Numerical Analysis I

Transcription

1 College of Science, Deprtment of Mthemtics Tibh University, Al Mdinh Al Munwwrh 010 / 1431 Numericl Anlysis I Mostf Zhri URL. emil. mzhri@tibhu.edu.s

2 CONTENTS CONTENTS Contents 1 The Concept of norms Definition of norms Exercises Numericl Derivtives 6.1 Definitions Exercises Error Terms Trunction Error Evlution of Functions Clcultion of Functions Clcultion of Polynomils Horner Scheme Exercises Numericl Integrtion Rectngle rule Trpezoidl rule Composite trpezium rule Prbol rule (Simpson s rule) Composite Simpson s rule Newton-Cotes formul Numericl Integrtion s Error Error of the Simpson s rule Solution of equtions f(x) = Newton Method Geometric Illustrtion Convergence Theorems Secnt Method Geometric Illustrtion URL. emil. zhri@gmx.net

3 CONTENTS CONTENTS 6.. Convergence Theorem Bisection Method Solution of System of liner Equtions Using the inverse mtrix Using Crmer Method Gussin elimintion URL. emil. zhri@gmx.net 3

4 REFERENCES REFERENCES References [1] Burden nd Fires, Numericl Anlysis, Eighth Edition, Brooks/Cole Publishing Compny, 005. [] K. Atkinson, An Introduction to Numericl Anlysis, John Wiley & Sons, Inc., Second Edition, [3] Buchnn nd Turner, Numericl Methods nd Anlysis, McGrw-Hill, N.Y., 199. [4] Jmes F. Epperson, n Introduction to Numericl Methods nd Anlysis, John Wiley N.Y., 00. [5] Iscson nd Keller, Anlysis of Numericl Methods, John Wiley N.Y., [6] L. F. Shmpine, R. C. Allen, Jr. & S. Pruess, Fundmentls of Numericl Computing, John Wiley & Sons, Inc., First Edition, [7] Rlston Anthony, A First course in Numericl Anlysis, McGrw-Hill, Inc., [8] Endre Süli nd Dvid F. MyersAn, Introduction to Numericl Anlysis University of Oxford, Cmbridge University Press, 003 [9] J. Stoer nd R. Bulirsch, Introduction to Numericl nlysis, Springer Verlg 1993 URL. emil. zhri@gmx.net 4

5 1 THE CONCEPT OF NORMS 1 The Concept of norms 1.1 Definition of norms The nlysis of the effects of rounding error on solutions of systems of liner equtions requires n pproprite mesure. This is provided by the concept of norm defined below. In order to motivte the xioms of norm, we note tht the set IR of rel numbers is liner spce, nd tht the bsolute vlue function: { x, if x 0; x := x, if x < 0. The bsolute vlue x of rel number x mesures the distnce between x nd 0 (the zero element of the liner spce IR). Definition 1. The nonnegtive rel-vlued function is sid to be norm on the spce IR n provided tht it stisfies the following xioms: 1. x = 0 if, nd only if, x = 0 in IR n ;. λx = λx x for ll λ IR nd ll x IR n ; 3. x + y x + y for ll x, y IR n (the tringle inequlity). Exmple 1. Prove tht the following mps stisfy the norm xioms for n = n ( n ) 1. x 1 = x i b. x = x i c. x = mx x i. i=1,...,n 1. Exercises i=1 Exercice 1. Compute the norms of x = (1, ) T, y = (1, 5, ) T, (T represents the trnsposed vector.) For n = 1 nd n = resp. interpret geometriclly on the rel line nd plne resp. the norm. b. c. Exercice. Often, the rel number π is pproximted by 3.14 nd e by.718, use your clcultor nd the norm bove to compute the error term given s ( ) ( ) e.718 e i = π 3.14 i, for i = 1,,, wht do you conclude. URL. emil. zhri@gmx.net 5 i=1

