensively: when α = 1 and β = 0, Gn is the classical Fibonaccci polynomial sequence. Likewise, when α = 2 and β = 0, one gets the classical Lucas polynomials. Hoggatt and Bicknell explicitly find the zeros of these two sequences in [2]. Moore [4] and Prodinger [5] studied the asymptotic behavior of the maximal roots when α = β = −1. Moore’s results have been generalized to more cases; in [6], Yu, Wang, and He study the case when α = β = −a for all positive integers a, while M´aty´as [3] examines the sequence for α = a 6= 0 and β = ±a. More recently, Wang and He fully generalized their results to any integers α and β with α 6= 0. 1
Let a be a positive real number and define the quasi-Fibonacci polynomial sequence Fna recursively by F0a (x) = −a, F1a (x) = x − a, and for n ≥ 2, a a Fna (x) = Fn−1 (x) + xFn−2 (x).
(1)
In [1], B. Alberts studies this sequence when a = 1 and finds that the maximum roots of even indexed polynomials converge monotonically to 2 from above, while the maximum roots of the odd indexed polynomials converge monotonically to 2 from below. In general, denote the maximum root of Fna (x) by rna . When no ambiguity will arise we will omit the superscript a . The paper is organized as follows: in section 2 we present technical results used to prove later theorems. In section 3 we examine the existence, boundedness, and monotonic behavior of the maximum roots. In sections 4 and 5 we study properties of the derivative needed to show convergence. Finally, main results are provided in section 6.
2
Technical results
Lemma 1. Denote the standard Fibonacci sequence by f0 = 0, f1 = 1, and for n ≥ 2, fn = fn−1 + fn−2 . Then Fn (1) = fn − a · fn+1
(2)
Fn (0) = −a
(3)
Fn (−1) ∈ {−a, −1 − a, −1, a, 1 + a, 1}
(4)
Proof. The proofs for Equations (2) and (3) follow directly from the initial conditions and recursion. For Equation (4), we notice that F0 (−1) = −a, F1 (−1) = −1 − a, F2 (−1) = −1, F3 (−1) = a, F4 (−1) = 1 + a, F5 (−1) = 1. An inductive argument shows this pattern holds for all values of n. Lemma 2. Let β = a2 + a. Then Fn (β) = (−a)n+1 . Proof. This is clear for F0 and F1 . Suppose it holds through n − 1. Then Fn (β) = Fn−1 (β) + β · Fn−2 (β) = (−a)n + β(−a)n−1 = (−a)n−1 (−a + β) = (−a)n−1 (a2 ) = (−a)n+1 . 2
Lemma 3. Fn (x) = (1 + 2x)Fn−2 (x) − x2 Fn−4 (x)
(5)
Proof. This follows directly from manipulation of the recursive formula. Theorem 4.
∞ X n−i
Fn (x) =
n−i −a i
i−1
i=0
xi
(6)
Proof. Notice that each sum is finite since for large enough i, i − 1 > n − i and i > n−i. Direct computation verifies the formula for F0 and F1 ; suppose we have shown the theorem through Fn−1 . Then Fn (x) = Fn−1 (x) + xFn−2 (x) ∞ ∞ X X n−1−i i n−1−i = − a x + x · i−1 i = =
i=0 ∞ X
n−1−i i−1
i=0 ∞ X
n−i i−1
−a
n−1−i i
n−i i
−a
xi +
n−2−i i−1
i=0 ∞ Xh n−1−j j−2 j=1
−a
−a
n−2−i i
n−1−j j−1
i
xi
xj + 0
xi
i=0
Corollary. F2n (x) =
n X 2n − i i=0
F2n+1 (x) =
n+1 X 2n + 1 − i i−1
i=0
3
i−1
2n − i −a xi i
(7)
2n + 1 − i −a xi i
Positive real roots
Lemma 5. If
n−i i−1
−a·
n−i i
If
n−i i−1
> 0, then
n−(i+1) (i+1)−1
−a·
n−i i
≤ 0, then
n−(i−1) (i−1)−1
3
−a·
n−(i+1) (i+1)
−a·
n−(i−1) (i−1)
> 0. < 0.
