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Given any positive integer $n$, let $t(n)$ be the smallest real number such that for any real numbers $x_1,\dots,x_n$, the following inequality holds $$ \sum\limits_{k=1}^n (x_1+\dots+x_k)^2 \leqslant t(n) \sum\limits_{i=1}^n x_i^2.$$ Please give an expression for $t(n)$.

Attempt:

The problem is equivalent to calculating the maximum of $\sum\limits_{k=1}^n (x_1+\dots+x_k)^2$ on the unit $n$-sphere $\sum\limits_{i=1}^n x_i^2=1$. Let

$$ A := \begin{pmatrix} n & n-1 & n-2 & \cdots & 1\\ n-1 & n-1 & n-2 & \cdots & 1\\ n-2 & n-2 & n-2 & \cdots & 1\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 1 & 1 & \cdots & 1 \end{pmatrix}$$

Then

$$\sum\limits_{k=1}^n (x_1+\cdots+x_k)^2=xAx^T$$

where $x=(x_1,\dots,x_n)$. According to Lagrange multipliers, we know the problem is equivalent to calculating the maximum eigenvalue of $A$. Then I'm stuck. The book says that the answer is

$$\frac{1}{4\sin^2\left(\frac{\pi}{4n+2}\right)}.$$

It is really amazing. I would appreciate it if someone could please give me some hints.

Rodrigo de Azevedo
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Jiongjie Wang
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2 Answers2

5

Hint. Note that the inverse of $A$ is $$B_n=\begin{pmatrix} 1 & -1 \\ -1 & 2 & -1 \\ & -1 & 2 & -1 \\ &&&\ddots\\ &&& -1 & 2 &-1\\ &&&& -1 & 2 \end{pmatrix}.$$ Then the characteristic polynomial $P_n$ of the tridiagonal matrix $B_n$ satisfies the recurrence relation: $$P_{n}(x)=P_{n-1}(x)(x-2)-P_{n-2}(x)$$ with $P_1(x)=x-1$ and $P_0(x)=1$.

See also the related question Eigenvalues of tridiagonal symmetric matrix with diagonal entries $2$ and subdiagonal entries $1$

Robert Z
  • 145,942
5

We will use the result of (Fonseca, 2004, On the eigenvalues of some tridiagonal matrices). For that, let's write the matrix in another way $$\sum\limits_{k=1}^n (x_1+\cdots+x_k)^2=\mathbf{yAy}^T$$ such that $$\mathbf{y}:=(x_\color{red}{n},x_\color{red}{{n-1}},...,x_\color{red}{1})^T$$ $$\mathbf{A}:=\begin{pmatrix} 1 & 1 & 1 & \cdots & 1\\ 1 & 2 & 2 & \cdots & 2\\ 1 & 2 & 3 & \cdots & 3\\ \vdots & \vdots & \vdots & \ddots & \vdots\\ 1 & 2 & 3 & \cdots & n \end{pmatrix}$$

The matrix $\mathbf{A}$ is the one defined in section 1. Then applying the theorem 2.1, the largest eigenvalue of $\mathbf{A}$ is $$\begin{align} \lambda_0&=\frac{1}{2}\frac{1}{1-\cos\left(\frac{\pi}{2n+1} \right)}\\ &=\frac{1}{2}\frac{1}{2\sin^2\left(\frac{\pi}{4n+2} \right)}\\ &=\color{red}{\frac{1}{4\sin^2\left(\frac{\pi}{4n+2} \right)}} \end{align}$$

NN2
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