6 NUMERICAL DERIVATIVES Exercice 3. For n = find α, β > 0 such tht α x 1 x β x. In this cse we sid tht the norms re equivlent. Is propriety is true for ll n IN (ns. yes). Show this for the following vector x = (3, 4) T. (T represents the trnsposed vector.) Numericl Derivtives.1 Definitions If you hve procedure which computes function f(x), nd now you wnt to compute its derivtive f (x). The definition of the derivtive, is given s f(x + h) f(x) lim = f f(x + h) f(x) (x). h 0 h h If you cn fford two function evlutions for ech derivtive clcultion, then it is significntly better to use the symmetrized form. Exercises f (x) f(x + h) f(x h) h Exercice 4. Let us consider f(x) = x nd by using the following pproximtion formul of the first derivtive Df h (x 0 ) f(x 0 + h) f(x 0 ), h compute Df h (1) for h = 0.1; 0.01; Compre your result to f (1), wht do you conclude. Exercice 5. Let us consider f(x) = sin(x) nd by using the following pproximtion formul of the first derivtive Df h (x 0 ) f(x 0 + h) f(x 0 ), h compute Df h (0) for h = 0.1; 0.01; Compre your result to f (0), wht do you conclude. URL. emil. zhri@gmx.net 6

7 3 ERROR TERMS Exercice 6. Let us consider f(x) = x nd by using the following pproximtion formul of the first derivtive Df h (x 0 ) f(x 0 + h) f(x 0 h), h compute Df h (1) for h = 0.1; 0.01; Compre your result to f (1), wht do you conclude. Exercice 7. Let us consider f(x) = sin(x) nd by using the following pproximtion formul of the first derivtive Df h (x 0 ) f(x 0 + h) f(x 0 h), h compute Df h (0) for h = 0.1; 0.01; Compre your result to f (0), wht do you conclude. Exercice 8. compre the results of the pproximtions bove. Exercice 9. By using the first pproximtion formul of the first derivtive, show tht f f(x + h) f(x) f(x h) (x). h 3 Error Terms 3.1 Trunction Error Using Tylor expnsion, we hve f(x 0 + h) = f(x 0 ) + f (x 0 )h + 1 f (α)h, with α [x 0 h, x 0 + h], which implies tht nd f (x 0 ) = f(x 0 + h) f(x 0 ) h } {{ } Approximtion 1 f (α)h. }{{} Error T erm f (x 0 ) f(x 0 + h) f(x 0 ) Kh where mx h = K. x [x 0,x 0 +h] URL. emil. zhri@gmx.net 7

8 4 EVALUATION OF FUNCTIONS We sy tht this pproximtion is first order. A p order pproximtion of the derivtive of f is given s f (x 0 ) f(x 0 + h) f(x 0 ) Kh p h where p 0. The generl Tylor expnsion is f (k) (x 0 ) f(x 0 + h) = h k, f(x 0 h) = ( 1) k f (k) (x 0 ) h k k! h! k=0 where f (k) (x 0 ) refers the k th derivtive of f t the point x 0. Hence f f(x + h) f(x h) [ h ] (x 0 ) = + h 3! f (3) (x 0 ) + h3 5! f (5) (x 0 ) +. }{{} Error T erm 4 Evlution of Functions 4.1 Clcultion of Functions The generl Tylor expnsion, the Trunction Error is given s f (k) (x 0 ) p f(x 0 + h) = h k f (k) (x 0 ) = h k f (k) (x 0 ) + h k k! k! k! k=0 k=0 k=p+1 }{{}}{{} Approximtion T runction Error Using the generl Tylor expnsion, we evlute the following functions e x k=0 = 1 + x + x! + x3 3! +... = Approximtion + Trunction Error x IR. ln(1 + x) = x x + x = Approximtion + Trunction Error x IR. sin(x) = x x3 3! + x5 5!... = Approximtion + Trunction Error x IR. cos(x) = 1 x! + x4 4!... = Approximtion + Trunction Error x IR. URL. emil. zhri@gmx.net 8

