3.2 Legendre Polynomials

Thus for \(n = 0, \, 1, \, 2, \, 3, \, 4, \, \cdots \) we have \[P_0(x) = 1, \,\hspace {0.5cm} P_1(x) = x, \,\hspace {0.5cm} P_2(x) = \frac {3}{2}x^2 - \frac {1}{2}, \,\hspace {0.5cm} P_3(x) = \frac {5}{2}x^3 - \frac {3}{2}x, \,\hspace {0.5cm} P_4(x) = \frac {35}{8}x^4 - \frac {30}{8}x^2 + \frac {3}{8}.\]

Using the Binomial expansion, we can rewrite the expression for the \(n^{\text {th}}\) Rodrigues’ \[(x^2 - 1)^n = \sum _{k = 0}^n \frac {(-1)^k\, n!}{k!\, (n - k)!}\, x^{2(n-k)},\] so that \[P_n(x) = \sum _{k = 0}^{[n/2]} \frac {(-1)^k\, (2n - 2k)!}{2^n\, k!\, (n - k)!\, (n - 2k)!}\, x^{n-2k},\] where \([n/2]\) denotes the largest integer \(\leq \frac {n}{2}\) from Rodrigues’ formula, it follows that \(P_n(x)\) is a polynomial of degree \(n\), and the coefficient of \(x^n\) is \[\frac {\binom {2n}{n}}{2^n}\] We can normalise \(P_n(x)\) by multiplying the coefficients by \(2^n/\binom {2n}{n}\), so that the coefficient of \(x^n\) is 1.

3.2.1 Generating Function

Another definition of Legendre polynomials is obtained by means of the generating function, \[g(x,r) = \frac {1}{(1 - 2rx + r^2)^{1/2}},\] by expanding it in as a power series for sufficiently small values of \(r\). The coefficient of \(r^n\) in this expansion turns out to be \(P_n(x)\), \(\, n = 0, \, 1, \, 2,\, \cdots \) that is \[g(x,r) = \sum _{n = 0}^{\infty } P_n(x) \, r^n.\]

To show this, consider expanding \((1 - z)^{-1/2}\) in a neighborhood of zero, where \begin {align*} z & = 2rx - r^2,\\ & = 1 + \frac {1}{2}z + \frac {3}{8}z^2 + \frac {5}{16}z^3 + \, \cdots \cdots \, , \hspace {0.3cm} |z| < 1\\ & = 1 + \frac {1}{2}(2rx - r^2) + \frac {3}{8}(2rx - r^2)^2 + \frac {5}{16}(2rx - r^2)^3 + \, \cdots \cdots \, \\ & = 1 + rx + \left (\frac {3}{2}x^2 - \frac {1}{2}\right )r^2 + \left (\frac {5}{2}x^3 - \frac {3}{2}x\right ) r^3 + \, \cdots \cdots \end {align*}

We observe that the coefficient of \(1, \, r, \, r^2\,\) and \(\, r^3\,\) are exactly, \(P_0(x), \, P_1(x),\, P_2(x)\) and \(P_3(x)\) respectively.

We generalise thus by the recurrence relation below.

We have \[g(x,r) = \frac {1}{(1 - 2rx + r^2)^{1/2}}.\] Differentiating this with respect to \(r\), we get, from \[(1 - 2rx + r^2)^{\frac {1}{2}}\, g(x, r) = 1\] \[(1 - 2rx + r^2)^{\frac {1}{2}}\, \frac {\partial }{\partial r}g(x, r) + \frac {1}{2}(1 - 2rx + r^2)^{-\frac {1}{2}}\, (-2x + 2r)\, g(x,r) = 0\] \[(1-2rx + r^2) \frac {\partial }{\partial r}g(x,r) - (x-r)g(x,r) = 0\] substituting \[g(x,r) = \sum _{n=0}^{\infty }P_n(x)\, r^n,\] get \[(1 - 2rx + r^2) \sum ^{\infty }_{n=1}n\, P_n(x)\, r^{n-1}\, -(x-r)\sum _{n=0}^{\infty } P_n(x)\, r^n = 0.\] The coefficient of \(r^n\) must be zero for each \(n\) and for all values of \(x\), we thus have \[(n+1)\, P_{n+1}(x) - (2n +1)xP_n(x) + nP_{n-1}(x) = 0,\] for \(n = 1, \, 2,\, 3,\, \cdots \)

