characterizing the boundary of the convergent region of $f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$












1














Let



$$f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$$



A domain colored portrait (with artifacts) for $f(z)$ on the unit disk looks like:





The gray and white regions are where the software package had difficulty with the sum, because it was divergent. Is there any good way of characterizing for which $|z|<1$ this series is convergent? I'm interested in the boundary of the red dagger shaped region to the right.



The code that was used to generate this is (using mpmath and matplotlib):



from mpmath import *

import pylab

def f(z):

    return fp.nsum(lambda n: z**(1/(z**n)), [1,inf])





fp.cplot(lambda z: f(z), [-1.0, 1.0], [-1.0, 1.0], points=800000, verbose=True)





complex-analysis







share|cite|improve this question
























  • The series converges for $$left| {{z^{1/z}}} right| < 1$$
    – Pedro Tamaroff♦
    Mar 6 '12 at 21:03






  • 1




    No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
    – Robert Israel
    Mar 6 '12 at 21:22


















1














Let



$$f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$$



A domain colored portrait (with artifacts) for $f(z)$ on the unit disk looks like:





The gray and white regions are where the software package had difficulty with the sum, because it was divergent. Is there any good way of characterizing for which $|z|<1$ this series is convergent? I'm interested in the boundary of the red dagger shaped region to the right.



The code that was used to generate this is (using mpmath and matplotlib):



from mpmath import *

import pylab

def f(z):

    return fp.nsum(lambda n: z**(1/(z**n)), [1,inf])





fp.cplot(lambda z: f(z), [-1.0, 1.0], [-1.0, 1.0], points=800000, verbose=True)





complex-analysis







share|cite|improve this question
























  • The series converges for $$left| {{z^{1/z}}} right| < 1$$
    – Pedro Tamaroff♦
    Mar 6 '12 at 21:03






  • 1




    No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
    – Robert Israel
    Mar 6 '12 at 21:22
















1












1








1







Let



$$f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$$



A domain colored portrait (with artifacts) for $f(z)$ on the unit disk looks like:





The gray and white regions are where the software package had difficulty with the sum, because it was divergent. Is there any good way of characterizing for which $|z|<1$ this series is convergent? I'm interested in the boundary of the red dagger shaped region to the right.



The code that was used to generate this is (using mpmath and matplotlib):



from mpmath import *

import pylab

def f(z):

    return fp.nsum(lambda n: z**(1/(z**n)), [1,inf])





fp.cplot(lambda z: f(z), [-1.0, 1.0], [-1.0, 1.0], points=800000, verbose=True)





complex-analysis







share|cite|improve this question















Let



$$f(z)= sum_{n=1}^{infty} z^{(1/z)^{n}}$$



A domain colored portrait (with artifacts) for $f(z)$ on the unit disk looks like:





The gray and white regions are where the software package had difficulty with the sum, because it was divergent. Is there any good way of characterizing for which $|z|<1$ this series is convergent? I'm interested in the boundary of the red dagger shaped region to the right.



The code that was used to generate this is (using mpmath and matplotlib):



from mpmath import *

import pylab

def f(z):

    return fp.nsum(lambda n: z**(1/(z**n)), [1,inf])





fp.cplot(lambda z: f(z), [-1.0, 1.0], [-1.0, 1.0], points=800000, verbose=True)





complex-analysis




complex-analysis






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 4 at 8:24









Glorfindel

3,41981830




3,41981830










asked Mar 6 '12 at 20:28









deoxygerbedeoxygerbe

3,06711736




3,06711736












  • The series converges for $$left| {{z^{1/z}}} right| < 1$$
    – Pedro Tamaroff♦
    Mar 6 '12 at 21:03






  • 1




    No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
    – Robert Israel
    Mar 6 '12 at 21:22




















  • The series converges for $$left| {{z^{1/z}}} right| < 1$$
    – Pedro Tamaroff♦
    Mar 6 '12 at 21:03






  • 1




    No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
    – Robert Israel
    Mar 6 '12 at 21:22


















The series converges for $$left| {{z^{1/z}}} right| < 1$$
– Pedro Tamaroff♦
Mar 6 '12 at 21:03




The series converges for $$left| {{z^{1/z}}} right| < 1$$
– Pedro Tamaroff♦
Mar 6 '12 at 21:03




1




1




No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
– Robert Israel
Mar 6 '12 at 21:22






No, it's $z^{(1/z^n)}$, not $(z^{1/z})^n$
– Robert Israel
Mar 6 '12 at 21:22












1 Answer
1






active

oldest

votes


















2














$|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
$log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.



