range of convergence of taylor approximation for $frac{1}{1-x}$












0












$begingroup$


My question concerns the function $f = frac{1}{1-x}$



The derivatives 1,2 ..(n+1) I believe are:




  • $f^{(1)} = frac{1}{(1-x)^2}$

  • $f^{(2)} = frac{2}{(1-x)^3}$

  • $f^{(n+1)} = frac{(n+1)!}{(1-x)^{n+2}}$


The taylor polynomial around $x=0$ would then be



$f(x) = sum_{k=0}^n x^k+ R_n(f,0)$



with the lagrange closing term $R_n(f,0)= frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$ and $c$ between $0$ and $x$



I hope so much is correct. Now a simple look at $lim_{nto infty} x^n$ confirms the idea that the terms will go to $0$ as $n to infty$ if and only if $|x|lt 1$ but somehow when I try to proof that using $lim_{nto infty} |R_n|$ I get the opposite!



$|R_n(f,0)|= lvert frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} rvert = lvert frac{frac{(n+1)!}{(1-c)^{n+2}}}{(n+1)!}x^{n+1}rvert = lvert frac{x^{n+1}}{(1-c)^{n+2}}rvert$



Taking the limit here:
$lim_{nto infty} lvert frac{x^{n+1}}{(1-c)^{n+2}}rvert = L$



Now this limit diverges for values of $|x|<1$



... I've been staring at it but did not find the mistake. Any pointers appreciated!






calculus convergence taylor-expansion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why does the limit of $left | frac {x^{n+1}} {(1-c)^{n+2}} right |$ diverge as $n rightarrow infty$ for $|x| < 1$?
    $endgroup$
    – Dbchatto67
    Jan 7 at 5:26












  • $begingroup$
    Hint $:$ $R_n (f,0) = sumlimits_{k=n+1}^{infty} x^k = frac {x^{n+1}} {1-x} rightarrow 0$ as $n rightarrow infty$ for $|x| < 1$.
    $endgroup$
    – Dbchatto67
    Jan 7 at 6:04












  • $begingroup$
    Yes, but can we derive that last line from the first ? Or is that not possible and exactly what the answer by jmerry then?
    $endgroup$
    – Chai
    Jan 7 at 7:35
















0












$begingroup$


My question concerns the function $f = frac{1}{1-x}$



The derivatives 1,2 ..(n+1) I believe are:




  • $f^{(1)} = frac{1}{(1-x)^2}$

  • $f^{(2)} = frac{2}{(1-x)^3}$

  • $f^{(n+1)} = frac{(n+1)!}{(1-x)^{n+2}}$


The taylor polynomial around $x=0$ would then be



$f(x) = sum_{k=0}^n x^k+ R_n(f,0)$



with the lagrange closing term $R_n(f,0)= frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$ and $c$ between $0$ and $x$



I hope so much is correct. Now a simple look at $lim_{nto infty} x^n$ confirms the idea that the terms will go to $0$ as $n to infty$ if and only if $|x|lt 1$ but somehow when I try to proof that using $lim_{nto infty} |R_n|$ I get the opposite!



$|R_n(f,0)|= lvert frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} rvert = lvert frac{frac{(n+1)!}{(1-c)^{n+2}}}{(n+1)!}x^{n+1}rvert = lvert frac{x^{n+1}}{(1-c)^{n+2}}rvert$



Taking the limit here:
$lim_{nto infty} lvert frac{x^{n+1}}{(1-c)^{n+2}}rvert = L$



Now this limit diverges for values of $|x|<1$



... I've been staring at it but did not find the mistake. Any pointers appreciated!






calculus convergence taylor-expansion







share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    Why does the limit of $left | frac {x^{n+1}} {(1-c)^{n+2}} right |$ diverge as $n rightarrow infty$ for $|x| < 1$?
    $endgroup$
    – Dbchatto67
    Jan 7 at 5:26












  • $begingroup$
    Hint $:$ $R_n (f,0) = sumlimits_{k=n+1}^{infty} x^k = frac {x^{n+1}} {1-x} rightarrow 0$ as $n rightarrow infty$ for $|x| < 1$.
    $endgroup$
    – Dbchatto67
    Jan 7 at 6:04












