MHB Solve Real Solutions of $(1-x_1)^2+\cdots+x_{2013}^2=\frac{1}{2014}$

[magneto1]

Find all the solutions in real numbers to the equation:
\[
(1-x_1)^2 +(x_1-x_2)^2+(x_2-x_3)^2 + \cdots + (x_{2012}-x_{2013})^2 + x_{2013}^2 = \frac{1}{2014},
\]
and show that you have all of the solutions that would satisfy the equation.

 

[Opalg]

[sp]and show that you have all of the solutions that would satisfy the equation.[/QUOTE]
Let $x_0 = 1$, $x_{2014} = 0$ and $y_i = x_{i-1} - x_i\ (1\leqslant i\leqslant 2014)$. Then $\sum y_i = x_0 - x_{2014} = 1-0 = 1$ (telescoping sum), and $\sum y_i^2 = \frac1{2014}.$ Let $z_i = 1\ (1\leqslant i\leqslant 2014)$. Then $1 = \sum y_i = \Bigl(\sum y_i\Bigr)^2 = \Bigl(\sum y_iz_i\Bigr)^2 \leqslant \sum y_i^2 \sum z_i^2 = \frac{2014}{2014} = 1$ (Cauchy–Schwarz). But equality holds in the Cauchy–Schwarz inequality only if $y_i = cz_i$ for some constant $c$. Thus all the $y_i$ are equal, and since their sum is $1$ they must all be equal to $\dfrac1{2014}.$ Therefore $x_i = 1 - \dfrac i{2014}\ (1\leqslant i\leqslant 2013)$, and that is the unique solution.[/sp]

 

[magneto1]

[sp]and show that you have all of the solutions that would satisfy the equation.

Let $x_0 = 1$, $x_{2014} = 0$ and $y_i = x_{i-1} - x_i\ (1\leqslant i\leqslant 2014)$. Then $\sum y_i = x_0 - x_{2014} = 1-0 = 1$ (telescoping sum), and $\sum y_i^2 = \frac1{2014}.$ Let $z_i = 1\ (1\leqslant i\leqslant 2014)$. Then $1 = \sum y_i = \Bigl(\sum y_i\Bigr)^2 = \Bigl(\sum y_iz_i\Bigr)^2 \leqslant \sum y_i^2 \sum z_i^2 = \frac{2014}{2014} = 1$ (Cauchy–Schwarz). But equality holds in the Cauchy–Schwarz inequality only if $y_i = cz_i$ for some constant $c$. Thus all the $y_i$ are equal, and since their sum is $1$ they must all be equal to $\dfrac1{2014}.$ Therefore $x_i = 1 - \dfrac i{2014}\ (1\leqslant i\leqslant 2013)$, and that is the unique solution.[/sp][/QUOTE]

Almost correct. There is actually another set of $x_i$ that would work.