Lecture Note 4: Numerical differentiation and integration. Xiaoqun Zhang Shanghai Jiao Tong University

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1 Lecture Note 4: Numericl differentition nd integrtion Xioqun Zng Sngi Jio Tong University Lst updted: November, 0

2 Numericl Anlysis. Numericl differentition.. Introduction Find n pproximtion of f (x 0 ), f (x 0 ),..., in terms of only f(x). Since we use Used in Secnt metod. > 0, forwrd difference formul < 0, bckwrd difference formul Use grp to illustrte tis. Not successful due to round-off error. Quntittive study of te error: f f(x 0 + ) f(x 0 ) (x 0 ) = lim, 0 f (x 0 ) f(x 0 + ) f(x 0 ) Assume tt x 0 (, b). Suppose tt f C [, b]. x = x 0 + [, b]. Use Tylor s expnsion. So f(x 0 + ) = f(x 0 ) + f (x 0 ) + f (ξ). f (x 0 ) = f(x 0 + ) f(x 0 ) Te error is of O() since f (ξ) is bounded. Error: 0., is O(0.) Error: 0.00, is O(0.00). f (ξ). : To get n ccurte f (x 0 ), must be very smll. However, smll denomintor exggerte te round-off error of f(x 0 + ) f(x 0 ). (Recll loss of significnt digits.) Error is of O( α ). α is te order of te formul. Error 0.0, order : = O(0.0), order : = O(0.). Error 0.000, order : = O(0.000), order : = O(0.0). To minimize te influence of te round-off error, iger order formuls re desired. Higer order formul. Ide: Involving more points tn only x 0, x 0 +.

3 Numericl Anlysis Metod: using polynomil P (x) interpoltion to pproximte f(x), nd use P (x 0 ) to pproximte f (x 0 ).. Construct polynomil P n (x) tt psses troug ll te given points. (Lgrnge or Newton s divided difference). P n (x) f(x), so P n(x) f (x).. Use P n(x 0 ) s te formul to pproximte f (x). Use grp to illustrte. Polynomil interpoltion (Newton s divided difference formul) n i f(x) = f[x 0,..., x i ] (x x j ) + f (n+) n (x x j ). (n + )! j=0 Two-point formul. (f (x 0 ) f(x0+) f(x0) ) Interpolte f(x) using two points x 0, x. j=0 f(x) = f[x 0 ] + f[x 0, x ](x x 0 ) + f (ξ(x)) (x x 0 )(x x ). Tke derivtive on bot nd side. f (x) =f[x 0, x ] ( + df (ξ(x)) dx (x x 0 )(x x ) + f (ξ(x)) Replce x by x 0 nd x by x 0 +, we obtin f (x 0 ) = f[x 0, x 0 + ] f (ξ(x)). Te sme s obtined from Tylor s expnsion. Higer order formul: tree-point formul. Interpolte f(x) using tree points x 0, x nd x. f(x) =f[x 0 ] + f[x 0, x ](x x 0 ) + f[x 0, x, x ](x x 0 )(x x ) + f (ξ(x)) (x x 0 )(x x )(x x ). Tke derivtive on bot nd sides, (x x 0 ) + f ) (ξ(x)) (x x ) f (x) = f[x 0, x ] + f[x 0, x, x ](x x 0 ) + f[x 0, x, x ](x x ) ( df (ξ(x)) + (x x 0 )(x x )(x x ) dx + f (ξ(x)) ) ((x x 0 )(x x ) + (x x )(x x ) + (x x 0 )(x x ))

4 Numericl Anlysis Let x = x 0, x = x 0, nd x = x 0 +. We ve f (x 0 ) = f[x 0, x ] + f[x 0, x, x ] f (ξ(x)) f[x 0, x ] + f[x 0, x, x ] = f(x 0 ) f(x 0 ) Centrl difference formul Error term = f(x 0 + ) f(x 0 ) f(x 0 + ) f(x 0 ). f (ξ(x)), O( ). Let x = x 0, x = x +, x = x +. We ve f (x 0 ) = f[x 0, x ] f[x 0, x, x ] + f (ξ(x)) f[x 0, x ] + f[x 0, x, x ] = f(x 0 + ) f(x 0 ) Anoter tree-point formul Error term = f(x 0) + 4f(x 0 + ) f(x 0 + ) f(x 0 ) + 4f(x 0 + ) f(x 0 + ). f (ξ(x)), O( ). f(x 0+) f(x 0 ) f(x0 ) f(x0) + f(x 0+) f(x 0+) f(x0+) f(x0) Te error of te centrl difference formul is lf of te oter. Te reson: x nd x re closer to x 0 in its derivtion. Similrly, we ve Five-point formul. f (x 0 ) = (f(x 0 ) 8f(x 0 )+8f(x+0+) f(x 0 +))+ 4 0 f (5) (ξ) Cn construct formule involving more points. 4