  • $begingroup$
    Yes, but can we derive that last line from the first ? Or is that not possible and exactly what the answer by jmerry then?
    $endgroup$
    – Chai
    Jan 7 at 7:35














0












0








0





$begingroup$


My question concerns the function $f = frac{1}{1-x}$



The derivatives 1,2 ..(n+1) I believe are:




  • $f^{(1)} = frac{1}{(1-x)^2}$

  • $f^{(2)} = frac{2}{(1-x)^3}$

  • $f^{(n+1)} = frac{(n+1)!}{(1-x)^{n+2}}$


The taylor polynomial around $x=0$ would then be



$f(x) = sum_{k=0}^n x^k+ R_n(f,0)$



with the lagrange closing term $R_n(f,0)= frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$ and $c$ between $0$ and $x$



I hope so much is correct. Now a simple look at $lim_{nto infty} x^n$ confirms the idea that the terms will go to $0$ as $n to infty$ if and only if $|x|lt 1$ but somehow when I try to proof that using $lim_{nto infty} |R_n|$ I get the opposite!



$|R_n(f,0)|= lvert frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} rvert = lvert frac{frac{(n+1)!}{(1-c)^{n+2}}}{(n+1)!}x^{n+1}rvert = lvert frac{x^{n+1}}{(1-c)^{n+2}}rvert$



Taking the limit here:
$lim_{nto infty} lvert frac{x^{n+1}}{(1-c)^{n+2}}rvert = L$



Now this limit diverges for values of $|x|<1$



... I've been staring at it but did not find the mistake. Any pointers appreciated!






calculus convergence taylor-expansion







share|cite|improve this question











$endgroup$




My question concerns the function $f = frac{1}{1-x}$



The derivatives 1,2 ..(n+1) I believe are:




  • $f^{(1)} = frac{1}{(1-x)^2}$

  • $f^{(2)} = frac{2}{(1-x)^3}$

  • $f^{(n+1)} = frac{(n+1)!}{(1-x)^{n+2}}$


The taylor polynomial around $x=0$ would then be



$f(x) = sum_{k=0}^n x^k+ R_n(f,0)$



with the lagrange closing term $R_n(f,0)= frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1}$ and $c$ between $0$ and $x$



I hope so much is correct. Now a simple look at $lim_{nto infty} x^n$ confirms the idea that the terms will go to $0$ as $n to infty$ if and only if $|x|lt 1$ but somehow when I try to proof that using $lim_{nto infty} |R_n|$ I get the opposite!



$|R_n(f,0)|= lvert frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} rvert = lvert frac{frac{(n+1)!}{(1-c)^{n+2}}}{(n+1)!}x^{n+1}rvert = lvert frac{x^{n+1}}{(1-c)^{n+2}}rvert$



Taking the limit here:
$lim_{nto infty} lvert frac{x^{n+1}}{(1-c)^{n+2}}rvert = L$



Now this limit diverges for values of $|x|<1$



... I've been staring at it but did not find the mistake. Any pointers appreciated!






calculus convergence taylor-expansion




calculus convergence taylor-expansion






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Jan 7 at 4:43







Chai

















asked Jan 6 at 18:59









ChaiChai

1276




1276








  • 1




    $begingroup$
    Why does the limit of $left | frac {x^{n+1}} {(1-c)^{n+2}} right |$ diverge as $n rightarrow infty$ for $|x| < 1$?
    $endgroup$
    – Dbchatto67
    Jan 7 at 5:26












  • $begingroup$
    Hint $:$ $R_n (f,0) = sumlimits_{k=n+1}^{infty} x^k = frac {x^{n+1}} {1-x} rightarrow 0$ as $n rightarrow infty$ for $|x| < 1$.
    $endgroup$
    – Dbchatto67
    Jan 7 at 6:04












  • $begingroup$
    Yes, but can we derive that last line from the first ? Or is that not possible and exactly what the answer by jmerry then?
    $endgroup$
    – Chai
    Jan 7 at 7:35














