Way to handle singularity,Division by zero
3 views (last 30 days)
Show older comments
Hi All, I am trying to find out the integral by algorithm but it is singular at right end limit.
U see beta is 0.5511 and at right hand limit 1 the term (x/((alpha*x*x)+1-alpha)).^2becomes 1 and further subtraction by 1 make it zero. I am trying (1-sqrt of eps) but I am not sure how much accurate the answer will be. With quadgk answer is 0.5.
but how sould I handle the singularity. I need an accurate answer to find value of alpha which in turn will be used to create some profiles. Thanks in advance. Amit Kadam
clc;
clear;
syms r;
Alpha =1.436269843278397;
H = 1;
R = 1;
Theta = 0;
Beta =o.5516
a = Beta;
b = 1; %(r./R);
npoints = 63;
h = (b - a) / (npoints);
node_r = sym(zeros(1,npoints));
for i=1:npoints
% The number of intervals must be even => number of points must be odd.
x = a - h + (h.*i);
if x==1
x =1-(sqrt(eps));
end
func_r= ( ( ( ( (x) ./ ( (Alpha.*(x).*(x) ) + 1 - Alpha ) ).^2 ) - 1 ) .^(- 0.5) );
node_r(i) = func_r;
end
simpson = (h / 3) .* (node_r(1) + node_r(end)) + ((4 * h) ./ 3) .* ..
0 Comments
Answers (1)
Doug Hull
on 10 Dec 2012
Could you calculate the denominator, and if the absolute value is within a certain tolerance you branch the code to a default value for the expression?
I am surprised if in floating point you ever get zero, you might get really close.
It is not clear which denominator is going to zero.
See Also
Categories
Find more on Numerical Integration and Differential Equations in Help Center and File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!