# test

what's new posted @ 2015年5月01日 13:38 in 未分类 , 993 阅读 Theoretical Computer Science Cheat Sheet
Deﬁnitions Series
f(n)=O(g(n)) iﬀ positive c, n
0
such that
0 f(n) cg(n) n n
0
.
n
i=1
i =
n(n +1)
2
,
n
i=1
i
2
=
n(n + 1)(2n +1)
6
,
n
i=1
i
3
=
n
2
(n +1)
2
4
.
In general:
n
i=1
i
m
=
1
m +1
(n +1)
m+1
1
n
i=1
(i +1)
m+1
i
m+1
(m +1)i
m
n1
i=1
i
m
=
1
m +1
m
k=0
m +1
k
B
k
n
m+1k
.
Geometric series:
n
i=0
c
i
=
c
n+1
1
c 1
,c=1,
i=0
c
i
=
1
1 c
,
i=1
c
i
=
c
1 c
, |c| < 1,
n
i=0
ic
i
=
nc
n+2
(n +1)c
n+1
+ c
(c 1)
2
,c=1,
i=0
ic
i
=
c
(1 c)
2
, |c| < 1.
Harmonic series:
H
n
=
n
i=1
1
i
,
n
i=1
iH
i
=
n(n +1)
2
H
n
n(n 1)
4
.
n
i=1
H
i
=(n +1)H
n
n,
n
i=1
i
m
H
i
=
n +1
m +1

H
n+1
1
m +1
.
f(n)=Ω(g(n)) iﬀ positive c, n
0
such that
f(n) cg(n) 0 n n
0
.
f(n)=Θ(g(n)) iﬀ f(n)=O(g(n)) and
f(n)=Ω(g(n)).
f(n)=o(g(n)) iﬀ lim
n→∞
f(n)/g(n)=0.
lim
n→∞
a
n
= a iﬀ >0, n
0
such that
|a
n
a| <, n n
0
.
sup S least b R such that b s,
s S.
inf S greatest b R such that b
s, s S.
lim inf
n→∞
a
n
lim
n→∞
inf{a
i
| i n, i N}.
lim sup
n→∞
a
n
lim
n→∞
sup{a
i
| i n, i N}.
n
k
Combinations: Size k sub-
sets of a size n set.
n
k
Stirling numbers (1st kind):
Arrangements of an n ele-
ment set into k cycles.
1.
n
k
=
n!
(n k)!k!
, 2.
n
k=0
n
k
=2
n
, 3.
n
k
=
n
n k
,
4.
n
k
=
n
k
n 1
k 1
, 5.
n
k
=
n 1
k
+
n 1
k 1
,
6.
n
m

m
k
=
n
k

n k
m k
, 7.
n
k=0
r + k
k
=
r + n +1
n
,
8.
n
k=0
k
m
=
n +1
m +1
, 9.
n
k=0
r
k

s
n k
=
r + s
n
,
10.
n
k
=(1)
k
k n 1
k
, 11.
n
1
=
n
n
=1,
12.
n
2
=2
n1
1, 13.
n
k
= k
n 1
k
+
n 1
k 1
,
n
k
Stirling numbers (2nd kind):
Partitions of an n element
set into k non-empty sets.
n
k
1st order Eulerian numbers:
Permutations π
1
π
2
...π
n
on
{1, 2,...,n} with k ascents.

n
k

2nd order Eulerian numbers.
C
n
Catalan Numbers: Binary
trees with n + 1 vertices.
14.
n
1
=(n 1)!, 15.
n
2
=(n 1)!H
n1
, 16.
n
n
=1, 17.
n
k
n
k
,
18.
n
k
=(n 1)
n 1
k
+
n 1
k 1
, 19.
n
n 1
=
n
n 1
=
n
2
, 20.
n
k=0
n
k
= n!, 21. C
n
=
1
n +1
2n
n
,
22.
n
0
=
n
n 1
=1, 23.
n
k
=
n
n 1 k
, 24.
n
k
=(k +1)
n 1
k
+(n k)
n 1
k 1
,
25.
0
k
=
1ifk =0,
0 otherwise
26.
n
1
=2
n
n 1, 27.
n
2
=3
n
(n + 1)2
n
+
n +1
2
,
28. x
n
=
n
k=0
n
k

x + k
n
, 29.
n
m
=
m
k=0
n +1
k
(m +1 k)
n
(1)
k
, 30. m!
n
m
=
n
k=0
n
k

k
n m
,
31.
n
m
=
n
k=0
n
k

n k
m
(1)
nkm
k!, 32.

