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Дата загрузки 29 Октябрь 2024
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Раздел Курсовые работы
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Benjamin Franklin

Дата регистрации 29 Октябрь 2024

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Chiziqsiz tenglamalar sistemasini yechish usullari. Nyuton usullari

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O‘ZBEKISTON RESPUBLIKASI OLIY TA’LIM, FAN VA INNOVATSIYALAR VAZIRLIGI MIRZO ULUG‘BEK NOMIDAGI O‘ZBEKISTON MILLIY UNIVERSITETI AMALIY MATEMATIKA VA INTELLEKTUAL TEXNOLOGIYALARI FAKULTETI AMALIY MATEMATIKA YO‘NALISHI SONLI USULLAR FANIDAN KURS ISH MAVZU: Chiziqsiz tenglamalar sistemasini yechish usullari. Nyuton usullari BAJARDI: ________________________ QABUL QILDI: ________________________ Toshkent 2024 Mavzu: Chiziqsiz tenglamalar sistemasini yechish usullari. Nyuton usullari Mundarija: 1. Kirish 2. Iteratsiyalar usuli (ketma-ket yaqinlashish usuli) 3. Oddiy iteratsiya usuli 4. Chiziqsiz tenglamalar sistemasini Mathcad dasturi yordamida taqribiy yechish 5. Foydalanilgan adabiyotlar ro’yxati Kirish Ma`lumki, t е nglamalar sist е masini yechish usullarini ikki guruhga bo‘linadi: aniq va it е ratsion. Aniq usullar yordamida sist е mani yechgan bilan aniq yechimni har doim ham topa olmasligimiz mumkin. Chunki b е rilgan sist е madagi ayrim qiymatlar taqriban olingan bo‘lishi, bundan tashqari, hisoblash jarayonida sonlarni yaxlitlashga to‘g’ri k е lishi mumkin. It е ratsion usullarda esa yechim ch е ksiz k е tma- k е tliklarning limiti sifatida olinadi. L е kin, bu usullarning o‘ziga xos tomonlaridan biri shundan iboratki, ular o‘z xatosini o‘zi tuzatib boradi. Agar aniq usullar bilan ishlayotganda biror qadamda xatoga yo‘l qo‘yilsa, bu xato oxirgi natijaga ham o‘z ta`sirini o‘tkazadi. Yaqinlashuvchi it е ratsion jarayonning biror qadamida yo‘l qo‘yilgan xato esa faqat bir n е cha it е ratsiya qadamini ortiqcha bajarishgagina olib k е ladi, xolos. Ya`ni, biror qadamda yo‘l qo‘yilgan xato k е yingi qadamlarda tuzatib boriladi. It е ratsion usullarning hisoblash sx е malari juda sodda bo‘lib, ularni dasturlash juda qulaydir. L е kin, har bir it е ratsion usulning qo‘llanish sohasi ch е garalangandir. Chunki, it е ratsiya jarayoni b е rilgan sist е ma uchun uzoqlashishi yoki, shuningd е k, s е kin yaqinlashishi mumkinki, amalda yechimni qoniqarli aniqlikda topib bo‘lmaydi. Shuning uchun ham, it е ratsion usullarda faqat yaqinlashish masalasigina emas, balki yaqinlashish t е zligi masalasi ham katta ahamiyatga egadir. Yaqinlashish t е zligi dastlabki yaqinlashish v е ktorining qulay tanlanishiga ham bog’liqdir. Aytib o‘tilgan mulohazalar chiziqli t е nglamalar sist е masini it е ratsion usullar yordamida yechishga t е gishli bo‘lib, chiziqsiz t е nglamalar sist е masini it е ratsion usullar yordamida yechishda bu jarayon birmuncha boshqacharoq k е chadi. Chiziqsiz t е nglamalar sist е masini yechishda eng qulay usullar bu it е ratsion usullardir. Chunki, chiziqsiz t е nglamalar sist е masini aniq usullar bilan yechish imkoniyati juda kam bo‘lganligi uchun, ularni yechishda taqribiy usullarni qo‘llashni tavsiya qilinadi. C h i z i q s i z t e n g l a m a l a r s i s t e m a s i n i y e c h i s h n i n g s o n l i u s u l l a r i Das t lab k i tu s h u nch a l ar K o ‘ p l a b a m aliy m asal a lar no ch iz iq li t e n gl a m alar siste m a s ini y echishga o l i b kelin a di. U m u m i y h o lda n no m a ’ li m li n t a no c h i z i qli alg e b r aik y o k i t r a n s e nde n t te n gla m alar siste m asi q u y i d ag ic h a y ozil ad i: f 1( x 1 , x 2 , … , x n ) = 0 f 2 ( x 1 , x 2 , … , x n ) = 0 … … … … … … … … … f n ( x 1 , x 2 , … , x n ) = 0 } ( 3.1 ) Ushbu ( 3 . 1 ) siste m ani ve k t or sha k lida q u y id a gi c ha y ozish m u m kin: f ( x ) = 0. ( 3 .1 ` ) bu y erda x = ( x 1 , x 2 , …, x n ) T – ar g u m entl a rni n g vektor ustuni; f=(f1,f2,… ,fn)T−¿ f u n k si y ala r n i n g v ek tor ustuni; ( … ) T – transpon irlas h o peratsi y asi bel g i s i. Bu siste m a y echi m in i t opi s hni geo m etr i k ta l qi n da 3. 1 - ras m dagi i kki no m a ’ l u m l i ikkita te n gla m alar siste m asining fazov i y t a s -v i r i m isolida tush u n t i r i s h m u m k i n. N o c h i z i q li t en gl am alar s i s te m asi y echi m in i i z l a sh – b u bitta noc h i z iq l i te n gla m a n i y echis h ga nisb a t an a n c ha m urakkab m a s ala. B i tta tengla m ani y echish uchun qo ‘ llani l g a n usullar n i noc h i z iqli t e n gla m alar s iste m a si ni y echishga u m u m lashtirishjuda k o ‘p h i - sobla sh la r ni tal a b qiladi y o ki uni a m a-l i y otda q o ‘ llab b o ‘l m a y di. X us u s an, b u or a li q ni t e n g ikkiga b o ‘ l is h u s u liga tegi s hli. Shu nga qa r a m a sd a n , nochi z i q li te n gla m a n i y echis h ning b ir q a t o r iter a t s ion usul l ar in i no c h i z iq l i te ng la m alar s is t e- m asini y echis h g a umu m lasht i r ish m u m kin. Iteratsiyalar usuli (ketma-ket yaqinlashish usuli) 3. 1 -ras m . I kk i n o m a ’ lu m li ikkita te ng l a m alar siste m asining fa zo viy tasviri Yuqoridagi (3.1) chiziqsiz tenglamalar sistemasi ushbu x 1 = φ 1 ( x 1 , x 2 , … , x n ) x 2 = φ 2 ( x 1 , x 2 , … , x n ) … … … … … … … … … … x n = φ n ( x 1 , x 2 , … , x n )} ( 3.15 ) ko‘rinishga keltirilgan bo‘lsin, bu yerda φ1,φ2,… ,φn - haqiqiy funksiyalar bo‘lib, ular bu sistema izolyatsiyalangan ( x 1¿ , x 2¿ , … , x n¿ ) yechimining biror atrofida aniqlangan va uzluksiz. Qulaylik uchun quyidagi vektorni kiritamiz: x=(x1,x2,… ,xn)va φ(x)=(φ1(x),φ2(x),… ,φn(x)). U holda (3.15) ni quyidagi vector shaklida yozish mumkin: x=φ(x)(3.16 ) (3.16) tenglamaning x¿=(x1¿,x2¿,… ,xn¿) vektor-ildizini topish uchun ko‘pincha quyidagi iteratsiyalar usulini qo‘llash juda qulay: x(k+1)=φ(x(k))yoki x1(k+1)= φ1(x1(k),x2(k),… ,xn(k)) x2(k+1)=φ2(x1(k),x2(k),… ,xn(k)) … … … … … … … … … … xn(k+1)=φn(x1(k),x2(k),… ,xn(k))} k = 0,1,2 , … , ( 3.17 ) bu y er d a y uqoridagi i n d e ks i te r a t s i y alar yaqi n l a s h ishi no m eri n i b i l d ir a d i ; x ( 0) ≈ x ¿ b oshl a ng ‘ i c h y aq in la s hi s h. Usulning b l ok - sxe m ali a l g ori t m i 3 . 6 - ras m d a t a svirla n g a n. A g ar (3.17) itera t si o n jara y on y aqinlashiv c h a n b o ‘ lsa, u holda ushbu ξ= limk→∞x(k)(3.18 ) l i m i tik q i y m at (3 .17) ten g la m aning il di z i bo ‘la d i. 3 . 6 -ras m . Chiziqsiz ten g l a- m alar si s te m as i ni y echi s h u c hun i ter a t si y alar usulining blok-sxe m ali a lg ori t m i . Haqiq a t d an h a m , agar ( 3 . 18) m uno s abat b a j arilg a n d esak, u holda ( 3 . 17) te ng likda k→ ∞ bo’yicha limitga o’tib, φ ( x ) funksiyalarning uzluksizligidan quyidagiga ega bo’lamiz: lim k → ∞ x ( k + 1 ) = φ ( lim k → ∞ x ( k) ) , y a ' ∋ ξ = φ ( ξ ) S h u n d ay q ilib, ξ - bu (3 .16) v e ktor tengla m a ni n g ildizi. A g ar, bund a n t a shqa r i, b a rc ha x(k)(k=0,1 ,… ) yaqinlashishlar biror Ω – sohaga tegishli bo’lsa, u holda ξ = x ¿ ekanligi yaqqol ko’rinadi. Soddaroq qilib aytganda, (3.17)iteratsion jarayon x(0)=(x1(0),x2(0),… ,xn(0)) – boshlan g ‘ ich y aqinl as hishdan b o sh l an i b , b i tta it e ra t si y ad a n k e y in barcha argu m entlar ort t i r - m as i ning m oduli be ri lg a n ε m iqdordan ki c hik b o ‘l m a gun c ha d a v o m ettir i l a di, y a ’ ni ‖x(k+1)− x(k)‖∞= max1≤i≤n{|xi(k+1)− xi(k)|}<ε Bu s ha r tga t e n g ku c hli bo ‘ l g a n q u y i d ag i s h artdan ham fo y d ala n ish m u m kin: ‖x(k+1)− x(k)‖2=|x(k+1)− x(k)|= max1≤i≤n{√xi(k+1)− xi(k)}<ε O d diy iter at si y a usuli d a st urla s h u c hun j uda q u la y , a m m o u qu y id ag i m u hi m ka m c hi likla r ga e g a : a) ‖ φ ' ( x ) ‖ ∞ ≤ q < 1 bu y erda φ' - v ek to r -funksi y a φ ni n g Y a k o b m atr i ts a s i , ‖∙‖∞ belgi bilan esa matritsa normasi kiritilgan: ‖ φ ' ( x ) ‖ ∞ = max 1 ≤ i ≤ n { ∑ j = 1n ∂ φ i ∂ x j } ; b) b ) ‖ φ ' ( x ) ‖ l ≤ q < 1 bu y erda φ ' − ¿ v e kto r - fu nksi y a φ n i n g Yakob m atrits a si, ‖ ∙‖ l belgi bilan esa matritsa normasi kiritilgan: ‖φ'(x)‖l= max1≤i≤n{∑j=1 n ∂φi ∂xj}; c) a g ar b oshl a ng‘ich y aq i nlashish an iq y echi m dan u z oqroq tanla n g a n b o ‘ lsa, a - sh a rtni n g baja r i l i sh ig a q ara m asd a n, usul n ing y aqinlashish i ga ka f ol a t y o ‘q; de m ak, boshla ng ‘i c h y aq i nla s hi s hni t a n l as hn i ng o ‘ zi ham s o dd a e m as ek a n; d) i ter a t sion j ara y on j u d a s e kin y aqinl as h a di. Od di y it e ra ts iya l ar u s u li Iter a tsi y alar u sul i ni dastl ab k i f ( x ) = 0 u m u m i y si ste m aga h am q o ‘ l l a sh m u m kin , bu y erda f ( x ) − ¿ v e ktor-fun k si y a b o ‘l i b , i zol y a t si y a la n gan x* - ve k to r - ildiz a trofi d a ani q l a n g a n va u z l uk si z. Mas a l a n, b u s iste m a n i qu y idagi k o ‘ r i ni s hda y oz a y l i k : x= x+Λ f(x) bu y erda Λ − ¿ x os m as m atr i tsa. Qu y id ag i bel g ila sh ni k i r i ta m i z: x+Λ f(x)=φ(x) ( 3 . 1 9) U h o lda qu y idagiga e ga bo ‘ la m i z: x = φ ( x ) ( 3 .20) A g ar f ( x ) fun k si y a f ' ( x ) − ¿ uzluksiz h osil a ga e g a b o ‘lsa, u h o l d a ( 3 .19) for m u-ladan q u y i d a g i t en gl i k ke l ib c hiq ad i: φ ' ( x ) = E + Λ f ' ( x( 0)) = 0 A g ar φ ' ( x ) o ‘zining normasi b o ‘ y icha k ichik b o ‘lsa, u holda ( 3 .20) t e ng l a m a u c hun ite r a t si y alar j a ra y oni t ez y aqinla s h a di. B u h ol a t n i e ’ ti b o r ga o li b , Λ m atritsani shun d ay ta n la y m iz k i , ushbu φ ' ( x( 0)) = E + Λ f ' ( x( 0)) = 0 te n gl i k b aja r i ls in. Bu y erdan, agar f'(x(0)) - xos m as b o ‘l sa, u h olda qu y idagi t e n gli k k a ega bo ‘ la m iz: Λ = − | f ' ( x( 0))| − 1 S h u n i t a ’ ki d l a sh m u m ink i , bu m az m unan N y ut o n m odi f ika ts ion jara y oni n ing ( 3 .19) tengla m aga q o ‘ l l a n i li shi de m akdir. Xususan, a gar det f'(x(0))=0 bo ‘lsa, u h ol d a boshqa x(0)−¿ bosh l a n g ‘ ich y aqi n la s hi s hni ta n l a sh l oz i m b o ‘ la d i. O d diy it era t si y a u s ul i nafaq a t haqiqiy ildizla r n i , b a lk i k o m pl e k s il d izlar n i h a m topish i m konini ber a di. O x i r g i holda ko m p l eks boshla n g ‘ i c h y aqinl a sh i shni ta n l a sh l o z im bo ‘ ladi. Iter