Problem 81 - Calculus Early Transcendentals

Question

Bessel functions arise in the study of wave propagation in circular geometries (for example, waves on a circular drum head). They are conveniently defined as power series. One of an infinite family of Bessel functions is $$J_{0}(x)=\sum_{k=0}^{\infty} \frac{(-1)^{k}}{2^{2 k}(k !)^{2}} x^{2 k}.$$ a. Write out the first four terms of \(J_{0}\) b. Find the radius and interval of convergence of the power series for \(J_{0}\) c. Differentiate \(J_{0}\) twice and show (by keeping terms through \(x^{6}\) ) that \(J_{0}\) satisfies the equation \(x^{2} y^{\prime \prime}(x)+x y^{\prime}(x)+x^{2} y(x)=0\)

Step-by-Step Solution

Answer

Question: Write out the first four terms of \(J_{0}(x)\) and find the radius and interval of convergence of the power series for \(J_{0}(x)\). Verify that \(J_{0}(x)\) satisfies the following second-order differential equation by differentiating it twice and keeping terms through \(x^{6}\): $$x^2 y^{\prime \prime}(x) + xy^{\prime}(x) + x^2 y(x) = 0$$ Answer: The first four terms of \(J_{0}(x)\) are: $$J_{0}(x) = 1-\frac{1}{2^2}x^2+\frac{1}{2^4\cdot2!^2}x^4-\frac{1}{2^6\cdot3!^2}x^6+\cdots$$ The radius of convergence is infinite and the interval of convergence is \((-\infty, \infty)\). By differentiating \(J_{0}(x)\) twice and substituting the resulting expressions into the given differential equation, we have verified that \(J_{0}(x)\) satisfies the second-order differential equation.

Step 1: a. First four terms of \(J_{0}(x)\)

To find the first four terms of \(J_{0}(x)\), we need to substitute the first four values of \(k\) into the given power series formula: $$J_{0}(x) = \sum_{k=0}^{\infty} \frac{(-1)^{k}}{2^{2 k}(k !)^{2}} x^{2 k}.$$ For \(k=0,1,2,3\), we have: $$J_{0}(x) = \frac{(-1)^{0}}{2^{0}(0!)^{2}} x^{0} + \frac{(-1)^{1}}{2^{2}(1!)^{2}} x^{2} + \frac{(-1)^{2}}{2^{4}(2!)^{2}} x^{4} + \frac{(-1)^{3}}{2^{6}(3!)^{2}} x^{6} + \cdots$$ Simplifying this, we get: $$J_{0}(x) = 1-\frac{1}{2^2}x^2+\frac{1}{2^4\cdot2!^2}x^4-\frac{1}{2^6\cdot3!^2}x^6+\cdots$$

Step 2: b. Radius and interval of convergence

To find the radius and interval of convergence for the power series, we'll use the Ratio Test. Applying this test to \(J_{0}(x)\), we compute the limit: $$\lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right| = \lim_{k \to \infty} \frac{\frac{(-1)^{k+1}}{2^{2(k+1)}((k+1)!)^{2}}x^{2(k+1)}}{\frac{(-1)^{k}}{2^{2k}(k!)^{2}}x^{2k}}$$ This limit simplifies to: $$\lim_{k \to \infty} \frac{(k!)^{2}2^{2k}}{(k+1)^{2}2^{2(k+1)}}x^{2} = \lim_{k \to \infty} \frac{2}{(k+1)^{2}}x^2$$ For the series to converge, this must satisfy: $$\lim_{k \to \infty} \frac{2}{(k+1)^{2}}x^2 < 1$$ Dividing both sides by 2, we get: $$\lim_{k \to \infty} \frac{1}{(k+1)^{2}}x^2 < \frac{1}{2}$$ Since \((k+1)^2\) goes to infinity as \(k\) goes to infinity, the limit will be zero for any finite value of \(x\). Therefore, the radius of convergence, \(R\), is infinite. Consequently, the interval of convergence is \((-\infty, \infty)\).

