Question

Pleasehelpmewiththisproblem! Thanks!

10. (a) By recalling that Pm(x) is a polynomial of degree m containing only the powers r", Х'n-2, X,"-4, . . . of x (Sec. 99), state why where the coefficients are constants, Apply the same argument to 2, etc., to conclude that x"is a finite linear combination of the polynomials nt PCx), P-20x), P4x),.... (b) With the aid of the result in part (a), point out why P (x)p(x) dx-0, where Pa(x) is a Legendre polynomial of degree n (n-1,2,... and p(x) is any polynomial whose degree is less than n. t have any one of the values n12

Answer 1

(a) We can write $P_m(x)=a_mx^m+a_{m-2}x^{m-2}+\cdots$ , we have

m-2 m-2 m-Z' m-Z'

Therefore,

$x^m=a_m^{-1}P_m(x)-a_m^{-1}(a_{m-2}x^{m-2}+\cdots)$

which is of the form in question (name coefficients as )

cm-24 mm-2 Cm

We can do this repeatedly, for
m2
, then
m-4
, et cetera, to get

$x^m=c_mP_m(x)+c_{m-2}P_{m-2}+\cdots$

(b) Since , we have

degp(r) < n

$p(x)=a_{n-1}x^{n-1}+\cdots+a_1x+a_0$

Using part a) above, this becomes a linear combination of the polynomials . Let us write

Pn-1, Pn-2, .. , Po(i)

$p(x)=b_{n-1}P_{n-1}(x)+b_{n-2}P_{n-2}(x)+\cdots+b_0P_0(x)$

Then

$\begin{align*}\int_{-1}^1p(x)P_n(x)~dx&=\int_{-1}^1P_n(x)(b_{n-1}P_{n-1}(x)+b_{n-2}P_{n-2}(x)+\cdots+b_0P_0(x))~dx\\ &=\int_{-1}^1P_n(x)b_{n-1}P_{n-1}~dx+\int_{-1}^1P_n(x)b_{n-2}P_{n-2}(x)~dx+\cdots+\int_{-1}^1P_n(x)b_0P_0(x)~dx\\ &=b_{n-1}\int_{-1}^1P_n(x)P_{n-1}~dx+b_{n-2}\int_{-1}^1P_n(x)P_{n-2}(x)~dx+\cdots+b_0\int_{-1}^1P_n(x)P_0(x)~dx\end{align*}$

Since the Legendre polynomials are orthogonal, each of the integrals above is zero. Hence,

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