9 4. Clcultion of Polynomils 4 EVALUATION OF FUNCTIONS 4. Clcultion of Polynomils Consider the polynomil function P given s P (x) := n x n + n 1 x n x + 0 ; n, n 1, , 0 IR, n 0. Cost of ny evlution using direct computtions re: Addition : + Multipliction : = n = n + n j=1 j = 1 n(n + 3). Totl number : = n ( (n + 3)) = Order(n ) The bove method is expensive in number of opertion, especilly for very smll nd very lrge x. To improve the number of opertions, one cn proceed s follow P (x) := n x n + n 1 x n x + ) 0 = ( n x n 1 + n 1 x n x + 0. = (... ( ) ( n x + n 1 ) x + }{{} n x )x + 0 } Step 1 {{ } } Step {{ } Step n Ech step requires two opertions one, ddition nd one multipliction. Hence, the totl number of opertion is n =Order(n): better thn Order(n ). To evlute P t the point x, one cn use the following lgorithm Algorithm 1. Step 1. Compute ( n x 0 + n 1 ) = A n, Step. Compute (A n x 0 + n ) = A n 1, Step 3. Compute (A n 1 x 0 + n ) = A n,.. Step n. Compute (A 1 x ) = A 0, Output A 0 = P (x 0 ), end. URL. emil. zhri@gmx.net 9

10 4.3 Horner Scheme 5 NUMERICAL INTEGRATION 4.3 Horner Scheme Consider the polynomil function P given s P n (x) := n x n + n 1 x n x + 0 ; n, n 1, , 0 IR, n 0. Horner s method gives P (x 0 ), P (x 0 ),..., P (n) (x 0 ): For j = 0, 1,..., n, define P n j (x) s { 4.4 Exercises P n j (x) = (j) n x n j + + (j) j+1 x + (j) j, where (j+1) n (j+1) k = (j) n = (j) k R j = 1 d j P n (x 0 ) j! dx j + x 0 (j+1) k+1, k = n 1,..., j. implies = (j+1) j nd p (j) n (x0) = j!r j. Exercice 10. Evlute the vlue of e using Tylor expnsion for p = 1,, 3, 4, 5. Exercice 11. Evlute the vlue of sin( π 17 ) using Tylor expnsion for p = 1, 3, 5. Exercice 1. Evlute the vlue of cos( π 11 ) using Tylor expnsion for p =, 4, 6. Exercice 13. Evlute the vlue of ln(1.314) using Tylor expnsion for p =, 3, 4, 5, 6. Exercice 14. Use the lgorithm (1) to evlute P ( ) nd P (5) for P (x) = 5x 4 x 3 + x + x Numericl Integrtion Consider the integrl of continuous rel-vlued function f defined on closed intervl [, b] of the rel line such tht the definite I = b URL. emil. zhri@gmx.net 10 f(x)dx

11 5.1 Rectngle rule 5 NUMERICAL INTEGRATION integrl is very hrd to reduce by mens of the usul methods of vrible substitution nd/or integrtion by prts. Let F be n nti-derivtive of f, i.e., its derivtive F = f. Then by the Fundmentl Theorem of Clculus: I = b f(x)dx = F (b) F (). However, unfortuntely, nti-derivtives re often not known. e.g. (f(x) = e x ). We hve to clculte integrls of such functions numericlly. Numericl integrtion formuls re bsed on formuls for the res of of simple geometricl shpes = Qudrture. 5.1 Rectngle rule The integrl of f cn be numericlly pproximted using the rectngle formul given s I = b f(x)dx (b )f(θ) for θ [, b]. This formul is exct, (i.e., without error) for constnt functions f(x) = c. 5. Trpezoidl rule Using n other pproximtion of f(θ) in the formul bove : Length verge of the Height. Hence the trpezoidl rule is given s ( ) b f() + f(b) I = f(x)dx (b ). 5.3 Composite trpezium rule Suppose tht f is function, defined nd continuous on nonempty closed intervl [, b] of the rel line. In order to construct n pproximtion to the integrl I, we now select n integer m nd divide the intervl [, b] into m equl subintervls, URL. emil. zhri@gmx.net 11

