$f(x)$ has a continuous derivative of $n+1$ order, and$$ f(a+h)=f(a)+hf'(a)+\frac{h^2}{2!}f''(a)+ \cdots +\frac{h^n}{n!} f^{(n)}(a+θh). \quad 0<θ<1, f^{(n + 1)}(a)≠0. $$ I know that for this question the Taylor formula should be used, but I don't know how to solve for it.


We have $$f(a+h)-f(a)=\displaystyle\sum_{i=1}^{n-1} {f^{(i)}(a)\over i!}\cdot h^i+{f^{(n)} (a+\theta h)\over n!}\cdot h^n$$$$=\sum_{i=1}^{n} {f^{(i)}(a)\over i!}\cdot h^i+{f^{(n+1)}(a+\xi h)\over (n+1)!}\cdot h^{n+1} ,$$ Here $\theta,\xi\in (0,1).$ Thus $$f^{(n)} (a+\theta h)=f^{(n)}(a)+\displaystyle{f^{(n+1)}(a+\xi h)\over (n+1)}\cdot h.$$ Notice that $f^{(n)}(a+\theta h)-f^{(n)}(a)=f^{(n+1)}(a+\eta h)\cdot \theta h,$ here $\eta\in (0,\theta),$ thus we have $$f^{(n+1)}(a+\eta h)\cdot \theta=\displaystyle{f^{(n+1)}(a+\xi h)\over (n+1)}.$$ Now let $h\to 0,$ and we obtain that $\theta\to \displaystyle {1\over n+1}.$

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    $\begingroup$ Does not touch the solution but shouldn't $i$ start from 1 rather than 0? $\endgroup$ – LRDPRDX Jan 15 '18 at 9:41
  • $\begingroup$ @BogdanSikach: You're right, thanks. I'll edit my answer. $\endgroup$ – painday Jan 15 '18 at 9:52

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