Radius of convergence $sum_{n=0}^{infty} a_n z^n$, where $a_n = 2a_{n-1} + 1$












1












$begingroup$


I want to show that the radius of convergence of $$g(z) = sum_{n=0}^{infty} a_n z^n$$ is $1/2$ when $a_0 = 1$ and $a_n = 2a_{n-1} + 1$ for $n geq 1$. Since $$g(z) = 1 + sum_{n=1}^{infty} a_n z^n = 1 + sum_{n=1}^{infty} (2a_{n-1} + 1)z^n = 2zg(z) + frac{1}{1-z}$$ $$iff g(z) = frac{1}{(1-2z)(1-z)}, $$



I am wondering if I can somehow use this closed form formula for $g(z)$. Since $1-2z = 0$ for $z=1/2$, that makes me suspect that somehow that is related to the radius of convergence, but I haven't been able to figure out how... Can anyone help explain it to me?






sequences-and-series complex-analysis power-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 21:33












  • $begingroup$
    Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
    $endgroup$
    – j.eee
    Dec 13 '18 at 21:45










  • $begingroup$
    If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
    $endgroup$
    – Daniel Schepler
    Dec 13 '18 at 22:28










  • $begingroup$
    That makes some sense. :) Thank you. @DanielSchepler
    $endgroup$
    – j.eee
    Dec 13 '18 at 22:41
















1












$begingroup$


I want to show that the radius of convergence of $$g(z) = sum_{n=0}^{infty} a_n z^n$$ is $1/2$ when $a_0 = 1$ and $a_n = 2a_{n-1} + 1$ for $n geq 1$. Since $$g(z) = 1 + sum_{n=1}^{infty} a_n z^n = 1 + sum_{n=1}^{infty} (2a_{n-1} + 1)z^n = 2zg(z) + frac{1}{1-z}$$ $$iff g(z) = frac{1}{(1-2z)(1-z)}, $$



I am wondering if I can somehow use this closed form formula for $g(z)$. Since $1-2z = 0$ for $z=1/2$, that makes me suspect that somehow that is related to the radius of convergence, but I haven't been able to figure out how... Can anyone help explain it to me?






sequences-and-series complex-analysis power-series







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 21:33












  • $begingroup$
    Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
    $endgroup$
    – j.eee
    Dec 13 '18 at 21:45










  • $begingroup$
    If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
    $endgroup$
    – Daniel Schepler
    Dec 13 '18 at 22:28










  • $begingroup$
    That makes some sense. :) Thank you. @DanielSchepler
    $endgroup$
    – j.eee
    Dec 13 '18 at 22:41














1












1








1





$begingroup$


I want to show that the radius of convergence of $$g(z) = sum_{n=0}^{infty} a_n z^n$$ is $1/2$ when $a_0 = 1$ and $a_n = 2a_{n-1} + 1$ for $n geq 1$. Since $$g(z) = 1 + sum_{n=1}^{infty} a_n z^n = 1 + sum_{n=1}^{infty} (2a_{n-1} + 1)z^n = 2zg(z) + frac{1}{1-z}$$ $$iff g(z) = frac{1}{(1-2z)(1-z)}, $$



I am wondering if I can somehow use this closed form formula for $g(z)$. Since $1-2z = 0$ for $z=1/2$, that makes me suspect that somehow that is related to the radius of convergence, but I haven't been able to figure out how... Can anyone help explain it to me?






sequences-and-series complex-analysis power-series







share|cite|improve this question









$endgroup$




I want to show that the radius of convergence of $$g(z) = sum_{n=0}^{infty} a_n z^n$$ is $1/2$ when $a_0 = 1$ and $a_n = 2a_{n-1} + 1$ for $n geq 1$. Since $$g(z) = 1 + sum_{n=1}^{infty} a_n z^n = 1 + sum_{n=1}^{infty} (2a_{n-1} + 1)z^n = 2zg(z) + frac{1}{1-z}$$ $$iff g(z) = frac{1}{(1-2z)(1-z)}, $$



I am wondering if I can somehow use this closed form formula for $g(z)$. Since $1-2z = 0$ for $z=1/2$, that makes me suspect that somehow that is related to the radius of convergence, but I haven't been able to figure out how... Can anyone help explain it to me?






sequences-and-series complex-analysis power-series




sequences-and-series complex-analysis power-series






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 13 '18 at 21:28









j.eeej.eee

787




787








  • 1




    $begingroup$
    Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 21:33












  • $begingroup$
    Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
    $endgroup$
    – j.eee
    Dec 13 '18 at 21:45










  • $begingroup$
    If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
    $endgroup$
    – Daniel Schepler
    Dec 13 '18 at 22:28










  • $begingroup$
    That makes some sense. :) Thank you. @DanielSchepler
    $endgroup$
    – j.eee
    Dec 13 '18 at 22:41














  • 1




    $begingroup$
    Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
    $endgroup$
    – SmileyCraft
    Dec 13 '18 at 21:33












  • $begingroup$
    Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
    $endgroup$
    – j.eee
    Dec 13 '18 at 21:45










  • $begingroup$
    If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
    $endgroup$
    – Daniel Schepler
    Dec 13 '18 at 22:28










  • $begingroup$
    That makes some sense. :) Thank you. @DanielSchepler
    $endgroup$
    – j.eee
    Dec 13 '18 at 22:41








1




1




$begingroup$
Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
$endgroup$
– SmileyCraft
Dec 13 '18 at 21:33






