1. Limited time only! Sign up for a free 30min personal tutor trial with Chegg Tutors
    Dismiss Notice
Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Homework Help: Riemann Sum

  1. Sep 14, 2008 #1
    About these formulas:

    [tex]\frac{n(n+1)}{2}, \frac{n(n+1)(2n+1)}{6} ,(\frac{n(n+1)}{2})^2[/tex]

    How were they evolved?

    And what about [tex]i^4, i^5, i^6,...[/tex]

    Thankyou everybody;;
  2. jcsd
  3. Sep 14, 2008 #2


    User Avatar
    Science Advisor

    Those are, of course, [itex]\sum_{i=1}^n i[/itex], [itex]\sum_{i=1}^n i^2[/itex], and [itex]\sum_{i=1}^n i^3[/itex], respectively. You can sum such things by using "Newton's Divided Difference" formula. It is a discrete version of Taylor's series. If f(n) is a function defined on the non-negative integers, [itex]\Delta f[/itex] is the "first difference", f(n)- f(n-1), [itex]\Delta^2 f[/itex] is the "second difference", [itex]\Delta f(n)- \Delta f(n-1)[/itex], [itex]\Delta^3 f[/itex] is the "third difference", etc. then
    [tex]f(n)= f(0)+ \Delta f(0) n+ \frac{\Delta^2 f(0)}{2!} n(n-1)+ \frac{\Delta^3 f(0)}{3!} n(n-1)(n-2)+ \cdot\cdot\cdot[/itex]
    In particular, for a sum of terms in involving nk, [itex]\Delta^{k+1} f[/itex] is always 0 so this gives a polynomial of order k+1.

    A "short cut" to finding [itex]\sum_{i=1}^n i^4[/itex] is to set it equal to An5+ Bn4+ Cn3+ Dn2+ En+ F and use the first 6 values to get eqations for A, B, C, D, E and F.
  4. Sep 14, 2008 #3
    Thakyou HallsofIvy for th explanation, but could you show me explicitly how to evolve he equation in [tex}i^4[/tex],for example.It would be very helpful.
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook