Is this determinant not zero?

Question

I would like to prove that for any $k\in \mathbb{N}$, there is a unique polynomial of $2k +1$ degree, $p(x)=a_{2k+1}x^{2k+1}+\dots+a_0$ such that $p(0), p'(0),\dots p^{(k)}(0)$ and $p(1), p'(1),\dots ,p^{(k)}(1)$ are given. For the cases $k=0,1,2$ it is fairly easy to prove that as it ends up to a linear system. But how about the general case ? So after some handling (developing the determinant by the rows with $p(0), p'(0),\dots p^{(k)}(0)$), I came up with the $(k+1)\times (k+1)$ determinant : $$ \begin{pmatrix} 1 & 1 & \cdots \cdots & 1\\ 2k+1 & 2k & \cdots \cdots& k+1 \\ (2k+1)\cdot2k & 2k\cdot(2k-1) & \cdots \cdots&(k+1)\cdot k \\ \vdots & \vdots & \vdots & \vdots &\\ \vdots & \vdots & \vdots & \vdots &\\ (2k+1)\cdot2k\cdots(k+2) & 2k\cdot(2k-1)\cdots(k+1) & \cdots \cdots&(k+1)\cdot k \cdots2\\ \end{pmatrix} $$

I do not know how to compute this determinant (is it a special one?)or at least show that it is not zero. Any suggestions are appreciated.

Answer 1

Let us consider the map: $u\in F_{\leq 2k+1}[X]\to F^{2k+2}$ sending $p$ to $(p(0),\ldots, p^{(k)}(0),p(1),\ldots,p^{k}(1))$, where $F$ is your prefered base field.

The map $u$ is linear, and you want to prove it is bijective. For, it is enough to prove its injectivity, since the source space and the target same have same dimension $2k+2$.

But $p\in\ker(u)$ means that $0$ and $1$ are roots of $p$ with multiplicity at least $k+1$ (maybe we need the characteristic of $F$ to be zero, here) , meaning that $p$ is a multiple of $x^{k+1}(x-1)^{k+1}$. This is not possible since the degree of $p$ is at most $2k+1$, unless $p=0$, and we are done.

Remark. As a consequence, you get that your determinant is indeed nonzero.

Answer 2

This is more of a constructive proof if you want to know how to actually compute these polynomials. But I also really like the simplicity of the algebraic proof mentioned by another user here, just by studying the kernel of a linear morphism between two vector spaces of the same dimension, and would consider that one as the best answer.

You can consider the functions

$$\begin{cases}L_{0, k}(x) = \frac{x^{k}}{k!} (1-x)^{k+1} \\ L_{1, k}(x) = \frac{(x-1)^{k}}{k!} x^{k+1} \end{cases}$$

So, these satisfy that

$$\begin{cases}L_{0, k}^{(n)}(1) = 0 \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \\ L_{0, k}^{(n)}(0) = \delta^n_k \hspace{1 cm} \forall \hspace{0.1 cm} n\leq k \end{cases} $$ $$\begin{cases} L_{1, k}^{(n)}(0) = 0 \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \\ L_{1, k}^{(n)}(1) = \delta^n_k \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \end{cases}$$

where $\delta^n_k$ is a Kronecker delta ($\delta^n_k=1$ if $n=k$, and $\delta^n_k=0$ if $n\neq k$)

Then, start constructing backwards. Given polynomials $\{L_{0, k}, L_{0, k-1},...,L_{0, N} \}$ and $\{L_{1, k}, L_{1, k-1},...,L_{1, N} \}$ (for some $1 \leq N \leq k$) of degree at most $2k+1$ such that

$$\begin{cases} L_{0, m}^{(n)}(0) = \delta^n_m \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \hspace{1 cm} \forall \hspace{0.1 cm} N \leq m \leq k \\ L_{0, m}^{(n)}(1) = 0 \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \hspace{1 cm} \forall \hspace{0.1 cm} N \leq m \leq k \end{cases}$$

$$\begin{cases} L_{1, m}^{(n)}(0) = 0 \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \hspace{1 cm} \forall \hspace{0.1 cm} N \leq m \leq k \\ L_{1, m}^{(n)}(1) = \delta^n_m \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \hspace{1 cm} \forall \hspace{0.1 cm} N \leq m \leq k \end{cases} $$

