example 6.5

er=95;                                                                  % titania
tand=.001;a=4.13e-3;L=8.255e-3;c=3e8;

x=[0:.001:15];y=besselj(0,x);                                   % order 0
[nx0,ny0]=intersections(x,y,x,zeros(1,numel(x))); 

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% finding zeros of J(n,x)

p01=nx0(1)                 % p01 = 2.404825587623834;

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% 1st approach

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L=8.255e-3;a=4.13e-3
f1=@(x) real(tan((er*x^2 - (p01/a)^2)^.5*L/2) - ...
((p01/a)^2-x^2)^.5./(er*x^2-(p01/a)^2)^.5)
s=0

for k=[10:.1:600]
    s=[s fzero(f1,k)];
end
s2=unique(s);
stem(s2)

k02=p01/a
k01=p01/(a*er^.5)

f1=c*k01/(2*pi)
f2=c*k02/(2*pi)

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k0_range=intersect(s2(s2>=k01),s2(s2<=k02))

min(k0_range)
max(k0_range)

k1=min(k0_range)
k2=max(k0_range)

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f1=c*k1/(2*pi)
f2=c*k2/(2*pi)

tand=.001;
Qd=1/tand

(1): Titania, AKA Titanium Dioxide TiO2  https://www.azom.com/article.aspx?ArticleID=1179

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comment: there are limitations when attempting to find Bessel function zeros with standard commands like solve and fslove in the following ways

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syms x;y=besselj(0,x); 

solve(y)

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fun=@(x) besselj(0,x)

fsolve(fun,1)

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