(8)
n−i i−1 n−i i
Proof. Suppose
n−i i−1
−a·
n−(i+1) (i+1)−1 n−(i+1) (i+1)
=
n−i i
> 0. Then
i n−2i+1
=
> a. So
(i + 1) i > >a n − 2(i + 1) + 1 n − 2i + 1
Our conclusion follows directly from this fact; the second result is proven similarly. Theorem 6. F2n+1 has exactly one nonnegative real root. When n ≤ a, F2n has no positive real roots, and when n > a, F2n has exactly one nonnegative real root. Proof. First notice that if a = 0 then the maximal root of each Fn is 0 for n ≥ 1. Suppose a > 0. We use Descartes’ Rule of Signs. Equation (8) shows that the constant term of F2n+1 is −a < 0 and the leading coefficient of F2n+1 is 1 > 0. Thus, by Lemma 5, there must be exactly one change in sign of the coefficients of F2n+1 . This implies that the polynomial has exactly one real positive root. Similarly, Equation (7) shows the constant term of F2n is −a < 0. However, when n ≤ a, the leading coefficient of F2n is n − a ≤ 0, so there are no sign changes and therefore no positive real roots. When n > a, n − a > 0, so there must be exactly one sign change and therefore exactly one positive real root. Corollary. Let rn denote the maximal real root of the polynomial Fn (if it exists). Then for all n, r2n+1 exists and r2n+1 < β. Similarly, for all n > a, r2n exists and β < r2n . Corollary. For all n and for all x ∈ (0, r2n+1 ), F2n+1 (x) < 0; for all x > r2n+1 , F2n+1 (x) > 0. Similarly, for all n > a, for all x ∈ (0, r2n ), F2n (x) < 0 and for all x > r2n , F2n (x) > 0.
4
Theorem 7. For all n, a = r1 < r3 < · · · < r2n+1 < · · · < β.
(9)
Furthermore, when N > a and n > N , β < · · · < r2n < · · · < r2N +2 < r2N .
(10)
Proof. Clearly r1 = a and F3 (r1 ) = F2 (a) < 0, so r1 < r3 . Suppose we have shown Equation (9) holds through r2n−1 . Then since r2n−3 < r2n−1 , Equation (5) gives us 2 F2n+1 (r2n−1 ) = (1 + 2r2n−1 )F2n−1 (r2n−1 ) − r2n−1 F2n−3 (r2n−1 ) 2 = −r2n−1 F2n−3 (r2n−1 ) < 0.
Thus, r2n+1 > r2n−1 . The proof of (10) is similar.
4
Behavior of derivatives
For the remainder of the paper, we will use Na (or just N ) to denote an integer satisfying the following four properties: (i) Na > a. (ii) r2Na +1 ≥ a2 . 0 0 (x) > a2Na −1 and F2N (x) > a2Na . (iii) For x ≥ β, F2N a −1 a 0 0 0 0 (iv) For x > r2Na +1 , F2N (x) ≥ 1, F2N (x) ≥ 1, and F2N (x) ≥ xF2N (x). a +1 a a +2 a
Numerical evidence suggests that such an Na exists for all a ≥ 0. For instance, when 0 ≤ a ≤ 1, it can be shown directly that Na = 2 satisfies the properties. Lemma 8. Let x ≥ β. Then for all n ≥ 2N − 1, Fn0 (x) > an .
