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Test2A solutions

# Test2A solutions - TEST#2A PGE 310 Name Solutions Signature...

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Unformatted text preview: TEST #2A (November 13, 2008) PGE 310 Name: Solutions Signature: UT EID: NOTES: 1. You have 1 hour and 15 minutes. Points will be deducted for tests turned in after 12:15 2. Write your name on all pages of the test 3. SHOW ALL WORK. Partial (or any credit) cannot be given for problems without work shown . Qualitative Questions (30 points) (a) I want to solve some nonlinear equation f(x) = 0 using x=1 and x=2 as guess values. If the ﬁrst 3 iterations of the Bisection and False-Position Method are given below, what would I get for “x” in the FIRST iteration of the Secant method? Iteration x_new (Bisection) x_new (False Position) 1 1.5 1.6667 2 1.75 1.72727 3 1 .875 1 .73 17 1.6667: The ﬁrst iteration of False Position is the same as the ﬁrst iteration of Secant (b) Consider the non-linear function below; f(x)=0. Using an initial guess of X = -0.5, On the curve, show GRAPHICALLY the ﬁrst step of the Newton- Raphson method. X (c) I have to solve a system of equations Ax=b several times (maybe hundreds). The matrix A never changes, but the vector b changes every time. Should I use Gauss Elimination or LU decomposition? Brieﬂy explain (and be speciﬁc) LU decomposition. If you have already created L and U in LU decomposition, you can backward substitute right away for d and then x. If you use Gauss Elimination, you will have to create an upper triangular matrix many times and thus increasing your computation (by a very large amount if you have a large A! l). ((1) Consider the following system of equations. As written, is convergence guaranteed for an iterative method like J acobi/Gauss Seidel. If not, rewrite the equations so that convergence is guaranteed. § § x1+3x2——6x3=4 )Il/ Bl 41,; _4x1+2x2+x3:6 1+ my? l : x1‘5x3 . . . of the absolute values of the residuals as opposed to the“sum of the squares” of the residuals. By summing the absolute value of the errors, you may have multiple solutions for a data set. (i) Experimental data is gathered for the growth of certain bacteria. A theoretical model for bacteria grth is given by the equation below. Rewrite in such a way that linear least squares regression can be used. _ (l/z‘)t P — Poe . ln(P) = ln(PO) + (l/r)t where ln(Po) = intercept = a0 (1/1) = slope = a1 2. Solve using 2 iterations of the secant method and guesses x1=1 and x; = 2 (15 points) f(x)=2x3+x'1/2—-4=0 f‘fzmﬂoa 1: 15cm: M + F; J1:- 0.707 «CW: 96 Ni :4 Ya: ;_ ‘9»*707<3~“U 9.70%”! 23A 9:“ \X: 3+; [email protected]}3’,gé"/ .ﬂy’zﬁ" MW W533» 15W — 12.707 Seam/ﬂ Meﬁxcxli ' Xm: X‘: ‘3“ (XIX Xk~f> ‘HX’J ~ \$(xkﬂ) ‘W.WMWW.W_V“-W.__W_a w._m._wwwwwmw...M.u‘.m.ﬂm.wm._v......_.m.-,.m “we. 3. Solve using Gauss Elimination. SHOW ALL WORK (15 points) x2 + 3x3 = 10 x1+2x2 +4x3 =16 2xl + 3x2 = 7 \ ; Limo 9L4; \ ; W/é OI 3 IO :7 013110 O! 8&5 0 O”5\"\5 4. Solve using ONE iteration of the multidimensional Newton-Raphson method. Use initial guesses of x = y = 1 (15 points) 2x+21n(y) = —3 5. Write a MATLAB function ﬁle call PolynomialFitm that went sent input vectors x and y of data, the coefﬁcients of a 2nd order polynomial that best ﬁts the data, a0, a1, and a2, are returned Assume that there already exists another function ﬁle in your directory called “GaussElimm” that when sent a matrix A and vector b, the solution to the system of equations, x, is returned. Do not use built-in functions max, average, sum, or MATLAB’s backslash operator. (25 points) function [a0,al,a2] = PolynomialFit(x,y) sumx = 0; sumxy = 0; sumx2 = O; sumx3 = O; sumx4 = 0; sumy = O; sumx2y = 0; n = length(x); for i = 1:11 sumx = sumx + x(i);_ sumx2 = sumx2 + (x(i))"2; sumx3 = sumx3 + (x(i))"3; sumx4 = sumx4 + (x(i))"4; l sumy = sumy + y(i); sumxy = sumxy + x(i)*y(i); sumx2y = sumx2y + (x(i)’\2)*y(i); end A(l,1) = n; A(1,2) = sumx;'A(l,3) = sumx2; A(2,l) = sumx; A(2,2) = sumx2; A(2,3) = sumx3; A(3,1) = sumx2; A(3,2) = sumx3; A(3,3) = sumx4; l' b(1,l) = sumy; b(2,l) = sumxy; b(3,1) = sumx2y; x = GaussElim(A,b) E a0 = x(l); l a1 = x(2); a2 = x(3); ...
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