Euler–Maclaurin formula

In mathematics, the Euler–Maclaurin formula provides a powerful connection between integrals (see calculus) and sums. It can be used to approximate integrals by finite sums, or conversely to evaluate finite sums and infinite series using integrals and the machinery of calculus. For example, many asymptotic expansions are derived from the formula, and Faulhaber's formula for the sum of powers is an immediate consequence.

The formula was discovered independently by Leonhard Euler and Colin Maclaurin around 1735 (and later generalized as Darboux's formula). Euler needed it to compute slowly converging infinite series while Maclaurin used it to calculate integrals.

The formula

If m and n are natural numbers and f(x) is a smooth (meaning: sufficiently often differentiable), analytic function of exponential type < 2π defined for all real numbers x in the interval $[m,n]$, then the integral

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

can be approximated by the sum (or vice versa)

$S={\frac {1}{2}}f(m)+f\left(m+1\right)+\cdots +f\left(n-1\right)+{\frac {1}{2}}f(n)$

(see trapezoidal rule). The Euler–Maclaurin formula provides expressions for the difference between the sum and the integral in terms of the higher derivatives ƒ^(k) at the end points of the interval m and n. Explicitly, for any natural number p, we have

$S-I=\sum _{k=2}^{p}{{\frac {B_{k}}{k!}}\left(f^{(k-1)}(n)-f^{(k-1)}(m)\right)}+R$

where B₁ = −1/2, B₂ = 1/6, B₃ = 0, B₄ = −1/30, B₅ = 0, B₆ = 1/42, B₇ = 0, B₈ = −1/30, … are the Bernoulli numbers, and R is an error term which is normally small for suitable values of p and depends on n, m, p and f. (The formula is often written with the subscript taking only even values, since the odd Bernoulli numbers are zero except for B₁.)

Note that

$-B_{1}\left(f(n)+f(m)\right)={\frac {1}{2}}\left(f(n)+f(m)\right).$

Hence, we may also write the formula as follows:

$\sum _{i=m}^{n}f(i)=\int _{m}^{n}f(x)\,dx-B_{1}\left(f(n)+f(m)\right)+\sum _{k=1}^{p}{\frac {B_{2k}}{(2k)!}}\left(f^{(2k-1)}(n)-f^{(2k-1)}(m)\right)+R$

or, more compactly,

$\sum _{i=m}^{n}f(i)=\sum _{k=0}^{2p}{\frac {1}{k!}}\left(B_{k}^{\ast }f^{(k-1)}(n)-B_{k}f^{(k-1)}(m)\right)+R$

with the convention of $f^{(-1)}(x)=\int f(x)\,dx$, i.e. the -1-th derivation of f is the integral of the function. This presentation also emphasizes the notation of the two kinds of Bernoulli numbers, called the first and the second kind. Here we denote the Bernoulli number of the second kind as $B_{k}^{\ast }:=(-1)^{k}B_{k}$ which differ from the first kind only for the index 1.

The remainder term

The remainder term R is most easily expressed using the periodic Bernoulli polynomials P_n(x). The Bernoulli polynomials B_n(x), n = 0, 1, 2, … are defined recursively as

${\begin{aligned}B_{0}(x)&=1\\B_{n}'(x)&=nB_{n-1}(x){\text{ and }}\int _{0}^{1}B_{n}(x)\,dx=0{\text{ for }}n\geq 1\end{aligned}}$

Then the periodic Bernoulli functions P_n are defined as

$P_{n}(x)=B_{n}\left(x-\lfloor x\rfloor \right){\text{ for }}x>0$

where $\scriptstyle \lfloor x\rfloor$ denotes the largest integer that is not greater than x. Then, in terms of P_n(x), the remainder term R can be written as

$R=(-1)\int _{m}^{n}f^{(2p)}(x){P_{2p}(x) \over (2p)!}\,dx$

or equivalently, integrating by parts, assuming ƒ^(2p) is differentiable again and recalling that the odd Bernoulli numbers are zero:

$R=\int _{m}^{n}f^{(2p+1)}(x){P_{(2p+1)}(x) \over (2p+1)!}\,dx$

When n > 0, it can be shown that

$\left|B_{n}\left(x\right)\right|\leq 2\cdot {\frac {n!}{\left(2\pi \right)^{n}}}\zeta \left(n\right)$

where ζ denotes the Riemann zeta function (see Lehmer; one approach to prove the inequality is to obtain the Fourier series for the polynomials B_n). The bound is achieved for even n when x is zero. Using this inequality, the size of the remainder term can be estimated using

$\left|R\right|\leq {\frac {2\zeta (2p)}{(2\pi )^{2p}}}\int _{m}^{n}\left|f^{(2p)}(x)\right|\ \,dx$

Applications

The Basel problem

The Basel problem asks to determine the sum

$1+{\frac {1}{4}}+{\frac {1}{9}}+{\frac {1}{16}}+{\frac {1}{25}}+\cdots =\sum _{n=1}^{\infty }{\frac {1}{n^{2}}}$

Euler computed this sum to 20 decimal places with only a few terms of the Euler–Maclaurin formula in 1735. This probably convinced him that the sum equals π² / 6, which he proved in the same year. Parseval's identity for the Fourier series of f(x) = x gives the same result.

Sums involving a polynomial

If f is a polynomial and p is big enough, then the remainder term vanishes. For instance, if f(x) = x³, we can choose p = 2 to obtain after simplification

$\sum _{i=0}^{n}i^{3}=\left({\frac {n(n+1)}{2}}\right)^{2}$

(see Faulhaber's formula).

Numerical integration

The Euler–Maclaurin formula is also used for detailed error analysis in numerical quadrature. It explains the superior performance of the trapezoidal rule on smooth periodic functions and is used in certain extrapolation methods. Clenshaw–Curtis quadrature is essentially a change of variables to cast an arbitrary integral in terms of integrals of periodic functions where the Euler–Maclaurin approach is very accurate (in that particular case the Euler–Maclaurin formula takes the form of a discrete cosine transform). This technique is known as a periodizing transformation.

Asymptotic expansion of sums

In the context of computing asymptotic expansions of sums and series, usually the most useful form of the Euler–Maclaurin formula is

$\sum _{n=a}^{b}f(n)\sim \int _{a}^{b}f(x)\,dx+{\frac {f(b)-f(a)}{2}}+\sum _{k=1}^{\infty }\,{\frac {B_{2k}}{(2k)!}}\left(f^{(2k-1)}(b)-f^{(2k-1)}(a)\right)$

where a and b are integers. Often the expansion remains valid even after taking the limits ${\scriptstyle a\to -\infty }$ or ${\scriptstyle b\to +\infty }$, or both. In many cases the integral on the right-hand side can be evaluated in closed form in terms of elementary functions even though the sum on the left-hand side cannot. Then all the terms in the asymptotic series can be expressed in terms of elementary functions. For example,

$\sum _{k=0}^{\infty }{\frac {1}{(z+k)^{2}}}\sim \underbrace {\int _{0}^{\infty }{\frac {1}{(z+k)^{2}}}\,dk} _{={\frac {1}{z}}}+{\frac {1}{2z^{2}}}+\sum _{t=1}^{\infty }{\frac {B_{2t}}{z^{2t+1}}}$

Here the left-hand side is equal to ${\scriptstyle \psi ^{(1)}(z)}$, namely the first-order polygamma function defined through ${\scriptstyle \psi ^{(1)}(z)={\frac {d^{2}}{dz^{2}}}\ln \Gamma (z)}$; the gamma function ${\scriptstyle \Gamma (z)}$ is equal to ${\scriptstyle (z-1)!}$ if ${\scriptstyle z}$ is a positive integer. This results in an asymptotic expansion for ${\scriptstyle \psi ^{(1)}(z)}$. That expansion, in turn, serves as the starting point for one of the derivations of precise error estimates for Stirling's approximation of the factorial function.

Examples

• $\sum _{k=1}^{n}{\frac {1}{k^{s}}}={\frac {1}{n^{s-1}}}+s\int _{1}^{n}{\frac {\lfloor x\rfloor }{x^{s+1}}}dx\qquad {\text{with }}\quad s\in \mathbb {R} \setminus \{1\}$

• $\sum _{k=1}^{n}{\frac {1}{k}}=\log n+1-\int _{1}^{n}{\frac {x-\lfloor x\rfloor }{x^{2}}}dx$

Proofs

Derivation by mathematical induction

We follow the argument given in (Apostol).

