Sequence with Undefined Terms?Calculate the sum of the first N terms of the sequenceMean of the terms of convergent sequenceTo show sequence $a_n$ is convergentCompute the limit of a recursively defined sequence in terms of its initial valuesGiven $a_1=1, a_2=2, a_n+1=n(a_n+a_n-1)$ find the general termSequence and series convergenceShow that the sequence $a_nleq a_2n+a_2n+1$ divergesHelp with Notation Concerning Bifurcation/Assimilation of Sequence ElementsInitial value changes, predict the character of the sequencehow to derive an $O(log(n))$ expression for a sequence with $3$ terms
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Sequence with Undefined Terms?
Calculate the sum of the first N terms of the sequenceMean of the terms of convergent sequenceTo show sequence $a_n$ is convergentCompute the limit of a recursively defined sequence in terms of its initial valuesGiven $a_1=1, a_2=2, a_n+1=n(a_n+a_n-1)$ find the general termSequence and series convergenceShow that the sequence $a_nleq a_2n+a_2n+1$ divergesHelp with Notation Concerning Bifurcation/Assimilation of Sequence ElementsInitial value changes, predict the character of the sequencehow to derive an $O(log(n))$ expression for a sequence with $3$ terms
$begingroup$
The sequence $a_n$ satisfies the relationship $a_n−1 a_n+1 + a_n = 1$ for all $n geq 2$. Let $a_1 = 1$ and $a_2 = 2$, what are the values of the next few terms? What can you say about the sequence? What happens for other starting values?
So for the initial case of $a_1 = 1$ and $a_2 = 2$, I find that for $n geq 7$, $a_n$ no longer has a defined value because $a_5 = 0$. But what can we say for the more general case where $a_1 = a$ and $a_2 = b$? Are there starting values that are simply forbidden for usage?
sequences-and-series
edited Mar 17 at 21:17
OmicronGamma
607
asked Mar 17 at 21:00
Omicron9GammaOmicron9Gamma
304
$endgroup$
add a comment |
$begingroup$
The sequence $a_n$ satisfies the relationship $a_n−1 a_n+1 + a_n = 1$ for all $n geq 2$. Let $a_1 = 1$ and $a_2 = 2$, what are the values of the next few terms? What can you say about the sequence? What happens for other starting values?
So for the initial case of $a_1 = 1$ and $a_2 = 2$, I find that for $n geq 7$, $a_n$ no longer has a defined value because $a_5 = 0$. But what can we say for the more general case where $a_1 = a$ and $a_2 = b$? Are there starting values that are simply forbidden for usage?
sequences-and-series
edited Mar 17 at 21:17
OmicronGamma
607
asked Mar 17 at 21:00
Omicron9GammaOmicron9Gamma
304
$endgroup$
$begingroup$
What is the relationship between $a_n$ and $y_n$?
$endgroup$
– clathratus
Mar 17 at 21:16
$begingroup$
If the sequence converges, it converges to one of the roots of $a^2+a=1$.
$endgroup$
– Yves Daoust
Mar 17 at 21:27
add a comment |
$begingroup$
The sequence $a_n$ satisfies the relationship $a_n−1 a_n+1 + a_n = 1$ for all $n geq 2$. Let $a_1 = 1$ and $a_2 = 2$, what are the values of the next few terms? What can you say about the sequence? What happens for other starting values?
So for the initial case of $a_1 = 1$ and $a_2 = 2$, I find that for $n geq 7$, $a_n$ no longer has a defined value because $a_5 = 0$. But what can we say for the more general case where $a_1 = a$ and $a_2 = b$? Are there starting values that are simply forbidden for usage?
sequences-and-series
edited Mar 17 at 21:17
OmicronGamma
607
asked Mar 17 at 21:00
Omicron9GammaOmicron9Gamma
304
$endgroup$
The sequence $a_n$ satisfies the relationship $a_n−1 a_n+1 + a_n = 1$ for all $n geq 2$. Let $a_1 = 1$ and $a_2 = 2$, what are the values of the next few terms? What can you say about the sequence? What happens for other starting values?
