Write the first four terms of the taylor series for the function $f(x, y) = ln(1 + x + y)$ around the point...











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(a) Write the first four terms of the taylor series for the function $f(x, y) = ln(1 + x + y)$ around the point $(x, y) = (0, 0)$



$f(0,0) = 0$



$f_x(x, y) = frac{1}{1+x+y}|_{0,0} = 1$



$f_y(x, y) = frac{1}{1+x+y}|_{0,0} = 1$



$f_{xx}(x, y) = frac{-1}{(1+x+y)}|_{0,0} = -1$



$f_{yy}(x, y) = frac{-1}{(1+x+y)^2}|_{0,0} = -1$



$f_{xy}(x, y) = frac{-1}{(1+x+y)^2}|_{0,0} = -1$



therefore the first four terms are



$= 0 + [x cdot 1 + y cdot 1] + frac{1}{2}(x^2(-1) + 2xy(-1) + y^2(-1)) = x + y - frac{1}{2}left[x^2 + 2xy + y^2right] + cdots$



would this be correct?






multivariable-calculus







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    (a) Write the first four terms of the taylor series for the function $f(x, y) = ln(1 + x + y)$ around the point $(x, y) = (0, 0)$



    $f(0,0) = 0$



    $f_x(x, y) = frac{1}{1+x+y}|_{0,0} = 1$



    $f_y(x, y) = frac{1}{1+x+y}|_{0,0} = 1$



    $f_{xx}(x, y) = frac{-1}{(1+x+y)}|_{0,0} = -1$



    $f_{yy}(x, y) = frac{-1}{(1+x+y)^2}|_{0,0} = -1$



    $f_{xy}(x, y) = frac{-1}{(1+x+y)^2}|_{0,0} = -1$



    therefore the first four terms are



    $= 0 + [x cdot 1 + y cdot 1] + frac{1}{2}(x^2(-1) + 2xy(-1) + y^2(-1)) = x + y - frac{1}{2}left[x^2 + 2xy + y^2right] + cdots$



    would this be correct?






    multivariable-calculus







    share|cite|improve this question
























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      (a) Write the first four terms of the taylor series for the function $f(x, y) = ln(1 + x + y)$ around the point $(x, y) = (0, 0)$



      $f(0,0) = 0$



      $f_x(x, y) = frac{1}{1+x+y}|_{0,0} = 1$



      $f_y(x, y) = frac{1}{1+x+y}|_{0,0} = 1$



      $f_{xx}(x, y) = frac{-1}{(1+x+y)}|_{0,0} = -1$



      $f_{yy}(x, y) = frac{-1}{(1+x+y)^2}|_{0,0} = -1$



      $f_{xy}(x, y) = frac{-1}{(1+x+y)^2}|_{0,0} = -1$



      therefore the first four terms are



      $= 0 + [x cdot 1 + y cdot 1] + frac{1}{2}(x^2(-1) + 2xy(-1) + y^2(-1)) = x + y - frac{1}{2}left[x^2 + 2xy + y^2right] + cdots$



      would this be correct?






      multivariable-calculus







      share|cite|improve this question













      (a) Write the first four terms of the taylor series for the function $f(x, y) = ln(1 + x + y)$ around the point $(x, y) = (0, 0)$



      $f(0,0) = 0$



      $f_x(x, y) = frac{1}{1+x+y}|_{0,0} = 1$



      $f_y(x, y) = frac{1}{1+x+y}|_{0,0} = 1$



      $f_{xx}(x, y) = frac{-1}{(1+x+y)}|_{0,0} = -1$



      $f_{yy}(x, y) = frac{-1}{(1+x+y)^2}|_{0,0} = -1$



      $f_{xy}(x, y) = frac{-1}{(1+x+y)^2}|_{0,0} = -1$



      therefore the first four terms are



      $= 0 + [x cdot 1 + y cdot 1] + frac{1}{2}(x^2(-1) + 2xy(-1) + y^2(-1)) = x + y - frac{1}{2}left[x^2 + 2xy + y^2right] + cdots$



      would this be correct?






