# cheat engine

Document Sample

```					                                                        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
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.
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
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
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
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
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
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
Theoretical Computer Science Cheat Sheet
Calculus Cont.                                                                         Finite Calculus
√
dx       1                                       dx                                                x2 ± a2     Diﬀerence, shift operators:
62.     √                      a
= a arccos |x| , a > 0,      63.        √          =                                                ,
x x  2 − a2                                       x2 x2 ± a2                                             a2 x              ∆f (x) = f (x + 1) − f (x),
√
x dx                                         x2 ± a2           (x2 + a2 )3/2                                           E f (x) = f (x + 1).
64.    √          = x2 ± a2 ,                65.                dx =                   ,
x2 ± a2                                        x4                 3a2 x3                                         Fundamental Theorem:
√

       1        2ax + b − b2 − 4ac                                                                 f (x) = ∆F (x) ⇔             f (x)δx = F (x) + C.
√
             ln            √           , if b2 > 4ac,
dx            b2 − 4ac     2ax + b + b2 − 4ac
66.                 =                                                                                                                   b                   b−1
ax2 + bx + c                                                                                                                       f (x)δx =             f (i).
√ 2
             arctan √
2ax + b
,          if b2 < 4ac,                                                          a
4ac − b2           4ac − b2                                                                                                      i=a
Diﬀerences:

 √ ln 2ax + b + 2√a ax2 + bx + c , if a > 0,

1                                                                                              ∆(cu) = c∆u,             ∆(u + v) = ∆u + ∆v,
dx          a
67.    √               =                                                                                                   ∆(uv) = u∆v + E v∆u,
ax2 + bx + c  √1 arcsin √
               −2ax − b
                         ,                  if a < 0,                                            ∆(xn ) = nxn−1 ,
−a           b2 − 4ac
∆(Hx ) = x−1 ,                            ∆(2x ) = 2x ,
2ax + b                                   4ax − b2                       dx
68.         ax2 + bx + c dx =                            ax2 + bx + c +                           √             ,          ∆(cx ) = (c − 1)cx ,               ∆     x
=    x
.
4a                                         8a                      ax2 + bx + c                                                     m          m−1
Sums:
√
x dx                        ax2 + bx + c    b                      dx                                             cu δx = c         u δx,
69.    √              =                              −                    √             ,
ax2 + bx + c                       a          2a                  ax2 + bx + c
(u + v) δx =          u δx +          v δx,
       √ √ 2
 −1
 √ ln 2 c ax + bx + c + bx + 2c ,                                                                    u∆v δx = uv −             E v∆u δx,

 c                                                                               if c > 0,
dx                            x                                                                                                 n+1
70.     √             =                                                                                                       xn δx =   x
x−1 δx = Hx ,
2 + bx + c    1                                                                                                             m+1 ,
x ax                         bx
 √ arcsin √ + 2c ,
                                                                                 if c < 0,                    c    x
x         x
−c     |x| b2 − 4ac                                                                                cx δx = c−1 ,             m δx = m+1 .
Falling Factorial Powers:
71.    x3     x2 + a2 dx = ( 1 x2 −
3
2 2
15 a )(x
2
+ a2 )3/2 ,                                                   xn = x(x − 1) · · · (x − n + 1), n > 0,

1
x0 = 1,
72.    xn sin(ax) dx = − a xn cos(ax) +                      n
a       xn−1 cos(ax) dx,
1
xn =                            ,            n < 0,
(x + 1) · · · (x + |n|)
1 n
73.     n
x cos(ax) dx =           ax      sin(ax) −        n
x n−1
sin(ax) dx,
a                                                                 xn+m = xm (x − m)n .
xn eax                                                                                          Rising Factorial Powers:
74.    xn eax dx =                −     n
xn−1 eax dx,
a          a                                                                                     xn = x(x + 1) · · · (x + n − 1),                 n > 0,

75.    xn ln(ax) dx = xn+1
ln(ax)
−
1
,                                                   x0 = 1,
n+1     (n + 1)2                                                                                    1
xn =                            ,            n < 0,
xn+1
m                                                                                 (x − 1) · · · (x − |n|)
76.    xn (ln ax)m dx =                 (ln ax)m −                         xn (ln ax)m−1 dx.
n+1            n+1                                                                     xn+m = xm (x + m)n .
