Sequence with non-positive terms

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Sequence with non-positive terms

G H J

Reveal topic tags

algebra proposed

L

G H BBookmark kLocked kLocked NReply

Source: Baltic Way 2014, Problem 2

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

2105 posts

#1 h Nov 11, 2014, 1:51 PM • 2 Y

Y by Adventure10, Mango247

Let $a_0, a_1, . . . , a_N$ be real numbers satisfying $a_0 = a_N = 0$ and $a_{i+1} - 2a_i + a_{i-1} = a^2_i$ for $i = 1, 2, . . . , N - 1.$ Prove that $a_i\leq 0$ for $i = 1, 2, . . . , N- 1.$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

15142 posts

#2 h Nov 11, 2014, 2:42 PM • 3 Y

Y by FlakeLCR, Adventure10, Mango247

The condition $a_{i+1} - 2a_i + a_{i-1} = a^2_i$ for $i = 1, 2, \ldots, N - 1$ is a herring; it is enough to know $a_{i+1} - 2a_i + a_{i-1} \geq 0$ .

Consider $x_i = \textrm{e}^{-a_i}$ ; then $x_1^2 = \textrm{e}^{-2a_i} \geq \textrm{e}^{-a_{i-1} - a_{i+1}} = x_{i-1}x_{i+1}$ , i.e. the sequence $(x_i)_{i=0}^{i=N}$ is log-concave, and as such, it is known (and easy to prove) it is unimodal, i.e. for some $0\leq j \leq N$ we have $x_0\leq x_1\leq \cdots \leq x_{j-1} \leq x_j \geq x_{j+1} \geq \cdots \geq x_N$ .
But that means $0 = a_0 \geq a_1\geq \cdots \geq a_{j-1} \geq a_j \leq a_{j+1} \leq \cdots \leq a_N = 0$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

9042 posts

#3 h Nov 12, 2014, 4:53 PM • 3 Y

Y by Kameawtamprooz, Adventure10, Mango247

mavropnevma wrote:

Consider $x_i = \textrm{e}^{-a_i}; \dotsc$

Why so complicated?
Just assume the maximum of the sequence would be at some $i$ not equal to $0$ or $N$ .
Then $a_i \ge a_{i+1}$ and $a_i \ge a_{i-1}$ which gives $0 \ge a_{i-1}+a_{i+1}-2a_i=a_i^2$ and thus $a_i=0$ which gives in every case the result that the maximum of the sequence is $0$ .

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

15142 posts

#4 h Nov 12, 2014, 5:20 PM • 4 Y

Y by Tintarn, FlakeLCR, Adventure10, Mango247

Tintarn wrote:

Why so complicated?

Just because I wanted to bring into picture the wide-spread literature on log-convex and log-concave sequences; of course, the whole thing is trivial.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

96 posts

#5 h Nov 8, 2018, 1:41 PM • 2 Y

Y by Adventure10, Mango247

$a^2_i+a^2_{i-1}+a^2_{i-2}+...+a^2_2+a^2_1 = -(a_i+a_1)$

LHS is nonnegative, then RHS must nonnegative too. We are done (cmiiw)

This post has been edited 2 times. Last edited by MADharm, Dec 6, 2018, 9:36 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

3760 posts

#6 h Nov 12, 2020, 8:22 PM

Y by

Suppose $a_i >0.$ Then, $2a_i+a_i^2=a_{i-1}+a_{i+1} >0.$ Then, either $a_{i-1}$ or $a_{i+1}$ is less than $0.$ We can continue using this process until we get $a_0$ or $a_N$ equal to $0,$ which is contradiction. Therefore, $a_i<0$ for all $1 \leq i \leq N-1.$

This post has been edited 2 times. Last edited by Williamgolly, Nov 13, 2020, 12:28 AM

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

1079 posts

#7 h Aug 13, 2021, 8:58 PM

Y by

Claim 1. Suppose $a_{i+1}> 0$ , $a_{i+1}-a_i>0$ for some $n-2\geq i\geq 0$ . Then, $a_{i+2}> 0$ and $a_{i+2}-a_{i+1}>0$ .
Proof.
We have $a_{i+2}=a_{i+1}^2+2a_{i+1}-a_{i}> 0$ and $a_{i+2}-a_{i+1}=a_{i+1}^2+a_{i+1}-a_{i}> 0.$

Claim 2. Suppose $a_{i-1}\leq 0$ for some $n-1\geq i\geq 1$ . Then, $a_i\leq 0$ .
Proof.
Suppose otherwise, i.e. $a_i>0$ and $a_{i-1}\leq 0$ , hence
$a_{i+1}=a_i^2+2a_i-a_{i-1}> 0$ and $a_{i+1}-a_i=a_i^2+a_i-a_{i-1}>0$ . By the first claim we have that $a_n>0$ , which is a contradiction.

Now by the second claim, we are done.

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

laikhanhhoang_3011

637 posts

laikhanhhoang_3011

#8 h Jan 10, 2023, 2:57 AM

Y by

s

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

35 posts

ray66

#9 h Apr 12, 2025, 3:40 AM

Y by

Because $a_{i+1}+a_{i-1} \ge 2a_i$ , we either have some increasing or decreasing sequence. But because $a_0$ and $a_n$ are both 0, the entire sequence has to be 0.

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a