1. Theoretical Computer Science Cheat Sheet
Definitions Series
f (n) = O(g(n)) iff ∃ positive c, n0 such that n n n
n(n + 1) n(n + 1)(2n + 1) n2 (n + 1)2
0 ≤ f (n) ≤ cg(n) ∀n ≥ n0 . i= , i2 = , i3 = .
i=1
2 i=1
6 i=1
4
f (n) = Ω(g(n)) iff ∃ positive c, n0 such that
In general:
f (n) ≥ cg(n) ≥ 0 ∀n ≥ n0 . n n
1
f (n) = Θ(g(n)) iff f (n) = O(g(n)) and im = (n + 1)m+1 − 1 − (i + 1)m+1 − im+1 − (m + 1)im
i=1
m+1 i=1
f (n) = Ω(g(n)).
n−1 m
1 m+1
f (n) = o(g(n)) iff limn→∞ f (n)/g(n) = 0. im = Bk nm+1−k .
i=1
m+1 k
k=0
lim an = a iff ∀ > 0, ∃n0 such that
n→∞ Geometric series:
|an − a| < , ∀n ≥ n0 . n ∞ ∞
cn+1 − 1 1 c
sup S least b ∈ R such that b ≥ s, ci = , c = 1, ci = , ci = , |c| < 1,
c−1 1−c 1−c
∀s ∈ S. i=0 i=0 i=1
n ∞
ncn+2 − (n + 1)cn+1 + c c
inf S greatest b ∈ R such that b ≤ ici = , c = 1, ici = , |c| < 1.
s, ∀s ∈ S. i=0
(c − 1)2 i=0
(1 − c)2
Harmonic series:
lim inf an lim inf{ai | i ≥ n, i ∈ N}. n n
n→∞ n→∞ 1 n(n + 1) n(n − 1)
Hn = , iHi = Hn − .
lim sup an lim sup{ai | i ≥ n, i ∈ N}. i=1
i i=1
2 4
n→∞ n→∞
n n
n i n+1 1
k Combinations: Size k sub- Hi = (n + 1)Hn − n, Hi = Hn+1 − .
sets of a size n set. i=1 i=1
m m+1 m+1
n n
Stirling numbers (1st kind): n n! n n n
k
Arrangements of an n ele- 1. = , 2. = 2n , 3. = ,
k (n − k)!k! k k n−k
k=0
ment set into k cycles.
n n n−1 n n−1 n−1
n 4. = , 5. = + ,
k Stirling numbers (2nd kind): k k k−1 k k k−1
Partitions of an n element n m n n−k
n
r+k r+n+1
set into k non-empty sets. 6. = , 7. = ,
m k k m−k k n
n k=0
1st order Eulerian numbers: n n
k k n+1 r s r+s
Permutations π1 π2 . . . πn on 8. = , 9. = ,
m m+1 k n−k n
{1, 2, . . . , n} with k ascents. k=0 k=0
n k−n−1 n n
n
2nd order Eulerian numbers. 10. = (−1) k
, 11. = = 1,
k k k 1 n
Cn Catalan Numbers: Binary n n−1 n n−1
trees with n + 1 vertices. 12. = 2n−1 − 1, , 13. =k +
2 k−1 k k
n n n n n
14. = (n − 1)!, 15. = (n − 1)!Hn−1 , 16. = 1, 17. ≥ ,
1 2 n k k
n
n n−1 n−1 n n n n 1 2n
18. = (n − 1) + , 19. = = , 20. = n!, 21. Cn = ,
k k k−1 n−1 n−1 2 k n+1 n
k=0
n n n n n n−1 n−1
22. = = 1, 23. = , 24. = (k + 1) + (n − k) ,
0 n−1 k n−1−k k k k−1
0 1 if k = 0, n n n+1
25. = 26. = 2n − n − 1, 27. = 3n − (n + 1)2n + ,
k 0 otherwise 1 2 2
n m n
n x+k n n+1 n n k
28. xn = , 29. = (m + 1 − k)n (−1)k , 30. m! = ,
k n m k m k n−m
k=0 k=0 k=0
n
n n n−k n n
31. = (−1)n−k−m k!, 32. = 1, 33. = 0 for n = 0,
m k m 0 n
k=0
n
n n−1 n−1 n (2n)n
34. = (k + 1) + (2n − 1 − k) , 35. = ,
k k k−1 k 2n
k=0
n n
x n x+n−1−k n+1 n k k
36. = , 37. = = (m + 1)n−k ,
x−n k 2n m+1 k m m
k=0 k k=0

2. Theoretical Computer Science Cheat Sheet
Identities Cont. Trees
n n n
n+1 n k k n−k 1 k x n x+k Every tree with n
38. = = n = n! , 39. = , vertices has n − 1
m+1 k m m k! m x−n k 2n
k k=0 k=0 k=0
edges.
