2、f?(?時(shí)??石??時(shí)>=貞?.i'3(.口—?r;)+?時(shí)?才;?(“2—>+.r;?.i'2(ill)(^2IH;)W
3、—(-r2~^2)—■/7<^4—?T;)?24(4〉=
4、卑町(孔)—咯町(才4〉
5、WI遼I[Ie;(才2)
6、+
7、e;(x4)
8、]012.計(jì)算/=(Q1屮?取吃~1?1?利用下列篩式計(jì)算?哪一個(gè)得到的結(jié)果戰(zhàn)好?—7^,(3-2施幾,99-70妊(>/2+1)$(3+2何3解(施一1)6=0.0050506…取72^1.4.(1)————1---?-=0052328(#+])?(】?4+W2.”(2)(3-2Q)3~(3-2Xl?l
9、)3^0.008(3)20.0051253(3+2施戶(3+2.8戶99一7072^99-70X1.4=1經(jīng)比較可知.以——「一計(jì)算得到的結(jié)果M好.(3+2Q第二章O1.當(dāng)r=l,1,2時(shí),/(.r)=0,3?1?求./(.門的:次插值多解U(x)=/(fcr0)4(.r)+f(xx)/
10、(x)4-/(.r2)/2(x)=0+(-3)(*—1)(j*—2)
11、.(j-—)(.i+1)(一1一1)(一1-2)(2-1)(2+1)◎2.給出/(,r)=ln.r數(shù)值表X0.10.50.6ln.r-0.916291一0.693147-0.510826JC0.70
12、.8Inx—0.356675一0.223144用線性插值及二次插值計(jì)算lnO.5l的近似值.分析利用Newum插值多項(xiàng)式.解依據(jù)插值誤差估計(jì)式選距肉0.51較近的點(diǎn)為插值卩點(diǎn)?并建立差商表?口=0.5一0.693117、〉1.823210、?門=0.6—0.0826〈J>0.201115>2.027325/?口=0.I-0.91G291^^“出NewtonMi值多項(xiàng)式.Vj.r)=-0.693117+1.823210(.r-0.5)、(.r)=Ni(m)+(—0.201115)(.r—0.5)(0.6)計(jì)算近似值.rj(0.51)=0.693117+
13、1.823210X(0.540.5)a-0.620219N?(0.51)=Nj(O.51)-0.201115X(0.5I-C).5)X(0.54—0.6)a-0.616839Oi?設(shè)巧為互異節(jié)點(diǎn)?求證:n)另m:/j(.r)=xk(k=0.1?,//);>-onii)工(?_-.r)%(x)=0(k=1.2,????")?證明i)函數(shù)〃及y;^//.r)均為被插値函數(shù)川的關(guān)于互異H點(diǎn){小;“的1超過〃次的捕值多項(xiàng)式?利川插值多項(xiàng)式的惟一性知兩者恒辱?ii)(?巧—?曠)%(才)==2加)2k”(一?『)…>-0>-0?—0inrkj—(、r-*'?
14、kn=ssi-o;-o(交換求和次序)(有關(guān)因子提出求和符號外)i=0(利用iWkJ及結(jié)論(1))=(.r—j'Y=0◎5?設(shè)/(?“€◎[>』]且/(=/(u)2—/(/>)^—^=0u—bb—a應(yīng)用插値余項(xiàng)公式右If(,r}—L(.r)I=—/^(^)(.—a)(.r—b)■?W丄mnx/"(€〉max
15、(.r—a〉(/—")I£』(“一"尸max
16、/"(”)O?所以17、/(j)^—(ba)-max/"(*).笫三章◎11?假設(shè)/(")在"]上連續(xù)?求J(工)的冬次赧隹一致逼近多項(xiàng)式?分析山閉區(qū)間連續(xù)函數(shù)n質(zhì)知?{(6:中?七w4??”]?使得max/(.?)—/(.])和min/(』)=/(』2)成工.證明M=max/(./)un=min/(?』)上€[jufW:&“」取3=£(m+“j?則右max
18、/(?/)—Po
19、=max!
20、M~PoI21、h-致遏近多項(xiàng)式.◎14.
22、求/")=*在[0.1]匕的帰住一次逼近多項(xiàng)式.分析