5 Numericl Anlysis Interpoltion error: Assume > 0. Summry of formuls f(x) P n (x) = f (n+) (ξ) (n + )! (x x )(x x )... (x x n ). Two-points: x 0 nd x 0 + Forwrd difference formul. Two-points: x 0 nd x 0 Bckwrd difference formul. Tree points: x 0, x 0, x 0 + Centrl difference formul. Tree points: x 0, x 0 +, x 0 + f(x 0 + ) f(x 0 ), error = O(). f(x 0 ) f(x 0 ), error = O(). f(x 0 + ) f(x 0 ), error = O( ). f(x 0 ) + 4f(x 0 + ) f(x 0 + ), error = O( ). Five points: x 0, x 0, x 0, x 0 +, x 0 + f(x 0 ) 8f(x 0 ) + 8f(x 0 + ) f(x 0 + ), error = O( 4 )... Higer order derivtive Expnd f t x 0 nd evlute t x 0, x 0 + : f(x 0 + ) = f(x 0 ) + f (x 0 ) + f (x 0 ) + f (x 0 ) + 4 f (4) (ξ ) 4 (.) f(x 0 ) = f(x 0 ) f (x 0 ) + f (x 0 ) f (x 0 ) + 4 f (4) (ξ ) 4 were x 0 < ξ < x 0 < ξ < x 0 +. Add te two equtions, we ve f(x 0 + ) + f(x 0 ) = f(x 0 ) + f (x 0 ) + 4 [f (4) (ξ ) + f (4) (ξ )] 4 5 (.)

6 Numericl Anlysis We obtin f (x 0 ) = [f(x 0 ) f(x 0 ) + f(x 0 + )] f (4) (ξ) for some x 0 < ξ < x Ricrdson s extrpoltion Use less ccurte formuls to generte ig ccurcy formul. Forwrd difference formul: f(x 0 + ) f(x 0 ) Assume tt f is smoot enoug. Expnd te terms involved in te formul by Tylor s expnsion. f(x 0 + ) = f(x 0 ) + f (x 0 ) + f (x 0 ) So te error of tis formul is f (x 0 ) = f(x 0 + ) f(x 0 ) Error is O(). + f (x 0 ) + f (x 0 ) + f (x 0 ) f (n)(x0) n +... n! f (n)(x0) n +... n! Ricrdson s extrpoltion: Using oter s to eliminte te term of O(). Replce by : f (x 0 ) = f(x 0) f(x 0 ) f (x 0 ) Summing togeter nd dividing by : f (x 0 ) = f(x 0 + ) f(x 0 ) Centrl difference formul. Error O( )! + f (x 0 ) + f (x 0 )...+ f (n)(x0) ( ) n +... n! + f (5) (x 0 ) To get more ccurcy formul: Replce by in te bove formul f (x 0 ) = f(x 0 + ) f(x 0 ) 4 One 4 T wo f (x 0 ) = (f(x 0 + ) f(x 0 )) Error O( 4 ) + 4f (x 0 ) + f (5) (x 0 ) f(x 0 + ) f(x 0 ) 4 f (x 0 ) = f(x 0 ) 8f(x 0 ) + 8f(x 0 + ) f(x 0 + ) + f (5) (x 0 ) f (5) (x 0 )

7 Numericl Anlysis. Numericl integrtion.. Elements of Numericl integrtion Te need often rises for evluting te definite integrl of function tt s no explicit ntiderivtive or wose ntiderivtive is not esy to obtin. We wnt to compute f(x)dx using only f(x), i.e. Clled numericl qudrture. f(x) n i f(x i ) Bsic ide: use piecewise polynomil to pproximte f(x). For ec subintervl, select set of distinct nodes x0,, x n from te intervl [, b]. Ten integrte te interpolting polynomil P n (x) = n f(x i)l i (x) nd its trunction error term over [, b] to obtin f(x) = = i P n (x) + (n + )! i f(x i ) + (n + )!,b,b Π n (x x i )f (n+) (ξ(x))dx Π n (x x i )f (n+) (ξ(x))dx were i (x) = L i(x) for ec i = 0,, n nd ξ(x) [, b]. Numericl qudrture: nd error Trpezoidl rule E n (f) = (n + )! f(x) n i f(x i ) Π n (x x i )f (n+) (ξ(x))dx Use liner polynomil nd pproximte f(x) by trpezoid re. Wen f is positive vlue function, te integrl is given by te trpezoid re (f(x 0 ) + f(x ))/. Used piecewise liner function to pproximte f(x) using x 0 =, x = b, = b. P n (x) = f(x 0 ) x x + f(x ) x x 0 x 0 x x x 0 7

