How to solve NaN values of integrand defined over a finite interval?
2 Ansichten (letzte 30 Tage)
Ältere Kommentare anzeigen
Hi, could someone know why quadv () returns NaN (see %r1), although my defined function is well behaved on [-theta; pi/2]?
see below the code:
f=@(k) k .* exp(-k);
x=1
al=1;
bet=0.5;
if al==1
c=exp(-(pi * x) /2 * bet);
c2=1 /(2*abs(bet));
theta0=pi/2
zet=0
v1=@(theta) 2/pi .* ((pi/2 + bet.*theta) ./cos(theta)) .* exp(1/bet .* (pi/2 + bet.*theta)...
.* tan(theta));
g=@(theta) c .* v1(theta);
%p=c2 * quadv(@(theta) f(g(theta)),-theta0,pi/2);
y= @(theta) c2 .* f(g(theta)) ; % values of integrand
%r1= quadv(@(theta) y(theta),-theta0,pi/2)
% check behavior of integrand
n= @(theta) f(g(theta));
a=[-theta0:0.01:pi/2]';
N=n(a);
P=[a;N];
fprintf('%6s %12s\n','x','f(x)');
fprintf('%6.2f %12.8f\n',P);
plot(a,P)
else
alm=al-1;
d=1/alm;
da=al .*d;
theta0=(1/al) .*atan(bet .*tan(pi .*al/2));
u=al .*theta0;
zet=-bet .*tan(pi*al/2)
v=@(theta) (cos(u) .^d) .*((cos(theta) ./sin(al .*(theta0+theta))) .^da) .*...
(cos(u+alm .*theta) ./cos(theta));
c=(x-zet) .^da;
c2=al ./(pi .*abs(al-1) .* (x-zet));
g=@(theta) c .* v(theta);
y= @(theta) f(g(theta)) ; % values of integrand
%y= @(theta) c2 .* f(g(theta)) ;
r2= c2 .* quadv(@(theta) y(theta),-theta0,pi/2)
end
% plot values
%a=[-theta0:0.01:pi/2]';
%Y=y(a);
%U=[a;Y];
%fprintf('%6s %12s\n','x','f(x)');
%fprintf('%6.2f %12.8f\n',U);
%plot(a,Y)
Also for curiosity I am wondering why from the screen x goes beyond the maximum limit pi/2. actually matlab outputs lots of zeros there.
Thanks
0 Kommentare
Antworten (0)
Siehe auch
Kategorien
Mehr zu Numerical Integration and Differential Equations finden Sie in Help Center und File Exchange
Community Treasure Hunt
Find the treasures in MATLAB Central and discover how the community can help you!
Start Hunting!