I'm trying to study polynomial interpolation for my numerical analysis course, but I've become absolutely stuck in proving deriving the second order divided difference for myself:

Assuming we have $y_2=p_2(x_2)=c_0+c_1(x_2-x_0)+c_2(x_2-x_0)(x_2-x_1)$, we know $$c_0=y_0$$ $$c_1=\frac{y_1-y_0}{x_1-x_0}=[y_0,y_1]$$ We're supposed to find that $$c_2=\frac{\frac{y_2-y_1}{x_2-x_1}-\frac{y_1-y_0}{x_1-x_0}}{x_2-x_0}=\frac{[y_1,y_2]-[y_0,y_1]}{x_2-x_0}$$ But if I do the naive calculation from $p_2(x_2)$, I get $$c_2=\frac{\frac{y_2-y_0}{x_2-x_0}-\frac{y_1-y_0}{x_1-x_0}}{x_2-x_1}=\frac{[y_0,y_2]-[y_0,y_1]}{x_2-x_1}$$

I've actually set these two expressions equal to each other and reduced to $0=0$, but I can't fathom how we transform one into the other. I'd like to see such a manipulation to understand the motivation to make an analogy between $n$th derivatives and $n$th divided differences. Does anyone know how to manipulate the expressions? I know it's pedantic, but I'm hung up on understanding how anyone made the logical jump to find the canonical expression.


1 Answer 1


what is $[y_0,y_2]$ ? If I remember correctly you can only have $[y_k,y_{k+1}]$ and this $c_2$ is $[y_0,y_1,y_2]$ , I mean you can define $[y_0,y_1,y_2]$ with $[y_0,y_2]$ if you want

but lets just show how to find $c_2$

$$ c_2 = \frac{\frac{y_2-y_0}{x_2-x_1}-[y_0,y_1]\frac{x_2-x_0}{x_2-x_1}}{x_2-x_0}=\frac{\frac{y_2-y_1}{x_2-x_1}+\frac{y_1-y_0}{x_2-x_1} -[y_0,y_1]\frac{x_2-x_1}{x_2-x_1} - [y_0,y_1]\frac{x_1-x_0}{x_2-x_1} }{x_2-x_0}=$$

$$=\frac{\frac{y_2-y_1}{x_2-x_1}+\frac{y_1-y_0}{x_2-x_1} -[y_0,y_1]\frac{x_2-x_1}{x_2-x_1} - \frac{y_1-y_0}{x_1-x_0}\frac{x_1-x_0}{x_2-x_1} }{x_2-x_0}=\frac{[y_1,y_2]-[y_0,y_1]}{x_2-x_0}$$

  • $\begingroup$ Ok, wow. Did not see that. I definitely tried keeping $x_2-x_1$ into the numerator while keeping $x_2-x_0$ in the denominator, but then I took a left turn and expanded everything getting lost in the morass of trying to cancel and factor. But this is much simpler. Thank you. $\endgroup$
    – CasaBonita
    Sep 4, 2015 at 19:30

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