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I want to show that $a_{3}=f[x_{0}, x_{1}, x_{2}, x_{3}]$, if $P_{n}(x)$ is the lagrange interpolation polynomial written in the form

$$P_{n}(x)=a_{0}+a_{1}(x-x_{0})+a_{2}(x-x_{0})(x-x_{1})+ \cdots+ a_{n}(x-x_{0})\cdots(x-x_{n-1})$$

from where $f[x_{0}, x_{1}, x_{2}, x_{3}]$ denoted the third divided difference, so we have that

$$ f[x_0, x_1, x_2, x_3] = \frac{f(x_3) - f(x_0) - f[x_0, x_1](x-x_0) - f[x_0, x_1, x_2](x-x_0)(x-x_1)}{(x-x_0)(x-x_1)(x-x_2)} $$

and $f[x_{0}, x_{1}]=\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}$ and

$$f[x_{0}, x_{1}, x_{2}]=\frac{\frac{f(x_{2})-f(x_{1})}{x_{2}-x_{1}}-\frac{f(x_{1})-f(x_{0})}{x_{1}-x_{0}}}{x_{2}-x_{0}}$$

I was thinking of replacing these values in the expression for $f[x_{0}, x_{1}, x_{2}, x_{3}]$ but I have not been able to fully obtain the result, any suggestions to simplify this expression? I will be grateful!

Wrloord
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1 Answers1

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You can view

$$a_3 = \frac{P(x) - a_0 - a_1(x - x_0) - a_2(x - x_0)(x - x_1) - \sum_{k\ge 4} a_k \prod_{j<k}(x - x_j)}{(x - x_0)(x - x_1)(x - x_2)}$$

which is also true by setting $x = x_3$.

Yimin
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