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Determine for which values the Bessel function of order one, J 1


(x)=∑ n=0
[infinity]

2 2n+1
(n!)(n+1)!
(−1) n
x 2n+1

, converges.

1 Answer

2 votes

Answer:

Step-by-step explanation:

The Bessel function of the first kind and order one, denoted as J1(x), can be represented by the following series expansion:

J1(x) = ∑[n=0 to ∞] ((-1)^n / n! * (n+1)! * (x/2)^(2n+1))

To determine the values for which this series converges, we need to examine the convergence properties of the terms in the series.

For a series to converge, the individual terms of the series must approach zero as n approaches infinity. Let's analyze the behavior of the terms in the series:

Let a_n = ((-1)^n / n! * (n+1)! * (x/2)^(2n+1))

We can consider the absolute value of the terms:

|a_n| = (1 / n! * (n+1)! * (x/2)^(2n+1))

Now, let's use the ratio test to investigate the convergence of the series:

lim[n->∞] |a_(n+1)| / |a_n| = lim[n->∞] ((1 / (n+1)!) * ((n!) / ((n+1)!) * (x/2)^(2n+3)) * ((n+1)! / n!) * (x/2)^(2n+1))

= lim[n->∞] ((x^2 / 4) / (n+1) * (n+1))

= (x^2 / 4) * lim[n->∞] (1 / (n+1))

The limit as n approaches infinity of (1 / (n+1)) is zero. Therefore, the ratio test gives us:

lim[n->∞] |a_(n+1)| / |a_n| = (x^2 / 4) * 0 = 0

Since the limit is less than 1 (in this case, it is zero), the ratio test indicates that the series converges for all values of x.

Hence, the Bessel function of order one, J1(x), converges for all real values of x.

User Andrew McNamee
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