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Trapezoidal rule

In calculus, the trapezoidal rule is a technique for numerical integration, i.e. approximating the definite integral:

History
A 2016 Science paper reports that the trapezoid rule was in use in Babylon before 50 BCE for integrating the velocity of Jupiter along the ecliptic. == Numerical implementation ==
Numerical implementation
Non-uniform grid When the grid spacing is non-uniform, one can use the formula \int_a^b f(x) \,dx \approx \sum_{k=1}^N \frac{f(x_{k-1}) + f(x_k)}{2} \Delta x_k, where \Delta x_k = x_k - x_{k-1}, or more a computationally efficient formula \int_a^b f(x) \,dx \approx \frac{1}{2} \biggl(f(x_0) \Delta_{+1} x_0 + f(x_N) \Delta_{-1} x_N + \sum_{k=1}^{N-1} f(x_k) \Delta_{\pm1} x_k\biggr), where \Delta_{+1} x_0 = x_1 - x_0, \Delta_{-1} x_N = x_N - x_{N-1}, \Delta_{\pm1} x_k = x_{k+1} - x_{k-1} are the corresponding forward, backward, and central differences. Uniform grid For a domain partitioned by N equally spaced points, considerable simplification may occur. Let \Delta x = \frac{b - a}{N} and x_k = a + k \Delta x for k = 0, 1, \ldots, N. The approximation to the integral becomes \begin{align} \int_a^b f(x) \,dx &\approx \frac{\Delta x}{2} \sum_{k=1}^N [f(x_{k-1}) + f(x_{k})] \\ &= \Delta x \biggl( \tfrac12f(x_0) + \tfrac12f(x_N) + \sum_{k=1}^{N-1} f(x_k) \biggr). \end{align} Sometimes this expression is written as \Delta x \!\! \mathop{\ \sum\vphantom\big)'}_{k=0}^{N} f(x_k), where the symbol indicates that the first and last terms are halved. ==Error analysis==
Error analysis
The error of the composite trapezoidal rule is the difference between the value of the integral and the numerical result: \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] There exists a number ξ between a and b, such that \text{E} = -\frac{(b-a)^3}{12N^2} f''(\xi) 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 \text{E} = -\frac{(b-a)^2}{12N^2} \big[ f'(b)-f'(a) \big] + O(N^{-3}). Further terms in this error estimate are given by the Euler–Maclaurin summation formula. Several techniques can be used to analyze the error, including: • Fourier seriesResidue calculusEuler–Maclaurin summation formulaPolynomial interpolation 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. Proof First suppose that h=\frac{b-a}{N} and a_k=a+(k-1)h. Let g_k(t) = \frac{1}{2} t[f(a_k)+f(a_k+t)] - \int_{a_k}^{a_k+t} f(x) \, dx be the function such that |g_k(h)| is the error of the trapezoidal rule on one of the intervals, [a_k, a_k+h] . Then {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), and {d^2g_k \over dt^2}={1\over 2}t\cdot f''(a_k+t). Now suppose that \left| f(x) \right| \leq \left| f(\xi) \right|, which holds if f is sufficiently smooth. It then follows that \left| f(a_k+t) \right| \leq f(\xi) which is equivalent to -f(\xi) \leq f(a_k+t) \leq f(\xi), or -\frac{f(\xi)t}{2} \leq g_k(t) \leq \frac{f(\xi)t}{2}. Since g_k'(0)=0 and g_k(0)=0, \int_0^t g_k''(x) dx = g_k'(t) and \int_0^t g_k'(x) dx = g_k(t). Using these results, we find -\frac{f''(\xi)t^2}{4} \leq g_k'(t) \leq \frac{f''(\xi)t^2}{4} and -\frac{f(\xi)t^3}{12} \leq g_k(t) \leq \frac{f(\xi)t^3}{12} Letting t = h we find -\frac{f(\xi)h^3}{12} \leq g_k(h) \leq \frac{f(\xi)h^3}{12}. Summing all of the local error terms we find \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. But we also have - \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} and \sum_{k=1}^N \frac{f(\xi)h^3}{12}=\frac{f(\xi)h^3N}{12}, so that -\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}. Therefore the total error is bounded by \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}. 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 f is p times continuously differentiable with period T, then \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)!