Help Understand Denominator Separation from Zero in Math Equations

[williamshipman]

Please help me to understand: "the denominator is numerically separated from zero"

Hi everyone,

I've come across this statement in a dissertation I'm reading and I don't have a clue as to what the author is speaking about. Can anyone give me an explanation. For reference, the equations are

\lambda_0 = \frac{C_T}{2 \eta \sqrt{\mu^2 + (\lambda_0 - mu_z)^2}}

and

C_T^{ideal} = \frac{a\sigma}{2}\left(\theta_0\left(\frac{1}{3}+\frac{\mu^2}{2}\right)+\frac{\mu_z-\lambda_0}{2}\right)

C_T equals C_T^{ideal} unless C_T^{ideal} is larger than C_T^{max} or smaller than C_T^{min}, in which case it saturates at the applicable limit.

The objective is to numerically solve the above equations for \lambda_0 and then calculate C_T. Problems occur when \mu is close to zero and \mu_z is close to \lambda_0 as the denominator of the first equation gets very small. Trying to solve this using Newton's method fails because the iterations don't converge to an answer. I'm hoping that if I understand what "numerically separating the denominator (of the first equation) from zero" means, then I might make some progress.

Thanks for the help.

 

[Tedjn]

Where does the author use that phrase "numerically separating..." or are you paraphrasing? I'm not enough of an expert to provide details, but it appears that the author is saying a denominator too small leads to round off error in the computation that will give inaccurate results. It is an important but purely practical matter.

I am also confused how one would calculate λ₀ numerically without knowing C_T, and then calculate C_T.

 

[arildno]

A numerically awkward difference (i.e, a difference between two numbers too close to have a barrier against loss of significant digits), can be made less awkward in various ways, for example by this strategy:

Let x be less than, but very close to 1.

Then, we have:
1-x=\frac{1-x^{2}}{1+x}=\frac{1-x^{4}}{(1+x)(1+x^{2})}
and so on.
Note that the retained difference within the expression will be an improvement relative to the first order, original difference 1-x

 

[williamshipman]

Thanks for the replies.

Tedjn: The quote comes straight from the author's work. You are right to say that a small denominator gives incorrect results, at least with the current iteration scheme. To solve the equations numerically, you can use something like Newton's method. One can substitute the equation for C_T into the first equation, giving you one equation to solve. Its possible to get an initial guess of \lambda_0, so the equations can be solved. Right now, I'm using Matlab's fsolve function, but I'd like something faster. So I tried the Newton's method iteration formula from the author's work, but with the noted problems. And I did check the derivation.

arildno: Thanks for this idea. So if I use this method to work out \lambda_0-\mu_z in the denominator of the first equation, is that then numerically separating the denominator from zero? I'll give this a go and see what happens.

Merry Christmas and thanks for helping me even though it is holiday time.

 

[arildno]

A bit of caution, though:
Newton's methon doesn't always work, as you probably know. It depends upon the local curvature (check it out).

Now, if you're unlucky, no amount of twiddling with that difference would help you, because your problem isn't really numerical, but rather with Newton's method as such.