Homework Help is Here – Start Your Trial Now!
Math
Advanced Math
THEOREM 5 Theorem on Orthogonal Polynomials The sequence of polynomials defined inductively as follows is orthogonal: Pn(x) = (x-an) Pn-1(x) — bn Pn-2(x) (n ≥ 2) with po(x) = 1, p₁(x) = x − a₁, and an = (xPn-1, Pn-1)/(Pn-1, Pn-1) bn = (xPn-1, Pn-2)/(Pn-2, Pn-2)

THEOREM 5 Theorem on Orthogonal Polynomials The sequence of polynomials defined inductively as follows is orthogonal: Pn(x) = (x-an) Pn-1(x) — bn Pn-2(x) (n ≥ 2) with po(x) = 1, p₁(x) = x − a₁, and an = (xPn-1, Pn-1)/(Pn-1, Pn-1) bn = (xPn-1, Pn-2)/(Pn-2, Pn-2)

Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
13th Edition
ISBN:9781133382119
Author:Swokowski
Publisher:Swokowski
Chapter3: Functions And Graphs
Section3.2: Graphs Of Equations
Problem 78E
See similar textbooks
Question

How do you find the legendre polynomial p3(x), p4(x), p5(x)?

**Legendre Polynomials Calculation** To obtain the Legendre polynomials, we use Theorem 5 with the inner product defined as: \[ \int_{-1}^{1} f(x)g(x) \, dx \] The initial steps in this calculation are as follows: - \( p_0(x) = 1 \) - Calculate \( a_1 \): \[ a_1 = \frac{\langle p_0, p_0 \rangle}{\langle p_0, p_0 \rangle} = 0 \] - Define \( p_1(x) = x \) - Calculate \( a_2 \): \[ a_2 = \frac{\langle p_1, p_1 \rangle}{\langle p_1, p_1 \rangle} = 0 \] - Calculate \( b_2 \): \[ b_2 = \frac{\langle p_1, p_0 \rangle}{\langle p_0, p_0 \rangle} = \frac{1}{3} \] - Determine \( p_2(x) \): \[ p_2(x) = x^2 - \frac{1}{3} \] These calculations outline the process of deriving the first few Legendre polynomials using the specified inner product.
Transcribed Image Text:**Legendre Polynomials Calculation** To obtain the Legendre polynomials, we use Theorem 5 with the inner product defined as: \[ \int_{-1}^{1} f(x)g(x) \, dx \] The initial steps in this calculation are as follows: - \( p_0(x) = 1 \) - Calculate \( a_1 \): \[ a_1 = \frac{\langle p_0, p_0 \rangle}{\langle p_0, p_0 \rangle} = 0 \] - Define \( p_1(x) = x \) - Calculate \( a_2 \): \[ a_2 = \frac{\langle p_1, p_1 \rangle}{\langle p_1, p_1 \rangle} = 0 \] - Calculate \( b_2 \): \[ b_2 = \frac{\langle p_1, p_0 \rangle}{\langle p_0, p_0 \rangle} = \frac{1}{3} \] - Determine \( p_2(x) \): \[ p_2(x) = x^2 - \frac{1}{3} \] These calculations outline the process of deriving the first few Legendre polynomials using the specified inner product.
**Theorem 5: Theorem on Orthogonal Polynomials** The sequence of polynomials defined inductively as follows is orthogonal: \[ p_n(x) = (x - a_n)p_{n-1}(x) - b_n p_{n-2}(x) \quad (n \geq 2) \] with \( p_0(x) = 1 \), \( p_1(x) = x - a_1 \), and \[ a_n = \frac{\langle x p_{n-1}, p_{n-1} \rangle}{\langle p_{n-1}, p_{n-1} \rangle} \] \[ b_n = \frac{\langle x p_{n-1}, p_{n-2} \rangle}{\langle p_{n-2}, p_{n-2} \rangle} \] **Explanation:** This theorem describes a method to build a sequence of orthogonal polynomials using an inductive approach. Starting with initial conditions \( p_0(x) \) and \( p_1(x) \), each subsequent polynomial \( p_n(x) \) is constructed based on its predecessors, minus terms involving coefficients \( a_n \) and \( b_n \). The orthogonality ensures that each polynomial in the sequence is distinct in its relation to others in terms of their polynomial spaces.
Transcribed Image Text:**Theorem 5: Theorem on Orthogonal Polynomials** The sequence of polynomials defined inductively as follows is orthogonal: \[ p_n(x) = (x - a_n)p_{n-1}(x) - b_n p_{n-2}(x) \quad (n \geq 2) \] with \( p_0(x) = 1 \), \( p_1(x) = x - a_1 \), and \[ a_n = \frac{\langle x p_{n-1}, p_{n-1} \rangle}{\langle p_{n-1}, p_{n-1} \rangle} \] \[ b_n = \frac{\langle x p_{n-1}, p_{n-2} \rangle}{\langle p_{n-2}, p_{n-2} \rangle} \] **Explanation:** This theorem describes a method to build a sequence of orthogonal polynomials using an inductive approach. Starting with initial conditions \( p_0(x) \) and \( p_1(x) \), each subsequent polynomial \( p_n(x) \) is constructed based on its predecessors, minus terms involving coefficients \( a_n \) and \( b_n \). The orthogonality ensures that each polynomial in the sequence is distinct in its relation to others in terms of their polynomial spaces.
Expert Solution
trending now

Trending now

This is a popular solution!

steps

Step by step

Solved in 6 steps with 5 images

SEE SOLUTIONCheck out a sample Q&A here
Blurred answer
Similar questions
  • SEE MORE QUESTIONS
Recommended textbooks for you
Algebra & Trigonometry with Analytic Geometry
Algebra & Trigonometry with Analytic Geometry
Algebra
ISBN:
9781133382119
Author:
Swokowski
Publisher:
Cengage
SEE MORE TEXTBOOKS