1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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

    statdad

    User Avatar
    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.

    [tex]
    \binom x y= \frac{x!}{y! (x-y)!}
    [/tex]

    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

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

    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

    statdad

    User Avatar
    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

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

    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

    [tex]
    t = x - y \quad \text{ so } \quad x = t+y
    [/tex]

    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

    [tex]
    \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!}
    [/tex]

    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:
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook




Similar Discussions: Maths summation algebra
  1. Summation ? (Replies: 4)

  2. Summation math help (Replies: 1)

  3. Summation Question (Replies: 7)

  4. Summation problem (Replies: 4)

Loading...