A doubt about Schlömilch remainder formula's proof I'm reading a proof of Schlömilch remainder formula in textbook Analysis I by Amann. > ![enter image description here]( While the remainder function is given by > ![enter image description here]( * * * **My questions:** 1. In the author's definition, > ![enter image description here]( ![enter image description here]( I could not see how the hypothesis "perfect interval" is applied in the proof. Please shed me some light! 2. We have $$R_{n}(f, a)(x)=g(x)-g(a) = 0 - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k !}(x-a)^{k} = - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k !}(x-a)^{k}$$ But $R_{n}(f, a)(x)$ given by Taylor theorem is $$f(x) - \sum_{k=0}^{n} \frac{f^{(k)}(a)}{k !}(x-a)^{k}$$ Could you please explain why there is no $f(x)$ as required in the formula? Thank you so much!

1. The assumption that $I$ is perfect is a fancy way of saying that $I$ contains a nonempty open interval, which is required to define the derivatives. It rules out the case that $I$ contains just a single point; it would be silly to define $C^n(I;\mathbb{R})$ in that case.

2. You are mistaken in writing $g(x) = 0$. The higher-order terms in the sum do zero out, but the $k=0$ term remains: $$ g(x) = \sum_{k=0}^n \frac{f^{(k)}(x)}{k!}(x-x)^k = f^{(0)}(x) = f(x). $$