How to show a particular recurrence $x_{n+1} = {4over 3}x_n - x_n^2$ is bounded?












0














Given a sequence:
$$
x_{n+1} = {4over 3}x_n - x_n^2
$$
And initial conditions:
$$
x_1 = {1over 6}
$$
I want to find $lim x_n$ as $ntoinfty$. The approach i want to use is prove $x_n$ is bounded and monotone, hence convergent and then find its fixed points one of which is going to be the limit.



I have difficulties showing $x_n$ is bounded. If I make an assumption that it's bounded then for $L = lim x_n$:
$$
3L^2 - L = 0 iff L = 0 text{or} L = {1over 3}
$$



Which in case $Lne 0$ matches the answer. Could someone please assist me on showing the bounds for $x_n$? I would also like to know whether the rest reasoning is fine. Thank you.






calculus limits recurrence-relations upper-lower-bounds







share|cite|improve this question




















  • 2




    There is at least a mistake in what you wrote. $x_{n+1} > x_n iff -x_{n+1} > -x_n$ is wrong. Inequality has to be reverted when multiplied by a negative number.
    – mathcounterexamples.net
    Jan 4 at 14:24












  • @mathcounterexamples.net you're right, thanks for pointing out. I've updated the post.
    – roman
    Jan 4 at 14:31






  • 1




    But then your following equivalences are wrong.
    – mathcounterexamples.net
    Jan 4 at 14:37










  • @mathcounterexamples.net oh, you are right once again. In such case i'm not sure how to show the sequence is monotone :(
    – roman
    Jan 4 at 14:53
















0














Given a sequence:
$$
x_{n+1} = {4over 3}x_n - x_n^2
$$
And initial conditions:
$$
x_1 = {1over 6}
$$
I want to find $lim x_n$ as $ntoinfty$. The approach i want to use is prove $x_n$ is bounded and monotone, hence convergent and then find its fixed points one of which is going to be the limit.



I have difficulties showing $x_n$ is bounded. If I make an assumption that it's bounded then for $L = lim x_n$:
$$
3L^2 - L = 0 iff L = 0 text{or} L = {1over 3}
$$



Which in case $Lne 0$ matches the answer. Could someone please assist me on showing the bounds for $x_n$? I would also like to know whether the rest reasoning is fine. Thank you.






calculus limits recurrence-relations upper-lower-bounds







share|cite|improve this question




















  • 2




    There is at least a mistake in what you wrote. $x_{n+1} > x_n iff -x_{n+1} > -x_n$ is wrong. Inequality has to be reverted when multiplied by a negative number.
    – mathcounterexamples.net
    Jan 4 at 14:24












  • @mathcounterexamples.net you're right, thanks for pointing out. I've updated the post.
    – roman
    Jan 4 at 14:31






  • 1




    But then your following equivalences are wrong.
    – mathcounterexamples.net
    Jan 4 at 14:37










  • @mathcounterexamples.net oh, you are right once again. In such case i'm not sure how to show the sequence is monotone :(
    – roman
    Jan 4 at 14:53














0












0








0


1





Given a sequence:
$$
x_{n+1} = {4over 3}x_n - x_n^2
$$
And initial conditions:
$$
x_1 = {1over 6}
$$
I want to find $lim x_n$ as $ntoinfty$. The approach i want to use is prove $x_n$ is bounded and monotone, hence convergent and then find its fixed points one of which is going to be the limit.



I have difficulties showing $x_n$ is bounded. If I make an assumption that it's bounded then for $L = lim x_n$:
$$
3L^2 - L = 0 iff L = 0 text{or} L = {1over 3}
$$



Which in case $Lne 0$ matches the answer. Could someone please assist me on showing the bounds for $x_n$? I would also like to know whether the rest reasoning is fine. Thank you.






calculus limits recurrence-relations upper-lower-bounds







share|cite|improve this question















Given a sequence:
$$
x_{n+1} = {4over 3}x_n - x_n^2
$$
And initial conditions:
$$
x_1 = {1over 6}
$$
I want to find $lim x_n$ as $ntoinfty$. The approach i want to use is prove $x_n$ is bounded and monotone, hence convergent and then find its fixed points one of which is going to be the limit.