  • 1




    $begingroup$
    Why does the limit of $left | frac {x^{n+1}} {(1-c)^{n+2}} right |$ diverge as $n rightarrow infty$ for $|x| < 1$?
    $endgroup$
    – Dbchatto67
    Jan 7 at 5:26












  • $begingroup$
    Hint $:$ $R_n (f,0) = sumlimits_{k=n+1}^{infty} x^k = frac {x^{n+1}} {1-x} rightarrow 0$ as $n rightarrow infty$ for $|x| < 1$.
    $endgroup$
    – Dbchatto67
    Jan 7 at 6:04












  • $begingroup$
    Yes, but can we derive that last line from the first ? Or is that not possible and exactly what the answer by jmerry then?
    $endgroup$
    – Chai
    Jan 7 at 7:35








1




1




$begingroup$
Why does the limit of $left | frac {x^{n+1}} {(1-c)^{n+2}} right |$ diverge as $n rightarrow infty$ for $|x| < 1$?
$endgroup$
– Dbchatto67
Jan 7 at 5:26






$begingroup$
Why does the limit of $left | frac {x^{n+1}} {(1-c)^{n+2}} right |$ diverge as $n rightarrow infty$ for $|x| < 1$?
$endgroup$
– Dbchatto67
Jan 7 at 5:26














$begingroup$
Hint $:$ $R_n (f,0) = sumlimits_{k=n+1}^{infty} x^k = frac {x^{n+1}} {1-x} rightarrow 0$ as $n rightarrow infty$ for $|x| < 1$.
$endgroup$
– Dbchatto67
Jan 7 at 6:04






$begingroup$
Hint $:$ $R_n (f,0) = sumlimits_{k=n+1}^{infty} x^k = frac {x^{n+1}} {1-x} rightarrow 0$ as $n rightarrow infty$ for $|x| < 1$.
$endgroup$
– Dbchatto67
Jan 7 at 6:04














$begingroup$
Yes, but can we derive that last line from the first ? Or is that not possible and exactly what the answer by jmerry then?
$endgroup$
– Chai
Jan 7 at 7:35




$begingroup$
Yes, but can we derive that last line from the first ? Or is that not possible and exactly what the answer by jmerry then?
$endgroup$
– Chai
Jan 7 at 7:35










2 Answers
2






active

oldest

votes


















2












$begingroup$

There's nothing wrong here. The error estimate in Taylor's theorem tells us that the series $1+x+x^2+cdots$ converges to $frac1{1-x}$ for $|x|<frac12$. This is true. The fact that the series also converges to that value for $frac12le |x|<1$ (because, you know, it's a geometric series that we can calculate the partial sums of exactly) is irrelevant.



The error estimate in Taylor's theorem is not always sharp; the derivative that goes into it can vary greatly, especially in larger intervals. It is normal for the error estimate from the theorem to lead us to an interval of convergence that isn't the largest possible.






share|cite|improve this answer









$endgroup$





















    0












    $begingroup$

    This is a very well know function that is used A LOT!



    Some hints to get you going



    (1) Find the general form for $f^n(x)$ at $x = 0$ to form your Maclaurin Series.



    (2) Apply the ratio test to find the region of convergence for $x$.



    Btw, you are definitely on the right track!






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thx! Yes I confirmed with the ratio test that in fact the convergence is as stated in the book. I just did not manage to get the same result with the lagrange term
      $endgroup$
      – Chai
      Jan 7 at 7:03











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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    There's nothing wrong here. The error estimate in Taylor's theorem tells us that the series $1+x+x^2+cdots$ converges to $frac1{1-x}$ for $|x|<frac12$. This is true. The fact that the series also converges to that value for $frac12le |x|<1$ (because, you know, it's a geometric series that we can calculate the partial sums of exactly) is irrelevant.



    The error estimate in Taylor's theorem is not always sharp; the derivative that goes into it can vary greatly, especially in larger intervals. It is normal for the error estimate from the theorem to lead us to an interval of convergence that isn't the largest possible.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      There's nothing wrong here. The error estimate in Taylor's theorem tells us that the series $1+x+x^2+cdots$ converges to $frac1{1-x}$ for $|x|<frac12$. This is true. The fact that the series also converges to that value for $frac12le |x|<1$ (because, you know, it's a geometric series that we can calculate the partial sums of exactly) is irrelevant.