Conversion:
x1 =                          x1                                   =                                x1                     xn = (−1)n (−x)n = (x − n + 1)n
x2 =                       x2 + x1                                 =                             x2 − x1                      = 1/(x + 1)−n ,
x3 =              x3 + 3x2 + x1                                    =                    x3 − 3x2 + x1                      xn = (−1)n (−x)n = (x + n − 1)n
x4 =           x + 6x3 + 7x2 + x1
4
=                  4
x − 6x3 + 7x2 − x1                       = 1/(x − 1)−n ,
x5 =   x5 + 15x4 + 25x3 + 10x2 + x1                                =           x5 − 15x4 + 25x3 − 10x2 + x1                       n
n k
n
n
xn =            x =                  (−1)n−k xk ,
k                    k
x1 =                         x1                                    x1 =                            x1                             k=1                   k=1
n
x2 =                        2
x + x1                                  x2 =                          x − x1
2
xn =
n
(−1)n−k xk ,
3                     3        2           1                        3                       3        2        1                        k
x =             x + 3x + 2x                                        x =                x − 3x + 2x                                 k=1
x4 =           4
x + 6x3 + 11x2 + 6x1                                 x4 =              4
x − 6x3 + 11x2 − 6x1                           n
n k
5      5          4            3               2        1           5          5        4            3            2   1   xn =           x .
x =    x + 10x + 35x + 50x + 24x                                   x =         x − 10x + 35x − 50x + 24x                                k
k=1
Theoretical Computer Science Cheat Sheet
Series
Taylor’s series:                                                                                                         Ordinary power series:
∞
(x − a)2                     (x − a)i (i)                                                                 ∞
f (x) = f (a) + (x − a)f (a) +          f (a) + · · · =             f (a).                                                   A(x) =            ai xi .
2                     i=0
i!
i=0
Expansions:
∞                                               Exponential power series:
1
= 1 + x + x2 + x3 + x4 + · · ·     =      xi ,                                                               ∞
xi
1−x                                                        i=0                                                        A(x) =            ai      .
∞                                                      i!
1                                                                                                                          i=0
= 1 + cx + c2 x2 + c3 x3 + · · ·                        =         ci xi ,
1 − cx                                                                              i=0
Dirichlet power series:
∞                                           ∞
1                                                                                                                                       ai
=1+x +x n         2n
+x   3n
+ ···            =           ni
x ,                       A(x) =                   .
1 − xn                                                                                                                           i=1
ix
i=0
∞                    Binomial theorem:
x
= x + 2x2 + 3x3 + 4x4 + · · ·                           =         ixi ,                             n
n n−k k
(1 − x)2                                                                             i=0                      (x + y)n =                 x  y .
∞                                                k
dn        1                                                                                                                k=0
xk                             = x + 2n x2 + 3n x3 + 4n x4 + · · · =                             in xi ,         Diﬀerence of like powers:
dxn       1−x                                                                         i=0
∞                                              n−1
1 2          1 3                                    xi               xn − y n = (x − y)               xn−1−k y k .
e   x
=1+x+         2x      +    6x          + ···            =            ,
i=0
i!                                        k=0
∞
xi           For ordinary power series:
ln(1 + x)                 = x − 1 x2 + 1 x3 − 1 x4 − · · ·
2      3      4                                   =     (−1)i+1 ,                                      ∞
i
i=1
∞
αA(x) + βB(x) =                 (αai + βbi )xi ,
1                                                                                      xi
= x + 1 x2 + 1 x3 + 1 x4 + · · ·
i=0
ln                             2      3      4                                   =            ,                                      ∞
1−x                                                                               i=1
i
∞                            xk A(x) =                ai−k xi ,
1       1       1                           x2i+1
sin x                   = x − 3! x3 + 5! x5 − 7! x7 + · · · =     (−1)i           ,                                                i=k
(2i + 1)!                                     k−1                      ∞
i=0
∞
A(x) −           ai x i
x2i
i=0
=            ai+k xi ,
cos x                   =1−    1 2        1 4
−      1 6
+ ···            (−1)i                         xk
2! x   +   4! x             6! x                 =
(2i)!
