Why does $sum_{ngeq0}(1-x)^n=frac1x$ have such a poor radius of convergence?












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I am confused as to why $$sum_{ngeq0}(1-x)^n=frac1x$$
only works for $xin (0,2)$. I get that it has a singularity at $x=0$, so that can't work, but there are no singularities for the rest of the positive real line. Why isn't there a power series representation of $frac1x$ which works for the whole positive real line? Are there any series representations of $1/x$ which work for $xin (0,infty)$? I can't find any. Thanks.






sequences-and-series convergence power-series taylor-expansion







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  • $begingroup$
    The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
    $endgroup$
    – Andreas Blass
    Nov 30 '18 at 18:27






  • 3




    $begingroup$
    $sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 18:28










  • $begingroup$
    $1/x$ has a pole at $x=0$.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 18:28






  • 1




    $begingroup$
    More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
    $endgroup$
    – user3482749
    Nov 30 '18 at 18:31
















5












$begingroup$


I am confused as to why $$sum_{ngeq0}(1-x)^n=frac1x$$
only works for $xin (0,2)$. I get that it has a singularity at $x=0$, so that can't work, but there are no singularities for the rest of the positive real line. Why isn't there a power series representation of $frac1x$ which works for the whole positive real line? Are there any series representations of $1/x$ which work for $xin (0,infty)$? I can't find any. Thanks.






sequences-and-series convergence power-series taylor-expansion







share|cite|improve this question











$endgroup$












  • $begingroup$
    The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
    $endgroup$
    – Andreas Blass
    Nov 30 '18 at 18:27






  • 3




    $begingroup$
    $sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 18:28










  • $begingroup$
    $1/x$ has a pole at $x=0$.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 18:28






  • 1




    $begingroup$
    More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
    $endgroup$
    – user3482749
    Nov 30 '18 at 18:31














5












5








5


1



$begingroup$


I am confused as to why $$sum_{ngeq0}(1-x)^n=frac1x$$
only works for $xin (0,2)$. I get that it has a singularity at $x=0$, so that can't work, but there are no singularities for the rest of the positive real line. Why isn't there a power series representation of $frac1x$ which works for the whole positive real line? Are there any series representations of $1/x$ which work for $xin (0,infty)$? I can't find any. Thanks.






sequences-and-series convergence power-series taylor-expansion







share|cite|improve this question











$endgroup$




I am confused as to why $$sum_{ngeq0}(1-x)^n=frac1x$$
only works for $xin (0,2)$. I get that it has a singularity at $x=0$, so that can't work, but there are no singularities for the rest of the positive real line. Why isn't there a power series representation of $frac1x$ which works for the whole positive real line? Are there any series representations of $1/x$ which work for $xin (0,infty)$? I can't find any. Thanks.






sequences-and-series convergence power-series taylor-expansion




sequences-and-series convergence power-series taylor-expansion






share|cite|improve this question















share|cite|improve this question













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edited Dec 1 '18 at 7:31









Asaf Karagila♦

302k32427757




302k32427757










asked Nov 30 '18 at 18:26









clathratusclathratus

3,541332




3,541332












  • $begingroup$
    The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
    $endgroup$
    – Andreas Blass
    Nov 30 '18 at 18:27






  • 3




    $begingroup$
    $sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 18:28










  • $begingroup$
    $1/x$ has a pole at $x=0$.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 18:28






  • 1




    $begingroup$
    More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
    $endgroup$
    – user3482749
    Nov 30 '18 at 18:31


















  • $begingroup$
    The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
    $endgroup$
    – Andreas Blass
    Nov 30 '18 at 18:27






  • 3




    $begingroup$
    $sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 18:28










  • $begingroup$
    $1/x$ has a pole at $x=0$.
    $endgroup$
    – Lord Shark the Unknown
    Nov 30 '18 at 18:28






  • 1




    $begingroup$
    More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
    $endgroup$
    – user3482749
    Nov 30 '18 at 18:31
















$begingroup$
The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
$endgroup$
– Andreas Blass
Nov 30 '18 at 18:27