The Bernoulli polynomials B_n(x), n = 0, 1, 2, … may be defined recursively as follows:

${\begin{aligned}B_{0}(x)&=1\\B_{n}'(x)&=nB_{n-1}(x){\text{ and }}\int _{0}^{1}B_{n}(x)\,dx=0{\text{ for }}n\geq 1\end{aligned}}$

The first several of these are

${\begin{aligned}B_{1}(x)&=x-{\frac {1}{2}}\\B_{2}(x)&=x^{2}-x+{\frac {1}{6}}\\B_{3}(x)&=x^{3}-{\frac {3}{2}}x^{2}+{\frac {1}{2}}x\\B_{4}(x)&=x^{4}-2x^{3}+x^{2}-{\frac {1}{30}}\\&\vdots \end{aligned}}$

The values B_n(0) are the Bernoulli numbers. Notice that for n ≥ 2 we have

$B_{n}(0)=B_{n}(1)=B_{n}\quad (n{\text{th Bernoulli number}})$

For n = 1,

$B_{1}(0)=-B_{1}(1)=B_{1}$

We define the periodic Bernoulli functions P_n by

$P_{n}(x)=B_{n}(x-\lfloor x\rfloor )$

where $\lfloor x\rfloor$ denotes the largest integer that is not greater than x. So P_n agree with the Bernoulli polynomials on the interval (0, 1) and are periodic with period 1. Thus,

$P_{n}(0)=P_{n}(1)=B_{n}$

Let k be an integer, and consider the integral

$\int _{k}^{k+1}f(x)\,dx=\int u\,dv$

where

${\begin{aligned}u&=f(x)\\du&=f'(x)\,dx\\dv&=P_{0}(x)\,dx\quad ({\text{since }}P_{0}(x)=1)\\v&=P_{1}(x)\end{aligned}}$

Integrating by parts, we get

${\begin{aligned}\int _{k}^{k+1}f(x)\,dx&=uv-\int v\,du&{}\\&={\Big [}f(x)P_{1}(x){\Big ]}_{k}^{k+1}-\int _{k}^{k+1}f'(x)P_{1}(x)\,dx\\[8pt]&=-B_{1}(f(k+1))-B_{1}(f(k))-\int _{k}^{k+1}f'(x)P_{1}(x)\,dx\end{aligned}}$

Summing the above from k = 0 to k = n − 1, we get

${\begin{aligned}&\int _{0}^{1}f(x)\,dx+\dotsb +\int _{n-1}^{n}f(x)\,dx\\&=\int _{0}^{n}f(x)\,dx\\&={\frac {f(0)}{2}}+f(1)+\dotsb +f(n-1)+{f(n) \over 2}-\int _{0}^{n}f'(x)P_{1}(x)\,dx\end{aligned}}$

Adding (ƒ(0) + ƒ(n))/2 to both sides and rearranging, we have

$\sum _{k=0}^{n}f(k)=\int _{0}^{n}f(x)\,dx+{f(0)+f(n) \over 2}+\int _{0}^{n}f'(x)P_{1}(x)\,dx\qquad (1)$

The last two terms therefore give the error when the integral is taken to approximate the sum.

Next, consider

$\int _{k}^{k+1}f'(x)P_{1}(x)\,dx=\int u\,dv$

where

${\begin{aligned}u&{}=f'(x)\\du&{}=f''(x)\,dx\\dv&{}=P_{1}(x)\,dx\\v&{}={\frac {1}{2}}P_{2}(x)\end{aligned}}$

Integrating by parts again, we get,

${\begin{aligned}uv-\int v\,du&=\left[{f'(x)P_{2}(x) \over 2}\right]_{k}^{k+1}-{1 \over 2}\int _{k}^{k+1}f''(x)P_{2}(x)\,dx\\&={B_{2} \over 2}(f'(k+1)-f'(k))-{1 \over 2}\int _{k}^{k+1}f''(x)P_{2}(x)\,dx\end{aligned}}$

