# Maclaurin Series Question ??

Question: Find maclaurin series using the first few terms for: $$\frac1{x^2}\left(\frac{\sin x}{\cos x}-x\right)$$

The answer is $x/6$.

My working out so far:

$$\frac{\sin x}{\cos x}= \frac{x-\frac{x^3}6+\cdots}{1-\frac{x^2}2+\cdots}$$

However, I cant seem to simplify it down to $x/6$?

• What do you mean x/6? Does the book just say x/6, or are there more terms after wards? – Crescendo Dec 9 '17 at 2:34
• says " x/6 ... " literally. – Mathematica Dec 9 '17 at 2:35
• @Crescendo Of course, there will be higher order terms. I suppose the question means to find the lowest order term – John Doe Dec 9 '17 at 2:41

Use the series expansion$$\frac1{1-y}=1+y+y^2+...$$ All you need are the first two terms. You have $$\frac{\sin x}{\cos x}=\frac{x-\frac{x^3}6+\cdots}{1-\frac{x^2}2+\cdots}$$ so take $y=\frac{x^2}2$. Then you get$$\frac{\sin x}{\cos x}=\frac{x-\frac{x^3}6+\cdots}{1-\frac{x^2}2+\cdots}=\left(x-\frac{x^3}6+\cdots\right)\left(1+\frac{x^2}2+\cdots\right)$$ Can you finish it from here?
• It is quite commonly known. It is the maclaurin expansion about $0$: $$(1-x)^{-1}=(1-0)^{-1}+(-1)(-1)(1-0)^{-2}\cdot x+(-2)(-1)(1-0)^{-3}\cdot(\frac{x^2}2)+\cdots \\=1+x+x^2+\cdots$$ – John Doe Dec 9 '17 at 3:55
• Also $\frac{1-y^n}{1-y}=1+y+y^2+...+y^{n-1}$ so $\frac{1}{1-y}=1+y+y^2+...+y^{n-1}+\frac{y^n}{1-y}$. Let $n \to \infty$ and use $y^n \to 0$. And if you integrate this you get the power series for $\ln(1-y)$. – marty cohen Dec 9 '17 at 5:07