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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
Can this sequence be bounded?
darij grinberg   66
N a few seconds ago by shendrew7
Source: German pre-TST 2005, problem 4, ISL 2004, algebra problem 2
Let $a_0$, $a_1$, $a_2$, ... be an infinite sequence of real numbers satisfying the equation $a_n=\left|a_{n+1}-a_{n+2}\right|$ for all $n\geq 0$, where $a_0$ and $a_1$ are two different positive reals.

Can this sequence $a_0$, $a_1$, $a_2$, ... be bounded?

Proposed by Mihai Bălună, Romania
66 replies
darij grinberg
Jan 19, 2005
shendrew7
a few seconds ago
Functional Equations Marathon March 2025
Levieee   5
N 9 minutes ago by mqoi_KOLA
1. before posting another problem please try your best to provide the solution to the previous solution because we don't want a backlog of many problems
2.one is welcome to send functional equations involving calculus (mainly basic real analysis type of proofs) as long it is of the form $\text{"find all functions:"}$
5 replies
Levieee
2 hours ago
mqoi_KOLA
9 minutes ago
Sequences and limit
lehungvietbao   14
N 17 minutes ago by eg4334
Source: Vietnam Mathematical OLympiad 2014
Let $({{x}_{n}}),({{y}_{n}})$ be two positive sequences defined by ${{x}_{1}}=1,{{y}_{1}}=\sqrt{3}$ and
\[ \begin{cases}  {{x}_{n+1}}{{y}_{n+1}}-{{x}_{n}}=0 \\   x_{n+1}^{2}+{{y}_{n}}=2 \end{cases} \] for all $n=1,2,3,\ldots$.
Prove that they are converges and find their limits.
14 replies
lehungvietbao
Jan 3, 2014
eg4334
17 minutes ago
IMO 2014 Problem 1
Amir Hossein   131
N 25 minutes ago by eg4334
Let $a_0 < a_1 < a_2 < \dots$ be an infinite sequence of positive integers. Prove that there exists a unique integer $n\geq 1$ such that
\[a_n < \frac{a_0+a_1+a_2+\cdots+a_n}{n} \leq a_{n+1}.\]
Proposed by Gerhard Wöginger, Austria.
131 replies
Amir Hossein
Jul 8, 2014
eg4334
25 minutes ago
No more topics!
Limit of a sequence involving square roots
hajimbrak   5
N May 15, 2018 by neel02
Source: Indian TST Day 1 Problem 3
Starting with the triple $(1007\sqrt{2},2014\sqrt{2},1007\sqrt{14})$, define a sequence of triples $(x_{n},y_{n},z_{n})$ by
$x_{n+1}=\sqrt{x_{n}(y_{n}+z_{n}-x_{n})}$
$y_{n+1}=\sqrt{y_{n}(z_{n}+x_{n}-y_{n})}$
$ z_{n+1}=\sqrt{z_{n}(x_{n}+y_{n}-z_{n})}$
for $n\geq 0$.Show that each of the sequences $\langle x_n\rangle _{n\geq 0},\langle y_n\rangle_{n\geq 0},\langle z_n\rangle_{n\geq 0}$ converges to a limit and find these limits.
5 replies
hajimbrak
Jul 11, 2014
neel02
May 15, 2018
Limit of a sequence involving square roots
G H J
G H BBookmark kLocked kLocked NReply
Source: Indian TST Day 1 Problem 3
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hajimbrak
209 posts
#1 • 5 Y
Y by the0myth0, Paragdey12, Adventure10, Mango247, and 1 other user
Starting with the triple $(1007\sqrt{2},2014\sqrt{2},1007\sqrt{14})$, define a sequence of triples $(x_{n},y_{n},z_{n})$ by
$x_{n+1}=\sqrt{x_{n}(y_{n}+z_{n}-x_{n})}$
$y_{n+1}=\sqrt{y_{n}(z_{n}+x_{n}-y_{n})}$
$ z_{n+1}=\sqrt{z_{n}(x_{n}+y_{n}-z_{n})}$
for $n\geq 0$.Show that each of the sequences $\langle x_n\rangle _{n\geq 0},\langle y_n\rangle_{n\geq 0},\langle z_n\rangle_{n\geq 0}$ converges to a limit and find these limits.
This post has been edited 1 time. Last edited by hajimbrak, Aug 9, 2014, 8:22 AM
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pco
23442 posts
#2 • 4 Y
Y by babu2001, Paragdey12, Adventure10, Mango247
hajimbrak wrote:
Starting with the triple $(1007\sqrt{2},2014\sqrt{2},1007\sqrt{14})$, define a sequence of triples $(x_{n},y_{n},z_{n})$ by
$x_{n+1}=\sqrt{x_{n}(y_{n}+z_{n}-x_{n})},  y_{n+1}=\sqrt{y_{n}(z_{n}+x_{n}-y_{n})},  z_{n+1}=\sqrt{z_{n}(x_{n}+y_{n}-z_{n})}$,
for $n\geq 0$.Show that each of these sequences converges to a limit and find these limits.
Sketch of proof :

