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The following taylor expansion of the function $f(x)$, requires $f(x)$ to have a derivative up to what order? $$ f(x)=f(0)+f'(0)x+f''(0)x^2/2+\mathcal{O}(x^3)$$

My solution: Based on the Taylor's Theorem

$f(x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k} + \ R_{k}(x)$

where $R_{k}(x)={\frac {f^{(k+1)}(\xi _{L})}{(k+1)!}}(x-a)^{k+1}$ and $ξ_L$ is between a and x. Therefore, the expansion in the question requires $f(x)$ to have up to the third derivative.

Is my solution correct? Is there a way that we need only up to the 2nd derivative?

Susan_Math123
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  • Relevant: http://math.stackexchange.com/questions/1809293/taylors-theorem-with-peanos-form-of-remainder (good to have that referenced) – Clement C. Jul 14 '16 at 22:35
  • @ClementC. Thanks. So, based on Peano's remainder, it just needs to have derivative up to the 2nd order? – Susan_Math123 Jul 14 '16 at 22:37
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    This is not exactly implied by what I linked. The above will show a remainder that is $o(x^2)$. But $o(x^2)$ is weaker than $O(x^3)$... – Clement C. Jul 14 '16 at 22:38

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tl;dr: as you expected, existence of the third derivative at $0$ is necessary and sufficient.

By your argument, differentiable thrice at $0$ is sufficient. Now, let us show it is necessary.

For that, let $f\colon\mathbb{R}\to\mathbb{R}$ be the even function defined by $f(x) = \lvert x\rvert^{5/2}$. It is a simple matter to check that it is twice differentiable; so that it $f\in\mathcal{D}^2$ were a sufficient condition, then we would have $|x|^{5/2} = O(x^3)$ around $0$ — which is clearly false.

As a side note: as mentioned in the comments, if the question were to find a condition on $f$ ensuring that $$ f(x) = f(0)+f'(0)x+\frac{f''(0)}{2}x^2 + o(x^2) $$ instead, then indeed $f$ being twice differentiable at $0$ would be enough. But the remainder being $O(x^3)$ is a strictly stronger condition than $o(x^2)$.

Community
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Clement C.
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  • Maybe I'm being overly picky, but I'm confused by the phrase "necessary and sufficient" here--there are $\mathcal O(x^3)$ functions that aren't thrice differentiable. (Consider $f(x)$ defined by $f(0) = 0$ and $f(x) = x^3 \sin(1/x)$ for $x$ nonzero; this isn't even twice differentiable at 0.) Do you just mean that $\mathcal D^3$ is sufficient and $\mathcal D^2$ isn't? – Ravi Fernando Jul 14 '16 at 23:32
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    @RaviFernando I am considering it in the context of the question: "how to choose $k$ such that $f$ $k$-differentiable at $0$ implies the desired expansion"? $k\geq 3$ is necessary, $k=3$ is sufficient. (Note that the statement to prove does not even make sense for $f$ not being twice differentiable at $0$, as it features the quantity $f''(0)$.) – Clement C. Jul 14 '16 at 23:34