I've been trying to calculate the maximum possible error for the Taylor Series (with $n=3$, $c=0$, $|x| \leq .1$) of the function $f(x)=(1+2x)^{-4}$ Here's my work:
$$f(x)=(1+2x)^{-4}$$ $$f'(x)=-8(1+2x)^{-5}$$ $$f''(x)=80(1+2x)^{-6}$$ $$f'''(x)=-960(1+2x)^{-7}$$ $$f''''(x)=13440(1+2x)^{-8}$$
Now we know, $$|R_n| \leq |\frac{f^{n+1}(z)(x-c)^{n+1}}{(n+1)!}|$$
Plugging values in,
$$R_3 \leq \frac{13440(1+2z)^{-8}(x)^{4}}{(4)!}$$ $$R_3 \leq \frac{13440x^{4}}{4!(1+2z)^{8}}$$ $$R_3 \leq \frac{560x^{4}}{(1+2z)^{8}}$$
Now, since $c \leq z \leq x$, we have
$$0 \leq z \leq .1$$
Thus,
$$(1+2z)^8 \leq 1.2^8$$
Now we have,
$$R_3 \leq \frac{560(.1)^4}{1.2^8} \approx .013$$
But my book has
$$R_3 \leq .34$$
Where did I go wrong? Thanks