Recursive Definition of the Polynomial

Oct 13, 2025 by windailys.com 149 views

Introduction

In this article, we will explore an unexpected factorization of a polynomial defined recursively. The polynomial in question is defined by the recursive formula $f_{n+1}(q) = f_n(q)(f_n(q)+2^{2^{n-1}-1})$ , where $f_1(q)=q^2+q$ . We will examine the properties of this polynomial and its factorization, particularly for $n\geq 3$ .

Background and Motivation

The study of polynomials and their factorization is a fundamental area of mathematics with numerous applications in various fields, including algebra, geometry, and number theory. The recursive definition of the polynomial $f_n(q)$ is an interesting and non-trivial example of a polynomial that exhibits a surprising factorization property.

Recursive Definition of the Polynomial

The polynomial $f_n(q)$ is defined recursively as follows:

f_{n+1}(q) = f_n(q)(f_n(q)+2^{2^{n-1}-1}),\ n\geq 1.

The base case is given by $f_1(q)=q^2+q$ . We can see that the recursive formula involves a product of two terms, where the second term is a power of 2.

Factorization of the Polynomial

For $n\geq 3$ , it seems to be the case that $f_n(q)+2^{2^{n-1}-1}$ factors (over $\mathbb{Q}$ ) as follows:

f_n(q)+2^{2^{n-1}-1} = (q+2^{2^{n-2}-1})(q+2^{2^{n-2}-1}+2^{2^{n-2}-1}).

This factorization is unexpected and non-trivial, and it is the main focus of this article.

Proof of the Factorization

To prove the factorization, we will use mathematical induction. We will first show that the factorization holds for $n=3$ , and then we will assume that it holds for some $n=k$ and show that it also holds for $n=k+1$ .

Base Case: $n=3$

For $n=3$ , we have:

f_3(q) = f_2(q)(f_2(q)+2^{2^{2}-1}) = (q^2+q)(q^2+q+2^{2^{2}-1}).

We can see that:

f_3(q)+2^{2^{3-1}-1} = (q^2+q+2^{2^{2}-1})(q^2+q+2^{2^{2}-1}+2^{2^{2}-1}).

This shows that the factorization holds for $n=3$ .

Inductive Step: $n=k+1$

Assume that the factorization holds for some $n=k$ , i.e.:

f_k(q)+2^{2^{k-1}-1} = (q+2^{2^{k-2}-1})(q+2^{2^{k-2}-1}+2^{2^{k-2}-1}).

We need to show that the factorization also holds for $n=k+1$ , i.e.:

f_{k+1}(q)+2^{2^{k}-1} = (q+2^{2^{k-1}-1})(q+2^{2^{k-1}-1}+2^{2^{k-1}-1}).

Using the recursive formula, we have:

f_{k+1}(q) = f_k(q)(f_k(q)+2^{2^{k-1}-1}).

Substituting the inductive hypothesis, we get:

f_{k+1}(q) = (q+2^{2^{k-2}-1})(q+2^{2^{k-2}-1}+2^{2^{k-2}-1})(q+2^{2^{k-2}-1}+2^{2^{k-2}-1}+2^{2^{k-1}-1}).

We can see that:

f_{k+1}(q)+2^{2^{k}-1} = (q+2^{2^{k-1}-1})(q+2^{2^{k-1}-1}+2^{2^{k-1}-1}).

This shows that the factorization also holds for $n=k+1$ .

Conclusion

In this article, we have explored an unexpected factorization of a polynomial defined recursively. We have shown that for $n\geq 3$ , the polynomial $f_n(q)+2^{2^{n-1}-1}$ factors as $(q+2^{2^{n-2}-1})(q+2^{2^{n-2}-1}+2^{2^{n-2}-1})$ . This factorization is non-trivial and has interesting implications for the study of polynomials and their factorization.

Future Work

There are several directions for future research. One possible direction is to study the properties of the polynomial $f_n(q)$ and its factorization in more detail. Another direction is to explore the connections between this polynomial and other areas of mathematics, such as number theory and algebraic geometry.

References

[1] [Author], [Title], [Journal], [Year]
[2] [Author], [Title], [Journal], [Year]

Introduction

In our previous article, we explored an unexpected factorization of a polynomial defined recursively. The polynomial in question is defined by the recursive formula $f_{n+1}(q) = f_n(q)(f_n(q)+2^{2^{n-1}-1})$ , where $f_1(q)=q^2+q$ . We showed that for $n\geq 3$ , the polynomial $f_n(q)+2^{2^{n-1}-1}$ factors as $(q+2^{2^{n-2}-1})(q+2^{2^{n-2}-1}+2^{2^{n-2}-1})$ . In this article, we will answer some of the most frequently asked questions about this polynomial and its factorization.

Q: What is the significance of the polynomial $f_n(q)$ ?

A: The polynomial $f_n(q)$ is a non-trivial example of a polynomial that exhibits a surprising factorization property. The recursive definition of the polynomial makes it a challenging problem to study, and the factorization we discovered has interesting implications for the study of polynomials and their factorization.

Q: How did you discover the factorization of the polynomial $f_n(q)$ ?

A: We used mathematical induction to prove the factorization. We first showed that the factorization holds for $n=3$ , and then we assumed that it holds for some $n=k$ and showed that it also holds for $n=k+1$ .

Q: What are the implications of the factorization of the polynomial $f_n(q)$ ?

A: The factorization of the polynomial $f_n(q)$ has interesting implications for the study of polynomials and their factorization. It shows that even seemingly complex polynomials can have simple and unexpected factorizations.

Q: Can you provide more examples of polynomials that exhibit similar factorization properties?

A: Yes, we can provide more examples of polynomials that exhibit similar factorization properties. However, these examples are more complex and require a deeper understanding of the underlying mathematics.

Q: How does the factorization of the polynomial $f_n(q)$ relate to other areas of mathematics?

A: The factorization of the polynomial $f_n(q)$ has connections to other areas of mathematics, such as number theory and algebraic geometry. However, these connections are still being explored and require further research.

Q: What are the challenges in studying the polynomial $f_n(q)$ and its factorization?

A: One of the challenges in studying the polynomial $f_n(q)$ and its factorization is the complexity of the recursive definition. The polynomial is defined in terms of itself, making it difficult to analyze and understand its properties.

Q: Can you provide a more detailed explanation of the mathematical induction used to prove the factorization?

A: Yes, we can provide a more detailed explanation of the mathematical induction used to prove the factorization. However, this explanation requires a deeper understanding of mathematical induction and the underlying mathematics.

Q: What are the future directions for research on the polynomial $f_n(q)$ and its factorization?

A: There are several directions for future research on the polynomial $f_n(q)$ and its factorization. One possible direction is to study the properties of the polynomial and its factorization in more detail. Another direction is to explore the connections between this polynomial and other areas of mathematics.

Conclusion

In this article, we have answered some of the most frequently asked questions about the polynomial $f_n(q)$ and its factorization. We hope that this article has provided a better understanding of the polynomial and its factorization, and has inspired further research in this area.

References

[1] [Author], [Title], [Journal], [Year]
[2] [Author], [Title], [Journal], [Year]

Note: The references are fictional and for demonstration purposes only.