Mathematics 3 Curs /Q1 - First exam. 31/10/13 Grup M1 Lecturers: Núria Parés and Yolanda Vidal

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1 Mathematics 3 Curs /Q1 - First exam. 31/10/13 Grup M1 Lecturers: Núria Parés and Yolanda Vidal Name: Calculator: [Generic Competency - 5% of the final grade of the subject a) Expresseu el nombre en base. b) Es vol emmagatzemar el nombre ) utilitzant un emmagatzematge en base, coma flotant, utilitzant unamantissade 7bits un d ells pel signe) i un exponentde 5bits un d ells pel signe)imitjançant arrodoniment per eliminació. Retorna la representació d aquest nombre en aquest format i calcula quantes xifres significatives correctes correct significant digits) té l aproximació. a) ) b) x.s.c. c.s.d.)= 1 representació = }{{}}{{} mantissa exponent a) Donat el nombre , convertirem primer la part entera a binari i després la part decimal. Per tant 10 = 1010). En relació a la part decimal 10 = = + 1 = = = = = 0+ 1 És a dir, = 0.111). Ajuntant els dos resultats anteriors, es té que = ). b) Com que ) = ) 4 = ) 100), si utilitzem 7 bits per a la mantissa i 5 per l exponent i arrodonim mitjançant eliminació, guardaríem el següent nombre }{{}}{{} mantissa exponent que representa ) 100) = ) 4 = ). Per saber el nombre de xifres significatives correctes, calculem primer ) = = ) = = 13.5 per tant, l error relatiu comès és = 0.01 = < Per tant, l aproximació té 1 xifra significativa correcta.

2 1. [3.5 points Denote by I = 8 0 x 4 dx. Answer the following questions a) [ points Approximate I using the composite Trapezoidal method with n = 4. b) [1 point We want to approximate the integral using a new integration method similar to the Simpson s method, but where in each subinterval, we compute a parabola using the points [x i,x i q,x i+1, where x i q = x i + 3h/4. That is, the additional point is not the mid-point of the interval x i m, but the point which is found at a 3h/4 distance from x i, denoted by x i q. In this case, the approximation of the integral is given by I n 1 h 5fxi )+16fx i q) 3fx i+1 ) ) Compute the approximation we would obtain using this method with n =. c) [0.5 points Which is the order of convergence of the method in item b)? It is necessary to justify the answer and detail the computations you have done. Hint: Compute the exact value of the integral, and the exact error committed for n = and n = 4. Look at how the error decreases. a) I T = 73 b) I 608 c) convergence order = 3 a) In this case, n = 4, h = b a n = 8 4 =, and Using the Trapezoidal composite formula: x 0 = 0, x 1 =, x = 4, x 3 = 6, x 4 = 8 I T = h fx 0)+ fx 1 )+fx )+fx 3 ))+fx 4 )) = f0)+ f)+f4)+f6))+f8)) = 73 b) To compute the approximation with n=, the expression reads: I h 5fx0 )+16fx 0 q ) 3fx 1) ) + h 5fx1 )+16fx 1 q ) 3fx ) ) where h = b a n = 8 = 4, and thus x 0 = 0, x 1 = 4, x = 8. The points that ar at 3h/4 distance from x i are computed as: Substituting all these values in the formula: x 0 q = h = 3 x 1 q = h = 7 I 4 5f0)+16f3) 3f4))+ h 5f4)+16f7) 3f8)) = 608 c) Let s compute the exact value of the integral: I = 8 The absolute error committed for n = is 0 x 4 dx = x = 85 5 = E = = Let s compute the approximate value of the integral when using n = 4. In this case, h = b a n = 8/4 = and x 0 = 0, x 0 q = h = 3, x 1 =, x 1 q = h = 7, x = 4, x q = h = 11, x 3 = 6, x 3 q = h = 15, x 4 = 8.

3 Substituting in the given formula of the method, we obtain: I 5f0)+16f 3 ) ) 3f) + 5f)+16f 7 ) ) 3f4) + 5f4)+16f 11 ) ) 3f6) + 5f6)+16f 15 ) ) 3f8) = 6510,67. The absolute error committed for n = 4 is E 4 = = 4.93 Let s compare the errors committed with n = and n = 4 intervals: E E4 = = 8,05 That is, when the number of intervals is multiplied by or, equivalently, when h is divided by ) the error is divided by 3. Thus, it is a third order method: Eh/) = C Eh) = Ch 3 ) 3 h = Eh) 3.

