I have two questions and I hope it is alright to ask them both at the same time. I'm currently trying to solve an online-exercise about differential calculus and the Taylor series and I'm having some trouble there.
Let $f: (a,b) \rightarrow \mathbb{R}$ be differentiable at the point $x_0$. Which of the following statements is true?
- $f(x) = f(x_0) + O(x-x_0) \;\; (x \rightarrow x_0)$
- $f(x) = f(x_0) + f'(x_0)(x-x_0) + O((x-x_0)^2) \;\; (x \rightarrow x_0)$
- $f(x) = f(x_0) + f'(x_0)(x-x_0) + o((x-x_0)) \;\; (x \rightarrow x_0)$
I am pretty sure the third statement is true, because that's basically our definition of the Taylor series. However, I'm not quite sure about the other two statements, though I believe either both or none of them is true. When proving the Fundamental Theorem of Algebra, we once "replaced" $o(r^k)$ by $O(r^{k+1})$ which already then I found hard to understand. Can one always do this? In that case, all three statements would be correct...
Let $f(x) = e^{\frac{-1}{|x|}} \cdot \cos(x^{-1})$ for $x \neq 0$ and $0$ for $x = 0$. $f$ has a local maximum or minimum at $x=0$.
Again, I am not sure at all. I plotted the function and came to the conclusion that it has a minimum at $x=0$. However, we usually have a minimum when $f'(x_0) = 0$ and $f^{(n)} (x_0) > 0$, where $n$ is an even number. This function is arbitrarily often differentiable and I believe at none of the derivatives $f^{(n)} (x_0) \neq 0$...
I'd be delighted if someone could help me out here.