Taylor polynomial of 1/(2+x-2y)

[Combinatus]

Homework Statement

Find the Taylor polynomial of degree 3 of $$\frac{1}{2+x-2y}$$ near (2,1).

Homework Equations

The Attempt at a Solution

I have already solved this problem by evaluating the R^2 Taylor series; I'm mostly curious about another aspect of the problem.

By substituting u = x-2y, it would seem that we can use the Maclaurin expansion of $$g(u) = \frac{1}{2+u}$$, and then substitute back to the original variables (since f(x,y) ~= g(u) when (x,y) ~= (2,1)) to get the relevant Taylor series for f(x,y).

I seem to be getting the wrong answer with this approach, but I'm curious why this is the case. Does this approach work under certain conditions for multivariable functions, or should it work in general if a relevant substitution can be made, indicating that I've made an arithmetical error somewhere?

Thanks.

 

[Mark44]

What did you get for your Maclaurin expansion for 1/(2 + u)? I think your approach might work, but I'm not sure of it. Keep in mind that you want to work with (1/2) *1/(1 + u/2).

 

[Dick]

It should work fine if you are doing it correctly.

 

[Combinatus]

What did you get for your Maclaurin expansion for 1/(2 + u)?

$$P_3(u) = \frac{1}{2} - \frac{1}{4}u + \frac{1}{8}u^2 - \frac{1}{16}u^3$$

So, substituting u = x - 2y, we get

$$P_3(x-2y) = \frac{1}{2} - \frac{x-2y}{4} + \frac{(x-2y)^2}{8} - \frac{(x-2y)^3}{16}$$...oh, nice, it does work out. I must have misplaced a sign somewhere when verifying the solution. The key to the problem states the third-degree Taylor polynomial of the initial function simply as

(1/2) - (1/4)*(x-2) + (1/2)*(y-1) + (1/8)*(x-2)^2 - (1/2)*(x-2)(y-1) + (1/2)*(y-1)^2 - (1/16)*(x-2)^3 + (3/8)*((x-2)^2)*(y-1) - (3/4)*(x-2)((y-1)^2) + (1/2)*((y-1)^3)), which equals what I obtained with the lengthy approach of evaluating the relevant partial derivatives. Wolframalpha seems to verify that the two polynomials are equal.

I'm not very comfortable with approaches that happen to work out that I don't really understand though.

 

[Dick]

The taylor series is an expansion of the function in powers of x and y. So is the expansion by series you did. It's not an accident they are equal. They are both doing the same thing. It's often easier to do it by series (if you can) than computing all of the high order derivatives.