= 5.8. Let f₁(x) = x³ x² +1, f₂(x) = x³ + x² + x and f3 (x) = x³ + 2x²+3x+1 be three polynomials in the vector space P3 and let W Span{f(x), f₂(x), ƒ3 (x)}. Does the polynomial g(x) = 4x³ + 11x² + 11x belong to the subspace W? If yes, write g(x) as a linear combination of the vectors f₁(x), f₂ (x), f(x).

Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
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### Polynomial Subspaces and Linear Combinations

#### Problem Statement:

Let \( f_1(x) = x^3 - x^2 + 1 \), \( f_2(x) = x^3 + x^2 + x \), and \( f_3(x) = x^3 + 2x^2 + 3x + 1 \) be three polynomials in the vector space \( P_3 \), and let \( W = \text{Span}\{f_1(x), f_2(x), f_3(x)\} \).

Does the polynomial \( g(x) = 4x^3 + 11x^2 + 11x \) belong to the subspace \( W \)? If yes, write \( g(x) \) as a linear combination of the vectors \( f_1(x), f_2(x), f_3(x) \).

### Solution:

To determine if \( g(x) \) belongs to the subspace \( W \), we need to express \( g(x) \) as a linear combination of \( f_1(x), f_2(x), f_3(x) \). In other words, we need to find scalars \( a \), \( b \), and \( c \) such that:

\[ g(x) = a \cdot f_1(x) + b \cdot f_2(x) + c \cdot f_3(x) \]

where:

\[ f_1(x) = x^3 - x^2 + 1 \]
\[ f_2(x) = x^3 + x^2 + x \]
\[ f_3(x) = x^3 + 2x^2 + 3x + 1 \]
\[ g(x) = 4x^3 + 11x^2 + 11x \]

### Steps to Solve:

1. **Write the linear combination:**
   \[ a(x^3 - x^2 + 1) + b(x^3 + x^2 + x) + c(x^3 + 2x^2 + 3x + 1) \]

2. **Combine like terms:**
   \[ (a + b + c)x^3 + (-a + b + 2c)x^2 + (b + 3c)x + a + c = 4
Transcribed Image Text:### Polynomial Subspaces and Linear Combinations #### Problem Statement: Let \( f_1(x) = x^3 - x^2 + 1 \), \( f_2(x) = x^3 + x^2 + x \), and \( f_3(x) = x^3 + 2x^2 + 3x + 1 \) be three polynomials in the vector space \( P_3 \), and let \( W = \text{Span}\{f_1(x), f_2(x), f_3(x)\} \). Does the polynomial \( g(x) = 4x^3 + 11x^2 + 11x \) belong to the subspace \( W \)? If yes, write \( g(x) \) as a linear combination of the vectors \( f_1(x), f_2(x), f_3(x) \). ### Solution: To determine if \( g(x) \) belongs to the subspace \( W \), we need to express \( g(x) \) as a linear combination of \( f_1(x), f_2(x), f_3(x) \). In other words, we need to find scalars \( a \), \( b \), and \( c \) such that: \[ g(x) = a \cdot f_1(x) + b \cdot f_2(x) + c \cdot f_3(x) \] where: \[ f_1(x) = x^3 - x^2 + 1 \] \[ f_2(x) = x^3 + x^2 + x \] \[ f_3(x) = x^3 + 2x^2 + 3x + 1 \] \[ g(x) = 4x^3 + 11x^2 + 11x \] ### Steps to Solve: 1. **Write the linear combination:** \[ a(x^3 - x^2 + 1) + b(x^3 + x^2 + x) + c(x^3 + 2x^2 + 3x + 1) \] 2. **Combine like terms:** \[ (a + b + c)x^3 + (-a + b + 2c)x^2 + (b + 3c)x + a + c = 4
### Linear Independence and Function Decomposition

**Example 5.8:**

Let \( g(x) = c_1 f_1(x) + c_2 f_2(x) + c_3 f_3(x) \), for all \( x \). Simplify this to get:

\[ (c_1 + c_2 + c_3 - 4)x^3 + (-c_1 + c_2 + 2c_3 - 11)x^2 + (c_2 + 3c_3 - 11)x + c_1 + c_3 = 0, \]

for all \( x \).

Since \( 1, x, x^2, x^3 \) are linearly independent, we must have:

\[
\begin{cases}
c_1 + c_2 + c_3 - 4 = 0, \\
-c_1 + c_2 + 2c_3 - 11 = 0, \\
c_2 + 3c_3 - 11 = 0, \\
c_1 + c_3 = 0.
\end{cases}
\]

Solving these equations, we get:

\[
c_1 = -\frac{7}{3}, \quad c_2 = 4, \quad c_3 = \frac{7}{3}.
\]

Hence, \( g(x) \) is in \( W \) and 

\[ g(x) = -\frac{7}{3}f_1(x) + 4f_2(x) + \frac{7}{3}f_3(x). \]

### Explanation of the Equations:

- The problem begins by expressing \( g(x) \) as a linear combination of functions \( f_1(x), f_2(x), \) and \( f_3(x) \).
- The resulting equation is simplified and represented in a polynomial form to consider the coefficients of \( x^3, x^2, x \), and the constant term.
- Since the polynomials are linearly independent, their coefficients must equal zero.

### Solution Process Breakdown:

1. Write \( g(x) \) as a linear combination: 
\[ g(x) = c_1 f_1(x) + c_2 f_2(x) + c_3 f_3(x) \]
Transcribed Image Text:### Linear Independence and Function Decomposition **Example 5.8:** Let \( g(x) = c_1 f_1(x) + c_2 f_2(x) + c_3 f_3(x) \), for all \( x \). Simplify this to get: \[ (c_1 + c_2 + c_3 - 4)x^3 + (-c_1 + c_2 + 2c_3 - 11)x^2 + (c_2 + 3c_3 - 11)x + c_1 + c_3 = 0, \] for all \( x \). Since \( 1, x, x^2, x^3 \) are linearly independent, we must have: \[ \begin{cases} c_1 + c_2 + c_3 - 4 = 0, \\ -c_1 + c_2 + 2c_3 - 11 = 0, \\ c_2 + 3c_3 - 11 = 0, \\ c_1 + c_3 = 0. \end{cases} \] Solving these equations, we get: \[ c_1 = -\frac{7}{3}, \quad c_2 = 4, \quad c_3 = \frac{7}{3}. \] Hence, \( g(x) \) is in \( W \) and \[ g(x) = -\frac{7}{3}f_1(x) + 4f_2(x) + \frac{7}{3}f_3(x). \] ### Explanation of the Equations: - The problem begins by expressing \( g(x) \) as a linear combination of functions \( f_1(x), f_2(x), \) and \( f_3(x) \). - The resulting equation is simplified and represented in a polynomial form to consider the coefficients of \( x^3, x^2, x \), and the constant term. - Since the polynomials are linearly independent, their coefficients must equal zero. ### Solution Process Breakdown: 1. Write \( g(x) \) as a linear combination: \[ g(x) = c_1 f_1(x) + c_2 f_2(x) + c_3 f_3(x) \]
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