a tsi y alar j ata y oni y aqinlashi s hi n ing y etar l i sha r ti q u y i d a g i c h a. Faraz qi l a y l i k, s h un d ay D ⊂ Rn y op i q s o ha m a v jud b o ‘ l s in k i , bu nda ixti y or i y x ∈ D u c h un φ(x)∈D bo ‘lsin. X ud d i shunda y , i x t i y oriy x 1 va x 2 ∈ D l ar uc hu n qu y idagi shart bajar il sin: ‖φ(x1)−φ(x2)‖≤q‖x1− x2‖,q<1 bu yerda ‖ ∙‖ − R n d ag i b i r o r n or m a. U ho lda osongina k o ‘rsati s h m u m ki n ki, D sohada ( 3 .16) tengla m aning x * y echi m i m avjud b o ‘l i b, (3.17) itera ts ion jara y on t a nla n g a n i x t i y oriy x 0 ∈ D u chun shu y ech i m g a y aqinl a shadi. Bunda qu y i da gi y aq i nlashish te z l i gini baholash o’rinli: ‖ x m − x ¿ ‖ ≤ c q m 1 − q , bu yerda c-biror o’zgarmas. Y u qorid ag i ( * ) shartni qano at la n tiruv c h i   fu n k si y a si qiluv c h an a k slan i sh d e b, ( 3 .18) tengla m aning y ech i m i esa   funk si y an in g qo‘zg‘a l m a s n u q t asi d e b a t a la d i. S h u n i ta ’ k idla y m i zk i , ‖xm+1− x¿‖=‖φ(xm)−φ(x¿)‖≤q‖xm− x¿‖ shuning uchun, odd i y ite r a t si y a us u lini n g yaqi n l a shish t ar ti bi 1 g a t e n g . A g ar  ( x ) fun k si y a D s o h ada uz l uk s iz va differen s ia l l a n uvchan b o ‘lsa, u holda ( * ) sha rt ning b a j a r i li shi u c hun i x ti y or i y x ∈ D l ar u c hun ‖ φ ' ( x ) ‖ ≤ q < 1 sh a r t n ing b a j arilis h i y etarl i . I z o h .  ( x ) f unksi y a f ( x ) funksi y a o r qali b i r q i y m at l i a ni q l a n m a y d i . ‖ φ ' ( x ) ‖ ≤ q < 1 shar t ni n g ba j ar i lis h i u c hun  ( x ) f un k si y ani q a nday t an l a s h l o zi m ? A g ar x ( 0) nuqta atrofida uzluksiz differensiallanuvchan va f ' ( x ) matritsa funksiya aynimagan bo ‘ lsa, unda u m u m i y h o l da q u y i d a gicha y o zi s h m u m kin: Xususiy hollarda φ ( x ) fu n ks i y a n i tanl as h v a u shbu ‖ φ ' ( x ) ‖ ≤ q < 1 s har tn ing b a j arilis h ini t ek shi r ish a n c ha sodda bo ‘lishi m u m k i n. Xususiy h o l . H i so b l a shla r ni a m al i y ot uchun qulay b o ‘lgan n =2 b o ‘ l g a n h o l da x=φ1(x,y) y=φ2(x,y)}(3.21 ) ko ‘r i b c hiqa y lik. ( 3 .15) s iste m a n i ko ‘r i ni s hda y oz i b o l a m iz. φ 1 ( x , y ) , φ 2 ( x , y ) funki y alar i tera s i y al o vc h i funksiyalar d e b y ur i t i l a d i . T a q r ibiy y ech i m n i topish a l g o ri t m i u s hbu xn+1= φ1(xn,yn) yn+1=φ2(xn,yn)}n= 0,1,2 ,… (3.22 )ko ‘r i ni s hda be r i la d i. Bu y erda ( x 0 , y 0 ) − ¿ bir i n c hi y aq in l a shish q i y m at l a r i. (3.22) i t e r a si on hi s obla s h j ara y oni y aqinl as huvchi b o ‘ la d i, a g ar d a ushbu | ∂φ1 ∂x|+| ∂φ1 ∂y|≤q1<1 | ∂φ1 ∂x|+| ∂φ1 ∂y|≤q2<1} (3.23 ) te n gsiz l i k lar baja r i l s a . 4 - m i so l . Q u y id ag i t e ng la m alar siste m a sini n g ite r a t si y al a nuv c hi φ 1 ( x , y ) v a φ 2 ( x , y ) f un k si y alar i ni ( x 0 , y 0 ) = ( 0 , 80; 0 , 55) bosh l a n g ‘ ich nu q t a da topi n g: { f 1 ( x , y ) = x 2 + y 2 − 1 = 0 f 2 ( x , y ) = x 3 − y Y echish. Bu si s te m a uchun φ 1 ( x , y ) v a φ2(x,y) f u n k si y ala r n i qu y idagi k o ‘ rini s hda iz l a y m iz: { φ 1 ( x , y ) = x + α ( x 2 + y 2 − 1 ) + β ( x 3 − y ) 2 ( x , y ) = y + γ ( x 2 + y 2 − 1 ) + δ ( x 3 − y ) α,β,γ,δ no m a ’ l i m ko rffisi y e ntla r ni topish u c h un y uqo r ida t a kl if e ti lgan siste m aga k iruvc h i x ususiy ho si l a lar v a ul a r ni n g (x0,y0) n uq t a d a g i qiy m at l arini hisoblaylik: ∂ f1 ∂x =2x;∂ f1(x0,y0) ∂x =1.6 ;∂ f1 ∂y=2y;∂ f1(x0,y0) ∂y =1.1 ; ∂ f 2 ∂ x = 3 x 2 ; ∂ f 2 ( x 0 , y 0 ) ∂ x = 1.92 ; ∂ f 2 ∂ y = − 1 ; ∂ f 2 ( x 0 , y 0 ) ∂ y = − 1 ; B u lar g a ko ‘ ra α , β , γ , δ n o m a ’ l i m ko rffisi y entlarga nisb a tan qu y idagi chiz iq li a lgeb r a i k t en g l a m alar s i s t e m asiga ke l a m i z: { 1 + 1,6 α + 1.92 β = 0 1.1 α − β = 0 1.6 γ + 1.92 δ = 0 1 + 1.1 γ − δ = 0 Buni y echib, q u y ida g ilarga e ga b o ‘ la m i z: α ≈ − 0.3 ; β ≈ − 0.5 ; γ ≈ − 0.3 ; δ ≈ 0.4 S h u n d ay qil i b, φ1(x,y) va φ2(x,y) f u n k s i y alar n i n g qu y id a gi i f o da la r iga k e la m iz: {φ 1 ( x , y ) = x − 0.3 ( x 2 + y 2 − 1 ) − 0.5 ( x 3 − y ) 2 ( x , y ) = y − 0.3 ( x 2 + y 2 − 1 ) + 0.4 ( x 3 − y ) E n di be r ilg a n tengl a m alar s is t e m a s ini taqribiy y ech i s h uchun y aqinlashuv c h a n ( 3 .22) iter a t s i y alar f o rmulasidan y oki qu y ida kelti r i l g a n Ze y del u s u li for m ulasid a n fo y dal a ni s h m u m k i n . Z eydel usuli O d diy iteratsi y a u sulining iter a t s i on j a ra y on y aqinl a shishi n i tez l a sh ti r itu ch i m odifikatsi y alar i dan bi r i Ze y d e l u s u l i bo ‘lib, b u usulning as osiy f o r m u l asi qu y id agi c ha i fo da l a n adi: x 1 ( k + 1 ) = φ 1 ( x 1 ( k) , x 2 ( k) , … , x n ( k) ) x 2 ( k + 1 ) = φ 2 ( x 1 ( k) , x 2 ( k) , … , x n ( k) ) … … … … … … … … … … … … … . x n ( k + 1 ) = φ 2 ( x 1 ( k) , x 2 ( k) , … , x n ( k) ) } k = 0,1,2 , … ( 3.25 ) Bu iteratsion jarayon bilan bir qatorda ushbu f1(ξ,x2(k),… ,xn(k))=0 f2(x1(k+1),ξ,… ,xn(k))= 0 … … … … … … … … … … … … … . fn(x1(k+1),… ,xn−1(k+1),ξ)=0 } (3.26 ) it e ra ts ion jara y onni q ar a b, y aqi n la s