n n k+1 n n+1 k
40. = (−1)n−k , 41. = (−1)m−k , Kraft inequal-
m k m+1 m k+1 m
k k ity: If the depths
m m
m+n+1 n+k m+n+1 n+k of the leaves of
42. = k , 43. = k(n + k) ,
m k m k a binary tree are
k=0 k=0
n n+1 k n n+1 k d1 , . . . , dn :
44. = (−1)m−k , 45. (n − m)! = (−1)m−k , for n ≥ m, n
m k+1 m m k+1 m 2−di ≤ 1,
k k
n m−n m+n m+k n m−n m+n m+k i=1
46. = , 47. = ,
n−m m+k n+k k n−m m+k n+k k and equality holds
k k
n +m k n−k n n +m k n−k n only if every in-
48. = , 49. = . ternal node has 2
+m m k +m m k
k k
sons.
Recurrences
Master method: Generating functions:
T (n) = aT (n/b) + f (n), a ≥ 1, b > 1 1 T (n) − 3T (n/2) = n 1. Multiply both sides of the equa-
3 T (n/2) − 3T (n/4) = n/2 tion by xi .
If ∃ > 0 such that f (n) = O(nlogb a− )
. . . 2. Sum both sides over all i for
then . . .
. . . which the equation is valid.
T (n) = Θ(nlogb a ).
3log2 n−1 T (2) − 3T (1) = 2 3. Choose a generating function
∞
If f (n) = Θ(nlogb a ) then G(x). Usually G(x) = i=0 xi gi .
T (n) = Θ(nlogb a log2 n). Let m = log2 n. Summing the left side 3. Rewrite the equation in terms of
we get T (n) − 3m T (1) = T (n) − 3m = the generating function G(x).
If ∃ > 0 such that f (n) = Ω(nlogb a+ ), T (n) − nk where k = log2 3 ≈ 1.58496.
4. Solve for G(x).
and ∃c < 1 such that af (n/b) ≤ cf (n) Summing the right side we get
for large n, then m−1 m−1 5. The coefficient of xi in G(x) is gi .
n i 3 i Example:
T (n) = Θ(f (n)). 3 =n 2 .
i=0
2i i=0
gi+1 = 2gi + 1, g0 = 0.
Substitution (example): Consider the
following recurrence Let c = 3 . Then we have
2
Multiply and sum:
i
Ti+1 = 22 · Ti2 , T1 = 2.
m−1
c −1m gi+1 xi = 2gi xi + xi .
i
n c =n i≥0 i≥0 i≥0
c−1
Note that Ti is always a power of two. i=0
We choose G(x) = i≥0 xi gi . Rewrite
Let ti = log2 Ti . Then we have = 2n(clog2 n − 1)
in terms of G(x):
ti+1 = 2i + 2ti , t1 = 1.
= 2n(c(k−1) logc n − 1) G(x) − g0
= 2G(x) + xi .
