Felix' Math Place » Taylor’s formula http://math.fontein.de Focussed on, but not limited to Computational Number Theory Sat, 30 Jul 2011 12:35:49 +0000 en hourly 1 http://wordpress.org/?v=3.1 The Hasse derivative, part II: Multivariate partial Hasse derivatives. http://math.fontein.de/2009/10/02/the-hasse-derivative-part-ii-multivariate-partial-hasse-derivatives/ http://math.fontein.de/2009/10/02/the-hasse-derivative-part-ii-multivariate-partial-hasse-derivatives/#comments Fri, 02 Oct 2009 19:47:27 +0000 Felix Fontein http://math.fontein.de/?p=474 suggestion by A. Maevskiy, we show how the Hasse derivative can be extended to partial Hasse derivative in arbitrary multivariate polynomial rings. We show multivariate versions of Taylor's Formula, of the Identity Theorem, and of the Generalized Leibnitz Rule.]]> Let R be a commutative unitary ring, and let x_i, i \in I be a family of indeterminates over R. We consider the polynomial ring S = R[\{ x_i \mid i \in I \}], obtained from R by adjoining all indeterminates x_i, i \in I. The two most used examples are I = \{ \bullet \}, where S = R[x], and I = \{ 1, \dots, n \}, where S = R[x_1, \dots, x_n]. (Note that if we use finitely many polynomials f_1, \dots, f_r, there exist a finite subset J \subset I such that f_1, \dots, f_r \in R[\{ x_j \mid j \in J \}], i.e. it suffices to consider finitely many variables.)

In classical analysis, one defines the partial derivative by fixing all variables but one, and then using the classical one-dimensional derivative. This can be done similarly in the case of Hasse derivatives. Given f \in S, i \in I and s \in R^I, we can define f_i(s) = f|_{x_j = s(j) \text{ for } j \neq i}, which is an element of R[x_i]. Then D^{(k)} f_i(s) is another element of R[x_i], whence we can define D^{(k)}_{x_i} f(s) := (D^{(k)} f_i(s))|_{x_i = s(i)} = (D^{(k)} f_i(s))(s(i)). This gives a function \displaystyle D^{(k)}_{x_i} f : R^I \to R, \quad s \mapsto D^{(k)}_{x_i} f(s), which can be considered as the k-th partial Hasse derivative of f with respect to x_i. Now we would like the derivative to be another element of S. In classical analysis, polynomial functions are in bijection to polynomials, and one can show using certain rules that the partial derivative of a polynomial is again a polynomial. In case R is an arbitrary ring, one has in general no longer the bijection between polynomials and polynomial functions.

In the case of derivatives over arbitrary rings, we have the advantage that we can naturally identify S with R_i[x_i], where R_i := R[\{ x_j \mid j \in I, j \neq i \}]; here, R_i is another commutative unitary ring. Now R_i[x_i] is a univariate polynomial ring, whence we have the usual Hasse derivative. Denote the k-th Hasse derivative D^{(k)} : R_i[x_i] \to R_i[x_i] by D^{(k)}_{x_i}; then we obtain an R-linear (in fact, R_i-linear) operator D^{(k)}_{x_i} : S \to S. In fact, if we evaluate D^{(k)} f at a point s \in R^I, we obtain the same value as D^{(k)} f(s) defined above. We will prove this in a minute; before that we will prove a more general result on D^{(k)}_{x_i}.

