A solution for problem 8.6.32a in Stewart Niels Joaquin The Bessel function of order 1 is defined by J 1 (x)= ∞ n=0 (-1) n x 2n+1 n!(n + 1)!2 2n+1 Show that J 1 satisfies the differential equation x 2 J 1 (x)+ xJ 1 (x)+(x 2 - 1)J 1 (x)=0 This solution is deliberately long and drawn-out so that you can see each step clearly. The computation is lengthy but not difficult to understand. First, we calculate the derivatives. J 1 (x)= ∞ n=0 (-1) n (2n + 1)x 2n n!(n + 1)!2 2n+1 J 1 (x)= ∞ n=0 (-1) n (2n + 1)(2n)x 2n-1 n!(n + 1)!2 2n+1 (We could let the second derivative start at n = 1, but I will defer changing indices for now.) Plugging our derivatives into the differential equation, we get x 2 J 1 (x)+ xJ 1 (x)+(x 2 - 1)J 1 (x) = x 2 ∞ n=0 (-1) n (2n + 1)(2n)x 2n-1 n!(n + 1)!2 2n+1 + x ∞ n=0 (-1) n (2n + 1)x 2n n!(n + 1)!2 2n+1 +(x 2 - 1) ∞ n=0 (-1) n x 2n+1 n!(n + 1)!2 2n+1 = ∞ n=0 (-1) n (2n + 1)(2n)x 2n+1 n!(n + 1)!2 2n+1 + ∞ n=0 (-1) n (2n + 1)x 2n+1 n!(n + 1)!2 2n+1 + ∞ n=0 (-1) n x 2n+3 n!(n + 1)!2 2n+1 - ∞ n=0 (-1) n x 2n+1 n!(n + 1)!2 2n+1 1
A solution for problem 8.6.32a in StewartNiels Joaquin
The Bessel function of order 1 is defined by
J1(x) =n=0
(1)nx2n+1n!(n+ 1)!22n+1
Show that J1 satisfies the differential equation
x2J 1 (x) + xJ1(x) + (x
2 1)J1(x) = 0
This solution is deliberately long and drawn-out so that you can
see eachstep clearly. The computation is lengthy but not difficult
to understand.First, we calculate the derivatives.
J 1(x) =n=0
(1)n(2n+ 1)x2nn!(n+ 1)!22n+1
J 1 (x) =n=0
(1)n(2n+ 1)(2n)x2n1n!(n+ 1)!22n+1
(We could let the second derivative start at n = 1, but I will
defer changingindices for now.) Plugging our derivatives into the
differential equation, weget
x2J 1 (x) + xJ1(x) + (x
2 1)J1(x)
= x2n=0
(1)n(2n+ 1)(2n)x2n1n!(n+ 1)!22n+1
+ xn=0
(1)n(2n+ 1)x2nn!(n+ 1)!22n+1
+ (x2 1)n=0
(1)nx2n+1n!(n+ 1)!22n+1
=n=0
(1)n(2n+ 1)(2n)x2n+1n!(n+ 1)!22n+1
+n=0
(1)n(2n+ 1)x2n+1n!(n+ 1)!22n+1
+n=0
(1)nx2n+3n!(n+ 1)!22n+1
n=0
(1)nx2n+1n!(n+ 1)!22n+1
1
We can combine these four summations into one.
=n=0
[(1)n(2n+ 1)(2n)x2n+1
n!(n+ 1)!22n+1+
(1)n(2n+ 1)x2n+1n!(n+ 1)!22n+1
+(1)nx2n+3
n!(n+ 1)!22n+1 (1)
nx2n+1
n!(n+ 1)!22n+1
]=
n=0
(1)n[(2n+ 1)(2n) + (2n+ 1) + x2 1
n!(n+ 1)!22n+1
]x2n+1
=n=0
(1)n[4n2 + 4n+ x2
n!(n+ 1)!22n+1
]x2n+1
Its a little difficult to work with that x2 term. Lets try to
take care of itby separating it.
=n=0
(1)n[
4n2 + 4nn!(n+ 1)!22n+1
+x2
n!(n+ 1)!22n+1
]x2n+1
=n=0
(1)n[
4n2 + 4nn!(n+ 1)!22n+1
]x2n+1 +
n=0
(1)n[
x2
n!(n+ 1)!22n+1
]x2n+1
=n=0
(1)n[
4n2 + 4nn!(n+ 1)!22n+1
]x2n+1 +
n=0
(1)n[
x2n+3
n!(n+ 1)!22n+1
]We want the x2n+3 in the second series to look like x2n+1. We
can do thisby starting the series at n = 1 and subtracting 1 from
each n. Furthermore,notice that the first term of the first series
is 0, so we can start this seriesat n = 1 as well without having to
change anything else.
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