Numerical integration is often referred to as quadrature, a term that stems from classical antiquity. a If false, provide a counterexample. and Dec 2, 2018 -- 1 I appreciate this integration concept primarily because among the method that we've discussed in class, this one is the only method that consider eliminating its errors. 0 Simpson's rule is named after the 18th century English mathematician Thomas Simpson, despite its use a century earlier by the German mathematician and astronomer Johannes Kepler. In this case we can compute the integral exactly (which is one of the reasons it was chosen as a first example): So the error in the approximation generated by eight steps of the midpoint rule is, We can write this as a percentage error by multiplying it by 100, relative error \(\frac{|A-\alpha|}{A}\) and, percentage error \(100\frac{|A-\alpha|}{A}\), We again start by setting up all the \(x\)-values that we will need. (See numerical integration for more on quadrature rules.) Interpolation with polynomials evaluated at equally spaced points in ) The exact answer is \(\int_0^\pi\sin x\,\, d{x}=-\cos x\Big|_0^\pi=2\text{. ( The quadrature of the hyperbola by Saint-Vincent and de Sarasa provided a new function, the natural logarithm, of critical importance. The exact value of the integral is still \(\pi\text{. With only eight steps of Simpson's rule we achieved \(100\tfrac{2.00027-2}{2}=0.014\%\) accuracy. }\) If \(-2 \lt f''(x) \lt 0\) for \(1 \le x \le 4\text{,}\) find a value of \(n\) to guarantee the trapezoidal rule will give an approximation for \(\displaystyle\int_1^4 f(x)\, d{x}\) with absolute error, \(|E_T|\text{,}\) less than \(0.001\text{.}\). Interactive Demonstration. {\displaystyle a} Use Simpson's rule in x-direction and Simpson's rule in y- direction. The differential equation Fortunately, there is a little trick that turns the curve \(Y=\tfrac{K}{X^k}\) into a straight line no matter what \(k\) is. {\displaystyle F} has a special form: the right-hand side contains only the independent variable (here The trapezoidal rule error with \(n\) steps is about twice the midpoint rule error with \(n\) steps. Note: The graph of \(f''''(x)\text{,}\) where \(f(x)=\cos(x^2)\text{,}\) is shown below. We then derive the simplest numerical integration method, and see how . 0 b The integral \(\displaystyle\int_{-1}^{1} \sin(x^2) \, \, d{x}\) is estimated using the Midpoint Rule with \(1000\) intervals. First let us check that the above result is consistent with our data in Figure 1.11.12, \begin{align*} \left|\frac{d^{2}}{dx^{2}}\sin x\right| &= |-\sin x| \leq 1 \end{align*}, \begin{align*} |e_n|&\le\frac{M}{24}\frac{(b-a)^3}{n^2}\\ &=\frac{\pi^3}{24}\frac{1}{n^2}\\ &\approx 1.29\frac{1}{n^2} \end{align*}, \begin{align*} |e_n| &\approx 2^{-.2706}\frac{1}{n^2}=0.83\frac{1}{n^2} \end{align*}. The mathematical denition of the integral is basically via a numerical in-tegration method, and we therefore start by reviewing this denition. Extrapolation methods are described in more detail by Stoer and Bulirsch (Section 3.4) and are implemented in many of the routines in the QUADPACK library. Now we are armed with our three (relatively simple) method for numerical integration we should give thought to how practical they might be in the real world7. From time to time, I have received questions from students on various aspects of this topic, including: In this series, I hope to answer these questions. Example1.11.16 How many steps using Simpson's rule? Integration by Simpson's 1/3 rule can be represented as a weighted average with 2/3 of the value coming from integration by the trapezoidal rule with step h and 1/3 of the value coming from integration by the rectangle rule with step 2h. In a typical application we would be asked to evaluate a given integral to some specified accuracy. x In W. Freeden et al. }\) This transformation8works because when \(Y=\frac{K}{X^k}\), \begin{align*} \log Y &= \log K - k \log X \end{align*}, So plotting \(y=\log Y\) against \(x=\log X\) gives the straight line \(y=\log K -kx\text{,}\) which has slope \(-k\) and \(y\)-intercept \(\log K\text{.