Let ui = ti /2i . Dividing both sides of = 2nk − 2n, x
i≥0
the previous equation by 2i+1 we get
ti+1 2i ti and so T (n) = 3n − 2n. Full history re-
k Simplify:
= i+1 + i . G(x) 1
2 i+1 2 2 currences can often be changed to limited = 2G(x) + .
history ones (example): Consider x 1−x
Substituting we find i−1
ui+1 = 1 + ui ,
2 u1 = 1 ,
2
Solve for G(x):
Ti = 1 + Tj , T0 = 1. x
G(x) = .
which is simply ui = i/2. So we find j=0
(1 − x)(1 − 2x)
i−1
that Ti has the closed form Ti = 2i2 . Note that
i Expand this using partial fractions:
Summing factors (example): Consider 2 1
the following recurrence Ti+1 = 1 + Tj . G(x) = x −
1 − 2x 1 − x
T (n) = 3T (n/2) + n, T (1) = 1. j=0
 
Subtracting we find
Rewrite so that all terms involving T i i−1 = x 2 2i xi − xi 
are on the left side Ti+1 − Ti = 1 + Tj − 1 − Tj i≥0 i≥0
T (n) − 3T (n/2) = n. j=0 j=0
= i+1
(2 − 1)x i+1
.
Now expand the recurrence, and choose = Ti . i≥0
a factor which makes the left side “tele-
scope” And so Ti+1 = 2Ti = 2i+1 . So gi = 2i − 1.

3. Theoretical Computer Science Cheat Sheet
√ √
1+ 5 ˆ 1− 5
π ≈ 3.14159, e ≈ 2.71828, γ ≈ 0.57721, φ= 2 ≈ 1.61803, φ= 2 ≈ −.61803
i 2i pi General Probability
1 2 2 Bernoulli Numbers (Bi = 0, odd i = 1): Continuous distributions: If
1 1 1 b
2 4 3 B0 = 1, B1 = B2 = −2,
B4 = 6, − 30 , Pr[a < X < b] = p(x) dx,
1 1 5
3 8 5 B6 = 42 , B8 = − 30 , B10 = 66 . a
4 16 7 Change of base, quadratic formula: then p is the probability density function of
√ X. If
5 32 11 loga x −b ± b2 − 4ac Pr[X < a] = P (a),
logb x = , .
6 64 13 loga b 2a
then P is the distribution function of X. If
7 128 17 Euler’s number e: P and p both exist then
1 1 11
8 256 19 e=1+ 2 + 24 + 120 + · · ·
+ 6
a
x n P (a) = p(x) dx.
9 512 23 lim 1 + = ex . −∞
n→∞ n Expectation: If X is discrete
10 1,024 29 1 n 1 n+1
1+ n <e< 1+ n .
11 2,048 31 E[g(X)] = g(x) Pr[X = x].
1 n e 11e 1 x
12 4,096 37 1+ =e− + 2
−O .
n 2n 24n n3 If X continuous then
13 8,192 41 ∞ ∞
Harmonic numbers:
14 16,384 43 E[g(X)] = g(x)p(x) dx = g(x) dP (x).
1, 3 , 11 , 25 , 137 , 49 , 363 , 761 , 7129 , . . .
2 6 12 60 20 140 280 2520
−∞ −∞
15 32,768 47 Variance, standard deviation:
16 65,536 53 ln n < Hn < ln n + 1, VAR[X] = E[X 2 ] − E[X]2 ,
17 131,072 59 1 σ = VAR[X].
Hn = ln n + γ + O .
18 262,144 61 n For events A and B:
19 524,288 67 Factorial, Stirling’s approximation: Pr[A ∨ B] = Pr[A] + Pr[B] − Pr[A ∧ B]
20 1,048,576 71 1, 2, 6, 24, 120, 720, 5040, 40320, 362880, ... Pr[A ∧ B] = Pr[A] · Pr[B],
21 2,097,152 73 iff A and B are independent.
√ n 1
n
22 4,194,304 79 n! = 2πn 1+Θ . Pr[A ∧ B]
e n Pr[A|B] =
23 8,388,608 83 Pr[B]
Ackermann’s function and inverse:

24 16,777,216 89 For random variables X and Y :
 2j i=1
25 33,554,432 97 a(i, j) = a(i − 1, 2) j=1 E[X · Y ] = E[X] · E[Y ],

26 67,108,864 101 a(i − 1, a(i, j − 1)) i, j ≥ 2 if X and Y are independent.
