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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*}

Jeroen van Riel
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  • (wrt. your question in the comments on my answer) - I think the general approach of comparing $\cos$ to $1-x^2/2$ probably works (and it's essentially the same thing I've done in my answer) but isn't it abuse of notation to use little/big o notation without a function? If $f(x) \in x+\mathcal{O}(x^2)$ means that $f(x)<x+kx^2$ for some constant $k$, then what does $\mathcal{O}(x^2)$ mean by itself? I'm not certain it's valid to use it as a single variable? – Jam Apr 10 '18 at 15:30
  • Also, would it not be easier to simply bound $\cos$ by a Taylor polynomial, and then compare that polynomial to $x^2/2 + x^3/6$ ? – Jam Apr 10 '18 at 15:32
  • I thought that writing $o(x^4)$ was equivalent to stating that the expression holds for "all function $f(x)$ such that $|f(x)| \leq |\epsilon x^4|$ for all constants $\epsilon > 0$". – Jeroen van Riel Apr 10 '18 at 15:56
  • You might be right, I don't actually know. – Jam Apr 10 '18 at 15:57
  • You propose 'simply bounding' $\cos$, which is probably what I'm looking for, but I am not sure how to approach this. How do I deal with this remainder function then? – Jeroen van Riel Apr 10 '18 at 16:03
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    By "bounding $\cos$", I meant with a method like the one I linked in my answer, where you consider the derivative of $(1-x^2/2)+\cos(x)$. But maybe you can bound the remainder with one of these formulae? https://en.wikipedia.org/wiki/Taylor%27s_theorem#Explicit_formulas_for_the_remainder I'm not actually completely sure about this, sorry :( – Jam Apr 10 '18 at 16:11
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    Thanks for your help so far! I will have a look at those. – Jeroen van Riel Apr 10 '18 at 16:15
  • Sure about the $+$ sign in $+\frac{x^3}6$ on the RHS? Since $1-\cos x\leqslant\frac{x^2}2$ is direct, adding a positive term to the upper bound $\frac{x^2}2$ is rather odd. – Did Apr 10 '18 at 16:27
  • The original problem (from an actual exam) is as stated in the title, so I thought this necessary to conclude. – Jeroen van Riel Apr 10 '18 at 16:29

2 Answers2

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For $x>0$,

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

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

Jam
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2

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*}

Jeroen van Riel
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