12 5.4 Prbol rule (Simpson s rule) 5 NUMERICAL INTEGRATION ech of width h = (b )/m, so tht I = b f(x)dx = m i=1 xi x i 1 f(x)dx, x i = + ih = + i (b ), i = 0, 1,..., m. m Ech of the integrls is then evluted by the trpezium rule, xi I i = f(x)dx h ( ) f(x i ) + f(x i 1 ). x i 1 summing these over i = 1,, ldots, m leds to the composite trpezium rule I = b f(x)dx m i=1 5.4 Prbol rule (Simpson s rule) 1 I i = h( f(x 0) + f(x 1 ) + + f(x m 1 ) + 1 ) f(x m). The Simpson method pproximte the function by Prbol p(x) = A(x ) + B(x ) + C through the points ( + b (, f()),, f( + b) ), (b, f(b)). By computing the constnts A, B nd C follows the pproximtion formul I = b f(x)dx 5.5 Composite Simpson s rule (b ) ( 6 f() + 4f ( + b ) + f(b) ). For Simpson s rule, let us suppose tht the intervl [, b] hs been divided into m intervls by the points x i = + ih, i = 0, 1,..., m with m nd h = (b )/m nd let us pply Simpson s rule on ech of the intervls [x i, x i ], URL. emil. zhri@gmx.net 1

13 5.6 Newton-Cotes formul 5 NUMERICAL INTEGRATION i = 1,,..., m, giving I = = b m i=1 f(x)dx = xi m i=1 I i x i f(x)dx m i=1 h ( ) f(x i ) + 4f(x i 1 ) + f(x i ). 6 This leds to the composite Simpson rule b f(x)dx h [ f(x 0 ) + 4f(x 1 ) + f(x ) + 4f(x 3 ) 3 ] f(x m ) + 4f(x m 1 ) + f(x m ) 5.6 Newton-Cotes formul Let f be rel-vlued function, defined nd continuous on the closed rel intervl [, b], nd suppose tht we hve to evlute the integrl b f(x)dx. Since polynomils re esy to integrte, the ide, roughly speking, is to pproximte the function f by its Lgrnge interpoltion polynomil p n of degree n, nd integrte p n insted. Thus, for positive integer n, we hve I = b 5.7 Numericl Integrtion s Error f(x)dx b p n (x)dx. Theorem 1. Let n be even nd let f be (n + )-times continuously differentible. Furthermore, let (b ) n j=0 c jf(x j ) be closed Newton-Cotes formul on the intervl [, b] The the qudrture error is given by where < θ < b. E n (f) = = b f(x)dx (b ) (b )n+3 f n+ (θ) (n + )! j=0 1 0 n c j f(x j ) n (t t j )dt, j=0 URL. emil. zhri@gmx.net 13

14 6 SOLUTION OF EQUATIONS F (X) = Error of the Simpson s rule Exmple. Simpson s rule (n = ) t 0 = 0, t 1 = 1/, t = 1 1 n 0 j=0 (t t j )dt = E (f) 1 0 t (t 1/)(t 1)dt = (b )5 880 mx θ b f (4) (θ). 6 Solution of equtions f(x) = 0 A simple exmple is x + bx + c = 0, involving qudrtic polynomil with rel coefficients, b, c, nd 0. The two solutions to this eqution, lbeled x 1 nd x, re found in terms of the coefficients of the polynomil from the fmilir formule x 1 = b + b 4c ; x = b b 4c. It trnspires tht for ech n 5 there exists polynomil eqution of degree n. with integer coefficients which cnnot be solved in terms of rdicls; such is, for exmple, x 5 4x = 0. Since there is no generl formul for the solution of polynomil equtions, no generl formul will exist for the solution of n rbitrry nonliner eqution of the form f(x) = 0 where f is continuous rel-vlued function. How cn we then decide whether or not such n eqution possesses solution in the set of rel numbers, nd how cn we find solution? Suppose tht f is rel-vlued function, defined nd continuous on bounded closed intervl [, b] of the rel line. It will be tcitly ssumed throughout the chpter tht < b, so tht the intervl is nonempty. We wish to find rel number α [, b] such tht f(α) = 0. If such α exists, it is clled solution to the eqution f(x) = 0. Theorem. Let f be rel-vlued function, defined nd continuous on bounded closed intervl [, b] of the rel line. Assume, further, tht f()f(b) 0;then, there exists α [, b] such tht f(α) = 0. Theorem 3 (Fixed Point Theorem). Suppose tht g is rel-vlued function, defined nd continuous on bounded closed intervl [, b] of the rel line, nd let URL. emil. zhri@gmx.net 14

