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Prerequisites. I know about integrals, Taylors polynomials, differentiation and i understand newtons binomial theorem. I list these terms for the sake of showing, that an answer with one of these mathematical concepts would be understandable to me.

So i get that differentiating, a polynomial with $n$ degrees yield a number that is zero, but how this function works i have no idea. I've seen the $f^{(n+1)}$ written as $M$ before, to denote max value but where this value comes from i have no idea. Like how would one know the upper bound of a natural logarithm function? For me nothing in this formula makes sense, and if anyone would be so helpful to clear this up it would be great. Thank you in advance.

$$R_n(x) = \frac{f^{(n+1)} (c)}{n!} (x - a)^\left(n+1\right) \,$$

Above is the hairy formula i'm referring to.

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    $\begingroup$ Does the mean value theorem make sense to you? It is basically the same ... just for higher order derivatives. $\endgroup$ – user251257 Nov 22 '16 at 0:38
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One possible trick is to consider the remainder as function of $a$, $$ f(x)=f(a)+f'(a)(x-a)+…+\frac{f^{(n)}(a)}{n!}(x-a)^n+r(a) $$ Derivation wrt. to $a$ then cancels a lot of terms so that the only remaining ones are $$ 0=\frac{f^{(n+1)}(a)}{n!}(x-a)^n+r'(a) $$ which expresses the fact that the approximation by the Taylor polynomial is indeed tangent of order $n+1$.

Integration and application of the mean value theorem of integration (with weights) then results in the different forms of the remainder, \begin{align} r(a)&=-\int_x^a\frac{f^{(n+1)}(s)}{n!}(x-s)^n\,ds \\[.6em] &=-\frac{f^{(n+1)}(c)}{n!}(x-c)^{n-m}\int_x^a(x-s)^m\,ds \\[.6em] &=\frac{f^{(n+1)}(c)}{(m+1)\,n!}(x-c)^{n-m}(x-a)^{m+1} \end{align} for some $c$ between $a$ and $x$.


Usually one is only interested in the provided approximation in a small interval around $a$. Thus only there one needs to bound the derivative in the remainder term to get accessible upper bounds for the error.

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