Editing Euler–Maclaurin formula (section)

==Applications==

===The Basel problem===
The [[Basel problem]] is to determine the sum
<math display=block> 1 + \frac14 + \frac19 + \frac1{16} + \frac1{25} + \cdots = \sum_{n=1}^\infty \frac{1}{n^2}. </math>

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 {{math|{{sfrac|''π''<sup>2</sup>|6}}}}, which he proved in the same year.<ref>{{cite book
 | last = Pengelley | first = David J.
 | arxiv = 1912.03527
 | contribution = Dances between continuous and discrete: Euler's summation formula
 | location = Washington, DC
 | mr = 2349549
 | pages = 169–189
 | publisher = Mathematical Association of America
 | series = MAA Spectrum
 | title = Euler at 300
 | year = 2007}}</ref>

===Sums involving a polynomial===
{{see also|Faulhaber's formula}}
If {{mvar|f}} is a [[polynomial]] and {{mvar|p}} is big enough, then the remainder term vanishes.  For instance, if {{math|''f''(''x'') {{=}} ''x''<sup>3</sup>}}, we can choose {{math|''p'' {{=}} 2}} to obtain, after simplification,
<math display=block>\sum_{i=0}^n i^3 = \left(\frac{n(n + 1)}{2}\right)^2.</math>

===Approximation of integrals===
The formula provides a means of approximating a finite integral.  Let {{math|''a'' < ''b''}} be the endpoints of the interval of integration.  Fix {{mvar|N}}, the number of points to use in the approximation, and denote the corresponding step size by {{math|''h'' {{=}} {{sfrac|''b'' − ''a''|''N'' − 1}}}}.  Set {{math|''x<sub>i</sub>'' {{=}} ''a'' + (''i'' − 1)''h''}}, so that {{math|''x''<sub>1</sub> {{=}} ''a''}} and {{math|''x<sub>N</sub>'' {{=}} ''b''}}.  Then:<ref name="Devries">{{cite book |last1=Devries |first1=Paul L. |last2=Hasbrun |first2=Javier E. |title=A first course in computational physics. |edition=2nd |publisher=Jones and Bartlett Publishers |year=2011 |page=156}}</ref>
<math display=block>
\begin{align}
I & = \int_a^b f(x)\,dx \\
&\sim h\left(\frac{f(x_1)}{2} + f(x_2) + \cdots + f(x_{N-1}) + \frac{f(x_N)}{2}\right) + \frac{h^2}{12}\bigl[f'(x_1) - f'(x_N)\bigr] - \frac{h^4}{720}\bigl[f'''(x_1) - f'''(x_N)\bigr] + \cdots
\end{align}
</math>

This may be viewed as an extension of the [[trapezoid rule]] by the inclusion of correction terms.  Note that this asymptotic expansion is usually not convergent; there is some {{mvar|p}}, depending upon {{mvar|f}} and {{mvar|h}}, such that the terms past order {{mvar|p}} increase rapidly.  Thus, the remainder term generally demands close attention.<ref name="Devries"/>

The Euler–Maclaurin formula is also used for detailed [[error analysis (mathematics)|error analysis]] in [[numerical quadrature]]. It explains the superior performance of the [[trapezoidal rule]] on smooth [[periodic function]]s and is used in certain [[Series acceleration|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 expansion]]s of sums and [[Series (mathematics)|series]], usually the most useful form of the Euler–Maclaurin formula is
<math display=block>\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),</math>

where {{mvar|a}} and {{mvar|b}} are integers.<ref>{{Cite book | editor1-last=Abramowitz | editor1-first=Milton | editor1-link=Milton Abramowitz | editor2-last=Stegun | editor2-first=Irene A. | editor2-link=Irene Stegun | title=[[Abramowitz and Stegun|Handbook of Mathematical Functions with Formulas, Graphs, and Mathematical Tables]] | publisher=[[Dover Publications]] | location=New York | isbn=978-0-486-61272-0 | year=1972 | pages = 16, 806, 886}}</ref> Often the expansion remains valid even after taking the limits {{math|''a'' → −∞}} or {{math|''b'' → +∞}} or both.  In many cases the integral on the right-hand side can be evaluated in [[Differential Galois theory|closed form]] in terms of [[elementary function]]s 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,
<math display=block>\sum_{k=0}^\infty \frac{1}{(z + k)^2} \sim \underbrace{\int_0^\infty\frac{1}{(z + k)^2}\,dk}_{= \dfrac{1}{z}} + \frac{1}{2z^2} + \sum_{t = 1}^\infty \frac{B_{2t}}{z^{2t + 1}}.</math>

Here the left-hand side is equal to {{math|''ψ''<sup>(1)</sup>(''z'')}}, namely the first-order [[Polygamma function#Series representation|polygamma function]] defined by
:<math>\psi^{(1)}(z) = \frac{d^2}{dz^2}\log \Gamma(z);</math>
the [[gamma function]] {{math|Γ(''z'')}} is equal to {{math|(''z'' − 1)!}} when {{mvar|z}} is a [[positive integer]]. This results in an asymptotic expansion for {{math|''ψ''<sup>(1)</sup>(''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====
If {{mvar|s}} is an integer greater than 1 we have:
<math display=block>\sum_{k=1}^n \frac{1}{k^s} \approx \frac 1{s-1}+\frac 12-\frac 1{(s-1)n^{s-1}}+\frac 1{2n^s}+\sum_{i=1}\frac{B_{2i}}{(2i)!}\left[\frac{(s+2i-2)!}{(s-1)!}-\frac{(s+2i-2)!}{(s-1)!n^{s+2i-1}}\right].</math>

Collecting the constants into a value of the [[Riemann zeta function]], we can write an asymptotic expansion:
<math display=block>\sum_{k=1}^n \frac{1}{k^s} \sim\zeta(s)-\frac 1{(s-1)n^{s-1}}+\frac 1{2n^s}-\sum_{i=1}\frac{B_{2i}}{(2i)!}\frac{(s+2i-2)!}{(s-1)!n^{s+2i-1}}.</math>

For {{mvar|s}} equal to 2 this simplifies to
<math display=block>\sum_{k=1}^n \frac{1}{k^2} \sim\zeta(2)-\frac 1n+\frac 1{2n^2}-\sum_{i=1}\frac{B_{2i}}{n^{2i+1}},</math>
or
<math display=block>\sum_{k=1}^n \frac{1}{k^2} \sim \frac{\pi^2}{6} -\frac{1}{n} +\frac{1}{2n^2} -\frac{1}{6n^3}+\frac{1}{30n^5}-\frac{1}{42n^7} + \cdots.</math>

When {{math|''s'' {{=}} 1}}, the corresponding technique gives an asymptotic expansion for the [[harmonic number]]s:
<math display=block>\sum_{k=1}^n \frac{1}{k} \sim \gamma + \log n + \frac{1}{2n} - \sum_{k=1}^\infty \frac{B_{2k}}{2kn^{2k}},</math>
where {{math|''γ'' ≈ 0.5772...}} is the [[Euler–Mascheroni constant]].