x 1 b 1 Consider the midpoint x 0 = 1 2

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1 1 chapte 2 : oot-finding def : Given a function f(), a oot is a numbe satisfying f() = 0. e : f() = 2 3 = ± 3 question : How can we find the oots of a geneal function f()? 2.1 bisection method idea : Find an inteval [a, b] such that f(a) and f(b) have opposite sign. Then f() has a oot in [a, b] (Intemediate Value Theoem, Math advanced calculus). f() 0 a0 b0 a 1 1 a2 2 b2 b 1 Conside the midpoint 0 = 1 2 (a + b). The oot is contained in eithe the left subinteval o the ight subinteval; to detemine which one, compute f( 0 ). Then epeat. e : f() = 2 3, f(1) = 2, f(2) = 1 f() has a oot in [1, 2], = n a n b n n f( n ) n Thus 1/17 bisection method (assume f(a) f(b) < 0 ) 1. n = 0, a 0 = a, b 0 = b 2. n = 1 2 (a n + b n ) : cuent estimate of the oot 3. if f( n ) f(a n ) < 0, then a n+1 = a n, b n+1 = n 4. else a n+1 = n, b n+1 = b n 5. set n = n + 1 and go to line 2

2 2 stopping citeion : hee ae thee options b n a n < ɛ, f( n ) < ɛ, n = n ma eo bound an n n 1 2 b n a n = ( 1 2 )2 b n 1 a n 1 = = ( 1 2 )(n+1) b 0 a 0 e : how many steps ae needed to ensue that the eo is less than 10 3? [a, b] = [1, 2], n ( 1 2 )(n+1) b 0 a n n fied-point iteation Suppose f() = 0 is equivalent to = g(). Then is a oot of f() if and only if is a fied point of g(). f() bn g() We ty to solve = g() by computing n+1 = g( n ) with some initial guess 0. This is called fied-point iteation. e : f() = 2 3 = 0 = g 1 () = 3, = g 2() = ( 2 3), = g 3 () = 1 2 (2 3) case 1 case 2 case 3 n n n n Case 1 and case 2 divege, but case 3 conveges (ecall : = ).

3 3 question : what detemines whethe fied-point iteation conveges o diveges? Conside two eamples. g() g() The 1st eample diveges and the 2nd eample conveges. thm Assume that 0 is sufficiently close to and let k = g (). Then fied-point iteation conveges if and only if k < 1. note : This is consistent with the two eamples above. pf (idea) n+1 = g() g( n ) g () n Taylo epansion : g( n ) = g() + g ()( n ) + n+1 k n k 2 n 1 k n+1 0 ok note 1. We showed that n+1 k n. This is called linea convegence and k is called the asymptotic eo constant. ecall : f() = 2 3, = 3 = Tues 1/22 g 1 () = 3 g 1() = 3 2 k = g 1() = 1 : diveges g 2 () = ( 2 3) g 2() = 1 2 k = g 2() = : diveges g 3 () = 1 2 (2 3) g 3() = 1 k = g 3() = : conveges 2. The bisection method also conveges linealy, with k = 1 2.

4 4 2.3 Newton s method idea : local linea appoimation f() tangent line n+1 n slope = f ( n ) = 0 f( n) n+1 = n f( n) n+1 n f ( n ) e f() = 2 3 n+1 = n 2 n 3 2 n note n n f( n ) n : apid convegence Newton s method is an eample of fied point iteation, n+1 = g( n ), whee the iteation function is g() = f() f (). g () = 1 f () 2 f() f () f () 2 g () = 1 f () 2 f() f () f () 2 = 0 This implies that Newton s method conveges faste than linealy; in fact it can be shown that n+1 C n 2, i.e. quadatic convegence. pf n+1 = g() g( n ) = g() ( g() + g ()( n ) + O(( n ) 2 ) ) ok

5 5 e : equation of state of chloine gas ideal gas law : P V = nrt, P : pessue, V : volume n : numbe of moles pesent R : univesal gas constant, R = atm lite/(mole K) van de Waals equation : ( P + n2 ) a (V nb) = nrt V 2, T : tempeatue a = 6.29 atm lite 2 /mole 2 (accounts fo intemolecula attactive foces) b = lite/mole (accounts fo size of gas molecules) Take n = 1 mole, P = 2 atm, T = 313 K, and find V by Newton s method with stating guess V 0 given by the ideal gas law. ( f(v ) = P + n2 ) ( a (V nb) nrt, f (V ) = P + n2 ) ( a 2n 2 ) a + (V nb) V 2 V 2 V 3 n V n atm less than V 0 given by ideal gas law We infe that V 0 has 2 coect digits and V 1 has 5 coect digits. How many coect digits does V 2 have? (hw) summay method ate of convegence cost pe step bisection linea, k = 1 2 f( n ) fied-point iteation linea, k = g () g( n ) Newton quadatic f( n ), f ( n ) note : Bisection is guaanteed to convege if the initial inteval contains a oot; the othe methods ae sensitive to the choice of 0. oot-finding fo nonlinea systems e : chemical eactions 2A + B C A + D C } : evesible eactions fo eactants A, B, D and poduct C a 0, b 0, d 0 : initial concentations (moles/lite) in chemical eacto (known) c 1, c 2 : equilibium concentations of C poduced by each eaction (unknown) k 1, k 2 : equilibium eaction constants (known)

6 6 These vaiables ae elated by the law of mass action. c 1 + c 2 (a 0 2c 1 c 2 ) 2 (b 0 c 1 ) = k 1 c 1 + c 2 (a 0 2c 1 c 2 )(d 0 c 2 ) = k 2 Hence to find c 1, c 2 we need to solve a system of nonlinea equations with 2 equations and 2 unknowns. Newton s method fo nonlinea systems Fist note the following altenative deivation of Newton s method fo the case of 1 equation and 1 unknown, f() = f( n+1 ) = f( n ) + f ( n )( n+1 n ) + n+1 = n f( n) f ( n ) Now conside a system of 2 equations and 2 unknowns. f(, y) = 0, g(, y) = 0 Given ( n, y n ), we want to find ( n+1, y n+1 ). f( n+1, y n+1 ) = f( n, y n ) + f ( n, y n )( n+1 n ) + f y ( n, y n )(y n+1 y n ) + g( n+1, y n+1 ) = g( n, y n ) + g ( n, y n )( n+1 n ) note ( ) f f y g g y (n,y n ) Jacobian mati + g y ( n, y n )(y n+1 y n ) + ( ) ( ) n+1 n f(n, y = n ) y n+1 y n g( n, y n ) 1. Given ( n, y n ), we can solve fo ( n+1, y n+1 ). Each step has the fom A = b, whee A is a given mati, b is a given vecto, and we must solve fo the vecto. 2. hw3 has an application to the chemical eaction system 5 Thus 1/24