Given a fixed $x_0$ can a reccurence relation give rise to two distinct sequences?
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If two sequences have the same initial term and satisfy the same reccurence relation $f(a_n+1, a_n) = c$ for some constant $c$, are the sequences necesarily the same?
I was trying to make up with an example. My guess was to construct something involving $x_n^2$ or $|x|$, the idea being that I can find two sequences whose corresponding terms are equal in magintude, but whose signs differ for some terms. However, simple examples such as $x_n+1 ^2 = x_n$ did not work.
For context, I was trying to solve this problem:
Let $x_n$ be a sequence f nonzero real numbers such that $x_n^2-x_n-1x_n+1 = 1$ for $n ge 1.$ Prove that there exists a real number $a$ such that $x_n+1 = ax_n - x_n-1$ for all $n ge 1$.
Assuming the conclusion, I found that we must have $x_n = x_0 + n$ for $n ge 1$. Thus for an actual proof, I was thinking to say:
"For any $x_0 in mathbbR$, the sequence $x_n+1 = x_0 + n$ with $n ge 1$ satisfies the reccurence relation. Moreover, since any solution of the relation is uniquely determined by its initial term $x_0$ and the relation, these are all the solutions. And with the explicit solution, it is easy to check that there is a constant $a$ such that $x_n+1 = ax_n - x_n-1$ for all $n ge 1$.
real-analysis sequences-and-series combinatorics recurrence-relations
edited Sep 11 at 4:17
Ross Millikan
286k23195363
asked Sep 11 at 3:26
Ovi
11.9k938106
add a comment |
up vote
1
down vote
favorite
If two sequences have the same initial term and satisfy the same reccurence relation $f(a_n+1, a_n) = c$ for some constant $c$, are the sequences necesarily the same?
I was trying to make up with an example. My guess was to construct something involving $x_n^2$ or $|x|$, the idea being that I can find two sequences whose corresponding terms are equal in magintude, but whose signs differ for some terms. However, simple examples such as $x_n+1 ^2 = x_n$ did not work.
For context, I was trying to solve this problem:
Let $x_n$ be a sequence f nonzero real numbers such that $x_n^2-x_n-1x_n+1 = 1$ for $n ge 1.$ Prove that there exists a real number $a$ such that $x_n+1 = ax_n - x_n-1$ for all $n ge 1$.
Assuming the conclusion, I found that we must have $x_n = x_0 + n$ for $n ge 1$. Thus for an actual proof, I was thinking to say:
"For any $x_0 in mathbbR$, the sequence $x_n+1 = x_0 + n$ with $n ge 1$ satisfies the reccurence relation. Moreover, since any solution of the relation is uniquely determined by its initial term $x_0$ and the relation, these are all the solutions. And with the explicit solution, it is easy to check that there is a constant $a$ such that $x_n+1 = ax_n - x_n-1$ for all $n ge 1$.
real-analysis sequences-and-series combinatorics recurrence-relations
edited Sep 11 at 4:17
Ross Millikan
286k23195363
asked Sep 11 at 3:26
Ovi
11.9k938106
depends on the recurrence relation, and whether its injective or not.
– Rushabh Mehta
Sep 11 at 3:31
@RushabhMehta What do you mean by an "injective recurrence relation"? Also, do you have an example?
– Ovi
Sep 11 at 3:32
Your edit has changed the problem considerably. Originally you did not restrict the recurrence to depending only on the immediately previous term and your example does not do that. marty cohen's answer is now directly applicable and mine is invalidated. Your example does not meet the condition of the first sentence. Please think about your question and do not change it after there are answers.
– Ross Millikan
Sep 11 at 4:13
@RossMillikan Sorry about that
– Ovi
Sep 11 at 4:14
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
If two sequences have the same initial term and satisfy the same reccurence relation $f(a_n+1, a_n) = c$ for some constant $c$, are the sequences necesarily the same?
