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1、where p(AT A) =max |X?i i(3) Let f (x ) = x 2 6 , the sequenceroot of equation f (x ) = 0isp = g (p )=nn - 1p3n -1 +2pn 一 1n-1/(p )n1f(p )n-1n-1p2 一 6n12pn - 1pn-1Numerical AnalysisAnswers to Test A (2008)1. Fill in the following blanks.(1) Let x *= 3.14159, then the number x = 3.141 approximate x*
2、with 3significant digits.(Solution. x 二 3.144 0.3 14m im 二 1,x * - x = 0.059 x 10 -2 n m - l = - 2 n l = 3.)(2) Suppose2 1A =-1 2 -Then | A=3.(p ( a ) = max 入 and 九 is an eigenvalue of a for each i. iii|a| = p (ATA),2and 兀 is an eigenvalue of A Ta for each i.) igenerated by Newton s method for findi
3、ng theis defined by(4) A natural cubic spline S on 0,2S (x) = 1 + 2 x - x3,if0 x 1,S (x) = 0S (x) = 2 + b (x - 1) + c (x - 1)2 + (x - 1)3, if1 x 0, x 0, show the sequence0(x2 + 3 ak+1converges toJa of order 3.Proof Letfor any x g (0, +8 )0x( x2 + 3a)g (x)=3x 2 + aIt is easy to prove that g(x) 0, g(x
4、)gC(0,+8), and3( x2 一 a 户 o , suppose thatlim xkk 一g=lim x 0 ,k +1k -gthen for x(x2 + 3 akk3 x2 + a k)一,one haswhich gives a = Ja .Andlimk fgXk +1一打I一叼1limkfg 3x2 + ak14aso the sequence x g converges to va of order 3.k k = 03. Use the following data to construct an interpolating polynomial P ( x) of
5、 degreethree so that p(x ) = f (x ) for i = 0,1,2 and p(x ) = f(x )x0f (x.)1广(x)01229Solution. First compute all the divided differences, and list the divided differences table as followsx=0f(x ) =100x=0f(x ) =1fx ,x = 00000x=1f(x)=2fx ,x =1fx ,x ,x =11101001x2=2f(x ) =92fx ,x = 712f x , x , x = 3 f
6、 x , x , x , x = 10 1 2 0 0 1 2Applying Newtons interpolation formula gives the interpolating polynomial p ( x)3as followsP (x) = f x + f x , x (x 一 x ) + f x , x , x (x 一 x )2 +300000010f x , x , x , x ( x 一 x )2 ( x 一1 2x)1=1 + x2 + x 2( x 一 1)=X3 + 1.4. The forward-difference formula can be expre
7、ssed asf (x ) = -f (x + h) 一 f (x ) - hf(x )h2一竺广(x ) + O (h 3),6Use Richardsons extrapolation to derive an O(h3) formula for广(x )0Solution. We have the O (h )approximationwheref，(x )0=M (h) = N (h) + k h + k h2 + o (h3)12N (h) = - f (x + h) 一 f (x ) , k =-丄 f( x ),h12Replacing h by in the formula g
8、ives that2(1)h11M (h) = N()+ k h + k h2 + o(h3),22 1 4 2(2)Subtracting Eq. (1) from2 time Eq. (2) gives1f，(x ) = N (h) 一 k h2 + o(h3),0 1 2 2(3)whereN (h) =1h2 N (_) 一 N (h),2Continuing this procedure gives1f，(x ) = N ( ) 一 k h2 + o(h3),0 1 2 8 2(4)Subtracting Eq. (3)from 4 time Eq.(4) and dividing
9、by 3 provides anO(h3)formula for fr( x )0f(x )=0N (h)+ o(h3)2Whereh4 N ( ) N (h)N (h)= 匸 234hh1h=2 N () N () (2 N () N (h)342328hh1=N (一) 2 N (一)+N (h)34231hh=f (x + h) 12 f (x +) + 32 f (x +) 21 f (x ).3h 0020405. Show the quadrature formulaJ1 f (x )dx a(皿) + 8f (0)+ 5f (航刀(1) Proof.Sincef(x)=1,J 1
10、 f ( x)dx = 2, 195 f (賦)+8 f (0)+5 f (丽=2,f (x) = x2 ,f ( x) = x3 ,f ( x) = x4 ,f ( x) = x5 ,J 1 f (x)dx =0,1J1f (x) dx=3J 1 f (x)dx =0,1J11f (x) dx=5J 1 f ( x)dx =0, 1丄5 f (皈)+ 8 f (0) + 5 f ( P0.6 ) = 0,9丄5 f &0.6) + 8 f (0) + 5 f (0.6 )=- 93+32丄5 f+ 8f (0) + 5f (0.6 ) = 0,910.36-5 f (W6) + 8 f (0
11、) + 5 f (70.6 )=- 9护f 帧+8 f (0)+5 f (-而=0,1gives the exact result whenever f (x) is a polynomial of degree 5 or less, and usesin x0 1+ xthe above formula to approximate the integral J1 dx -so for f (x) =1, x, x2,x3,x4,x5, it follows1f (x) dx=95 f(纭)+8 f (0)+5 f (八莎.J1 f(x)dx1But for f ( x) = x6 ,-5 f (、：0.6 ) + 8 f (0) + 5 f ( ：0.6 ) = 0.24 ,9therefore the formula gives the exact result whenever f ( x) is a polynomial of degree 5 or less.(2) Solution Let x = t1,
