Editing Legendre polynomials (section)

===Recurrence relations===
As discussed above, the Legendre polynomials obey the three-term recurrence relation known as Bonnet's recursion formula given by
<math display="block"> (n+1) P_{n+1}(x) = (2n+1) x P_n(x) - n P_{n-1}(x)</math>
and
<math display="block"> \frac{x^2-1}{n} \frac{d}{dx} P_n(x) = xP_n(x) - P_{n-1}(x) </math>
or, with the alternative expression, which also holds at the endpoints
<math display="block"> \frac{d}{dx} P_{n+1}(x) = (n+1)P_n(x) + x \frac{d}{dx}P_{n}(x) \,.</math>

Useful for the integration of Legendre polynomials is
<math display="block">(2n+1) P_n(x) = \frac{d}{dx} \bigl( P_{n+1}(x) - P_{n-1}(x) \bigr) \,.</math>

From the above one can see also that
<math display="block">\frac{d}{dx} P_{n+1}(x) = (2n+1) P_n(x) + \bigl(2(n-2)+1\bigr) P_{n-2}(x) + \bigl(2(n-4)+1\bigr) P_{n-4}(x) + \cdots</math>
or equivalently
<math display="block">\frac{d}{dx} P_{n+1}(x) = \frac{2 P_n(x)}{\left\| P_n \right\|^2} + \frac{2 P_{n-2}(x)}{\left\| P_{n-2} \right\|^2} + \cdots</math>
where {{math|{{norm|''P<sub>n</sub>''}}}} is the norm over the interval {{math|−1 ≤ ''x'' ≤ 1}}
<math display="block">\| P_n \| = \sqrt{\int_{-1}^1 \bigl(P_n(x)\bigr)^2 \,dx} = \sqrt{\frac{2}{2 n + 1}} \,.</math>More generally, all orders of derivatives are expressible as a sum of Legendre polynomials:<ref>{{Cite journal |last=Doha |first=E. H. |date=1991-01-01 |title=The coefficients of differentiated expansions and derivatives of ultraspherical polynomials |url=https://dx.doi.org/10.1016/0898-1221%2891%2990089-M |journal=Computers & Mathematics with Applications |volume=21 |issue=2 |pages=115–122 |doi=10.1016/0898-1221(91)90089-M |issn=0898-1221}}</ref><math display="block">\begin{aligned}
&\begin{aligned}
& \frac{d^q}{dx^q} P_{q+2 j}(x)=\frac{2^{q-1}}{(q-1)!} \sum_{i=0}^j(4 i+1) \frac{(q+j-i-1)!\Gamma\left(q+j+i+\frac{1}{2}\right)}{(j-i)!\Gamma(j+i+3 / 2)} P_{2 i}(x) \\
& \quad=\frac{1}{2^{q-2}(q-1)!} \sum_{i=0}^j(4 i+1) \frac{(q+j-i-1)!(2 q+2 j+2 i-1)!}{(j-i)!(2 j+2 i+2)!} \frac{(j+i+1)!}{(q+j+i-1)!} P_{2 i}(x)
\end{aligned}\\
&\begin{aligned}
& \frac{d^q}{dx^q} P_{q+2 j+1}(x)=\frac{2^{q-1}}{(q-1)!} \sum_{i=0}^j(4 i+3) \frac{(q+j-i-1)!\Gamma(q+j+i+3 / 2)}{(j-i)!\Gamma(j+i+5 / 2)} P_{2 i+1}(x) \\
& \quad=\frac{1}{2^{q-2}(q-1)!} \sum_{i=0}^j(4 i+3) \frac{(q+j-i-1)!(2 q+2 j+2 i+1)!}{(j-i)!(2 j+2 i+4)!} \frac{(j+i+2)!}{(q+j+i)!} P_{2 i+1}(x)
\end{aligned}
\end{aligned}</math>