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 ,
being a unitary ring, define
. This satisfies the same rules as the usual derivative, for example
-linearity and the product rule
. One can also define
recursively by
,
for
. Unfortunately, one looses certain properties; for example, if
is of finite characteristic
, the polynomial
satisfies
, but is not constant as
(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
for
; 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.
Let and
. Define
The function is called the
-th Hasse derivative.
The Hasse derivative shares several properties with the usual derivative, but not all of them; for example, in general. But we have the following properties:
Let and
, and
.
- We have that
is
-linear, i.e.
and
.
- We have
; in particular,
.
- We have
.
-
(Leibniz Rule) We have
more generally, for
, we have
where the sum goes over all such tuples
.
-
(Faà di Bruno's Formula) We have
where the sum goes over all tuples
with
; here,
is a multinomial coefficient having the value
-
(Taylor Formula) We have
- (Identity Theorem) If we have
for all
, then
.
For that reason, one can define , so that we can write Taylor's formula in a more tempting form as
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.
- This follows from the definition of
.
-
Write
; then
-
By 1., it is suffices to show this for
,
(for smaller
, both sides will be zero). We have
and
But since
these terms are equal.
-
Note that if we fix
, we get an
-linear function
,
. Hence, it suffices to show this for arbitrary
and
,
. By the same argument, for
, we get an
-linbear function
,
; therefore, it suffices to consider
,
. But now,
Hence, it suffices to show
. By reorganizing the binomial coefficients, one transforms this into the equality
which is Vandermonde's Identity and, hence, true.
The more general equation is shown by induction on
. For
, we have
. Now, assume that the equation is true for all
for one
. Then, for any
,
by the Leibniz rule and by the induction hypothesis; here, the second sum goes over all such tuples
. But this equals
, what we had to show.
-
Again, by 1., it suffices to show this for
,
. Now, by the second part of 4.,
where the sum goes over all such
. The formula we want is now obtained by sorting the summands by the different powers of
appearing,
.
Consider the map
there
,
. Now, if
satisfies
, then
and
. Now, for a fixed
with
and
, the number
equals the multinomial coefficient
whence we get that the above formula for
equals
where the sum goes over all tuples
with
and
.
Now note that
and
. Therefore, we get
If one now replaces
by
, we obtain the formula from the statement.
-
By 1., it suffices to show this for
,
. Now
by the Binomial Theorem, what we had to show.
- This follows directly from 6.
Comments.
Hi! Perhaps it may be useful to place here some facts about mixed partial Hasse derivatives of polynomials from
. 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.
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.
I believe your Faa di Bruno formula has a small error. In particular, I believe it should be:
where the sum ranges over all
where
.
In particular, in the derivation above for
you end with a
term, but this term is not equal to
, but is rather equal to
. 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.
Yes, you are right. Thank you very much!