Let f(x_1,x_2,...,x_n) be an analytic function of n…
Let $f(x_1,x_2,\dots,x_n)$ be an analytic function of $n$ variables. How many distinct partial derivatives of order $r$ does it have?
Explanation of Method
For an analytic (in fact, for any sufficiently smooth) function, mixed partial derivatives do not depend on the order of differentiation. So a partial derivative of order $r$ is determined only by how many times we differentiate with respect to each variable: numbers $r_1,\dots,r_n\ge0$ with $r_1+\dots+r_n=r$. Counting these is exactly the stars-and-bars problem from Problem 58.
Step 1: The order of differentiation does not matter
For smooth functions Schwarz's theorem says $\dfrac{\partial^2 f}{\partial x\,\partial y}=\dfrac{\partial^2 f}{\partial y\,\partial x}$. Repeating this, any partial derivative of order $r$ can be rearranged into the form
\[ \frac{\partial^{\,r} f}{\partial x_1^{r_1}\,\partial x_2^{r_2}\cdots\partial x_n^{r_n}}, \qquad r_1+r_2+\dots+r_n = r,\quad r_i\ge0. \]
So distinct derivatives correspond to distinct vectors $(r_1,\dots,r_n)$.
Step 2: Count the vectors — stars and bars
Write $r$ stars (one per differentiation) and $n-1$ bars separating the variables. For example, for $n=3$, $r=4$:
| Picture | $(r_1,r_2,r_3)$ | Derivative |
|---|---|---|
| $\ast\ast\,|\,\ast\,|\,\ast$ | $(2,1,1)$ | $\dfrac{\partial^4 f}{\partial x_1^2\,\partial x_2\,\partial x_3}$ |
| $|\,\ast\ast\ast\ast\,|$ | $(0,4,0)$ | $\dfrac{\partial^4 f}{\partial x_2^4}$ |
Each arrangement of $r$ stars and $n-1$ bars in a row of $r+n-1$ places gives one derivative, and vice versa:
\[ \boxed{\text{number of distinct derivatives of order } r = \binom{n+r-1}{r} = \frac{(n+r-1)!}{r!\,(n-1)!}.} \]
Step 3: Checks
| $n$ | $r$ | $\binom{n+r-1}{r}$ | List |
|---|---|---|---|
| $1$ | any | $1$ | $f^{(r)}$ only |
| $2$ | $2$ | $3$ | $f_{xx},\ f_{xy},\ f_{yy}$ |
| $3$ | $2$ | $6$ | $f_{xx},f_{yy},f_{zz},f_{xy},f_{xz},f_{yz}$ |
| $2$ | $3$ | $4$ | $f_{xxx},f_{xxy},f_{xyy},f_{yyy}$ |
Without the symmetry of mixed derivatives there would be $n^r$ ordered derivatives; symmetry reduces this to $\binom{n+r-1}{r}$.
Result
| Answer | |
|---|---|
| Distinct partial derivatives of order $r$ | $\binom{n+r-1}{r}=\binom{n+r-1}{n-1}$ |