1O9OqeO

f(n) = O(g(n))
f(n) = Ω(g(n))
Definitions
iff ∃ positive c, n 0 such that
0 ≤ f(n) ≤ cg(n) ∀n ≥ n 0 .
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 a n = a iff ∀ϵ > 0, ∃n 0 such that
n→∞
|a n − a| < ϵ, ∀n ≥ n 0 .
sup S least b ∈ R such that b ≥ s,
∀s ∈ S.
inf S
lim inf
n→∞ a n
greatest b ∈ R such that b ≤
s, ∀s ∈ S.
lim inf{a i | i ≥ n, i ∈ N}.
n→∞
Theoretical Computer Science Cheat Sheet
n∑
i =
i=1
In general:
n∑
i=1
i=1
n(n + 1)
,
2
i m = 1
m + 1
n−1
∑
i m = 1
m + 1
i=0
n∑
i 2 =
i=1
[
(n + 1) m+1 − 1 −
m∑
( m + 1
k=0
k
Series
n(n + 1)(2n + 1)
,
6
n∑
i=1
i 3 = n2 (n + 1) 2
.
4
n∑ (
(i + 1) m+1 − i m+1 − (m + 1)i m)]
i=1
)
B k n m+1−k .
Geometric series:
n∑
c i = cn+1 − 1
c − 1 , c ≠ 1, ∑
∞ c i = 1
1 − c , ∑
∞ c i =
i=0
i=0
i=1
n∑
ic i = ncn+2 − (n + 1)c n+1 + c
∞∑
(c − 1) 2 , c ≠ 1, ic i =
Harmonic series:
n∑ 1
H n =
i ,
n ∑
i=0
n(n + 1) n(n − 1)
iH i = H n − .
2
4
n∑
( ( ) (
i n + 1
H i =
m)
m + 1
c , |c| < 1,
1 − c
c
, |c| < 1.
(1 − c)
2
lim sup a n lim sup{a i | i ≥ n, i ∈ N}.
i=1 i=1
n→∞
n→∞
( n∑
H i = (n + 1)H n − n,
H n+1 − 1 )
n
k)
Combinations: Size k subsets
of a size n set.
m + 1
.
i=1
i=1
[ n
]
(
k
Stirling numbers (1st kind): n n!
Arrangements of an n element
set into k cycles.
( n
1. =
k)
(n − k)!k! , 2. ∑
n ( ( ( )
n n n
= 2
k)
n , 3. = ,
k)
n − k
k=0
4. =
k)
n ( )
( ( ) ( )
{ n − 1
n n − 1 n − 1
n
}
, 5. = + ,
k
Stirling numbers (2nd kind):
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
n∑
( ( )
k n + 1
n∑
( )( ) ( )
k
1st order Eulerian numbers:
r s r + s
Permutations π 1 π 2 . . . π n on 8. = , 9.
= ,
m)
m + 1
k n − k n
{1, 2, . . . , n} with k ascents.
k=0
( ( )
k=0
{ } {
⟨ n
⟩
n k − n − 1
n n
10. = (−1)
k)
k , 11. = = 1,
k
2nd order Eulerian numbers.
k
1 n}
C { }
{ } { } { }
n Catalan Numbers: Binary
n n n − 1 n − 1
trees with n + 1 vertices. 12. = 2 n−1 − 1, 13. = k + ,
2 k k k − 1
[ ]
[ ]
[ [ ] { }
n n n n n
14. = (n − 1)!, 15. = (n − 1)!H n−1 , 16. = 1, 17. ≥ ,
1 2 n]
k k
[ ] [ ] [ ] { } [ ] ( n n − 1 n − 1
n n n
n∑
[ ] n
18. = (n − 1) + , 19. = = , 20. = n!, 21. C n =
k k k − 1
n − 1 n − 1 2)
1 ) 2n
,
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. = 2 n − n − 1, 27. = 3 n − (n + 1)2 n + ,
k 0 otherwise
1 2 2
n∑
⟨ ⟩( )
⟨ n x + k
n
m∑
( )
{ n + 1
n
n∑
⟨ ⟩( )
n k
28. x n =
, 29. =
(m + 1 − k)
k n
m⟩
n (−1) k , 30. m! =
,
k
m}
k n − m
k=0
k=0
k=0
⟨ n
n∑
{ }( )
⟨ ⟩
⟨ ⟩
n n − k
n n
31. =
(−1)
m⟩
n−k−m k!, 32. = 1, 33. = 0 for n ≠ 0,
k m
0
n
k=0
⟨ ⟩ ⟨ ⟩
⟨ ⟩
n n − 1
n − 1
n∑
⟨ ⟩ n
34. = (k + 1) + (2n − 1 − k) , 35.
= (2n)n
k
k
k − 1
k 2 n ,
k=0
{ } x
n∑
⟨ ⟩( )
{ }
n x + n − 1 − k
n + 1
36. =
, 37.
= ∑ ( ){ n k
n∑
{ k
= (m + 1)
x − n k 2n
m + 1 k m}
m}
n−k ,
k=0
k
k=0

38.
40.
42.
44.
46.
48.
Theoretical Computer Science Cheat Sheet
Identities Cont.
[ ] n + 1
= ∑ [ ]( n k
n∑
[ k
n∑
] [ ]
= n
m + 1 k m)
m]
n−k 1 k x
n∑
⟨ ⟩( )
n x + k
= n!
