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Prove that:

$\begin{equation} \begin{vmatrix} x_0^{2n+1}&x_0^{2n}&\cdots&x_0&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (2n+1)x_0^{2n}&2nx_0^{2n-1}&\cdots&1&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_n^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix}=(-1)^{n}\prod_{n\geq i>j\geq 0}(x_i-x_j)^4 \end{equation}$

where $(x_i^{2n+1})'=(2n+1)x_i^{2n}$,$\cdots$,$x_i'=1$,$1'=0$.

3 Answers 3

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Each column contains the values and first derivatives of a power of $x$ evaluated at $x_0$ to $x_n$. Thus multiplying a column vector of coefficients by this matrix yields a column vector of values and first derivatives of the corresponding polynomial of degree up to $2n+1$ evaluated at $x_0$ to $x_n$. Since this linear evaluation map from the vector space of polynomials of degree up to $2n+1$ to the space of $(2n+2)$-dimensional vectors is an isomorphism if the $x_i$ are distinct (Hermite interpolation establishes a surjection in one direction, and this is a bijection because the dimensions coincide), the product of the matrix with a vector is non-zero unless the vector is zero or two of the $x_i$ coincide; thus the determinant is non-zero unless two of the $x_i$ coincide. Considering the determinant as a polynomial of degree $4n$ in, say, $x_0$, it must be a constant times linear factors $x_0-x_i$ for its roots. By symmetry, each of these factors must appear the same number of times, and thus $4$ times. Applying this argument for each $x_j$ shows that the determinant must be a constant times the product given in the question. To evaluate the constant, consider the term proportional to $x_n^{4n}x_{n-1}^{4(n-1)}\dotso x_{1}^4x_0^0$, which arises from $n+1$ determinants of $2\times2$ matrices containing only one $x_i$ and evaluating to $-x_i^{4i}$; then you just have to check that an odd permutation reorders the matrix such that these $2\times2$ matrices are on the diagonal.

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    Extraordinary....2012-12-15
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Having seen the answer given by joriki,I think it is the time to share my own answer(I appreciate his answer,and need some time to digest...).

Let \begin{equation} f(x)= \begin{vmatrix} x^{2n+1}&x^{2n}&\cdots&x&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (2n+1)x^{2n}&2nx^{2n-1}&\cdots&1&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix} \end{equation}

It is easy to verify that $f(x)$ is a polynomial of degree $4n$.And \begin{equation} f(x_1)=\cdots =f(x_n)=0 \end{equation}

So $(x-x_1)(x-x_2)\cdots (x-x_n)|f(x)$.And

\begin{align*} f'(x)= \begin{vmatrix} x^{2n+1}&x^{2n}&\cdots&x&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ 2n(2n+1)x^{2n-1}&(2n-1)2nx^{2n-2}&\cdots&0&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix} \end{align*} So $f'(x_1)=\cdots f'(x_n)=0$.So $(x-x_1)^2(x-x_2)^2\cdots (x-x_n)^2|f(x)$.And

\begin{align*} f''(x)= \begin{vmatrix} (2n+1)x^{2n}&2nx^{2n-1}&\cdots&1&0\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ 2n(2n+1)x^{2n-1}&(2n-1)2nx^{2n-2}&\cdots&0&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix}+ \begin{vmatrix} x^{2n+1}&x^{2n}&\cdots&x&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (2n-1)2n(2n+1)x^{2n-2}&(2n-2)(2n-1)2nx^{2n-3}&\cdots&0&0\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (2n+1)x_{n}^{2n}&2nx_n^{2n-1}&\cdots&1&0\\ \end{vmatrix} \end{align*}

It is easy to verify that $f''(x_1)=\cdots f''(x_n)=0$.So

\begin{equation} (x-x_1)^3(x-x_2)^3\cdots (x-x_n)^3|f(x) \end{equation} And it is also easy to figure out $f'''(x)$,so it is easy to verify that

\begin{equation} f'''(x_1)=\cdots =f'''(x_n)=0 \end{equation} So \begin{equation} (x-x_1)^4(x-x_2)^4\cdots (x-x_n)^4|f(x) \end{equation}

