Chapter 8: Problem 31

(a) Show that $$ \begin{aligned} D\left(e^{a x} y\right) &=e^{a x}(D+a) y \\ D^{2}\left(e^{a x} y\right) &=e^{a x}(D+a)^{2} y \end{aligned} $$ and so on, that is, for any positive integral $n_{y}$ $$ D^{n}\left(e^{a x} y\right)=e^{a x}(D+a)^{n} y $$ Thus show that if $L(D)$ is any polynomial in the operator $D$, then $$L(D)\left(e^{a x} y\right)=e^{\alpha x} L(D+a) y$$ This is called the exponential shift.

Short Answer

Expert verified

The exponential shift is shown by differentiating step-by-step, leading to the formula: $ L(D)(e^{ax} y)) = e^{ax} L(D+a)y $

Step by step solution

Differentiating the Product

Start by differentiating the product of functions using the product rule: \[ D(e^{ax}y) = e^{ax}D(y) + yD(e^{ax}) \]

Apply Derivative of Exponential Function

Compute the derivative of the exponential function: \[ D(e^{ax}) = ae^{ax} \] So, \[ D(e^{ax}y) = e^{ax}D(y) + y(ae^{ax}) \]

Group Terms

Factor out the common term $e^{ax}$ from the right-hand side: \[ D(e^{ax}y) = e^{ax}(D(y) + ay) \] This simplifies to: \[ D(e^{ax}y) = e^{ax}(D + a)y \]

Higher-Order Derivatives

Use the result from Step 3 to compute the second derivative: \[ D^{2}(e^{ax}y) = D(e^{ax}(D + a)y) \] Applying the product rule again: \[ D^{2}(e^{ax}y) = D(e^{ax}) \times (D + a)y + e^{ax}D((D + a)y) \]

Simplify Second Derivative

Simplify using the same approach: \[ D^{2}(e^{ax}y) = ae^{ax}(D + a)y + e^{ax}[(D + a)D(y)] \] Factor out $e^{ax}$: \[ D^{2}(e^{ax}y) = e^{ax}(a(D + a) + (D + a)D)y = e^{ax}(D + a)^{2}y \]

Generalize to Higher Derivatives

By induction, show for any positive integer $n$, \[ D^{n}(e^{ax}y) = e^{ax}(D + a)^n y \]

Applying Polynomial Operator

Let $L(D)$ be any polynomial in operator $D$. Write $L(D)$ as a sum of terms with coefficients: \[ L(D) = c_0 + c_1 D + c_2 D^2 + \ ... + c_n D^n \] Then apply each term to $e^{ax}y$ using the results from previous steps.

Finalize the Exponential Shift

Based on step 7: \[ L(D)(e^{ax}y) = c_0(e^{ax} y) + c_1(D(e^{ax} y)) + c_2(D^2(e^{ax} y)) + \ ... + c_n(D^n(e^{ax} y)) \], So, \[ L(D)(e^{ax} y) = e^{ax} (c_0 + c_1 (D+a) + c_2 (D+a)^2 + \ ... + c_n (D+a)^n ) y = e^{ax} L(D+a) y \]

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Key Concepts

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

Operator Algebra

Operator algebra is key to understanding how we manipulate and apply operators like differentiation. In mathematics, particularly in calculus, an operator is a symbol or function that signifies a mathematical operation. For example, the differentiation operator, denoted as $D$, takes a function and returns its derivative.

Understanding operator algebra helps us simplify complex expressions. For instance, when it comes to exponential functions and differentiation, knowing that $D(e^{ax}) = ae^{ax}$ allows us to combine and work with different parts of a function effectively. This is essential when dealing with higher-order derivatives or polynomial operators.

Product Rule

The product rule is a fundamental technique used in differentiation. It is crucial when differentiating the product of two functions. The product rule states:

For two functions $f(x)$ and $g(x)$: \ $ (fg)' = f'g + fg' $

This rule ensures that we account for the derivative of each function in the product. When differentiating $ e^{ax}y $, we apply the product rule:

\[ D(e^{ax}y) = e^{ax} D(y) + y D(e^{ax}) \]
Since $ D(e^{ax}) = ae^{ax} $, the expression simplifies:
\[ D(e^{ax}y) = e^{ax} (D + a)y \]
This step shows how the product rule is applied and helps us in overall simplification.

Polynomial Operators

Polynomial operators involve operators that act like polynomials. In our context, $L(D)$ is a polynomial operator where $D$ represents the differentiation operator. For example:

\[ L(D) = c_0 + c_1 D + c_2 D^2 + \ ... + c_n D^n \]

This means $L(D)$ is a sum of terms, each involving the differentiation operator raised to some power. By applying our earlier results, we can transform polynomial operators using the exponential shift theorem:

\[ L(D)(e^{ax}y) = e^{ax} L(D + a)y \]
This simplifies our computations significantly, as it shifts the polynomial operator effectively. It allows us to handle complex expressions and combinations of functions without performing tedious individual derivatives each time.

Induction

Mathematical induction is a powerful tool for proving statements involving integers. It consists of two steps:

Base Case: Verify the statement for the initial value (usually n=1).
Induction Step: Assume the statement is true for some integer k and prove it for k+1.

For instance, to show \[ D^n(e^{ax}y) = e^{ax}(D+a)^n y \] for all positive integers, we use induction:
Base Case: For $n=1$, we have already shown: \[ D(e^{ax}y) = e^{ax}(D+a)y \]
Induction Step: Assume it holds for $n=k$: \[ D^k(e^{ax}y) = e^{ax}(D+a)^k y \] Now, for $n=k+1$:
Applying differentiation to $D^k(e^{ax}y)$ gives: \[ D^{k+1}(e^{ax}y) = D(e^{ax}(D+a)^k y) \] Using the product rule again: \[ D(e^{ax}(D+a)^k y) = e^{ax}(D+a)^{k+1} y \]
Thus, by induction, the statement holds for all positive integers.