, 39. =
,
k![
m
x − n k 2n
{
k
k=0
k=0
k=0
n
=
m}
∑ ( ){ }
[ n k + 1
n
(−1) n−k , 41. =
k m + 1
m]
∑ [ ]( )
n + 1 k
(−1) m−k ,
k + 1 m
k
k
{ }
m + n + 1
m∑
{ }
[ ]
n + k
m + n + 1
m∑
[ ] n + k
= k , 43.
= k(n + k) ,
m
k
m
k
(
k=0
k=0
n
=
m)
∑ { }[ ]
( n + 1 k n
(−1) m−k , 45. (n − m)! =
k + 1 m
m)
∑ [ ]{ }
n + 1 k
(−1) m−k , for n ≥ m,
k + 1 m
{ }
k
k
n
= ∑ ( )( )[ ] [ ]
m − n m + n m + k
n
, 47. = ∑ ( )( ){ }
m − n m + n m + k
,
n − m m + k n + k k
n − m m + k n + k k
{ }(
k
)
k
n l + m
= ∑ { }{ }( [ ]( )
k n − k n n l + m
, 49.
=
l + m l
l m k)
∑ [ ][ ]( k n − k n
.
l + m l
l m k)
k
k
Trees
Every tree with n
vertices has n − 1
edges.
Kraft inequality:
If the depths
of the leaves of
a binary tree are
d 1 , . . . , d n :
n∑
2 −d i
≤ 1,
i=1
and equality holds
only if every internal
node has 2
sons.
Master method:
T (n) = aT (n/b) + f(n), a ≥ 1, b > 1
If ∃ϵ > 0 such that f(n) = O(n log b a−ϵ )
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 ∃ϵ > 0 such that f(n) = Ω(n log b a+ϵ ),
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 2i · 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 2i
=
2i+1 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 i2i−1 .
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 “telescope”
Recurrences
1 ( T (n) − 3T (n/2) = n )
3 ( T (n/2) − 3T (n/4) = n/2 )
. . .
3 log n−1( 2
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
m−1
∑
m−1
n ∑ (
2 i 3i = n 3
) i
2 .
i=0
i=0
Let c = 3 2
. Then we have
m−1
∑
( c
n c i m )
− 1
= n
c − 1
i=0
= 2n(c log 2 n − 1)
= 2n(c (k−1) log c n − 1)
= 2n k − 2n,
and so T (n) = 3n k − 2n. Full history recurrences
can often be changed to limited
history ones (example): Consider
∑i−1
T i = 1 + T j , T 0 = 1.
Note that
j=0
T i+1 = 1 +
Subtracting we find
T i+1 − T i = 1 +
= T i .
i∑
T j .
j=0
i∑ ∑i−1
T j − 1 −
j=0
And so T i+1 = 2T i = 2 i+1 .
j=0
T j
Generating functions:
1. Multiply both sides of the equation
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 xi 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) is g i .
Example:
g i+1 = 2g i + 1, g 0 = 0.
Multiply and sum:
∑
i≥0
g i+1 x i = ∑ i≥0
2g i x i + ∑ i≥0
x i .
We choose G(x) = ∑ i≥0 xi g i . Rewrite
in terms of G(x):
G(x) − g 0
= 2G(x) + ∑ x i .
x
i≥0
Simplify:
G(x)
= 2G(x) + 1
x
1 − x .
Solve for G(x):
x
G(x) =
(1 − x)(1 − 2x) .
Expand this ( using partial fractions:
2
G(x) = x
1 − 2x − 1 )
1 − x
⎛
⎞
= x ⎝2 ∑ 2 i x i − ∑ x i ⎠
i≥0 i≥0
So g i = 2 i − 1.
= ∑ i≥0(2 i+1 − 1)x i+1 .

Theoretical Computer Science Cheat Sheet
π ≈ 3.14159, e ≈ 2.71828, γ ≈ 0.57721, ϕ = 1+√ 5
2
≈ 1.61803, ˆϕ =
1− √ 5
2
≈ −.61803
i 2 i p i General Probability
1 2 2 Bernoulli Numbers (B i = 0, odd i ≠ 1):
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
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 .
Change of base, quadratic formula:
log b x = log a x
log a b ,
Euler’s number e:
−b ± √ b 2 − 4ac
.
2a
e = 1 + 1 2 + 1 6 + 1 24 + 1
120
(
+ · · ·
lim 1 + x n
= e
n→∞ n) x .
( )
1 +
1 n (
n < e < 1 +
1 n+1
n)
.
( )
1 +
1 n e
n = e −
2n + 11e ( ) 1
24n 2 − O 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,
( 1
H n = ln n + γ + O .
n)
Factorial, Stirling’s approximation:
1, 2, 6, 24, 120, 720, 5040, 40320, 362880, . . .
n! = √ ( ) n ( ( n 1
2πn 1 + Θ .
e n))
Ackermann’s ⎧ function and inverse:
⎨ 2 j i = 1
a(i, j) = a(i − 1, 2) j = 1
⎩
a(i − 1, a(i, j − 1)) i, j ≥ 2
α(i) = min{j | a(j, j) ≥ i}.
28 268,435,456 107 Binomial distribution:
(
29 536,870,912 109
n
Pr[X = k] = p
k)
k q n−k , q = 1 − p,
30 1,073,741,824 113
n∑
(
31 2,147,483,648 127
n
E[X] = k p
k)
k q n−k = np.
32 4,294,967,296 131
k=1
Pascal’s Triangle
Poisson distribution:
Pr[X = k] = e−λ λ k
1
, E[X] = λ.
k!
