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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?

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  • $\begingroup$ 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/… $\endgroup$ – NoChance Nov 26 '18 at 19:25
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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.

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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.

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  • $\begingroup$ would I then set y=0? such that for example $y'(0)=-2x$ ? $\endgroup$ – seraphimk Nov 26 '18 at 19:41
  • $\begingroup$ @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. $\endgroup$ – Arthur Nov 26 '18 at 20:54
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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 $$0=-a_1 x+\left(a_1^2-a_2+1\right) x^2+(2 a_1 a_2-a_3) x^3+\left(a_2^2+2 a_1 a_3-a_4\right) x^4+(2 a_2 a_3+2 a_1 a_4-a_5) x^5+\left(a_3^2+2 a_2 a_4+2 a_1 a_5\right) x^6+(2 d_3 d_4+2 d_2 d_5+2 d_1 d_6-d_7) x^7+\cdots$$ 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 $$y=\sum_{k=1}^\infty C_k\, x^{2k}=\sum_{k=1}^\infty \frac{(2n)!}{n! \, (n+1)!}\, x^{2k}$$

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