MATH 251 Final Multiple Previouses, Fall 2015

CMPS 251: Numerical Computing February 2nd, 2006 Final Exam Instructors: F. Abu Salem and G. Turkiyyah Name: Duration: 120 minutes Notes: 1. You are allowed to use no more than two A4 pages of notes (front and back). 2. This exam consists of seven questions. Make sure you have all of them. Question 1 (8pts) (a) Let f be an infinitely differentiable function in a closed interval I = [α, β]. Let h be any value such that x + h ∈ I and consider a, b ∈ R. Use Taylor series to show that f ′ can be approximated using the formula f ′(x) ≈ 1 (b − a)h [bf(x + h) − f(x + ah) − (b − 1)f(x)] (1) (5pts) (b) Establish the error term for Equation (1) above. (3pts) Solution: F-1 Solution cont’d: F-2 Question 2 (9pts) Consider the following system of equations:    x1 + αx2 − 3x3 = 7 12x1 + 12αx2 + x3 = 121 αx1 + 4x2 + 2x3 = 6 where α ∈ R. For which values of α does the above system possess 1. no solution? (3pts) 2. infinitely many solutions? (3pts) 3. a unique solution (to be determined if it exists)? (3pts) Solution: F-3 Solution cont’d: F-4 Question 3 (21pts) A symmetric pentadiagonal system is a pentadiagonal system in which the subdiagonals are the same as the superdiagonals. (a) Develop a pseudo-code to solve a symmetric pentadiagonal system using scaled partial pivoting. (7pts) (b) Determine the operational count (run-time in terms of the input size) of your pseudo- code. (7pts) (c) Is this more efficient than the pseudo-code for solving general pentadiagonal systems? Justify your answer in terms of operational and/or spatial costs, whenever applicable. (7pts) Solution: F-5 Solution cont’d: F-6 Question 4 (18pts) (a) What is the minimum number of sub-intervals required to insure that the error in approximating � 2 1 ln x3dx using the trapezoid rule is less than 5 × 10−5? (5pts) (b) Consider the integral � 1 0 exdx. • Use the trapezoid rule to find the value of the integral with 2, 4, and 8 sub-intervals. (5pts) • Use Romberg integration to obtain the best approximation of the integral from the results above. (5pts) • What is the order of the approximation of the Romberg rule above? (i.e., what is the value of k if the error is written as O(hk) where h is the interval size). (3pts) Solution: F-7 Solution cont’d: F-8 Question 5 (20pts) (a) Consider the trapezoid rule for the interval [0, 1] using n + 1 equally spaced points. Explain why the composite trapezoid rule can be expressed as � 1 0 f(x)dx ≈ h 2 � f(0) + 2 n−1 � i=1 f � i n � + f(1) � where h is the step size. (5pts) (b) Rewriting the above expression as � 1 0 f(x)dx ≈ n � i=0 Aif � i n � how would the Ai’s be defined? (5pts) (c) Consider the composite trapezoid rule for double integration of bivariate functions over the unit square [0, 1] × [0, 1]: � 1 0 � 1 0 f(x, y)dxdy ≈ n � i=0 n � j=0 AiAjf � i n, j n � Use the above expression to develop a pseudo-code approximating � � e−(x2+y2)dxdy over the unit square. (10pts) Solution: F-9 Solution cont’d: F-10 Question 6 (12pts) Consider an iterative solution of the following 2 × 2 set of linear equations: �3 1 1 3 � �x1 x2 � = �1 1 � (a) Determine the Jacobi iteration matrix. (4pts) (b) Will Jacobi converge? Why or why not? (4pts) (c) Perform one iteration of Jacobi with the starting value x0 = (0, 0). (4pts) Solution: F-11 Solution cont’d: F-12 Question 7 (12pts) (a) Determine the values of a and b that make the function f(x) = � x2 x ∈ [0, 1] 1 2(x − 1)2 + a(x − 1) + b x ∈ [1, 3] a quadratic spline. (4pts) (b) In class, we considered the problem of interpolating n + 1 data points with co- ordinates (xi, yi), i = 1, · · · , n + 1, using a quadratic spline Q(x). The continuity and smoothness conditions at x allowed us to express the n spline segements as: qi(x) = zi+1 − zi 2(xi+1 − xi)(x − xi)2 + zi(x − xi) + yi, x ∈ [xi, xi+1] where zi+1 = 2 yi+1−yi xi+1−xi − zi, zi = Q′(xi). In order to determine the parameters zi uniquely we needed one additional condition. In class, we imposed z1 = 0 and used it to compute the remaining zi’s. • Explain how you would compute the unknown coefficients zi if the extra condition becomes alternatively z1 = z2. (4pts) • What is the geometric interpretation of this condition? (4pts) Solution: F-13 Solution cont’d: F-14 AMERICAN UNIVERSITY OF BEIRUT Mathematics Department-FAS MATH 251 FINAL EXAM Fall 2013-2014 Closed Book, 2hours STUDENT NAME ID NUMBER Problem Out of Grade 1 20 2 20 3 20 4 25 5 15 TOTAL 100 1 1. (20 points) Consider the following Initial value Problem: (IV P) � dy dt = 2et − y ; t ∈ [1.00, 2.00] y(1) = 0 To solve (IVP) (a) (5 pts)(IVP) is first solved on [1.00 , 1.25] using 1 step of the discrete scheme of the first order Runge Kutta method -(Euler’s method) : (RK1). • Write the formulae of this scheme: (RK1) � ................................................................ yi+1 = .................................................. • Use 1 step ONLY of this scheme to approximate y(1.25) with 5 significant digits and rounding to the closest. i ti yi k1 yi+1 0 1 . . . (b) (7 pts) (IVP) is then solved on [1.25 , 1.75] using 2 steps of the discrete scheme of the 2nd order Runge Kutta method - Trape- zoid Rule (Heun’s method): (RK2.T). • Write the formulae of this scheme: (RK2.T)    ............................................................... ................................................................ yi+1 = .................................................... • (7 pts)Use 2 steps ONLY of this scheme to approximate y(1.50) and y(1.75) with 5 significant digits and rounding to 2 the closest.. i ti yi k1 k2 yi+1 1 1.25 . . . . 2 . . . . . (c) (5 pts) (IVP) is then solved on [1.75 ,2.00] using the 4th order discrete Runge Kutta scheme given by: (RK4)            k1 = f(ti, yi) k2 = f(ti + h 2, yi + h 2k1) k3 = f(ti + h 2, yi + h 2k2) k4 = f(ti + h, yi + hk3). yi+1 = yi + h 6(k1 + 2k2 + 2K3 + k4) Use 1 step ONLY of this scheme to approximate y(2.00) with 5 significant digits and rounding to the closest. i ti yi k1 k2 k3 k4 yi+1 3 1.75 . . . . (d) (1pt) The discrete solution of (IVP) is Y4 = {y0 = 0, ......................................................................................................} 3 2. (20 points) Consider the following set of data: Dn = {(xi, yi)|i = 0, ..., n where yi = f(xi) , and xi+1−xi = h, 0 < h ≤ 1} (a) (7 points) Write first the Central difference formula ψh ≡ ψh(f(xi)) that approximates the first derivative f ′(xi), ∀i = 1, 2, ..., n − 1. Assuming then that f is differentiable up to any or- der, derive the expression of the infinite error series ϵ(h) = c1hα1 + c2hα2 + c3hα3 + .... given that: f ′(xi) = ψh + ϵ(h), by determining the values of the exponents {α1, α2, α3, ...}. • ψh ≡ ψh(f(xi)) = • Derive the Error series ϵ(h) : ϵ(h) = 4 (b) (6 points) Based on the Central difference formula, derive Richard- son extrapolation operators of orders 1 and 2 and the order of their error series. • 1st order Richardson extrapolation operator ψ1 h(f(xi)) : ψ1 h ≡ ψ1 h(f(xi)) = Corresponding Error = O(...........) • 2nd order Richardson extrapolation operator ψ2 h(f(xi)): ψ2 h ≡ ψ2 h(f(xi)) = Corresponding Error = O(...........) 