Return to Answer

Wadim, isn't that 95% of the proof? First let me correct your first displayed equation (thanks to fherzig for pointing this out): It is not sufficient for the proof, but $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ is, where $t$ and $s$ are two independent variables.

Let me rename your $P_i$ as $Q_{2i}$ and your $\hat{P_i}$ as $Q_{2i+1}$, so that your equation $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now let $Q$ be the polynomial $Q\left(t\right)=t^{n-1}$ (I suspect the below proof works with any polynomial $Q$ of degree $n-1$ (not less!), but I'm not completely sure and don't have the time to check) and let $Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$ for every $i\in\mathbb Z$. Being a polynomial in $i$ of degree $2\left(2\left(n-1\right)+1\right)<4n$ (for fixed $t$ and $s$), the term $Q_i\left(t\right)Q_i\left(s\right)$ satisfies $$ \sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0, $$ since the $4n$-th finite difference of a polynomial of degree $< 4n$ is zero. Due to the symmetry of the function $i\mapsto Q_i\left(t\right)Q_i\left(s\right)$ around $i=2n$, and due to $Q_{2n}\left(t\right)=0$, this becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now it remains to prove that each of the families $\left(Q_1,Q_3,...,Q_{2n-1}\right)$ and $\left(Q_0,Q_2,...,Q_{2n-2}\right)$ spans the space of all polynomials in $t$ of degree $< n$. This is a particular case of a more general fact: If $x_1$, $x_2$, ..., $x_n$ are $n$ pairwise distinct reals, then the polynomials $\left(t-x_1\right)^{n-1}$, $\left(t-x_2\right)^{n-1}$, ..., $\left(t-x_n\right)^{n-1}$ are linearly independent. In order to prove this, assume that they are linearly dependent, take their derivatives of all possible orders, evaluate at $t=0$, and get a contradiction because Vandermonde's determinant is nonzero.

Wadim, isn't that 95% of the proof? First let me correct your first displayed equation (thanks to fherzig for pointing this out): It is not sufficient for the proof, but $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ is, where $t$ and $s$ are two independent variables.

Let me rename your $P_i$ as $Q_{2i}$ and your $\hat{P_i}$ as $Q_{2i+1}$, so that your equation $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now let $Q$ be the polynomial $Q\left(t\right)=t^{n-1}$. (With some work, the proof below works just as well if $Q$ is any polynomial of degree $n-1$ (not less!), but let me use $t^{n-1}$ for simplicity's sake.) Let $Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$ for every $i\in\mathbb Z$. For any fixed $t$ and $s$, the term $Q_i\left(t\right)Q_i\left(s\right)$ is a polynomial in $i$ of degree $2\left(2\left(n-1\right)+1\right)<4n$, and thus satisfies $$ \sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0, $$ since the $4n$-th finite difference of a polynomial of degree $< 4n$ is zero. Due to the symmetry of the function $i\mapsto Q_i\left(t\right)Q_i\left(s\right)$ around $i=2n$, and due to $Q_{2n}\left(t\right)=0$, this becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now it remains to prove that each of the families $\left(Q_1,Q_3,...,Q_{2n-1}\right)$ and $\left(Q_0,Q_2,...,Q_{2n-2}\right)$ spans the space of all polynomials in $t$ of degree $< n$. This is a particular case of a more general fact: If $x_1$, $x_2$, ..., $x_n$ are $n$ distinct reals, then the polynomials $\left(t-x_1\right)^{n-1}$, $\left(t-x_2\right)^{n-1}$, ..., $\left(t-x_n\right)^{n-1}$ are linearly independent. In order to prove this, assume that they are linearly dependent, take their derivatives of all possible orders, evaluate at $t=0$ (or alternatively, just take their coefficients), and get a contradiction because Vandermonde's determinant is nonzero.

