Question

Let \(f_{1}(x)=x\) and \(f_{2}(x)=|x|\) for \(x \in[-1,1]\). Show that these two functions are linearly independent in the given interval, and that their Wronskian vanishes. Is this a violation of Theorem 14.4.3?

Short answer

Yes, the functions \(f_{1}(x)=x\) and \(f_{2}(x)=|x|\) are linearly independent in the interval \([-1,1]\), despite their Wronskian being zero at \(x = 0\). This observation does not violate Theorem 14.4.3 as the theorem allows zero Wronskian at isolated points, and \(x = 0\) is an isolated point in the given interval.

Step-by-step solution

Prove Linear Independence

Show that \(f_{1}(x)\) and \(f_{2}(x)\) are linearly independent. This can be achieved by showing that the equation \(c_{1}f_{1}(x) + c_{2}f_{2}(x) = 0\) for \(x \in [-1,1]\) has no solution for \(c_1\) and \(c_2\) other than \(c_1 = 0\) and \(c_2 = 0\). After substitution, you get two separate cases: \(c_1x + c_2x = 0\) for \(x\geq0\) and \(c_1x - c_2x = 0\) for \(x<0\). Solving for these cases, you find that only \(c_1 = 0\) and \(c_2 = 0\) is valid for \(x\) in \([-1,1]\). Therefore, the functions are linearly independent.

Calculate the Wronskian

The Wronskian of \(f_{1}(x)\) and \(f_{2}(x)\) is the determinant of a matrix with the first row as \(f_{1}(x)\) and \(f_{2}(x)\) and the second row as the derivatives of these functions. The derivative of \(f_{1}(x)\) is 1 and of \(f_{2}(x)\) is undefined at 0, therefore, the determinant which is the Wronskian is either 0 or undefined at \(x = 0\). So, the Wronskian indeed vanishes for these two functions within the given interval \([-1,1]\).

Interpretation Relating to Theorem 14.4.3

Theorem 14.4.3 states that if two functions are linearly independent, their Wronskian is not equal to zero except possibly at isolated points. However, the zero Wronskian observed here exists at a single point, \(x = 0\), which is an isolated point in the interval \([-1,1]\). Thus, this is not a violation but is in line with Theorem 14.4.3, even though it's an exceptional case.

Key concepts

Linear Independence

In mathematics, two functions are defined as linearly independent if there are no constants other than zero that can be multiplied with them to sum to zero. In simpler terms, one function is not a multiple of the other. For functions like \(f_{1}(x) = x\) and \(f_{2}(x) = |x|\), we check for linear independence by observing whether the only solution to \(c_{1}f_{1}(x) + c_{2}f_{2}(x) = 0\) is \(c_{1} = 0\) and \(c_{2} = 0\). In our case:

• For \(x \geq 0\), both functions reduce to \(c_{1}x + c_{2}x = 0\), which leads to \(c_{1} + c_{2} = 0\).
• For \(x < 0\), they form \(c_{1}x - c_{2}x = 0\), leading to \(c_{1} - c_{2} = 0\).

The only solution to these system of equations is \(c_{1} = 0\) and \(c_{2} = 0\), thus proving their linear independence. Both functions travel in distinct paths across the interval, affirming each other's independence.

Wronskian

The Wronskian is a fascinating tool in mathematics to help determine if functions are linearly independent. The Wronskian of two functions \(f_{1}(x)\) and \(f_{2}(x)\) is formed by placing these functions and their derivatives in a 2x2 matrix, then calculating the determinant. For our functions \(f_{1}(x) = x\) and \(f_{2}(x) = |x|\):

• The derivative of \(f_{1}(x)\) is 1.
• \(f_{2}(x)\) is not differentiable at \(x = 0\), making room for exceptional cases.

The determinant, or the Wronskian, is computed as \( \text{det} \begin{bmatrix} x & |x| \1 & \text{sign}(x) \end{bmatrix} \), which results in zero or undefined at \(x = 0\). Thus, the Wronskian vanishes except at this isolated point, marking a special circumstance for these functions.

Theorem 14.4.3

Theorem 14.4.3 is a key principle that relates linear independence to the Wronskian. This theorem states that if two functions are linearly independent, then their Wronskian is not zero except potentially at isolated points. If, for example, you have functions that meet this theorem's conditions, their Wronskian may only vanish at certain specific points, not across a continuous interval. In our case with \(f_{1}(x) = x\) and \(f_{2}(x) = |x|\), we confirm they are linearly independent; yet, the Wronskian vanishes at \(x = 0\). According to the theorem, since \(x = 0\) is an isolated point in the domain \([-1,1]\), this behavior is in adherence to Theorem 14.4.3, and not a violation. It’s essential to recognize how this illustrates a fascinating exception handled by mathematical theory.