Mixing
sequence converges to different value
$begingroup$
Im trying to study the sequence
$$ x_{i+1} = 111- (1130-3000/x_{i-1})/x_i $$
with $x_1=11/2$ and $x_2 = 61/11$
Now, I implemented it in Matlab to see the behavoiur:
x(1) = 11/2;
x(2) = 61/11;
for k=2:20
x(k+1) = 111-(1130-3000/x(k-1))/x(k);
disp(x(k))
end
but, this seems to be approaching $6$ first but suddenly it jumps to 100 at about $n=8 $ or so. Is my code wrong? or this sequence does have this behaviour?
calculus sequences-and-series matlab
edited Jan 13 at 9:05
idriskameni
641319
asked Jan 13 at 8:47
Jimmy SabaterJimmy Sabater
2,625322
$endgroup$
add a comment |
$begingroup$
Im trying to study the sequence
$$ x_{i+1} = 111- (1130-3000/x_{i-1})/x_i $$
with $x_1=11/2$ and $x_2 = 61/11$
Now, I implemented it in Matlab to see the behavoiur:
x(1) = 11/2;
x(2) = 61/11;
for k=2:20
x(k+1) = 111-(1130-3000/x(k-1))/x(k);
disp(x(k))
end
but, this seems to be approaching $6$ first but suddenly it jumps to 100 at about $n=8 $ or so. Is my code wrong? or this sequence does have this behaviour?
calculus sequences-and-series matlab
edited Jan 13 at 9:05
idriskameni
641319
asked Jan 13 at 8:47
Jimmy SabaterJimmy Sabater
2,625322
$endgroup$
add a comment |
$begingroup$
Im trying to study the sequence
$$ x_{i+1} = 111- (1130-3000/x_{i-1})/x_i $$
with $x_1=11/2$ and $x_2 = 61/11$
Now, I implemented it in Matlab to see the behavoiur:
x(1) = 11/2;
x(2) = 61/11;
for k=2:20
x(k+1) = 111-(1130-3000/x(k-1))/x(k);
disp(x(k))
end
but, this seems to be approaching $6$ first but suddenly it jumps to 100 at about $n=8 $ or so. Is my code wrong? or this sequence does have this behaviour?
calculus sequences-and-series matlab
edited Jan 13 at 9:05
idriskameni
641319
asked Jan 13 at 8:47
Jimmy SabaterJimmy Sabater
2,625322
$endgroup$
Im trying to study the sequence
$$ x_{i+1} = 111- (1130-3000/x_{i-1})/x_i $$
with $x_1=11/2$ and $x_2 = 61/11$
Now, I implemented it in Matlab to see the behavoiur:
x(1) = 11/2;
x(2) = 61/11;
for k=2:20
x(k+1) = 111-(1130-3000/x(k-1))/x(k);
disp(x(k))
end
but, this seems to be approaching $6$ first but suddenly it jumps to 100 at about $n=8 $ or so. Is my code wrong? or this sequence does have this behaviour?
calculus sequences-and-series matlab
calculus sequences-and-series matlab
edited Jan 13 at 9:05
idriskameni
641319
asked Jan 13 at 8:47
Jimmy SabaterJimmy Sabater
2,625322
edited Jan 13 at 9:05
idriskameni
641319
asked Jan 13 at 8:47
Jimmy SabaterJimmy Sabater
2,625322
edited Jan 13 at 9:05
idriskameni
641319
edited Jan 13 at 9:05
idriskameni
641319
edited Jan 13 at 9:05
idriskameni
641319
641319
asked Jan 13 at 8:47
Jimmy SabaterJimmy Sabater
2,625322
asked Jan 13 at 8:47
Jimmy SabaterJimmy Sabater
2,625322
asked Jan 13 at 8:47
Jimmy SabaterJimmy Sabater
2,625322
2,625322
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Nothing abnormal, because the corresponding "fixed point" 3rd degree equation
$$x = f(x) text{with} f(x):=111-(1130-3000/x)/x$$
has 3 real roots $5, 6, 100$ : thus, this sequence is just "jumping" from the vicinity of a fixed point ($6$) to another one ($100$).
