**You are now following this question**

- You will see updates in your followed content feed.
- You may receive emails, depending on your communication preferences.

# difficult inequality to solve

2 views (last 30 days)

Show older comments

##### 0 Comments

### Accepted Answer

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

##### 19 Comments

Niels
on 25 Jan 2017

nope

it does not:

>> f=@(x)abs(((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))

range=[-solution solution]

f =

function_handle with value:

@(x)abs(((cos(x)+1./2.*x.^2)-1).*x.^4)-1/24

solution =

1.0042

range =

-1.0042 1.0042

Walter Roberson
on 25 Jan 2017

The abs() makes a difference if you are expecting complex numbers for input.

Walter Roberson
on 26 Jan 2017

Are you looking for real valued solutions or complex valued solutions?

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

### More Answers (0)

### See Also

### Tags

### Community Treasure Hunt

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

Start Hunting!**An Error Occurred**

Unable to complete the action because of changes made to the page. Reload the page to see its updated state.

Select a Web Site

Choose a web site to get translated content where available and see local events and offers. Based on your location, we recommend that you select: .

You can also select a web site from the following list:

## How to Get Best Site Performance

Select the China site (in Chinese or English) for best site performance. Other MathWorks country sites are not optimized for visits from your location.

### Americas

- América Latina (Español)
- Canada (English)
- United States (English)

### Europe

- Belgium (English)
- Denmark (English)
- Deutschland (Deutsch)
- España (Español)
- Finland (English)
- France (Français)
- Ireland (English)
- Italia (Italiano)
- Luxembourg (English)

- Netherlands (English)
- Norway (English)
- Österreich (Deutsch)
- Portugal (English)
- Sweden (English)
- Switzerland
- United Kingdom (English)

### Asia Pacific

- Australia (English)
- India (English)
- New Zealand (English)
- 中国
- 日本Japanese (日本語)
- 한국Korean (한국어)