Use MATLAB to show that the sum of the infinite series 1/(2n + 1)(2n + 2) conver
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Use MATLAB to show that the sum of the infinite series 1/(2n + 1)(2n + 2) converges to ln2. Do this for computing the sum for n = 50 n = 500 n = 5000 For each ease compute tilt percent error using the equation: Percent error = Exact Value - Approximate Value Exact Value/ Exact Value times 100% Hint: You need to create a vector n ranging between 0 and 50.500 and 5000. then use element by element calculations, and a built - in function to the get the approximations 3) Given the following simultaneous equations: 4r - 5t + 6u - 8y = - 13 7s - 3x + 5y + 2z = 6 2r - 3s - 6u - 5v + 9y = - 12 6t + 2u + 3v + 4x + 5y + z = 0 3r + 6s + 8t + 5u + 2v + 3x + 4y + 4z = 2 v - x - y - z = 8 5r + 6t + 4v - 6y = 6 4s - 2u - 6x + 7z = 17 Using MATLAB. solve the simultaneous equations three ways by: Using the inverse computed by using exponentiation Using the inverse computed by using a built - in function Using matrix division (left and right) In your report, verify that solutions found are correct by substituting the solutions into each equation and verifying that the both sides of each equation arc equal.Explanation / Answer
clc n=[0:50]; %sum(1./((2*n+1).*(2*n+2))) we can also use this and avoid defining a new %function below funct=1./((2*n+1).*(2*n+2)); sumofinf=sum(funct) percenterror50=(log(2)-sumofinf)/log(2) fprintf('The sum of the infinite series with n=50 is %f ', sumofinf) fprintf('The percent error is %f ', percenterror50) %I'm not sure whether you need to write it out nicely, but if you do, just %copy/paste the above w/ the different variables below %repeat for n=500 and 1000 %note we can also use a for loop i.e for n=[0:x]; for x=[50 500 1000]... n=[0:500]; %sum(1./((2*n+1).*(2*n+2))) we can also use this and avoid defining a new %function below funct=1./((2*n+1).*(2*n+2)); sumofinf=sum(funct) percenterror500=(log(2)-sumofinf)/log(2) n=[0:1000]; %sum(1./((2*n+1).*(2*n+2))) we can also use this and avoid defining a new %function below funct=1./((2*n+1).*(2*n+2)); sumofinf=sum(funct) percenterror1000=(log(2)-sumofinf)/log(2) %part 3 %first write out matrix, r seems to be the smallest alphabet A=[4 0 -5 6 0 0 -8 0; 0 7 0 0 0 -3 5 2; 2 -3 0 -6 -5 0 9 0; 0 0 6 2 3 4 5 1; 3 6 8 5 2 3 4 4; 0 0 0 0 1 -1 -1 -1; 5 0 6 0 4 0 -6 0; 0 4 0 -2 0 -6 0 7]; b=[-13; 6; -12; 0; 2; 8; 6; 17]; %a: inverse using exponentiation: solna=(A^-1)*b %note you can make it look better by using fprintf or disp as done before %b: using built in function: solnb=inv(A)*b %c: using matrix division: solnc=A %verifying the answers hold true: %Sub into the eqn: bof1shouldbe=A(1,:)*solnb %You can do all of the above, or just use one eq. to give all the answers: bshouldbe=A*solnb %You can rewrite them better if required
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