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 $n ge 1$.










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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














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 $n ge 1$.










share|cite|improve this question























  • 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












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 $n ge 1$.










share|cite|improve this question















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






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edited Sep 11 at 4:17









Ross Millikan

286k23195363




286k23195363










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
















  • 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










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.






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  • 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


















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1
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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.






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  • 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










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2 Answers
2






active

oldest

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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.






share|cite|improve this answer




















  • 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















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.






share|cite|improve this answer




















  • 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













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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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

















  • 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











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.






share|cite|improve this answer




















  • 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














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.






share|cite|improve this answer




















  • 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












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.






share|cite|improve this answer












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.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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
















  • 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

















 

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