Subfield: Algebra


Question: Let $n$ be a positive integer and let $a_{1}, \ldots, a_{n-1}$ be arbitrary real numbers. Define the sequences $u_{0}, \ldots, u_{n}$ and $v_{0}, \ldots, v_{n}$ inductively by $u_{0}=u_{1}=v_{0}=v_{1}=1$, and

$$
u_{k+1}=u_{k}+a_{k} u_{k-1}, \quad v_{k+1}=v_{k}+a_{n-k} v_{k-1} \quad \text { for } k=1, \ldots, n-1
$$

Prove that $u_{n}=v_{n}$.

Solution:
We prove by induction on $k$ that

$$
u_{k}=\sum_{\substack{0<i_{1}<\ldots<i_{t}<k \\ i_{j+1}-i_{j} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}}
\tag{1}
$$

Note that we have one trivial summand equal to 1 (which corresponds to $t=0$ and the empty sequence, whose product is 1 ).

For $k=0,1$ the sum on the right-hand side only contains the empty product, so (1) holds due to $u_{0}=u_{1}=1$. For $k \geqslant 1$, assuming the result is true for $0,1, \ldots, k$, we have

$$
\begin{aligned}
u_{k+1} & =\sum_{\substack{0<i_{1}<\ldots<i_{t}<k, i_{j+1}-i_{j} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}}+\sum_{\substack{0<i_{1}<\ldots<i_{t}<k-1, i_{j+1}-i_{j} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}} \cdot a_{k} \\
& =\sum_{\substack{0<i_{1}<\ldots<i_{t}<k+1, i_{j+1}-i_{j} \geqslant 2, k \notin\left\{i_{1}, \ldots, i_{t}\right\}}} a_{i_{1}} \ldots a_{i_{t}}+\sum_{\substack{0<i_{1}<\ldots<i_{t}<k+1, i_{j+1}-i_{j} \geqslant 2, k \in\left\{i_{1}, \ldots, i_{t}\right\}}} a_{i_{1}} \ldots a_{i_{t}} \\
& =\sum_{\substack{0<i_{1}<\ldots<i_{t}<k+1, i_{j+1}-i_{j} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}},
\end{aligned}
$$

as required.

Applying (1) to the sequence $b_{1}, \ldots, b_{n}$ given by $b_{k}=a_{n-k}$ for $1 \leqslant k \leqslant n$, we get

$$
v_{k}=\sum_{\substack{0<i_{1}<\ldots<i_{t}<k, i_{j+1}-i_{j} \geqslant 2}} b_{i_{1}} \ldots b_{i_{t}}=\sum_{\substack{n>i_{1}>\ldots>i_{t}>n-k \\ i_{j}-i_{j+1} \geqslant 2}} a_{i_{1}} \ldots a_{i_{t}}
\tag{2}
$$

For $k=n$ the expressions (1) and (2) coincide, so indeed $u_{n}=v_{n}$.
Define recursively a sequence of multivariate polynomials by

$$
P_{0}=P_{1}=1, \quad P_{k+1}\left(x_{1}, \ldots, x_{k}\right)=P_{k}\left(x_{1}, \ldots, x_{k-1}\right)+x_{k} P_{k-1}\left(x_{1}, \ldots, x_{k-2}\right),
$$

so $P_{n}$ is a polynomial in $n-1$ variables for each $n \geqslant 1$. Two easy inductive arguments show that

$$
u_{n}=P_{n}\left(a_{1}, \ldots, a_{n-1}\right), \quad v_{n}=P_{n}\left(a_{n-1}, \ldots, a_{1}\right)
$$



so we need to prove $P_{n}\left(x_{1}, \ldots, x_{n-1}\right)=P_{n}\left(x_{n-1}, \ldots, x_{1}\right)$ for every positive integer $n$. The cases $n=1,2$ are trivial, and the cases $n=3,4$ follow from $P_{3}(x, y)=1+x+y$ and $P_{4}(x, y, z)=$ $1+x+y+z+x z$.

