Theoretical Computer Science Cheat Sheet
Definitions
iff ∃ positive c, n0 such that
0 ≤ f (n) ≤ cg(n) ∀n ≥ n0 .
f (n) = O(g(n))
f (n) = Ω(g(n))
iff ∃ positive c, n0 such that
f (n) ≥ cg(n) ≥ 0 ∀n ≥ n0 .
f (n) = Θ(g(n))
iff f (n) = O(g(n)) and
f (n) = Ω(g(n)).
f (n) = o(g(n))
iff limn→∞ f (n)/g(n) = 0.
iff ∀ǫ > 0, ∃n0 such that
|an − a| < ǫ, ∀n ≥ n0 .
lim an = a
n→∞
least b ∈ R such that b ≥ s,
∀s ∈ S.
sup S
greatest b ∈ R such that b ≤
s, ∀s ∈ S.
inf S
lim inf{ai | i ≥ n, i ∈ N}.
lim inf an
Series
n
X
i=
i=1
n(n + 1)
,
2
n
X
i2 =
i=1
n
X
n(n + 1)(2n + 1)
,
6
i3 =
i=1
n2 (n + 1)2
.
4
In general:
n
n
X
X
1
(i + 1)m+1 − im+1 − (m + 1)im
(n + 1)m+1 − 1 −
im =
m+1
i=1
i=1
n−1
m
X
1 X m+1
im =
Bk nm+1−k .
m
+
1
k
i=1
k=0
Geometric series:
n
X
cn+1 − 1
ci =
,
c−1
i=0
n
X
i=0
ici =
c 6= 1,
∞
X
ci =
i=0
ncn+2 − (n + 1)cn+1 + c
,
(c − 1)2
Harmonic series:
n
X
1
Hn =
,
i
i=1
n
X
1
,
1−c
c 6= 1,
∞
X
ci =
i=1
∞
X
ici =
i=0
c
,
1−c
c
,
(1 − c)2
|c| < 1,
|c| < 1.
n(n + 1)
n(n − 1)
Hn −
.
2
4
lim sup an
lim sup{ai | i ≥ n, i ∈ N}.
i=1
n→∞
n→∞
n n
X
X
1
i
n+1
n
Combinations:
Size
k
subHi =
Hn+1 −
Hi = (n + 1)Hn − n,
.
k
m
m+1
m+1
sets of a size n set.
i=1
i=1
n
n X
Stirling numbers (1st kind):
n
n
n
n!
n
k
n
,
2.
=2 ,
3.
=
,
1.
=
Arrangements of an n ele(n − k)!k!
k
k
n−k
k
k=0
ment set into k cycles.
n
n
n−1
n−1
n n−1
n
,
5.
4.
=
=
+
,
Stirling
numbers
(2nd
kind):
k k−1
k
k
k
k−1
k
n Partitions of an n element
X
n
m
n
n−k
r+k
r+n+1
6.
=
,
7.
=
,
set into k non-empty sets.
m
k
k
m−k
k
n
n
k=0
n n 1st order Eulerian numbers:
X
X
k
k
n+1
r
s
r+s
8.
=
,
9.
=
,
Permutations π1 π2 . . . πn on
m
m+1
k
n−k
n
k=0
k=0
{1, 2, . . . , n} with k ascents.
k−n−1
n
n
n
n ,
11.
=
= 1,
10.
= (−1)k
2nd order Eulerian numbers.
k
k
1
n
k
Cn
Catalan Numbers: Binary
n
n
n−1
n−1
n−1
12.
=
2
−
1,
13.
=
k
+
,
trees with n + 1 vertices.
2
k
k
k−1
n
n
n
n
n
14.
= (n − 1)!,
15.
= (n − 1)!Hn−1 ,
16.
= 1,
17.
≥
,
1
2
n
k
k
n X
2n
1
n
n−1
n−1
n
n
n
n
,
18.
= (n − 1)
+
,
19.
=
=
,
20.
= n!,
21. Cn =
n+1 n
k
k
k−1
n−1
n−1
2
k
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
n
n
n
0
n+1
1 if k = 0,
26.
= 2n − n − 1,
27.
25.
= 3n − (n + 1)2n +
=
,
1
2
k
2
0 otherwise
X
X
n m n X
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
X
n n
n
n−k
n
n
n−k−m
31.
=
(−1)
k!,
32.
= 1,
33.
= 0 for n 6= 0,
m
k
m
0
n
k=0
n X
n
n−1
(2n)n
n−1
n
,
34.
= (k + 1)
+ (2n − 1 − k)
,
35.
=
2n
k
k
k−1
k
k=0
X
X X
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
n→∞
n→∞
k=0
iHi =
k
k=0
Theoretical Computer Science Cheat Sheet
38.
40.
42.
44.
46.
48.
Identities Cont.
n
X
1 k
,
nn−k = n!
k! m
m
Trees
n X
x
n
x+k
n+1
39.
=
,
=
=
x−n
k
2n
m+1
k m
k=0
k=0
k
k=0
X X
n
n
k+1
n
n+1
k
n−k
=
(−1)
,
41.
=
(−1)m−k ,
m
k
m+1
m
k+1 m
k
k
X
X
m
m
n+k
n+k
m+n+1
m+n+1
k(n + k)
,
k
,
43.
=
=
k
k
m
m
k=0
k=0
X
X
n
n+1
k
n
n+1
k
=
(−1)m−k , 45. (n − m)!
=
(−1)m−k , for n ≥ m,
m
k+1
m
m
k+1
m
k k X X m − nm + n m + k n
m−n m+n m+k
n
=
,
47.
=
,
n−m
m+k
n+k
k
n−m
m+k
n+k
k
k X k
X n
ℓ+m
k
n−k
n
n
ℓ+m
k n−k n
=
,
49.
=
.
ℓ+m
ℓ
ℓ
m
k
ℓ+m
ℓ
ℓ
m
k
X n k n X
k
Every tree with n
vertices has n − 1
edges.
k
Kraft
inequality: If the depths
of the leaves of
a binary tree are
d1 , . . . , dn :
n
X
2−di ≤ 1,
i=1
and equality holds
only if every internal node has 2
sons.
k
Recurrences
Master method:
T (n) = aT (n/b) + f (n),
a ≥ 1, b > 1
If ∃ǫ > 0 such that f (n) = O(n
then
T (n) = Θ(nlogb a ).
log b a−ǫ
)
If f (n) = Θ(nlogb a ) then
T (n) = Θ(nlogb a log2 n).
If ∃ǫ > 0 such that f (n) = Ω(nlogb a+ǫ ),
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
i
Ti+1 = 22 · Ti2 , T1 = 2.
Note that Ti is always a power of two.
Let ti = log2 Ti . Then we have
ti+1 = 2i + 2ti , t1 = 1.
Let ui = ti /2i . Dividing both sides of
the previous equation by 2i+1 we get
ti+1
2i
ti
=
+ i.
i+1
i+1
2
2
2
Substituting we find
ui+1 = 21 + ui ,
u1 = 12 ,
which is simply ui = i/2. So we find
i−1
that Ti has the closed form Ti = 2i2 .
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”
1 T (n) − 3T (n/2) = n
3 T (n/2) − 3T (n/4) = n/2
..
..
..
.
.
.
log2 n−1
3
T (2) − 3T (1) = 2
Let m = log2 n. Summing the left side
we get T (n) − 3m T (1) = T (n) − 3m =
T (n) − nk where k = log2 3 ≈ 1.58496.
Summing the right side we get
m−1
m−1
X i
X n
i
3
3
=
n
.
2
i
2
i=0
i=0
Let c = 23 . Then we have
m
m−1
X
c −1
i
c =n
n
c−1
i=0
= 2n(c(k−1) logc n − 1)
= 2nk − 2n,
and so T (n) = 3n − 2n. Full history recurrences can often be changed to limited
history ones (example): Consider
i−1
X
Tj , T0 = 1.
Ti = 1 +
j=0
Ti+1 = 1 +
Tj .
j=0
Subtracting we find
i−1
i
X
X
Tj
Tj − 1 −
Ti+1 − Ti = 1 +
j=0
= Ti .
And so Ti+1 = 2Ti = 2i+1 .
i≥0
i≥0
We choose G(x) = i≥0 xi gi . Rewrite
in terms of G(x):
X
G(x) − g0
= 2G(x) +
xi .
x
P
i≥0
k
i
X
Multiply
X and sum:
X
X
gi+1 xi =
2gi xi +
xi .
i≥0
= 2n(clog2 n − 1)
Note that
Generating functions:
1. Multiply both sides of the equation by xi .
2. Sum both sides over all i for
which the equation is valid.
3. Choose a generatingPfunction
i
G(x). Usually G(x) = ∞
i=0 x gi .
3. Rewrite the equation in terms of
the generating function G(x).
4. Solve for G(x).
5. The coefficient of xi in G(x) is gi .
Example:
gi+1 = 2gi + 1, g0 = 0.
j=0
Simplify:
G(x)
1
= 2G(x) +
.
x
1−x
Solve for G(x):
G(x) =
x
.
(1 − x)(1 − 2x)
Expand this using partial fractions:
1
2
−
G(x) = x
1 − 2x 1 − x


