## How do I count how many iterations my root is in a smaller interval?

Matthew Worker

### Matthew Worker (view profile)

さんによって質問されました 2017 年 11 月 11 日

### Rena Berman (view profile)

さんによって 編集されました 2018 年 3 月 21 日
Mischa Kim

### Mischa Kim (view profile)

さんの 回答が採用されました
In my code I need to display in the command line how many of the iterations my root was in the smaller interval. I am not entirely sure what this means, but I'm hoping someone does and can give me a hand to work out what I'm missing. At the moment the command line displays the same number of iterations for both the iterations to the root and how many iterations the root was in the smaller interval.
Code removed. Question has been answered.

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## 1 件の回答

### Mischa Kim (view profile)

2017 年 11 月 11 日
採用された回答

In each iteration you have two intervals, [a,c] and [c,b]. Compute the size of both intervals and use the last if-else condition in your code to increment itc.

Mischa Kim

### Mischa Kim (view profile)

2017 年 11 月 11 日
I used
f = @(x) sqrt(x) - cos(x);
[a,b,it,itc] = WB(f,0,15,0.25,1e-5)
function [a,b,it,itc] = WB(f,a,b,w,tol)
it=0;
itc=0;
fa=f(a);
fb=f(b);
if fa*fb>0
error('Not guaranteed a root in interval [a,b]');
end
if abs(fa)<abs(fb)
fprintf('The root is probably closer to a than b\n');
else
abs(fa)>abs(fb);
fprintf('The root is probably closer to b than a\n');
end
while (b-a)/2>tol
c = (1-w)*a+w*b;
% fc = f(c); % not really needed
% if fc == 0
% return
% end
x = abs(a-c);
y = abs(c-b);
if f(a)*f(c)<0
b = c;
if x<y
itc = itc + 1;
end
else
a = c;
if y<x
itc = itc + 1;
end
end
it = it+1;
end
fprintf('Final interval: [%f, %f] with %d iterations and %d iterations where the root was in the smaller interval\n',a,b,it,itc)
end

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