5 (c) (7 pts) For the purpose of approximating f ′(2.0), we consider the following table Dn = {(xi, yi)|i = 0, 1, ..., n} associated with a function f(x), i.e., yi = f(xi), i = 1, ..., n i xi yi = f(xi) 0 1.6 3.803929 × 10+01 1 1.7 4.745912 × 10+01 2 1.8 5.880257 × 10+01 3 1.9 7.240824 × 10+01 4 2.0 8.866867 × 10+01 5 2.1 1.080384 × 10+02 6 2.2 1.310432 × 10+02 7 2.3 1.582903 × 10+02 8 2.4 1.904805 × 10+02 Fill in the empty slots in the following table adequately, starting with h0 = 0.4. Express all the results obtained with 5 significant digits and rounding to the closest. h ψh(.) ψ1 h(.) ψ2 h(.) h0 = 0.4 XXXXX XXXXX h0/2 XXXXX h0/4 Best approximation to f ′(2.0) : 6 3. (20 points) Consider the function erf(x) = 2 √π � x 0 e−t2 dt Let I = erf(1). (1)-(5 pts) Determine the minimum number of partition points needed to approximate I by means of the Composite Simpson’s rule, if the absolute error in this approximation verifies: |ϵS| < 1 210−2 7 (2)-(5 pts) Approximate I by means of the Composite Simpson’s Rule using the number of equi-spaced partition points found in part (1) above . Express your answer with a precision p = 5 with rounding to the closest. 8 (3)- (5 pts)Assume h = b−a 2i , i = 0, 1, 2, .... Given that: I = S(h) + s1h4 + s2h6 + ..... + sjh2j+2 + .... Derive the first order Romberg extrapolation formula based on Simp- son’s rule rule. S1(h) = (4)- (5 pts) Apply this formula to the previous result obtained in (a-3) in order to improve the approximation of I . Do not compute the final answer: Write only the detailed formula that should be used with the adequate numerical values of h. 9 4. (25 points) Consider the following set of data Dn = {(xi, yi) for i = 0, ..., n|x0 < x1 < ... < xn and yi = f(xi)} where the nodes are not necessarily equally spaced. (a) (7 pts) Based on the set Dn, derive the general equations of the Non Natural Quadratic spline function S(x), that approxi- mate the function f(x). 10 (b) (6 pts) Use the results obtained in (a) to write the equations of the Non Natural Quadratic Spline S(x) that interpolates the set of data D2 = {(0, 1), (0.25, 4/5), (0.5, 2/3), (0.75, 4/7), (1, 1/2)} and fill in the following table: N.B.Express the results in fractions. i xi yi zi [xi, xi+1] 0 0.00 1 . . 1 0.25 4/5 . . 2 0.50 2/3 . . 3 0.75 4/7 . . 4 1.00 1/2 . . S(x) =                11 (c) (7 pts)Use the equations of the Quadratic Spline S(x) obtained in (b), to approximate the integral I = � 1 0 f(x)dx 12 (d) (5 pts)Let Ik = � xk+1 xk f(x)dx ≈ Qk = � xk+1 xk Sk(x)dx where Sk(x) is the equation of the Quadratic Spline on the interval [xk, xk+1], ∀ k = 0, 1, ..., n − 1. Let also I = n−1 � k=0 Ik ≈ Q(h) = n−1 � k=0 Qk Given that on each interval [xk, xk+1], Maxxk<x<xk+1|f(x) − Sk(x)| ≤ ckh3 prove that |I − Q(h)| ≤ n−1 � k=0 |Ik − Qk| ≤ C(b − a)h3 Hint: Prove first that |Ik − Qk| ≤ ch4 13 5. (15 points) We seek the approximation of r = a3/4, where 1 ≤ a < 2, by seeking the positive root of a polynomial function p(x). (a) (2 pts)For 1 ≤ a < 2, show that 1 ≤ r < 2. (b) (3 pts)Write Newton’s iterative formula: rn+1 = g(rn). Show then graphically that lim n→∞ rn = r, ∀r0 > 0. 14 (c) (5 pts)For r0 = 2, show that the resulting Newton’s sequence {rn} is decreasing. Hint Show that rn+1 − rn = 1 4r3n(a3 − r4 n), n ≥ 0 (d) (5 pts) Show that the convergence of rn to r is quadratic, by finding C such that: |rn+1 − r| ≤ C|r − rn|2 15 AMERICAN UNIVERSITY OF BEIRUT Mathematics Department-FAS MATH 251 FINAL EXAM Fall 2014-2015 Closed Book, 2hours STUDENT NAME ID NUMBER SECTION Problem Out of Grade 1 15 2 25 3 20 4 20 5 10 TOTAL 100 1 1. (15 points) Consider the following Initial value Problem: (IV P) � dy dt = 2y + et ; t ∈ [0, 1.00] y(0) = 2 (a) (5 points)(IVP) is first solved on [0 , 0.25] using 1 step of the dis- crete scheme of the 2nd order Runge Kutta method (RK2.M): Midpoint rule that is also referred to as Modified Euler method. • Write the formulae of this scheme: (RK2.M)    ................................................................ ................................................................ yi+1 = .................................................. • Use 1 step ONLY of this scheme to approximate y(0.25) with 5 significant digits and rounding to the closest. i ti yi k1 k2 1 0 . . . 2 . . XXX XXX (b) (7 points) (IVP) is then solved on [0.25 , 0.75] using 2 steps of the discrete scheme of the 2nd order Runge Kutta method (RK2.T): Trapezoid Rule also referred to as Heun’s method. • Write the formulae of this scheme: (RK2.T)    ............................................................... ................................................................ yi+1 = .................................................... • (7 pts)Use 2 steps ONLY of this scheme to approximate y(0.50) and y(0.75) with 5 significant digits and rounding to 2 the closest. i ti yi k1 k2 1 0.25 . . . 2 . . . . 3 . . XXX XXX (c) (3 points) (IVP) is then solved on [0.75 ,1.00] using one step of the 4th order discrete Runge Kutta scheme given by: (RK4)            k1 = f(ti, yi) k2 = f(ti + h 2, yi + h 2k1) k3 = f(ti + h 2, yi + h 2k2) k4 = f(ti + h, yi + hk3). yi+1 = yi + h 6(k1 + 2k2 + 2K3 + k4) Use 1 step ONLY of this scheme to approximate the value of y(1.00) with 5 significant digits and rounding to the closest. i ti yi k1 k2 k3 k4 4 0.75 . . . . . 5 . . XXX XXX XXX XXX 3 2. (25 points) (a) (7 points)Consider the definite integral: I = � 1 0 sin(x2) dx. Approximate I by the Composite trapezoidal Rule with 5 equally spaced partition points. Express the result with 5 significant digits and rounding to the closest. Find also an upper bound to the absolute value of the error in this approximation. N.B Angles are evaluated in radians. • I ≈ • Upper bound to the absolute Error: 4 (b) (8 points)Let I = � b a f(x) dx, with f ∈ Ck[a, b], ∀k > 0. Consider the following set of data: Dn = {(xi, yi)|i = 0, 1, ..., n = 2m, where a = x0, b = xn, and yi = f(xi)} where the x-coordinates are equally spaced with h = b−a n and n = 2i, i ∈ N. i. Write the Romberg extrapolation formulae of orders 1,2 then k > 2, based on the Composite Trapezoidal rule (Do not de- rive the formulae): • R(1)(h) = • R(2)(h) = ........ • k ≥ 2, R(k)(h) = ii. Prove that the 1st Romberg operator applied to the composite Trapezoid rule is equivalent to Simpson’s rule, that is prove that: R(1)(h) = S(h) = 2 3M(h) + 1 3T(2h). 5 (c) (10 points) In order to estimate the integral I = � b a f(x)dx using Romberg integration up to order k ≥ 1, given the MATLAB function that approximates I using the Composite Trapezoidal rule: function I =CompositeTrapez(f,a,b,h) % Input: function f of one variable % 2 real numbers a, b (a<b) % h positive number with n=(b-a)/h integer >=1 % Output: The composite trapezoidal rule approximation I n=(b - a)/h; sum=f(a)+f(b); for i=1:n-1 sum=sum+2*f(a+i*h); end I=sum*(h/2); Complete then on the next page the MATLAB function that ap- plies k iterations of the Romberg process based on the Composite Trapezoidal rule. The results are displayed in a lower triangular matrix R of size (k + 1) × (k + 2). (Call for the previous function if adequate) 6 function R = RombergCompositeTrapezoid(f,a, b, k) % Input: a function f, the limits of integration a and b and % k: the number of iterations % Output: a lower triangular matrix R of size (k+1) x (k+2). R = zeros(k+1,k+2) ; % The first column of R consists of all values of h, starting % with the largest value h=(b-a) and ending at h=(b-a)/2^k R(1,1)=b-a ; % The elements of the second column of R are all values of % T(h)-approximations of I using the Composite Trapezoidal rule % All elements of the remaining columns of R from the 3rd till % the last one are the Romberg approximations of all orders % Display the best approximation to I 7 3. (20 points) Let f be a differentiable function up to any order, i.e. f ∈ Ck ∀ k. (a) (5 points) The Central difference formula that approximates the THIRD order derivative f ′′′(x), is given by: ψh(f(x)) = 6[x − 2h, x − h, x + h, x + 2h] where [x−2h, x−h, x+h, x+2h] is a third order divided difference. Show that ψh(f(x)) = f(x+2h)−f(x−2h)−2[f(x+h)−f(x−h)] 2h3 8 (b) (5 points) Derive the expression of the infinite error series ϵ(h) = c1hα1 + c2hα2 + c3hα3 + .... given that: f ′′′(x) = ψh(f(x)) + ϵ(h), by determining the values of the exponents {α1, α2, α3, ...