Taylor Series expansion of an implicitly defined function $x^2 +y^2=y, y(0)=0$












4















Find the first 6 terms of the Taylor series for $y$ in powers of $x$ of the following implicitly defined function;



$$x^2 +y^2=y, y(0)=0$$




I'm a bit stuck in how to proceed do i need to implicitly differentiate the function such that $y'=frac{-2x}{(1-2y)}$ and again so as to find $y'',y^{(3)},...,y^{(6)}$ and then plug these into the Taylor expansion and set $y=0$ or $y=x$?



or do I define say $f(x,y):=x^2+y^2-y=0$ and do a multivariate expansion?






taylor-expansion







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  • the right-hand side "y" could represent f(x,y) and not the variable "y". Without this we don't have a relation or a function to apply T.S. for. Also see: math.stackexchange.com/questions/69610/…
    – NoChance
    Nov 26 '18 at 19:25
















4















Find the first 6 terms of the Taylor series for $y$ in powers of $x$ of the following implicitly defined function;



$$x^2 +y^2=y, y(0)=0$$




I'm a bit stuck in how to proceed do i need to implicitly differentiate the function such that $y'=frac{-2x}{(1-2y)}$ and again so as to find $y'',y^{(3)},...,y^{(6)}$ and then plug these into the Taylor expansion and set $y=0$ or $y=x$?



or do I define say $f(x,y):=x^2+y^2-y=0$ and do a multivariate expansion?






taylor-expansion







share|cite|improve this question






















  • the right-hand side "y" could represent f(x,y) and not the variable "y". Without this we don't have a relation or a function to apply T.S. for. Also see: math.stackexchange.com/questions/69610/…
    – NoChance
    Nov 26 '18 at 19:25














4












4








4








Find the first 6 terms of the Taylor series for $y$ in powers of $x$ of the following implicitly defined function;



$$x^2 +y^2=y, y(0)=0$$




I'm a bit stuck in how to proceed do i need to implicitly differentiate the function such that $y'=frac{-2x}{(1-2y)}$ and again so as to find $y'',y^{(3)},...,y^{(6)}$ and then plug these into the Taylor expansion and set $y=0$ or $y=x$?



or do I define say $f(x,y):=x^2+y^2-y=0$ and do a multivariate expansion?






taylor-expansion







share|cite|improve this question














Find the first 6 terms of the Taylor series for $y$ in powers of $x$ of the following implicitly defined function;



$$x^2 +y^2=y, y(0)=0$$




I'm a bit stuck in how to proceed do i need to implicitly differentiate the function such that $y'=frac{-2x}{(1-2y)}$ and again so as to find $y'',y^{(3)},...,y^{(6)}$ and then plug these into the Taylor expansion and set $y=0$ or $y=x$?



or do I define say $f(x,y):=x^2+y^2-y=0$ and do a multivariate expansion?






taylor-expansion




taylor-expansion






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share|cite|improve this question











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share|cite|improve this question










asked Nov 26 '18 at 19:06









seraphimk

1429




1429












  • the right-hand side "y" could represent f(x,y) and not the variable "y". Without this we don't have a relation or a function to apply T.S. for. Also see: math.stackexchange.com/questions/69610/…
    – NoChance
    Nov 26 '18 at 19:25


















  • the right-hand side "y" could represent f(x,y) and not the variable "y". Without this we don't have a relation or a function to apply T.S. for. Also see: math.stackexchange.com/questions/69610/…
    – NoChance
    Nov 26 '18 at 19:25
















the right-hand side "y" could represent f(x,y) and not the variable "y". Without this we don't have a relation or a function to apply T.S. for. Also see: math.stackexchange.com/questions/69610/…
– NoChance
Nov 26 '18 at 19:25




the right-hand side "y" could represent f(x,y) and not the variable "y". Without this we don't have a relation or a function to apply T.S. for. Also see: math.stackexchange.com/questions/69610/…
– NoChance
Nov 26 '18 at 19:25










3 Answers
3






active

oldest

votes


















1














hint



If$$y=a_1x+a_2x^2+...a_6x^6+...$$



then



$$y^2=a_1^2x^2+a_2^2x^4+a_3^2x^6+2a_1a_2x^3+2a_1a_3x^4+2a_1a_4x^5+2a_1a_5x^6+2a_2a_3x^5+2a_2a_4x^6+...$$



on the other hand



$$y-y^2=x^2$$



thus by identification,



$a_1=0$



$a_2=1$



can you take it.