hi s h v e kto r i k o m po ne n t al ar i ni shu te n gla m alar siste m asidan topish m u m kin. Bu t e ngl a m alar s iste m as i n ing h ar bi r i d a bitta ξ no -ma ’ l u m q atnash a di. A n a shu ξ1 l a r n i n g qi y m at l ari (26) t e n gl a m alar siste m asini n g y an g i birin c hi x1(k+1)= ξ1 y aq i nl a s hishi qiymati t o ‘p la m i b o ‘lib xi z m at q ila d i . N avbatd a gi ξ2 l ar es a i kkin c hi y aqinla s h i sh n i n g qiy m atlar t o ‘ p l a m ini beradi, y a ’ ni x 2 ( k + 1 ) = ξ 2 va h o k a z o. B u u sulni n g qula y li g i shund a ki, uni b itta t en gla m a y echi m in i topish g a q o ‘l la s h j u da o d di y , a m m o bu u s u l a m ali y otda ju d a katta h aj m d agi hisobl a shla r ni baja r i s hni t al a b q ilishi m u mkin. Xususiy h o l . I kki no m a ’ l u m li ikk i ta n oc h iz iq li t en gla m alar si s te m as i ni t a q rib i y y echish uchun ba ’ zi h ol l a r da (3 .22) ite r asi o n h i so bl a s h jara y o n i o ‘ r n iga q u y i d a g i « Z e y d el jara y oni » dan fo y dal a nish j u da qula y : xn+1=φ1(xn,yn) yn+1=φ2(xn+1,yn)}n=0,1,2,3 ,… Misol. Quyidagi sistemani oddiy iteratsiya va Zeydel usullari bilan yeching:{ x 1 − x 22 3 = 1 x 2 + 1 x 1 + 1 = − 1 → { x 1 = 1 + x 22 3 = φ 1 ( x 1 , x 2 ) x 2 = − 1 − 1 x 1 + 1 = φ 2 ( x 1 , x 2 ) | ∂ φ 1 ∂ x 1 | +| ∂ φ 1 ∂ x 2 | = | 2 ∙ x 2 3 | < 1 ; | ∂ φ 2 ∂ x 1 | +| ∂ φ 2 ∂ x 2 | = | 1( x 1 + 1 ) 2| < 1 ; x1>0,x1←1,−1.5 <x2<1.5 ;x1(0)=1.5 ,x2(0)=0 Yechish : oddiy iteratsiya usuli: x 1 ( 1) = 1 ; x 2 ( 1) = − 1 − 1 2.5 = − 1.4 ; x1(2)=1+1.96 3 =1.65 ;x2(2)=− 1− 1 2=−1.5 ; x 1 ( 3) = 1 + 2.25 3 = 1.75 ; x 23 = − 1 − 1 2.65 = − 1.38 εn=2= max (|1.61 −1.75 |,|−1.39 +1.36 |)=0.14 ;x1¿≈1.61 ,x2¿≈−1.39 P a r a m etr l ar n i qo‘ z g ‘atish usu l i Bu usulning g ‘ o y asi qu y i d ag i c ha. Da s tlab ( 3.1) t en gl a m alar s is te m asi bi l an b ir qato r da a v va l d a n uning y ec h i m i m a ’ l u m b o ‘ l g a n q u y i d ag i bi r or ten g l a m alar s i s t e m asi qar a l a di: h 1 ( 0)( x 1 , x 2 , … , x n ) = 0 h 2 ( 0)( x 1 , x 2 , … , x n ) = 0 … … … … … … … … … … … h n ( 0)( x 1 , x 2 , … , x n ) = 0 } ( 3.27 ) Bunga m isol qi l ib c hi zi qli a l g eb r a i k te n gl a m alar s iste m asini o l i sh m u m kin . ( 3 . 2 7) te n gla m alar sis t e m as i nin i ng c h a p tara f i ni sh u n d ay o‘zgartir a m izki, u bi r or K s onga ni s ba t an (3.1) t e ngl a m aning c h a p ta r afiga q o ‘ y ilib, uni qu y idagi k o ‘r i nishga k e l t i r s i n: bu y erda k = 0 ,1,…, K . A g ar K ning q i y m ati k a t ta r o q t anla n sa b u f un k si y al a r q i y m atlari n ing k e t m a -k e t o‘z g ari s hi k i c h r a y ib bor a di. Har b i r o‘zga r t irishd a n k e y in para m etrla r i q o ‘zg ‘ a ti lg a n ushbu h 1( k + 1 )( x 1 , x 2 , … , x n ) = 0 h 2 ( k + 1 )( x 1 , x 2 , … , x n ) = 0 … … … … … … … … … … … h n ( k + 1 )( x 1 , x 2 , … , x n ) = 0 } ( 3.29 ) te n gla m alar siste m asi ite r a t si o n u s u l lar bi l a n y ech i b b oril a d i . ( 3 .27) tengla m alar s i s t e m asini n g y ech i m i ( 3 . 2 9 ) uc hu n k = 0 da bos h l an g ‘ i c h y aqi n la s h deb fo y da l anil a d i. ( 3 . 29) tengl a m alar siste m a si ni n g y echi m i ( 3 .27) si s t e m a y echi m id an k am farq q ilga n l ig i u c h u n itera ts ion jara y on n ing y aq i nl a s hishi ta’ m in l a n a di deb h is o bla sh i m i z m u m k i n. Sh und a n k e y in o l ing a n y ech i m ( 3.29 ) siste m aning k = 1 d a gi b o sh l an g ‘ ich y aqinl as hi s hi d eb qar a l ad i v a h oka z o. O x iriga b o rib, k = K - 1 b o ‘ lganda h o s i l b o ‘ lg a n ( 3 . 2 9 ) tengla m alar s iste m asi dastla bk i ( 3 .1) te n gla m alar siste m asiga e kvi v a l e n t b o ‘ lib q ol a d i. S h u n d ay qi l i b, pa ram etrla r ni q o ‘ zgat i sh us u lini n g bos h l a n g ‘ich y aqin l a s hish i ni ta n l as h m asa l a s i y echil ad i. Bu u suln i ng noq u la y ligi shund a ki, ( 3 .2 7 ) te n gla m alar s iste m asini y echila di gan te n gla m alar s iste m as i ga a y lanti r i s h k a t ta h i so b qada m larini ( h at to 10 d a n 1 0 0 g a c ha) ta l ab q ilishi m u m ki n. S h uning ushun, bu usulning q o ‘l l a nil is hi mashin a da j u da k a t ta hisob v a qt i ni talab qili sh i m u m kin. Bu usulning afz a ll igi s h u n da k i, ( 3 .2 8) si s t e m a m uvaffaqi y atli y echi l g a n d a ( 3 .29) siste m a n i n g y echi m i b i r n e c ha i ter a t s i y a q ad a m -la r dagina topil is hi m u m k i n. 3 . 9 . P i k ar it e r a t s i yalari B i r qator h o lla r da (3 . 