Question

Prove that the functions $L_n\left(x\right)$are orthogonal on$(0,\infty )$with respect to the weight function$e^{-x}$

Hint: Write the differential equation$\left(22.21\right)$as$e^x\frac{d}{dx}\left(x{e}^{-x}{y}^{\text{'}}\right)+ny=0$, and see Sections$7$and$19$ .

Short answer

The required equation is${\int }_0^{\infty }{e}^{-x}{L}_m\left(x\right)L_n\left(n\right)dx=0$.

Step-by-step solution

Concept of Differential equation and Orthogonal functions 

Differential equations are equations that connect one or more derivatives of a function. This implies that their answer is a function.

Two mathematical functions whose definite integral of their product is 0 given proper bounds are known as orthogonal functions.

Step 2:Determining Orthogonal function

Consider the differential equation $x{y}^{\text{'}\text{'}}+(1-x)y^{\text{'}}+ny=0$

This can be written as $e^x\frac{d}{dx}\left(x{e}^{-x}{y}^{\text{'}}\right)+ny=0$.

Write $1$for $L_n\left(x\right)$and $L_m\left(x\right)$

$\begin{array}{l}{e}^x\frac{d}{dx}\left(x{e}^{-x}{L}_n^{\text{'}}\left(x\right)\right)+n{L}_n\left(x\right)=0\dots \\ e^x\frac{d}{dx}\left(x{e}^{-x}{L}_m^{\text{'}}\left(x\right)\right)+m{L}_m\left(x\right)=0.\end{array}$

Multiply $2$by $L_n\left(x\right)$and $3$by $L_m\left(x\right)$and subtract

$\left\{L_m\left(x\right)e^x\frac{d}{dx}\left(x{e}^{-x}{L_n}^{\text{'}}\left(x\right)\right)-L_n\left(x\right)e^x\frac{d}{dx}\left(x{e}^{-x}{L}_m^{\text{'}}\left(x\right)\right)\right\}+(n-m)L_m\left(x\right)L_n\left(n\right)=0$

Step 2: Integrating the limits

Integrate in the limits $(0,\infty )$

$\begin{array}{c}{\int }_0^{\infty }\frac{d}{dx}\left[x{e}^{-x}\left(L_m\left(x\right){L_n}^{\text{'}}\left(x\right)-L_n\left(x\right)L_m^{\text{'}}\left(x\right)\right)\right]dx+(n-m){\int }_0^{\infty }{e}^{-x}{L}_m\left(x\right)L_n\left(n\right)dx=0\\ {\left[x{e}^{-x}\left(L_m\left(x\right)L_n^{\text{'}}\left(x\right)-L_n\left(x\right)L_m^{\text{'}}\left(x\right)\right)\right]}_0^{\infty }+(n-m){\int }_0^{\infty }{e}^{-x}{L}_m\left(x\right)L_n\left(n\right)dx=0\\ 1+{\int }_0^{\infty }{e}^{-x}{L}_m\left(x\right)L_n\left(n\right)dx=0\end{array}$

The integration term above will be equal to zero because for lower limit,$x=0$ and for higher $e^{-x}=0$.

Hence,${\int }_0^{\infty }{e}^{-x}{L}_m\left(x\right)L_n\left(n\right)dx=0$.