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Use Solve with a condition of a limit x is much less than D?

Posted 8 years ago

Greetings, I am working on a very complicated equation that needs to be solved. To simplify the function I use a limit in which x is much less than D or ( x<<D). I cannot find much less/greater condition in Mathematica. I need to solve that equation symbolically and I know that I should get four solutions. By taking a limit manually I got something like Ax^4-Bx^2+C=0. Then using solve I was able to get four solution, but with coefficients that were very long. (In manual calculations, I was able to get shorter coefficients) How do I solve equation with much less condition? Just for the sake of example let say I want to solve

f(x)= sqrt( D^2 - ( x - z )^2 ) + x * cos(a) + x + z

after limit

f(x)= sqrt( D^2 - ( z )^2 ) + x * cos(a) + x + z

Is there a command in Mathematica to selectively simplify/factor/expand some terms instead all terms in equations?

POSTED BY: Adam Szewczyk
9 Replies

That square root looks to be malformed. Also I don't see how this type of elimination will be helpful in terms of getting a smaller solution to the biquadratic.

POSTED BY: Daniel Lichtblau

(1) It would help to have actual code provided in the question.

(2) Might want to look at ref guide page for Series.

POSTED BY: Daniel Lichtblau
Posted 8 years ago

Hi Daniel, Thanks for reply.
1) I attached the code. 2) I used series to do x<<D (much lesser than) approximation. In the code, I wrote both methods to solve the quartic equation. Both give me quartic equation solution in which 3rd and 1st power coefficients are zeros. However the coefficient values and solutions are different. First method using series approximation, gave me very lengthy solution. The second method by hand gave me simpler solution but still it is a long one. My question is: Which method is consistent approximation? Are there commands to do condition (much lesser than)? Any other options of calculating the limits such that x variable is still present in equation? For last method, FullSimplify does the job in simplifying the results.

Attachments:
POSTED BY: Adam Szewczyk
POSTED BY: Daniel Lichtblau
Posted 8 years ago

Thanks Daniel, I will start working on it and see what happens.

POSTED BY: Adam Szewczyk
Posted 8 years ago

Hi Daniel, Unfortunately, I waited 5 minutes and no results.

POSTED BY: Adam Szewczyk
Posted 8 years ago

Hi Daniel,

I think I might solve it if I get rid of square roots in it. I attempted to expand the f5 function in order to get rid of square roots because for my situation Det[f5]==0. I attempted by using Eliminate command as well as GroebnerBasis command. For some reasons it does not work...it takes more than 15 minutes.

eli=ReplaceAll[Sqrt[f5,D^2 - (x - c)^2]->u]
Eliminate[{eli==0,u^2==D^2 - (x - c)^2},u]
GroebnerBasis[{eli,u^2-D^2 + (x - c)^2},x,u]

Am I right with the code?

POSTED BY: Adam Szewczyk

Hey Adam, have you tried FullSimplify ?

POSTED BY: Robert Poenaru
Posted 8 years ago

Hi Robert, Yes I tried FullSimplify. The system was busy for a long time, that I usually aborted the calculations after 5 minutes. Perhaps next time I should try a bit longer.

POSTED BY: Adam Szewczyk
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