27 134,217,728 103 α(i) = min{j | a(j, j) ≥ i}. E[X + Y ] = E[X] + E[Y ],
E[cX] = c E[X].
28 268,435,456 107 Binomial distribution:
n k n−k Bayes’ theorem:
29 536,870,912 109 Pr[X = k] = p q , q = 1 − p,
k Pr[B|Ai ] Pr[Ai ]
30 1,073,741,824 113 Pr[Ai |B] = n .
n j=1 Pr[Aj ] Pr[B|Aj ]
31 2,147,483,648 127 n k n−k
E[X] = k p q = np. Inclusion-exclusion:
32 4,294,967,296 131 k n n
k=1
Poisson distribution: Pr Xi = Pr[Xi ] +
Pascal’s Triangle
e−λ λk i=1 i=1
1 Pr[X = k] = , E[X] = λ. n k
k!
11 Normal (Gaussian) distribution: (−1)k+1 Pr Xij .
k=2 ii <···<ik j=1
121 1 2 2
p(x) = √ e−(x−µ) /2σ , E[X] = µ. Moment inequalities:
1331 2πσ
1
14641 The “coupon collector”: We are given a Pr |X| ≥ λ E[X] ≤ ,
random coupon each day, and there are n λ
1 5 10 10 5 1 1
different types of coupons. The distribu- Pr X − E[X] ≥ λ · σ ≤ .
1 6 15 20 15 6 1 tion of coupons is uniform. The expected λ2
1 7 21 35 35 21 7 1 Geometric distribution:
number of days to pass before we to col-
lect all n types is Pr[X = k] = pq k−1 , q = 1 − p,
1 8 28 56 70 56 28 8 1
∞
1 9 36 84 126 126 84 36 9 1 nHn . 1
E[X] = kpq k−1 = .
p
1 10 45 120 210 252 210 120 45 10 1 k=1

4. Theoretical Computer Science Cheat Sheet
Trigonometry Matrices More Trig.
Multiplication: C
n
(0,1)
C = A · B, ci,j = ai,k bk,j . a
b b h
(cos θ, sin θ) k=1
C θ Determinants: det A = 0 iff A is non-singular.
A
(-1,0) (1,0) A c B
det A · B = det A · det B, Law of cosines:
c a
c2 = a2 +b2 −2ab cos C.
n
(0,-1)
B det A = sign(π)ai,π(i) .
π i=1 Area:
Pythagorean theorem:
C 2 = A2 + B 2 . 2 × 2 and 3 × 3 determinant:
a b A = 1 hc,
2
Definitions: = ad − bc, = 1 ab sin C,
c d 2
sin a = A/C, cos a = B/C,
a b c c2 sin A sin B
csc a = C/A, sec a = C/B, b c a c a b = .
d e f =g −h +i 2 sin C
sin a A cos a B e f d f d e Heron’s formula:
tan a = = , cot a = = . g h i
cos a B sin a A aei + bf g + cdh √
Area, radius of inscribed circle: =
− ceg − f ha − ibd. A = s · sa · sb · sc ,
1 AB
Permanents: s = 1 (a + b + c),
2 AB, . 2
A+B+C n
sa = s − a,
Identities: perm A = ai,π(i) .
π i=1
sb = s − b,
1 1
sin x = , cos x = , Hyperbolic Functions sc = s − c.
csc x sec x
1 Definitions: More identities:
tan x = , sin2 x + cos2 x = 1,
cot x ex − e−x e +e x −x
1 − cos x
sinh x = , cosh x = , sin x = ,
1 + tan2 x = sec2 x, 1 + cot2 x = csc2 x, 2
−x
2 2 2
e −e
x
1
tanh x = x , csch x = , 1 + cos x
sin x = cos π
2 −x , sin x = sin(π − x), e + e−x sinh x cos x =
2 ,
1 1 2
sech x = , coth x = .
