Final Exam—Math 331, Section 201—April 17, 2007
page 2 of ??
Formulas
On the exam itself, I’ll add a few more words to help indicate what the notations mean. But
these will be the formulas included with the exam.
ops
If {an } ←→ f (x) then, for h ≥ 1,
!
f (x) − a0 − a1 x − · · · − ah−1 xh−1
{an+h } ←→
xh
ops
"
n1 +···+nk =n
egf
If {bn } ←→ g(x) then, for h ≥ 1,
an1 an2 · · · ank
#
ops
←→ f (x)k
egf
{bn+h } ←→ Dh g(x)
" $α %
n
"
n
$
n
" $n + k %
α
x = (1 + x)
n
√
%
1
2n n 1 − 1 − 4x
x =
n
+
1
n
2x
n
√
%k
$
" k(2n + k − 1)!
1 − 1 − 4x
n
x =
n!(n + k)!
2x
n≥0
&
xn
h(n, k) =
n!
""
Nr =
Nr =
t≥0
r
(
n
= log
1
1−x
#
1
1 − yxr
r≥1
x=1
E(x) = N (x − 1)
N (⊃ S)
S : #S=r
" $t%
" xn
*++
)
σ 2 = (log F )# + (log F )## +
F # (1)
F (1)
"
D(x)k
k!
p(n, k)xn y k =
n≥0 k≥0
µ=
'!
1
(1 − x)k+1
n
$
" 2n%
1
xn = √
n
1 − 4x
n
n≥1
H(x, y) = eyD(x)
xn =
Et
Et =
"
r≥0
r−t
(−1)
$ %
r
Nr
t
Let C be a (strangely shaped) chessboard contained in an n × n board. If rk is the number of
ways of placing k non-attacking rooks on C, then the number of permutations of {1, 2, . . . , n}
that hit C in exactly j squares is
"
[xj ]
(n − k)!rk (x − 1)k .
k
Final Exam—Math 331, Section 201—April 17, 2007
page 3 of ??
Lagrange Inversion: If f , φ, and u are power series in t with φ(0) (= 0 and u = tφ(u), then
)
* 1
)
*
[tn ] f (u(t)) = [un−1 ] f # (u)φ(u)n .
n
$
%#
r
" !"
a−j
(−1)j a−j n + j − 1
n
P P (f ; z0 ) =
=
z
.
(z − z0 )j
j−1
z0n+j
j=1
j=1
n≥0
!"
#
$,
s
-n %
1
n
n
+ε
[z ]f (z) = [z ]
P P (f ; zk ) + O
#
R
k=0
$
%
m
)
* "
.
/
n−β−j−1
n
β
[z ] (1 − z) v(z) =
vj
+ O n−m−β−2
n
j=0
r
"
If f (z) is admissible, a(r) = rf # (r)/f (r), b(r) = ra# (r), a(rn ) = n, then
f (rn )
.
[z n ]f (z) ∼ 0
rnn 2πb(rn )
If p(z) is a polynomial with [z 1 ]p(z) (= 0, then ep(z) is admissible.