I'm going to try asking this question in two different ways, because either may be easier to understand. The first one is more specific, and the second one is more general.

Specific Question

Let's say that I have a function $f:\mathbb{R} \mapsto \mathbb{R} $, and its taylor polynomial about point $P$ to degree $n$ is $P^f_n$. Now let's say that I have another function $g: \mathbb{R}^n \mapsto \mathbb{R}$. The domains and such of the functions assures that I can nest: $$z(\vec{x}) = f(g(\vec{x}))$$ given that $\vec{x} \in D(g)$. If I do the same thing to the taylor polynomials, about which point will the new polynomial be expanded? To clarify, consider: $$P(\vec{x}) = P_n^f(g(\vec{x}))$$ To my knowledge, this nested polynomial is a taylor expansion of $z$ around a point, but if this is true, which one?

General Question

If I have a composite function $z:\mathbb{R}^n \mapsto \mathbb{R}$ and I want to find its taylor polynomial of degree $n$ around point $\vec{P}$, how can I accomplish this using the taylor polynomials of the outer function? (And maybe even inner functions, if needed, such as with $z = \sin(\sin(x) + \sin(y))$.) What are the exact demands that assures me that the result will be around the point $\vec{P}$, rather than some other point? I'm mainly curious about the demands, because in the past, I've come across cases where not being certain of the demands, leading me to a completely different answer from what I expected. Basically, all I want is to be certain that a nested function is expanded around an asked point, when having the outer- and inner functions' taylor expansions.

Complicated example (to visualize)

Let's say that I have a slightly complicated function, such as:

$$z(x, y) = \sin((x-1)^5(y+2)(x+3)(y-18)^{13})$$

and I want to find its taylor expansion around $(1,-2)$ to a certain degree. I know the taylor expansion for $\sin x$ around $x = 0$, so one can argue that I can just substitute $(x-1)^5(y+2)(x+3)(y-18)^{13}$ in that formula, giving me the result for the expansion of $z$ around $(1, -2)$, because: $$(x, y) = (1, -2) \implies (x-1)^5(y+2)(x+3)(y-18)^{13} = 0$$ However, this is a one-way arrow, meaning that there are other points around which it's $0$. So couldn't you just argue that the obtained expansion, if you use this method, is around point $(-3, 18)$ as well? As you can tell, I'm somewhat confused about what's really going on.

  • $\begingroup$ From what I've figured out, the nested taylor polynomial doesn't even have to be a taylor polynomial. As long as $g(\vec{x})$ is close to a value $x_0$ when $\vec{x}$ is close to $\vec{x_0}$, you can nest it with another function's expansion around $x_0$ and get a taylor polynomial around $\vec{x_0}$. However, it may require some rewriting. $\endgroup$
    – Max
    Commented May 27, 2017 at 9:20


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