Taylor expansion - imaginary coefficients?

[Froskoy]

Homework Statement

Find the first two non-zero terms in the Taylor expansion of $\frac{x}{\sqrt{x^2-a^2}}$ where a is a real constant

Homework Equations

$$f(x)=f(x_0)+f^{\prime}(x_0)(x-x_0)+\frac{f^{\prime\prime}(x_0)}{2!}(x-x_0)^2+...+\frac{f^{(n)}(x_0)}{n!}(x-x_0)^n$$

The Attempt at a Solution

If a=0 then f(x)=1 (in the case a=0,x=0, use l'Hopital's rule to find the limit of the ratio as $x$ approaches 0 - is this correct?)

then $$f(0)=0<br /> <br /> f^{\prime}(0)=\frac{-ai}{a}<br /> <br /> f^{\prime\prime}(0)=\frac{-i}{2a^{13}}$$

Is it OK that there are imaginary terms here? I guess $f(x)$ is imaginary if $a^2>x^2$ anyway?

 

[clamtrox]

If you really need to expand it at x=0, then your function is imaginary, so of course you expect to get imaginary coefficients. Usually though in expressions like this, x is limited to values larger than a.

I'm not sure if your coefficients are correct though. Maybe it's easier to expand $g(x)=(x^2-a^2)^{-1/2}$

 

[Froskoy]

Thanks! I hadn't thought of expanding $\frac{1}{\sqrt{x^2-a^2}}$ and then multiplying my $x$ - that's really cool - thanks!