What is the region of convergence of $x_n=left(frac{x_{n-1}}{n}right)^2-a$, where $a$ is a constant?












5












$begingroup$


The following recurrence relation came up in some research I was working on:



$$x_n=left(frac{x_{n-1}}{n}right)^2-a$$



Or equivalently the map:



$$zmapstofrac{z^2}{n^2}-a$$



Where $n$ is the iteration number. Specifically, I'm interested in the size of the convergence region across the real line. Some stuff I know about this map:




  • For $a = 1$, it's easy, the "size on the real line" is $[-3,3]$.


I do have an infinite radical expansion for the size of the convergence region on the real line (see Solving the infinite radical $sqrt{6+sqrt{6+2sqrt{6+3sqrt{6+...}}}}$):



$$sqrt{a+2sqrt{a+3sqrt{a+...}}}$$



That's why it's easy for $a=1$ -- it's just the Ramanujan radical, and equals 3. It's also easy for $a=0$ -- it's $expleft(-mathrm{PolyLog}^{(1,0)}(0,1/2)right)$ as per Wolfram Alpha.



Has anyone seen this map before? Here's the region of convergence on the complex plane, plotted numerically (for $a=6$):



enter image description here










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For your title, instead asking "have you seen this recurrence before" to which most will say "no" and skip the question, you rather ask "what is the region of convergence for ...." and get more help.
    $endgroup$
    – DanielV
    Dec 24 '18 at 19:34






  • 1




    $begingroup$
    if the sequence converges, then it must converge to $-a.$ I am not sure if any other trivial observation can be made.
    $endgroup$
    – dezdichado
    Dec 24 '18 at 22:58
















5












$begingroup$


The following recurrence relation came up in some research I was working on:



$$x_n=left(frac{x_{n-1}}{n}right)^2-a$$



Or equivalently the map:



$$zmapstofrac{z^2}{n^2}-a$$



Where $n$ is the iteration number. Specifically, I'm interested in the size of the convergence region across the real line. Some stuff I know about this map:




  • For $a = 1$, it's easy, the "size on the real line" is $[-3,3]$.


I do have an infinite radical expansion for the size of the convergence region on the real line (see Solving the infinite radical $sqrt{6+sqrt{6+2sqrt{6+3sqrt{6+...}}}}$):



$$sqrt{a+2sqrt{a+3sqrt{a+...}}}$$



That's why it's easy for $a=1$ -- it's just the Ramanujan radical, and equals 3. It's also easy for $a=0$ -- it's $expleft(-mathrm{PolyLog}^{(1,0)}(0,1/2)right)$ as per Wolfram Alpha.



Has anyone seen this map before? Here's the region of convergence on the complex plane, plotted numerically (for $a=6$):



enter image description here










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    For your title, instead asking "have you seen this recurrence before" to which most will say "no" and skip the question, you rather ask "what is the region of convergence for ...." and get more help.
    $endgroup$
    – DanielV
    Dec 24 '18 at 19:34






  • 1




    $begingroup$
    if the sequence converges, then it must converge to $-a.$ I am not sure if any other trivial observation can be made.
    $endgroup$
    – dezdichado
    Dec 24 '18 at 22:58














5












5








5


1



$begingroup$


The following recurrence relation came up in some research I was working on:



$$x_n=left(frac{x_{n-1}}{n}right)^2-a$$



Or equivalently the map:



$$zmapstofrac{z^2}{n^2}-a$$



Where $n$ is the iteration number. Specifically, I'm interested in the size of the convergence region across the real line. Some stuff I know about this map:




  • For $a = 1$, it's easy, the "size on the real line" is $[-3,3]$.


I do have an infinite radical expansion for the size of the convergence region on the real line (see Solving the infinite radical $sqrt{6+sqrt{6+2sqrt{6+3sqrt{6+...}}}}$):



$$sqrt{a+2sqrt{a+3sqrt{a+...}}}$$



That's why it's easy for $a=1$ -- it's just the Ramanujan radical, and equals 3. It's also easy for $a=0$ -- it's $expleft(-mathrm{PolyLog}^{(1,0)}(0,1/2)right)$ as per Wolfram Alpha.



