Near point $(0,0)$ find taylor formula for $f(x,y)=\ln(1+x+y)$ Near point $(0,0)$ find taylor formula for $f(x,y)=\ln(1+x+y)$
$f(x)=f(x_0)+\sum_{k=1}^n \frac{1}{k!}d^kf(x_0)+o(|x-x_0|^n)$
$$d^kf(0,0)=\sum_{n=1}^k\binom{k}{n}\frac{\partial^kf}{\partial x^n\partial y^{k-n}}(0,0)x^ny^{k-n}$$
$\frac{\partial^kf}{\partial x^n \partial y^{k-n}}=\frac{(-1)^k k!}{(1+x+y)^k}$
$d^kf(0,0)=\sum_{n=0}^k \binom{k}{n} (-1)^{k-1}k!x^n y^{k-n} = (-1)^{k-1} k!(x+y)^n$
Plugging in the formula I don't get the answer. Answer in the book is $\sum_{n=1} ^m \frac{(-1)^{n-1}(x+y)^n}{n} + o((x^2+y^2)^\frac{m}{2})$
 A: The key to finding this summation is the fact that since $f_x(x,y)=f_y(x,y)$, all partial derivatives of the same order are equal (e.g. $f_{xyxy}=f_{xxxx}$). This means the summation can be:$$\sum_{n=0}^\infty \frac{d^nf}{dx^n}\frac{(x+y)^n}{n!}$$
$f_x(x, y)=\frac{1}{1+x+y}$, $f_{xx}(x,y)=-\frac{1}{(1+x+y)^2}$, $f_{xxx}(x,y)=\frac{2}{(1+x+y)^3}$. This pattern continues as such: $$\frac{d^nf}{dx^n}=-\frac{(-1)^{n}(n-1)!}{(1+x+y)^n}$$Substituting this in and simplifying leaves $\sum_{n=0}^\infty \frac{(-1)^{n-1}(x+y)^n}{n(1+x+y)^n}$.
A: The Taylor formula $f(x_0) + Df(x_0)(x-x_0) + D^2f(x_0)(x-x_0,x-x_0) + \ldots + D^nf(x_0)(x-x_0)^n$ is unique in the sense that if $|f - P_n|/|x-x_0|^n\to 0$ at $x\to x_0$ where $P_n$ is a Polynomial $A_0 + A_1(x-x_0) + A_2(x-x_0,x-x_0) + \ldots + A_n(x-x_0)^n$ (with $A_i$ being an $i$-multivariate functional) then $A_i = D^if(x_0)$.
This is useful, as it implies that if we can find any such form we have found the taylor formula. Now consider: $f(x,y)$ can we written as $\hat f(\phi(x,y))$ where $\hat f(x) = \log(1+x)$ and $\phi(x,y) = x+y$.
Since we know the taylor formua for $\hat f$:
$$ \hat f(x) = \sum_{k=1}^n (-1)^{k-1}/k x^k + R_n(x) $$
we immediately get
$$ f(x,y) = \sum_{k=1}^n \frac{(-1)^{k-1}}k (x+y)^k + R_n(x+y) $$
Note that $(x+y)^k=A_k(x,y)^k$ since $\phi$ is linear (to be precise: $A_k((x_1,y_1),\ldots,(x_k,y_k)) = (x_1+y_1)\cdot\ldots\cdot(x_k+y_k)$). Also we have
$$ |R_n(x+y)| / \|(x,y)\|^n \sim |R_n(x+y)| / \|(x,y)\|_1 \leq |R_n(x+y)| / |x+y| \xrightarrow{x+y\to 0} 0 $$
This suffices to show that the above is indeed our Taylor formula. More generally: If $f = \hat f\circ \phi$ with $\phi:\mathbb R^d\to\mathbb R$ linear and $T_{n,x_0}\hat f(x) = \sum_{k=0}^n a_k (x-x_0)^k$ then
$$ \sum_{k=0}^n a_k\phi(\mathbf x-\mathbf x_0)^k$$
will be the Taylor formula of degree $n$ for $f$. This also follows from
$$ d^k_v (f\circ\phi)(\mathbf x) = d^{k-1}_v(f'\circ\phi(\mathbf x)\cdot d_v(\mathbf x)) = \ldots = f^{(k)}\circ\phi(\mathbf x)\cdot d_v\phi(\mathbf x)^k $$
This follows directly as if $\phi$ is linear then $d_v\phi = \phi(v)$ is a constant.