X
X
= x 2
2i xi −
xi 
i≥0
=
X
(2
i≥0
So gi = 2i − 1.
i+1
i≥0
i+1
− 1)x
.
Theoretical Computer Science Cheat Sheet
π ≈ 3.14159,
e ≈ 2.71828,
γ ≈ 0.57721,
φ=
√
1+ 5
2
i
2i
pi
General
1
2
2
4
2
3
3
4
8
16
5
7
Bernoulli Numbers (Bi = 0, odd i 6= 1):
1
,
B0 = 1, B1 = − 21 , B2 = 61 , B4 = − 30
1
1
5
B6 = 42 , B8 = − 30 , B10 = 66 .
5
6
32
64
11
13
7
8
128
256
17
19
9
10
512
1,024
23
29
11
12
2,048
4,096
31
37
13
14
8,192
16,384
41
43
15
32,768
47
16
17
65,536
131,072
53
59
18
19
262,144
524,288
61
67
20
21
1,048,576
2,097,152
71
73
22
23
4,194,304
8,388,608
79
83
24
25
16,777,216
33,554,432
89
97
26
27
67,108,864
134,217,728
101
103
28
268,435,456
107
29
30
536,870,912
1,073,741,824
109
113
31
32
2,147,483,648
4,294,967,296
127
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
≈ 1.61803,
1
1
+ 61 + 24
+ 120
+ ···
n
x
lim 1 +
= ex .
n→∞
n
n
n+1
1 + n1 < e < 1 + n1
.
11e
1
e
n
1 + n1 = e −
.
−O
+
2
2n 24n
n3
Harmonic numbers:
1, 23 ,
1
2
11 25 137 49 363 761 7129
6 , 12 , 60 , 20 , 140 , 280 , 2520 , . . .
ln n < Hn < ln n + 1,
1
.
Hn = ln n + γ + O
n
Factorial, Stirling’s approximation:
1, 2, 6, 24, 120, 720, 5040, 40320, 362880,
√
1− 5
2
≈ −.61803
Probability
Change of base, quadratic formula:
√
−b ± b2 − 4ac
loga x
,
.
logb x =
loga b
2a
Euler’s number e:
e=1+
φ̂ =
...
n √
1
n
2πn
1+Θ
.
e
n
Ackermann’s
 function and inverse:
i=1
 2j
a(i, j) = a(i − 1, 2)
j=1