1 1
Normal (Gaussian) distribution:
1 2 1
p(x) = √ 1 e −(x−µ)2 /2σ 2 , E[X] = µ.
1 3 3 1
2πσ
1 4 6 4 1
The “coupon collector”: We are given a
1 5 10 10 5 1
random coupon each day, and there are n
different types of coupons. The distribution
of coupons is uniform. The expected
1 6 15 20 15 6 1
1 7 21 35 35 21 7 1
number of days to pass before we to collect
all n types is
1 8 28 56 70 56 28 8 1
1 9 36 84 126 126 84 36 9 1
nH n .
1 10 45 120 210 252 210 120 45 10 1
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)] = ∑ g(x) Pr[X = x].
x
If X continuous then
E[g(X)] =
∫ ∞
−∞
g(x)p(x) dx =
Variance, standard deviation:
∫ ∞
−∞
VAR[X] = E[X 2 ] − E[X] 2 ,
g(x) dP (x).
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[B|A i ] Pr[A i ]
Pr[A i |B] = ∑ n
j=1 Pr[A j] Pr[B|A j ] .
Inclusion-exclusion:
[ ∨
n ] n∑
Pr X i = Pr[X i ] +
i=1
i=1
n∑
(−1) k+1
k=2
Moment inequalities:
∑
i i

Theoretical Computer Science Cheat Sheet
Trigonometry Matrices More Trig.
Multiplication:
(0,1)
C = A · B,
n∑
c i,j = a i,k b k,j .
b
(cos θ, sin θ)
k=1
C
A
θ
Determinants: det A ≠ 0 iff A is non-singular.
(-1,0) (1,0)
det A · B = det A · det B,
c a
(0,-1)
B
det A = ∑ n∏
sign(π)a i,π(i) .
Pythagorean theorem:
π i=1
C 2 = A 2 + B 2 .
2 × 2 and 3 × 3 determinant:
Definitions:
∣ a b
c d ∣ = ad − bc,
sin a = A/C, cos a = B/C,
csc a = C/A, sec a = C/B,
a b c
∣ ∣ ∣ ∣∣∣ tan a = sin a
cos a = A B , cos a
cot a =
sin a = B d e f
A . ∣ g h i ∣ = g b c
∣∣∣ e f ∣ − h a c
∣∣∣ d f ∣ + i a b
d e ∣
aei + bfg + cdh
Area, radius of inscribed circle:
=
− ceg − fha − ibd.
1
2 AB, AB
A + B + C .
Permanents:
Identities:
perm A = ∑ n∏
a i,π(i) .
sin x = 1
csc x , cos x = 1
π
sec x ,
i=1
Hyperbolic Functions
tan x = 1
cot x , sin2 x + cos 2 x = 1,
Definitions:
sinh x = ex − e −x
, cosh x = ex + e −x
,
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 ± tan y
tan(x ± y) =
1 ∓ tan x tan y ,
cot x cot y ∓ 1
cot(x ± y) =
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,
tan 2x =
cos 2x = 1 − tan2 x
1 + tan 2 x ,
2 tan x
1 − tan 2 x , cot 2x = cot2 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 sin x, e iπ = −1.
v2.02 c⃝1994 by Steve Seiden
sseiden@acm.org
http://www.csc.lsu.edu/~seiden
2
2
tanh x = ex − e −x
1
e x , csch x =
+ e−x 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 cosh2 x 2
= cosh x + 1.
θ sin θ cos θ tan θ
0 0 1 0
π
6
π
4
π
3
π
1
2
√
2
2
√
3
2
√ √
3 3
2 3
√
2
2
1
√
1
2 3
2
1 0 ∞
. . . in mathematics
you don’t understand
things, you
just get used to
them.
– J. von Neumann
b
C
h
a
A c
Law of cosines:
B
c 2 = a 2 +b 2 −2ab cos C.
Area:
A = 1 2 hc,
= 1 2ab sin C,
= c2 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:
√
1 − cos x
sin x 2 = ,
2
√
1 + cos x
cos x 2 = ,
2
√
1 − cos x
tan x 2 = 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 = eix − e −ix
,
2i
cos x = eix + e −ix
,
2
tan x = −i eix − e −ix
e ix + e −ix ,
= −i e2ix − 1
e 2ix + 1 ,
sinh ix
sin x = ,
i
cos x = cosh ix,
tanh ix
tan x = .
i

Number Theory
The Chinese remainder theorem: There exists
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 pe i
i is the prime factorization
of x then
n∏
ϕ(x) = p ei−1
i (p i − 1).
i=1
Euler’s theorem: If a and b are relatively
prime then
1 ≡ a ϕ(b) mod b.
Fermat’s theorem:
1 ≡ a p−1 mod p.
The Euclidean algorithm: if a > b are integers
then
gcd(a, b) = gcd(a mod b, b).
If ∏ n
then
i=1 pe i
i
is the prime factorization of x
S(x) = ∑ d|x
d =
n∏
i=1
p e i+1
i − 1
p i − 1 .
Perfect Numbers: x is an even perfect number
iff x = 2 n−1 (2 n −1) and 2 n −1 is prime.
Wilson’s theorem: n is a prime iff
(n − 1)! ≡ −1 mod n.
Möbius ⎧inversion:
1 if i = 1.
⎪⎨
0 if i is not square-free.
µ(i) =
⎪⎩ (−1) r if i is the product of
r distinct primes.
