# Taylor polynomial with Lagrange remainder

In my course there's a paragraph: Taylor polynomial with Lagrange remainder,
The paragraph starts with a theorem (I left out the constraints):

$$( \exists \theta \in ]0,1[)(f(a +h) = T_{f,a,n}(a + h) + \frac{h^{n+1}}{(n+1)!}D^{n+1}f(a + \theta h))$$

With $f$ an $R-R$ function and $a$ and $a + h$ defined over an interval $I$.
And the only other thing in the paragraph is a proof of the above theorem.

Before I'm starting with the proof, could someone please explain the above theorem in human language?

I understand what a Taylor polynomial is, and what it's good for, but I can't turn this into anything that makes sense...

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Please read en.wikipedia.org/wiki/Taylor_series –  anonymous Dec 25 '10 at 16:37

I think what you've got there is the $n$th-degree Taylor polynomial for a function $f$ centered at $a$ and being evaluated at $x=a+h$. Basically, the error term $\frac{h^{n+1}}{(n+1)!}D^{n+1}f(a + \theta h))$ looks like the last term in the $(n+1)$st degree Taylor polynomial would, except that instead of evaluating the $(n+1)$st derivative at $a$, it's evaluated somewhere between $a$ and $a+h$ ($\theta$ measures how much of the way from $a$ to $a+h$).
This formula is not meant to be a terribly complicated way to compute $f(a+h)$ (e.g. $\log(1+h)$) $exactly$, but it tells you the following: If you compute the value $T_{f,a,n}(a+h)$ (which is a simple polynomial value) instead you commit an error which is small because (i) $h^{n+1}$ might be very small, (ii) ${1\over(n+1)!}$ is very small and (iii) hopefully the $(n+1)^{\rm st}$ derivative of $f$ is bounded by a constant throughout, so that it doesn't matter where the point $a+\theta h$ guaranteed by the theorem lies.