Why doesn't the Stone-Weierstrass theorem imply that every function has a power series expansion?

The name of the pictureThe name of the pictureThe name of the pictureClash Royale CLAN TAG#URR8PPP











up vote
5
down vote

favorite
2












I know that every function doesn't have a power series expansion.
Yet what I don't understand is that for every $C^infty$ functions there is a sequence of polynomial $(P_n)$ such that $P_n$ converges uniformly to $f$. That's to say :



$$forall x in [a,b], f(x) = lim_n to infty sum_k = 0^infty a_k,nx^k$$



But then because it converges uniformly why can't I say that :



$$forall x in [a,b], f(x) = sum_k = 0^infty lim_n to infty a_k,nx^k$$



And so $f$ has a power series expansion with coefficients: $lim_n to infty a_k,nx^k$.






calculus real-analysis power-series







share|cite|improve this question



















  • 5




    Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
    – Kavi Rama Murthy
    3 hours ago











  • Note that it's also possible to "have a power series" for a function (like the Taylor series) and yet have it converge to the function only at a single point. The "bathtub" function $f(x) = begincases0 & x = 0 \ e^-x^2 & x ne 0endcases$ is a good example: its Taylor series $T$ at $x = 0$ exists and has all coefficients zero, but $f(x) = T(x)$ only for $x = 0$.
    – John Hughes
    8 mins ago














up vote
5
down vote

favorite
2












I know that every function doesn't have a power series expansion.
Yet what I don't understand is that for every $C^infty$ functions there is a sequence of polynomial $(P_n)$ such that $P_n$ converges uniformly to $f$. That's to say :



$$forall x in [a,b], f(x) = lim_n to infty sum_k = 0^infty a_k,nx^k$$



But then because it converges uniformly why can't I say that :



$$forall x in [a,b], f(x) = sum_k = 0^infty lim_n to infty a_k,nx^k$$



And so $f$ has a power series expansion with coefficients: $lim_n to infty a_k,nx^k$.






calculus real-analysis power-series







share|cite|improve this question



















  • 5




    Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
    – Kavi Rama Murthy
    3 hours ago











  • Note that it's also possible to "have a power series" for a function (like the Taylor series) and yet have it converge to the function only at a single point. The "bathtub" function $f(x) = begincases0 & x = 0 \ e^-x^2 & x ne 0endcases$ is a good example: its Taylor series $T$ at $x = 0$ exists and has all coefficients zero, but $f(x) = T(x)$ only for $x = 0$.
    – John Hughes
    8 mins ago












up vote
5
down vote

favorite
2









up vote
5
down vote

favorite
2






2





I know that every function doesn't have a power series expansion.
Yet what I don't understand is that for every $C^infty$ functions there is a sequence of polynomial $(P_n)$ such that $P_n$ converges uniformly to $f$. That's to say :



$$forall x in [a,b], f(x) = lim_n to infty sum_k = 0^infty a_k,nx^k$$



But then because it converges uniformly why can't I say that :



$$forall x in [a,b], f(x) = sum_k = 0^infty lim_n to infty a_k,nx^k$$



And so $f$ has a power series expansion with coefficients: $lim_n to infty a_k,nx^k$.






calculus real-analysis power-series







share|cite|improve this question















I know that every function doesn't have a power series expansion.
Yet what I don't understand is that for every $C^infty$ functions there is a sequence of polynomial $(P_n)$ such that $P_n$ converges uniformly to $f$. That's to say :



$$forall x in [a,b], f(x) = lim_n to infty sum_k = 0^infty a_k,nx^k$$



But then because it converges uniformly why can't I say that :



$$forall x in [a,b], f(x) = sum_k = 0^infty lim_n to infty a_k,nx^k$$



And so $f$ has a power series expansion with coefficients: $lim_n to infty a_k,nx^k$.






calculus real-analysis power-series




calculus real-analysis power-series






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited 16 mins ago









Andrew Uzzell

1,2111127




1,2111127










asked 3 hours ago









auhasard

929




929







  • 5




    Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
    – Kavi Rama Murthy
    3 hours ago











  • Note that it's also possible to "have a power series" for a function (like the Taylor series) and yet have it converge to the function only at a single point. The "bathtub" function $f(x) = begincases0 & x = 0 \ e^-x^2 & x ne 0endcases$ is a good example: its Taylor series $T$ at $x = 0$ exists and has all coefficients zero, but $f(x) = T(x)$ only for $x = 0$.
    – John Hughes
    8 mins ago












  • 5




    Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
    – Kavi Rama Murthy
    3 hours ago











  • Note that it's also possible to "have a power series" for a function (like the Taylor series) and yet have it converge to the function only at a single point. The "bathtub" function $f(x) = begincases0 & x = 0 \ e^-x^2 & x ne 0endcases$ is a good example: its Taylor series $T$ at $x = 0$ exists and has all coefficients zero, but $f(x) = T(x)$ only for $x = 0$.
    – John Hughes
    8 mins ago







