Find $a,b,c\in \mathbb{R}$ that minimize the integral $\int_0^\infty |t^3 -at^2 -bt-c|^2e^{-t}$dt

Question

Hint:Use orthogonality of $(P_n)_{n=0}^\infty$ in $H=L_{2,\rho}(\mathbb{R}_+)$ with $\rho(t)=e^{-t}$

Where $P_n$ Laguerre Polynomials satisfying the recurrence relation:

$(n+1)L_{n+1}(t)=(2n+1-t)L_n(t)-nL_{n-1}$ and $L_0=1, L_1(t)=1-t$

I did something similar here http://math.stackexchange.com/questions/415885/question-about-finding-minimum-hilbert-spaces/415906#415906 (actually, it is same, just here are different coefficients) — Cortizol, Apr 24 '14 at 16:27

user108539 · Accepted Answer · 2014-04-24T16:51:55.040

I first calculated $L_2=1-2t+{t^2}/2 $ and $L_3=1-3t+3t^2/2-t^3/6$ and $||L_0||=1, ||L_1||=1, ||L_2||=2 $ with the help of $\int_0^\infty t^ne^{-t}dt=\Gamma(n+1)=n!$ (note that $\widetilde{L_2}=L_2/2=1/2-t+t^2/4$

Now I need to write $t^3$ in terms of $L_0,L_1,\widetilde{L_2},L_3$, i.e. $t^3=6L_0-18L_1+36L_2-6L_3$

Then $<t^3-P_2,t^3-P_2>$ where $P_2=at^2+bt+c$ which will give $<-6L_3,-6L_3>=36<L_3,L_3>=36||L_3||^2=36.\int (L_3)^2e^{-t}dt$ which is desired minimum value and we can find a,b,c by opening the terms of $t^3$ except $L_3$

Is this correct?

Find $a,b,c\in \mathbb{R}$ that minimize the integral $\int_0^\infty |t^3 -at^2 -bt-c|^2e^{-t}$dt

1 Answers1