MHB Is f in the vector space of cubic spline functions?

AI Thread Summary
The discussion centers on determining if the function f(x) = ||x|^3 - |x + 1/3|^3| belongs to the vector space of cubic spline functions S_{X,3} defined on the interval [-1,1]. It is established that f(x) is piecewise defined and consists of polynomials of degree less than or equal to 3. However, concerns arise regarding the continuity of f, particularly at the point x = -1/3, where the derivative is expected to be discontinuous. Participants suggest expanding the inner expression to eliminate absolute values and better analyze continuity across the defined intervals. The conclusion leans towards the idea that f may not meet the criteria for inclusion in S_{X,3}.
mathmari
Gold Member
MHB
Messages
4,984
Reaction score
7
Hey! :o

Let $S_{X,3}$ be the vector space of cubic spline functions on $[-1,1]$ in respect to the points $$X=\left \{x_0=-1, x_1=-\frac{1}{2}, x_2=0, x_3=\frac{1}{2}, x_4\right \}$$ I want to check if the function $$f(x)=\left ||x|^3-\left |x+\frac{1}{3}\right |^3\right |$$ is in $S_{X,3}$.

We have that \begin{align*}f(x)&=\left ||x|^3-\left |x+\frac{1}{3}\right |^3\right |\\ & =\begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^3 , & |x|^3-\left |x+\frac{1}{3}\right |^3\geq 0 \\-|x|^3+\left |x+\frac{1}{3}\right |^3 , & |x|^3-\left |x+\frac{1}{3}\right |^3<0\end{cases} \\ & =\begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^3 , & |x|^3\geq \left |x+\frac{1}{3}\right |^3 \\-|x|^3+\left |x+\frac{1}{3}\right |^3 , & |x|^3<\left |x+\frac{1}{3}\right |^3\end{cases} \\ & = \begin{cases}|x|^3-\left |x+\frac{1}{3}\right |^3 , & |x|\geq \left |x+\frac{1}{3}\right | \\-|x|^3+\left |x+\frac{1}{3}\right |^3 , & |x|<\left |x+\frac{1}{3}\right |\end{cases}\end{align*}

The function is piecewise a polynomial of degree smaller or equal to $3$, right?

Now we have to check if $f$ is continuous on $[-1,1]$.

How could we continue to get the definition of $f$ ? (Wondering)
 
Mathematics news on Phys.org
Yes. And those pieces should be between those $x_i$.
That is, between each $x_i$ and $x_{i+1}$ we should have a polynomial of degree $\le 3$.
It doesn't look like we will get that, since the derivative at $x=-\frac 13$ will be discontinuous. (Worried)

mathmari said:
Now we have to check if $f$ is continuous on $[-1,1]$.

How could we continue to get the definition of $f$ ?

How about we start with the inner expression $|x|^3-\left |x+\frac{1}{3}\right|$ and expand it for the cases:
\begin{cases} -1 \le x<-\frac 13 \\ -\frac 13\le x < 0 \\ 0 \le x \le 1\end{cases}
Then all the absolute signs should disappear. (Thinking)
 
Suppose ,instead of the usual x,y coordinate system with an I basis vector along the x -axis and a corresponding j basis vector along the y-axis we instead have a different pair of basis vectors ,call them e and f along their respective axes. I have seen that this is an important subject in maths My question is what physical applications does such a model apply to? I am asking here because I have devoted quite a lot of time in the past to understanding convectors and the dual...
Fermat's Last Theorem has long been one of the most famous mathematical problems, and is now one of the most famous theorems. It simply states that the equation $$ a^n+b^n=c^n $$ has no solutions with positive integers if ##n>2.## It was named after Pierre de Fermat (1607-1665). The problem itself stems from the book Arithmetica by Diophantus of Alexandria. It gained popularity because Fermat noted in his copy "Cubum autem in duos cubos, aut quadratoquadratum in duos quadratoquadratos, et...
Thread 'Imaginary Pythagorus'
I posted this in the Lame Math thread, but it's got me thinking. Is there any validity to this? Or is it really just a mathematical trick? Naively, I see that i2 + plus 12 does equal zero2. But does this have a meaning? I know one can treat the imaginary number line as just another axis like the reals, but does that mean this does represent a triangle in the complex plane with a hypotenuse of length zero? Ibix offered a rendering of the diagram using what I assume is matrix* notation...

Similar threads

Replies
4
Views
2K
Replies
4
Views
1K
Replies
4
Views
11K
Replies
1
Views
11K
Back
Top