}. (c) (5 points) Let ψ(h) ≡ ψh(f(x)). Based on the Central difference formula, derive Richardson extrapolation operator of first order ψ(1)(h), and the order in h of its error. 9 (d) (5 points) For the purpose of approximating the third derivative, f ′′′(2), consider the following table D8 = {(xi, yi)|i = 0, 1, ..., 8} associated with a function f(x), i.e., yi = f(xi), i = 0, ..., 8. i xi yi = f(xi) 0 1.6 0.47000 1 1.7 0.53063 2 1.8 0.58779 3 1.9 0.64185 4 2.0 0.69315 5 2.1 0.74194 6 2.2 0.78846 7 2.3 0.83291 8 2.4 0.87547 Fill in the empty slots in the following table adequately, starting with h0 = 0.2. Express all the results obtained with 5 significant digits and rounding to the closest. h ψ(h) ψ(1)(h) h0 = 0.2 XXXXX 0.1 Best approximation to f ′′′(2) : 10 4. (20 points) Consider the following set of data Dn = {(xi, yi) for i = 0, ..., n|x0 < x1 < ... < xn and yi = f(xi)} where the nodes are not necessarily equally spaced, with: xi+1 − xi = hi+1 ∀i = 0, 1, ..., n − 1 (a) (10 points) Based on the set Dn, derive the general equations of the Natural Cubic spline function S(x), that approximate the function f(x). 11 (b) (10 points) Use the results obtained in (a) to write the equations of the Natural Cubic Spline S(x) that interpolates the set of data D3 = {(1 2, 0), (3 2, 1), (5 2, 0), (3, 1)} N.B. Express the computations in fractions and fill in the follow- ing table to simplify the presentation of your results.. i xi yi [., .] [., ., .] wi zi 0 1 2 0 . . ... 1 3 2 1 ... . . ... 2 5 2 0 ... . . ... 3 3 1 . . S(x) =                        12 5. (20 points) Let a ∈ R, a > 0. Based on Newton’s method, we seek an approximation of r = 1 a1/4 by seeking the positive root of a NON polynomial function f(x), that leads to an iterative formula with no division by the iterate. (a) (5 points) Determine such a function f(x) and plot its graph. 13 (b) (5 points) Write Newton’s iterative formula: rn+1 = g(rn) then determine the restrictions on the choice of the initial condition if there are any. Justify your answer. (c) (5 points) Show that for f ∈ C2, the iterates of Newton’s sequence verify the following relation: rn+1 − r = 1 2 f”(cn) f ′(rn) (r − rn)2, ∀ n ≥ 0, where cn is between r and rn. 14 ... (d) (5 points) In what follows, let a = 2 and let also the initial condi- tion r0 = 0.5 < r. Check graphically that: 0 < rn < rn+1 < r then use these estimates to show that: |r − rn+1| ≤ 5|r − rn|2 15 1 AMERICAN UNIVERSITY OF BEIRUT Faculty of Arts and Sciences ‐ Mathematics Department. MATH 251 Final Exam (2hrs) SPRING 2007 – 2008 STUDENT NAME . ID NUMBER . Problem 1 / 19 . Problem 2 / 22 . Problem 3 / 15 . Problem 4 / 27 . Problem 5 / 17 . TOTAL / 100 2 1 . (a) ‐ Loss of significant figures may result in the computation of the functions for certain values of x. Explain and specify these values then propose alternative functions that would remedy the loss of significant figures. Note that : (10 points) (b) ‐ Let f(x) = g(x) – h(x). Using the results in (a), find a way to calculate accurately f(x) for values of x close to zero. Find then . (3 points) 3 (c) –Solving the quadratic equation , using a machine that carries only 8 decimal digits causes a problem. Investigate this example, observe the difficulty, and propose a remedy to approximate the 2 solutions . (6 points) 2 . The Reciprocal of