5




5




Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
– Kavi Rama Murthy
3 hours ago





Uniform convergence does not even guarantee the existence of $lim_nto infty a_k,n$. If $p(x)=sum a_kx^k$ then $a_i=p^(i) (0)/(i!)$ but Uniform convergence of continuous functions does not give convergence of derivatives.
– Kavi Rama Murthy
3 hours ago













Note that it's also possible to "have a power series" for a function (like the Taylor series) and yet have it converge to the function only at a single point. The "bathtub" function $f(x) = begincases0 & x = 0 \ e^-x^2 & x ne 0endcases$ is a good example: its Taylor series $T$ at $x = 0$ exists and has all coefficients zero, but $f(x) = T(x)$ only for $x = 0$.
– John Hughes
8 mins ago




Note that it's also possible to "have a power series" for a function (like the Taylor series) and yet have it converge to the function only at a single point. The "bathtub" function $f(x) = begincases0 & x = 0 \ e^-x^2 & x ne 0endcases$ is a good example: its Taylor series $T$ at $x = 0$ exists and has all coefficients zero, but $f(x) = T(x)$ only for $x = 0$.
– John Hughes
8 mins ago










3 Answers
3






active

oldest

votes

















up vote
3
down vote













$$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.






share|cite|improve this answer




















  • Yes, but because I have the uniform convergence the switch of limits should work?
    – auhasard
    3 hours ago






  • 1




    @auhasard It should? Why?
    – 5xum
    3 hours ago

















up vote
0
down vote













What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?






share|cite|improve this answer




















  • While the title doesn't say it, the second sentence of the question itself limits the discussion to $C^infty$ functions; even then, it's a quite interesting question.
    – John Hughes
    11 mins ago


















up vote
0
down vote













Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.






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',
    convertImagesToLinks: true,
    noModals: false,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    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%2f2951249%2fwhy-doesnt-the-stone-weierstrass-theorem-imply-that-every-function-has-a-power%23new-answer', 'question_page');

    );

    Post as a guest

















    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes








    up vote
    3
    down vote













    $$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
    and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.






    share|cite|improve this answer




















    • Yes, but because I have the uniform convergence the switch of limits should work?
      – auhasard
      3 hours ago






    • 1




      @auhasard It should? Why?
      – 5xum
      3 hours ago














    up vote
    3
    down vote













    $$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
    and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.






    share|cite|improve this answer




















    • Yes, but because I have the uniform convergence the switch of limits should work?
      – auhasard
      3 hours ago






    • 1




      @auhasard It should? Why?
      – 5xum
      3 hours ago












    up vote
    3
    down vote










    up vote
    3
    down vote









    $$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
    and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.






    share|cite|improve this answer












    $$lim_ntoinftyleft(lim_ktoinfty a_n,kright)$$ is, in general, not the same as $$lim_ktoinftyleft(lim_ntoinfty a_n,kright)$$
    and in order to switch the order of your infinite sum (which is in its definition a limit) and your limit, you would need something like that.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 3 hours ago









    5xum

    85.2k388154




    85.2k388154











    • Yes, but because I have the uniform convergence the switch of limits should work?
      – auhasard
      3 hours ago






    • 1




      @auhasard It should? Why?
      – 5xum
      3 hours ago
















    • Yes, but because I have the uniform convergence the switch of limits should work?
      – auhasard
      3 hours ago






    • 1




      @auhasard It should? Why?
      – 5xum
      3 hours ago















    Yes, but because I have the uniform convergence the switch of limits should work?
    – auhasard
    3 hours ago




    Yes, but because I have the uniform convergence the switch of limits should work?
    – auhasard
    3 hours ago




    1




    1




    @auhasard It should? Why?
    – 5xum
    3 hours ago




    @auhasard It should? Why?
    – 5xum
    3 hours ago










    up vote
    0
    down vote













    What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?






    share|cite|improve this answer




















    • While the title doesn't say it, the second sentence of the question itself limits the discussion to $C^infty$ functions; even then, it's a quite interesting question.
      – John Hughes
      11 mins ago















    up vote
    0
    down vote













    What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?






    share|cite|improve this answer




















    • While the title doesn't say it, the second sentence of the question itself limits the discussion to $C^infty$ functions; even then, it's a quite interesting question.
      – John Hughes
      11 mins ago













    up vote
    0
    down vote










    up vote
    0
    down vote









    What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?






    share|cite|improve this answer












    What about the absolute-value function $xmapsto|x|$ on the interval $[-1,1]$. Can you give us a power series for that?