1) siste m a m axsu s k o‘rinishga e ga bo ‘ lib, u v e kto r - m atrit s a ko ‘ rini s hida q u y i d agi cha y ozil a di: A x − f ( x ) = 0 , ( 3 .30) bu y erda A – be ri lg a n a y ni m agan m atrit s a; f – noc h i z iq l i v ek t or -f u nksi y a. Bun day te n gla m alar s iste m asiga, m asa l an, n oc h i z i q li c h e gar a v iy m asalalarni che k li a y i r m a l ar usuli bil a n y echishda kel i n a di. ( 3 .30) siste m a u c hun q u y id ag i i ter a ts i on p r osedura o ‘ rinli: x ( k + 1 ) = A − 1 f( x( k)) ( 3.31 ) v a u P i k ar iterat s i y alari d e b a t a la d i. I ter at si o n a lgo r it m ni i xch a m y ozish m a qs a d id a ( 3 .31) for m ul ada A - 1 teskari m atrits a d a n fo y da l a n ildi. Asl i da e s a iter a tsi y aning har bir q a da m ida qu y idagi c h i zi qli a lge brai k t en gla m alar siste m a s i y e ch il a d i : Ax(k+1)= f(x(k)) Pikar iteratsiyalarini quyidagi umumlashgan iteratsiyon jarayonning xususiy holi deb qarash mumkin: x ( k + 1 ) = x ( k) − BF ( x( k)) ,( 3.32 ) bu yerda B – berilgan aynimagan matritsa. Bu yerdan ko‘rinadiki, agar F(x)= Ax − f(x)va B= A−1 bo‘lsa, u holda (3.32) tenglik (3.31) ga aylanadi. Agar B matritsa boshqacharoq tanlansa, u holda boshqa bir necha algoritmlar yuzaga keladi, xususan, Nyuton usuli algoritmlari va ko‘p o‘chovli kesuvchilar usuli. B r o y d en us u li N y uto n -Rafs o n u s uli juda y axs h i y aqinl ash is h ni b e r a d i, a m m o Yakob m atrits a s-ining tesk a risi ni h i soblash juda k o ‘p m ashina v a qt i ni oladi. B unday ho l l arda Ya k ob m atritsasi n i h iso b la s h o ‘ r n iga biror b o s hqa y aqi n la sh i sh n i tu z ish uslubi k v a zin yu ton usullar ( a l go r it la r ) deb ham atal a di. A n a shu n d ay us ulla r d a n b i r i bu 1 965 y i l d a t a klif et il g a n Bro y den usuli bo ‘lib, u N y u to n - R a f s on u s u li n i n g tako m ill a shtirilgan v a r i a nti ha m d a u y uqo rida t a’ kidl a ngan ka m chilikd a n h o li. Bu u s u l n i n g qu y ida g i i k k ita m u -hi m farqla r i m avjud: 1 ) itera ts i y alarni n g h ar b i r qada m ida Y ak ob m atrits a s i t o ‘ g‘r i s i y o k i te s ka r i s i hisobl a n m a y di, o ‘zgaruv c hilar c h e t l a shi s h ini s onli b a h ola s h u chun q o ‘sh i m cha f u n k si y alar h iso b l a nil m a y di, faqatgina sx e m a o ‘ zgar m as m atritsasini n g m a v j ud li g ini topish d agi f unk s i y al a rdangina fo y d a la ni la d i; 2 ) y echi m n i ng y aqinl as hi s hini k o ‘ r s a t u v c hi s o ‘n ish koef f itsi y e n ti har bi r it e r-atsi y ada hisobl a na d i, bu o‘z navbatida, Br o y den u sulining y ut u g ‘i b o ‘l i b, N y uto n -Rafson usuli y aqin l a sh i sh n i ka f olat b era ol m a y di . Bundan t ash q ari, bu ko eff i t s i y ent hatto, h a li y ech i m t opil m agan b o ‘lsa h a m , h iso b la s h xa to lig in i bahol a sh i m konini b e-ra d i. Bro y den us u lini n g m az m u ni q u y ida g ich a : N y uto n -Rafs o n f or m ulasi b o ‘ y icha navb a t d a gi x ( k + 1 ) y aqin l as h ish n i olish u c hun k - y aq i n l a sh i sh g a t uz a t m a v e ktor q o ‘sh i la di , y a ’ ni: Δ x ( k + 1 ) =( − W ( k)) − 1 f ( k) ; x ( k + 1 ) = x ( k ) + Δ x ( k + 1 ) Bro y den u sul i da bu t u zat m an i ng h am m a s i d a n e m a s , b a l k i u n ing b i r qis m i d a n f o y da l a nil m a y di: x(k+1)= x(k)+λ(k)Δx(k+1) bu y erda λ ( k) - ska l y ar ko e ffitsi y ent shunday tanl a n a di k i, f ( k + 1 ) v ek tor y o k i uni n g m aksi m al qiy m at q abul q i luv ch i ele m enti nor m asi m ini m u m lashtiril a di ( y o k i k a- m a y t i rila d i). Agar N y uto n - R afson u s u l i ni n g y aqinlashi s hi t a’ m i n l a n g a n b o ‘lsa, u holda λ ( k) > 1 ni tanl as h h isobi g a Br o y den u suli ju d a ka t ta y aqinlash i shga e r i s hadi. A k si n c ha, a g ar N y u t o n - R a f so n u sulini n g y aq i nl a shis h i ta’ m in m alangan b o ‘ lsa, u holda λ(k)<1 ni t a n l a s h h i so b i g a b u y aqinl a shish ta ’ m i n l a n a d i. Bro y den usu l i Ya ko b m atr i t s a si va u n ing te s ka r is i ni hi s obl a sh bilan b o g ‘ liq bo ‘l g a n N y uto n -Ra f son usuli n i n g q i y in c hil i gi n i bart a raf qil a di. Bunga iter a tsi y aning har b i r qada m ida Yakob m atritsasini n g o ‘ r n iga qu y idagi f o r m ula bilan b e r i l g a n y aqinla s hi s hni hiso b la s h e v az i ga e r i s hila d i: H (k+1)= H (k)− [λ(k)Δx(k)+H(k)(f(k+1)− f(k))](Δx(k))TH (k) (Δx(k))TH (k)(f(k+1)− f(k)) (3.33 ) Bro y den us u lini n g a l g o ri t m i q u y id a gi c h a : 1. x (0) bos h l a n g ‘i c h y aqinlashi s h t a nla nad i. 2. Yakob m atr it s a s i W ( 0 ) ni ng t es k a r isini his o b l a s h bilan H (0) m atrits an ing bo sh lang ‘ i c h q iy m ati hisob l a n ad i. 3. f (k) ¿ f x(k) , k = 0 ,1,2, … his o bl a n a di. 4.Δx(k)¿H (k) f (k) hisobla n a d i. 5. λ(k) ko e ffit s i y ent shun d a y tanlana di ki, ‖ f ( k + 1 ) ‖ < ‖ f ( k ) ‖ bo ‘ lsin. 6. x(k+1)= x(k)+λ(k)Δx(k+1) hisobla n a d i. 7. ‖f(k+1)‖ m atr i t sa n or m a s ini n g y aqinl as hi s hi te k shi r iladi. 8. ( f ( k + 1 ) − f ( k ) ) his oblanadi. 9. H (k+1) m atr i tsa ( 33 ) f or m u la b o ‘ y icha hi sob l a n a d i . 