Wadim, isn't that 95% of the proof? First let me correct your first displayed equation (thanks to fherzig for pointing this out): It is not sufficient for the proof, but $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ is, where $t$ and $s$ are two independent variables.

Let me rename your $P_i$ as $Q_{2i}$ and your $\hat{P_i}$ as $Q_i$, so that your equation $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now let $Q$ be the polynomial $Q\left(t\right)=t^{n-1}$ (I suspect the below proof works with any polynomial $Q$ of degree $n-1$ (not less!), but I'm not completely sure and don't have the time to check) and let $Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$ for every $i\in\mathbb Z$. Being a polynomial in $i$ of degree $2\left(2\left(n-1\right)+1\right)=4n$ (for fixed $t$ and $s$), the term $Q_i\left(t\right)Q_i\left(s\right)$ satisfies $$ \sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0, $$ since the $\left(4n+1\right)$-th finite difference of a polynomial of degree $4n$ is zero. Due to the symmetry of the function $i\mapsto Q_i\left(t\right)Q_i\left(s\right)$ around $i=2n$, and due to $Q_{2n}\left(t\right)=0$, this becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now it remains to prove that each of the families $\left(Q_1,Q_3,...,Q_{2n-1}\right)$ and $\left(Q_0,Q_2,...,Q_{2n-2}\right)$ spans the space of all polynomials in $t$ of degree $< n$. This is a particular case of a more general fact: If $x_1$, $x_2$, ..., $x_n$ are $n$ pairwise distinct reals, then the polynomials $\left(t-x_1\right)^{n-1}$, $\left(t-x_2\right)^{n-1}$, ..., $\left(t-x_n\right)^{n-1}$ are linearly independent. In order to prove this, assume that they are linearly dependent, take their derivatives of all possible orders, evaluate at $t=0$, and get a contradiction because Vandermonde's determinant is nonzero.

Wadim, isn't that 95% of the proof? First let me correct your first displayed equation (thanks to fherzig for pointing this out): It is not sufficient for the proof, but $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ is, where $t$ and $s$ are two independent variables.

Let me rename your $P_i$ as $Q_{2i}$ and your $\hat{P_i}$ as $Q_{2i+1}$, so that your equation $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now let $Q$ be the polynomial $Q\left(t\right)=t^{n-1}$ (I suspect the below proof works with any polynomial $Q$ of degree $n-1$ (not less!), but I'm not completely sure and don't have the time to check) and let $Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$ for every $i\in\mathbb Z$. Being a polynomial in $i$ of degree $2\left(2\left(n-1\right)+1\right)<4n$ (for fixed $t$ and $s$), the term $Q_i\left(t\right)Q_i\left(s\right)$ satisfies $$ \sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0, $$ since the $4n$-th finite difference of a polynomial of degree $< 4n$ is zero. Due to the symmetry of the function $i\mapsto Q_i\left(t\right)Q_i\left(s\right)$ around $i=2n$, and due to $Q_{2n}\left(t\right)=0$, this becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now it remains to prove that each of the families $\left(Q_1,Q_3,...,Q_{2n-1}\right)$ and $\left(Q_0,Q_2,...,Q_{2n-2}\right)$ spans the space of all polynomials in $t$ of degree $< n$. This is a particular case of a more general fact: If $x_1$, $x_2$, ..., $x_n$ are $n$ pairwise distinct reals, then the polynomials $\left(t-x_1\right)^{n-1}$, $\left(t-x_2\right)^{n-1}$, ..., $\left(t-x_n\right)^{n-1}$ are linearly independent. In order to prove this, assume that they are linearly dependent, take their derivatives of all possible orders, evaluate at $t=0$, and get a contradiction because Vandermonde's determinant is nonzero.

