Let \(\alpha\) be the first positive zero of \(J_{1}(x)\) and let \(\beta_{n}\) be the zeros of \(J_{0}(x) .\) In terms of \(\alpha\) and \(\beta_{n},\) find the values of \(x\) at the maximum and minimum points of the function \(y=x J_{1}(\alpha x) .\) By computer or tables, find the needed zeros and compute the coordinates of the maximum and minimum points on the graph of \(y(x)\) for \(x\) between 0 and \(5 .\) Computer plot \(y\) from \(x=0\) to 5 and compare your computed maximum and minimum points with what the plot shows.

Short Answer

Expert verified
Critical points of \( y = x J_{1}(\alpha x) \) occur at \( x = \frac{\beta_{n}}{\alpha} \), where \( \alpha \approx 3.8317 \) and \( \beta_{n} \) are zeros of \( J_{0}(x) \).

Step by step solution

01

- Understand the Functions Involved

The function given is \( y = x J_{1}(\alpha x) \), where \( J_{1}(x) \) is the Bessel function of the first kind and \( \alpha \) is the first positive zero of \( J_{1}(x) \). \( \beta_{n} \) are the zeros of \( J_{0}(x) \).
02

- Define Critical Points

Critical points of the function \( y \) occur where its derivative equals zero. Therefore, we need to find \( \frac{dy}{dx} = 0 \) for \( y = x J_{1}(\alpha x) \).
03

- Compute the Derivative

Using the product rule, \( \frac{dy}{dx} = J_{1}(\alpha x) + x \alpha J_{1}'(\alpha x) \). Set the derivative to zero: \[ J_{1}(\alpha x) + \alpha x J_{1}'(\alpha x) = 0. \]
04

- Simplify the Equation

Using the recurrence relation of Bessel functions, \[ J_{0}(x) = J_{1}'(x) + \frac{1}{x} J_{1}(x), \] we can substitute \( J_{1}'(\alpha x) \) in the derivative equation to get: \[ J_{1}(\alpha x) + \alpha x (J_{0}(\alpha x) - \frac{1}{\alpha x} J_{1}(\alpha x)) = 0. \] Simplifying, we get: \[ J_{0}(\alpha x) \alpha x = (1 - \alpha x) J_{1}(\alpha x). \]
05

- Find Zeros

To satisfy the equation, we need \[ J_{0}(\beta_{n}x) = 0. \] Therefore, \( \alpha x \) must be a zero of \( J_{0}(x) \), implying \( \alpha x = \beta_{n} \). Therefore, the values of \( x \) at the critical points are \( x = \frac{\beta_{n}}{\alpha} \).
06

- Determine the Values

Using tables or computer software, find the first positive zeros \( \beta_{n} \) of \( J_{0}(x) \). For example, \( \alpha \approx 3.8317 \) (the first zero of \( J_{1}(x) \)) and \( \beta_{1} \approx 2.4048, \beta_{2} \approx 5.5201, \beta_{3} \approx 8.6537 \). Consequently, the values of \( x \) for the first few critical points are approximately \( \frac{2.4048}{3.8317}, \frac{5.5201}{3.8317}, \frac{8.6537}{3.8317} \).
07

- Compute Coordinates

The y-coordinates of the maximum and minimum points can be obtained by substituting the critical \( x \) values back into the function \( y = x J_{1}(\alpha x) \).
08

- Plot and Compare

Plot the function \( y = x J_{1}(\alpha x) \) from \( x = 0 \) to 5 using computer software. Compare the computed critical points with the maxima and minima observed in the plot to verify the results.