So for the initial case of $a_1 = 1$ and $a_2 = 2$, I find that for $n geq 7$, $a_n$ no longer has a defined value because $a_5 = 0$. But what can we say for the more general case where $a_1 = a$ and $a_2 = b$? Are there starting values that are simply forbidden for usage?
sequences-and-series
sequences-and-series
edited Mar 17 at 21:17
OmicronGamma
607
asked Mar 17 at 21:00
Omicron9GammaOmicron9Gamma
304
edited Mar 17 at 21:17
OmicronGamma
607
asked Mar 17 at 21:00
Omicron9GammaOmicron9Gamma
304
edited Mar 17 at 21:17
OmicronGamma
607
edited Mar 17 at 21:17
OmicronGamma
607
edited Mar 17 at 21:17
OmicronGamma
607
607
asked Mar 17 at 21:00
Omicron9GammaOmicron9Gamma
304
asked Mar 17 at 21:00
Omicron9GammaOmicron9Gamma
304
asked Mar 17 at 21:00
Omicron9GammaOmicron9Gamma
304
304
$begingroup$
What is the relationship between $a_n$ and $y_n$?
$endgroup$
– clathratus
Mar 17 at 21:16
$begingroup$
If the sequence converges, it converges to one of the roots of $a^2+a=1$.
$endgroup$
– Yves Daoust
Mar 17 at 21:27
add a comment |
$begingroup$
What is the relationship between $a_n$ and $y_n$?
$endgroup$
– clathratus
Mar 17 at 21:16
$begingroup$
If the sequence converges, it converges to one of the roots of $a^2+a=1$.
$endgroup$
– Yves Daoust
Mar 17 at 21:27
$begingroup$
What is the relationship between $a_n$ and $y_n$?
$endgroup$
– clathratus
Mar 17 at 21:16
$begingroup$
What is the relationship between $a_n$ and $y_n$?
$endgroup$
– clathratus
Mar 17 at 21:16
$begingroup$
If the sequence converges, it converges to one of the roots of $a^2+a=1$.
$endgroup$
– Yves Daoust
Mar 17 at 21:27
$begingroup$
If the sequence converges, it converges to one of the roots of $a^2+a=1$.
$endgroup$
– Yves Daoust
Mar 17 at 21:27
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
The argument is that if the sequence converges to a fixed point, call it $x$, then at that fixed point
$$a_n-1=a_n=a_n+1=x$$
Consequently,
$$a_n-1 a_n+1+a_n=1$$
becomes
$$x times x+x=1$$
i.e.
$$x^2+x-1=0$$
This has solutions,
$$x=frac-1 pm sqrt52$$
So, essentially your iteration is attracted or repelled from these two values of $x$ regardless of the initial values of $a$ and $b$ but, as you have observed, a lot can go wrong with an iteration as it heads towards or away from it's fixed points, and picking the 'right' $a$ and $b$ might enable the iteration to avoid such problems.
A division by zero, a negative square root or the log of a negative number are typical reasons why an iteration may 'crash'.
answered Mar 17 at 22:02
Martin HansenMartin Hansen
707114
$endgroup$
$begingroup$
Is there any way to find generally what starting values are banned? After writing out the general case of $a_1 = a$ and $a_2 = b$, we see that $a, b neq 1$, but it seems that there might be other values that these two elements cannot take on.
$endgroup$
– OmicronGamma
Mar 17 at 23:15
1
$begingroup$
I don't know of a way in general although perhaps for a relatively simple iteration such as this, it would be possible to map out a surface with $a$ on the $x$-axis and $b$ on the $y$-axis and use a computer to look at points to see if they converge, diverge or hit a division by zero error. Of course, with complex numbers, mapping out the single starting value is what gave rise to Julia Sets and Fractal geometry where the boundary between what converges and what diverges is so amazingly complicated; another issue with your iteration is it could get stuck in a periodic loop.
