Chapter 12: Q6P (page 590)

To show the following equation shown in the problem
$2 \frac{d}{d x} J_{a} (x) = J_{a - 1} (x)$ .

Short Answer

Expert verified

The answer is, $2 \frac{d}{dx} J_{a} (x) = J_{a - 1} (x)$ .

Step by step solution

Step 1: Concept of Bessel Function:

The Bessel functions of the first kind $J_{n} (x)$ are defined as the solutions to the Bessel differential equation.

$x^{2} \frac{d^{2} y}{d x^{2}} + x \frac{d y}{d x} + (x^{2} - n^{2}) y = 0$

Solve the equation 2ddxJa(x)=Ja-1(x)-Ja+1(x) about the Bessel Function:

The given equation for the following equation is shown as,

$2 \frac{d}{dx} J_{a} (x) = J_{a - 1} (x) - J_{a + 1} (x)$

By the use of the following proposition about the zeros of Bessel functions solve the equation as follows:

\frac{d}{d x} J_{α} (x) = \frac{d}{d x} (x^{a} (x^{- α} J_{α} (x))) = α x^{α - 1} (J_{α} (x)) + x^{α} \frac{d}{d x} (x^{α} J_{α} (x)) = - α x^{α - 1} J_{a} (x) + x^{- α} (x^{α} J_{α - 1} (x)) = - α x^{α - 1} J_{α} (x) + J_{a - 1} (x)

Calculation of the equation ddx(xαx-αJα(x) :

By adding the two expressions as follows:

$\frac{d}{d x} J_{a} (x) = \frac{d}{d x} (x^{α} (x^{- α} J_{α}) (x)) = α x^{α - 1} (J_{α} (x)) + x^{α} \frac{d}{d x} (x^{α} J_{α} (x)) = (J_{α - 1} (x) - J_{α + 1} (x)) / 2 2 \frac{d}{d x} J_{α} (x) = J_{a - 1} (x) - J_{a + 1} (x)$

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To show the following equation shown in the problem
$2 \frac{d}{d x} J_{a} (x) = J_{a - 1} (x)$ .

Short Answer

Step by step solution

Step 1: Concept of Bessel Function:

Solve the equation 2ddxJa(x)=Ja-1(x)-Ja+1(x) about the Bessel Function:

Calculation of the equation ddx(xαx-αJα(x) :

One App. One Place for Learning.

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To show the following equation shown in the problem2ddxJa(x)=Ja-1(x).

Short Answer

Step by step solution

Step 1: Concept of Bessel Function:

Solve the equation 2ddxJa(x)=Ja-1(x)-Ja+1(x) about the Bessel Function:

Calculation of the equation ddx(xαx-αJα(x) :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Circular Motion and Gravitation

Wave Optics

Atoms and Radioactivity

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Space Physics

Study anywhere. Anytime. Across all devices.

To show the following equation shown in the problem
$2 \frac{d}{d x} J_{a} (x) = J_{a - 1} (x)$ .