- #1

mathmari

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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 $S_{X,3}$.

- $f_1(x):=|x|^3$
- $f_2(x)=\left (x-\frac{1}{3}\right )_+^3$

- $f_3(x)=-x+x^3+3x^5$

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

I have done the following:

- $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)

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

What exactly does the $+$ mean? (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? (Wondering)