# Taylor’s Theorem confusion

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?

• As an aside, I would rather derive Taylor theorem by integrating the inequality $\inf\le f^{(n)}(t)\le\sup~n$ times ;-P – Simply Beautiful Art Jul 14 at 22:36

$$\beta$$ 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.