      multivariable-calculus




      multivariable-calculus






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      asked Nov 16 at 0:33









      Tree Garen

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          $frac {d}{du} ln (1+u) = frac 1{1+u}\
          frac {1}{1+u} = sum_limits{n=0}^{infty} (-1)^{n} u^n\
          int_1^{u} sum_limits{n=0}^{infty} (-1)^{n}u^n = sum_limits{n=1}^{infty} (-1)^{n+1}frac {u^n}{n} = ln (1+u)$



          $u = x + y\
          ln(1+x+y) = sum_limits{n=1}^{infty} (-1)^{n+1}frac {(x+y)^n}{n}$



          The first four terms are $x + y - frac{x^2}{2} - xy - frac {y^2}{2} $



          Looks like we match.






          share|cite|improve this answer























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            $frac {d}{du} ln (1+u) = frac 1{1+u}\
            frac {1}{1+u} = sum_limits{n=0}^{infty} (-1)^{n} u^n\
            int_1^{u} sum_limits{n=0}^{infty} (-1)^{n}u^n = sum_limits{n=1}^{infty} (-1)^{n+1}frac {u^n}{n} = ln (1+u)$



            $u = x + y\
            ln(1+x+y) = sum_limits{n=1}^{infty} (-1)^{n+1}frac {(x+y)^n}{n}$



            The first four terms are $x + y - frac{x^2}{2} - xy - frac {y^2}{2} $



            Looks like we match.






            share|cite|improve this answer



























              up vote
              0
              down vote



              accepted










              $frac {d}{du} ln (1+u) = frac 1{1+u}\
              frac {1}{1+u} = sum_limits{n=0}^{infty} (-1)^{n} u^n\
              int_1^{u} sum_limits{n=0}^{infty} (-1)^{n}u^n = sum_limits{n=1}^{infty} (-1)^{n+1}frac {u^n}{n} = ln (1+u)$



              $u = x + y\
              ln(1+x+y) = sum_limits{n=1}^{infty} (-1)^{n+1}frac {(x+y)^n}{n}$



              The first four terms are $x + y - frac{x^2}{2} - xy - frac {y^2}{2} $



              Looks like we match.






              share|cite|improve this answer

























                up vote
                0
                down vote



                accepted







                up vote
                0
                down vote



                accepted






                $frac {d}{du} ln (1+u) = frac 1{1+u}\
                frac {1}{1+u} = sum_limits{n=0}^{infty} (-1)^{n} u^n\
                int_1^{u} sum_limits{n=0}^{infty} (-1)^{n}u^n = sum_limits{n=1}^{infty} (-1)^{n+1}frac {u^n}{n} = ln (1+u)$



                $u = x + y\
                ln(1+x+y) = sum_limits{n=1}^{infty} (-1)^{n+1}frac {(x+y)^n}{n}$



                The first four terms are $x + y - frac{x^2}{2} - xy - frac {y^2}{2} $



                Looks like we match.






                share|cite|improve this answer














                $frac {d}{du} ln (1+u) = frac 1{1+u}\
                frac {1}{1+u} = sum_limits{n=0}^{infty} (-1)^{n} u^n\
                int_1^{u} sum_limits{n=0}^{infty} (-1)^{n}u^n = sum_limits{n=1}^{infty} (-1)^{n+1}frac {u^n}{n} = ln (1+u)$



                $u = x + y\
                ln(1+x+y) = sum_limits{n=1}^{infty} (-1)^{n+1}frac {(x+y)^n}{n}$



                The first four terms are $x + y - frac{x^2}{2} - xy - frac {y^2}{2} $



                Looks like we match.







                share|cite|improve this answer














                share|cite|improve this answer



                share|cite|improve this answer








                edited Nov 16 at 1:04

























                answered Nov 16 at 0:41









                Doug M

                42.7k31752




                42.7k31752






























                     

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