cos x = − cos(π − x), tan x = cot π
2 −x , cosh x tanh x 1 − cos x
tan x =
2 ,
Identities: 1 + cos x
cot x = − cot(π − x), csc x = cot x − cot x,
2 1 − cos x
cosh2 x − sinh2 x = 1, tanh2 x + sech2 x = 1, =
sin x
,
sin(x ± y) = sin x cos y ± cos x sin y,
sin x
coth2 x − csch2 x = 1, sinh(−x) = − sinh x, = ,
cos(x ± y) = cos x cos y sin x sin y, 1 + cos x
cosh(−x) = cosh x, tanh(−x) = − tanh x, 1 + cos x
tan x ± tan y cot x = ,
tan(x ± y) = , 2 1 − cos x
1 tan x tan y sinh(x + y) = sinh x cosh y + cosh x sinh y,
1 + cos x
cot x cot y 1 = ,
cot(x ± y) = , cosh(x + y) = cosh x cosh y + sinh x sinh y, sin x
cot x ± cot y sin x
2 tan x sinh 2x = 2 sinh x cosh x, = ,
sin 2x = 2 sin x cos x, sin 2x =
1 + tan2 x
, 1 − cos x
cosh 2x = cosh2 x + sinh2 x, eix − e−ix
cos 2x = cos2 x − sin2 x, cos 2x = 2 cos2 x − 1, sin x = ,
2i
1 − tan2 x cosh x + sinh x = ex , cosh x − sinh x = e−x , eix + e−ix
cos 2x = 1 − 2 sin2 x, cos 2x = ,
1 + tan2 x cos x = ,
(cosh x + sinh x)n = cosh nx + sinh nx, n ∈ Z, 2
2 tan x cot2 x − 1 eix − e−ix
tan 2x = 2 , cot 2x = , 2 sinh2 x = cosh x − 1, 2 cosh2 x = cosh x + 1. tan x = −i ix ,
1 − tan x 2 cot x 2 2 e + e−ix
sin(x + y) sin(x − y) = sin2 x − sin2 y, e2ix − 1
θ sin θ cos θ tan θ . . . in mathematics = −i 2ix ,
e +1
cos(x + y) cos(x − y) = cos2 x − sin2 y. you don’t under- sinh ix
0 0 √
1 √
0 sin x = ,
π 1 3 3 stand things, you i
Euler’s equation: 6 2 2 3
√ √ just get used to
eix = cos x + i sin x, eiπ
= −1. π 2 2
cos x = cosh ix,
4 2 2 1 them.
v2.02 c 1994 by Steve Seiden π
√
3 1
√ – J. von Neumann tan x =
tanh ix
.
3 2 2 3 i
sseiden@acm.org π
2 1 0 ∞
http://www.csc.lsu.edu/~seiden

5. Theoretical Computer Science Cheat Sheet
Number Theory Graph Theory
The Chinese remainder theorem: There ex- Definitions: Notation:
ists a number C such that: Loop An edge connecting a ver- E(G) Edge set
tex to itself. V (G) Vertex set
C ≡ r1 mod m1 c(G) Number of components
Directed Each edge has a direction.
. .
. . .
. Simple Graph with no loops or G[S] Induced subgraph
. . .
multi-edges. deg(v) Degree of v
C ≡ rn mod mn ∆(G) Maximum degree
Walk A sequence v0 e1 v1 . . . e v .
if mi and mj are relatively prime for i = j. Trail A walk with distinct edges. δ(G) Minimum degree
Path A trail with distinct χ(G) Chromatic number
Euler’s function: φ(x) is the number of
vertices. χE (G) Edge chromatic number
positive integers less than x relatively
n Connected A graph where there exists Gc Complement graph
prime to x. If i=1 pei is the prime fac-
i
a path between any two Kn Complete graph
torization of x then
n
vertices. Kn1 ,n2 Complete bipartite graph
φ(x) = pi i −1 (pi − 1).
e
Component A maximal connected
r(k, ) Ramsey number
i=1
subgraph. Geometry
Euler’s theorem: If a and b are relatively Tree A connected acyclic graph.
prime then Projective coordinates: triples
Free tree A tree with no root.
1 ≡ aφ(b) mod b. (x, y, z), not all x, y and z zero.
