Multivariable Maclaurin Series











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Find Maclaurin series for



a) $cos(x + y)$



b) $frac{log(1 + x)}{(1 + y)}$.



Honestly, I'm really just confused about the process of finding a Taylor series expansion for multivariate equation.






sequences-and-series multivariable-calculus







share|cite|improve this question




























    up vote
    0
    down vote

    favorite












    Find Maclaurin series for



    a) $cos(x + y)$



    b) $frac{log(1 + x)}{(1 + y)}$.



    Honestly, I'm really just confused about the process of finding a Taylor series expansion for multivariate equation.






    sequences-and-series multivariable-calculus







    share|cite|improve this question


























      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      Find Maclaurin series for



      a) $cos(x + y)$



      b) $frac{log(1 + x)}{(1 + y)}$.



      Honestly, I'm really just confused about the process of finding a Taylor series expansion for multivariate equation.






      sequences-and-series multivariable-calculus







      share|cite|improve this question















      Find Maclaurin series for



      a) $cos(x + y)$



      b) $frac{log(1 + x)}{(1 + y)}$.



      Honestly, I'm really just confused about the process of finding a Taylor series expansion for multivariate equation.






      sequences-and-series multivariable-calculus




      sequences-and-series multivariable-calculus






      share|cite|improve this question















      share|cite|improve this question













      share|cite|improve this question




      share|cite|improve this question








      edited Nov 13 '14 at 9:27









      Swapnil Tripathi

      3,04621736




      3,04621736










      asked Nov 13 '14 at 8:59









      Katie

      1




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          2 Answers
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          0
          down vote













          HINT:
          Consider the general formula, try relating it to single variable formula of Taylor's series.



          $f(x,y)=f(x_0,y_0)$
          $$+frac{partial f(x_0,y_0)}{partial x}(x-x_0)+frac{partial f(x_0,y_0)}{partial y}(y-y_0)quad Rightarrow Order 1$$$
          $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$



          And repeat this for higher orders.



          For Maclaurin's the formula is much more simpler as $x_0=y_0=0$



          Do you still have a problem evaluating the thing? Tell me where you're struck.






          share|cite|improve this answer





















          • In the 2nd order why $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$$ instead of $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2bigg) +frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0) quad Rightarrow Order 2$$?
            – KYHSGeekCode
            Sep 2 at 14:42


















          up vote
          0
          down vote













          Try this expression:



          $$f(x,y)=sum^infty_{k=0}sum^k_{i=0} frac{(x-a)^{k-i}(y-b)^i}{(k-i)!i!} left[frac{partial^kf}{partial x^{k-i} partial y^i}right]_{(a=0,b=0)}$$






          share|cite|improve this answer





















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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes








            up vote
            0
            down vote













            HINT:
            Consider the general formula, try relating it to single variable formula of Taylor's series.



            $f(x,y)=f(x_0,y_0)$
            $$+frac{partial f(x_0,y_0)}{partial x}(x-x_0)+frac{partial f(x_0,y_0)}{partial y}(y-y_0)quad Rightarrow Order 1$$$
            $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$



            And repeat this for higher orders.



            For Maclaurin's the formula is much more simpler as $x_0=y_0=0$



            Do you still have a problem evaluating the thing? Tell me where you're struck.






            share|cite|improve this answer





















            • In the 2nd order why $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$$ instead of $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2bigg) +frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0) quad Rightarrow Order 2$$?
              – KYHSGeekCode
              Sep 2 at 14:42















            up vote
            0
            down vote













            HINT:
            Consider the general formula, try relating it to single variable formula of Taylor's series.



            $f(x,y)=f(x_0,y_0)$
            $$+frac{partial f(x_0,y_0)}{partial x}(x-x_0)+frac{partial f(x_0,y_0)}{partial y}(y-y_0)quad Rightarrow Order 1$$$
            $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$



            And repeat this for higher orders.



