Q. 65 - Calculus

Question

The second-order differential equation 

$x^2{y}^{\text{'}\text{'}}+x{y}^{\text{'}}+\left(x^2-p^2\right)=0$

where p is a nonnegative integer, arises in many applications in physics and engineering, including one model for the vibration of a beaten drum. The solution of the differential equation is called the Bessel function of order p, denoted by $J_p\left(x\right)$. It may be shown that $J_p\left(x\right)$ is given by the following power series in x: 

$J_p\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{k!(k+p)!2^{2k+p}}{x}^{2k+p}$

Find and graph the first four terms in the sequence of partial sums of $J_o\left(x\right)$. 

Step-by-Step Solution

Answer

The four terms are $1,1-\frac{1}{4}{x}^2,1-\frac{1}{4}{x}^2+\frac{1}{64}{x}^4,1-\frac{1}{4}{x}^2+\frac{1}{64}{x}^4-\frac{1}{2304}{x}^6$

And the graph is 

Step 1. Given information

An expression is given as $J_p\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{k!(k+p)!2^{2k+p}}{x}^{2k+p}$

Step 2. Finding four terms

The Bessel function is given in the order of p. So the value of $J_o\left(x\right)$ is

$$\begin{gathered} J_0\left(x\right)=\sum _{k=0}^{\infty }\frac{(-1{)}^k}{k!(k+0)!2^{2k+0}}{x}^{2k+0} \\ =\sum _{k=0}^{\infty }\frac{(-1{)}^k}{(k!{)}^2{2}^{2k}}{x}^{2k} \end{gathered}$$

Therefore the first four terms of partial sums are as,

$1,1-\frac{1}{4}{x}^2,1-\frac{1}{4}{x}^2+\frac{1}{64}{x}^4,1-\frac{1}{4}{x}^2+\frac{1}{64}{x}^4-\frac{1}{2304}{x}^6$

And the graph is