If
then
G(a) = ∑ d|a
F (a) = ∑ d|a
F (d),
( a
µ(d)G .
d)
Prime numbers:
ln ln n
p n = n ln n + n ln ln n − n + n
( )
ln n
n
+ O ,
ln n
π(n) =
n
ln n + n
(ln n) 2 + 2!n
(ln n)
( )
3
n
+ O
(ln n) 4 .
Theoretical Computer Science Cheat Sheet
Definitions:
Graph Theory
Loop An edge connecting a vertex
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 l v l .
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.
Tree A connected acyclic graph.
Free 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 removal
increases the number
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
which are adjacent.
Clique A set of vertices, all of
which are adjacent.
Ind. set A set of vertices, none of
which are adjacent.
Vertex cover A set of vertices which
cover all edges.
Planar graph A graph which can be embeded
in the plane.
Plane graph An embedding of a planar
graph.
∑
deg(v) = 2m.
v∈V
If G is planar then n − m + f = 2, so
f ≤ 2n − 4, m ≤ 3n − 6.
Any planar graph has a vertex with degree
≤ 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 n1 ,n 2
Complete bipartite graph
r(k, l) 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:
√
(x1 − x 0 ) 2 + (y 1 − y 0 ) 2 ,
[
|x1 − x 0 | p + |y 1 − y 0 | p] 1/p
,
[
lim |x1 − x 0 | p + |y 1 − y 0 | p] 1/p
.
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)
l 2
l 1
(x 2 , y 2 )
(x 1 , y 1 )
cos θ = (x 1, y 1 ) · (x 2 , y 2 )
.
l 1 l 2
Line through two points (x 0 , y 0 )
and (x 1 , y 1 ):
x y 1
x 0 y 0 1
∣ x 1 y 1 1 ∣ = 0.
Area of circle, volume of sphere:
A = πr 2 , V = 4 3 πr3 .
If I have seen farther than others,
it is because I have stood on the
shoulders of giants.
– Issac Newton

Wallis’ identity:
π = 2 · 2 · 2 · 4 · 4 · 6 · 6 · · ·
1 · 3 · 3 · 5 · 5 · 7 · · ·
π
Brouncker’s continued fraction expansion:
π
4 = 1 + 1 2
3
2 + 2
2+ 52
2+
2+···
72
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 (
1 − 1
3 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 − · · ·
Partial Fractions
Let N(x) and D(x) be polynomial functions
of x. We can break down
N(x)/D(x) using partial fraction expansion.
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 following
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:
m−1
N(x)
(x − a) m D(x) = ∑ A k (x)
(x − a) m−k +N′ D(x) ,
where
A k = 1 k!
k=0
[ ( )]
d
k N(x)
dx k .
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
Theoretical Computer Science Cheat Sheet
Derivatives:
1. d(cu)
dx
4. d(un )
dx
7. d(cu )
dx
9.
11.
13.
15.
17.
19.
21.
23.
25.
27.
29.
d(sin u)
dx
d(tan u)
dx
d(sec u)
dx
Calculus
= cdu
d(u + v)
, 2. = du
dx dx
du
= nun−1
dx ,
du
= (ln c)cu
dx
d(arcsin u)
dx
d(arctan u)
dx
d(arcsec u)
dx
d(sinh u)
dx
d(tanh u)
dx
d(sech u)
dx
d(arcsinh u)
dx
d(arctanh u)
dx
dx + dv
dx ,
d(u/v)
5. = v( du
dx
dx
d(uv)
3.
dx
= u dv
dx + v du
dx ,
) (
− u
dv
)
dx
v 2 , 6. d(ecu )
dx
, 8.
d(ln u)
dx
cu du
= ce
dx ,
= 1 du
u dx ,
= cos u du
d(cos u)
, 10. = − sin u du
dx dx
dx ,
= sec 2 u du
d(cot u)
, 12. = csc 2 u du
dx dx
dx ,
= tan u sec u du
d(csc u)
, 14. = − cot u csc u du
dx dx
dx ,
=
1 du
d(arccos u)
√ , 16. = −1 du
√
1 − u
2 dx dx 1 − u
2 dx ,
= 1 du
1 + u 2 dx
1
=
u √ du
1 − u 2 dx
, 18.
d(arccot u)
dx
, 20.
d(arccsc u)
dx
=
= −1
1 + u 2 du
dx ,
−1
u √ du
1 − u 2 dx ,
= cosh u du
d(cosh u)
, 22. = sinh u du
dx dx
dx ,
= sech 2 u du
d(coth u)
, 24. = − csch 2 u du
dx dx
dx ,
= − sech u tanh u du
d(csch u)
, 26. = − csch u coth u du
dx dx
dx ,
1 du
d(arccosh u) 1 du
= √ , 28. = √
1 + u
2 dx dx u2 − 1 dx ,
= 1 du
d(arccoth u)
1 − u 2 , 30. = 1 du
dx dx u 2 − 1 dx ,
d(arcsech u) −1
31. =
dx u √ du
d(arccsch u)
, 32. =
1 − u 2 dx dx
Integrals:
∫ ∫
∫
∫
1. cu dx = c u dx, 2. (u + v) dx =
3.
6.
8.
10.
12.
14.
−1
|u| √ du
1 + u 2 dx .