Proposition.
Let \varphi : R \to R' be a homomorphism of unitary commutative rings R and R' (i.e. it additionally satisfies \varphi(1_R) = 1_{R'}). If x is an indeterminate over R and R', and \varphi^* : R[x] \to R'[x] the natural continuation of \varphi with \varphi^*(x) = x and \varphi^*(r) = \varphi(r) for all r \in R, then for f \in R[x], we have \displaystyle D^{(k)} \varphi^*(f) = \varphi^*(D^{(k)} f).
Proof.
This follows directly from the definition of the Hasse derivative: if f = \sum_{i=0}^n a_i x^i \in R[x] and k \in \N, we have \varphi^*(D^{(k)} f) ={} & \varphi^*\biggl( \sum_{i=k}^n \binom{i}{k} a_i x^{i - k} \biggr) \\ {}={} & \sum_{i=k}^n \binom{i}{k} \varphi(a_i) x^{i - k} = D^{(k)} \varphi^*(f).
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Corollary.
For all f \in S and s \in R^I, we have (D^{(k)}_{x_i} f)(s) = D^{(k)} f(s).
Proof.
Let \varphi : R_i \to R be the substitution homomorphism f \mapsto f|_{x_j = s(j) \text{ for } j \neq i}. We interpret f as an element of R_i[x_i]; note that the above defined f_i equals \varphi^*(f). Then, by the proposition, \displaystyle D^{(k) }\varphi^*(f) = \varphi^*(D^{(k)} f) = \varphi^*(D^{(k)}_{x_i} f). Now \varphi^*(D^{(k)}_{x_i} f)|_{x_i = s(i)} = (D^{(k)}_{x_i} f)(s), whence \displaystyle (D^{(k)}_{x_i} f)(s) = (D^{(k)} \varphi^*(f))|_{x_i = s(i)} = (D^{(k)} f_i)|_{x_i = s(i)} = D^{(k)}_{x_i} f(s) by definition of D^{(k)}_{x_i} f(s).
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Hence, the following definition makes sense:

Definition.
The R_i-linear opeator D^{(k)}_{x_i} : S \to S is called the k-th partial Hasse derivative with respect to x_i.

As in the case of usual partial derivatives (in the case of “nice” functions, i.e. the partial derivatives are continuous in a neighborhood of the point we are interested in), the Hasse derivatives commute:

Proposition.
Let i, j \in I, k, \ell \in \N and f \in S. Then D^{(k)}_{x_i} D^{(\ell)}_{x_j} f = D^{(\ell)}_{x_j} D^{(k)}_{x_i} f.

For this, consider the notation R_{i,j} = R[\{ t \in I \mid t \neq i, j \}]; then S = R_{i,j}[x_i, x_j] = R_{i,j}[x_j][x_i] = R_{i,j}[x_i][x_j].

Proof.
By R_{i,j}-linearity of D^{(k)}_{x_i} and D^{(\ell)}_{x_j}, it suffices to show the result for f = x_i^s x_j^t with s, t \in \N. Now D^{(k)}_{x_i} D^{(\ell)}_{x_j} f ={} & D^{(k)}_{x_i} \binom{t}{\ell} x_i^s x_j^{t - \ell} = \binom{s}{k} \binom{t}{\ell} x_i^{s - k} x_j^{t - \ell} \\ {}={} & D^{(\ell)}_{x_j} \binom{s}{k} x_i^{s - k} x_j^t = D^{(\ell)}_{x_j} D^{(k)}_{x_i} f.
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Now, let us define the following notation. We let \N^{(I)} be the set of functions I \to \N which are zero for all but finitely many elements of I. In case I is finite, \N^{(I)} = \N^I in the usual sense. The set \N^{(I)} will be the set of multiindices we use. For s \in \N^{(I)}, define D^{(s)} : S \to S by D^{(s)} := D^{(s(i_1))}_{x_{i_1}} \circ \dots \circ D^{(s(i_t))}_{x_{i_t}}, where \{ i_1, \dots, i_t \} \subseteq I is the support of s. By the above proposition, D^{(s)} is independent of the order of i_1, \dots, i_t, and as D^{(0)}_{x_i} = \id_S, we can also include elements outside the support. We sometimes use the more suggestive notation \displaystyle D^{(s)} = \prod_{i \in I} D^{(s(i))}_{x_i}, which is hence also justified. Moreover, for s \in \N^{(I)} and \lambda \in R^I, define (x - \lambda)^s := \prod_{j=1}^t (x_j - \lambda(j))^{s(j)} \in S. Again, this is well-defined. Using this notation, we obtain Taylor’s formula:

