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Changing the Limits of Summation

  1. Jan 3, 2015 #1
    1. The problem statement, all variables and given/known data

    Dear Mentors and PF helpers,

    Here's my question, I see these on my textbook but couldn't really understand how they derived this short cut.
    Please show me how they got to these. Thank you for your time.

    image.jpg

    2. Relevant equations

    These is what I understand from now.

    image.jpg

    3. The attempt at a solution
     
  2. jcsd
  3. Jan 3, 2015 #2

    ShayanJ

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    Gold Member

    So we have the sum [itex] \displaystyle \sum_{r=m}^n T_r [/itex]. We may change the index of summation from r to s=r+k. Then because r starts from m, s should start from m+k and because r ends at n, s should end at n+k. This means that we changed our summation to [itex] \displaystyle \sum_{s=m+k}^{n+k}T_{s-k} [/itex]. But you should note that this sum doesn't depend on s as the first sum didn't depend on r. r and s are just dummy indices because after you do the sums, there'll remain no sign of them. So I can safely rename s to r. But because the sums are just numbers and because the changes I did to the first sum doesn't change the result, I'll have:
    [itex] \displaystyle \sum_{r=m}^n T_r=\sum_{r=m+k}^{n+k}T_{r-k} [/itex].
     
  4. Jan 3, 2015 #3
    Dear Shyan,

    Do you mind giving me a better view with these words, can give your explanation with examples to explain these.

    Thank you for your time
     
  5. Jan 3, 2015 #4

    ShayanJ

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    Gold Member

    [itex] \displaystyle \sum_{r=m}^n T_r=T_m+T_{m+1}+T_{m+2}+\dots+T_{n-2}+T_{n-1}+T_n [/itex]

    [itex] \displaystyle \sum_{r=m+k}^{n+k} T_{r-k}=T_{m+k-k}+T_{m+k+1-k}+T_{m+k+2-k}+\dots+T_{n+k-2-k}+T_{n+k-1-k}+T_{n+k-k}=\\ T_m+T_{m+1}+T_{m+2}+\dots+T_{n-2}+T_{n-1}+T_n[/itex]

    Is it clear enough?
     
  6. Jan 3, 2015 #5
    Thank you very much Shyan.;)
     
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