a(i − 1, a(i, j − 1)) i, j ≥ 2
n! =
α(i) = min{j | a(j, j) ≥ i}.
Binomial distribution:
n k n−k
Pr[X = k] =
p q
,
q = 1 − p,
k
n
X
n k n−k
[X]
=
k
p q
= np.
E
k
k=1
Poisson distribution:
e−λ λk
, E[X] = λ.
Pr[X = k] =
k!
Normal (Gaussian) distribution:
2
2
1
p(x) = √
e−(x−µ) /2σ , E[X] = µ.
2πσ
The “coupon collector”: We are given a
random coupon each day, and there are n
different types of coupons. The distribution of coupons is uniform. The expected
number of days to pass before we to collect all n types is
nHn .
Continuous distributions: If
Z b
Pr[a < X < b] =
p(x) dx,
a
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
Z a
p(x) dx.
P (a) =
−∞
Expectation: If X is discrete
X
g(x) Pr[X = x].
E[g(X)] =
x
If X continuous
Z ∞ then
Z
[g(X)]
=
g(x)p(x)
dx
=
E
−∞
∞
g(x) dP (x).
−∞
Variance, standard deviation:
VAR[X] = E[X 2 ] − E[X]2 ,
p
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[B]
For random variables X and Y :
E[X · Y ] = E[X] · E[Y ],
if X and Y are independent.
Pr[A|B] =
E[X + Y ] = E[X] + E[Y ],
E[cX] = c E[X].
Bayes’ theorem:
Pr[B|Ai ] Pr[Ai ]
Pr[Ai |B] = Pn
.
j=1 Pr[Aj ] Pr[B|Aj ]
Inclusion-exclusion:
n
n
i X
h_
Pr[Xi ] +
Xi =
Pr
i=1
i=1
n
X
k=2
(−1)k+1
X
ii <···<ik
Pr
k
h^
j=1
i
Xij .
Moment inequalities:
1
Pr |X| ≥ λ E[X] ≤ ,
λ
h
i
1
Pr X − E[X] ≥ λ · σ ≤ 2 .
λ
Geometric distribution:
Pr[X = k] = pq k−1 ,
q = 1 − p,
∞
X
1
kpq k−1 = .
E[X] =
p
k=1
Theoretical Computer Science Cheat Sheet
Trigonometry
Matrices
More Trig.
C
Multiplication:
(0,1)
b
C
A
θ
(-1,0)
c
(1,0)
a
cos a = B/C,
sec a = C/B,
cos a
B
cot a =
= .
sin a
A
circle:
AB
.
A+B+C
cos x =
1 + cot2 x = csc2 x,
sin x = sin(π − x),
tan x = cot π2 − x ,
cos x = − cos(π − x),
cot x = − cot(π − x),
1
,
sec x
sin2 x + cos2 x = 1,
1 + tan2 x = sec2 x,
sin x = cos π2 − x ,
csc x = cot x2 − 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
,
1 ∓ tan x tan y
cot x cot y ∓ 1
cot(x ± y) =
,
cot x ± cot y
cos 2x = cos2 x − sin2 x,
cos 2x = 1 − 2 sin2 x,
2 × 2 and 3 × 3 determinant:
a b
c d = ad − bc,
a b c d e f = g b c − ha c + ia
e f d f d
g h i aei + bf g + cdh
=
− ceg − f ha − ibd.
Permanents:
n
XY
ai,π(i) .
perm A =
b e
−x
e +e
,
2
1
csch x =
,
sinh x
1
coth x =
.
tanh x
cosh x =
Identities:
cosh2 x − sinh2 x = 1,
tanh2 x + sech2 x = 1,
coth2 x − csch2 x = 1,
sinh(−x) = − sinh x,
tanh(−x) = − tanh x,
cosh(x + y) = cosh x cosh y + sinh x sinh y,
2 tan x
,
1 + tan2 x
cos 2x = 2 cos2 x − 1,
sin 2x =
cos 2x =
1 − tan2 x
,
1 + tan2 x
cos(x + y) cos(x − y) = cos2 x − sin2 y.
e
iπ
= −1.
c
v2.02 1994
by Steve Seiden
sseiden@acm.org
http://www.csc.lsu.edu/~seiden
sinh 2x = 2 sinh x cosh x,
cosh 2x = cosh2 x + sinh2 x,
cosh x + sinh x = ex ,
cosh x − sinh x = e−x ,
(cosh x + sinh x)n = cosh nx + sinh nx,
2 sinh2 x2
= cosh x − 1,
θ
sin θ
cos θ
0
0
π
6
π
4
π
3
π
2
1
2
√
2
2
√
3
2
√
3
2
√
2
2
1
2
1
0
1
tan θ
0
√
3
3
1
√
3
∞
2 cosh2 x2
A
c
B
Law of cosines:
c2 = a2 +b2 −2ab cos C.
= 21 ab sin C,
π i=1
x
a
A = 21 hc,
Hyperbolic Functions
Definitions:
ex − e−x
sinh x =
,
2
x
−x
e −e
,
tanh x = x
e + e−x
1
sech x =
,
cosh x
h
Area:
sinh(x + y) = sinh x cosh y + cosh x sinh y,
2 tan x
cot2 x − 1
,
tan 2x =
cot 2x =
2 ,
2 cot x
1 − tan x
sin(x + y) sin(x − y) = sin2 x − sin2 y,
Euler’s equation:
eix = cos x + i sin x,
b
det A · B = det A · det B,
n
XY
sign(π)ai,π(i) .
det A =
cosh(−x) = cosh x,
tan(x ± y) =
sin 2x = 2 sin x cos x,
ai,k bk,j .
k=1
π i=1
Definitions:
sin a = A/C,
csc a = C/A,
sin a
A
tan a =
= ,
cos a
B
Area, radius of inscribed
Identities:
1
,
sin x =
csc x
1
tan x =
,
cot x
ci,j =
Determinants: det A 6= 0 iff A is non-singular.
(0,-1)
B
Pythagorean theorem:
C 2 = A2 + B 2 .
1
2 AB,
C = A · B,
(cos θ, sin θ)
n
X
n ∈ Z,
= cosh x + 1.
. . . in mathematics
you don’t understand things, you
just get used to
them.
– J. von Neumann
c2 sin A sin B
.
2 sin C
Heron’s formula:
=
√
A = s · sa · sb · sc ,
s = 12 (a + b + c),
sa = s − a,
sb = s − b,
sc = s − c.
More identities:
r
1 − cos x
x
,
sin 2 =
2
r
1 + cos x
cos x2 =
,
2
r
1 − cos x
tan x2 =
,
1 + cos x
1 − cos x
,
=
sin x
sin x
=
,
1 + cos x
r
1 + cos x
cot x2 =
,
1 − cos x
1 + cos x
=
,
sin x
sin x
,
=
1 − cos x
eix − e−ix
sin x =
,
2i
eix + e−ix
,
cos x =
2
eix − e−ix
,
tan x = −i ix
e + e−ix
e2ix − 1
= −i 2ix
,
e +1
sinh ix
,
sin x =
i
cos x = cosh ix,
tanh ix
tan x =
.
i
Theoretical Computer Science Cheat Sheet
Number Theory
The Chinese remainder theorem: There exists a number C such that:
C ≡ r1
.. ..
. .
mod m1
..
.
C ≡ rn mod mn
if mi and mj are relatively prime for i 6= j.
Euler’s function: φ(x) is the number of
positive integers
Qnless than x relatively
prime to x. If i=1 pei i is the prime factorization of x then
n
Y
piei −1 (pi − 1).
φ(x) =
i=1
Euler’s theorem: If a and b are relatively
prime then
1 ≡ aφ(b) mod b.
Fermat’s theorem:
1 ≡ ap−1 mod p.
The Euclidean algorithm: if a > b are integers then
gcd(a, b) = gcd(a mod b, b).
Qn
If i=1 pei i is the prime factorization of x
then
n
Y
X
piei +1 − 1
.
d=
S(x) =
pi − 1
i=1
d|x
Perfect Numbers: x is an even perfect number iff x = 2n−1 (2n −1) and 2n −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) =
r