n
0

=1, 33.

n
n

= 0 for n =0,
34.

n
k

=(k +1)

n 1
k

+(2n 1 k)

n 1
k 1

, 35.
n
k=0

n
k

=
(2n)
n
2
n
,
36.
x
x n
=
n
k=0

n
k

x + n 1 k
2n
, 37.
n +1
m +1
=
k
n
k

k
m
=
n
k=0
k
m
(m +1)
nk
, Theoretical Computer Science Cheat Sheet
Identities Cont. Trees
38.
n +1
m +1
=
k
n
k

k
m
=
n
k=0
k
m
n
nk
= n!
n
k=0
1
k!
k
m
, 39.
x
x n
=
n
k=0

n
k

x + k
2n
,
40.
n
m
=
k
n
k

k +1
m +1
(1)
nk
, 41.
n
m
=
k
n +1
k +1

k
m
(1)
mk
,
42.
m + n +1
m
=
m
k=0
k
n + k
k
, 43.
m + n +1
m
=
m
k=0
k(n + k)
n + k
k
,
44.
n
m
=
k
n +1
k +1

k
m
(1)
mk
, 45. (n m)!
n
m
=
k
n +1
k +1

k
m
(1)
mk
, for n m,
46.
n
n m
=
k
m n
m + k

m + n
n + k

m + k
k
, 47.
n
n m
=
k
m n
m + k

m + n
n + k

m + k
k
,
48.
n
+ m

+ m
=
k
k

n k
m

n
k
, 49.
n
+ m

+ m
=
k
k

n k
m

n
k
.
Every tree with n
vertices has n 1
edges.
Kraft inequal-
ity: If the depths
of the leaves of
a binary tree are
d
1
,...,d
n
:
n
i=1
2
d
i
1,
and equality holds
only if every in-
ternal node has 2
sons.
Recurrences
Master method:
T (n)=aT (n/b)+f (n),a 1,b > 1
If >0 such that f(n)=O(n
log
b
a
)
then
T (n)=Θ(n
log
b
a
).
If f(n)=Θ(n
log
b
a
) then
T (n)=Θ(n
log
b
a
log
2
n).
If >0 such that f(n)=Ω(n
log
b
a+
),
and c<1 such that af(n/b) cf(n)
for large n, then
T (n)=Θ(f(n)).
Substitution (example): Consider the
following recurrence
T
i+1
=2
2
i
· T
2
i
,T
1
=2.
Note that T
i
is always a power of two.
Let t
i
= log
2
T
i
. Then we have
t
i+1
=2
i
+2t
i
,t
1
=1.
Let u
i
= t
i
/2
i
. Dividing both sides of
the previous equation by 2
i+1
we get
t
i+1
2
i+1
=
2
i
2
i+1
+
t
i
2
i
.
Substituting we ﬁnd
u
i+1
=
1
2
+ u
i
,u
1
=
1
2
,
which is simply u
i
= i/2. So we ﬁnd
that T
i
has the closed form T
i
=2
i2
i1
.
Summing factors (example): Consider
the following recurrence
T (n)=3T (n/2) + n, T(1) = 1.
Rewrite so that all terms involving T
are on the left side
T (n) 3T (n/2) = n.
Now expand the recurrence, and choose
a factor which makes the left side “tele-
scope”
1
T (n) 3T (n/2) = n
3
T (n/2) 3T (n/4) = n/2
