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7/27/2019 Some asymptotics on Bessel functions
1/2
A note about Bessel functions of the first kind
Jack DAurizio
July 11, 2013
The starting point of our study is the integral:
In() =1
0
cos(nx) e cosxdx.
By considering the Taylor series of ey and the Fourier series of (cos x)m, we have that:
In() =+=0
(/2)n+2
!(n + )!.
Moreover, integrating by parts and using Briggs formulas we have:
In =
2n(In1 In+1) ,
d
dIn =
1
2(In1 + In+1) .
From the second identity, we have that the ratio between In and In1 is the continued fraction:
InIn1
= 12n + 12
(n+1)+...
,
that obviously gives:
In (/2)n
n!I0.
By defining:
Jn =n!
(/2)nIn,
so J0 = I0, we have that Jn is decreasing as n goes from 0 to +, and since:
Jn =+
=0
(/2)2
!2 n+
exp 24(n + 1)
we have that limn+ Jn = 1.
Jn = Jn1 2
4n(n + 1)Jn+1.
Jn 2
4I0 + (I0 1)n
2
4 + (I0 1)n,
1
Jn 1
1
I0 1+
4n
2.
M=0
1
n+
=
n
n 1
1
1
M+nM+1
1n+
= 1 nn + 1
1k=0
1n+1+k
k
1
7/27/2019 Some asymptotics on Bessel functions
2/2
M=0
1
!2(M )!2=
(2M)!
M!4
M=0
1
!2(M )!2n+
= n!(2M+ n)!(M+ n)!2M!2M=0
1
!2(M )!2n+n
n+M
n
= n!2(2M+ 2n)!(M+ 2n)!(M+ n)!2M!
+=0
(/2)2
!2 n+
= +=0
(/2)2
!2
(/2)2+2
(+ 1)!2
n
n 1
1
1n++1
Jn [1;4n
2, (I0 1)]
JnJn1 = [0; 1,
4
2 n(n + 1),n + 2
n ,4
2 n(n + 3), . . .]
Jn+1 = n + 1
/2d
dJn
Conosciamo lespressione della derivata logaritmica rispetto ad di Jn e vogliamo provare che
J0 Jn =
nk=1
(/2)2
k(k + 1)Jk+1
4n
2(Jn 1)(J0 1).
O semplicemente provare la convessita (rispetto ad n) della funzione 1Jn1
. Abbiamo anche:
J0 Jn = nn + 1
+=1
(/2)2
!2
1k=0
1n+1+kk
= nn + 1+k=1
1n+kk1
+=k
(/2)2
!2
che sembra estremamente promettente. In particolare si ha:
(Jn 1) 1
n+11
(J0 1) ,(Jn 1)
1n+22
J0 1 + n2
(/2)2,
(Jn 1) 1
n+33 (J0 1) +
1
n+11 1
n+33
(/2)2 +
1
n+22 1
n+33
(/2)4
2!2
Jn 1 1
n+KK
(J0 1) + K1=1
1
n+
1n+KK
(/2)2
!2
Inoltre, al crescere di K le disuguaglianze ottenute sono progressivamente piu strette,
k k1 =n(k 1)!
(n + 1) . . . (n + k)
+=k
(/2)2
!2
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