Writing a complicated equation in

I'm trying to write a code for the function attached. The only parameter which isn't a constant is h_c. The problem of course is that h_c appears in both sides of the equation. The only mathematical way I found to overcome this problem is by using the Lambert W function, but I'm not sure how to make it work in matlab. Is there a way to write this type of equation and have matlab solve it?

[Walter Roberson]

syms b f h_c theta theta_1 v
syms Z2 integer
h_c = b*(exp(-(lambertw(Z2, (8*pi*exp((8*pi*cos(f*theta_1)*(v + 1))/(v*cos(theta)^2 - 1))*cos(f*theta_1)*(v + 1))/(v*cos(theta)^2 - 1))*(v*cos(theta)^2 - 1) - 8*pi*cos(f*theta_1)*(v + 1))/(v*cos(theta)^2 - 1)) - 1)

for every integer Z2

This assumes that the equation was

h_c == b*(1-v*cos(theta)^2)/(8*pi*(1+v)*cos(theta_1*f))*log(h_c/b+1)

as it is not completely clear whether the f is inside or outside of the cos()

[Rebeca Miyar]

the f is outside, Thank you so much though!

[Walter Roberson]

Are you certain that the f is outside? The convention is that the arguments to function extend over multiplications and exponentiation, so like

cos theta f^2

would be cos(theta*f^2) not (cos(theta)*f)^2

This is because you could have written f cos theta if you had intended the cos to not include the f

f is often used for frequency, and multiplying an angle by a frequency before taking cos is done often, which would tend to argue that the f is inside the cos .

[Torsten]

My guess is that "f" stands for frequency, thus should be inside the "cos" function.

[Rebeca Miyar]

the f is for lattice mismatch, and is outside of the cos

Answer 1

[David Goodmanson]

Hello Rebeca,

Matlab does have a lambertw function, so you can use it to solve this.

First of all, since b is known, one can solve for x = h_c / b, in which case

x = C*(log(x)+1)

where C is the product of all the factors in front on the rhs of the equation, except not including the b in the numerator.

You didn't say whether C can be of either sign. Plotting both sides of the equation together for various values of C shows that for negative C there is one real solution, and there are no real solutions for 0<=C<1. The two curves are tangent for C=1 at the point x=1. For C>1 there are two real solutions, one with x<1 and one with x>1. You may well have worked out the algebra, and the two solutions are

C = 2;  % example
e = exp(1);
f = @(x) C*(log(x)+1);
x1 = (-C)*lambertw( 0,-1/(C*e))
x2 = (-C)*lambertw(-1,-1/(C*e))
x = .1:.001:7;
plot(x,x,x,f(x),x1,f(x1),'o',x2,f(x2),'o')

Two get both solutions you need two branches of the lambertw function, 0 and 1.

For negative C, there is just the one eal solution x1.

To find roots numerically , the solver fzero is available

C = 2;
g = @(x) C*(log(x)+1) - x;   % find zeros
e = exp(1);
x1f = fzero(g,[1/e,1])
x2f = fzero(g,[1,2*C^2])

This gives the same results as before. It helps to supply fzero with intervals that contain one root (that way you are sure to get both roots), and for this problem the hardest part was coming up with suitable intervals for x1 and x2. But for many fzero situations the interval is pretty clear.