Lesson 2 Quadrature rules 1 Quadrature is the computation of - PowerPoint PPT Presentation

Lesson 2 Quadrature rules 1


 
 • Quadrature is the computation of integrals Z b f ( x ) d x a • Only in very special cases can integrals be computed exactly, otherwise, we must compute them numerically (i.e., approximately) • The goal of standard quadrature methods is to express the integral as a finite sum n � w n,k f ( x n,k ) k =1 so that the approximation converges: � b � � n � � � lim w n,k f ( x n,k ) − f ( x ) d x � = 0 � � � � n →∞ a � i =1 2


 SIMPLEST METHOD: RIGHTPOINT RULE • The standard definition for an integral from Z 1 introductory calculus is cos x d x 0 � b n = 3 n � � b − a a + k b − a � 1.0 f ( x ) d x = lim f n n - n →∞ a k =1 0.8 f ( h ) 0.6 0.4 0.2 0.0 0.2 0.4 0.6 0.8 1.0 h = b − a n


 
 SIMPLEST METHOD: RIGHTPOINT RULE • The standard definition for an integral from Z 1 introductory calculus is cos x d x 0 � b n = 3 n � � b − a a + k b − a � 1.0 f ( x ) d x = lim f n n n →∞ a k =1 0.8 • Idea: use the sum with finite n 
 f ( h ) 0.6 � b n x k = a + k b − a 0.4 � f ( x ) d x ≈ Q R n = h f ( x k ) for n a k =1 0.2 • This is equivalent to approximating the integral 0.0 0.2 0.4 0.6 0.8 1.0 by small rectangles h = b − a n


 LEFTPOINT RULE Z 1 cos x d x 0 • Alternatively, we can evaluate f on n = 3 1.0 the left: 
 0.8 � b n f (0) � f ( x ) d x ≈ Q L 0.6 n = h f ( x k ) a k =1 0.4 x k = a + ( k − 1) b − a for 0.2 n 0.0 0.2 0.4 0.6 0.8 1.0 h = b − a n

TRAPEZOIDAL RULE • Instead of rectangles, we can use trapezoids Z 1 cos x d x • We then have 
 0 n = 3 4 1.0 � b � x k +1 n − 1 � f ( x k ) + ( x − x k ) f ( x k +1 ) − f ( x k ) � � 0.8 f ( x ) d x ≈ d x x k +1 − x k a x k k =1 0.6 � � 0.4 0.2 0.0 0.2 0.4 0.6 0.8 1.0 for h = b − a and x k = a + ( k − 1) h n − 1 6

TRAPEZOIDAL RULE • Instead of rectangles, we can use trapezoids Z 1 cos x d x • We then have 
 0 n = 3 4 1.0 � b � x k +1 n − 1 � f ( x k ) + ( x − x k ) f ( x k +1 ) − f ( x k ) � � 0.8 f ( x ) d x ≈ d x x k +1 − x k a x k k =1 0.6 n − 1 � � f ( x k ) h + hf ( x k +1 ) − f ( x k ) � = 2 0.4 k =1 n − 1 0.2 � 0.0 0.2 0.4 0.6 0.8 1.0 for h = b − a and x k = a + ( k − 1) h n − 1 7

TRAPEZOIDAL RULE • Instead of rectangles, we can use trapezoids Z 1 cos x d x • We then have 
 0 n = 3 4 1.0 � b � x k +1 n − 1 � f ( x k ) + ( x − x k ) f ( x k +1 ) − f ( x k ) � � 0.8 f ( x ) d x ≈ d x x k +1 − x k a x k k =1 0.6 n − 1 � � f ( x k ) h + hf ( x k +1 ) − f ( x k ) � = 2 0.4 k =1 n − 1 = h 0.2 � [ f ( x k ) + f ( x k +1 )] 2 k =1 0.0 0.2 0.4 0.6 0.8 1.0 for h = b − a and x k = a + ( k − 1) h n − 1 8

