April 15 Math 2335 sec 001 Spring 2014

Transcription

1 April 15 Math 2335 sec 001 Spring 2014 Trapezoid and Simpson s Rules I(f ) = b a f (x) dx Trapezoid Rule: If [a, b] is divided into n equally spaced subintervals of length h = (b a)/n, then I(f ) T n (f ) where T n (f ) = h 2 [f (x 0) + 2f (x 1 ) + 2f (x 2 ) + + 2f (x n 1 ) + f (x n )] () April 15, / 64

2 Trapezoid and Simpson s Rule I(f ) = b a f (x) dx Simpson s Rule: If [a, b] is divided into an even number n equally spaced subintervals of length h = (b a)/n, then I(f ) S n (f ) where S n (f ) = h 3 [f (x 0) + 4f (x 1 ) + 2f (x 2 ) + 4f (x 3 ) + 2f (x 4 ) f (x n 2 ) + 4f (x n 1 ) + f (x n )]. () April 15, / 64

3 Section 5.2: Error in T n and S n If the proper integral I(f ) = b a f (x) dx is approximated using T n or S n with n equally spaced subintervals, then the error formula (for f twice continuously differentiable) is En T (f ) = I(f ) T n (f ) = h2 12 (b a)f (b a)3 (c) = 12n 2 f (c) En S (f ) = I(f ) S n (f ) = h4 180 (b a)f (4) (b a)5 (c) = 180n 4 f (4) (c) for some c between a and b. () April 15, / 64

4 Example The length L of the curve defined by y = g(x) for a x b is given by b ( ) dy 2 L = 1 + dx. dx a In general, such integrals can not be evaluated in terms of elementary functions. () April 15, / 64

5 Example: Catenary A heavy, flexible cable fixed at two ends subject to a uniform force form a catenary. Figure: Telephone cables are examples of a catenary. () April 15, / 64

6 Example: Catenary Figure: The equation for a catenary is a hyperbolic cosine: cosh(t) = et +e t 2. () April 15, / 64

7 Example Write an integral for the length of the catenary with equation y = 3 + cosh(x), 1 x It is useful to note that d dx cosh(t) = et e t 2 = sinh(t). () April 15, / 64

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9 Example Estimate the number of subintervals n needed to approximate your integral using the Trapezoid rule T n and then the Simpson s rule S n to obtain the length with an error It is useful to know that for f (x) = and f (4) (x) for 1 x sinh 2 (x), both f (x) () April 15, / 64

10 () April 15, / 64

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12 Section 5.3: Gaussian Quadrature Consider the integral I(f ) = 1 1 f (x) dx 3. Both the Trapezoid and Simpson s rule start with approximating f by a polynomial to obtain the rule. Both rules have a certain form. T n (f ) = h 2 [f (x 0) + 2f (x 1 ) + + 2f (x n 1 ) + f (x n )] = w 0 f (x 0 ) + w 1 f (x 1 ) + + w n 1 f (x n 1 ) + w n f (x n ) = numbers function values. 3 We can consider more general limits [a, b] later. () April 15, / 64

13 Gaussian Quadrature Here we are going to approximate the integral I(f ) by the new rule called Gaussian Quadrature. The integration formula will be given by I n (f ) = n w j f (x j ) where the numbers {w 1,..., w n } are called the weights and {x 1,..., x n } are called the nodes. j=1 Main Idea: The weights and nodes are chosen so that I n (p) = I(p) exactly, for p(x) any polynomial of degree as high as possible. () April 15, / 64

14 Gaussian Quadrature: n = 1 Case When n = 1, the formula becomes 1 I 1 (f ) = w j f (x j ) = w 1 f (x 1 ). j=1 There is one weight w 1 and one node x 1. We need to determine two things, w 1 and x 1, so we can impose two conditions. We ll insist that the formula is exact for all polynomials of degree 1 i.e. all polynomials of the form p(x) = p 0 + p 1 x. () April 15, / 64

15 Building Block of Polynomials We note that a polynomial p(x) = p 0 + p 1 x + p 2 x p n x n is a linear combination (sum of constant multiples of) the basic building blocks 1,, x, x 2,, x n Because integrating is a linear transformation i.e. (αf (x) + βg(x)) dx = α f (x) dx + β g(x) dx we will use these building blocks to obtain our weights and nodes. 4 4 An alternative approach using the theory of orthogonal polynomials can be used (see pg. 223 of our text). () April 15, / 64

