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1)上根的分布及证明》这篇文章中我们阐述了Legendre多项式在

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[−1,1]上的根分布情况并给出了证明。

本文将证明Legendre多项式在

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I_{mn}m!n!2^m2^n\int_{-1}^{1}p_{m}(x)p_{n}(x)

\mathrm{d}x

Imn​m!n!2m2n∫−11​pm​(x)pn​(x)dx∫−11​dxmdm​(x2−1)m⋅dxndn​(x2−1)ndx​​

用分部积分法对

Imn​​∫−11​dxmdm​(x2−1)md(dxn−1dn−1​(x2−1)n)dxmdm​(x2−1)m⋅dxn−1dn−1​(x2−1)n

​−11​−∫−11​dxn−1dn−1​(x2−1)n⋅dxm1dm1​(x2−1)mdx​​​​

kn时

dxmdm​(x2−1)m⋅dxn−1dn−1​(x2−1)n

(2)

I_{mn}-\int_{-1}^{1}\frac{\mathrm

d^{n-1}}{\mathrm{d}

Imn​−∫−11​dxn−1dn−1​(x2−1)n⋅dxm1dm1​(x2−1)mdx​​

继续用分部积分法对

Imn​(−1)n∫−11​dxmndmn​(x2−1)m⋅(x2−1)ndx​​

\displaystyle

C_{2n}^{i}[(x1)^n]^{(i)}[(x-1)^n]^{(2n-i)}C_{2n}^{n}[(x1)^n]^{(n)}[(x-1)^n]^{(n)}(2n)!

\end{equation}

dxmndmn​(x2−1)mi0∑2n​C2ni​[(x1)n](i)[(x−1)n](2n−i)C2nn​[(x1)n](n)[(x−1)n](n)(2n)!​​

(6)

\mathrm{d}\left(\dfrac{(1x)^{n1}}

{n1}\right)\nonumber

\left.\dfrac{(2n)!}{n1}(1-x)^n(1x)^{n1}\right|_{-1}^{1}\dfrac{(2n)!n}{n1}\int_{-1}^{1}(1-x)^{n-1}(1x)^{n1}\mathrm{d}x

\nonumber

\dfrac{(2n)!n}{n1}\int_{-1}^{1}(1-x)^{n-1}(1x)^{n1}\mathrm{d}x

\nonumber

\dfrac{(2n)!n(n-1)}{(n1)(n2)}\int_{-1}^{1}(1-x)^{n-2}(1x)^{n2}\mathrm{d}x

\nonumber

(n!)^2\int_{-1}^{1}(1x)^{2n}\mathrm{d}x

\dfrac{(n!)^2

Inn​​(2n)!(−1)n∫−11​(x−1)n(x1)ndx(2n)!∫−11​(1−x)nd(n1(1x)n1​)n1(2n)!​(1−x)n(1x)n1

​−11​n1(2n)!n​∫−11​(1−x)n−1(1x)n1dxn1(2n)!n​∫−11​(1−x)n−1(1x)n1dx(n1)(n2)(2n)!n(n−1)​∫−11​(1−x)n−2(1x)n2dx...(n!)2∫−11​(1x)2ndx2n1(n!)222n1​​​​​

(7)

\dfrac{I_{nn}}{(n!)2^{2n}}\dfrac{2}{2n1}0

\end{equation}

∫−11​pm​(x)pn​(x)dx(n!)22nInn​​2n12​0​​

(5),(8)

\\\displaystyle\dfrac{2}{2n1}0,

mn\end{cases}