The Hasse derivative.

In real and complex analysis, one has a powerful tool, namely the Taylor expansion, which expands an analytic function into a power series. In algebra, one can define the derivative of a polynomial aswell; for $f = \sum_{i=0}^n a_i x^i \in R[x]$ , being a unitary ring, define $f' := \sum_{i=1}^n i a_i x^i \in R[x]$ . This satisfies the same rules as the usual derivative, for example -linearity and the product rule . One can also define $f^{(k)}$ recursively by $f^{(0)} = f$ , $f^{(k + 1)} = (f^{(k)})'$ for $k \in \N$ . Unfortunately, one looses certain properties; for example, if is of finite characteristic , the polynomial $f = x^m \in R[x]$ satisfies , but is not constant as $f(0) = 0 \neq 1 = f(1)$ (assuming is not the zero ring). In particular, the Identity Theorem does not work. This example also shows one problem with a possible Taylor expansion: for that, one needs to compute $\frac{f^{(i)}(a)}{i!}$ for $i = 0, \dots, \deg f$ ; but in , whence has no inverse in ! Hence, a Taylor expansion in the classical sense cannot be defined. A “fix” for this problem is offered by Hasse derivatives: they are defined to make both the Identity Theorem and Taylor expansions work again.

Definition.

Let $f = \sum_{i=0}^n a_i x^i \in R[x]$ and $k \in \N$ . Define

$D^{(k)} f := \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} \in R[x].$

The function $D^{(k)} : R[x] \to R[x]$ is called the -th Hasse derivative.

The Hasse derivative shares several properties with the usual derivative, but not all of them; for example, $D^{(k)} D^{(\ell)} \neq D^{(k+\ell)}$ in general. But we have the following properties:

Theorem.

Let $f, g \in R[x]$ and $\lambda \in R$ , and $k, \ell \in \N$ .

We have that $D^{(k)}$ is -linear, i.e. $D^{(k)} (f + g) = D^{(k)} f + D^{(k)} g$ and $D^{(k)}(\lambda f) = \lambda D^{(k)} f$ .
We have $k! \cdot D^{(k)} f = f^{(k)}$ ; in particular, $D^{(1)} f = f'$ .
We have $D^{(k)} D^{(\ell)} f = \binom{k + \ell}{\ell} D^{(k+\ell)} f$ .
(Leibniz Rule) We have

$D^{(k)}(f g) = \sum_{i=0}^k D^{(i)} f \cdot D^{(k-i)} g;$

more generally, for $f_1, \dots, f_t \in R[x]$ , we have

$D^{(k)} \prod_{i=1}^t f_i = \sum_{m_1 + \dots + m_t = k} \prod_{i=1}^t D^{(m_i)} f_i,$

where the sum goes over all such tuples $(m_1, \dots, m_t) \in \N^t$ .
(Faà di Bruno's Formula) We have

$D^{(k)} (f \circ g) = \sum \binom{c_1 + \dots + c_k}{c_1, \dots, c_k} (D^{(c_1 + \dots + c_k)} f) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j},$

where the sum goes over all tuples $(c_1, \dots, c_k) \in \N^k$ with $\sum_{i=1}^k i c_i = k$ ; here, $\binom{c_1 + \dots + c_k}{c_1, \dots, c_k}$ is a multinomial coefficient having the value

$\frac{(c_1 + \dots + c_k)!}{c_1! \cdots c_k!}.$
(Taylor Formula) We have

$f = \sum_{i=0}^{\deg f} (D^{(i)} f)(\lambda) (x - \lambda)^i.$
(Identity Theorem) If we have $(D^{(i)} f)(\lambda) = 0$ for all $i \ge 0$ , then .

For that reason, one can define $\frac{f^{(k)}}{k!} := D^{(k)} f$ , so that we can write Taylor's formula in a more tempting form as

$f = \sum_{i=0}^n \frac{f^{(i)}}{i!}(\lambda) (x - \lambda)^i,$

which almost equals the classical form.

Note that the Leibniz rule, Faà di Bruno's formula and Taylor's formula take simpler forms than their classical counterparts; this is due to the fact that the additional factorial terms or binomial coefficients already hide in the definition of the Hasse derivative.

Proof.

