What are the first four terms of the Maclaurin series of $2x\cos^2(x)$, using the series for $\cos x$?

Question

How do I find Maclaurin series of $2x\cos^2(x)$ using the series for $\cos x$? \begin{align} \cos x&\approx 1- \frac {x^2}2 \\ \cos^2(x)&\approx 1-2x^2+\frac{x^4}4\\ 2x\cos^2(x)&\approx 2x -4x^3+ \frac{x^5}5 \end{align}

Answer 1

$$\cos(2x) = 2\cos^2 x -1$$

\begin{align} \cos^2x &= \frac12 ( 1+\cos (2x))\\ &= \frac12\left( 1+ \sum_{k=0}^\infty \frac{(-1)^k(2x)^{2k}}{(2k)!}\right)\end{align}

\begin{align} 2x\cos^2x &= \left( x+ \sum_{k=0}^\infty \frac{(-1)^k2^{2k}x^{2k+1}}{(2k)!}\right)\end{align}

Remark about your attempt:

Note that you did not start with sufficient terms to obtain $4$ terms.

Also, remember that $(a-b)^2 = a^2-\color{blue}2ab+b^2$, also $\frac24 \ne \frac15$ in your last line.

Answer 2

You should write by little-o (or also Big-O) notation to take into account the remainder

$$\cos x = 1- \frac {x^2}2+o(x^2)$$

or also (since the next term is a 4th order)

$$\cos x = 1- \frac {x^2}2+o(x^3)$$

then

$$\cos^2 x=\left(1- \frac {x^2}2+o(x^3)\right)^2=1-2\frac {x^2}2+\frac {x^4}4+o(x^3)=1-x^2+o(x^3)\implies2x \cos^2x=2x-2x^3+o(x^4)$$

thus to obtain more terms we should expand further $\cos x$ that is

$$\cos x = 1- \frac {x^2}2+\frac {x^4}{24}- \frac {x^6}{720}+o(x^7)$$

Answer 3

Instead of squaring, it's often easier to use the identity $2 \cos^2 x = 1 + \cos 2x$.

$$\begin{align}2 \cos^2 x &= 1 + \cos 2x\\ &= 1 + \left(1 - \frac{(2x)^2}{2!}+\frac{(2x)^4}{4!}-\frac{(2x)^6}{6!} + O(x^8)\right) \\ &=2 - 2x^2 +\frac 23 x^4-\frac4{45} x^6 + O(x^8)\\ 2 x \cos^2 x &= 2x - 2x^3 +\frac 23 x^5-\frac4{45} x^7 + O(x^9). \end{align}$$

This is ultimately just a less general form of Siong Thye Goh's answer, but may be clearer to follow for some.