One can get a good idea of what happens on the following figure with the graphical representation of $f$ intersecting the line bissector of the first quadrant at points $(5,5), (6,6), (100,100)$. On the same figure, we have placed points $(y_k,y_k)$ and $(y_k,y_{k+1})$ of the "cousin sequence" $(y_n)$ defined by the first order recurrence relationship :
$$y_1=11/2 text{and} y_{k+1} = 111-(1130-3000/y_{k})/y_{k}$$
Please note how the staircase pattern connects the two areas of $(6,6)$ and $(100,100)$.
Sequence $(y_n)$ is such that $y_n/x_n$ tends to $1$ when $n to infty$.
Remark : sequence ($x_n$) has become a classical (counter)example ; see for example Müller series : solving a recursive sequence with different initial conditions and an interesting discussion ; I think the origin of sequence $(x_n)$ can be traced to the book by Jean-Michel Muller "Arithmétique des ordinateurs" (Eyrolles Ed., 1989) : see https://perso.ens-lyon.fr/jean-michel.muller/chapitre1.pdf and http://www.mat.unb.br/~ayala/EVENTS/munoz2006.pdf
answered Jan 13 at 9:03
Jean MarieJean Marie
29.8k42051
$endgroup$
add a comment |
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1 Answer
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1 Answer
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active
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votes
$begingroup$
Nothing abnormal, because the corresponding "fixed point" 3rd degree equation
$$x = f(x) text{with} f(x):=111-(1130-3000/x)/x$$
has 3 real roots $5, 6, 100$ : thus, this sequence is just "jumping" from the vicinity of a fixed point ($6$) to another one ($100$).
One can get a good idea of what happens on the following figure with the graphical representation of $f$ intersecting the line bissector of the first quadrant at points $(5,5), (6,6), (100,100)$. On the same figure, we have placed points $(y_k,y_k)$ and $(y_k,y_{k+1})$ of the "cousin sequence" $(y_n)$ defined by the first order recurrence relationship :
$$y_1=11/2 text{and} y_{k+1} = 111-(1130-3000/y_{k})/y_{k}$$
Please note how the staircase pattern connects the two areas of $(6,6)$ and $(100,100)$.
Sequence $(y_n)$ is such that $y_n/x_n$ tends to $1$ when $n to infty$.
Remark : sequence ($x_n$) has become a classical (counter)example ; see for example Müller series : solving a recursive sequence with different initial conditions and an interesting discussion ; I think the origin of sequence $(x_n)$ can be traced to the book by Jean-Michel Muller "Arithmétique des ordinateurs" (Eyrolles Ed., 1989) : see https://perso.ens-lyon.fr/jean-michel.muller/chapitre1.pdf and http://www.mat.unb.br/~ayala/EVENTS/munoz2006.pdf
answered Jan 13 at 9:03
Jean MarieJean Marie
29.8k42051
$endgroup$
add a comment |
$begingroup$
Nothing abnormal, because the corresponding "fixed point" 3rd degree equation
$$x = f(x) text{with} f(x):=111-(1130-3000/x)/x$$
has 3 real roots $5, 6, 100$ : thus, this sequence is just "jumping" from the vicinity of a fixed point ($6$) to another one ($100$).
One can get a good idea of what happens on the following figure with the graphical representation of $f$ intersecting the line bissector of the first quadrant at points $(5,5), (6,6), (100,100)$. On the same figure, we have placed points $(y_k,y_k)$ and $(y_k,y_{k+1})$ of the "cousin sequence" $(y_n)$ defined by the first order recurrence relationship :
$$y_1=11/2 text{and} y_{k+1} = 111-(1130-3000/y_{k})/y_{k}$$
Please note how the staircase pattern connects the two areas of $(6,6)$ and $(100,100)$.
Sequence $(y_n)$ is such that $y_n/x_n$ tends to $1$ when $n to infty$.
