Mathematics 3 Curs /Q1 - First exam. 30/10/14 Group M1 Lecturer: Yolanda Vidal

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1 Mathematics Curs /Q1 - First exam. 0/10/14 Group M1 Lecturer: Yolada Vidal Name: Calculator: 1. [ poits] You are desigig a spherical tak to hold water for a small village i a developig coutry. The volume of liquid it ca hold ca be computed as V πh 2 R h where V volume [m ], h depth of water i tak [m], ad R the tak radius [m]. If R m, what depth must the tak be filled to so that it holds 0m? I order to solve this problem follow the steps: a) [0.5 poit] Reformulate the problem ito a root fidig problem. b) [1.5 poits] Do three iteratios of the Newto method. Choose the iitial guess by yourself explaiig the reasos why you selected that value. c) [1 poits] Does x 2 verify the stoppig (covergece) criteria with tol x ? Solutio a) Takig V πh 2 R h ad substitutig V 0 ad R we obtai which ca be arraged as 0 πh 2 9 h 9πh 2 πh 90 0 }{{} f(h) b) A iitial guess should be selected i the iterval [0, 6] which are the possible values of h. Here we select h h h h c) r 2 h h2 h e 04 < Yes, it verifies the stoppig criteria.

2 2. [.5 poits] The data below represets the bacterial growth i a liquid culture over a umber of days. Day 0 16 Amout a) [1 poit] Fid the pure iterpolatig polyomial usig all the data poits. Use Lagrage polyomials. b) [1 poit] Fid the liear Splie that iterpolates the give data. Use the shape fuctios Φ i. c) We wat to approximate with the least squares criterio the give data usig a iterpolat of the form Solutio p(x) e β x. c.1) [1 poit]write dow the equatio to solve i order to fid the value of β. Substitute the values of the give data ad simplify the most you ca the equatio. c.2) [0.5 poit]solve the equatio obtaied i item c.1) usig Newto root fidig method with two iteratios ad iitial guess equal to 0.. a) Let s defie: x 0 0, x 1, x 2 16 the the pure iterpolatig polyomial usig Lagrage polyomials is p 2 (x) f(x 0 )L 0 (x) + f(x 1 )L 1 (x) + f(x 2 )L 2 (x) 67L 0 (x) + 9L 1 (x) + 149L 2 (x) where L 0 (x) L 1 (x) L 2 (x) (x )(x 16) ( )( 16) (x 0)(x 16) ( 0)( 16) (x )(x 16) 12 x(x 16) 64 (x 0)(x ) x(x ) (16 0)(16 ) 12 b) Usig the Lagrage polyomial the Φ 0 (x) basis fuctio is defied as L 0 0 x 0 x Usig the Lagrage polyomials the Φ 1 (x) basis fuctio is defied as Φ 0 (x) { x x [0, ] L 0 1 x 0 0 x L 1 0 x x Φ 1 (x) x x [0, ] 16 x x [, 16] Usig the Lagrage polyomial the Φ 2 (x) basis fuctio is defied as L 1 1 x 16 x Φ 2 (x) Fially, the liear Splie ca be writte as: { x x [, 16] S(x) f(x 0 )Φ 0 (x) + f(x 1 )Φ 1 (x) + f(x 2 )Φ 2 (x) 67Φ 0 (x) + 9Φ 1 (x) + 149Φ 2 (x)

3 c) The error term, i this case, is E(β) [ f(xi ) e β ] 2 x i, ad we wat to solve the followig miimizatio problem Thus, the β parameter is determied imposig E β mi E(β) β 2e β x i [ f(xi ) e β x i ] 0 After some algebra it reads, therefore, substitutig the give data, ad usig the atural logarithm we fid out the value of β, that is ( x 2 i ) e β x i f(x i ) 20e β 16 e β l e β l β Note: Item c.2 has ot bee take ito accout as the equatio i item c.1 was aalytically solvable.