DAG Directed acyclic graph.
Eulerian Graph with a trail visiting (x, y, z) = (cx, cy, cz) ∀c = 0.
Fermat’s theorem:
each edge exactly once. Cartesian Projective
1 ≡ ap−1 mod p.
Hamiltonian Graph with a cycle visiting (x, y) (x, y, 1)
The Euclidean algorithm: if a > b are in- each vertex exactly once. y = mx + b (m, −1, b)
tegers then Cut A set of edges whose re- x=c (1, 0, −c)
gcd(a, b) = gcd(a mod b, b). moval increases the num- Distance formula, Lp and L∞
n
If i=1 pei is the prime factorization of x ber of components. metric:
i
then Cut-set A minimal cut. (x1 − x0 )2 + (y1 − y0 )2 ,
pi i +1 − 1
n e Cut edge A size 1 cut. 1/p
S(x) = d= . k-Connected A graph connected with |x1 − x0 |p + |y1 − y0 |p ,
pi − 1
d|x i=1 the removal of any k − 1 lim |x1 − x0 |p + |y1 − y0 | p 1/p
.
Perfect Numbers: x is an even perfect num- vertices. p→∞
ber iff x = 2n−1 (2n −1) and 2n −1 is prime. k-Tough ∀S ⊆ V, S = ∅ we have Area of triangle (x0 , y0 ), (x1 , y1 )
Wilson’s theorem: n is a prime iff k · c(G − S) ≤ |S|. and (x2 , y2 ):
(n − 1)! ≡ −1 mod n. k-Regular A graph where all vertices 1 x1 − x0 y1 − y0
2 abs x − x .
have degree k. 2 0 y2 − y0
M¨bius 
o inversion: k-Factor A k-regular spanning
1 if i = 1. Angle formed by three points:
 subgraph.
0 if i is not square-free.
µ(i) = Matching A set of edges, no two of

 (−1)r if i is the product of (x2 , y2 )
r distinct primes. which are adjacent.
2
Clique A set of vertices, all of
If θ
which are adjacent.
G(a) = F (d), (0, 0) 1 (x1 , y1 )
Ind. set A set of vertices, none of
d|a
which are adjacent. (x1 , y1 ) · (x2 , y2 )
cos θ = .
then Vertex cover A set of vertices which 1 2
a
F (a) = µ(d)G . cover all edges. Line through two points (x0 , y0 )
d
d|a Planar graph A graph which can be em- and (x1 , y1 ):
Prime numbers: beded in the plane. x y 1
ln ln n Plane graph An embedding of a planar x0 y0 1 = 0.
pn = n ln n + n ln ln n − n + n
ln n graph. x1 y1 1
n Area of circle, volume of sphere:
+O , deg(v) = 2m.
ln n A = πr2 , V = 4 πr3 .
v∈V 3
n n 2!n
π(n) = + + If G is planar then n − m + f = 2, so
ln n (ln n)2 (ln n)3 If I have seen farther than others,
f ≤ 2n − 4, m ≤ 3n − 6. it is because I have stood on the
n
+O . Any planar graph has a vertex with de- shoulders of giants.
