Chebyshev Polynomials Series offer a fascinating glimpse into the elegant structure underlying polynomial expansions. We’ll explore the relationship between Chebyshev polynomials of the second kind and the series expansion of ##y^k + \frac{1}{y^k}##, where ##y + \frac{1}{y} = m##. This seemingly simple equation unlocks a world of intricate patterns in polynomial coefficients, revealing connections to binomial coefficients and trigonometric identities. Understanding this relationship is key to simplifying complex calculations and gaining valuable insights into various mathematical and scientific applications.

Therefore, we’ll unpack this mathematical journey step-by-step. We will first establish the connection between the given expression and trigonometric functions, then introduce the Chebyshev Polynomials Series as the elegant solution. Furthermore, we’ll demonstrate how these polynomials, with their unique properties and recurrence relations, provide a powerful tool for expressing ##y^k + \frac{1}{y^k}## solely in terms of m. Finally, we’ll examine the relationship between the resulting polynomial coefficients and binomial coefficients, adding another layer of depth to this intriguing mathematical exploration.

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This blog post delves into a fascinating mathematical relationship between Chebyshev polynomials of the second kind and the series expansion of ##y^k + \frac{1}{y^k}##, where ##y + \frac{1}{y} = m##. We will explore this relationship, providing detailed explanations and examples, and highlighting the connection with binomial coefficients. Let’s embark on this mathematical journey!

Problem Statement

Given that ##y + \frac{1}{y} = m##, find expressions for ##y^k + \frac{1}{y^k}## in terms of ##m## for various integer values of ##k##. Analyze the patterns in the coefficients of the resulting polynomials and their relationship with Chebyshev polynomials of the second kind and binomial coefficients. Understanding this relationship provides insights into the elegant structure underlying these polynomial expansions. The problem requires a blend of algebraic manipulation, trigonometric identities, and the properties of Chebyshev polynomials.

This problem is not just a mathematical curiosity; it has applications in various fields, including physics and engineering. The ability to express powers of ##y## and ##\frac{1}{y}## in terms of ##m## simplifies calculations and provides a more manageable form for further analysis. This is particularly useful when dealing with trigonometric functions or oscillatory systems. The Chebyshev polynomials, known for their remarkable properties, provide a powerful tool for solving this problem.

Solution

Understanding the Problem

The core of the problem lies in expressing ##y^k + \frac{1}{y^k}## solely in terms of ##m##, where ##m = y + \frac{1}{y}##. This requires finding a recursive relationship or a direct formula that eliminates ##y##. The use of Chebyshev polynomials of the second kind, denoted as ##U_k(x)##, provides an elegant solution. These polynomials possess a unique structure that perfectly fits this problem. Their definition and recurrence relations are crucial to understanding the solution. The connection to trigonometric functions, specifically ##\cos(k\theta)##, will also be essential.

The Chebyshev polynomials of the second kind are defined by the recurrence relation ##U_{k+1}(x) = 2xU_k(x) – U_{k-1}(x)##, with initial conditions ##U_0(x) = 1## and ##U_1(x) = 2x##. These polynomials have numerous applications in approximation theory, numerical analysis, and signal processing. Their orthogonality properties and their ability to represent trigonometric functions make them a powerful tool in various mathematical problems. This particular problem demonstrates their effectiveness in simplifying expressions involving powers of trigonometric functions.

Solving the Problem

Step 1: Trigonometric Representation

Let ##y = e^{i\theta}##. Then ##\frac{1}{y} = e^{-i\theta}##. Substituting into ##y + \frac{1}{y} = m##, we get:

### e^{i\theta} + e^{-i\theta} = m ###

Using Euler’s formula, ##e^{i\theta} = \cos(\theta) + i\sin(\theta)## and ##e^{-i\theta} = \cos(\theta) – i\sin(\theta)##, this simplifies to ##2\cos(\theta) = m##, or ##\cos(\theta) = \frac{m}{2}##.

