Remainder in Lagrange form of $ln (1-x)$












1















Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=ln(1-x)$.



Let $T_{2,f,0}$ be the MacLaurin polynomial of second degree for $f$
and $r_2$ it's remainder.



Find $r_2(x)$ and evaluate it in $x=-1$ i.e $r_2(frac{1}{2})$.




My first thought was to use Lagrange's form of remainder...



$r_2(x)=frac{f^{'''}(c)}{3!}x^3=frac{frac{-2}{(1-c)^3}}{3!}=frac{-2}{3(1-c)^3}x^3$



Now, I don't understand that




"evaluate it in $x=-1$ i.e $r_2(frac{1}{2})$"




I anyway evaluated it in $x=-1$ and $x=frac{1}{2}$:



When $x=1$: $r_2=frac{2}{3(1-c)^3}$



When $x=frac{1}{2}$ $r_2=frac{-1}{12(1-c)^3}$



Which none of them gives me the correct answer, which apparently is



$r_2=frac{1}{3(1-c)^3}$ with $c in (-1,0)$.



Do I have to find the remainder in another way?






calculus real-analysis derivatives taylor-expansion







share|cite|improve this question



























    1















    Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=ln(1-x)$.



    Let $T_{2,f,0}$ be the MacLaurin polynomial of second degree for $f$
    and $r_2$ it's remainder.



    Find $r_2(x)$ and evaluate it in $x=-1$ i.e $r_2(frac{1}{2})$.




    My first thought was to use Lagrange's form of remainder...



    $r_2(x)=frac{f^{'''}(c)}{3!}x^3=frac{frac{-2}{(1-c)^3}}{3!}=frac{-2}{3(1-c)^3}x^3$



    Now, I don't understand that




    "evaluate it in $x=-1$ i.e $r_2(frac{1}{2})$"




    I anyway evaluated it in $x=-1$ and $x=frac{1}{2}$:



    When $x=1$: $r_2=frac{2}{3(1-c)^3}$



    When $x=frac{1}{2}$ $r_2=frac{-1}{12(1-c)^3}$



    Which none of them gives me the correct answer, which apparently is



    $r_2=frac{1}{3(1-c)^3}$ with $c in (-1,0)$.



    Do I have to find the remainder in another way?






    calculus real-analysis derivatives taylor-expansion







    share|cite|improve this question

























      1












      1








      1








      Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=ln(1-x)$.



      Let $T_{2,f,0}$ be the MacLaurin polynomial of second degree for $f$
      and $r_2$ it's remainder.



      Find $r_2(x)$ and evaluate it in $x=-1$ i.e $r_2(frac{1}{2})$.




      My first thought was to use Lagrange's form of remainder...



      $r_2(x)=frac{f^{'''}(c)}{3!}x^3=frac{frac{-2}{(1-c)^3}}{3!}=frac{-2}{3(1-c)^3}x^3$



      Now, I don't understand that




      "evaluate it in $x=-1$ i.e $r_2(frac{1}{2})$"




      I anyway evaluated it in $x=-1$ and $x=frac{1}{2}$:



      When $x=1$: $r_2=frac{2}{3(1-c)^3}$



      When $x=frac{1}{2}$ $r_2=frac{-1}{12(1-c)^3}$



      Which none of them gives me the correct answer, which apparently is



      $r_2=frac{1}{3(1-c)^3}$ with $c in (-1,0)$.



      Do I have to find the remainder in another way?






      calculus real-analysis derivatives taylor-expansion







      share|cite|improve this question














      Let $f:mathbb{R} to mathbb{R}$ be a function, $f(x)=ln(1-x)$.



      Let $T_{2,f,0}$ be the MacLaurin polynomial of second degree for $f$
      and $r_2$ it's remainder.



      Find $r_2(x)$ and evaluate it in $x=-1$ i.e $r_2(frac{1}{2})$.




      My first thought was to use Lagrange's form of remainder...



      $r_2(x)=frac{f^{'''}(c)}{3!}x^3=frac{frac{-2}{(1-c)^3}}{3!}=frac{-2}{3(1-c)^3}x^3$



      Now, I don't understand that




      "evaluate it in $x=-1$ i.e $r_2(frac{1}{2})$"




      I anyway evaluated it in $x=-1$ and $x=frac{1}{2}$:



      When $x=1$: $r_2=frac{2}{3(1-c)^3}$



      When $x=frac{1}{2}$ $r_2=frac{-1}{12(1-c)^3}$



      Which none of them gives me the correct answer, which apparently is



      $r_2=frac{1}{3(1-c)^3}$ with $c in (-1,0)$.



