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Domain of the Bessel Function

  1. Oct 4, 2009 #1
    1. The problem statement, all variables and given/known data
    Find domain of [tex]\sum_{n= 0}^\infty \frac{(-1)^{n}x^{2n}}{2^{2n}(n!)^{2}}[/tex]


    2. Relevant equations


    3. The attempt at a solution
    I set it all up but I can't really seem to simplify it.

    [tex]\frac{(-1)^{n+1}x^{2(n+1)}}{2^{2n+2}(n+1)!^{2}}\bullet\frac{2^{2n}(n!)^{2}}{(-1)^{n}x^{2n}} [/tex]
     
  2. jcsd
  3. Oct 4, 2009 #2

    Dick

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    Why can't you simplify it? What's 2^(2n)/(2^(2n+2))? What's n!/(n+1)!?
     
  4. Oct 4, 2009 #3
    You mean polynomial division?
     
  5. Oct 4, 2009 #4

    Dick

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    No, I mean the law of exponents for the first one and realizing (n)!/(n+1)!=(1*2*3*...*n)/(1*2*3*...*n*(n+1)) for the second one.
     
  6. Oct 4, 2009 #5
    Explain what the mean by showing me on this much simpler one.

    [tex]\sum_{n= 0}^\infty n!x^{n}[/tex]

    Setting it up I get [tex]\frac{(1+n!)x^{n+1}}{n!x^{n}}[/tex]

    Do I cross out the factors?
     
  7. Oct 4, 2009 #6

    Dick

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    You meant to write (1+n)!, I hope, not (1+n!). Yes, you just cancel the common factors in the numerator and denominator. What do you get?
     
  8. Oct 4, 2009 #7
    So, I got x+1 but I must have messed something up.
     
  9. Oct 4, 2009 #8

    Dick

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    Apparently, but not showing how you got that doesn't make it easy to help. What are x^(n+1)/x^n and (n+1)!/n!?????
     
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