Help with Taylor polynomial

Question

I need to find the Taylor polynomial of order $2n$ of the function $$f(x,y)=\frac{1}{1+xy}$$ on $(x_0,y_0)=(0,0)$. Can anyone give me a hand please?

Sorry, i make a mistake, their should say "polynomial" instead of "series, i have corrected now.

score 3 · Answer 1 · answered May 17 '13 at 03:08

3

Hint, for $|u|<1$, $$\frac{1}{1-u}=1+u+u^2+u^3+\ldots.$$

answered May 17 '13 at 03:08

Jonathan

thanks for the answer, but i correct now, i mean Taylor polynomial, or the order $2n$ doesn´t make any sense. – Dimitri May 17 '13 at 03:27
@Dimitri so only keep a fraction of the terms... – obataku May 17 '13 at 05:00

score 0 · Answer 2 · edited Apr 13 '17 at 12:20

A related technique. Here is the general formula for the Taylor series of a function in two variables around the point $(x,y)=(0,0)$

$$ f(x,y)=\sum_{k=0}^{\infty}\sum_{m=0}^{\infty}\frac{x^m y^k}{m! k!}\frac{\partial^{m+k}}{\partial x^m \partial y^k }f(x,y)|_{x=0,y=0} . $$

So, to get the order $2n$, you can consider the truncated series

$$ f(x,y)=\sum_{k=0}^{n}\sum_{m=0}^{n}\frac{x^m y^k}{m! k!}\frac{\partial^{m+k}}{\partial x^m \partial y^k }f(x,y)|_{x=0,y=0}. $$

You may want to show that

$$ \frac{\partial^{m+k}}{\partial x^m \partial y^k }\frac{1}{1-xy}\Big|_{x=0,y=0}=0\,\quad \mathrm{ when }\quad m\neq n. $$

Otherwise, use Jonathan's technique to get the required series.

2 Answers2