to find root of transcendental equation

As from below program i need to confirm The roots are α = {2.7859,5.5215,7.9573} . how to find

a=0:0.01:9;
y1=sqrt(64-a.^2);
y2=-1.*a.*cot(a);
plot(a,y1)
grid on
hold on
plot(a, y2)
ylim([0, 12])

[Dyuman Joshi]

Numerically you won't be able to find the exact values, more so as the data has 2 digit precision and the values you want to obtain have 4 digit precision.

As you can see below, for tolerance of 0.1 for step-size 0.01, you get 2 out of 3 intersection points -

a=0:0.01:9;

y1=sqrt(64-a.^2);
y2=-1.*a.*cot(a);
a(abs(y1-y2)<0.1)

ans = 1×2

5.5200 7.9600

For step-size 1e-4 and tolerance 1e-3, you get 4 different roots

a=0:0.0001:9;

y1=sqrt(64-a.^2);
y2=-1.*a.*cot(a);
a(abs(y1-y2)<0.001)

ans = 1×4

2.7859 5.5214 5.5215 7.9573

Even solve() is unable to find a solution symbolically, and returns a numeric solution via vpasolve. Note that vpasolve only returns a single solution for non-polynomial equations

syms x
assume(x>=0 & x<=9)
f1(x)=sqrt(64-x.^2);
f2(x)=-1.*x.*cot(x);

sol = solve(f1-f2==0, x)

Warning: Unable to solve symbolically. Returning a numeric solution using vpasolve.

sol = 

5.521446429976302205509075939559

You can call vpasolve() multiple times by setting 'Random' as true to obtain the solutions -

for k=1:5
    s = double(vpasolve(f1-f2==0, x, [0 9], 'Random', true))
end

s = 5.5214

s = 2.7859

s = 5.5214

s = 7.9573

s = 7.9573

plot(a,y1)

Warning: Imaginary parts of complex X and/or Y arguments ignored.

grid on
hold on
plot(a, y2)
ylim([0, 12])

Answer 1

[Matt J]

As from below program i need to confirm The roots are α = {2.7859,5.5215,7.9573}

If you only need to confirm the roots, just plug them in and check:

dy=@(a) sqrt(64-a.^2) + a.*cot(a);

dy([2.7859,5.5215,7.9573])

ans = 1×3

1.0e-03 * 0.0550 -0.7282 0.3823