DeepAI

# Balancing polynomials, Fibonacci numbers and some new series for π

We evaluate some types of infinite series with balancing and Lucas-balancing polynomials in closed form. These evaluations will lead to some new curious series for π involving Fibonacci and Lucas numbers. Our findings complement those of Castellanos from 1986 and 1989.

• 1 publication
• 4 publications
10/24/2022

### Gosper's algorithm and Bell numbers

Computers are good at evaluating finite sums in closed form, but there a...
07/16/2021

### Tropical Laurent series, their tropical roots, and localization results for the eigenvalues of nonlinear matrix functions

Tropical roots of tropical polynomials have been previously studied and ...
06/06/2021

### A Generalization of Classical Formulas in Numerical Integration and Series Convergence Acceleration

Summation formulas, such as the Euler-Maclaurin expansion or Gregory's q...
01/30/2020

### The Lagrangian remainder of Taylor's series, distinguishes O(f(x)) time complexities to polynomials or not

The purpose of this letter is to investigate the time complexity consequ...
02/15/2022

### On Polynomial Ideals And Overconvergence In Tate Algebras

In this paper, we study ideals spanned by polynomials or overconvergent ...
08/01/2021

### Multidimensional Padé approximation of binomial functions: Equalities

Let ω_0,…,ω_M be complex numbers. If H_0,…,H_M are polynomials of degree...
12/12/2016

### Geographical Load Balancing across Green Datacenters

"Geographic Load Balancing" is a strategy for reducing the energy cost o...

## 1 Motivation and Preliminaries

Castellanos [1, 2] has found, among other things, the following curious series for :

 π4=∞∑n=0(−1)n2n+1F(2n+1)(2m+1)5n(2F2m+1+√F22m+1+4/5)2n+1,m≥0. (1)

When the expression simplifies to ([1, Eq. (46)] and [2, Eq. (3.4)])

 π4=√5∞∑n=0(−1)n2n+1F2n+1(23+√5)2n+1,

which has the equivalent form

 π4=√5∞∑n=0(−1)n2n+1F2n+1α4n+2. (2)

Another series involving the squares of is ([2, Eqs. (3.2) and (3.3)])

 π20=∞∑n=0(−1)n2n+1F2(2n+1)(2m+1)(t+√t2+1)2n+1,m≥0, (3)

with

 t=5F22m+14(1+√1+(1625(52F22m+1−1)/F42m+1),

which, for , simplifies to

 π20=∞∑n=0(−1)n2n+1F22n+1(3+√10)2n+1. (4)

In the identities above, (respectively ) are the famous Fibonacci (Lucas) numbers, defined for by the recursion () with initial conditions ( and ), and is the golden ratio. Castellanos’ series pose the question, whether more series of these types exist? Here, we give a positive answer to this question. Working with balancing and Lucas-balancing polynomials we derive more series for involving Fibonacci and Lucas numbers and exhibiting such a structure.

Recall that, for any integer and , balancing polynomials and Lucas-balancing polynomials are defined by the second-order homogeneous linear recurrence [4]

 un(x)=6xun−1(x)−un−2(x), (5)

but with different initial terms. Balancing polynomials start with and , while for Lucas-balancing polynomials we set , . These polynomials have been introduced as a natural extension of the popular balancing and Lucas-balancing numbers and , respectively, and must be seen as a special member of the Horadam sequence. Obviously, and . The first few polynomials are

 B0(x)=0,B1(x)=1,B2(x)=6x,B3(x)=36x2−1, B4(x)=216x3−12x,B5(x)=1296x4−108x2+1,

and

 C0(x)=1,C1(x)=3x,C2(x)=18x2−1,C3(x)=108x3−9x, C4(x)=648x4−72x2+1,C5(x)=3888x5−540x3+15x.