share|cite|improve this answer































    0














    Your first approach is the correct one. Although, to find $y''(0)$, I think it's easier to differentiate $2x+2yy'=y'$ and then solve for $y''$ than to differentiate $frac{-2x}{1-2y}$. And so on for $y'''(0)$ etc.






    share|cite|improve this answer





















    • would I then set y=0? such that for example $y'(0)=-2x$ ?
      – seraphimk
      Nov 26 '18 at 19:41










    • @seraphimk Almost. You set $x=0$, and use $y(0)=0$ to get $y'(0)=0$. Next you differentiate $2x+2yy'=y'$, insert $x=0$ and use $y(0)=0$ and $y'(0)=0$ to find $y''(0)$. And so on.
      – Arthur
      Nov 26 '18 at 20:54





















    0














    Just added for your curiosity.



    If you use hamam_Abdallah's hint, writing
    $$y=sum_{k=1}^n a_k,x^k$$ and replacing, you will get
    $$0=-a_1 x+left(a_1^2-a_2+1right) x^2+(2 a_1 a_2-a_3) x^3+left(a_2^2+2 a_1
    a_3-a_4right) x^4+(2 a_2 a_3+2 a_1 a_4-a_5) x^5+left(a_3^2+2 a_2
    a_4+2 a_1 a_5right) x^6+(2 d_3 d_4+2 d_2 d_5+2 d_1 d_6-d_7) x^7+cdots$$     Since this holds for all $x$, set each coefficient equal to $0$ (do it for one at the time). You will quickly notice that $a_{2k+1}=0$ and that, for $a_{2k}$ the sequence is ${1,1,2,5,14,42,132,429,1430,cdots}$. These are Catalan numbers which you will find in many counting problems.



    This makes
    $$y=sum_{k=1}^infty C_k, x^{2k}=sum_{k=1}^infty frac{(2n)!}{n! , (n+1)!}, x^{2k}$$






    share|cite|improve this answer





















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      3 Answers
      3






      active

      oldest

      votes








      3 Answers
      3






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes









      1














      hint



      If$$y=a_1x+a_2x^2+...a_6x^6+...$$



      then



      $$y^2=a_1^2x^2+a_2^2x^4+a_3^2x^6+2a_1a_2x^3+2a_1a_3x^4+2a_1a_4x^5+2a_1a_5x^6+2a_2a_3x^5+2a_2a_4x^6+...$$



      on the other hand



      $$y-y^2=x^2$$



      thus by identification,



      $a_1=0$



      $a_2=1$



      can you take it.






      share|cite|improve this answer




























        1














        hint



        If$$y=a_1x+a_2x^2+...a_6x^6+...$$



        then



        $$y^2=a_1^2x^2+a_2^2x^4+a_3^2x^6+2a_1a_2x^3+2a_1a_3x^4+2a_1a_4x^5+2a_1a_5x^6+2a_2a_3x^5+2a_2a_4x^6+...$$



        on the other hand



        $$y-y^2=x^2$$



        thus by identification,



        $a_1=0$



        $a_2=1$



        can you take it.






        share|cite|improve this answer


























          1












          1








          1






          hint



          If$$y=a_1x+a_2x^2+...a_6x^6+...$$



          then



          $$y^2=a_1^2x^2+a_2^2x^4+a_3^2x^6+2a_1a_2x^3+2a_1a_3x^4+2a_1a_4x^5+2a_1a_5x^6+2a_2a_3x^5+2a_2a_4x^6+...$$



          on the other hand



          $$y-y^2=x^2$$



          thus by identification,



          $a_1=0$



          $a_2=1$



          can you take it.






          share|cite|improve this answer














          hint



          If$$y=a_1x+a_2x^2+...a_6x^6+...$$



          then



          $$y^2=a_1^2x^2+a_2^2x^4+a_3^2x^6+2a_1a_2x^3+2a_1a_3x^4+2a_1a_4x^5+2a_1a_5x^6+2a_2a_3x^5+2a_2a_4x^6+...$$



          on the other hand



          $$y-y^2=x^2$$



          thus by identification,



          $a_1=0$



          $a_2=1$



          can you take it.







