Question

Instructions: Submit your script in a file named hwk08.m to the dropbox before 11:59 pm on the due date NOTE: This assignment is neither quick nor simple. You will be well served to start on it early, and to ask for help if you need it. Being a more substantial assignment than earlier hwk, it is worth- s. When you ask your calculator for the value of a function for a specified argument, (e.g., sin(22), cos(74), In(6.5)), it almost certainly uses a series expansion of that function to compute the value. This assignment is interested in studying the exponential function for which the series expansion can be expessed as where: n! is the factorial function, 1!-1, 2!-2-1, 3!-3-2-1 special case, o!-1 The equality holds for an infinite number of terms. For a finite number of terms, the value of the series is an approximation for the value of e* for which the relative error is approxValue- trueValue trueValue n! realtive Error Craft a script that: Prompts the user for a 2-element vector named myData. The 1st element specifies the x value. The 2nd element specifies the relative error convergence tolerance Determines how many terms of the series must be included to compute a series-value that satisfies the specified convergence tolerance for the relative error Display to the CW, in an easily understood format; - in the 1st line, the value of x and the value of e* computed using the MATLAB function in the 2nd line, the value computed by the series computation, the relative error in that value, and the number of terms required to satisfy the convergence criterion Computing the relative error is a problem for very small values of e because it requires division by a number close zero. Include in the script a safety check that immediately stops the computations if e< 1x10-6 and displays in the CW an appropriate message In a single figure window displays - the value of e* for-3 sxs 3, computed by the MATLAB function - the value of ex for-3 sxs 3, computed using 1 term, 2 terms, 3 terms, and 4 terms, " of the series approximation - all 5 plots should be side-by-side, starting with the plot for the MATLAB function at the left, with successive plots showing values for successively higher number of terms in the series, until the final plot shows the values when 4 terms are included

Answer 1

###########Programming code############

myData = input("\Enter x and Convergence tolerance in vector form [x,tol]: ");
x = myData(1);
tol = myData(2);
if exp(x) < 10^-6
error("The value is very small, terminating program")
end
% Zeroth term
n = 0; % Term number
result = (x^n)/factorial(n);
rel_Error = (result - exp(x))/(exp(x));

while abs(rel_Error) > tol % Repeating until convergence
n = n+1; % Incrementing terms counter
result = result + (x^n)/factorial(n);
rel_Error = (result - exp(x))/(exp(x));
end
fprintf("\nUsing in-built function, exp(%.3f) = %.6f",x,exp(x))
fprintf("\nUsing Series computation, exp(%.3f) = %.6f, relative error = %.6f, Number of terms:%d\n",x,result,abs(rel_Error),n);
% Plots
subplot(1,5,1)
xp = linspace(-3,3);
plot(xp,exp(xp))
title("e^x (Matlab Function)")

ex_1 = zeros(1,length(xp)); % Initializing all for faster execution
ex_2 = zeros(1,length(xp));
ex_3 = zeros(1,length(xp));
ex_4 = zeros(1,length(xp));

for k = 1:length(xp) % calculating result for all xp
ex_1(k) = 1 + (xp(k)^1); % 1 term
ex_2(k) = ex_1(k) + (xp(k)^2)/ factorial(2); % 2 term
ex_3(k) = ex_2(k) + (xp(k)^3)/ factorial(3); % 3 term
ex_4(k) = ex_3(k) + (xp(k)^4)/ factorial(4); % 4 term
end
subplot(1,5,2)
plot(xp,ex_1) % Plotting 1 term approximation
title("1 term")
subplot(1,5,3)
plot(xp,ex_2) % 2 terms
title("2 terms")
subplot(1,5,4)
plot(xp,ex_3) % 3 terms
title("3 terms")
subplot(1,5,5)
plot(xp,ex_4) % 4 terms
title("4 terms")

Command Window:
Command Window Enter x and Convergence tolerance in vector form [x, tol] [2 10A-5] Using in-built function, exp (2 . 000) = 7 . 389056 Using Series computation, exp (2.000) = 7.388995, relative error = 0.000008, Number of terms : 10

3 terms 4 terms 1 term 2 terms 18 e* (Matlab Function) 25 16 12 20 b 10 12 15 10 10 4 2 0 24 4 2 0 24 4 2 024