(ln n)4
gree ≤ 5. – Issac Newton

6. Theoretical Computer Science Cheat Sheet
π Calculus
Wallis’ identity: Derivatives:
2 · 2 · 4 · 4 · 6 · 6···
π =2· d(cu) du d(u + v) du dv d(uv) dv du
1 · 3 · 3 · 5 · 5 · 7··· 1. =c , 2. = + , 3. =u +v ,
dx dx dx dx dx dx dx dx
Brouncker’s continued fraction expansion:
12 d(un ) du d(u/v) v du − u dv
d(ecu ) du
π 4. = nun−1 , 5. = dx dx
, 6. = cecu ,
4 =1+ 32 dx dx dx v2 dx dx
2+ 52
2+
2+ 72 d(cu ) du d(ln u) 1 du
2+···
7. = (ln c)cu , 8. = ,
dx dx dx u dx
Gregrory’s series:
1 1 1 1
4 =1− 3 + − − ···
π
+ d(sin u) du d(cos u) du
5 7 9 9. = cos u , 10. = − sin u ,
dx dx dx dx
Newton’s series:
d(tan u) du d(cot u) du
1 1 1·3 11. = sec2 u , 12. = csc2 u ,
dx dx dx dx
6 = 2 + 2 · 3 · 23 + 2 · 4 · 5 · 25 + · · ·
π
d(sec u) du d(csc u) du
Sharp’s series: 13. = tan u sec u , 14. = − cot u csc u ,
dx dx dx dx
1 1 1 1 d(arcsin u) 1 du d(arccos u) −1 du
π
= √ 1− 1 + 2 − 3 +··· 15. =√ , 16. =√ ,
6 3 ·3 3 ·5 3 ·7 dx 1−u 2 dx dx 1 − u2 dx
3
Euler’s series: d(arctan u) 1 du d(arccot u) −1 du
17. = , 18. = ,
dx 1 + u2 dx dx 1 + u2 dx
π2 1 1 1 1 1
6 = 12 + 22 + 32 + 42 + 52 + ··· d(arcsec u) 1 du d(arccsc u) −1 du
π2 1 1 1 1 1
19. = √ , 20. = √ ,
u 1−u 2 dx u 1−u 2 dx
8 = 12 + 32 + 52 + 72 + 92 + ··· dx dx
π2 1 1 1 1 1 d(sinh u) du d(cosh u) du
12 = 12 − 22 + 32 − 42 + 52 − ··· 21. = cosh u , 22. = sinh u ,
dx dx dx dx
Partial Fractions d(tanh u) du d(coth u) du
23. = sech2 u , 24. = − csch2 u ,
Let N (x) and D(x) be polynomial func- dx dx dx dx
tions of x. We can break down d(sech u) du d(csch u) du
N (x)/D(x) using partial fraction expan- 25. = − sech u tanh u , 26. = − csch u coth u ,
dx dx dx dx
sion. First, if the degree of N is greater
than or equal to the degree of D, divide d(arcsinh u) 1 du d(arccosh u) 1 du
27. =√ , 28. =√ ,
dx 1+u 2 dx dx u 2 − 1 dx
N by D, obtaining
N (x) N (x) d(arctanh u) 1 du d(arccoth u) 1 du
= Q(x) + , 29. = , 30. = 2 ,
D(x) D(x) dx 1 − u2 dx dx u − 1 dx
where the degree of N is less than that of d(arcsech u) −1 du d(arccsch u) −1 du
31. = √ , 32. = √ .
D. Second, factor D(x). Use the follow- dx u 1 − u2 dx dx |u| 1 + u2 dx
ing rules: For a non-repeated factor: Integrals:
N (x) A N (x)
= + ,
(x − a)D(x) x−a D(x) 1. cu dx = c u dx, 2. (u + v) dx = u dx + v dx,
where
N (x) 1 1
A= . 3. xn dx = xn+1 , n = −1, 4. dx = ln x, 5. ex dx = ex ,
D(x) x=a
n+1 x
For a repeated factor: dx dv du
6. = arctan x, 7. u dx = uv − v dx,
N (x)
m−1
Ak N (x) 1 + x2 dx dx
= + ,
(x − a)m D(x) (x − a)m−k D(x) 8. sin x dx = − cos x, 9. cos x dx = sin x,
k=0
where
1 dk N (x) 10. tan x dx = − ln | cos x|, 11. cot x dx = ln | cos x|,
Ak = .
k! dxk D(x) x=a
12. sec x dx = ln | sec x + tan x|, 13. csc x dx = ln | csc x + cot x|,
The reasonable man adapts himself to the
world; the unreasonable persists in trying
to adapt the world to himself. Therefore 14. arcsin x dx = arcsin x +
a a a2 − x2 , a > 0,
all progress depends on the unreasonable.
– George Bernard Shaw

7. Theoretical Computer Science Cheat Sheet
Calculus Cont.