Because $f(x)$ is a polynomial of degree $4n$,so $f(x)=a(x-x_1)^4(x-x_2)^4\cdots (x-x_n)^4$.According to symmetry ammong $x_0,x_1,\cdots,x_n$ in this determinant,it is easy to verify that

\begin{equation} f(x_0)=c\prod_{n\geq i>j\geq 0}(x_i-x_j)^4 \end{equation} Now we determine the constant $c$.We do it by induction.It is easy to verify that When $n=1$,

\begin{equation} \det \begin{pmatrix} x_0^3&x_0^2&x_0&1\\ x_1^3&x_1^2&x_1&1\\ 3x_0^2&2x_0&1&0\\ 3x_1^2&2x_1&1&0\\ \end{pmatrix}=-(x_0-x_1)^4 \end{equation}Then when $n=2$,Let's see the determinant

\begin{equation} \det\begin{pmatrix} x_0^5&x_0^4&x_0^3&x_0^2&x_0&1\\ x_1^5&x_1^4&*x_1^3&*x_1^2&*x_1&*1\\ x_2^5&x_2^4&*x_2^3&*x_2^2&*x_2&*1\\ 5x_0^4&4x_0^3&3x_0^2&2x_0&1&0\\ 5x_1^4&4x_1^3&*3x_1^2&*2x_1&*1&*0\\ 5x_2^4&4x_2^3&*3x_2^2&*2x_2&*1&*0\\ \end{pmatrix} \end{equation}

The element marked * also form a determinant,we know that the constant of this determinant is -1,so the constant term of the determinant \begin{equation} \det\begin{pmatrix} x_0^5&x_0^4&x_0^3&x_0^2&x_0&1\\ x_1^5&x_1^4&*x_1^3&*x_1^2&*x_1&*1\\ x_2^5&x_2^4&*x_2^3&*x_2^2&*x_2&*1\\ 5x_0^4&4x_0^3&3x_0^2&2x_0&1&0\\ 5x_1^4&4x_1^3&*3x_1^2&*2x_1&*1&*0\\ 5x_2^4&4x_2^3&*3x_2^2&*2x_2&*1&*0\\ \end{pmatrix} \end{equation} is $-1\times (4-5)=1$ ……

So by induction,we know that $c=(-1)^n$.


In my answer I make use of the following result

If \begin{equation} f(x)=\begin{vmatrix} f_{11}(x)&f_{12}(x)&\cdots&f_{1n}(x)\\ f_{21}(x)&f_{22}(x)&\cdots&f_{2n}(x)\\ \vdots&\vdots& &\vdots\\ f_{n1}(x)&f_{n2}(x)&\cdots&f_{nn}(x)\\ \end{vmatrix} \end{equation} Then \begin{equation} f'(x)=\sum_{i=1}^n \begin{vmatrix} f_{11}(x)&f_{12}(x)&\cdots&f_{1n}(x)\\ \vdots&\vdots & &\vdots\\ f'_{i1}(x)&f'_{i2}(x)&\cdots&f'_{in}(x)\\ \vdots&\vdots&&\vdots\\ f_{n1}(x)&f_{n2}(x)&\cdots&f_{nn}(x)\\ \end{vmatrix} \end{equation}

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    I see; sorry, I didn't mean to be critical; as I said, it's a very nice answer. I hope my summary can contribute to its understanding.2012-12-15
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There is a sign error in the right-hand side; the term $(-1)^n$ should be $(-1)^{(n+1)(n+2)/2}$.

By the usual formula for the determinant of the Vandermonde matrix,

\begin{equation} \begin{vmatrix} x_0^{2n+1}&x_0^{2n}&\cdots&x_0&1\\ x_1^{2n+1}&x_1^{2n}&\cdots&x_1&1\\ \vdots&\vdots&\cdots&\vdots\\ x_n^{2n+1}&x_n^{2n}&\cdots&x_n&1\\ (x_0+h)^{2n+1}&(x_0+h)^{2n}&\cdots&x_0+h&1\\ \vdots&\vdots&\cdots&\vdots&\cdots\\ (x_n+h)^{2n+1}&(x_n+h)^{2n}&\cdots&x_n+h&1\\ \end{vmatrix}=\prod_{i$n+1$ rows of the matrix, subtract it from the row $n+1$ rows lower down. This does not change the determinant of the matrix. Then divide the right-hand side by $h^{n+1}$, and divide the left-hand side by $h^{n+1}$ by dividing each of the last $n+1$ rows of the matrix by $h$. Setting $h:=0$ then gives the identity.