Question

4.2 The Maclaurin series expansion for cos x is 6 .8 Starting with the simplest version, cos -I, add terms one at a time to estimate cos(π/3). After each new term is added, compute the true and approximate percent relative errors. Use your pocket calculator to determine the true value. Add terms until the absolute value of the approximate error estimate falls below an error crite- rion conforming to two significant figures.

Answer 1

Solution Code: alize the epsilon value %initialize the x value that is pi/3 x = pi/3; %assign the approximation value is zero app 0,; %set the term, n= 1 n-1; %assign the true is cos (x) t cos (x); %print the statement 1NIt TrueltltApproximationltAbs_Error tApprox_Error\n') %check condition in while while(1) %assign old approximation is in app aapp find the approximation value appappt (-1) 1 (n-1) *x* (2*n-2)/factorial (2*n-2) %increment the term n=n+1; %formula for find the abs error ae = abs ( (t - app) /t)*100; %formula for find the approx. error ap-e r r= abs( (app- appold) /app) *100; %print the given result in table format fprintf. ' %4d\t%8 .4f\t% 8.4f\ t%8.4f\ t%8.4f\n', n- 1,t,app, ae, ap_err)

%check if appro. error is <-epsilon value %then it break If ap err <= epsilon, break, end end Sample Output: $octave -qf no-window-system demo.m Approximation Abs_Error True 0.5000 0.5000 0.5000 0.5000 0.5000 Approx_Error 1.0000 0.4517 0.5018 0.5000 0.5000 warning: function ./demo.m shadows a core library function 100.0000 9.6623 0.3592 0.0071 0.0001 100.0000 121.3914 9.9856 0.3664 0.0072 4

copyable Code:

%initialize the epsilon value

epsilon=0.5*0.1;

%initialize the x value that is pi/3

x = pi/3;

%assign the approximation value is zero

app= 0;

%set the term, n = 1

n = 1;

%assign the true is cos (x)

t = cos(x);

%print the statement

fprintf('   N True Approximation Abs_Error Approx_Error ')

%check condition in while

while(1)

    %assign old approximation is in app

    appOld= app;

    %find the approximation value

    app = app+(-1)^(n-1)*x^(2*n-2)/factorial(2*n-2);

    %increment the term

    n = n+1;

    %formula for find the abs error

    ae = abs((t - app)/t)*100;

    %formula for find the approx. error

    ap_err = abs((app - appOld)/app)*100;

    %print the given result in table format

    fprintf('%4d %8.4f %8.4f %8.4f %8.4f ',n-1,t,app,ae,ap_err);

    %check if appro.error is <= epsilon value

    %then it break

    if ap_err <= epsilon,

        break,

    end

end