Derivatives and trees
Michael Anshelevich
January 20, 2005
All know
(F G)′ = F ′ G + F G′
(product rule)
(F ◦ G)′ = F ′(G)G′
1
−1
′
(F ) = ′ −1
F (F )
(chain rule)
(inverse function rule)
What about
(F G)′′′′
(F ◦ G)′′′′
(F −1)′′′′?
Use Taylor series (invented by Newton)
Once have the answer, can check it even if don’t have
Taylor expansions.
1
F ′(0)
F ′′(0) 2
F (3)(0) 3
F (x) = F (0) +
x+
x +
x + ...
1!
2!
3!
= a0 + a1x + a2x2 + a3x3 + · · · ,
F (n)(0)
an =
.
n!
Problem: find the coefficients in the power series expansions of
F G,
F ◦ G,
F −1.
2
P RODUCT RULE .
F (x) = a0 + a1x + a2x2 + . . .
G(x) = b0 + b1x + b2x2 + . . .
(a0 + a1x + a2x2 + . . .)(b0 + b1x + b2x2 + . . .)
= a0b0 + (a0b1 + a1b0)x + (a0b2 + a1b1 + a2b0)x2 + . . .
F (x)G(x) =
∞
X
anxn
n=0
=
∞ X
∞
X
anbk x
∞
X
b k xk
=
∞
X
k=0
n+k
n=0 k=0
m
X
anbm−nxm.
m=0 n=0
Thus
F (x)G(x) =
∞
X
cmxm,
m=0
cm =
m
X
anbm−n.
n=0
3
(m)
(F G)
= m!cm = m!
m
X
F (n) G(m−n)
n=0 n! (m − n)!
m
X
m!
=
F (n)G(m−n)
n=0 n!(m − n)!
m X
m (n) (m−n)
(m)
(F G)
=
F
G
.
n=0 n
Binomial coefficients:
m X
m n m−n
m
(s + t) =
s t
.
n=0 n
4
C HAIN RULE
Assume a0 = b0 = 0,
F (x) = a1x + a2x2 + a3x3 + . . .
G(x) = b1x + b2x2 + b3x3 . . .
F (G(x)) = a1G + a2G2 + a3G3 + . . .
X
X
X
u
u
= a1
bux + a2
bux
bv xv + . . .
= a1
u
X
+ a3
buxu + a2
X
u
X
v
bu1 bu2 xu1+u2
bu1 bu2 bu3 xu1+u2+u3 + . . . .
c1 = a1b1,
c2 = a1b2 + a2b1b1,
c3 = a1b3 + a2b1b2 + a2b2b1 + a3b1b1b1,
cm =
m
X
n=1
an
X
bu1 bu2 . . . bun .
u1 ,u2,...,un≥1
u1+u2 +...+un=m
5
(F ◦ G)(m) = m!
m
X
F (n) X G(u1) G(u2)
n!
n=1
m
X
=
F
n=1
m
X
(n)
=
u1 !
G(un)
...
u2 !
un !
X 1
F (n)
n=1
m!
G(u1)G(u2) . . . G(un)
n! u1!u2! . . . un!
X 1
m
G(u1) . . . G(un).
n! u1, u2, . . . , un
Multinomial coefficients:
(s1 + s2 + . . . + sn)m
=
X
u1 ,u2,...,un≥0
u1+u2 +...+un=m
m
u u
n
s11 s22 . . . su
n .
u1 , u2 , . . . , u n
Ugly! Reformulate.
6
Back to product rule.
cm =
m
X
anbm−n.
n=0
1
2
3
(m)
(F G)
m n
n
n+1
m
m X
m (n) (m−n)
=
F
G
.
n
n=0
= # of n-element subsets in an m-element set.
(F G)(m) =
X
F |S|Gm−|S|.
S⊂[m]
7
F’’’’
F’’’ G’
F’’ G’’
F’ G’’’
G’’’’
8
What is the combinatorial structure for the chain rule?
cm =
m
X
n=1
an
X
bu1 bu2 . . . bun .
u1 ,u2,...,un≥1
u1+u2 +...+un=m
bushes.
m leaves
u_1
u_2
u_n
u_3
n
9
For a shrubbery P
A(P) = a# bushes
Y
b# leaves on a bush
bushes
cm =
X
A(P)
shrubberies with m leaves
a_4 b_1 b_1 b_1 b_1
a_3 b_1 b_1 b_2
a_2 b_1 b_3
a_2 b_2 b_2
a_1 b_4
10
(m)
(F ◦ G)
=
X
=
ordered shrubberies
X
.
partitions
F’’’’ G’ G’ G’ G’
F’’ G’ G’’’
F’’ G’’ G’’
F’’’ G’ G’ G’’
F’ G’’’’
11
(F −1)(n).
F −1 ◦ F = x
(F −1 ◦ F )′ = 1,
(F −1 ◦ F )(n) = 0.
1 = a1b1
0 = a1b2 + a2b1b1
0 = a1b3 + a2b1b2 + a2b2b1 + a3b1b1b1.
Let b1 = 1.
a1 = 1
a2 = −a1b2
a3 = −a1b3 − a2b1b2 − a2b2b1
12
am = −
m−1
X
n=1
X
an
bu1 bu2 . . . bun .
u1 ,u2,...,un≥1
u1 +u2+...+un=m
In its turn
an = −
n−1
X
k=1
ak
X
v1 ,v2 ,...,vk
bv1 bv2 . . . bvk .
13
am = −
m−1
X
n=1
X
an
bu1 bu2 . . . bun .
u1 ,u2,...,un≥1
u1 +u2+...+un=m
m leaves
(−1)
u_1
u_n
u_3
u_2
v_1
v_2
v_k
(−1)
k
Decorated tree, on each level at least one branch splits.
14
B(T ) = bk bv1 . . . bvk bu1 . . . bun .
am =
X
(−1)# levelsB(T )
splitting trees with m leaves
−b_2 b_2 b_2
b_2 b_3
b_2 b_2 b_2
b_3 b_2
−b_4
15
Used combinatorics to find derivatives.
Can also use derivatives for combinatorics.
What is the sum of binomial coefficients?
Many ways. One of them:
Let F (x) = G(x) = ex. Then
F (n)(0) = G(n)(0) = 1.
Use the product rule:
(m)
(F G)
=
m X
m
n=0
n
F
(n)
G
(m−n)
=
m X
m
n=0
n
.
But (F G)(x) = e2x and
(F G)(m) (0) = 2m.
16
How many partitions of a set of n elements are there?
Called Bell numbers, no easy formula. But:
Let F (x) = eax − 1, G(x) = ebx − 1.
F (n)(0) = an,
G(n)(0) = bn.
Use the chain rule:
(F ◦ G)(m) =
X
a# classesbm.
partitions
But
a(ebx −1)
F ◦G=e
− 1.
Thus
# partitions of m with k classes
bx
= coefficient of ak bm in ea(e −1) − 1.
bx
# partitions of m = coefficient of bm in ee −1 − 1.
17