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Consider the subspace \(W\) of \(\mathscr{D}\), given by \(W=\operatorname{span}\left(e^{2 x}, e^{-2 x}\right)\) (a) Show that the differential operator \(D\) maps \(W\) into itself. (b) Find the matrix of \(D\) with respect to \(\mathcal{B}=\left\\{e^{2 x}, e^{-2 x}\right\\}\) (c) Compute the derivative of \(f(x)=e^{2 x}-3 e^{-2 x}\) indirectly, using Theorem \(6.26,\) and verify that it agrees with \(f^{\prime}(x)\) as computed directly

Short Answer

Expert verified
(a) D maps W into itself. (b) Matrix is \(\begin{bmatrix} 2 & 0 \\ 0 & -2 \end{bmatrix}\). (c) Derivative is \(2e^{2x} + 6e^{-2x}\), consistent with direct computation.

Step by step solution

01

Understand the Problem

We are given a subspace \(W\) of a differential function space and the differential operator \(D\) which is \(\frac{d}{dx}\). We need to show that applying \(D\) to any element in \(W\) still gives an element in \(W\). Additionally, we are asked to find the matrix representation of \(D\) with respect to a basis \(\mathcal{B}\), and finally, use a theorem to indirectly compute the derivative of a function using the matrix, verifying it with the direct computation.
02

Show D Maps W into Itself

Consider the basis functions \(e^{2x}\) and \(e^{-2x}\). Applying the differential operator \( D = \frac{d}{dx} \), we compute:\[ D(e^{2x}) = \frac{d}{dx}(e^{2x}) = 2e^{2x} \] and \[ D(e^{-2x}) = \frac{d}{dx}(e^{-2x}) = -2e^{-2x}. \] Both results are linear combinations of the elements in \(W\), confirming \(W\) is closed under \(D\).
03

Find the Matrix of D with Respect to \(\mathcal{B}\)

Construct the matrix \(A\) of \(D\) with respect to \(\mathcal{B} = \{e^{2x}, e^{-2x}\}\). Since \(D(e^{2x}) = 2e^{2x} + 0 \cdot e^{-2x}\), the first column of \(A\) is \(\begin{bmatrix} 2 \ 0 \end{bmatrix}\). Similarly, \(D(e^{-2x}) = 0 \cdot e^{2x} - 2e^{-2x}\), so the second column is \(\begin{bmatrix} 0 \ -2 \end{bmatrix}\). Hence, the matrix of \(D\) is:\[ A = \begin{bmatrix} 2 & 0 \ 0 & -2 \end{bmatrix}. \]
04

Compute Derivative Indirectly Using Theorem 6.26

Express \(f(x) = e^{2x} - 3e^{-2x}\) in terms of \(\mathcal{B}\) as \( \begin{bmatrix} 1 \ -3 \end{bmatrix} \). Applying the matrix \(A\) to this vector gives:\[ A \begin{bmatrix} 1 \ -3 \end{bmatrix} = \begin{bmatrix} 2 & 0 \ 0 & -2 \end{bmatrix} \begin{bmatrix} 1 \ -3 \end{bmatrix} = \begin{bmatrix} 2 \cdot 1 + 0 \cdot (-3) \ 0 \cdot 1 + (-2) \cdot (-3) \end{bmatrix} = \begin{bmatrix} 2 \ 6 \end{bmatrix}. \] This suggests \(f'(x) = 2e^{2x} + 6e^{-2x}\).
05

Verify by Direct Computation

Compute \(f'(x)\) directly:\[ f'(x) = \frac{d}{dx}(e^{2x} - 3e^{-2x}) = 2e^{2x} + 6e^{-2x}. \] This matches our previous indirect computation result, confirming correctness.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Differential Operators
Differential operators are fundamental in calculus and are primarily used to express the process of differentiation. In simpler terms, these operators apply derivatives to functions. The differential operator \(D\), for instance, represents taking the derivative with respect to \(x\), denoted as \(\frac{d}{dx}\). This operator can act on a wide range of functions, transforming them by computing rates of change. For example, when you apply it to a function \(e^{2x}\), it evaluates to \(2e^{2x}\) because the rate of change of \(e^{2x}\) with respect to \(x\) is twice the function itself.

Differential operators, like \(D\), help in determining how functions behave as their values change. They can map functions from one space to another, and in this case, we're investigating if applying \(D\) to a subspace \(W\) results in another function that still lies within that subspace, essentially checking the closure under \(D\). Consequently, this operation assists in exploring the stability and properties of function spaces in the context of linear algebra.
Matrix Representation
Matrix representation is an instrumental tool in linear algebra that involves representing linear transformations in a concise format. When considering transformations, such as a differential operator \(D\), on a given subspace, matrices provide a structured way to encapsulate these transformations.

For example, in the case of the operator \(D\), the matrix representation allows us to see how each basis function transforms. If \(\mathcal{B} = \{e^{2x}, e^{-2x}\}\) is our basis, a matrix \(A\) helps in understanding how these functions change under \(D\). With this basis, finding \(A\) involves computing how each function in \(\mathcal{B}\) is influenced by \(D\), organizing these results into columns. For \(D(e^{2x})=2e^{2x}\), the matrix entry becomes \(\begin{bmatrix} 2 \ 0 \end{bmatrix}\), meaning \(e^{2x}\) maps to itself scaled by 2, and similarly for \(e^{-2x}\).

This process facilitates easy calculations when applying the transformation to any linear combination of the basis functions, which is helpful in more complex calculations and theoretical explorations, such as indirect computation of derivatives.
Basis Functions
Basis functions are the building blocks of vector spaces in linear algebra. They provide a means to constructively represent any element of the space through linear combinations. In our given subspace \(W\), the basis functions are \(\{e^{2x}, e^{-2x}\}\). These functions are invaluable because any function in \(W\) can be expressed as a combination of these basis elements, much like representing vectors in terms of coordinate axes.

Recognizing basis functions is crucial since they simplify complex functions into comprehensible components. For instance, if you have a function like \(f(x) = e^{2x} - 3e^{-2x}\), it can be broken down into its basis components \(1 \times e^{2x}\) and \(-3 \times e^{-2x}\).

This decomposition can be elegantly expressed in vector terms relative to the basis, making further operations, such as applying the matrix representation of a differential operator \(D\), straightforward. The basis provides a scaffold for understanding the structure of a vector space, facilitating easier manipulation and transformation of its elements, ensuring that operations like differentiation preserve the structure contained within \(W\).

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