Chapter 12: Q7P (page 590)

To show that, $\lim_{x \to 0} \frac{(x)}{x} = \frac{1}{2}$ .

Short Answer

Expert verified

Hence, $\lim_{x \to 0} (\frac{J_{1} (x)}{x}) = \frac{1}{2}$

Step by step solution

Concept of the Property of Jp(x) :

Infinite series:

$J_{p} (x) = \sum_{n = 0}^{\infty} \frac{{(- 1)}^{n}}{⌜ (n + 1) ⌜ (n + 1 + p)} {(\frac{x}{2})}^{2 n + p}$

Calculation of the property of Jα :

The statement is,

$\lim_{x \to 0} (\frac{J_{1} (x)}{x})$ .

According to the property of $J_{α}$ :

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

Now substituting, 1 for $α$ .

$J_{1 + 1} (x) + J_{1 - 1} (x) = \frac{2 (1)}{x} J_{1} (x) J_{2} (x) + J_{0} (x) = \frac{2}{x} J_{1} (x)$

This equation can be written as follows:

$\frac{x}{2} (J_{2} (x) + J_{0} (x) = J_{1} (x)) J_{0} (0) = 1 and J_{α} (0), α > 0$ .

Once again from the property of $J_{α}$ ;

Hence,

$J_{2} (0) = 0$

Calculation of the limx→0J1(x)x :

As per statement:

$\lim_{x \to 0} (\frac{J_{1} (x)}{x}) = \lim_{x \to 0} (\frac{\frac{x}{2} (J_{2} (x) + J_{0} (x))}{x}) = \lim_{x \to 0} (\frac{x}{2 x} (J_{2} (x) + J_{0} (x))) = \lim_{x \to 0} (\frac{J_{2} (0) + J_{0} (x)}{2}) = (\frac{0 + 1}{2})$ x

Therefore, :

$\lim_{x \to 0} (\frac{J_{1} (x)}{x}) = \frac{1}{2}$

Hence, $\lim_{x \to 0} (\frac{J_{1} (x)}{x}) = \frac{1}{2}$ is proved.

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To show that, $\lim_{x \to 0} \frac{(x)}{x} = \frac{1}{2}$ .

Short Answer

Step by step solution

Concept of the Property of Jp(x) :

Calculation of the property of Jα :

Calculation of the limx→0J1(x)x :

One App. One Place for Learning.

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To show that,limx→0(x)x=12 .

Short Answer

Step by step solution

Concept of the Property of Jp(x) :

Calculation of the property of Jα :

Calculation of the limx→0J1(x)x :

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Quantum Physics

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Study anywhere. Anytime. Across all devices.

To show that, $\lim_{x \to 0} \frac{(x)}{x} = \frac{1}{2}$ .