This relation connects three successive Legendre polynomials, e.g for \[P_2(x) = \frac {3}{2}x^2 - \frac {1}{2}\, , \hspace {0.5cm} P_3(x) = \frac {5}{2}x^3 - \frac {3}{2}x,\] then \[P_4(x) = \frac {1}{4}\left [7x\, P_3(x) - 3P_2(x)\right ] = \frac {35}{8}x^4 - \frac {30}{8}x^2 + \frac {3}{8}.\]

Further Recurrence Relations

From \[g(x,r) = \frac {1}{(1 - 2rx + r^2)^{\frac {1}{2}}},\] get \[(1 - 2rx + r^2)^{\frac {1}{2}}\, g(x,r) = 1.\] Differentiating with respect to \(x\), get \[(1 - 2rx + r^2)\frac {\partial }{\partial x}g(x,r) - r\, g(x,r) = 0.\] But \[g(x,r) = \sum _{n=0}^{\infty } P_n(x)\, r^n,\] so replacing, get \[(1 - 2rx + r^2)\sum _{n=0}^{\infty } P_n'(x) r^n - \sum _{n=0}^{\infty } P_n(x)\, r^{n+1} = 0\] which implies \begin {equation} P_{n+1}'(x) - 2xP_n'(x) + P_{n-1}'(x) - P_n(x) = 0, \hspace {0.5cm} n=1,\, 2,\, 3,\, \cdots \end {equation}

Now using \[(n+1)P_{n+1}(x) - (2n+1)xP_n(x) + n P_{n-1}(x) = 0\] derived in 4.2.1, and differentiating with respect to \(x\), get \begin {equation} (n+1)\, P_{n+1}'(x) - (2n+1) \, P_n(x) - (2n + 1)\, x\, P'_n(x) + nP'_{n-1}(x) =0. \end {equation}

Eliminating \(P'_{n-1}(x)\) from (4.1) and (4.2), get multiplying equation (4.1) by \(n\) we obtain \[n P_{n+1}'(x) - 2x nP'_n(x) + nP'_{n-1}(x) -n P_n(x) = 0\]

\[(n+1) P'_{n+1}(x) - (2n+1)P_n(x) - (2n+1)xP_n'(x) + n P'_{n-1}(x) = 0\] subtracting equation (4.1) from (4.2) \[P'_{n+1}(x) - (n+1)P_n(x) - xP'_n(x) = 0\] or \begin {equation} P'_{n+1}(x) - xP_n'(x) = (n+1)P_n(x) \end {equation}

Similarly, eliminating \(P'_{n+1}(x)\) from (4.1) and (4.2), we get \begin {equation} xP'_n(x) - P'_{n-1} (x) = nP_n(x) \end {equation}

Adding (4.3) and (4.4), get \[P'_{n+1}(x) - P'_{n-1}(x) = (2n+1) P_n(x), \, \hspace {0.5cm} n = 1, \, 2,\, 3, \, \cdots \] Replacing \(n\) by \(n-1\) in (4.3), get \begin {equation} P'_n(x) - xP'_{n-1}(x) = nP_{n-1}(x). \end {equation}

Then we eliminate \(P'_{n-1}(x)\) from (4.4) and (4.5), get (by multiplying equation 4.4 by \(x\)) \[x^2P'_n(x) - xP'_{n-1}(x) = xnP_n(x)\] \[P'_n(x) - x P'_{n-1}(x) = n P_{n-1}(x)\] implies that \begin {equation} (x^2-1)P_n'(x) = nP_{n-1}(x) - n xP_n(x). \end {equation}

