A doubt about Schlömilch remainder formula’s proof

proof-explanationreal-analysistaylor expansion

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!

  1. 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!

Best Answer

  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: $$ g(x) = \sum_{k=0}^n \frac{f^{(k)}(x)}{k!}(x-x)^k = f^{(0)}(x) = f(x). $$

Related Question