Question

Show that \(e^{x^{2} / 2} D\left[e^{-x^{2} / 2} f(x)\right]=(D-x) f(x) .\) Now set $$f(x)=(D-x) g(x)=e^{x^{2} / 2} D\left[e^{-x^{2} / 2} g(x)\right]$$ to get $$(D-x)^{2} g(x)=e^{x^{2} / 2} D^{2}\left[e^{-x^{2} / 2} g(x)\right]$$. Continue this process to show that $$(D-x)^{n} F(x)=e^{x^{2} / 2} D^{n}\left[e^{-x^{2} / 2} F(x)\right]$$ for any \(F(x) .\) Then let \(F(x)=e^{-x^{2} / 2}\) to get (22.11).

Short answer

Short

Step-by-step solution

Prove the Initial Statement

We start by proving that

Define \f(x) as a Differential Operator

Let's define i step, w

Apply the Definition and Simplify

Now, substituteln

Differentiate Further

Apply the differentiatioind

Generalize for ⁿ

Generalize this process for any

Key concepts

Exponential functions

Exponential functions are fundamental in mathematics and are defined as functions of the form \(e^x\), where \(e\) is Euler's number, approximately equal to 2.71828. These functions grow rapidly and are widely used in various fields such as physics, engineering, and economics.

A key property of exponential functions is that their rate of change is proportional to their current value. This property makes them very suitable for modeling growth and decay processes.

In our exercise, we encounter the term \(e^{x^2 / 2}\). This is an exponential function where the exponent is a quadratic expression. Such generalizations of exponential functions appear frequently in mathematics and have specific applications.

When dealing with these types of exponential functions, especially in the context of differential operators, understanding their properties deeply helps in solving complex problems. The key to this exercise is manipulating these exponential terms correctly to demonstrate the required identities.

Differentiation

Differentiation is a fundamental concept in calculus that deals with the rate at which a function is changing at any given point. The operator \(D\) is commonly used to denote differentiation with respect to \(x\).

In the given exercise, we need to differentiate functions embedded within exponential terms. The challenge comes from applying the product rule and the chain rule correctly. The product rule states that \( (uv)' = u'v + uv' \), while the chain rule is used for compositions of functions and is written as \( (f(g(x)))' = f'(g(x)) g'(x) \).

For instance, we need to deal with differentiating expressions of the form \(e^{-x^2 / 2} g(x)\). Applying the product rule will involve computing the derivatives of both \(e^{-x^2 / 2}\) and \(g(x)\), and then combining these results appropriately. This allows us to work through the core parts of the problem step by step.

Understanding how to handle differentiation in these contexts is crucial for solving higher-level mathematical problems.

Mathematical proof

Mathematical proof is a logical argument demonstrating the truth of a given statement based on previously established statements and axioms.

In our exercise, we need to prove the initial identity stated in the problem. To do this, we start by manipulating the differential operators and exponential terms step by step.

First, we prove that \(e^{x^2 / 2} D[e^{-x^2 / 2} f(x)] = (D-x) f(x)\). This requires applying the product rule of differentiation carefully and simplifying the resulting terms.

We then define \(f(x) = (D-x) g(x)\) and show how this can be generalized to higher derivatives. By carefully proving the identity for the base case, we create a foundation from which the generalization can be carried out. Each subsequent step builds on the previous ones, ensuring that our argument is watertight.

Mathematical proofs require rigor and a detailed understanding of the underlying concepts. In this exercise, the approach we take ensures that every step is logically sound and leads us to the final result.

Generalization in mathematics

Generalization is a powerful tool in mathematics that allows us to extend results from specific cases to more general ones.

In the context of our exercise, after proving the initial case, we generalize the result to higher-order derivatives. This is done by repeatedly applying the previous steps and showing that the pattern holds for all \(n\).

Specifically, we demonstrate that \((D - x)^2 g(x) = e^{x^2 / 2} D^2 [e^{-x^2 / 2} g(x)]\) and then generalize this to \((D - x)^n F(x) = e^{x^2 / 2} D^n [e^{-x^2 / 2} F(x)]\) for any function \(F(x)\).

Generalization requires careful handling of the terms and ensuring that each step follows logically from the previous ones. By recognizing patterns and using inductive reasoning, we extend the results to broader contexts, which is a central technique in higher mathematics.