Bisection method relative error
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Hello everyone, I don't use MATLAB very well. I have a question. If you can help, I'd appreciate.I have a function below that I have to find its roots using bisection method. I want the for loop to stop on the point where relative error is lower than %0.05. I couldn't understand how I can define n.
f=@(x) log(x)-cos(x)-exp(-x);
x1=1;
x2=2;
xmid=(x1+x2)/2
for i=1:n;
if (f(xmid)*f(x2))<0
x1=xmid;
else
x2=xmid;
end
xmid=(x1+x2)/2;
end
fprintf('The root is: %3.8g\n',xmid)
采纳的回答
Mohammed Hamaidi
2022-3-17
编辑：Mohammed Hamaidi
2022-3-18
Hi
Just use "while" loop with your condition as follows:
f=@(x) log(x)-cos(x)-exp(-x);
x1=1;
x2=2;
xmid=(x1+x2)/2;
while (x2-x1)>0.0005
if (f(xmid)*f(x2))<0
x1=xmid;
else
x2=xmid;
end
xmid=(x1+x2)/2;
end
fprintf('The root is: %3.8g\n',xmid)
4 个评论
Sazcl
2022-3-17
Thanks for your time.
But as far as I know, error tolerance is calculated by (upper limit-lower limit)/(2^iteration number) when the exact root is not given in the question.
Actually your code gives the right answer but I don't think it's what the question asks.
Jan
2022-3-17
@Sazcl: If you do have the mathematical definition of "relative error", it should be easy to insert it in the posted code. Simply use it as condition in the WHILE command.
To avoid an infinite loop, add a counter, which stops the loop after a certain limit, e.g.:
kMax = 1e6;
k = 0;
while <InsertYourCondition> && k < kMax
...
k = k + 1;
end
if k == kMax
error('No convergence')
end
But think twice: Under which circumstances is this possible?
By the way, f(xmid)*f(x2) < 0 does not catch the cases in which xmid or x2 is exactly the root. What a pity, if the root way found and the iteration goes on.
Sazcl
2022-3-17
Jan
2022-3-18
If this answer solves the problem, please accept it.
更多回答（1 个）
John
2023-7-31
0 个投票
function [p, pN] = Bisection_371(a,b,N, tol)
if f(a)*f(b) > 0
disp("IVT does not guarantee a root in [a,b]")
elseif f(a)*f(b) == 0
disp("The root is either a or b")
else
for n = 1:N
p = (a+b)/2;
pN(n) = p;
if f(p) == 0 || (b-a)/2 < tol
break
elseif f(p)*f(a) < 0
b = p;
else
a = p;
end
end
end
end
%f = @(x)x^2 - 1;
function y = f(x)
y = x^2 - 1;
end
1 个评论
Jan
2023-8-2
For numerical reasons it is rather unlikely that the condiotion f(p) == 0 is met exactly. Use a tolerance instead.
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