the cubic root of a positive number (i.e. ), can be computed by an iterative formula that does not use division by the iterate i – Establish this formula by applying Newton’s method to some appropriate function f(x). Determine also any necessary restrictions on the choice of the initial value of the iterative procedure. (9 points) 4 ii – Let a = 2. Plot the graph of the function f, and give a graphic justification of the necessity of this restriction on . (4 points) iii – For a suitable choice of , approximate by computing the first 3 iterates by Newton’s formula. (5 points) 5 iv – Prove that Newton’s method is quadratic. (4 points) 3. Based on a set of data , where , and nh=2, let i. Give the formula for the composite Trapezoid Rule T(h) to approximate the integral I, and derive the expression of I – T(h) in terms of the powers of h. (8 points) 6 ii. Obtain an upper bound on the absolute error when we compute by means of the composite Trapezoid Rule using 51 equally spaced points. (7 points) 4. Based on the following table of values for a function f(x) i x(i) f(x(i)) 0 0.000 1.0000000 1 0.125 0.9844964 2 0.250 0.9394131 3 0.375 0.8688151 4 0.500 0.7788008 5 0.625 0.6766338 6 0.750 0.5697828 7 0.875 0.4650432 8 1.000 0.3678794 7 Consider the approximation the integral , using Romberg integration based on the composite trapezoid rule T(h). i. Derive Romberg subsequent integration formulae, , and obtain expressions of the errors , and . (12 points) 8 ii. Prove that the Simpson’s composite rule S(h) is equivalent to R1 (h). (5 points) iii. Using the table and the results above, fill the empty slots in the following table. (10 points) h M(h) T(h) R1(h) R2(h) R3(h) 1 0.5 0.25 0.125 9 10 5. (a) – Fill in the missing statements in the following Matlab program (8 points) function p = Neville(X, Y, x) % Input arguments: a vector X representing the distinct xi data % : a vector Y representing the corresponding yi = f(xi )data % : x is a vector of numbers at which we want to interpolate % Output arguments: p = p(x) the value of the interpolation polynomial at the X data % Assume length (X) = length (Y) > 0 % Place the base statement of the recursion n = length(X); if n = 1 ………………………………………………………………………………………………………………………. else % Place the statements of the recursion X1 = X(1 : n­1) ; Y1 = Y(1 : n­1); X2 = X(2 : n) ; Y2 = Y(2 : n); p1 = ………………………………………………………………………………………………………………… ; p2 = ………………………………………………………………………………………………………………… ; p = ………………………………………………………………………………………………………………….; end 11 (b) – Show the various steps of the execution process of this program, when n = 3 , in the form of a diagram: (9 points) p1 = ………. p2 = ………. 12 AMERICAN UNIVERSITY OF BEIRUT Mathematics Department-FAS MATH 251 FINAL EXAM SPRING 2010-2011 Closed Book, 2hours STUDENT NAME ID NUMBER Problem Out of Grade 1 15 2 20 3 20 4 15 5 15 6 15 TOTAL 100 1 1. (15 points) For which values of x is there a difficulty in computing f(x) ? What is the problem ? What remedy do you suggest ? • (7 points)f(x) = x + √ x2 + 4 • (8 points)f(x) = tan(x) − 1 (N.B. Determine only the smallest positive value of x that causes a problem.) 2 2. (20 points) Consider the function f(x) = x2 − a where 1 < a ≤ 2. (a) (6 points)Write Newton’s formula: rn = g(rn−1) that computes the positive root of the function f. Is there any restriction on the initial condition r0? Justify your answer. (b) (2 points)Find the minimum number of flops necessary in com- puting one iteration rn = g(rn−1). 3 (c) (6 points)Prove that: (rn − r) = 1 2rn−1 (rn−1 − r)2 then show that for n ≥ 1: |rn − r| = 1 2(rn−1 − r)2 < 1 2n(r0 − r)2n (d) (6 points)Let r0 = 3/2. Find an estimate on the least value of n for which: |rn − r| < 2−50 Hint: Using the result in (c), show first that |rn − r| < 2(−2n) 4