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I'm reading through Jackson's Classical Electrodynamics book and am working through the derivation of the Legendre polynomials. He uses this ##\alpha## term that seems to complicate the derivation more and is throwing me for a bit of a loop. Jackson assumes the solution is of the form:

Inserting this into the differential equation we need to solve,

,

we obtain:

He then says:

,

which I agree with. The issue I'm having is with the next statement:

First, in saying the two relations are equivalent, I assume he means that if ##a_0 \neq 0## then we have ##P(x)_{a_0 \neq 0} = P_{\alpha=0}(x) + P_{\alpha=1}(x)##

##= a_0 +a_1*x^1+a_2*x^2 +...## ##+## ##a_0'*x^1+a_1'*x^2+a_2'*x^3 +...##

whereas if ##a_1 \neq 0## then we have ##P(x)_{a_1 \neq 0} = P_{\alpha=0}(x) + P_{\alpha=-1}(x)##

##= a_0 +a_1*x^1+a_2*x^2 +...## ##+## ##a_0''*x^{-1}+a_1''+a_2''*x^1+...##

and, setting ##a''##=0 to avoid the potential blowing up at x=0, ##P(x)_{a_0 \neq 0}## is basically the same as ##P(x)_{a_1 \neq 0}##, since each can be expressed as an infinite series in the form of ##c_0+c_1*x^1+c_2*x^2+...##. Therefore, we can choose either ##a_0 \neq 0## or ##a_1 \neq 0##.

But then he says we can't have both ##a_0 \neq 0## and ##a_1 \neq 0##, which I don't quite follow. If they were both 0, we'd need ##\alpha \equiv 0 ## (to satisfy his eq 3.13) and the solution would be of the form ##a_0+a_1*x^1+a_2*x^2+...##. What's wrong with that at this point in the derivation?

I recognize thatwe will see that either ##a_0## (and ##a_1'## ) or ##a_1 ##(and ##a_0'##) must equal 0 for the solution to be finite, but that's at a later stage in the derivation. At this stage in the derivation, what's wrong with still assuming ##a_0 \neq 0##eventually##a_1 \neq 0##??and

I also recognize that I may be mucking things up due to staring at this for far too long..

Thanks!

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# A Legendre Polynomials Jackson Derivation

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