HONORS CALCULUS I, FALL 2015 - BESSEL FUNCTIONS
For each nonnegative integer r, we define the Bessel function of order r to be
Jr (x) =
∞
x r X
2
(−1)n x 2n
.
n!(n + r)! 2
n=0
1. Bessel Function of Order 0
We first study the r = 0 case:
J0 (x) =
∞
X
(−1)n x2n
.
22n (n!)2
n=0
This is a power series of the form
∞
X
an (x − x0 )n ,
n=0
where x0 = 0, a2n−1 = 0 for all n ≥ 1, and
a2n =
(−1)n
.
22n (n!)2
1.1. Computation of the Radius of Convergence. Set
bn (x) =
(−1)n x2n
22n (n!)2
for each n ≥ 0, so that
J0 (x) =
∞
X
bn (x).
n=0
Observe that, for a fixed x ∈ R,
bn+1 (x) (−1)n+1 x2n+2 /22n+2 ((n + 1)!)2 lim
= lim n→∞ bn (x) n→∞
(−1)n x2n /22n (n!)2
−x2
=0
= lim 2
2
n→∞ 2 (n + 1) regardless of the choice of |x|. The ratio test implies that J0 converges absolutely
for all x ∈ R.
Date: October 2, 2015.
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HONORS CALCULUS I, FALL 2015 - BESSEL FUNCTIONS
1.2. A Cautionary Remark. We remark that it is impractical to use the powerseries radius of convergence formulas
−1
−1 p
|an+1 |
R = lim n |an |
= lim
n→∞
n→∞ |an |
directly, as the odd terms a2n+1 are zero. We can nevertheless use the ratio test or
the root test directly to the series
∞
X
bn (x)
n=0
for each fixed x.
1.3. Root Test? We could, of course, apply the root test to
∞
X
bn (x)
n=0
as well. Let’s see what we get:
s
lim
p
n
n→∞
n
|bn (x)| = lim
n→∞
|(−1)n x2n |
|x|2
.
= lim
2n
2
n→∞ 4(n!)2/n
2 (n!)
It would appear that we need to know how to compute
lim (n!)2/n .
n→∞
To this end, we observe that a simple arrangement yields the following:
(n!)2 = (n · 1)((n − 1) · 2)((n − 2) · 3) · · · (3 · (n − 2))(2 · (n − 1))(1 · n).
In other words,
(n!)2 =
n
Y
(n + 1 − k)k.
k=1
Now, for each 1 ≤ k ≤ n,
(n + 1 − k)k − n = nk − n + k − k 2 = n(k − 1) + k(k − 1) = (n + k)(k − 1) ≥ 0,
and so
(n + 1 − k) ≥ n.
It follows that
(n!)2 =
n
Y
(n + 1 − k)k ≥
k=1
n
Y
n = nn ,
k=1
whence it follows that
lim (n!)2/n ≥ lim n = ∞.
n→∞
n→∞
Returning to the task at hand, we see that
p
lim n |bn (x)| = lim
n→∞
|x|2
=0
n→∞ 4(n!)2/n
regardless of the choice of x. The root test therefore implies that J0 converges
absolutely for all x.
HONORS CALCULUS I, FALL 2015 - BESSEL FUNCTIONS
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1.4. Bessel’s Differential Equation of Order 0. We now show that
x2 J000 (x) + xJ00 (x) + x2 J0 (x) = 0
for all x ∈ R. To this end, we observe that
J00 (x) =
=
∞
∞
X
X
(−1)n (2n)x2n−1
(−1)n x2n−1
=
22n (n!)2
22n−1 n!(n − 1)!
n=1
n=1
∞
X
(−1)n+1 x2n+1
(−1)n+1 x2(n+1)−1
=
.
22(n+1)−1 (n + 1)!((n + 1) − 1)! n=0 22n+1 n!(n + 1)!
n=0
∞
X
Observe that J00 (x) does not have a constant term, and so the computation of J000 (x)
does not eliminate the n = 0 term:
∞
X
(−1)n+1 (2n + 1)x2n
J000 (x) =
.
22n+1 n!(n + 1)!
n=0
Now, for each x ∈ R,
x2 J000 (x) + xJ00 (x) + x2 J0 (x) =
∞
X
(−1)n x2n+2
[2(n + 1) − 1 − (2n + 1)] = 0,
22n+1 n!(n + 1)!
n=0
as was to be shown.
2. Bessel Functions of Order r
In this section, we shall compute the radius of convergence of Jr for a fixed
positive integer r. We shall also show that y = Jr (x) is a solution to Bessel’s
differential equation
x2 y 00 + xy 0 + (x2 − r2 )y = 0.
r
2.1. Computation of the Radius of Convergence. The x2 term is irrelevant
in computing the radius of convergence: it suffices to investigate
∞
X
(−1)n x 2n
.
n!(n + r)! 2
n=0
We set
bn =
(−1)n x 2n
n!(n + r)! 2
for each n ≥ 0 and observe that
bn+1 (−1)n+1 x2n+2 /22+2n (n + 1)!(n + r + 1)! =
bn (−1)n x2n /22n n!(n + r)!
