# Where to stop a taylor expansion of a function of more than one variable?

I tried to taylor expand a function of three variables $f(x,y,z)$ around $y=0,z=0$ but I don't find a way to decide where to stop the expansion.

Usually when one expands a function of one variable $g(x)$ around $x=0$

$g(0)+\partial_x g(0) \cdot x + \frac 1 2 \partial^2_x g(0) \cdot x^2 + \mathcal O (x^3)$

one can immediately say that contributions of (say) $x^3$ or higher can be neglected if $x^3$ is already smaller than the desired precision.

But in the case of three variables I get an expansion like

$f(x,0,0) + \partial_yf(x,0,0)\cdot y + \partial_zf(x,0,0)\cdot z+\frac 1 2 \big(\partial_y^2f(x,0,0)\cdot y^2 + 2\partial_y\partial_z f(x,0,0)\cdot yz + \partial^2_z f(x,0,0)\cdot z^2\big) + ...$

but I don't have any clue where to stop. I think the reason is that there are still functions on $x$ to which I can't say anything about their order in opposite to the one dimensional case where just constants appear.

So for example when I say that both $y$ and $z$ are very small but $z$ is way smaller than $y$ and I keep a term like $\partial^2_z f(x,0,0)\cdot z^2$ then there is no reason to neglect the term (say) $\partial_y^7\partial_zf(x,0,0)\cdot y^7z$ since the latter can be bigger than the former for some $x$.

Can someone help me out?

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The Taylor expansion of degree $k$ will involve all terms $x^ay^bz^c$ with $0\le a+b+c\le k$, and then the error will be $O\big((x^2+y^2+z^2)^{(k+1)/2}\big)$. You need to work simultaneously with all three variables.
thanks for your answer. hm, but then it is pointless for my situation. because in my case $y$ and $z$ are very small (almost 0) and $x$ shall be general. so is there no way to handle that? – thyme May 26 '14 at 14:50
Well, you can handle it by fixing $x=x_0$ for any value of $x_0$ and looking near the point $(x_0,0,0)$. What I said then applies to the polynomial whose terms are of the form $(x-x_0)^ay^bz^c$ ... – Ted Shifrin May 26 '14 at 14:59
Hm, no this is no option for me. The dependence on $x$ is so important that I can not expand it (I would at least need terms with $x^6$)... but what I did is the following: I just ignored the x dependency and pretended that I have a funcion just of $y$ and $z$. I then just guessed the important orders and the result agrees perfectly with the exact version of the function. I think there must be a way to understand that mathematically... – thyme May 27 '14 at 7:56
Sure. You can do that. Mathematically, we think of a family of functions $f_x(y,z)$, where we view $x$ as a parameter. – Ted Shifrin May 27 '14 at 11:42
Ok, thanks. But I now have a problem with the error term $\mathcal O\big((y^2+z^2)^{(k+1)/2}\big)$, because my variables have physical units. $y$ is an angle in radiant and $z$ is a length in meter, so adding up $y^2+z^2$ makes no sense. I already saw at some stage before that it is problematic to expand a function in variables with units... but what should I do? I tried to expand in unitless variables but it is a bit artificial then... of course at the end the forced parameter (which I introduced to make a variable unitless) cancels out... – thyme May 27 '14 at 13:05