Be negative!

Algebra Level 5

Consider the function f ( x 1 , … , x n ) = ∑ k = 1 n x k 2 − x 1 x 3 − ∑ k = 2 n − 1 x k x k + 1 \large f(x_1,\ldots ,x_n)=\sum_{k=1}^{n}x_k^2-x_1x_3-\sum_{k=2}^{n-1}x_kx_{k+1}

where n ≥ 3 n\geq 3 and x 1 , … , x n x_1,\ldots ,x_n are real numbers. Find the smallest value of n n such that f f attains some negative values.

If you come to the conclusion that no such n n exists, enter 666.


The answer is 666.

Otto Bretscher by Otto Bretscher , 65, USA

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1 solution

Otto Bretscher
Jan 18, 2016

We can complete the squares, f ( x 1 , . . . , x n ) = ( x 1 − x 3 2 ) 2 + ( x 2 − x 3 2 ) 2 + 1 2 ∑ k = 3 n − 1 ( x k − x k + 1 ) 2 + [ 2 x n 2 f(x_1,...,x_n)=\left(x_1-\frac{x_3}{2}\right)^2+\left(x_2-\frac{x_3}{2}\right)^2+\frac{1}{2}\sum_{k=3}^{n-1}\left(x_k-x_{k+1}\right)^2+\frac{[}{2}x_n^2 , to see that this function does not attain negative values for any n n . The answer is 666 \boxed{666} .

2 Helpful 0 Interesting 0 Brilliant 0 Confused

Moderator note:

Good observation of completing the square to find the minimium.

Alternatively, we could consider a variable, fix the rest, and treat this as a quadratic equation.

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