$endgroup$
– Martin Hansen
Mar 18 at 0:17
add a comment |
$begingroup$
Considering $$a_n−1, a_n+1 + a_n = 1qquad textwithqquad a_1=A qquad textandqquad a_2=B$$ and computing, we get a cyclic system
$$leftcolorbluealpha ,beta ,frac1-beta alpha ,fracalpha +beta -1alpha
beta ,frac1-alpha beta ,colorgreenalpha ,beta ,frac1-beta alpha
,fracalpha +beta -1alpha beta ,frac1-alpha beta , cdotsright$$
answered Mar 18 at 10:23
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
The argument is that if the sequence converges to a fixed point, call it $x$, then at that fixed point
$$a_n-1=a_n=a_n+1=x$$
Consequently,
$$a_n-1 a_n+1+a_n=1$$
becomes
$$x times x+x=1$$
i.e.
$$x^2+x-1=0$$
This has solutions,
$$x=frac-1 pm sqrt52$$
So, essentially your iteration is attracted or repelled from these two values of $x$ regardless of the initial values of $a$ and $b$ but, as you have observed, a lot can go wrong with an iteration as it heads towards or away from it's fixed points, and picking the 'right' $a$ and $b$ might enable the iteration to avoid such problems.
A division by zero, a negative square root or the log of a negative number are typical reasons why an iteration may 'crash'.
answered Mar 17 at 22:02
Martin HansenMartin Hansen
707114
$endgroup$
$begingroup$
Is there any way to find generally what starting values are banned? After writing out the general case of $a_1 = a$ and $a_2 = b$, we see that $a, b neq 1$, but it seems that there might be other values that these two elements cannot take on.
$endgroup$
– OmicronGamma
Mar 17 at 23:15
1
$begingroup$
I don't know of a way in general although perhaps for a relatively simple iteration such as this, it would be possible to map out a surface with $a$ on the $x$-axis and $b$ on the $y$-axis and use a computer to look at points to see if they converge, diverge or hit a division by zero error. Of course, with complex numbers, mapping out the single starting value is what gave rise to Julia Sets and Fractal geometry where the boundary between what converges and what diverges is so amazingly complicated; another issue with your iteration is it could get stuck in a periodic loop.
$endgroup$
– Martin Hansen
Mar 18 at 0:17
add a comment |
$begingroup$
The argument is that if the sequence converges to a fixed point, call it $x$, then at that fixed point
$$a_n-1=a_n=a_n+1=x$$
Consequently,
$$a_n-1 a_n+1+a_n=1$$
becomes
$$x times x+x=1$$
i.e.
$$x^2+x-1=0$$
This has solutions,
$$x=frac-1 pm sqrt52$$
So, essentially your iteration is attracted or repelled from these two values of $x$ regardless of the initial values of $a$ and $b$ but, as you have observed, a lot can go wrong with an iteration as it heads towards or away from it's fixed points, and picking the 'right' $a$ and $b$ might enable the iteration to avoid such problems.
A division by zero, a negative square root or the log of a negative number are typical reasons why an iteration may 'crash'.
answered Mar 17 at 22:02
Martin HansenMartin Hansen
707114
$endgroup$
$begingroup$
Is there any way to find generally what starting values are banned? After writing out the general case of $a_1 = a$ and $a_2 = b$, we see that $a, b neq 1$, but it seems that there might be other values that these two elements cannot take on.
$endgroup$
– OmicronGamma
Mar 17 at 23:15
1
$begingroup$
I don't know of a way in general although perhaps for a relatively simple iteration such as this, it would be possible to map out a surface with $a$ on the $x$-axis and $b$ on the $y$-axis and use a computer to look at points to see if they converge, diverge or hit a division by zero error. Of course, with complex numbers, mapping out the single starting value is what gave rise to Julia Sets and Fractal geometry where the boundary between what converges and what diverges is so amazingly complicated; another issue with your iteration is it could get stuck in a periodic loop.
