Evening. I am having difficulty solving the the problems that have been included below. For the first problem, I essentially followed what the hint suggested that I do and I still cannot "see" the solution. I am honestly not sure how to even go about solving the second problem. For the third problem, I differentiated the generating function for the legendre polynomial, which gave a solution that was somewhat similiar to the "Henyey-Greenstein phase function" but deviates from it slightly (the answer expression that I have right now is(adsbygoogle = window.adsbygoogle || []).push({});

[tex]2[r^2-1](\frac{d}{dx})[1+r^2-2cos(\theta)]^{-1/2}[/tex].

Finally, for the fourth problem, I really don't know where to begin. Any and all assistance/guidance/hints would be very much appreciated. Thank you.

Regards

I. Show that the nth Legendre polynomial is given by

[tex]P_n(x)=\sum_{k=0}^n\left(\stackrel{n}{k}\right)\left(\stackrel{n+k}{k}\right)\left(\stackrel{x-1}{2}\right)^k.[/tex]

Hint: Write [tex](x^2-1)^n=(x-1)^n[(x-1)+2]^n[/tex], apply the binomial expansion to the term in [\cdots], and differentiate n-times.

II. Find the degree three Legendre approximation of the function

[tex]f(x)=\left\{\stackrel{0 (-1\leq x<0)}{1 (0\leq x<1)}.[/tex]

III. Use the formula

[tex]\frac{1}{\sqrt{1+r^2-2rx}}=\sum_{n=0}^\infty P_n(x)r^n[/tex]

to derive the formula for the ``Henyey-Greenstein phase function''

[tex]\frac{1-r^2}{(1+r^2-2rcos(\theta))^{3/2}}=\sum_{n=0}^\infty (2n+1)P_n(cos(\theta))r^n[/tex].

IV. Let [tex]c_n[/tex] be the leading term of [tex]P_n[/tex] and set

[tex]\tilde{P}_n=c^{-1}P_n=x^n+[/tex] (lower powers of x).

Prove that if [tex]Q=x^n+\cdots[/tex] is any polynomial of degree n with leading coefficient

one, then

[tex]<Q,Q>\geq<\tilde{P}_n,\tilde{P}_n>[/tex]

with equality only if [tex]Q=\tilde{P}_n[/tex].

(Hint: Write [tex]Q=\tilde{P}_n+h[/tex], where h is a linear

combination of [tex]P_0,P_1,\ldots P_{n-1}[/tex], and note that [tex]<Pn,h>=0[/tex].)

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# Homework Help: A Few Diff. Eq. Problems (please help)

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