Question

Problem 2. The Laguerre polynomials L (x) are orthogonal with respect to the inner product (u,u)=/o e-ru (x) u(x) dx and standardized so that L" (0) = 1 . In addition to their importance in numerical analysis, the La- guerre polynomials are notable for their use in electron orbitals in atoms Use Gram-Schmidt orthogonalization along with the standardization L(0)1 to find the Laguerre polynomials of degree ns4 The Laguerre polynomials obey the TTRR (n + 1) L" +1 (x) = (2n + 1-x)L" (x)-nL"-1 (x) with initial values Lo (x) and L1 (x)-1-x. Use the TTRR to find Ln (x) through degree 4. You should of course find the same polynomials as in part 1

Answer 1

We start with the polynomials and use Gram-Schmidt. Note that, using integration by parts we get

$\begin{align*}\int_0^\infty x^ne^{-x}~dx&=n!\end{align*}$

for $n\geq 0$ .

Let $u_0\equiv 1$ .

Let

$\begin{align*}v_1&=x-{\frac{\int_0^\infty xe^{-x}~dx}{\int_0^\infty e^{-x}~dx}}\\ &=x-\int_0^\infty xe^{-x}~dx\\ &=x-1\end{align*}$

To standardize, we let $\begin{align*}u_1&=1-x\end{align*}$ .

Let

$\begin{align*}v_2&=x^2-{\frac{\int_0^\infty x^2e^{-x}~dx}{\int_0^\infty e^{-x}~dx}}-{\frac{\int_0^\infty x^2(1-x)e^{-x}~dx}{\int_0^\infty (1-x)^2e^{-x}~dx}}(1-x)\end{align*}$

Since

$\begin{align*}\int_0^\infty x^2(1-x)e^{-x}~dx&=\int_0^\infty (2x-3x^2)e^{-x}~dx\\ &=\int_0^\infty (2-6x)e^{-x}~dx\\ &=2-6\\ &=-4\\ \int_0^\infty (1-x)^2e^{-x}~dx&=\int_0^\infty (1-2x+x^2)e^{-x}~dx\\ &=1-2\int_0^\infty xe^{-x}~dx+\int_0^\infty x^2e^{-x}~dx\\ &=1-2+2\\ &=1\end{align*}$

we get

$\begin{align*}v_2&=x^2-{\frac{\int_0^\infty x^2e^{-x}~dx}{\int_0^\infty e^{-x}~dx}}-{\frac{\int_0^\infty x^2(1-x)e^{-x}~dx}{\int_0^\infty (1-x)^2e^{-x}~dx}}(1-x)\\ &=x^2-2+4(1-x)\\ &=x^2-4x+2\end{align*}$

To standardize, we let

$\begin{align*}u_2&=0.5x^2-2x+1\end{align*}$

Let

$\begin{align*}v_3&=x^3-{\frac{\int_0^\infty x^3e^{-x}~dx}{\int_0^\infty e^{-x}~dx}}-{\frac{\int_0^\infty x^3(1-x)e^{-x}~dx}{\int_0^\infty (1-x)^2e^{-x}~dx}}(1-x)-{\frac{\int_0^\infty x^3(0.5x^2-2x+1)e^{-x}~dx}{\int_0^\infty (0.5x^2-2x+1)^2e^{-x}~dx}}(0.5x^2-2x+1)\end{align*}$

Since

$\begin{align*}\int_0^\infty x^3(1-x)e^{-x}~dx&=\int_0^\infty (x^3-x^4)e^{-x}~dx\\ &=3!-4!\\ &=-18\\ \int_0^\infty x^3(0.5x^2-2x+1)e^{-x}~dx&=0.5\cdot 5!-2\cdot 4!+3!\\ &=18\\ \int_0^\infty (0.5x^2-2x+1)^2e^{-x}~dx&=\int_0^\infty (0.25x^4+4x^2+1-2x^3+x^2-4x)e^{-x}~dx\\ &=0.5\cdot 4!+5\cdot 3!+1-2\cdot 3!-4\\ &=27\end{align*}$