10. H i soblash j ara y oni 3 -q a da m dan tak r orlan a di. Misol. Ushbu f 1 ( x , y ) = x 5 + y 3 − xy − 1 = 0 f 2 ( x , y ) = x 2 y + y − 2 = 0 } te n gla m alar siste m asining nolinc h i y aqinlashish ini X 0 = ( x 0 , y 0 ) deb o li b , uning a n iq y ech i m i X= (x, y) = (1; 1) ni B r o y den usuli y orda m i d a a n i q l a n g . Y echish. Miso l ning y ech i m i jara y o ni n i , iter a tsi y alar d a g i X k = ( x k , y k ) y aqinlashishlarni quyidagi jadval shaklida ifodalaylik Jadv a ldan k o ‘ rin a diki, bu m iso l ning nati jas iga N y u t on u suli bil an 8 qada m da e r-ishilg a n e d i. B u e sa Br o y den usulining i t e ra ts i y a tezligi N y u t o n u suli n ik i d a n p a st ek a nl i gini k o ‘ r s at ad i. S h u n g a qara m asda n , Br o y den u s uli o m m a bop b o ‘l i b, h i so b l a s h a m ali y otida k e ng qo ‘ ll an ilib kel i n m oqda. Tez k or tushish usuli (gra di yen t lar u s uli) Y u qorid ag i (3 . 1) t e n g l a m alar siste m a si ni q aray m iz. Faraz qil a y li k , fi f u nksi y a-lar o‘zla r i ni n g u m u m i y an i qlan i sh s o h asida ha qi q i y v a uzluksiz differ e nsi a l-lanuv c h a n. Qu y idagi funksi y ani qaray m iz: U ( x ) = ∑ i = 1n [ f i ( x )] 2 = ( f ( x ) , f ( x ) ) Ravch a nki, (3 . 1) sist e m aning har b i r y echi m i U ( x ) f unk si y ani n o l ga a y lantir a di; va a ks i ncha U ( x ) funksi y a n ol g a t e ng b o ‘ la d ig a n x 1 , x 2 , … , x n s o n lar ( 3 . 1) sis t e m an-ing i l dizla r i. Faraz qi l a m i z k i, ( 3 .1) si ste m a i zo l y atsi y ala n g a n y echi m ga g ina e g a b o ‘lib, b u y echim U ( x ) f u nksi y an in g qat ’ iy m i nim u m n u qtasi bo ‘ls i n. Bu b il a n m asala n o ‘ lc h ovli fa z oda U ( x ) funks i y ani n g m ini m u m i n i top i shga olib kelin ad i. Faraz qila y lik, x - (1 ) siste m aning v ektor - i l dizi, x ( 0) esa u ning boshl a ng‘ich y aqin l a s hi s hi b o ‘lsin. x ( 0 ) n uqta o rq a li U ( x ) funksi y aning sath si r t in i o ‘tka z a m iz. A g ar x ( 0 ) n uqta x il d i z ga y etarl i c ha y aqin b o ‘l sa, u h olda biz ni ng fa r az i m iz b o‘ y icha s a t h sirti U (x)=U (x(0)) ellipsoidga o’xshash. x(0) nuqtadan boshlab U ( x ) = U ( x ( 0) ) sirt normali bo‘ylab harakatlanib, bu harakatni shu normal qaysidir boshqa bir U (x)=U (x(1)) sath sirtiga biror x ( 1) nuqtada urinmaguncha davom ettiramiz. Keyin yana x ( 1) nuqtadan harakatni U (x)=U (x(1)) sath sirti bo‘ylab davom ettirib, bu harakatni shu normal boshqa bir yangi U ( x ) = U ( x ( 2) ) sath sirtiga biror x(2) nuqtada urinmaguncha davom ettiramiz va hokazo. Vaholanki, U ( x ( 0)) > U ( x ( 1)) > U ( x( 2)) > … bo‘lsa, u holda biz shu yo‘l bilan harakatlanib, U(x) ning (3.1) sistemaning izlanayotgan x ildiziga mos keluvchi eng kichik qiymatli nuqtasiga tezkor yaqinlashib boramiz. Ushbu ∇ U( x ) = [ ∂ U ∂ x 1 … ∂ U ∂ x n ] tuzilgan U(x) funksiyaning gradiyenti belgilashini kiritib, O M 0M 1,O M 1M 2,… vektorli uchburchaklardan quyidagini yozamiz (xususiy holda 3.10-rasmga qarang): x(p+1)= x(p)− λp∇U (x(p)),(p=0,1,2 ,… ) Endi  p ko‘paytuvchin aniqlash qoldi. Buning uchun quyidagi skalyar funksiyani qaraymiz: Φ ' ( λ ) = d dλ U [ x ( p) − λ p ∇ U ( x ( p))] = 0 ( 3.40 ) ( 3 .40) t e ngla m aning e n g ki c hik m u s bat i l di z i bizga λp n ing q i y m atini b er a d i . E n di λp s o n n i taq ri biy t opish u sulini q a r ab c h iqa y l i k. Faraz q il a m i zk i , λ−¿ ki c h ik para m etr b o ‘ li b , u ni n g kv a dr a ti va u n dan y uq o r i da r a j ala r ini e ’ tiborga ol m aslik m u m -kin. U ho l da qu y idagiga k ela m iz: Φ ( λ ) = ∑ i = 1n { f i [ x( p) − λ p ∇ U ( x( p))]} 2 fi f u nksi y ani λ n ing c h izi ql i had l ar i g acha a n iq l ikd ag i d ara j ala r iga y o y i b , qu y id a giga ega bo ‘ la m iz: Φ (λ)=∑i=1 n {fi[x(p)− λ∂ fi(x(p)) ∂x ∇ U (x(p))]} 2 Bu yerda ∂ fi ∂x=[ ∂ fi ∂x1 ,∂ fi ∂x2 ,… ,∂ fi ∂xn] Bulardan 3.10-rasm. Ikki o’zgaruvchili funksiya holida gradiyent usulining geometrik talqini Φ '( λ ) = − 2 ∑ i = 1n [ f i ( x( p)) − λ ∂ f i ( x( p)) ∂ x ∇ U ( x( p))] × ∂ f i ( x( p)) ∂ x ∇ U ( x( p)) = 0 Natijada λ p = ∑ i = 1n f i ( x( p)) ∂ f i ( x( p)) ∂ x ∇ U ( x ( p)) ∑ i = 1n [ ∂ f i ( x( p)) ∂ x ∇ U ( x( p))] 2 = ( f i ( x( p))) , W ( x( p)) ∇ U ( x ( p)) ( W ( x ( p)) ∇ U ( x( p))) ,( W ( x( p)) ∇ U ( x ( p))) Bu yerda W ( x ) = ∂ f i / ∂ x – vektor funksiya f ning Yakob matritsasi. Shularga ko’ra quyidagi natijaga kelamiz: μ p = 2 λ p = ( f p , W p W p' f ( p)) ( W p W p' f ( p) , W p W p' f ( p)) ( 3.41 ) bu yerda soddalik uchun quyidagicha yozuv qabul qilingan: f p = f ( x ( p)) ; W p = W ( x ( p ) ) bularga ko’ra x ( p + 1 ) = x ( p) − μ p W p' f ( p) p = 0,1,2 , … ¿ ( 3.42 ) Gradiyent tushish algoritmining blok- sxemasi 3.11-rasmda tasvirlangan. Misol. Tezkor tushish usuli (gradiyent usuli) yordamida ushu x + x 2 − 2 yz = 0.1 y − y 2 + 3 xz = − 0.2 z + z 2 + 2 xy = 0.3 } Tenglamalar sistemasini koordinata boshi atrofida yotgan ildizlarini taqribiy hisoblang. Yechish. Berilganlarga ko’ra f = [ x + x 2 − 2 yz − 0.1 y − y 2 + 3 xz + 0.2 z + z 2 + 2 xy − 0.3 ] , W = [ 1 + 2 x − 2 z − 2 y 3 z 1 − 2 y 3 x 2 y 2 x 1 + 2 z ] , x ( 0) = [ 0 0 0 ] (3.41) va (3.42) formulalar bo‘yicha quyidagi birinchi yaqinlashishni olamiz: μ0= (f(0),f(0)) (f(0),f(0)) =1,x(1)= x(0)−1∙E f(0)= [ 0.1 − 0.2 0.3 ]Xuddi shunday x ( 2) - ikkinchi yaqinlashishni aniqlaymiz. Bu yerdan: f ( 0) = [ 0.13 0.05 0.05 ] , W 1 = [ 1.2 − 0.6 0.4 0.9 1.4 0.3 − 0.4 0.2 1.6 ] , W 1' f ( 1) = [ 0.181 0.002 0.147 ] W 1W 1'f(1)= [ 0.181 0.002 0.147 ] ,μ= 0.13 ∙0.2748 +0.05 ∙0.2098 +0.05 ∙0.1632 0.2748 2+0.2098 2+0.1632 2 =0.3719 x(2)= [ 0.1 −0.2 0.3 ] −0.37119 ∙ [ 0.181 0.002 0.147 ] = [ 0.0327 −0.2007 0.2453 ] Natijaning qanchalik to’g’ri va aniq ekanligini tekshirish uchun tafovut hisoblanadi. N yuton u suli ( 3 .1 ` ) t e n gla m alar siste m asini y echish u c hun ket m a - k et y aqi n la sh i s h usulid a n f o y da l a na m i z. Faraz q ila y lik, ( 3 .1 ` ) vektor tengla m aning iz o l y atsi y al a ng a n x = ( x 1 , x 2 , …, x n ) i l di z la r idan bi t ta s i bo ‘lgan us h b u k -inchi y aq i nl a sh i sh x ( k) = ( x 1 ( k) , x 2 ( k) , … , x n ( k) ) top il g a n b o ‘lsin. U h olda ( 3 . 