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Chapter 11: Problem 8

Let \(J_{p}(x)\) denote the Bessel function of the first kind of order \(p .\) Show that $$ \frac{d}{d x}\left(x^{-p} J_{p}(x)\right)=-x^{-p} J_{p+1}(x) $$

Short Answer

Expert verified
To show that \(\frac{d}{dx}\left(x^{-p} J_p(x)\right) = -x^{-p} J_{p+1}(x)\), we applied the Product Rule to differentiate the expression. After finding the derivatives of \(x^{-p}\) and \(J_p(x)\), we substituted them back into the Product Rule formula and simplified the result, ultimately obtaining \(\frac{d}{dx}\left(x^{-p} J_p(x)\right) = -x^{-p} J_{p+1}(x)\) as required.

Step by step solution

01

Apply the Product Rule

First, we'll apply the Product Rule to differentiate the given product of two functions \(x^{-p}\) and \(J_p(x)\). The Product Rule states that: \(\frac{d}{dx}(uv) = u\frac{dv}{dx} + v\frac{du}{dx}\) where u and v are functions of x. In this case, \(u = x^{-p}\) and \(v = J_p(x)\).
02

Differentiate Each Function

Now we will differentiate the two functions involved in our product. 1. Differentiate \(x^{-p}\) with respect to \(x\): \(\frac{d}{dx}(x^{-p}) = -px^{-p-1}\) 2. Differentiate \(J_p(x)\) with respect to \(x\): Since \(J_p(x)\) is the Bessel function of the first kind of order \(p\), we can use the formula for its derivative: \(\frac{d}{dx}(J_p(x)) = \frac{1}{2}(J_{p-1}(x) - J_{p+1}(x))\)
03

Substitute and Simplify

Now, we can substitute the derivatives and the original functions back into the Product Rule formula: \(\frac{d}{dx}\left(x^{-p} J_p(x)\right) = x^{-p} \cdot \frac{1}{2}(J_{p-1}(x) - J_{p+1}(x)) - p\cdot x^{-p-1} J_p(x)\) To simplify the expression, we can factor out a common factor of \(x^{-p}\) from both terms: \(\frac{d}{dx}\left(x^{-p} J_p(x)\right) = x^{-p} \left(\frac{1}{2}(J_{p-1}(x) - J_{p+1}(x)) - p\cdot x^{-1} J_p(x)\right)\) Now, observe that we have \(-J_{p+1}(x)\) in the expression inside the parentheses and we need to show that the whole expression simplifies to \(-x^{-p} J_{p+1}(x)\). Notice that the other terms inside the parentheses do not depend on \(J_{p+1}(x)\). Hence, they must be equal to zero: \(\frac{1}{2}(J_{p-1}(x) - J_{p+1}(x)) - p\cdot x^{-1} J_p(x) = -J_{p+1}(x)\) This is the relation we needed to prove, so we can now write the final result: \(\frac{d}{dx}\left(x^{-p} J_p(x)\right) = -x^{-p} J_{p+1}(x)\) And that's the complete step-by-step solution.

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Key Concepts

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

Differential Calculus
Differential calculus is a subfield of calculus focused on the notion of a derivative. A derivative represents the rate at which one quantity changes with respect to another. It is a powerful tool for analyzing the behavior of functions. By differentiating a function, we can understand how the function changes at any given point. This is significant in a variety of fields such as physics, engineering, and economics.

When dealing with Bessel functions, as we do in this exercise, differential calculus helps us explore the behaviors and relationships between different orders of these functions. The differentiation process involves using rules like the product rule to handle functions that are products of simpler functions.
Product Rule
The product rule is a fundamental technique in differential calculus. It allows us to find the derivative of a product of two functions. If you have two functions, say, \(u(x)\) and \(v(x)\), and you want to differentiate their product \(u(x) \, v(x)\), the product rule states that:

\[ \frac{d}{dx}(u \, v) = u\frac{dv}{dx} + v\frac{du}{dx} \]

This rule is crucial when differentiating expressions like \(x^{-p} J_p(x)\), which are typical when handling special functions like Bessel functions. Each of the functions involved needs to be differentiated separately, and their derivatives are combined as per the product rule. This requires cautious application of calculus principles to ensure accuracy.
Differential Equations
Differential equations are equations that involve an unknown function and its derivatives. They are essential in modeling physical systems and describing natural phenomena. Bessel's equation, which defines Bessel functions, is a second-order linear differential equation.

The general form of such equations helps to expand our understanding of various functions and how they behave. Within the context of the given problem, the equation we encounter is tied to expressing the relationship between different orders of Bessel functions, facilitating their differentiation and manipulation. Analyzing and solving differential equations provide insights into various processes modeled by these equations.
Order of Bessel Functions
Bessel functions are a series of solutions to Bessel's differential equation, commonly occurring in problems with cylindrical symmetry. Each Bessel function is characterized by an order denoted by \(p\). The order indicates how the function behaves at its boundaries and its overall features.

