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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 $a \rightarrow \infty$.

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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 7 at 12:39
if you meant to take the limit of the expression as a->Infinity, then type eq = (x^(17/6))/(a^(17/6)) + (x^2)/a - 1; Limit[eq, a -> Infinity] this gives -1. But @m_goldberg said, you are using wrong terms all over the place. You mention Solve, then write an equation, then talk about dependence, which implies yet another function, then talk about limits. So it is not clear. In Math you need to use clear and correct terms so not to confuse others. – Nasser Feb 7 at 12:53
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A remarkably similar question appears at mathematica.stackexchange.com/questions/19047/…. – whuber Feb 7 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 mathematica.stackexchange.com/questions/19047... – Alexei Boulbitch Feb 8 at 9:57
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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 8 at 10:06

1 Answer

up vote 5 down vote

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).

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