Using Hermite interpolation determine the fourth degree polynomial $p(x)$ for which \begin{equation} p(0) = p'(0) = 0, \quad p(1) = p'(1) = 1, \quad p(2) = 1 \end{equation}
I can only seem to find content on how to apply Hermite interpolation to data points with odd degree polynomial. Some guidance would be greatly appreciated.
Let $P_0(x) =x^2, P_1(x)=(x-1)^2, P_2(x)=x-2$.
Let $A_0(x) = 0, A_1(x) = 1+(x-1)=x, A_2(x)=1$.
We want to find $p(x)$ such that $$p(x) \equiv A_i(x) \pmod{P_i(x)}$$
We let $$Q(x)=x^2(x-1)^2(x-2)$$
Denote $Q_i(x) = \frac{Q(x)}{P_i(x)}.$ We have $$Q_0(x) =(x-1)^2(x-2)$$ $$Q_1(x) = x^2(x-2)$$ $$Q_2(x)=x^2(x-1)^2$$
Let $$\sum_{i=0}^2S_i(x)Q_i(x) = 1$$
where $degree(S_i)<degree(P_i)$
$$S_0(x)(x-1)^2(x-2)+S_1(x)x^2(x-2)+S_2(x)x^2(x-1)^2=1\tag{1}$$
Let $x=2$, we conclude that $S_2(x) = \frac14$.
Let $x=1$, we have $S_1(1)=-1$. We let $S_1(x)=ax+b$, we have $a+b=-1$.
Differentiating $(1)$ once,
$$\frac{d}{dx}[S_0(x)(x-1)^2(x-2)+S_2(x)x^2(x-1)^2]+\frac{d}{dx}[S_1(x)(x^3-2x^2)]=0$$
$$\frac{d}{dx}[S_0(x)(x-1)^2(x-2)+S_2(x)x^2(x-1)^2]+a(x^3-2x^2)+(ax+b)(3x^2-4x)=0$$
Let $x=1$, we have $-a-(a+b)=0$, that is $b=-2a$. That is $a=1$ and $b=-2$.
$S_1(x)=x-2$.
Now let's compute $A_1(x)S_1(x) \pmod{P_1(x)}$ to reduce the degree:
\begin{align} A_1(x) S_1(x) &\equiv x(x-2) \pmod{(x-1)^2} \\ &\equiv x^2-2x \pmod{(x-1)^2} \\ &\equiv x^2-2x+1-1 \pmod{(x-1)^2} \\ &\equiv -1 \pmod{(x-1)^2}. \end{align}
Let $B_i \equiv A_iS_i \pmod{P_i}$, The desired polynomial is $\sum_{i=0}^2 B_i(x)Q_i(x)$, that is
\begin{align}-Q_1(x)+ \frac14Q_2(x)&=-x^2(x-2)+\frac14x^2(x-1)^2 \\ &=x^2 \left[ -(x-2) + \frac14(x-1)^2\right]\end{align}
Relevant reading from wikipedia