I was trying to make up with an example. My guess was to construct something involving $x_n^2$ or $|x|$, the idea being that I can find two sequences whose corresponding terms are equal in magintude, but whose signs differ for some terms. However, simple examples such as $x_n+1 ^2 = x_n$ did not work.
For context, I was trying to solve this problem:
Let $x_n$ be a sequence f nonzero real numbers such that $x_n^2-x_n-1x_n+1 = 1$ for $n ge 1.$ Prove that there exists a real number $a$ such that $x_n+1 = ax_n - x_n-1$ for all $n ge 1$.
Assuming the conclusion, I found that we must have $x_n = x_0 + n$ for $n ge 1$. Thus for an actual proof, I was thinking to say:
"For any $x_0 in mathbbR$, the sequence $x_n+1 = x_0 + n$ with $n ge 1$ satisfies the reccurence relation. Moreover, since any solution of the relation is uniquely determined by its initial term $x_0$ and the relation, these are all the solutions. And with the explicit solution, it is easy to check that there is a constant $a$ such that $x_n+1 = ax_n - x_n-1$ for all $n ge 1$.
real-analysis sequences-and-series combinatorics recurrence-relations
edited Sep 11 at 4:17
Ross Millikan
286k23195363
asked Sep 11 at 3:26
Ovi
11.9k938106
If two sequences have the same initial term and satisfy the same reccurence relation $f(a_n+1, a_n) = c$ for some constant $c$, are the sequences necesarily the same?
I was trying to make up with an example. My guess was to construct something involving $x_n^2$ or $|x|$, the idea being that I can find two sequences whose corresponding terms are equal in magintude, but whose signs differ for some terms. However, simple examples such as $x_n+1 ^2 = x_n$ did not work.
For context, I was trying to solve this problem:
Let $x_n$ be a sequence f nonzero real numbers such that $x_n^2-x_n-1x_n+1 = 1$ for $n ge 1.$ Prove that there exists a real number $a$ such that $x_n+1 = ax_n - x_n-1$ for all $n ge 1$.
Assuming the conclusion, I found that we must have $x_n = x_0 + n$ for $n ge 1$. Thus for an actual proof, I was thinking to say:
"For any $x_0 in mathbbR$, the sequence $x_n+1 = x_0 + n$ with $n ge 1$ satisfies the reccurence relation. Moreover, since any solution of the relation is uniquely determined by its initial term $x_0$ and the relation, these are all the solutions. And with the explicit solution, it is easy to check that there is a constant $a$ such that $x_n+1 = ax_n - x_n-1$ for all $n ge 1$.
real-analysis sequences-and-series combinatorics recurrence-relations
real-analysis sequences-and-series combinatorics recurrence-relations
edited Sep 11 at 4:17
Ross Millikan
286k23195363
asked Sep 11 at 3:26
Ovi
11.9k938106
edited Sep 11 at 4:17
Ross Millikan
286k23195363
asked Sep 11 at 3:26
Ovi
11.9k938106
edited Sep 11 at 4:17
Ross Millikan
286k23195363
edited Sep 11 at 4:17
Ross Millikan
286k23195363
edited Sep 11 at 4:17
Ross Millikan
286k23195363
286k23195363
asked Sep 11 at 3:26
Ovi
11.9k938106
asked Sep 11 at 3:26
Ovi
11.9k938106
asked Sep 11 at 3:26
Ovi
11.9k938106
11.9k938106
depends on the recurrence relation, and whether its injective or not.
– Rushabh Mehta
Sep 11 at 3:31
@RushabhMehta What do you mean by an "injective recurrence relation"? Also, do you have an example?
– Ovi
Sep 11 at 3:32
Your edit has changed the problem considerably. Originally you did not restrict the recurrence to depending only on the immediately previous term and your example does not do that. marty cohen's answer is now directly applicable and mine is invalidated. Your example does not meet the condition of the first sentence. Please think about your question and do not change it after there are answers.