15 6.1 Newton Method 6 SOLUTION OF EQUATIONS F (X) = 0 g(x) [, b] for ll x [, b]. Then, there exists α [, b] such tht α = g(α); the rel number α is clled fixed point of the function g. Proof. Note tht (use the theorem bove) ( g()) (b g(b) 0. }{{}}{{} 0 0 Exmple. Consider the function f defined by f(x) = e x x 1 for x [1, ]. Clerly, f(1) < 0 nd f() > 0. Thus we deduce the existence of α [1, ] such tht f(α) = 0. Suppose tht g is rel-vlued function, defined nd continuous on bounded closed intervl [, b] of the rel line, nd ssume tht g(x) [, b] for ll x [, b]. Given tht x 0 [, b], the recursion defined by x k+1 = g(x k ), k = 0, 1,,... is clled simple itertion; the numbers x k, k 0, re referred to s itertes. If the sequence (x k ) converges, the limit must be fixed point of the function g, since g is continuous on closed intervl. Indeed, writing α = lim k x k, we hve tht α = lim k x k+1 = lim k g(x k ) = g ( lim k x k ) = g(α), A sufficient condition for the convergence of the sequence (x k ) is provided by our next result which represents refinement of BrouwerŠs Fixed Point Theorem, under the dditionl ssumption tht the mpping g is contrction. Definition. (Contrction) Suppose tht g is rel-vlued function, defined nd continuous on bounded closed intervl [, b] of the rel line. Then, g is sid to be contrction on [, b] if there exists constnt L such tht 0 < L < 1 nd g(x) g(y) L x y x, y [, b]. 6.1 Newton Method Newton ws born on 4 Jnury 1643 in Woolsthorpe, Lincolnshire, Englnd nd died on 31 Mrch 177 in London, Englnd. Newton mde revolutionry dvnces in mthemtics, physics, stronomy nd optics; his contributions to the foundtions URL. emil. zhri@gmx.net 15

16 6.1 Newton Method 6 SOLUTION OF EQUATIONS F (X) = 0 of clculus were mrred by priority disputes with Leibniz. Newton ws ppointed to the Lucsin chir t Cmbridge t the ge of 7. In 1705, two yers fter becoming president of the Royl Society ( position to which he ws re-elected ech yer until his deth), Newton ws knighted by Queen Anne; he ws the first scientist to be honored in this wy. Newtons Philosophie nturlis principi mthemtic is one of the most importnt scientific books ever written. Let f be (rel) continuous nd differentible function. To solve the nonliner eqution f(x) = 0, Newton suggested the following lgorithm defined by Geometric Illustrtion x k+1 = x k f(x k) ; k = 0, 1,, 3,... (1) f (x k ) The geometric interprettion of the lgorithm 1 is illustrted in Figure 1: the tngent to the curve y = f(x) t the point (x k, f(x k )) is the line with the eqution y f(x k ) = f (x k )(x x k ); it meets the x-xis t the point (x k+1, 0). α x x 1 x 0 Figure 1: Newton Method Convergence Theorems Definition 3. Suppose tht α = lim k x k. We sy tht the sequence (x k ) converges to α with t lest order q > 1, if there exist sequence (α k ) of positive rel URL. emil. zhri@gmx.net 16

