Question: In Equations (6.35) through (6.42), we provide several identities for Legendre polynomials. Derive the results in Equations (6.36) through (6.42) as described in the text.
In Equations (6.35) through (6.42), we provide several identities for Legendre polynomials. Derive the results in Equations (6.36) through (6.42) as described in the text. Namely,
a. Differentiating Equation (6.35) with respect to \(x\), derive Equation \((6.36)\).
b. Derive Equation (6.37) by differentiating \(g(x, t)\) with respect to \(x\) and rearranging the resulting infinite series.
c. Combining the previous result with Equation (6.35), derive Equations \((6.38)\) and (6.39).
d. Adding and subtracting Equations (6.38) and (6.39), obtain Equations \((6.40)\) and \((6.41)\).
e. Derive Equation (6.42) using some of the other identities.
![(n+1)Pn+1(x) == (2n+1)xPn(x)-nPn-1(x) (6.35) (n+1)P+1(x) == (2n+1)[Pn(x)+xP(x)] - nPn-1(x) (6.36) Pn(x) =](https://dsd5zvtm8ll6.cloudfront.net/images/question_images/1716/5/3/3/98166503add13c751716577146833.jpg)
(n+1)Pn+1(x) == (2n+1)xPn(x)-nPn-1(x) (6.35) (n+1)P+1(x) == (2n+1)[Pn(x)+xP(x)] - nPn-1(x) (6.36) Pn(x) = P+1(x) - 2x P(x) + Pn-1(x) (6.37) P-1(x) = xP(x) - nPn(x) (6.38) = P+1(x) Pn+1(x) + Pn-1(x) Ph+1(x) - P-1(x) (x-1)P(x) = = = 2x Pn(x) + Pn(x). (2n+1)Pn(x). nxPn(x)-nPn-1(x) xP(x)+(n+1)P(x) (6.39) (6.40) (6.41) (6.42)
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