Answer to Question #304168 in Calculus for itz_perfect

Question #304168

The function J1 defined by

J1(x) = ∑((((−1)^n).x^(2n+1))/(n! (n + 1)! 2)).2n+1

n=0


,


is called the Bessel function of order 1.

a) Find its domain.

b) Graph the first several partial sums on a common screen using programming software.


1
Expert's answer
2022-03-02T16:29:58-0500

a) the domain of the function is the interval of convergence of power series. Let an=(1)nx2n+1n!(n+1)!22n+1a_n=\frac{(-1)^{n}x^{2n+1}}{n!(n+1)!2^{2n+1}} and find the limnan+1anlim_{n\to\infin}|\frac{a_{n+1}}{a_n}|

(1)n+1x2(n+1)+1(n+1)!((n+1)+1)!22(n+1)+1(1)nx2n+1n!(n+1)!22n+1=x2n+3(n+1)!(n+2)!22n+3×n!(n+1)!22n+1x2n+1|\frac{\frac{(-1)^{n+1}x^{2(n+1)+1}}{(n+1)!((n+1)+1)!2^{2(n+1)+1}}}{\frac{(-1)^{n}x^{2n+1}}{n!(n+1)!2^{2n+1}}}|=|\frac{x^{2n+3}}{(n+1)!(n+2)!2^{2n+3}}\times\frac{n!(n+1)!2^{2n+1}}{x^{2n+1}}|

=x2(n+1)n!(n+1)(n+2)!22×n!(n+1)!=|\frac{x^2}{(n+1)n!(n+1)(n+2)!2^2}\times n!(n+1)!|

=x24(n+1)(n+2)=|\frac{x^2}{4(n+1)(n+2)}|

0 as n \to0\ as\ n\ \to\infin

By ratio test, Since limnan+1an=01  xlim_{n\to\infin|\frac{a_{n+1}}{a_{n}}|}=0\le1\ \forall\ x , The given series converges ,With this ,write write the interval which is exactly the domain of the function

(,)(-\infin,\infin)

b)

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