.
.
.
.
.
.
.
.
.
3
log
2
n1
T (2) 3T (1) = 2
Let m = log
2
n. Summing the left side
we get T (n) 3
m
T (1) = T (n) 3
m
=
T (n) n
k
where k = log
2
3 1.58496.
Summing the right side we get
m1
i=0
n
2
i
3
i
= n
m1
i=0
3
2
i
.
Let c =
3
2
. Then we have
n
m1
i=0
c
i
= n
c
m
1
c 1
=2n(c
log
2
n
1)
=2n(c
(k1) log
c
n
1)
=2n
k
2n,
and so T (n)=3n
k
2n. Full history re-
currences can often be changed to limited
history ones (example): Consider
T
i
=1+
i1
j=0
T
j
,T
0
=1.
Note that
T
i+1
=1+
i
j=0
T
j
.
Subtracting we ﬁnd
T
i+1
T
i
=1+
i
j=0
T
j
1
i1
j=0
T
j
= T
i
.
And so T
i+1
=2T
i
=2
i+1
.
Generating functions:
1. Multiply both sides of the equa-
tion by x
i
.
2. Sum both sides over all i for
which the equation is valid.
3. Choose a generating function
G(x). Usually G(x)=
i=0
x
i
g
i
.
3. Rewrite the equation in terms of
the generating function G(x).
4. Solve for G(x).
5. The coeﬃcient of x
i
in G(x)isg
i
.
Example:
g
i+1
=2g
i
+1,g
0
=0.
Multiply and sum:
i0
g
i+1
x
i
=
i0
2g
i
x
i
+
i0
x
i
.
We choose G(x)=
i0
x
i
g
i
. Rewrite
in terms of G(x):
G(x) g
0
x
=2G(x)+
i0
x
i
.
Simplify:
G(x)
x
=2G(x)+
1
1 x
.
Solve for G(x):
G(x)=
x
(1 x)(1 2x)
.
Expand this using partial fractions:
G(x)=x
2
1 2x
1
1 x
= x
2
i0
2
i
x
i
i0
x
i
=
i0
(2
i+1
1)x
i+1
.
So g
i
=2
i
1. Theoretical Computer Science Cheat Sheet
π 3.14159, e 2.71828, γ 0.57721, φ =
1+
5
2
1.61803,
ˆ
φ =
1
5
2
≈−.61803
i 2
i
p
i
General Probability
1 2 2
Bernoulli Numbers (B
i
=0,oddi = 1):
B
0
=1,B
1
=
1
2
, B
2
=
1
6
, B
4
=
1
30
,
B
6
=
1
42
, B
8
=
1
30
, B
10
=
5
66
.
log
b
x =
log
a
x
log
a
b
,
b ±
b
2
4ac
2a
.
Euler’s number e:
e =1+
1
2
+
1
6
+
1
24
+
1
120
+ ···
lim
n→∞
1+
x
n
n
= e
x
.
1+
1
n
n
<e<
1+
1
n
n+1
.
1+
1
n
n
= e
e
2n
+
11e
24n
2
O
1
n
3
.
Harmonic numbers:
1,
3
2
,
11
6
,
25
12
,
137
60
,
49
20
,
363
140
,
761
280
,
7129
2520
,...
ln n<H
n
< ln n +1,
H
n
=lnn + γ + O
1
n
.
Factorial, Stirling’s approximation:
1, 2, 6, 24, 120, 720, 5040, 40320, 362880, ...
n!=
2πn
n
e
n
1+Θ
1
n

