Let $f(x) = x^2\cdot (x-1)^2 \cdot (x-2)^2 \cdot (x-3)^2$. What is the piecewise cubic Hermite interpolant of $f$ on the grid $x_0 = 0$, $x_1 = 1$, $x_2 = 2$, $x_3 = 3$. Let $g(x) = ax^3 + bx^2 + cx +d$ for some parameters $a, b, c, d$ write down the piecewise cubic Hermite interpolation of g on the same grid.

I realize that $f$ and $f'$ are 0 at each node, so essentially the cubic polynomial that interpolates $f$ is just $g$. But I'm not sure of how to actually split $g$ into a piecewise cubic since it's already a cubic function.

  • $\begingroup$ Please, check the formulation of the very first formula. Is it something like $f(x)=x^2\cdot (x-1)^2\cdot (x-2)^2 \cdot (x-3)^2$? $\endgroup$ – Karel Macek Nov 4 '14 at 13:21
  • $\begingroup$ Oh yes sorry about that $\endgroup$ – Meggany Nov 4 '14 at 14:27

Unfortunately, the only cubic polynomial in each interval taking zero values and zero derivatives at the end-points is the zero polynomial.

We look for a polynomial for the first interval $[x_0,x_1]$ in this form $$p(x)=a+b(x-x_0)+c(x-x_0)^2+d(x-x_0)^2(x-x_1)$$ with its first derivative being $$p^\prime(x)=b+2c(x-x_0)+2d(x-x_0)(x-x_1)+d(x-x_0)^2$$

Zero value and zero derivative at $x=x_0$ means $a=b=0$. Same conditions at $x=x_1$ lead to $c=d=0$.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.