4 . [.5 poits] Deote by I 0 e x2 dx. Aswer the followig questios a) [1 poit] Approximate I usig the composite Simpso method with 2. b) [1 poit] Approximate I usig the composite Trapezoidal method with 4. c) [0.5 poits] Is it possible to obtai the exact value of the proposed itegral usig Simpso or Trapezoidal composite methods? It is ecessary to justify the aswer. d) [1 poit] Whe usig the Trapezoidal rule, how may itervals,, are ecessary to approximate I with a absolute error (i absolute value) smaller tha 0.001? Solutio Hit: E T C (h) f (ξ) (b a) 12 h2 a) As a 0, b, ad 2 the h b a Simpso s rule 4 ad x 0 0, x 0,1 2, x 1 4, x 1,2 6, x 2. Applyig I h 6 (f(x 0) + 2f(x 1 ) + f(x 2 ) + 4(f(x 0,1 ) + f(x 1,2 ))) 4 (f(0) + 2f(4) + f() + 4(f(2) + f(6))) b) As a 0, b, ad 4 the h b a Trapezoidal rule 2 ad x 0 0, x 1 2, x 2 4, x 6, x 4. Applyig I h 2 (f(x 0) + 2(f(x 1 ) + f(x 2 ) + f(x )) + f(x 4 )) 2 (f(0) + 2(f(2) + f(4) + f(6)) + f()) c) No, because the fuctio to be itegrated is ot a polyomial. d) We are lookig for the value such that E T C (h) f (b a) (ξ) 12 ( b a ) 2 f (ξ) () 12 ( ) 1 2 < (1) Thus, the value f (ξ) must be bouded. Let s start computig the secod derivative of the fuctio, So, f(x) e x2 f (x) 2xe x2 f (x) 2e x2 + 4x 2 e x2 (4x 2 2)e x2 f (x) (4x 2 2) e x2 O oe had, the fuctio (4x 2 2) is icreasig i the iterval [0, ]. Therefore, its maximum is achieved at x. O the other had, the fuctio e x2 is decreasig i the iterval [0, ]. Therefore its maximum is achieved at x 0. Thus, the secod derivative ca be bouded by whe ξ [0, ]. f (ξ) 254 Usig the obtaied boud for the secod derivative ad isolatig from equatio 1 we obtai: ( ) 254 () 1 < Performig the calculatios, 292 <. Thus, at least 29 itervals are ecessary to obtai a error smaller tha

5 GENERIC COMPETENCY - 5% of the fial grade of the subject [CG. 1] U ombre e format miifloat usa bits de memòria. Utilitzeu 4 bits (u d ells pel sige) per a l expoet i 4 bits (u d ells pel sige) per a la matissa. Calculeu els ombres màxims i míims (e valor absolut) que es pode emmagatzemar e aquest format. El ombre més gra (e valor absolut) que es pot represetar és que represeta el ombre }{{}}{{} matissa expoet +(0.111) 2 2 +(111)2 ( ) El ombre més petit (e valor absolut) que es pot represetar és que represeta el ombre }{{}}{{} matissa expoet +(0.100) 2 2 (111) ( ) [CG. 2] Represeteu el ombre 5.75 e base 2 i coma flotat utilitzat ua matissa de 6 dígits (u d ells pel sige) i u expoet de 4 dígits (u d ells pel sige). Utilitzeu arrodoimet per elimiació. Preem la part etera i aem dividit per 2: 5/2 26 i residu 1 26/2 1 i residu 0 1/2 6 i residu 1 6/2 i residu 0 /2 1 i residu 1 1/2 0 i residu 1 Per tat el ombre 5 e base 2 s escriu com: (110101) 2. Preem ara la part decimal i aem multiplicat per 2 (a cada pas agafem la part decimal del resultat aterior i torem a multiplicar per 2): Per tat el ombre 0.75 e base 2 s escriu com: (0.011) 2. Així teim, (5.75) 10 ( ) (6) (110)2. Com omés teim 6 dígits per la matisa i u d ells és pel sige, el ombre que s emmagatzema a l ordiador és: (110)2.