Determination of error in interpolating polynomial

  • #1
MAXIM LI
6
2
TL;DR Summary
Help needed
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):

$$ \Pi_{n, m} = \{ p(x, y) = \sum_{i = 0}^n \sum_{j = 0}^m a_{i,j} x^i y^j : a_{i, j} \in \mathbb{R} \quad \forall 0 \leq i \leq n, 0 \leq j \leq m \} $$
 
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  • #2
On a quick view, I would look for the multivariate Taylor and see how far I get, or a multidimensional version of an intermediate value theorem possibly applied to a double induction along ##n## and ##m.##

However, I haven't done any of it, it's just ideas.
 
  • #3
Help?
 
  • #4
MAXIM LI said:
Help?
I had hoped that you would have started the calculations, especially as ##f(\xi,y)## occurs on the right side of ##\dfrac{\partial^{n+1} f(\xi,y)}{\partial^{n+1} x}## which makes the term
$$
\dfrac{\partial^{n+1} f(\xi,y)}{\partial^{n+1} x}\prod_{i=0}^n(x-x_i)
$$
confusing and I have to guess whether this means
$$
f(\xi,y)\cdot\dfrac{\partial^{n+1} }{\partial^{n+1} x}\prod_{i=0}^n(x-x_i)
$$

Anyway, your formula is a direct consequence of the multivariate Taylor expansion, here the bivariate version in two variables. Of course, I could type what my textbook says and use very likely a different notation than your textbook uses, so let me instead directly ask you what you know about Taylor's theorem for functions ##f\, : \,\mathbb{R}^2\longrightarrow \mathbb{R}\;##?
 
  • #5
fresh_42 said:
I had hoped that you would have started the calculations, especially as ##f(\xi,y)## occurs on the right side of ##\dfrac{\partial^{n+1} f(\xi,y)}{\partial^{n+1} x}## which makes the term
$$
\dfrac{\partial^{n+1} f(\xi,y)}{\partial^{n+1} x}\prod_{i=0}^n(x-x_i)
$$
confusing and I have to guess whether this means
$$
f(\xi,y)\cdot\dfrac{\partial^{n+1} }{\partial^{n+1} x}\prod_{i=0}^n(x-x_i)
$$

Anyway, your formula is a direct consequence of the multivariate Taylor expansion, here the bivariate version in two variables. Of course, I could type what my textbook says and use very likely a different notation than your textbook uses, so let me instead directly ask you what you know about Taylor's theorem for functions ##f\, : \,\mathbb{R}^2\longrightarrow \mathbb{R}\;##?

I've found a explanation in this book in chapter 6 section 6, but it doesn't explain the concrete steps and I'm managing to figure them out.
 
  • #6
I do not have this book and it needs a lot of notation to write down what I have. It all is a rephrasing of the Taylor expansion for two variables
$$
f((x+\xi,y+\eta))=\sum_{|\alpha|\leq k} \dfrac{D^\alpha f(x,y)}{\alpha!} (\xi,\eta)^\alpha +\sum_{|\alpha|= k+1} \dfrac{D^\alpha f((x,y)+\vartheta (\eta,\xi) )}{\alpha!} (\xi,\eta)^\alpha
$$
From there it is a few steps to
$$
f((x+\xi,y+\eta))=f(x,y)+\bigl\langle \nabla f((x,y))\, , \,(\xi,\eta) \bigr\rangle +\dfrac{1}{2}\sum_{i,j=1}^2 \dfrac{\partial^i }{\partial x^i}\dfrac{\partial^j}{\partial x^j}f((x,y))\cdot (\xi^i,\eta^j)+o\left(\|(\xi,\eta)\|^2\right)
$$
Now set ##\xi=x_i-x## and ##\eta=y_i-y## and the derivatives become the coefficients ##a_{i,j}## of the three homogenous polynomials of degree ##0,1,## and ##2.##
The actual formulas and the steps between them are in my book which is unfortunately not in English. Hence my recommendation is to search for the multi- or at least two-dimensional version of Taylor's theorem and what happens if you only consider the first three terms up to degree two.

https://teaching.smp.uq.edu.au/scims/Num_analysis/Taylor.html
is an example I found on searching "multidimensional Taylor series".
 
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