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

Questions
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1

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?

2

Hey Adam, have you tried FullSimplify ?

3

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.

4

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

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

5

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.
6

I can make a few suggestions.

(1) Avoid use of capitalized symbols, expecially those tht might already be built-in symbols. D is an example.

(2) You might rename x to xx wherever it appears with the big variable (I’ll call it bigD), and expand as a series in xx at the origin. Then substitute xx->x. I think the expansion would be to order zero, else total degree of resulting Taylor polynomials will exceed 4.

(3) Notice that this gives a biquadratic in x so the solution will be done by iterating solutions for quadratics. This gives those complicated formulas. One can get a more concise form via Root objects (prametrized in terms of the variables other than x). To do this, add a “junk” linear term to avoid the biquadratic, and set Quartics->False in Solve to avoid the Cardano-Tartaglia form of result. Then remove that junk term. If expr is that big quartic, then the code to do this would be:

Solve[expr+junk*x==0,x,Quartics->False]/.junk->0
7

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

8

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

9

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?

10

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.