Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

\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)

share|cite|improve this question
What is the question? (In any case, there is no Riemann sum in there.) – Did Apr 13 '14 at 7:54
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 $$ A(n_1,\cdots,n_d)=\frac{\partial^m f(a_1, \dots,a_d)}{\partial^{n_1} x_1\cdots \partial^{n_d} x_d}\cdot\prod_{j=1}^d (x_j - a_j)^{n_j}. $$

share|cite|improve this answer
@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

share|cite|improve this answer
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.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.