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered 40 mins ago









    hartkp

    68134




    68134











    • While the title doesn't say it, the second sentence of the question itself limits the discussion to $C^infty$ functions; even then, it's a quite interesting question.
      – John Hughes
      11 mins ago

















    • While the title doesn't say it, the second sentence of the question itself limits the discussion to $C^infty$ functions; even then, it's a quite interesting question.
      – John Hughes
      11 mins ago
















    While the title doesn't say it, the second sentence of the question itself limits the discussion to $C^infty$ functions; even then, it's a quite interesting question.
    – John Hughes
    11 mins ago





    While the title doesn't say it, the second sentence of the question itself limits the discussion to $C^infty$ functions; even then, it's a quite interesting question.
    – John Hughes
    11 mins ago











    up vote
    0
    down vote













    Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.






    share|cite|improve this answer
























      up vote
      0
      down vote













      Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.






      share|cite|improve this answer






















        up vote
        0
        down vote










        up vote
        0
        down vote









        Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.






        share|cite|improve this answer












        Short answer. The sequence of polynomials guaranteed by the Stone Weierstrass theorem may not be constructible by appending terms of higher and higher order. The early coefficients can vary as the sequence grows. So you don't have the sequence of partial sums of a power series.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered 37 mins ago









        Ethan Bolker

        37.2k54299




        37.2k54299



























             

            draft saved


            draft discarded















































             


            draft saved


            draft discarded














            StackExchange.ready(
            function ()
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2951249%2fwhy-doesnt-the-stone-weierstrass-theorem-imply-that-every-function-has-a-power%23new-answer', 'question_page');

            );

            Post as a guest
































































            Comments

            Post a Comment

            Popular posts from this blog

            What does second last employer means? [closed]

            Image
            Clash Royale CLAN TAG #URR8PPP .everyoneloves__top-leaderboard:empty,.everyoneloves__mid-leaderboard:empty margin-bottom:0; up vote -3 down vote favorite Say, for example I have worked in the following companies, Company A => 2010 - 2011 Company B => 2011 - 2012 Company C => 2012 - Till Date Company D => Have offer Now assume Company D is asking to submit documents from my second last employer. So from the list of companies given, which company would be qualified as second last employer? software-industry employer share | improve this question asked Aug 10 '14 at 5:16 Rajaprabhu Aravindasamy 559 1 6 13 closed as off-topic by Jim G., jmort253 ♦ Aug 11 '14 at 1:58 This question appears to be off-topic. The users who voted to close gave this specific reason: "Questions seeking advice on company-specific regulations, agreements, or policies should be directed to your manager or HR department. Questions that address...
            Read more

            List of Gilmore Girls characters

            Image
            The season three cast—(Back row): Lane, Michel, Paris, Emily, Richard, Sookie, Miss Patty, Kirk; Front row: Jess, Luke, Lorelai, Rory, Dean This is a list of characters for the comedy-drama television series Gilmore Girls . Contents 1 Main characters 1.1 Lorelai Gilmore 1.2 Rory Gilmore 1.3 Sookie St. James 1.4 Lane Kim 1.5 Michel Gerard 1.6 Luke Danes 1.7 Emily Gilmore 1.8 Richard Gilmore 1.9 Paris Geller 1.10 Dean Forester 1.11 Jess Mariano 1.12 Kirk Gleason 1.13 Jason Stiles 1.14 Logan Huntzberger 2 Major recurring characters 2.1 Christopher Hayden 2.2 Jackson Belleville 2.3 Mrs. Kim 2.4 Tristin DuGray 2.5 Max Medina 2.6 Taylor Doose 2.7 Dave Rygalski 2.8 Marty 3 Recurring characters 3.1 Stars Hollow 3.2 Chilton 3.3 Yale 3.4 Other 4 Notable guest stars 5 Unseen characters 6 Notes 7 References Main characters Actor Character Appearances Season 1 Season 2 Season 3 Season 4 Season 5 Season 6 Season 7 A Year in the Life Lauren Graham Lorelai Gilmore...
            Read more

            One-line joke

            A one-liner is a joke that is delivered in a single line. A good one-liner is said to be pithy - concise and meaningful. [1] Comedians and actors use this comedic method as part of their act, e.g. Jimmy Carr, Tommy Cooper, Rodney Dangerfield, Norm Macdonald, Ken Dodd, Stewart Francis, Zach Galifianakis, Mitch Hedberg, Anthony Jeselnik, Milton Jones, Shappi Khorsandi, Jay London, Mark Linn-Baker, Demetri Martin, Groucho Marx, Dan Mintz, Emo Philips, Tim Vine, Steven Wright and Henny Youngman and Arnold Schwarzenegger. [ citation needed ] Many fictional characters are also known to deliver one-liners, including James Bond, who usually includes pithy and laconic quips after disposing of a villain. [ citation needed ] Examples "A baby seal walks into a club." "A dyslexic man walks into a bra." (Anonymous) "There are three types of people, those who can count and those who can't." "Venison's dear, isn't it?" (Jimmy Carr) "An escala...
            Read more