8 Numericl Anlysis x x 0 (f(x 0 ) x x x 0 x + f(x ) x x 0 x x 0 dx = f(x 0) x 0 x (x x ) x x=x 0 + f(x ) x x 0 (x x 0) x x=x 0 = f(x 0)(x 0 x ) + f(x )(x x 0 ) Remrks: = (f(x 0) + f(x ))(x x 0 ) = (f(x 0) + f(x )). Te error term for te Trpezoidl rule involves f, so te rule gives te exct result wen pplied to ny function wose second derivtive is identiclly zero, tt is, ny polynomil of degree one or less. Simpson s rule Assume tt te polynomil on [x 0, x ] is P (x). Write P (x) = αx + βx + γ x x 0 P (x)dx = x x 0 (αx + βx + γ)dx = ( αx + βx + γx) x x=x 0 = α(x x 0) + β(x x 0) + γ(x x 0 ) = (α(x x 0 )(x + x x 0 + x 0) + β(x x 0 )(x + x 0 ) + γ(x x 0 )) = x x 0 (α(x + x x 0 + x 0) + β(x + x 0 ) + γ) = (α(x + x x 0 + x 0) + β(x + x 0 ) + γ) Let us prove tt α(x + x x 0 + x 0) + β(x + x 0 ) + γ = f(x 0 ) + 4f(x ) + f(x ) Since P (x) is te interpoltion of f t x 0, x nd x, we ve f(x 0 ) = αx 0 + βx 0 + γ ( ) ( ) f(x ) = αx x0 + x x0 + x + βx + γ = α + β + γ f(x ) = αx + βx + γ 8

9 Numericl Anlysis So f(x 0 ) + 4f(x ) + f(x ) ( ) =α(x x0 + x x ) + βα(x x 0 + x + x ) + γ =... x.. Composite formul x 0 P (x) = (f(x 0) + 4f(x ) + f(x )). We wnt to find f(x)dx using piecewise polynomil to void (Runge s penomenon). Cut [, b] into n subintervls using eqully spced nodes. x 0 =, x, x,..., x n = b. Do integrl on ec subintervl. Composite Trpezoidl formul: let = (b )/n nd x j = + j, j = 0,, n. I n (f) := (f(x 0 )/ + f(x ) + f(x ) f(x n ) + f(x n )/) := n [f() + f(x j ) + f(b)] j= Composite Simpson rule: coose n even integer n nd divide [, b] into n subintervls wit = (b )/n nd x j = + j. Group te nodes every tree points, nd we coose te unique polynomil of degree wic goes troug te points. Do Integrl on ec sub-intervl. I n (f) := (f(x 0) + 4f(x ) + f(x ) + 4f(x ) + f(x 4 ) f(x n ) + 4f(x n ) + f(x n )). Remrk := n/ n/ (f(x 0) + f(x j ) + 4 f(x j ) + f(x n )). j= j=. Wy not use one polynomil nd ten integrl? (Runge s penomenon). lim n + I n (f) = f(x)dx. How fst does it converge?. Trpezoid: order. Simpson: order 4. I n(f) f(x)dx C I n(f) f(x)dx C4 9