– Ross Millikan
Sep 11 at 4:13
@RossMillikan Sorry about that
– Ovi
Sep 11 at 4:14
add a comment |
depends on the recurrence relation, and whether its injective or not.
– Rushabh Mehta
Sep 11 at 3:31
@RushabhMehta What do you mean by an "injective recurrence relation"? Also, do you have an example?
– Ovi
Sep 11 at 3:32
Your edit has changed the problem considerably. Originally you did not restrict the recurrence to depending only on the immediately previous term and your example does not do that. marty cohen's answer is now directly applicable and mine is invalidated. Your example does not meet the condition of the first sentence. Please think about your question and do not change it after there are answers.
– Ross Millikan
Sep 11 at 4:13
@RossMillikan Sorry about that
– Ovi
Sep 11 at 4:14
depends on the recurrence relation, and whether its injective or not.
– Rushabh Mehta
Sep 11 at 3:31
depends on the recurrence relation, and whether its injective or not.
– Rushabh Mehta
Sep 11 at 3:31
@RushabhMehta What do you mean by an "injective recurrence relation"? Also, do you have an example?
– Ovi
Sep 11 at 3:32
@RushabhMehta What do you mean by an "injective recurrence relation"? Also, do you have an example?
– Ovi
Sep 11 at 3:32
Your edit has changed the problem considerably. Originally you did not restrict the recurrence to depending only on the immediately previous term and your example does not do that. marty cohen's answer is now directly applicable and mine is invalidated. Your example does not meet the condition of the first sentence. Please think about your question and do not change it after there are answers.
– Ross Millikan
Sep 11 at 4:13
Your edit has changed the problem considerably. Originally you did not restrict the recurrence to depending only on the immediately previous term and your example does not do that. marty cohen's answer is now directly applicable and mine is invalidated. Your example does not meet the condition of the first sentence. Please think about your question and do not change it after there are answers.
– Ross Millikan
Sep 11 at 4:13
@RossMillikan Sorry about that
– Ovi
Sep 11 at 4:14
@RossMillikan Sorry about that
– Ovi
Sep 11 at 4:14
add a comment |
2 Answers
2
active
oldest
votes
up vote
1
down vote
accepted
It depends on the recurrence relation. If the recurrence only relates $x_n+1$ to $x_n$ and you can solve for $x_n+1$ then one term determines the whole series. If it relates things two steps back, like the Fibonacci recurrence $x_n+1=x_n+x_n-1$ you need two starting terms to determine the relation. The usual Fibonacci series has $x_0=0,x_1=1$, but if you start with $x_0=0,x_1=0$ all the terms are $0$. If you can solve the relation for the latest term, you need as many starting terms as the earliest one on the right, so $x_n+1=x_n+x_n-4$ would need five starting terms.
answered Sep 11 at 3:33
Ross Millikan
286k23195363
Hmm I overlooked that in the question; could you kindly take a look at the yellow box and the text below and let me know if you think the argument is correct?
– Ovi
Sep 11 at 3:41
That was really the motivation for the question.
– Ovi
Sep 11 at 3:43
The argument is not correct, nor is what you found. We can rewrite the recurrence as $x_n+1=frac x_n^2-1x_n-1$ If you start with $x_0=2,x_1=6$, for example, you do not have $x_n+1=x_0+n$ but the thing you are supposed to prove is correct with $a=13/4$
– Ross Millikan
Sep 11 at 3:52
Could you please elaborate? I don't see how riting $x_n+1=frac x_n^2-1x_n-1$ invalidates either the argument or that $x_n = x_0 + n$ is a solution
– Ovi
Sep 11 at 4:01
Writing it that way shows explicitly how to compute $x_n+1$ from previous terms. It shows you need two previous terms to compute it, so you need two starting terms. You can't compute $x_1$ only knowing $x_0$. Once you know $x_0$ and $x_1$ you can compute all the later terms. It is true that $x_n=x_0+n$ is a solution (though in the question you had $x_n+1$) but that is not the only solution. Try $x_0=1,x_1=3$ and see what happens.