 (−1) if i is the product of
r distinct primes.
If
G(a) =
X
F (d),
d|a
then
F (a) =
X
d|a
a
µ(d)G
.
d
Prime numbers:
ln ln n
pn = n ln n + n ln ln n − n + n
ln n
n
,
+O
ln n
n
2!n
n
π(n) =
+
+
ln n (ln n)2
(ln n)3
n
.
+O
(ln n)4
Graph Theory
Definitions:
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 v0 e1 v1 . . . eℓ vℓ .
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 6= ∅ 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.
X
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
χE (G) Edge chromatic number
Gc
Complement graph
Kn
Complete graph
Kn1 ,n2 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 6= 0.
Cartesian
Projective
(x, y)
(x, y, 1)
y = mx + b (m, −1, b)
x=c
(1, 0, −c)
Distance formula, Lp and L∞
metric:
p
(x1 − x0 )2 + (y1 − y0 )2 ,
1/p
,
|x1 − x0 |p + |y1 − y0 |p
1/p
lim |x1 − x0 |p + |y1 − y0 |p
.
p→∞
Area of triangle (x0 , y0 ), (x1 , y1 )
and (x2 , y2 ):
x1 − x0 y1 − y0 1
.
2 abs x − x
y2 − y0 2
0
Angle formed by three points:
(x2 , y2 )
ℓ2
θ
(x1 , y1 )
ℓ1
(0, 0)
(x1 , y1 ) · (x2 , y2 )
cos θ =
.
ℓ1 ℓ2
Line through two points (x0 , y0 )
and (x1 , y1 ):
x y 1
x0 y0 1 = 0.
x1 y1 1 Area of circle, volume of sphere:
A = πr2 ,
V = 34 πr3 .
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:
2 ·2 · 4 ·4 · 6 ·6···
π =2·
1 ·3 · 3 ·5 · 5 ·7···
Brouncker’s continued fraction expansion:
12
π
4 = 1+
32
2+
52
2+
2+
Gregrory’s series:
1
π
4 =1− 3 +
1
5
−
1
7
72
2+···
+
1
9
− ···
Derivatives:
1.
du
d(cu)
=c ,
dx
dx
4.
d(un )
du
= nun−1 ,
dx
dx
7.
d(cu )
du
= (ln c)cu ,
dx
dx
9.
d(sin u)
du
= cos u ,
dx
dx
Newton’s series:
π
6
1
1
1·3
= +
+
+ ···
3
2 2·3·2
2 · 4 · 5 · 25
Sharp’s series:
π
6
1 1
1
1
= √ 1− 1
+ 2
− 3
+···
3 ·3 3 ·5 3 ·7
3
Euler’s series:
π2
6
π2
8
π2
12
=
=
=
1
12
1
12
1
12
+
+
−
1
22
1
32
1
22
+
+
+
1
32
1
52
1
32
+
+
−
1
42
1
72
1
42
+
+
+
1
52
1
92
1
52
+ ···
+ ···
− ···
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)
N ′ (x)
= Q(x) +
,
D(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:
A
N ′ (x)
N (x)
=
+
,
(x − a)D(x)
x−a
D(x)
where
N (x)
A=
D(x)
.
k=0
where
d(u + v)
du dv
=
+
,
dx
dx dx
v du
d(u/v)
dx − u
5.
=
dx
v2
13.
d(sec u)
du
= tan u sec u ,
dx
dx
14.
du
d(arcsec u)
1
= √
,
2
dx
u 1 − u dx
d(sinh u)
du
21.
= cosh u ,
dx
dx
23.
d(tanh u)
du
= sech2 u ,
dx
dx
25.
d(sech u)
du
= − sech u tanh u ,
dx
dx
20.
26.
du
−1
d(arcsech u)
= √
,
dx
u 1 − u2 dx
Integrals:
Z
Z
1.
cu dx = c u dx,
Z
sin x dx = − cos x,
1 dk N (x)
.
Ak =
k! dxk D(x) x=a
10.
32.
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
12.
14.
n 6= −1,
2.
4.
dx
= arctan x,
1 + x2
Z
Z
Z
d(cos u)
du
= − sin u ,
dx
dx
du
d(arccsc u)
−1
= √
,
2
dx
u 1 − u dx
d(cosh u)
du
22.
= sinh u ,
dx
dx
d(coth u)
du
= − csch2 u ,
dx
dx
d(csch u)
du
= − csch u coth u ,
dx
dx
Z
Z
du
d(arccsch u)
−1
√
=
.
dx
|u| 1 + u2 dx
(u + v) dx =
11.
sec x dx = ln | sec x + tan x|,
p
a2 − x2 ,
Z
u dx +
Z
v dx,
Z
1
dx = ln x,
5.
ex dx = ex ,
x
Z
Z
dv
du
7.
u dx = uv − v dx,
dx
dx
Z
9.
cos x dx = sin x,
tan x dx = − ln | cos x|,
arcsin xa dx = arcsin xa +
d(cot u)
du
= csc2 u ,
dx
dx
28.
31.
8.
d(ln u)
1 du
=
,
dx
u dx
du
d(arccosh u)
1
= √
,
2
dx
u − 1 dx
d(arccoth u)
1 du
30.
= 2
,
dx
u − 1 dx
1
du
d(arcsinh u)
=√
,
2
dx
1 + u dx
d(arctanh u)
1 du
29.
=
,
dx
1 − u2 dx
6.
d(ecu )
du
= cecu ,
dx
dx
d(csc u)
du
= − cot u csc u ,
dx
dx
24.
27.
Z
6.
16.
19.
3.
,
d(arccos u)
−1 du
= √
,
dx
1 − u2 dx
d(arccot u)
−1 du
18.
=
,
dx
1 + u2 dx
d(arcsin u)
du
1
=√
,
2
dx
1 − u dx
1 du
d(arctan u)
=
,
17.
dx
1 + u2 dx
1
x dx =
xn+1 ,
n+1
12.
15.
n
dv
dx
10.
d(tan u)
du
= sec2 u ,
dx
dx
Z
d(uv)
dv
du
=u
+v ,
dx
dx
dx
3.
8.
11.
x=a
For a repeated factor:
m−1
X
N (x)
N ′ (x)
Ak
+
=
,
m
m−k
(x − a) D(x)
(x − a)
D(x)
2.
13.
a > 0,
Z
Z
cot x dx = ln | cos x|,
csc x dx = ln | csc x + cot x|,
Theoretical Computer Science Cheat Sheet
Calculus Cont.
15.
17.
19.
21.
23.
25.
26.
29.
33.
36.
Z
Z
Z
Z
Z
Z
Z
Z
Z
Z
arccos
x
a dx
= arccos
sin2 (ax)dx =
1
2a
x
a
p
− a2 − x2 ,
a > 0,
16.
Z
arctan xa dx = x arctan xa −
ax − sin(ax) cos(ax) ,
18.
Z
sec2 x dx = tan x,
cos2 (ax)dx =
1
2a
20.
a
2
ln(a2 + x2 ),
a > 0,
ax + sin(ax) cos(ax) ,
Z
csc2 x dx = − cot x,
Z
sech2 x dx = tanh x,
Z
Z
Z
sinn−1 x cos x n − 1
cosn−1 x sin x n − 1
+
+
sinn−2 x dx,
22.
cosn x dx =
cosn−2 x dx,
n
n
n
n
Z
Z
Z
tann−1 x
cotn−1 x
n
n−2
n
tan x dx =
− tan
x dx, n 6= 1,
24.
cot x dx = −
− cotn−2 x dx, n 6= 1,
n−1
n−1
Z
tan x secn−1 x n − 2
+
secn x dx =
secn−2 x dx, n 6= 1,
n−1
n−1
Z
Z
Z
cot x cscn−1 x n − 2
cscn x dx = −
+
cscn−2 x dx, n 6= 1,
27.
sinh x dx = cosh x,
28.
cosh x dx = sinh x,
n−1
n−1
Z
Z
Z
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 x2 ,
sinn x dx = −
2
sinh x dx =
arcsinh
x
a dx
1
4
sinh(2x) −
= x arcsinh
x
a
1
2 x,
p
− x2 + a2 ,
34.
Z
a > 0,
2
cosh x dx =
1
4
sinh(2x) +
37.
Z
1
2 x,
35.
arctanh xa dx = x arctanh xa +
a
2
ln |a2 − x2 |,

