Taylor expansion of ln(1-x) Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Basic Taylor expansion questionStuck on Taylor expansion questionTaylor expansion of the Error functionUsing substitution while using taylor expansionTaylor expansion of a matrix to scalar functionTaylor expansion of $log(x - x^2)$ at 0?Taylor expansion of $(1-x)(1-y)$.Taylor Expansion of Eigenvector PerturbationTaylor expansion of $ln(1 + frac2^xn)$How to see the following Taylor expansion?

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Taylor expansion of ln(1-x)



Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Basic Taylor expansion questionStuck on Taylor expansion questionTaylor expansion of the Error functionUsing substitution while using taylor expansionTaylor expansion of a matrix to scalar functionTaylor expansion of $log(x - x^2)$ at 0?Taylor expansion of $(1-x)(1-y)$.Taylor Expansion of Eigenvector PerturbationTaylor expansion of $ln(1 + frac2^xn)$How to see the following Taylor expansion?










3












$begingroup$


I was just wondering where the minus sign in the first term of the Taylor expansion of $ ln(1-x) $ comes from? In wikipedia page and everywhere else $ln(1-x)$ is given by
$$
ln(1-x) = -x-dots
$$

But assuming $x$ is small and expand around $1$, I got
$$
ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
$$

Using the definition of Taylor expansion $f(z) approx f(a) + fracdf(z)dzbiggvert_z=a(z-a) $, where here $z=1-x$, $f(z) = ln(1-z)$ and $a=1$.



I know you can get $ln(1-x) approx -x$ by e.g. substitute $xrightarrow -x$ into the expansion of $ln(1+x)$ and through other methods etc. But I still don't quite get how you can get the minus sign from Taylor expansion alone. Thanks.










share|cite|improve this question







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Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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    3












    $begingroup$


    I was just wondering where the minus sign in the first term of the Taylor expansion of $ ln(1-x) $ comes from? In wikipedia page and everywhere else $ln(1-x)$ is given by
    $$
    ln(1-x) = -x-dots
    $$

    But assuming $x$ is small and expand around $1$, I got
    $$
    ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
    $$

    Using the definition of Taylor expansion $f(z) approx f(a) + fracdf(z)dzbiggvert_z=a(z-a) $, where here $z=1-x$, $f(z) = ln(1-z)$ and $a=1$.



    I know you can get $ln(1-x) approx -x$ by e.g. substitute $xrightarrow -x$ into the expansion of $ln(1+x)$ and through other methods etc. But I still don't quite get how you can get the minus sign from Taylor expansion alone. Thanks.










    share|cite|improve this question







    New contributor




    Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.







    $endgroup$














      3












      3








      3





      $begingroup$


      I was just wondering where the minus sign in the first term of the Taylor expansion of $ ln(1-x) $ comes from? In wikipedia page and everywhere else $ln(1-x)$ is given by
      $$
      ln(1-x) = -x-dots
      $$

      But assuming $x$ is small and expand around $1$, I got
      $$
      ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
      $$

      Using the definition of Taylor expansion $f(z) approx f(a) + fracdf(z)dzbiggvert_z=a(z-a) $, where here $z=1-x$, $f(z) = ln(1-z)$ and $a=1$.



      I know you can get $ln(1-x) approx -x$ by e.g. substitute $xrightarrow -x$ into the expansion of $ln(1+x)$ and through other methods etc. But I still don't quite get how you can get the minus sign from Taylor expansion alone. Thanks.










      share|cite|improve this question







      New contributor




      Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.







      $endgroup$




      I was just wondering where the minus sign in the first term of the Taylor expansion of $ ln(1-x) $ comes from? In wikipedia page and everywhere else $ln(1-x)$ is given by
      $$
      ln(1-x) = -x-dots
      $$

      But assuming $x$ is small and expand around $1$, I got
      $$
      ln(1-x) approx ln(1) + fracd(ln(1-x))dxbiggvert_x=0[(1-x)-1] approx 0 + frac11-xbiggvert_x=0(-1)(-x) = x.
      $$

      Using the definition of Taylor expansion $f(z) approx f(a) + fracdf(z)dzbiggvert_z=a(z-a) $, where here $z=1-x$, $f(z) = ln(1-z)$ and $a=1$.



      I know you can get $ln(1-x) approx -x$ by e.g. substitute $xrightarrow -x$ into the expansion of $ln(1+x)$ and through other methods etc. But I still don't quite get how you can get the minus sign from Taylor expansion alone. Thanks.







      calculus






      share|cite|improve this question







      New contributor




      Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      share|cite|improve this question







      New contributor




      Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.









      share|cite|improve this question




      share|cite|improve this question






      New contributor




      Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      asked 3 hours ago









      LepnakLepnak

      182




      182




      New contributor




      Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      New contributor





      Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.




















          2 Answers
          2






          active

          oldest

          votes


















          1












          $begingroup$

          If one considers
          $$
          f(x)=ln (1-x),qquad |x|<1,
          $$
          one has
          $$
          f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
          $$
          giving, by the Taylor expansion,
          $$
          f(x)=0-x-fracx^22+O(x^3)
          $$
          as $x to 0$.






          share|cite|improve this answer











          $endgroup$












          • $begingroup$
            Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
            $endgroup$
            – Lepnak
            2 hours ago










          • $begingroup$
            The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
            $endgroup$
            – Minus One-Twelfth
            2 hours ago











          • $begingroup$
            Hmm I think I see what I did wrong. Thanks for all your answers.
            $endgroup$
            – Lepnak
            1 hour ago


















          2












          $begingroup$

          $$y=ln(1-x)$$
          $$y'=-frac11-x=-sum_n=0^inftyx^n$$
          so
          $$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$






          share|cite|improve this answer











          $endgroup$













            Your Answer








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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            1












