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I can follow the proof but what I do not get is that $g(t)=0$ If we replace $f(t)$ in $26$ by $f(t)$ defined $25$. So $g(t)=0$ for all $t$. Why does that make sense?

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  • $\begingroup$ As an aside, I would rather derive Taylor theorem by integrating the inequality $\inf\le f^{(n)}(t)\le\sup~n$ times ;-P $\endgroup$ – Simply Beautiful Art Jul 14 at 22:36
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$\beta$ is some value between $a$ and $b$, and $M$ is the value which satisfies $(25)$ for that $\beta$. Hence $g(t)$ is zero when $t=\beta$, but not necessarily otherwise.

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