This follows from the definition of $D^{(k)}$ .
Write $f = \sum_{i=0}^n a_i x^i$ ; then

$k! \cdot D^{(k)} f ={} & k! \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} = \sum_{i=k}^n k! \binom{i}{k} a_i x^{i-k} \\ {}={} & \sum_{i=k}^n a_i \cdot i (i - 1) (i - 2) \cdots (i - k + 1) x^{i - k} \\ {}={} & \sum_{i=k}^n a_i (x^i)^{(k)} = f^{(k)}.$
By 1., it is suffices to show this for , $i \ge k + \ell$ (for smaller , both sides will be zero). We have

$D^{(k)} D^{(\ell)} f = D^{(k)} \binom{i}{\ell} x^{i - \ell} = \binom{i - \ell}{k} \binom{i}{\ell} x^{i - k - \ell}$

and

$D^{(k + \ell)} f = \binom{i}{k + \ell} x^{i - k - \ell}.$

But since

$\binom{k + \ell}{\ell} \binom{i}{k + \ell} ={} & \frac{(k + \ell)!}{\ell! k!} \frac{i!}{(k + \ell)! (i - k - \ell)!} \\ {}={} & \frac{i!}{\ell! k! (i - k - \ell!)} \\ {}={} & \frac{(i - \ell)!}{k! (i - k - \ell)!} \frac{i!}{\ell! (i - \ell)!} = \binom{i - \ell}{k} \binom{i}{\ell},$

these terms are equal.
Note that if we fix , we get an -linear function $R[x] \to R[x]$ , $g \mapsto D^{(k)} (f g)$ . Hence, it suffices to show this for arbitrary $f \in R[x]$ and , $m \in \N$ . By the same argument, for , we get an -linbear function $R[x] \to R[x]$ , $f \mapsto D^{(k)} (f x^m)$ ; therefore, it suffices to consider , $n \in \N$ . But now,

$D^{(k)} (x^n x^m) ={} & D^{(k)} x^{n + m} = \binom{n + m}{k} x^{n + m - k} \\ \text{and} \quad D^{(i)} x^n \cdot D^{(k-i)} x^m ={} & \binom{n}{i} x^{n - i} \binom{m}{k - i} x^{m - (k - i)} \ {}={} & \binom{n}{i} \binom{m}{k - i} x^{n + m - k}.$

Hence, it suffices to show $\sum_{i=0}^k \binom{n}{i} \binom{m}{k - i} = \binom{n + m}{k}$ . By reorganizing the binomial coefficients, one transforms this into the equality

$\sum_{i=0}^k \binom{k}{i} \binom{n + m - k}{n - i} = \binom{n + m}{n},$

which is Vandermonde's Identity and, hence, true.

The more general equation is shown by induction on . For , we have $\sum_{m_1 + \dots + m_t = k} \prod_{i=1}^t D^{(m_i)} f_i = D^{(k)} f_1$ . Now, assume that the equation is true for all for one $t \ge 1$ . Then, for any ,

$& D^{(k)} \prod_{i=1}^{t+1} f_i = D^{(k)} \biggl( f_{t+1} \cdot \prod_{i=1}^t f_i \biggr) \ {}={} & \sum_{m_{t+1} = 0}^k D^{(m_{t+1})} f_{t+1} \cdot D^{(k-m_{t+1})} \biggl( \prod_{i=1}^t f_i \biggr) \ {}={} & \sum_{m_{t+1} = 0}^k D^{(m_{t+1})} f_{t+1} \cdot \sum_{m_1 + \dots + m_t = k - m_{t+1}} \prod_{i=1}^t D^{(m_i)} f_i$

by the Leibniz rule and by the induction hypothesis; here, the second sum goes over all such tuples $(m_1, \dots, m_t) \in \N^t$ . But this equals $\sum_{m_1+\dots+m_t+m_{t+1}=k} \prod_{i=1}^{t+1} D^{(m_i)} f_i$ , what we had to show.
Again, by 1., it suffices to show this for , $n \in \N$ . Now, by the second part of 4.,

$D^{(k)}(g^n) = D^{(k)}(g \cdots g) = \sum_{m_1 + \dots + m_n = k} \prod_{i=1}^n D^{(m_i)} g,$

where the sum goes over all such $(m_1, \dots, m_n) \in \N^n$ . The formula we want is now obtained by sorting the summands by the different powers of $D^{(i)} g$ appearing, $0 \le i \le k$ .

