Question

Find the domain of the Bessel function of order 0 defined by [infinity]J0(x) = Σ (−1)^nx^2n/ 2^2n(n!)^2 n = 0

Answer 1

The domain of J_0(x) is all real numbers, i.e., (-∞, ∞).

The Bessel function of order 0, denoted by J_0(x), is defined by the infinite series:

J_0(x) = Σ (-1)^n x^(2n) / (2^2n(n!)^2)

To determine the domain of J_0(x), we need to investigate the convergence of this series. We can use the ratio test to do this.

The ratio test states that if the limit of the absolute value of the ratio of successive terms is less than 1, then the series converges; if the limit is greater than 1, the series diverges; and if the limit is 1, the test is inconclusive.

Let's apply the ratio test to the series for J_0(x):

lim |a_(n+1) / a_n| = lim |x^2| / (2^2(n+1)!^2) = 0

Since the limit is 0, which is less than 1, the series converges for all real values of x. Therefore, the domain of J_0(x) is all real numbers, i.e., (-∞, ∞).

Answer 2

Following are the given series for all x:

Explanation:

Given equation:

`\bold{J_0(x)=\sum_(n=0)^(\infty)(((-1)^(n)(x^(2n))))/((2^(2n))(n!)^2)}\\`

Let the value a so, the value of
`a_n` and the value of
`a_(n+1)`is:

`\to a_n=((-1)^2n x^(2n))/(2^(2n)(n!)^2)`

`\to a_((n+1))=((-1)^(n+1) x^(2(n+1)))/(2^(2(n+1))((n+1))!^2)`

To calculates its series we divide the above value:

`\left | (a_(n+1))/(a_n)\right |= (((-1)^(n+1) x^(2(n+1)))/(2^(2(n+1))((n+1))!^2))/(((-1)^2n x^(2n))/(2^(2n)(n!)^2))\\\\`

`= \left | ((-1)^(n+1) x^(2(n+1)))/(2^(2(n+1))((n+1))!^2) \cdot \frac {2^(2n)(n!)^2}{(-1)^2n x^(2n)} \right |`

`= \left | ( x^(2n+2))/(2^(2n+2)(n+1)!^2) \cdot \frac {2^(2n)(n!)^2}{x^(2n)} \right |`

`= \left | ( x^(2n+2))/(2^(2n+2)(n+1)^2 (n!)^2) \cdot \frac {2^(2n)(n!)^2}{x^(2n)} \right |\\\\= \left | (x^(2n)\cdot x^2)/(2^(2n) \cdot 2^2(n+1)^2 (n!)^2) \cdot \frac {2^(2n)(n!)^2}{x^(2n)} \right |\\\\`

`= (x^2)/(2^2(n+1)^2)\longrightarrow 0 <1` for all x

The final value of the converges series for all x.