      Do I have to find the remainder in another way?






      calculus real-analysis derivatives taylor-expansion




      calculus real-analysis derivatives taylor-expansion






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 25 at 19:41









      parishilton

      18110




      18110






















          1 Answer
          1






          active

          oldest

          votes


















          1














          I think that it should be




          Find $r_2(x)$ and evaluate it in $x=-1$, i.e $r_2(-1)$.




          By the the way, your computation is almost correct, you just missed a factor $2$.



          The remainder of second degree is
          $$r_2(x)=f(x)-T_{2,f,0}(x)=frac{f'''(c)}{3!}x^3=frac{-2}{3!(1-c)^3}x^3=frac{-x^3}{3(1-c)^3}$$
          where $c$ is some point between $x$ and $0$ and
          $$f'(x)=-(1-x)^{-1},,,f''(x)=-(1-x)^{-2},,,f'''(x)=-2(1-x)^{-3}.$$
          Therefore if $x=-1$ then $cin (-1,0)$ and
          $$r_2(-1)=frac{1}{3(1-c)^3}.$$






          share|cite|improve this answer























          • Thank you so much!!!
            – parishilton
            Nov 25 at 20:08











          Your Answer





          StackExchange.ifUsing("editor", function () {
          return StackExchange.using("mathjaxEditing", function () {
          StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
          StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
          });
          });
          }, "mathjax-editing");

          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
          });


          }
          });














          draft saved

          draft discarded


















          StackExchange.ready(
          function () {
          StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013290%2fremainder-in-lagrange-form-of-ln-1-x%23new-answer', 'question_page');
          }
          );

          Post as a guest















          Required, but never shown

























          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          I think that it should be




          Find $r_2(x)$ and evaluate it in $x=-1$, i.e $r_2(-1)$.




          By the the way, your computation is almost correct, you just missed a factor $2$.



          The remainder of second degree is
          $$r_2(x)=f(x)-T_{2,f,0}(x)=frac{f'''(c)}{3!}x^3=frac{-2}{3!(1-c)^3}x^3=frac{-x^3}{3(1-c)^3}$$
          where $c$ is some point between $x$ and $0$ and
          $$f'(x)=-(1-x)^{-1},,,f''(x)=-(1-x)^{-2},,,f'''(x)=-2(1-x)^{-3}.$$
          Therefore if $x=-1$ then $cin (-1,0)$ and
          $$r_2(-1)=frac{1}{3(1-c)^3}.$$






          share|cite|improve this answer























          • Thank you so much!!!
            – parishilton
            Nov 25 at 20:08
















          1














          I think that it should be




          Find $r_2(x)$ and evaluate it in $x=-1$, i.e $r_2(-1)$.




          By the the way, your computation is almost correct, you just missed a factor $2$.



          The remainder of second degree is
          $$r_2(x)=f(x)-T_{2,f,0}(x)=frac{f'''(c)}{3!}x^3=frac{-2}{3!(1-c)^3}x^3=frac{-x^3}{3(1-c)^3}$$
          where $c$ is some point between $x$ and $0$ and
          $$f'(x)=-(1-x)^{-1},,,f''(x)=-(1-x)^{-2},,,f'''(x)=-2(1-x)^{-3}.$$
          Therefore if $x=-1$ then $cin (-1,0)$ and
          $$r_2(-1)=frac{1}{3(1-c)^3}.$$






          share|cite|improve this answer























          • Thank you so much!!!
            – parishilton
            Nov 25 at 20:08














          1












          1








          1






          I think that it should be




          Find $r_2(x)$ and evaluate it in $x=-1$, i.e $r_2(-1)$.




          By the the way, your computation is almost correct, you just missed a factor $2$.



