MATLAB: Solve a numerical equation

equationnumericalnumerical equationsolvezero

Hi everyone,

I have a problem solving a numerical equation. I want the result near zero for:

((1/6)*abs(sin(x*6/2)/sin(x/2))) = 1/sqrt(2)

It should be around x = 0.4695;

I tried the following:

myfunc = @(x) ((1/6)*abs(sin(x*6/2)/sin(x/2)))-1/sqrt(2);
x0 = 0;
fzero(myfunc,x0);

But somehow doesn't work for me. Can somebody help me please and tell me why it doesn't work?

Thanks.

Best Answer

Hi,

use another x0:

myfunc = @(x) ((1/6)*abs(sin(x*6/2)/sin(x/2)))-1/sqrt(2);
x0 = 1;
fzero(myfunc,x0);

This will make it work.

Best regards

Stephan

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FMINCON is not a root finder. Minimizing fun will not necessarily give you zero. You need to use fsolve, or lsqnonlin depending on the x,y constraints you are trying to implement.

Also, it is overkill to use MultiStart for a 2-variable problem. Simply make a surf plot of your objective function and see where it gets close to zero. Or use min(abs(z)) to search the surface samples.

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First, consider if you are trying to solve the wrong problem.

In your question, you state it as e^2*sin(x) - x = 0

However, in your code, you write exp(2*sin(x)) - x. You do realize there is a difference? What I don't know is if you have miswritten your question, or is it your code?

fun = @(x) exp(2)*sin(x) - x;
fplot(fun,[0,3])
yline(0);
fzero(fun,3)
ans =
       2.7589

So the function you claim to want to solve has a zero around x==2.76, which is inconsistent with your claim of where the root lies. Next, look at the function you wrote code for:

fun = @(x) exp(2*sin(x)) - x;
fplot(fun,[-5,3])
yline(0);

Indeed, this does seem to have a root near 2.6357. So, just possibly, you really do want to solve the problem exp(2*sin(x))-x==0.

fzero(fun,3)
ans =
       2.6357

But now, look at the plot! What happens when you start Newton's method at x==1? THINK! Where will the first iteration go? On which side of that hump is x==1?

The point is, Newton's method tries to drive the functino to zero. But if you start the iterations BELOW x==1.5028, which direction will Newton's method push you?

fminbnd(@(x) -fun(x),1,3)
ans =
       1.5028

Think about the meaning of the iterations of Newton's method. What is the goal? What will happen?

Best Answer

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