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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.

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In the future, it would be better to post this as two separate questions. But don't change it now. – Nate Eldredge Nov 21 '10 at 18:10
I will, sorry about that. – Huy Nov 21 '10 at 18:27
up vote 1 down vote accepted

For your first question, the first statement is true; it follows from the third.

The second statement is not always true; it essentially requires the function to be twice differentiable. Try a function that is only once differentiable at $x_0$ and see what happens.

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How do I get to the first statement from the second statement? Is the relation $o(x^n) = O(x^{n+1})$ true then? – Huy Nov 21 '10 at 18:17
I think you misunderstand me. You cannot get from the first statement to the second. The first statement is always true, but the second statement is not. Also, a function like $x^{n+1/2}$ is $o(x^n)$ but not $O(x^{n+1})$. – Nate Eldredge Nov 21 '10 at 19:58
Actually, I meant how to get to the first statement from the third, but I got this by now. Thanks for the help! – Huy Nov 22 '10 at 19:28

For your second question, there are points arbitrarily close to zero, such as $\frac{1}{2k\pi}$ for k a natural number, where f(x)>0 and also points $\frac{1}{(2k+1)\pi}$where f(x)<0. So zero is neither a maximum nor a minimum.

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For your first question, the third statement is the usual definition of the derivative. For the second statement, take the absolute value function and integrate it to get a function $f(x)$; check what happens at $x_0=0$.

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