Binomial Theorem Application in Cauchy's and Sellmeier's Equations

[Luminous Blob]

I am trying to do a question from Eugene Hecht's Optics book, which goes something like this:

Given the following equations:

Cauchy's Equation:

$$n = C_1 + \frac{C_2}{\lambda^2} + \frac{C_3}{\lambda^4} + ...$$

Sellmeier's Equation:

$$n^2 = 1 + \sum_{j} \frac{A_j\lambda^2}{\lambda^2-\lambda_0_j^2}$$

where the $$A_j$$ terms are constants and each $$\lambda_0_j$$ is the vacuum wavelength associated with a natural frequency $$v_0_j$$, such that $$\lambda_0_jv_0_j = c$$.

Show that where $$\lambda >> \lambda_0_j$$, Cauchy's Equation is an approximation of Sellmeier's Equation.

Now it also gives a hint which is as follows:

Write the above expression with only the first term in the sum; expand it by the binomial theorem; take the square root of $$n^2$$ and expand again.

From the hint, I gather that it means to rewrite Sellmeier's Equation as:

$$n^2 = 1 + \frac{A\lambda^2}{\lambda^2 - \lambda_0^2}$$

From there though, I have no idea how to apply the binomial theorem to expand it. I just don't see how anything in that equation has the form $$(x+y)^n$$, except for where n = 1.

If anyone can explain to me how to apply the binomial theorem to the equation, or if I've misunderstood what the hint means, it would be much appreciated.

 

[Galileo]

You can use the binomial theorem to expand $(1+x)^{1/2}$ when x<<1.

 

[Luminous Blob]

So you mean first take the square root of both sides, then expand it using the binomial theorem , letting $$x = \frac{A\lambda^2}{\lambda^2 - \lambda_0^2}$$, rather than first applying the binomial theorem, then taking the square root of both sides and then expanding again like the hint suggests?

 

[Dr Transport]

Rewrite $$\frac{A_j\lambda^2}{\lambda^2-\lambda_0_j^2}$$ as

$$\frac{A_j}{\lambda^2}\frac{1}{1-\frac{\lambda_0_j^2}{\lambda^2}}$$ and expand the second part as

$$\frac{1}{1-x^2} \approx 1 - x^2 + x^4 - x^6 \ldots$$ where $$x = \frac{\lambda}{\lambda_0_j}$$

 

[Luminous Blob]

Aah, I didn't think to do that. Thanks, that was a great help.