# series expansion

I have a fourth order equation

equation (22)

and I must solve it with respect to delta0 (i.e.: Solve[equation,delta0]) to have a solution for small value of mu1 and mu2. So is it possible to expand in series of delta0 (second order) and take only first order terms in mu (i.e.: not mu1*mu2 but only mu1+mu2 ).

The equation is (23).

Edit

Solve[Subscript[[Mu], 1]*Subscript[[Mu], 2]^2 (1+d/2-d^2/8)^4+2Subscript[[Mu], 1]^2
Subscript[[Mu], 2] (1+d/2-d^2/8)^3+(1+d/2-d^2/8)^2
(Subscript[[Mu], 1]^3+Subscript[[Mu], 2]^2-(Subscript[[Mu], 1]+
Subscript[[Mu], 2])^3-3^(4/3) Subscript[[Mu], 1]*Subscript[[Mu], 2]
(Subscript[[Mu], 1]+Subscript[[Mu], 2])^(5/3))+2Subscript[[Mu],1]*
Subscript[[Mu], 2]^2 (1+d/2-d^2/8)+Subscript[[Mu], 1]^2 Subscript[[Mu], 2]==0,d]


... so i can't match this equation witch the one i can found in the pape

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Welcome to Mathematica.SE, Federico! Please consider registering your account so you can use the same account from multiple machines. You can use LaTeX markup on this site so there is no need to show equations as images. To help people answer the question, please let us know what you have tried. –  Verbeia Oct 11 '12 at 12:46
Welcome! Please write down those equations in Mathematica code to relief potential answerers from the burden –  belisarius Oct 11 '12 at 12:49
I'm not sure I understand the question. $\delta_0$ is expanded to $\sqrt{1+\Delta_c}\approx1+0.5\Delta_C-\frac{\Delta_C^2}{8}$ in the line below (22) with the truncated solution for $\Delta_C$ then given as $2.4(\mu_1+\mu_2)^\frac{1}{3}$. Is that the expansion you are looking for? Are you looking for a step by step or the result? –  fizzics Oct 11 '12 at 13:15
yes fizzics if you see the equation (23) i try to find this espression but i can't understand how, with mathematica. –  federico Oct 11 '12 at 13:43
i put here my attempt to solve the equation. i put in the equation the truncated expansion of 'δ_0' and i try to solve it but i found a very complex expression. –  federico Oct 11 '12 at 13:53