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?

  • 2
    $\begingroup$ The symbol "$\approx$" leaves space for many different interpretations... Looking at the expressions it seems that since $\varepsilon$ is very small it is reasonable to admit that $x \approx y$ (from (1)). You'll get (2) if you replace $y$ by $x$ in the RHS of (1).\ $\endgroup$ Commented Jul 23, 2021 at 12:42

2 Answers 2


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.

  • $\begingroup$ I tried as you say, but the second-order partial derivatives are also zero... So the expansion remains $y(x,z) \approx x$, without the $\varepsilon$ $\endgroup$
    – Invenietis
    Commented Jul 23, 2021 at 17:09
  • $\begingroup$ If all second order derivatives are zero, you move to the third order terms. In fact, if you look at the expression you want to obtain, there are only first and third order terms. $\endgroup$ Commented Jul 23, 2021 at 23:44

While the accepted answer is focused on performing taylor expansions in $x$ and $z$, it is worth noting that we can also perform an expansion in $\varepsilon$ - this is the basis of perturbation analysis, as we know $\varepsilon$ to be a small number.

To achieve this, we begin by assuming that $y$ can be expressed in the form $y=y_0+\varepsilon y_1+\varepsilon^2 y_2 + O(\varepsilon^3)$ (where $O(\varepsilon^3)$ is big-O notation, meaning terms of that size or smaller - that is, higher powers of $\varepsilon$), which gives us $$ x=y_0+\varepsilon y_1+\varepsilon^2 y_2 + O(\varepsilon^3) + \varepsilon \left(a(y_0 + \varepsilon y_1 + O(\varepsilon^2))^3+bz^2(y_0+\varepsilon y_1 + O(\varepsilon^2))\right) $$ which, on expanding and simplifying, can be written as $$ x=y_0+\varepsilon y_1 + \varepsilon^2 y_2 + \varepsilon ay_0^3+\varepsilon^2 3ay_0^2y_1+\varepsilon bz^2y_0+\varepsilon^2 bz^2y_1 + O(\varepsilon^3) $$ or, collecting the various powers of $\varepsilon$, $$ x=y_0 + \varepsilon (y_1+ay_0^3+bz^2y_0) + \varepsilon^2 (y_2 + 3ay_0^2y_1+bz^2y_1) + O(\varepsilon^3) $$ Now, we just match powers. For $\varepsilon^0$, we have $x=y_0$, and thus $$y_0=x$$

For $\varepsilon^1$, we have $y_1+ay_0^3+bz^2y_0 = 0$. And since $y_0=x$, we have $$y_1=-ax^3+bxz^2$$

This gives $$ y = x - \varepsilon (ax^3 + bxz^2) + O(\varepsilon^2) $$ and we could then examine the $\varepsilon^2$ terms to get $y_2 + 3ay_0^2y_1+bz^2y_1=0$, or $$ y_2 = (3ax^2 + bz^2)(ax^3+bxz^2) $$ which means we have $$ y = x - \varepsilon (ax^3 + bxz^2) + \varepsilon^2 (3ax^2 + bz^2)(ax^3+bxz^2) + O(\varepsilon^3) $$ It is worth noting that we assume that $\varepsilon$ is sufficiently small for the given $x$ and $z$ values.


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