Considering that $(x_n,y_n,z_n)$ are the three positive sidelengths or a triangle, we can rather easily show that :
1) $(x_{n+1},y_{n+1},z_{n+1})$ are the sides of a new triangle with same area
2) $|x_n^2-y_n^2|$, $|x_n^2-z_n^2|$ and $|y_n^2-z_n^2|$ are decreasing sequences
3) $x_n,y_n,z_n$ have same limits, sides of an equilateral triangle of same area

Area of starting triangle is $\frac{\sqrt{(x_0+y_0+z_0)(x_0+y_0-z_0)(x_0-y_0+z_0)(-x_0+y_0+z_0)}}4$

Area of limit equilateral triangle is $\frac{\sqrt 3}4l^2$

Hence required limit is $l=\sqrt[4]{\frac{(x_0+y_0+z_0)(x_0+y_0-z_0)(x_0-y_0+z_0)(-x_0+y_0+z_0)}3}$

With $(x_0,y_0,z_0)=(1007\sqrt 2,2014\sqrt 2,1007\sqrt{14})$, this formula gives $\boxed{l=2014}$
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utkarshgupta
2280 posts
#3 • 2 Y
Y by Adventure10, Mango247
Any motivation for that invariant ?
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WizardMath
2487 posts
#4 • 2 Y
Y by Adventure10, Mango247
scale down by 1007
we get sides of a triangle
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WizardMath
2487 posts
#5 • 2 Y
Y by Adventure10, Mango247
Adding some motivation to my 'cryptic' solution from more than 2 years ago :D
We basically want to remove radicals from this expression, so squaring is very natural. The existence of the square root implicitly assumes that we have $y_n + z_n \ge x_n$. Also, the thing under the radical suspiciously looks like the product of length of tangent from a vertex and some other thing. This motivates the introduction of a triangle with sides $x_n, y_n, z_n$. Usually we need some kind of invariant in multivariable limits. So, we try the squared area of the triangle since it is a polynomial in $x_n, y_n, z_n$, more precisely, $16 \Delta^2 = 2\sum_{cyc} b^2c^2 - \sum_{cyc} a^4$. This surprisingly comes out to be invariant. Also to show the convergence, it seems better to choose the difference between squares.

Also this problem can be done using this paper.
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neel02
66 posts
#6 • 2 Y
Y by Adventure10, Mango247
Exceptionally easy for P3; Idea only since I am in hurry.
Central observation is if we treat the 3 values as lengths of a triangle then the ares is an invariant !
Another important observation is the subtended angles converges to 60 deg. Now bad calculations but a smart ans. ! :gleam:
This post has been edited 2 times. Last edited by neel02, May 15, 2018, 5:32 PM
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