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Sum of series

  1. May 26, 2009 #1

    Mentallic

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    This problem has intrigued me for some time now. I don't have much prerequisite knowledge on the topic.

    1. The problem statement, all variables and given/known data
    Find the sum of [itex]1+4+9+16+...+n^2[/itex] for n=x and thus, find the sum after 50 terms.

    3. The attempt at a solution
    Well, I have studied sums for the simple arithmetic and geometric progressions, but this one clearly doesn't follow those patterns. Please, for any helpers, don't just throw formulas at me, I would like this to be solved from scratch if need be (formulas being proven before being used). Thanks for any help.
     
  2. jcsd
  3. May 26, 2009 #2
    There is a simple trick and a more general approach.

    Simple trick:

    If you sum [(n+1)^3 - n^3 ] from n = 0 to N, the terms in the summation will cancel except the (n+1)^3 for n = N and the n^3 for n = 0, so the summation is (N+1)^3. But we also have that:

    (n+1)^3 - n^3 = 3 n^2 + 3 n +1

    So, it is also 3 times the desired summation plus 3 times the summation of the arithmetic series plus N.
     
  4. May 28, 2009 #3

    Mentallic

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    Sorry I couldn't completely follow:

    But skipping ahead for the moment:

    So somehow (by your previous statement)
    [tex](n+1)^3-n^3=\sum_1^n{n^2}=3n^2+3n+1[/tex]

    and [itex]n^2[/itex] is just the last term in the series, or taking the nth term in the sequence, then simply n2

    and n is obviously the nth term in the sequence.

    Ok so testing for n=4,
    [tex]\sum_1^4{n^2}=1+4+9+16=30[/tex]

    and [tex]3n^2+3n+1=3(4)^2+3(4)+1=61\neq 30[/tex]

    Well it's not working. I did something wrong :biggrin:
     
  5. May 28, 2009 #4
    You have some sign mistakes.
     
  6. May 28, 2009 #5

    Mentallic

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    Can you please point them out?

    I'm checked my work, and the arithmetic seems correct.

    The "formula" [itex]3n^2+3n+1[/itex] was never going to work anyway. I must be misunderstanding how to solve this.
     
  7. May 28, 2009 #6
    n3-(n-1)3 = n3-(n3-3n2+3n-1)=3n2-3n+1
    On the left you go to
    (n-1)3-(n-2)3
    ...
    23-13

    The sum of the left sides is n3-1

    Now expand the right sides and sum those.
    Solve for [tex]\sum n^{2}[/tex] on the right side and you have your formula
     
  8. May 28, 2009 #7

    Mentallic

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    I'm going to need it re-explained as to why this is being done? How this is actually working. Sorry to be a hassle.
     
  9. May 28, 2009 #8
    No hassle at all. The idea is to sum from 1 to n or from 12 to n2 in this case. So here I listed the sums in descending order. It could just have easily been written in ascending order:
    23-13 = 7
    33-23 = 19
    ...
    (n-1)3-(n-2)3 = ...
    n3-(n-1)3 = ...

    The sum on the left is simple because almost all terms cancel out. The sum on the right is interesting in that the cube terms are removed leaving a sum of terms with exponents that are less than cubic. When the right side is regrouped and added we find so many squared terms, so many linear terms, and a constant.

    Replace the [tex]\sum n[/tex] in the formula with the known formula for that expression. Now all you have left in the equation is a [tex]\sum n^{2}[/tex] term. Solve for that term to get the formula.
     
  10. May 28, 2009 #9
    My math professor referred to this as a technique. He said tricks are used once, techniques are used 2 or more times.
     
  11. May 28, 2009 #10
    I would call this a trick because the fact that we consider summing (n+1)^3 - n^3 is itself not derived using formal rules. We just do it because "it works".
     
