You can also use the new in M12 function AsymptoticSolve
. First, it will be simpler to use a new variable x2 = x^2
:
eqn = x2 == (Subscript[a, 4] K^6)/(9 x2^3) + (Subscript[a, 3] K^4)/(7x2^2) +
(Subscript[a, 2] K^2)/(5 x2) + Subscript[a, 1]/3;
eqn //TeXForm
$\text{x2}=\frac{a_4 K^6}{9 \text{x2}^3}+\frac{a_3 K^4}{7 \text{x2}^2}+\frac{a_2 K^2}{5
\text{x2}}+\frac{a_1}{3}$
At K=0
, x^2
is given by $a_1/3$, so the call to AsymptoticSolve
is:
x2 /. First @ AsymptoticSolve[eqn, {x2, Subscript[a,1]/3}, {K, 0, 6}] //TeXForm
$\frac{3 \left(1134 a_2^3-2025 a_1 a_3 a_2+875 a_1^2 a_4\right) K^6}{875 a_1^5}+\frac{9
\left(25 a_1 a_3-21 a_2^2\right) K^4}{175 a_1^3}+\frac{3 a_2 K^2}{5 a_1}+\frac{a_1}{3}$
in agreement with my other answer.
Before you have access to M12, you can use CloudEvaluate
:
$VersionNumber
x2 /. CloudEvaluate @ First @ System`AsymptoticSolve[
eqn,
{x2, Subscript[a,1]/3},
{K, 0, 6}
] //TeXForm
11.3
$\frac{3 \left(1134 a_2^3-2025 a_1 a_3 a_2+875 a_1^2 a_4\right) K^6}{875 a_1^5}+\frac{9
\left(25 a_1 a_3-21 a_2^2\right) K^4}{175 a_1^3}+\frac{3 a_2 K^2}{5 a_1}+\frac{a_1}{3}$
NestList
), still it doesn't make much sense to calculate successive iterations symbolically. What are you actually trying to do? $\endgroup$a1, a2, a3 and a4
too. I want to keep these parameters. $\endgroup$x^2
. Just useSolve
$\endgroup$Solve[x^2 == (a[4] k^6)/(9 x^6) + (a[3] k^4)/(7 x^4) + (a[2] k^2)/(5 x^2) + a[1]/3, x]
and gaze at the horror that are the symbolic solutions... Also, stay away fromSubscript
at the beginning; use indexed variables, i.e.a[1]
instead ofSubscript[a, 1]
. Also stay away from uppercase variables; for instance, yourK
has a conflict with a built in symbol. $\endgroup$x^2
in the form:x^2=A+B K^2+C K^4+D K^6
$\endgroup$