.
Ackermann’s function and inverse:
a(i, j)=
2
j
i =1
a(i 1, 2) j =1
a(i 1,a(i, j 1)) i, j 2
α(i) = min{j | a(j, j) i}.
Continuous distributions: If
Pr[a<X<b]=
b
a
p(x) dx,
then p is the probability density function of
X.If
Pr[X<a]=P (a),
then P is the distribution function of X.If
P and p both exist then
P (a)=
a
−∞
p(x) dx.
Expectation: If X is discrete
E
[g(X)] =
x
g(x)Pr[X = x].
If X continuous then
E
[g(X)] =
−∞
g(x)p(x) dx =
−∞
g(x) dP(x).
Variance, standard deviation:
VAR[X]=
E
[X
2
]
E
[X]
2
,
σ =
VAR[X].
For events A and B:
Pr[A B] = Pr[A] + Pr[B] Pr[A B]
Pr[A B] = Pr[A] · Pr[B],
iﬀ A and B are independent.
Pr[A|B]=
Pr[A B]
Pr[B]
For random variables X and Y :
E
[X · Y ]=
E
[X] ·
E
[Y ],
if X and Y are independent.
E
[X + Y ]=
E
[X]+
E
[Y ],
E
[cX]=c
E
[X].
Bayes’ theorem:
Pr[A
i
|B]=
Pr[B|A
i
]Pr[A
i
]
n
j=1
Pr[A
j
]Pr[B|A
j
]
.
Inclusion-exclusion:
Pr
n
i=1
X
i
=
n
i=1
Pr[X
i
]+
n
k=2
(1)
k+1
i
i
<···<i
k
Pr
k
j=1
X
i
j
.
Moment inequalities:
Pr
|X|≥λ
E
[X]
1
λ
,
Pr
X
E
[X]
λ · σ
1
λ
2
.
Geometric distribution:
Pr[X = k]=pq
k1
,q=1p,
E
[X]=
k=1
kpq
k1
=
1
p
.
2 4 3
3 8 5
4 16 7
5 32 11
6 64 13
7 128 17
8 256 19
9 512 23
10 1,024 29
11 2,048 31
12 4,096 37
13 8,192 41
14 16,384 43
15 32,768 47
16 65,536 53
17 131,072 59
18 262,144 61
19 524,288 67
20 1,048,576 71
21 2,097,152 73
22 4,194,304 79
23 8,388,608 83
24 16,777,216 89
25 33,554,432 97
26 67,108,864 101
27 134,217,728 103
28 268,435,456 107
Binomial distribution:
Pr[X = k]=
n
k
p
k
q
nk
,q=1p,
E
[X]=
n
k=1
k
n
k
p
k
q
nk
= np.
Poisson distribution:
Pr[X = k]=
e
λ
λ
k
k!
,
E
[X]=λ.
Normal (Gaussian) distribution:
p(x)=
1
2πσ
e
(xµ)
2
/2σ
2
,
E
[X]=µ.
The “coupon collector”: We are given a
random coupon each day, and there are n
diﬀerent types of coupons. The distribu-
tion of coupons is uniform. The expected
number of days to pass before we to col-
lect all n types is
nH
n
.
29 536,870,912 109
30 1,073,741,824 113
31 2,147,483,648 127
32 4,294,967,296 131
Pascal’s Triangle
1
11
121
1331
14641
1 5 10 10 5 1
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
1 10 45 120 210 252 210 120 45 10 1 Theoretical Computer Science Cheat Sheet
Trigonometry Matrices More Trig.
A
c
θ
B
a
b
C
(0,-1)
(0,1)
(-1,0) (1,0)
(cos θ, sin θ)
Pythagorean theorem:
C
2
= A
2
+ B
2
.
Deﬁnitions:
sin a = A/C, cos a = B/C,
csc a = C/A, sec a = C/B,
tan a =
sin a
cos a
=
A
B
, cot a =
cos a
sin a
=
B
A
.
1
2
AB,
AB
A + B + C
.
Identities:
sin x =
1
csc x
, cos x =
1
sec x
,
tan x =
1
cot x
, sin
2
x + cos
2
x =1,
1 + tan
2
x = sec
2
x, 1 + cot
2
x = csc
2
x,
sin x = cos
π
2
x
, sin x = sin(π x),
cos x = cos(π x), tan x = cot
π
2
x
,
cot x = cot(π x), csc x = cot
x
2
cot x,
sin(x ± y) = sin x cos y ± cos x sin y,
cos(x ± y) = cos x cos y sin x sin y,
tan(x ± y)=
tan x ± tan y
1 tan x tan y
,
cot(x ± y)=
cot x cot y 1