Now we proceed by induction, assuming that $n \geqslant 5$ and the claim hold for all smaller cases. Using $F(a, b)$ as an abbreviation for $P_{|a-b|+1}\left(x_{a}, \ldots, x_{b}\right)$ (where the indices $a, \ldots, b$ can be either in increasing or decreasing order),

$$
\begin{aligned}
F(n, 1) & =F(n, 2)+x_{1} F(n, 3)=F(2, n)+x_{1} F(3, n) \\
& =\left(F(2, n-1)+x_{n} F(2, n-2)\right)+x_{1}\left(F(3, n-1)+x_{n} F(3, n-2)\right) \\
& =\left(F(n-1,2)+x_{1} F(n-1,3)\right)+x_{n}\left(F(n-2,2)+x_{1} F(n-2,3)\right) \\
& =F(n-1,1)+x_{n} F(n-2,1)=F(1, n-1)+x_{n} F(1, n-2) \\
& =F(1, n),
\end{aligned}
$$

as we wished to show.
Using matrix notation, we can rewrite the recurrence relation as

$$
\left(\begin{array}{c}
u_{k+1} \\
u_{k+1}-u_{k}
\end{array}\right)=\left(\begin{array}{c}
u_{k}+a_{k} u_{k-1} \\
a_{k} u_{k-1}
\end{array}\right)=\left(\begin{array}{cc}
1+a_{k} & -a_{k} \\
a_{k} & -a_{k}
\end{array}\right)\left(\begin{array}{c}
u_{k} \\
u_{k}-u_{k-1}
\end{array}\right)
$$

for $1 \leqslant k \leqslant n-1$, and similarly

$$
\left(v_{k+1} ; v_{k}-v_{k+1}\right)=\left(v_{k}+a_{n-k} v_{k-1} ;-a_{n-k} v_{k-1}\right)=\left(v_{k} ; v_{k-1}-v_{k}\right)\left(\begin{array}{cc}
1+a_{n-k} & -a_{n-k} \\
a_{n-k} & -a_{n-k}
\end{array}\right)
$$

for $1 \leqslant k \leqslant n-1$. Hence, introducing the $2 \times 2$ matrices $A_{k}=\left(\begin{array}{cc}1+a_{k} & -a_{k} \\ a_{k} & -a_{k}\end{array}\right)$ we have

$$
\left(\begin{array}{c}
u_{k+1} \\
u_{k+1}-u_{k}
\end{array}\right)=A_{k}\left(\begin{array}{c}
u_{k} \\
u_{k}-u_{k-1}
\end{array}\right) \quad \text { and } \quad\left(v_{k+1} ; v_{k}-v_{k+1}\right)=\left(v_{k} ; v_{k-1}-v_{k}\right) A_{n-k} .
$$

for $1 \leqslant k \leqslant n-1$. Since $\left(\begin{array}{c}u_{1} \\ u_{1}-u_{0}\end{array}\right)=\left(\begin{array}{l}1 \\ 0\end{array}\right)$ and $\left(v_{1} ; v_{0}-v_{1}\right)=(1 ; 0)$, we get

$$
\left(\begin{array}{c}
u_{n} \\
u_{n}-u_{n-1}
\end{array}\right)=A_{n-1} A_{n-2} \cdots A_{1} \cdot\left(\begin{array}{l}
1 \\
0
\end{array}\right) \quad \text { and } \quad\left(v_{n} ; v_{n-1}-v_{n}\right)=(1 ; 0) \cdot A_{n-1} A_{n-2} \cdots A_{1} \text {. }
$$

It follows that

$$
\left(u_{n}\right)=(1 ; 0)\left(\begin{array}{c}
u_{n} \\
u_{n}-u_{n-1}
\end{array}\right)=(1 ; 0) \cdot A_{n-1} A_{n-2} \cdots A_{1} \cdot\left(\begin{array}{l}
1 \\
0
\end{array}\right)=\left(v_{n} ; v_{n-1}-v_{n}\right)\left(\begin{array}{l}
1 \\
0
\end{array}\right)=\left(v_{n}\right) .
$$