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As far as I can tell, if you start with the integral form of the remainder of a Taylor polynomial then you can derive the Lagrange form by an application of the mean value theorem for integrals. From Spivak's Calculus:

If $f^{\left({n + 1}\right)}$ is continuous on $[a, x]$, then

$$\displaystyle R_{n,a} \left({x}\right) = \int_a^x \dfrac {f^{\left({n + 1}\right)}\left({t}\right)}{n!} \left({x - t}\right)^n \, \mathrm d t$$

Let m and M be the minimum and maximum of $\dfrac {f^{\left({n + 1}\right)}}{n!}$ on $[a, x]$, then $R_{n,a}(x)$ satisfies

$$m\int_a^x \left({x - t}\right)^n \, \mathrm d t \le R_{n,a} \left({x}\right) \le M\int_a^x \left({x - t}\right)^n \, \mathrm d t$$

so we can write

$$\displaystyle R_{n,a} \left({x}\right) = \alpha \cdot \dfrac {\left({x - a}\right)^{n+1}} {{n+1}!}$$

Now here is my question. Shouldn't this imply that there is a number $x^* \in \left[{a \,.\,.\, x}\right]$ such that

$$\displaystyle R_{n,a} \left({x}\right) = \dfrac {f^{\left({n + 1}\right)} \left({x^*}\right)} {\left({n + 1}\right)!} \left({x - a}\right)^{n + 1}$$

in accordance with the mean value theorem for integrals?

Yet all sources I reviewed, including Spivak, state that $x^* \in \left({a \,.\,.\, x}\right)$.

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    $\begingroup$ For sure it is not wrong to consider the closed interval... It would be nice though to come up with an example where $x^*=a$. $\endgroup$ – PierreCarre Feb 4 at 16:26
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If for instance $f(x)$ is a 2nd degree polynomial, the 1st order Taylor approximation around $x=0$ tells you that $$ f(x)=f(0) + f'(0)x + \frac{f''(x^*)}{2!} x^2. $$

But of course, since the second order approximation is exact, we have that $x^*=0$. So, for sure, it can happen that $x^*$ is in boundary.

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