2
2
|x|
−x
=
= .
4(n + 1)(n + r + 1) 4(n + 1)(n + r + 1)
Since
bn+1 =0<1
lim n→∞
bn regardless of our choice of x, the ratio test implies that
∞
X
(−1)n x 2n
n=0
n!(n + r)!
2
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HONORS CALCULUS I, FALL 2015 - BESSEL FUNCTIONS
converges absolutely on (−∞, ∞). It follows that the radius of convergence of Jr
is ∞.
2.2. Bessel’s Differential Equation of Order r. Now,
Jr (x) =
∞
X
(−1)n x 2n+r
,
n!(n + r)! 2
n=0
and r ≥ 1, and so Jr has no constant term. Therefore,
∞
X
(−1)n (2n + r) x 2n+r−1
.
2(n!)(n + r)! 2
n=0
Jr0 (x) =
If r = 1, then
J10 (x) =
∞
X
(−1)n (2n + 1) x 2n
,
2(n!)(n + 1)! 2
n=0
and so J10 (x) has a constant term. Therefore,
J100 (x) =
∞
X
(−1)n (2n + 1)(2n) x 2n−1
4(n!)(n + 1)!
2
n=1
If r > 1, then
Jr0 (x) =
∞
X
(−1)n (2n + r) x 2n+r−1
2(n!)(n + r)! 2
n=0
does not have a constant term, and so
Jr00 (x) =
∞
X
(−1)n (2n + r)(2n + r − 1) x 2n+r−2
.
4(n!)(n + r)!
2
n=0
Note, however, that r = 1 implies that
(−1)n (2n + r)(2n + r − 1) x 2n+r−2
4(n!)(n + r)!
2
=
(−1)n (2n + 1)(2n) x 2n−1
.
4(n!)(n + 1)!
2
When n = 0, we obtain the following:
(−1)n (2 · 0 + 1)(2 · 0) x 2·0−1
= 0.
4(0!)(0 + 1)!
2
It follows that
Jr00 (x)
∞
X
(−1)n (2n + r)(2n + r − 1) x 2n+r−2
=
4(n!)(n + r)!
2
n=0
in the r = 1 case as well, despite the presence of the x1 term. In other words, there
is no need to consider the r = 1 case and the r > 1 case separately.
Let us now verify that Jr (x) satisfies Bessel’s differential equation
x2 Jr00 (x) + xJr0 (x) + (x2 − r2 )Jr (x) = 0.
HONORS CALCULUS I, FALL 2015 - BESSEL FUNCTIONS
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Since
x2 Jr00 (x) =
xJr0 (x) =
(x2 − r2 )Jr (x) =
∞
X
(−1)n (2n + r)(2n + r − 1) x 2n+r
n!(n + r)!
2
n=0
∞
X
(−1)n (2n + r) x 2n+r
n!(n + r)!
2
n=0
∞
X
(−1)n (x2 − r2 ) x 2n+r
,
n!(n + r)!
2
n=0
we see that
x2 Jr00 (x) + xJr0 (x) + (x2 − r2 )Jr (x)
∞
X
(−1)n x 2n+r
=
(2n + r)(2n + r − 1) + (2n + r) + (x2 − r2 )
n!(n + r)! 2
n=0
∞
X
=
(−1)n x 2n+r
4n2 + 4nr + x2
n!(n
+
r)!
2
n=0
=
∞
∞
X
(−1)n 4n(n + r) x 2n+r X (−1)n x2 x 2n+r
+
.
n!(n + r)!
2
n!(n + r)! 2
n=0
n=0
Now, when n = 0,
(−1)0 4 · 0 · (0 + r) x 2·0+r
(−1)n 4n(n + r) x 2n+r
=
,
n!(n + r)!
2
0!(0 + r)!
2
and so
∞
∞
X
(−1)n 4n(n + r) x 2n+r X (−1)n 4n(n + r) x 2n+r
=
.
n!(n + r)!
2
n!(n + r)!
2
n=1
n=0
It now follows that
x2 Jr00 (x) + xJr0 (x) + (x2 − r2 )Jr (x)
∞
∞
X
(−1)n 4n(n + r) x 2n+r X (−1)n x2 x 2n+r
+
=
n!(n + r)!
2
n!(n + r)! 2
n=1
n=0
=
=
∞
X
∞
x 2n+r X
(−1)n 4
(−1)n 4 x 2n+r+2
+
(n − 1)!(n + r − 1)! 2
n!(n + r)! 2
n=1
n=0
∞
X
∞
x 2n+r X
x 2n+r
(−1)n 4
(−1)n−1 4
+
= 0,
(n − 1)!(n + r − 1)! 2
(n − 1)!(n + r − 1)! 2
n=1
n=1
as was to be shown.