Multivariable calculus: How do I find the Taylor series for a function about a certain point?

Question

Suppose $$f(x,y) = x^2 + xy + y^3$$

I have to find the Taylor series for this function about the point $(1,-1)$. How should one solve this?

Now usually, I'm used to solving questions that ask you to find the Taylor polynomial of a certain degree of a function near a given point. But I feel like this one is different. It doesn't ask you to find the taylor polynomial of a certain degree. It just asks you to find the taylor series about a given point. I suppose there's a different way to do this.

EDIT: Here's a new example:

f(x,y) = 1/(2 + xy^2)

Suppose you had to find the taylor series for this function about a certain point, say (0,0).

In this case, calculating every partial derivative would be quite painstaking. Is it possible to find the taylor series for this function without calculating all the partial derivatives?

Answer 1

If it exists, the Taylor series of $f$ at $a$ looks like:

$$\sum_{n=0}^{\infty}\frac1{n!} D^nf(a)(x-a)^n$$

where e.g. $D^2 f(a)(x-a)^2 = [((D(Df))(a))(x-a)](x-a)$.

The easiest way to compute the derivative map is through partial differentiation. So there isn't any easier way to do your problem without going to partial derivatives. If these have no nice closed form them the Taylor series is simply too ugly to compute.

Hope this helps

Answer 2

Since $f$ is a polynomial it is for each point $(x_0,y_0)$ its own Taylor expansion at $(x_0,y_0)$ in disguise. Given $(x_0,y_0):=(1,-1)$ write $x:=1+\xi$, $y:=-1+\eta$ and obtain $$\eqalign{\hat f(\xi,\eta)&=f(1+\xi,-1+\eta)=(1+\xi)^2+(1+\xi)(-1+\eta)+(-1+\eta)^3\cr &=-1+(\xi+4\eta)+(\xi^2+\xi\eta-3\eta^2)+\eta^3\ .\cr}$$ Of course you can rewrite that in the form $$f(x,y)=-1+\bigl((x-1)+4(y+1)\bigr)+\bigl((x-1)^2+(x-1)(y+1)-3(y-1)^2\bigr)+(y+1)^3\ .$$