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How would I go about showing that the third coefficient of the 2nd order Newton's interpolating polynomial is :

$$a3=D^2y1= \frac{Dy2-Dy1}{x3-x1}= \frac{\frac{y_3-y_2}{x_3-x_2}-\frac{y_2-y_1}{x_2-x_1}}{x_3-x_1}$$

I dont know how to use math equations here, but I hope putting this in word and pressing the INSERT EQUATION you can see what I'm trying to prove ...

Whenever I try doing it the normal way (placing $x=x_0$, $x=x_1$, $x=x_2$ etc..), I get : $$\frac{\frac{y_3-y_1}{x_3-x_1}-\frac{y_2-y_1}{x_2-x_1}}{x_3-x_2}$$

So how can I prove that both of these equations are the same !!?

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I've partially edited your question to try to better display the expressions in question. If I've got it wrong, feel free to re-edit or rever it. You might want to take a look at the code I used and/or try using codecogs.com/latex/eqneditor.php –  Isaac Mar 10 '11 at 21:02

2 Answers 2

up vote 1 down vote accepted

$$\begin{align} \frac{\frac{y_3-y_2}{x_3-x_2}-\frac{y_2-y_1}{x_2-x_1}}{x_3-x_1} & =\frac{y_3-y_2}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_2-x_1)(x_3-x_1)} \\ & =\frac{y_3-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_2-x_1)(x_3-x_1)} \\ & = \frac{y_3-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_3-x_1)}\left(\frac{1}{x_3-x_2}+\frac{1}{x_2-x_1}\right) \\ & = \frac{y_3-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_3-x_1)}\frac{x_3-x_1}{(x_3-x_2)(x_2-x_1)} \\ & = \frac{y_3-y_1}{(x_3-x_2)(x_3-x_1)}-\frac{y_2-y_1}{(x_3-x_2)(x_2-x_1)} \\ &= \frac{\frac{y_3-y_1}{x_3-x_1}-\frac{y_2-y_1}{x_2-x_1}}{x_3-x_2}\end{align}$$

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I think if, for each of the two expressions, you get a common denominator with the two fractions in the numerator of the overall fraction, you'll be able to rewrite each expression into an identical form. That is, rewrite each of the two expressions as a simplified polynomial expression divided by $(x_3-x_1)(x_3-x_2)(x_2-x_1)$.

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