Linear Algebra and Its Applications, 5th Edition
Authors: David C. Lay, Steven R. Lay, Judi J. McDonald
ISBN-13: 978-0321982384
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See our solution for Question 6E from Chapter 6.7 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 6E
Chapter:
Problem:
Exercises 3–8 refer to P2 with the inner product given by evaluation at –1, 0, and 1. (See Example 2.) Compute , for p and q in Exercise 4. Reference Example 2: Reference Exercises 4: Exercises 3–8 refer to P2 with the inner product given by evaluation at –1, 0, and 1. (See Example 2.) Reference Example 2:
Step-by-Step Solution
Given Information
We are given that: \[ p ( t ) = 3 t - t ^ { 2 } \text { and } q ( t ) = 3 + 2 t ^ { 2 } \] We have to find $\| p \|$ and $\| q \|$ at $t = - 1,0 ,$ and 1
Step-1:
The polynomials at $t _ { 0 } = - 1 , \quad t _ { 1 } = 0 , \quad t _ { 2 } = 1$ \[ \begin{array} { l } { p \left( t _ { 0 } \right) = p ( - 1 ) = - 4 q \left( t _ { 0 } \right) = q ( - 1 ) = 5 } \\ { p \left( t _ { 1 } \right) = p ( 0 ) = 0 \text { And } q \left( t _ { 1 } \right) = q ( 0 ) = 3 } \\ { p \left( t _ { 2 } \right) = p ( 1 ) = 2 \quad q \left( t _ { 2 } \right) = q ( 1 ) = 5 } \end{array} \]
Step-2: The norms
Norm of p: \[ \begin{aligned} \| p \| & = \sqrt { \langle p , p \rangle } \\ & = \sqrt { p ( - 1 ) p ( - 1 ) + p ( 0 ) p ( 0 ) + p ( 1 ) p ( 1 ) } \\ & = \sqrt { ( - 4 ) ( - 4 ) + ( 0 ) ( 0 ) + ( 2 ) ( 2 ) } \\ & = \sqrt { 20 } \end{aligned} \] Norm of q \[ \begin{aligned} \| q \| & = \sqrt { \langle q , q \rangle } \\ & = \sqrt { q ( - 1 ) q ( - 1 ) + q ( 0 ) q ( 0 ) + q ( 1 ) q ( 1 ) } \\ & = \sqrt { ( 5 ) ( 5 ) + ( 3 ) ( 3 ) + ( 5 ) ( 5 ) } \\ & = \sqrt { 59 } \end{aligned} \]
We are given that: \[ p ( t ) = 3 t - t ^ { 2 } \text { and } q ( t ) = 3 + 2 t ^ { 2 } \] We have to find $\| p \|$ and $\| q \|$ at $t = - 1,0 ,$ and 1
Step-1:
The polynomials at $t _ { 0 } = - 1 , \quad t _ { 1 } = 0 , \quad t _ { 2 } = 1$ \[ \begin{array} { l } { p \left( t _ { 0 } \right) = p ( - 1 ) = - 4 q \left( t _ { 0 } \right) = q ( - 1 ) = 5 } \\ { p \left( t _ { 1 } \right) = p ( 0 ) = 0 \text { And } q \left( t _ { 1 } \right) = q ( 0 ) = 3 } \\ { p \left( t _ { 2 } \right) = p ( 1 ) = 2 \quad q \left( t _ { 2 } \right) = q ( 1 ) = 5 } \end{array} \]
Step-2: The norms
Norm of p: \[ \begin{aligned} \| p \| & = \sqrt { \langle p , p \rangle } \\ & = \sqrt { p ( - 1 ) p ( - 1 ) + p ( 0 ) p ( 0 ) + p ( 1 ) p ( 1 ) } \\ & = \sqrt { ( - 4 ) ( - 4 ) + ( 0 ) ( 0 ) + ( 2 ) ( 2 ) } \\ & = \sqrt { 20 } \end{aligned} \] Norm of q \[ \begin{aligned} \| q \| & = \sqrt { \langle q , q \rangle } \\ & = \sqrt { q ( - 1 ) q ( - 1 ) + q ( 0 ) q ( 0 ) + q ( 1 ) q ( 1 ) } \\ & = \sqrt { ( 5 ) ( 5 ) + ( 3 ) ( 3 ) + ( 5 ) ( 5 ) } \\ & = \sqrt { 59 } \end{aligned} \]