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.










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












    $begingroup$


    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.










    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.










      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






      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.
      Check out our Code of Conduct.









      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.
      Check out our Code of Conduct.





      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.






      Lepnak is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      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
          $$
          f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
          $$
          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
            2 hours ago












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



















          2












          $begingroup$

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











          $endgroup$














            Your Answer








            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',
            autoActivateHeartbeat: false,
            convertImagesToLinks: true,
            noModals: true,
            showLowRepImageUploadWarning: true,
            reputationToPostImages: 10,
            bindNavPrevention: true,
            postfix: "",
            imageUploader: {
            brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
            contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
            allowUrls: true
            },
            noCode: true, onDemand: true,
            discardSelector: ".discard-answer"
            ,immediatelyShowMarkdownHelp:true
            });


            }
            });






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










            draft saved

            draft discarded


















            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3193068%2ftaylor-expansion-of-ln1-x%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown

























            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
              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
            $$
            f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
            $$
            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
              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
            $$
            f(0)=0,quad f'(x)=-frac{1}{1-x},quad f'(0)=-1,quad f''(x)=-frac{1}{(1-x)^2},quad f''(0)=-1,
            $$
            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
              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












            $begingroup$

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






















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










                    draft saved

                    draft discarded


















                    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.












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
















                    Thanks for contributing an answer to Mathematics Stack Exchange!


                    • Please be sure to answer the question. Provide details and share your research!

                    But avoid



                    • Asking for help, clarification, or responding to other answers.

                    • Making statements based on opinion; back them up with references or personal experience.


                    Use MathJax to format equations. MathJax reference.


                    To learn more, see our tips on writing great answers.




                    draft saved


                    draft discarded














                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3193068%2ftaylor-expansion-of-ln1-x%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown





















































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown

































                    Required, but never shown














                    Required, but never shown












                    Required, but never shown







                    Required, but never shown







                    Popular posts from this blog

                    Knooppunt Holsloot

                    Altaar (religie)

                    Gregoriusmis