          share|cite|improve this answer














          share|cite|improve this answer



          share|cite|improve this answer








          edited Nov 26 '18 at 19:31

























          answered Nov 26 '18 at 19:11









          hamam_Abdallah

          37.9k21634




          37.9k21634























              0














              Your first approach is the correct one. Although, to find $y''(0)$, I think it's easier to differentiate $2x+2yy'=y'$ and then solve for $y''$ than to differentiate $frac{-2x}{1-2y}$. And so on for $y'''(0)$ etc.






              share|cite|improve this answer





















              • would I then set y=0? such that for example $y'(0)=-2x$ ?
                – seraphimk
                Nov 26 '18 at 19:41










              • @seraphimk Almost. You set $x=0$, and use $y(0)=0$ to get $y'(0)=0$. Next you differentiate $2x+2yy'=y'$, insert $x=0$ and use $y(0)=0$ and $y'(0)=0$ to find $y''(0)$. And so on.
                – Arthur
                Nov 26 '18 at 20:54


















              0














              Your first approach is the correct one. Although, to find $y''(0)$, I think it's easier to differentiate $2x+2yy'=y'$ and then solve for $y''$ than to differentiate $frac{-2x}{1-2y}$. And so on for $y'''(0)$ etc.






              share|cite|improve this answer





















              • would I then set y=0? such that for example $y'(0)=-2x$ ?
                – seraphimk
                Nov 26 '18 at 19:41










              • @seraphimk Almost. You set $x=0$, and use $y(0)=0$ to get $y'(0)=0$. Next you differentiate $2x+2yy'=y'$, insert $x=0$ and use $y(0)=0$ and $y'(0)=0$ to find $y''(0)$. And so on.
                – Arthur
                Nov 26 '18 at 20:54
















              0












              0








              0






              Your first approach is the correct one. Although, to find $y''(0)$, I think it's easier to differentiate $2x+2yy'=y'$ and then solve for $y''$ than to differentiate $frac{-2x}{1-2y}$. And so on for $y'''(0)$ etc.






              share|cite|improve this answer












              Your first approach is the correct one. Although, to find $y''(0)$, I think it's easier to differentiate $2x+2yy'=y'$ and then solve for $y''$ than to differentiate $frac{-2x}{1-2y}$. And so on for $y'''(0)$ etc.







              share|cite|improve this answer












              share|cite|improve this answer



              share|cite|improve this answer










              answered Nov 26 '18 at 19:10









              Arthur

              110k7105186




              110k7105186












              • would I then set y=0? such that for example $y'(0)=-2x$ ?
                – seraphimk
                Nov 26 '18 at 19:41










              • @seraphimk Almost. You set $x=0$, and use $y(0)=0$ to get $y'(0)=0$. Next you differentiate $2x+2yy'=y'$, insert $x=0$ and use $y(0)=0$ and $y'(0)=0$ to find $y''(0)$. And so on.
                – Arthur
                Nov 26 '18 at 20:54




















              • would I then set y=0? such that for example $y'(0)=-2x$ ?
                – seraphimk
                Nov 26 '18 at 19:41










              • @seraphimk Almost. You set $x=0$, and use $y(0)=0$ to get $y'(0)=0$. Next you differentiate $2x+2yy'=y'$, insert $x=0$ and use $y(0)=0$ and $y'(0)=0$ to find $y''(0)$. And so on.
                – Arthur
                Nov 26 '18 at 20:54


















              would I then set y=0? such that for example $y'(0)=-2x$ ?
              – seraphimk
              Nov 26 '18 at 19:41




              would I then set y=0? such that for example $y'(0)=-2x$ ?
              – seraphimk
              Nov 26 '18 at 19:41












              @seraphimk Almost. You set $x=0$, and use $y(0)=0$ to get $y'(0)=0$. Next you differentiate $2x+2yy'=y'$, insert $x=0$ and use $y(0)=0$ and $y'(0)=0$ to find $y''(0)$. And so on.
              – Arthur
              Nov 26 '18 at 20:54






              @seraphimk Almost. You set $x=0$, and use $y(0)=0$ to get $y'(0)=0$. Next you differentiate $2x+2yy'=y'$, insert $x=0$ and use $y(0)=0$ and $y'(0)=0$ to find $y''(0)$. And so on.
              – Arthur
              Nov 26 '18 at 20:54













              0














              Just added for your curiosity.



