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MAXIM LI

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Professor showed this result in the lecture without giving any proof (after proving the existence of the interpolating polynomial in two variables). I've been trying to prove it myself or find a book where is proved but I failed. This is the theorem:

Let

$$ x_0 < x_1 < \cdots < x_n \in [a, b], \quad y_0 < y_1 < \cdots < y_m \in [c, d],$$

$$ M = \{ (x_i, y_j) : 0 \leq i \leq n, 0 \leq j \leq m \}, \quad f \in \mathcal{C}^{m + n + 2}([a,b] \times [c,d]), $$

$$ p \in \Pi_{n, m} : p(x_i, y_j) = f(x_i, y_j) \quad \forall 0 \leq i \leq n, 0 \leq j \leq m. $$

Then, for all ##(x, y) \in (x_0, x_n) \times (y_0, y_m)## there exist ##\xi, \xi' \in (x_0, x_n), \eta, \eta' \in (y_0, y_m)## such that

$$ f(x, y) - p(x, y) = \frac{1}{(n + 1)!} \frac{\partial^{n + 1} f(\xi, y)}{\partial x^{n + 1}} \prod_{i = 0}^n (x - x_i) $$

$$ + \frac{1}{(m + 1)!} \frac{\partial^{m + 1} f(x, \eta)}{\partial y^{m + 1}} \prod_{j = 0}^m (y - y_j) $$

$$ - \frac{1}{(n + 1)! (m + 1)!} \frac{\partial^{n + m + 2} f(\xi', \eta')}{\partial x^{n + 1} \partial y^{m + 1}} \prod_{i = 0}^n (x - x_i) \prod_{j = 0}^m (y - y_j) $$

I appreciate any kind of help, even if it is only from where could I start the proof.

Edit 1 (clarification):

$$ \Pi_{n, m} = \{ p(x, y) = \sum_{i = 0}^n \sum_{j = 0}^m a_{i,j} x^i y^j : a_{i, j} \in \mathbb{R} \quad \forall 0 \leq i \leq n, 0 \leq j \leq m \} $$

Let

$$ x_0 < x_1 < \cdots < x_n \in [a, b], \quad y_0 < y_1 < \cdots < y_m \in [c, d],$$

$$ M = \{ (x_i, y_j) : 0 \leq i \leq n, 0 \leq j \leq m \}, \quad f \in \mathcal{C}^{m + n + 2}([a,b] \times [c,d]), $$

$$ p \in \Pi_{n, m} : p(x_i, y_j) = f(x_i, y_j) \quad \forall 0 \leq i \leq n, 0 \leq j \leq m. $$

Then, for all ##(x, y) \in (x_0, x_n) \times (y_0, y_m)## there exist ##\xi, \xi' \in (x_0, x_n), \eta, \eta' \in (y_0, y_m)## such that

$$ f(x, y) - p(x, y) = \frac{1}{(n + 1)!} \frac{\partial^{n + 1} f(\xi, y)}{\partial x^{n + 1}} \prod_{i = 0}^n (x - x_i) $$

$$ + \frac{1}{(m + 1)!} \frac{\partial^{m + 1} f(x, \eta)}{\partial y^{m + 1}} \prod_{j = 0}^m (y - y_j) $$

$$ - \frac{1}{(n + 1)! (m + 1)!} \frac{\partial^{n + m + 2} f(\xi', \eta')}{\partial x^{n + 1} \partial y^{m + 1}} \prod_{i = 0}^n (x - x_i) \prod_{j = 0}^m (y - y_j) $$

I appreciate any kind of help, even if it is only from where could I start the proof.

Edit 1 (clarification):

$$ \Pi_{n, m} = \{ p(x, y) = \sum_{i = 0}^n \sum_{j = 0}^m a_{i,j} x^i y^j : a_{i, j} \in \mathbb{R} \quad \forall 0 \leq i \leq n, 0 \leq j \leq m \} $$

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