5
0 2N −1 0 Proof. Since N satisfies condition (iii), F2N and F2N (x) > −1 (x) > a 2N a . Suppose we have shown the lemma holds for all integers between 2N −1 0 and 2n. Then since F2n−1 > 0 on [β, ∞), F2n−1 (x) ≥ F2n−1 (β), so 0 0 0 F2n+1 (x) = F2n (x) + xF2n−1 (x) + F2n−1 (x) 2n 2n−1 ≥a +β·a + a2n = a2n+1 + 3 · a2n > a2n+1 . 0 follows similarly. The proof for F2n+1
Lemma 9. For all n > N and x > r2n−1 , 0 0 (x) F2n−1 (x) F2n+1 0 ≥ ≥ · · · ≥ F2N +1 (x) ≥ 1 n−N n−N −1 x x and 0 0 F2n+2 (x) F (x) 0 ≥ 2n ≥ · · · ≥ F2N (x) ≥ 1. n−N +1 n−N x x 0 Proof. Let x > r2n−1 . Since N satisfies condition (iv), F2N +1 (x) ≥ 1, 0 0 0 F2N (x) ≥ 1, and F2N +2 (x) ≥ xF2N (x). Suppose we have shown the lemma holds through n − 1. Let x ≥ r2n−1 > r2n−3 . Then 0 0 0 F2n+1 (x) = F2n (x) + xF2n−1 (x) + F2n−1 (x) 0 ≥ xn−N + xF2n−1 (x) + 0 0 ≥ xF2n−1 (x),
so
0 0 F2n+1 (x) F2n−1 (x) ≥ . n−N n−N −1 x x 0 0 Similarly, since F2n (x) ≥ xF2n−2 (x), Equation (5) shows: 0 0 0 (x) + 2F2n (x) − 2xF2n−2 (x) F2n+2 (x) = (1 + 2x)F2n (x) − x2 F2n−2 0 0 ≥ F2n (x) + (2x − x)F2n (x) + 2F2n−1 (x) 0 ≥ xn−N + xF2n (x) + 0 0 ≥ xF2n (x),
so
0 0 F2n+2 (x) F2n (x) ≥ . xn−N +1 xn−N
6
Remark Given some n0 that satisfies condition (iii), each n ≥ n0 also satisfies condition (iii). Likewise, if n1 satisfies condition (iv), each n ≥ n1 also satisfies condition (iv). 0 F2n (x) = ∞. 2n n→∞ a Proof. Let x ≥ β. Notice from Lemma 8 that when n > N and x ≥ β,
Lemma 10. When x ≥ β, lim
0 0 (x) (x) F2n−2 1 F2n 0 0 2 0 − = F (x) + xF (x) + F (x) − a F 2n−2 2n−1 2n−2 2n−2 a2n a2n−2 a2n 1 0 (x) − a2n−1 ≥ 2n a2n−1 + (β − a2 )F2n−2 a 1 1 ≥ 2n a · a2n−2 = a a Then n X 1 ∞ = lim n→∞ a i=N +1 n 0 X (x) F2i0 (x) F2i−2 ≤ lim − n→∞ a2i a2i−2 i=N +1 0 0 F2n (x) F2N (x) = lim − n→∞ a2n a2N F 0 (x) = lim 2n2n n→∞ a Lemma 11. Let {x2n+1 } be a sequence of numbers satisfying x2n+1 ≥ r2n−1 . Then F 0 (x2n+1 ) lim 2n+12n+1 = ∞. n→∞ a
Proof. Since N satisfies condition (ii), r2N +1 ≥ a2 . Then for n > N , Lemma 9 and a similar argument to the previous proof show that when x > r2n−1 ≥ r2N +1 ≥ a2 , 2n 0 0 (x) F2n+1 (x) F2n−1 1 a 1 − ≥ 2n+1 = 2N +1 . 2n+1 2n−1 2N a a a a a Thus, by an analogous argument to the previous proof, we find 0 F2n (x) = ∞. n→∞ a2n
lim
7
5
Existence of Na
