Editing Euler–Maclaurin formula (section)

==The formula==
If {{mvar|m}} and {{mvar|n}} are [[natural number]]s and {{math|''f''(''x'')}} is a [[Real number|real]] or [[Complex number|complex]] valued [[continuous function]] for [[real number]]s {{mvar|x}} in the [[Interval (mathematics)|interval]] {{math|[''m'',''n'']}}, then the integral
<math display=block>I = \int_m^n f(x)\,dx</math>
can be approximated by the sum (or vice versa)
<math display=block>S = f(m + 1) + \cdots + f(n - 1) + f(n)</math>
(see [[Riemann sum#Right rule|rectangle method]]). The Euler–Maclaurin formula provides expressions for the difference between the sum and the integral in terms of the higher [[derivative]]s {{math|''f''{{isup|(''k'')}}(''x'')}} evaluated at the endpoints of the interval, that is to say {{math|''x'' {{=}} ''m''}} and {{math|''x'' {{=}} ''n''}}.

Explicitly, for {{mvar|p}} a positive [[integer]] and a function {{math|''f''(''x'')}} that is {{mvar|p}} times [[continuously differentiable]] on the interval {{math|[''m'',''n'']}}, we have
<math display=block>S - I = \sum_{k=1}^p {\frac{B_k}{k!} \left(f^{(k - 1)}(n) - f^{(k - 1)}(m)\right)} + R_p,</math>
where {{mvar|B<sub>k</sub>}} is the {{mvar|k}}th [[Bernoulli numbers|Bernoulli number]] (with {{math|''B''<sub>1</sub> {{=}} {{sfrac|1|2}}}}) and {{mvar|R<sub>p</sub>}} is an [[numerical integration|error term]] which depends on {{mvar|n}}, {{mvar|m}}, {{mvar|p}}, and {{mvar|f}} and is usually small for suitable values of {{mvar|p}}.

The formula is often written with the subscript taking only even values, since the odd Bernoulli numbers are zero except for {{math|''B''<sub>1</sub>}}.  In this case we have<ref name=":0" /><ref name="DLMF">{{cite web|url=http://dlmf.nist.gov/2.10|title=Digital Library of Mathematical Functions: Sums and Sequences|publisher=[[National Institute of Standards and Technology]]}}</ref>
<math display=block>\sum_{i=m}^n f(i) =
    \int^n_m f(x)\,dx + \frac{f(n) + f(m)}{2} +
    \sum_{k=1}^{\left\lfloor \frac{p}{2}\right\rfloor} \frac{B_{2k}}{(2k)!} \left(f^{(2k - 1)}(n) - f^{(2k - 1)}(m)\right) + R_p,
</math>
or alternatively
<math display=block>\sum_{i=m+1}^n f(i) =
    \int^n_m f(x)\,dx + \frac{f(n) - f(m)}{2} +
    \sum_{k=1}^{\left\lfloor \frac{p}{2}\right\rfloor} \frac{B_{2k}}{(2k)!} \left(f^{(2k - 1)}(n) - f^{(2k - 1)}(m)\right) + R_p.
</math>

===The remainder term===
{{see also|Bernoulli polynomials}}

The remainder term arises because the integral is usually not exactly equal to the sum.  The formula may be derived by applying repeated [[integration by parts]] to successive intervals {{math|[''r'', ''r'' + 1]}} for {{math|''r'' {{=}} ''m'', ''m'' + 1, …, ''n'' − 1}}.  The boundary terms in these integrations lead to the main terms of the formula, and the leftover integrals form the remainder term.

The remainder term has an exact expression in terms of the periodized Bernoulli functions {{math|''P<sub>k</sub>''(''x'')}}.  The Bernoulli polynomials may be defined recursively by {{math|''B''<sub>0</sub>(''x'') {{=}} 1}} and, for {{math|''k'' ≥ 1}},
<math display=block>\begin{align}
  B_k'(x) &= kB_{k - 1}(x), \\
  \int_0^1 B_k(x)\,dx &= 0.
\end{align}</math>
The periodized Bernoulli functions are defined as
<math display=block>P_k(x) = B_k\bigl(x - \lfloor x\rfloor\bigr),</math>
where {{math|⌊''x''⌋}} denotes the largest integer less than or equal to {{mvar|x}}, so that {{math|''x'' − ⌊''x''⌋}} always lies in the interval {{math|[0,1)}}.

With this notation, the remainder term {{mvar|R<sub>p</sub>}} equals
<math display="block">R_{p} = (-1)^{p+1}\int_m^n f^{(p)}(x) \frac{P_p(x)}{p!}\,dx. </math>

When {{math|''k'' > 0}}, it can be shown that for {{math|0 ≤ ''x'' ≤ 1}},
<math display=block>\bigl|B_k(x)\bigr| \le \frac{2 \cdot k!}{(2\pi)^k}\zeta(k),</math>
where {{mvar|ζ}}  denotes the [[Riemann zeta function]]; one approach to prove this inequality is to obtain the Fourier series for the polynomials {{math|''B<sub>k</sub>''(''x'')}}. The bound is achieved for even {{mvar|k}} when {{mvar|x}} is zero.  The term {{math|''ζ''(''k'')}} may be omitted for odd {{mvar|k}} but the proof in this case is more complex (see Lehmer).<ref name="Lehmer">{{cite journal|last1=Lehmer|first1=D. H.|author-link=D. H. Lehmer |title=On the maxima and minima of Bernoulli polynomials | date=1940 | journal=[[The American Mathematical Monthly]]|volume=47|issue=8|pages=533–538 |doi=10.2307/2303833|jstor=2303833}}</ref> Using this inequality, the size of the remainder term can be estimated as
<math display=block>\left|R_p\right| \leq \frac{2 \zeta(p)}{(2\pi)^p}\int_m^n \left|f^{(p)}(x)\right|\,dx.</math>

===Low-order cases===

The Bernoulli numbers from {{math|''B''<sub>1</sub>}} to {{math|''B''<sub>7</sub>}} are {{math|{{sfrac|1|2}}, {{sfrac|1|6}}, 0, −{{sfrac|1|30}}, 0, {{sfrac|1|42}}, 0}}. Therefore, the low-order cases of the Euler–Maclaurin formula are:
<math display=block>\begin{align}
\sum_{i=m}^n f(i) - \int_m^n f(x)\,dx &= \frac{f(m)+f(n)}{2} + \int_m^n f'(x)P_1(x)\,dx \\
&=\frac{f(m)+f(n)}{2} + \frac{1}{6}\frac{f'(n) - f'(m)}{2!} - \int_m^n f''(x)\frac{P_2(x)}{2!}\,dx \\
&=\frac{f(m)+f(n)}{2} + \frac{1}{6}\frac{f'(n) - f'(m)}{2!} + \int_m^n f'''(x)\frac{P_3(x)}{3!}\,dx \\
&=\frac{f(m)+f(n)}{2} + \frac{1}{6}\frac{f'(n) - f'(m)}{2!} - \frac{1}{30}\frac{f'''(n) - f'''(m)}{4!}-\int_m^n f^{(4)}(x) \frac{P_4(x)}{4!}\, dx \\
&=\frac{f(m)+f(n)}{2} + \frac{1}{6}\frac{f'(n) - f'(m)}{2!} - \frac{1}{30}\frac{f'''(n) - f'''(m)}{4!} + \int_m^n f^{(5)}(x)\frac{P_5(x)}{5!}\,dx \\
&=\frac{f(m)+f(n)}{2} + \frac{1}{6}\frac{f'(n) - f'(m)}{2!} - \frac{1}{30}\frac{f'''(n) - f'''(m)}{4!} + \frac{1}{42}\frac{f^{(5)}(n) - f^{(5)}(m)}{6!} - \int_m^n f^{(6)}(x)\frac{P_6(x)}{6!}\,dx \\
&=\frac{f(m)+f(n)}{2} + \frac{1}{6}\frac{f'(n) - f'(m)}{2!} - \frac{1}{30}\frac{f'''(n) - f'''(m)}{4!} + \frac{1}{42}\frac{f^{(5)}(n) - f^{(5)}(m)}{6!} + \int_m^n f^{(7)}(x)\frac{P_7(x)}{7!}\,dx.
\end{align}</math>