10 Numericl Anlysis 4. If 00 points, = b 00 = 0. Error 0 4 or Error nlysis Lemm (Weigted men vlue teorem) Suppose tt g(x), w(x) C[, b] nd w(x) doesn t cnge sign in [, b]. Ten, tere exists θ (, b) suc tt g(x)w(x)dx = g(θ) Error nlysis for Trpezoidl rule w(x)dx I n (f) = (f(x 0 )/ + f(x ) + f(x ) f(x n ) + f(x n )/) Teorem Suppose f C [, b]. Ten ξ n [, b] suc tt Proof. f(x)dx I n (f) = (b )f (ξ n ) Step Let P (x) be te unique polynomil tt goes troug x i nd x i+. Error is Ten f(x) P (x) = f (ξ(x)) (x x i )(x x i+ ), ξ(x) (x i, x i+ ). E i = = xi+ x i xi+ x i (f(x) P (x))dx f (ξ(x)) (x x i )(x x i+ )dx (x x i )(x x i+ ) is continuous nd doesn t cnge te sign in [x i, x i+ ] f (ξ(x)) is continuous. By te weigted men vlue teorem, tere exists η i [x i, x i+ ] suc tt E i = f (η i ) xi+ x i (x x i )(x x i+ )dx. We ve Finlly, xi+ x i (x x i )(x x i+ )dx = (x i+ x i ). E i = f (η i ) 0

11 Numericl Anlysis Step Te totl error is n E = E i = n Notice tt n = b. Ten, Clim: Tere exists η [, b] suc tt n proof: f (η i ) = n ( ) n f (η i ) n ( ) n (b ) f(x) I n (x) = f (η i ) n n f (η i ) = f (η). min f (x) f (η i ) mx f (x) x [,b] x [,b] n n min f (x) f (η i ) n mx f (x) x [,b] x [,b] min f (x) n f (η i ) mx x [,b] n f (x) x [,b] Since f is continuous on [, b]. it must ttin ll vlue between its minimum nd mximum t, sy c nd d. Let g(x) = f (x) n f (η i ). n Ten, g(c) 0, g(d) 0. Applying intermedite vlue teorem, we get g (η) = 0. Hence f (η) = n f (η i ). n Corollry Te error is f(x)dx I n (f) M(b ).

12 Numericl Anlysis Error nlysis for Composite Simpson s rule. I n (f) = n/ (f(x i ) + 4f(x i+ ) + f(x i+ )) Teorem Suppose f C 4 [, b]. Ten ξ n [, b] suc tt f(x)dx I n (f) = (b )f (4) (ξ n ) 4 80 Proof. Let E i be te error in te intervl [x i, x i+ ]. Ten, E i = xi+ x i (f(x) P (x))dx = xi+ x i f (ξ(x)) (x x i )(x x i+ )(x x i+ )dx, were ξ(x) [x i, x i+ ]. Let g(x) = f (ξ(x)) nd w(x) = (x x i )(x x i+ )(x x i+ ). We cnnot use te Weigted Men Vlue Teorem, becuse w(x) > 0, wen x i < x < x i+, w(x) < 0, wen x i+i < x < x i+. Clim: η i [x i, x i+ ] s.t. E i = 5 90 f (4) (η i ). Proof. Will not prove it. So, te totl error is 5 n/ 90 f (4) (η i ) = n Need to nlyze n 5 n/ f (4) (η i ) 90 n n/ f (4) (η i ). We ve m f (4) (η i ) M, m n Since f (4) is continuous, η [, b] s.t. n/ = n/ f (4) (η i ) = f (4) (η). (b )4 80 n f (4) (η i ) M. n/ f (4) (η i ). Corollry Te error is f(x)dx I n (f) M(b ) 4. 80

13 Numericl Anlysis Degree of precision Trpezoidl rule is O( ). ( (b )f (ξ) ). Simpson s rule is O( 4 ). ( (b )f (4) (ξ) 80 4 ). Simpson s rule is so muc better tn te Trpezoidl rule. Definition: A numericl integrtion formul I n (f) is sid ve degree of precision m if. I n (f) = f(x)dx for ll f(x) = xk, k m.. I n (f) f(x)dx for f(x) = xm+. I n (f + g) = I n (f) + I n (g), I n (λf) = λi n (f). Tis is equivlent to: A numericl integrtion formul I n (f) is sid ve degree of precision m if. I n (f) = f(x)dx for ll polynomil f(x) of degree m.. I n (f) f(x)dx for some polynomil f(x) of degree m +. It is expected tt Trpezoidl rule, te degree is, nd Simpson s rule, te degree is. In fct, Trp. rule s degree of precision, nd Simpson s rule s degree of precision. Let prove tt Simpson s rule is exct if f(x) is polynomil of degree. Te reson is tt te error x P (x) on [x 0, x ] cncels tt on [x, x ]. Proof. Step Let prove tt Simpson s rule is exct if f(x) = x. We only need to prove it in subintervl [x i, x i+ ], i.e., (f(x i) + 4f(x i+ ) + f(x i+ ))dx = xi+ x i x dx. In fct Note tt xi+ x i x dx = x4 4 xi+ x=x i = 4 (x4 i+ x 4 i) x 4 i+ x 4 i = (x i+ x i )(x i+ + x i+x i + x i+ x i + x i) nd x i+ x i =.

