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What does this expression mean?

$$f\in C^2[a,b]$$

More specifically, I don't know what $C$ means.

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    $\begingroup$ as has been noted, the "$C$" stands for "continuous", the $2$ for "second derivative"; so functions with domain $[a,b]$ that have second derivatives that are continuous. $\endgroup$ Nov 26, 2010 at 5:20

2 Answers 2

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$f∈C^2[a,b]$ means that $f : [a,b] \rightarrow \mathbb{R}$ is a function that is twice differentiable with each derivative continuous. That is, $f'$ and $f''$ both exist and are both continuous.

$C^0$ means the function is continuous, $C^1$ means the first derivative is continuous, $C^2$ means the second derivative is continuous, In general, $C^n$ means the $n^{th}$ derivative is continuous.

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    $\begingroup$ Did you mean $f:[a,b] \rightarrow \mathbb{R}$? $\endgroup$
    – Shai Covo
    Nov 25, 2010 at 20:44
  • $\begingroup$ In other words, $C^2$ means that the function and its first derivative is continuous in $[a,b]$? $\endgroup$ Nov 26, 2010 at 0:25
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    $\begingroup$ $C^0$ means the function is continuos, $C^1$ means the first derivative is continuos, $C^2$ means the second derivative is continuos, In general, $C^{n}$ means the $n^{th}$ derivative is continuos. $\endgroup$
    – user17762
    Nov 26, 2010 at 0:44
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    $\begingroup$ @tomm89: It means that $f$, $f'$, and $f''$ are continuous on $[a,b]$ (equivalently, $f''$ is continuous on $[a,b]$, as Sivaram noted). $\endgroup$
    – Shai Covo
    Nov 26, 2010 at 0:46
  • $\begingroup$ @Shai: yes, thank you. Fixed. $\endgroup$ Nov 26, 2010 at 1:21
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$f:[a,b]\to \mathbb{R}$ or $\to \mathbb{C}$. And both the first and second derivatives of $f$ are continuous.

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