Number of Partial Derivatives of Order n

From ProofWiki
Jump to navigation Jump to search

Theorem

Let $u = \map f {x_1, x_2, \ldots, x_m}$ be a function of the $m$ independent variables $x_1, x_2, \ldots, x_m$.


There are $m^n$ partial derivatives of $u$ of order $n$.


Proof

The proof proceeds by induction.

For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:

There are $m^n$ partial derivatives of $u$ of order $n$.


$\map P 0$ is the degenerate case where $f$ is not partially differentiated at all:

$\map f {x_1, x_2, \ldots, x_m}$

and it is apparent that there is only $1 = m^0$ such.


Thus $\map P 0$ is seen to hold.


Basis for the Induction

$\map P 1$ is the case:

There are $m^1 = m$ partial derivatives of $u$ of order $1$.

These can be instanced as:

\(\ds \map {f_1} {x_1, x_2, \ldots, x_m}\) \(=\) \(\ds \dfrac {\partial u} {\partial {x_1} }\)
\(\ds \map {f_2} {x_1, x_2, \ldots, x_m}\) \(=\) \(\ds \dfrac {\partial u} {\partial {x_2} }\)
\(\ds \) \(\vdots\) \(\ds \)
\(\ds \map {f_m} {x_1, x_2, \ldots, x_m}\) \(=\) \(\ds \dfrac {\partial u} {\partial {x_m} }\)

and it is apparent that there are $m$ such.

Thus $\map P 1$ is seen to hold.


This is the basis for the induction.


Induction Hypothesis

Now it needs to be shown that if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.


So this is the induction hypothesis:

There are $m^k$ partial derivatives of $u$ of order $k$.


from which it is to be shown that:

There are $m^{k + 1}$ partial derivatives of $u$ of order $k + 1$.


Induction Step

This is the induction step:

Let $g$ be one of the partial derivatives of $u$ of order $k$.

Then by the basis for the induction, there are $m$ partial derivatives of $g$.

By the induction hypothesis, there are $m^k$ partial derivatives of $u$ of order $k$.

Thus by the Product Rule for Counting, there are $m \times m^n$ partial derivatives of $u$ of order $k + 1$.

That is, a total of $m^{k + 1}$ partial derivatives of $u$ of order $k + 1$.

So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.


Therefore:

$\forall n \in \Z_{\ge 0}:$ there are $m^n$ partial derivatives of $u$ of order $n$.

$\blacksquare$


Examples

Partial Derivatives of Order $1$ of Function of $2$ Variables

Let $u = \map f {x, y}$ be a real function of $2$ variables.

There are $2$ partial derivatives of $u$ of order $1$.


Partial Derivatives of Order $2$ of Function of $2$ Variables

Let $u = \map f {x, y}$ be a real function of $2$ variables.

There are $4$ partial derivatives of $u$ of order $2$.


Partial Derivatives of Order $n$ of Function of $2$ Variables

Let $u = \map f {x, y}$ be a real function of $2$ variables.

There are $2^n$ partial derivatives of $u$ of order $n$.


Sources

Retrieved from "https://proofwiki.org/w/index.php?title=Number_of_Partial_Derivatives_of_Order_n&oldid=466898"