How to find the quadratic approximation of a quotient?

Question

How would I find the quadratic approximation of some quotient like tan(x)?

Can I rewrite it as $\tan(x) = \frac{\sin(x)}{\cos(x)}$, and then find the approximation for sine and cosine separately?

Answer 1

You can do it; this is series composition.

Suppose that you want the Taylor series of $$y=\frac{f(x)}{g(x)}$$ built around $x=0$ up to second order. You have $$f(x)=f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right)$$ $$g(x)=g(0)+x g'(0)+\frac{1}{2} x^2 g''(0)+O\left(x^3\right)$$ $$y=\frac{f(0)+x f'(0)+\frac{1}{2} x^2 f''(0)+O\left(x^3\right) }{g(0)+x g'(0)+\frac{1}{2} x^2 g''(0)+O\left(x^3\right) }$$ Now, use the long division to get $$y=\frac{f(0)}{g(0)}+\frac{ \left(g(0) f'(0)-f(0) g'(0)\right)}{g(0)^2}x+\frac{ \left(g(0)^2 f''(0)-2 g(0) f'(0) g'(0)-f(0) g(0) g''(0)+2 f(0) g'(0)^2\right)}{2 g(0)^3}x^2+O\left(x^3\right)$$

Answer 2

I’m not entirely sure if this is what you’re asking, but I suppose you could use the first few terms of the Taylor series

$$\tan x = x + \frac{x^3}{3} + \frac{2x^5}{15} +\cdots $$

for $|x|<\pi/2$. Does that help?

Answer 3

$$\begin{align} f(x) &= \tan x \\ f’(x) &= \sec^2x \\ f’’(x) &= 2\sec(x)\tan(x) \\ \end{align}$$

You should just memorize $f’(x)$, and $f’’(x)$ comes from chain rule.

Assuming we expand the approximation around $x=a$, we’ll use

$$f(x)\approx f(0)+f’(0)\,(x-a)+\frac{f’’(0)\,(x-a)^2}{2}$$

Use

$$\begin{align} \tan0&=0\\ \sec^20&=1\\ 2\sec(0)\tan(0)&=0\\ \end{align}\\$$

Giving you the rather lousy $f(x)\approx x$, hence my recommendation in my other answer.