Has anyone seen this map before? Here's the region of convergence on the complex plane, plotted numerically (for $a=6$):



enter image description here










share|cite|improve this question











$endgroup$




The following recurrence relation came up in some research I was working on:



$$x_n=left(frac{x_{n-1}}{n}right)^2-a$$



Or equivalently the map:



$$zmapstofrac{z^2}{n^2}-a$$



Where $n$ is the iteration number. Specifically, I'm interested in the size of the convergence region across the real line. Some stuff I know about this map:




  • For $a = 1$, it's easy, the "size on the real line" is $[-3,3]$.


I do have an infinite radical expansion for the size of the convergence region on the real line (see Solving the infinite radical $sqrt{6+sqrt{6+2sqrt{6+3sqrt{6+...}}}}$):



$$sqrt{a+2sqrt{a+3sqrt{a+...}}}$$



That's why it's easy for $a=1$ -- it's just the Ramanujan radical, and equals 3. It's also easy for $a=0$ -- it's $expleft(-mathrm{PolyLog}^{(1,0)}(0,1/2)right)$ as per Wolfram Alpha.



Has anyone seen this map before? Here's the region of convergence on the complex plane, plotted numerically (for $a=6$):



enter image description here







sequences-and-series convergence recurrence-relations






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 29 '18 at 6:47







Abhimanyu Pallavi Sudhir

















asked Dec 24 '18 at 19:02









Abhimanyu Pallavi SudhirAbhimanyu Pallavi Sudhir

909719




909719








  • 1




    $begingroup$
    For your title, instead asking "have you seen this recurrence before" to which most will say "no" and skip the question, you rather ask "what is the region of convergence for ...." and get more help.
    $endgroup$
    – DanielV
    Dec 24 '18 at 19:34






  • 1




    $begingroup$
    if the sequence converges, then it must converge to $-a.$ I am not sure if any other trivial observation can be made.
    $endgroup$
    – dezdichado
    Dec 24 '18 at 22:58














  • 1




    $begingroup$
    For your title, instead asking "have you seen this recurrence before" to which most will say "no" and skip the question, you rather ask "what is the region of convergence for ...." and get more help.
    $endgroup$
    – DanielV
    Dec 24 '18 at 19:34






  • 1




    $begingroup$
    if the sequence converges, then it must converge to $-a.$ I am not sure if any other trivial observation can be made.
    $endgroup$
    – dezdichado
    Dec 24 '18 at 22:58








1




1




$begingroup$
For your title, instead asking "have you seen this recurrence before" to which most will say "no" and skip the question, you rather ask "what is the region of convergence for ...." and get more help.
$endgroup$
– DanielV
Dec 24 '18 at 19:34




$begingroup$
For your title, instead asking "have you seen this recurrence before" to which most will say "no" and skip the question, you rather ask "what is the region of convergence for ...." and get more help.
$endgroup$
– DanielV
Dec 24 '18 at 19:34




1




1




$begingroup$
if the sequence converges, then it must converge to $-a.$ I am not sure if any other trivial observation can be made.
$endgroup$
– dezdichado
Dec 24 '18 at 22:58




$begingroup$
if the sequence converges, then it must converge to $-a.$ I am not sure if any other trivial observation can be made.
$endgroup$
– dezdichado
Dec 24 '18 at 22:58










1 Answer
1






active

oldest

votes


















2












$begingroup$

Reminded me of this particular Julia set. It's the one generated from the Mandelbrot set's (0,0) point. Picture



Mandelbrot set on the left, Julia on the right.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Interesting -- it's not quite the origin, though (0 just gives an ellipse). I'm not sure if they're related -- going further to the left or right gives something quite different from my regions.
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:08












  • $begingroup$
    It does seem that different values of $a$ correspond to different real $c$ generators for the Julia set. But I think the similarity is just approximate -- it stops working for large $|c|$, etc. (where you get little Mandelbrots from the Julia set but bats from mine).
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:38