I have difficulties showing $x_n$ is bounded. If I make an assumption that it's bounded then for $L = lim x_n$:
$$
3L^2 - L = 0 iff L = 0 text{or} L = {1over 3}
$$



Which in case $Lne 0$ matches the answer. Could someone please assist me on showing the bounds for $x_n$? I would also like to know whether the rest reasoning is fine. Thank you.






calculus limits recurrence-relations upper-lower-bounds




calculus limits recurrence-relations upper-lower-bounds






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 15:10







roman

















asked Jan 4 at 14:17









romanroman

2,02221221




2,02221221








  • 2




    There is at least a mistake in what you wrote. $x_{n+1} > x_n iff -x_{n+1} > -x_n$ is wrong. Inequality has to be reverted when multiplied by a negative number.
    – mathcounterexamples.net
    Jan 4 at 14:24












  • @mathcounterexamples.net you're right, thanks for pointing out. I've updated the post.
    – roman
    Jan 4 at 14:31






  • 1




    But then your following equivalences are wrong.
    – mathcounterexamples.net
    Jan 4 at 14:37










  • @mathcounterexamples.net oh, you are right once again. In such case i'm not sure how to show the sequence is monotone :(
    – roman
    Jan 4 at 14:53














  • 2




    There is at least a mistake in what you wrote. $x_{n+1} > x_n iff -x_{n+1} > -x_n$ is wrong. Inequality has to be reverted when multiplied by a negative number.
    – mathcounterexamples.net
    Jan 4 at 14:24












  • @mathcounterexamples.net you're right, thanks for pointing out. I've updated the post.
    – roman
    Jan 4 at 14:31






  • 1




    But then your following equivalences are wrong.
    – mathcounterexamples.net
    Jan 4 at 14:37










  • @mathcounterexamples.net oh, you are right once again. In such case i'm not sure how to show the sequence is monotone :(
    – roman
    Jan 4 at 14:53








2




2




There is at least a mistake in what you wrote. $x_{n+1} > x_n iff -x_{n+1} > -x_n$ is wrong. Inequality has to be reverted when multiplied by a negative number.
– mathcounterexamples.net
Jan 4 at 14:24






There is at least a mistake in what you wrote. $x_{n+1} > x_n iff -x_{n+1} > -x_n$ is wrong. Inequality has to be reverted when multiplied by a negative number.
– mathcounterexamples.net
Jan 4 at 14:24














@mathcounterexamples.net you're right, thanks for pointing out. I've updated the post.
– roman
Jan 4 at 14:31




@mathcounterexamples.net you're right, thanks for pointing out. I've updated the post.
– roman
Jan 4 at 14:31




1




1




But then your following equivalences are wrong.
– mathcounterexamples.net
Jan 4 at 14:37




But then your following equivalences are wrong.
– mathcounterexamples.net
Jan 4 at 14:37












@mathcounterexamples.net oh, you are right once again. In such case i'm not sure how to show the sequence is monotone :(
– roman
Jan 4 at 14:53




@mathcounterexamples.net oh, you are right once again. In such case i'm not sure how to show the sequence is monotone :(
– roman
Jan 4 at 14:53










2 Answers
2






active

oldest

votes


















3














Let's define $$e_n={1over 3}-x_n$$then by substituting we obtain $$e_{n+1}={2over 3}e_n+e_n^2=e_nleft({2over 3}+e_nright)$$with $e_1={1over 6}$. Also note that if $e_n>0$ then $e_{n+1}>0$ therefore since $e_1>0$ we have $e_n>0$. Also if $e_nle {1over 6}$ we obtain $$e_{n+1}=e_nleft({2over 3}+e_nright)le {5over 6}e_n$$ finally $$0<e_{n+1}le {5over 6}e_n$$which means that $e_nto 0$ and $x_nto {1over 3}$ therefore $x_n$ is bounded.






share|cite|improve this answer





















  • Could you please elaborate on how you came up with such substitution? For me it looks like a lucky guess, i would like to grasp the technique
    – roman
    Jan 4 at 14:36






  • 1




    Sure! Since after finding $L=0$ and $L={1over 3}$ we guessed that $x_n$ should converge ${1over 3}$ then we are able to define the distance of $x_n$ to its final limit as $e_n=x_n-{1over 3}$ or $e_n={1over 3}-x_n$, not much different which one to choose and prove that the distance converges to $0$. A proof on convergence to $0$ is rather simpler than proving the convergence to any other point ---at least for me :)
    – Mostafa Ayaz
    Jan 4 at 14:40



