$begingroup$
The domain of convergence of a real power series is always either a finitely long interval or the whole real line.
$endgroup$
– Andreas Blass
Nov 30 '18 at 18:27




3




3




$begingroup$
$sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
$endgroup$
– Lord Shark the Unknown
Nov 30 '18 at 18:28




$begingroup$
$sum r^n$ converges iff $-1<r<1$. If $r=1-x$ that's the same as $0<x<2$.
$endgroup$
– Lord Shark the Unknown
Nov 30 '18 at 18:28












$begingroup$
$1/x$ has a pole at $x=0$.
$endgroup$
– Lord Shark the Unknown
Nov 30 '18 at 18:28




$begingroup$
$1/x$ has a pole at $x=0$.
$endgroup$
– Lord Shark the Unknown
Nov 30 '18 at 18:28




1




1




$begingroup$
More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
$endgroup$
– user3482749
Nov 30 '18 at 18:31




$begingroup$
More specifically: their intervals of convergence are symmetric around the point of evaluation, which is $1$ here, so the Taylor series does not converge beyond $x = 2$ (and you can check that it doesn't converge at $x = 2$ either).
$endgroup$
– user3482749
Nov 30 '18 at 18:31










4 Answers
4






active

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9












$begingroup$

The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
at all points farther away from the centre than that point, even though the function may be analytic at those other points.



On the other hand, you could take the series
$$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
Of course, this is not a power series in the usual sense.






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$endgroup$













  • $begingroup$
    What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
    $endgroup$
    – clathratus
    Nov 30 '18 at 21:36








  • 2




    $begingroup$
    @clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
    $endgroup$
    – adfriedman
    Nov 30 '18 at 22:51












  • $begingroup$
    @adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
    $endgroup$
    – clathratus
    Nov 30 '18 at 23:20








  • 2




    $begingroup$
    No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
    $endgroup$
    – adfriedman
    Nov 30 '18 at 23:29












  • $begingroup$
    @adfriedman Thank you for the clarification. I understand now.
    $endgroup$
    – clathratus
    Dec 1 '18 at 2:31



















4












$begingroup$

The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
exponents, including Taylor series, does not have singularities in the finite complex plane
and consequently cannot reproduce them.



A function $ f(z)$ , which is analytic in a domain of the complex plane,
can be developed (by definition) in power series around a point in that domain, and the radius
of convergence of the series will be equal to the distance of that point from the nearest
singularity, of course excluded.



Thus, for $1/z$ which has a simple pole at $z=0$ :

- when developed at $z=1$ will have a convergence radius of $ 1$;

- to get a larger radius( $R$), you shall develop it around $z_0=R$ ;

- there is no possibility to develop it so as to encompass positive and negative real values of it.



The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.






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$endgroup$





















    2












    $begingroup$

    The geometric series $sum_{ngeq0}(1-x)^n$ converges for



    $$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$



    and we have



    $$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$






    share|cite|improve this answer









    $endgroup$





















      2












      $begingroup$

      The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:





      • ${a}$;


      • $(a-r,a+r)$, for some $rin(0,+infty)$;


      • $(a-r,a+r]$, for some $rin(0,+infty)$;


      • $[a-r,a+r)$, for some $rin(0,+infty)$;


      • $[a-r,a+r]$, for some $rin(0,+infty)$;


      • $mathbb R$.


      Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.






      share|cite|improve this answer









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        4 Answers
        4






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        4 Answers
        4






        active

        oldest

        votes









        active

        oldest

        votes






        active

        oldest

        votes









        9












        $begingroup$

        The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
        at all points farther away from the centre than that point, even though the function may be analytic at those other points.



        On the other hand, you could take the series
        $$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
        which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
        Of course, this is not a power series in the usual sense.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
          $endgroup$
          – clathratus
          Nov 30 '18 at 21:36








        • 2




          $begingroup$
          @clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
          $endgroup$
          – adfriedman
          Nov 30 '18 at 22:51












        • $begingroup$
          @adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
          $endgroup$
          – clathratus
          Nov 30 '18 at 23:20








        • 2




          $begingroup$
          No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
          $endgroup$
          – adfriedman
          Nov 30 '18 at 23:29












        • $begingroup$
          @adfriedman Thank you for the clarification. I understand now.
          $endgroup$
          – clathratus
          Dec 1 '18 at 2:31
















        9












        $begingroup$

        The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
        at all points farther away from the centre than that point, even though the function may be analytic at those other points.