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1 Answer
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1 Answer
1






active

oldest

votes









active

oldest

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active

oldest

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2












$begingroup$

Reminded me of this particular Julia set. It's the one generated from the Mandelbrot set's (0,0) point. Picture



Mandelbrot set on the left, Julia on the right.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Interesting -- it's not quite the origin, though (0 just gives an ellipse). I'm not sure if they're related -- going further to the left or right gives something quite different from my regions.
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:08












  • $begingroup$
    It does seem that different values of $a$ correspond to different real $c$ generators for the Julia set. But I think the similarity is just approximate -- it stops working for large $|c|$, etc. (where you get little Mandelbrots from the Julia set but bats from mine).
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:38


















2












$begingroup$

Reminded me of this particular Julia set. It's the one generated from the Mandelbrot set's (0,0) point. Picture



Mandelbrot set on the left, Julia on the right.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Interesting -- it's not quite the origin, though (0 just gives an ellipse). I'm not sure if they're related -- going further to the left or right gives something quite different from my regions.
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:08












  • $begingroup$
    It does seem that different values of $a$ correspond to different real $c$ generators for the Julia set. But I think the similarity is just approximate -- it stops working for large $|c|$, etc. (where you get little Mandelbrots from the Julia set but bats from mine).
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:38
















2












2








2





$begingroup$

Reminded me of this particular Julia set. It's the one generated from the Mandelbrot set's (0,0) point. Picture



Mandelbrot set on the left, Julia on the right.






share|cite|improve this answer









$endgroup$



Reminded me of this particular Julia set. It's the one generated from the Mandelbrot set's (0,0) point. Picture



Mandelbrot set on the left, Julia on the right.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 24 '18 at 22:30









FrostFrost

1776




1776












  • $begingroup$
    Interesting -- it's not quite the origin, though (0 just gives an ellipse). I'm not sure if they're related -- going further to the left or right gives something quite different from my regions.
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:08












  • $begingroup$
    It does seem that different values of $a$ correspond to different real $c$ generators for the Julia set. But I think the similarity is just approximate -- it stops working for large $|c|$, etc. (where you get little Mandelbrots from the Julia set but bats from mine).
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:38




















  • $begingroup$
    Interesting -- it's not quite the origin, though (0 just gives an ellipse). I'm not sure if they're related -- going further to the left or right gives something quite different from my regions.
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:08












  • $begingroup$
    It does seem that different values of $a$ correspond to different real $c$ generators for the Julia set. But I think the similarity is just approximate -- it stops working for large $|c|$, etc. (where you get little Mandelbrots from the Julia set but bats from mine).
    $endgroup$
    – Abhimanyu Pallavi Sudhir
    Dec 25 '18 at 4:38


















$begingroup$
Interesting -- it's not quite the origin, though (0 just gives an ellipse). I'm not sure if they're related -- going further to the left or right gives something quite different from my regions.
$endgroup$
– Abhimanyu Pallavi Sudhir
Dec 25 '18 at 4:08






$begingroup$
Interesting -- it's not quite the origin, though (0 just gives an ellipse). I'm not sure if they're related -- going further to the left or right gives something quite different from my regions.
$endgroup$
– Abhimanyu Pallavi Sudhir
Dec 25 '18 at 4:08














$begingroup$
It does seem that different values of $a$ correspond to different real $c$ generators for the Julia set. But I think the similarity is just approximate -- it stops working for large $|c|$, etc. (where you get little Mandelbrots from the Julia set but bats from mine).
$endgroup$
– Abhimanyu Pallavi Sudhir
Dec 25 '18 at 4:38






$begingroup$
It does seem that different values of $a$ correspond to different real $c$ generators for the Julia set. But I think the similarity is just approximate -- it stops working for large $|c|$, etc. (where you get little Mandelbrots from the Julia set but bats from mine).
$endgroup$
– Abhimanyu Pallavi Sudhir
Dec 25 '18 at 4:38




















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