Applying any method and show all the steps to determine whether f,(x) = cos 2x f,(x) =1 f,(x) = cosx are linearly independent.

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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**Problem Statement:**

Determine whether the functions \( f_1(x) = \cos 2x \), \( f_2(x) = 1 \), and \( f_3(x) = \cos^2 x \) are linearly independent by applying any method and showing all steps.

**Explanation:**

To determine the linear independence of the given functions, start by checking if there exist constants \( c_1, c_2, \) and \( c_3 \), not all zero, such that:

\[ c_1 f_1(x) + c_2 f_2(x) + c_3 f_3(x) = 0 \]

for all \( x \).

**Steps:**

1. **Substitute the given functions:**

   \[ c_1 \cos 2x + c_2 \cdot 1 + c_3 \cos^2 x = 0 \]

2. **Use trigonometric identities:**

   Recall the identity: \( \cos 2x = 2\cos^2 x - 1 \).

   Substitute back:

   \[ c_1 (2\cos^2 x - 1) + c_2 + c_3 \cos^2 x = 0 \]

3. **Simplify the equation:**

   \[ (2c_1 + c_3) \cos^2 x - c_1 + c_2 = 0 \]

   Set the coefficients of like terms equal to zero:

   \[
   \begin{align*}
   2c_1 + c_3 &= 0 \\
   -c_1 + c_2 &= 0
   \end{align*}
   \]

4. **Solve the system of equations:**

   From \( 2c_1 + c_3 = 0 \), we get \( c_3 = -2c_1 \).

   From \(-c_1 + c_2 = 0\), we get \( c_2 = c_1 \).

   Thus the solution is of the form \( c_1(1, 1, -2) \).

5. **Conclusion:**

   Since the constants \( c_1, c_2, \) and \( c_3 \) are not all zero, the functions \( f_1(x), f
Transcribed Image Text:**Problem Statement:** Determine whether the functions \( f_1(x) = \cos 2x \), \( f_2(x) = 1 \), and \( f_3(x) = \cos^2 x \) are linearly independent by applying any method and showing all steps. **Explanation:** To determine the linear independence of the given functions, start by checking if there exist constants \( c_1, c_2, \) and \( c_3 \), not all zero, such that: \[ c_1 f_1(x) + c_2 f_2(x) + c_3 f_3(x) = 0 \] for all \( x \). **Steps:** 1. **Substitute the given functions:** \[ c_1 \cos 2x + c_2 \cdot 1 + c_3 \cos^2 x = 0 \] 2. **Use trigonometric identities:** Recall the identity: \( \cos 2x = 2\cos^2 x - 1 \). Substitute back: \[ c_1 (2\cos^2 x - 1) + c_2 + c_3 \cos^2 x = 0 \] 3. **Simplify the equation:** \[ (2c_1 + c_3) \cos^2 x - c_1 + c_2 = 0 \] Set the coefficients of like terms equal to zero: \[ \begin{align*} 2c_1 + c_3 &= 0 \\ -c_1 + c_2 &= 0 \end{align*} \] 4. **Solve the system of equations:** From \( 2c_1 + c_3 = 0 \), we get \( c_3 = -2c_1 \). From \(-c_1 + c_2 = 0\), we get \( c_2 = c_1 \). Thus the solution is of the form \( c_1(1, 1, -2) \). 5. **Conclusion:** Since the constants \( c_1, c_2, \) and \( c_3 \) are not all zero, the functions \( f_1(x), f
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