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\begin{align} f(x_1,\dots,x_d) &= \sum_{n_1=0}^\infty \sum_{n_2=0}^\infty \cdots \sum_{n_d = 0}^\infty \frac{(x_1-a_1)^{n_1}\cdots (x_d-a_d)^{n_d}}{n_1!\cdots n_d!}\,\left(\frac{\partial^{n_1 + \cdots + n_d}f}{\partial x_1^{n_1}\cdots \partial x_d^{n_d}}\right)(a_1,\dots,a_d) \\ &= f(a_1, \dots,a_d) \\ &\quad + \sum_{j=1}^d \frac{\partial f(a_1, \dots,a_d)}{\partial x_j} (x_j - a_j) \\ &\quad + \sum_{j=1}^d \sum_{k=1}^d \frac{1}{2!} \frac{\partial^2 f(a_1, \dots,a_d)}{\partial x_j \partial x_k} (x_j - a_j)(x_k - a_k) \\ &\quad + \sum_{j=1}^d\sum_{k=1}^d\sum_{l=1}^d \frac{1}{3!} \frac{\partial^3 f(a_1, \dots,a_d)}{\partial x_j \partial x_k \partial x_l} (x_j - a_j)(x_k - a_k)(x_l - a_l) \\ &\quad + \dots \end{align}

I have read through wikipedia, and when I saw the above formula, I didn't know how the second equality is justified. Anybody can help me please? (The first equality is assumed to be true by myself thus doesn't need to be proved)

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  • $\begingroup$ What is the question? (In any case, there is no Riemann sum in there.) $\endgroup$
    – Did
    Apr 13 '14 at 7:54
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The second RHS is an enumeration of the first RHS according to the value of $m=n_1+\cdots+n_d$. For $m=0$, one gets one term, which is $f(a_1, \dots,a_d)$. For $m=1$, one gets $d$ terms, which are the products $\frac{\partial f(a_1, \dots,a_d)}{\partial x_j}\cdot(x_j - a_j)$ for each $1\leqslant j\leqslant d$. More generally, for each $m\geqslant0$, one gets $d^m$ terms, hence the multiple sums from $1$ to $d$ with $m$ sums.

To "sum" the above, one uses the identity $$ \sum_{n_1=0}^\infty \sum_{n_2=0}^\infty \cdots \sum_{n_d = 0}^\infty A(n_1,\cdots,n_d) =\sum_{m=0}^\infty\sum_{\begin{array}{c}(n_1,\cdots,n_d)\\ n_1+\cdots+n_d=m\end{array}} A(n_1,\cdots,n_d), $$ with $$ A(n_1,\cdots,n_d)=\frac{\partial^m f(a_1, \dots,a_d)}{\partial^{n_1} x_1\cdots \partial^{n_d} x_d}\cdot\prod_{j=1}^m (x_j - a_j)^{n_j}. $$

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  • $\begingroup$ @pxc3110 The first term in your third expression corresponds to $m=0$. There is only one way to find non-negative integers $n_1+n_2+\cdots+n_d = 0$, which is where all the $n_i = 0$. That is why the first term only has one term in it. And the zeroth derivative of $f$ is simply $f$. $\endgroup$ Apr 13 '14 at 15:27
  • $\begingroup$ Also when Did says "first and second RHS" he means "second and third expression." Did has answered your question. You just haven't realized it yet. $\endgroup$ Apr 13 '14 at 15:32
  • $\begingroup$ Is there a more compact notation for this which does not depend on the coordinates. For example in terms of ∇ operator? $\endgroup$ Jul 8 '17 at 6:22
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Let me just take an analogy. When you make a first order Taylor expansion of $f(x)$, you basically write the equation of a straight line saying that the model is linear with respect to $x$. If you do the same with $g(x,y)$ and you want the model to be linear with respect to both $x$ and $y$, you need to write that $$g(x,y) \simeq a +b (x-x_0)+c(y-y_0)+d(x-x_0)(y-y_0)$$

If you prefer, say that for a given value of $y$,$\text{ } g(x,y)$ is linear with respect to $x$; this write $$g(x,y)=a(y)+b(y) \times (x-x_0)$$ and now consider that $a(y)$ and $b(y)$ are expanded as Taylor series around $y_0$. So $$a(y)=\alpha_0+\alpha_1 (y-y_0)$$ $$b(y)=\beta_0+\beta_1 (y-y_0)$$ and replace in the previous expansion for $g(x,y)$.

You can generalize this for as many variables as you wish. I let you finding the analogy between the coefficients and the derivatives.

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For simplicity, I demonstrate the two-dimensional case, up to the second order. The multi-dimensional case is similar.

Suppose that the function $f=f(x,y)$ is defined on a convex open subset of $\mathbb{R}^{2}$, and suppose $f$ is sufficiently smooth. Let $(x_{0},y_{0})$ be a point in the domain of $f$. Let $h,k\in\mathbb{R}$ such that $(x_{0}+h,y_{0}+k)$ is also point in the domain of $f$. Define a function $g(t)=f(x_{0}+th,y_{0}+tk)$. Note that $g$ is well-defined on an open interval containing $[0,1]$ and $g$ is sufficiently differentiable. Consider the Taylor expansion of the one-variable function $g$: $$ f(x_{0}+h,y_{0}+k)-f(x_{0},y_{0})=g(1)-g(0)=g'(0)(1-0)+\frac{1}{2!}g''(0)(1-0)^{2}+\ldots $$ By chain rule, $g'(t)=f_{x}h+f_{y}k,$ $g''(t)=(f_{xx}h+f_{xy}k)h+(f_{yx}h+f_{yy}k)k=f_{xx}h^{2}+2f_{xy}hk+f_{yy}k^{2}$, where the derivatives $f_{x},f_{y}, f_{xx}$ etc are evaluated at the point $(x_{0}+th,y_{0}+tk)$. Plug-in $t=0$, we have $g'(0)=f_{x}(x_{0},y_{0})h+f_{y}(x_{0},y_{0})k$ and $g''(0)=\ldots$ (I don't want to type). The result follows.

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  • $\begingroup$ I think the derivatives should be computed at $(x_0,y_0)$. Is there a more compact notation for this which does not depend on the coordinates. For example in terms of ∇ operator? $\endgroup$ Jul 8 '17 at 7:06
  • $\begingroup$ @ H.R. If we write in full, it should be $g'(t)=f_{x}(x_{0}+th,y_{0}+tk)h+f_{y}(x_{0}+th,y_{0}+tk)k$. Only when we put $t=0$, the derivatives $f_x$, $f_y$ etc are evaluated at the point $(x_0,y_0)$ $\endgroup$ Jul 8 '17 at 7:10
  • $\begingroup$ (+1) That's right. What about my second question? Is there a compact coordinate free notation? $\endgroup$ Jul 8 '17 at 7:11

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