∫
u dx +
v dx,
∫
x n dx = 1
∫ ∫
1
n + 1 xn+1 , n ≠ −1, 4.
x dx = ln x, 5. e x dx = e x ,
∫
∫
dx
1 + x 2 = arctan x, 7. u dv ∫
dx dx = uv − v du
dx dx,
∫
∫
sin x dx = − cos x, 9. cos x dx = sin x,
∫
∫
∫
tan x dx = − ln | cos x|, 11.
sec x dx = ln | sec x + tan x|, 13.
arcsin x a dx = arcsin x a + √ a 2 − x 2 , a > 0,
∫
∫
cot x dx = ln | cos x|,
csc x dx = ln | csc x + cot x|,

15.
17.
19.
21.
23.
25.
26.
29.
33.
36.
∫
∫
∫
∫
∫
∫
∫
∫
∫
∫
Theoretical Computer Science Cheat Sheet
Calculus Cont.
arccos x a dx = arccos x a − √ a 2 − x 2 , a > 0, 16.
∫
arctan x a dx = x arctan x a − a 2 ln(a2 + x 2 ), a > 0,
∫
sin 2 ( )
(ax)dx = 1
2a ax − sin(ax) cos(ax) , 18. cos 2 )
(ax)dx =
2a( 1 ax + sin(ax) cos(ax) ,
∫
sec 2 x dx = tan x, 20. csc 2 x dx = − cot x,
sin n x dx = − sinn−1 x cos x
n
∫
−
tan n x dx = tann−1 x
n − 1
sec n x dx = tan x secn−1 x
n − 1
+ n − 1
n
∫
sin n−2 x dx, 22.
tan n−2 x dx, n ≠ 1, 24.
+ n − 2 ∫
n − 1
sec n−2 x dx, n ≠ 1,
∫
∫
cos n x dx = cosn−1 x sin x
n
cot n x dx = − cotn−1 x
n − 1
+ n − 1 ∫
cos n−2 x dx,
n
∫
−
cot n−2 x dx, n ≠ 1,
csc n x dx = − cot x cscn−1 x
+ n − 2 ∫
∫
∫
csc n−2 x dx, n ≠ 1, 27. sinh x dx = cosh x, 28. cosh x dx = sinh x,
n − 1 n − 1
∫
∫
∫
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
,
∫
∫
sinh 2 x dx = 1 4 sinh(2x) − 1 2 x,
34. cosh 2 x dx = 1 4 sinh(2x) + 1 2 x,
35. sech 2 x dx = tanh x,
arcsinh x a dx = x arcsinh x a − √ x 2 + a 2 , a > 0, 37.
∫
arctanh x a dx = x arctanh x a + a 2 ln |a2 − x 2 |,
38.
39.
40.
42.
43.
46.
48.
50.
52.
54.
56.
58.
60.
⎧
∫
⎨ x arccosh x
arccosh x a dx = a − √ x 2 + a 2 , if arccosh x a
> 0 and a > 0,
⎩
x arccosh x a + √ x 2 + a 2 , if arccosh x a
< 0 and a > 0,
∫
dx
(
√
a2 + x = ln x + √ )
a 2 + x 2 , a > 0,
2
∫
∫
∫
dx
√a2
a 2 + x 2 = 1 a arctan x a , a > 0, 41. √
− x 2 dx = x 2 a2 − x 2 + a2
2 arcsin x a , a > 0,
(a 2 − x 2 ) 3/2 dx = x 8 (5a2 − 2x 2 ) √ a 2 − x 2 + 3a4
8 arcsin x a , a > 0,
∫
∫
dx
√
a2 − x = arcsin x 2 a , a > 0, 44. dx
a 2 − x 2 = 1 ∣ ∣∣∣
2a ln a + x
a − x∣ ,
∫ √a2 √ ∣
± x 2 dx = x ∣∣x
2 a2 ± x 2 ± a2
2 ln √ ∣ ∫
∣∣
+ a2 ± x 2 , 47.
∫
dx
ax 2 + bx = 1 ∣ ∣∣∣
a ln x
a + bx∣ ,
∫
45. dx
(a 2 − x 2 ) = x
3/2 a 2√ a 2 − x , 2
dx
∣ ∣∣x
√
x2 − a = ln √ ∣ ∣∣ + x2 − a 2 , a > 0,
2
∫
49. x √ 2(3bx − 2a)(a + bx)3/2
a + bx dx =
15b 2 ,
∫ √
a + bx
dx = 2 √ ∫
∫
1
a + bx + a
x
x √ a + bx dx,
51. x
√ dx = 1
√ √ ∣ √ ln
a + bx − a∣∣∣
a + bx 2
∣√ √ , a > 0,
a + bx + a
∫ √
a2 − x 2
dx = √ a
x
2 − x 2 a + √ a
− a ln
2 − x 2
∫
∣ x ∣ ,
53. x √ a 2 − x 2 dx = − 1 3 (a2 − x 2 ) 3/2 ,
∫
x 2√ a 2 − x 2 dx = x 8 (2x2 − a 2 ) √ ∫
∣
a 2 − x 2 + a4
8 arcsin x a , a > 0, 55. dx
∣∣∣∣
√
a2 − x = − 1 2 a ln a + √ a 2 − x 2
x ∣ ,
∫
∫
x dx
√
a2 − x = −√ a 2 − x 2 x 2 dx
, 57. √ 2 a2 − x = − √ x
2 2 a2 − x 2 + a2
2 arcsin x a,
a > 0,
∫ √
a2 + x 2
dx = √ a
x
2 + x 2 a + √ a
− a ln
2 + x 2
∫ √ ∣ x ∣ , 59. x2 − a 2
dx = √ x
x
2 − a 2 − a arccos a
|x| , a > 0,
∫
x √ ∫
∣ ∣
x 2 ± a 2 dx = 1 3 (x2 ± a 2 ) 3/2 dx
∣∣∣
, 61.
x √ x 2 + a = 1 2 a ln x ∣∣∣
a + √ ,
a 2 + x 2

Theoretical Computer Science Cheat Sheet
Calculus Cont.
