Multivariable Differentials Problems












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I am having a lot of trouble with this question:



For $j=1,...,n$, define the function $f_{j}$ on $R^{n}$ {0} by $f_{j}(x)=x_{j}$/|x|. Show that $sum_{i=1}^n x_{j}df_{j}$=0.






multivariable-calculus







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  • 1




    $begingroup$
    Let's see your calculation.
    $endgroup$
    – Ted Shifrin
    May 26 '14 at 23:39
















0












$begingroup$


I am having a lot of trouble with this question:



For $j=1,...,n$, define the function $f_{j}$ on $R^{n}$ {0} by $f_{j}(x)=x_{j}$/|x|. Show that $sum_{i=1}^n x_{j}df_{j}$=0.






multivariable-calculus







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    Let's see your calculation.
    $endgroup$
    – Ted Shifrin
    May 26 '14 at 23:39














0












0








0





$begingroup$


I am having a lot of trouble with this question:



For $j=1,...,n$, define the function $f_{j}$ on $R^{n}$ {0} by $f_{j}(x)=x_{j}$/|x|. Show that $sum_{i=1}^n x_{j}df_{j}$=0.






multivariable-calculus







share|cite|improve this question









$endgroup$




I am having a lot of trouble with this question:



For $j=1,...,n$, define the function $f_{j}$ on $R^{n}$ {0} by $f_{j}(x)=x_{j}$/|x|. Show that $sum_{i=1}^n x_{j}df_{j}$=0.






multivariable-calculus




multivariable-calculus






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share|cite|improve this question











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asked May 26 '14 at 23:34









user153009user153009

359211




359211








  • 1




    $begingroup$
    Let's see your calculation.
    $endgroup$
    – Ted Shifrin
    May 26 '14 at 23:39














  • 1




    $begingroup$
    Let's see your calculation.
    $endgroup$
    – Ted Shifrin
    May 26 '14 at 23:39








1




1




$begingroup$
Let's see your calculation.
$endgroup$
– Ted Shifrin
May 26 '14 at 23:39




$begingroup$
Let's see your calculation.
$endgroup$
– Ted Shifrin
May 26 '14 at 23:39










2 Answers
2






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oldest

votes


















1












$begingroup$

Here’s a direct calculation:



For each j, $df_j = sum_{i=1}^nfrac{partial f_j}{partial x_i}dx_i$.

By the product rule, $frac{partial f_j}{partial x_j} = frac{1}{|x|}-frac{x_j^2}{|x|^3}$. Meanwhile, if i≠j, then $frac{partial f_j}{partial x_i} = -frac{x_ix_j}{|x|^3}$.



Thus, $df_j = frac{dx_j}{|x|}-sum_{i=1}^nfrac{x_ix_j}{|x|^3}dx_i$.



Multiplying by $x_j$ and summing on $j$ gives $sum_{j=1}^nx_jdx_j = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{j=1}^nsum_{i=1}^nfrac{x_ix_j^2}{|x|^3}dx_i$



$= sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nsum_{j=1}^nfrac{x_ix_j^2}{|x|^3}dx_i = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nfrac{x_i}{|x|}dx_i = 0$.






share|cite|improve this answer











$endgroup$





















    0












    $begingroup$

    We have $f_i=frac{x_i}{|x|}$ thus
    $$sum f_i^2=1$$



    The differential is therefore



    $$sum 2 f_i df_i =0$$



    or
    $$sum frac{x_i}{|x|} df_i =0$$ now multipy by $|x|$ to get
    $$sum x_i df_i =0$$



    A direct calculation should also work but this is more conceptual.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Thank you but how did you know to differentiate the inside term like that (is there a rule/concept that allows you to do that).
      $endgroup$
      – user153009
      May 27 '14 at 0:37













    Your Answer





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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    Here’s a direct calculation:



    For each j, $df_j = sum_{i=1}^nfrac{partial f_j}{partial x_i}dx_i$.

    By the product rule, $frac{partial f_j}{partial x_j} = frac{1}{|x|}-frac{x_j^2}{|x|^3}$. Meanwhile, if i≠j, then $frac{partial f_j}{partial x_i} = -frac{x_ix_j}{|x|^3}$.



    Thus, $df_j = frac{dx_j}{|x|}-sum_{i=1}^nfrac{x_ix_j}{|x|^3}dx_i$.



    Multiplying by $x_j$ and summing on $j$ gives $sum_{j=1}^nx_jdx_j = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{j=1}^nsum_{i=1}^nfrac{x_ix_j^2}{|x|^3}dx_i$



    $= sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nsum_{j=1}^nfrac{x_ix_j^2}{|x|^3}dx_i = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nfrac{x_i}{|x|}dx_i = 0$.






    share|cite|improve this answer











    $endgroup$


















      1












      $begingroup$

      Here’s a direct calculation:



      For each j, $df_j = sum_{i=1}^nfrac{partial f_j}{partial x_i}dx_i$.

      By the product rule, $frac{partial f_j}{partial x_j} = frac{1}{|x|}-frac{x_j^2}{|x|^3}$. Meanwhile, if i≠j, then $frac{partial f_j}{partial x_i} = -frac{x_ix_j}{|x|^3}$.



      Thus, $df_j = frac{dx_j}{|x|}-sum_{i=1}^nfrac{x_ix_j}{|x|^3}dx_i$.