1 ` ) vektor t en gla m aning a n iq i ld iz i ni ushbu x= x(k)+ε(k)(3.2 ) ko ‘r i ni s hda ifod a l a sh m u m kin, bu y erda ε ( k ) = ( ε 1 ( k) , ε 2 ( k) , … , ε n ( k) ) x a t o likni tuz a tuv c hi had ( ildi z ni n g x a toli g i). ( 3 . 2) i f oda n i ( 3 . 1 ` ) ga qo ‘ y i b , q u y i d ag i t en g l a m a n i hosil qila m iz: f ( x( k) + ε ( k)) = 0 ( 3.3 ) Faraz q i la y l i k , f ( x ) - bu x va x ( k ) la r ni o ‘ z i c hi ga o lg a n bi r or qovariq D soha d a uzluksiz differensiallanuvchan funksiya bo’lsin. (3 . 3) t engla m aning o ‘ng taraf in i ε (k) -kic h ik v e ktor dara j ala r i b o ‘ y icha q a to rg a y o y a m iz v a bu qatorn i ng c hi zi qli hadl a ri bil a ng i na c he k l a na m iz: f ( x( k) + ε ( k)) = f ( x ( k)) + f ' ( x ( k)) ε( k) ( 3.4 ) ( 3 . 4) f or m ul a d a n kelib chiqadiki, f’(x) hosila deb x1,x2,… ,xn−¿ o‘zga r uvc h ilar g a n isbat a n f1,f2,… ,fn−¿ funks i y alar siste m asini n g qu y i d a gi Yakob m atritsasi tu s hu n i l ad i: y ok i u ni qi s qa c ha v e k t o r sha k lida y oz s ak, f'(x)=W (x)=[ ∂ f1 ∂xj],i,j=1,n (3.4) sistema bu xatolikni tuzatuvchi had ε i ( k)( i = 1 , n ) larga nisbatan W(x) matritsali chiziqli sistema. Bundan (3.4) formulani quyidagicha yozish mumkin: f(x(k))+W (x(k))ε(k)= 0 Bu yerdan, W (x(k))  maxsus bo‘lmagan matritsa deb faraz qilib, quyidagiga ega bo‘lamiz: ε(k)=−W −1(x(k))∙f(x(k)) Natijada ushbu x ( k + 1 ) = x ( k) − W − 1 ( x ( k)) ∙ f ( x( k)) ( 3.5 ) Nyut o n usuli f o r m ula s iga kela m i z, b unda x ( 0 ) n olin c hi y aqinl as hi s h si f a t ida i z l a -na y otgan il d izning qo‘pol q i y m atini oli s h m u m ki n. A m ali y otda ( 3 . 1 ` ) noc h izi qli t e ngla m alar si ste m a s ini bu u s u l b i lan y echi s h u c hun hi s obla s hlar ( 3.5) for m ula bo ‘ y icha qu y ida g i sha r t b a j a ril g unga q a d ar d avo m et t i r iladi: | x ( k + 1 ) − x ( k)| ∞ < ε (3 . 6 ) Y uqoridagila r d a n k e lib c hiqib, N y u t on usul i n - ing a lgo r i t m i ni q u y idagic h a y oz a m i z: 1. x ( 0) − ¿ bo s hlang‘ich y aqinlashi s h a n i q l a n a di. 2. Ildizni n g qi y m a t i ( 3 .5) f o r m ul a b o ‘y icha ani q l a shti rila di. 3. Ag a r ( 3 . 6 ) sha r t bajarilsa, u h olda m as a la y echilgan b o ‘l a di v a x( k +1) − ¿ (3.1 ` ) ve ktor te n gla m a ni n g i ldi z i d e b q a bul q ilin a di, a k s holda e s a 2 - q a da m g a o ‘ tila d i. H i s o bl a shla r da ( 3 .1 ` ) c hi zi qsiz tengl a m alar siste m asining f ( x ) f unk s i y alari v a u l a r n ing h o sil a la r i m atritsasi W ( x ) a niq beri l g a n g e y m i z, u h o lda bu siste m ani y echis h n ing blo k -sxe m asi 3 . 2- r a s m da g i f ( x ) v e kt o r- f un k s i y a x il d izi at r o f ida i k ki m a r - ta g a c ha u z l uksiz differe n s i a l l a nuv c h i , Y a k o b m atr i t -sasi W ( x ) m ax s us b o ‘ l m agan (a y ni m a g an), ko ‘p o ‘ lc ho vli N y ut on usuli kvad r a t ik y aqi n la sh is h g a ega: | x ( k + 1 ) − x | < C | x ( k) − x | 2 S h u n i t a ’ ki d la y m i zki, us ulning y aqin l a s hish i ni t a ’ m in l ash u c hun b oshl a n g ‘i c h y aqinlashi s hni m u v a f fa q i y at l i t a n l a s h m u hi m ah a m i y atga e ga. Tengla m alar sonin in g oshishi va u l a r n ing m ur a kka b li g i o r ti s hi bi l an y aqinl as h i sh soh as i tora y ib b oradi. Xususiy h ol . H is o bl as h a m ali y otida n = 2 bo‘lgan hol k o ‘p u chr a y di. Bun i , m a s a-la n , f ( z ) = 0 no ch iz iq li t e ngla m a ni ng ko m pleks il d i z la r i n i t opis h da h am k o ‘ rish m u m kin. H a q i q at a n ha m , a g ar ushbu f 1 ( x , y ) = ℜ ( f ( x + jy )) va f 2 ( x , y ) = ℑ ( f ( x + jy )) f u n k si y ala r n i kiri t sa k , z - ko m pleks i ld i z n ing x – h a qi q iy q i s m i v a y – m avh um qis m i q u y i d ag i i k k i n o m a ’ lu m li ik k i ta n oc h izi qli t e ngla m alar si ste m a s i n i t a q r i b i y yechishdan hosil bo’ladi: { f1(x,y)=0 f2(x,y)=0 bu t a qri b iy h i sobl a shni N y ut o n usuli y orda m ida   a ni q l ik bil a n b aj ara y l i k. D sohaga t e gish l i X 0  (x0, y0) - n oli n c h i y aqin l a s hi s hni t a nl a b o la m i z. 3 . 2 -ras m . Chi z iq siz te n gla m alar siste m asini y echish uchun N y uton u s u- lini n g a lg orit m i . ( 3.4) d a n qu y ida g i c hiz i qli al g eb r a i k t e n g l a m alar sis t e m as i ni tuz i b ol a m i z: ∂ f 1 ∂ x( x − x 0 ) + ∂ f 1 ∂ y ( y − y 0 ) = − f 1 ( x 0 , y 0 ) ; ∂ f 2 ∂ x ( x − x 0 ) + ∂ f 2 ∂ y ( y − y 0 ) = − f 2 ( x 0 , y 0 ) ( 3.8 ) Qu y ida g i b el gil a shla r ni ki r ita m iz: x − x 0 = Δ x 0 , y − y 0 = Δ y 0 . ( 3.9 ) ( 3 . 