Proving uniqueness of an interpolating polynomial with high degree The values of a function $f$ and its first and second derivatives $f',f''$ are known at $2$ distinct points $a,b$. Assume that a polynomial $p$ of degree $5$ exists which satisfies $p^n(a)=f^n(a)$ and $p^n(b)=f^n(b)$ for $n\in\{0,1,2\}$.
How can I prove the uniqueness of such an interpolating polynomial? Usually for this type of question I would write the conditions out in terms of the coefficients of $p$. So I would end up with $6$ equations in terms of $a,b$ and the coefficients of $p$.
I could then write these equations in matrix form $A\vec{x}=\vec{b}$, where $\vec{x}=(p_0,p_1,p_2,p_3,p_4,p_5)$ are the coefficients of $p$, and $\vec{b}=(f(a),f'(a),f''(a),f(b),f'(a),f''(a))$. So if I could show that $\det(A)$ is non-zero, then I would be done. But in this case it is quite tedious to find $\det(A)$ by hand, not only because I have a $6x6$ matrix but also because there are many steps required to convert it to upper triangular form. So, is there a quicker way to go about this problem?
 A: We may assume wlog your two points are $a=0$ and $b=1$.  If
there is more than one such polynomial, their difference is a polynomial $p$ of degree $\le 5$ with $p(0) = p'(0) = p''(0) = 0$ and $p(1) = p'(1) = p''(1) = 0$.
From the conditions at $0$, 
$p(x) = a_3 x^3 + a_4 x^4 + a_5 x^5$ for some $a_3, a_4, a_5$.  From the conditions at $1$,
$$ \eqalign{a_3 + a_4 + a_5 &= 0\cr
            3 a_3 + 4 a_4 + 5 a_5 &= 0\cr
            6 a_3 + 12 a_4 + 20 a_5 &= 0\cr}$$
Subtracting the appropriate multiples of the first equation from the other two,
$$ \eqalign{a_4 + 2 a_5 &= 0\cr
            6 a_4 + 14 a_5 &= 0\cr}$$
and since $14/2 \ne 6$, ...
EDIT:  Alternative:
Start with $g(x) = f(a) + f'(a) (x-a) + f''(a) (x-a)^2/2$ which is a polynomial of degree $\le 2$ with the given values of $g(a), g'(a), g''(a)$.  Add an appropriate multiple of $(x-a)^3$ to get the desired value of $g(b)$ in a polynomial of degree $\le 3$.  Add an appropriate multiple of $(x-a)^3(x-b)$ (noting that the first derivative of this at $b$ is nonzero) to get the desired value of $g'(b)$.  Add an appropriate multiple of $(x-a)^3(x-b)^2$ to get the desired value of $g''(b)$.  Since the map $g \mapsto [g(a), g'(a), g''(a), g(b), g'(b), g''(b)]$ from 
polynomials of degree $\le 5$ to $\mathbb R^6$ is linear and onto, and both vector spaces have the same dimension, it must also be one-to-one, i.e. the interpolation is unique.  This solution can be generalized. 
A: Let $$p(t) = \sum_{j=0}^5 c_j t^j$$ denote a polynomial of degree at most 5. Then $p$ solves your problem if and only if the vector $c = (c_0,c_1,\dotsc,c_5)$ solve a linear system $$Ac=f$$ of dimension $6$. It is straightforward to assemble the matrix $A$, but it is not necessary. Instead we will show that $A$ is nonsingular. From this it will follow that there is a unique polynomial that solves your problem. 
It is enough to show that $Ac=0$ has the unique solution $c=0$. Now assume that $Ac = 0$, then the corresponding polynomial $p$ has zeros of order 3 at each of the points $a$ and $b$. It follows that $(x-a)^3(x-b)^3$ divides $p$, i.e., $$p(x) = q(x)(x-a)^3(x-b)^3$$ for some polynomial $q$. By considering the degree, the only choice is $q=0$, i.e., the constant zero. This shows that $c=0$ is the only solution of $Ac=0$. This completes the proof.
 I invoked a result about multiple roots of polynomials. I will sketch how this is established. Suppose $p$ and $p'$ share a root at $a$. I claim that $(x-a)^2$ divides $p$. By assumption, we have $$p(x) = q(x)(x-a) \quad \text{and} \quad p'(x) = q'(x)(x-a) + q(x).$$ Since $p'(a) = 0$ we have $q(a) = 0$, hence $$q(x) = r(x)(x-a)  \quad \text{and} \quad p(x) = r(x)(x-a)^2.$$

The techniques used here apply to all interpolation problems where $f^{(i)}(x_j)$ is specified for all $0 \leq i \leq m_j$ and $j=0,1,2,\dotsc,n$. The crucial feature is that we do not skip a derivative at any of the nodes, i.e., if $f^{(2)}(x_0)$ is specified, then we also know $f(x_0)$ and $f'(x_0)$.
