Show that $|\sin(0.1) – 0.1| \leq 0.001$
I know that's a basic exercise on taylor polynomial but I have made a mistake somewhere that I don't find out. Anyway, here's my attempt :
Because the function $f: \mathbb{R} \rightarrow \mathbb{R}$, $x \rightarrow \sin(x)$ is $1$ time derivable,
by the Taylor polynomial formula we find:
\begin{equation*}
\sin(x) = x + R^1_0 \sin(x)
\end{equation*}
Therefore,
\begin{equation*}
|sin(0.1) – 0.1| = |R^1_0 \sin(0.1)|
\end{equation*}
Because $f$ est 2 times derivable,
by the Lagrange remainder formula, $\exists c \in ]0, 0.1[$ such that
\begin{align*}
R^1_0 \sin(0.1)
&= f^{(2)}(c) \frac{(0.1)^2}{2!} \\
&= – sin(c) \frac{0.01}{2} \\
|R^1_0 \sin(0.1)| &= |sin(c) \cdot 0.005|
\end{align*}
Because $|sin(x)| \leq 1$, $\forall x \in \mathbb{R}$
\begin{align*}
|R^1_0 \sin(0.1)|
&\leq |0.005|
\end{align*}
However, $0.005 > 0.001$ so I'm wondering where I did a mistake ?
Best Answer
You also know that $\vert \sin x \vert \le \vert x \vert$. Hence
$$\vert \sin(c) \vert \frac{0.01}{2} \le 0.1 \frac{0.01}{2} = 0.0005 < 0.001$$