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Prove of sigma notation

  1. Oct 11, 2012 #1
    1. The problem statement, all variables and given/known data
    Prove that:

    [tex]\sum_{n=1}^{14} 10n = \sum_{n=1}^{7} (20n+70)[/tex]

    2. Relevant equations
    properties of sigma notation


    3. The attempt at a solution
    I know several properties of sigma notation but none that I know can be used to prove this. I don't know how to change 10 n to 20 n

    Thanks
     
  2. jcsd
  3. Oct 11, 2012 #2

    micromass

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    What properties do you know of sigma notation?
     
  4. Oct 11, 2012 #3

    tiny-tim

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    hint: there's exactly twice as many terms in one sum as in the other sum :wink:
     
  5. Oct 11, 2012 #4

    Ray Vickson

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    Sigma notation is a shorthand that helps you to write things in a more compact form, and is very useful once you have grasped what is being asked. However, if you have not yet understood what is being asked, or what has been written, the sigma notation is just getting in the way. It would be better in this case to write out in detail both sides of what was written above, but NOT using sigma notation. That way, you can understand what, exactly, you are being asked to do. After you understand these matters better, THEN you can start to switch back to using sigma notation, to save space, etc.

    RGV
     
  6. Oct 12, 2012 #5
    [tex]\sum_{n=1}^{r}c = cr ; c = constant[/tex]
    [tex]\sum_{n=1}^{r}c.p_n=c\sum_{n=1}^{r}p_n[/tex]
    [tex]\sum_{n=1}^{r} (p_n+q_n)=\sum_{n=1}^{r}p_n+\sum_{n=1}^{r}q_n[/tex]
    [tex]\sum_{n=1}^{r}p(n)=\sum_{n=1+s}^{r+s}p(n-s)[/tex]
    [tex]\sum_{n=1}^{r}p_n=\sum_{n=1}^{m}p_n+\sum_{n=m+1}^{r}p_n[/tex]

    Hm...sorry I don't know how to use your hint...:shy:

    Not sure what you mean but let me try

    Expand the LHS:
    10 + 20 + ... + 140
    = (20 + 40 + 60 + 80 + ... + 140) + (10 + 30 + 50 + ... + 130)
    [tex]=\sum_{n=1}^{7}20n + 70.7[/tex]
    [tex]=\sum_{n=1}^{7}(20n+70)[/tex]

    But I seriously doubt the validity of my work....

    Thanks
     
  7. Oct 12, 2012 #6

    tiny-tim

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    hi songoku! :smile:
    there's 14 terms on the left, and 7 on the right …

    so try putting the 14 terms into 7 pairs of 2 terms, so that each pair adds to the correct amount :wink:

    (eg, {1,2},{3,4}etc or {1,14}{2,13}etc or … :wink:)
     
  8. Oct 12, 2012 #7
    hi tiny tim :smile:

    So it is basically the same as what I have done in post #5?

    Is there another way to prove it without expanding the sigma notation? I mean just using the properties such as changing the upper and lower bound or other properties

    Thanks
     
  9. Oct 12, 2012 #8

    tiny-tim

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    hi songoku! :smile:
    you mean …
    i don't see how you got that last line :confused:

    try writing the LHS on one line (in numbers, in full), and the RHS on the next line, and then just connecting up anything that adds up :smile:
    yes, but you need to find what to do first before you can put in into proper maths :wink:
     
  10. Oct 13, 2012 #9
    hi tiny-tim :smile:
    [tex]\sum_{n=1}^{14}10n[/tex]
    [tex]= (20 + 40 + 60 + 80 + ... + 140) + (10 + 30 + 50 + ... + 130)[/tex]
    [tex]=\sum_{n=1}^{7}(20n) + (10+130) + (30+110)+(50+90)+70[/tex]
    [tex]=\sum_{n=1}^{7}(20n) + 140 + 140 + 140 + 70[/tex]
    [tex]=\sum_{n=1}^{7}(20) + 70 + 70 + 70 + 70 + 70 + 70 + 70[/tex]
    [tex]=\sum_{n=1}^{7}(20n) + 7.70[/tex]
    [tex]=\sum_{n=1}^{7}(20n+70)[/tex]

    Is this valid? And how to prove it without expanding?

    Thanks
     
  11. Oct 13, 2012 #10

    tiny-tim

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    hi songoku! :smile:

    yes, that's certainly valid :smile:

    to put that into symbolic language, try replacing the index n (from 1 to 14) by m (from 1 to 7), so that you use 2m for the even numbers and 2m±1 for the odd numbers :wink:

    (another way to see the result, which i was thinking of at the start, is to match:
    10 + 20 + 30 + 40 + 50 + 60 + 70 +
    80 + 90 + 100+110+120+130+140

    90 +110 + 130+150+170+190+210 :wink:)​
     
  12. Oct 17, 2012 #11
    Hi tiny-tim :smile:

    Sorry for late reply. Thanks a lot for all the help
     
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