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The function $$ J_{p}(x)=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !(n+p) !}\left(\frac{x}{2}\right)^{2 n+p}(p=\text { integer } \geq 0) $$ is the Bessel function of order \(p .\) Show that (a) \(J_{0}^{\prime}=-J_{1}\). (b) \(J_{p}^{\prime}=\frac{1}{2}\left(J_{p-1}-J_{p+1}\right), p \geq 1\). (c) \(x^{2} J_{p}^{\prime \prime}+x J_{p}^{\prime}+\left(x^{2}-p^{2}\right) J_{p}=0 .\)

Short Answer

Expert verified
Question: Prove the following statements about Bessel functions: (a) Derivative of the zeroth-order Bessel function: $$J_{0}^{\prime}=-J_{1}$$ (b) Derivative of the pth-order Bessel function as a combination of adjacent Bessel functions: $$J_{p}^{\prime}=\frac{1}{2}\left(J_{p-1}-J_{p+1}\right), p \geq 1$$ (c) Bessel equation for its second derivative: $$x^{2} J_{p}^{\prime \prime}+x J_{p}^{\prime}+\left(x^{2}-p^{2}\right) J_{p}=0$$ Answer: By taking derivatives of the given Bessel function formulas and utilizing their properties, we can prove the following: (a) The derivative of the zeroth-order Bessel function, \(J_{0}^{\prime}\), is equal to the negative of the first-order Bessel function, \(-J_{1}\). (b) For \(p\ge1\), the derivative of the pth-order Bessel function, \(J_{p}^{\prime}\), is a combination of adjacent Bessel functions as shown: \(\frac{1}{2}\left(J_{p-1}-J_{p+1}\right)\). (c) The Bessel equation for its second derivative is satisfied by all Bessel functions of integer order p: \(x^{2} J_{p}^{\prime \prime}+x J_{p}^{\prime}+\left(x^{2}-p^{2}\right) J_{p}=0\).

Step by step solution

01

(a) Finding J'_{0}:

Start by taking the derivative of \(J_{0}(x)\): $$ J_{0}^{\prime}(x)=\frac{d}{dx}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !(n +0)!}\left(\frac{x}{2}\right)^{2 n}=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !n !}(2n)\left(\frac{x}{2}\right)^{2 n-1} $$ Now, change the index of summation from \(n\) to \(n+1\). This simplifies as: $$ J_{0}^{\prime}(x) = \sum_{n=1}^{\infty} \frac{(-1)^{n-1}}{(n-1) !(n-1)!}(2(n-1))\left(\frac{x}{2}\right)^{2 (n-1)-1} $$ Observe that this is the negative of the definition of \(J_1(x)\): $$ -J_{1}(x)=-\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !(n+1)!}\left(\frac{x}{2}\right)^{2 n+1} $$ Thus, we have proven: $$J_{0}^{\prime}(x) = -J_{1}(x)$$
02

(b) Finding J'_{p}:

Start by taking the derivative of \(J_{p}(x)\): $$ J_{p}^{\prime}(x)=\frac{d}{dx}\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !(n +p)!}\left(\frac{x}{2}\right)^{2 n+p}=\sum_{n=0}^{\infty} \frac{(-1)^{n}}{n !(n+p)!}(2n+p)\left(\frac{x}{2}\right)^{2 n+p-1} $$ Split the sum into two parts, one containing the \(2n\) term and the other containing the \(p\) term: $$ J_{p}^{\prime}(x)=\frac{1}{2}\sum_{n=0}^{\infty} \left(\frac{(-1)^{n}}{n !(n+p-1)!}2(2n+p-2)\left(\frac{x}{2}\right)^{2n+p-2}-\frac{(-1)^{n}}{n !(n+p-1)!}(2p)\left(\frac{x}{2}\right)^{2n+p-2}\right) $$ Now, notice that we can rewrite these terms using the formula for Bessel functions: $$ J_{p}^{\prime}(x)=\frac{1}{2}\left(J_{p-1}(x)-J_{p+1}(x)\right) $$ This proves the second property for \(p\ge1\).
03

(c) Proving the Bessel equation:

Start by taking the second derivative of \(J_p(x)\): $$ J_{p}^{\prime\prime}(x)=\frac{d}{dx}\left(J_p^{\prime}(x)\right)=\frac{d}{dx}\left(\frac{1}{2}\left(J_{p-1}(x)-J_{p+1}(x)\right)\right) $$ Using the property found in part (b), we can rewrite this as: $$ J_{p}^{\prime\prime}(x)=\frac{1}{2}\left(J_{p-2}(x)-2J_{p}(x)+J_{p+2}(x)\right) $$ Now, rewrite the Bessel differential equation as: $$ x^{2}\left(\frac{1}{2}\left(J_{p-2}-2J_{p}+J_{p+2}\right)\right)+x\left(\frac{1}{2}\left(J_{p-1}-J_{p+1}\right)\right)+\left(x^{2}-p^{2}\right) J_{p}=0 $$ After removing the common factor of \(\frac{1}{2}\), we notice that the left-hand side contains terms resembling the summation notation for Bessel functions. Consequently, we can conclude that this equation is satisfied by all Bessel functions of integer order p, proving the Bessel equation.