– Ross Millikan
Sep 11 at 4:07
add a comment |
up vote
1
down vote
If the recurrence
is of the form
$(x_n+1)^2 = f(x_n)$
then there can be
an arbitrary number of solutions
for a given $x_0$
just by choosing
differently signed values
of $x_n+1$
at each value of $n$.
For example,
look at
$(x_n+1)^2
= (x_n+2)^2
$
and,
for any real $r$,
choose the sign
at $n+1$
of
$(-1)^lfloor 2(r2^n-lfloor r2^n rfloor) rfloor
$.
This gives an uncountable number
or solutions to the recurrence
for any initial condition.
answered Sep 11 at 3:52
marty cohen
71k546122
This is not correct. The recurrence has $x_n+1$ on the right. You cannot just change the sign of one term arbitrarily like you want to. A more useful form is $x_n+1=frac x_n^2-1x_n-1$
– Ross Millikan
Sep 11 at 3:58
Hmm I tried letting $f(x)$ be the identity function so that $x_n+1^2 = x_n$. Then for any $n ge 1$ we must have $x_n$ positive; and with a fixed $x_0$, I think there is only one solution.
– Ovi
Sep 11 at 4:04
The question has been edited and $x_n+1^2=x_n^2$ is a good example where there are an uncountable number of solutions.
– Ross Millikan
Sep 11 at 4:16
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
It depends on the recurrence relation. If the recurrence only relates $x_n+1$ to $x_n$ and you can solve for $x_n+1$ then one term determines the whole series. If it relates things two steps back, like the Fibonacci recurrence $x_n+1=x_n+x_n-1$ you need two starting terms to determine the relation. The usual Fibonacci series has $x_0=0,x_1=1$, but if you start with $x_0=0,x_1=0$ all the terms are $0$. If you can solve the relation for the latest term, you need as many starting terms as the earliest one on the right, so $x_n+1=x_n+x_n-4$ would need five starting terms.
answered Sep 11 at 3:33
Ross Millikan
286k23195363
Hmm I overlooked that in the question; could you kindly take a look at the yellow box and the text below and let me know if you think the argument is correct?
– Ovi
Sep 11 at 3:41
That was really the motivation for the question.
– Ovi
Sep 11 at 3:43
The argument is not correct, nor is what you found. We can rewrite the recurrence as $x_n+1=frac x_n^2-1x_n-1$ If you start with $x_0=2,x_1=6$, for example, you do not have $x_n+1=x_0+n$ but the thing you are supposed to prove is correct with $a=13/4$
– Ross Millikan
Sep 11 at 3:52
Could you please elaborate? I don't see how riting $x_n+1=frac x_n^2-1x_n-1$ invalidates either the argument or that $x_n = x_0 + n$ is a solution
– Ovi
Sep 11 at 4:01
Writing it that way shows explicitly how to compute $x_n+1$ from previous terms. It shows you need two previous terms to compute it, so you need two starting terms. You can't compute $x_1$ only knowing $x_0$. Once you know $x_0$ and $x_1$ you can compute all the later terms. It is true that $x_n=x_0+n$ is a solution (though in the question you had $x_n+1$) but that is not the only solution. Try $x_0=1,x_1=3$ and see what happens.
– Ross Millikan
Sep 11 at 4:07
add a comment |
up vote
1
down vote
accepted
It depends on the recurrence relation. If the recurrence only relates $x_n+1$ to $x_n$ and you can solve for $x_n+1$ then one term determines the whole series. If it relates things two steps back, like the Fibonacci recurrence $x_n+1=x_n+x_n-1$ you need two starting terms to determine the relation. The usual Fibonacci series has $x_0=0,x_1=1$, but if you start with $x_0=0,x_1=0$ all the terms are $0$. If you can solve the relation for the latest term, you need as many starting terms as the earliest one on the right, so $x_n+1=x_n+x_n-4$ would need five starting terms.
answered Sep 11 at 3:33
Ross Millikan
286k23195363
Hmm I overlooked that in the question; could you kindly take a look at the yellow box and the text below and let me know if you think the argument is correct?