              If you use hamam_Abdallah's hint, writing
              $$y=sum_{k=1}^n a_k,x^k$$ and replacing, you will get
              $$0=-a_1 x+left(a_1^2-a_2+1right) x^2+(2 a_1 a_2-a_3) x^3+left(a_2^2+2 a_1
              a_3-a_4right) x^4+(2 a_2 a_3+2 a_1 a_4-a_5) x^5+left(a_3^2+2 a_2
              a_4+2 a_1 a_5right) x^6+(2 d_3 d_4+2 d_2 d_5+2 d_1 d_6-d_7) x^7+cdots$$               Since this holds for all $x$, set each coefficient equal to $0$ (do it for one at the time). You will quickly notice that $a_{2k+1}=0$ and that, for $a_{2k}$ the sequence is ${1,1,2,5,14,42,132,429,1430,cdots}$. These are Catalan numbers which you will find in many counting problems.



              This makes
              $$y=sum_{k=1}^infty C_k, x^{2k}=sum_{k=1}^infty frac{(2n)!}{n! , (n+1)!}, x^{2k}$$






              share|cite|improve this answer


























                0














                Just added for your curiosity.



                If you use hamam_Abdallah's hint, writing
                $$y=sum_{k=1}^n a_k,x^k$$ and replacing, you will get
                $$0=-a_1 x+left(a_1^2-a_2+1right) x^2+(2 a_1 a_2-a_3) x^3+left(a_2^2+2 a_1
                a_3-a_4right) x^4+(2 a_2 a_3+2 a_1 a_4-a_5) x^5+left(a_3^2+2 a_2
                a_4+2 a_1 a_5right) x^6+(2 d_3 d_4+2 d_2 d_5+2 d_1 d_6-d_7) x^7+cdots$$                 Since this holds for all $x$, set each coefficient equal to $0$ (do it for one at the time). You will quickly notice that $a_{2k+1}=0$ and that, for $a_{2k}$ the sequence is ${1,1,2,5,14,42,132,429,1430,cdots}$. These are Catalan numbers which you will find in many counting problems.



                This makes
                $$y=sum_{k=1}^infty C_k, x^{2k}=sum_{k=1}^infty frac{(2n)!}{n! , (n+1)!}, x^{2k}$$






                share|cite|improve this answer
























                  0












                  0








                  0






                  Just added for your curiosity.



                  If you use hamam_Abdallah's hint, writing
                  $$y=sum_{k=1}^n a_k,x^k$$ and replacing, you will get
                  $$0=-a_1 x+left(a_1^2-a_2+1right) x^2+(2 a_1 a_2-a_3) x^3+left(a_2^2+2 a_1
                  a_3-a_4right) x^4+(2 a_2 a_3+2 a_1 a_4-a_5) x^5+left(a_3^2+2 a_2
                  a_4+2 a_1 a_5right) x^6+(2 d_3 d_4+2 d_2 d_5+2 d_1 d_6-d_7) x^7+cdots$$                   Since this holds for all $x$, set each coefficient equal to $0$ (do it for one at the time). You will quickly notice that $a_{2k+1}=0$ and that, for $a_{2k}$ the sequence is ${1,1,2,5,14,42,132,429,1430,cdots}$. These are Catalan numbers which you will find in many counting problems.



                  This makes
                  $$y=sum_{k=1}^infty C_k, x^{2k}=sum_{k=1}^infty frac{(2n)!}{n! , (n+1)!}, x^{2k}$$






                  share|cite|improve this answer












                  Just added for your curiosity.



                  If you use hamam_Abdallah's hint, writing
                  $$y=sum_{k=1}^n a_k,x^k$$ and replacing, you will get
                  $$0=-a_1 x+left(a_1^2-a_2+1right) x^2+(2 a_1 a_2-a_3) x^3+left(a_2^2+2 a_1
                  a_3-a_4right) x^4+(2 a_2 a_3+2 a_1 a_4-a_5) x^5+left(a_3^2+2 a_2
                  a_4+2 a_1 a_5right) x^6+(2 d_3 d_4+2 d_2 d_5+2 d_1 d_6-d_7) x^7+cdots$$                   Since this holds for all $x$, set each coefficient equal to $0$ (do it for one at the time). You will quickly notice that $a_{2k+1}=0$ and that, for $a_{2k}$ the sequence is ${1,1,2,5,14,42,132,429,1430,cdots}$. These are Catalan numbers which you will find in many counting problems.



                  This makes
                  $$y=sum_{k=1}^infty C_k, x^{2k}=sum_{k=1}^infty frac{(2n)!}{n! , (n+1)!}, x^{2k}$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 27 '18 at 8:01









                  Claude Leibovici

                  119k1157132




                  119k1157132






























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