Help to solv a equation with iteration

I need help, how do i make a code in matlab for iterate the following eq:

x=0.46-0.5*cos(3.14*(y/d))+0.004*cos(2*3.14*(y/d))

where y is the only unknown constant? i know the value of x and d.

[Azzi Abdelmalek]

What do you mean?

Answer 1

[Youssef Khmou]

hi Clauss,

there are many ways to solve the equation, but with iterations i have few ideas : there is a method called Bisection :

 d=1.3; % I SUPPOSED d=1.3
 f=@(y) 0.46-0.5*cos(3.14*(y/d))+0.004*cos(2*3.14*(y/d))
 format long
 eps_abs = 1e-5;
 eps_step = 1e-5;
 a = 70.0; % Initial Guess to your function such that f(a)>0.
 b = 78.0; % Initial Guess to your function such that f(b)<0.

 while (b - a >= eps_step || ( abs( f(a) ) >= eps_abs && abs( f(b) )  >= eps_abs ) )
    c = (a + b)/2;
    if ( f(c) == 0 )
       break;
    elseif ( f(a)*f(c) < 0 )
       b = c;
    else
       a = c;
    end
end

[Claus Madsen]

thanks for the answer. Im quit new in matlab, is this the most simple way to iterate? And it do not work if i change the value of d?

[Claus Madsen]

From what i can understand this solve the eq for f(c)=0 but as i wrote in my problem, i know what x is, and its not 0 but 0.2?

Answer 2

[Roger Stafford]

You can solve that equation without doing any iteration, Claus. Using the formula

 cos(2*A) = 2*cos(A)^2-1

your equation is equivalent to

 .008*cos(3.14*y/d)^2-.5*cos(3.14*y/d)+.456-x = 0

and this is a quadratic equation in cos(3.14*y/d) which you can solve for. You will then have the unknown y expressed as d/3.14 times the arccosine of the two possible roots of this quadratic. (I presume 3.14 is your approximation for pi.) Provided either of these roots lie between -1 and +1, this will give you an infinite number of possible real-valued solutions for y.