MATLAB: Roots of nonlinear eqn

MATLABnonlinear equationone unknownroots

I tried so many ways to find this question's roots, but I couldn't find it. Could you help me to find it,please?

y = -1200*x.^2*sin(1200*x) + x.*cos(1200*x).^2 – 2*x.*cos(1200*x) + x.*sin(1200*x).^2 + x

Best Answer

Well, you did come back and make an effort. There are infinitely many roots.

First though, fully use the . operators to get rid of those annoying warnings. You missed at least one.

fun = @(x) -1200*x.^2.*sin(1200*x) + x.*cos(1200*x).^2 - 2*x.*cos(1200*x) + x.*sin(1200*x).^2 + x;
ezplot(fun,[0,.01]),grid on

fzero(fun,0.005)
ans =
       0.00523598775598288

WTP?

If you cannot figure out how to find a decent estimate of that first root, then try a Taylor series approximation. Use an expansion point someplace in that vicinity.

Yapprox = taylor(y,x,'order',15,'expansion',0.003);
vpa(vpasolve(Yapprox),15)
ans =
                           0.00523598465330524
                           0.00751869453910503
                           0.00953318487432745
                           0.00987708713167847
   - 0.0015385284492191 + 0.00427836249361709i
   - 0.0015385284492191 - 0.00427836249361709i
 - 0.000770867501573772 - 0.00128010168392448i
 - 0.000770867501573772 + 0.00128010168392448i
  0.000154479476404463 + 0.000558172154407523i
  0.000154479476404463 - 0.000558172154407523i
  0.000389259331551612 - 0.000176673046991433i
  0.000389259331551612 + 0.000176673046991433i
    0.00934267887256653 + 0.00394573409659824i
    0.00934267887256653 - 0.00394573409659824i

That first root listed is a splendidly good approximation to the first root. In fact, it even gave a pretty good approximation for the second root.

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MATLAB: How to find the roots of the Equation

This question gets asked so often, that it must be part of a homework or something.

fplot(@(x) tan(x/2)+tanh(x/2),[-10,10])
grid
yline(0);

There are infinitely many real roots. You can see three of them in the figure. They will surely not have any simple algebraic form you can write down. There may be complex roots, in fact, that is entirely possible. You could investigate that behavior by plotting things very carefully in the complex plane. How?

Hr = fcontour(@(xr,xi) real(tan((xr + i*xi)/2)+tanh((xr + i*xi)/2)),[-10,10,-10,10]);
Hr.LevelList = 0;
Hr.LineColor = 'b';
hold on
Hi = fcontour(@(xr,xi) imag(tan((xr + i*xi)/2)+tanh((xr + i*xi)/2)),[-10,10,-10,10]);
Hi.LevelList = 0;
Hi.LineColor = 'r';
grid

So it appears there are solutions in the complex plane. However, they exist only when x is purely real, OR purely imaginary. Those solutions lie at the intersections of the red and blue curves as I have plotted them. It appears that no solutions exist with simultaneously non-zero real and imaginary parts. With some effort, you could probably prove that claim, but it hardly seems worth the effort.

Actually finding any of the solutions for real or imaginary x will involve nothing more than use of a root finder, suze as fzero or vpasolve. Again, the solution that you find will depend on the start point or the starting interval. Here are three arbitrarily found real solutions.

syms x
vpasolve(tan(x/2)+tanh(x/2),x)
ans =
0
vpasolve(tan(x/2)+tanh(x/2),x,5)
ans =
4.730040744862704026024048100833884819898
vpasolve(tan(x/2)+tanh(x/2),x,12)
ans =
10.99560783800167090666903251910589241754

I seem to recall, from the last time I answered this question, that the roots approach a periodic behavior for large or small x. There will be some multiple of pi involved as I recall. This too can surely be shown, with some effort that may not be worth the effort invested for what is surely a homework assignment.

MATLAB: Positive roots only for cubic eqn

 ROOTS = roots([1, -5, -2, 24]);
 Wanted = ROOTS(ROOTS > 0)

Best Answer

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MATLAB: How to find the roots of the Equation

MATLAB: Positive roots only for cubic eqn

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