we get

$\begin{align*}v_3&=x^3-{\frac{\int_0^\infty x^3e^{-x}~dx}{\int_0^\infty e^{-x}~dx}}-{\frac{\int_0^\infty x^3(1-x)e^{-x}~dx}{\int_0^\infty (1-x)^2e^{-x}~dx}}(1-x)-{\frac{\int_0^\infty x^3(0.5x^2-2x+1)e^{-x}~dx}{\int_0^\infty (0.5x^2-2x+1)^2e^{-x}~dx}}(0.5x^2-2x+1)\\ &=x^3-6+18(1-x)-1.5(0.5x^2-2x+1)\\ &=x^3-0.75x^2-15x-10.5\end{align*}$

To standardize, we let

$\begin{align*}u_3&=-{\frac 2{21}}(x^3-0.75x^2-15x-10.5)\end{align*}$

Finally, let

$\begin{align*}v_4&=x^4-{\frac{\int_0^\infty x^4e^{-x}~dx}{\int_0^\infty e^{-x}~dx}}-{\frac{\int_0^\infty x^4(1-x)e^{-x}~dx}{\int_0^\infty (1-x)^2e^{-x}~dx}}(1-x)\\ &~~~~~~-{\frac{\int_0^\infty x^4(0.5x^2-2x+1)e^{-x}~dx}{\int_0^\infty (0.5x^2-2x+1)^2e^{-x}~dx}}(0.5x^2-2x+1)\\ &~~~~~~~~~~+10.5{\frac{\int_0^\infty x^4(x^3-0.75x^2-15x-10.5)e^{-x}~dx}{\int_0^\infty (x^3-0.75x^2-15x-10.5)^2e^{-x}~dx}}(x^3-0.75x^2-15x-10.5)\\ &=x^4-24+96(1-x)-8(0.5x^2-2x+1)\\ &~~~~~~~~~~~~+10.5{\frac{\int_0^\infty x^4(x^3-0.75x^2-15x-10.5)e^{-x}~dx}{\int_0^\infty (x^3-0.75x^2-15x-10.5)^2e^{-x}~dx}}(x^3-0.75x^2-15x-10.5)\end{align*}$

Since

$\begin{align*}\int_0^\infty x^4(x^3-0.75x^2-15x-10.5)e^{-x}~dx&=7!-0.75\cdot 6!-15\cdot 5!-10.5\cdot 4!\\ &=440\\ \int_0^\infty (x^3-0.75x^2-15x-10.5)^2e^{-x}~dx&=\int_0^\infty (x^6+9x^4/16+225x^2+110.25-1.5x^5-30x^4-21x^3+22.5x^3+15.75x^2+315x)e^{-x}~dx\\ &=6!+4!\cdot 9/16+450+110.25-1.5\cdot 5!-30\cdot4!+1.5\cdot 3!+31.5+315\\ &=720+13.5+450+110.25-180-720+9+346.5\\ &=740.25\end{align*}$

we get

$\begin{align*}v_3&=x^3-{\frac{\int_0^\infty x^3e^{-x}~dx}{\int_0^\infty e^{-x}~dx}}-{\frac{\int_0^\infty x^3(1-x)e^{-x}~dx}{\int_0^\infty (1-x)^2e^{-x}~dx}}(1-x)-{\frac{\int_0^\infty x^3(0.5x^2-2x+1)e^{-x}~dx}{\int_0^\infty (0.5x^2-2x+1)^2e^{-x}~dx}}(0.5x^2-2x+1)\\ &=x^3-6+18(1-x)-1.5(0.5x^2-2x+1)\\ &=x^3-0.75x^2-15x-10.5\end{align*}$

Thus,

$\begin{align*}v_4&=x^4-24+96(1-x)-8(0.5x^2-2x+1)\\ &~~~~~~~~~~~~+{\frac{4620}{740.25}}(x^3-0.75x^2-15x-10.5)\end{align*}$

To standardize, we divide by its constant term.

Then, $\begin{align*}u_0,u_1,u_2,u_3,u_4\end{align*}$ are the required polynomials.