Infinite Loop help Riemann sums
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For some reason it's get stuck in an infinite loop. The programs says line 8 is the problem so the "first = first" line so I'm assuming its the bound problem? I don't see the problem any suggestions?
Note: my functions needs to give me an approx that has an difference from the previous estimate less than the tol that's why I'm running a while and for loop.
3 个评论
Walter Roberson
2013-2-24
I suggest you step through it line by line with the debugger.
Azzi Abdelmalek
2013-2-24
It's difficult to say if we don not know a,b,tol and your function f.
Kenny
2013-2-24
采纳的回答
Youssef Khmou
2013-2-24
编辑：Youssef Khmou
2013-2-24
hi Kenny, your program works fine for most of cases
Note : no need to declare syms x ,
I tried to to make some change according to your model , like number of samples n, no need to increase through the loop, the theorem states that the Sum converges when N->Inf :
function y = myintegralapprox(f,a,b,tol)
tol=100*tol;
n = 1000;
c = tol + 1;
first = 0;
second = 0;
counter=1;
while (c > tol)
for m=a+(((b-a)/n)/2):((b-a)/n):(b-((b-a)/n)/2)
first = first + ((b-a)/n)*(f(m));
end
counter=counter+1;
if counter>1e+6
break;
end
y = first;
c = abs(first - second);
second = first;
end
I made some tests comparing you func with "quad" function :
>>myintegralapprox(@(x) 1/(1+exp(x)),0,pi,.1)
>>quad(@(x) 1./(1+exp(x)),0,pi)
>>myintegralapprox(@(x) cos(x),0,pi,.1)
>>quad(@(x) cos(x),0,pi)
>>myintegralapprox(@(x) exp(-x^2),0,pi,.1)
>>quad(@(x) exp(-x.^2),0,pi,.1)
>>myintegralapprox(@(x) (x^3)+(x^2)+1,0,pi,1)
>>quad(@(x) (x.^3)+(x.^2)+1,0,pi)
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