Example. By applying the Remainder Estimate to the function $f(x)=$sin $x$, with $a=0$ and $n=3$, calculate sin $0.1$ to four decimal places.
Recall. Let $f$ be $(n+1)$-times differentiable on an open interval containing the points $a$ and $x$.
$\left| R_{n}\left( x\right) \right| \leq \dfrac {M} {\left( n+1\right) !}\left| x-a\right| ^{n+1}$.
Strictly speaking, we should use more complicated notation to indicate that the upper bound $M$ depends on $n$, $a$ and $f$.
Soulution of example.
By the Taylor's theorem, we obtain $P_3 (x)= x-\dfrac {1} {6} x^3$.
Also, $f^{ıv} (x)=$sin $x$, so that
$\left| f^{ıv} (c)\right| \leq \left| \sin(c) \right| \leq 1$ for all $c$.
$(*)$ Taking $M=1$ in the remainder estimate, we deduce that
$\left| R_3 (0.1)\right| \leq \dfrac {1} {2}\times 10^{-5}$.
Now,
$\sin(0.1) \simeq P_3 (0.1)=0.1-\dfrac {1} {6}\times 10^{-3}=0.00001666...=0.0998333...$. Hence, $\sin(0.1)=0.0998 (to four decical places). We are done.
My question is Why did we do this statement $(*)$, how did we use $(*)$? So, I couldn't see in the proof that we couldn't use this statement $(*)$.
That $M$ is a bound on the absolute value of the fourth derivative of $\sin(x)$ between $0$ and $x$. That fourth derivative is again $\sin(x)$ which is uniformly bounded in absolute value by $1$. So $|R_3(x)|$ in this case is bounded above by $\frac{1}{24} |x|^4$.