Determine the interval [-a,a] on which the degree-4

Determine the interval $[-a,a]$ on which the degree-4 Maclaurin polynomial approximates $y=\cos x$ with error less than $\varepsilon=10^{-5}$.

Step 1: Degree-4 polynomial and its remainder

\[ \cos x = 1-\frac{x^2}{2!}+\frac{x^4}{4!}+R_5(x), \qquad P_4(x)=1-\frac{x^2}{2}+\frac{x^4}{24}. \tag{1} \]

Since $f^{(5)}(x)=\sin x$, the Lagrange remainder is $R_5(x)=\dfrac{\sin\xi}{120}x^5$, $\xi$ between $0$ and $x$; as $|\sin\xi|\le1$, \[|R_5(x)| \le \frac{|x|^5}{120}. \tag{2}\]

Step 2: Impose the error condition

The bound is largest at $|x|=a$, so requiring $|R_5(x)|\lt 10^{-5}$ throughout $[-a,a]$ means

\[ \frac{a^5}{120} \lt 10^{-5} \;\Longrightarrow\; a^5 \lt 1.2\times10^{-3} \;\Longrightarrow\; a \lt \bigl(1.2\times10^{-3}\bigr)^{1/5}. \tag{3} \]

Step 3: Numerical evaluation

\[ \bigl(1.2\times10^{-3}\bigr)^{1/5} \approx 0.2605. \tag{4} \]

Checking the boundary: at $a=0.26$, $a^5/120\approx9.9\times10^{-6}\lt 10^{-5}$ $\checkmark$; at $a=0.27$, $a^5/120\approx1.19\times10^{-5}\gt 10^{-5}$ (too large).

Step 4: Verification

\[x=0.26\text{: }P_4(0.26)\approx0.96639370,\ \cos(0.26)\approx0.96639352,\ |{\rm error}|\approx1.8\times10^{-7}\lt 10^{-5}\ \checkmark\]