      Multiplying by $x_j$ and summing on $j$ gives $sum_{j=1}^nx_jdx_j = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{j=1}^nsum_{i=1}^nfrac{x_ix_j^2}{|x|^3}dx_i$



      $= sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nsum_{j=1}^nfrac{x_ix_j^2}{|x|^3}dx_i = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nfrac{x_i}{|x|}dx_i = 0$.






      share|cite|improve this answer











      $endgroup$
















        1












        1








        1





        $begingroup$

        Here’s a direct calculation:



        For each j, $df_j = sum_{i=1}^nfrac{partial f_j}{partial x_i}dx_i$.

        By the product rule, $frac{partial f_j}{partial x_j} = frac{1}{|x|}-frac{x_j^2}{|x|^3}$. Meanwhile, if i≠j, then $frac{partial f_j}{partial x_i} = -frac{x_ix_j}{|x|^3}$.



        Thus, $df_j = frac{dx_j}{|x|}-sum_{i=1}^nfrac{x_ix_j}{|x|^3}dx_i$.



        Multiplying by $x_j$ and summing on $j$ gives $sum_{j=1}^nx_jdx_j = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{j=1}^nsum_{i=1}^nfrac{x_ix_j^2}{|x|^3}dx_i$



        $= sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nsum_{j=1}^nfrac{x_ix_j^2}{|x|^3}dx_i = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nfrac{x_i}{|x|}dx_i = 0$.






        share|cite|improve this answer











        $endgroup$



        Here’s a direct calculation:



        For each j, $df_j = sum_{i=1}^nfrac{partial f_j}{partial x_i}dx_i$.

        By the product rule, $frac{partial f_j}{partial x_j} = frac{1}{|x|}-frac{x_j^2}{|x|^3}$. Meanwhile, if i≠j, then $frac{partial f_j}{partial x_i} = -frac{x_ix_j}{|x|^3}$.



        Thus, $df_j = frac{dx_j}{|x|}-sum_{i=1}^nfrac{x_ix_j}{|x|^3}dx_i$.



        Multiplying by $x_j$ and summing on $j$ gives $sum_{j=1}^nx_jdx_j = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{j=1}^nsum_{i=1}^nfrac{x_ix_j^2}{|x|^3}dx_i$



        $= sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nsum_{j=1}^nfrac{x_ix_j^2}{|x|^3}dx_i = sum_{j=1}^nfrac{x_j}{|x|}dx_j-sum_{i=1}^nfrac{x_i}{|x|}dx_i = 0$.







        share|cite|improve this answer














        share|cite|improve this answer



        share|cite|improve this answer








        edited Jan 22 at 1:54

























        answered May 27 '14 at 0:56









        Sarah T.Sarah T.

        1933




        1933























            0












            $begingroup$

            We have $f_i=frac{x_i}{|x|}$ thus
            $$sum f_i^2=1$$



            The differential is therefore



            $$sum 2 f_i df_i =0$$



            or
            $$sum frac{x_i}{|x|} df_i =0$$ now multipy by $|x|$ to get
            $$sum x_i df_i =0$$



            A direct calculation should also work but this is more conceptual.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you but how did you know to differentiate the inside term like that (is there a rule/concept that allows you to do that).
              $endgroup$
              – user153009
              May 27 '14 at 0:37


















            0












            $begingroup$

            We have $f_i=frac{x_i}{|x|}$ thus
            $$sum f_i^2=1$$



            The differential is therefore



            $$sum 2 f_i df_i =0$$



            or
            $$sum frac{x_i}{|x|} df_i =0$$ now multipy by $|x|$ to get
            $$sum x_i df_i =0$$



            A direct calculation should also work but this is more conceptual.






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Thank you but how did you know to differentiate the inside term like that (is there a rule/concept that allows you to do that).
              $endgroup$
              – user153009
              May 27 '14 at 0:37
















            0












            0








            0





            $begingroup$

            We have $f_i=frac{x_i}{|x|}$ thus
            $$sum f_i^2=1$$



            The differential is therefore



            $$sum 2 f_i df_i =0$$



            or
            $$sum frac{x_i}{|x|} df_i =0$$ now multipy by $|x|$ to get
            $$sum x_i df_i =0$$



            A direct calculation should also work but this is more conceptual.






            share|cite|improve this answer









            $endgroup$



            We have $f_i=frac{x_i}{|x|}$ thus
            $$sum f_i^2=1$$



            The differential is therefore



            $$sum 2 f_i df_i =0$$



            or
            $$sum frac{x_i}{|x|} df_i =0$$ now multipy by $|x|$ to get
            $$sum x_i df_i =0$$



            A direct calculation should also work but this is more conceptual.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered May 27 '14 at 0:28









            Rene SchipperusRene Schipperus

            32.4k11960




            32.4k11960












            • $begingroup$
              Thank you but how did you know to differentiate the inside term like that (is there a rule/concept that allows you to do that).
              $endgroup$
              – user153009
              May 27 '14 at 0:37




















            • $begingroup$
              Thank you but how did you know to differentiate the inside term like that (is there a rule/concept that allows you to do that).
              $endgroup$
              – user153009
              May 27 '14 at 0:37


















            $begingroup$
            Thank you but how did you know to differentiate the inside term like that (is there a rule/concept that allows you to do that).
            $endgroup$
            – user153009
            May 27 '14 at 0:37






            $begingroup$
            Thank you but how did you know to differentiate the inside term like that (is there a rule/concept that allows you to do that).
            $endgroup$
            – user153009
            May 27 '14 at 0:37




















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