# Proof

• By hypothesis, $I\subset\mathbb R$ is an interval and $f:I\to\mathbb R$ is a $(n+1)$ times continuously differentiable function.
• Let $a,x\in I.$
• By the Taylor's formula, we have $$f(x)=f(a)+\sum_{k=1}^n \frac{f^{\{k\}}(a)}{k!}(x-a)^k+R_{n+1}(x)$$ with the remainder term $$R_{n+1}(x)=\frac 1{n!}\int_a^x (x-t)^nf^{\{n+1\}}(t)dt.$$
• Since the $(n+1)$-th derivative is continuous, the mean value theorem for Riemann integrals yields that there is a value $\xi$ between $a$ and $x$ such that \begin{align}R_{n+1}(x)&=\frac 1{n!}\int_a^x (x-t)^nf^{\{n+1\}}(t)dt\nonumber\\ &=f^{\{n+1\}}(\xi)\int_a^x \frac {(x-t)^n}{n!}dt\nonumber\\ &=-f^{\{n+1\}}(\xi)\frac{(x-t)^{n+1}}{(n+1) !}\;\Rule{1px}{4ex}{2ex}^{x}_{a}\nonumber\\ &=\frac{f^{\{n+1\}}(\xi)}{(n+1) !}(x-a)^{n+1}\nonumber\\ \end{align}

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### References

#### Bibliography

1. Forster Otto: "Analysis 1, Differential- und Integralrechnung einer VerĂ¤nderlichen", Vieweg Studium, 1983