6 views (last 30 days)

Niels
on 25 Jan 2017

Edited: Niels
on 25 Jan 2017

since your function is even it is symmetric and it is monotone for x>0 or x<0

use fzero

>> f=@(x)((cos(x)+1./2.*x.^2)-1).*x.^4 - 1/24 % subtract the value to transform it into an issue of roots

>> solution=abs(fzero(f,0))

solution =

1.0042

>> range=[-solution solution]

range =

-1.0042 1.0042

Walter Roberson
on 26 Jan 2017

There are an infinity of complex solutions, symmetric in positive and negative real components, and symmetric in positive and negative imaginary components. The boundaries on the imaginary components are +/- -.9958714409867068 approximately and the boundaries on the real components are +/- -1.004205445912837 approximately.

The area is not circular in real and imaginary components, but it is approximately circular.

For any given real component inside the given range, there are two imaginary components that lead to solution; likewise for any given imaginary component within the range, there are two real components that lead to solution.

When I talk about infinity of solutions, I am referring just to the boundary; because if the inequality, everything within the boundary is included too.

The boundary is the solutions for

(1/2)*sqrt((2*cosh(2*b)+2*cos(2*a)+(4*a^2-4*b^2-8)*cos(a)*cosh(b)-8*a*b*sin(a)*sinh(b)+a^4+(2*b^2-4)*a^2+b^4+4*b^2+4)*(a^2+b^2)^4)-1/24

where a is the real component and b is the imaginary component.

Walter Roberson
on 26 Jan 2017

Find the treasures in MATLAB Central and discover how the community can help you!

Start Hunting!
## 0 Comments

Sign in to comment.