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Consider the Legendre differential equation $$ (1-x^2) y'' - 2xy' + n(n+1)y = 0 $$ Then its solution is given by $$ y = c_1 P_n (x) + \text{an infinite series} $$ In fact $y = c_1 P_n (x) + c_2 Q_n (x) $ where $P_n$ is Legendre polynomials and $Q_n $ is Legendre function of the second kind. Here I want to prove that 'an infinite series' above can be written by $c_2 Q_n (x)$ for some constant $c_2$.

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  • $\begingroup$ You should see this. $\endgroup$ Commented Jul 29, 2012 at 4:15
  • $\begingroup$ @J.M. Thank you J.M., this proof is what I wanted. $\endgroup$
    – Ann
    Commented Jul 29, 2012 at 4:31

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The infinite series you are talking about stems from the recurrence relation that can be derived when solving the Legendre differential equation: plugging in $$y = \sum_{j=0}^{\infty}a_jx^j $$ into $$[(1-x^2)y']' + n(n+1)y = 0$$ you get $$\frac{a_{k+2}}{a_k} = \frac{k(k+1) - n(n+1)}{(k+1)(k+2)}$$ Obviously, this leads to separate solutions with all even or all odd powers of $x$. When $n$ is even, then starting with $a_0 \not= 0$, the series terminates and you get (a multiple of) $P_n$. The same holds for odd $n$ starting with $a_1 \not= 0$. But when n is even and we start with $a_1 \not= 0$, then the series does not terminate. For instance, for $n=0$, we get $$\frac{a_{k+2}}{a_k} = \frac{k(k+1)}{(k+1)(k+2)} = \frac{k}{k+2}$$ which clearly leads to (taking $a_1=1$) $$y= \sum_{k=0}^{\infty}\frac{1}{2k+1}x^{2k+1}$$ Now, as $$ln(1+x) = \sum_{j=0}^{\infty}\frac{(-1)^{j+1}}{j}x^j$$ it is easy to derive \begin{equation} \begin{split} ln(\frac{1+x}{1-x}) &= ln(1+x) - ln(1-x) \\ &=\sum_{j=0}^{\infty}\frac{(-1)^{j+1}}{j}x^j + \sum_{j=0}^{\infty}\frac{x^j}{j} \\ &=\sum_{j=0}^{\infty}\frac{(-1)^{j+1} + 1}{j}x^j \\ &= 2x + 2\frac{x^3}{3} + 2\frac{x^5}{5} + ... \end{split} \end{equation} which is twice what we are looking for. So $$Q_0(x) = \frac{1}{2} ln(\frac{1+x}{1-x})$$ Similarly, $$Q_1(x) = \frac{x}{2} ln(\frac{1+x}{1-x}) - 1$$ And the rest of the $Q_n$ can then be determined by using the recurrence relation $$(n+1)Q_{n+1}(x) = (2n+1)xQ_n(x) - nQ_{n-1}(x)$$

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