Question

How to Interpret the Euler-Maclaurin formula:
(It uses summation, Cartesian products and Bernoulli Polynomials (I think)).

I don't know how to use it to get an approximately-closed form of the summation shown on the left side of the image with a given bounds.

http://imgur.com/Z2XGkgO

Answer 1

@Kainui

Answer 2

Shouldn't the 0.4 be 0.5 ?

Answer 3

\[
\left(\prod_{i=1}^{2k_1} (2.5-i) \right) (b^{2.5-2k} - a^{2.5-2k})
\]
is the \((2k-1)\)-th derivative of \(b^{1.5} - a^{1.5}\).

Answer 4

*or deriviation, one of the two

Answer 5

\(\int b^{3/2} = \frac{b^{5/2}}{5/2} = 0.4 b^{2.5}\).
You're right.

So what is the question exactly?

Answer 6

Using normal integeration is not enough to estimate the curve due to error, which increases as range (b-a) increases. The second part is supposed to help eliminate the erorr by a huge factor.

Answer 7

As far as I'm aware.

Answer 8

*elimated the error. It supposingly can only give -0.07 absolete errors even with a range of 200,000. I'm not a math person,this is what I was told.

Answer 9

If the function \(f\) is many times differentiable, many terms of the sum of the \(f^{(2k-1)}(b)-f^{(2k-1)}(a)\) can be computed. If you go to infinity, the formula is an equality. Otherwise, there's a rest \(R\). WIkipedia: the remainder term can be written as
\[
R = \int_a^b f^{(2p)}(x) {P_{2p}(x) \over (2p)!}\,dx
\]
And Using this inequality, the size of the remainder term can be estimated using
\[
\left|R\right|\leq\frac{2 \zeta (2p)}{(2\pi)^{2p}}\int_m^n\left|f^{(2p)}(x)\right|\ \, dx
\]

You might see here an analogy with the Taylor series.