Given the data: 
Find an interpolating polynomial using Newton's method.
I am a bit confused on how to do this. I think it should be fairly simple.. Here is what I have so far:
$p_0(x) = -1$
$p_1(x) = p_0(x) + c(x_1 - x_0) \implies -1 + C(2 - 1)$
which $= -1 + C = \frac{-1}{3} \implies C = \frac{2}{3}$
so $p_1(x) = -1 + \frac{2}{3} x$ .. Is this the correct approach? & Also if it is, what is the next step?
would it be:
$p_2(x) = p_1(x) + C(x_2 - x_0)(x_2 - x_1)$
$p_2(x) = -1 + \frac{2}{3} x + C(2.5 - 1)(2.5 - 2)$ ? Then solve for $C$?
Thanks in advance
Following the method from our previous discussion, we use divided differences to find $p_4(x)=f[x_0]+f[x_1](x-x_0)+f[x_2](x-x_0)(x-x_1)+f[x_3](x-x_0)(x-x_1)(x-x_2)$
We know that $p(1)=-1=f[x_0]$, $p(2)=-1/3$, $p(2.5)=3/32$, $p(3)=4/3$, and $p(4)=25$.
$$p[1,2]=\frac{-1/3+1}{1}=\frac{2}{3}=f[x_1]$$ $$p[2,2.5]=\frac{3/32+1/3}{0.5}=\frac{41}{48}$$ $$p[2.5,3]=\frac{4/3-3/32}{0.5}=\frac{119}{48}$$ $$p[3,4]=\frac{25-4/3}{1}=\frac{71}{3}$$ $$p[1,2,2.5]=\frac{41/48-2/3}{1.5}=1/8=f[x_2]$$ $$p[2,2.5,3]=\frac{119/48-41/48}{1}=\frac{13}{8}$$ $$p[2.5,3,4]=\frac{71/3-119/48}{1.5}=\frac{113}{8}$$ $$p[1,2,2.5,3]=\frac{13/8-1/8}{2}=\frac{3}{4}=f[x_3]$$ $$p[2,2.5,3,4]=\frac{113/8-13/8}{2}=\frac{25}{4}$$ $$p[1,2,2.5,3,4]=\frac{25/4-3/4}{3}=\frac{11}{6}=f[x_3]$$
Thus, $p_4(x)=-1+\frac{2}{3}(x-1)+\frac{1}{8}(x-1)(x-2)+\frac{3}{4}(x-1)(x-2)(x-2.5)+\frac{11}{6}(x-1)(x-2)(x-2.5)(x-3)$
Verifying...
$p_4(1)=-1\,\checkmark$
$p_4(2)=-1+\frac{2}{3}(1)=-\frac{1}{3}\,\checkmark$
$p_4(2.5)=-1+\frac{2}{3}(1.5)+\frac{1}{8}(1.5)(0.5)=\frac{3}{32}\,\checkmark$
$p_4(3)=-1+\frac{2}{3}(2)+\frac{1}{8}(2)(1)+\frac{3}{4}(2)(1)(0.5)=\frac{4}{3}\,\checkmark$
$p_3(4)=-1+\frac{2}{3}(3)+\frac{1}{8}(3)(2)+\frac{3}{4}(3)(2)(1.5)+\frac{11}{6}(3)(2)(1.5)(1)=25\,\checkmark$