Remark : sequence ($x_n$) has become a classical (counter)example ; see for example Müller series : solving a recursive sequence with different initial conditions and an interesting discussion ; I think the origin of sequence $(x_n)$ can be traced to the book by Jean-Michel Muller "Arithmétique des ordinateurs" (Eyrolles Ed., 1989) : see https://perso.ens-lyon.fr/jean-michel.muller/chapitre1.pdf and http://www.mat.unb.br/~ayala/EVENTS/munoz2006.pdf
answered Jan 13 at 9:03
Jean MarieJean Marie
29.8k42051
$endgroup$
add a comment |
$begingroup$
Nothing abnormal, because the corresponding "fixed point" 3rd degree equation
$$x = f(x) text{with} f(x):=111-(1130-3000/x)/x$$
has 3 real roots $5, 6, 100$ : thus, this sequence is just "jumping" from the vicinity of a fixed point ($6$) to another one ($100$).
One can get a good idea of what happens on the following figure with the graphical representation of $f$ intersecting the line bissector of the first quadrant at points $(5,5), (6,6), (100,100)$. On the same figure, we have placed points $(y_k,y_k)$ and $(y_k,y_{k+1})$ of the "cousin sequence" $(y_n)$ defined by the first order recurrence relationship :
$$y_1=11/2 text{and} y_{k+1} = 111-(1130-3000/y_{k})/y_{k}$$
Please note how the staircase pattern connects the two areas of $(6,6)$ and $(100,100)$.
Sequence $(y_n)$ is such that $y_n/x_n$ tends to $1$ when $n to infty$.
Remark : sequence ($x_n$) has become a classical (counter)example ; see for example Müller series : solving a recursive sequence with different initial conditions and an interesting discussion ; I think the origin of sequence $(x_n)$ can be traced to the book by Jean-Michel Muller "Arithmétique des ordinateurs" (Eyrolles Ed., 1989) : see https://perso.ens-lyon.fr/jean-michel.muller/chapitre1.pdf and http://www.mat.unb.br/~ayala/EVENTS/munoz2006.pdf
answered Jan 13 at 9:03
Jean MarieJean Marie
29.8k42051
$endgroup$
Nothing abnormal, because the corresponding "fixed point" 3rd degree equation
$$x = f(x) text{with} f(x):=111-(1130-3000/x)/x$$
has 3 real roots $5, 6, 100$ : thus, this sequence is just "jumping" from the vicinity of a fixed point ($6$) to another one ($100$).
One can get a good idea of what happens on the following figure with the graphical representation of $f$ intersecting the line bissector of the first quadrant at points $(5,5), (6,6), (100,100)$. On the same figure, we have placed points $(y_k,y_k)$ and $(y_k,y_{k+1})$ of the "cousin sequence" $(y_n)$ defined by the first order recurrence relationship :
$$y_1=11/2 text{and} y_{k+1} = 111-(1130-3000/y_{k})/y_{k}$$
Please note how the staircase pattern connects the two areas of $(6,6)$ and $(100,100)$.
Sequence $(y_n)$ is such that $y_n/x_n$ tends to $1$ when $n to infty$.
Remark : sequence ($x_n$) has become a classical (counter)example ; see for example Müller series : solving a recursive sequence with different initial conditions and an interesting discussion ; I think the origin of sequence $(x_n)$ can be traced to the book by Jean-Michel Muller "Arithmétique des ordinateurs" (Eyrolles Ed., 1989) : see https://perso.ens-lyon.fr/jean-michel.muller/chapitre1.pdf and http://www.mat.unb.br/~ayala/EVENTS/munoz2006.pdf
answered Jan 13 at 9:03
Jean MarieJean Marie
29.8k42051
edited Jan 14 at 0:09
answered Jan 13 at 9:03
Jean MarieJean Marie
29.8k42051
answered Jan 13 at 9:03
Jean MarieJean Marie
29.8k42051
answered Jan 13 at 9:03
Jean MarieJean Marie
29.8k42051
29.8k42051
add a comment |
add a comment |
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