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

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?

• 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

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.

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.

• 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

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