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

[mathmari]

Hey!

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)

 

[I like Serena]

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

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)

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

What exactly does the $+$ mean?

I don't know. I haven't seen such a subscript + before.
Can it be a typo? (Wondering)

[*] $f_3(x)=-x+x^3+3x^5$

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)

 

[S.G. Janssens]

Could the subscript $+$ be the positive part of the expression between parentheses?

 

[mathmari]

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)

 

[I like Serena]

Looks like we need to divide it further into sub cases for [-1,-1/3), [-1/3, 0), [0, 1], don't we? (Wondering)