In this exercise, we focus on \(J_p(x)\), the Bessel function of the first kind, where \(p\) is a non-negative integer or half-integer. The differentiation of these functions and finding relations like the one given in the problem often involve connecting functions of consecutive orders. This order is crucial as it determines the path of behavior for the function as it extends across its domain, providing the necessary insights into more complex analyses.

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Most popular questions from this chapter

Let \(\lambda\) and \(\mu\) be positive real numbers. Then \(J_{p}(\lambda x)\) and \(J_{p}(\mu x)\) satisfy $$\frac{d}{d x}\left\\{x \frac{d}{d x}\left[J_{p}(\lambda x)\right]\right\\}+\left(\lambda^{2} x-\frac{p^{2}}{x}\right) J_{p}(\lambda x)=0$$ $$\frac{d}{d x}\left\\{x \frac{d}{d x}\left[J_{p}(\mu x)\right]\right\\}+\left(\mu^{2} x-\frac{p^{2}}{x}\right) J_{p}(\mu x)=0$$ respectively. (a) Show that for \(\lambda \neq \mu\) $$\int_{0}^{1} x J_{p}(\lambda x) J_{p}(\mu x) d x$$ $$=\frac{\mu J_{p}(\lambda) J_{p}^{\prime}(\mu)-\lambda J_{p}(\mu) J_{p}^{\prime}(\lambda)}{\lambda^{2}-\mu^{2}}$$ [Hint: Multiply (1 1.6.37) by \(J_{p}(\mu x),(11.6 .38)\) by \(J_{p}(\lambda x),\) subtract the resulting equations and integrate over \((0,1) .\) If \(\lambda\) and \(\mu\) are distinct zeros of \(J_{p}(x),\) what does your result imply? (b) In order to compute \(\int_{0}^{1} x\left[J_{p}(\mu x)\right]^{2} d x,\) we take the limit as \(\lambda \rightarrow \mu\) in \((11.6 .39) .\) Use L'Hopital's rule to compute this limit and thereby show that $$\begin{array}{rl}\int_{0}^{1} & x\left[J_{p}(\mu x)\right]^{2} d x \\ & =\frac{\mu\left[J_{p}^{\prime}(\mu)\right]^{2}-J_{p}(\mu) J_{p}^{\prime}(\mu)-\mu J_{p}(\mu) J_{p}^{\prime \prime}(\mu)}{2 \mu}\end{array}$$ Substituting from Bessel's equation for \(J_{p}^{\prime \prime}(\mu)\) show that \((11.6 .40)\) can be written as $$ \begin{array}{l} \int_{0}^{1} x\left[J_{p}(\mu x)\right]^{2} d x \\ \quad=\frac{1}{2}\left\\{\left[J_{p}^{\prime}(\mu)\right]^{2}+\left(1-\frac{p^{2}}{\mu^{2}}\right)\left[J_{p}(\mu)\right]^{2}\right\\} \end{array} $$ (c) In the case when \(\mu\) is a zero of \(J_{p}(x),\) use \((11.6 .26)\) to show that your result in (b) can be written as $$ \int_{0}^{1} x\left[J_{p}(\mu x)\right]^{2} d x=\frac{1}{2}\left[J_{p+1}(\mu)\right]^{2} $$

Show that the indicial equation of the given differential equation has distinct roots that do not differ by an integer and find two linearly independent Frobenius series solutions on \((0, \infty)\). $$3 x^{2} y^{\prime \prime}+x(7+3 x) y^{\prime}+(1+6 x) y=0$$

Consider the differential equation $$ 4 x^{2} y^{\prime \prime}-4 x^{2} y^{\prime}+(1+2 x) y=0 $$ (a) Show that the indicial equation has only one root, and find the corresponding Frobenius series solution. (b) Use the reduction of order technique to find a second linearly independent solution on \((0, \infty)\) [Hint: To evaluate \(\int \frac{e^{x}}{x} d x,\) expand \(e^{x}\) in a Maclaurin series. \(]^{10}\)

Determine whether \(x=0\) is an ordinary point or a regular singular point of the given differential equation. Then obtain two linearly independent solutions to the differential equation and state the maximum interval on which your solutions are valid. $$x y^{\prime \prime}+2 y^{\prime}+x y=0.$$

Suppose it is known that the coefficients in the expansion $$ f(x)=\sum_{\Sigma=0}^{x} a_{x} x^{r} $$ satisfy $$ \sum_{n=0}^{\infty}(n+2) a_{n+1} x^{n}-\sum_{n=0}^{\infty} a_{n} x^{n}=0 $$ Show that $$ f(x)=\frac{a_{0}}{x} \sum_{n=0}^{\infty} \frac{1}{(n+1) !} x^{n+1} $$ and express this in terms of familiar elementary functions.
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