We will now investigate the existence of an Na which satisfies the base cases for each of the proofs in the previous sections. The properties and conditions mentioned below are listed at the beginning of section 4. Notice that if n0 satisfies any given condition, then each n ≥ n0 also satisfies that condition. Furthermore, direct computation of Na for some values of a (see table below) suggests that as a increases, the minimum possible value of Na increases. This implies that the value Na would work for all values less than or equal to a; e.g., since N25 = 102 satisfies the conditions when a = 30, 102 should satisfy the conditions for all a ∈ [0, 30]. a 1 2 3 5 10 20 30
n satisfying (ii) 1 2 4 7 18 44 72
n satisfying (iii) 2 4 7 12 26 59 95
n satisfying (iv) 2 4 6 12 28 63 102
Na 2 4 7 12 28 63 102
Lemma 12. Suppose there exists some N satisfying condition (iv) for a given value of a. Then there exists k ≥ 0 such that N + k satisfies condition (iii). Proof. By Lemma 9, for all k ≥ 0 and x > r2(N +k)−1 , 0 k F2(N +k)+1 (x) ≥ x
and
0 k F2(N +k) (x) ≥ x
Let k be large enough such that (a2 + a)k−1 ≥ a2N +1 (a2 )k−1
and
(a2 + a)k ≥ a2N . (a2 )k
Then for all x ≥ β, 0 k−1 F2(N ≥ β k−1 = (a2 + a)k−1 ≥ a2(N +k)−1 . +k)−1 (x) ≥ x
Likewise, for all x ≥ β, 0 k 2 k 2(N +k) F2(N . +k) (x) ≥ x ≥ (a + a) ≥ a
Thus, N + k satisfies condition (iii). 8
6
Main results
Theorem 13. Let a be an integer and n > 1. If n is odd or greater than 2a, then rn is irrational. Proof. Suppose r is a rational root of Fn . Then in reduced form, r = pq where p is a factor of −a. So |p| ≤ a, which means r ≤ a. However, from Theorem 7 we know r1 = a, r2n+1 > r1 , and when n2 > a, r2n > β > a. Theorem 14. The sequence r2n converges to β monotonically from above, and the sequence r2n+1 converges to β monotonically from below. Proof. Recall that Fn (β) = (−a)n+1 , r2n > β for n > a, and r2n+1 < β. By the Mean Value Theorem, for each n > a there exists c2n ∈ [β, r2n ] such that F2n (r2n ) − F2n (β) 0 = F2n (c2n ). r2n − β Then −F2n (β) n→∞ n→∞ r2n − β a2n lim (r2n − β) = a lim 0 =0 n→∞ n→∞ F2n (c2n ) lim r2n = β. 0 lim F2n (c2n ) = lim
n→∞
By combining a similar argument and Lemma 11, we obtain our second result.
7
Acknowledgements
I would like to thank Michigan State University and the Lyman Briggs College for their support of our REU, as well as my faculty mentor, Dr. Aklilu Zeleke. I would also like to thank his graduate assistants, Justin Droba, Rani Satyam and Richard Shadrach. Project sponsored by the National Security Agency under Grant Number H98230-13-1-0259. Project sponsored by the National Science Foundation under Grant Number DMS 1062817.
