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
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          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











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          1 Answer
          1






          active

          oldest

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          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


















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