Write the first four terms of the taylor series for the function $f(x, y) = ln(1 + x + y)$ around the point $(x, y) = (0, 0)$

Question

(a) Write the first four terms of the taylor series for the function $f(x, y) = ln(1 + x + y)$ around the point $(x, y) = (0, 0)$

$f(0,0) = 0$

$f_x(x, y) = \frac{1}{1+x+y}|_{0,0} = 1$

$f_y(x, y) = \frac{1}{1+x+y}|_{0,0} = 1$

$f_{xx}(x, y) = \frac{-1}{(1+x+y)}|_{0,0} = -1$

$f_{yy}(x, y) = \frac{-1}{(1+x+y)^2}|_{0,0} = -1$

$f_{xy}(x, y) = \frac{-1}{(1+x+y)^2}|_{0,0} = -1$

therefore the first four terms are

$= 0 + [x \cdot 1 + y \cdot 1] + \frac{1}{2}(x^2(-1) + 2xy(-1) + y^2(-1)) = x + y - \frac{1}{2}\left[x^2 + 2xy + y^2\right] + \cdots$

would this be correct?

Answer 1

$\frac {d}{du} \ln (1+u) = \frac 1{1+u}\\ \frac {1}{1+u} = \sum_\limits{n=0}^{\infty} (-1)^{n} u^n\\ \int_1^{u} \sum_\limits{n=0}^{\infty} (-1)^{n}u^n = \sum_\limits{n=1}^{\infty} (-1)^{n+1}\frac {u^n}{n} = \ln (1+u)$

$u = x + y\\ \ln(1+x+y) = \sum_\limits{n=1}^{\infty} (-1)^{n+1}\frac {(x+y)^n}{n}$

The first four terms are $x + y - \frac{x^2}{2} - xy - \frac {y^2}{2} $

Looks like we match.