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Convergent sequence of points

1

$begingroup$

Let $nge 0$ and $A_0, A_1, A_2$ 3 points of the plane. We define recursively $A_n+3$ as the barycenter of the system $(A_n;2);(A_n+1;1);(A_n+2;1)$. Show that the sequence $(A_n)$ converges and find $A_infty$.

My try: I define $z_n$ as the affix of $A_n$. This provide the relation: $4z_n+3-z_n+2-z_n+1-2z_n=0$. Solving this gives: $z_n=lambda+murho_1^n+nurho_2^n$. Since $ |rho_1,2|<1$ the sequence converges and the limit will be $lambda$ which I can find.

Is there a synthetic proof, using geometry tools only?

linear-algebra sequences-and-series geometry complex-numbers alternative-proof

share|cite|improve this question

edited Apr 2 at 17:30

HAMIDINE SOUMARE

asked Apr 2 at 16:57

HAMIDINE SOUMAREHAMIDINE SOUMARE

2,8851420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you can show that the barycenter lies within the triangle formed by the previous 3 points, and also show that there is fixed ratio $r < 1$ such that the longest side of the new triangle is less than $r$ times the longest side of the old, then you can prove that you have a nested sequence of compact sets with diameter going to $0$, which garantees a unique limit point.
  $endgroup$
  – Paul Sinclair
  Apr 3 at 0:32
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $nge 0$ and $A_0, A_1, A_2$ 3 points of the plane. We define recursively $A_n+3$ as the barycenter of the system $(A_n;2);(A_n+1;1);(A_n+2;1)$. Show that the sequence $(A_n)$ converges and find $A_infty$.

My try: I define $z_n$ as the affix of $A_n$. This provide the relation: $4z_n+3-z_n+2-z_n+1-2z_n=0$. Solving this gives: $z_n=lambda+murho_1^n+nurho_2^n$. Since $ |rho_1,2|<1$ the sequence converges and the limit will be $lambda$ which I can find.

Is there a synthetic proof, using geometry tools only?

linear-algebra sequences-and-series geometry complex-numbers alternative-proof

share|cite|improve this question

edited Apr 2 at 17:30

HAMIDINE SOUMARE

asked Apr 2 at 16:57

HAMIDINE SOUMAREHAMIDINE SOUMARE

2,8851420

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  If you can show that the barycenter lies within the triangle formed by the previous 3 points, and also show that there is fixed ratio $r < 1$ such that the longest side of the new triangle is less than $r$ times the longest side of the old, then you can prove that you have a nested sequence of compact sets with diameter going to $0$, which garantees a unique limit point.
  $endgroup$
  – Paul Sinclair
  Apr 3 at 0:32
  

  
  
  
  

add a comment | 

$begingroup$

Let $nge 0$ and $A_0, A_1, A_2$ 3 points of the plane. We define recursively $A_n+3$ as the barycenter of the system $(A_n;2);(A_n+1;1);(A_n+2;1)$. Show that the sequence $(A_n)$ converges and find $A_infty$.

My try: I define $z_n$ as the affix of $A_n$. This provide the relation: $4z_n+3-z_n+2-z_n+1-2z_n=0$. Solving this gives: $z_n=lambda+murho_1^n+nurho_2^n$. Since $ |rho_1,2|<1$ the sequence converges and the limit will be $lambda$ which I can find.

Is there a synthetic proof, using geometry tools only?

linear-algebra sequences-and-series geometry complex-numbers alternative-proof

share|cite|improve this question

edited Apr 2 at 17:30

HAMIDINE SOUMARE

asked Apr 2 at 16:57

HAMIDINE SOUMAREHAMIDINE SOUMARE

2,8851420

$endgroup$

share|cite|improve this question

edited Apr 2 at 17:30

HAMIDINE SOUMARE

asked Apr 2 at 16:57

HAMIDINE SOUMAREHAMIDINE SOUMARE

2,8851420

share|cite|improve this question

edited Apr 2 at 17:30

HAMIDINE SOUMARE

asked Apr 2 at 16:57

HAMIDINE SOUMAREHAMIDINE SOUMARE

2,8851420

asked Apr 2 at 16:57

HAMIDINE SOUMAREHAMIDINE SOUMARE

2,8851420

asked Apr 2 at 16:57

HAMIDINE SOUMAREHAMIDINE SOUMARE

2,8851420