# 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).

## 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.

Mentor
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).

Homework Helper
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.

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