trigonometric equation solution help

Hi everyone,

I need to solve the following equation:

with k as unknown, in the range ]0.5 , 1]

I try with this,

syms err space k

vq_equation = err^2 == (space^2 * (2 - 2*cos(4*pi*k))/(8*pi^2*k^2*(4*k^2+1)).^2);

assume(k>0.5 & k<=1)

a = solve(vq_equation,k);

but the result is

Warning: Unable to find explicit solution. For options, see help.

> In sym/solve (line 317)

Have you got any idea?

Answer 1

[Torsten]

When I plot the function for several values of err and space, I only see a zero at k=0 and k=0.5.

Can you give values for err and space such there are more zeros in (0.5,1) ?

[Giuseppe Avallone]

Hi,

by plotting err as a function of k for several space i found different values in the range ]0.5 , 1].

So, to answer your question, you can run this (i plot the two sides of the equation and the intersection is in that range).

 err = 0.00614737147202947;
 space = deg2rad(5);
 k = linspace(0.3,1.0,1000);
 memb1 = err^2*(8*pi^2*k.^2.*(4*k.^2-1)).^2;
 memb2 = space^2 * (2 - 2*cos(4*pi*k));

 figure
 hold on
 grid on
 plot(k',[memb1', memb2']);
 legend('memb1','memb2')

[Torsten]

An analytical solution for k using MATLAB's "solve" is not probable.

I rewrote the equation as

a^2*(8*pi^2*k^2*(4*k^2-1))^2 - 2*(1-cos(4*pi*k) = 0

with the dimensionless number a = err/space and solved for k in the interval [0.5:1] for a given value of a.

The result is plotted.

A = 0:0.001:0.2;
nk = 1000;
k_min = 0.50001;
k_max = 1.0;
dk = (k_max-k_min)/nk;

for i = 1:numel(A)
    a = A(i);
    f = @(k) a^2*(8*pi^2*k.^2.*(4*k.^2-1)).^2 - 2*(1-cos(4*pi*k));
    k = k_min;
    fl = f(k);
    flag = 0;
    while k < k_max
        k = k + dk;
        fr = f(k);
        if fl*fr <= 0
            sol_k(i) = k-dk/2.0;
            flag = 1;
            break
        end  
        fl = fr;
    end
    if flag == 0 
        sol_k(i) = NaN;
    end
    i
end   
plot(A,sol_k)

Answer 2

[David Goodmanson]

Hi Giuseppe

Although Matlab can't come up with an analytic solution with symbolics, you can still obtain a numerical solution that does not involve doing a numerical solve for each desired value of err^2. (err^2 is denoted by e2 here). For a vector of values e2_new, the code below works much like an analytic expression would.

At k = 1/2, e2 has the form 0/0. It's obvious from the plot that the value of e2 is s^2/2 there. It's possible to prove it, but I am just as happy not to have to.

In place of spline( ...) you can use interp1(...,'spline') which is more prominently documented than spline is these days.

s = 3
k = linspace(1/2,1,1e5);
e2 = fun(k,s);
plot(k,e2)    % the allowed range of e2 is 0 to s^2/2.

% pick a few values of e2 (in the allowed range)
e2_new = [2 2.5 3 3.2 4 4.1];
k_new = spline(e2,k,e2_new);      % the solution

% check to see that the function works, difference is small
fun(k_new,s) - e2_new

       ans = 1.0e-13 *
      -0.0022    0.0266         0   -0.0133   -0.0844    0.4263

function e2 = fun(k,s)
e2 = s^2*(2-2*cos(4*pi*k))./(8*pi^2*k.^2.*(4*k.^2-1).^2);
e2(abs(k-1/2)<1e-10) = s^2/2;
end