If we want $$p(x_i)=a_i, \qquad x_1 < \dotsb < x_{n+1},$$ then there is a unique polynomial of degree $\leq n$ that accomplishes this (Lagrange interpolation). If we want $$p(x_i)=a_i, \qquad x_1<\dotsb < x_n\\ p'(\xi)=\alpha,$$ then there is a unique polynomial of degree $\leq n$ that accomplishes this, except we can run into problems (for instance, suppose $a_i=0$ for all $i$ and $\xi$ is a zero of $\frac{d}{dx}(x\mapsto \prod_{i=1}^n (x-x_i))$, but $\alpha \neq 0$. Then this is impossible.) If we want $$p(x)=a \\ p'(\xi_i)=\alpha_i,\qquad \xi_1< \dotsb < \xi_n$$ then there is a unique polynomial of degree $\leq n$ that accomplishes this: take $q$ such that $q(x_i)=a_i$, and let $p(x):=\left(\int_0^xq\right) + c$ for the appropriate constant $c$.

We've taken care of cases $(n+1,0)$, $(n,1)$, and $(1,n)$. It seems clear that cases like $(0,n+1)$ are in general not possible. What about intermediate cases, or prescribing values of the second or greater derivatives? What would be the method here? What is the general rule?

Edit: @RobertIsrael, I wanted to try to modify your linear algebraic characterization to show that the normal Hermite interpolation exists uniquely. I.e. I want to show that a matrix such as $$ \left[ \begin{matrix} 1&\alpha_1&\alpha_1^2&\alpha_1^3&\alpha_1^4 \\ 0&1&2\alpha_1 & 2\alpha_1^2 & 4\alpha_1^3 \\ 0&0&2&6\alpha_1 &12 \alpha_1^2 \\ 1 & \alpha_2 & \alpha_2^2 & \alpha_2^3 & \alpha_2^4\\ 0&1&2\alpha_2 &3\alpha_2^2 &4\alpha_2^3 \end{matrix} \right] $$ is invertible. I could try using your characterization of the determinant and show that $$ \frac{\partial^4}{\partial x_2 \partial x_3^2 \partial x_5}\left((x_1, \dotsc, x_5)\mapsto \prod_{1\leq i < j <5}(x_i - x_j)\right) \left. \right|_{\substack{x_1=x_2=x_3=\alpha_1\\ x_4=x_5=\alpha_2}} \neq 0. $$ But so far I've been unable to do so in a way that extends more generally. I'm also trying to do column manipulations on matrices of the above type and come up with an induction proof (as with showing the determinant of the Vandermonde matrix) to show that they must be invertible.

Any ideas?

Edit 2: I have been able to show this, at another post: There is a unique polynomial interpolating $f$ and its derivatives

  • $\begingroup$ Makes me think of Hermite interpolation. Go thou and search. $\endgroup$ Sep 7, 2014 at 4:25
  • $\begingroup$ @martycohen It made me think of Hermite interpolation, too. I went, searched, but have not found an answer to Eric's question. $\endgroup$
    – user147263
    Sep 7, 2014 at 5:59

1 Answer 1


Suppose you want $p(x_i) = a_i$, $i = 1 \ldots, k$ and $p'(x_i) = \alpha_i$, $i = k+1 \ldots n$, where $p(x) = \sum_{j=0}^{n-1} u_j x^j$. Then you have an $n \times n$ linear system to solve, with coefficient matrix $A = A(x_1,\ldots,x_n)$ where $A_{ij} = x_i^{j-1}$ for $1 \le i \le k$ and $(j-1) x_i^{j-2}$ for $k+1 \le i \le n$. Whenever $\det(A) \ne 0$, there is a unique solution.

Let $V(x_1,\ldots,x_n)$ be the Vandermonde matrix with entries $x_i^{j-1}$, $i,j=1\ldots n$. Note that rows $k+1$ to $n$ are the derivatives with respect to $x_{k+1}, \ldots,x_n$ respectively of the corresponding rows of $V(x_1,\ldots,x_n)$. By multilinearity of determinant, it follows that

$$ \det(A) = \dfrac{\partial^{n-k}}{\partial x_{k+1} \ldots \partial x_n} \det \left(V(x_1,\ldots,x_n)\right)$$

It is well-known that $\det(V(x_1,\ldots,x_n)) = \prod_{i < j} (x_j - x_i)$, which is a homogeneous polynomial of total degree $n(n-1)/2$. So $\det(A)$ is a homogeneous polynomial of total degree $n(n-1)/2 - (n-k)$. Since $n > k$, this will be nonzero except on a set of dimension at most $n-1$. This set can indeed be nontrivial, even when all $x_i$ are distinct. For example, taking $k=3$, $n=5$, $x_1 = 1$, $x_2 = 2$, $x_3 = 3$, $x_4 = 4$, here is a picture of the points in the $x_5-x_6$ plane where $\det(A) = 0$:

enter image description here

  • $\begingroup$ Thanks for the helpful answer. How are we defining $$\frac{\partial^{n-k}}{\partial x_{k+1}\dotsb \partial x_n}A?$$ (If it were taken entrywise we'd end up with all entries zero.) I see how it works out with the determinant; I'm just curious about that definition. $\endgroup$
    – Eric Auld
    Sep 6, 2014 at 2:20
  • $\begingroup$ OK, I unaccepted it. Can you tell me what was wrong with you answer? Because it made sense to me, except for the question I ask above. $\endgroup$
    – Eric Auld
    Sep 7, 2014 at 8:16
  • $\begingroup$ OK, I revisited the question, and came up with a better answer. $\endgroup$ Sep 16, 2014 at 3:42

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