Riemann Sum

  • Thread starter Brunll
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  • #1
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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;;
 

Answers and Replies

  • #2
HallsofIvy
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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;;
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.
 
  • #3
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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.
 

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