Talk:Euler-Maclaurin formula

From Wikipedia, the free encyclopedia

[edit] Motivation for the existence

I don't understand why $Δ = e D - I$ , or even what I is (the integral used earlier in the article?). Can someone explain? Also, is there a reference for this argument (other than "Legendre")? Fredrik Johansson 20:05, 17 May 2006 (UTC)

No, I is not that integral. I is the identity operator on functions, i.e.

$(If)(x) = f(x);\,$

D is the differentiation operator, i.e.

$(Df)(x) = f'(x);\,$

and Δ is the forward difference operator, i.e.

$(\Delta f)(x) = f(x+1) - f(x).\,$

Using the power series

$e^x = 1 + x + {x^2 \over 2} + {x^3 \over 6} + {x^4 \over 24} + \cdots,$

we have

$e^D = I + D + {D^2 \over 2} + {D^3 \over 6} + \cdots.$

Therefore

$(e^D - I)f(x) = f'(x) + {f''(x) \over 2} + {f'''(x) \over 6} + \cdots.$

If f happens to be a polynomial function, then all but finitely many of these terms vanish, and a bit of algebra shows that this sum adds up to

$f(x+1) - f(x),\,$

i.e. it adds up to

$\Delta f(x).\,$

To what extent this works when f is not a polynomial, is a more difficult question. Michael Hardy 22:58, 17 May 2006 (UTC)

Thanks, Michael. Fredrik Johansson 23:02, 17 May 2006 (UTC)

---I should add that the Differentiation operator acting on that space of polynomials is just an emulation of the differentiation operation but uses no calculus since is just defined as :: $D [a 0 x n + a 1 x n - 1 + a 2 x n - 2 + ... + a n - 1 x + a n] = n * a 0 x n - 1 + (n - 1) * a 1 x n - 2 + ... + 2 * a n - 2 x + a n - 1$ and isn't necessarily defined for an arbitrary function.

[edit] Some Formula

The article doesn't say exactly how to find the following Sn, here is the process. The idea is from Trapezium rule

Sn=Sum i^2 = n*(n+1)*(2n+1)/6

from b^3-a^3=(b-a)*(b^2+ab+a^2) b=i, a=i-1,

i^3-(i-1)^3=(i^2+i*(i-1)+(i-1)^2)=2*i^2-i+(i-1)^2

Sum (i^3-(i-1)^3)=n^3=3Sn-n^2-n*(n+1)/2, you find Sn

Sn=Sum i^3 = n^2*(n+1)^2/4

from b^4-a^4=(b^2-a^2)(b^2+a^2)=(b-a)(b+a)(b^2+a^2)

b=i, a=i-1, i^4-(i-1)^4=4*i^3-6*i^2+4^i-1

Sum (i^4-(i-1)^4)=n^4=4Sn-n*(n+1)*(2n+1)+2n*(n+1)-n, you find Sn

With the same method, we can find all others. Please help put the text in a nice format.

RESPONSE

I'd be happy to help put that in a nice format, but it's a bit unclear. What are the sums ranging from, and what does "from" refer to? If you want to do it yourself, use the simple commands like

<math> and </math> \sum_{0}^{\infty} would give sum from 0 to infinity

Lavaka 19:34, 6 September 2006 (UTC)

[edit] unclear variable

In the "Remainder Term" section, where did the variable $N$ come from? I can only assume it's supposed to be $p$ ? Lavaka 19:34, 6 September 2006 (UTC)

Retrieved from "http://en.wikipedia.org../../../e/u/l/Talk%7EEuler-Maclaurin_formula_4a47.html"

Talk:Euler-Maclaurin formula

From Wikipedia, the free encyclopedia

[edit] Motivation for the existence

[edit] Some Formula

[edit] unclear variable

Views

Navigation

Search