TRAPEZOIDAL RULE • Instead of rectangles, we can use trapezoids Z 1 cos x d x • We then have 
 0 n = 3 4 1.0 � b � n − 1 � f ( a ) f ( x k ) + f ( b ) � f ( x ) d x ≈ Q T n = h + 0.8 2 2 a k =2 0.6 0.4 0.2 0.0 0.2 0.4 0.6 0.8 1.0 for h = b − a and x k = a + ( k − 1) h n − 1 9

Experiments of quadrature error 10

• We’ve introduced three quadrature rules – Left point rule, right point rule and trapezoidal rule • Each use precisely n function evaluations • How fast will they converge to the true integral as n increases? • Let’s try some experiments 11

� 1 Error approximating � x � x = � − 1 0 0.4 0.3 0.2 0.1 1000 n 200 400 600 800 Right-point rule, Left-point rule, Trapezoidal rule 12

� 1 Error approximating � x � x = � − 1 0 Log-scale plot 0.4 0.01 0.3 10 - 4 0.2 10 - 6 0.1 1000 n 5000 n 200 400 600 800 1000 2000 3000 4000 Right-point rule, Left-point rule, Trapezoidal rule 13

� 1 Error approximating � x � x = � − 1 0 Log-log-scale plot 0.4 0.01 0.3 10 - 4 0.2 10 - 6 0.1 1000 n 5000 n 200 400 600 800 100 200 500 1000 2000 Right-point rule, Left-point rule, Trapezoidal rule 14

� 1 Error approximating � x � x = � − 1 0 1 Log-log-scale plot n 0.4 0.01 0.3 1 10 - 4 n 2 0.2 10 - 6 0.1 1000 n n 200 400 600 800 100 200 500 1000 2000 5000 Right-point rule, Left-point rule, Trapezoidal rule 15

� x 0.01 10 - 4 10 - 6 n 100 200 500 1000 2000 5000 16

��� 500 x � x 0.1 0.01 0.001 10 - 4 10 - 5 10 - 6 10 - 7 n n 100 200 500 1000 2000 5000 100 200 500 1000 2000 5000 16

��� 500 x � x 0.1 0.01 0.001 10 - 4 10 - 5 10 - 6 10 - 7 n n 100 200 500 1000 2000 5000 100 200 500 1000 2000 5000 | x − . 1 | 0.1 0.001 10 - 5 10 - 7 n 100 200 500 1000 2000 5000 16

��� 500 x � x 0.1 0.01 0.001 10 - 4 10 - 5 10 - 6 10 - 7 n n 100 200 500 1000 2000 5000 100 200 500 1000 2000 5000 | x − . 1 | ��� ( x − . 1) 0.1 0.1 0.001 0.001 10 - 5 10 - 5 10 - 7 10 - 7 n n 100 200 500 1000 2000 5000 100 200 500 1000 2000 5000 16

� x 0.01 10 - 4 10 - 6 n 100 200 500 1000 2000 5000 17

� x ��� 10 π x 1 0.01 0.001 10 - 6 10 - 4 10 - 9 10 - 12 10 - 6 10 - 15 n 100 n 100 200 500 1000 2000 5000 5 10 20 50 17

� x ��� 10 π x 1 0.01 0.001 10 - 6 10 - 4 10 - 9 10 - 12 10 - 6 10 - 15 n 100 n 100 200 500 1000 2000 5000 5 10 20 50 ��� 50 π x 1 0.001 10 - 6 10 - 9 10 - 12 10 - 15 n 5 10 20 50 100 17

� x ��� 10 π x 1 0.01 0.001 10 - 6 10 - 4 10 - 9 10 - 12 10 - 6 10 - 15 n 100 n 100 200 500 1000 2000 5000 5 10 20 50 ��� 50 π x ��� ��� 2 π x 1 0.1 0.001 10 - 4 10 - 6 10 - 7 10 - 9 10 - 10 10 - 12 10 - 13 10 - 15 n n 5 10 20 50 100 5 10 20 50 100 17