17 6. Secnt Method 6 SOLUTION OF EQUATIONS F (X) = 0 numbers converging to 0, nd µ > 0, such tht α k+1 x k α α k, k = 0, 1,,..., nd lim k α q = µ. k with α k = x k α for k = 0, 1,,..., then the sequence (x k ) is sid to converge to α with order q. In prticulr, if q =, then we sy tht the sequence (x k ) converges to α qudrticlly. Theorem 4. [Convergence of Newton s method] Suppose tht f is continuous rel-vlued function with continuous second derivtive f, defined on the closed intervl I δ = [α δ, α + δ], δ > 0, such tht f(α) = 0 nd f (α) 0. Suppose further tht there exists positive constnt A such tht f (x) f (y) A, x, y I δ. If f(α) x 0 h, where h is the smller of δ nd 1/A, then the sequence (x k ) defined by Newton method (1) converges qudrticlly to α. Theorem 5. Suppose tht the function f stisfies the conditions of Theorem 4 nd lso tht there exists rel number X, X > α, such tht in the intervl J = [α, X] both f nd f re positive. Then, the sequence (x k ) defined by Newton s method converges qudrticlly to α from ny strting vlue x 0 J. 6. Secnt Method Let f be (rel) continuous nd differentible function, using Newtons method to solve nonliner eqution f(x) = 0 requires explicit knowledge of the first derivtive f of the function f. Unfortuntely, in mny prcticl situtions f is not explicitly vilble or it cn only be obtined t high computtionl cost. In such cses, the vlue f (x k ) cn be pproximted by difference quotient; tht is, f (x k ) f(x k) f(x k 1 ) x k x k 1. () The lgorithm of the secnt method is defined by ( x k x ) k 1 x k+1 = x k f(x k ) ; k = 1,, 3,... f(x k ) f(x k 1 ) URL. emil. zhri@gmx.net 17

18 6. Secnt Method 6 SOLUTION OF EQUATIONS F (X) = 0 where x 0 nd x 1 re given strting vlues. It is implicitly ssumed here tht f(x k ) f(x k 1 ) 0, k > Geometric Illustrtion The method is illustrted in Figure. The new iterte x k+1 is obtined from x k 1 nd x k by drwing the chord joining the points B(x k 1, f(x k 1 )) nd A(x k, f(x k )), nd using s x k+1 the point t which this chord intersects the x-xis. If x k 1 nd x k re close together nd f is differentible, x k+1 is pproximtely the sme s the vlue supplied by Newton s method, which uses the tngent t the point A. A B C α Figure : Secnt Method. 6.. Convergence Theorem Theorem 6. Suppose tht f is rel-vlued function, defined nd continuously differentible on n intervl I = [α h, α + h], h > 0, with center point α. Suppose further tht f(α) = 0, f (α) 0. Then, the sequence (x k ) defined by the secnt method converges t lest linerly to α provided tht x 0 nd x 1 re sufficiently close to α. The experiment shows the fster convergence of Newton s method, but it must be remembered tht ech itertion of Newton s method requires the clcultion of both f(x k ) nd f (x k ), while ech itertion of the secnt method requires the clcultion of f(x k ) only (s f(x k 1 ) hs lredy been computed). In our exmples URL. emil. zhri@gmx.net 18

19 6.3 Bisection Method 6 SOLUTION OF EQUATIONS F (X) = 0 the computtions re quite trivil, but in prcticl sitution the clcultion of ech vlue of f(x k ) nd f (x k ) my demnd substntil mount of work, nd then ech itertion of Newton s method is likely to involve t lest twice s much work s one itertion of the secnt method. 6.3 Bisection Method Suppose tht f is rel-vlued function defined nd continuous on bounded closed intervl [, b] of the rel line nd such tht f(α) = 0 for some α [, b]. A very simple itertive method for the solution of the nonliner eqution f(x) = 0 cn be constructed by beginning with n intervl [ 0, b 0 ] which is known to contin the required solution α (e.g., one my choose [ 0, b 0 ] s the intervl [, b] itself, with 0 = nd b 0 = b), nd successively hlving its size. More precisely, we proceed s follows. Let k > 0, nd suppose tht it is known tht f( k ) nd f(b k ) hve opposite signs; we then conclude tht the intervl ( k, b k ) contins solution of f(x) = 0. Consider the midpoint c k of the intervl ( k, b k ) defined by c k = 1 ( k + b k ) nd evlute f(c k ). If f(c k ) is zero, then we hve locted solution α of f(x) = 0, nd the itertion stops. Else, we define the new intervl ( k+1, b k+1 ) by { (k, c ( k+1, b k+1 ) := k ), if f( k )f(c k ) < 0 (c k, b k ), if f(c k )f(b k ) < 0 nd repet this procedure. This my t first seem to be very crude method, but it hs some importnt dvntges. The nlysis of convergence is trivil; the size of the intervl contining î is hlved t ech itertion, so the sequence (c k ) defined by the bisection method converges linerly, with rte ρ = log 10 (). Even Newton s method my often converge more slowly thn this in the erly stges, when the strting vlue is fr from the desired solution. Moreover, the convergence nlysis ssumes only tht the function f is continuous, nd requires no bounds on the derivtives, nor even their existence. Once we cn find n intervl [ 0, b 0 ] such tht f( 0 ) nd f(b 0 ) hve opposite signs, we cn gurntee convergence to solution, nd tht fter k itertions the solution α will lie in n intervl of length (b 0 0 )/ k. The bisection method is therefore very robust, though Newton s method will lwys win once the current iterte is sufficiently close to α. URL. emil. zhri@gmx.net 19

