I was given a problem like this:
Suppose $f$ has derivatives of all orders and all continuous. Suppose the sequence $\{a_n\}$ is bounded and $|f^{(n)}(x)| \le a_n$ for all $x$. Use the Cauchy Remainder Formula to show that $f$ equals its Taylor Expansion.
So far I have been covered Lagrange Remainder and Cauchy Remainder Theorem. My intuition is showing the remainder part $R_n(x)$ of Cauchy equal $R_n(x)$ from Lagrange. But I don't know how to apply the info that $|f^{(n)}(x)| \le a_n$ for all $x$. Can someone please help with this? Thanks
Best Answer
That the Cauchy and Lagrange forms of the remainder are equal is not something you need to show. Since they are both expressions for the same value, they are equal (however- the unknown constants in them are not the same value).
The problem is to prove this by the Cauchy version. For simplicity I will assume it is about $0$:
$$R_n(x) = \frac{f^{(n+1)}(x_n)}{n!}(x - x_n)^nx$$ for some $x_n$ between $0$ and $x$.
user143462 didn't spot the more egregious poor wording in the problem statement. We are told that $|f^{(n)}(x)| \le a_n$ for all $x$ and That $a_n$ itself is bounded. Since we know nothing else about $a_n$, this just means that there is some $M$ such that $|f^{(n)}(x)| < M$ for all $x$ and $n$. This is all we need. Stating it in terms of some sequence is a red herring.
Note that since $x_n$ is between $x$ and $0$, we have $|x - x_n| < |x|$. So $$|R_n(x)| = \frac{|f^{(n+1)}(x_n)|}{n!}|x - x_n|^n|x| < \frac M{n!}|x|^n|x| = \frac{Mx^{n+1}}{n!}$$ which goes to $0$ as $n \to \infty$.