14 Numericl Anlysis So xi+ x i On te oter nd, x dx = (x i+ + x i+x i + x i+ x i + x i). f(x i )+4f(x i+ ) + f(x i+ ) = x i + 4( x i + x i+ ) + x i+ = x i + (x i + x i+ ) + x i+ Terefore, = x i + (x i+ + x i+x i + x i+ x i + x i) + x i+ = (x i+ + x i+x i + x i+ x i + x i) (f(x i) + 4f(x i+ ) + f(x i+ )) = (x i+ + x i+x i + x i+ x i + x i) Summing ll togeter, I n (x ) = x dx. Higer order metod (Closed Newton-Cote formul) Trpezoid: points; Simpson s rule: points. Higer order: wit more points. 4 points (/8 rule): interpolte polynomil of degree. x x 0 f(x)dx x x 0 P (x)dx = 8 (f(x 0) + f(x ) + f(x ) + f(x )). Error O( 4 ). f(x)dx 8 n/ (f(x i ) + f(x i+ ) + f(x i+ ) + f(x i+ )). 5 points (Boole s rule): interpolte polynomil of degree 4. x4 x 0 f(x)dx x4 x 0 P (x)dx = 45 (7f(x 0)+f(x )+f(x )+f(x )+7f(x 4 )). 4

15 Numericl Anlysis f(x)dx 8 Te error is O( ). n/4 Summry: Closed Newton-Cotes Formuls 45 (7f(x 4i)+f(x 4i+ )+f(x 4i+ )+f(x 4i+ )+7f(x 4i+4 )) Trpezoidl Simpson s /8 Boole? Num. of Points 4 5 Deg. of inter. poly. 4 5 Deg. of Precision. 5 5 Order. of Error(composite formul) O( ) O( 4 ) O( 4 ) O( ) O( ) Even polynomils (odd number of points) re better. Te left prt of te errors x n+ P n (x)) cncels te rigt prt!. Romberg integrtion Te min ide of Romber integrtion is to use te composite Trpezoidl rule to give preliminry pproximtion nd ten pplies te Ricrdson verge process to get iger order trunction error. Te Romberg integrtion lgoritm produces tringulr rry of numbers, ll of wic re numericl estimtes of te definite integrl f(x)dx. Te rry is denoted s R, R, R, R, R, R, R n, R n, R n, R n,n Let n be positive number. Te first column R n, contins estimtes of te integrl obtined by te recursive trpezoid formul wit decresing vlues of te step size n ( wit n equl subintervls). R, = (f() + f(b)) ( = b ) R, = (f() + f(b) + f( + )) = b b (f() + f(b) + f( + 4 ) = (R, + f( + )) ( = b ) R, = (R, + (f( + ) + f( + ))) 5 ( = b )

16 Numericl Anlysis In generl R k, = k (R k, + k f( + (i ) k )) i= for ec k =,,, n, k = b k. It cn be sown tt if f C [, b], te composite Trpezoidl rule cn be written wit n lterntive error term in te form f(x)dx R k, = Replce k+ = k /, giving f(x)dx R k+, = i= i= K i i k = K k + i= K i i k+ = 4 K k + Using Ricrdson s verging tecnique, we get f(x)dx [R k+, + R k+, R k, ] = Higer order formul (second column) for ec k =,,, n i= K i i k i= K i i k 4 i K i 4i ( 4 i ) i k = K 4 4 k + R k, = R k, + R k, R k, i= K i 4i ( 4 i ) i k Continue to pply te Ricrdson extrpoltion procedure to tese vlues, we ve, for ec k =,,, n, nd j =,, k, n O( j k pproximtion formul defined by Remrks R k,j = R k,j + R k,j R k,j 4 j Order of computing R,, R,, R,, R,, R,, R, etc. Stopping criterion for n: bot R n,n R n,n nd R n,n R n,n re smller tn tolernce.

