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I have an equation I want to solve, so I entered the following into Mathematica. 

(x^(17/6))/(a^(17/6)) - (x^2)/a -1 == 0

where I assume $a>>1$ and $x$ is the unknown. How do I compute the dependence of $x$ on $a$ as $a \rightarrow \infty$.

Descartes
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  • It is still not clear what you were expecting Mathematica to do with your equation. Just entering an equation doesn't make any magic happen. In order to help you, we need to have much more detail on what you did and what result you expected. – m_goldberg Feb 07 '13 at 12:39
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    A remarkably similar question appears at http://mathematica.stackexchange.com/questions/19047/how-can-i-solve-an-equation-with-symbolic-coefficients. – whuber Feb 07 '13 at 17:11
  • The question asked at the site given in a previous comment has another nick. If it is not another post of the same person using two nicks, it looks like a homework. In that case we should not answer, should we? I have, however, already given the solution that should be simply translated into the notations used in this post. In particular one should rescale the equation, rename the parameter c=a^(-5/16) and the unknown variable: z=x*a^(6/17). Large a corresponds to small c. Solution at small c is given at http://mathematica.stackexchange.com/questions/19047... – Alexei Boulbitch Feb 08 '13 at 09:57
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    I should add that in the past (before Mathematica) such simple problems we did by hands, in few seconds. General approach one may find, for instance, in the book Arthur Erdélyi, Asymptotic expansions. New York, N.Y., Dover 1956 or in any other book on the same subject. – Alexei Boulbitch Feb 08 '13 at 10:06

1 Answers1

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Find the first two terms of the expression expanded as a series at a=infinity. Trial and error quickly indicates that expanding to order 2 suffices.

ee = (x^(17/6))/(a^(17/6)) + (x^2)/a - 1;

ser = Normal[Series[ee, {a, Infinity, 2}]]

(* Out[112]= -1 + x^2/a *)

This tells you that x will be approximately sqrt(a). You can check this for plausibility by using FindRoot with a set to 1000, say. The solution is around 31.622, which is fairly close to sqrt(1000).

Daniel Lichtblau
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