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Chapter 12: Q9P (page 584)

Use problem 7 to show that
P_l^m (x) = (-1)^m (l+m)!/(l-m)! (1-x²)/2^l! d^l-m/dx^l-m(x²-1)^l

Short Answer

Expert verified

The answer is stated below.

Step by step solution

Relation between Plm(x) and Pl(x):

The relation between P_l^m(x) and P_l(x) is as follow.

P_l^m(x) =1/2^l l! (1-x²)^m/2d^l+m/dx^l+m(x²-1)^l ..... (1)

Put the values in the equation (1):

Putting the above given equation in equation (1) and obtain the solution as follows:

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Most popular questions from this chapter

To study the approximations in the table, a computer plot on the same axes the given function together with its small x approximation and its asymptotic approximation. Use an interval large enough to show the asymptotic approximation agreeing with the function for large x. If the small x approximation is not clear, plot it alone with the function over a small interval $j1(x)$

Expand the following functions in Legendre series.

Hint: Solve the recursion relation (5.8e)for P_l(x)and show that ∫_a¹ P_l(x) dx=1/2l+1 [P_l-1 (a)-P_l+1 (a)].

Prove $J_{p} (x) J'_{- P} (x) - J_{- p} (x) J'_{p} (x) = - \frac{2}{π x} s i n p π$ as follows:

Write Bessel's equation (12.1) with $y = J_{p}$ and with $y = J_{- p}$ ; multiply the $J_{p}$ equation by $J_{- p}$ and the $J_{- p}$ equation by $J_{p}$ and subtract to get $\frac{d}{d x} ['' x (J_{p} {J_{- p}}^{'} - J_{- p} J_{p}^{'})^{''}]^{''} = 0$ . Then $J_{p} {J_{- p}}^{'} - J_{- p}^{'} J_{p}^{'} = \frac{c}{x}$ . To find , use equation for each of the four functions and pick out the $1 / x$ terms in the products.

To study the approximations in the table, a computer plot on the same axes the given function together with its small x approximation and its asymptotic approximation. Use an interval large enough to show the asymptotic approximation agrees with the function for large x . If the small x approximation is not clear, plot it alone with the function over a small interval $N2(x)$ .

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Use problem 7 to show thatPlm (x) = (-1)m (l+m)!/(l-m)! (1-x2)/2l! dl-m/dxl-m(x2-1)l

Short Answer

Step by step solution

Relation between Plm(x) and Pl(x):

Put the values in the equation (1):

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Most popular questions from this chapter

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Use problem 7 to show that
P_l^m (x) = (-1)^m (l+m)!/(l-m)! (1-x²)/2^l! d^l-m/dx^l-m(x²-1)^l