Taylor expansion of ln(1-x)












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
$$
ln(1-x) approx ln(1) + frac{d(ln(1-x))}{dx}biggvert_{x=0}[(1-x)-1] approx 0 + frac{1}{1-x}biggvert_{x=0}(-1)(-x) = x.
$$
Using the definition of Taylor expansion $f(z) approx f(a) + frac{df(z)}{dz}biggvert_{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.






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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
    $$
    ln(1-x) approx ln(1) + frac{d(ln(1-x))}{dx}biggvert_{x=0}[(1-x)-1] approx 0 + frac{1}{1-x}biggvert_{x=0}(-1)(-x) = x.
    $$
    Using the definition of Taylor expansion $f(z) approx f(a) + frac{df(z)}{dz}biggvert_{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.







    $endgroup$















      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
      $$
      ln(1-x) approx ln(1) + frac{d(ln(1-x))}{dx}biggvert_{x=0}[(1-x)-1] approx 0 + frac{1}{1-x}biggvert_{x=0}(-1)(-x) = x.
      $$
      Using the definition of Taylor expansion $f(z) approx f(a) + frac{df(z)}{dz}biggvert_{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.







      $endgroup$




      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
      $$
      ln(1-x) approx ln(1) + frac{d(ln(1-x))}{dx}biggvert_{x=0}[(1-x)-1] approx 0 + frac{1}{1-x}biggvert_{x=0}(-1)(-x) = x.
      $$
      Using the definition of Taylor expansion $f(z) approx f(a) + frac{df(z)}{dz}biggvert_{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




      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.











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      New contributor




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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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      asked 2 hours ago









      LepnakLepnak

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          2 Answers
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          active

          oldest

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          1












          $begingroup$

          If one considers
          $$
          f(x)=ln (1-x),qquad |x|<1,
          $$          one has
          $$
          f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
          $$           giving, by the Taylor expansion,
          $$
          f(x)=0-x-frac{x^2}2+O(x^3)
          $$          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
            1 hour ago












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



















          2












          $begingroup$

          $$y=ln(1-x)$$
          $$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
          so
          $$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$






          share|cite|improve this answer











          $endgroup$














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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
            $$
            f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
            $$             giving, by the Taylor expansion,
            $$
            f(x)=0-x-frac{x^2}2+O(x^3)
            $$            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
              1 hour 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
            $$
            f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
            $$             giving, by the Taylor expansion,
            $$
            f(x)=0-x-frac{x^2}2+O(x^3)
            $$            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
              1 hour 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
            $$
            f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
            $$             giving, by the Taylor expansion,
            $$
            f(x)=0-x-frac{x^2}2+O(x^3)
            $$            as $x to 0$.






            share|cite|improve this answer











            $endgroup$



            If one considers
            $$
            f(x)=ln (1-x),qquad |x|<1,
            $$            one has
            $$
            f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
            $$             giving, by the Taylor expansion,
            $$
            f(x)=0-x-frac{x^2}2+O(x^3)
            $$            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
              1 hour 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
              1 hour 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
            1 hour 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
            1 hour 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












            $begingroup$

            $$y=ln(1-x)$$
            $$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
            so
            $$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$






            share|cite|improve this answer











            $endgroup$


















              2












              $begingroup$

              $$y=ln(1-x)$$
              $$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
              so
              $$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$






              share|cite|improve this answer











              $endgroup$
















                2












                2








                2





                $begingroup$

                $$y=ln(1-x)$$
                $$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
                so
                $$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$






                share|cite|improve this answer











                $endgroup$



                $$y=ln(1-x)$$
                $$y'=-frac{1}{1-x}=-sum_{n=0}^{infty}x^n$$
                so
                $$ln(1-x)=-sum_{n=0}^{infty}frac{x^{n+1}}{n+1}=-sum_{n=1}^{infty}frac{x^{n}}{n}$$







                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






















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