          The remainder of second degree is
          $$r_2(x)=f(x)-T_{2,f,0}(x)=frac{f'''(c)}{3!}x^3=frac{-2}{3!(1-c)^3}x^3=frac{-x^3}{3(1-c)^3}$$
          where $c$ is some point between $x$ and $0$ and
          $$f'(x)=-(1-x)^{-1},,,f''(x)=-(1-x)^{-2},,,f'''(x)=-2(1-x)^{-3}.$$
          Therefore if $x=-1$ then $cin (-1,0)$ and
          $$r_2(-1)=frac{1}{3(1-c)^3}.$$






          share|cite|improve this answer














          I think that it should be




          Find $r_2(x)$ and evaluate it in $x=-1$, i.e $r_2(-1)$.




          By the the way, your computation is almost correct, you just missed a factor $2$.



          The remainder of second degree is
          $$r_2(x)=f(x)-T_{2,f,0}(x)=frac{f'''(c)}{3!}x^3=frac{-2}{3!(1-c)^3}x^3=frac{-x^3}{3(1-c)^3}$$
          where $c$ is some point between $x$ and $0$ and
          $$f'(x)=-(1-x)^{-1},,,f''(x)=-(1-x)^{-2},,,f'''(x)=-2(1-x)^{-3}.$$
          Therefore if $x=-1$ then $cin (-1,0)$ and
          $$r_2(-1)=frac{1}{3(1-c)^3}.$$







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 25 at 20:05

























          answered Nov 25 at 19:47









          Robert Z

          93.1k1061132




          93.1k1061132












          • Thank you so much!!!
            – parishilton
            Nov 25 at 20:08


















          • Thank you so much!!!
            – parishilton
            Nov 25 at 20:08
















          Thank you so much!!!
          – parishilton
          Nov 25 at 20:08




          Thank you so much!!!
          – parishilton
          Nov 25 at 20:08


















          draft saved

          draft discarded




















































          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.





          Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


          Please pay close attention to the following guidance:


          • 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.


          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%2f3013290%2fremainder-in-lagrange-form-of-ln-1-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

          Ellipse (mathématiques)

          Image
          Pour les articles homonymes, voir Ellipse. L'ellipse est le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante. Section du cône ou projection du cercle. En géométrie, une ellipse est une courbe plane fermée obtenue par l’intersection d’un cône de révolution avec un plan, à condition que celui-ci coupe l'axe de rotation du cône ou du cylindre : c'est une conique d'excentricité strictement comprise entre 0 et 1. On peut également la définir comme le lieu des points dont la somme des distances à deux points fixes, dits foyers, est constante (sa construction par la méthode du jardinier est très simple). Dans la vie courante, l’ellipse est la forme qu'on perçoit en regardant un cercle en perspective, ou la figure formée par l’ombre d'un disque sur une surface plane. On retrouve aussi, en première approximation [ 1 ] , des ellipses dans les trajectoires des corps célestes (planètes, comètes ou satellites arti...
          Read more

          Quarter-circle Tiles

          Image
          up vote 0 down vote favorite I have $99$ identical square tiles, each with a quarter-circle drawn on it like this: [asy] size(1.5cm); draw(Arc((2,0),1,90,180),red+1); draw((0,0)--(2,0)--(2,2)--(0,2)--(0,0)); [/asy] When I arrange the tiles in a $9times 11$ rectangular grid, each with a random orientation, what is the expected value of the number of full circles I form? I think this problem has to do with finding the chance any given 2x2 square has a circle, but I can't find it. expected-value share | cite | improve this question asked Nov 20 at 15:03 6minecraftninja 1 2 ...
          Read more

          Mont Emei

          Image
          Mont Emei.mw-parser-output .entete.map{background-image:url("//upload.wikimedia.org/wikipedia/commons/7/7a/Picto_infobox_map.png")} Vue du sommet Wanfo du mont Emei Géographie Altitude 3 099  m , Wanfo Coordonnées 29° 31′ 11″ nord, 103° 19′ 57″ est Administration Pays   Chine Province Sichuan Ville-préfecture Leshan Géolocalisation sur la carte : Sichuan Mont Emei Géolocalisation sur la carte : Chine Mont Emei modifier   Paysage panoramique du mont Emei, incluant le paysage panoramique du grand Bouddha de Leshan *   Patrimoine mondial de l'UNESCO Passerelle en bois sur le versant ouest Pays   Chine Subdivision Sichuan Type Mixte Critères (iv) (vi) (x) Superficie 15 400 ha Numéro d’identification 779 Zone géographique Asie et Pacifique  ** Année d’inscription 1996 (20 e session) * Descriptif officiel UNESCO ** Classification géog...
          Read more