  12. May 28, 2009 #11

    HallsofIvy

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    You could use "Newton's divided difference method"
    The sum you give has successive "partial sums" 1, 1+ 4= 5, 1+ 4+ 9= 14, 1+ 4+ 9+ 16= 30
    Set the numbers up by
    0 1 3 2
    1 4 5 2
    5 9 7
    14 16
    30


    Where the first column is the list of partial sums. The second column are the succesive differences from the first column ,third column is the differences from the second column, and the fourth is the differences form the third column. We could continue, but obviously the second column is those squares and the third consecutive is successive odd numbers ([itex](n+1)^2- n^2= n^2+ 2n+ 1- n^2= 2n+1[/itex]) so the fourth column is all "2"s and all succeeding columns are "0"s.

    Now "Newton's divided difference formula" says the the formula for f(n) giving the first column is [itex]f(0)+ \Delta f(0) n+ \frac{\Deltaj^2 f(0)}{2}n(n-1)+ \cdot\cdot\cdot [/itex]. That's a lot like "McLaurin series" using the "differences" rather than derivatives and "factorial" powers rather than ordinary powers. Here this would be [itex]0+ 1(n)+ (3/2!)n(n-1)+ (2/3!)n(n-1)(n-2)[/itex].


    We can factor an "n" out of that: [itex]n(1+ (3/2)(n-1)+ (1/3)(n^2- 3n+ 2)= n\frac{6+ 9n- 9+ 2n^2- 6n+ 4}{6}[/itex][itex]= n\frac{1+ 3n+ 2n^2}{6}= \frac{n(n+1)(2n+1)}{6}[/itex].

    Notice, by the way, that "Newton's divided difference formula" implies that a sum of [itex]n^{th}[/itex] powers of n is an [itex]n+1[/itex] degree polynomial. Since here we are summing squares, the sum must be a cubic polynomial, of the form [itex]an^3+ bn^2+ cn+ d[/itex]. When n= 1, [itex]a+ b+ c+ d= 1[/itex]. When n= 2, [itex]8a+ 4b+ 2c+ d= 5[/itex]. When n= 3, [itex]27a+ 9b+ 3c+ d= 14. When n= 4, [itex]64a+ 16b+ 4c+ d= 30[/itex]. That gives four equations to solve for a, b, c, d and should give the same answer as above. In fact, notice that the first thing you can do is subtract two succesive equations to eliminate d, then subtract to eliminate, c, etc. You will basically reproduce the "difference table" above!
     
  13. May 28, 2009 #12
    Count Iblis the reason this is a technique and not a trick is that it allows a whole series of summation formulas to be created. The technique can be used to get the formula for sums of 1+2+ .. +n and anything else. This is not a trick exclusive to the sum of squares problem.
     
  14. May 28, 2009 #13
  15. May 28, 2009 #14
    My favorite derivation is to use the formula:

    [tex]\sum_{k=0}^{N}x^{k}=\frac{1-x^{N+1}}{1-x}[/tex]

    If we put x = exp(t), we get:

    [tex]\sum_{k=0}^{N}\exp(kt)=\frac{1-\exp\left[(N+1)t\right]}{1-\exp(t)}[/tex]

    Then expand both sides in powers of t to see that the sum of powers are given by the Bernoulli polynomials.
     
  16. May 29, 2009 #15

    Mentallic

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    Yes, now I remember. I used to be able to solve problems like this using the method that Hallsofivy showed. Completely slipped my mind about that method, and I never really understood why it worked, just that it was a handy tool to exploit.


    Oh uh... this maths is rightfully out of my league.

    [tex]\sum_{k=0}^{N}x^{k}=\frac{1-x^{N+1}}{1-x}[/tex]

    This is following the same pattern as that with the cubic expansions just previously for x2.

    [tex]\frac{1-x^{N+1}}{1-x}=1-x+x^2-x^3...+(-x)^n[/tex]

    (if I got my expansion right that is)



    Thanks for that :smile: Yet, I still don't understand why the [itex]n^3-(n-1)^3[/itex] is valid?
     
  17. May 29, 2009 #16
    I'm not saying that the formula is valid I'm simply saying evaluate it. When you do you get an expression without a cubic term.
     
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