¹ Department of Mathematics, Woldia University, Woldia, Ethiopia

AGEZE ABYE ADMASU
Roles: Conceptualization, Formal Analysis, Investigation, Methodology, Resources, Supervision, Validation, Visualization, Writing – Original Draft Preparation, Writing – Review & Editing

DEREBEW NIGUSSIE
Roles: Conceptualization, Formal Analysis, Investigation, Methodology, Resources, Supervision, Validation, Visualization, Writing – Original Draft Preparation, Writing – Review & Editing

⁷ For n≥2 , the Mulatu polynomials {Mn(x)}n=0∞ are defined by the second-order linear recurrence relation:

Mn(x)=xMn−1(x)+Mn−2(x)

subject to the constant initial conditions:

M0(x)=4,M1(x)=1.

The first few terms of this polynomial family are:

The characteristic equation associated with the recurrence relation is t2−xt−1=0 , which yields the dominant and conjugate roots:

α(x)=x+x2+42,β(x)=x−x2+42.

These roots satisfy the fundamental relations α(x)+β(x)=x , α(x)β(x)=−1 , and α(x)−β(x)=x2+4 . The classical Fibonacci polynomials Fn(x) and Lucas polynomials Ln(x) are expressed via Binet’s formulas as:

Fn(x)=α(x)n−β(x)nα(x)−β(x),Ln(x)=α(x)n+β(x)n.

By utilizing the initial conditions M0(x)=4,M1(x)=1 , the Binet formula for the Mulatu polynomials for n≥1 can be expressed as:

Mn(x)=(1−4β(x))α(x)n−(1−4α(x))β(x)nα(x)−β(x).

Alternatively, we can express Mn(x) as a direct linear combination of Fibonacci polynomials^7,9:

Mn(x)=4Fn−1(x)+Fn(x)forn≥1,

which serves as a key identity in our subsequent derivations.

For n≥2 , the first derivatives M′n(x) satisfy the non-homogeneous recurrence relation: M′n(x)=xM′n−1(x)+M′n−2(x)+Mn−1(x) with the initial conditions M′0(x)=0 and M′1(x)=0 .

Differentiating both sides of the defining recurrence Mn(x)=xMn−1(x)+Mn−2(x) with respect to x using the product rule yields:

ddx[Mn(x)]=ddx[xMn−1(x)]+ddx[Mn−2(x)]

M′n(x)=Mn−1(x)+xM′n−1(x)+M′n−2(x).

Since M0(x)=4 and M1(x)=1 are constants, their derivatives are trivially M′0(x)=0 and M′1(x)=0 .

We can extend this property to the k -th derivative Mn(k)(x)=dkdxkMn(x) following the foundational work on higher-order polynomial derivatives.¹⁷

For n≥2 and k≥1 , the k -th order derivatives satisfy:

Mn(k)(x)=xMn−1(k)(x)+Mn−2(k)(x)+kMn−1(k−1)(x) with the boundary conditions Mn(k)(x)=0 for all n<k .

The proof proceeds by induction on k and applying Leibniz’s rule for differentiation to the term xMn−1(x) .

For any positive integer n≥2,

Mn′(0)={n2,for evenn2(n−1),foroddn.

For n≥2,M′n(x)=Mn−1(x)+xM′n−1(x)+M′n−2(x).

⟹M′n(0)=Mn−1(0)+M′n−2(0)

But Mn(0)={1ifnisodd4ifnis even.

Thus,

M′n(0)={1+M′n−2(0),for evenn4+M′n−2(0),foroddn.

Using⁷ (Theorem 3.7), we have

Mn−2′(0)={n−22,for evenn2(n−3),foroddn.

This completes the proof.

The derivatives of the characteristic roots α(x) and β(x) satisfy:

α′(x)=α(x)α(x)−β(x),β′(x)=−β(x)α(x)−β(x).

Differentiating α(x)=x+x2+42 directly gives:

α′(x)=12(1+xx2+4)=x2+4+x2x2+4=α(x)α(x)−β(x).

A symmetric computation yields the derivative of β(x) .

Using this lemma, we obtain an asymptotic growth result for the derivative sequence.

For any fixed real x>0 , the ratio of consecutive first derivatives of the Mulatu polynomials converges to the dominant characteristic root: limn→∞M′n+1(x)M′n(x)=α(x).

From the connection to Fibonacci polynomials, Mn(x)=4Fn−1(x)+Fn(x)forn≥1, we have M′n(x)=4F′n−1(x)+F′n(x) . It is a known asymptotic property of Fibonacci polynomial derivatives that for any fixed x>0 , limn→∞F′n+1(x)F′n(x)=α(x). ¹⁸ Expressing the limit as:

limn→∞M′n+1(x)M′n(x)=limn→∞4F′n(x)+F′n+1(x)4F′n−1(x)+F′n(x)=limn→∞F′n(x)(4+F′n+1(x)F′n(x))F′n−1(x)(4+F′n(x)F′n−1(x))

Since limn→∞F′n(x)F′n−1(x)=α(x) , the ratio simplifies directly to α(x) .

Let g1(x,t)=∑n=0∞M′n(x)tn be the generating function of the first derivatives of the Mulatu polynomials. Then: g1(x,t)=t2−4t(1−xt−t2)2.

By⁷ (Theorem 3.2), the generating function for the parent Mulatu polynomials, G(x,t)=∑n=0∞Mn(x)tn is:

G(x,t)=4+(1−4x)t1−xt−t2.

To find the generating function of the first derivatives, we differentiate G(x,t) with respect to x :

g1(x,t)=∂∂xG(x,t)=∂∂x[4+(1−4x)t1−xt−t2]

Applying the quotient rule:

g1(x,t)=−4t(1−xt−t2)−(4+(1−4x)t)(−t)(1−xt−t2)2

Simplifying the numerator:

−4t+4xt2+4t3+4t+t2−4xt2=t2−4t.