– Ovi
Sep 11 at 3:41
That was really the motivation for the question.
– Ovi
Sep 11 at 3:43
The argument is not correct, nor is what you found. We can rewrite the recurrence as $x_n+1=frac x_n^2-1x_n-1$ If you start with $x_0=2,x_1=6$, for example, you do not have $x_n+1=x_0+n$ but the thing you are supposed to prove is correct with $a=13/4$
– Ross Millikan
Sep 11 at 3:52
Could you please elaborate? I don't see how riting $x_n+1=frac x_n^2-1x_n-1$ invalidates either the argument or that $x_n = x_0 + n$ is a solution
– Ovi
Sep 11 at 4:01
Writing it that way shows explicitly how to compute $x_n+1$ from previous terms. It shows you need two previous terms to compute it, so you need two starting terms. You can't compute $x_1$ only knowing $x_0$. Once you know $x_0$ and $x_1$ you can compute all the later terms. It is true that $x_n=x_0+n$ is a solution (though in the question you had $x_n+1$) but that is not the only solution. Try $x_0=1,x_1=3$ and see what happens.
– Ross Millikan
Sep 11 at 4:07
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
It depends on the recurrence relation. If the recurrence only relates $x_n+1$ to $x_n$ and you can solve for $x_n+1$ then one term determines the whole series. If it relates things two steps back, like the Fibonacci recurrence $x_n+1=x_n+x_n-1$ you need two starting terms to determine the relation. The usual Fibonacci series has $x_0=0,x_1=1$, but if you start with $x_0=0,x_1=0$ all the terms are $0$. If you can solve the relation for the latest term, you need as many starting terms as the earliest one on the right, so $x_n+1=x_n+x_n-4$ would need five starting terms.
answered Sep 11 at 3:33
Ross Millikan
286k23195363
It depends on the recurrence relation. If the recurrence only relates $x_n+1$ to $x_n$ and you can solve for $x_n+1$ then one term determines the whole series. If it relates things two steps back, like the Fibonacci recurrence $x_n+1=x_n+x_n-1$ you need two starting terms to determine the relation. The usual Fibonacci series has $x_0=0,x_1=1$, but if you start with $x_0=0,x_1=0$ all the terms are $0$. If you can solve the relation for the latest term, you need as many starting terms as the earliest one on the right, so $x_n+1=x_n+x_n-4$ would need five starting terms.
answered Sep 11 at 3:33
Ross Millikan
286k23195363
answered Sep 11 at 3:33
Ross Millikan
286k23195363
answered Sep 11 at 3:33
Ross Millikan
286k23195363
answered Sep 11 at 3:33
Ross Millikan
286k23195363
286k23195363
Hmm I overlooked that in the question; could you kindly take a look at the yellow box and the text below and let me know if you think the argument is correct?
– Ovi
Sep 11 at 3:41
That was really the motivation for the question.
– Ovi
Sep 11 at 3:43
The argument is not correct, nor is what you found. We can rewrite the recurrence as $x_n+1=frac x_n^2-1x_n-1$ If you start with $x_0=2,x_1=6$, for example, you do not have $x_n+1=x_0+n$ but the thing you are supposed to prove is correct with $a=13/4$
– Ross Millikan
Sep 11 at 3:52
Could you please elaborate? I don't see how riting $x_n+1=frac x_n^2-1x_n-1$ invalidates either the argument or that $x_n = x_0 + n$ is a solution
– Ovi
Sep 11 at 4:01
Writing it that way shows explicitly how to compute $x_n+1$ from previous terms. It shows you need two previous terms to compute it, so you need two starting terms. You can't compute $x_1$ only knowing $x_0$. Once you know $x_0$ and $x_1$ you can compute all the later terms. It is true that $x_n=x_0+n$ is a solution (though in the question you had $x_n+1$) but that is not the only solution. Try $x_0=1,x_1=3$ and see what happens.