Theorem (Taylor formula).
For f \in S and \lambda \in R^I, we have \displaystyle f = \sum_{s \in \N^{(I)}} (D^{(s)} f)(\lambda) (x - \lambda)^s.
Proof.
By the R-linearity of the equality, it suffices to consider f = x_{i_1}^{e_1} \cdots x_{i_t}^{e_t} with i_1, \dots, i_t \in I pairwise distinct and e_1, \dots, e_t \in \N. Define I' := \{ i_1, \dots, i_t \}. If s \in \N^{(I)} satisfies s(i) \neq 0 for i \not\in I', then D^{(s)} f = 0. Hence, \displaystyle \sum_{s \in \N^{(I)}} (D^{(s)} f)(\lambda) (x - \lambda)^s = \sum_{s \in \N^{(I')}} (D^{(s)} f)(\lambda) (x - \lambda)^s; moreover, one can restrict to the subring S' = R[x_{i_1}, \dots, x_{i_t}], as every appearing object lies in that ring. Hence, it suffices to show the result for I = I', i.e. for a finite index set I. We can assume I = \{ 1, \dots, n \}.
But now we can prove this by induction on n. If we write s = (s', s_n) with s' \in \N^{n-1}, s_n \in \N and \lambda = (\lambda', \lambda_n) with \lambda' \in R^{n-1}, \lambda_n \in R for s \in \N^n and \lambda \in R^n, then \displaystyle D^{(s)} f = D^{(s_n)} D^{(s')} f \quad \text{and} \quad (x - \lambda)^s = (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n}. Hence, & \sum_{s \in \N^n} (D^{(s)} f)(\lambda) (x - \lambda)^s \\ {}={} & \sum_{s \in \N^n} (D^{(s_n)} D^{(s')} f)(\lambda) (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n} \\ {}={} & \sum_{s_n \in \N} \sum_{s' \in \N^{n-1}} (D^{(s_n)} D^{(s')} f)(\lambda) (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n} \\ {}={} & \sum_{s_n \in \N} \sum_{s' \in \N^{n-1}} \bigl(D^{(s_n)} (D^{(s')} f)(\lambda') \bigr)(\lambda_n) (x - \lambda')^{s'} (x_n - \lambda_n)^{s_n} \\ {}={} & \sum_{s_n \in \N} D^{(s_n)} \biggl( \sum_{s' \in \N^{n-1}} (D^{(s')} f)(\lambda') (x - \lambda')^{s'} \biggr)(\lambda_n) (x_n - \lambda_n)^{s_n}. Now, by induction hypothesis (n - 1 with the base ring R[x_n]), \displaystyle \sum_{s' \in \N^{n-1}} (D^{(s')} f)(\lambda') (x - \lambda')^{s'} = f, whence \displaystyle \sum_{s \in \N^n} (D^{(s)} f)(\lambda) (x - \lambda)^s = \sum_{s_n \in \N} D^{(s_n)} f(\lambda_n) (x_n - \lambda_n)^{s_n}. But by the classical univariate case which we already proved, this equals f itself.
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Again, we get the Identity Theorem as a direct corollary:

Corollary (Identity Theorem).
Let f \in S and \lambda \in R^I. If D^{(s)} f(\lambda) = 0 for all s \in \N^{(I)}, then f = 0.
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In a similar manner as Taylor’s formula, we can also adapt other one-dimensional results to the multivariate case.

For example, for s, t \in \N^{(I)}, let us define \binom{s + t}{s} := \prod_{j=1}^u \binom{s(i_j) + t(i_j)}{s(i_j)} if both s and t have no non-zero values outside \{ i_1, \dots, i_u \}. Note that \binom{s + t}{s} = \binom{s + t}{t}. We now obtain the following:

Proposition.
For s, t \in \N^{(I)}, we have D^{(s)} \circ D^{(t)} = \binom{s + t}{s} D^{(s + t)}. In particular, D^{(s)} \circ D^{(t)} = D^{(t)} \circ D^{(s)}.
Proof.
Let R' denote the ring R[x_j \mid j \neq i_1, \dots, i_u\}], where s and t have no non-zero values outside \{ i_1, \dots, i_u \}; then we have reduced to the ring R'[x_{i_1}, \dots, x_{i_u}], i.e. to the case of I being finite, say I = \{ 1, \dots, n \}. Write s = (s_1, \dots, s_n) and t = (t_1, \dots, t_n). As D^{(k)}_{x_i} and D^{(\ell)}_{x_j} commute for i \neq j, we obtain \displaystyle D^{(s)} \circ D^{(t)} = \prod_{i=1}^n (D^{(s_i)}_{x_i} \circ D^{(t_i)}_{x_i}). Using the corresponding univariate result, we have D^{(s_i)}_{x_i} \circ D^{(t_i)}_{x_i} = \binom{s_i + t_i}{s_i} D^{(s_i + t_i)}_{x_i}. Hence, \displaystyle D^{(s)} \circ D^{(t)} = \prod_{i=1}^n \biggl( \binom{s_i + t_i}{s_i} D^{(s_i + t_i)}_{x_i} \biggr) = \binom{s + t}{s} D^{(s + t)}.
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We also obtain the Leibniz rule:

Proposition (Generalized Leibniz Rule).
For f_1, \dots, f_m \in S and s \in \N^{(I)}, we have \displaystyle D^{(s)} \prod_{i=1}^m f_i = \sum_{s_1 + \dots + s_m = s} \prod_{i=1}^m D^{(s_i)} f_i; here, the sum ranges over all (s_1, \dots, s_m) \in (\N^{(I)})^m with s_1 + \dots + s_m = s. As a special case, for f, g \in S, we have \displaystyle D^{(s)}(f g) = \sum_{s' + s'' = s} D^{(s')}(f) D^{(s'')}(g).
Proof.
Using the standard argument, we reduce to the case I = \{ 1, \dots, n \} and we use the standard decomposition \N^n = \N^{n-1} \times \N. Hence, \displaystyle \sum_{s_1 + \dots + s_m = s} \prod_{i=1}^m D^{(s_i)} f_i = \sum_{s'_1 + \dots + s_m' = s'} \sum_{s_{1,n} + \dots + s_{m,n} = s_n} \prod_{i=1}^m D^{(s_i')} D^{(s_{i,n})} f_i, where the second sum ranges over all (s'_1, \dots, s_ m') \in (\N^{n-1})^m and the third sum ranges over all (s_{1,n}, \dots, s_{m,n}) \in \N^m. Now & \sum_{s'_1 + \dots + s_m' = s'} \sum_{s_{1,n} + \dots + s_{m,n} = s_n} \prod_{i=1}^m D^{(s_i')} D^{(s_{i,n})} f_i \\ {}={} & \sum_{s'_1 + \dots + s_m' = s'} D^{(s_i')} \sum_{s_{1,n} + \dots + s_{m,n} = s_n} \prod_{i=1}^m D^{(s_{i,n})} f_i, and applying the univariate result to the base ring R[x_1, \dots, x_{n-1}], we obtain \sum_{s_1 + \dots + s_m = s} \prod_{i=1}^m D^{(s_i)} f_i ={} & \sum_{s'_1 + \dots + s_m' = s'} D^{(s_i')} D^{(s_n)}_{x_n} \prod_{i=1}^m f_i \\ {}={} & D^{(s_n)}_{x_n} \sum_{s'_1 + \dots + s_m' = s'} D^{(s_i')} \prod_{i=1}^m f_i. By induction hypothesis, this equals \displaystyle D^{(s_n)}_{x_n} D^{(s')} \prod_{i=1}^m f_i = D^{(s)} \prod_{i=1}^m f_i.
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]]> http://math.fontein.de/2009/10/02/the-hasse-derivative-part-ii-multivariate-partial-hasse-derivatives/feed/ 2 The Hasse derivative. http://math.fontein.de/2009/08/12/the-hasse-derivative/ http://math.fontein.de/2009/08/12/the-hasse-derivative/#comments Wed, 12 Aug 2009 05:43:38 +0000 Felix Fontein http://math.fontein.de/?p=277 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.

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

Theorem.
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 \displaystyle 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 \displaystyle 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. (Faà di Bruno’s Formula) We have \displaystyle D^{(k)} (f \circ g) = \sum \binom{n}{c_0, c_1, \dots, c_k} (D^{(c_0)} f) \circ g \cdot \prod_{j=1}^k (D^{(j)} g)^{c_j}, where the sum goes over all tuples (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; here, \binom{n}{c_0, c_1, \dots, c_k} is a multinomial coefficient having the value \displaystyle \frac{n!}{c_0! \cdot c_1! \cdots c_k!}.
  6. (Taylor Formula) We have \displaystyle 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 \displaystyle 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.
  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 \displaystyle 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 \displaystyle 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 \displaystyle \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., \displaystyle 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 \displaystyle \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 \abs{\varphi^{-1}(c_0, \dots, c_k)} equals the multinomial coefficient \displaystyle \binom{n}{c_0, c_1, \dots, c_k}, whence we get that the above formula for D^{(k)}(g^n) equals \displaystyle \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.
  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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