Confusion surrounding Big-$O$ notation in multidimensional Taylor series

Question

Let $f: \mathbb R^{2}\to \mathbb R, f(x,y)=y-e^{x}\sin(y)$

I was asked to find the the Taylor polynomial about $(1,\pi)$ until the terms of third order.

Solution: setting $z=(1+x,\pi+y)$, we get $f(z)=y+\pi-e^{x+1}\sin(y+\pi)=\pi+y+ee^{x}\sin(y)$

and then taking the Taylor polynomials of that we get:

$= \pi + y +e(1+x+\frac{x^{2}}{2!}+\frac{x^{3}}{3!}+\mathcal{O(\vert\vert x,y\vert\vert^{4})})(y-\frac{y^{3}}{3!}+\mathcal{O}(\vert\vert x,y\vert\vert^{5}))$ (*)

And since $\mathcal{O}(g)\mathcal{O}(f)=\mathcal{O}(g\times f)$ and for $h=\mathcal{O}(f_{m})$ and $g=\mathcal{O}(f_{n})$, it follows $g,h=\mathcal{O}(\max\{f_{n},f_{m}\})$

Am I correct in saying that (*)$=...+\mathcal{O}(\vert\vert x,y\vert\vert^{9})$

Somehow our lecturer got to (*)$=...+\mathcal{O}(\vert\vert (x,y)\vert\vert^{4})$, and I have no idea how he got that.

Answer 1

Let us forget about the $\pi$, the $y$ and the $e$ to focus on what matters, the big product.

Using distributivity, it becomes a sum $$\ldots+\text{ $\ldots\mathcal{O}(\| x,y\|^{4})$ $+$ $\ldots\mathcal{O}(\| x,y\|^{5})$ $+\ldots\mathcal{O}(\| x,y\|^{9})$.}$$ where each $\ldots$ stands for a polynomial.

Now the fundamental question is what precision do we have when we write this?

• The part of degree $2$ contains $y+xy$ for instance, and none of the $\mathcal O$ is going to bring you any terms of degree $1$ or $2$ to spoil that.
• The part of degree $6$ contains $\frac{x^3}{3!}\frac{y^3}{3!}$. But it is not really relevant to write that part down, because we also have $y\mathcal{O}(\| x,y\|^{4})$ and the $\mathcal{O}(\| x,y\|^{4})$ can contain unknown terms of degree $5$, which together with the factor $y$ may alter or even cancel your $\frac{x^3}{3!}\frac{y^3}{3!}\ldots $ or not! There is no way you can know (unless you go further in the series for each factor).

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A related problem: Say you know the value of a number $x\sim 3.1415 $ up to four digits and another one $y\sim 2.71828$ up to five digits. Is this enough to write down $xy$ with a precision of nine digits?