Differentiating (4.6) with respect to \(x\) and then using (4.4) to eliminate \(P'_{n-1}(x)\), we get \[\left [(1-x^2)P_n'(x)\right ]' + n(n+1) P_n(x) = 0\] which shows that \(U=P_n(x)\) is a particular integral of the second order differential equations.

\[\left [(1-x^2)U'\right ]' + n(n+1) U = 0.\]

3.2.2 Orthogonality of Legendre Polynomials

Orthogonality of Legendre polynomials on the interval \([-1,1]\) follows from the Legendre differential equations \[\left [(1-x^2)\, P'_n(x) \right ]' + n(n+1)\, P_n(x) = 0\hspace {0.5cm} n = 0,\, 1, \, 2,\, \cdots \] We have \begin {equation} \left [(1 - x^2)\, P_n'(x)\right ]'\, P_m(x) + n(n+1)\, P_n(x)\, P_m(x) = 0 \end {equation} and \begin {equation} \left [(1-x^2)\, P'_m(x)\right ]'\, P_n(x) + m(m+1)\, P_m(x)\, P_n(x) = 0 \end {equation} subtracting (4.7) from (4.8), get \[\left \{(1-x^2)\, \left [P'_m(x)\, P_n(x) - P_n'(x)\, P_m(x)\right ]\right \}' + (m-n)(m+n-1)P_m(x)P_n(x) = 0.\] Integrating this last equation over \([-1,1]\) and noting that the first integral vanishes, get \[(m-n)(m+n+1)\int _{-1}^1P_m(x)\, P_n(x)\, dx = 0\] i.e \[\int _{-1}^1P_m(x)\, P_n(x)\, dx = 0 \hspace {0.7cm} \text {when}\,\,\, m\neq n.\]

Further, it can be shown that \[\int _{-1}^1P_n^2(x)\, dx = \frac {2}{2n + 1}\, , \hspace {0.6cm} n = 0, \, 1, \, 2, \, \cdots \]

To show this, one uses a recurrence relation \begin {equation} (n+1)P_{n+1}(x) - (2n+1)xP_n(x) + nP_{n-1}(x) = 0, \hspace {0.5cm} n = 1, \, 2,\, \cdots \end {equation} Reduce \(n\) by \(n-1\) in (4.9) and multiply the result by \((2n+1)P_n\).

Then from this equation we subtract (4.9) multiplied by \((2n-1)P_{n-1}(x)\).

Finally integrate the result over \([-1,1]\), noting orthogonality to get \begin {align*} \int _{-1}^1P_n^2(x)\, dx & = \frac {2n - 1}{2n+1}\int _{-1}^1P^2_n(x)\, dx\\ & = \frac {3}{2n+1}\int _{-1}^1P^2_1(x)\, dx \hspace {0.6cm} \text {letting} \,\,\, n-1 = 1 \implies n = 2\\ & =\frac {2}{2n+1}. \end {align*}

3.2.3 Expansion of a Function using Legendre

Suppose that \(f(x)\) is a function defined on \([-1,1]\) such that \[\int _{-1}^1f(x)\, P_n(x)\, dx\hspace {0.6cm} \text {exists for }\,\, \, \, n = 0,\, 1,\, 2, \, \cdots \]

Consider the series \begin {equation} f(x) = \sum _{i=0}^{\infty } a_iP_i(x). \end {equation} Multiplying both sides of (4.10) by \(P_n(x)\) and integrating over \([-1,1]\), get \[\int _{-1}^1f(x)\, P_n(x)\, dx = \int _{-1}^1a_n\, P^2_n(x)\, dx\, , \hspace {0.3cm} n = 0,\, 1,\, 2, \, \cdots \] Thus \begin {align*} a_n & = \left [\int _{-1}^1P^2_n(x)\, dx \right ]^{-1}\, \int _{-1}^1f(x)\, P_n(x)\, dx\\\\ & = \frac {2n + 1}{2}\int ^1_{-1} f(x) \, P_n(x)\, dx. \end {align*}

Recall Fourier series \[f(x) = \sum _{n=0}^{\infty }\left [a_n\cos nx + b_n \sin nx\right ].\]

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