x p
 x arccosh − x2 + a2 , if arccosh xa > 0 and a > 0,
a
38.
arccosh xa dx =
p
 x arccosh x + x2 + a2 , if arccosh x < 0 and a > 0,
a
a
Z
p
dx
√
= ln x + a2 + x2 , a > 0,
39.
a2 + x2
Z
Z p
p
2
dx
1
x
= a arctan a , a > 0,
40.
41.
a2 − x2 dx = x2 a2 − x2 + a2 arcsin xa , a > 0,
2
2
a +x
Z
p
4
42.
(a2 − x2 )3/2 dx = x8 (5a2 − 2x2 ) a2 − x2 + 3a8 arcsin xa , a > 0,
Z
Z
Z
a + x
dx
dx
1
dx
x
x
,
√
= arcsin a , a > 0,
,
45.
43.
=
ln
44.
= √
2
2
2
2
3/2
2
2
2
a −x
2a
a−x
(a − x )
a −x
a a2 − x2
Z p
Z
p
p
p
2
dx
√
a2 ± x2 dx = x2 a2 ± x2 ± a2 ln x + a2 ± x2 ,
46.
= ln x + x2 − a2 , a > 0,
47.
x2 − a2
Z
Z
√
2(3bx − 2a)(a + bx)3/2
1 x dx
,
=
ln
,
49.
x a + bx dx =
48.
2
ax + bx
a
a + bx
15b2
√
√ Z
Z
Z √
√
a + bx − a a + bx
1
x
1
√
√
dx,
51.
dx = √ ln √
50.
dx = 2 a + bx + a
√ , a > 0,
x
x a + bx
a + bx
a + bx + a 2
Z p
Z √ 2
a + √a2 − x2 p
a − x2
2
2
53.
x a2 − x2 dx = − 13 (a2 − x2 )3/2 ,
dx = a − x − a ln 52.
,
x
x
Z
Z
a + √a2 − x2 p
p
4
dx
√
54.
x2 a2 − x2 dx = x8 (2x2 − a2 ) a2 − x2 + a8 arcsin xa , a > 0,
= − a1 ln 55.
,
2
2
x
a −x
Z
Z
2
p
p
2
x dx
x dx
x
√
√
= − a2 − x2 ,
= − x2 a2 − x2 + a2 arcsin a,
57.
a > 0,
56.
2
2
2
2
a −x
a
−
x
√
√
√
Z
Z
a + a2 + x2 p
p
a2 + x2
x2 − a2
a
dx = a2 + x2 − a ln dx = x2 − a2 − a arccos |x|
59.
58.
, a > 0,
,
x
x
x
Z
Z p
dx
x
,
√
√
61.
60.
x x2 ± a2 dx = 31 (x2 ± a2 )3/2 ,
= a1 ln x x2 + a2
a + a2 + x2 Z
Theoretical Computer Science Cheat Sheet
Calculus Cont.
62.
Z
64.
Z
66.
Z
67.
68.
Z
Z
√
dx
dx
x2 ± a2
1
a
√
√
= a arccos |x| , a > 0,
=∓
,
63.
a2 x
x x2 − a2
x2 x2 ± a2
√
Z
p
x2 ± a2
x dx
(x2 + a2 )3/2
√
65.
= x2 ± a2 ,
dx
=
∓
,
x4 3a2 x3
x2 ± a2

