MHB Checking if $f_1, f_2, f_3 Belong to $S_{X,3}$

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The discussion revolves around determining whether specific functions belong to the vector space of cubic spline functions, denoted as $S_{X,3}$. The functions under consideration include $f_1(x) = |x|^3$, which is continuous and has a derivative that needs verification for continuity; $f_2(x) = (x - \frac{1}{3})_+^3$, where the meaning of the $+$ sign is questioned; and $f_3(x) = -x + x^3 + 3x^5$, which is excluded from $S_{X,3}$ due to its degree exceeding 3. Additionally, $f_4(x) = \sum_{n=0}^3 a_n x^n$ is confirmed to be in $S_{X,3}$ as it meets the criteria of being of degree 3 and $C^2$. The conversation also touches on the need to analyze further functions and their intervals for continuity and differentiability.
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Hey! :o

Let $S_{X,3}$ be the vector space of the cubic splines functions on $[-1, 1]$ with the points \begin{equation*}X=\left \{x_0=-1, \ x_1=-\frac{1}{2},\ x_2=0,\ x_3=\frac{1}{2}, \ x_4=1\right \}\end{equation*}

I want to check if the following function are in $S_{X,3}$.
  1. $f_1(x):=|x|^3$
  2. $f_2(x)=\left (x-\frac{1}{3}\right )_+^3$
  3. $f_3(x)=-x+x^3+3x^5$
  4. $f_4(x)=\sum_{n=0}^3a_nx^n$, $a_n\in \mathbb{R}, n=0, \ldots , 3$
We have to check at each case if the function are of degree at most $3$ and are $C^2$, or not? (Wondering)

I have done the following:

  1. $f_1(x):=|x|^3=|x|^3=\begin{cases}
    x^3 \ \ \ ,& x\geq 0\\
    -x^3 \ ,& x<0
    \end{cases}$

    This function is continuous at every point, i.e. at $[-1, 0), (0, 1]$ and at $x=0$.

    Then we have to check if the derivative id continuous. How can we calculate the derivative? (Wondering)
  2. $f_2(x)=\left (x-\frac{1}{3}\right )_+^3$

    What exactly does the $+$ mean? (Wondering)
  3. $f_3(x)=-x+x^3+3x^5$

    This function is not in $S_{X,3}$, since it is of order $5$ instead of at most $3$.
  4. $f_4(x)=\sum_{n=0}^3a_nx^n$, $a_n\in \mathbb{R}, n=0, \ldots , 3$

    This function is $C^2$ and of degree $3$.

    From that it follows that $f_4\in S_{X,3}$, right? (Wondering)
 
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mathmari said:
[*] $f_1(x):=|x|^3=|x|^3=\begin{cases}
x^3 \ \ \ ,& x\geq 0\\
-x^3 \ ,& x<0
\end{cases}$

This function is continuous at every point, i.e. at $[-1, 0), (0, 1]$ and at $x=0$.

Then we have to check if the derivative id continuous. How can we calculate the derivative?

Hey mathmari!

Isn't the derivative:
$$f_1'(x)=\begin{cases}
3x^2 \ \ \ ,& x\geq 0\\
-3x^2 \ ,& x<0
\end{cases}$$
(Wondering)

mathmari said:
[*] $f_2(x)=\left (x-\frac{1}{3}\right )_+^3$

What exactly does the $+$ mean?

I don't know. I haven't seen such a subscript + before.
Can it be a typo? (Wondering)
mathmari said:
[*] $f_3(x)=-x+x^3+3x^5$

This function is not in $S_{X,3}$, since it is of order $5$ instead of at most $3$.

[*] $f_4(x)=\sum_{n=0}^3a_nx^n$, $a_n\in \mathbb{R}, n=0, \ldots , 3$

This function is $C^2$ and of degree $3$.

From that it follows that $f_4\in S_{X,3}$, right?

Yep. (Nod)
 
Could the subscript $+$ be the positive part of the expression between parentheses?
 
I see! Thank you! (Happy) What about the following function?

$f(x)=\left ||x|^3-\left |x+\frac{1}{3}\right |^2\right |=\begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^2 , & |x|^3-\left |x+\frac{1}{3}\right |^2>0\\ |x|^3-\left |x+\frac{1}{3}\right |^2 , & |x|^3-\left |x+\frac{1}{3}\right |^2<0\end{cases}=\begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^2 , & |x|^3>\left (x+\frac{1}{3}\right )^2\\ |x|^3-\left |x+\frac{1}{3}\right |^2 , & |x|^3<\left (x+\frac{1}{3}\right )^2\end{cases}$ How can we check what subintervals of $[-1,1]$ we have here? (Wondering)
 
Looks like we need to divide it further into sub cases for [-1,-1/3), [-1/3, 0), [0, 1], don't we? (Wondering)
 
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