Question

Usingverify $(22.19)$and $\left(22.20\right)$also find$L_3\left(x\right)$and$L_4\left(x\right)$.

Short answer

The value of the $L_3\left(x\right)$and $L_4\left(x\right)$is given by :

$\begin{array}{l}{L}_3\left(x\right)=1-x+\frac{3x^2}{2}-\frac{x^3}{6}\\ L_4\left(x\right)=1-4x+3x^2-\frac{2x^3}{3}+\frac{x^4}{24}\end{array}$

Step-by-step solution

Concept of Differential equation

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

Step 2:Determining equation with the help of Leibniz’ rule

The equation is given as:

$\begin{array}{l}{L}_n\left(x\right)=\sum _{m=0}^n{(-1)}^m{n}_{c_m}\frac{x^m}{m!}\\ L_n\left(x\right)=\sum _{m=0}^n{(-1)}^m{n}_{c_m}\frac{x^m}{m!}\\ L_0\left(x\right)=1\\ L_1\left(x\right)=\sum _{m=0}^1{(-1)}^m{1}_{c_m}\frac{x^m}{m!}\end{array}$

So, the value obtained of $L_1\left(x\right)$is $1-x$.

$\begin{array}{l}{L}_2\left(x\right)=\sum _{m=0}^2{(-1)}^m{2}_{c_m}\frac{x^m}{m!}\\ L_2\left(x\right)=1-2x+\frac{x^2}{2!}\\ L_3\left(x\right)=\sum _{m=0}^3{(-1)}^m{3}_{c_m}\frac{x^m}{m!}\\ L_3\left(x\right)=1-x+\frac{3x^2}{2}-\frac{x^3}{6}\end{array}$

Similarly find the series for $L_4\left(x\right)$.

$L_4\left(x\right)=1-4x+3x^2-\frac{2x^3}{3}+\frac{x^4}{24}$

The results for the given sequence are:

$\begin{array}{l}{L}_3\left(x\right)=1-x+\frac{3x^2}{2}-\frac{x^3}{6}\\ L_4\left(x\right)=1-4x+3x^2-\frac{2x^3}{3}+\frac{x^4}{24}\end{array}$.