Consider the map

$\varphi : \N^n \to \N^{k+1}, \quad m = (m_1, \dots, m_n) \mapsto (c_0(m), \dots, c_k(m)),$

there $c_i(m) := \abs{\{ j \in \{ 1, \dots, n \} \mid m_j = i \}}$ , $0 \le i \le k$ . Now, if $m = (m_1, \dots, m_n) \in \N^n$ satisfies $\sum_{i=1}^n m_i = k$ , then $\sum_{j=0}^k j c_j(m) = k$ and $\sum_{j=0}^k c_j(m) = n$ . Now, for a fixed $(c_0, \dots, c_k) \in \N^{k+1}$ with $\sum_{i=0}^k c_i = n$ and $\sum_{i=0}^k i c_i = k$ , the number $\abs{\varphi^{-1}(c_0, \dots, c_k)}$ equals the multinomial coefficient

$\binom{n}{c_0, c_1, \dots, c_k} = \frac{n!}{c_0! \cdot c_1! \cdots c_n!},$

whence we get that the above formula for $D^{(k)}(g^n)$ equals

$\sum \binom{n}{c_0, c_1, \dots, c_k} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j},$

where the sum goes over all tuples $(c_0, c_1, \dots, c_k) \in \N^k$ with $\sum_{i=0}^k i c_i = k$ and $\sum_{i=0}^k c_i = n$ .

Now note that $\binom{n}{n - c_0} g^{c_0} = \binom{n}{n - c_0} x^{c_0} \circ g = D^{(n - c_0)}(x^n) \circ g$ and $\binom{n}{n - c_0} = \frac{n!}{c_0! (n - c_0)!}$ . Therefore, we get

$& \sum \binom{n}{c_0, c_1, \dots, c_k} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j} \\ {}={} & \sum \frac{n!}{c_0! \cdot c_1! \cdots c_n!} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j} \\ {}={} & \sum \frac{n!}{c_0! \cdot (n - c_0)!} \frac{(n - c_0)!}{c_1! \cdots c_n!} g^{c_0} \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j} \ {}={} & \sum \frac{(n - c_0)!}{c_1! \cdots c_n!} D^{(n - c_0)}(x^n) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}.$

If one now replaces by $c_1 + \dots + c_n$ , we obtain the formula from the statement.
By 1., it suffices to show this for , $n \in \N$ . Now

$\sum_{i=0}^{\deg f} (D^{(i)} f)(\lambda) (x - \lambda)^k ={} & \sum_{i=0}^n \biggl(\binom{n}{i} x^{n - i}\biggr)(\lambda) (x - \lambda)^i \\ {}={} & \sum_{i=0}^n \binom{n}{i} \lambda^{n-i} (x - \lambda)^i \\ {}={} & ((x - \lambda) + \lambda)^n = x^n$

by the Binomial Theorem, what we had to show.
This follows directly from 6.

Comments.

Alexey Maevskiy wrote on September 22, 2009:

Hi! Perhaps it may be useful to place here some facts about mixed partial Hasse derivatives of polynomials from $R[x_1,\ldots,x_n]$ . As for me I worked with polynomials from and had to spend a lot of time proving some needed results on its Hasse derivatives. In particular, one of the popular fact is Taylor Formula.

Felix Fontein wrote on October 2, 2009:

Hi, sorry for not replying earlier, I was not available the last weeks. This sounds interesting, and I will write up something about that. Thanks for the suggestion!

Maybe a question for you: are there other facts about (mixed partial) Hasse derivatives which you think are worth showing here?

I wrote a first article on partial Hasse derivatives, which can be found here.

Michael Forbes wrote on November 6, 2012:

I believe your Faa di Bruno formula has a small error. In particular, I believe it should be:

$\displaystyle D^{(k)} (f \circ g) = \sum \binom{c_1+\cdots+c_k}{ c_1, \dots, c_k} (D^{(c_1+\cdots+c_k)} f) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}$

where the sum ranges over all where $\sum_{i=1}^k i c_i = k$ .

In particular, in the derivation above for you end with a $g^{c_0}$ term, but this term is not equal to $D^{c_0}(f)(g)$ , but is rather equal to $D^{n-c_0}(f)(g)/\binom{n}{c_0}$ . When you do this substitution, you get what I list above.

Incidentally, if you look on Wikipedia for this formula (for normal partial derivatives) it agrees with what I put above, when you ignore the factorials.

Felix Fontein wrote on November 7, 2012:

Yes, you are right. Thank you very much!

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Felix' Math Place

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The Hasse derivative.

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