# Solve equation perturbatively

1. Jul 8, 2010

### norman86

Hi all! I want to perturbatively solve this equation in $$\beta$$ at second order in $$\alpha$$

$$\frac{\beta^{2}}{4}-\frac{3}{8}\beta^{4}\alpha+\frac{2}{3}\beta^{6}\alpha^{2}=\frac{1}{4}$$

I rewrite this formula in this way

$$\beta^{2}=1+\frac{3}{2}\beta^{4}\alpha-\frac{8}{3}\beta^{6}\alpha^{2}$$

When I try to solve it perturbatively, I obtain

$$\beta^{2}=1+\frac{3}{2}1^{2}\alpha-\frac{8}{3}\alpha^{2}\left(1+\frac{3}{2}1^{2}\alpha\right)^{3}$$

The result is

$$\beta^{2}=1+\frac{3}{2}\alpha-\frac{8}{3}\alpha^{2}$$

The square root of which gives

$$\beta=1+\frac{3}{4}\alpha-\frac{155}{96}\alpha^{2}$$

I know that the correct result is

$$\beta=1+\frac{3}{4}\alpha+\frac{61}{96}\alpha^{2}$$

Clearly, there is something wrong in the second order of my calculus.

Can anyone please tell me where I'm mistaking? Thanks a lot.

2. Jul 8, 2010

### arildno

Hmm..
Not quite sure what you are doing here, I'm afraid.
Here's how I would do it.
Assume a power series expansion as the following:
$$\beta=1+k_{1}\alpha+k_{2}\alpha^{2}+++$$
We are to determine the k's in orders of alpha.
Inserting in your equation, and retaining terms of order 0 and 1, we get:
$$1+2k_{1}\alpha=1+\frac{3}{2}*1\to{k}_{1}=\frac{3}{4}$$

To order 2, we have:
$$k_{1}^{2}\alpha_{2}+2k_{2}\alpha^{2}=\frac{3}{2}*4k_{1}\alpha^{2}-\frac{8}{3}*1*\alpha^{2}$$
Or, we get the equation for k_2:
$$\frac{9}{16}+2k_{2}=\frac{9}{2}-\frac{8}{3}$$
Which yields:
$$2k_{2}=\frac{216-128-27}{48}=\frac{61}{48}$$
whereby the result follows.

3. Jul 8, 2010

### norman86

You are right. That's the right way to do it. Thank you very much!