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I have questions on a similar topic here, here, and here, but this is a different question.

It is well-known that a Hermite interpolation polynomial (where we sample the function and its derivatives at certain points) exists uniquely. That means that the sort of "modified Vandermonde 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 for $\alpha_1 \neq \alpha_2$, because $$ \left[ \begin{matrix} 1&\alpha_1&\alpha_1^2&\alpha_1^3&\alpha_1^4 \\ 0&1&2\alpha_1 & 2\alpha_1^2 & 4\alpha_1^3 \\ 0&0&2&6\alpha_1 &12 \alpha_1^2 \\ 1 & \alpha_2 & \alpha_2^2 & \alpha_2^3 & \alpha_2^4\\ 0&1&2\alpha_2 &3\alpha_2^2 &4\alpha_2^3 \end{matrix} \right] \left[ \begin{matrix} c_0 \\ c_1 \\ c_2 \\ c_3 \\ c_4 \end{matrix} \right] = \left[ \begin{matrix} f(\alpha_1) \\ f'(\alpha_1)\\ f''(\alpha_1) \\ f(\alpha_2) \\ f'(\alpha_2) \end{matrix} \right] $$ is the equation for the coefficients of the Hermite polynomial $c_4 x^4 + c_3 x^3 + \dotsb + c_0$ that agrees with $f$ at $\alpha_1$ up to order 2 and at $\alpha_2$ up to order 1. This will be the unique polynomial of degree $\leq 4$ with this property.

This matrix has another interesting application, which I'll place below my question.

I'm wondering: how can we show this matrix, and others of the same form, are invertible? The normal way to show uniqueness of the Hermite interpolation is through "divided differences", and I'd like a proof that doesn't rely on them.

Things I've tried:

  • Playing with column reduction, similar to the way that we show that the Vandermonde matrix has determinant $\prod_{i<j}(x_i - x_j)$. It got messy.
  • Note that the determinant of the matrix is $$ \frac{\partial^4}{\partial x_2 \partial x_3^2 \partial x_5}\left((x_1, \dotsc, x_5)\mapsto \prod_{1\leq i < j \leq 5}(x_i - x_j)\right) \left. \right|_{\substack{x_1=x_2=x_3=\alpha_1\\ x_4=x_5=\alpha_2}}. $$ Through computation I could show this is nonzero in this case, but I haven't found a way to extend this result more generally.

I claim also that the invertibility of a matrix like this one (or, more precisely, its transpose) is exactly what we need in order to show that there is a unique solution to an initial value problem $f^{(n)}=\sum a_i f^{(i)}$ when we have initial data at zero, and when $x^n - \sum a_i x^i$ has repeated roots. The above case would be if $\alpha_1$ were a thrice-repeated root, and $\alpha_2$ were a twice-repeated root. Then the system would be

$$ \left[ \begin{matrix} 1&\alpha_1&\alpha_1^2&\alpha_1^3&\alpha_1^4 \\ 0&1&2\alpha_1 & 2\alpha_1^2 & 4\alpha_1^3 \\ 0&0&2&6\alpha_1 &12 \alpha_1^2 \\ 1 & \alpha_2 & \alpha_2^2 & \alpha_2^3 & \alpha_2^4\\ 0&1&2\alpha_2 &3\alpha_2^2 &4\alpha_2^3 \end{matrix} \right]^T \left[ \begin{matrix} c_1 \\ c_2 \\ c_3 \\ c_4 \\ c_5 \end{matrix} \right] = \left[ \begin{matrix} f(0) \\ f'(0)\\ f''(0) \\ f^{(3)}(0) \\ f^{(4)}(0) \end{matrix} \right], $$

where we want a solution of the form $c_1 e^{\alpha_1 x} + c_2 x e^{\alpha_1 x} + c_3 x^2 e^{\alpha_1 x} + c_4 e^{\alpha_2 x} + c_5 x e^{\alpha_2 x}$.

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One can prove that derivatives of the form we mention of $x_1, \dotsc, x_k \mapsto \prod\limits_{1\leq i<j\leq n}(x_i-x_j)$ are nonzero. Note that the operation of taking the derivative naturally divides the $x_i$ into groups, e.g. in the above matrix there are two groups, $x_1, x_2, x_3$ and $x_4, x_5$. So it helps to divide $\prod\limits_{1\leq i<j\leq n}(x_i-x_j)$ into groups accordingly: $p_1\cdot p_2 \cdot \dotsc \cdot p_k \cdot p_{k+1}$, where the first $k$ factors contain only the factors from one group, and $p_{k+1}$ is all the factors left over. Similarly divide the differential operators into groups $\partial^{\beta_1}\dotsb \partial ^{\beta_k}$ where $\partial^{\beta_i}$ contains all the operators pertaining to group $i$. For each group we have a unique real number $\alpha_j$ that represents the point at which we are sampling (see the example above, where there are two groups). Recall that if $x_{j}, x_{j+1}, \dotsc, x_{j+l}$ is group $m$, then $\partial ^{\beta_m}$ is of the form $$ \partial^{\beta_ i} = \frac{\partial}{\partial x_{j+1}} \frac{\partial^2}{\partial x_{j+2}^2} \dotsb \frac{\partial^l}{\partial x_{j+l}^l}\left.\right|_{x_{q}=\alpha_m, \,\,j\leq q \leq j+l }. $$

Now when we take $$(\partial^{\beta_1}\dotsb \partial^{\beta_k})(p_1\cdot p_2\cdot \dotsc \cdot p_k \cdot p_{k+1}),$$

we have the product rule, where we apply the operators to the factors in every combination and then sum up, with certain coefficients. I claim that if we do not apply all of the operators in $\partial ^{\beta_k}$ to group $k$, group $k$ will evaluate to zero, since we will have two identical rows in the corresponding matrix. This causes things to simplify greatly, and we have $$ (\partial^{\beta_1}\dotsb \partial^{\beta_k})(p_1\cdot p_2\cdot \dotsc \cdot p_k \cdot p_{k+1}) = C \partial^{\beta_1}p_1 \dotsb \partial^{\beta_k}p_k\cdot p_{k+1}, $$ where $C$ is a constant depending on the size of the various groups. This expression is nonzero, since $\partial^{\beta_i}p_i$ corresponds to an upper-triangular matrix with positive diagonal entries, and $p_{k+1}$ is just a product of nonzero terms (since $\alpha_i \neq \alpha_j$ for $i\neq j$).

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