cot x ± cot y
,
sin 2x = 2 sin x cos x, sin 2x =
2 tan x
1 + tan
2
x
,
cos 2x = cos
2
x sin
2
x, cos 2x = 2 cos
2
x 1,
cos 2x =1 2 sin
2
x, cos 2x =
1 tan
2
x
1 + tan
2
x
,
tan 2x =
2 tan x
1 tan
2
x
, cot 2x =
cot
2
x 1
2 cot x
,
sin(x + y) sin(x y) = sin
2
x sin
2
y,
cos(x + y) cos(x y) = cos
2
x sin
2
y.
Euler’s equation:
e
ix
= cos x + i sinx, e
= 1.
Multiplication:
C = A · B, c
i,j
=
n
k=1
a
i,k
b
k,j
.
Determinants: det A =0iﬀA is non-singular.
det A · B = det A ·detB,
det A =
π
n
i=1
sign(π)a
i,π(i)
.
2 × 2 and 3 × 3 determinant:
ab
cd
abc
def
ghi
= g
bc
ef
h
ac
df
+ i
ab
de
=
aei + bfg + cdh
ceg fha ibd.
Permanents:
perm A =
π
n
i=1
a
i,π(i)
.
A
a
c
h
b
B
C
Law of cosines:
c
2
= a
2
+b
2
2ab cos C.
Area:
A =
1
2
hc,
=
1
2
ab sin C,
=
c
2
sin A sin B
2 sin C
.
Heron’s formula:
A =
s · s
a
· s
b
· s
c
,
s =
1
2
(a + b + c),
s
a
= s a,
s
b
= s b,
s
c
= s c.
More identities:
sin
x
2
=
1 cos x
2
,
cos
x
2
=
1 + cos x
2
,
tan
x
2
=
1 cos x
1 + cos x
,
=
1 cos x
sin x
,
=
sin x
1 + cos x
,
cot
x
2
=
1 + cos x
1 cos x
,
=
1 + cos x
sin x
,
=
sin x
1 cos x
,
sin x =
e
ix
e
ix
2i
,
cos x =
e
ix
+ e
ix
2
,
tan x = i
e
ix
e
ix
e
ix
+ e
ix
,
= i
e
2ix
1
e
2ix
+1
,
sin x =
sinh ix
i
,
cos x = cosh ix,
tan x =
tanh ix
i
.
Hyperbolic Functions
Deﬁnitions:
sinh x =
e
x
e
x
2
, cosh x =
e
x
+ e
x
2
,
tanh x =
e
x
e
x
e
x
+ e
x
, csch x =
1
sinh x
,
sech x =
1
cosh x
, coth x =
1
tanh x
.
Identities:
cosh
2
x sinh
2
x =1, tanh
2
x + sech
2
x =1,
coth
2
x csch
2
x =1, sinh(x)=sinh x,
cosh(x) = cosh x, tanh(x)=tanh x,
sinh(x + y) = sinh x cosh y + cosh x sinh y,
cosh(x + y) = cosh x cosh y + sinh x sinh y,
sinh 2x = 2 sinh x cosh x,
cosh 2x = cosh
2
x + sinh
2
x,
cosh x + sinh x = e
x
, cosh x sinh x = e
x
,
(cosh x + sinh x)
n
= cosh nx + sinh nx, n Z,
2 sinh
2
x
2
= cosh x 1, 2 cosh
2
x
2
= cosh x +1.
θ sin θ cos θ tan θ
00 1 0
π
6
1
2
3
2
3
3
π
4
2
2
2
2
1
π
3
3
2
1
2
3
π
2
10
...in mathematics
you don’t under-
stand things, you
just get used to
them.
– J. von Neumann
v2.02
c
1994 by Steve Seiden
sseiden@acm.org
http://www.csc.lsu.edu/~seiden Theoretical Computer Science Cheat Sheet
Number Theory Graph Theory
The Chinese remainder theorem: There ex-
ists a number C such that:
C r
1
mod m
1
.
.
.
.
.
.
.
.
.
C r
n
mod m
n
if m
i
and m
j
are relatively prime for i = j.
Euler’s function: φ(x) is the number of
positive integers less than x relatively
prime to x.If
n
i=1
p
e
i
i
is the prime fac-
torization of x then
φ(x)=
n
i=1
p
e
i
1
i
(p
i
1).
Euler’s theorem: If a and b are relatively
prime then
1 a
φ(b)
mod b.
Fermat’s theorem:
1 a
p1
mod p.
The Euclidean algorithm: if a>bare in-
tegers then
gcd(a, b) = gcd(a mod b, b).
If
n
i=1
p
e
i
i
is the prime factorization of x
then
S(x)=
d|x
d =
n
i=1
p
e
i
+1
i
1
p
i
1
.
Perfect Numbers: x is an even perfect num-
ber iﬀ x =2