$begingroup$
Can't you just use the ratio test? en.wikipedia.org/wiki/Ratio_test
$endgroup$
– SmileyCraft
Dec 13 '18 at 21:33














$begingroup$
Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
$endgroup$
– j.eee
Dec 13 '18 at 21:45




$begingroup$
Perhaps, but in the solutions manual they seem to do it through this formula, although they aren't very explicit about how they do it. If it is possible then it seems like the easier way. :)
$endgroup$
– j.eee
Dec 13 '18 at 21:45












$begingroup$
If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
$endgroup$
– Daniel Schepler
Dec 13 '18 at 22:28




$begingroup$
If you could show $g(z)$ has a positive radius of convergence, then the sum would have to be equal to this formula. Then, since this function extends to a complex function analytic off ${ 1/2, 1 }$, that would imply that $g$ has a Taylor series expansion with radius of convergence $frac{1}{2}$ - which then must have coefficients $a_n$. However, I don't see any way to bootstrap that initial step without a method which just directly proves the radius of convergence anyway...
$endgroup$
– Daniel Schepler
Dec 13 '18 at 22:28












$begingroup$
That makes some sense. :) Thank you. @DanielSchepler
$endgroup$
– j.eee
Dec 13 '18 at 22:41




$begingroup$
That makes some sense. :) Thank you. @DanielSchepler
$endgroup$
– j.eee
Dec 13 '18 at 22:41










1 Answer
1






active

oldest

votes


















0












$begingroup$

You are nearly done:
$$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
=sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
sumlimits_{n=0}(2^{n+1}-1)z^n$$
from infinite geometric progression, or
$$a_n=2^{n+1}-1$$
This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.






share|cite|improve this answer











$endgroup$













    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%2f3038586%2fradius-of-convergence-sum-n-0-infty-a-n-zn-where-a-n-2a-n-1-1%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









    0












    $begingroup$

    You are nearly done:
    $$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
    =sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
    sumlimits_{n=0}(2^{n+1}-1)z^n$$
    from infinite geometric progression, or
    $$a_n=2^{n+1}-1$$
    This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.






    share|cite|improve this answer











    $endgroup$


















      0












      $begingroup$

      You are nearly done:
      $$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
      =sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
      sumlimits_{n=0}(2^{n+1}-1)z^n$$
      from infinite geometric progression, or
      $$a_n=2^{n+1}-1$$
      This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.






      share|cite|improve this answer











      $endgroup$
















        0












        0








        0





        $begingroup$

        You are nearly done:
        $$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
        =sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
        sumlimits_{n=0}(2^{n+1}-1)z^n$$
        from infinite geometric progression, or
        $$a_n=2^{n+1}-1$$
        This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.






        share|cite|improve this answer











        $endgroup$



        You are nearly done:
        $$frac{1}{(1-2z)(1-z)}=frac{2}{1-2z}-frac{1}{1-z}
        =sumlimits_{n=0}2^{n+1}z^n-sumlimits_{n=0}z^n=\
        sumlimits_{n=0}(2^{n+1}-1)z^n$$
        from infinite geometric progression, or
        $$a_n=2^{n+1}-1$$
        This method is called method of generating functions for solving recurrences, here is another good resource. As it was mentioned in the comments, ratio test shows $r=frac{1}{2}$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Dec 13 '18 at 23:39

























        answered Dec 13 '18 at 22:52









        rtybasertybase

        11k21533




        11k21533






























            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3038586%2fradius-of-convergence-sum-n-0-infty-a-n-zn-where-a-n-2a-n-1-1%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

            Probability when a professor distributes a quiz and homework assignment to a class of n students.

            Image
            0 0 $begingroup$ Need help with this problem. Suppose our lazy professor collects a quiz and a homework assignment from a class of n students one day, then distributes both the quizzes and the homework assignments back to the class in a random fashion for grading. Each student receives one quiz and one homework assignment to grade. (a) What is the probability that every student receives someone else's quiz to grade, and someone else's homework to grade? (b) What is the probability that no student receives both their own quiz and their own homework assignment to grade? In this case, some students may receive their own quiz, and others may receive their own homework assignment. (c) Compute the limiting probability as n approaches infinity in each case. ...
            Read more

            Aardman Animations

            Image
            Aardman Animations, Ltd. Aardman Animations Tipo privada Atividade Animação em stop-motion , Animação em CGI Fundação 1972 Fundador(es) Peter Lord David Sproxton Sede Bristol, Inglaterra  Reino Unido Proprietário(s) Dreamworks Pessoas-chave Peter Lord David Sproxton Nick Park Divisões Aardman Features Aardman Digital Aardman Commercials Aardman Broadcast Aardman International Aardman Rights Aardman Effects Aardman 3-D Systems Aardman Nathan Love AardBoiled Website oficial aardman.com ...
            Read more

            Are they similar matrix

            Image
            0 $begingroup$ Do $begin{bmatrix} 0&i&0\0&0&1\0&0&0 end{bmatrix} $ and $begin{bmatrix} 0&0&0\-i&0&0\0&1&0 end{bmatrix} $ are similar.Is this True/false Clearly both are nilpotent and one is conjucate transpose of other but how to know if they are similar.i'm stuck. Please help me linear-algebra vector-spaces eigenvalues-eigenvectors generalizedeigenvector share | cite | improve this question edited Dec 10 '18 at 6:38 Vasanth Kris asked Dec 10 '18 at 6:19 ...
            Read more