1














Hint: find explicit pattern and prove that it is bounded






share|cite|improve this answer

















  • 1




    Do you know what the explicit pattern is? maybe you can share it in your answer
    – caverac
    Jan 4 at 14:32












  • Are you sure closed form exists?
    – roman
    Jan 4 at 14:33











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









3














Let's define $$e_n={1over 3}-x_n$$then by substituting we obtain $$e_{n+1}={2over 3}e_n+e_n^2=e_nleft({2over 3}+e_nright)$$with $e_1={1over 6}$. Also note that if $e_n>0$ then $e_{n+1}>0$ therefore since $e_1>0$ we have $e_n>0$. Also if $e_nle {1over 6}$ we obtain $$e_{n+1}=e_nleft({2over 3}+e_nright)le {5over 6}e_n$$ finally $$0<e_{n+1}le {5over 6}e_n$$which means that $e_nto 0$ and $x_nto {1over 3}$ therefore $x_n$ is bounded.






share|cite|improve this answer





















  • Could you please elaborate on how you came up with such substitution? For me it looks like a lucky guess, i would like to grasp the technique
    – roman
    Jan 4 at 14:36






  • 1




    Sure! Since after finding $L=0$ and $L={1over 3}$ we guessed that $x_n$ should converge ${1over 3}$ then we are able to define the distance of $x_n$ to its final limit as $e_n=x_n-{1over 3}$ or $e_n={1over 3}-x_n$, not much different which one to choose and prove that the distance converges to $0$. A proof on convergence to $0$ is rather simpler than proving the convergence to any other point ---at least for me :)
    – Mostafa Ayaz
    Jan 4 at 14:40
















3














Let's define $$e_n={1over 3}-x_n$$then by substituting we obtain $$e_{n+1}={2over 3}e_n+e_n^2=e_nleft({2over 3}+e_nright)$$with $e_1={1over 6}$. Also note that if $e_n>0$ then $e_{n+1}>0$ therefore since $e_1>0$ we have $e_n>0$. Also if $e_nle {1over 6}$ we obtain $$e_{n+1}=e_nleft({2over 3}+e_nright)le {5over 6}e_n$$ finally $$0<e_{n+1}le {5over 6}e_n$$which means that $e_nto 0$ and $x_nto {1over 3}$ therefore $x_n$ is bounded.






share|cite|improve this answer





















  • Could you please elaborate on how you came up with such substitution? For me it looks like a lucky guess, i would like to grasp the technique
    – roman
    Jan 4 at 14:36






  • 1




    Sure! Since after finding $L=0$ and $L={1over 3}$ we guessed that $x_n$ should converge ${1over 3}$ then we are able to define the distance of $x_n$ to its final limit as $e_n=x_n-{1over 3}$ or $e_n={1over 3}-x_n$, not much different which one to choose and prove that the distance converges to $0$. A proof on convergence to $0$ is rather simpler than proving the convergence to any other point ---at least for me :)
    – Mostafa Ayaz
    Jan 4 at 14:40














3












3








3






Let's define $$e_n={1over 3}-x_n$$then by substituting we obtain $$e_{n+1}={2over 3}e_n+e_n^2=e_nleft({2over 3}+e_nright)$$with $e_1={1over 6}$. Also note that if $e_n>0$ then $e_{n+1}>0$ therefore since $e_1>0$ we have $e_n>0$. Also if $e_nle {1over 6}$ we obtain $$e_{n+1}=e_nleft({2over 3}+e_nright)le {5over 6}e_n$$ finally $$0<e_{n+1}le {5over 6}e_n$$which means that $e_nto 0$ and $x_nto {1over 3}$ therefore $x_n$ is bounded.






share|cite|improve this answer












Let's define $$e_n={1over 3}-x_n$$then by substituting we obtain $$e_{n+1}={2over 3}e_n+e_n^2=e_nleft({2over 3}+e_nright)$$with $e_1={1over 6}$. Also note that if $e_n>0$ then $e_{n+1}>0$ therefore since $e_1>0$ we have $e_n>0$. Also if $e_nle {1over 6}$ we obtain $$e_{n+1}=e_nleft({2over 3}+e_nright)le {5over 6}e_n$$ finally $$0<e_{n+1}le {5over 6}e_n$$which means that $e_nto 0$ and $x_nto {1over 3}$ therefore $x_n$ is bounded.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 14:31