n1
(2
n
1) and 2
n
1 is prime.
Wilson’s theorem: n is a prime iﬀ
(n 1)! ≡−1modn.
M¨obius inversion:
µ(i)=
1ifi =1.
0ifi is not square-free.
(1)
r
if i is the product of
r distinct primes.
If
G(a)=
d|a
F (d),
then
F (a)=
d|a
µ(d)G
a
d
.
Prime numbers:
p
n
= n ln n + nln ln n n + n
ln ln n
ln n
+ O
n
ln n
,
π(n)=
n
ln n
+
n
(ln n)
2
+
2!n
(ln n)
3
+ O
n
(ln n)
4
.
Deﬁnitions:
Loop An edge connecting a ver-
tex to itself.
Directed Each edge has a direction.
Simple Graph with no loops or
multi-edges.
Walk A sequence v
0
e
1
v
1
...e
v
.
Trail A walk with distinct edges.
Path A trail with distinct
vertices.
Connected A graph where there exists
a path between any two
vertices.
Component A maximal connected
subgraph.
T ree A connected acyclic graph.
F ree tree A tree with no root.
DAG Directed acyclic graph.
Eulerian Graph with a trail visiting
each edge exactly once.
Hamiltonian Graph with a cycle visiting
each vertex exactly once.
Cut A set of edges whose re-
moval increases the num-
ber of components.
Cut-set A minimal cut.
Cut edge A size 1 cut.
k-Connected A graph connected with
the removal of any k 1
vertices.
k-Tough S V,S = we have
k ·c(G S) ≤|S|.
k-Regular A graph where all vertices
have degree k.
k-Factor A k-regular spanning
subgraph.
Matching A set of edges, no two of
Clique A set of vertices, all of
Ind. set A set of vertices, none of
Vertex cover A set of vertices which
cover all edges.
Planar graph A graph which can be em-
beded in the plane.
Plane graph An embedding of a planar
graph.
vV
deg(v)=2m.
If G is planar then n m + f =2,so
f 2n 4,m 3n 6.
Any planar graph has a vertex with de-
gree 5.
Notation:
E(G) Edge set
V (G) Vertex set
c(G) Number of components
G[S] Induced subgraph
deg(v) Degree of v
∆(G) Maximum degree
δ(G) Minimum degree
χ(G) Chromatic number
χ
E
(G) Edge chromatic number
G
c
Complement graph
K
n
Complete graph
K
n
1
,n
2
Complete bipartite graph
r
(k, ) Ramsey number
Geometry
Projective coordinates: triples
(x, y, z), not all x, y and z zero.
(x, y, z)=(cx, cy, cz) c =0.
Cartesian Projective
(x, y)(x, y, 1)
y = mx + b (m, 1,b)
x = c (1, 0, c)
Distance formula, L
p
and L
metric:
(x
1
x
0
)
2
+(y
1
y
0
)
2
,
|x
1
x
0
|
p
+ |y
1
y
0
|
p
1/p
,
lim
p→∞
|x
1
x
0
|
p
+ |y
1
y
0
|
p
1/p
.
Area of triangle (x
0
,y
0
), (x
1
,y
1
)
and (x
2
,y
2
):
1
2
abs
x
1
x
0
y
1
y
0
x
2
x
0
y
2
y
0
.
Angle formed by three points:
θ
(0, 0)
(x
1
,y
1
)
(x
2
,y
2
)
2
1
cos θ =
(x
1
,y
1
) · (x
2
,y
2
)
1
2
.
Line through two points (x
0
,y
0
)
and (x
1
,y
1
):
xy1
x
0
y
0
1
x
1
y
1
1
=0.
Area of circle, volume of sphere:
A = πr
2
,V=
4
3
πr
3
.
If I have seen farther than others,
it is because I have stood on the
shoulders of giants.
– Issac Newton Theoretical Computer Science Cheat Sheet
π Calculus
Wallis’ identity:
π =2·
2 · 2 · 4 · 4 ·6 ·6 ···