Step 2: Expressing ##y^k + \frac{1}{y^k}## in terms of ##\cos(k\theta)##

Now consider ##y^k + \frac{1}{y^k}##:

### y^k + \frac{1}{y^k} = e^{ik\theta} + e^{-ik\theta} = 2\cos(k\theta) ###

This shows a direct relationship between the expression we want to find and the cosine of a multiple angle. This is where the Chebyshev polynomials come into play.

Step 3: Introducing Chebyshev Polynomials

The Chebyshev polynomials of the second kind are defined such that ##U_k(\cos(\theta)) = \frac{\sin((k+1)\theta)}{\sin(\theta)}##. However, a more useful property for our problem is that ##2\cos(k\theta) = 2U_k(\cos(\theta))##. Since ##\cos(\theta) = \frac{m}{2}##, we have:

### y^k + \frac{1}{y^k} = 2\cos(k\theta) = 2U_k(\frac{m}{2}) ###

This equation elegantly expresses ##y^k + \frac{1}{y^k}## in terms of ##m## using the Chebyshev polynomial of the second kind.

Step 4: Example for k=5

Let’s consider the case where ##k=5##. The 5th Chebyshev polynomial of the second kind is:

### U_5(x) = 16x^5 – 20x^3 + 5x ###

Therefore:

### y^5 + \frac{1}{y^5} = 2U_5(\frac{m}{2}) = 2(16(\frac{m}{2})^5 – 20(\frac{m}{2})^3 + 5(\frac{m}{2})) = m^5 – 5m^3 + 5m ###

This demonstrates how to obtain the expression for a specific value of ##k##.

Step 5: Generalization and Binomial Coefficients

The coefficients of the resulting polynomials are related to binomial coefficients. While a rigorous proof is beyond the scope of this blog post, it can be shown that the general expression is given by:

### y^k + \frac{1}{y^k} = \sum_{j=0}^{\lfloor k/2 \rfloor} (-1)^j \binom{k}{2j} m^{k-2j} ###

The expression above reveals a deep connection between the powers of ##(y + \frac{1}{y})## and the Chebyshev polynomials of the second kind. The binomial coefficients dictate the weights of the various powers of ##m##, leading to the elegant structure observed in the resulting polynomials.

Final Solution

The expression for ##y^k + \frac{1}{y^k}## in terms of ##m = y + \frac{1}{y}## is given by ##2U_k(\frac{m}{2})##, where ##U_k(x)## is the ##k##th Chebyshev polynomial of the second kind. This can also be expressed using binomial coefficients as shown above.

The final solution highlights the power of Chebyshev polynomials in solving problems involving trigonometric functions and polynomial expansions. The connection to binomial coefficients adds another layer of mathematical richness to this problem.

Below are few additional problems similar to the above.

Problem 1: Find ##y^3 + \frac{1}{y^3}## if ##y + \frac{1}{y} = 2##

Using the formula, ##y^3 + \frac{1}{y^3} = m^3 – 3m = 2^3 – 3(2) = 2##.

Problem 2: Find ##y^4 + \frac{1}{y^4}## if ##y + \frac{1}{y} = 3##

Using the formula, ##y^4 + \frac{1}{y^4} = m^4 – 4m^2 + 2 = 3^4 – 4(3^2) + 2 = 47##.

Problem 3: Express ##y^6 + \frac{1}{y^6}## in terms of ##m = y + \frac{1}{y}##

Using the formula, ##y^6 + \frac{1}{y^6} = m^6 – 6m^4 + 9m^2 – 2##.

Problem 4: If ##y + \frac{1}{y} = 1##, find ##y^2 + \frac{1}{y^2}##

Using the formula, ##y^2 + \frac{1}{y^2} = m^2 – 2 = 1^2 – 2 = -1##.

Problem 5: Determine the coefficient of ##m^2## in the expansion of ##y^7 + \frac{1}{y^7}##

The coefficient is given by ##(-1)^2\binom{7}{4} = 35##.

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