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I know that the Laguerre polynomials are orthogonal with respect to $\exp(-x)$, but is there an orthogonal set with respect to $\exp(-ax)$? Is it possible to extend the Laguerre polynomials to obtain orthogonality/orthonormality with respect to this slightly changed weight?

What I would like to do is to find polynomials $\{P_i(x)\}$ such that $\langle P_i(x)\exp(-a_i\; x) | P_j(x)\exp(-a_j\;x)\rangle$ is orthonormal.

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Hint: if $a > 0$, do a change of variables in $\int_0^\infty f(x) g(x) e^{-ax}\ dx$ –  Robert Israel Mar 29 '12 at 23:26
    
Maybe if you first try Gram--Schmidt to get a set of orthogonal polynomials, then notice how they're related to the standard Laguerre polynomials with $a=1$, you can quickly conjecture the answer to your question and then prove it. –  Michael Hardy Mar 29 '12 at 23:55
    
You need to provide an interval as well. Say: orthogonal with respect to $\exp(-ax)$ on $[0,+\infty)$ or something like that. –  GEdgar Mar 30 '12 at 0:13
    
@GEdgar : You're right that Laubert515 needs to put that there. I overlooked it since "everybody [?] knows" that that's what you do when you think about Laguerre polynomials. –  Michael Hardy Mar 30 '12 at 2:56
    
I have tried gram-schmidt and it is nearly impossible. I can get the coefficients of the first 2 functions in about an hour, but the third took 40+ hours on mathematica with a 3.1 GHz machine. –  Laurbert515 Mar 30 '12 at 11:27

1 Answer 1

up vote 2 down vote accepted

The thing is, OP really should remember that specifying just the weight function doesn't cut it, seeing that the inner product involves a definite integral, and thus a support interval is very much needed to completely specify the inner product involved in constructing the orthogonal polynomials involved. To use the Laguerre polynomials as an example, the weight function is $\exp(-x)$, and the support interval is $[0,\infty)$, corresponding to the inner product

$$\langle f,g\rangle=\int_0^\infty \exp(-u)f(u)g(u)\mathrm du$$

and we have the Laguerre polynomials $L_n(x)$ satisfying $\langle L_j,L_k\rangle=\delta_{j,k}$.

Now... as already hammered to you in the comments, since

$$\int_0^\infty \exp(-au)f(u)g(u)\mathrm du=\frac1{a}\int_0^\infty \exp(-v)f(v/a)g(v/a)\mathrm dv$$

assuming that your support interval is indeed $[0,\infty)$, building up your "modified" Laguerre polynomials shouldn't be too hard.

Otherwise, there is always the Stieltjes procedure for generating orthogonal polynomials from a given inner product. I already described it at length here, and I'd rather not repeat myself. All I'll say is that the procedure generates monic orthogonal polynomials; you'll have to figure it out on your own if you require a different normalization for your polynomials. In this regard,

$$\int_0^\infty \exp(-au)u^k\mathrm du=\frac{k!}{a^{k+1}}$$

will be a useful identity for you.


If the support interval is not $[0,\infty)$, but some finite interval $[r,s]$; that is, the inner product involved is

$$\langle f,g\rangle_{a,[r,s]}=\int_r^s \exp(-au)f(u)g(u)\mathrm du$$

the Stieltjes procedure is still useful, but one now needs the identity

$$\int_r^s \exp(-au)u^k\mathrm du=\frac{\Gamma(k+1,ar)-\Gamma(k+1,as)}{a^{k+1}}$$

where $\Gamma(k,a)$ is the upper incomplete gamma function. In Mathematica, one could compactly express the right hand side of that last identity as Gamma[k + 1, a r, a s]/(a^(k+1)), making use of Mathematica's support for incomplete gamma functions with arbitrary integration limits.

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