...and orthogonality relation.(adsbygoogle = window.adsbygoogle || []).push({});

The book says

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \delta_{mn} \frac{2}{2n+1}[/tex]

So I sat and tried derieving it. First, I gather an inventory that might be useful:

[tex](1-x^2)P_n''(x) - 2xP_n'(x) + n(n+1) = 0[/tex]

[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]

[tex]P_n(-x) = (-1)^n P_n(x)[/tex]

[tex]P_n'(-x) = -(-1)^n P_n'(x)[/tex]

And started the job. Now...

[tex][(1-x^2)P_n'(x)]' = -n(n+1)P_n(x)[/tex]

multiplying both sides with [tex]P_m(x)[/tex] and integrating from -1 to 1

[tex]\int_{-1}^{1} [(1-x^2)P_n'(x)]' P_m(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]

applying a partial to the lefthand side

[tex](1-x^2)P_n'(x) P_m(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_n'(x) P_m'(x) dx = \int_{-1}^{1} -n(n+1)P_n(x) P_m(x) dx[/tex]

switching n and m's, ve get another equation like it

[tex](1-x^2)P_m'(x) P_n(x) |_{-1}^{1} - \int_{-1}^{1} (1-x^2)P_m'(x) P_n'(x) dx = \int_{-1}^{1} -m(m+1)P_m(x) P_n(x) dx[/tex]

substituing these

[tex][-n(n+1) + m(m+1)]\int_{-1}^{1} P_n(x) P_m(x) dx = [(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))] |_{-1}^{1}[/tex]

now, from the inventory

[tex]P_n(-x)P_m'(-x) = -P_n(x) P_m'(x)[/tex]

the righthand side becomes

[tex]\lim_{x \to 1} 2[(1-x^2)(P_n'(x) P_m(x) - P_m'(x) P_n(x))[/tex]

and in it's final shape

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2(1-x^2) \frac {(P_n'(x) P_m(x) - P_m'(x) P_n(x))}{-n(n+1) + m(m+1)}[/tex]

For [tex]n \neq m[/tex], there's no problem, it's straightly 0. But for [tex]n = m[/tex]...

And everything starts going wrong. By stating

[tex](1-x^2)P_n'(x) = \int -n(n+1)P_n(x) dx[/tex]

i'm making a little change

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

Now I plug [tex]P_n(x) = P_m(x)[/tex], [tex]x=1[/tex] and [tex]P_n(1) = 1[/tex] and state

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = \lim_{x \to 1} 2 \frac {P_m(x) (\int -n(n+1)P_n(x) dx) - P_n(x) (\int -m(m+1)P_m(x) dx))}{-n(n+1) + m(m+1)}[/tex]

[tex]\int_{-1}^{1} P_n(x) P_m(x) dx = 2 (\int P_n(x) dx)_{x=1}[/tex]

which is the nonsense of the day.

Where's the error?

**Physics Forums - The Fusion of Science and Community**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Legendre polynomials

Loading...

Similar Threads for Legendre polynomials | Date |
---|---|

I Legendre Differential Equation | Oct 12, 2016 |

I Legendre polynomials and Rodrigues' formula | Sep 3, 2016 |

Associated Legendre polynomials | Nov 22, 2014 |

Legendre Equation & Polynomials | Jun 28, 2014 |

Proof - Legendre polynomials | Feb 10, 2012 |

**Physics Forums - The Fusion of Science and Community**