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Chapter 12: Q7P (page 590)

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

Short Answer

Expert verified

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

Step by step solution

01

Concept of the Property of Jp(x) :

Infinite series:

$J_{p} (x) = {鈭�}_{n = 0}^{鈭�} \frac{{(- 1)}^{n}}{鈱� (n + 1) 鈱� (n + 1 + p)} {(\frac{x}{2})}^{2 n + p}$

02

Calculation of the property of Jα :

The statement is,

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

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

$J_{a + 1} (x) + J_{伪 - 1} (x) = \frac{2 a}{x} J_{a} (x)$ 鈥︹€� (1)

Now substituting, 1 for $伪$ .

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

This equation can be written as follows:

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

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

Hence,

$J_{2} (0) = 0$

03

Calculation of the limx→0J1(x)x :

As per statement:

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

Therefore, :

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

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

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Most popular questions from this chapter

To calculate the given system of equation.

$\frac{d}{dx} [{(x)}^{p} J_{p} (x)] = {(x)}^{p} J_{p - 1} (x)$

Using (17.3) and (15.1) to (15.5), find the recursion relations for $I_{p} (x)$ . In particular, show that $I_{0}^{'} = I_{1}$ .

To show the first few terms of $J_{0} (x), J_{1} (x), J_{- 1} (x), J_{2} (x), J_{- 2} (x)$ . Show that $J_{- 1} (x) = - J_{1} (x) {andJ}_{- 2} (x) = J_{2} (x)$ .

Plot

$l_{0} (x), l_{1} (x), l_{2} (x)$ , from x = 0 to 2 .
$K_{0} (x), K_{1} (x), K_{2} (x)$ , from x = 0.1 to 2 .
role="math" localid="1659270507695" $Ai (x)$ from x = -10 to 10 .
role="math" localid="1659270512971" $Bi (x)$ from x = -10 to 1 .

Find the best (in the least squares sense) second-degree polynomial approximation to each of the given functions over the interval -1<x<1.

|x|

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