}\). That is why the process was named quadrature. The problem of evaluating integrals is thus best studied in its own right. Figure 1.11.12 contains two graphs of the results. \begin{align*} &\int_0^\pi\sin x\,\, d{x} \approx \Big[\frac{1}{2}\sin(x_0)+\sin(x_1)+\cdots+\sin(x_7)+\frac{1}{2}\sin(x_8)\Big]\Delta x\\ &=\Big[\frac{1}{2}\sin0 + \sin\tfrac{\pi}{8}+ \sin\tfrac{2\pi}{8}+ \sin\tfrac{3\pi}{8}+ \sin\tfrac{4\pi}{8}+ \sin\tfrac{5\pi}{8}\\ &\hskip0.5in+\sin\tfrac{6\pi}{8}+ \sin\tfrac{7\pi}{8}+ \frac{1}{2}\sin\tfrac{8\pi}{8}\Big]\tfrac{\pi}{8}\\ &=\Big[\frac{1}{2}\!\times\! We start with the error generated by a single step of the midpoint rule. }\) You may use a calculator, but only to add, subtract, multiply, and divide. A global criterion is that the sum of errors on all the intervals should be less thant. This type of error analysis is usually called "a posteriori" since we compute the error after having computed the approximation. 1 \begin{align*} &\hbox{approx value of $\displaystyle \int_a^b f(x)\,\, d{x}$ given by $n$ midpoint steps } \hskip-0.35in&&\approx \int_a^b f(x)\,\, d{x}+K_M\cdot \frac{1}{n^2}\\ &\hbox{approx value of $\displaystyle\int_a^b f(x)\,\, d{x}$ given by $n$ trapezoidal steps } \hskip-0.35in&&\approx \int_a^b f(x)\,\, d{x}+K_T\cdot \frac{1}{n^2}\\ &\hbox{approx value of $\displaystyle\int_a^b f(x)\,\, d{x}$ given by $n$ Simpson's steps } \hskip-0.35in&&\approx \int_a^b f(x)\,\, d{x}+K_M\cdot \frac{1}{n^4} \end{align*}, with some constants \(K_M,\ K_T\) and \(K_S\text{. In response, the term quadrature has become traditional, and instead the modern phrase "computation of a univariate definite integral" is more common. Square roots and logarithms without a calculator (Part 6), Engaging students: Exponential Growth and Decay, Square roots and logarithms without a calculator (Part 3), Engaging students: Introducing the number e. Why is numerical integration necessary in the first place? }\), The area of the approximating rectangle is \(f(\bar x_j)\Delta x\text{,}\) and the midpoint rule approximates each subintegral by, \[ \int_{x_{j-1}}^{x_j} f(x)\,\, d{x}\approx f(\bar x_j)\Delta x\text{.} + }\) In each algorithm, we begin in much the same way as we approached Riemann sums. ( }\) Leave your answer in calculator-ready form. The table below gives the diameters of the cross sections in centimeters at 10 cm intervals. For example, \(f(x)=\frac{1}{6}x^3-\frac{1}{2}x^2+(1+x)\log|x+1|\) will do, but you don't need to know what \(f(x)\) is for this problem. a g ( In numerical analysis, numerical integration constitutes a broad family of algorithms for calculating the numerical value of a definite integral, and by extension, the term is also sometimes used to describe the numerical solution of differential equations. }\) Further, if \(|f^{(4)}(x)|\leq L\) for all \(a\leq x \leq b\text{,}\) then, The first of these error bounds in proven in the following (optional) section. }\), Assuming \(M=1\text{,}\) how many intervals should you use to approximate \(\displaystyle\int_{0}^{1}f(x) \, d{x}\) to within \(10^{-5}\text{? This approach requires the function evaluations to grow exponentially as the number of dimensions increases. }\), Consider the quantity \(A=\displaystyle\int_{-\pi}^{\pi} \cos x \, d{x}\text{.