How to solve transcendental equations in matlab

I am practicing solving transcendental equations in matlab I have the given equation and I want to solve for x

tan(((0.9)*pi/w)*sqrt(1.468^2-x.^2))- sqrt((x.^2-1.458^2))/(sqrt(1.468^2-x.^2))==0

plotting w varies from 1.55 to 5 I tried all day but I cannot find solution, I tried with fzero(fun,xo) without success any suggestion?

[Walter Roberson]

Could you confirm that the equation has two copies of 1.468 and one copy of 1.458 ? The two numbers differ in only 1 digit, so it would be good to check that there has not been a typing error.

Are the three related? For example could you have written

G = 1.468;
tan(((0.9)*pi/w)*sqrt(G^2-x.^2))- sqrt((x.^2-(G-1/100)^2))/(sqrt(G^2-x.^2))==0

[Anton Semechko]

Walter, I am fairly certain there is another type in this equation. The "/" between the last two square roots should be "./".

[Walter Roberson]

Please do not close questions that have an Answer.

[alburary daniel]

A feel confuse because a will trying to solve problemas using newton rapson method

[Walter Roberson]

Do not close questions that have an Answer. The volunteers spent time responding to you, and their responses may help other people with similar questions.

I know that I spent about 2 hours investigating this myself, and I am disappointed that my contribution would just disappear as if I had never done the work.

Answer 1

[Anton Semechko]

This is a complex-valued function. Just from visual inspection, it is apparent that the real part of this function has infinitely many solutions. The imaginary part, however, appears to have a finite number of zeros. Are those the solutions you are looking for?

x=linspace(-2*pi,2*pi,1E4);
w=1.55;
f=tan(((0.9)*pi/w)*sqrt(1.468^2-x.^2)) - sqrt((x.^2-1.458^2))./(sqrt(1.468^2-x.^2));

figure('color','w')
plot(x,real(f))
hold on
plot(x,imag(f))

set(gca,'YLim',[-10 10])

[alburary daniel]

in fact, I really trying to plot w versus b, where variable w runs from 1.5 to 5

[Anton Semechko]

what is b?

[Walter Roberson]

My tests show that it does have real solutions, provided that it is correct that the middle value is 1.458 and not 1.468 like the other two.

For example, for w = 2, x = +/- 1.45854467945914

[alburary daniel]

Let me explain, I trying to obtain specific values of b taking specific value of w, for example, if I take a particular value of w, the above equation that present to you shoud be give me a value of beta as solution, in this case I want plot beta vs w, where w will run from 1.5 to 5. In this case as is a transcedental equation I tried to find the roots for every value of w and ploting the roots vrs w, which in this case the roots will be the solution of b for every w evaluated in the transcedental equation

a=0.9;
n1=1
n2=1.5

tan(((a)*pi/w)*sqrt(n2^2-b.^2))- sqrt((b.^2-1.n1^2))/(sqrt(n2^2-b.^2))==0

by the way I forget that n2<x<n1 should be satisfy

[Anton Semechko]

Here you go:

% Equation
a=0.9;
n1=1;
n2=1.5;
f=@(x,w) tan((a*pi/w)*sqrt(n2^2-x.^2)) - sqrt((x.^2-n1^2))./(sqrt(n2^2-x.^2));

% Look for zeros for w=[1.5 5]
W=linspace(1.5,5,1E2);
X=linspace(n1+1E-12,n2-1E-12,1E3);
Z=0.5*(n1+n2)*ones(size(W));

Opt=optimset('Display','off');
figure('color','w')
for i=1:numel(W)

      % Initialize search
      F=f(X,W(i));
      [F_min,id]=min(abs(F));
      Z(i)=X(id);

      % Refine solution
      f_i=@(x) f(x,W(i));
      [Zi,Fi]=fzero(f_i,Z(i),Opt);    
      if abs(Fi)<abs(F_min), Z(i)=Zi; end

      plot(Z(i),W(i),'.k','MarkerSize',20)
      hold on
 end
 plot(Z,W,'-b')

xlabel('b','FontSize',25,'FontWeight','bold')
ylabel('w','FontSize',25,'FontWeight','bold')
set(gca,'FontSize',20)
grid on

[alburary daniel]

but in this case I want that x-axis will be W not b, so change the result rigth in the code? by the way what do Z? can you explain that sentences

[Anton Semechko]

If you want to plot w along horizontal axis, change

plot(Z(i),W(i),'.k','MarkerSize',20)  to  plot(W(i),Z(i),'.k','MarkerSize',20),

plot(Z,W,'-b') to plot(W,Z,'-b')

x- and y-labels would have to be modified accordingly.

Variable Z is a vector that contains solutions to f(x,w)=0, so that abs(f(Z(i),W(i)))<tol, where tol=1E-16. W is vector of w values that go from 1.5 to 5.

[Anton Semechko]

Here is modified version of the code and visualization of computed results:

% Equation
a=0.9;
n1=1;
n2=1.5;
f=@(x,w) tan((a*pi/w)*sqrt(n2^2-x.^2)) - sqrt((x.^2-n1^2))./(sqrt(n2^2-x.^2));

% Look for zeros for w=[1.5 5]
W=linspace(1.5,5,1E2);
X=linspace(n1+1E-12,n2-1E-12,1E3);
Z=0.5*(n1+n2)*ones(size(W));

Opt=optimset('Display','off');
figure('color','w')
for i=1:numel(W)

      % Initialize search
      F=f(X,W(i));
      [F_min,id]=min(abs(F));
      Z(i)=X(id);

      % Refine solution
      f_i=@(x) f(x,W(i));
      [Zi,Fi]=fzero(f_i,Z(i),Opt);    
      if abs(Fi)<abs(F_min), Z(i)=Zi; end

      plot(W(i),Z(i),'.k','MarkerSize',20)
      hold on
  end
  plot(W,Z,'-b')

xlabel('w','FontSize',25,'FontWeight','bold')
ylabel('b','FontSize',25,'FontWeight','bold')
set(gca,'FontSize',20)
grid on

Answer 2

[Walter Roberson]

Stop closing your questions that people have already answered!!

We are not free private consultants: public questions and public answers are the price we "charges". We answer publicly as volunteers so that everyone can benefit. When you close questions after people have answered, then you are taking advantage of what you learned but you are grabbing it away to prevent anyone else from benefiting from the time and knowledge that other people have contributed.