Find the maximum and minimum of $\sum_{i=1}^{n-1}x_ix_{i+1}$ subject to $\sum_{i=1}^nx_i^2=1$. Find the maximum and minimum of 
$$
\sum_{i=1}^{n-1}x_ix_{i+1}
$$
subject to 
$$
\sum_{i=1}^nx_i^2=1
$$
for all $n\in\mathbb{N}-\{1,0\}$.
 A: Answer: max is $\displaystyle\cos{\frac{\pi}{n+1}}$,   min is $\displaystyle -\cos{\frac{\pi}{n+1}}$.

It can be used Lagrange multiplier method. 
Let $\mathbf{x}=(x_1,x_2, \ldots,x_n)$.
We have function $f(\mathbf{x}) = x_1 x_2 + x_2x_3+\cdots + x_{n-1}x_n \rightarrow \max$,  
and we have condition $g(\mathbf{x}) = x_1^2 + x_2^2+\cdots+x_n^2-1 = 0$.
We create other function
$$
F(\mathbf{x},\lambda) = f(\mathbf{x}) + \lambda \cdot g(\mathbf{x}).
$$
If $F(\mathbf{x},\lambda)$ has extremum somewhere, then all partial derivatives must be 0 here.
So, we have (n+1) conditions:
$
\left\{ 
\begin{array}{r}
x_2 + 2\lambda x_1 = 0, \\
x_1 + x_3 + 2\lambda x_2 = 0, \\
x_2 + x_4 + 2\lambda x_3 = 0, \\
\cdots \\
x_{n-2}+x_n+2\lambda x_{n-1} = 0, \\
x_{n-1}+2\lambda x_n = 0; \\ 
x_1^2 + x_2^2+\cdots+x_n^2-1 = 0.
\end{array}
\right.
$
We can consider $\lambda$ as parameter for first $n$ equations. Then the system of first $n$ linear equations has 3-diagonal matrix
$M_n(\lambda) = \left(
\begin{array}{cccccc}
2\lambda & 1 & 0 &  \cdots & 0 & 0 \\
1 & 2\lambda & 1 &  \ddots & 0 & 0 \\
0 & 1 & 2\lambda &   \ddots & 0 & 0 \\
\vdots & \ddots & \ddots & \ddots & \ddots & \vdots \\
0 & 0 & 0 & \ddots  & 2\lambda & 1\\
0 & 0 & 0 & \cdots  & 1 & 2\lambda \\
\end{array}
\right).
$
We see that $\det M_1(\lambda) = 2\lambda$, $\det M_2(\lambda) = 4\lambda^2-1$, 
$\det M_{n+1}(\lambda) = 2\lambda \det M_n(\lambda) - \det M_{n-1}(\lambda)$. 
It is recurrent formula for Chebyshev polynomials of the second kind (by definition). 

So, for $\lambda_j = \cos(j\pi/(n+1))$ we have  $\det M_n(\lambda_j) = \frac{\sin(j\pi)}{\sin(j\pi/(n+1))}=0$, where $j=1,\ldots,n$.

It is easy to see that for $\lambda = \lambda_j$, $j=1,\ldots,n$, we have 
$\mathbf{x}_j =  (x_{j1}, \ldots, x_{jn} )$, where $x_{jk} = C_j \sin \frac{(n+1-j)k\pi}{n+1} $, $k=1,\ldots,n$.
Condition $g(\mathbf{x}_j)=0$  implies $C_j = \sqrt{2/(n+1)}$.
$f(\mathbf{x}_j) = \cos \frac{(n+1-j)\pi}{n+1}$.

So, 
$\max\limits_{g(\mathbf{x})=0} f(\mathbf{x}) = 
\max\limits_{j=1,\ldots,n} \cos\frac{(n+1-j)\pi}{n+1} = \cos \frac{\pi}{n+1}$. (It is attained when $j = n$).
$\min\limits_{g(\mathbf{x})=0} f(\mathbf{x}) = 
\min\limits_{j=1,\ldots,n} \cos\frac{(n+1-j)\pi}{n+1} = \cos \frac{n\pi}{n+1} = -\cos \frac{\pi}{n+1}$. (It is attained when $j = 1$).
$\max\limits_{g(\mathbf{x})=0} |f(\mathbf{x})| = \cos \frac{\pi}{n+1}$. (obviously).
$\min\limits_{g(\mathbf{x})=0} |f(\mathbf{x})| = 
\min\limits_{j=1,\ldots,n} \left|\cos\frac{(n+1-j)\pi}{n+1} \right| = $
$
\left\{ 
\begin{array}{cl}
0, & \mbox{when } n \mbox{ is odd},  (\mbox{ when } j=(n+1)/2) \\
\cos \frac{n\pi}{2n+2}, & \mbox{when } n \mbox{ is even}, (\mbox{ when } j=n/2 + 1).
\end{array}
\right.
$

Examples:
$n=2$: $f_{max} = \cos{\frac{\pi}{3}}$, 
$$(x_1,x_2) = \left( \sqrt{\frac{2}{3}} \sin{\frac{\pi}{3}},  \sqrt{\frac{2}{3}} \sin{\frac{2\pi}{3}}\right);$$
$n=3$: $f_{max} = \cos{\frac{\pi}{4}}$, 
$$(x_1,x_2,x_3) = \left( 
\sqrt{\frac{2}{4}} \sin{\frac{\pi}{4}},
\sqrt{\frac{2}{4}} \sin{\frac{2\pi}{4}},
\sqrt{\frac{2}{4}} \sin{\frac{3\pi}{4}}
\right);$$
$n=4$: $f_{max} = \cos{\frac{\pi}{5}}$, 
$$(x_1,x_2,x_3,x_4) = \left( 
\sqrt{\frac{2}{5}} \sin{\frac{\pi}{5}},
\sqrt{\frac{2}{5}} \sin{\frac{2\pi}{5}},
\sqrt{\frac{2}{5}} \sin{\frac{3\pi}{5}},
\sqrt{\frac{2}{5}} \sin{\frac{4\pi}{5}}
\right);$$
$n=5$: $f_{max} = \cos{\frac{\pi}{6}}$,  
$$(x_1,x_2,x_3,x_4,x_5) = \left( 
\sqrt{\frac{2}{6}} \sin{\frac{\pi}{6}},
\sqrt{\frac{2}{6}} \sin{\frac{2\pi}{6}},
\sqrt{\frac{2}{6}} \sin{\frac{3\pi}{6}},
\sqrt{\frac{2}{6}} \sin{\frac{4\pi}{6}},
\sqrt{\frac{2}{6}} \sin{\frac{5\pi}{6}}
\right);$$
$\cdots$
A: Hint: Use Lagrange multiplier method.
A: We can write the original expression as 
$-1/r + (1/r)(\sum_{i=1}^{n}x_i^2+r\sum_{i=1}^{n-1}x_ix_{i+1})$
And completing the squares we get a sequence $\{a_n\}$ with
$\sum_{i=1}^{n}x_i^2+r\sum_{i=1}^{n-1}x_ix_{i+1}=\sum_{k=1}^{n-1}(\sqrt{a_k}x_k+\frac{r}{2\sqrt{a_k}}x_{k+1})^2$.
From 
Sequence $a_k=1-\frac{\lambda^2}{4a_{k-1}},\ k=2,3,\ldots,n$. 
we can pick $r=\pm\frac1{\cos{\frac{\pi}{n+1}}}$ for maximum and minimum. It remains to verify we can attain equality.
