Let B = {1, x, x², x³} be an ordered basis for P3. Find the coordinate vector of ƒ(x) = −4x³ + 1x² + 9x − 2 relative to B. fB = B
Let B = {1, x, x², x³} be an ordered basis for P3. Find the coordinate vector of ƒ(x) = −4x³ + 1x² + 9x − 2 relative to B. fB = B
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem Statement:**
Let \( B = \{1, x, x^2, x^3\} \) be an ordered basis for \( P_3 \). Find the coordinate vector of
\[ f(x) = -4x^3 + 1x^2 + 9x - 2 \]
relative to \( B \).
**Solution:**
To express the polynomial \( f(x) = -4x^3 + 1x^2 + 9x - 2 \) in terms of the basis \( B = \{1, x, x^2, x^3\} \), we identify the coefficients of each term in the polynomial. These coefficients correspond to the coordinates of the vector relative to the basis \( B \).
The polynomial can be rewritten in terms of the basis as:
\[ f(x) = -4(x^3) + 1(x^2) + 9(x) - 2(1) \]
From this expression, the coordinate vector \( f_B \) is:
\[ f_B = \begin{bmatrix} -2 \\ 9 \\ 1 \\ -4 \end{bmatrix} \]
This vector represents the coefficients of the polynomial \( f(x) \) with respect to the basis \( B \), ordered according to \( \{1, x, x^2, x^3\} \).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa35463d6-bfa8-46b5-8d12-2e8ead64b10c%2F6a168446-8657-40cf-baf8-bc0de5c1cb2c%2Fe8dgr5u_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
Let \( B = \{1, x, x^2, x^3\} \) be an ordered basis for \( P_3 \). Find the coordinate vector of
\[ f(x) = -4x^3 + 1x^2 + 9x - 2 \]
relative to \( B \).
**Solution:**
To express the polynomial \( f(x) = -4x^3 + 1x^2 + 9x - 2 \) in terms of the basis \( B = \{1, x, x^2, x^3\} \), we identify the coefficients of each term in the polynomial. These coefficients correspond to the coordinates of the vector relative to the basis \( B \).
The polynomial can be rewritten in terms of the basis as:
\[ f(x) = -4(x^3) + 1(x^2) + 9(x) - 2(1) \]
From this expression, the coordinate vector \( f_B \) is:
\[ f_B = \begin{bmatrix} -2 \\ 9 \\ 1 \\ -4 \end{bmatrix} \]
This vector represents the coefficients of the polynomial \( f(x) \) with respect to the basis \( B \), ordered according to \( \{1, x, x^2, x^3\} \).
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