Matching values and derivatives needs repeated-node divided differences

Let $f(x)=x^4$. Osculatory (Hermite) interpolation at the nodes $0$ and $1$ seeks the unique polynomial $H$ of degree at most $3$ satisfying \[ H(0)=f(0),\quad H'(0)=f'(0),\quad H(1)=f(1),\quad H'(1)=f'(1). \] 1. Form the Newton divided-difference table for the repeated abscissae $z=(0,0,1,1)$, using the rule that a first divided difference on a repeated node equals the derivative. Compute every entry, and write the Newton–Hermite form \[ H(x)=f[z_0]+f[z_0,z_1](x-z_0)+f[z_0,z_1,z_2](x-z_0)(x-z_1)+f[z_0,z_1,z_2,z_3](x-z_0)(x-z_1)(x-z_2). \] Simplify $H$ to the monomial basis. 2. Independently assemble $H$ in the two-node Hermite basis \[ \begin{aligned} \alpha_0(x)&=(1+2x)(1-x)^2,& \beta_0(x)&=x(1-x)^2,\\ \alpha_1(x)&=(3-2x)x^2,& \beta_1(x)&=(x-1)x^2, \end{aligned} \] and verify that the two constructions agree. Check the four interpolation conditions directly on your simplified $H$. 3. State the Hermite remainder theorem for this data: there is $\xi_x$ strictly between the extreme nodes and $x$ (when $x\notin\{0,1\}$) such that \[ f(x)-H(x)=\frac{f^{(4)}(\xi_x)}{4!}\,x^2(x-1)^2. \] Evaluate both sides exactly at $x=1/2$, and conclude that the remainder identity is an equality of polynomials in this example. 4. Explain why a degree-at-most-$1$ Lagrange interpolant of the values $f(0),f(1)$ alone cannot meet the derivative conditions, and why the nodal factor here is $x^2(x-1)^2$ rather than the degree-$4$ equispaced nodal product of a Runge/Chebyshev value-interpolation comparison. Do not replace the osculatory table by value interpolation on four distinct nodes, and do not treat this as an equispaced-versus-Chebyshev remainder audit.

The first-layer entries of the table are $f'(0)=0$, $(f(1)-f(0))/(1-0)=1$, and $f'(1)=4$.