untitled.m

基本的迭代求和办法

for m=3:15;
for j=2:m-1
 a(j-1)=((j-1)*cos((pi/6)-((j-1)*pi)/(6*(m-1))));
 
end
sum=a(j-1);
j=1;
while j<m-2
    j=j+1;
    sum=sum+a(j-1);

end
sum

a=(m-1)/2+sum;
b=a^2;
c=27*128*(sin(pi/(12*(m-1)))^2);
d1=(pi^2)*[(4+sqrt(3))*((m-1)^2)];
d2=1+(11-6*sqrt(3))/(13*(m-1)^2);
d=d1*d2;
THD=sqrt(d/(c*b)-1)
end
%THD