,                                         i=0
∞
i=0
∞                                                      ci ai xi ,
−1                        1 3        1 5          1 7                                  x2i+1                   A(cx) =
tan          x           =x−    3x     +   5x       −   7x           + ···       =     (−1)          , i
i=0
i=0
(2i + 1)                                 ∞
∞
n(n−1) 2                                             n i                      A (x) =              (i + 1)ai+1 xi ,
(1 + x)n                 = 1 + nx +        2  x             + ···                =             x,
i=0
i                                      i=0
∞
∞
1                                                                                        i+n i                   xA (x) =               iai xi ,
= 1 + (n + 1)x +            n+2
2          x2 + · · · =                    x,
(1 − x)n+1                                                                            i=0
i                                     i=1
∞            i                                 ∞
x                                                                                    Bi x                                             ai−1 i
= 1 − 1x +
2
1 2
12 x        −    1
720 x
4
+ ··· =                   ,                  A(x) dx =                    x,
ex   −1                                                                             i=0
i!                                      i=1
i
∞
1     √                                                                                          1  2i i         A(x) + A(−x)
∞
(1 − 1 − 4x)                = 1 + x + 2x2 + 5x3 + · · ·                             =                x,                     =                  a2i x2i ,
2x                                                                                         i=0
i+1 i                  2                 i=0
∞
1                                                                                            2i i                                    ∞
√                                              2
= 1 + x + 2x + 6x + · · ·      3
=              x,          A(x) − A(−x)
1 − 4x                                                                                          i                           =                  a2i+1 x2i+1 .
√                                                                                  i=0                          2
n                                                                ∞                                             i=0
1      1 − 1 − 4x                                             4+n                                     2i + n i
√                                    = 1 + (2 + n)x +             2          x2 + · · · =                       x,     Summation: If bi = j=0 ai then
i
1 − 4x        2x                                                                                i=0
i
∞                                     1
1    1                              3 2        11 3             25 4                                                  B(x) =      A(x).
ln                         =x+    2x     +    6 x      +      12 x      + ··· =              Hi xi ,                        1−x
1−x 1−x                                                                                                         Convolution:
2
i=1
∞                                             
1        1                                                                                      Hi−1 xi                       ∞             i
ln                       = 1 x2 + 3 x3 +           11 4
24 x     + ···                =                 ,                             
2       1−x                     2      4                                                          i             A(x)B(x) =                       aj bi−j  xi .
i=2
∞                                  i=0         j=0
x                             2             3            4
= x + x + 2x + 3x + · · ·                               =         Fi x ,  i
1 − x − x2                                                                            i=0                   God made the natural numbers;
∞
Fn x                                            2                 3                                         all the rest is the work of man.
= Fn x + F2n x + F3n x + · · ·                          =         Fni x .     i
1 − (Fn−1 + Fn+1 )x − (−1)n x2                                                                     i=0
– Leopold Kronecker
Theoretical Computer Science Cheat Sheet
Series                                                                                                        Escher’s Knot
Expansions:
∞                                                      −n                       ∞
1         1                                               n+i i                   1                                   i
ln               =           (Hn+i − Hn )               x,                                       =                xi ,
(1 − x)n+1    1−x               i=0
i                      x                           i=0
n
∞                                                                                  ∞
n i                                                                            i n!xi
x   n
=                 x,                                     (e − 1)
x         n
=                     ,
i=0
i                                                                      i=0
n  i!
∞                                                                                  ∞
(−4)i B2i x2i
n
1                                  i n!xi
ln                     =                      ,                                  x cot x               =                          ,
1−x                     i=0
n i!                                                                   i=0
(2i)!
∞                  2i 2i                                                           ∞
i−1 2 (2        − 1)B2i x2i−1                                              1
tan x                =           (−1)                                 ,         ζ(x)                 =               ,
i=1
(2i)!                                               i=1
ix
∞                                                                                  ∞
1                        µ(i)                                                ζ(x − 1)                            φ(i)
=          ,                                                                     =                 ,
ζ(x)                  i=1
ix                                                   ζ(x)                        i=1
ix
1
ζ(x)                =            ,                                                                                           Stieltjes Integration
p
1 − p−x
∞
If G is continuous in the interval [a, b] and F is nondecreasing then
2                               d(i)                                                                                                       b
ζ (x)               =                     where d(n) =             d|n   1,
i=1
xi                                                                                                            G(x) dF (x)
a
∞
S(i)                                          exists. If a ≤ b ≤ c then
ζ(x)ζ(x − 1)            =                      where S(n) =            d|n   d,                             c                                   b                                c
i=1
xi
G(x) dF (x) =                       G(x) dF (x) +                    G(x) dF (x).