The closed forms, known as Binet’s formulas, for balancing and Lucas-balancing polynomials are given by

 Bn(x)=λn1(x)−λn2(x)λ1(x)−λ2(x),Cn(x)=λn1(x)+λn2(x)2, (6)

where and . We note the following properties: and for . The polynomials possess a simple connection to Chebyshev polynomials [4] via

 Bn(x)=Un−1(3x)andCn(x)=Tn(3x),

where and are Chebyshev polynomials of the first and second kind, respectively. Some other interesting properties have been discovered in the articles [5, 6, 7, 8, 9]. Balancing and Lucas-balancing polynomials are also linked in various ways to Fibonacci and Lucas numbers. Two such connections, which will be exploited in the text, are

 Bn(L2m6)=F2mnF2m,Cn(L2m6)=L2mn2, (7)

and

 B2n+1(√56F2m+1)=L(2m+1)(2n+1)L2m+1,C2n+1(√56F2m+1)=√52F(2m+1)(2n+1). (8)

These relations can be deduced from the corresponding relations between Fibonacci numbers and Chebyshev polynomials (see [2] and [4]).

In the present article, we study special infinite series involving and , respectively. We evaluate these series in closed form. Based on these results, new infinite series for with Fibonacci and Lucas numbers will be stated as an immediate consequence. These new and curious series must be seen as complements of Castellanos’ results from 1986 and 1989.

## 2 Complementing the work of Castellanos (Part 1)

The next theorem is our starting point.

###### Theorem 1.

For each real with and with , we have

 ∞∑n=0(−1)n2n+1B2n+1(x)z2n+1=12√9x2−1arctan(2√9x2−1zz2+1) (9)

and

 ∞∑n=0(−1)n2n+1C2n+1(x)z2n+1=12arctan(6xzz2−1). (10)
###### Proof.

Recall the Taylor series for the arctangent function

 arctan(z)=∞∑n=0(−1)n2n+1z2n+1,|z|≤1. (11)

Hence, for all and with ,

 ∞∑n=0(−1)n2n+1B2n+1(x)z2n+1 = 1λ1(x)−λ2(x)(arctan(λ1(x)z)−arctan(λ2(x)z)) = 12√9x2−1arctan(2√9x2−1zz2+1),

where in the last step we applied the identity

 arctan(a)±arctan(b)=arctan(a±b1∓ab). (12)

This proves the first identity. The proof of the second identity is very similar and omitted. ∎

###### Corollary 2.

The following series representation for holds

 π8=∞∑n=0(−1)n2n+1C2n+1(3+√10)2n+1. (13)
###### Remark 3.

Observe the similarity of equation (13) to Castellanos’ series (4). To be particular, the two series provide an interesting example for series whose members’ denominators and the sum are the same but the numerators are different. Namely,

 ∞∑n=0(−1)n2n+1F22n+1(3+√10)2n+1=π20=∞∑n=0(−1)n2n+125C2n+1(3+√10)2n+1.

Such series seem to be rare. Another example given by Mező [10] (and involving Lucas numbers) are the series

 ∞∑n=1Lnn2n=2⋅ln(2)=∞∑n=12n2n.

### 2.1 Series with even Fibonacci (Lucas) coefficients

Setting , we first note that

 λ1(L2m6)=L2m+√L22m−42=α2m,

with and where we have used that . We can now use (7) to obtain the relations, valid for all ,

 ∞∑n=0(−1)n2n+1F2m(2n+1)z2n+1=1√5arctan(√5F2mzz2+1), (14)

and

 ∞∑n=0(−1)n2n+1L2m(2n+1)z2n+1=arctan(L2mzz2−1). (15)

Especially, for ,

 ∞∑n=0(−1)n2n+1F2m(2n+1)L2n+12m=1√5arctan(√5F4mL22m+1), (16)

and

 ∞∑n=0(−1)n2n+1L2m(2n+1)L2n+12m=arctan(L22mL22m−1). (17)

Next, from we see that

 α2m<2F2m+F2m−1=F2m+2.

Hence, inserting , we also get the relations

 ∞∑n=0(−1)n2n+1F2m(2n+1)F2n+12m+2=1√5arctan(√5F2mF2m+2F2m+1F2m+3), (18)

and

 ∞∑n=0(−1)n2n+1L2m(2n+1)F2n+12m+2=arctan(L2mF2m+2F2mF2m+4), (19)

where we have used the Catalan identities

 F22m+2+1=F2m+3F2m+1andF22m+2−1=F2m+4F2m.