Step 3: c. Verifying the differential equation

We are asked to differentiate \(J_{0}(x)\) twice and show that it satisfies the following differential equation, keeping terms through \(x^{6}\): $$x^2 y^{\prime \prime}(x) + xy^{\prime}(x) + x^2 y(x) = 0$$ First, we find the first derivative of \(J_{0}(x)\) using the first four terms we found in part (a): $$y^\prime(x) = \frac{d}{dx}\left(1-\frac{1}{2^2}x^2+\frac{1}{2^4\cdot2!^2}x^4-\frac{1}{2^6\cdot3!^2}x^6+\cdots\right)$$ $$y^\prime(x)=-\frac{1}{2^2}(2x)+\frac{1}{2^4\cdot2!^2}(4x^3)-\frac{1}{2^6\cdot3!^2}(6x^5)+\cdots$$ Now we find the second derivative of \(J_{0}(x)\): $$y^{\prime\prime}(x) = \frac{d^2}{dx^2}\left(1-\frac{1}{2^2}x^2+\frac{1}{2^4\cdot2!^2}x^4-\frac{1}{2^6\cdot3!^2}x^6+\cdots\right)$$ $$y^{\prime\prime}(x)=-\frac{1}{2^2}(2)+\frac{1}{2^4\cdot2!^2}(12x^2)-\frac{1}{2^6\cdot3!^2}(30x^4)+\cdots$$ Now, let's substitute \(y(x), y^\prime(x),\) and \(y^{\prime\prime}(x)\) into the given differential equation and check if it is satisfied: $$x^2\left(-\frac{1}{2^2}(2)+\frac{1}{2^4\cdot2!^2}(12x^2)-\frac{1}{2^6\cdot3!^2}(30x^4)\right)+x\left(-\frac{1}{2^2}(2x)+\frac{1}{2^4\cdot2!^2}(4x^3)-\frac{1}{2^6\cdot3!^2}(6x^5)\right)+x^2\left(1-\frac{1}{2^2}x^2+\frac{1}{2^4\cdot2!^2}x^4-\frac{1}{2^6\cdot3!^2}x^6\right) = 0$$ On simplifying, we see that all terms cancel out, leaving us with: $$0 = 0$$ Thus, the given second-order differential equation is satisfied by \(J_{0}(x)\), and we have completed all tasks.

Key Concepts

Power Series

Power series are an infinite series of the form \[ \sum_{k=0}^{\infty} a_k x^k \] which can be thought of as a polynomial with infinitely many terms. A specific case of a power series, like the Bessel function \(J_0(x)\), is defined by a formula for its coefficients, such as \[J_0(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{2k}(k!)^2} x^{2k}. \] When dealing with power series, each "term" of the series is calculated based on a set pattern of coefficients and exponents of \(x\).

• The coefficient \(\frac{(-1)^k}{2^{2k}(k!)^2}\) determines the influence of each term, involving an alternating sign (as indicated by \((-1)^k\)) and factorial terms.
• A power series can be used to define functions that are not expressible by simple algebraic means, providing accurate approximations for complex curves.

Understanding the convergence of these series is crucial, as it tells us where the series reliably represents the function.

Radius of Convergence

The radius of convergence indicates the interval over which a power series converges, meaning where it has a finite sum. For the Bessel function power series \[J_0(x) = \sum_{k=0}^{\infty} \frac{(-1)^k}{2^{2k}(k!)^2} x^{2k},\] the radius of convergence is determined using the Ratio Test. This involves comparing consecutive terms in the series to see if they decrease in size.

• To find this, we compute the limit: \[ \lim_{k \to \infty} \left|\frac{a_{k+1}}{a_k}\right|, \] where \(a_k\) is the general term of the series.
• If this limit approaches a finite value less than 1, the series converges within a radius \(R\).
• In cases like the Bessel functions, where the limit approaches zero for any \(x\), \(R\) is infinite, indicating convergence for all \(x\).

In essence, if a series has an infinite radius of convergence, it can represent the function everywhere along the x-axis.

Differential Equations

Differential equations are relations between a function and its derivatives. They are frequently used to model physical systems and phenomena. In this context, Bessel functions solve a particular kind of differential equation that arises in circular wave systems. For example, the differential equation in question is \[ x^2 y^{\prime\prime}(x) + x y^{\prime}(x) + x^2 y(x) = 0, \] and the task is to verify that the Bessel function \(J_0(x)\) satisfies this equation.

• To solve or verify solutions of differential equations, we often differentiate the series (like \(J_0(x)\)) and plug the results back into the equation.
• The process involves careful differentiation and substitution, showing that all terms of the series cancel to satisfy the equation.
• Successfully applying \(J_0(x)\) shows it is a valid solution, essential for applications in physics, like the vibrational modes in circular membranes.

Differential equations form a central framework in mathematical modeling, providing substantial insight into the behavior of natural and engineered systems.