substituting f(x) = 1 :[1 +1]['idx=b-asubstituting f(x) = x :Cbxdx = b-α?b[a+b]2asubstituting f(x) = x2 :xdx=b2"[a? +b”]主3上页so . algebraic precision =1下页返圆
上页 下页 返回 substituting f(x) = 1： = − b a 1 dx b a = [1 1] 2 + b−a substituting f(x) = x ： 2 2 2 b a b a x dx − = = [ ] 2 a b b a + − substituting f(x) = x 2 ： 3 2 3 3 b a b a x dx − = [ ] 2 2 2 a b b a + − so， algebraic precision = 1
Example 2 we construct numerical integrationformula like f。" f(x)dx ~ Aof(O)+Af(h)+ A,f(2h) , tryingto improve its algebraic precision as high as possibleand find out the corresponding order.Solution: let formula can be correctly established onf(x) =1,x,x2[3h=A+A+A29SOh2=0+Ah+A2h,29h3=0+Ah2+4h2A.上页49wecanwork outA =0,下页44返圆
上页 下页 返回 Example 2 we construct numerical integration formula like , trying to improve its algebraic precision as high as possible, and find out the corresponding order. 3 0 1 2 0 ( ) (0) ( ) (2 ) A f x dx A f A f h A f h + + Solution: let formula can be correctly established on 2 f x x x ( ) 1, , = so 0 1 2 2 1 2 3 2 2 1 2 3 , 9 0 2 , 2 9 0 4 . h A A A h A h A h h A h h A = + + = + + = + + we can work out 0 3 , 4 A h = 1 A = 0, 2 9 . 4 A h =
So the form of integration formula is°" (x)dx ~ 3 r(0) + h (2h),According to the process of construction, this formula has atf(x)=x3least 2 algebraic precision, when481 h4 the right sidethe left side= 18h4left side ≠ right side, so it proves that the formula f(x)= xcan not be established correctly. So the formula can onlyhave 2 algebraic precision.上页下页返圆
上页 下页 返回 So the form of integration formula is 3 0 3 9 ( ) (0) (2 ). 4 4 h h h f x dx f f h + According to the process of construction, this formula has at least 2 algebraic precision, when the left side the right side left side right side, so it proves that the formula can not be established correctly. So the formula can only have 2 algebraic precision. 3 f x x ( ) = 81 4 4 = h 4 =18h 3 f x x ( ) =
Example3.Try to improve the algebraic precision of parameters as high aspossible in below integration formula.hI = I" f(x)dx ~=[f(0)+ f(h)+ ah'[F'(O)- f'(h)= I,2solution:1I,=h1dx = hwhen f(x) = 1 I ==Joh?h2when f(x) = x I =x'dx12220上页下页返圆
上页 下页 返回 Example 3. Try to improve the algebraic precision of parameters as high as possible in below integration formula. 2 2 0 [ (0) ( )] [ (0) ( )] 2 ( ) f f h ah f f h I h I f x dx h = + + − = = h I dx 0 1 solution: 2 2 2 h I = = h I2 = h = h I x dx 0 1 2 2 h = = =
when f(x) = x2h3h3x"dx+ ah'[0 -2h]= (-2a)h1321let I = 12a12上页下页返圆
上页 下页 返回 = h I x dx 0 2 3 3 h = [0 2 ] 2 2 3 2 ah h h I = + − 3 2 ) 2 1 = ( − a h 12 1 a =