Hi bohan shen,
The first six roots are shown below. First, since J1 is odd and J0,J2 are even, if z is a root then so is -z. The contour plot of abs(f(z)) shows roots in quadrant IV, meaning there also roots in quadrant II. No roots in quadrants I or III.
Roots are determined by Newton's method. Each initial estimate has to be within an enclosed contour. From the contour plot the estimate of 3 looks a bit sketchy (2-.6i would have been a sure thing) but it worked anyway.
clear i
xx = 0:.01:20;
yy = -1:.01:1;
[x y] = meshgrid(xx,yy);
z = x+i*y;
f = @(z) i*besselj(1,z) - (z/2).*(besselj(0,z) - besselj(2,z));
dfdz = @(z) ((i-1)/2)*(besselj(0,z) - besselj(2,z)) ...
+ (z/2).*((3/2)*besselj(1,z) -(1/2)*besselj(3,z));
contour(x,y,abs(f(z)))
grid on
w0 = [3;6;9;12;15;18];
w = w0;
for k = 1:10
w = w - f(w)./dfdz(w);
end
w
f(w)
w =
2.0811 - 0.6681i
5.3355 - 0.1967i
8.5372 - 0.1193i
11.7063 - 0.0863i
14.8637 - 0.0677i
18.0156 - 0.0557i
Best Answer