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

**arbolis** · May 18th 2009, 04:12 PM

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?

**Jester** · May 18th 2009, 04:28 PM

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

$<br /> \frac{1}{1+u} = 1 - u + u^2 - u^3 \cdots<br />$

**arbolis** · May 18th 2009, 05:03 PM

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).

**Jester** · May 18th 2009, 05:18 PM

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

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

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

**arbolis** · May 18th 2009, 06:08 PM

Originally Posted by danny arrigo

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

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

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

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.)

**arbolis** · May 19th 2009, 06:15 PM

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.

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