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Discrete Math Cheat Sheet by

Equations commonly used in Discrete Math

Complex Numbers

j2 = -1
j3 = -j
j4 = 1
z = a + bj
z = r(sin θ + jsinθ)
z = re
tan-1 b/a = θ
cos-1 a/r = θ
sin-1 b/r = θ
(a + bj)* = a - bj
|z|=r=­sqrt(a2 + b2)
|z|x = |zx|
arg(z)x = x arg (z)
arg(z) = θ + 2kπ
(cos θ + jsin θ)k
= cos kθ + jsin kθ
= (e)k = ejkθ
< DeMoivre's Theorum
* means conjugate
j = i = sqrt(-1) = imaginary unit
Find roots example:
z2 = -4j
Convert to expone­ntial form first:
z2 = 4e-jπ/2

|z2| = r2 = sqrt(02 + 42) = 4
|z| = r = 2
k = (0, 1 ...n where n = expon' of z) = 0, 1
arg(z2) = 2 arg(z) = -π/2 + 2kπ
arg(z) = -π/4 + kπ

Substitute values of k (0, 1) for z = |z|ejarg(z) = 2e-jπ/4, 2ej3π/4
 

Discrete Probab­ility & Sets & Whatever

Probab­ility
1. P(x) = nCx . px . (1-p)n-x
2. P(x) = (XCk)((N-X)C(n-k))/NCn

Set Theory
A = B when A subset of B & B subset of A
A - B = A n B'
A u (A n B) = A
A n (A u B) = A
A u A' = U
A n A' = nullset or {}

Power set of S is the set of ALL SUBSETS of S e.g. S = {1,2} , P(S) = { {}, {1}, {2}, {1,2}}

|A| = n, |P(A)| = 2n

Sets A and B are disjoint iff A n B = {}

Cardin­ality of union: |A u B| = |A| + |B| - |A n B|

Proof by induction:
Show that when p(k) is true, p(k + 1) follows.
1. Binomial Distri­bution
n = trials, x = successes, p = probab­ility of success

2. Hyperg­eom­etric Distri­bution
N = deck size, n = draws, X = copies of card, k = successes
 

Matrix Manipu­lations

AT: Transpose of A - Switch Rows with Columns (R1 becomes C1, R2 becomes C2 etc.)

-A = -1 . A

A-1: Inverse of A
A-1 . I = I = A . I
A-1A=I

Augment Identity matrix to matrix and perform Guass-­Jordon elimin­ation on both to get change Identity matrix to the Inverse.

EROs:
Switch Rows
Scale Row (Multiply entire row)
Add multiple of different row to another

A matrix A is in row echelon form if
1. The nonzero rows in A lie above all zero rows (when there is at least a nonzero row and a zero row).
2. The first nonzero entry in a nonzero row (called a pivot) lies to the right of the pivot in the row immedi­ately above it.
                                   
 

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