Q: 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:

Question 1

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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

Accepted Answer

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	heta)##, will also be essential.

Question 2

Solving the Problem

Step 1: Trigonometric Representation

Accepted Answer

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

Question 3

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

Accepted Answer

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

Question 4

Step 3: Introducing Chebyshev Polynomials

Accepted Answer

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

Question 5

Step 4: Example for k=5

Accepted Answer

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

Question 6

Step 5: Generalization and Binomial Coefficients

Accepted Answer

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:

Question 7

Final Solution

Accepted Answer

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.

Question 8

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

Accepted Answer

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

Question 9

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

Accepted Answer

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

Question 10

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

Accepted Answer

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

Question 11

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

Accepted Answer

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

Question 12

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

Accepted Answer

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

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Trigonometric Functions

sinθ (Sine of an angle)

cosθ (Cosine of an angle)

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