√
2ax + b − b2 − 4ac 
1


√
√
ln , if b2 > 4ac,

2
2
dx
b − 4ac
2ax + b + b − 4ac =
ax2 + bx + c 

2ax + b
2

√
arctan √
,
if b2 < 4ac,
4ac − b2
4ac − b2

√ p
1


√ ln 2ax + b + 2 a ax2 + bx + c , if a > 0,

a
dx
√
=
−2ax − b
1
ax2 + bx + c 

√
,
if a < 0,
arcsin √
−a
b2 − 4ac
Z
p
2ax + b p 2
4ax − b2
dx
√
ax2 + bx + c dx =
ax + bx + c +
,
2
4a
8a
ax + bx + c
Z
√
ax2
dx
b
+ bx + c
√
,
−
2
a
2a
ax + bx + c
√ √

−1 2 c ax2 + bx + c + bx + 2c 


Z
, if c > 0,
 √c ln x
dx
√
70.
=
2

x ax + bx + c 
bx + 2c
1

 √ arcsin √
,
if c < 0,
−c
|x| b2 − 4ac
Z
p
2 2
a )(x2 + a2 )3/2 ,
71.
x3 x2 + a2 dx = ( 13 x2 − 15
69.
72.
73.
74.
75.
76.
Z
Z
Z
Z
Z
Z
x dx
√
=
2
ax + bx + c
Z
xn sin(ax) dx = − a1 xn cos(ax) +
n
x cos(ax) dx =
xn eax dx =
1 n
ax
xn eax
−
a
xn ln(ax) dx = xn+1
xn (ln ax)m dx =
sin(ax) −
n
a
Z
n
a
n
a
Z
Z
xn−1 cos(ax) dx,
Difference, shift operators:
∆f (x) = f (x + 1) − f (x),
E f (x) = f (x + 1).
Fundamental Theorem:
X
f (x) = ∆F (x) ⇔
f (x)δx = F (x) + C.
b
X
f (x)δx =
a
Differences:
∆(cu) = c∆u,
b−1
X
f (i).
i=a
∆(u + v) = ∆u + ∆v,
∆(uv) = u∆v + E v∆u,
∆(xn ) = nxn−1 ,
∆(Hx ) = x−1 ,
∆(cx ) = (c − 1)cx ,
∆(2x ) = 2x ,
x
x
∆ m
= m−1
.
Sums:
P
P
cu δx = c u δx,
P
P
P
(u + v) δx = u δx + v δx,
P
P
u∆v δx = uv − E v∆u δx,
P −1
n+1
P n
x δx = Hx ,
x δx = xm+1 ,
P
P x
x
x
x
c
,
c δx = c−1
m δx = m+1 .
Falling Factorial Powers:
xn = x(x − 1) · · · (x − n + 1),
n−1
x
x = 1,
1
,
(x + 1) · · · (x + |n|)
xn+m = xm (x − m)n .
Rising Factorial Powers:
sin(ax) dx,
1
ln(ax)
−
n+1
(n + 1)2
n+1
x
m
(ln ax)m −
n+1
n+1
Z
x1 =
x2 =
x1
x2 + x1
=
=
x3 =
x4 =
x3 + 3x2 + x1
4
x + 6x3 + 7x2 + x1
=
=
x5 =
x5 + 15x4 + 25x3 + 10x2 + x1
=
x1 =
x2 =
x1
x + x1
x1 =
x2 =
x3 =
x4 =
x3 + 3x2 + 2x1
x4 + 6x3 + 11x2 + 6x1
x3 =
x4 =
x5 =
x5 + 10x4 + 35x3 + 50x2 + 24x1
x5 =
n > 0,
0
xn =
xn−1 eax dx,
2
Finite Calculus
n < 0,
xn = x(x + 1) · · · (x + n − 1),
n > 0,
x0 = 1,
,
xn =
xn (ln ax)m−1 dx.
x1
x2 − x1
x3 − 3x2 + x1
4
x − 6x3 + 7x2 − x1
x5 − 15x4 + 25x3 − 10x2 + x1
x1
x − x1
2
x3 − 3x2 + 2x1
x4 − 6x3 + 11x2 − 6x1
x5 − 10x4 + 35x3 − 50x2 + 24x1
1
,
(x − 1) · · · (x − |n|)
n < 0,
xn+m = xm (x + m)n .
Conversion:
xn = (−1)n (−x)n = (x − n + 1)n
= 1/(x + 1)−n ,
xn = (−1)n (−x)n = (x + n − 1)n
= 1/(x − 1)−n ,
n n X
n k X n
xn =
x =
(−1)n−k xk ,
k
k
k=1
k=1
n X
n
(−1)n−k xk ,
xn =
k
k=1
n X
n k
n
x .
x =
k
k=1
