Question

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

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

- 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 \( J_{0}(x) \).

- 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 \( y = x J_{1}(\alpha x) \).

- 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: \[ J_{1}(\alpha x) + \alpha x J_{1}'(\alpha x) = 0. \]

- 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: \[ J_{0}(\alpha x) \alpha x = (1 - \alpha x) J_{1}(\alpha x). \]

- 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 \( x = \frac{\beta_{n}}{\alpha} \).

- 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 \( \frac{2.4048}{3.8317}, \frac{5.5201}{3.8317}, \frac{8.6537}{3.8317} \).

- Compute Coordinates

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

- 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.

Key concepts

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 $$ y = x J_{1}(alpha x) $$.