Some asymptotics on Bessel functions

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    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

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    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

    Pagina 2 di 2