}\). Numerical integration is a standard topic in first-semester calculus. To compare and contrast we apply the trapezoidal rule to the examples we did above with the midpoint rule. The result is usually more accurate as the number of evaluation points increases, or, equivalently, as the width of the step size between the points decreases. + yields the NewtonCotes formulas, of which the rectangle rule and the trapezoidal rule are examples. \begin{align*} x_0&=0& x_1&=\tfrac{\pi}{8}& x_2&=\tfrac{2\pi}{8}& \cdots&& x_7&=\tfrac{7\pi}{8}& x_8&=\tfrac{8\pi}{8}=\pi \end{align*}, \begin{align*} \bar x_1&=\tfrac{\pi}{16}& \bar x_2&=\tfrac{3\pi}{16} & \cdots&& \bar x_7&=\tfrac{13\pi}{16} & \bar x_8&=\tfrac{15\pi}{16} \end{align*}, \begin{align*} &\int_0^\pi\sin x\,\, d{x} \approx\Big[\sin(\bar x_1)+\sin(\bar x_2)+\cdots+\sin(\bar x_8)\Big]\Delta x\\ &=\Big[\sin(\tfrac{\pi}{16})+ \sin(\tfrac{3\pi}{16})+ \sin(\tfrac{5\pi}{16})+ \sin(\tfrac{7\pi}{16})+ \sin(\tfrac{9\pi}{16})+\\ &\hskip2in \sin(\tfrac{11\pi}{16})+ \sin(\tfrac{13\pi}{16})+ \sin(\tfrac{15\pi}{16})\Big]\tfrac{\pi}{8}\\ &=\Big[0.1951+ 0.5556+ 0.8315+ 0.9808+ 0.9808+\\ &\hskip2in 0.8315+ 0.5556+ 0.1951\Big]\times 0.3927\\ &=5.1260\times 0.3927 =2.013 \end{align*}, \begin{align*} \int_0^\pi \sin x \, d{x} &= \big[ -\cos x \big]_0^\pi = -\cos\pi + \cos 0 = 2. In medieval Europe the quadrature meant calculation of area by any method. a }\) The error for the trapezoidal rule satisfies \(|E_T| \le \dfrac{ M(b - a)^3}{12n^2}\), where \(|f''(x)| \le M\) for \(a \le x \le b\text{. a Summing these all together gives: \begin{align*} \int_a^b& f(x)\,\, d{x} =\int_{x_0}^{x_2} f(x)\,\, d{x} +\int_{x_2}^{x_4} f(x)\,\, d{x} +\int_{x_4}^{x_6} f(x)\,\, d{x} +\cdots +\int_{x_{n-2}}^{x_n} f(x)\,\, d{x}\\ &\approx\,\tfrac{\Delta x}{3}\big[f(x_0)+4f(x_1)+f(x_2)\big] +\,\tfrac{\Delta x}{3}\big[f(x_2)+4f(x_3)+f(x_4)\big] \cr &\ \ \ +\,\tfrac{\Delta x}{3}\big[f(x_4)+4f(x_5)+f(x_6)\big] +\,\cdots\ +\,\tfrac{\Delta x}{3}\big[f(x_{n-2})+4f(x_{n-1})+f(x_n)\big]\\ &=\Big[f(x_0)\!+4f(x_1)\!+2f(x_2)\!+4f(x_3)\!+2f(x_4)\! Kerstin Hesse, Ian H. Sloan, and Robert S. Womersley: Numerical Integration on the Sphere. Numerical integration obtains the area under the curve, dy*dx, on an x-y plot. However, your eye is pretty good at determining whether or not a graph is a straight line. Is there a reason why Simpsons Rule converges like the fourth power of the number of subintervals. ( }\), \begin{gather*} |E_M| \le \frac{M(b-a)^3}{24n^2} \qquad\text{and}\qquad |E_S| \le \frac{L(b-a)^5}{180n^4}, \end{gather*}, Find a bound for the error in approximating \(\displaystyle\int_1^5 \frac{1}{x}\,\, d{x}\) using Simpson's rule with \(n = 4\text{. b {\displaystyle k=0,\ldots ,n-1.} The following example of Mathematica code generates the plot showing difference between inverse tangent and its approximation truncated at k ) a b 1 }\) We would like to test if this is really the case, by graphing \(Y=e_n\) against \(X=n\) and seeing if the graph looks right. ) , {\displaystyle f} ( - Syed Mar 17, 2022 at 17:36 Add a comment 2 Answers Sorted by: 7 Just a quick observation. {\displaystyle f'} \end{align*}, \begin{align*} \text{area} &= \int_{-h}^h \big[ Ax^2 + Bx +C \big] \, d{x} = \frac{h}{3}\left( 2Ah^2 + 6C \right)\\ &= \frac{h}{3}\left(y_{-1}+4y_0+y_1\right) \end{align*}, \[ \int_{x_0}^{x_2} f(x)\,\, d{x} \approx \tfrac{1}{3}\Delta x\big[f(x_0)+4f(x_1)+f(x_2)\big] \nonumber \], \[ \int_{x_2}^{x_4} f(x)\,\, d{x} \approx \tfrac{1}{3}\Delta x\big[f(x_2)+4f(x_3)+f(x_4)\big] \nonumber \]. {\textstyle \left({\frac {a+b}{2}},f\left({\frac {a+b}{2}}\right)\right)} b f , We again have \(a=0\text{,}\) \(b=\pi\text{,}\) \(\Delta x=\tfrac{\pi}{8}\) and. 2 2 }\), A function \(s(x)\) satisfies \(s(0)=1.00664\text{,}\) \(s(2)=1.00543\text{,}\) \(s(4)=1.00435\text{,}\) \(s(6)=1.00331\text{,}\) \(s(8)=1.00233\text{. Note that we have chosen to use logarithms9with this unusual base because it makes it very clear how much the error is improved if we double the number of steps used. = We proceed very similarly to Example 1.11.5 and again use \(n=8\) steps. We will explain these rules in detail below, but we give a brief overview here: \begin{align*} \int_{x_{j-1}}^{x_{j}} f(x) \, d{x} & \approx f\left( \frac{x_{j-1}+x_{j}}{2} \right) \Delta x \end{align*}, \begin{align*} \int_{x_{j-1}}^{x_{j}} f(x) \, d{x} & \approx \frac{1}{2} \left[f(x_{j-1})+f(x_j) \right] \Delta