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I am asking a simple conceptual question. I saw in many Mathematics and Mathematical physics text books that the Legendre polynomials and Laguerre polynomials "falling from the sky"! I mean, I didn't get the concept behind and how those polynomials derived. The textbooks rather says a second order differential equation and says the solution is in this form. I wish to know why people assuming this kind of differential equations and also what is the physical concept behind. It would be rather interesting to know how basically the solutions for those differential equations derived.

Or please tell me a text book/reference where I can find those concepts

Thanks a lot.

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This book has an accessible introduction to these concepts from the point of view of representing and approximating probability distributions. I found it suitable for an engineer like myself. D. Xiu Numerical Methods for Stocahstic Computations: A Spectral Method Approach Princeton University Press, 2010. ISBN-10: 0691142122, ISBN-13: 978-0691142128 –  Sriram S Feb 11 '13 at 1:33
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The basic concept: when we work with a Hilbert space of functions, we want to have an orthonormal basis that consists of particularly simple functions. The trigonometric (or complex exponential) basis works very well for $L^2([a,b])$. But algebraic polynomials can serve this purpose too. Since polynomials are not integrable over unbounded intervals, one includes a weight such as $w(x)=e^{-x}$ on $[0,\infty)$ or $w(x)=e^{-x^2}$ on $\mathbb R$. Then the monomials $1,x,x^2,\dots$ are elements of the weighted Lebesgue space $L^2_w = \{f:\int |f|^2 w<\infty\}$. Applying the Gram-Schmidt process to the sequence of monomials produces an orthonormal sequence, which in many cases turns out to be complete, i.e., an orthonormal basis. With the weights I mentioned above we get the Laguerre and Hermite polynomials, respectively.

Orthogonal polynomials have a large number of interesting properties, many of which are easily proved by induction. Recurrence relations, for example: polynomials of different degrees turn out to be related via linear combinations involving derivatives. Also, they usually satisfy a simple differential equation with degree $n$ as parameter, which can be used as a shortcut to quickly define the polynomials (without going through Gram-Schmidt, etc).

Further reading: you can start with the short Wikipedia article Orthogonal polynomials and proceed to the references listed there, out of which I recommend the classical book by Szegő, Orthogonal polynomials.

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Well, the best way I can think of to introduce a real physical problem where Laguerre ànd Legendre (even of the associated form!) differential equations play a role, is the (exact, that is, without taking spin into consideration) solution of the combined angular momentum / magnetic moment / energy eigenvalue-problem for the hydrogen atom.

I cannot tell the whole story here, but I can sketch the process.

When solving the Schrodinger equation involved, and transforming it to spherical coordinates, all applicable quantum numbers roll into your lap and on route you will encounter the Laguerre and associated Legendre differential equations.

Now for the approach of solving these equations. Let's have a look at the Laguerre equation first.

$$xy'' + (1+\alpha-x)y' + ny = 0$$

If you try $y = \sum_{j=0}^{\infty}a_jx^j$, you will find that $$\frac{a_{j+1}}{a_j} = -\frac{n-j}{(j+1)(j+1+\alpha)}$$ So, when moving from one coefficient to the next, the sign changes, it is divided by $j+1$, multiplied by $n-j$ and divided by $j+1+\alpha$. When j reaches n, the coefficients vanish. That means that we will not be far off if we try $$a_j = \frac{(-1)^j}{j!}\binom{n+\alpha}{n-j}$$ In fact, this is exactly what we need.

Now, which function will this be? The factor $\frac{(-1)^j}{j!}$ points at $e^{-x}$. The factor $\binom{n+\alpha}{n-j}$ points at the $j^{th}$ derivative of $x^{n+\alpha}$, and we saw above that the series ends for j=n. But when we try $[e^{-x}x^{n+\alpha}]^{(n)}$, firstly we have to compensate for all the $e^{-x}$ factors by multiplying the result by $e^x$ , and secondly we would get terms for the running index j that have a factor $x^{n + \alpha - (n-j)} = x^{j + \alpha}$ and we would like to have $x^j$. So we have to compensate at the end here too, by multiplying by $x^{-\alpha}$. This reasoning leads to trying $$y = e^xx^{-\alpha}[e^{-x}x^{n+\alpha}]^{(n)}$$ Where $f^{(n)}$ stands for the $n^{th}$ derivative of $f$. And funnily enough, it's spot on again.

Over to the Legendre equation, first the ordinary one.

$$[(1-x^2)y']' + n(n+1)y = 0$$

Trying $y = \sum_{j=0}^{\infty}a_jx^j$ again, we find that $$\frac{a_{j+2}}{a_j} = \frac{j(j+1) - n(n+1)}{(j+1)(j+2)}$$ As can be seen in one of my other posts (About the Legendre differential equation) there are solutions that are polynomials ($P_n$) and solutions that are represented by an unending series expansion ($Q_n$). The $P_n$ are either odd-powered or even powered polynomials, depending on $n$ which is also the degree of $P_n$.

The $Q_n$ contain a factor $ln(\frac{1+x}{1-x})$, see (About the Legendre differential equation) again.

Finally, the associated Legendre equation. $$[(1-x^2)y']' + \{l(l+1) - \frac{m^2}{1-x^2}\}y = 0$$ It is natural to try $y = (1-x^2)^{\alpha}P_l$. It is easily checked that we get the desired term $\frac{m^2}{1-x^2}y$ if we take $\alpha = m/2$. Further analysis shows that if we wish to arrive at all the correct terms, we have to use $P_l^{(m)}$ instead of $P_l$. Note that I still use the notation $f^{(m)}$ as denoting the $m^{th}$ derivative of $f$. The notation I use for the solutions of the associated Legendre equation is $$P_{l,m} = (1-x^2)^{m/2}P_l^{(m)}$$

I hope this helps :-).

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