– Ross Millikan
Sep 11 at 4:07
add a comment |
Hmm I overlooked that in the question; could you kindly take a look at the yellow box and the text below and let me know if you think the argument is correct?
– Ovi
Sep 11 at 3:41
That was really the motivation for the question.
– Ovi
Sep 11 at 3:43
The argument is not correct, nor is what you found. We can rewrite the recurrence as $x_n+1=frac x_n^2-1x_n-1$ If you start with $x_0=2,x_1=6$, for example, you do not have $x_n+1=x_0+n$ but the thing you are supposed to prove is correct with $a=13/4$
– Ross Millikan
Sep 11 at 3:52
Could you please elaborate? I don't see how riting $x_n+1=frac x_n^2-1x_n-1$ invalidates either the argument or that $x_n = x_0 + n$ is a solution
– Ovi
Sep 11 at 4:01
Writing it that way shows explicitly how to compute $x_n+1$ from previous terms. It shows you need two previous terms to compute it, so you need two starting terms. You can't compute $x_1$ only knowing $x_0$. Once you know $x_0$ and $x_1$ you can compute all the later terms. It is true that $x_n=x_0+n$ is a solution (though in the question you had $x_n+1$) but that is not the only solution. Try $x_0=1,x_1=3$ and see what happens.
– Ross Millikan
Sep 11 at 4:07
Hmm I overlooked that in the question; could you kindly take a look at the yellow box and the text below and let me know if you think the argument is correct?
– Ovi
Sep 11 at 3:41
Hmm I overlooked that in the question; could you kindly take a look at the yellow box and the text below and let me know if you think the argument is correct?
– Ovi
Sep 11 at 3:41
That was really the motivation for the question.
– Ovi
Sep 11 at 3:43
That was really the motivation for the question.
– Ovi
Sep 11 at 3:43
The argument is not correct, nor is what you found. We can rewrite the recurrence as $x_n+1=frac x_n^2-1x_n-1$ If you start with $x_0=2,x_1=6$, for example, you do not have $x_n+1=x_0+n$ but the thing you are supposed to prove is correct with $a=13/4$
– Ross Millikan
Sep 11 at 3:52
The argument is not correct, nor is what you found. We can rewrite the recurrence as $x_n+1=frac x_n^2-1x_n-1$ If you start with $x_0=2,x_1=6$, for example, you do not have $x_n+1=x_0+n$ but the thing you are supposed to prove is correct with $a=13/4$
– Ross Millikan
Sep 11 at 3:52
Could you please elaborate? I don't see how riting $x_n+1=frac x_n^2-1x_n-1$ invalidates either the argument or that $x_n = x_0 + n$ is a solution
– Ovi
Sep 11 at 4:01
Could you please elaborate? I don't see how riting $x_n+1=frac x_n^2-1x_n-1$ invalidates either the argument or that $x_n = x_0 + n$ is a solution
– Ovi
Sep 11 at 4:01
Writing it that way shows explicitly how to compute $x_n+1$ from previous terms. It shows you need two previous terms to compute it, so you need two starting terms. You can't compute $x_1$ only knowing $x_0$. Once you know $x_0$ and $x_1$ you can compute all the later terms. It is true that $x_n=x_0+n$ is a solution (though in the question you had $x_n+1$) but that is not the only solution. Try $x_0=1,x_1=3$ and see what happens.
