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The Legendre equation is given by $$(1-x^2)y''-2xy'+l(l+1)y=0,$$ which has the solution $$y(x)=a_0\Big[1-\frac{l(l+1)}{2!}x^2+\frac{l(l+1)(l-2)(l+3)}{4!}x^4+...\Big]+a_1\Big[x-\frac{(l-1)(l+2)}{3!}x^3+\frac{(l-1)(l+2)(l-3)}{5!}x^5+...$$ We then substitute integer values of $l$ to obtain the Legendre polynomials, and we get the following: $$l=0 \implies y(x)=a_0,$$ $$l=1 \implies y(x)=y(x)=a_1x,$$ $$l=2 \implies y(x)=a_0(1-3x^2),$$ $$l=3 \implies y(x)=a_1(x-\frac{5}{3}x^3).$$ This I understand. However, then in the book I am reading it goes on to say: $$P_0(x)=1,$$ $$P_1(x)=x,$$ $$P_2(x)=\frac{1}{2}(3x^2-1),$$ $$P_3(x)=-\frac{3}{2}(x-\frac{5}{3}x^3)=\frac{5}{2}x^3-\frac{3}{2}x.$$

This is the bit I am struggling with. I don't understand what operation has been done to make each $P_l(x)$ different to its respective $y(x)$ above. For example, for $l=2$ $y(x)$ has been multiplied by $-1/2$. Why is this and what is happening?

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The Legendre polynomials are normally normalised so that $P_n(1)=1$ (probably for historical reasons). It's mentioned in the introduction of the Wikipedia article. This makes them easier to work with using a generating function, for example.

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