$endgroup$
– Martin Hansen
Mar 18 at 0:17
add a comment |
$begingroup$
The argument is that if the sequence converges to a fixed point, call it $x$, then at that fixed point
$$a_n-1=a_n=a_n+1=x$$
Consequently,
$$a_n-1 a_n+1+a_n=1$$
becomes
$$x times x+x=1$$
i.e.
$$x^2+x-1=0$$
This has solutions,
$$x=frac-1 pm sqrt52$$
So, essentially your iteration is attracted or repelled from these two values of $x$ regardless of the initial values of $a$ and $b$ but, as you have observed, a lot can go wrong with an iteration as it heads towards or away from it's fixed points, and picking the 'right' $a$ and $b$ might enable the iteration to avoid such problems.
A division by zero, a negative square root or the log of a negative number are typical reasons why an iteration may 'crash'.
answered Mar 17 at 22:02
Martin HansenMartin Hansen
707114
$endgroup$
The argument is that if the sequence converges to a fixed point, call it $x$, then at that fixed point
$$a_n-1=a_n=a_n+1=x$$
Consequently,
$$a_n-1 a_n+1+a_n=1$$
becomes
$$x times x+x=1$$
i.e.
$$x^2+x-1=0$$
This has solutions,
$$x=frac-1 pm sqrt52$$
So, essentially your iteration is attracted or repelled from these two values of $x$ regardless of the initial values of $a$ and $b$ but, as you have observed, a lot can go wrong with an iteration as it heads towards or away from it's fixed points, and picking the 'right' $a$ and $b$ might enable the iteration to avoid such problems.
A division by zero, a negative square root or the log of a negative number are typical reasons why an iteration may 'crash'.
answered Mar 17 at 22:02
Martin HansenMartin Hansen
707114
edited Mar 17 at 22:18
answered Mar 17 at 22:02
Martin HansenMartin Hansen
707114
answered Mar 17 at 22:02
Martin HansenMartin Hansen
707114
answered Mar 17 at 22:02
Martin HansenMartin Hansen
707114
707114
$begingroup$
Is there any way to find generally what starting values are banned? After writing out the general case of $a_1 = a$ and $a_2 = b$, we see that $a, b neq 1$, but it seems that there might be other values that these two elements cannot take on.
$endgroup$
– OmicronGamma
Mar 17 at 23:15
1
$begingroup$
I don't know of a way in general although perhaps for a relatively simple iteration such as this, it would be possible to map out a surface with $a$ on the $x$-axis and $b$ on the $y$-axis and use a computer to look at points to see if they converge, diverge or hit a division by zero error. Of course, with complex numbers, mapping out the single starting value is what gave rise to Julia Sets and Fractal geometry where the boundary between what converges and what diverges is so amazingly complicated; another issue with your iteration is it could get stuck in a periodic loop.
$endgroup$
– Martin Hansen
Mar 18 at 0:17
add a comment |
$begingroup$
Is there any way to find generally what starting values are banned? After writing out the general case of $a_1 = a$ and $a_2 = b$, we see that $a, b neq 1$, but it seems that there might be other values that these two elements cannot take on.
$endgroup$
– OmicronGamma
Mar 17 at 23:15
1
$begingroup$
I don't know of a way in general although perhaps for a relatively simple iteration such as this, it would be possible to map out a surface with $a$ on the $x$-axis and $b$ on the $y$-axis and use a computer to look at points to see if they converge, diverge or hit a division by zero error. Of course, with complex numbers, mapping out the single starting value is what gave rise to Julia Sets and Fractal geometry where the boundary between what converges and what diverges is so amazingly complicated; another issue with your iteration is it could get stuck in a periodic loop.
$endgroup$
– Martin Hansen
Mar 18 at 0:17
$begingroup$
Is there any way to find generally what starting values are banned? After writing out the general case of $a_1 = a$ and $a_2 = b$, we see that $a, b neq 1$, but it seems that there might be other values that these two elements cannot take on.
$endgroup$
– OmicronGamma
Mar 17 at 23:15
$begingroup$
Is there any way to find generally what starting values are banned? After writing out the general case of $a_1 = a$ and $a_2 = b$, we see that $a, b neq 1$, but it seems that there might be other values that these two elements cannot take on.
