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IS this coincidence? Integral adn summation

  1. Apr 13, 2012 #1
    The sum of 1 + 2 + 3...........n = n(n+1) / 2 - highest power term is n^2
    sum of 1^2 + 2^2 + 3^2...........n^2 - n(n+1)(2n+1) / 6 - highest power term is n^3
    sum of 1^3 + 2^3 + 3^3............n^3 - it has highest power term of n^4
    similarly 1^k +2^k ...............n^k - it has highest power term of n^(k+1)

    Is it a coincidence that ∫x^k dx = x^(k+1) / (k+1) - power is k+1 ?
     
  2. jcsd
  3. Apr 13, 2012 #2
    Do you understand how the Reimann integral is derived? (i.e. its relationship to summation)
     
  4. Apr 13, 2012 #3

    chiro

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    Hey jd12345.

    For the summation part there is a formula for a finite n which is known as the Bernoulli polynomials that basically allow you to derive the closed form solution for x^p where p >= 0 and p is a integer and x ranges from 1 to n.
     
  5. Apr 13, 2012 #4

    Stephen Tashi

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    Suppose [itex] F(n) = \sum_{i=1}^n f(n) [/itex]

    then [itex] F(n+1) - F(n) = f(n+1) [/itex]

    Denote [itex] F(n+1) - F(n) [/itex] by [itex] \triangle F(n) [/itex].

    Then [itex] \triangle F(n) [/itex] resembles a finite version of the derivative of [itex] F(x) [/itex]

    To sum [itex] f(n) [/itex] you must solve [itex] \triangle F(n) = f(n+1) [/itex] for [itex] F(n) [/itex]. i.e. you must find the anti-[itex]\triangle[/itex] of f(n+1). So this resembles integration.

    The study of stuff like this is called "The Calculus Of Finite Differences". (George Boole himself wrote an interesting book about it.)
     
  6. Apr 14, 2012 #5
    Ok - thanks Stephen
     
  7. Apr 14, 2012 #6

    AlephZero

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    Incidentally, if you want to see neater analogy between sums of series and integrals, find the formulas for
    1 + 2 + 3 + ...
    1.2 + 2.3 + 3.4 + ...
    1.2.3 + 2.3.4 + 3.4.5 + ....
    Or in general
    $$\sum_{n=0}^N \frac{(n+k)!}{ n!}$$
    for k = 1, 2, 3, .....
     
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