Apply the Maclaurin formula to f(x)=sin x, estimate

Apply the Maclaurin formula to $f(x)=\sin x$, estimate the remainder, and determine how many terms are needed so that the error is less than $10^{-3}$ for $x\in[0,\tfrac\pi4]$.

Step 1: Derivatives of $\sin x$ at $x=0$

The derivatives cycle with period 4: $\sin x,\cos x,-\sin x,-\cos x,\ldots$; at $x=0$, even-order derivatives vanish and odd-order ones alternate $\pm1$: \[f^{(2k+1)}(0)=(-1)^k. \tag{1}\]

Step 2: Maclaurin polynomial and Lagrange remainder

\[ \sin x = x-\frac{x^3}{3!}+\frac{x^5}{5!}-\frac{x^7}{7!}+\cdots+(-1)^n\frac{x^{2n+1}}{(2n+1)!}+R_{2n+2}(x). \tag{2} \]

The next derivative, $f^{(2n+2)}(x)=\pm\sin x$ or $\pm\cos x$, is bounded by $1$ in absolute value, so

\[ |R_{2n+2}(x)| \le \frac{|x|^{2n+2}}{(2n+2)!}. \tag{3} \]

Step 3: Bound on $[0,\pi/4]$

The maximum of $|x|$ on $[0,\pi/4]$ is $x=\pi/4\approx0.7854$. Testing successive $n$: \[n=1:\ \frac{(\pi/4)^4}{4!}\approx\frac{0.3802}{24}\approx0.0158\quad(\text{too large}). \tag{4}\]

\[ n=2:\ \frac{(\pi/4)^6}{6!}\approx\frac{0.2346}{720}\approx0.000326\ \lt \ 10^{-3}.\ \checkmark \tag{5} \]

Step 4: Verification

\[x=\pi/4\text{: }P(x)\approx0.70663,\ \sin(\pi/4)\approx0.70711,\ |{\rm error}|\approx4.8\times10^{-4}\lt 10^{-3}\ \checkmark\]