8) s iste m ani x 0 , y 0 la r ga n i sb a t a n, m asalan, Kra m er usu l i y orda m ida y ech a m i z. K r a m er f o r m ul a l ar in i qu y idagi c ha y oza m iz: Δ x 0 = Δ 1 J , Δ y 0 = Δ 2 J ( 3.10 ) ( 3 .10) bu y erda ( 8) siste m aning a sos i y deter m i nanti q u y idagich a : J = | ∂ f 1 ( x 0 , y 0 ) ∂ x ∂ f 1 ( x 0 , y 0 ) ∂ y ∂ f 2 ( x 0 , y 0 ) ∂ x ∂ f 2 ( x 0 , y 0 ) ∂ y | ≠ 0 ( 3.11 ) (3.8) sistemaning yordamchi determinantlari esa quyidagicha: Δ1= | − f1(x0,y0) ∂ f1(x0,y0) ∂y − f2(x0,y0) ∂f2(x0,y0) ∂y | ;Δ2= | ∂ f1(x0,y0) ∂x − f1(x0,y0) ∂ f2(x0,y0) ∂x − f2(x0,y0)| Δx0, Δy0 la r ni n g to pilgan qiy m atlar i n i ( 3 .9) ga q o ‘ y ib , (3 . 8) s i st e m aning X 1 = ( x 1 , y 1 ) - bi r inc h i y aqin l as hi s hi ko m ponentala r ini t opa m iz: x 1 = x 0 + Δ x 0 ; y 1 = y 0 + Δ y 0 ( 3.12 ) Qu y ida g i sha r tn i ng b a j aril i shi n i t e k shira m iz: max (|Δx0|,|Δy0|)≤ε A g ar bu sha r t ba j ar ilsa, u h olda X1 = (x , y1) birinchi y aqinlashi s hni ( 3.8) siste m an- ing taq r ibiy y ech i m i d eb, hisobl a sh n i t o ‘ xt a ta m iz. Agar ( 3 .13 ) sh a rt ajar i l m asa, u u holda x0− x1,y0− y1 d eb olib, y angi (3 . 8) c hi zi qli algebraik t en gla m alar s is t em asini tuza m iz. Uni y echib, X 2 =(x2, y2) - i kkin c hi y aqin l ash i sh n i topa m i z. T opilg a n y echi m ni ε g a n i s b a t a n (3 .13) b o ‘ y icha tekshira m iz. Agar b u s hart baja r ilsa, u h o l da ( 3.8) siste m aning ta q ribiy y ech i m i d eb X 2 =(x2, y2) ni q abul qila m iz. Agar ( 3 .13) shart b a j ar i l m asa, u h o l da x1= x2,y1= y2 d eb oli b , X 3 =(x3, y3) n i t o p ish u c hun y an g i ( 3 . 8 ) si s te m a n i tuza m iz v a hoka z o. B u siste m ani y ech is h n ing blo k -sxe m asi 3 .3-ras m d a tas v irlang a n. 3 . 3 -ras m . Ik k i n o m a ’ l u m li ikkita noc h i z i q l i ten g la m alar siste m a si ni ta q rib i y y echishning N y u t on u su l i b l o k -sx e m a si . Chiziqsiz tenglamalar sistemasini Mathcad dasturi yordamida taqribiy yechish.{ x1+ x12 10 +ln x2 3 = 4.6 e−x1+√x2+x2=3.1 chiziqsiz t е nglamalar sist е masini oddiy it е ratsiya usuli bilan 0,001 aniqlikda yeching. Е chish: Avvalo sist е maning ko‘rinishini o‘zgartirib olamiz, ya`ni ularni x 1 va x 2 larga nisbatan yechib olamiz: { x1= 4.6 − x12 10 − ln x2 3 x2=3.1 −e−x1−√x2 ; { φ1(x1,x2)= 4.6 − x12 10 − ln x2 3 φ2(x1,x2)=3.1 −e−x1−√x2Endi qidirilayotgan o‘zgaruvchilar bo‘yicha hususiy hosilalar olinadi: ∂φ1 ∂x1 = − x1 5 ,∂φ1 ∂x2 = −1 3x2 ,∂φ2 ∂x1 = e−x1,∂φ2 ∂x2 = −1 2√x2 Aytaylik, boshlang’ich yaqinlashish x1 va x2 lar bo‘yicha [1,4] k е smada bo‘lsin. U holda hosilalar uchun quyidagi t е ngsizliklar o‘rinli bo‘ladi: | ∂ φ 1 ∂ x 1 | ≤ 4 5 = 0.8 = p 1 | ∂ φ 1 ∂ x 2 | ≤ 1 12 = 0.083 = q 1 | ∂φ2 ∂x1|≤ 1 e4≈0.0069 = p2| ∂φ2 ∂x2|≤1 4= 0.25 =q2 Dеmak, qaralayotgan kvadratda: p1+p2=0.8 +0.0 069 =0.8069 <1 q 1 + q 2 = 0.083 + 0.25 = 0.333 < 1 yaqinlashish shartlari bajariladi. U holda dastlabki yaqinlashish sifatida x 1 ( 0) = 3.5 x 2 ( 0) = 1.7 ni olib, k е yingi yaqinlashishlarni oddiy it е ratsiya usuliga mos dastur ta`minoti yordamida aniqlanadi. Buning uchun MathCAD dasturining ishchi oynasiga quyidagi buyruqlar kiritiladi.  1 x1 x2( ) 4.6 x1 2 10 ln x2( ) 3 2 x1 x2 ( ) 3.1 e x1  x2  iter x1 y1 ( ) k 0 x0 x1 y0 y1 x1  1 x0 y0( ) y1  2 x0 y0( ) x x1 x0 y y1 y0 k k 1 break max x y( )( ) if1while x1 y1      iter 3.5 2.2 0.001 ( ) 3.315 1.743        Quyida Nyuton usulining Mathcad hisoblash tizimida tuzilgan dasturiy modul keltirilganF x( ) x0 x1  x1  3  x0  5  1  x0  2 x1  x1  2           J x( ) x1 5 x0  4   2 x0  x1  x0 3 x1  2   x0  2 1           x 2 2        Chiziqsiz tenglamalar sistemasini yechishning Nyuton usuli NSys_N x f J ( )  x lsolve J x( ) f x( )( ) x x  x xreturn  x T  x if1while NSys_N x F J 10 4    1 1        F 1 1             0 0        J 1 1             4 2 2 2        ORIGIN 1  F x y ( ) 2 x2  x y  5 y  1  x log x( )3  y2          J x y( ) x F x y( ) 1d d x F x y( ) 2d d y F x y( ) 1d d y F x y( ) 2d d            D x y( ) J x y( ) 1  Nyuton X ( )  D X 1 X 2   x 1 X 1 x 2 X 2 Y  F x 1 x 2   X x Y break max x X  ( ) if1while X X0 3.5 2.1        X Nyuton X0 10 5   X 2.667 1.986        Foydalanilgan adabiyotlar ro’yxati 1. , Mahmudov A.A, Baxramov S.B « Hisoblash usullari» fanidan o’quv- uslubiy. Toshkent 2008 2. Ismatullaev G'.P., Jo'raev G'.U. “Hisoblash usullaridan metodik qo'llanma” Toshkent 2007 3. N.N. Kalitkin , “Chislennniye metodi” . , О ‘q.q о ‘l . , M., Nauka ., 1978 . 4. M.I. Isroilov ., “Hisoblash metodlari” ., Toshkent . , O´qituvchi , 1988 . 5. A.A. Samarskiy , A.V. Gulin , “Chislenniye metodi” . О ‘q. q о ‘llanma, M . , Nauka , 1989.

Mavzu: Chiziqsiz tenglamalar sistemasini yechish usullari. Nyuton usullari

Mundarija:

  1. Kirish
  2. Iteratsiyalar usuli (ketma-ket yaqinlashish usuli)
  3. Oddiy iteratsiya usuli
  4. Chiziqsiz tenglamalar sistemasini Mathcad dasturi yordamida taqribiy yechish

Foydalanilgan adabiyotlar ro’yxati

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