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Chapter 12: Q4P (page 599)

Show that $J_{P} (x) N'_{P} (x) - J'_{P} (x) N_{P} (x) = \frac{J'_{P} (x) J_{- p} (x) - J_{p} (x) j'_{- p} (x)}{蝉颈苍辫蟺} = \frac{2}{蟺虫}$ .

Short Answer

Expert verified

The resultant answer is $J_{P} (x) N'_{P} (x) - J'_{P} (x) N_{P} (x) = \frac{2}{蟺虫}$ .

Step by step solution

01

Concept of Equation (13.3):

Equation (13.3):

$N_{p} = \frac{\cos {蟺辫闯}_{P} (x) - J_{- p} (x)}{蝉颈苍蟺辫} N'_{p} = \frac{\cos 蟺辫闯'_{P} (x) - J'_{- p} (x)}{蝉颈苍蟺辫}$

02

Consider the equation and simplify it:

By considering the equation and simplifying it, obtain:

$J_{P} (x) N'_{P} (x) - J'_{P} (x) N_{P} (x) = \frac{\cos {蟺辫闯}_{P} (x) J'_{p} (x) - J_{P} (x) J'_{- p} (x)}{蝉颈苍蟺辫} - \frac{\cos 蟺辫闯'_{P} (x) J_{p} (x) - J'_{P} (x) J_{- p} (x)}{蝉颈苍蟺辫} = \frac{J'_{P} J_{- P} - J'_{- P} J_{P}}{蝉颈苍蟺辫} = \frac{\frac{2 \sin x p}{x x}}{蟺虫}$ Hence, it is proved that $J_{P} (x) N'_{P} (x) - J'_{P} (x) N_{P} (x) = \frac{2}{蟺虫}$ .

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Verify the recursion relations $(22.24)$ as follows:

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

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