Shape preserving piecewise polynomial interpolation — Examples

By [F3], $\left(\genfrac{}{}{0px}{}{2}{1}\right)=2$; applying the recursion in [F4] twice shows its canonical real value is $2$. Expanding $\frac{1}{10}(1-t{)}^2-\frac{4}{5}t(1-t)+\frac{1}{10}{t}^2$ gives $t^2-t+1\mathrm{/}10$, so [F1] identifies its Bernstein controls as $c_0=1\mathrm{/}10$, $c_1=-2\mathrm{/}5$, $c_2=1\mathrm{/}10$.

Direct substitution gives $p(0)=p(1)=1\mathrm{/}10>0$, while $p(1\mathrm{/}2)=1\mathrm{/}4-1\mathrm{/}2+1\mathrm{/}10=-3\mathrm{/}20<0$. Thus positive samples at the two endpoints do not control the interpolant between them.

The middle Bernstein control is negative, so the sufficient hypothesis in [F2] is absent; the negative value in step 1.2 is consistent with the control-hull theorem.

Each displayed piece has endpoint values $S_0(0)=1$, $S_0(1)=0$, $S_1(0)=S_1(1)=0$, $S_2(0)=0$, and $S_2(1)=1$, so the pieces interpolate the given table. By [F4], their first derivatives are $S_0^{\mathrm{\prime }}=-6\mathrm{/}5+3t^2\mathrm{/}5$, $S_1^{\mathrm{\prime }}=-3\mathrm{/}5+6t\mathrm{/}5$, and $S_2^{\mathrm{\prime }}=3\mathrm{/}5+6t\mathrm{/}5-3t^2\mathrm{/}5$; the derivative values match at both shared knots. Their second derivatives are $S_0^{\mathrm{\prime }\mathrm{\prime }}=6t\mathrm{/}5$, $S_1^{\mathrm{\prime }\mathrm{\prime }}=6\mathrm{/}5$, and $S_2^{\mathrm{\prime }\mathrm{\prime }}=6\mathrm{/}5-6t\mathrm{/}5$; these match at shared knots, and $S_0^{\mathrm{\prime }\mathrm{\prime }}(0)=S_2^{\mathrm{\prime }\mathrm{\prime }}(1)=0$. By [F1], this is a natural cubic spline.

For $0<t<1$, $S_1(t)=-3t(1-t)\mathrm{/}5<0$, although every tabulated value is nonnegative. Thus the natural cubic spline undershoots the data range on the middle cell.

The secants are $d_0=-1$, $d_1=0$, and $d_2=1$. By [F2], the interior slopes at nodes $1$ and $2$ are both zero because each neighboring pair includes a zero secant; the Hermite piece joining the two zero values with zero slopes is constant zero. By [F3], the first and last PCHIP pieces remain between $1$ and $0$, and between $0$ and $1$, respectively. Hence PCHIP is nonnegative throughout the interval and does not reproduce the natural spline's undershoot.

The secants are $d_0=1$, $d_1=0$, and $d_2=1$. The interior rule in [F1] gives $m_1=m_2=0$. The endpoint rule gives ${\widehat{m}}_0=((3)(1)-0)\mathrm{/}2=3\mathrm{/}2$ and, by reflection, ${\widehat{m}}_3=3\mathrm{/}2$; neither clip nor cap changes these values.

On the middle cell both endpoint values are $1$ and both slopes are zero, so the Hermite formula is the constant polynomial $1$. On the first cell it is $-t^3\mathrm{/}2+3t\mathrm{/}2$, and on the last cell it is $1+3t^2\mathrm{/}2-t^3\mathrm{/}2$, for normalized local coordinate $t\in [0,1]$.

The shape theorem [F2] makes the first and last pieces monotone from $0$ to $1$ and from $1$ to $2$ and makes the middle piece constant. Thus the interpolant is globally nondecreasing, has no overshoot, and retains the flat interval exactly.

The secants on $[-2h,-h]$, $[-h,0]$, $[0,h]$, and $[h,2h]$ are $-3h$, $-h$, $h$, and $3h$. At $x=-h$, the equal-width weights are $3h,3h$, so [F1] gives $m(-h)=6h\mathrm{/}(3h\mathrm{/}(-3h)+3h\mathrm{/}(-h))=-3h\mathrm{/}2$. At $x=0$ the adjacent secants have opposite signs, so [F1] gives $m(0)=0$.

With $t=(x+h)\mathrm{/}h$ on $[-h,0]$, [F2] gives $P_h(x)\mathrm{/}{h}^2=H_{00}(t)-\frac{3}{2}{H}_{10}(t)=1-\frac{3}{2}t+\frac{1}{2}{t}^3$, whereas $f(x)\mathrm{/}{h}^2=(1-t{)}^2$. Therefore $(P_h(x)-f(x))\mathrm{/}{h}^2=\frac{1}{2}t(1-t{)}^2$.

On this cell, step 2.1 gives the error $E(x)=\mathrm{\mid }{P}_h(x)-f(x)\mathrm{\mid }=\frac{h^2}{2}t(1-t{)}^2$, since $0\le t\le 1$. This expression is a polynomial in $x$ through $t=(x+h)\mathrm{/}h$, so [F4] makes it differentiable, hence continuous, on the cell. As $h>0$, the interval $[-h,0]$ is nonempty and compact by [F5]; [F6] therefore gives an attained maximum of $E$ there. Since $t=1\mathrm{/}3$ corresponds to a point of the cell and step 2.1 gives $E=2h^2\mathrm{/}27$ at that point, the maximum is at least $2h^2\mathrm{/}27$. Any uniform bound $\mathrm{\mid }{P}_h-f\mathrm{\mid }\le C{h}^3$ on these grids with $C$ independent of $n$ would imply $2\mathrm{/}27\le Ch=C\mathrm{/}n$ for all sufficiently large $n$, which is impossible as $n\to \mathrm{\infty }$. The hypothesis $\mathrm{\mid }{f}^{\mathrm{\prime }}\mathrm{\mid }\ge c>0$ in [F3] also fails because [F4] gives $f^{\mathrm{\prime }}(0)=0$.