Is There A Minimal-degree Integer Polynomial F ( X ) F(x) F ( X ) Such That F ( Cos ⁡ ( 2 Π / 13 ) ) = 13 F(\cos(2\pi/13)) = \sqrt{13} F ( Cos ( 2 Π /13 )) = 13 ​ ?

Is there a minimal-degree integer polynomial $f(x)$ such that $f(\cos(2\pi/13)) = \sqrt{13}$?

In the realm of constructive mathematics, we often delve into the intricacies of algebraic numbers and their properties. The given problem revolves around the existence of a minimal-degree integer polynomial $f(x)$ that satisfies the equation $f(\cos(2\pi/13)) = \sqrt{13}$. To tackle this problem, we need to understand the algebraic structure of $\cos(2\pi/13)$ and its relationship with the number $\sqrt{13}$.

Algebraic Structure of $\cos(2\pi/13)$

It is known that $\cos(2\pi/13)$ is an algebraic number of degree $6$ over $\mathbb{Q}$. This means that there exists a polynomial $g(x) \in \mathbb{Q}[x]$ of degree $6$ such that $g(\cos(2\pi/13)) = 0$. In other words, $\cos(2\pi/13)$ is a root of the polynomial $g(x)$.

Gaussian Period Theory

From Gaussian period theory, we have the following equation:

$$\sqrt{13} = -2 \cos \frac{2 \pi}{13}$$

This equation provides a connection between the number $\sqrt{13}$ and the value of $\cos(2\pi/13)$. We can use this equation to derive a polynomial that has $\sqrt{13}$ as a root.

Deriving a Polynomial with $\sqrt{13}$ as a Root

Let's consider the polynomial $h(x) = x^2 - 13$. This polynomial has $\sqrt{13}$ as a root, since $h(\sqrt{13}) = 0$. We can also rewrite the equation from Gaussian period theory as:

$$\sqrt{13} = -2 \cos \frac{2 \pi}{13}$$

This implies that:

$$h(-2 \cos \frac{2 \pi}{13}) = 0$$

Substituting $x = -2 \cos \frac{2 \pi}{13}$ into the polynomial $h(x)$, we get:

$$h(-2 \cos \frac{2 \pi}{13}) = (-2 \cos \frac{2 \pi}{13})^2 - 13 = 0$$

This equation shows that the polynomial $h(x)$ has $-2 \cos \frac{2 \pi}{13}$ as a root.

Constructing a Minimal-Degree Polynomial

Since $\cos(2\pi/13)$ is an algebraic number of degree $6$ over $\mathbb{Q}$, we can construct a polynomial $f(x)$ of degree $6$ that has $\cos(2\pi/13)$ as a root. We can then use this polynomial to derive a polynomial that has $\sqrt{13}$ as a root.

Let's consider the polynomial $f(x) = x^6 - 2x^3 + 1$. This polynomial has $\cos(2\pi/13)$ as a root, since $f(\cos(2\pi/13)) = 0$. We can also rewrite the equation from Gaussian period theory as:

$$\sqrt{13} = -2 \cos \frac{2 \pi}{13}$$

This implies that:

$$f(-2 \cos \frac{2 \pi}{13}) = 0$$

Substituting $x = -2 \cos \frac{2 \pi}{13}$ into the polynomial $f(x)$, we get:

$$f(-2 \cos \frac{2 \pi}{13}) = (-2 \cos \frac{2 \pi}{13})^6 - 2(-2 \cos \frac{2 \pi}{13})^3 + 1 = 0$$

This equation shows that the polynomial $f(x)$ has $-2 \cos \frac{2 \pi}{13}$ as a root.

In conclusion, we have shown that there exists a minimal-degree integer polynomial $f(x)$ such that $f(\cos(2\pi/13)) = \sqrt{13}$. The polynomial $f(x)$ is of degree $6$ and has $\cos(2\pi/13)$ as a root. We can use this polynomial to derive a polynomial that has $\sqrt{13}$ as a root.

• [1] Gaussian period theory
• [2] Algebraic structure of $\cos(2\pi/13)$

• Investigate the properties of the polynomial $f(x)$
• Derive a polynomial that has $\sqrt{13}$ as a root using the polynomial $f(x)$
• Explore the applications of the polynomial $f(x)$ in constructive mathematics
  Q&A: Is there a minimal-degree integer polynomial $f(x)$ such that $f(\cos(2\pi/13)) = \sqrt{13}$?

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A: The number $\cos(2\pi/13)$ is an algebraic number of degree $6$ over $\mathbb{Q}$. This means that there exists a polynomial $g(x) \in \mathbb{Q}[x]$ of degree $6$ such that $g(\cos(2\pi/13)) = 0$. In other words, $\cos(2\pi/13)$ is a root of the polynomial $g(x)$.

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A: The equation $\sqrt{13} = -2 \cos \frac{2 \pi}{13}$ provides a connection between the number $\sqrt{13}$ and the value of $\cos(2\pi/13)$. We can use this equation to derive a polynomial that has $\sqrt{13}$ as a root.

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A: Yes, we can derive a polynomial with $\sqrt{13}$ as a root using the equation $\sqrt{13} = -2 \cos \frac{2 \pi}{13}$. Let's consider the polynomial $h(x) = x^2 - 13$. This polynomial has $\sqrt{13}$ as a root, since $h(\sqrt{13}) = 0$. We can also rewrite the equation from Gaussian period theory as:

$$\sqrt{13} = -2 \cos \frac{2 \pi}{13}$$

This implies that:

$$h(-2 \cos \frac{2 \pi}{13}) = 0$$

Substituting $x = -2 \cos \frac{2 \pi}{13}$ into the polynomial $h(x)$, we get:

$$h(-2 \cos \frac{2 \pi}{13}) = (-2 \cos \frac{2 \pi}{13})^2 - 13 = 0$$

This equation shows that the polynomial $h(x)$ has $-2 \cos \frac{2 \pi}{13}$ as a root.

$$$$

A: Yes, we can construct a minimal-degree polynomial $f(x)$ that has $\cos(2\pi/13)$ as a root. Let's consider the polynomial $f(x) = x^6 - 2x^3 + 1$. This polynomial has $\cos(2\pi/13)$ as a root, since $f(\cos(2\pi/13)) = 0$. We can also rewrite the equation from Gaussian period theory as:

$$\sqrt{13} = -2 \cos \frac{2 \pi}{13}$$

This implies that:

$$f(-2 \cos \frac{2 \pi}{13}) = 0$$

Substituting $x = -2 \cos \frac{2 \pi}{13}$ into the polynomial $f(x)$, we get:

$$f(-2 \cos \frac{2 \pi}{13}) = (-2 \cos \frac{2 \pi}{13})^6 - 2(-2 \cos \frac{2 \pi}{13})^3 + 1 = 0$$

This equation shows that the polynomial $f(x)$ has $-2 \cos \frac{2 \pi}{13}$ as a root.

A: The result implies that there exists a minimal-degree integer polynomial $f(x)$ such that $f(\cos(2\pi/13)) = \sqrt{13}$. This has significant implications in constructive mathematics, as it provides a new way to construct polynomials with specific roots.

A: Some potential applications of this result include:

• Constructing polynomials with specific roots
• Investigating the properties of algebraic numbers
• Developing new algorithms for solving polynomial equations

A: Some potential future directions for research include:

• Investigating the properties of the polynomial $f(x)$
• Deriving a polynomial that has $\sqrt{13}$ as a root using the polynomial $f(x)$
• Exploring the applications of the polynomial $f(x)$ in constructive mathematics