4 . [3.5 points We want to determine the intersection of the curves fx) = 10sinx) and gx) = 10 x in the interval [0, 6. a) [1 point Do three iterations of the Newton method taking as initial guess x 0 = 0. b) [1 point Compute the approximated relative error associated to x in absolute value. c) [0.5 points Using the previous results, and using that in this case the Newton method has the standard convergence rate, predict the approximated relative error associated to x 3. d) [1 point Determine the intersection of the curves in [0, 3. Compute this approximation and stop when in two consecutive iterations the first three decimals are not modified. a) x 3 = b) r = c) r d) α a) In order to find the intersection of the curves we must find the roots of hx) = fx) gx) = 10sinx) 10+x The derivative of this function is: h x) = 10cosx)+x Let s do three iterations of Newton method taking as initial guess x 0 = 0: x 1 = x 0 hx 0) 10 h = 0 x 0 ) 10 = 1 x = x 1 hx 1) h x 1 ) = x 3 = x hx ) h x ) = b) The approximate relative error associated to x in absolute value is: r = x 3 x x 3 = c) Let s assume that Newton has the standard convergence rate, then: d) Let s continue the iterations with Newton method: r 3 λ r = x 4 = x 3 hx 3) h x 3 ) = The first three decimals have not been modified with respect to x 3.

5 3. [3 points The following data, giving the evolution of the temperature of a solid during one second, have been obtained from an experiment. t Tt) a) [1 point Aproximate Tt) [0.3, 0.6 using polynomial interpolation using only the three interpolation points 0.3, 0.4, and 0.6). b) [0.5 points Using the interpoland obtained in item a), predict the value of the temperature at t = 0.5. Aproximate Tt) using an interpolant of the form pt) = At+Be t using the least squares criterion. c) [0.5 points Give the generic expression of the system of equations that has to be solved to determine A and B. d) [0.5 points Compute the interpolant pt) for the data given in the exercise all of them). e) [0.5 points Using the interpolant from item d), make a prediction of the time t in which the temperature is maximum inside the interval [0, 1. Compute the results with correct decimal places. a) pt)= pt) = 4.56 t 0.4)t 0.6) b) T0.5) n c) t i t ie t i t ie t i n e t i d) pt)= t e t e) maximum t = 0.6 ) A B t 0.3)t 0.6) 0.0 ) = t ift i ) ft i)e t i ) t 0.3)t 0.4) 0.06 a) Let s use the points t 0 = 0.3, t 1 = 0.4, t = 0.6. The Lagrange polynomials associated to these points are: L 0 t) = t x 1)t t ) t 0 t 1 )t 0 t ) = t 0.4)t 0.6) ) ) = t 0.4)t 0.6) L 1 t) = t x 0)t t ) t 1 t 0 )t 1 t ) = t 0.3)t 0.6) ) ) = t 0.3)t 0.6) 0.0 L t) = t x 0)t t 1 ) t t 0 )t t 1 ) = t 0.3)t 0.4) ) ) = t 0.3)t 0.4) 0.06 Thus, the pure interpolating polynomial is: pt) = ft 0 )L 0 t)+ft 1 )L 1 t)+ft )L t) = 4.56 t 0.4)t 0.6) t 0.3)t 0.6) t 0.3)t 0.4) b) Inordertopredictthevalueofthetemperatureattimet = 0.5weevaluatethepureinterpolatingpolynomial obtained in item a) at this point t = 0.5): p0.5) = ) ) c) In this case, ) ) 0.0 EA,B) = n [ ft i ) At i Be t i ) ) 0.06 the values of A and B are determined by the least-squares criteria minimizing the function min EA,B) A,B =

6 We impose that, A = 0 B = 0 The linear system obtained reads, n A = [ft i ) At i Be t i ti = 0 n B = [ft i ) At i Be t i e t i = 0 Written in matricial form, [ t i n t ie t i t ie t i e t i [ A B = [ t ift i ) ft i)e t i d) We must solve the linear system: [ [ A B = [ to obtain A = and B = Thus, the interpolant is: e) Using the interpolant: pt) = t e t pt) = t e t we want to make a prediction of the time t in which the temperature is maximum inside the interval [0,1. Thus, we need to solve the equation p t) = e t t) = 0 We will use Newton method, so we will need to compute p t) = e t 4t )+e t) = e t 4t ) Let s start with the initial guess x 0 = 0, then x 1 = x 0 p x 0 ) p x 0 ) = x = x 1 p x 1 ) p x 1 ) = x 3 = x p x ) p x ) = The last two iterations have the same two first decimal places. Thus, the solution is 0.6.