∫
∫
√
dx
62.
x √ x 2 − a = 1 2 a arccos a
|x| , a > 0, 63. dx
x 2√ x 2 ± a = ∓ x2 ± a 2
2 a 2 ,
x
∫
x dx
64. √
x2 ± a = √ ∫ √
x2
x 2 ± a 2 ± a
, 65.
2
2 x 4 dx = ∓ (x2 + a 2 ) 3/2
3a 2 x 3 ,
∣
1 ∣∣∣∣
∫
⎧⎪
dx ⎨ √
66.
ax 2 + bx + c = b2 − 4ac ln 2ax + b − √ b 2 − 4ac
2ax + b + √ b 2 − 4ac∣ , if b2 > 4ac,
⎪ ⎩
67.
68.
69.
∫
2
√
4ac − b
2 arctan
2ax + b √
4ac − b
2 ,
if b2 < 4ac,
1
∣
⎧⎪
dx ⎨ √a ln ∣2ax + b + 2 √ a √ ax 2 + bx + c∣ , if a > 0,
√
ax2 + bx + c = ⎪ 1 ⎩ √ arcsin √ −2ax − b , if a < 0,
−a b2 − 4ac
∫ √ax2
+ bx + c dx = 2ax + b √ ∫
4ax − b2
ax2 + bx + c +
4a
8a
∫
√
x dx
√
ax2 + bx + c = ax2 + bx + c
− b ∫
a 2a
dx
√
ax2 + bx + c ,
dx
√
ax2 + bx + c ,
−1
2
∫
⎧⎪ √ c √ ax
dx ⎨ √ ln
2 + bx + c + bx + 2c
, if c > 0,
c
70.
x √ ax 2 + bx + c = ∣
x
∣
⎪ 1
bx + 2c
⎩ √ arcsin −c |x| √ , if c < 0,
b 2 − 4ac
∫
71. x 3√ x 2 + a 2 dx = ( 1 3 x2 − 2
15 a2 )(x 2 + a 2 ) 3/2 ,
∫
∫
72. x n sin(ax) dx = − 1 a xn cos(ax) + n a
x n−1 cos(ax) dx,
∫
∫
73. x n cos(ax) dx = 1 a xn sin(ax) − n a
x n−1 sin(ax) dx,
∫
74. x n e ax dx = xn e ax ∫
− n
a
a
x n−1 e ax dx,
∫
( )
ln(ax)
75. x n ln(ax) dx = x n+1 n + 1 − 1
(n + 1) 2 ,
∫
76. x n (ln ax) m dx = xn+1
n + 1 (ln ax)m −
m ∫
x n (ln ax) m−1 dx.
n + 1
x 1 = x 1 = x 1
x 2 = x 2 + x 1 = x 2 − x 1
x 3 = x 3 + 3x 2 + x 1 = x 3 − 3x 2 + x 1
x 4 = x 4 + 6x 3 + 7x 2 + x 1 = x 4 − 6x 3 + 7x 2 − x 1
x 5 = x 5 + 15x 4 + 25x 3 + 10x 2 + x 1 = x 5 − 15x 4 + 25x 3 − 10x 2 + x 1
x 1 = x 1 x 1 = x 1
x 2 = x 2 + x 1 x 2 = x 2 − x 1
x 3 = x 3 + 3x 2 + 2x 1 x 3 = x 3 − 3x 2 + 2x 1
x 4 = x 4 + 6x 3 + 11x 2 + 6x 1 x 4 = x 4 − 6x 3 + 11x 2 − 6x 1
x 5 = x 5 + 10x 4 + 35x 3 + 50x 2 + 24x 1 x 5 = x 5 − 10x 4 + 35x 3 − 50x 2 + 24x 1
Finite Calculus
Difference, shift operators:
∆f(x) = f(x + 1) − f(x),
E f(x) = f(x + 1).
Fundamental Theorem:
f(x) = ∆F (x) ⇔ ∑ f(x)δx = F (x) + C.
b∑ ∑b−1
f(x)δx = f(i).
a
Differences:
∆(cu) = c∆u,
∆(uv) = u∆v + E v∆u,
∆(x n ) = nx n−1 ,
i=a
∆(u + v) = ∆u + ∆v,
∆(H x ) = x −1 , ∆(2 x ) = 2 x ,
∆(c x ) = (c − 1)c x , ∆ ( ) (
x
m = x
m−1)
.
Sums:
∑ ∑ cu δx = c u δx,
∑ (u + v) δx =
∑ u δx +
∑ v δx,
∑ u∆v δx = uv −
∑
E v∆u δx,
∑ x n δx = xn+1
m+1 , ∑ x −1 δx = H x ,
∑ c x δx = cx
c−1 , ∑ ( x
) (
m δx = x
m+1)
.
Falling Factorial Powers:
x n = x(x − 1) · · · (x − n + 1), n > 0,
x 0 = 1,
x n 1
=
(x + 1) · · · (x + |n|) , n < 0,
x n+m = x m (x − m) n .
Rising Factorial Powers:
x n = x(x + 1) · · · (x + n − 1), n > 0,
x 0 = 1,
x n 1
=
(x − 1) · · · (x − |n|) , n < 0,
x n+m = x m (x + m) n .