– Ross Millikan
Sep 11 at 4:07
Writing it that way shows explicitly how to compute $x_n+1$ from previous terms. It shows you need two previous terms to compute it, so you need two starting terms. You can't compute $x_1$ only knowing $x_0$. Once you know $x_0$ and $x_1$ you can compute all the later terms. It is true that $x_n=x_0+n$ is a solution (though in the question you had $x_n+1$) but that is not the only solution. Try $x_0=1,x_1=3$ and see what happens.
– Ross Millikan
Sep 11 at 4:07
add a comment |
up vote
1
down vote
If the recurrence
is of the form
$(x_n+1)^2 = f(x_n)$
then there can be
an arbitrary number of solutions
for a given $x_0$
just by choosing
differently signed values
of $x_n+1$
at each value of $n$.
For example,
look at
$(x_n+1)^2
= (x_n+2)^2
$
and,
for any real $r$,
choose the sign
at $n+1$
of
$(-1)^lfloor 2(r2^n-lfloor r2^n rfloor) rfloor
$.
This gives an uncountable number
or solutions to the recurrence
for any initial condition.
answered Sep 11 at 3:52
marty cohen
71k546122
This is not correct. The recurrence has $x_n+1$ on the right. You cannot just change the sign of one term arbitrarily like you want to. A more useful form is $x_n+1=frac x_n^2-1x_n-1$
– Ross Millikan
Sep 11 at 3:58
Hmm I tried letting $f(x)$ be the identity function so that $x_n+1^2 = x_n$. Then for any $n ge 1$ we must have $x_n$ positive; and with a fixed $x_0$, I think there is only one solution.
– Ovi
Sep 11 at 4:04
The question has been edited and $x_n+1^2=x_n^2$ is a good example where there are an uncountable number of solutions.
– Ross Millikan
Sep 11 at 4:16
add a comment |
up vote
1
down vote
If the recurrence
is of the form
$(x_n+1)^2 = f(x_n)$
then there can be
an arbitrary number of solutions
for a given $x_0$
just by choosing
differently signed values
of $x_n+1$
at each value of $n$.
For example,
look at
$(x_n+1)^2
= (x_n+2)^2
$
and,
for any real $r$,
choose the sign
at $n+1$
of
$(-1)^lfloor 2(r2^n-lfloor r2^n rfloor) rfloor
$.
This gives an uncountable number
or solutions to the recurrence
for any initial condition.
answered Sep 11 at 3:52
marty cohen
71k546122
This is not correct. The recurrence has $x_n+1$ on the right. You cannot just change the sign of one term arbitrarily like you want to. A more useful form is $x_n+1=frac x_n^2-1x_n-1$
– Ross Millikan
Sep 11 at 3:58
Hmm I tried letting $f(x)$ be the identity function so that $x_n+1^2 = x_n$. Then for any $n ge 1$ we must have $x_n$ positive; and with a fixed $x_0$, I think there is only one solution.
– Ovi
Sep 11 at 4:04
The question has been edited and $x_n+1^2=x_n^2$ is a good example where there are an uncountable number of solutions.
– Ross Millikan
Sep 11 at 4:16
add a comment |
up vote
1
down vote
up vote
1
down vote
If the recurrence
is of the form
$(x_n+1)^2 = f(x_n)$
then there can be
an arbitrary number of solutions
for a given $x_0$
just by choosing
differently signed values
of $x_n+1$
at each value of $n$.
For example,
look at
$(x_n+1)^2
= (x_n+2)^2
$
and,
for any real $r$,
choose the sign
at $n+1$
of
$(-1)^lfloor 2(r2^n-lfloor r2^n rfloor) rfloor
$.
This gives an uncountable number
or solutions to the recurrence
for any initial condition.
answered Sep 11 at 3:52
marty cohen
71k546122
If the recurrence
is of the form
$(x_n+1)^2 = f(x_n)$
then there can be
an arbitrary number of solutions
for a given $x_0$
just by choosing
differently signed values
of $x_n+1$
at each value of $n$.
