Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Find the sum S(n)

  1. Dec 17, 2004 #1
    How can I find the a function to express the sum of a function, like
    [tex]1^2+2^2+3^2+....+n^2=?[/tex]
    What's the expression?
    Anyway, how can I use latex?
    Thanks for advice.
     
    Last edited: Dec 17, 2004
  2. jcsd
  3. Dec 17, 2004 #2
    Isn't that sum just [tex]\sum_{n=1}^\infty n^2[/tex]?

    Read this thread for info about LaTeX.
     
  4. Dec 17, 2004 #3

    matt grime

    User Avatar
    Science Advisor
    Homework Helper

    It's the sum from 1 to n of r^2, not an infinite sum, especially an infinite sum that obviously diverges.

    Seach these forums for it as it comes up a lot.
     
  5. Dec 17, 2004 #4
    With what keywords?
     
  6. Dec 17, 2004 #5
  7. Dec 17, 2004 #6
    Thanks. They are useful
     
  8. Dec 17, 2004 #7

    HallsofIvy

    User Avatar
    Staff Emeritus
    Science Advisor

    For a problem like that, "Newton's finite Difference" method, a variation on Taylor's polynomial works nicely.

    Make a list of values and repeated differences:
    n y(n) dy d2y d3y d4y
    0 0 1 3 2 0
    1 1 4 5 2
    2 5 9 7
    3 14 16
    4 30

    Where the number in each "difference" column is the difference between two successive numbers in the previous column. Of course, since the y value itself is gotten by adding squares, the first difference is just the square itself. The second differences are just odd numbers and the third differences are the same: 2. All succeeding differences are 0.

    Newton's polynomial now is y+ (dy)n+ (d2y/2) n(n-1)+ (d3y/3!)n(n-1)(n-2)+ ... with the ith term being (diy)/n! n(n-1)..(n-i+1)

    Here, that gives 0+ 1n+ (3/2)n(n-1)+ (2/6)n(n-1)(n-2)= (n/6)(6+ 9n- 9+ 2n[sup[/sup]-6n+ 4)= (1/6)n(2n2+3n+1)= (1/6)n(2n+1)(n+1).

    One can show that the sum of "kth" powers is a polynomial of degree k+1.
     
    Last edited: Dec 17, 2004
  9. Dec 17, 2004 #8
    I wish I have learnt calculus.
     
  10. Dec 18, 2004 #9
    One way to do it.

    [tex] \sum_{k=1}^{n} k^2 [/tex]

    [tex] \sum_{k=1}^{n} k^2 -k + k -1 + 1[/tex]

    [tex] \left( \sum_{k=1}^{n} k^2 - k - {1 \over 2} + 1 \right) + \left( \sum_{k=1}^{n} k - {1 \over 2} \right) [/tex]

    [tex] \left( \sum_{k=1}^{n} k^2 - k - {1 \over 2} + {1 \over 3 } + {2 \over 3} \right) + \left( {1 \over 2} \sum_{k=1}^{n} 2 \left(k - {1 \over 2} \right) \right) [/tex]

    [tex] \left( \sum_{k=1}^{n} k^2 - k + {1 \over 3} \right) + \left( \sum_{k=1}^{n} {2 \over 3} - {1 \over 2} \right) + \left( {1 \over 2} \sum_{k=1}^{n} 2k - 1 \right) [/tex]

    [tex] \left( {1 \over 3} \sum_{k=1}^{n} 3k^2 - 3k + 1 \right) + \left( \sum_{k=1}^{n} {1 \over 6} \right) + \left( {1 \over 2} \sum_{k=1}^{n} 2k - 1 \right) [/tex]

    [tex] \left( {1 \over 3} \sum_{k=1}^{n} k^3 - (k-1)^3 \right) + \left( \sum_{k=1}^{n} {1 \over 6} \right) + \left( {1 \over 2} \sum_{k=1}^{n} 2k - 1 \right) [/tex]

    [tex]{k^3 \over 3} + {n \over 6} + {n^2 \over 2} [/tex]

    If you're wondering how [tex] \sum_{k=1}^{n} 2k - 1 = n^2[/tex]

    [tex] \sum_{k=1}^{n} 2k - 1 [/tex]

    [tex] \sum_{k=1}^{n} k^2 - k^2 + 2k - 1 [/tex]

    [tex] \sum_{k=1}^{n} k^2 - (k^2 - 2k + 1) [/tex]

    [tex] \sum_{k=1}^{n} k^2 - (k-1)^2 [/tex]

    [tex] k^2 [/tex]

    This last step is called telescoping property of sums, it was also
    used in the last step of the first proof, for the leftmost term.

    [tex] \sum_{k=1}^{n} (a_k - a_{k-1}) = a_n - a_0 [/tex]
     
    Last edited: Dec 18, 2004
  11. Dec 18, 2004 #10
    After reading the web site last night, I got it.
     
  12. Dec 18, 2004 #11
    1^2+2^2+3^2+4^2+5^2.......x^2

    all you have to do is look at increasing three digit numbers, and the amount of combinations in an increasing three digit number is 1+2+3+4+5+6+7+(1+2+3+4+5+6)+(1+2+3+4+5).......

    but there is any easier way to express this, it could be 9(8)(7)/6 because an increasing number is only one out of six.

    now look 100(1)+/99(2)/+98(3)+/97(4)/+96(5)+/95(6)/.........+/1(100)/= 102(101)(100)/6

    and 99(1)+97(2)+95(3)+93(5)...............+1(50)

    so the terms in / / are going to be repeated again, and you divide that by two so you get the series below, and you get (102)(101)(100)/24 as the sum which is (2n+2)(2n+1)(2n)/24 which is equal to the sum of the nth squares.
    kind of a cool way. to find the sum of cubes, use four digit numbers.
     
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?



Similar Discussions: Find the sum S(n)
  1. Sum of n-th roots (Replies: 9)

  2. Find sum (Replies: 4)

Loading...