Why does knowing the $n$th derivative of a function at a point allow you to know the functions value everywhere else? I apologise if this question has already been asked, but I couldn't find anything answering my specific question.
My question is why does knowing the $n$ derivative at a point allow you to approximate (or know exactly, for an infinite Taylor series for functions like $\sin(x)$) a function at places (far) away from that point.
Specifically if you consider the Sine functions Taylor Series:
$$ \sin(x)=\sum_{i=0}^{\infty}\frac{(-1)^nx^{2n+1}}{{2n+1}!} $$
We could derive this series by considering the $n$th derivative of $\sin(x)$ at $x=0$ which turns out to only be $4$ different values (after the first four it repeats). It seems strange to me that knowing infinitely well what the derivative of the Sine function is at $x=0$ allows one to know the value for the function at $x=1000$ (or any number).
It seems that encoded in the derivative(s) of a function (at a point) must be information that entirely defines that function (for functions like the Sine function). Though there's most likely a flaw in my thinking somewhere, I've only started looking into series. Is there any intuitive way of understanding why knowing infinitely well a function at a point can define a function anywhere?
 A: Functions with this property are "nice" in a strong sense; specifically they are called real analytic. Generally "analytic" means "representable as a power series", although usually "analytic" without specification is used to describe a complex analytic function.
This turns out to be a strictly stronger property than merely having infinitely many derivatives, because there are functions which are smooth but grow slower than any polynomial near a specified point. One such function is 
$$f(x) = \begin{cases} e^{-1/x^2}, & x \neq 0 \\
0, & x = 0 \end{cases}$$
This function is not identically zero, clearly. Perhaps surprisingly, it is infinitely differentiable, even at zero. Yet one can prove that its Taylor polynomials at zero are all identically zero. So this function is not real analytic.
This is an especially nasty example because the Taylor series converges for any $x$ that you substitute, yet it only converges to the function $f$ at $0$ itself, never anywhere else. There are functions "in between" where the Taylor series does agree on a neighborhood but not globally. For instance:
$$f(x)=\frac{1}{1+x^2}$$
is equal to its Taylor series at $0$ on the interval $(-1,1)$. Yet beyond this interval the series fails to converge at all, even though $f$ is defined and smooth over the whole line. These functions are called real analytic on the domain where the series converges. 
Properly studying this phenomenon requires complex analysis, at least in my opinion. For instance, for the previous function, one can identify the fact that the Taylor series converges only on $(-1,1)$ with the fact that the Taylor series of the corresponding complex function converges only on the disk of radius $1$ centered at zero. It could never converge on any larger disk because it blows up at $z=i$.
Perhaps surprisingly, the properties of Taylor polynomials (not series) can be understood using only real-variable methods. Taylor's theorem (with its various error estimates) is all there is to say on the subject.
A: Here's some intuition (not a rigorous argument).
\begin{align}
f(x) &= f(x_0) + \int_{x_0}^x f'(s) \, ds \\
\tag{$\clubsuit$} &\approx f(x_0) + \int_{x_0}^x \underbrace{f'(x_0)}_{\text{zeroth-order approximation to } f'(s)} \, ds \\
&= f(x_0) + f'(x_0)(x - x_0).
\end{align}
But we can get a better approximation by using a better approximation for $f'(s)$ in step $(\clubsuit)$:
\begin{align}
f(x) &= f(x_0) + \int_{x_0}^x f'(s) \, ds \\
\tag{$\spadesuit$}& \approx f(x_0) + \int_{x_0}^x \underbrace{f'(x_0) + f''(x_0)(s - x_0)}_{\text{first-order approximation to } f'(s)} \, ds \\
&= f(x_0) + f'(x_0)(x - x_0) + \frac12 f''(x_0)(x - x_0)^2.
\end{align}
This approximation seems likely to be superior to the first-order approximation derived above.
By using higher-order approximations to $f'(s)$ in step ($\spadesuit$), such as
\begin{equation}
f'(s) \approx f'(x_0) + f''(x_0)(s - x_0) + \frac12 f'''(x_0) (s - x_0)^2,
\end{equation}
we get higher-order Taylor approximations to $f(x)$, and it seems like these approximations should be increasingly accurate.  So we can hope that the Taylor series for $f(x)$ will be exact.
(Of course, this doesn't always happen, but we at least have a reason for hoping it will happen.)