2n−1
2       |B2n | 2n                                                          a                                   a                                b
ζ(2n)               =                  π ,              n ∈ N,                 If the integrals involved exist
(2n)!
b                                                             b                                  b
∞
x                                       i−1 (4
i
− 2)B2i x2i                                  G(x) + H(x) dF (x) =                                      G(x) dF (x) +                      H(x) dF (x),
=            (−1)                          ,                    a                                                             a                                  a
sin x                      i=0
(2i)!
b                                                             b                                  b
√                 n         ∞
1 − 1 − 4x                            n(2i + n − 1)! i                                           G(x) d F (x) + H(x) =                                         G(x) dF (x) +                      G(x) dH(x),
=                          x,                                   a                                                             a                                  a
2x                        i=0
i!(n + i)!                                              b                                        b                                                 b
∞       i/2         iπ                                                c · G(x) dF (x) =                       G(x) d c · F (x) = c                                 G(x) dF (x),
x                                2      sin    4   i                                   a                                        a                                                    a
e sin x              =                              x,                                           b                                                                                    b
i!
i=1
G(x) dF (x) = G(b)F (b) − G(a)F (a) −                                                F (x) dG(x).
√                   ∞                                                                  a                                                                                    a
1−      1−x                              (4i)!
=               √                xi ,                      If the integrals involved exist, and F possesses a derivative F at every
x                   i=0
16i 2(2i)!(2i + 1)!
point in [a, b] then
2             ∞
arcsin x                                 4i i!2                                                                          b                                      b
=                             x2i .                                                                    G(x) dF (x) =                          G(x)F (x) dx.
x                       i=0
(i + 1)(2i + 1)!                                                                  a                                      a

Cramer’s Rule                                                                                                                                   Fibonacci Numbers
00 47 18 76 29 93 85 34 61 52
If we have equations:                                                                86 11 57 28 70 39 94 45 02 63
a1,1 x1 + a1,2 x2 + · · · + a1,n xn = b1                                                                                                               1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89, . . .
95 80 22 67 38 71 49 56 13 04
a2,1 x1 + a2,2 x2 + · · · + a2,n xn = b2                                    59 96 81 33 07 48 72 60 24 15
Deﬁnitions:
.          .                 .                                         73 69 90 82 44 17 58 01 35 26                                              Fi = Fi−1 +Fi−2 , F0 = F1 = 1,
.
.          .
.                 .
.                                         68 74 09 91 83 55 27 12 46 30                                                         F−i = (−1)i−1 Fi ,
an,1 x1 + an,2 x2 + · · · + an,n xn = bn                                    37 08 75 19 92 84 66 23 50 41                                                                  1           ˆ
14 25 36 40 51 62 03 77 88 99
Fi =     √
5
φi − φi ,
Let A = (ai,j ) and B be the column matrix (bi ). Then                               21 32 43 54 65 06 10 89 97 78                                              Cassini’s identity: for i > 0:
there is a unique solution iﬀ det A = 0. Let Ai be A
with column i replaced by B. Then
42 53 64 05 16 20 31 98 79 87
Fi+1 Fi−1 − Fi2 = (−1)i .
det Ai                                                                                                                               Additive rule:
xi =         .                                                  The Fibonacci number system:
det A                                                     Every integer n has a unique                                                   Fn+k = Fk Fn+1 + Fk−1 Fn ,
representation                                                                 F2n = Fn Fn+1 + Fn−1 Fn .
Improvement makes strait roads, but the crooked                                        n = Fk1 + Fk2 + · · · + Fkm ,                                            Calculation by matrices:
roads without Improvement, are roads of Genius.                                      where ki ≥ ki+1 + 2 for all i,                                                                                           n
Fn−2 Fn−1           0 1
– William Blake (The Marriage of Heaven and Hell)                                    1 ≤ i < m and km ≥ 2.                                                                      =                                 .
Fn−1     Fn         1 1

```
DOCUMENT INFO
Shared By:
Categories:
Tags:
Stats:
 views: 22 posted: 4/25/2012 language: pages: 10
How are you planning on using Docstoc?