16 Gaussian Quadrature: n = 1 Case 1 1 f (x) dx I n (f ) = w 1 f (x 1 ) The formula should be exact for p(x) = 1. () April 15, / 64

17 Gaussian Quadrature: n = 1 Case 1 1 f (x) dx I n (f ) = w 1 f (x 1 ) The formula should be exact for p(x) = x. () April 15, / 64

18 Gaussian Quadrature: 1 1 f (x) dx I 1(f ) = 2f (0) Use I 1 (f ) to approximate 1 value π 2. 1 dx 1+x 2. Compare the result to the true () April 15, / 64

19 Gaussian Quadrature: n = 2 Case When n = 2, the formula becomes 2 I 1 (f ) = w j f (x j ) = w 1 f (x 1 ) + w 2 f (x 2 ). j=1 There are two weights {w 1, w 2 } and two nodes {x 1, x 2 }. We have four things to determine, so we can impose four conditions. () April 15, / 64

20 Gaussian Quadrature: n = 2 Case We ll insist that the formula is exact for p(x) = 1, x, x 2, and x dx = 2 = w 1 + w 2 x dx = 0 = w 1 x 1 + w 2 x 2 x 2 dx = 2 3 = w 1x w 2x 2 2 x 3 dx = 0 = w 1 x w 2x 3 2 () April 15, / 64

21 Gaussian Quadrature: n = 2 Case We have four equations in four unknowns w 1 + w 2 = 2 (1) w 1 x 1 + w 2 x 2 = 0 (2) w 1 x w 2x 2 2 = 2 3 (3) w 1 x w 2x 3 2 = 0 (4) () April 15, / 64

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24 ( ) Gaussian Quadrature: I 2 (f ) = f f The weights and nodes for I 2 are (always!) w 1 = w 2 = 1, and x 1 = 1 3, x 2 = 1 3. ( ) 1 3 Use I 2 (f ) to approximate 1 value π 2. 1 dx 1+x 2. Compare the result to the true () April 15, / 64

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26 Gaussian Quadrature for n > 2 We would insist that the formula I n (p) is exact for p(x) = 1, x, x 2,..., x 2n 1. 5 We ll get a nonlinear system of 2n equations w 1 + w w n = 2 w 1 x 1 + w 2 x w n x n = 0 w 1 x w 2x w nx 2 n = 2 3 w 1 x 2n w 2 x 2n w n x 2 n 2n 2 = w 1 x 2n w 2 x 2n w n x 2n 1 n = n 1 5 This is 2n conditions for the 2n unknowns {w 1,..., w n} and {x 1,..., x n}. () April 15, / 64

27 Gaussian Quadrature for n > 2 Fortunately, solutions for various n values are known. Table 5.7 (pg. 223) in Atkinson and Han shows the weights and nodes for n = 2, 3,..., 8. For example, the weights and nodes for I 3 (f ) are w 1 = w 2 = 5 9, and w 3 = x 1 = 5, x 2 = 3 5, and x 3 = 0. () April 15, / 64

28 Example for n = 3 Use I 3 (f ) to approximate 1 value π 2. 1 dx 1+x 2. Compare the result to the true () April 15, / 64

29 () April 15, / 64

30 Other Intervals Suppose we wish to evaluate b a f (x) dx. We d like a change of variables x t so that 1 t 1 when a x b. We already know that t = 2 b a (x a) 1, i.e. x = (t + 1) + a b a 2 does the trick. This gives dx = b a 2 dt so that where b a f (x) dx = b a g(t) dt ( ) b a g(t) = f (t + 1) + a. 2 () April 15, / 64

31 Example Use I 2 (f ) to approximate 1 0 x dx () April 15, / 64

32 () April 15, / 64

33 Comparison of Methods I 2 (f ) = f 1 0 ( 1 3 ) + f x dx = 2 3 = ( ) 1 3 = , f (t) = 1 2 ] 1 3 [ T 4 (f ) = [ S 4 (f ) = The errors are ] t = = I I 2 = , I T 4 = , and I S 4 = () April 15, / 64