Way to handle singularity,Division by zero
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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) .* ..
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Doug Hull
le 10 Déc 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.
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