= Solution
The <cubic spline> is even, so the symmetry of the normalized <B-splines> gives $N_{j,4}(-x)=N_{6-j,4}(x)$. It therefore suffices to extract three <coefficients>. The two <polynomial> pieces join with matching values and first two <derivatives> at zero, so the function belongs to the stated <cubic spline> space. At zero its right-hand <derivatives> are
$$
f(0)=1,\qquad f'(0)=0,\qquad f''(0)=-27,\qquad f'''(0+)=81.
$$
Use the supplied <De Boor–Fix spline coefficient functional> with the unnormalized knot <polynomial> $\psi_j(x)=\prod_{r=j+1}^{j+3}(t_r-x)$. In this sign convention the <coefficient> is
$$
c_j=\frac{\psi_jf'''-\psi_j'f''+\psi_j''f'-\psi_j'''f}{6}.
$$
For $j=1$, $\psi_1=(-1-x)^3$. Taking the right-hand limit at $-1$ kills the first three terms and gives $c_1=f(-1)=1$. For the next two <coefficients> take the right-hand limit at zero. Their knot <polynomials> and <derivatives> there are
$$
\begin{aligned}
\psi_2(x)&=-x(1+x)^2,&(\psi_2,\psi_2',\psi_2'',\psi_2''')(0)&=(0,-1,-4,-6),\\
\psi_3(x)&=x-x^3,&(\psi_3,\psi_3',\psi_3'',\psi_3''')(0)&=(0,1,0,-6).
\end{aligned}
$$
Consequently $c_2=(-27+6)/6=-7/2$ and $c_3=(27+6)/6=11/2$. Although the third <derivative> jumps at zero, its multiplier $\psi_j(0)$ vanishes for these two <functionals>; taking either one-sided limit gives the same <coefficients>. Symmetry gives $c_4=c_2$ and $c_5=c_1$. Thus
$$
\boxed{(c_1,c_2,c_3,c_4,c_5)=(1,-7/2,11/2,-7/2,1).}
$$
For a normalized <B-spline> <basis>, define the synthesis <linear map> $Tc=\sum_jc_jN_{j,4}$ from the maximum <coefficient> <norm> into the <spline> space with its <supremum norm>. Nonnegativity and partition of unity give $\|T\|=1$, including equality on the all-ones <vector>. The <coefficient condition number of a normalized B-spline basis> is
$$
\kappa(\mathcal S)=\|T\|\|T^{-1}\|=\sup_{0\ne s\in\mathcal S}\frac{\|c(s)\|_{\ell^\infty}}{\|s\|_\infty}.
$$
On $[0,1]$, $f'(x)=(27/2)x(3x-2)$. The only interior extremum is at $2/3$, where $f=-1$; at zero and one, $f=1$. Reflection therefore gives $\|f\|_\infty=1$. Its largest absolute <coefficient> is $11/2$, proving
$$
\boxed{\kappa(\mathcal S)\ge\frac{11}{2}.}
$$
Back to article page