        On the other hand, you could take the series
        $$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
        which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
        Of course, this is not a power series in the usual sense.






        share|cite|improve this answer











        $endgroup$













        • $begingroup$
          What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
          $endgroup$
          – clathratus
          Nov 30 '18 at 21:36








        • 2




          $begingroup$
          @clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
          $endgroup$
          – adfriedman
          Nov 30 '18 at 22:51












        • $begingroup$
          @adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
          $endgroup$
          – clathratus
          Nov 30 '18 at 23:20








        • 2




          $begingroup$
          No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
          $endgroup$
          – adfriedman
          Nov 30 '18 at 23:29












        • $begingroup$
          @adfriedman Thank you for the clarification. I understand now.
          $endgroup$
          – clathratus
          Dec 1 '18 at 2:31














        9












        9








        9





        $begingroup$

        The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
        at all points farther away from the centre than that point, even though the function may be analytic at those other points.



        On the other hand, you could take the series
        $$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
        which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
        Of course, this is not a power series in the usual sense.






        share|cite|improve this answer











        $endgroup$



        The radius of convergence of a power series in $z-a$ is the largest $r$ such that the sum of the series is analytic in the disk ${z: |z-a| < r}$ in the complex plane. Thus a singularity at some point stops the series from converging
        at all points farther away from the centre than that point, even though the function may be analytic at those other points.



        On the other hand, you could take the series
        $$ frac{1}{x} = -1 + sum_{n=1}^infty left(frac{1+x}{1+2x}right)^n$$
        which converges for all $x > 0$ (in fact, everywhere outside a circle of radius $1/3$ centred at $-1/3$ in the complex plane).
        Of course, this is not a power series in the usual sense.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Nov 30 '18 at 18:47

























        answered Nov 30 '18 at 18:35









        Robert IsraelRobert Israel

        319k23209460




        319k23209460












        • $begingroup$
          What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
          $endgroup$
          – clathratus
          Nov 30 '18 at 21:36








        • 2




          $begingroup$
          @clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
          $endgroup$
          – adfriedman
          Nov 30 '18 at 22:51












        • $begingroup$
          @adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
          $endgroup$
          – clathratus
          Nov 30 '18 at 23:20








        • 2




          $begingroup$
          No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
          $endgroup$
          – adfriedman
          Nov 30 '18 at 23:29












        • $begingroup$
          @adfriedman Thank you for the clarification. I understand now.
          $endgroup$
          – clathratus
          Dec 1 '18 at 2:31


















        • $begingroup$
          What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
          $endgroup$
          – clathratus
          Nov 30 '18 at 21:36








        • 2




          $begingroup$
          @clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
          $endgroup$
          – adfriedman
          Nov 30 '18 at 22:51












        • $begingroup$
          @adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
          $endgroup$
          – clathratus
          Nov 30 '18 at 23:20








        • 2




          $begingroup$
          No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
          $endgroup$
          – adfriedman
          Nov 30 '18 at 23:29












        • $begingroup$
          @adfriedman Thank you for the clarification. I understand now.
          $endgroup$
          – clathratus
          Dec 1 '18 at 2:31
















        $begingroup$
        What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
        $endgroup$
        – clathratus
        Nov 30 '18 at 21:36






        $begingroup$
        What type of series is $$-1+sum_{ngeq0}bigg(frac{1+x}{1+2x}bigg)^n$$ and how do you find it? Is this type of series generally better at approximating rational functions?
        $endgroup$
        – clathratus
        Nov 30 '18 at 21:36






        2




        2




        $begingroup$
        @clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
        $endgroup$
        – adfriedman
        Nov 30 '18 at 22:51