            $begingroup$

            If one considers
            $$
            f(x)=ln (1-x),qquad |x|<1,
            $$
            one has
            $$
            f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
            $$
            giving, by the Taylor expansion,
            $$
            f(x)=0-x-fracx^22+O(x^3)
            $$
            as $x to 0$.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
              $endgroup$
              – Lepnak
              2 hours ago










            • $begingroup$
              The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
              $endgroup$
              – Minus One-Twelfth
              2 hours ago











            • $begingroup$
              Hmm I think I see what I did wrong. Thanks for all your answers.
              $endgroup$
              – Lepnak
              1 hour ago















            1












            $begingroup$

            If one considers
            $$
            f(x)=ln (1-x),qquad |x|<1,
            $$
            one has
            $$
            f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
            $$
            giving, by the Taylor expansion,
            $$
            f(x)=0-x-fracx^22+O(x^3)
            $$
            as $x to 0$.






            share|cite|improve this answer











            $endgroup$












            • $begingroup$
              Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
              $endgroup$
              – Lepnak
              2 hours ago










            • $begingroup$
              The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
              $endgroup$
              – Minus One-Twelfth
              2 hours ago











            • $begingroup$
              Hmm I think I see what I did wrong. Thanks for all your answers.
              $endgroup$
              – Lepnak
              1 hour ago













            1












            1








            1





            $begingroup$

            If one considers
            $$
            f(x)=ln (1-x),qquad |x|<1,
            $$
            one has
            $$
            f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
            $$
            giving, by the Taylor expansion,
            $$
            f(x)=0-x-fracx^22+O(x^3)
            $$
            as $x to 0$.






            share|cite|improve this answer











            $endgroup$



            If one considers
            $$
            f(x)=ln (1-x),qquad |x|<1,
            $$
            one has
            $$
            f(0)=0,quad f'(x)=-frac11-x,quad f'(0)=-1,quad f''(x)=-frac1(1-x)^2,quad f''(0)=-1,
            $$
            giving, by the Taylor expansion,
            $$
            f(x)=0-x-fracx^22+O(x^3)
            $$
            as $x to 0$.







            share|cite|improve this answer














            share|cite|improve this answer



            share|cite|improve this answer








            edited 2 hours ago

























            answered 2 hours ago









            Olivier OloaOlivier Oloa

            109k17178294




            109k17178294











            • $begingroup$
              Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
              $endgroup$
              – Lepnak
              2 hours ago










            • $begingroup$
              The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
              $endgroup$
              – Minus One-Twelfth
              2 hours ago











            • $begingroup$
              Hmm I think I see what I did wrong. Thanks for all your answers.
              $endgroup$
              – Lepnak
              1 hour ago
















            • $begingroup$
              Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
              $endgroup$
              – Lepnak
              2 hours ago










            • $begingroup$
              The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
              $endgroup$
              – Minus One-Twelfth
              2 hours ago











            • $begingroup$
              Hmm I think I see what I did wrong. Thanks for all your answers.
              $endgroup$
              – Lepnak
              1 hour ago















            $begingroup$
            Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
            $endgroup$
            – Lepnak
            2 hours ago




            $begingroup$
            Thanks for the answer but what about the $(z-a)$ part in the Taylor expansion $f(z) = f(a)+f^prime(a)(z-a)$? Substitute $z=1-x$ and $a=1$ gives a $-x$ though?
            $endgroup$
            – Lepnak
            2 hours ago












            $begingroup$
            The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
            $endgroup$
            – Minus One-Twelfth
            2 hours ago





            $begingroup$
            The Taylor series centred at $0$ is $$f(x)=f(0)+f'(0)x +cdots.$$ Use $f(0)$ and $f'(0)$ from Olivier Oloa's answer and you should get the right answer. In your OP, you are actually expanding $f(x)$ around $0$, not around $1$ (where $f(x)=ln (1-x)$). So $a=0$. By the way, if you substitute $z=1-x$ where $f(z)=ln (1-z)$, you would get $ln(1-(1-x))=ln x$, rather than $ln(1-x)$ (which is what you want). So no need to do this substitution.
            $endgroup$
            – Minus One-Twelfth
            2 hours ago













            $begingroup$
            Hmm I think I see what I did wrong. Thanks for all your answers.
            $endgroup$
            – Lepnak
            1 hour ago




            $begingroup$
            Hmm I think I see what I did wrong. Thanks for all your answers.
            $endgroup$
            – Lepnak
            1 hour ago











            2












            $begingroup$

            $$y=ln(1-x)$$
            $$y'=-frac11-x=-sum_n=0^inftyx^n$$
            so
            $$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$






            share|cite|improve this answer











            $endgroup$

















              2












              $begingroup$

              $$y=ln(1-x)$$
              $$y'=-frac11-x=-sum_n=0^inftyx^n$$
              so
              $$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$






              share|cite|improve this answer











              $endgroup$















                2












                2








                2





                $begingroup$

                $$y=ln(1-x)$$
                $$y'=-frac11-x=-sum_n=0^inftyx^n$$
                so
                $$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$






                share|cite|improve this answer











                $endgroup$



                $$y=ln(1-x)$$
                $$y'=-frac11-x=-sum_n=0^inftyx^n$$
                so
                $$ln(1-x)=-sum_n=0^inftyfracx^n+1n+1=-sum_n=1^inftyfracx^nn$$







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited 2 hours ago

























                answered 2 hours ago









                E.H.EE.H.E

                16.8k11969




                16.8k11969




















                    Lepnak is a new contributor. Be nice, and check out our Code of Conduct.









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                    Lepnak is a new contributor. Be nice, and check out our Code of Conduct.











                    Lepnak is a new contributor. Be nice, and check out our Code of Conduct.














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