Mostafa AyazMostafa Ayaz

14.3k3937




14.3k3937












  • Could you please elaborate on how you came up with such substitution? For me it looks like a lucky guess, i would like to grasp the technique
    – roman
    Jan 4 at 14:36






  • 1




    Sure! Since after finding $L=0$ and $L={1over 3}$ we guessed that $x_n$ should converge ${1over 3}$ then we are able to define the distance of $x_n$ to its final limit as $e_n=x_n-{1over 3}$ or $e_n={1over 3}-x_n$, not much different which one to choose and prove that the distance converges to $0$. A proof on convergence to $0$ is rather simpler than proving the convergence to any other point ---at least for me :)
    – Mostafa Ayaz
    Jan 4 at 14:40


















  • Could you please elaborate on how you came up with such substitution? For me it looks like a lucky guess, i would like to grasp the technique
    – roman
    Jan 4 at 14:36






  • 1




    Sure! Since after finding $L=0$ and $L={1over 3}$ we guessed that $x_n$ should converge ${1over 3}$ then we are able to define the distance of $x_n$ to its final limit as $e_n=x_n-{1over 3}$ or $e_n={1over 3}-x_n$, not much different which one to choose and prove that the distance converges to $0$. A proof on convergence to $0$ is rather simpler than proving the convergence to any other point ---at least for me :)
    – Mostafa Ayaz
    Jan 4 at 14:40
















Could you please elaborate on how you came up with such substitution? For me it looks like a lucky guess, i would like to grasp the technique
– roman
Jan 4 at 14:36




Could you please elaborate on how you came up with such substitution? For me it looks like a lucky guess, i would like to grasp the technique
– roman
Jan 4 at 14:36




1




1




Sure! Since after finding $L=0$ and $L={1over 3}$ we guessed that $x_n$ should converge ${1over 3}$ then we are able to define the distance of $x_n$ to its final limit as $e_n=x_n-{1over 3}$ or $e_n={1over 3}-x_n$, not much different which one to choose and prove that the distance converges to $0$. A proof on convergence to $0$ is rather simpler than proving the convergence to any other point ---at least for me :)
– Mostafa Ayaz
Jan 4 at 14:40




Sure! Since after finding $L=0$ and $L={1over 3}$ we guessed that $x_n$ should converge ${1over 3}$ then we are able to define the distance of $x_n$ to its final limit as $e_n=x_n-{1over 3}$ or $e_n={1over 3}-x_n$, not much different which one to choose and prove that the distance converges to $0$. A proof on convergence to $0$ is rather simpler than proving the convergence to any other point ---at least for me :)
– Mostafa Ayaz
Jan 4 at 14:40











1














Hint: find explicit pattern and prove that it is bounded






share|cite|improve this answer

















  • 1




    Do you know what the explicit pattern is? maybe you can share it in your answer
    – caverac
    Jan 4 at 14:32












  • Are you sure closed form exists?
    – roman
    Jan 4 at 14:33
















1














Hint: find explicit pattern and prove that it is bounded






share|cite|improve this answer

















  • 1




    Do you know what the explicit pattern is? maybe you can share it in your answer
    – caverac
    Jan 4 at 14:32












  • Are you sure closed form exists?
    – roman
    Jan 4 at 14:33














1












1








1






Hint: find explicit pattern and prove that it is bounded






share|cite|improve this answer












Hint: find explicit pattern and prove that it is bounded







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 4 at 14:24









VirtualUserVirtualUser

54712




54712








  • 1




    Do you know what the explicit pattern is? maybe you can share it in your answer
    – caverac
    Jan 4 at 14:32












  • Are you sure closed form exists?
    – roman
    Jan 4 at 14:33














  • 1




    Do you know what the explicit pattern is? maybe you can share it in your answer
    – caverac
    Jan 4 at 14:32












  • Are you sure closed form exists?
    – roman
    Jan 4 at 14:33








1




1




Do you know what the explicit pattern is? maybe you can share it in your answer
– caverac
Jan 4 at 14:32






Do you know what the explicit pattern is? maybe you can share it in your answer
– caverac
Jan 4 at 14:32














Are you sure closed form exists?
– roman
Jan 4 at 14:33




Are you sure closed form exists?
– roman
Jan 4 at 14:33


















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