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?

  • 1
    $\begingroup$ After a certain point the partial derivatives vanish. So there's no series here just a few computations.. $\endgroup$ – Cauchy Jul 7 '17 at 11:00
  • $\begingroup$ But is there a way to do it that doesn't involve calculating all the partial derivatives? $\endgroup$ – User95 Jul 7 '17 at 11:08
  • $\begingroup$ Just calculate the partial derivatives as usual is probably the easiest way. $\endgroup$ – mathreadler Jul 7 '17 at 11:08
  • $\begingroup$ You could try dividing the polynomial by $(y+1)$ and $(x-1)$ respectively to kind of refactor it, although in general trying to factor multivariate polynomials can become tricky. $\endgroup$ – mathreadler Jul 7 '17 at 11:10
  • $\begingroup$ I solved it by calculating all the partial derivatives and got the correct answer as expected. However, the question I wrote in the description was just an example. I've come across similar questions where the partial derivatives just go on for a long time. You can't expect someone calculate every partial derivative in that case. It would be painstaking. That's why I'm suspecting that there might be different ways to solve this....way that don't involve calculating every partial derivative. $\endgroup$ – User95 Jul 7 '17 at 11:26

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 $$f(x,y)=-1+\bigl((x-1)+4(y+1)\bigr)+\bigl((x-1)^2+(x-1)(y+1)-3(y-1)^2\bigr)+(y+1)^3\ .$$


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


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