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 10E from Chapter 6.7 from Lay's Linear Algebra and Its Applications, 5th Edition.
Problem 10E
Chapter:
Problem:
Let P3 have the inner product as in Exercise 9, with p0, p1, and q the polynomials described there. Find the best approximation to by polynomials in Span . Reference: Let P3 have the inner product given by evaluation at a. Compute the orthogonal projection of p2 onto the subspace spanned by p0 and p1. b. Find a polynomial q that is orthogonal to p0 and p1, such that is an orthogonal basis for Span Scale the polynomial q so that its vector of values at .
Step-by-Step Solution
Given Information
We are given with polynomials: \[ p _ { 0 } ( t ) = 1 , p _ { 1 } ( t ) = t , \text { and } q = t ^ { 2 } - 5 \] We have to find the best approximation to $p ( t ) = t ^ { 3 }$ by polynomials in $\operatorname { Span } \left\{ p _ { 0 } , p _ { 1 } , q \right\}$
Step-1:
The polynomials at $t _ { 0 } = - 3 , t _ { 1 } = - 1 , t _ { 2 } = 1 ,$ and $t _ { 3 } = 3$ \[\begin{array}{l} p\left( {{t_0}} \right) = p( - 3) = - 27\quad \\ {p_0}\left( {{t_0}} \right) = {p_0}( - 3) = 1\quad \\ {p_1}\left( {{t_0}} \right) = {p_1}( - 3) = - 3\quad \\ q\left( {{t_0}} \right) = q( - 3) = 4\\ \\ p\left( {{t_1}} \right) = p( - 1) = - 1\\ {p_0}\left( {{t_1}} \right) = {p_0}( - 1) = 1\\ {p_1}\left( {{t_1}} \right) = {p_1}( - 1) = - 1\\ q\left( {{t_1}} \right) = q( - 1) = - 4\\ \\ p\left( {{t_2}} \right) = p(1) = 1\quad \\ {p_0}\left( {{t_2}} \right) = {p_0}(1) = 1\quad \\ {p_1}\left( {{t_2}} \right) = {p_1}(1) = 1\quad \\ q\left( {{t_2}} \right) = q(1) = - 4\\ \\ p\left( {{t_3}} \right) = p(3) = 27\\ {p_0}\left( {{t_3}} \right) = 1\\ {p_1}\left( {{t_3}} \right) = {p_1}(3) = 3\\ q\left( {{t_3}} \right) = q(3) = 4 \end{array}\]
Step-2: The inner products
\[ \langle p , q \rangle = p \left( t _ { 0 } \right) q \left( t _ { 0 } \right) + p \left( t _ { 1 } \right) q \left( t _ { 1 } \right) + p \left( t _ { 2 } \right) q \left( t _ { 2 } \right) + p \left( t _ { 3 } \right) q \left( t _ { 3 } \right) \] The inner product $\left\langle p , p _ { 0 } \right\rangle$ \[ \begin{aligned} \left\langle p \cdot p _ { 0 } \right\rangle & = p ( - 3 ) p _ { 0 } ( - 3 ) + p ( - 1 ) p _ { 0 } ( - 1 ) + p ( 1 ) p _ { 0 } ( 1 ) + p ( 3 ) p _ { 0 } ( 3 ) \\ & = ( - 27 ) ( 1 ) + ( - 1 ) ( 1 ) + ( 1 ) ( 1 ) + ( 27 ) ( 1 ) \\ & = 0 \end{aligned} \] Find the inner product $\langle p , q \rangle$ \[ \begin{aligned} \langle p , q \rangle & = p ( - 3 ) q ( - 3 ) + p ( - 1 ) q ( - 1 ) + p ( 1 ) q ( 1 ) + p ( 3 ) q ( 3 ) \\ & = ( - 27 ) ( 4 ) + ( - 1 ) ( - 4 ) + ( 1 ) ( - 4 ) + ( 27 ) ( 4 ) \\ & = 0 \end{aligned} \] Find the inner product $\left\langle p _ { 0 } , p _ { 0 } \right\rangle$ \[ \begin{aligned} \left\langle p _ { 0 } \cdot p _ { 0 } \right\rangle & = p _ { 0 } ( - 3 ) p _ { 0 } ( - 3 ) + p _ { 0 } ( - 1 ) p _ { 0 } ( - 1 ) + p _ { 0 } ( 1 ) p _ { 0 } ( 1 ) + p _ { 0 } ( 3 ) p _ { 0 } ( 3 ) \\ & = ( 1 ) ( 1 ) + ( 1 ) + ( 1 ) + ( 1 ) + ( 1 ) + ( 1 ) \\ & = 4 \end{aligned} \] Find the inner product $\left\langle p _ { 1 } , p _ { 1 } \right\rangle$ \[ \begin{aligned} \left\langle p _ { 1 } , p _ { 1 } \right\rangle & = p _ { 1 } ( - 3 ) p _ { 1 } ( - 3 ) + p _ { 1 } ( - 1 ) p _ { 1 } ( - 1 ) + p _ { 1 } ( 1 ) p _ { 1 } ( 1 ) + p _ { 1 } ( 3 ) p _ { 1 } ( 3 ) \\ & = ( - 3 ) ( - 3 ) + ( - 1 ) ( - 1 ) + ( 1 ) + ( 1 ) + ( 3 ) ( 3 ) \\ & = 20 \end{aligned} \] Find the inner product $\langle q , q \rangle$ \[ \begin{aligned} \langle q , q \rangle & = q ( - 3 ) q ( - 3 ) + q ( - 1 ) q ( - 1 ) + q ( 1 ) q ( 1 ) + q ( 3 ) q ( 3 ) \\ & = ( 4 ) ( 4 ) + ( - 4 ) ( - 4 ) + ( - 4 ) ( - 4 ) + ( 4 ) ( 4 ) \\ & = 64 \end{aligned} \]
Step-3: The best approximation
\[ \begin{aligned} \hat { p } & = \dfrac { \left\langle p , p _ { 0 } \right\rangle } { \left\langle p _ { 0 } , p _ { 0 } \right\rangle } p _ { 0 } + \dfrac { \left\langle p , p _ { 1 } \right\rangle } { \left\langle p _ { 1 } , p _ { 1 } \right\rangle } p _ { 1 } + \dfrac { \langle p , q \rangle } { \langle q , q \rangle } q \\ & = \dfrac { 0 } { 4 } ( 1 ) + \dfrac { 164 } { 20 } ( t ) + \dfrac { 0 } { 64 } \left( t ^ { 2 } - 5 \right) \\ & = \dfrac { 41 } { 5 } t \end{aligned} \] Therefore,
\[\hat p = \dfrac{{41}}{5}t\]
We are given with polynomials: \[ p _ { 0 } ( t ) = 1 , p _ { 1 } ( t ) = t , \text { and } q = t ^ { 2 } - 5 \] We have to find the best approximation to $p ( t ) = t ^ { 3 }$ by polynomials in $\operatorname { Span } \left\{ p _ { 0 } , p _ { 1 } , q \right\}$
Step-1:
The polynomials at $t _ { 0 } = - 3 , t _ { 1 } = - 1 , t _ { 2 } = 1 ,$ and $t _ { 3 } = 3$ \[\begin{array}{l} p\left( {{t_0}} \right) = p( - 3) = - 27\quad \\ {p_0}\left( {{t_0}} \right) = {p_0}( - 3) = 1\quad \\ {p_1}\left( {{t_0}} \right) = {p_1}( - 3) = - 3\quad \\ q\left( {{t_0}} \right) = q( - 3) = 4\\ \\ p\left( {{t_1}} \right) = p( - 1) = - 1\\ {p_0}\left( {{t_1}} \right) = {p_0}( - 1) = 1\\ {p_1}\left( {{t_1}} \right) = {p_1}( - 1) = - 1\\ q\left( {{t_1}} \right) = q( - 1) = - 4\\ \\ p\left( {{t_2}} \right) = p(1) = 1\quad \\ {p_0}\left( {{t_2}} \right) = {p_0}(1) = 1\quad \\ {p_1}\left( {{t_2}} \right) = {p_1}(1) = 1\quad \\ q\left( {{t_2}} \right) = q(1) = - 4\\ \\ p\left( {{t_3}} \right) = p(3) = 27\\ {p_0}\left( {{t_3}} \right) = 1\\ {p_1}\left( {{t_3}} \right) = {p_1}(3) = 3\\ q\left( {{t_3}} \right) = q(3) = 4 \end{array}\]