Conversion:
x n = (−1) n (−x) n = (x − n + 1) n
= 1/(x + 1) −n ,
x n = (−1) n (−x) n = (x + n − 1) n
= 1/(x − 1) −n ,
n∑
{ } n
n∑
{ } n
x n = x k = (−1) n−k x k ,
k k
x n =
x n =
k=1
n∑
k=1
n∑
k=1
k=1
[ n
k
]
(−1) n−k x k ,
[ n
k
]
x k .

Taylor’s series:
Expansions:
f(x) = f(a) + (x − a)f ′ (a) +
1
1 − x
1
1 − cx
Theoretical Computer Science Cheat Sheet
(x − a)2
f ′′ (a) + · · · =
2
Series
∞∑ (x − a) i
f (i) (a).
i!
i=0
= 1 + x + x 2 + x 3 + x 4 + · · · =
= 1 + cx + c 2 x 2 + c 3 x 3 + · · · =
1
1 − x n = 1 + x n + x 2n + x 3n + · · · =
∞∑
x i ,
i=0
∞∑
c i x i ,
i=0
∞∑
x ni ,
x
∞∑
(1 − x) 2 = x + 2x 2 + 3x 3 + 4x 4 + · · · = ix i ,
i=0
n∑
{ n k!z
k
∞∑
k}
(1 − z) k+1 = x + 2 n x 2 + 3 n x 3 + 4 n x 4 + · · · = i n x i ,
k=0
i=0
∞∑
e x = 1 + x + 1 2 x2 + 1 x i
6 x3 + · · · =
i! ,
i=0
∞∑
ln(1 + x) = x − 1 2 x2 + 1 3 x3 − 1 4 x4 i+1 xi
− · · · = (−1)
i ,
i=1
1
∞∑
ln
= x + 1
1 − x
2 x2 + 1 3 x3 + 1 x i
4 x4 + · · · =
i ,
i=1
∞∑
sin x = x − 1 3! x3 + 1 5! x5 − 1 7! x7 + · · · = (−1) i x 2i+1
(2i + 1)! ,
cos x = 1 − 1 2! x2 + 1 4! x4 − 1 6! x6 + · · · =
tan −1 x = x − 1 3 x3 + 1 5 x5 − 1 7 x7 + · · · =
(1 + x) n = 1 + nx + n(n−1)
2
x 2 + · · · =
1
(1 − x) n+1 = 1 + (n + 1)x + ( )
n+2
2 x 2 + · · · =
x
e x − 1
= 1 − 1 2 x + 1
12 x2 − 1
720 x4 + · · · =
1
2x (1 − √ 1 − 4x) = 1 + x + 2x 2 + 5x 3 + · · · =
1
√ 1 − 4x
= 1 + 2x + 6x 2 + 20x 3 + · · · =
( √
1 1 − 1 − 4x
√ 1 − 4x 2x
1
1 − x ln 1
1 − x
(
ln
1
2
1
1 − x
) n
= 1 + (2 + n)x + ( )
4+n
2 x 2 + · · · =
= x + 3 2 x2 + 11
6 x3 + 25
12 x4 + · · · =
) 2
= 1 2 x2 + 3 4 x3 + 11
24 x4 + · · · =
x
1 − x − x 2 = x + x 2 + 2x 3 + 3x 4 + · · · =
F n x
1 − (F n−1 + F n+1 )x − (−1) n x 2 = F n x + F 2n x 2 + F 3n x 3 + · · · =
i=0
i=0
∞∑
i=0
(−1) i x2i
(2i)! ,
∞∑
(−1) i x 2i+1
(2i + 1) ,
∞∑
( ) n
x i ,
i
i=0
i=0
∞∑
( i + n
i
i=0
∞∑ B i x i
,
i!
i=0
∞∑
(
1 2i
i + 1 i
i=0
∞∑
( ) 2i
x i ,
i
i=0
∞∑
( 2i + n
i
∞∑
H i x i ,
i=0
i=1
∞∑ H i−1 x i
,
i
∞∑
F i x i ,
i=2
i=0
∞∑
F ni x i .
i=0
)
x i ,
)
x i ,
)
x i ,
Ordinary power series:
∞∑
A(x) = a i x i .
i=0
Exponential power series:
∞∑ x i
A(x) = a i
i! .
i=0
Dirichlet power series:
∞∑ a i
A(x) =
i x .
i=1
Binomial theorem:
n∑
( n
(x + y) n = x
k)
n−k y k .
k=0
Difference of like powers:
n−1
∑
x n − y n = (x − y) x n−1−k y k .
k=0
For ordinary power series:
∞∑
αA(x) + βB(x) = (αa i + βb i )x i ,
x k A(x) =
i=0
∞∑
a i−k x i ,
i=k
A(x) − ∑ k−1
i=0 a ix i
x k =
∫
A(cx) =
A ′ (x) =
xA ′ (x) =
A(x) dx =
A(x) + A(−x)
2
A(x) − A(−x)
2
=
=
∞∑
a i+k x i ,
i=0
∞∑
c i a i x i ,
i=0
∞∑
(i + 1)a i+1 x i ,
i=0
∞∑
ia i x i ,
i=1
∞∑
i=1
a i−1
x i ,
i
∞∑
a 2i x 2i ,
i=0
∞∑
a 2i+1 x 2i+1 .
i=0
Summation: If b i = ∑ i
j=0 a i then
B(x) = 1
1 − x A(x).