$endgroup$
– OmicronGamma
Mar 17 at 23:15
1
1
$begingroup$
I don't know of a way in general although perhaps for a relatively simple iteration such as this, it would be possible to map out a surface with $a$ on the $x$-axis and $b$ on the $y$-axis and use a computer to look at points to see if they converge, diverge or hit a division by zero error. Of course, with complex numbers, mapping out the single starting value is what gave rise to Julia Sets and Fractal geometry where the boundary between what converges and what diverges is so amazingly complicated; another issue with your iteration is it could get stuck in a periodic loop.
$endgroup$
– Martin Hansen
Mar 18 at 0:17
$begingroup$
I don't know of a way in general although perhaps for a relatively simple iteration such as this, it would be possible to map out a surface with $a$ on the $x$-axis and $b$ on the $y$-axis and use a computer to look at points to see if they converge, diverge or hit a division by zero error. Of course, with complex numbers, mapping out the single starting value is what gave rise to Julia Sets and Fractal geometry where the boundary between what converges and what diverges is so amazingly complicated; another issue with your iteration is it could get stuck in a periodic loop.
$endgroup$
– Martin Hansen
Mar 18 at 0:17
add a comment |
$begingroup$
Considering $$a_n−1, a_n+1 + a_n = 1qquad textwithqquad a_1=A qquad textandqquad a_2=B$$ and computing, we get a cyclic system
$$leftcolorbluealpha ,beta ,frac1-beta alpha ,fracalpha +beta -1alpha
beta ,frac1-alpha beta ,colorgreenalpha ,beta ,frac1-beta alpha
,fracalpha +beta -1alpha beta ,frac1-alpha beta , cdotsright$$
answered Mar 18 at 10:23
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Considering $$a_n−1, a_n+1 + a_n = 1qquad textwithqquad a_1=A qquad textandqquad a_2=B$$ and computing, we get a cyclic system
$$leftcolorbluealpha ,beta ,frac1-beta alpha ,fracalpha +beta -1alpha
beta ,frac1-alpha beta ,colorgreenalpha ,beta ,frac1-beta alpha
,fracalpha +beta -1alpha beta ,frac1-alpha beta , cdotsright$$
answered Mar 18 at 10:23
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
add a comment |
$begingroup$
Considering $$a_n−1, a_n+1 + a_n = 1qquad textwithqquad a_1=A qquad textandqquad a_2=B$$ and computing, we get a cyclic system
$$leftcolorbluealpha ,beta ,frac1-beta alpha ,fracalpha +beta -1alpha
beta ,frac1-alpha beta ,colorgreenalpha ,beta ,frac1-beta alpha
,fracalpha +beta -1alpha beta ,frac1-alpha beta , cdotsright$$
answered Mar 18 at 10:23
Claude LeiboviciClaude Leibovici
125k1158135
$endgroup$
Considering $$a_n−1, a_n+1 + a_n = 1qquad textwithqquad a_1=A qquad textandqquad a_2=B$$ and computing, we get a cyclic system
$$leftcolorbluealpha ,beta ,frac1-beta alpha ,fracalpha +beta -1alpha
beta ,frac1-alpha beta ,colorgreenalpha ,beta ,frac1-beta alpha
,fracalpha +beta -1alpha beta ,frac1-alpha beta , cdotsright$$
answered Mar 18 at 10:23
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 18 at 10:23
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 18 at 10:23
Claude LeiboviciClaude Leibovici
125k1158135
answered Mar 18 at 10:23
Claude LeiboviciClaude Leibovici
125k1158135
125k1158135
add a comment |
add a comment |
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$begingroup$
What is the relationship between $a_n$ and $y_n$?
$endgroup$
– clathratus
Mar 17 at 21:16
$begingroup$
If the sequence converges, it converges to one of the roots of $a^2+a=1$.
$endgroup$
– Yves Daoust
Mar 17 at 21:27