(using $f[x_1, ... , x_n]$ to denote the forward difference operator)

I have a polynomial $P(x)$ interpolating $5$ points $x_0, ... , x_4$ and $2$ derivative values $x_0, x_3$ across an evenly spaced grid such that $x_i = x_0 + ih$. I constructed a Hermite polynomial to interpolate the data, and obtained a polynomial of degree $9$ from $f[z_0, z_1, ... , z_n]$ where $z_1 = z_0 = x_0$ and $n = 10$, however, I also have the constraint that the polynomial must be of degree $6$.

My initial idea was to alter the Lagrangian basis s.t. the kronecker delta only interpolates the $2$ derivative points. If I were to take this approach, I would in effect be doing the forward difference similar to $f[z_0, z_1, x_1, x_2, z_3, z_4, x_4]$ (where $z_3 = z_4 = x_3$, etc.) correct?

Is this a valid operation? While looking for a result over the past few days I've been finding a lot of sites referencing Birkhoff interpolation which is a concept that I do not fully understand, and while I don't think it's relevant, perhaps I'm incorrect? Thanks in advance! (thank you for cleaning this up)

  • $\begingroup$ $x_0,\dotsc,x_5$ are $6$ points. $\endgroup$ – joriki May 1 '13 at 2:22
  • $\begingroup$ Sorry about that, fixed :) $\endgroup$ – user60977 May 1 '13 at 2:48

The interpolating polynomial is linear in the data, so all you need is $7$ basis polynomials corresponding to one of the given values being $1$ and the others being $0$. The three polynomials for $x_1$, $x_2$ and $x_4$, are straightforward; for instance

$$ \begin{align} p_4(x) &= \frac{(x-x_0)^2(x-x_1)(x-x_2)(x-x_3)^2}{(x_4-x_0)^2(x_4-x_1)(x_4-x_2)(x_4-x_3)^2} \\ &= \frac1{96h^7}(x-x_0)^2(x-x_1)(x-x_2)(x-x_3)^2\;. \end{align} $$

For the polynomials for $x_0$ and $x_3$ (two each), you can make the ansatz e.g.

$$ p_3(x)=(ax+b)(x-x_0)^2(x-x_1)(x-x_2)(x-x_4) $$

and then choose $a$ and $b$ such that the function value at $x_3$ is $1$ and the derivative $0$, and vice versa.

  • $\begingroup$ Thank you, this helped a lot :) $\endgroup$ – user60977 May 1 '13 at 20:55

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