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\begin{align} T(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) + \sum_{j=1}^d \frac{\partial f(a_1, \dots,a_d)}{\partial x_j} (x_j - a_j) \\ &\quad {} + \frac{1}{2!} \sum_{j=1}^d \sum_{k=1}^d \frac{\partial^2 f(a_1, \dots,a_d)}{\partial x_j \partial x_k} (x_j - a_j)(x_k - a_k) \\ &\quad {} + \frac{1}{3!} \sum_{j=1}^d\sum_{k=1}^d\sum_{l=1}^d \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) + \dots \end{align}

I have read through, and when I saw this(under the title "Taylor series in several variables"), I didn't know how the second equality is justified. Anybody help me please? (The first equality is assumed to be true by myself thus doesn't need to be prove)

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What is the question? (In any case, there is no Riemann sum in there.) – Did Apr 13 '14 at 7:54

3 Answers 3

up vote 2 down vote accepted

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 $$ A(n_1,\cdots,n_d)=\frac{\partial^m f(a_1, \dots,a_d)}{\partial^{n_1} x_1\cdots \partial^{n_d} x_d}\cdot\prod_{j=1}^d (x_j - a_j)^{n_j}. $$

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@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 $f$. – Stephen Montgomery-Smith Apr 13 '14 at 15:27
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. – Stephen Montgomery-Smith Apr 13 '14 at 15:32

Not quite sure on the details, but maybe you can think of expanding using vectors/matrices?'s_theorem#Taylor.27s_theorem_for_multivariate_functions

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Thanks, I'll check'em out later. – pxc3110 Apr 13 '14 at 14:30
@pxc3110, I just checked out the link in the OP. I think the vector/matrix version that comes right after it is precisely the rhs of the second equality.…… – BCLC Apr 13 '14 at 14:35

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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