Orthogonal polynomials with respect to $e^{-|x|} \mathrm{d} x$ on the entire real line?

Question

The Laguerre polynomials https://en.wikipedia.org/wiki/Laguerre_polynomials form a system of orthogonal polynomials with respect to the measure $e^{ -x} \mathrm{d} x$ on $(0,\infty)$.

Is anything known about the orthogonal polynomials with respect to the measure $e^{ -|x|} \mathrm{d} x$ on $(-\infty,\infty)$?

Thank you!

Answer 1

I'll quote from T. J. Sullivan's Introduction to Uncertainty Quantification (SpringerLink). Let $\mathfrak P$ be the set of polynomials (as a subset of $L^2(\mu)$, where $\mu$ is a measure). Then by Gram-Schmidt on monomials, we can get the existence of orthogonal polynomials for $\mu=\exp(-|x|)dx$-

Theorem 8.5. If the $L^{2}(\mu)$ inner product is positive definite on $\mathfrak{P},$ then there exists an infinite sequence of orthogonal polynomials for $\mu$.

The book also mentions that $\int \exp(a|x|)d\mu(x)<\infty$ for some $a>0$ implies the completeness of the above constructed set of orthogonal polynomials; thus this works for $\mu=\exp(-|x|)dx$.

Then, to find the polynomials, you can use the three-term recurrence-

Theorem 8.9. Let $\mathcal{Q}=\left\{q_{n} \mid n \in \mathcal{N}\right\}$ be the monic orthogonal polynomials for a measure $\mu .$ Then $$ \begin{aligned} q_{n+1}(x) &=\left(x-\alpha_{n}\right) q_{n}(x)-\beta_{n} q_{n-1}(x) \\ q_{0}(x) &=1 \\ q_{-1}(x) &=0 \end{aligned} $$ where $$ \begin{aligned} \alpha_{n} &:=\frac{\left\langle x q_{n}, q_{n}\right\rangle_{L^{2}(\mu)}}{\left\langle q_{n}, q_{n}\right\rangle_{L^{2}(\mu)}}, & \text { for } n \geq 0 \\ \beta_{n} &:=\frac{\left\langle q_{n}, q_{n}\right\rangle_{L^{2}(\mu)}}{\left\langle q_{n-1}, q_{n-1}\right\rangle_{L^{2}(\mu)}}, & \text { for } n \geq 1, \\ \beta_{0} &:=\left\langle q_{0}, q_{0}\right\rangle_{L^{2}(\mu)} \equiv \int_{\mathbb{R}} \mathrm{d} \mu. \end{aligned} $$ Hence, the orthonormal polynomials $\left\{p_{n} \mid n \in \mathcal{N}\right\}$ for $\mu$ satisfy $$ \sqrt{\beta_{n+1}} p_{n+1}(x)=\left(x-\alpha_{n}\right) p_{n}(x)-\sqrt{\beta_{n}} p_{n-1}(x) $$ $$ \begin{aligned} p_{0}(x) &=\beta_{0}^{-1 / 2} \\ p_{-1}(x) &=0. \end{aligned} $$

For instance, $q_{0,-1}$ are given, and to find $q_1$, we need to compute $\alpha_0$ and $\beta_0$, which are $$\beta_0 = \int_{\mathbb R} d\mu = \int_{-\infty}^\infty \exp(-|x|)dx = 2,$$ and $$ \alpha_{0} =\frac{\left\langle x q_{0}, q_{0}\right\rangle_{L^{2}(\mu)}}{\left\langle q_{0}, q_{0}\right\rangle_{L^{2}(\mu)}} = \frac12\int x\exp(-|x|)dx = 0 $$ so the first two in the list are $$q_0=1,\quad q_1=x.$$ Then you can continue inductively. $\langle xq_1,q_1\rangle_{L^2(\mu)}$ is zero (hence so is $\alpha_1$). In fact, a quick induction argument (once you have the three-term recurrence, and since $\mu$ is even; see comments) shows that $q_{2k}$ is an even polynomial, and $q_{2k+1}$ is an odd one. This also implies $\alpha_n\equiv 0$. A quick calculation shows that $\beta_1 = \frac12\int x^2 d\mu = 2,$ and so $$q_2=x q_1 -2 q_0 = x^2-2.$$ A less easy calculation gives $ \quad \beta_2 = 10,$ which means that $$ q_3 = xq_2 -20 q_1 = x^3 - 12x.$$ And it continues. Maybe there's a nice closed form for the recurrence, but I haven't tried.

