Explanation Given: The equation is, ∫ − 1 1 P n ( x ) q ( x ) d x = 0 ... ( 1 ) for any polynomial q ( x ) of degree at most n − 1 . Calculation: The polynomials P 0 , P 1 , ... , P n − 1 are linearly independent. Then, q ( x ) = a 0 P 0 ( x ) + a 1 P 1 ( x ) + a 2 P 2 ( x ) + ... + a n − 1 P n − 1 ( x ) Substitute the above expression for q ( x ) in equation ( 1 ) . ∫ − 1 1 P n ( x ) q ( x ) d x = ∫ − 1 1 P n ( x ) [ a 0 P 0 ( x ) + a 1 P 1 ( x ) + a 2 P 2 ( x ) + ... + a n − 1 P n − 1 ( x ) ] d x = ∫ − 1 1 [ a 0 P n ( x ) P 0 ( x ) + a 1 P n ( x ) P 1 ( x ) + a 2 P n ( x ) P 2 ( x ) + ... + a n − 1 P n ( x ) P n − 1 ( x ) ] d x = a 0 ∫ − 1 1 P n ( x ) P 0 ( x ) d x + a 1 ∫ − 1 1 P n ( x ) P 1 ( x ) d x + To determine (b) To prove: If Q n ( x ) is a polynomial of degree n such that ∫ − 1 1 Q n ( x ) P k ( x ) d x = 0 for k = 0 , 1 , ... , n − 1 , then Q n ( x ) = c P n ( x ) for some constant c.

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

7th Edition

ISBN: 9780134768717

Author: R. Kent Nagle, Edward B. Saff, Arthur David Snider

Publisher: PEARSON

See similar textbooks

Videos

Question

Chapter 8.8, Problem 31E

To determine

(a)

To show:

The orthogonality condition for Legendre polynomials implies that ∫−11Pn(x)q(x)dx=0.

To determine

(b)

To prove:

If Qn(x) is a polynomial of degree n such that ∫−11Qn(x)Pk(x)dx=0 for k=0,1,...,n−1, then Qn(x)=cPn(x) for some constant c.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 8.8, Problem 30E

Chapter 8.8, Problem 33E

Students have asked these similar questions

Consider the alphabet {a, b, c}.• Design a regular expression that recognizes all strings over {a, b, c} that have at least three nonconsec-utive c characters (two characters are non-consecutive if there is at least one character between them)and at least one a character.• Explain how your regular expression recognizes the string cbbcccac by clearly identifying which partsof the string match to the components of your regular expression

Complex Analysis 2 z3+3 Q1: Evaluate cz(z-i)² the Figure. First exam 2024-2025 dz, where C is the figure-eight contour shown in

Construct a state-level description (i.e., a state diagram with transitions) for aTuring machine that decides the language {a^(n)b^(2n)c^(n) | n ∈ N}.

Chapter 8 Solutions

Fundamentals Of Differential Equations And Boundary Value Problems Plus Mylab Math With Pearson Etext -- Title-specific Access Card Package (7th ... Fundamentals Of Differential Equations)

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, subject and related others by exploring similar questions and additional content below.

(a) To show: The orthogonality condition for Legendre polynomials implies that ∫ − 1 1 P n ( x ) q ( x ) d x = 0 .

Solution Summary: The author explains that the orthogonality condition for Legendre polynomials implies that displaystyle

Videos

(a)

To show:

The orthogonality condition for Legendre polynomials implies that ∫−11Pn(x)q(x)dx=0.

(b)

To prove:

If Qn(x) is a polynomial of degree n such that ∫−11Qn(x)Pk(x)dx=0 for k=0,1,...,n−1, then Qn(x)=cPn(x) for some constant c.

Want to see the full answer?

Chapter 8 Solutions

(a) To show: The orthogonality condition for Legendre polynomials implies that ∫ − 1 1 P n ( x ) q ( x ) d x = 0 .

Solution Summary: The author explains that the orthogonality condition for Legendre polynomials implies that displaystyle

Videos

(a) To show: The orthogonality condition for Legendre polynomials implies that ∫−11Pn(x)q(x)dx=0.

(b) To prove: If Qn(x) is a polynomial of degree n such that ∫−11Qn(x)Pk(x)dx=0 for k=0,1,...,n−1, then Qn(x)=cPn(x) for some constant c.

Want to see the full answer?

Chapter 8 Solutions

(a)

To show:

The orthogonality condition for Legendre polynomials implies that ∫−11Pn(x)q(x)dx=0.

(b)

To prove:

If Qn(x) is a polynomial of degree n such that ∫−11Qn(x)Pk(x)dx=0 for k=0,1,...,n−1, then Qn(x)=cPn(x) for some constant c.