While loop runs infinitely

2 次查看(过去 30 天)
I'm trying to create a function that takes two parameters: x and threshold. X is the angle and the threshold is the percent error I need to get when approximating sine with the taylor series. However my while loop runs infinitely and I'm very confused as to how do go about fixing it.
Here's what I have so far.
function [approx, terms] = approx_sine(x, threshold)
approx = x; % Initial approximation
terms = 0; % Number of additional terms added to improve the approximation;
% Write your code here using a while loop to improve the above approximation
while threshold <= abs((sin(x)-approx)/sin(x))
terms = terms + 2;
approx = (-1)^(terms+1)*(x.^terms)/ factorial(terms)+x;
end
end

采纳的回答

Thorsten
Thorsten 2015-12-1
编辑:Thorsten 2015-12-1
The code for approximation is wrong. There are some errors in your formula: term takes values 0, 2, 4, ... but should take values 1,3,5,7; the (-1)^n changes sign with every new term, but since your term is always even, (-1)^(term+1) is always odd; and finally you add the n'th approximation term to the initial approximation x, instead of adding it to the most recent approximation.
This works
function [approx, n] = approxsine(x, threshold)
approx = x; % Initial approximation
n = 0; % Number of additional terms added to improve the approximation;
% Write your code here using a while loop to improve the above approximation
while threshold <= abs((sin(x)-approx)/sin(x))
n = n + 1;
approx = approx + (-1)^(n)*x.^(2*n+1)/factorial(2*n+1);
end
end
  2 个评论
Image Analyst
Image Analyst 2015-12-1
And I'm a very strong advocate of failsafes, especially in while loops to prevent just this problem of infinite loops. Check the loop counter and if it's more than you ever realistically expect, bail out with an error message:
n = 1;
nMax = 1000000
while threshold <= abs((sin(x)-approx)/sin(x)) && n < nMax
n = n + 1;
approx = approx + (-1)^(n)*x.^(2*n+1)/factorial(2*n+1);
end
if n >= nMax
errorMessage = sprintf('No solution found after %d iterations!', n-1);
uiwait(errordlg(errorMessage));
end
Note the additional check of "&& n < nMax" I added to the while line.
Torsten
Torsten 2015-12-1
Since sin(x) < = 1, better use absolute instead of relative error within error estimate:
while threshold <= abs(sin(x)-approx)
Best wishes
Torsten.

请先登录,再进行评论。

更多回答(1 个)

Walter Roberson
Walter Roberson 2015-12-1
Remove the line
threshold = abs((sin(x)-approx)/sin(x))
It is overwriting the threshold that you passed in.
  2 个评论
Anthony Siddique
Anthony Siddique 2015-12-1
I'm still having the same problem. The while loop won't resolve.
Walter Roberson
Walter Roberson 2015-12-1
>> limit((-1)^(terms+1)*x^terms/factorial(terms), terms = infinity);
0
Your approx is (-1)^(terms+1)*x^terms/factorial(terms) + x . As I indicate above, as terms increases, the first part of that tends to 0 for all x. With the first part tending to 0, only the second part will start to matter, the "+ x" part. Your approx goes to x.
Your termination test then becomes
threshold <= abs((sin(x)-x)/sin(x))
or threshold <= abs(1 - x/sin(x))
For any given threshold there is a limited range of x that can satisfy this, and that range of x does not increase smoothly with the threshold.

请先登录,再进行评论。

类别

Help CenterFile Exchange 中查找有关 Mathematics 的更多信息

产品

Community Treasure Hunt

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!

Translated by