} \left(f^{(2k - 1)}(T) - f^{(2k - 1)}(0)\right) - (-1)^p h^p \int_0^T\tilde{B}_{p}(x/T)f^{(p)}(x) \, dx, where h := T/N, and \tilde{B}_p is the periodic extension of the p-th Bernoulli polynomial. Due to the periodicity, the derivatives at the endpoint cancel, and we see that the error is O(h^p). A similar effect is available for peak-like functions, such as Gaussian, Exponentially modified Gaussian and other functions with derivatives at integration limits that can be neglected. The evaluation of the full integral of a Gaussian function by trapezoidal rule with 1% accuracy can be made using just 4 points. Simpson's rule requires 1.8 times more points to achieve the same accuracy. "Rough" functions For functions that are not in C2, 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 N than the O(N^{-2}) 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. == Applicability and alternatives ==
Applicability and alternatives
The trapezoidal rule is one of a family of formulas for numerical integration called Newton–Cotes formulas, of which the midpoint rule is similar to the trapezoid rule. Simpson's rule is another member of the same family, and in general has faster convergence than the trapezoidal rule for functions which are twice continuously differentiable, though not in all specific cases. However, for various classes of rougher functions (ones with weaker smoothness conditions), the trapezoidal rule has faster convergence in general than Simpson's rule. Moreover, the trapezoidal rule tends to become extremely accurate when periodic functions are integrated over their periods, which can be analyzed in various ways. Convergence usually is exponential or faster. A similar effect is available for peak functions. For non-periodic functions, however, methods with unequally spaced points such as Gaussian quadrature and Clenshaw–Curtis quadrature are generally far more accurate; Clenshaw–Curtis quadrature can be viewed as a change of variables to express arbitrary integrals in terms of periodic integrals, at which point the trapezoidal rule can be applied accurately. ==Numerical examples==
Numerical examples
Approximating the natural logarithm of 3 Since \int_1^3 \frac{1}{x} \,dx = \ln 3 - \ln 1 = \ln 3, we can use the trapezoidal rule to approximate the integral, thereby generating an approximation of \ln 3. Applying the rule with n = 3 segments gives \ln 3 = \int_1^3 \frac{1}{x} \,dx \approx \frac{1}{3} \left(1 + \frac{6}{5} + \frac{6}{7} + \frac{1}{3}\right) = \frac{356}{315} \approx 1.13015873, which has absolute error of 0.031546 and a relative error of 2.87148\%. Applying the rule with n = 6 segments gives \ln 3 = \int_1^3 \frac{1}{x} \,dx \approx \frac{1}{6} \left(1 + \frac{3}{2} + \frac{6}{5} + 1 + \frac{6}{7} + \frac{3}{4} + \frac{1}{3}\right) = \frac{2789}{2520} \approx 1.10674603, which has absolute error of 8.13 \times 10^{-3} and a relative error of 0.74036\%. Approximating the integral of a product The following integral is given: \int_{0.1}^{1.3} 5xe^{- 2x} \,dx. Solution {{ordered list |list-style-type=lower-alpha \int_a^b f(x) \,dx \approx \frac{b - a}{2n} \left[ f(a) + 2\sum_{i=1}^{n-1} f(a + ih) + f(b) \right]. \begin{align} n &= 3, \\ a &= 0.1, \\ b &= 1.3, \\ h &= \frac{b - a}{n} = \frac{1.3 - 0.1}{3} = 0.4. \end{align} Using the composite trapezoidal rule formula, \int_a^b f(x) \,dx \approx \frac{b - a}{2n} \left[ f(a) + 2\left\{ \sum_{i=1}^{n-1} f(a + ih) \right\} + f(b) \right]. \begin{align} I &\approx \frac{1.3 - 0.1}{6} \left[ f(0.1) + 2\sum_{i=1}^{3-1} f(0.1 + 0.4i) + f(1.3) \right] \\ I &\approx \frac{1.3 - 0.1}{6} \left[ f(0.1) + 2\sum_{i=1}^2 f(0.1 + 0.4i) + f(1.3) \right] \\ &= 0.2 [f(0.1) + 2f(0.5) + 2f(0.9) + f(1.3)] \\ &= 0.2 \left[5 \times 0.1 \times e^{-2(0.1)} + 2(5 \times 0.5 \times e^{-2(0.5)}) + 2(5 \times 0.9 \times e^{-2(0.9)}) + 5 \times 1.3 \times e^{-2(1.3)}\right]\\ &= 0.84385. \end{align} \int_{0.1}^{1.3} 5xe^{-2x} \,dx = 0.89387, so the true error is \begin{align} E_t &= (\text{true value}) - (\text{approximate value}) \\ &= 0.89387 - 0.84385 \\ &= 0.05002. \end{align} \begin{align} |\varepsilon_t| &= \left| \frac{\text{true error}}{\text{true value}} \right| \times 100\% \\ &= \left| \frac{0.05002}{0.89387} \right| \times 100\% \\ &= 5.5959\%. \end{align} }} ==See also==
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