To derive Castellanos-like expressions for with Fibonacci (Lucas) coefficients, we use (14) and (15), and relate them to special arctan arguments. We start with

 √5F2mzz2+1=1andL2mzz2−1=1.

Solving gives

 z1/2=√5F2m±√5F22m−42.

But it can be checked easily that for all

 0

with and as . Similarly, solving the equation gives

 z1/2=L2m±√L22m+42

and

 z2=L2m−√L22m+42<0<α2m

Note also that while as . Hence, inserting in (15) and simplifying proves the next result.

###### Theorem 4.

For each integer the following expression for is valid

 π4=∞∑n=0(−1)n2n+1L2m(2n+1)(2L2m+√L22m+4)2n+1. (20)

Equation (20) is the first Lucas number counterpart of (1). It is worth to note, that the case yields

 π8=∞∑n=0(−1)n2n+1(11+√2)2n+1,

which can be obtained directly from (11) with and using the arctan identity

 2arctan(x)=arctan(2x1−x2),x2<1.

When , then

 π4=∞∑n=0(−1)n2n+1L4n+2(23+√13)2n+1.

In a similar manner, working with other arguments of the arctan function, we can derive the following presumably new series for involving even Lucas numbers.

###### Theorem 5.

For each integer the following expressions for are valid

 π6=∞∑n=0(−1)n2n+1L2m(2n+1)(2√3L2m+√3L22m+4)2n+1, (21)
 π12=∞∑n=0(−1)n2n+1L2m(2n+1)(2(2−√3)L2m+√L22m+4(2−√3)2)2n+1, (22)

and

 π5=∞∑n=0(−1)n2n+1L2m(2n+1)(2√5−2√5L2m+√L22m+4(5−2√5))2n+1. (23)

### 2.2 Series with odd Fibonacci (Lucas) coefficients

Setting , we observe that

 λ1(√5F2m+16)=√5F2m+1+L2m+12=α2m+1,

where again we have used . Proceeding as before, inserting the value of into (9) and (10) we obtain the following expressions via (8), valid for all ,

 ∞∑n=0(−1)n2n+1F(2m+1)(2n+1)z2n+1=1√5arctan(√5F2m+1zz2−1), (24)

and

 ∞∑n=0(−1)n2n+1L(2m+1)(2n+1)z2n+1=arctan(L2m+1zz2+1). (25)

Especially, for ,

 (26)

and

 ∞∑n=0(−1)n2n+1L(2m+1)(2n+1)L2n+12m+1=arctan(L22m+1L22m+1−1). (27)

Once more, deriving Castellanos-like expressions for is fairly simple. We start with

 √5F2m+1zz2−1=1andL2m+1zz2+1=1.

Solving gives

 z1/2=√5F2m+1±√5F22m+1+42.

But it can be checked easily that for all . Inserting the value in (24) gives Castellanos’ series (1). Similarly, solving the equation gives

 z1/2=L2m+1±√L22m+1−42

We note that, for the zeros are complex numbers with . For , the zeros are real and . Hence, no such series for will exist.

In the same manner other arctan arguments can be analyzed. Interestingly, the careful analysis shows that no series involving odd Lucas numbers will come to appearance. The next theorem contains additional odd-indexed Fibonacci series for

that we found.

###### Theorem 6.

For each integer the following expressions for are valid

 π12=√5∞∑n=0(−1)n2n+1F(2m+1)(2n+1)(2(2−√3)√5F2m+1+√5F22m+1+4(2−√3)2)2n+1, (28)
 π6=√5∞∑n=0(−1)n2n+1F(2m+1)(2n+1)(2√15F2m+1+√15F22m+1+4)2n+1, (29)

and

 π5=√5∞∑n=0(−1)n2n+1F(2m+1)(2n+1)(2(√5−2√5)√5F2m+1+√5F22m+1+4(5−2√5))2n+1. (30)

The special case of (28) when appears in one of Castellanos’ papers [1, Eq. (48)]. The generalization and the other two series seem to be new.

