# Thread: Taylor formula for function of 2 variables

1. ## Taylor formula for function of 2 variables

Hi! Could anyone help me with this question? I'm unable to sleep as I haven't found the solution!

Find the first and second-degree Taylor polynomials of
f(x, y) = e^(-x^2 - y^2)
at (0, 0). I know I've got to use the Taylor formula for function of 2 variables but as my understanding is weak, I'm unable to do so.

Any help is much appreciated!!

Thank you!!

2. ## Could this be the solution?

First-degree = 1
Second-degree = 1-x^2-y^2

Has anyone managed to obtain something similar?

Thanks!! =)

3. Originally Posted by keenlearner
Find the first and second-degree Taylor polynomials of
f(x, y) = e^(-x^2 - y^2)
at (0, 0). I know I've got to use the Taylor formula for function of 2 variables but as my understanding is weak, I'm unable to do so.
Two ways to do this. The first is more straightforward, but it uses the Taylor formula for one variable rather than two. You know that $e^t = 1 + t + t^2/2 +$ (higher powers of t). Substitute $t = -x^2-y^2$ and you get $f(x,y) = 1 - x^2 - y^2 +$ (higher powers). Here, there is a constant term, no terms of degree 1, and two terms of degree 2. So the first-degree Taylor polynomial of f(x,y) is just 1. The second-degree Taylor polynomial is $1 - x^2 - y^2$.

The second method (presumably the one you are supposed to be using) is to use the two-variable Taylor expansion

$f(x,y) =$ $f(0,0) + \Bigl(x\tfrac{\partial f}{\partial x}(0,0) + y\tfrac{\partial f}{\partial x}(0,0)\Bigr) + \tfrac1{2!}\Bigl(x^2\tfrac{\partial^2 f}{\partial x^2}(0,0) + 2xy\tfrac{\partial^2 f}{\partial x\partial y}(0,0) + y^2\tfrac{\partial^2 f}{\partial x^2}(0,0)\Bigr) +$ (higher powers).

If you do the partial differentiations, evaluate them at (0,0) and put them into that formula, then you should get the same answer as by the first method.