Question

State Euler’s Maclaurin’s formula for numerical integration.

Answer 1

If {\displaystyle m} m and {\displaystyle n} n are natural numbers and {\displaystyle f(x)} f(x) is a complex or real valued continuous function for real numbers {\displaystyle x} x in the interval {\displaystyle [m,n]} [m,n], then the integral

{\displaystyle I=\int _{m}^{n}f(x)\,dx} I = \int_m^n f(x)\,dx

can be approximated by the sum (or vice versa)

{\displaystyle S=f(m+1)+\cdots +f(n-1)+f(n)} {\displaystyle S=f(m+1)+\cdots +f(n-1)+f(n)}

(see rectangle method). The Euler–Maclaurin formula provides expressions for the difference between the sum and the integral in terms of the higher derivatives {\displaystyle f^{(k)}(x)} {\displaystyle f^{(k)}(x)} evaluated at the end points of the interval, that is to say when {\displaystyle x=m} {\displaystyle x=m} and {\displaystyle x=n} {\displaystyle x=n}.

Explicitly, for {\displaystyle p} p a positive integer and a function {\displaystyle f(x)} f(x) that is {\displaystyle p} p times continuously differentiable in the interval {\displaystyle [m,n]} [m,n], we have

{\displaystyle S-I=\sum _{k=1}^{p}{{\frac {B_{k}}{k!}}(f^{(k-1)}(n)-f^{(k-1)}(m))}+R_{p},} {\displaystyle S-I=\sum _{k=1}^{p}{{\frac {B_{k}}{k!}}(f^{(k-1)}(n)-f^{(k-1)}(m))}+R_{p},}

where {\displaystyle B_{k}} B_{k} is the {\displaystyle k} kth Bernoulli number (with {\displaystyle B_{1}=1/2} {\displaystyle B_{1}=1/2}) and {\displaystyle R_{p}} R_{p} is an error term which depends on {\displaystyle n} n, {\displaystyle m} m, {\displaystyle p} p, and {\displaystyle f} f and is usually small for suitable values of {\displaystyle p} p.

The formula is often written with the subscript taking only even values, since the odd Bernoulli numbers are zero except for {\displaystyle B_{1}} B_{1}. In this case we have[1][2]

{\displaystyle \sum _{i=m}^{n}f(i)=\int _{m}^{n}f(x)\,dx+{\frac {f(n)+f(m)}{2}}+\sum _{k=1}^{\lfloor p/2\rfloor }{\frac {B_{2k}}{(2k)!}}(f^{(2k-1)}(n)-f^{(2k-1)}(m))+R_{p},} {\displaystyle \sum _{i=m}^{n}f(i)=\int _{m}^{n}f(x)\,dx+{\frac {f(n)+f(m)}{2}}+\sum _{k=1}^{\lfloor p/2\rfloor }{\frac {B_{2k}}{(2k)!}}(f^{(2k-1)}(n)-f^{(2k-1)}(m))+R_{p},}

Answer 2

The original form of Euler Maclaurin Formula takes the form:

$\\ \sum_{n = a + 1} ^{b} f(n) = \int_{a} ^{b} f(x)dx + \sum_{k = 1} ^{p} \frac{B_{k} }{k!} ( {f}^{(k - 1)} (b) - {f}^{(k - 1)} (a)) + \frac{( - 1) ^{p +1 } }{p!} \int _{ a } ^{b} {f}^{(p)} (x)B_{p}(x - \lfloor x \rfloor)dx$

where, p≥0

Bₙ and Bₙ(x) refers to Bernoulli number and polynomial respectively.

Usually, the Last Remainder term is small and many times turns out to be zero, so it may be ignored.

You can use this for approximating integrals numerically or sums.