Find a function f so that Taylor expansion is always accurate to this degree Find a function $f$ from R to N such that with $T$ be the Taylor expansion of $\sin(x)$ around $0$.
$ | \sin (x) - T_{f(x)}x$| $\leq 1$
The hint is to use $n! \leq 3 \sqrt{n} {(\frac{n}{e})}^n$
 A: So Taylor's theorem* says that if $f$ is smooth (i.e. infinitely differentiable), then $\sin(x) = \sum_{k = 0}^{n} (-1)^{k} \frac{x^{2k + 1}}{(2k + 1)!} + R_{n}(x)$, where $R_{n}(x) = \frac{\sin^{(2n + 3)}(c)}{(2n + 3)!}x^{2k n + 3}$, where $c$ is dependent on $n, x$. But we do know that $|\sin^{(2n + 3)}(c)| \leq 1$, so we have that
\begin{align*}
\left| \sin(x) - \sum_{k = 0}^{n} (-1)^{k} \frac{x^{2k + 1}}{(2k + 1)!} \right| & \leq \frac{|x|^{2n + 3}}{(2n + 3)!} \\
\end{align*}
But more generally, if you have a function that's equal to its (infinite) Taylor series expansion, then by the definition of a limit such an $f(x)$ exists.
Provided here is a way to show $f(x)$ does in fact exist. I don't know how to hide it like a spoiler, so I'm just gonna put it here as a link. If anyone wants to show me how to do the spoiler deal I'd be very appreciative.
http://www.texpaste.com/n/b9vrly1z
*I misspoke, as a commenter pointed out. What I'm more particularly using is the Lagrange form of the remainder. Conveniently, this can still be found on the Wikipedia page on Taylor's theorem.
A: So you want to know the degree $n-1$, depending on $x$, where the error is less than $1$. This number is your desired $f(x)$. With this degree, you have the Lagrange error bound of $\frac{|x|^n}{n!}$. It is actually not at all clear how to use your hint here, because you really want a lower bound on $n!$, but your hint gives an upper bound.
One lower bound turns out to be $n^n/e^n$; see http://www.johndcook.com/blog/2011/06/10/stirling-approximation/ So you have an error bound of $\left ( \frac{|x| e}{n} \right )^n$, and you want that to be less than $1$. Solve this inequality by taking a logarithm.
