<s>[INST] find upper bound M taylor inequality Let $f(x) = \frac{1}{1-x}$
Find an upper bound M for $|f^{(n+1)}(x)|$ on the interval $[-1/2,1/2]$
I found derivative of the function and plugged in the points to see which one gives the largest value.
$f^{'}(x)=\frac{1}{(1-x)^2}$
$f^{'}(1/2)=4$
The solution was:  $2^{n+2}(n+1)!$
I know the formula $|R_n(x)|= \frac{M|x-a|^{n+1}}{(n+1)!}$ but not sure how they got the M. Any help?
 [/INST] That is because you need to find the largest value of the $n$th derivative of $f$ over the interval [-0.5,0.5]. Given $f(x) = \dfrac{1}{1-x}$,
\begin{align*}
f'(x) & = \frac{1}{(1-x)^2}.\\
f''(x) & = \frac{1\cdot 2}{(1-x)^3}.\\
f'''(x) & = \frac{1\cdot 2\cdot 3}{(1-x)^4}.\\
\vdots & \qquad \vdots \\
f^{(n)}(x) & = \frac{1\cdot 2 \cdot\ldots\cdot n}{(1-x)^{n+1}} = \frac{n!}{(1-x)^{n+1}}.
\end{align*}
Thus, the maximum of $f^{(n)}(x)$ over the interval $[-0.5,0.5]$ is given by
$$ \max_{x\in [-0.5,0.5]} f^{(n)}(x) = \max_{x\in [-0.5,0.5]} \frac{n!}{(1-x)^{n+1}} = \frac{n!}{(1-0.5)^{n+1}} = 2^{n+1}n!.$$
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