Verifications: this link shows that $\langle q_0,q_2\rangle_{L^2(\mu)} = 0$. $\langle q_1,q_2\rangle_{L^2(\mu)} = 0 = \langle q_0,q_3\rangle_{L^2(\mu)}$ and also $\langle q_2 ,q_3\rangle_{L^2(\mu)} = 0$ by the parity of the polynomials. This link shows that $\langle q_1 ,q_3\rangle_{L^2(\mu)} = 0$.

Answer 2

It's a known result that symmetric weights lead to definite parity polynomials. For if $w(x)=w(-x)\ge0$ for $-a\lt x\lt a$ then if $$\int_{-a}^aw(x)P_n(x)x^mdx=0$$ For all $0\le m\le n-1$, then $$\begin{align}\int_{-a}^aw(x)P_n(x)x^mdx&=\int_a^{-a}w(-x)P_n(-x)(-x)^m(-dx)\\ &=(-1)^m\int_{-a}^aw(x)P_n(-x)x^mdx=0\end{align}$$ Also for $0\le m\le n-1$. And since orthogonal polynomials are unique up to a constant factor because if $\pi_n(x)$ and $\rho_n(x)$ are monic polynomials such that $$\int_{-a}^aw(x)\pi_n(x)x^mdx=\int_{-a}^aw(x)\rho_n(x)x^mdx=0$$ for all $0\le m\le n-1$, then $\left(\pi_n(x)-\rho_n(x)\right)^*$ is a polynomial of degree at most $n-1$ so $$\int_{-a}^aw(x)\left(\pi_n(x)-\rho_n(x)\right)^*\left(\pi_n(x)-\rho_n(x)\right)dx=0$$ So the integrand, being inherently nonnegative, must be identically zero. Thus $P_n(-x)$ is a constant times $P_n(x)$ and the polynomials $\frac12\left(P_n(x)+P_n(-x)\right)$ and $\frac12\left(P_n(x)-P_n(-x)\right)$ are orthogonal polynomials with definite parity and exactly one of them is nonzero and of degree $n$. With this insight in hand we can crank out the first few orthogonal polynomials knowing that polynomials of one parity need only be orthogonal to polynomials of the same parity over the interval $[0,\infty)$:

