Solving Quadratic Problem: Find \omega

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In summary, the conversation revolves around solving the equation \sigma=(\omega + i \nu k^2)+\frac{\alpha^2}{\omega + i \eta k^2} and finding the complex variable \omega as a function of the other variables under certain conditions. The book suggests two approximations for \omega and the person is trying to understand how they were derived.
  • #1
dhris
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Hi, I'm hoping someone out there is going to see something in this problem that I don't because I really don't get it:

Consider the equation:

[tex]
\sigma=(\omega + i \nu k^2)+\frac{\alpha^2}{\omega + i \eta k^2}
[/tex]

It doesn't really matter what the variables mean, (i^2=-1 of course) but what I really need is to figure out [tex] \omega [/tex], which is complex, as a function of the rest (under a certain approximation). The book I found this in claims that under the following conditions:

[tex]
|\sigma|>>|\alpha|
[/tex]

as well as some vague statement about [tex] \nu, \eta [/tex] being small, the two roots of the quadratic are:

[tex]
\omega \approx -i \nu k^2 + \sigma + \frac{\alpha^2}{\sigma + i(\eta-\nu)k^2}
[/tex]

and

[tex]
\omega \approx -i \eta k^2 - \frac{\alpha^2}{\sigma}
[/tex]

I don't know how they came up with this, but it would be really great to find out. Anybody have any ideas?

Thanks,
dhris
 
Last edited:
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  • #2
Well, what is the exact solution for [itex]\omega[/itex]; maybe dwelling upon that will indicate how to come up with those approximations.
 
  • #3
Thanks, that's what I was doing. I couldn't see how they applied the approximation though, but figured it out soon after I posted. Why does it always happen that way?

dhris
 

1. What is a quadratic problem?

A quadratic problem involves finding the solution(s) to an equation that contains a quadratic term (x^2) and may also contain a linear term (x) and a constant term. The solutions to a quadratic problem are typically in the form of x-values that make the equation true.

2. What is \omega in the context of solving a quadratic problem?

In the context of solving a quadratic problem, \omega represents the variable or unknown value in the equation. It is typically represented as x or another letter and is the value that needs to be solved for in order to find the solution to the problem.

3. How do you find \omega in a quadratic problem?

The most common method for finding \omega in a quadratic problem is by using the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. This formula allows you to plug in the values for a, b, and c (from the standard form of a quadratic equation: ax^2 + bx + c) and solve for the two possible values of x. Other methods for solving quadratic problems include factoring and completing the square.

4. What is the discriminant and how is it related to solving a quadratic problem?

The discriminant is the part of the quadratic formula under the square root symbol: b^2 - 4ac. It is used to determine the nature of the solutions to a quadratic problem. If the discriminant is positive, there will be two distinct real solutions. If the discriminant is zero, there will be one real solution. And if the discriminant is negative, there will be no real solutions, but instead two complex solutions. The discriminant can also be used to determine whether a quadratic equation can be factored or not.

5. Can you provide an example of solving a quadratic problem to find \omega?

Sure! Let's say we have the equation 3x^2 + 4x - 5 = 0. To find the value of x, we can use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / 2a. Plugging in the values from our equation, we get x = (-4 ± √(4^2 - 4(3)(-5))) / 2(3) = (-4 ± √(16 + 60)) / 6 = (-4 ± √76) / 6. This simplifies to x = (-4 ± 2√19) / 6. So, the two possible values of x (or \omega) are (-4 + 2√19) / 6 and (-4 - 2√19) / 6.

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