Assume $f$ is real analytic in several variables. The Taylor series of $f$ can be represented in two forms (cf. e.g., wikipedia). The first form is power series form: $$T(x_1,\dots,x_d)=\sum\limits_{(n_1,\dots,n_d)\in\Bbb N^d}\frac{1}{n_1!\cdots n_d!}\frac{\partial^{n_1+\cdots+n_d}f(a_1,\dots,a_d)}{\partial x_1^{n_1}\cdots\partial x_d^{n_d}}(x_1-a_1)^{n_1}\cdots(x_d-a_d)^{n_d}.\tag{1}$$
The second form is grouping summands by the order (number of variables) of partial derivatives: $$T(x_1,\dots,x_d)=f(a_1,\dots,a_d)+\sum\limits_{j=1}^d\frac{\partial f(a_1,\dots,a_d)}{\partial x_j}(x_j-a_j)\\+\frac{1}{2!}\sum\limits_{j=1}^d\sum\limits_{k=1}^d\frac{\partial^2f(a_1,\dots,a_d)}{\partial x_j\partial x_k}(x_j-a_j)(x_k-a_k)+\cdots\\+\frac{1}{n!}\sum\limits_{i_1=1}^d\cdots\sum\limits_{i_n=1}^d\frac{\partial^n f(a_1,\dots,a_d)}{\partial x_{i_1}\cdots\partial x_{i_n}}(x_{i_1}-a_{i_1})\cdots(x_{i_n}-a_{i_n})+\cdots.\tag{2}$$
I can derive the coefficients in power series form $(1)$ by differentiating term-by-term. But I encountered difficulty in deriving coefficients in the second form. Following is my attempt. The general $n$-th term in the second form is like this (please correct me if I am wrong): $$\sum\limits_{i_1=1}^d\cdots\sum\limits_{i_n=1}^d c_{(i_1,\cdots,i_n)}(x_{i_1}-a_{i_1})\cdots(x_{i_n}-a_{i_n}).\tag{3}$$ $i_1,\dots,i_n$ can take duplicated values in $1..d$. Assuming in $n$ indices $i_1,\dots,i_n$, there are $n_1\ 1$'s, $n_2\ 2$'s, $\cdots$, $n_d\ d$'s, we have actually constructed a histogram mapping $\varphi: (i_1,\cdots,i_n)\to (n_1,\dots,n_d)$ with the restriction that $n_1+\cdots+n_d=n$. Based on different histogram $d$-tuple $(n_1,\dots,n_d)$, we can further subgroup the summands in sum $(3)$ as follows. For a specific histogram distribution $(n_1,\dots,n_d)$, by merging same factors, the polynomial part of $(3)$ becomes $(x_1-a_1)^{n_1}\cdots(x_d-a_d)^{n_d}$. So, $(3)$ can be rewritten in accordance with this subgrouping as $$\sum\limits_{\begin{array}{c}(n_1,\dots,n_d)\in\Bbb N^d\\n_1+\cdots+n_d=n\end{array}}c_{(n_1,\dots,n_d)}(x_1-a_1)^{n_1}\cdots(x_d-a_d)^{n_d}.$$ This looks much like the power series form. If we take partial derivative $\frac{\partial^{n}f}{\partial x_1^{n_1}\cdots\partial x_d^{n_d}}$ and plugging in $(a_1,\dots,a_d)$, all other terms not corresponding to this $(n_1,\dots,n_d)$ tuple will vanish either because the exponent is less or because the factor is evaluated to zero, and not surprisingly we get the coefficient in exactly the same form as ones in the power series form: $c_{(n_1,\dots,n_d)}=\frac{1}{n_1!\cdots n_d!}\frac{\partial^n f(a_1,\dots,a_d)}{\partial x_1^{n_1}\cdots\partial x_d^{n_d}}=\frac{1}{n_1!\cdots n_d!}\frac{\partial^n f(a_1,\dots,a_d)}{\partial x_{i_1}\cdots\partial x_{i_n}}$ since $f$ is $C^\infty$ and therefore changing the order in which variables are taken does not change the result of mixed partial derivative.
But the denominator before the mixed partial derivative is supposed to be $n!$ from the ground truth in $(2)$ as opposed to $n_1!\cdots n_d!$. I know I must have committed some error by messing up something during the derivation, but I cannot spot it. So, could you please help me figure out the error so that I can correct it and eventually derive the coefficients in the second form having expected $n!$ in the denominator. Thank you.
Best Answer
For any $ (n_1,\dots,n_d)\in\Bbb N^d $ let us count the coefficient at $$\frac{\partial^{n_1+\cdots+n_d}f(a_1,\dots,a_d)}{\partial x_1^{n_1}\cdots\partial x_d^{n_d}}(x_1-a_1)^{n_1}\cdots(x_d-a_d)^{n_d}$$ in the series (1) and (2). In the series (1) the coefficient is $\frac{1}{n_1!\cdots n_d!}.$ In the series (2) it is $\frac{1}{(n_1+\dots n_d)!}$ multiplied by the multinomial coefficient ${n_1+\dots n_d\choose n_1\dots n_k}$, that is the same as in (1).