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Math Help - Bessel function p=0

  1. #1
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    Bessel function p=0

    Hi there. Well, I'm stuck with this problem, which says:
    When p=0 the Bessel equation is: x^2y''+xy'+x^2y=0

    Show that its indicial equation only has one root and find the Frobenius solution correspondingly. (Answer: y=\sum \frac{(-1)^n}{ 2^{2n}(n!)^2 }x^{2n}

    Well, this is what I did:

    At first I normalized the equation:
    y''+\frac{y'}{x}+y=0
    Then
    P(x)=\frac{1}{ x} \rightarrow xP(x)=1
    Q(x)=1 \rightarrow x^2Q(x)=x^2
    So x=0 is regular singular point.


    Then the indicial equation is: \alpha(\alpha-1)+p_0\alpha+q_0=0 \rightarrow \alpha^2-alpha+\alpha=0 \rightarrow \alpha=0

    y=\sum_{n = 0}^\infty a_n x^n \rightarrow y'=\sum_{n = 1}^\infty a_n n x^{n-1} \rightarrow y''=\sum_{n = 2}^\infty a_n n(n-1) x^{n-2}

    Then x^2y''+xy'+x^2y=\sum_{n = 2}^\infty a_n n(n-1) x^n+ \sum_{n = 1}^\infty a_n n x^n+\sum_{n = 0}^\infty a_n x^{n+2} =
    =\sum_{n = 2}^\infty a_n n(n-1) x^n+ \sum_{n = 1}^\infty a_n n x^n+\sum_{n = 0}^\infty a_{n-2}x^n

    So from here I took
    a_n n(n-1)+a_{n-2}=0

    a_n=\frac{-a_{n-2}}{n(n-1)},n=2k \rightarrow a_{2k}=\frac{-a_{2k-2}}{2k(2k-1)}

    Then I've made some iterations, but I can't find the form that the problem gives as the answer, some of the iterations:

    a_2=\frac{-a_0}{2 },a_4=\frac{a_0}{4.3.2 },a_6=\frac{-a_0}{6.5.4.3.2 }

    So the answer I seem to get is a_{2k}=\frac{(-1)^k}{(2k)!}a_0

    But I should get \sum \frac{(-1)^n}{ 2^{2n}(n!)^2 }a_0 or something like that, which is how the answer the problem gives looks like.

    I'm probably doing something wrong, but I couldn't figure it out what it is.

    Bye there, thanks for helping!
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  2. #2
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    Quote Originally Posted by Ulysses View Post
    Hi there. Well, I'm stuck with this problem, which says:
    When p=0 the Bessel equation is: x^2y''+xy'+x^2y=0

    Show that its indicial equation only has one root and find the Frobenius solution correspondingly. (Answer: y=\sum \frac{(-1)^n}{ 2^{2n}(n!)^2 }x^{2n}

    Well, this is what I did:

    At first I normalized the equation:
    y''+\frac{y'}{x}+y=0
    Then
    P(x)=\frac{1}{ x} \rightarrow xP(x)=1
    Q(x)=1 \rightarrow x^2Q(x)=x^2
    So x=0 is regular singular point.


    Then the indicial equation is: \alpha(\alpha-1)+p_0\alpha+q_0=0 \rightarrow \alpha^2-alpha+\alpha=0 \rightarrow \alpha=0

    y=\sum_{n = 0}^\infty a_n x^n \rightarrow y'=\sum_{n = 1}^\infty a_n n x^{n-1} \rightarrow y''=\sum_{n = 2}^\infty a_n n(n-1) x^{n-2}

    Then x^2y''+xy'+x^2y=\sum_{n = 2}^\infty a_n n(n-1) x^n+ \sum_{n = 1}^\infty a_n n x^n+\sum_{n = 0}^\infty a_n x^{n+2} =
    =\sum_{n = 2}^\infty a_n n(n-1) x^n+ \sum_{n = 1}^\infty a_n n x^n+\sum_{n = 0}^\infty a_{n-2}x^n

    So from here I took
    a_n n(n-1)+a_{n-2}=0
    You missed the term from the middle summation.
    a_n n( n - 1) + a_n n + a_{n - 2} = 0

    -Dan
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  3. #3
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    Thanks! its done now.
    Last edited by Ulysses; May 7th 2011 at 03:39 PM. Reason: Done!
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