Wadim, isn't that 95% of the proof? Let me rename your $P_i$ as $Q_{2i}$ and your $\hat{P_i}$ as $Q_i$, so that your equation $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)^2 =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)^2 $$ becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)^2=0. $$ Now let $Q$ be the polynomial $Q\left(t\right)=t^{n-1}$ (I suspect the below proof works with any polynomial $Q$ of degree $n-1$ (not less!), but I'm not completely sure and don't have the time to check) and let $Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$ for every $i\in\mathbb Z$. Being a polynomial in $i$ of degree $2\left(2\left(n-1\right)+1\right)=4n$ (for fixed $t$), the square $Q_i\left(t\right)^2$ satisfies $$ \sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)^2=0, $$ since the $\left(4n+1\right)$-th finite difference of a polynomial of degree $4n$ is zero. Due to the symmetry of the function $i\mapsto Q_i\left(t\right)^2$ around $i=2n$, and due to $Q_{2n}\left(t\right)=0$, this becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)^2=0. $$ Now it remains to prove that each of the families $\left(Q_1,Q_3,...,Q_{2n-1}\right)$ and $\left(Q_0,Q_2,...,Q_{2n-2}\right)$ spans the space of all polynomials in $t$ of degree $< n$. This is a particular case of a more general fact: If $x_1$, $x_2$, ..., $x_n$ are $n$ pairwise distinct reals, then the polynomials $\left(t-x_1\right)^{n-1}$, $\left(t-x_2\right)^{n-1}$, ..., $\left(t-x_n\right)^{n-1}$ are linearly independent. In order to prove this, assume that they are linearly dependent, take their derivatives of all possible orders, evaluate at $t=0$, and get a contradiction because Vandermonde's determinant is nonzero.

Wadim, isn't that 95% of the proof? First let me correct your first displayed equation (thanks to fherzig for pointing this out): It is not sufficient for the proof, but $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ is, where $t$ and $s$ are two independent variables.

Let me rename your $P_i$ as $Q_{2i}$ and your $\hat{P_i}$ as $Q_i$, so that your equation $$ \sum_{i=0}^{n-1}\binom{4n}{2i}P_i(t)P_i(s) =\sum_{i=0}^{n-1}\binom{4n}{2i+1}\hat P_i(t)\hat P_i(s) $$ becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now let $Q$ be the polynomial $Q\left(t\right)=t^{n-1}$ (I suspect the below proof works with any polynomial $Q$ of degree $n-1$ (not less!), but I'm not completely sure and don't have the time to check) and let $Q_i\left(t\right)=\left(2n-i\right)Q\left(t-\left(2n-i\right)^2\right)$ for every $i\in\mathbb Z$. Being a polynomial in $i$ of degree $2\left(2\left(n-1\right)+1\right)=4n$ (for fixed $t$ and $s$), the term $Q_i\left(t\right)Q_i\left(s\right)$ satisfies $$ \sum_{i=0}^{4n}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0, $$ since the $\left(4n+1\right)$-th finite difference of a polynomial of degree $4n$ is zero. Due to the symmetry of the function $i\mapsto Q_i\left(t\right)Q_i\left(s\right)$ around $i=2n$, and due to $Q_{2n}\left(t\right)=0$, this becomes $$ \sum_{i=0}^{2n-1}\left(-1\right)^i\binom{4n}{i}Q_i(t)Q_i(s)=0. $$ Now it remains to prove that each of the families $\left(Q_1,Q_3,...,Q_{2n-1}\right)$ and $\left(Q_0,Q_2,...,Q_{2n-2}\right)$ spans the space of all polynomials in $t$ of degree $< n$. This is a particular case of a more general fact: If $x_1$, $x_2$, ..., $x_n$ are $n$ pairwise distinct reals, then the polynomials $\left(t-x_1\right)^{n-1}$, $\left(t-x_2\right)^{n-1}$, ..., $\left(t-x_n\right)^{n-1}$ are linearly independent. In order to prove this, assume that they are linearly dependent, take their derivatives of all possible orders, evaluate at $t=0$, and get a contradiction because Vandermonde's determinant is nonzero.