While loop isn't working
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Hello,
My function is as follows:
function root = NM(f, f1, x_0, epsilon, iteration)
x = zeros(1, 50);
x(1) = x_0;
count = 0;
while abs(f(x)) < epsilon
count = count + 1;
if count > iteration
break
end
x = x - f(x)/f1(x);
end
root = x;
end
Here's how I tried to test my function:
clear
syms t
x_0 = 2*pi/3;
epsilon = 0.0001;
iteration = 30;
v = @(t) sin(t);
vprime = matlabFunction(diff(v(t)));
root1 = NM(v, vprime, x_0, epsilon, iteration)
The loop does not seem to work, as the result I got was the initial value 2pi/3. What have I done wrong?
4 个评论
dpb
2020-8-23
I see no reference to function NM anywhere in the test script??? You call something named NewtonsMethod instead with at least one undefined argument f.
As for the while construct presuming you do somehow have a way in which you did call NM, use the debugger and step through the function looking carefully at the logical test after also reading very carefully the definition of TRUE for if and while logical tests in MATLAB.
StillANovice
2020-8-23
Adam Danz
2020-8-23
You could probably see what's wrong by investigating the variables in debug mode and it will probably be much faster than the time it will take to explain things to use and wait for a response.
采纳的回答
TADA
2020-8-23
You have two mistakes
1. while loop should work with a scalar condition, but as far as I know if you use a logical vector it will only enter the loop if all values are true:
% endless loop - as expected
while true
disp(rand(1));
end
% never enters loop - as expected
while false
disp(rand(1));
end
% endless loop
while [true true]
disp(rand(1));
end
% never enters loop
while [false true]
disp(rand(1));
end
% never enters loop
while [true false]
disp(rand(1));
end
So when you declared x as a vector 1x50 you booby trapped your loop. just initialize x to the x_0 value
x = x_0;
2. In your loop condition you checked if the value is smaller than the tolerance value instead of larger, so naturally it will never enter the loop unless by luck you started from zero:
while abs(f(x)) > epsilon T % should work
2 个评论
StillANovice
2020-8-23
your preallocation was fine, it could be improved by using the input of number of iterations as vector size (that is if the vector size correlated to number of iterations):
x = zeros(1, iteration);
The problem wasn't the preallocation though, the problem was how you used it later on
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