Then summing from k = 0 to k = n − 1, and then replacing the last integral in (1) with what we have thus shown to be equal to it, we have

$\sum _{k=0}^{n}f(k)=\int _{0}^{n}f(x)\,dx+{f(0)+f(n) \over 2}+{\frac {B_{2}}{2}}(f'(n)-f'(0))-{1 \over 2}\int _{0}^{n}f''(x)P_{2}(x)\,dx.$

By now the reader will have guessed that this process can be iterated. In this way we get a proof of the Euler–Maclaurin summation formula by mathematical induction, in which the induction step relies on integration by parts and on the identities for periodic Bernoulli functions.

Derivation by functional analysis

The Euler–MacLaurin formula can be understood as a curious application of some ideas from Banach spaces and functional analysis.

First we restrict the problem to the domain of the unit interval [0, 1]. Let $\scriptstyle B_{n}(x)$ be the Bernoulli polynomials. A set of functions dual to the Bernoulli polynomials are given by

${\tilde {B}}_{n}(x)={\frac {(-1)^{n+1}}{n!}}\left(\delta ^{(n-1)}(x-1)-\delta ^{(n-1)}(x)\right)$

where δ is the Dirac delta function. The above is a formal notation for the idea of taking derivatives at a point; thus one has

$\int _{0}^{1}{\tilde {B}}_{n}(x)f(x)\,dx={\frac {1}{n!}}\left(f^{(n-1)}(1)-f^{(n-1)}(0)\right)$

for n > 0 and some arbitrary but differentiable function ƒ(x) on the unit interval. For the case of n = 0, one defines $\scriptstyle {\tilde {B}}_{0}(x)\;=\;1$. The Bernoulli polynomials, along with their duals, form an orthogonal set of states on the unit interval: one has

$\int _{0}^{1}{\tilde {B}}_{m}(x)B_{n}(x)\,dx=\delta _{mn}$

and

$\sum _{n=0}^{\infty }B_{n}(x){\tilde {B}}_{n}(y)=\delta (x-y)$

The Euler–MacLaurin summation formula then follows as an integral over the latter. One has

${\begin{aligned}f(x)&=\int _{0}^{1}\sum _{n=0}^{\infty }B_{n}(x){\tilde {B}}_{n}(y)f(y)\,dy\\&=\int _{0}^{1}f(y)\,dy+\sum _{n=1}^{N}B_{n}(x){\frac {1}{n!}}\left(f^{(n-1)}(1)-f^{(n-1)}(0)\right)-{\frac {1}{(N+1)!}}\int _{0}^{1}B_{N+1}(x-y)f^{(N)}(y)\,dy\end{aligned}}$

Then setting x = 0 and rearranging terms, one obtains an expression for ƒ(0). Note that the Bernoulli numbers are defined as B_n = B_n(0), and that these vanish for odd n greater than 1.

Then, using the periodic Bernoulli function P_n defined above and repeating the argument on the interval [1,2], one can obtain an expression of ƒ(1). This way one can obtain expressions for ƒ(n), n = 0, 1, 2, ..., N, and adding them up gives the Euler–MacLaurin formula. Note that this derivation does assume that ƒ(x) is sufficiently differentiable and well-behaved; specifically, that ƒ may be approximated by polynomials; equivalently, that ƒ is a real analytic function of exponential type $2\pi$. Written in explicit terms,

${\begin{aligned}\sum _{j=0}^{n-1}f(j)&=\int _{0}^{n}f(x)\,dx+\sum _{i=1}^{p}{B_{i} \over i!}\left(f^{(i-1)}(n)-f^{(i-1)}(0)\right)-(-1)^{p}\int _{0}^{n}{B_{p}(x-\lfloor x\rfloor ) \over p!}f^{(p)}(x)dx\\\sum _{j=1}^{n}f(j)&=\int _{0}^{n}f(x)\,dx+\sum _{i=1}^{p}(-1)^{i}{B_{i} \over i!}\left(f^{(i-1)}(n)-f^{(i-1)}(0)\right)-(-1)^{p}\int _{0}^{n}{B_{p}(x-\lfloor x\rfloor ) \over p!}f^{(p)}(x)dx\\\sum _{j=0}^{n}f(j)&=\int _{0}^{n}f(x)\,dx+\sum _{i=1}^{p}{1 \over i!}\left(B_{i}f^{(i-1)}(n)-B_{i}^{\star }f^{(i-1)}(0)\right)-(-1)^{p}\int _{0}^{n}{B_{p}(x-\lfloor x\rfloor ) \over p!}f^{(p)}(x)dx\end{aligned}}$

where $B_{i}^{\star }=(-1)^{i}B_{i}$ are the second kind of Bernoulli numbers and $B_{i}(x)$ indicate the periodic Bernoulli polynomials. This general formula holds for even and odd p ≥ 1.

The Euler–MacLaurin summation formula can thus be seen to be an outcome of the representation of functions on the unit interval by the direct product of the Bernoulli polynomials and their duals. Note, however, that the representation is not complete on the set of square-integrable functions. The expansion in terms of the Bernoulli polynomials has a non-trivial kernel. In particular, sin(2πnx) lies in the kernel; the integral of sin(2πnx) is vanishing on the unit interval, as is the difference of its derivatives at the endpoints. This is the essentially the reason for the restriction to exponential type of less than 2π: the function sin(2πnz) grows faster than e^2π|z| along the imaginary axis! Essentially, Euler-MacLaurin summation can be applied whenever Carlson's theorem holds; the Euler-MacLaurin formula is essentially a result obtaining from the study of finite differences and Newton series.

See also

• Cesàro summation
• Euler summation
• Gauss–Kronrod quadrature formula
• Darboux's formula

Notes

43 year old Petroleum Engineer Harry from Deep River, usually spends time with hobbies and interests like renting movies, property developers in singapore new condominium and vehicle racing. Constantly enjoys going to destinations like Camino Real de Tierra Adentro.

References

• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534, pp. 16, 806, 886
• 
  
  I had like 17 domains hosted on single account, and never had any special troubles. If you are not happy with the service you will get your money back with in 45 days, that's guaranteed. But the Search Engine utility inside the Hostgator account furnished an instant score for my launched website. Fantastico is unable to install WordPress in a directory which already have any file i.e to install WordPress using Fantastico the destination directory must be empty and it should not have any previous installation files. When you share great information, others will take note. Once your hosting is purchased, you will need to setup your domain name to point to your hosting. Money Back: All accounts of Hostgator come with a 45 day money back guarantee. If you have any queries relating to where by and how to use Hostgator Discount Coupon, you can make contact with us at our site. If you are starting up a website or don't have too much website traffic coming your way, a shared plan is more than enough. Condition you want to take advantage of the worldwide web you prerequisite a HostGator web page, -1 of the most trusted and unfailing web suppliers on the world wide web today. Since, single server is shared by 700 to 800 websites, you cannot expect much speed.
  
  
  
  Hostgator tutorials on how to install Wordpress need not be complicated, especially when you will be dealing with a web hosting service that is friendly for novice webmasters and a blogging platform that is as intuitive as riding a bike. After that you can get Hostgator to host your domain and use the wordpress to do the blogging. Once you start site flipping, trust me you will not be able to stop. I cut my webmaster teeth on Control Panel many years ago, but since had left for other hosting companies with more commercial (cough, cough) interfaces. If you don't like it, you can chalk it up to experience and go on. First, find a good starter template design. When I signed up, I did a search for current "HostGator codes" on the web, which enabled me to receive a one-word entry for a discount. Your posts, comments, and pictures will all be imported into your new WordPress blog.
• Xavier Gourdon and Pascal Sebah, Introduction on Bernoulli's numbers, (2002)
• D.H. Lehmer, "On the Maxima and Minima of Bernoulli Polynomials", American Mathematical Monthly, volume 47, pages 533–538 (1940)
• 20 year-old Real Estate Agent Rusty from Saint-Paul, has hobbies and interests which includes monopoly, property developers in singapore and poker. Will soon undertake a contiki trip that may include going to the Lower Valley of the Omo.
  
  My blog: http://www.primaboinca.com/view_profile.php?userid=5889534