Question about showing the equation defines an implicit function and give the Taylor expansion I have a question while doing the exercise, I want to be sure if my step is correct, the exercise is:
Show that the equation
$$x^3 - y^3 -3xy + 1 = 0$$
defines an implicit function $\phi:x \to y$ in a neighbourhood of $0$ such that $\phi (0) =1$. Give the third order Taylor expansion of $\phi$ in this neighbourhood of $0$.
Here is my attempt:
I have first calculated the derivative of $\frac{dF}{dx}$ and $\frac{dF}{dy}$, and then find $dF=(3x^2-3y  \ \ \ -3y^2-3x)$, then we have $det(\frac{dF}{dy})(0)=(-3y^2-3x)(0)=0$ but $\phi(0)=1 \ne 0$
now we have $$\frac{d\phi}{dy} = -[\frac{dF}{dy}(x,\phi(x)]^{-1}[\frac{dF}{dx}(x,\phi(x)]=\frac{3x^2-3y}{3y^2-3y}$$
Then Taylor expansion for $\phi(x)$ at $0$ is:
$$\phi(0) + \phi'(0)(x) +\frac{\phi''(0)(x^2)}{2}+\frac{\phi'''(0)(x^3)}{6}+ ...$$
which gives me:
$$1+0+0+0+...=1$$
I doubt my step is not correct, can somebody helps me to check, thank you very much in advance.
 A: You are asked to show that $\phi$ exists in the first place - the implicit function theorem comes to your aid here. It is sufficient to show that there exists $(a,b)$ such that $f(a,b)=0$, and that $D_yf(a,b)$ is invertible - in the one-dimensional case, this just means that $D_yf(a,b)\neq0$. Such an $(a,b)$ is found with $(0,1)$, as hinted by your problem statement - and is also easily verified. At $(0,1)$, $D_yf=-3y^2-3x=-3\neq0$, so the implicit function theorem guarantees that there is an open neighbourhood of $a$ and $b$ in which $f(x,\phi(x))=0$ for some $\phi\in C^k$, where $C^k$ is the continuity class of $f$. Our $f$ here is smooth, and thus so is $\phi$.
Now we have shown $\phi$ exists, we need its higher derivatives at $0$. $f(x,\phi(x))$ describes a contour (it is always $0$ by construction), so its derivative w.r.t $x$ should be everywhere $0$.
$$\begin{align}0&=D_xf(x,\phi(x))\\&=D_x[x^3-\phi^3(x)-3x\cdot\phi(x)+1]\\&=3x^2-3\phi^2(x)\cdot\phi'(x)-3\cdot\phi(x)-3x\cdot\phi'(x)\\\implies \tag{1}x^2&=\phi^2(x)\cdot\phi'(x)+\phi(x)+x\cdot\phi'(x)\end{align}$$
Evaluating at $x=0$, remembering that $\phi(0)=1$ by construction, we have $\phi'(0)=-1$. We can take higher derivatives of $(1)$ as $\phi$ is smooth:
$$2\cdot\phi(x)\cdot(\phi'(x))^2+\phi^2(x)\cdot\phi''(x)+2\phi'(x)+x\cdot\phi''(x)=2x$$
Which at $x=0$ gives: $2\cdot(-1)^2+1\cdot\phi''(0)+2(-1)=0$ and thus $\phi''(0)=0$.
Taking third derivatives of $(1)$ gives: $$2(\phi'(x))^3+4\phi(x)\phi'(x)\phi''(x)+2\phi(x)\phi'(x)\phi''(x)+\phi^2(x)\phi'''(x)+3\phi''(x)+x\phi'''(x)=2$$
Ugly, but at $x=0$, and using previous results, one has:
$$2(-1)^3+0+0+1\cdot\phi'''(0)+0+0=2\\\therefore\phi'''(0)=4$$
So the Taylor expansion of $\phi$ near $0$ is:
$$\phi(x)\approx 1-x+\frac{2}{3}x^3$$
Which a quick sketch on Desmos can confirm!
Not necessary for the question, but it is interesting to keep going as the third-order expansion is really not very accurate...
Taking fourth derivatives of $(1)$ (I evaluate at $x=0$, so all $\phi''(x)$ terms will be deleted for convenience) gives:
$$6\phi(x)\phi'(x)\phi'''(x)+2\phi(x)\phi'(x)\phi'''(x)+\phi^2(x)\phi''''(x)+3\phi'''(x)+\phi'''(x)=0,x=0\\\therefore-24-8+\phi''''(0)+16=0$$
So one has $\phi''''(0)=16$. The approximation:
$$\phi(x)\approx1-x+\frac{2}{3}x^3+\frac{2}{3}x^4$$
Is much more accurate, and is nice to look at.
Hope this helped - there's a lot of fiddly product and chain rule business here, I can elaborate on any confusing steps.
EDIT:
During lesson time, out of a lack of things to do, I pressed on - it was mindlessly tedious, but the following 10th order approximation I have visually shown (I rigorously showed it of course, but with so many product derivative terms it is possible some of these numbers are slightly out) to be far superior to the 4th order one:
$$\phi(x)\approx1-x+\frac{2}{3}x^3+\frac{2}{3}x^4-\frac{10}{9}x^6-\frac{14}{9}x^7+\frac{274}{81}x^9+\frac{3118}{567}x^{10}$$
Which is accurate for $|x|\lt0.4$, and extremely accurate as you get closer to $0$.