20 7 SOLUTION OF SYSTEM OF LINEAR EQUATIONS If the initil intervl [ 0, b 0 ] contins more thn one solution, the limit of the bisection method will depend on the positions of these solutions. Figure 3 illustrtes possible sitution. c 1 B 0 α c0 b 0 Figure 3: Bisection Method. 7 Solution of System of liner Equtions 7.1 Using the inverse mtrix Let A = ( ij ) be n n n rel mtrix nd b given column vector of IR n. The ims of the following section is to find vector x IR n such tht Ax = b. Denoting by x i the i th entry of the vector x, we cn lso write Ax = b in the following expnded form: 1,1 x 1 + 1, x + + 1,n x n = b 1,,1 x 1 +, x + +,n x n = b,... n,1 x 1 + n, x + + n,n x n = b n. The inverse mtrix A 1 of nonsingulr mtrix A IR n n stisfies A 1 A = AA 1 = I, where I is the n n identity mtrix. We now multiply both sides of the eqution Ax = b on the left by A 1 to deduce tht A 1 Ax = A 1 b x = A 1 b. URL. emil. zhri@gmx.net 0

21 7.1 Using the inverse mtrix 7 SOLUTION OF SYSTEM OF LINEAR EQUATIONS Where A 1 = 1 det(a) Adj(A)T, Adj(A) T = (( 1) i+j det(a ij )) 1 i,j n, nd the mtrix A ij is the mtrix A without the i th row nd j th column. Exmple. Consider the three dimensionl liner system Ax = b : Where A = x 1 + 3x x 3 = 7 5x 1 x + x 3 = 3 x 1 + x + x 3 = The Adjcent mtrix is given s ( ) 1 1 det ( 1 1 ) Adj(A) = 3 1 det ( 1 1 ) 3 1 det 1 1 Hence Adj(A) = ; b = 7 3 ; x = ( ) 5 1 det ( 1 1 ) 1 det ( 1 1 ) 1 det 5 1 nd A 1 = 1 x 1 x x 3 ( ) 5 1 det ( 1 1 ) 3 det ( 1 1 ; ) 3 det Which implies tht x 1 1 x = A 1 b = 3. x 3 1 Exercice 15. Solve the following liner system using the inverse mtrix method x 1 + 3x x 3 = 1 x 1 x + x 3 = 0 3x 1 + x + x 3 = 3 URL. emil. zhri@gmx.net 1

22 7. Using Crmer Method 7 SOLUTION OF SYSTEM OF LINEAR EQUATIONS (The solution is x 1 = 1; x = ; x 3 = 4.) 7. Using Crmer Method An lterntive pproch to the solution of the liner system Ax = b, clled Crmer s rule, proceeds by expressing the ith entry of x s x i = det(a i), i = 1,, 3,..., n. det(a) where det(a) is the determinnt of A, nd det(a i ) is the n n determinnt obtined by replcing the i th column of A by the entries of b. This method require the nonsingulrity od A, i.e. det(a) 0. Thus, solving Ax = b is due to evlute n + 1 determinnts A, A 1,..., A n. Exmple 3. Solve the following liner system using Crmer s method Where A = The solution is given s Where A 1 = x 1 + x x 3 = 1 x 1 x + x 3 = x 1 + x + x 3 = 0 ; b = 1 0 x 1 = det(a 1) det(a) = 6 4 = 3, x = det(a ) det(a) = 4 = 1, x 3 = det(a 3) det(a) = 4 4 = 1. ; A = ; x = ; A 3 = x 1 x x URL. emil. zhri@gmx.net