        $begingroup$
        @clathratus It's just a geometric series where $y=frac{1+x}{1+2x}$. It is a special case for $a=2$ of: $$frac{1}{x}+a = frac{1+ax}{x}\ = frac{1+ax}{1+ax - (1+ax-x)}\ = frac{1}{1 - frac{1+(a-1)x}{1+ax}}\ =sum_{n=0}^infty left(frac{1+(a-1)x}{1+ax}right)^n $$ Note that the series given in the posted answer does not include the term $n=0$ as you wrote in your comment.
        $endgroup$
        – adfriedman
        Nov 30 '18 at 22:51














        $begingroup$
        @adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
        $endgroup$
        – clathratus
        Nov 30 '18 at 23:20






        $begingroup$
        @adfriedman I see. So the $$sum_{n=0}^{infty}$$ in your comment is supposed to be different too?
        $endgroup$
        – clathratus
        Nov 30 '18 at 23:20






        2




        2




        $begingroup$
        No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
        $endgroup$
        – adfriedman
        Nov 30 '18 at 23:29






        $begingroup$
        No, I wrote what I meant. you could rewrite mine as $$frac{1}{x} = 1-a + sum_{n=1}^infty left(frac{1+(a-1)x}{1+ax}right)^n$$ to be consistent if you'd like. (The $n=0$ term is just $1$, so take your pick on whether you want to index it or not)
        $endgroup$
        – adfriedman
        Nov 30 '18 at 23:29














        $begingroup$
        @adfriedman Thank you for the clarification. I understand now.
        $endgroup$
        – clathratus
        Dec 1 '18 at 2:31




        $begingroup$
        @adfriedman Thank you for the clarification. I understand now.
        $endgroup$
        – clathratus
        Dec 1 '18 at 2:31











        4












        $begingroup$

        The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
        exponents, including Taylor series, does not have singularities in the finite complex plane
        and consequently cannot reproduce them.



        A function $ f(z)$ , which is analytic in a domain of the complex plane,
        can be developed (by definition) in power series around a point in that domain, and the radius
        of convergence of the series will be equal to the distance of that point from the nearest
        singularity, of course excluded.



        Thus, for $1/z$ which has a simple pole at $z=0$ :

        - when developed at $z=1$ will have a convergence radius of $ 1$;

        - to get a larger radius( $R$), you shall develop it around $z_0=R$ ;

        - there is no possibility to develop it so as to encompass positive and negative real values of it.



        The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.






        share|cite|improve this answer











        $endgroup$


















          4












          $begingroup$

          The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
          exponents, including Taylor series, does not have singularities in the finite complex plane
          and consequently cannot reproduce them.



          A function $ f(z)$ , which is analytic in a domain of the complex plane,
          can be developed (by definition) in power series around a point in that domain, and the radius
          of convergence of the series will be equal to the distance of that point from the nearest
          singularity, of course excluded.



          Thus, for $1/z$ which has a simple pole at $z=0$ :

          - when developed at $z=1$ will have a convergence radius of $ 1$;

          - to get a larger radius( $R$), you shall develop it around $z_0=R$ ;

          - there is no possibility to develop it so as to encompass positive and negative real values of it.



          The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.






          share|cite|improve this answer











          $endgroup$
















            4












            4








            4





            $begingroup$

            The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
            exponents, including Taylor series, does not have singularities in the finite complex plane
            and consequently cannot reproduce them.



            A function $ f(z)$ , which is analytic in a domain of the complex plane,
            can be developed (by definition) in power series around a point in that domain, and the radius
            of convergence of the series will be equal to the distance of that point from the nearest
            singularity, of course excluded.



            Thus, for $1/z$ which has a simple pole at $z=0$ :

            - when developed at $z=1$ will have a convergence radius of $ 1$;

            - to get a larger radius( $R$), you shall develop it around $z_0=R$ ;

            - there is no possibility to develop it so as to encompass positive and negative real values of it.



            The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.






            share|cite|improve this answer











            $endgroup$



            The basic answer to you question is that a "normal" power series, i.e. with non-negative integral
            exponents, including Taylor series, does not have singularities in the finite complex plane
            and consequently cannot reproduce them.



            A function $ f(z)$ , which is analytic in a domain of the complex plane,
            can be developed (by definition) in power series around a point in that domain, and the radius
            of convergence of the series will be equal to the distance of that point from the nearest
            singularity, of course excluded.



