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Changing order of a double sum

  1. Feb 8, 2009 #1
    [SOLVED] Changing order of a double sum

    Hello everybody,

    I am a bit confused here, any help would be greatly appreciated..
    I have this double sum:

    [tex]\sum_{i=0}^{n-1}a_i \sum_{j=0}^{i} {i \choose j} b^{i-j}x^j[/tex]

    How can I take x out of the inner sum?

    Thank you very much in advance...
     
    Last edited: Feb 9, 2009
  2. jcsd
  3. Feb 8, 2009 #2
    Would it be helpful to use the binomial theorem?

    [tex]\sum_{j=0}^{i} {i \choose j} b^{i-j}x^j = (b + x)^i[/tex]
     
  4. Feb 8, 2009 #3
    Thanks for taking the time to answer!

    Well, no, I started from there, I want to write this as a polynomial of x in the usual way, that is, in the form:

    a_n*x^n+....+a_1*x+a_0

    I want to have only x there...
     
  5. Feb 8, 2009 #4
    It is possible, is it not?!
     
  6. Feb 9, 2009 #5
    Whoops, I thought I had hit the submit button hours ago, but apparently I didn't.

    For changing order of sums, the Iverson bracket
    http://xrl.us/befjqx
    is a useful tool.

    [tex]
    \sum_{i=0}^{n-1}a_i \sum_{j=0}^{i} \binom{i}{j} b^{i-j} x^j
    [/tex]
    [tex]
    =
    \sum_{i,j} [0 \le j \le i][0 \le i \le n-1] a_i \binom{i}{j} b^{i-j} x^j
    [/tex]
    [tex]
    =
    \sum_{i,j} [0 \le j \le i \le n-1] a_i \binom{i}{j} b^{i-j} x^j
    [/tex]
    [tex]
    =
    \sum_{j,i} [0 \le j \le n-1][j \le i \le n-1] a_i \binom{i}{j} b^{i-j} x^j
    [/tex]
    [tex]
    =
    \sum_{j=0}^{n-1} x^j \sum_{i=j}^{n-1} a_i \binom{i}{j} b^{i-j}.
    [/tex]
     
  7. Feb 9, 2009 #6
    Thanks so much for the help!!
     
  8. Feb 9, 2009 #7
    What a nice tool! I was struggling for so much time trying to change that variables!
     
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