23 7.3 Gussin elimintion 7 SOLUTION OF SYSTEM OF LINEAR EQUATIONS Exercice 16. Solve the following liner system using the Crmer method 5x 1 + 3x x 3 = 1 x 1 3x + x 3 = 1 3x 1 + x + 1x 3 = 0 (The solution is x 1 = 0; x = 0.5; x 3 = 0.5.) 7.3 Gussin elimintion To solve liner systems of lgebric equtions ws developed by Crl Friedrich Guss (Germn scientist) nd ws first published in his Theori motus corporum coelestium in sectionibus conicis solem mbientium (1809). Guss ws concerned with the study of the steroid Plls, nd derived set of six liner equtions with six unknowns, lso giving systemtic method for its solution. The method proceeds by successively eliminting the elements below the digonl of the mtrix of the liner system until the mtrix becomes tringulr, when the solution of the system is very esy. This technique is now known under the nme Gussin elimintion. Let us consider the following system 1,1 x 1 + 1, x ,n x n = b 1,1 x 1 +, x ,n x n = b 3,1 x 1 + 3, x ,n x n = b n,1 x + n, x n,n x n = b n The first gussin elimintions consists to eliminte the under mtrix entries using the following lgorithm { i,j (1) i,j = i,j i,1 1,j b i b (1) i = b i b 1 1,j 1,1 ; Suppose 1,1 0, it follows tht 1,1 ; i, j =,..., n i =,..., n URL. emil. zhri@gmx.net 3

24 7.3 Gussin elimintion 7 SOLUTION OF SYSTEM OF LINEAR EQUATIONS 1,1 x 1 + 1, x ,n x n = b (1), x (1),n x n = b (1) Suppose (1), 0, it follows tht nd 0 + (1) 3, x (1) 3,n x n = b (1) (1) n, x (1) n,nx n = b (1) n (1) i,j () i,j = (1) i,j (1) i,1 b (1) i b () i = b (1) i b (1) (1) 1,j (1) 1,1 (1) 1,j (1) 1,1 ; i, j = 3,..., n ; i = 3,..., n 1,1 x 1 + 1, x ,n x n = b (1), x... + (1),n x n = b (1) () 3,3 x n... + () 3,n x n = b () () n,3 x... + () n,nx n = b () n nd so on, the finl step results 1,1 x 1 + 1, x ,n x n = b (1), x... + (1),n x n = b (1) () 3,3 x n... + () 3,n x n = b () (n 1) n,n x n = b (n 1) n The system bove cn be written in consistency form (n) 1,1 x 1 + (n) 1, x... + (n) 1,n x n = b (n) (n), x... + (n),n x n = b (n) (n) 3,3 x n... + (n) 3,n x n = b (n) (n) n,nx n = b (n) n URL. emil. zhri@gmx.net 4

25 7.3 Gussin elimintion 7 SOLUTION OF SYSTEM OF LINEAR EQUATIONS The, the solution of the liner system is given by x n = b(n) n (n) n,n x i = b (n) i n j=i+1 (n) i,j x j (n) i,i for i = n 1,..., 1. Exmple 4. Let us consider the following three dimensionl liner system 1 1 x x = 4 6 x 3 10 Using the Guss elimintion lgorithm, we get (1) x()-(1) 0.5x()-(1) (1) () (3) (4) (5) (6) x 3 = 4 4 = 1, x = (3 1)/3 = 3, x 1 = ( )/ = 3. Exercice 17. Solve the following three dimensionl liner system using Guss elimintion 1 5 x x = 0 x 3 1 (The solution is x 1 = 7 4 ; x = 5 4 ; x 3 = 1 4 ;) URL. emil. zhri@gmx.net 5