            For Maclaurin's the formula is much more simpler as $x_0=y_0=0$



            Do you still have a problem evaluating the thing? Tell me where you're struck.






            share|cite|improve this answer





















            • In the 2nd order why $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$$ instead of $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2bigg) +frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0) quad Rightarrow Order 2$$?
              – KYHSGeekCode
              Sep 2 at 14:42













            up vote
            0
            down vote










            up vote
            0
            down vote









            HINT:
            Consider the general formula, try relating it to single variable formula of Taylor's series.



            $f(x,y)=f(x_0,y_0)$
            $$+frac{partial f(x_0,y_0)}{partial x}(x-x_0)+frac{partial f(x_0,y_0)}{partial y}(y-y_0)quad Rightarrow Order 1$$$
            $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$



            And repeat this for higher orders.



            For Maclaurin's the formula is much more simpler as $x_0=y_0=0$



            Do you still have a problem evaluating the thing? Tell me where you're struck.






            share|cite|improve this answer












            HINT:
            Consider the general formula, try relating it to single variable formula of Taylor's series.



            $f(x,y)=f(x_0,y_0)$
            $$+frac{partial f(x_0,y_0)}{partial x}(x-x_0)+frac{partial f(x_0,y_0)}{partial y}(y-y_0)quad Rightarrow Order 1$$$
            $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$



            And repeat this for higher orders.



            For Maclaurin's the formula is much more simpler as $x_0=y_0=0$



            Do you still have a problem evaluating the thing? Tell me where you're struck.







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 13 '14 at 13:25









            Swapnil Tripathi

            3,04621736




            3,04621736












            • In the 2nd order why $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$$ instead of $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2bigg) +frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0) quad Rightarrow Order 2$$?
              – KYHSGeekCode
              Sep 2 at 14:42


















            • In the 2nd order why $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$$ instead of $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2bigg) +frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0) quad Rightarrow Order 2$$?
              – KYHSGeekCode
              Sep 2 at 14:42
















            In the 2nd order why $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$$ instead of $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2bigg) +frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0) quad Rightarrow Order 2$$?
            – KYHSGeekCode
            Sep 2 at 14:42




            In the 2nd order why $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2+frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0)bigg) quad Rightarrow Order 2$$ instead of $$+frac{1}{2} bigg(frac{partial^2 f(x_0,y_0)}{partial x^2}(x-x_0)^2+frac{partial^2 f(x_0,y_0)}{partial y^2}(y-y_0)^2bigg) +frac{partial^2 f(x_0,y_0)}{partial xpartial y}(x-x_0)(y-y_0) quad Rightarrow Order 2$$?
            – KYHSGeekCode
            Sep 2 at 14:42










            up vote
            0
            down vote













            Try this expression:



            $$f(x,y)=sum^infty_{k=0}sum^k_{i=0} frac{(x-a)^{k-i}(y-b)^i}{(k-i)!i!} left[frac{partial^kf}{partial x^{k-i} partial y^i}right]_{(a=0,b=0)}$$






            share|cite|improve this answer

























              up vote
              0
              down vote













              Try this expression:



              $$f(x,y)=sum^infty_{k=0}sum^k_{i=0} frac{(x-a)^{k-i}(y-b)^i}{(k-i)!i!} left[frac{partial^kf}{partial x^{k-i} partial y^i}right]_{(a=0,b=0)}$$






              share|cite|improve this answer























                up vote
                0
                down vote










                up vote
                0
                down vote









                Try this expression:



                $$f(x,y)=sum^infty_{k=0}sum^k_{i=0} frac{(x-a)^{k-i}(y-b)^i}{(k-i)!i!} left[frac{partial^kf}{partial x^{k-i} partial y^i}right]_{(a=0,b=0)}$$






                share|cite|improve this answer












                Try this expression:



                $$f(x,y)=sum^infty_{k=0}sum^k_{i=0} frac{(x-a)^{k-i}(y-b)^i}{(k-i)!i!} left[frac{partial^kf}{partial x^{k-i} partial y^i}right]_{(a=0,b=0)}$$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Sep 2 at 14:53









                KYHSGeekCode

                302112




                302112






























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