help in fzero function
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Here x, y,teta are vectors. When I call the function fzero it shows this message:
fzero(@(t)shkruajek(x,y,1,3,teta,0.1,4,t),0)
Exiting fzero: aborting search for an interval containing a sign change
because NaN or Inf function value encountered during search.
(Function value at -2.54232e+306 is NaN.)
Check function or try again with a different starting value.
NaN
can you help me to determine where is the problem here??
function [ek]=shkruajek(x,y,i,j,teta,speed,R,t)
ek=sqrt(((x(i)+ speed*cos(teta(i)*t)) - (x(j) + speed*sin(cos(teta(j)*t))))^ 2+ ((y(i) + speed*sin(teta(i)*t)) - (y(j) + speed*sin(teta(j)*t))).^ 2)-R;
end
a=fzero(@(t)shkruajek(x,y,i,j,teta,speed,R,t),0);
4 个评论
John Chilleri
2017-1-13
Is it possible your function doesn't have a root?
ana take
2017-1-13
ana take
2017-1-13
Walter Roberson
2017-1-13
Note that only the newer versions of MATLAB support fzero on complex-valued functions.
采纳的回答
Walter Roberson
2017-1-13
0 个投票
You only provided a single bound, 0, to fzero. It makes guesses about where to look, and it happens to probe at -2.54232e+306 . Something in the calculation there is resulting in NaN being produced.
My guess is that at -2.54232e+306 you have multiple sub-expressions come out as infinite, and that you end up with infinity minus infinity (or negative infinity plus positive infinity). infinity minus infinity is NaN.
You should be considering giving two element bounding vector to fzero instead of just a single location.
更多回答（1 个）
John Chilleri
2017-1-13
编辑：John Chilleri
2017-1-13
1 个投票
Hello,
In addition to what Walter Roberson said as good practice, it would appear that your function does not have a root.
I say this because your t is only found within sine and cosine functions, and after fzero has tested one period then no other value of t will make a difference, so the fact that it does not find a root within one period and rather continues to eventually error at -2.5 e306 suggests there is no root.
Hope this helps!
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