## 3 Complementing the work of Castellanos (Part 2)

Replacing in (9) by and , respectively, and combining the terms according to the Binet form gives

 ∞∑n=0(−1)n2n+1B22n+1(x)z2n+1=14(9x2−1)(arctan(2√9x2−1λ1(x)zz2+λ21(x))−arctan(2√9x2−1λ2(x)zz2+λ22(x)))

Now, we can apply (12) and simplify to end with the following identity involving squares of balancing polynomials

 ∞∑n=0(−1)n2n+1B22n+1(x)z2n+1=14(9x2−1)arctan(4(9x2−1)z(z2−1)z4+6(12x2−1)z2+1),z>λ21(x). (31)

Similarly, the same replacement in (10) gives

 ∞∑n=0(−1)n2n+1C22n+1(x)z2n+1=14arctan(36x2z(z2−1)z4−2(36x2−1)z2+1),z>λ21(x). (32)

To get a series for, say, we set in (31) and are left with the quadric

 z4−32z3+66z2+32z+1=0. (33)

The roots are

 8−√47−4√7−√47,8−√47+4√7−√47, 8+√47−4√7+√47,8+√47+4√7+√47.

None of the roots satisfies the condition for convergence and therefore we can conclude that no such series with exists. Setting in the above Lucas-balancing identity we obtain

 π16=∞∑n=0(−1)n2n+1C22n+1z∗2n+1,

where is a positive root of the polynomial equation

 z4−36z3−70z2+36z+1=0. (34)

Here, also all four roots of equation (34) are real

 9−7√2−3√20−14√2,9−7√2+3√20−14√2, 9+7√2−3√20+14√2,9+7√2+3√20+14√2

and the only (biggest) root that satisfies the condition is

 z∗=9+7√2+3√20+14√2=37,8254….

Hence,

 π16=∞∑n=0(−1)n2n+1C22n+1(9+7√2+3√20+14√2)2n+1. (35)

From (31) and (32), with , we arrive at

 ∞∑n=0(−1)n2n+1F22m(2n+1)z2n+1=15arctan(5F22mz(z2−1)z4+2(5F22m+1)z2+1) (36)

and

 ∞∑n=0(−1)n2n+1L22m(2n+1)z2n+1=arctan(L22mz(z2−1)z4−2(L22m−1)z2+1). (37)

To get a new Castellanos’ type series for it is necessary to study the equations

 z4−L22mz3−2(L22m−1)z2+L22mz+1=0 (38)

and

 z4−5F22mz3+2(5F22m+1)z2+5F22mz+1=0. (39)

Note that, if in (38) we set then the quartic takes the form

 z4+az3+2(a+1)z2−az+1=0. (40)

The four roots are given by

 z1/2=−a4−14√a2−8a−16±12√a2−4a+a√a2−8a−162 (41)

and

 z3/4=−a4+14√a2−8a−16±12√a2−4a−a√a2−8a−162. (42)

Finally, if we set and then

 z1/2=~t±√~t2+1andz3/4=t±√t2+1.

Among the four roots we choose the biggest root that satisfies the condition for convergence and arrive at the main theorem.

###### Theorem 7.

For each we have the following series for involving squared even-indexed Lucas numbers

 π4=∞∑n=0(−1)n2n+1L22m(2n+1)(t+√t2+1)2n+1 (43)

with

 t=t(m)=L22m+√L42m+8L22m−164. (44)

Theorem 7 is a wonderful Lucas analogue of Castellanos’ equation (3). For , and

 π16=∞∑n=0(−1)n2n+11(1+√2+√4+2√2)2n+1. (45)

For , and

 π16=∞∑n=0(−1)n2n+1L24n+242n(9+√137+3√26+2√137)2n+1. (46)

The analysis of the quartic (39) is very similar. With the relevant equation becomes

 z4+bz3+2(1−b)z2−bz+1=0. (47)

The four roots are given by

 z1/2=−b4−14√b2+8b−16±12√b2+4b+b√b2+8b−162

and

 z3/4=−b4+14√b2+8b−16±12√b2+4b−b√b2+8b−162.

These roots can be expressed as

 z1/2=~s±√~s2+1andz3/4=s±√s2+1

with and . For and we have only complex roots: