Editing Trapezoidal rule (section)

==Error analysis==
[[File:Trapezium2.gif|right|thumb|An animation showing how the trapezoidal rule approximation improves with more strips for an interval with <math>a=2</math> and <math>b=8</math>. As the number of intervals <math>N</math> increases, so too does the accuracy of the result.]]
The error of the composite trapezoidal rule is the difference between the value of the integral and the numerical result:
<math display="block"> \text{E} = \int_a^b f(x)\,dx - \frac{b-a}{N} \left[ {f(a) + f(b) \over 2} + \sum_{k=1}^{N-1} f \left( a+k \frac{b-a}{N} \right) \right]</math>

There exists a number ''ξ'' between ''a'' and ''b'', such that<ref>{{harvtxt|Atkinson|1989|loc=equation (5.1.7)}}</ref>
<math display="block"> \text{E} = -\frac{(b-a)^3}{12N^2} f''(\xi)</math>

It follows that if the integrand is [[concave up]] (and thus has a positive second derivative), then the error is negative and the trapezoidal rule overestimates the true value. This can also be seen from the geometric picture: the trapezoids include all of the area under the curve and extend over it. Similarly, a [[concave-down]] function yields an underestimate because area is unaccounted for under the curve, but none is counted above. If the interval of the integral being approximated includes an [[inflection point]], the sign of the error is harder to identify.

An asymptotic error estimate for ''N'' → ∞ is given by
<math display="block"> \text{E} = -\frac{(b-a)^2}{12N^2} \big[ f'(b)-f'(a) \big] + O(N^{-3}). </math>
Further terms in this error estimate are given by the Euler–Maclaurin summation formula.

Several techniques can be used to analyze the error, including:<ref name="w0223">{{Harv|Weideman|2002|loc=p. 23, section 2}}</ref>
#[[Fourier series]]
#[[Residue calculus]]
#[[Euler–Maclaurin summation formula]]<ref>{{harvtxt|Atkinson|1989|loc=equation (5.1.9)}}</ref><ref>{{harvtxt|Atkinson|1989|loc=p. 285}}</ref>
#[[Polynomial interpolation]]<ref>{{harvtxt|Burden|Faires|2011|p=194}} </ref>

It is argued that the speed of convergence of the trapezoidal rule reflects and can be used as a definition of classes of smoothness of the functions.<ref name="rs90" />

=== Proof ===
First suppose that <math>h=\frac{b-a}{N}</math> and <math>a_k=a+(k-1)h</math>. Let <math display="block"> g_k(t) = \frac{1}{2}  t[f(a_k)+f(a_k+t)] - \int_{a_k}^{a_k+t} f(x) \, dx</math> be the function such that <math> |g_k(h)| </math> is the error of the trapezoidal rule on one of the intervals, <math> [a_k, a_k+h] </math>. Then
<math display="block"> {dg_k \over dt}={1 \over 2}[f(a_k)+f(a_k+t)]+{1\over2}t\cdot f'(a_k+t)-f(a_k+t),</math>
and
<math display="block"> {d^2g_k \over dt^2}={1\over 2}t\cdot f''(a_k+t).</math>

Now suppose that <math> \left| f''(x) \right| \leq \left| f''(\xi) \right|, </math> which holds if <math> f </math> is sufficiently smooth. It then follows that
<math display="block"> \left| f''(a_k+t) \right| \leq f''(\xi)</math>
which is equivalent to
<math> -f''(\xi) \leq f''(a_k+t) \leq f''(\xi)</math>, or <math> -\frac{f''(\xi)t}{2} \leq g_k''(t) \leq \frac{f''(\xi)t}{2}.</math>

Since <math> g_k'(0)=0</math> and <math> g_k(0)=0</math>,
<math display="block"> \int_0^t g_k''(x)  dx = g_k'(t)</math> and <math display="block"> \int_0^t g_k'(x) dx = g_k(t).</math>

Using these results, we find
<math display="block"> -\frac{f''(\xi)t^2}{4} \leq g_k'(t) \leq \frac{f''(\xi)t^2}{4}</math>
and
<math display="block"> -\frac{f''(\xi)t^3}{12} \leq g_k(t) \leq \frac{f''(\xi)t^3}{12}</math>

Letting <math> t = h </math> we find
<math display="block"> -\frac{f''(\xi)h^3}{12} \leq g_k(h) \leq \frac{f''(\xi)h^3}{12}.</math>

Summing all of the local error terms we find
<math display="block"> \sum_{k=1}^{N} g_k(h) = \frac{b-a}{N} \left[ {f(a) + f(b) \over 2} + \sum_{k=1}^{N-1} f \left( a+k \frac{b-a}{N} \right) \right] - \int_a^b f(x)dx.</math>

But we also have
<math display="block"> - \sum_{k=1}^N \frac{f''(\xi)h^3}{12} \leq \sum_{k=1}^N g_k(h) \leq  \sum_{k=1}^N \frac{f''(\xi)h^3}{12}</math>
and
<math display="block"> \sum_{k=1}^N \frac{f''(\xi)h^3}{12}=\frac{f''(\xi)h^3N}{12},</math>

so that

<math display="block"> -\frac{f''(\xi)h^3N}{12} \leq \frac{b-a}{N} \left[ {f(a) + f(b) \over 2} + \sum_{k=1}^{N-1} f \left( a+k \frac{b-a}{N} \right) \right]-\int_a^bf(x)dx \leq \frac{f''(\xi)h^3N}{12}.</math>

Therefore the total error is bounded by

<math display="block"> \text{error} = \int_a^b f(x)\,dx - \frac{b-a}{N} \left[ {f(a) + f(b) \over 2} + \sum_{k=1}^{N-1} f \left( a+k \frac{b-a}{N} \right) \right] = \frac{f''(\xi)h^3N}{12}=\frac{f''(\xi)(b-a)^3}{12N^2}.</math>

=== Periodic and peak functions ===
The trapezoidal rule converges rapidly for periodic functions. This is an easy consequence of the [[Euler-Maclaurin summation formula]], which says that
if <math>f</math> is <math>p</math> times continuously differentiable with period <math>T</math>
<math display="block">\sum_{k=0}^{N-1} f(kh)h =
    \int_0^T f(x)\,dx  +
    \sum_{k=1}^{\lfloor p/2\rfloor} \frac{B_{2k}}{(2k)!} (f^{(2k - 1)}(T) - f^{(2k - 1)}(0)) - (-1)^p h^p \int_0^T\tilde{B}_{p}(x/T)f^{(p)}(x) \, dx
</math>
where <math>h:=T/N</math> and <math>\tilde{B}_{p}</math> is the periodic extension of the <math>p</math>th Bernoulli polynomial.<ref>{{cite book|title=Numerical Analysis, volume 181 of Graduate Texts in Mathematics|first=Rainer|last=Kress |year=1998 |publisher=Springer-Verlag}}</ref> Due to the periodicity, the derivatives at the endpoint cancel and we see that the error is <math>O(h^p)</math>.

A similar effect is available for peak-like functions, such as [[Gaussian function|Gaussian]], [[Exponentially modified Gaussian distribution|Exponentially modified Gaussian]] and other functions with derivatives at integration limits that can be neglected.<ref>{{Cite journal |last=Goodwin|first=E. T. |date=1949 |title=The evaluation of integrals of the form <math>\textstyle\int_{-\infty}^\infty{f(x)e^{-x^2}dx}</math> | journal=[[Mathematical Proceedings of the Cambridge Philosophical Society]] |language=en |volume=45 |issue=2 |pages=241–245 |doi=10.1017/S0305004100024786 |bibcode=1949PCPS...45..241G |issn=1469-8064}}</ref> The evaluation of the full integral of a Gaussian function by trapezoidal rule with 1% accuracy can be made using just 4 points.<ref name=":0">{{Cite journal| last1=Kalambet|first1=Yuri |last2=Kozmin|first2=Yuri |last3=Samokhin|first3=Andrey |date=2018 |title=Comparison of integration rules in the case of very narrow chromatographic peaks |journal=Chemometrics and Intelligent Laboratory Systems|volume=179 |pages=22–30 |doi=10.1016/j.chemolab.2018.06.001|issn=0169-7439}}</ref> [[Simpson's rule]] requires 1.8 times more points to achieve the same accuracy.<ref name=":0" /><ref name="w02" />

=== "Rough" functions ===
For functions that are not in [[Smoothness|''C''<sup>2</sup>]], the error bound given above is not applicable. Still, error bounds for such rough functions can be derived, which typically show a slower convergence with the number of function evaluations <math>N</math> than the <math>O(N^{-2})</math> behaviour given above. Interestingly, in this case the trapezoidal rule often has sharper bounds than [[Simpson's rule]] for the same number of function evaluations.<ref name="cun02" />