Step-2: The inner products
\[ \langle p , q \rangle = p \left( t _ { 0 } \right) q \left( t _ { 0 } \right) + p \left( t _ { 1 } \right) q \left( t _ { 1 } \right) + p \left( t _ { 2 } \right) q \left( t _ { 2 } \right) + p \left( t _ { 3 } \right) q \left( t _ { 3 } \right) \] The inner product $\left\langle p , p _ { 0 } \right\rangle$ \[ \begin{aligned} \left\langle p \cdot p _ { 0 } \right\rangle & = p ( - 3 ) p _ { 0 } ( - 3 ) + p ( - 1 ) p _ { 0 } ( - 1 ) + p ( 1 ) p _ { 0 } ( 1 ) + p ( 3 ) p _ { 0 } ( 3 ) \\ & = ( - 27 ) ( 1 ) + ( - 1 ) ( 1 ) + ( 1 ) ( 1 ) + ( 27 ) ( 1 ) \\ & = 0 \end{aligned} \] Find the inner product $\langle p , q \rangle$ \[ \begin{aligned} \langle p , q \rangle & = p ( - 3 ) q ( - 3 ) + p ( - 1 ) q ( - 1 ) + p ( 1 ) q ( 1 ) + p ( 3 ) q ( 3 ) \\ & = ( - 27 ) ( 4 ) + ( - 1 ) ( - 4 ) + ( 1 ) ( - 4 ) + ( 27 ) ( 4 ) \\ & = 0 \end{aligned} \] Find the inner product $\left\langle p _ { 0 } , p _ { 0 } \right\rangle$ \[ \begin{aligned} \left\langle p _ { 0 } \cdot p _ { 0 } \right\rangle & = p _ { 0 } ( - 3 ) p _ { 0 } ( - 3 ) + p _ { 0 } ( - 1 ) p _ { 0 } ( - 1 ) + p _ { 0 } ( 1 ) p _ { 0 } ( 1 ) + p _ { 0 } ( 3 ) p _ { 0 } ( 3 ) \\ & = ( 1 ) ( 1 ) + ( 1 ) + ( 1 ) + ( 1 ) + ( 1 ) + ( 1 ) \\ & = 4 \end{aligned} \] Find the inner product $\left\langle p _ { 1 } , p _ { 1 } \right\rangle$ \[ \begin{aligned} \left\langle p _ { 1 } , p _ { 1 } \right\rangle & = p _ { 1 } ( - 3 ) p _ { 1 } ( - 3 ) + p _ { 1 } ( - 1 ) p _ { 1 } ( - 1 ) + p _ { 1 } ( 1 ) p _ { 1 } ( 1 ) + p _ { 1 } ( 3 ) p _ { 1 } ( 3 ) \\ & = ( - 3 ) ( - 3 ) + ( - 1 ) ( - 1 ) + ( 1 ) + ( 1 ) + ( 3 ) ( 3 ) \\ & = 20 \end{aligned} \] Find the inner product $\langle q , q \rangle$ \[ \begin{aligned} \langle q , q \rangle & = q ( - 3 ) q ( - 3 ) + q ( - 1 ) q ( - 1 ) + q ( 1 ) q ( 1 ) + q ( 3 ) q ( 3 ) \\ & = ( 4 ) ( 4 ) + ( - 4 ) ( - 4 ) + ( - 4 ) ( - 4 ) + ( 4 ) ( 4 ) \\ & = 64 \end{aligned} \]
Step-3: The best approximation
\[ \begin{aligned} \hat { p } & = \dfrac { \left\langle p , p _ { 0 } \right\rangle } { \left\langle p _ { 0 } , p _ { 0 } \right\rangle } p _ { 0 } + \dfrac { \left\langle p , p _ { 1 } \right\rangle } { \left\langle p _ { 1 } , p _ { 1 } \right\rangle } p _ { 1 } + \dfrac { \langle p , q \rangle } { \langle q , q \rangle } q \\ & = \dfrac { 0 } { 4 } ( 1 ) + \dfrac { 164 } { 20 } ( t ) + \dfrac { 0 } { 64 } \left( t ^ { 2 } - 5 \right) \\ & = \dfrac { 41 } { 5 } t \end{aligned} \] Therefore,
\[\hat p = \dfrac{{41}}{5}t\]