Taylor Series Remainder Use Taylor's Theorem to estimate the error in approximating $\sinh 2x$ by $2x + 4/3x^3$ on the interval $[-0.5,0.5]$.
For this question, I use the Taylor's remainder formular, $$
R_n(x)= \frac{f^{(n+1)}(z)(x-a)^{n+1}}{(n+1)!}$$
and I get $R_4(0.5) = 0.012\,85$.
Is this correct?
 A: Generally, you want to use:
$$\displaystyle |R_{n}(x)| \le \frac{M_{n+1}}{(n+1)!}|x-a|^{n+1}$$
where,  $\displaystyle R_{n}(x) = f(x) -T_n(x)$ is the remainder term and $T_n(x)$ is the Taylor polynomial of degree $n$ for $f(x)$, centered at $x = a$.
For this problem, we have an odd function, $f(x) = \sinh 2x$ on the interval $\displaystyle \left[-\frac{1}{2},\frac{1}{2}\right]$, so we can take:
$$\displaystyle T_4(x) = P_0(x) + P_1(x) + P_2(x) + P_3(x) + P_4(x) = f(0) + f'(0)(x-0) + \frac{f''(0)}{2!}(x-0)^2 + \frac{f'''(0)}{3!}(x-0)^3 + \frac{f^{(4)}(0)}{4!}(x-0)^3= 0 + 2x + 0 + \frac{8 x^3}{6} + 0 = 2x + \frac{4}{3} x^3$$
We also have: $f^{(5)}(x) = 32 \cosh x$, so
Max $\displaystyle |f^{(5)}(x)|$ on $\displaystyle -\frac{1}{2} \le x \le \frac{1}{2}$ occurs on the endpoints, that is $\displaystyle x = \pm \frac{1}{2}$, so:
$$\displaystyle M_5 = \max_{-\frac{1}{2} \le x \le \frac{1}{2}} \left|f^{(5)}(x)\right| = 32 \cosh(1) \approx 49.37858$$
So, the upper error bound is given by:
$$\displaystyle |f(x) -T_4(x)| = |R_4(x)| \le \frac{M_5}{5!}|x-0|^5 = \frac{49.37858}{5!}\left|\frac{1}{2}\right|^5 = 0.0128590052083333$$
