Chapter 12: Q4P (page 599)

Show that $J_{P} (x) N'_{P} (x) - J'_{P} (x) N_{P} (x) = \frac{J'_{P} (x) J_{- p} (x) - J_{p} (x) j'_{- p} (x)}{sinpπ} = \frac{2}{πx}$ .

Short Answer

Expert verified

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

Step by step solution

Concept of Equation (13.3):

Equation (13.3):

$N_{p} = \frac{\cos {πpJ}_{P} (x) - J_{- p} (x)}{sinπp} N'_{p} = \frac{\cos πpJ'_{P} (x) - J'_{- p} (x)}{sinπp}$

Consider the equation and simplify it:

By considering the equation and simplifying it, obtain:

$J_{P} (x) N'_{P} (x) - J'_{P} (x) N_{P} (x) = \frac{\cos {πpJ}_{P} (x) J'_{p} (x) - J_{P} (x) J'_{- p} (x)}{sinπp} - \frac{\cos πpJ'_{P} (x) J_{p} (x) - J'_{P} (x) J_{- p} (x)}{sinπp} = \frac{J'_{P} J_{- P} - J'_{- P} J_{P}}{sinπp} = \frac{\frac{2 \sin x p}{x x}}{πx}$ Hence, it is proved that $J_{P} (x) N'_{P} (x) - J'_{P} (x) N_{P} (x) = \frac{2}{πx}$ .

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Show that $J_{P} (x) N'_{P} (x) - J'_{P} (x) N_{P} (x) = \frac{J'_{P} (x) J_{- p} (x) - J_{p} (x) j'_{- p} (x)}{sinpπ} = \frac{2}{πx}$ .

Short Answer

Step by step solution

Concept of Equation (13.3):

Consider the equation and simplify it:

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Show thatJP(x)N'P(x)-J'P(x)NP(x)=J'P(x)J-p(x)-Jp(x)j'-p(x)sinpπ=2πx.

Short Answer

Step by step solution

Concept of Equation (13.3):

Consider the equation and simplify it:

One App. One Place for Learning.

Most popular questions from this chapter

Recommended explanations on Physics Textbooks

Work Energy and Power

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Study anywhere. Anytime. Across all devices.

Show that $J_{P} (x) N'_{P} (x) - J'_{P} (x) N_{P} (x) = \frac{J'_{P} (x) J_{- p} (x) - J_{p} (x) j'_{- p} (x)}{sinpπ} = \frac{2}{πx}$ .