Warning! • Only use these methods for periodic functions!! • Much better methods exist for non-periodic smooth functions • We will see this later in the course • The better method is not Simpson’s rule 18

Convergence of quadrature rules 19

• We want to prove the observed behaviour � Left and right point rules converge like O (1/ n ) � Trapezoidal rule converges like O � 1/ n 2 � � All rules converge a lot faster for periodic and smooth functions • The key is integration by parts 20

� � � � Right-hand rule • Assume for simplicity that f is smooth (infinitely differentiable in [ a, b ] ) • We start with the error in one panel: � x k +1 � x k +1 ( x − x k ) � [ f ( x ) − f ( x k +1 )] � x [ f ( x ) − f ( x k +1 )] � x = x k x k � 21

� � � Right-hand rule • Assume for simplicity that f is smooth (infinitely differentiable in [ a, b ] ) • We start with the error in one panel: � x k +1 � x k +1 ( x − x k ) � [ f ( x ) − f ( x k +1 )] � x [ f ( x ) − f ( x k +1 )] � x = x k x k � x k +1 = [( x − x k )( f ( x ) − f ( x k +1 ))] x k +1 ( x − x k ) f � ( x ) � x − x k x k � 22

� � Right-hand rule • Assume for simplicity that f is smooth (infinitely differentiable in [ a, b ] ) • We start with the error in one panel: � x k +1 � x k +1 ( x − x k ) � [ f ( x ) − f ( x k +1 )] � x [ f ( x ) − f ( x k +1 )] � x = x k x k � x k +1 = [( x − x k )( f ( x ) − f ( x k +1 ))] x k +1 ( x − x k ) f � ( x ) � x − x k x k � x k +1 = − 1 ( x − x k ) 2 � � f � ( x ) � x � 2 x k � 23

� Right-hand rule • Assume for simplicity that f is smooth (infinitely differentiable in [ a, b ] ) • We start with the error in one panel: � x k +1 � x k +1 ( x − x k ) � [ f ( x ) − f ( x k +1 )] � x [ f ( x ) − f ( x k +1 )] � x = x k x k � x k +1 = [( x − x k )( f ( x ) − f ( x k +1 ))] x k +1 ( x − x k ) f � ( x ) � x − x k x k � x k +1 = − 1 ( x − x k ) 2 � � f � ( x ) � x � 2 x k � x k +1 = − 1 + 1 � x k +1 ( x − x k ) 2 f � ( x ) ( x − x k ) 2 f �� ( x ) � x � 2 x k 2 x k � x 2 24

Right-hand rule • Assume for simplicity that f is smooth (infinitely differentiable in [ a, b ] ) • We start with the error in one panel: � x k +1 � x k +1 ( x − x k ) � [ f ( x ) − f ( x k +1 )] � x [ f ( x ) − f ( x k +1 )] � x = x k x k � x k +1 = [( x − x k )( f ( x ) − f ( x k +1 ))] x k +1 ( x − x k ) f � ( x ) � x − x k x k � x k +1 = − 1 ( x − x k ) 2 � � f � ( x ) � x � 2 x k � x k +1 = − 1 + 1 � x k +1 ( x − x k ) 2 f � ( x ) ( x − x k ) 2 f �� ( x ) � x � 2 x k 2 x k � x k +1 = − h 2 2 f � ( x k +1 ) + 1 ( x − x k ) 2 f �� ( x ) � x 2 x k 25

� � ��� � ��� • We can bound the first term � 1 � 2 h 2 � ≤ h 2 � � � b − a � � � 2 f � ( x k +1 ) x | f � ( x ) | = x | f � ( x ) | = O � � 2 ��� n 2 n � • we can also bound the second term • Summing over all panels we get 26