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], R 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 K-linearity and the product rule (f g)' = f' g + f g'. 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 R is of finite characteristic m > 0, the polynomial f = x^m \in R[x] satisfies f' = 0, but is not constant as f(0) = 0 \neq 1 = f(1) (assuming R 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 m = 0 in R, whence m! has no inverse in R! 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 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 k-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:


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

  1. We have that D^{(k)} is R-linear, i.e. D^{(k)} (f + g) = D^{(k)} f + D^{(k)} g and D^{(k)}(\lambda f) = \lambda D^{(k)} f.
  2. We have k! \cdot D^{(k)} f = f^{(k)}; in particular, D^{(1)} f = f'.
  3. We have D^{(k)} D^{(\ell)} f = \binom{k + \ell}{\ell} D^{(k+\ell)} f.
  4. (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.

  5. 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!}.
  6. (Taylor Formula) We have

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

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.

  1. This follows from the definition of D^{(k)}.
  2. 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)}.
  3. By 1., it is suffices to show this for f = x^i, i \ge k + \ell (for smaller i, 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}


    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.

  4. Note that if we fix f, we get an R-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 g = x^m, m \in \N. By the same argument, for g = x^m, we get an R-linbear function R[x] \to R[x], f \mapsto D^{(k)} (f x^m); therefore, it suffices to consider f = x^n, 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 t. For t = 1, 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 k for one t \ge 1. Then, for any k,

    & 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.

  5. Again, by 1., it suffices to show this for f = x^n, 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 n - c_0 by c_1 + \dots + c_n, we obtain the formula from the statement.

  6. By 1., it suffices to show this for f = x^n, 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.

  7. This follows directly from 6.


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