! plaguerre.f90
module plaguerre
   use ISO_FORTRAN_ENV, only: wp => REAL128, ik => INT64
   implicit none
   contains
      subroutine coeffs(n,det,num)
         integer, intent(in) :: n
         real(wp), intent(out) :: det
         real(wp), intent(out) :: num(n/2+1)
         real(wp) A(n/2,n/2)
         real(wp) b(n/2)
         integer i, k, m
         real(wp) mag
         integer(ik) d
         do i = 1, n/2
            A(i,1) = 1
            do k = 2, n/2
               A(i,k) = (2*i+2*k-4+2*mod(n,2))*(2*i+2*k-5+2*mod(n,2))*A(i,k-1)
            end do
            b(i) = -(2*i+2*k-4+2*mod(n,2))*(2*i+2*k-5+2*mod(n,2))*A(i,k-1)
         end do
         det = 1
         do k = 1, n/2
            do i = k, n/2
               mag = norm2(A(i,k:))
               A(i,k:) = A(i,k:)/mag
               det = det*mag
               b(i) = b(i)/mag
            end do
            m = k-1+maxloc(abs(A(k:,k)),1)
            if(k /= m) then
               A([k,m],k:) = A([m,k],k:)
               b([k,m]) = b([m,k])
               det = -det
            end if
            A(k,k+1:) = A(k,k+1:)/A(k,k)
            det = det*A(k,k)
            b(k) = b(k)/A(k,k)
            do i = k+1, n/2
               A(i,k+1:) = A(i,k+1:)-A(i,k)*A(k,k+1:)
               b(i) = b(i)-A(i,k)*b(k)
            end do
         end do
         do k = n/2, 1, -1
            b(k) = b(k)-dot_product(A(k,k+1:),b(k+1:))
         end do
         num = [det*b,det]
         if(any(abs(num) > huge(0_ik))) stop
         write(*,'(*(g0))',advance='no') '$$P_{',n,'}(x) ='
         if(n == 0) write(*,'(*(g0))',advance='no') 1
         do i = 1, n/2+1
            d = gcd(nint(abs(num(i)),ik),nint(abs(det),ik))
            if(i == 1) then
               if(num(i)/det < 0) write(*,'(*(g0))',advance='no') '-'
               call rational(nint(abs(num(i)/d),ik),nint(abs(det/d),ik))
               if(mod(n,2) == 1) write(*,'(*(g0))',advance='no') 'x'
            else
               write(*,'(*(g0))',advance='no') merge('-','+',num(i)/det < 0)
               call rational(nint(abs(num(i)/d),ik),nint(abs(det/d),ik))
               write(*,'(*(g0))',advance='no') 'x^{',2*i-2+mod(n,2),'}'
            end if
         end do
         write(*,'(*(g0))') '$$'
      end subroutine coeffs
      recursive function gcd(a,b) result(c)
         integer(ik) c
         integer(ik), value :: a, b
         if(b == 0) then
            c = a
            return
         else
            c = gcd(b, modulo(a,b))
         end if
      end function gcd
      subroutine rational(n, d)
         integer(ik), value :: n, d
         if(d > 1) then
            write(*,'(*(g0))',advance='no') '\frac{',n,'}{',d,'}'
         else if( n > 1) then
            write(*,'(*(g0))',advance='no') n
         end if
      end subroutine rational
end module plaguerre

program test
   use plaguerre
   implicit none
   integer n
   real(wp) det
   integer, parameter :: MAXN = 12
   real(wp) num(MAXN/2+1)
   do n = 0, MAXN
      call coeffs(n,det,num)
   end do
end program test

$$P_{0}(x) =1$$ $$P_{1}(x) =x$$ $$P_{2}(x) =-2+x^{2}$$ $$P_{3}(x) =-12x+x^{3}$$ $$P_{4}(x) =\frac{216}{5}-\frac{168}{5}x^{2}+x^{4}$$ $$P_{5}(x) =520x-\frac{220}{3}x^{3}+x^{5}$$ $$P_{6}(x) =-\frac{392400}{149}+\frac{382680}{149}x^{2}-\frac{20010}{149}x^{4}+x^{6}$$ $$P_{7}(x) =-\frac{5342400}{109}x+\frac{992880}{109}x^{3}-\frac{24360}{109}x^{5}+x^{7}$$ $$P_{8}(x) =\frac{339308928}{1073}-\frac{1917474048}{5365}x^{2}+\frac{4668384}{185}x^{4}-\frac{9216032}{26825}x^{6}+x^{8}$$ $$P_{9}(x) =\frac{48561690240}{6011}x-\frac{10820234880}{6011}x^{3}+\frac{364446432}{6011}x^{5}-\frac{3017160}{6011}x^{7}+x^{9}$$

Of course the coefficients of the $3$-term recurrence relations will depend on the normalization chosen, here as monic polynomials. I haven't checked these very thoroughly so some more verification might be in order.

EDIT: $$P_{10}(x) =-\frac{1842353840275200}{29286409}+\frac{2318870138659200}{29286409}x^{2}-\frac{199736268379200}{29286409}x^{4}+\frac{3777252968160}{29286409}x^{6}-\frac{20526107490}{29286409}x^{8}+x^{10}$$ $$P_{11}(x) =-\frac{5535810553996800}{2686261}x+\frac{1408500020188800}{2686261}x^{3}-\frac{58597576329600}{2686261}x^{5}+\frac{679907186640}{2686261}x^{7}-\frac{2547041860}{2686261}x^{9}+x^{11}$$