Solution1

Unset MECE 390 Assignment 1 Question 1 – Part I. Show the Taylor Series Approximation of orders 1, 2, and 3 to ? =6 - 4/ ? . This is to be done manually (i.e. do NOT perform the differentiation numerically) and submitted as part of your report. NOTE: Don't use the decimal format of numbers, use fractions and whole numbers instead. Your answers should contain the lowercase terms ? and 𝑎 . P 1 (x) = (4*(x-a))/a^2-4/a+6 P 2 (x) = -((4*(x-a)^2)/a^3)+(4*(x-a))/a^2-4/a+6 P 3 (x) = (4*(x-a)^3)/a^4-(4*(x-a)^2)/a^3+(4*(x-a))/a^2-4/a+6 Store each approximation as a separate function in MATLAB. You should have four different functions that you can recall for the next questions in the lab, one for the primary function and one for each of the three approximations. Question 1 – Part II. Write a MATLAB code (M-file) that calculates each of the above approximations about 𝑎 =1.0 at the values ? =0.85 and ? =1.15. Plot the results along with the exact function ? ( ? ). Plot the actual function in a solid line, and the approximations asked for below as single points using a marker (i.e. 'o'). You may plot the lines for the approximations, but they must appear as dashed or dotted lines, and a legend must be included. function taylor_Approx() %output=funcname(inputs) % in this section put some some text describing the matlab function % Prepared by :NAME ID:## Date:## % Constants lb=0.85; %Beg. of interval ub=1.15; %End of interval a = 1; % Exact solution f_exact=@(x) 6-4./x; % given function (for example only): 5*x^2+2*exp(2*x) % Taylor Series Approximation p1= @(x)(4.*(x-a))./a^2-4/a+6; %First order approx. p2= @(x) -((4.*(x-a).^2)./a^3)+(4.*(x-a))./a^2-4/a+6; %Second order p3= @(x)(4.*(x-a).^3)./a^4-(4.*(x-a).^2)./a^3+(4.*(x-a))/a^2-4/a+6; %3rd order % Plotting hold on fplot(f_exact,[lb,ub]) fplot(p1,[lb,ub]) fplot(p2,[lb,ub])

fplot(p3,[lb,ub]) grid on xlabel('x') ylabel('f(x)') legend('Exact','First Order','Second Order','Third Order') title('Taylor Expansion') % ERROR Calculation and Plot E1 = @(x)(f_exact(x)-p1(x));% True (absolute) error E2 = @(x)(f_exact(x)-p2(x));% True (absolute) error E3 = @(x)(f_exact(x)-p3(x));% True (absolute) error figure; hold on fplot(E1,[lb,ub]) fplot(E2,[lb,ub]) fplot(E3,[lb,ub]) xlabel('x') ylabel('True Error') legend('E1','E2','E3') title('Error Comparison') grid on end Using your code, answer the following questions. NOTE: Insert at least three digits of numbers, e.g. 0.00123 for 0.0012345. ? ( ? =0.85) = 1.29412 𝑃 1 ( ? =0.85) = 1.4 𝑃 2 ( ? =0.85) = 1.31 𝑃 3 ( ? =0.85) = 1.2965 ? ( ? =1.15) = 2.52174 𝑃 1 ( ? =1.15) = 2.6 𝑃 2 ( ? =1.15) = 2.51 𝑃 3 ( ? =1.15) = 2.5235 The red curve in figure 1 is the plot of …! Explain your reasoning. a) f b) P 1 c) P 2 d) P 3

Figure 2 Question 1 – Part IV. Comment on the Following - How does the error change with the order of the approximation? Is this behavior expected? o By increasing the order of approximation, the error reduces, and that can be shown by writing the error estimation formula the order of approximation converges to infinity the error converges to zero - For a given order of approximation, how does the error change with respect to the base point 𝑎 ? o Error increases as we get farther away from the base point the explanation can be shown based on the theory of the taylor series

Question 2 The damped vibration response of a shock absorber in a vehicle can be represented by the following equation: ? = 𝐴(? −? 1 ? + ? −? 2 ? ) where ? 1 , ? 2 are found to be -1.58 ? −1 and -6.32 ? −1 respectively, and 𝐴 = 4.21 cm. Discuss the approach to using the Taylor series approximation of to determine the mass position at ? = 0.5 seconds. You ? ? = 1 + ? 1! + ? 2 2! + ? 3 3! + … must discuss how the method (i.e. Taylor Series Approximation) can be implemented as well as the sources of error in your estimation. Note: you do NOT need to provide a numerical response to the question in your report, and even if you do there is no grade assigned to this value. points that should be discussed: - substitute e^(-r1*t) with 1+(r1*t)/1! + (r1*t)^2/2! + … - same goes for e^(-r2*t) - a for loop in Matlab could solve for y - sources of error are the inherent error that comes with Taylor series approximation(truncation error) and rounding error that happens in Matlab calculations