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Maths summation algebra

  1. Jan 3, 2010 #1
    1. The problem statement, all variables and given/known data

    Ok I have the answer to a question, all the working is given, however, I'm having trouble following it.

    2. Relevant equations

    http://img695.imageshack.us/img695/426/answer.jpg [Broken]

    3. The attempt at a solution

    I am completely lost, could someone please explain the steps that have been taken, it would really help me.
    Last edited by a moderator: May 4, 2017
  2. jcsd
  3. Jan 3, 2010 #2


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    Homework Helper

    First line to second: use the definition of [tex] \binom x y [/tex], then factor out terms that do not depend on x.

    \binom x y= \frac{x!}{y! (x-y)!}

    and factor from the sum any term that does not depend on the index of summation, [tex] x [/tex]

    Second line to third: shift the origin of summation from [tex] y [/tex] to 0 by replacing the index of summation by [tex] x = y[/tex]. After this the sum becomes

    \sum_{x=0}^\infty \lambda^{x+y} \frac{0.9^x}{x!} = \lambda^y \sum_{x=0}^\infty \frac{(0.9\, \lambda)^x}{x!}

    You should be able to fill in the final step yourself.
  4. Jan 3, 2010 #3
    Thankyou very much for your explanation, I've pretty much got my head around it. Am I right i thinking the second line to the third all of the [tex] x [/tex] change to [tex] x+y [/tex] and is that a sort of rule when using summation? Thanks
  5. Jan 3, 2010 #4


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    Homework Helper

    I wouldn't say a rule, but a common bit of work. It's similar to making a substitution in a definite integral.

    The original sum in line 2 is

    \sum_{x=y}^\infty \frac{\lambda^x (0.9)^{x-y}}{(x-y)!}

    The form of the summand is similar to the infinite series for an exponential, but the starting value isn't zero. Suppose I use a new index
    of summation, defined as

    t = x - y \quad \text{ so } \quad x = t+y

    Since the original sum begins at [tex] x = y[/tex], the rewritten form begins at [tex] t = x - x = 0 [/tex]. In terms of the new variable the sum looks like

    \sum_{t = 0}^\infty \frac{\lambda^{t+y} (0.9)^t}{t!} = \lambda^y \sum_{t=0}^\infty \frac{\lambda^t (0.9)^t}{t!}

    Writing this new form with summation index equal to [tex] x [/tex] gives the form mentioned above.
    Last edited: Jan 3, 2010
  6. Jan 3, 2010 #5
    Thanks for your time, you've really helped me, I'm confident I understand this now. Thanks again!!! :biggrin:
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