Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 6: Problem 2

Use the Gram-Schmidt process to find the first four orthogonal polynols satisfying the following: a. Interval: \((-\infty, \infty)\) Weight Function: \(e^{-x^{2}}\). b. Interval: \((0, \infty)\) Weight Function: \(e^{-x}\).

Short Answer

Expert verified
The first four orthogonal polynomials satisfying the conditions for part (a) are \(P_{0}(x) = 1\), \(P_{1}(x) = x\), \(P_{2}(x) = x^2 - 1\), and \(P_{3}(x) = x^3 - 3x\). For part (b), those are \(Q_{0}(x) = 1\), \(Q_{1}(x) = x - 1\), \(Q_{2}(x) = x^2 - 4x + 2\), and \(Q_{3}(x) = x^3 - 9x^2 + 18x - 6\).

Step by step solution

01

Part (a): Gram-Schmidt process with interval: (-∞, ∞) and weight function: \(e^{-x^{2}}\)

Consider the first four monomials \(1,x,x^{2},x^{3}\). Use the Gram Schmidt Process:1. The first polynomial \(P_{0}(x) = 1\)2. Next, use the idea of Gram Schmidt Process to find \(P_{1}(x), P_{2}(x), P_{3}(x)\). i. Orthogonalize: Subtract off the part of \(x, x^{2}, x^{3}\) that is in the direction of \(P_{0}(x)\) ii. Normalize: Divide by its norm.3. Compute them using integrals, we get: \(P_{1}(x) = x\), \(P_{2}(x) = x^2 - 1\), \(P_{3}(x) = x^3 - 3x\)
02

Part (b): Gram-Schmidt process with interval (0, ∞) and weight function \(e^{-x}\)

Again, consider the first four monomials \(1, x, x^{2}, x^{3}\). 1. The first polynomial \(Q_{0}(x) = 1\)2. Then apply Gram Schmidt Process to find \(Q_{1}(x),Q_{2}(x),Q_{3}(x)\) following the same process as in part (a)3. Compute them using integrals, we get: \(Q_{1}(x) = x - 1\), \(Q_{2}(x) = x^2 - 4x + 2\), \(Q_{3}(x) = x^3 - 9x^2 + 18x - 6\)

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Bessel functions \(J_{p}(\lambda x)\) are solutions of \(x^{2} y^{\prime \prime}+x y^{\prime}+\left(\lambda^{2} x^{2}-p^{2}\right) y=\) 0. Assume that \(x \in(0,1)\) and that \(J_{p}(\lambda)=0\) and \(J_{p}(0)\) is finite.s a. Show that this equation can be written in the form $$ \frac{d}{d x}\left(x \frac{d y}{d x}\right)+\left(\lambda^{2} x-\frac{p^{2}}{x}\right) y=0 $$ This is the standard Sturm-Liouville form for Bessel's equation. b. Prove that $$ \int_{0}^{1} x J_{p}(\lambda x) J_{p}(\mu x) d x=0, \quad \lambda \neq \mu $$ by considering $$ \int_{0}^{1}\left[J_{p}(\mu x) \frac{d}{d x}\left(x \frac{d}{d x} J_{p}(\lambda x)\right)-J_{p}(\lambda x) \frac{d}{d x}\left(x \frac{d}{d x} J_{p}(\mu x)\right)\right] d x $$ Thus, the solutions corresponding to different eigenvalues \((\lambda, \mu)\) are orthogonal. c. Prove that $$ \int_{0}^{1} x\left[J_{p}(\lambda x)\right]^{2} d x=\frac{1}{2} J_{p+1}^{2}(\lambda)=\frac{1}{2} J_{p}^{\prime 2}(\lambda) $$

The Hermite polynomials, \(H_{n}(x)\), satisfy the following: i. \(=\int_{-\infty}^{\infty} e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n} n ! \delta_{n, m}\) ii. \(H_{n}^{\prime}(x)=2 n H_{n-1}(x)\). iii. \(H_{n+1}(x)=2 x H_{n}(x)-2 n H_{n-1}(x)\) iv. \(H_{n}(x)=(-1)^{n} e^{x^{2}} \frac{d^{n}}{d x^{n}}\left(e^{-x^{2}}\right) .\) ng these, show that a. \(H_{n}^{\prime \prime}-2 x H_{n}^{\prime}+2 n H_{n}=0\). [Use properties ii. and iii.] b. \(\int_{-\infty}^{\infty} x e^{-x^{2}} H_{n}(x) H_{m}(x) d x=\sqrt{\pi} 2^{n-1} n !\left[\delta_{m, n-1}+2(n+1) \delta_{m, n+1}\right] .\) [Use properties i. and iii.]

Prove the double factorial identities: $$ \begin{gathered} (2 n) ! !=2^{n} n ! \\ (2 n-1) ! !=\frac{(2 n) !}{2^{n} n !} \end{gathered} $$

Use integration by parts to show \(\Gamma(x+1)=x \Gamma(x)\)

Show that a Sturm-Liouville operator with periodic boundary condion \([a, b]\) is self-adjoint if and only if \(p(a)=p(b)\). [Recall that periodic dary conditions are given as \(u(a)=u(b)\) and \(\left.u^{\prime}(a)=u^{\prime}(b) .\right]\).
See all solutions

Recommended explanations on Combined Science Textbooks

Synergy

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free