Homework Answers
For part a the MATLAB code is:
clc
m=100;
series1=0;
for i=1:m
p=(((-1)^(i+1))*((pi/2)^(2*i-1)))/factorial(2*i-1);
series1=series1+p;
end
series1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
Ans of sin(pi/2) for m=100 is =1.000
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
For part b the MATLAB code is:
clc
m=5;
series1=0;
for i=1:m
p=(((-1)^(i+1))*((pi)^(2*i-1)))/factorial(2*i-1);
series1=series1+p;
end
series1
%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%%
ans of sin(pi) for m=5 is 0.0069.
ans of sin(pi) for m=100 is 3.3281e-16 which is approximately equal to zero.
So, m=100 is close to the exact value.
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