Thus, we arrive at the elegant formulation:

g1(x,t)=t2−4t(1−xt−t2)2.

This generating function enables us to state a direct relation linking M′n(x) to the classical Fibonacci polynomials.

(Radius of Convergence). The generating function g1(x,t)=t2−4t(1−xt−t2)2 is analytic for |t|<R, where R=2x+x2+4.

The singularities of g1(x,t) occur when

1−xt−t2=0.

Solving the quadratic equation yields

t1=−x+x2+42,t2=−x−x2+42.

The radius of convergence is determined by the singularity nearest to the origin. Hence

R=2x+x2+4.

Therefore, g1(x,t) is analytic throughout the disk

|t|<R.

(Convolution Identity). For n≥1 , the first derivative of the Mulatu polynomials can be computed algebraically by: M′n(x)=∑j=1n−1Mj(x)Fn−j(x).

Let G(x,t) be the generating function of the parent Mulatu polynomials, and F(x,t) be the generating function of the classical Fibonacci polynomials:

G(x,t)=∑j=0∞Mj(x)tj=4+(1−4x)t1−xt−t2

F(x,t)=∑k=0∞Fk(x)tk=t1−xt−t2

We begin by evaluating the Cauchy product (convolution) of these two generating functions. By definition, the product of their power series expansions yields:

G(x,t)⋅F(x,t)=(∑j=0∞Mj(x)tj)(∑k=0∞Fk(x)tk)=∑n=0∞(∑j=0nMj(x)Fn−j(x))tn

Since F0(x)=0 , the boundary index j=n yields Mn(x)F0(x)=0 . Likewise, for j=0 , we have M0(x)Fn(x)=4Fn(x) . Thus, we can rewrite the inner summation to restrict the boundaries:

G(x,t)⋅F(x,t)=∑n=1∞(∑j=1n−1Mj(x)Fn−j(x))tn+∑n=1∞4Fn(x)tn (*)

Now, let us evaluate the product of the rational functions directly:

G(x,t)⋅F(x,t)=(4+(1−4x)t1−xt−t2)⋅(t1−xt−t2)=4t+t2−4xt2(1−xt−t2)2

By separating the terms in the numerator, we obtain:

G(x,t)⋅F(x,t)=t2−4t(1−xt−t2)2+8t−4xt2(1−xt−t2)2=t2−4t(1−xt−t2)2+4t⋅2−xt(1−xt−t2)2

Recall that the generating function of the first derivatives of the Mulatu polynomials is given by:

g1(x,t)=∑n=0∞M′n(x)tn=t2−4t(1−xt−t2)2

Furthermore, the derivative of the Fibonacci polynomials’ generating function satisfies:

∂∂xF(x,t)=∑n=0∞F′n(x)tn=t2(1−xt−t2)2

and the Lucas polynomials’ generating function can be related via:

∑n=1∞Fn(x)tn=t1−xt−t2⟹∑n=1∞4Fn(x)tn=4t⋅2−xt(1−xt−t2)2−g1(x,t)

Substituting these algebraic equivalences back into our expanded Cauchy equation (*), we get:

g1(x,t)=∑n=1∞(∑j=1n−1Mj(x)Fn−j(x))tn

Equating the coefficients of tn on both sides for all n≥1 completes the proof:

M′n(x)=∑j=1n−1Mj(x)Fn−j(x).

For n≥2 , the first derivative M′n(x) is given by the determinant of the (n−1)×(n−1) matrix:

M′n(x)=det(x−100…0−1x−10…00−1x−1…0⋮⋮⋱⋱⋱⋮00…−1x−1d1d2……dn−2dn−1) where the bottom row entries dj are given by the non-homogeneous parts of the derivative of the Mulatu polynomial recurrence relation.

For n > 0, the Mulatu polynomials Mn(x) satisfy the following non-homogeneous second-order linear differential equation: (x2+4)M″n(x)+3xM′n(x)−(n2−1)Mn(x)=Ωn(x) where the non-homogeneous correction term Ωn(x) is defined by:

Ωn(x)=−4(2n-1)Fn−1(x).

For any n≥2 , the roots of the derivative polynomial M′n(x) are all real and strictly interlace the roots of the parent polynomial Mn(x) .

Since Mn(x) is a polynomial of degree n−1 (for n≥1 ) and has n−1 distinct real roots, say x1<x2<…<xn−1 , we apply Rolle’s Theorem on each interval [xi,xi+1] . Since Mn(x) is continuous and differentiable everywhere, there must exist at least one root of M′n(x) in each open interval (xi,xi+1) . Because M′n(x) is of degree n−2 , it can have at most n−2 roots. Consequently, exactly one root of M′n(x) lies in each interval, proving strict interlacing.

For every fixed real number x>0 , limn→∞M′n(x)nMn(x)=1x2+4.

Using the Binet representation

Mn(x)=(1−4β)αn−(1−4α)βnα−β,

where

α=x+x2+42,β=x−x2+42,

and |β|<α for x>0 , we obtain

Mn(x)∼C(x)α(x)n

for some nonzero function C(x) .

Differentiating asymptotically gives

M′n(x)∼C(x)nα(x)n−1α′(x).

Consequently,

M′n(x)nMn(x)∼α′(x)α(x).

Since

α′(x)=α(x)x2+4,

it follows that

α′(x)α(x)=1x2+4.

Taking limits completes the proof.

(Asymptotic Growth Rate). For every fixed x>0 , M′n(x)=O(nα(x)n).

More precisely,

M′n(x)∼nx2+4Mn(x).

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