For example,
look at
$(x_n+1)^2
= (x_n+2)^2
$
and,
for any real $r$,
choose the sign
at $n+1$
of
$(-1)^lfloor 2(r2^n-lfloor r2^n rfloor) rfloor
$.
This gives an uncountable number
or solutions to the recurrence
for any initial condition.
answered Sep 11 at 3:52
marty cohen
71k546122
answered Sep 11 at 3:52
marty cohen
71k546122
answered Sep 11 at 3:52
marty cohen
71k546122
answered Sep 11 at 3:52
marty cohen
71k546122
71k546122
This is not correct. The recurrence has $x_n+1$ on the right. You cannot just change the sign of one term arbitrarily like you want to. A more useful form is $x_n+1=frac x_n^2-1x_n-1$
– Ross Millikan
Sep 11 at 3:58
Hmm I tried letting $f(x)$ be the identity function so that $x_n+1^2 = x_n$. Then for any $n ge 1$ we must have $x_n$ positive; and with a fixed $x_0$, I think there is only one solution.
– Ovi
Sep 11 at 4:04
The question has been edited and $x_n+1^2=x_n^2$ is a good example where there are an uncountable number of solutions.
– Ross Millikan
Sep 11 at 4:16
add a comment |
This is not correct. The recurrence has $x_n+1$ on the right. You cannot just change the sign of one term arbitrarily like you want to. A more useful form is $x_n+1=frac x_n^2-1x_n-1$
– Ross Millikan
Sep 11 at 3:58
Hmm I tried letting $f(x)$ be the identity function so that $x_n+1^2 = x_n$. Then for any $n ge 1$ we must have $x_n$ positive; and with a fixed $x_0$, I think there is only one solution.
– Ovi
Sep 11 at 4:04
The question has been edited and $x_n+1^2=x_n^2$ is a good example where there are an uncountable number of solutions.
– Ross Millikan
Sep 11 at 4:16
This is not correct. The recurrence has $x_n+1$ on the right. You cannot just change the sign of one term arbitrarily like you want to. A more useful form is $x_n+1=frac x_n^2-1x_n-1$
– Ross Millikan
Sep 11 at 3:58
This is not correct. The recurrence has $x_n+1$ on the right. You cannot just change the sign of one term arbitrarily like you want to. A more useful form is $x_n+1=frac x_n^2-1x_n-1$
– Ross Millikan
Sep 11 at 3:58
Hmm I tried letting $f(x)$ be the identity function so that $x_n+1^2 = x_n$. Then for any $n ge 1$ we must have $x_n$ positive; and with a fixed $x_0$, I think there is only one solution.
– Ovi
Sep 11 at 4:04
Hmm I tried letting $f(x)$ be the identity function so that $x_n+1^2 = x_n$. Then for any $n ge 1$ we must have $x_n$ positive; and with a fixed $x_0$, I think there is only one solution.
– Ovi
Sep 11 at 4:04
The question has been edited and $x_n+1^2=x_n^2$ is a good example where there are an uncountable number of solutions.
– Ross Millikan
Sep 11 at 4:16
The question has been edited and $x_n+1^2=x_n^2$ is a good example where there are an uncountable number of solutions.
– Ross Millikan
Sep 11 at 4:16
add a comment |
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depends on the recurrence relation, and whether its injective or not.
– Rushabh Mehta
Sep 11 at 3:31
@RushabhMehta What do you mean by an "injective recurrence relation"? Also, do you have an example?
– Ovi
Sep 11 at 3:32
Your edit has changed the problem considerably. Originally you did not restrict the recurrence to depending only on the immediately previous term and your example does not do that. marty cohen's answer is now directly applicable and mine is invalidated. Your example does not meet the condition of the first sentence. Please think about your question and do not change it after there are answers.
– Ross Millikan
Sep 11 at 4:13
@RossMillikan Sorry about that
– Ovi
Sep 11 at 4:14