Theoretical Computer Science Cheat Sheet
Series
Taylor’s series:
∞
X (x − a)i
(x − a)2 ′′
f (a) + · · · =
f (i) (a).
f (x) = f (a) + (x − a)f (a) +
2
i!
i=0
Expansions:
∞
X
1
xi ,
= 1 + x + x2 + x3 + x4 + · · ·
=
1−x
i=0
∞
X
1
ci xi ,
= 1 + cx + c2 x2 + c3 x3 + · · ·
=
1 − cx
i=0
∞
X
1
n
2n
3n
xni ,
=
1
+
x
+
x
+
x
+
·
·
·
=
1 − xn
i=0
∞
X
x
2
3
4
ixi ,
= x + 2x + 3x + 4x + · · ·
=
(1 − x)2
i=0
n ∞
X
X
n
k!z k
n 2
n 3
n 4
in xi ,
=
x
+
2
x
+
3
x
+
4
x
+
·
·
·
=
k (1 − z)k+1
i=0
k=0
∞
X
xi
=
ex
= 1 + x + 21 x2 + 16 x3 + · · ·
,
i!
i=0
∞
X
xi
(−1)i+1 ,
=
ln(1 + x)
= x − 12 x2 + 31 x3 − 41 x4 − · · ·
i
i=1
∞
i
Xx
1
ln
= x + 21 x2 + 31 x3 + 41 x4 + · · ·
,
=
1−x
i
i=1
∞
X
x2i+1
1 3
1 5
1 7
sin x
= x − 3!
,
(−1)i
x + 5!
x − 7!
x + ··· =
(2i + 1)!
i=0
∞
X
x2i
1 4
1 6
1 2
(−1)i
x + 4!
x − 6!
x + ··· =
cos x
= 1 − 2!
,
(2i)!
i=0
∞
X
x2i+1
(−1)i
,
=
tan−1 x
= x − 31 x3 + 51 x5 − 71 x7 + · · ·
(2i + 1)
i=0
∞ X
n i
2
(1 + x)n
= 1 + nx + n(n−1)
x,
x
+
·
·
·
=
2
i
i=0
∞ X
1
i+n i
n+2 2
x,
= 1 + (n + 1)x + 2 x + · · · =
i
(1 − x)n+1
i=0
∞
X
Bi xi
x
1
1 2
1
4
=
1
−
,
x
+
x
−
x
+
·
·
·
=
2
12
720
x
e −1
i!
i=0
∞
X
√
1
1
2i i
2
3
=
x,
(1 − 1 − 4x)
= 1 + x + 2x + 5x + · · ·
2x
i+1 i
i=0
∞ X
1
2i i
√
x,
= 1 + 2x + 6x2 + 20x3 + · · ·
=
i
1 − 4x
i=0
√
n
∞ X
1 − 1 − 4x
2i + n i
1
4+n 2
√
x,
= 1 + (2 + n)x + 2 x + · · · =
2x
i
1 − 4x
i=0
∞
X
1
1
3 2
11 3
25 4
ln
= x + 2 x + 6 x + 12 x + · · · =
Hi xi ,
1−x 1−x
i=1
2
∞
X
1
Hi−1 xi
1
11 4
ln
= 21 x2 + 43 x3 + 24
,
x + ···
=
2
1−x
i
i=2
∞
X
x
2
3
4
Fi xi ,
=
x
+
x
+
2x
+
3x
+
·
·
·
=
1 − x − x2
i=0
∞
X
Fn x
2
3
Fni xi .
=
F
x
+
F
x
+
F
x
+
·
·
·
=
n
2n
3n
1 − (Fn−1 + Fn+1 )x − (−1)n x2
i=0
′
Ordinary power series:
∞
X
ai xi .
A(x) =
i=0
Exponential power series:
∞
X
xi
ai .
A(x) =
i!
i=0
Dirichlet power series:
∞
X
ai
A(x) =
.
ix
i=1
Binomial theorem:
n X
n n−k k
(x + y)n =
x
y .
k
k=0
Difference of like powers:
n−1
X
xn − y n = (x − y)
xn−1−k y k .
k=0
For ordinary power series:
∞
X
(αai + βbi )xi ,
αA(x) + βB(x) =
i=0
xk A(x) =
∞
X
ai−k xi ,
i=k
A(x) −
Pk−1
i=0
i
ai x
xk
A(cx) =
=
∞
X
ai+k xi ,
i=0
∞
X
ci ai xi ,
i=0
A′ (x) =
∞
X
(i + 1)ai+1 xi ,
i=0
xA′ (x) =
∞
X
iai xi ,
i=1
Z
A(x) dx =
∞
X
ai−1
i=1
A(x) + A(−x)
=
2
A(x) − A(−x)
=
2
∞
X
i
xi ,
a2i x2i ,
i=0
∞
X
a2i+1 x2i+1 .
i=0
Pi
Summation: If bi = j=0 ai then
1
B(x) =
A(x).
1−x
Convolution:


∞
i
X
X

aj bi−j  xi .
A(x)B(x) =
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
ln
n+1
(1 − x)
1−x
xn
ln
1
1−x
n
tan x
1
ζ(x)
ζ(x)
ζ 2 (x)
Escher’s Knot
−n
∞ X
i
n+i i
1
xi ,
=
(Hn+i − Hn )
x,
=
n
i
x
i=0
i=0
∞ ∞ X
X
n i
i n!xi
x
n
,
=
x,
(e − 1)
=
i!
i
n
i=0
i=0
∞ ∞
X
X
i n!xi
(−4)i B2i x2i
=
,
x cot x
=
,
n i!
(2i)!
i=0
i=0
∞
∞
2i 2i
2i−1
X
X
1
i−1 2 (2 − 1)B2i x
(−1)
=
,
,
ζ(x)
=
x
(2i)!
i
i=1
i=1
∞
∞
X
X
µ(i)
φ(i)
ζ(x − 1)
,
,
=
=
x
i
ζ(x)
ix
i=1
i=1
Y
1
=
,
Stieltjes Integration
1 − p−x
p
If G is continuous in the interval [a, b] and F is nondecreasing then
∞
X
Z b
P
d(i)
=
where d(n) = d|n 1,
G(x) dF (x)
xi
∞
X
i=1
ζ(x)ζ(x − 1)
ζ(2n)
x
sin x
√
n
1 − 1 − 4x
2x
x
e sin x
=
∞
X
S(i)
xi
√
1−x
x
2
arcsin x
x
1−
where S(n) =
i=1
2n−1
d|n d,
|B2n | 2n
=
π , n ∈ N,
(2n)!
∞
X
(4i − 2)B2i x2i
(−1)i−1
=
,
(2i)!
i=0
2
=
=
∞
X
n(2i + n − 1)!
i=0
∞
X
i=1
s
a
P
=
∞
X
i=0
=
∞
X
i=0
i!(n + i)!
2
i/2
sin
i!
iπ
4
xi ,
i
x,
(4i)!
√
xi ,
i
16 2(2i)!(2i + 1)!
4i i!2
x2i .
(i + 1)(2i + 1)!
Cramer’s Rule
If we have equations:
a1,1 x1 + a1,2 x2 + · · · + a1,n xn = b1
a2,1 x1 + a2,2 x2 + · · · + a2,n xn = b2
..
..
..
.
.
.
an,1 x1 + an,2 x2 + · · · + an,n xn = bn
Let A = (ai,j ) and B be the column matrix (bi ). Then
there is a unique solution iff det A 6= 0. Let Ai be A
with column i replaced by B. Then
det Ai
xi =
.
det A
Improvement makes strait roads, but the crooked
roads without Improvement, are roads of Genius.
– William Blake (The Marriage of Heaven and Hell)
exists. If a ≤ b ≤ c then
Z c
Z
G(x) dF (x) =
a
b
G(x) dF (x) +
a
G(x) dF (x).
b
Z
G(x) dF (x) +
a
a
b
G(x) d F (x) + H(x) =
a
Z
c
b
If the integrals involved exist
Z
Z b
G(x) + H(x) dF (x) =
Z
Z
b
c · G(x) dF (x) =
a
Z
a
b
Z
a
b
Z
b
H(x) dF (x),
a
b
Z
G(x) dF (x) +
a
G(x) d c · F (x) = c
G(x) dF (x) = G(b)F (b) − G(a)F (a) −
b
G(x) dH(x),
a
Z b
Z
G(x) dF (x),
a
b
F (x) dG(x).
a
If the integrals involved exist, and F possesses a derivative F ′ at every
point in [a, b] then
Z b
Z b
G(x) dF (x) =
G(x)F ′ (x) dx.
a
00 47 18 76 29 93 85 34 61 52
86 11 57 28 70 39 94 45 02 63
95 80 22 67 38 71 49 56 13 04
59 96 81 33 07 48 72 60 24 15
73 69 90 82 44 17 58 01 35 26
68 74 09 91 83 55 27 12 46 30
37 08 75 19 92 84 66 23 50 41
14 25 36 40 51 62 03 77 88 99
21 32 43 54 65 06 10 89 97 78
42 53 64 05 16 20 31 98 79 87
a
Fibonacci Numbers
1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . .
Definitions:
Fi = Fi−1 +Fi−2 ,
F0
i−1
= F1 = 1,
F−i = (−1) Fi ,
Fi = √15 φi − φ̂i ,
Cassini’s identity: for i > 0:
Fi+1 Fi−1 − Fi2 = (−1)i .
The Fibonacci number system:
Every integer n has a unique
representation
Additive rule:
Fn+k = Fk Fn+1 + Fk−1 Fn ,
n = Fk1 + Fk2 + · · · + Fkm ,
where ki ≥ ki+1 + 2 for all i,
1 ≤ i < m and km ≥ 2.
Calculation by matrices:
n
Fn−2 Fn−1
0 1
=
.
1 1
Fn−1
Fn
F2n = Fn Fn+1 + Fn−1 Fn .