# how do I find the taylor polynomial of multivariable functions?

I know taylor polynomial for single variable functions but I am having trouble understanding how to find taylor polynomials for multivariable functions. I know how to find partial derivatives as well as the chain rule for multivariable functions.

Can someone please explain how I can find the taylor polynomial of a multivariable function? for simplicity lets do f(x,y)=x^5-y^4

Above is text from my textbook that I read but it doesnt make any sense to me especially step 3 and 4

If you know of a site or source that can better explain how to find multivariable function's taylor series I would appreciate that so that this way you dont spend so much time in typing explanation for me.

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One of my uncertainties is exactly what is tripping you up in trying to understand (3) and (4). All four equations are the exact same: the function on the left-hand side is being considered as solely a function of either $x$ or $y$, and you take the expansion of that function and put it on the right. My assumption is that you didn't realize you could take partial derivatives of something like $f_x(x_0,y)$ with respect to $y$ - is that accurate? – anon Mar 4 '12 at 9:43
Instead of doing such long equations in the book I prefer that would write the steps as to what to do to find the taylor polynomial. for example, given this function, take its partial derivatives with respect to x and add them up, take the partial derivatives with respect to y and add those up, then multiply the sum of the derivatives...I am having trouble following the steps because of so much arithmatic, if I was told in simple steps of what to do then i can do those rather than follow up with all the lines trying to figure out what happened. – Raynos Mar 4 '12 at 13:41
The assumption you mentioned, you mean if you have fx(0,0) then u can differentiate it with respect to y because it would have a y in it? if there is no y in there the expression is 0 because x would be a constant – Raynos Mar 4 '12 at 13:44

We can fix $y$ and then consider $f(x,y)$ to be strictly a function of $x$ (say $h(x)=f(x,y)$ if you want, which makes sense because $y$ is fixed somewhere). Doing this yields the Taylor expansion seen in equation $(1)$. We would say that, by the usual formula, we have

$$h(x)=h(x_0)+h'(x_0)(x-x_0)+h''(x_0)(x-x_0)^2+\cdots \tag{0}$$

Now $h(x_0)=f(x_0,y)$ and $h'(x_0)=f_x(x_0,y)$ and $h''(x_0)=f_{xx}(x_0,y)$ and so forth. Now notice that this works regardless of what we fixed $y$ at to begin with, so it must hold for all available $y$.

We will switch gears now; since the formula holds for all $y$ we no longer need to consider $y$ fixed.

The Taylor expansion in the variable $x$ seen in $(0)$ involves a number of terms like $f_{xx}(x_0,y)$. Now we know that $x_0$ is fixed but $y$ is not, so this is a function of $y$! Say $g(y)=f_{xx}(x_0,y)$. Then we can speak of its Taylor expansion just as well, $g(y)=g(y_0)+g'(y)(y-y_0)+g''(y)(y-y_0)+\cdots$.

It turns out that $g(y_0)=f_{xx}(x_0,y_0)$ and $g'(y_0)=f_{xxy}(x_0,y_0)$ and $g''(y_0)=f_{xxyy}(x_0,y_0)$ and so on, because taking partial derivatives commutes with "evaluating at $y=y_0$" (or $x=x_0$, as in the last part). This means that differentiation and plugging things in can be done in any order here.

Doing a Taylor series expansion in the variable $y$ for $f_{xx}(x_0,y)$, as seen in $(4)$, can be done with any of the terms in $(1)$, like $f(x_0,y)$ and $f_x(x_0,y)$ seen in $(2)$ and $(3)$.

I have assumed that $f$ is sufficiently nice in this answer.

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