9
References [1] Alberts, B.: On the properties of a quasi-Fibonacci polynomial sequence. June 2011. – SURIEM [2] Hoggatt, Verner Jr. ; Bicknell, Marjorie: Roots of Fibonacci polynomials. In: Fibonacci Quart. 11 (1973), Nr. 3, S. 271–274. – ISSN 0015–0517 ´ tya ´ s, Ferenc: Bounds for the zeros of Fibonacci-like polynomials. [3] Ma In: Acta Acad. Paedagog. Agriensis Sect. Mat. (N.S.) 25 (1998), S. 15–20 (1999). – ISSN 1216–6014 [4] Moore, Gregory: The limit of the golden numbers is 23 . In: Fibonacci Quart. 32 (1994), Nr. 3, S. 211–217. – ISSN 0015–0517 [5] Prodinger, Helmut: The asymptotic behavior of the golden numbers. In: Fibonacci Quart. 34 (1996), Nr. 3, S. 224–225. – ISSN 0015–0517 [6] Yu, Hongquan ; Wang, Yi ; He, Mingfeng: On the limit of generalized golden numbers. In: Fibonacci Quart. 34 (1996), Nr. 4, S. 320–322. – ISSN 0015–0517
10
1
Introduction
Consider a Fibonacci type polynomial sequence defined recursively by G0 (x) = α, G1 (x) = x + β, and for n ≥ 2 Gn = xGn−1 + Gn−2 . Here α and β are integers with α 6= 0. The properties of these polynomials have been studied extensively: when α = 1 and β = 0, Gn is the classical Fibonaccci polynomial sequence. Likewise, when α = 2 and β = 0, one gets the classical Lucas polynomials. Hoggatt and Bicknell explicitly find the zeros of these two sequences in [2]. Moore [4] and Prodinger [5] studied the asymptotic behavior of the maximal roots when α = β = −1. Moore’s results have been generalized to more cases; in [6], Yu, Wang, and He study the case when α = β = −a for all positive integers a, while M´aty´as [3] examines the sequence for α = a 6= 0 and β = ±a. More recently, Wang and He fully generalized their results to any integers α and β with α 6= 0. 1
Let a be a positive real number and define the quasi-Fibonacci polynomial sequence Fna recursively by F0a (x) = −a, F1a (x) = x − a, and for n ≥ 2, a a Fna (x) = Fn−1 (x) + xFn−2 (x).
(1)
In [1], B. Alberts studies this sequence when a = 1 and finds that the maximum roots of even indexed polynomials converge monotonically to 2 from above, while the maximum roots of the odd indexed polynomials converge monotonically to 2 from below. In general, denote the maximum root of Fna (x) by rna . When no ambiguity will arise we will omit the superscript a . The paper is organized as follows: in section 2 we present technical results used to prove later theorems. In section 3 we examine the existence, boundedness, and monotonic behavior of the maximum roots. In sections 4 and 5 we study properties of the derivative needed to show convergence. Finally, main results are provided in section 6.
2
Technical results
Lemma 1. Denote the standard Fibonacci sequence by f0 = 0, f1 = 1, and for n ≥ 2, fn = fn−1 + fn−2 . Then Fn (1) = fn − a · fn+1
(2)
Fn (0) = −a
(3)
Fn (−1) ∈ {−a, −1 − a, −1, a, 1 + a, 1}
(4)
Proof. The proofs for Equations (2) and (3) follow directly from the initial conditions and recursion. For Equation (4), we notice that F0 (−1) = −a, F1 (−1) = −1 − a, F2 (−1) = −1, F3 (−1) = a, F4 (−1) = 1 + a, F5 (−1) = 1. An inductive argument shows this pattern holds for all values of n. Lemma 2. Let β = a2 + a. Then Fn (β) = (−a)n+1 . Proof. This is clear for F0 and F1 . Suppose it holds through n − 1. Then Fn (β) = Fn−1 (β) + β · Fn−2 (β) = (−a)n + β(−a)n−1 = (−a)n−1 (−a + β) = (−a)n−1 (a2 ) = (−a)n+1 . 2
Lemma 3. Fn (x) = (1 + 2x)Fn−2 (x) − x2 Fn−4 (x)
(5)
Proof. This follows directly from manipulation of the recursive formula. Theorem 4.