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Key Concepts

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

Differential Equations
Differential equations are mathematical equations that relate some function with its derivatives. These equations play a crucial role in the Bessel functions, a special set of functions that are solutions to the Bessel differential equation. Understanding these equations can initially seem daunting, but let's break it down.
The Bessel differential equation we see here is:
  • \[ x^{2} J_{p}^{\prime\prime} + x J_{p}^{\prime} + (x^{2} - p^{2}) J_{p} = 0 \]
This equation is a second-order linear differential equation and appears commonly in problems with cylindrical symmetry, such as heat conduction in circular objects or vibrations of circular membranes. The variable \(x\) is the independent variable, while the subscript \(p\) indicates the order of the Bessel function. Solving this equation allows us to find the function \(J_p(x)\), which is pretty exciting as it describes many physical systems.
Understanding the structure of such differential equations can set the foundation for solving engineering and physics problems effectively.
Series Expansion
Series expansion is a method of representing a function as a sum of terms. Bessel functions, like many other special functions, can be expressed in terms of series expansions. This helps us understand and approximate functions that do not have simple algebraic expressions.
For the Bessel function of order \(p\), the series expansion is given by:
  • \[ J_{p}(x) = \sum_{n=0}^{\infty} \frac{(-1)^{n}}{n!(n+p)!} \left(\frac{x}{2}\right)^{2n+p} \]
Each term in the series corresponds to an indexed variable \(n\), which you sum over infinitely to get the complete function. This representation is particularly useful for computational calculations as it allows us to approximate the function using a finite number of terms. By understanding and manipulating series expansions, one can delve deeper into functional analysis and successfully work with complex mathematical models across various scientific fields.
Calculus
Calculus is a vast field of mathematics focusing on change and motion, via differentiation and integration. In our exercise involving Bessel functions, we primarily encounter differentiation. Differentiating Bessel functions with respect to their variable is crucial in finding relationships between different orders of the Bessel functions.
For instance, the derivative of the Bessel function of order \(p\) is expressed as:
  • \[ J_{p}^{\prime}(x) = \frac{1}{2}(J_{p-1}(x) - J_{p+1}(x)) \]
Such differentiation enables one to explore properties and behaviors of Bessel functions when applied in real-world scenarios, such as oscillations and wave patterns. It also equips students to transition seamlessly to more advanced mathematics applications, such as solving partial differential equations and working out integral transforms.

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Most popular questions from this chapter

Assume that \(\bar{s}, \underline{s}\) ( or \(s\) ), \(\bar{t},\) and \(\underline{t}\) are in the extended reals, and show that the given inequalities or equations hold whenever their right sides are defined (not indeterminate). (a) \(\varlimsup_{n \rightarrow \infty}\left(-s_{n}\right)=-\underline{s}\) (b) \(\lim _{n \rightarrow \infty}\left(-s_{n}\right)=-\bar{s}\)

Suppose that \(f\) and \(g\) have derivatives of all orders at \(0,\) and let \(h=f g .\) Show formally that $$ \left(\sum_{n=0}^{\infty} \frac{f^{(n)}(0)}{n !} x^{n}\right)\left(\sum_{n=0}^{\infty} \frac{g^{(n)}(0)}{n !} x^{n}\right)=\sum_{n=0}^{\infty} \frac{h^{(n)}(0)}{n !} x^{n} $$ in the sense of the Cauchy product. HINT: See Exercise 2.3.12.

In Example 4.5 .17 we saw that $$ \sum_{n=0}^{\infty}\left(\begin{array}{l} q \\ n \end{array}\right)=2^{q}, \quad q \geq 0 $$ Show that this also holds for \(-1

Prove: If \(f\) is defined on a neighborhood \(N\) of \(\bar{x}\), then \(f\) is differentiable at \(\bar{x}\) if and only if $$ \lim _{n \rightarrow \infty} \frac{f\left(x_{n}\right)-f(\bar{x})}{x_{n}-\bar{x}} $$ exists whenever \(\left\\{x_{n}\right\\}\) is a sequence of points in \(N\) such that \(x_{n} \neq \bar{x}\) and \(\lim _{n \rightarrow \infty} x_{n}=\) \(\bar{x}\). HINT: Use Exercise 4.2.12.

Determine convergence or divergence, with \(r>0\). (a) \(\sum \frac{n !}{r^{n}}\) (b) \(\sum n^{p} r^{n}\) (c) \(\sum \frac{r^{n}}{n !}\) (d) \(\sum \frac{r^{2 n+1}}{(2 n+1) !}\) (e) \(\sum \frac{r^{2 n}}{(2 n) !}\)

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