Understanding Taylor's formula I have been taught the following Theorem:
$$\textbf{Taylors Formula}$$
Let $f:(a,b)\rightarrow \mathbb{R}$ be $(n+1)$-times continuously differentiable, and let $x_0 \in (a,b)$. Then for any $x \in (a,b)$, $x\neq x_0$ exists some $\xi$ strictly between $x_0$ and $x$ s.t.
$$f(x):=\sum_{k=0}^n \frac{f^{(k)}}{k!} (x-x_0)^k +R_n$$,
where:
$$R_n:=\frac{f^{(n+1)}(\xi)}{(n+1)!} (x-x_0)^{n+1}.$$
$$\textbf{My Questions}$$
First of all, I understand that a taylor series may be used to represent some function, when it comes to application to real life problems. I understand that Taylor's formula differs to the Taylor series by the addition of $R_n$, the remainder term. However, I don't understand why this remainder term is necessary when it comes to the real life problems/ what it actually means intuitively.
Secondly, I am unsure as to how we know that $\xi$ is bound between $x_0$ and $x$, I am thinking that perhaps an answer to my first question may allow me to work that out anyway.
If anyone could clear these points up for me, It'd be much appreciated
 A: Second question first.  We know that $\xi$ is between $x_0$ and $x_1$ because we get its existence (for $x$ and $x_0$ fixed) essentially by applying the mean value theorem (which says that a continuous function is somewhere equal to its average over an interval).
For the remainder term, if you have a function which is not polynomial, but you wish to approximate it by a polynomial, then taking the first $n$ terms of the Taylor series gives you a good approximation, but it won't be perfect, and no approximation is safe to use if you don't know how good an approximation it is.  The theorem tells you that your approximation's error can be bounded by a $C(x-x_0)^{n+1}$, for some constant $C$, which tells you that your approximation will be good if $|x-x_0|\ll 1$.
Depending on how $f^{(n+1)}$ behaves, however, the constant in front might be so large that the approximation will only be good if $|x-x_0|$ is very small, and so depending on your application, it will tell you that your approximation isn't good enough and you need to use something different.
But to give an example of an application, say you wanted to approximate $\sin(0.2)$ to 6 decimal places.  By knowing the error term, you can figure out how many terms in the Taylor series you need to take before you are accurate enough for your needs.
A: Taylor's formula provides a more general a result.
Taylor's formula can be used to prove if the Taylor series converges.
Not every function is infinitely differentiable, so your infinite Taylor series will not exist, has some error, and the Taylor formula will give you the error bound.
It is never practical to calculate an infinite series, so if you round of a the $n^{th}$ term, what is the consequence?
"Real life problems" always have error.  The question becomes whether the error is within tolerance or not.
A: A fundamental problem in applying mathematics to reality is that most functions $f:(a,b) \to \mathbb{R}$ that actually describe some phenomenon in reality, are such that they cannot be written as combinations of elementary functions in mathematics ($x\mapsto x^a$, $x\mapsto e^x$, $x \mapsto \log x$...).
However, that does not mean that all is lost, because for most purposes, it is enough to find a function $g:(a,b) \to \mathbb{R}$ such that $g$ is "close" to $f$ in some way that satisfies our "purpose". For example, if the only thing we care about is to evaluate $f$ at some small interval $(x_0 - \delta, x_0 + \delta) \subset (a,b)$, then all that matters is to find $g$ such that $g$ is an elementary function and that $|g(x) - f(x)| \leq \epsilon$, $x \in (x_0 - \delta, x_0 + \delta)$ where $\epsilon$ is our tolerance for error.
Taylor's formula serves exactly that purpose. What it says is that if $f:(a,b) \to \mathbb{R}$ is any function that is $(n+1)$-differentiable, then you may let $g(x) = \sum_{k=0}^n \frac{f^{(k)}(x_0)}{k!}(x - x_0)^k$, an explicit polynomial (provided that you can at least calculate the ${f^{(k)}(x_0)}$), and that you have an almost explicit way of calculating the error $|g(x) - f(x)|$, that is $\frac{ f^{(n+1)}(\xi) }{ (n+1)!}(x - x_0)^{n+1}$.
The error is not explicit because in general there is no way to know what $\xi$ is (knowing $\xi$ is equivalent to knowing $f$), but at least you can bound it, that is, you know that $\xi$ is between $x$ and $x_0$ for $x \in (x_0 - \delta, x_0 +\delta)$, hence it must be in $(x_0 - \delta, x_0 +\delta)$ itself. If you can find a bound on $f^{(n+1)}(x)$, $x \in (x_0 - \delta, x_0 +\delta)$, you can confidently say that $g(x)$ is at most ... away from $f(x)$, for $x \in (x_0 - \delta, x_0 +\delta)$ and that's good enough at least for most engineering, say.
To recap, if all you care about is to evaluate $f(x)$ is some small interval and you are able to tolerate some error, what the Taylor formula allows you to do is to reduce that problem from evaluating an infinite number of values $\{ f(x) : x \in (x_0 - \delta,x_0 + \delta) \}$ (which is impossible, period), to just a finite number of values $\{ f^{(k)}(x_0) : k \in \{0,\dots,n\} \}$ and a bound on $f^{(n+1)}(x)$, $x \in (x_0 - \delta,x_0 + \delta)$ (which is "possible").
A: The magnitude of remainder term $|R_n|$ is an indicator of the error in real-life problems. In those problems, the answer is found approximately (as you increase the number of terms, you are getting closer to the exact answer). In taylor series, we don't have the term $R_n$ because Taylor Series exists if $$\lim_{n \to \infty} R_n = 0$$
A: The Taylor series can only be used for real analytic functions which
are as subset of the smooth functions, so it has limited applicability
for many problems.
Taylor's formula can be used for a larger collection of functions.
One way to look at Taylor's formula is as a generalisation of the mean value theorem,
if $f$ is differentiable on $(a,b)$ then we can write
$f(b)-f(a) = f'(\xi) (b-a)$ where $\xi \in (a,b)$. The $\xi \in (a,b)$ (and the intuition)
comes from Rolle's theorem. This is a '$0$th' order Taylor formula
and the remainder term is $f'(\xi) (b-a)$. 
The Taylor formula is useful because it gives an exact expression (albeit the $\xi \in (a,b)$ is unknown) that can be used to form useful estimates.
To illustrate, if we take $f(x) = \log (1+x)$ with $x \ge 0$, Taylor's
formula gives
$f(x)=0 + 1 \cdot x - { 1\over 2(1+\xi)^2} x^2 $ with $\xi \in [0,x]$ .
This gives
$0  \le x-\log(1+x) \le {x^2 \over 2}$.