1 · 3 · 3 · 5 ·5 ·7 ···
Brouncker’s continued fraction expansion:
π
4
=1+
1
2
2+
3
2
2+
5
2
2+
7
2
2+···
Gregrory’s series:
π
4
=1
1
3
+
1
5
1
7
+
1
9
−···
Newton’s series:
π
6
=
1
2
+
1
2 · 3 · 2
3
+
1 · 3
2 · 4 · 5 · 2
5
+ ···
Sharp’s series:
π
6
=
1
3
1
1
3
1
· 3
+
1
3
2
· 5
1
3
3
· 7
+ ···
Euler’s series:
π
2
6
=
1
1
2
+
1
2
2
+
1
3
2
+
1
4
2
+
1
5
2
+ ···
π
2
8
=
1
1
2
+
1
3
2
+
1
5
2
+
1
7
2
+
1
9
2
+ ···
π
2
12
=
1
1
2
1
2
2
+
1
3
2
1
4
2
+
1
5
2
−···
Derivatives:
1.
d(cu)
dx
= c
du
dx
, 2.
d(u + v)
dx
=
du
dx
+
dv
dx
, 3.
d(uv)
dx
= u
dv
dx
+ v
du
dx
,
4.
d(u
n
)
dx
= nu
n1
du
dx
, 5.
d(u/v)
dx
=
v
du
dx
u
dv
dx
v
2
, 6.
d(e
cu
)
dx
= ce
cu
du
dx
,
7.
d(c
u
)
dx
= (ln c)c
u
du
dx
, 8.
d(ln u)
dx
=
1
u
du
dx
,
9.
d(sin u)
dx
= cos u
du
dx
, 10.
d(cos u)
dx
= sin u
du
dx
,
11.
d(tan u)
dx
= sec
2
u
du
dx
, 12.
d(cot u)
dx
= csc
2
u
du
dx
,
13.
d(sec u)
dx
= tan u sec u
du
dx
, 14.
d(csc u)
dx
= cot u csc u
du
dx
,
15.
d(arcsin u)
dx
=
1
1 u
2
du
dx
, 16.
d(arccos u)
dx
=
1
1 u
2
du
dx
,
17.
d(arctan u)
dx
=
1
1+u
2
du
dx
, 18.
d(arccot u)
dx
=
1
1+u
2
du
dx
,
19.
d(arcsec u)
dx
=
1
u
1 u
2
du
dx
, 20.
d(arccsc u)
dx
=
1
u
1 u
2
du
dx
,
21.
d(sinh u)
dx
= cosh u
du
dx
, 22.
d(cosh u)
dx
= sinh u
du
dx
,
23.
d(tanh u)
dx
= sech
2
u
du
dx
, 24.
d(coth u)
dx
= csch
2
u
du
dx
,
25.
d(sech u)
dx
= sech u tanh u
du
dx
, 26.
d(csch u)
dx
= csch u coth u
du
dx
,
27.
d(arcsinh u)
dx
=
1
1+u
2
du
dx
, 28.
d(arccosh u)
dx
=
1
u
2
1
du
dx
,
29.
d(arctanh u)
dx
=
1
1 u
2
du
dx
, 30.
d(arccoth u)
dx
=
1
u
2
1
du
dx
,
31.
d(arcsech u)
dx
=
1
u
1 u
2
du
dx
, 32.
d(arccsch u)
dx
=
1
|u|
1+u
2
du
dx
.
Integrals:
1.
cu dx = c
u dx, 2.
(u + v) dx =
udx+
v dx,
3.
x
n
dx =
1
n +1
x
n+1
,n= 1, 4.
1
x
dx =lnx, 5.
e
x
dx = e
x
,
6.
dx
1+x
2
= arctan x, 7.
u
dv
dx
dx = uv
v
du
dx
dx,
8.
sin xdx= cos x, 9.
cos xdx= sin x,
10.
tan xdx= ln |cos x|, 11.
cot xdx=ln|cos x|,
12.
sec xdx=ln|sec x + tan x|, 13.
csc xdx=ln|csc x + cotx|,
14.
arcsin
x
a
dx = arcsin
x
a
+
a
2
x
2
,a>0,
Partial Fractions
Let N(x) and D(x) be polynomial func-
tions of x. We can break down
N(x)/D(x) using partial fraction expan-
sion. First, if the degree of N is greater
than or equal to the degree of D, divide
N by D, obtaining
N(x)
D(x)
= Q(x)+
N
(x)
D(x)
,
where the degree of N
is less than that of
D. Second, factor D(x). Use the follow-
ing rules: For a non-repeated factor:
N(x)
(x a)D(x)
=
A
x a
+
N
(x)
D(x)
,
where
A =
N(x)
D(x)
x=a
.
For a repeated factor:
N(x)
(x a)
m
D(x)
=
m1
k=0
A
k
(x a)
mk
+
N
(x)
D(x)
,
where
A
k
=
1
k!
d
k
dx
k
N(x)
D(x)

x=a
.
The reasonable man adapts himself to the
world; the unreasonable persists in trying
to adapt the world to himself. Therefore
all progress depends on the unreasonable.
– George Bernard Shaw  (输入验证码)
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