            Thus, for $1/z$ which has a simple pole at $z=0$ :

            - when developed at $z=1$ will have a convergence radius of $ 1$;

            - to get a larger radius( $R$), you shall develop it around $z_0=R$ ;

            - there is no possibility to develop it so as to encompass positive and negative real values of it.



            The Laurent series for $1/z$ at $z_0=0$ is infact $1/z$, i.e. the function itself.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited Dec 1 '18 at 16:53

























            answered Nov 30 '18 at 21:29









            G CabG Cab

            18.1k31237




            18.1k31237























                2












                $begingroup$

                The geometric series $sum_{ngeq0}(1-x)^n$ converges for



                $$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$



                and we have



                $$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$






                share|cite|improve this answer









                $endgroup$


















                  2












                  $begingroup$

                  The geometric series $sum_{ngeq0}(1-x)^n$ converges for



                  $$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$



                  and we have



                  $$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$






                  share|cite|improve this answer









                  $endgroup$
















                    2












                    2








                    2





                    $begingroup$

                    The geometric series $sum_{ngeq0}(1-x)^n$ converges for



                    $$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$



                    and we have



                    $$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$






                    share|cite|improve this answer









                    $endgroup$



                    The geometric series $sum_{ngeq0}(1-x)^n$ converges for



                    $$|1-x|<1 iff -1<1-x<1 iff 0<x<2$$



                    and we have



                    $$sum_{ngeq0}(1-x)^n=frac{1}{1-(1-x)}=frac1x$$







                    share|cite|improve this answer












                    share|cite|improve this answer



                    share|cite|improve this answer










                    answered Nov 30 '18 at 18:29









                    gimusigimusi

                    1




                    1























                        2












                        $begingroup$

                        The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:





                        • ${a}$;


                        • $(a-r,a+r)$, for some $rin(0,+infty)$;


                        • $(a-r,a+r]$, for some $rin(0,+infty)$;


                        • $[a-r,a+r)$, for some $rin(0,+infty)$;


                        • $[a-r,a+r]$, for some $rin(0,+infty)$;


                        • $mathbb R$.


                        Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.






                        share|cite|improve this answer









                        $endgroup$


















                          2












                          $begingroup$

                          The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:





                          • ${a}$;


                          • $(a-r,a+r)$, for some $rin(0,+infty)$;


                          • $(a-r,a+r]$, for some $rin(0,+infty)$;


                          • $[a-r,a+r)$, for some $rin(0,+infty)$;


                          • $[a-r,a+r]$, for some $rin(0,+infty)$;


                          • $mathbb R$.


                          Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.






                          share|cite|improve this answer









                          $endgroup$
















                            2












                            2








                            2





                            $begingroup$

                            The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:





                            • ${a}$;


                            • $(a-r,a+r)$, for some $rin(0,+infty)$;


                            • $(a-r,a+r]$, for some $rin(0,+infty)$;


                            • $[a-r,a+r)$, for some $rin(0,+infty)$;


                            • $[a-r,a+r]$, for some $rin(0,+infty)$;


                            • $mathbb R$.


                            Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.






                            share|cite|improve this answer









                            $endgroup$



                            The set of those real numbers at which a power series $displaystylesum_{n=0}^infty a_n(x-a)^n$ belongs to one of these types:





                            • ${a}$;


                            • $(a-r,a+r)$, for some $rin(0,+infty)$;


                            • $(a-r,a+r]$, for some $rin(0,+infty)$;


                            • $[a-r,a+r)$, for some $rin(0,+infty)$;


                            • $[a-r,a+r]$, for some $rin(0,+infty)$;


                            • $mathbb R$.


                            Since, $(0,+infty)$ doesn't appear here, no power series exists whose sum is $frac1x$ in $(0,+infty)$ and which diverges otherwise.







                            share|cite|improve this answer












                            share|cite|improve this answer



                            share|cite|improve this answer










                            answered Nov 30 '18 at 18:32









                            José Carlos SantosJosé Carlos Santos

                            153k22123226




                            153k22123226






























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