∞ X n−i
Fn (x) =
n−i −a i
i−1
i=0
xi
(6)
Proof. Notice that each sum is finite since for large enough i, i − 1 > n − i and i > n−i. Direct computation verifies the formula for F0 and F1 ; suppose we have shown the theorem through Fn−1 . Then Fn (x) = Fn−1 (x) + xFn−2 (x) ∞ ∞ X X n−1−i i n−1−i = − a x + x · i−1 i = =
i=0 ∞ X
n−1−i i−1
i=0 ∞ X
n−i i−1
−a
n−1−i i
n−i i
−a
xi +
n−2−i i−1
i=0 ∞ Xh n−1−j j−2 j=1
−a
−a
n−2−i i
n−1−j j−1
i
xi
xj + 0
xi
i=0
Corollary. F2n (x) =
n X 2n − i i=0
F2n+1 (x) =
n+1 X 2n + 1 − i i−1
i=0
3
i−1
2n − i −a xi i
(7)
2n + 1 − i −a xi i
Positive real roots
Lemma 5. If
n−i i−1
−a·
n−i i
If
n−i i−1
> 0, then
n−(i+1) (i+1)−1
−a·
n−i i
≤ 0, then
n−(i−1) (i−1)−1
3
−a·
n−(i+1) (i+1)
−a·
n−(i−1) (i−1)
> 0. < 0.
(8)
n−i i−1 n−i i
Proof. Suppose
n−i i−1
−a·
n−(i+1) (i+1)−1 n−(i+1) (i+1)
=
n−i i
> 0. Then
i n−2i+1
=
> a. So
(i + 1) i > >a n − 2(i + 1) + 1 n − 2i + 1
Our conclusion follows directly from this fact; the second result is proven similarly. Theorem 6. F2n+1 has exactly one nonnegative real root. When n ≤ a, F2n has no positive real roots, and when n > a, F2n has exactly one nonnegative real root. Proof. First notice that if a = 0 then the maximal root of each Fn is 0 for n ≥ 1. Suppose a > 0. We use Descartes’ Rule of Signs. Equation (8) shows that the constant term of F2n+1 is −a < 0 and the leading coefficient of F2n+1 is 1 > 0. Thus, by Lemma 5, there must be exactly one change in sign of the coefficients of F2n+1 . This implies that the polynomial has exactly one real positive root. Similarly, Equation (7) shows the constant term of F2n is −a < 0. However, when n ≤ a, the leading coefficient of F2n is n − a ≤ 0, so there are no sign changes and therefore no positive real roots. When n > a, n − a > 0, so there must be exactly one sign change and therefore exactly one positive real root. Corollary. Let rn denote the maximal real root of the polynomial Fn (if it exists). Then for all n, r2n+1 exists and r2n+1 < β. Similarly, for all n > a, r2n exists and β < r2n . Corollary. For all n and for all x ∈ (0, r2n+1 ), F2n+1 (x) < 0; for all x > r2n+1 , F2n+1 (x) > 0. Similarly, for all n > a, for all x ∈ (0, r2n ), F2n (x) < 0 and for all x > r2n , F2n (x) > 0.
4
Theorem 7. For all n, a = r1 < r3 < · · · < r2n+1 < · · · < β.
(9)
Furthermore, when N > a and n > N , β < · · · < r2n < · · · < r2N +2 < r2N .
(10)
Proof. Clearly r1 = a and F3 (r1 ) = F2 (a) < 0, so r1 < r3 . Suppose we have shown Equation (9) holds through r2n−1 . Then since r2n−3 < r2n−1 , Equation (5) gives us 2 F2n+1 (r2n−1 ) = (1 + 2r2n−1 )F2n−1 (r2n−1 ) − r2n−1 F2n−3 (r2n−1 ) 2 = −r2n−1 F2n−3 (r2n−1 ) < 0.
Thus, r2n+1 > r2n−1 . The proof of (10) is similar.