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Key Concepts

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

Critical Points
In calculus, critical points are where a function's derivative is zero or undefined. They help identify maximums, minimums, or points of inflection. For the function given, which is $$ y = x J_{1}(alpha x) $$, critical points occur where its derivative $$ \frac{dy}{dx} $$ equals zero. This means setting the derivative, obtained using the product rule, to zero and solving for $$ x $$: \[ J_{1}(alpha x) + alpha x J_{1}'(alpha x) = 0. \] Understanding critical points helps in sketching the function accurately and identifying key features.
Zeros of Bessel Functions
Bessel functions, named after Friedrich Bessel, are solutions to Bessel's differential equation and appear in many physical situations. The zeros of these functions are crucial in various applications, such as signal processing. Specifically, for this exercise, the first positive zero of the Bessel function $$ J_{1}(x) $$, denoted as $$ alpha $$, and the zeros of $$ J_{0}(x) $$, denoted as $$ beta_n $$, are necessary for finding the critical points of the given function. The zero of $$ J_{1}(x) $$ means $$ J_{1}(alpha) = 0 $$, and similarly, the zeros of $$ J_{0}(x) $$ follow $$ J_{0}(beta_n) = 0 $$. These zeros help in solving the derived equation for $$ y = x J_{1}(alpha x) $$.
Recurrence Relation
Bessel functions have useful recurrence relations that connect them to their derivatives and other Bessel functions. For $$ J_{n}(x) $$, one important recurrence relation is: \[ J_{n-1}(x) - J_{n+1}(x) = \frac{2n}{x} J_{n}(x). \] Another key relation used in this exercise is: \[ J_{0}(x) = J_{1}'(x) + \frac{1}{x} J_{1}(x). \] This relation helps simplify the derivative calculation of the given function. By substituting $$ J_{1}'(alpha x) $$ using this recurrence relation, we get a more manageable equation that leads us to find the critical points of the function $$ y = x J_{1}(alpha x) $$.

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Most popular questions from this chapter

By a method similar to that we used to show that the \(P_{l^{\prime}}\) s are an orthogonal set of functions on \((-1,1),\) show that the solutions of \(y_{n}^{\prime \prime}=-n^{2} y_{n}\) are an orthogonal set on \((-\pi, \pi) .\) Hint: You should know what functions the solutions \(y_{n}\) are; do not use the functions themselves, but you may use their values and the values of their derivatives at \(-\pi\) and \(\pi\) to evaluate the integrated part of your equation.

Solve the following differential equations by the method of Frobenius (generalized power series). Remember that the point of doing these problems is to learn about the method (which we will use later), not just to find a solution. You may recognize some series [as we did in (11.6)] or you can check your series by expanding a computer answer. $$2 x y^{\prime \prime}+y^{\prime}+2 y=0$$

Solve the following eigenvalue problem (see end of Section 2 and Problem 11): Given the differential equation $$y^{\prime \prime}+\left(\frac{\lambda}{x}-\frac{1}{4}-\frac{l(l+1)}{x^{2}}\right) y=0$$ where \(l\) is an integer \(\geq 0,\) find values of \(\lambda\) such that \(y \rightarrow 0\) as \(x \rightarrow \infty\), and find the corresponding eigenfunctions. Hint: let \(y=x^{l+1} e^{-x / 2} v(x),\) and show that \(v(x)\) satisfies the differential equation $$x v^{\prime \prime}+(2 l+2-x) v^{\prime}+(\lambda-l-1) v=0$$ Compare (22.26) to show that if \(\lambda\) is an integer \(>l,\) there is a polynomial solution \(v(x)=L_{\lambda-l-1}^{2 l+1}(x)\).

By power series, solve the Hermite differential equation $$y^{\prime \prime}-2 x y^{\prime}+2 p y=0$$ You should find an \(a_{0}\) series and an \(a_{1}\) series as for the Legendre equation in Section 2 Show that the \(a_{0}\) series terminates when \(p\) is an even integer, and the \(a_{1}\) series terminates when \(p\) is an odd integer. Thus for each integer \(n\), the differential equation (22.14) has one polynomial solution of degree \(n\). These polynomials with \(a_{0}\) or \(a_{1}\) chosen so that the highest order term is \((2 x)^{n}\) are the Hermite polynomials. Find \(H_{0}(x), H_{1}(x),\) and \(H_{2}(x) .\) Observe that you have solved an eigenvalue problem (see end of Section 2 ), namely to find values of \(p\) for which the given differential equation has polynomial solutions, and then to find the corresponding solutions (eigenfunctions).

Solve the following differential equations by the method of Frobenius (generalized power series). Remember that the point of doing these problems is to learn about the method (which we will use later), not just to find a solution. You may recognize some series [as we did in (11.6)] or you can check your series by expanding a computer answer. $$x^{2} y^{\prime \prime}+2 x y^{\prime}-6 y=0$$

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