define

$$\begin{cases} G_{0, N-1}(x) = \frac{x^{N-1}}{(N-1)!} (1-x)^{k+1} \\ G_{1, N-1}(x) = \frac{(x-1)^{N-1}}{(N-1)!} x^{k+1} \end{cases}$$

and

$$\begin{cases} L_{0, N-1}(x) = G_{0, N-1}(x) - \sum_{n=N}^k G_{0, N-1}^{(n)}(0) \cdot L_{0, n}(x) \\ L_{1, N-1}(x) = G_{1, N-1}(x) - \sum_{n=N}^k G_{1, N-1}^{(n)}(1) \cdot L_{1, n}(x) \end{cases}$$

It is immediate that the new set of polynomials $\{L_{0, k}, L_{0, k-1},...,L_{0, N}, L_{0, N-1} \}$ and $\{L_{1, k}, L_{1, k-1},...,L_{1, N}, L_{1, N-1} \}$ satisfies

$$\begin{cases} L_{0, m}^{(n)}(0) = \delta^n_m \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \hspace{1 cm} \forall \hspace{0.1 cm} N-1 \leq m \leq k \\ L_{0, m}^{(n)}(1) = 0 \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \hspace{1 cm} \forall \hspace{0.1 cm} N-1 \leq m \leq k \end{cases}$$

$$\begin{cases} L_{1, m}^{(n)}(0) = 0 \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \hspace{1 cm} \forall \hspace{0.1 cm} N-1 \leq m \leq k \\ L_{1, m}^{(n)}(1) = \delta^n_m \hspace{1 cm} \forall \hspace{0.1 cm} n \leq k \hspace{1 cm} \forall \hspace{0.1 cm} N-1 \leq m \leq k \end{cases} $$

and have degree $2k+1$ at most.

You can iterate this process until you get $\{L_{0, k}, L_{0, k-1},...,L_{0, 1}, L_{0, 0} \}$ and $\{L_{1, k}, L_{1, k-1},...,L_{1, 1}, L_{1, 0} \}$ satisfying

$$\begin{cases} L_{0, m}^{(n)}(0) = \delta^n_m \hspace{1 cm} \forall \hspace{0.1 cm} n,m \leq k\\ L_{0, m}^{(n)}(1) = 0 \hspace{1 cm} \forall \hspace{0.1 cm} n,m \leq k \end{cases}$$

$$\begin{cases} L_{1, m}^{(n)}(0) = 0 \hspace{1 cm} \forall \hspace{0.1 cm} n,m \leq k \\ L_{1, m}^{(n)}(1) = \delta^n_m \hspace{1 cm} \forall \hspace{0.1 cm} n,m \leq k \end{cases} $$

Then doing linear combinations of these polynomials you can impose any derivative of any order at $0$ and $1$, defining

$$f(x) = \sum_{n=0}^k p^{(n)}(0)L_{0, n}(x) + \sum_{n=0}^k p^{(n)}(1)L_{1, n}(x)$$

where $\{p^{(n)}(0)\}_{n=0,...,k}$ and $\{p^{(n)}(1)\}_{n=0,...,k}$ are the given values you wanted.

To see the uniqueness, you can consider two polynomials $f$ and $g$ of degree at most $2k+1$, such that $f^{(n)}(0) = g^{(n)}(0)$ and $f^{(n)}(1) = g^{(n)}(1)$ for all $n\leq k$. Now if we consider $h=f-g$, we have that $h^{(n)}(0)=0$ and $h^{(n)}(1)=0$ for all $n \leq k$, and $deg(h) \leq 2k+1$. So we have to see that a polynomial $h$ satisfying these conditions has to be $0$.

Such a polynomial should be divisible by $x^{k+1}$ and by $(x-1)^{k+1}$, which are coprime (no common factors in their unique factorization), which implies that such a polynomial has to be divisible by $x^{k+1}.(x-1)^{k+1}$, and therefore the degree is at least $2k+2$ unless $h=0$.