4
Behavior of derivatives
For the remainder of the paper, we will use Na (or just N ) to denote an integer satisfying the following four properties: (i) Na > a. (ii) r2Na +1 ≥ a2 . 0 0 (x) > a2Na −1 and F2N (x) > a2Na . (iii) For x ≥ β, F2N a −1 a 0 0 0 0 (iv) For x > r2Na +1 , F2N (x) ≥ 1, F2N (x) ≥ 1, and F2N (x) ≥ xF2N (x). a +1 a a +2 a
Numerical evidence suggests that such an Na exists for all a ≥ 0. For instance, when 0 ≤ a ≤ 1, it can be shown directly that Na = 2 satisfies the properties. Lemma 8. Let x ≥ β. Then for all n ≥ 2N − 1, Fn0 (x) > an .
5
0 2N −1 0 Proof. Since N satisfies condition (iii), F2N and F2N (x) > −1 (x) > a 2N a . Suppose we have shown the lemma holds for all integers between 2N −1 0 and 2n. Then since F2n−1 > 0 on [β, ∞), F2n−1 (x) ≥ F2n−1 (β), so 0 0 0 F2n+1 (x) = F2n (x) + xF2n−1 (x) + F2n−1 (x) 2n 2n−1 ≥a +β·a + a2n = a2n+1 + 3 · a2n > a2n+1 . 0 follows similarly. The proof for F2n+1
Lemma 9. For all n > N and x > r2n−1 , 0 0 (x) F2n−1 (x) F2n+1 0 ≥ ≥ · · · ≥ F2N +1 (x) ≥ 1 n−N n−N −1 x x and 0 0 F2n+2 (x) F (x) 0 ≥ 2n ≥ · · · ≥ F2N (x) ≥ 1. n−N +1 n−N x x 0 Proof. Let x > r2n−1 . Since N satisfies condition (iv), F2N +1 (x) ≥ 1, 0 0 0 F2N (x) ≥ 1, and F2N +2 (x) ≥ xF2N (x). Suppose we have shown the lemma holds through n − 1. Let x ≥ r2n−1 > r2n−3 . Then 0 0 0 F2n+1 (x) = F2n (x) + xF2n−1 (x) + F2n−1 (x) 0 ≥ xn−N + xF2n−1 (x) + 0 0 ≥ xF2n−1 (x),
so
0 0 F2n+1 (x) F2n−1 (x) ≥ . n−N n−N −1 x x 0 0 Similarly, since F2n (x) ≥ xF2n−2 (x), Equation (5) shows: 0 0 0 (x) + 2F2n (x) − 2xF2n−2 (x) F2n+2 (x) = (1 + 2x)F2n (x) − x2 F2n−2 0 0 ≥ F2n (x) + (2x − x)F2n (x) + 2F2n−1 (x) 0 ≥ xn−N + xF2n (x) + 0 0 ≥ xF2n (x),
so
0 0 F2n+2 (x) F2n (x) ≥ . xn−N +1 xn−N
6
Remark Given some n0 that satisfies condition (iii), each n ≥ n0 also satisfies condition (iii). Likewise, if n1 satisfies condition (iv), each n ≥ n1 also satisfies condition (iv). 0 F2n (x) = ∞. 2n n→∞ a Proof. Let x ≥ β. Notice from Lemma 8 that when n > N and x ≥ β,
Lemma 10. When x ≥ β, lim
0 0 (x) (x) F2n−2 1 F2n 0 0 2 0 − = F (x) + xF (x) + F (x) − a F 2n−2 2n−1 2n−2 2n−2 a2n a2n−2 a2n 1 0 (x) − a2n−1 ≥ 2n a2n−1 + (β − a2 )F2n−2 a 1 1 ≥ 2n a · a2n−2 = a a Then n X 1 ∞ = lim n→∞ a i=N +1 n 0 X (x) F2i0 (x) F2i−2 ≤ lim − n→∞ a2i a2i−2 i=N +1 0 0 F2n (x) F2N (x) = lim − n→∞ a2n a2N F 0 (x) = lim 2n2n n→∞ a Lemma 11. Let {x2n+1 } be a sequence of numbers satisfying x2n+1 ≥ r2n−1 . Then F 0 (x2n+1 ) lim 2n+12n+1 = ∞. n→∞ a
Proof. Since N satisfies condition (ii), r2N +1 ≥ a2 . Then for n > N , Lemma 9 and a similar argument to the previous proof show that when x > r2n−1 ≥ r2N +1 ≥ a2 , 2n 0 0 (x) F2n+1 (x) F2n−1 1 a 1 − ≥ 2n+1 = 2N +1 . 2n+1 2n−1 2N a a a a a Thus, by an analogous argument to the previous proof, we find 0 F2n (x) = ∞. n→∞ a2n
lim