15. arccos x dx = arccos x −
a a a2 − x2 , a > 0, 16. arctan x dx = x arctan x −
a a
a
2 ln(a2 + x2 ), a > 0,
17. sin2 (ax)dx = 1
2a ax − sin(ax) cos(ax) , 18. cos2 (ax)dx = 1
2a ax + sin(ax) cos(ax) ,
19. sec2 x dx = tan x, 20. csc2 x dx = − cot x,
sinn−1 x cos x n − 1 cosn−1 x sin x n − 1
21. sinn x dx = − + sinn−2 x dx, 22. cosn x dx = + cosn−2 x dx,
n n n n
tann−1 x cotn−1 x
23. tann x dx = − tann−2 x dx, n = 1, 24. cotn x dx = − − cotn−2 x dx, n = 1,
n−1 n−1
tan x secn−1 x n − 2
25. secn x dx = + secn−2 x dx, n = 1,
n−1 n−1
cot x cscn−1 x n − 2
26. cscn x dx = − + cscn−2 x dx, n = 1, 27. sinh x dx = cosh x, 28. cosh x dx = sinh x,
n−1 n−1
29. tanh x dx = ln | cosh x|, 30. coth x dx = ln | sinh x|, 31. sech x dx = arctan sinh x, 32. csch x dx = ln tanh x ,
2
33. sinh2 x dx = 1
4 sinh(2x) − 1 x,
2 34. cosh2 x dx = 1
4 sinh(2x) + 1 x,
2 35. sech2 x dx = tanh x,
36. arcsinh x dx = x arcsinh x −
a a x2 + a2 , a > 0, 37. arctanh x dx = x arctanh x +
a a
a
2 ln |a2 − x2 |,
 x
 x arccosh − x2 + a2 , if arccosh x > 0 and a > 0,
a
38. x
arccosh a dx = a
x
 x arccosh + x2 + a2 , if arccosh x < 0 and a > 0,
a a
dx
39. √ = ln x + a2 + x2 , a > 0,
a2 + x2
dx 1 a2
40. = a arctan x , a > 0, 41. a2 − x2 dx = x
2 a2 − x2 + 2 arcsin x , a > 0,
a2 + x2 a a
3a4
42. (a2 − x2 )3/2 dx = x (5a2 − 2x2 ) a2 − x2 +
8 8 arcsin x ,
a a > 0,
dx dx 1 a+x dx x
43. √ = arcsin x , a > 0, 44. = ln , 45. = √ ,
a2 − x2 a a2 −x 2 2a a−x (a2 −x2 )3/2
a2 a2 − x2
a2 dx
46. a2 ± x2 dx = x
2 a2 ± x2 ± 2 ln x + a2 ± x2 , 47. √ = ln x + x2 − a2 , a > 0,
x2 − a2
dx 1 x √ 2(3bx − 2a)(a + bx)3/2
48. = ln , 49. x a + bx dx = ,
ax2 + bx a a + bx 15b2
√ √ √
a + bx √ 1 x 1 a + bx − a
50. dx = 2 a + bx + a √ dx, 51. √ dx = √ ln √ √ , a > 0,
x x a + bx a + bx 2 a + bx + a
√ √
a2 − x2 a + a2 − x2
52. dx = a2 − x2 − a ln , 53. x a2 − x2 dx = − 1 (a2 − x2 )3/2 ,
3
x x
√
2 2 2 a4 dx 1 a+ a2 − x2
54. x a2 − x2 dx = x
8 (2x −a ) a2 − x2 + 8 arcsin x
a, a > 0, 55. √ = − a ln ,
a2 − x2 x
x dx x2 dx 2
56. √ = − a2 − x2 , 57. √ = − x a2 − x2 + a arcsin a,
2 2
x
a > 0,
a2 − x2 a2 − x2
√ √ √
a2 + x2 2 + x2 − a ln
a + a2 + x2 x2 − a2
58. dx = a , 59. dx = x2 − a2 − a arccos |x| ,
a
a > 0,
x x x
dx x
60. x x2 ± a2 dx = 1 (x2 ± a2 )3/2 ,
3 61. √ = 1
ln √ ,
x x2 + a2 a
a+ a2 + x2