Talk:Euler–Maclaurin formula

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[edit] Motivation for the existence

I don't understand why Δ = eD − 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[a0xn + a1xn − 1 + a2xn − 2 + ... + an − 1x + an] = n * a0xn − 1 + (n − 1) * a1xn − 2 + ... + 2 * an − 2x + an − 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)

[edit] Remainder term

The 'p+1' st derivative inside the integral looks suspicious to me - is it correct? Crackling 13:46, 25 March 2007 (UTC)

[edit] Definition of Bernoulli numbers

These are defined here (at least twice) as Bn(1), which is inconsistent with some other Wikipedia pages. They are also defined, in this article, as Bn(0). Some clarification would be helpful. Crackling 13:46, 25 March 2007 (UTC)

[edit] Periodic Bernoulli functions Pn(x)

These functions are not defined for integer x, but the evaluation of the integration by parts requires this. Explicitly it seems to assume that P1(k) is 1/2, but B1(0) is -1/2, so I am rather confused. Extending the definition to integral x would be helpful. Crackling 13:46, 25 March 2007 (UTC)

Use one-sided limits. Michael Hardy 18:30, 23 September 2007 (UTC)

[edit] Invalid proof???

I don't see how the proof on this page is correct. Perhaps I am missing something, but f(x)P1(x) evaluated from k to k+1 does not appear to equal (f(k)+f(k+1))/2 Instead, it appears to equal (-f(k+1)+f(k))/2. This invalidates the proof. Can someone show me that I am wrong by adding the intermediate steps?

Thanks. —Preceding unsigned comment added by Lasher9999 (talk • contribs)

As x approaches k FROM ABOVE, P1(x) approaches −1/2. As x approaches k + 1 FROM BELOW, P1(x) approaches +1/2. Thus we have
\Bigg[f(x)P_1(x)\Bigg]_k^{k+1} = f(k+1)P_1((k+1)-) - f(k)P_1(k+) \,


= f(k+1)\frac12 - f(k)\left(\frac{-1}{2}\right) = \frac{f(k+1) + f(k)}{2}. \,
Michael Hardy 18:27, 23 September 2007 (UTC)

[edit] Integratable???

I assume from the discussion that the integral of P_1(x) times f(x) cannot be found exactly but only approximated by a never ending series of higher orders of bernoulli polynomials via integration by parts, but I am not sure why. P_1(x) is defined at the end points and has a very simple form over the interval of k to k+1. It seems like it should be integrable. Thanks.

User:Lasher9999 —Preceding signed but undated comment was added at 00:38, 2 October 2007 (UTC)