17 Numericl Anlysis.4 Gussin Qudrture In order to find f(x)dx. f(x)dx I n(f). Generl form: I n (f) = w 0 f(x 0 ) + w f(x ) + w f(x ) w n f(x n ) In Newton-Cotes formuls, eqully spced. Te degree of precision is t most m if m points used. How to furter improve te ccurcy? Use non-eqully spced points! Gussin qudrture: coose x 0, x,..., x n nd w 0, w,..., w n simultneously so tt I n (f) s degree of precision s ig s possible. We ve n + unknowns, wic cn be determined by n + equtions for monomils, x, x,..., x n+. We will see tt n + coices llow us to rec degree of precision n +. Exmple: two points, n =. We wnt f(x)dx. I n (f) = w 0 f(x 0 ) + w f(x ). We wnt to coose x 0, x, w 0 nd w s.t. (Here n + =.) I n (f) = f(x)dx, f(x) = x k, k. So we ve 4 unknowns, 4 equtions. w 0 + w = dx = w 0 x 0 + w x = xdx = 0 w 0 x 0 + w x = x dx = w 0 x 0 + w x = x dx = 0 (.) Solve tis system of nonliner equtions to get x 0, x, w 0 nd w. Ten I n (f) = w 0 f(x 0 ) + w f(x ) will be exct for every polynomil of degree. Solution of te eqution.. () implies w x = w 0 x 0. 7

18 Numericl Anlysis. So (4) implies w 0 x 0 w 0 x 0 x = 0 w 0 x 0 = w 0 x 0 x x 0 = x x = ±x 0.. we wnt to coose different points, so x x 0. Ten x = x So w x 0 = w 0 x 0. Ten w = w () implies w =. w 0 =, w =.. () implies x 0 + x =. x = x 0 =, x =. Formul I (f) = f( ) + f( ) Let try ex dx. I (e x ) = e + e =.48 Exmple: tree points, n =. We wnt f(x)dx. I n (f) = w 0 f(x 0 ) + w f(x ) + w f(x ). We wnt to coose x 0, x, x, w 0, w nd w s.t. (Here n + = 5.) I n (f) = f(x)dx, f(x) = x k, k 5. So we ve w 0 + w + w = dx = w 0 x 0 + w x + w x = xdx = 0 w 0 x 0 + w x + w x = x dx = w 0 x 0 + w x + w x = x dx = 0 w 0 x w x 4 + w x 4 = x4 dx = 5 w 0 x w x 5 + w x 5 = x5 dx = 0 (.4) unknowns, equtions. Solve tis system of nonliner equtions to get x 0, x, w 0 nd w. Ten I n (f) = w 0 f(x 0 ) + w f(x ) will be exct for every polynomil of degree. Solution of te eqution. 8

19 Numericl Anlysis. ()(4)() implies In mtrix form, w 0 x 0 + w x + w x = xdx = 0 w 0 x 0 + w x + w x = x dx = 0 w 0 x w x 5 + w x 5 = x5 dx = 0 x 0 x x w 0x 0 w x = 0 0 x 4 0 x 4 x 4 w x 0 (.5) We ve two cses () If w 0x 0 w x = 0, ten one of te following olds. i) x i = 0 for t lest w x two indices. Ten x i = x j. Impossible. ii) x i = 0 (sy, x 0 = 0) for only one index. So w = w = 0. Ten I n (f) = w 0 f(0) No solution. (b) Ten te mtrix is nonsingulr. It implies x 0 = x. So x 0 = x.. Ten w 0 + w = w (w 0 w )x 0 = w x (w 0 + w )x 0 = w x (w 0 w )x 0 = w x (.) (w 0 + w )x 4 0 = 5 w x 4 (w 0 w )x 5 0 = w x 5. ()(4)() implies (w 0 w ) x 0 x 0 = w x x x 5 x 5 Ten eiter x = x 0 wic is impossible or bot nds sides re zero vectors. So w 0 = w nd w = 0 (impossible) or w 0 = w nd x = 0 4. So 5. ()() implies x 0 = 5. Ten w 0 = w w 0 x 0 = w 0 x 4 0 = 5 (.7) x 0 = 5, x = 9 5.

20 Numericl Anlysis. () implies w 0 = Finlly, w 0 = 5 9, w = 8 9, w = 5 9, x 0 = 5, x = 0, x = 5. Formul I (f) = 5 9 f( 5 ) f(0) f( 5 ) Let try ex dx. I (e x ) = 5 9 e e e 5 =.50 ex dx = e x =.504. Tble.: Gussin Qudrture Nodes nd Weigts n Degree of Precision Nodes x 0 x n Weigts w 0 w i