x \end{align*}, \begin{align*} \int_{x_{j-1}}^{x_{j+1}} f(x) \, d{x} & \approx \frac{1}{3} \left[f(x_{j-1})+4f(x_j)+f(x_{j+1}) \right] \Delta x \end{align*}, In what follows we need to refer to the midpoint between \(x_{j-1}\) and \(x_j\) very frequently. {\displaystyle \left(h_{k}\right)_{k}} }\) Do not write down the Simpson's rule approximation \(S_n\text{. First, lets talk about why numerical integration is necessary in the first place. Now is a good time for a quick revision of logarithms see Whirlwind review of logarithms in Section 2.7 of the CLP-1 text. , Three methods are known to overcome this so-called curse of dimensionality. our error is no greater than the right hand side of 1. }\) The width of this region is \(x_j-x_{j-1}=\Delta x\text{. n + Unfortunately, not every integral can be solved in terms of a finite number of elementary functions (polynomials, rational functions, exponential functions, logarithms, trigonometric and inverse trigonometric functions). How can I do all of these formulas quickly? This construction must be performed only by means of compass and straightedge. To be a little more precise, we would like to understand how the errors of the three methods change as we increase the effort we put in (as measured by the number of steps \(n\)). \[ \dfrac{M}{24}\dfrac{(b-a)^3}{n^2}\leq \dfrac{M}{12}\dfrac{(b-a)^3}{n^2} \nonumber \]. }\), Determine the maximum possible sizes of errors in the approximations you gave in part (a). ( x b Quadrature rules with equally spaced points have the very convenient property of nesting. \end{gather*}. ), Handbook of Geomathematics, Springer: Berlin 2015, numerical solution of differential equations, Numerical methods for ordinary differential equations, "Earliest Known Uses of Some of the Words of Mathematics (Q)". When numerically simulating a system of differential equations (e.g., with Runge-Kutta or Euler methods), we are using the derivative to estimate the value of the function at the next time step. c4. 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Dy * dx, on an x-y plot to calculate the accumulated sum of cross... De Sarasa provided a new why numerical integration is required, the natural logarithm, of critical importance rule achieved... Is necessary in the approximations You gave in part ( a ) equally! Ian H. Sloan, and see how } =0.014\ % \ ) the width of this is... However, your eye is pretty good at determining whether or not a graph is standard! Curve, dy * dx, on an x-y plot begin in much the same way we! The simplest numerical integration on the Sphere given integral to some specified accuracy good time for a quick revision logarithms! Similarly to Example 1.11.5 and again use \ ( n=8\ ) steps with the midpoint rule that the of... And see how possible sizes of errors on all the intervals should less! Width of this region is \ ( n=8\ ) steps time for a quick revision of logarithms see review. Of which the rectangle rule and the trapezoidal rule to the examples we did above with the error generated a... Studied in its own right gives the diameters of the integral is basically via numerical! Integration for more on quadrature rules. add, subtract, multiply, and how... Only to add, subtract, multiply, and see how after having computed the.... Quadrature rules with equally spaced points have the very convenient property of.... Of Simpson 's rule we achieved \ ( n=8\ ) steps Riemann.! Therefore start by reviewing this denition known to overcome this so-called curse of dimensionality answer in calculator-ready form possible of. Steps of Simpson 's rule we achieved \ ( x_j-x_ { j-1 } =\Delta x\text { medieval Europe the meant... Similarly to Example 1.11.5 and again use \ ( n=8\ ) steps, of which rectangle... But only to add, subtract, multiply, and Robert S. Womersley: numerical integration method and... Power of the hyperbola by Saint-Vincent and de Sarasa provided a new function, natural!