      The error estimate in Taylor's theorem is not always sharp; the derivative that goes into it can vary greatly, especially in larger intervals. It is normal for the error estimate from the theorem to lead us to an interval of convergence that isn't the largest possible.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        There's nothing wrong here. The error estimate in Taylor's theorem tells us that the series $1+x+x^2+cdots$ converges to $frac1{1-x}$ for $|x|<frac12$. This is true. The fact that the series also converges to that value for $frac12le |x|<1$ (because, you know, it's a geometric series that we can calculate the partial sums of exactly) is irrelevant.



        The error estimate in Taylor's theorem is not always sharp; the derivative that goes into it can vary greatly, especially in larger intervals. It is normal for the error estimate from the theorem to lead us to an interval of convergence that isn't the largest possible.






        share|cite|improve this answer









        $endgroup$



        There's nothing wrong here. The error estimate in Taylor's theorem tells us that the series $1+x+x^2+cdots$ converges to $frac1{1-x}$ for $|x|<frac12$. This is true. The fact that the series also converges to that value for $frac12le |x|<1$ (because, you know, it's a geometric series that we can calculate the partial sums of exactly) is irrelevant.



        The error estimate in Taylor's theorem is not always sharp; the derivative that goes into it can vary greatly, especially in larger intervals. It is normal for the error estimate from the theorem to lead us to an interval of convergence that isn't the largest possible.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 7 at 5:51









        jmerryjmerry

        16k1633




        16k1633























            0












            $begingroup$

            This is a very well know function that is used A LOT!



            Some hints to get you going



            (1) Find the general form for $f^n(x)$ at $x = 0$ to form your Maclaurin Series.



            (2) Apply the ratio test to find the region of convergence for $x$.



            Btw, you are definitely on the right track!






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thx! Yes I confirmed with the ratio test that in fact the convergence is as stated in the book. I just did not manage to get the same result with the lagrange term
              $endgroup$
              – Chai
              Jan 7 at 7:03
















            0












            $begingroup$

            This is a very well know function that is used A LOT!



            Some hints to get you going



            (1) Find the general form for $f^n(x)$ at $x = 0$ to form your Maclaurin Series.



            (2) Apply the ratio test to find the region of convergence for $x$.



            Btw, you are definitely on the right track!






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thx! Yes I confirmed with the ratio test that in fact the convergence is as stated in the book. I just did not manage to get the same result with the lagrange term
              $endgroup$
              – Chai
              Jan 7 at 7:03














            0












            0








            0





            $begingroup$

            This is a very well know function that is used A LOT!



            Some hints to get you going



            (1) Find the general form for $f^n(x)$ at $x = 0$ to form your Maclaurin Series.



            (2) Apply the ratio test to find the region of convergence for $x$.



            Btw, you are definitely on the right track!






            share|cite|improve this answer









            $endgroup$



            This is a very well know function that is used A LOT!



            Some hints to get you going



            (1) Find the general form for $f^n(x)$ at $x = 0$ to form your Maclaurin Series.



            (2) Apply the ratio test to find the region of convergence for $x$.



            Btw, you are definitely on the right track!







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Jan 7 at 5:10









            DavidGDavidG

            1




            1












            • $begingroup$
              Thx! Yes I confirmed with the ratio test that in fact the convergence is as stated in the book. I just did not manage to get the same result with the lagrange term
              $endgroup$
              – Chai
              Jan 7 at 7:03


















            • $begingroup$
              Thx! Yes I confirmed with the ratio test that in fact the convergence is as stated in the book. I just did not manage to get the same result with the lagrange term
              $endgroup$
              – Chai
              Jan 7 at 7:03
















            $begingroup$
            Thx! Yes I confirmed with the ratio test that in fact the convergence is as stated in the book. I just did not manage to get the same result with the lagrange term
            $endgroup$
            – Chai
            Jan 7 at 7:03




            $begingroup$
            Thx! Yes I confirmed with the ratio test that in fact the convergence is as stated in the book. I just did not manage to get the same result with the lagrange term
            $endgroup$
            – Chai
            Jan 7 at 7:03


















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