# [SOLVED] General form of a Taylor's polynomial approx. to a multivariable function

• May 18th 2009, 04:12 PM
arbolis
[SOLVED] General form of a Taylor's polynomial approx. to a multivariable function
The problem states : Find the Taylor's polynomial of degree $\displaystyle 2n$ of the function $\displaystyle f(x,y)=\frac{1}{1+xy}$ in the point $\displaystyle (0,0)$.
My attempt: I've calculated the polynomial of degree $\displaystyle 0$ and $\displaystyle 2$. ( I doubt my choice of degree $\displaystyle 0$ is a good one... but as there's no restriction over $\displaystyle n$, I can guess it is not a bad choice.)
I got that $\displaystyle P_0(0,0)=1$ and $\displaystyle P_2(0,0)=1$.
So $\displaystyle P_{2n}=1$, $\displaystyle \forall n \geq 0$.
I can't believe my result. Calculating $\displaystyle P_4(0,0)$ seems a really long work! Have I done what I've done right?
• May 18th 2009, 04:28 PM
Jester
Quote:

Originally Posted by arbolis
The problem states : Find the Taylor's polynomial of degree $\displaystyle 2n$ of the function $\displaystyle f(x,y)=\frac{1}{1+xy}$ in the point $\displaystyle (0,0)$.
My attempt: I've calculated the polynomial of degree $\displaystyle 0$ and $\displaystyle 2$. ( I doubt my choice of degree $\displaystyle 0$ is a good one... but as there's no restriction over $\displaystyle n$, I can guess it is not a bad choice.)
I got that $\displaystyle P_0(0,0)=1$ and $\displaystyle P_2(0,0)=1$.
So $\displaystyle P_{2n}=1$, $\displaystyle \forall n \geq 0$.
I can't believe my result. Calculating $\displaystyle P_4(0,0)$ seems a really long work! Have I done what I've done right?

Not really sure what your $\displaystyle P_0(0,0)$ and $\displaystyle P_2(0,0)$ is but you'll notice that the function is

$\displaystyle f = \frac{1}{1+u}$ where $\displaystyle u = xy$ so you can use

$\displaystyle \frac{1}{1+u} = 1 - u + u^2 - u^3 \cdots$
• May 18th 2009, 05:03 PM
arbolis
Thanks for the quick response.
Oh... I didn't realize that I could use $\displaystyle \frac{1}{1+u} = 1 - u + u^2 - u^3 \cdots$.

A stupid question from my part : from your formula, is $\displaystyle P_0(x,y)=1$ and $\displaystyle P_1(x,y)=1-xy$, $\displaystyle P_2(x,y)=1-xy+x^2y^2$, etc...?
If so then I was right by saying that $\displaystyle P_{2n}(0,0)=1$. (I don't think I was right since it would imply that $\displaystyle P_n(0,0)=1$, no matter what $\displaystyle n$ is).
• May 18th 2009, 05:18 PM
Jester
Quote:

Originally Posted by arbolis
Thanks for the quick response.
Oh... I didn't realize that I could use $\displaystyle \frac{1}{1+u} = 1 - u + u^2 - u^3 \cdots$.

A stupid question from my part : from your formula, is $\displaystyle P_0(x,y)=1$ and $\displaystyle P_1(x,y)=1-xy$, $\displaystyle P_2(x,y)=1-xy+x^2y^2$, etc...?
If so then I was right by saying that $\displaystyle P_{2n}(0,0)=1$. (I don't think I was right since it would imply that $\displaystyle P_n(0,0)=1$, no matter what $\displaystyle n$ is).

OK, I see your form. The Taylor series in 2 spatial variables is

$\displaystyle z = f(a,b) + \frac{\partial f}{\partial x}(a,b)(x-a) + \frac{\partial f}{\partial y}(a,b)(y-b)$

$\displaystyle + \frac{\partial^2 f}{\partial x^2}(a,b)\frac{(x-a)^2}{2!} + 2 \frac{\partial^2 f}{\partial x \partial y}(a,b)\frac{(x-a)(y-b)}{2!}+ \frac{\partial^2 f}{\partial^2 y}(a,b)\frac{(y-b) ^2}{2!} \cdots$.
• May 18th 2009, 06:08 PM
arbolis
Quote:

Originally Posted by danny arrigo
OK, I see your form. The Taylor series in 2 spatial variables is

$\displaystyle z = f(a,b) + \frac{\partial f}{\partial x}(a,b)(x-a) + \frac{\partial f}{\partial y}(a,b)(y-b)$

$\displaystyle + \frac{\partial^2 f}{\partial x^2}(a,b)\frac{(x-a)^2}{2!} + 2 \frac{\partial^2 f}{\partial x \partial y}(a,b)\frac{(x-a)(y-b)}{2!}+ \frac{\partial^2 f}{\partial^2 y}(a,b)\frac{(y-b) ^2}{2!} \cdots$.

This is exactly what I've done on my draft and I got $\displaystyle P_2(0,0)=1$ because all the terms were equal to $\displaystyle 0$, except the first (that is $\displaystyle f(0,0)$ which is worth 1.)
• May 19th 2009, 06:15 PM
arbolis
Oh! I just realized my error...
$\displaystyle P_2(x,y)=1-xy+x^2y^2$ is right and $\displaystyle P_{2n}=\sum _{i=1}^{n} (-1)^n (xy)^n$ thanks to you. It's obvious that any of these polynomials evaluated in $\displaystyle (0,0)$ give 1!
I don't know what I was thinking, I thought that the whole polynomial was worth 1 and that there were a polynomial for every values of x and y. Senseless.