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

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

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

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

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

$
\frac{1}{1+u} = 1 - u + u^2 - u^3 \cdots
$

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

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

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

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

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

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

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

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

$+ \frac{\partial^2 f}{\partial x^2}(a,b)\frac{(x-a)^2}{2!} + 2 \frac{\partial^2 f}{\partial x \partial y}(a,b)\frac{(x-a)(y-b)}{2!}+ \frac{\partial^2 f}{\partial^2 y}(a,b)\frac{(y-b) ^2}{2!} \cdots$.
This is exactly what I've done on my draft and I got $P_2(0,0)=1$ because all the terms were equal to $0$, except the first (that is $f(0,0)$ which is worth 1.)

6. Oh! I just realized my error...
$P_2(x,y)=1-xy+x^2y^2$ is right and $P_{2n}=\sum _{i=1}^{n} (-1)^n (xy)^n$ thanks to you. It's obvious that any of these polynomials evaluated in $(0,0)$ 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.