Convergent sequence of points Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)a recursive and dificult sequenceHow to prove the sequence $a_0, a_1, ldots$ converges iff $a_0, a, a_1, a ldots$ converges?Sequence $a_k=1-fraclambda^24a_k-1, k=2,3,ldots,n$.Determine the convergence of a sequence given by $a_n= fraca_n-1 + a_n-22$A sequence 〈a_n〉 is defined recursively ,and find infinite seriestwo convergent sequencesFinding the limit of a recursively defined sequence (recurrence relation). Specifically and generallyEvaluating the recurrence $a_n+1=a_ncdot a_n-1-1$Why is there no difference operator?Alternative proof that the 3-cycles generate the alternating group $A_n$
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Convergent sequence of points
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)a recursive and dificult sequenceHow to prove the sequence $a_0, a_1, ldots$ converges iff $a_0, a, a_1, a ldots$ converges?Sequence $a_k=1-fraclambda^24a_k-1, k=2,3,ldots,n$.Determine the convergence of a sequence given by $a_n= fraca_n-1 + a_n-22$A sequence 〈a_n〉 is defined recursively ,and find infinite seriestwo convergent sequencesFinding the limit of a recursively defined sequence (recurrence relation). Specifically and generallyEvaluating the recurrence $a_n+1=a_ncdot a_n-1-1$Why is there no difference operator?Alternative proof that the 3-cycles generate the alternating group $A_n$
$begingroup$
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
$endgroup$
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
$endgroup$
$begingroup$
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
$endgroup$
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
linear-algebra sequences-and-series geometry complex-numbers alternative-proof
edited Apr 2 at 17:30
HAMIDINE SOUMARE
asked Apr 2 at 16:57
HAMIDINE SOUMAREHAMIDINE SOUMARE
2,8851420
2,8851420
$begingroup$
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 |
$begingroup$
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
$begingroup$
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
$begingroup$
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 |
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$begingroup$
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