# Some question about definition of splines and periodic splines

I have some questions about the definition of splines and in particular periodic splines. So in non periodic case splines are piecewise polynomials of degree $<=m$ which are $m-1$ times differentiable.(I took $m-1$ time differentiability for simplicity) By this definition we can call linear or quadratic polynomials splines of degree 3.

For periodic case we need boundary conditions on some interval$(a,b)$ $s^{(q)}(a)=s^{(q)}(b),q=1...m-1$. Does it mean that periodic splines are piecewise polynomials of degree $m$(not smaller then $m$) or constants?

Is there analogue for truncated polynomials basis in periodic case?

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What do you mean by "constant" here? Normally I would consider a polynomial constant only if it has degree -1 or 0, but that doesn't fit the context. –  Peter Taylor Jan 21 '11 at 11:32
i mean polynomials of degree 0 –  Katja Jan 21 '11 at 11:34
So the question is asking whether periodic splines of order greater than 0 can exist? (Incidentally, what are $a$ and $b$? And by "periodic" do you mean that there is a period $P$ such that $\forall x: f(x-P) = f(x)$?) –  Peter Taylor Jan 21 '11 at 13:14
It sounds like you are asking whether there exists a non-constant periodic spline of degree less than $m$. Is that correct? And do you mean that all segments should be polynomials of degree less than $m$? –  Rahul Jan 21 '11 at 13:28
Rahul Narain, a non periodic spline of degree less then $m$ exists. For example, a periodic spline of degree $m-1$. My question if there is a spline of degree $m-1$, which belongs to subspace of splines degree $m$. –  Katja Jan 21 '11 at 13:35

Wlog we can take the knot at $x=0$, and then we're looking for two polynomials of degree $\le m$ which join there with $C(m-1)$ continuity.

If the two polynomials both have degree $m-1$ then, since they have $m$ derivatives (including the 0th derivative) in common they are equal. (Whether you consider a polynomial to be a spline on the basis that it's piecewise polynomial with all the pieces having the same function is up to you).

However, it's possible for one to be of degree $m$ and the other of degree $m-1$. Consider the case $m=2$. Our polynomials are $f(x) = ax^2 + bx + c$ and $g(x) = dx + e$. $f(0) = g(0) \Rightarrow c = e$, $f^\prime(0) = g^\prime(0) \Rightarrow b = d$. $a$ is unconstrained.

### Example

$f(x) = \left\{\begin{array}{l} (x-1)^2 & 0 <= x < 2 \\ 2x - 3 & 2 <= x < 3 \\ 4 - (x-4)^2 & 3 <= x < 5 \\ 13 - 2x & 5 <= x < 6 \\ f(x-6) & x >= 6 \\ f(x+6) & x < 0 \end{array}\right.$

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don't boundary conditions make one more constrain? –  Katja Jan 27 '11 at 13:17
@Katja, periodicity adds some constraints, but they don't necessarily conflict with these constraints. Do you want a constructive example of a periodic spline of order 2 containing a piecewise linear (order 1) segment? –  Peter Taylor Jan 27 '11 at 13:27
yes, I am looking for such an example. I am still not sure it exists. But I guess it is more computational problem. Thank you for the hint –  Katja Jan 27 '11 at 14:14
@Katja, added example –  Peter Taylor Jan 27 '11 at 14:29