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clc
clear
a=0
b=2
function p=fA(c,n)
p=c*cos((n*c*%pi)/2)
endfunction
function q=fB(c,n)
q=c*sin((n*c*%pi)/2)
endfunction
for n=1:5
for t=20:180:200
a=0
ST=0
ST1=0
h=(b-a)/t
for i=1:t+1
A(1,i)=a
a=a+h
end
for i=2:t
ST=ST+2*fA(A(1,i),n)
ST1=ST1+2*fB(A(1,i),n)
end
TZA(t,n)=(h/2)*(fA((0),n)+fA((2),n)+ST)
TZB(t,n)=(h/2)*(fB((0),n)+fB((2),n)+ST1)
end
end
for t=20:180:200
for n=1:5
a=0
ST1=0
ST2=0
ST3=0
ST4=0
h=(b-a)/t
for i=1:t+1
A(1,i)=a
a=a+h
end
for i=2:2:t-2
ST1=ST1+2*fA(A(1,i+1),n)
ST2=ST2+4*fA(A(1,i),n)
ST3=ST3+2*fB(A(1,i+1),n)
ST4=ST4+4*fB(A(1,i),n)
end
ST2=ST2+4*fA(A(1,t),n)
ST4=ST4+4*fB(A(1,t),n)
TSA3(t,n)=(h/3)*(fA(0,n)+fA(2,n)+ST1+ST2)
TSB3(t,n)=(h/3)*(fB(0,n)+fB(2,n)+ST3+ST4)
end
end
for t=20:180:200
printf('Comparison of numerical integration of %f subdivisions of [0 2]',t)
for n=1:5
T=[n,TZA(t,n),TZB(t,n),TSA3(t,n),TSB3(t,n)]
disp(T)
end
end
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