= Solution
Substitute $y(x)=\sum_{j\geq0}c_jx^j$ into the <Legendre differential equation>. Matching the coefficient of $x^j$ gives
$$
\boxed{c_{j+2}=\frac{j(j+1)-n(n+1)}{(j+2)(j+1)}c_j.}
$$
Choose the parity of $n$: start with nonzero $c_0$ for even $n$, or nonzero $c_1$ for odd $n$, and set the other parity to zero. The numerator is nonzero for every preceding index of that parity below $n$, so $c_n\ne0$, but at $j=n$ it vanishes. All higher coefficients are zero. This constructs a nonzero polynomial solution of degree exactly $n$.
Its value at $x=1$ cannot be zero. If it had a zero there of multiplicity $k\geq1$, write $y=(x-1)^kR$ with $R(1)\ne0$. The lowest-order term of $(1-x^2)y''-2xy'$ would be $-2k^2R(1)(x-1)^{k-1}$, whereas $n(n+1)y$ starts at order $k$. The differential equation could not hold. Hence the polynomial can be scaled to satisfy $P_n(1)=1$.
The resulting <Legendre polynomials> begin
$$
\boxed{P_0(x)=1,\qquad P_1(x)=x,\qquad P_2(x)=\tfrac12(3x^2-1).}
$$
Each satisfies both the differential equation and the specified normalization.
Back to article page
