Given this implicit equation, where $a$ and $b$ are constants, and assuming that $\varepsilon$ is very small,
$$x =y+\varepsilon(ay^3+byz^2) \tag{1}$$
I'm tryng to approximate the expression of the variable $y$ in terms of the variables $x$ and $z$, $y(x,z)$, by writting a Taylor series of $\varepsilon$ up to first order.
The result I'm supposed to get is
$$y\approx x-\varepsilon(ax^3+bxz^2) \tag{2}$$
How a Taylor series expansion should be approached for an implicit function, such as the one in this case?
Best Answer
You can differentiate (1) and get partial derivatives of $y$ with respect to $x$ and $z$. With those, you can use Taylor's formula for functions with two variables to approximate $y(x,z)$. For instance, differentiating with respect to $x$, you get $$ x = y + \varepsilon(a y^2 + byz^2) \Rightarrow 1 = \frac{\partial y}{\partial x}+\varepsilon (a \frac{\partial y}{\partial x}\cdot y+b z^2 \frac{\partial y}{\partial x}) $$
Substituting $x=z=0$ and knowing that $y(0,0)=0$, you obtain $\frac{\partial y}{\partial x}(0,0) = 1$. Similarly, you get $\frac{\partial y}{\partial z}(0,0)=0$. The first order approximation would be
$$ y(x,z) \approx y(0,0) + x\frac{\partial y}{\partial x}(0,0) + z \frac{\partial y}{\partial z}(0,0) = x. $$
Now you just need to go on.