Convolution:
⎛ ⎞
∞∑ i∑
A(x)B(x) = ⎝ a j b i−j
⎠ x i .
i=0
j=0
God made the natural numbers;
all the rest is the work of man.
– Leopold Kronecker

Theoretical Computer Science Cheat Sheet
Series
Expansions:
1
(1 − x) n+1 ln 1
∞∑
( ) ( ) −n n + i
1
= (H n+i − H n ) x i ,
=
1 − x
i
x
i=0
∞∑
[ ] n
x n = x i , (e x − 1) n =
i
∞∑
i=0
{ i
n}
x i ,
∞∑
{ i n!x
i
,
n}
i!
i=0
i=0
( ) n
1
∞∑
[ i n!x
i
∞∑ (−4)
ln
=
, x cot x =
1 − x
n] i B 2i x 2i
,
i!
(2i)!
i=0
i=0
∞∑
tan x = (−1) i−1 22i (2 2i − 1)B 2i x 2i−1
∞∑ 1
, ζ(x) =
(2i)!
i x ,
i=1
i=1
1
∞∑ µ(i)
=
ζ(x)
i x , ζ(x − 1)
∞∑
=
ζ(x)
i=1
ζ(x) = ∏ p
∞∑
ζ 2 (x) =
ζ(x)ζ(x − 1) =
i=1
∞∑
i=1
1
1 − p −x ,
d(i)
x i where d(n) = ∑ d|n 1,
S(i)
x i where S(n) = ∑ d|n d,
ζ(2n) = 22n−1 |B 2n |
π 2n , n ∈ N,
(2n)!
x
∞∑
= (−1) i−1 (4i − 2)B 2i x 2i
,
sin x
(2i)!
2x
i=0
( √ ) n
1 − 1 − 4x
∞∑
=
e x sin x =
√
1 − √ 1 − x
=
x
( ) 2 arcsin x
=
x
i=0
∞∑
i=1
∞∑
i=0
∞∑
i=0
n(2i + n − 1)!
x i ,
i!(n + i)!
2 i/2 sin iπ 4
i!
x i ,
(4i)!
16 i√ 2(2i)!(2i + 1)! xi ,
4 i i! 2
(i + 1)(2i + 1)! x2i .
Cramer’s Rule
If we have equations:
a 1,1 x 1 + a 1,2 x 2 + · · · + a 1,n x n = b 1
i=1
ϕ(i)
i x , Stieltjes Integration
Escher’s Knot
If G is continuous in the interval [a, b] and F is nondecreasing then
exists. If a ≤ b ≤ c then
∫ c
a
G(x) dF (x) =
∫ b
a
∫ b
a
G(x) dF (x)
G(x) dF (x) +
∫ c
If the integrals involved exist
∫ b
∫
( ) b
G(x) + H(x) dF (x) = G(x) dF (x) +
a
∫ b
a
∫ b
a
∫ b
G(x) d ( F (x) + H(x) ) =
a
c · G(x) dF (x) =
∫ b
a
a
∫ b
a
b
G(x) dF (x) +
G(x) d ( c · F (x) ) = c
G(x) dF (x) = G(b)F (b) − G(a)F (a) −
G(x) dF (x).
∫ b
a
∫ b
a
∫ b
a
∫ b
a
H(x) dF (x),
G(x) dH(x),
G(x) dF (x),
F (x) dG(x).
If the integrals involved exist, and F possesses a derivative F ′ at every
point in [a, b] then
∫ b
∫ b
G(x) dF (x) = G(x)F ′ (x) dx.
00 47 18 76 29 93 85 34 61 52
13, 21, 34, 55, 89, . . .
86 11 57 28 70 39 94 45 02 63
det A . Additive rule:
95 80 22 67 38 71 49 56 13 04
59 96 81 33 07 48 72 60 24 15
1, 1, 2, 3, 5, 8,
Definitions:
73 69 90 82 44 17 58 01 35 26 F i = F i−1 +F i−2 , F 0 = F 1 = 1,
.
68 74 09 91 83 55 27 12 46 30
F −i = (−1) i−1 F i ,
37 08 75 19 92 84 66 23 50 41
F
14 25 36 40 51 62 03 77 88 99
i = √ 1 5
(ϕ i − ˆϕ i) ,
21 32 43 54 65 06 10 89 97 78 Cassini’s identity: for i > 0:
42 53 64 05 16 20 31 98 79 87
F i+1 F i−1 − Fi 2 = (−1) i .
a 2,1 x 1 + a 2,2 x 2 + · · · + a 2,n x n = b 2
. .
a n,1 x 1 + a n,2 x 2 + · · · + a n,n x n = b n
Let A = (a i,j ) and B be the column matrix (b i ). Then
there is a unique solution iff det A ≠ 0. Let A i be A
with column i replaced by B. Then
x i = det A i
Improvement makes strait roads, but the crooked
roads without Improvement, are roads of Genius.
– William Blake (The Marriage of Heaven and Hell)
The Fibonacci number system:
Every integer n has a unique
representation
n = F k1 + F k2 + · · · + F km ,
where k i ≥ k i+1 + 2 for all i,
1 ≤ i < m and k m ≥ 2.
a
a
Fibonacci Numbers
F n+k = F k F n+1 + F k−1 F n ,
F 2n = F n F n+1 + F n−1 F n .
Calculation by matrices:
( ) ( ) n
Fn−2 F n−1 0 1
= .
F n−1 F n 1 1