$1-\cos (x) \leq \frac{x^2}{2} + \frac{x^3}{6}$, for $x > 0$, using Taylor expansion

Question

I want to solve this problem using Taylor expansions.

I tried

\begin{align*} 1 - \cos (x) = 1 - \left( \sum_{k=0}^{n} (-1)^k \frac{x^{2k}}{(2k)!} +R_{2n}(x) \right) \;, \end{align*}

where $R_{2n}(x)$ is the remainder function.

For $n=2$, this yields

\begin{align*} 1 - \cos (x) = 1 - \left( 1 - \frac{x^2}{2} + \frac{x^4}{24} + R_4(x) \right) = \frac{x^2}{2} - \frac{x^4}{24} - o(x^4) \;. \end{align*}

Is it now possible to just state

\begin{align*} |o(x^4)| \leq \frac{x^4}{24} \; , \end{align*} and hence \begin{align*} \frac{x^2}{2}-\frac{x^4}{24}-o(x^4)\leq \frac{x^2}{2} \leq \frac{x^2}{2} + \frac{x^3}{6} \; ? \end{align*}

Answer 1

I think I found an appropriate method.

The formula for the remainder term of an $n$-degree Taylor polynomial around $a$ is given by

\begin{align*} R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x-a)^{n+1} \; , \end{align*}

for some constant $c$ in the open interval between $a$ and $x$, so $c\in(a,x)$ or $c\in(x,a)$.

The remainder term for the second-order Taylor polynomial $T_2(x)$ of $\cos x$ is given by\begin{align*} R_2(x) = \frac{-\cos c}{6} x^3 \; , \end{align*}

which can be bounded: $-x^3/6 \leq R_2(x) \leq x^3/6$.

From this, the claim now immediately follows

\begin{align*} 1 - \cos x = 1 - (T_2(x) + R_2(x)) = 1 - (1 - \frac{x^2}{2} + R_2(x)) = \frac{x^2}{2} - R_2(x) \leq \frac{x^2}{2} + \frac{x^3}{6} \;. \end{align*}

Answer 2

For $x>0$,

$$\cos x > 1 -\frac{x^2}2$$

$$\begin{aligned}1-(\cos x) &< 1-\left(1 -\frac{x^2}2\right)\\ 1-\cos x&< \frac{x^2}2 +\frac{x^3}6\end{aligned}$$