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}<br />

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 &gt;&gt; \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.