So the thickness of your red region is really just an artifact of the numerical methods being used.






share|cite|improve this answer





















    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f117219%2fcharacterizing-the-boundary-of-the-convergent-region-of-fz-sum-n-1-inft%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    $|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
    $log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.



    So the thickness of your red region is really just an artifact of the numerical methods being used.






    share|cite|improve this answer


























      2














      $|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
      $log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.



      So the thickness of your red region is really just an artifact of the numerical methods being used.






      share|cite|improve this answer
























        2












        2








        2






        $|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
        $log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.



        So the thickness of your red region is really just an artifact of the numerical methods being used.






        share|cite|improve this answer












        $|z^{(1/z)^n}| = exp(text{Re}(log(z) (1/z)^{n}))$. Now if $z$ is not a positive real number,
        $log(z) (1/z)^n$ will be in some sector ${z: -pi/2+epsilon le arg(z) le pi/2 - epsilon}$ infinitely often, and since $|log(z) (1/z)^n| to infty$ as $n to infty$ the terms of your series won't go to $0$, and the series will diverge. On the other hand, for $0 < z < 1$ the series converges.



        So the thickness of your red region is really just an artifact of the numerical methods being used.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Mar 6 '12 at 21:20









        Robert IsraelRobert Israel

        319k23208457




        319k23208457






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.





            Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


            Please pay close attention to the following guidance:


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f117219%2fcharacterizing-the-boundary-of-the-convergent-region-of-fz-sum-n-1-inft%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            An IMO inspired problem

            Image
            14 4 This problem from IMO 1988 is said to be one of the most elegant ones in functional equations. Problem : The function $f$ is defined on the set of all positive integers as follows: begin{align} f(1) = 1, f(3) &= 3, f(2n) = f(n), \ f(4n+1) &= 2f(2n+1) - f(n), \ f(4n+3) &= 3 f(2n+1) - 2f(n) end{align} Find the number of $n$ with $f(n) = n, 1 leq n leq 1988$. The main idea towards the solution is realizing that these conditions stand for the fact that $f(n)$ just reverses the digits in the binary representation of the number $n$. So, essentially the solution is finding the number of binary pallindromes $leq 1988_{10}$. My question is the following : How to reformulate the problem so that $f(n)$ reverses the digits of $n$ in its ternary representation. Or even better, can we reformulate it for a
            Read more

            Management

            Image
            Management From Wikipedia, the free encyclopedia Jump to navigation Jump to search "Manager" redirects here. For other uses, see Management (disambiguation) and Manager (disambiguation). An organization chart for the United States Coast Guard shows the hierarchy of managerial roles in that organization. Business administration Management of a business Accounting Management accounting Financial accounting Financial audit Business entities Cooperative Corporation Limited liability company Partnership Sole proprietorship State-owned enterprise Corporate governance Annual general meeting Board of directors Supervisory board Advisory board Audit committee Corporate law Commercial law Constitutional documents Contract Corporate crime Corporate liability Insolvency law International trade law Mergers and acquisitions Economics Commodity Pu
            Read more

            Has there ever been an instance of an active nuclear power plant within or near a war zone?

            Image
            36 3 The question Has there ever been a military combat going on around an active (running or in outage but not decommissioned) commercial nuclear power plant? Or dangerously near it? If yes, how did the plant and its personnel fare in such situation? Why I ask Let's assume for the scope of this question modern commercial nuclear power plants (CANDU, PWR, BWR of generation II plus all of gen III and newer) are usually reasonably safe to operate given that the engineering has been done right and there is a whole cohort of very well trained personnel on-site at all times, well rested and with considerable resources on their hands. Things hard to get with an armed conflict like the recent Syrian war raging around. I can't seem to find any reference or a comprehensive article about this, hence this question.
            Read more