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5
Existence of Na
We will now investigate the existence of an Na which satisfies the base cases for each of the proofs in the previous sections. The properties and conditions mentioned below are listed at the beginning of section 4. Notice that if n0 satisfies any given condition, then each n ≥ n0 also satisfies that condition. Furthermore, direct computation of Na for some values of a (see table below) suggests that as a increases, the minimum possible value of Na increases. This implies that the value Na would work for all values less than or equal to a; e.g., since N25 = 102 satisfies the conditions when a = 30, 102 should satisfy the conditions for all a ∈ [0, 30]. a 1 2 3 5 10 20 30
n satisfying (ii) 1 2 4 7 18 44 72
n satisfying (iii) 2 4 7 12 26 59 95
n satisfying (iv) 2 4 6 12 28 63 102
Na 2 4 7 12 28 63 102
Lemma 12. Suppose there exists some N satisfying condition (iv) for a given value of a. Then there exists k ≥ 0 such that N + k satisfies condition (iii). Proof. By Lemma 9, for all k ≥ 0 and x > r2(N +k)−1 , 0 k F2(N +k)+1 (x) ≥ x
and
0 k F2(N +k) (x) ≥ x
Let k be large enough such that (a2 + a)k−1 ≥ a2N +1 (a2 )k−1
and
(a2 + a)k ≥ a2N . (a2 )k
Then for all x ≥ β, 0 k−1 F2(N ≥ β k−1 = (a2 + a)k−1 ≥ a2(N +k)−1 . +k)−1 (x) ≥ x
Likewise, for all x ≥ β, 0 k 2 k 2(N +k) F2(N . +k) (x) ≥ x ≥ (a + a) ≥ a
Thus, N + k satisfies condition (iii). 8
6
Main results
Theorem 13. Let a be an integer and n > 1. If n is odd or greater than 2a, then rn is irrational. Proof. Suppose r is a rational root of Fn . Then in reduced form, r = pq where p is a factor of −a. So |p| ≤ a, which means r ≤ a. However, from Theorem 7 we know r1 = a, r2n+1 > r1 , and when n2 > a, r2n > β > a. Theorem 14. The sequence r2n converges to β monotonically from above, and the sequence r2n+1 converges to β monotonically from below. Proof. Recall that Fn (β) = (−a)n+1 , r2n > β for n > a, and r2n+1 < β. By the Mean Value Theorem, for each n > a there exists c2n ∈ [β, r2n ] such that F2n (r2n ) − F2n (β) 0 = F2n (c2n ). r2n − β Then −F2n (β) n→∞ n→∞ r2n − β a2n lim (r2n − β) = a lim 0 =0 n→∞ n→∞ F2n (c2n ) lim r2n = β. 0 lim F2n (c2n ) = lim
n→∞
By combining a similar argument and Lemma 11, we obtain our second result.
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Acknowledgements
I would like to thank Michigan State University and the Lyman Briggs College for their support of our REU, as well as my faculty mentor, Dr. Aklilu Zeleke. I would also like to thank his graduate assistants, Justin Droba, Rani Satyam and Richard Shadrach. Project sponsored by the National Security Agency under Grant Number H98230-13-1-0259. Project sponsored by the National Science Foundation under Grant Number DMS 1062817.
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