Formulas - MATH 1
Functions:
Polar coordinates (r, 𝜃):
Cartesian coordinates:
𝑟 = √𝑥 2 + 𝑦 2
𝑥 = 𝑟 𝑐𝑜𝑠𝜃
𝑦
𝑡𝑎𝑛𝜃 = 𝑥
Rational functions
𝑓(𝑥) =
𝑦 = 𝑟 𝑠𝑖𝑛𝜃
𝑝(𝑥)
𝑞(𝑥)
where 𝑝 and 𝑞 are polynomials
Strictly proper rational function: 𝑑𝑒𝑔𝑟𝑒𝑒(𝑝) < 𝑑𝑒𝑔𝑟𝑒𝑒(𝑞)
Improper rational function: 𝑑𝑒𝑔𝑟𝑒𝑒(𝑝) > 𝑑𝑒𝑔𝑟𝑒𝑒(𝑞)
Hyperbolic functions
cosh(𝑥) = 12(𝑒 𝑥 + 𝑒 −𝑥 )
1
1
sinh(𝑥) = 2(𝑒 𝑥 − 𝑒 −𝑥 )
tanh(𝑥) =
1
sinh(𝑥)
cosh(𝑥)
𝑥(𝑡)
𝑦(𝑡)
Parametric representation
Curve given by {
Implicit function
Relation between x and y on the form 𝑓(𝑥, 𝑦) = 0
,𝑡 ∈ ℝ
Complex Numbers:
Cartesian form of 𝑧 ∈ ℂ
𝑧 = 𝑥 + 𝑦𝑗
Imaginary unit, 𝑗
𝑗 2 = −1
Real part
Imaginary part
𝑥 = 𝑅𝑒(𝑧)
𝑦 = 𝐼𝑚(𝑧)
Modulus
|𝑧| = √𝑥 2 + 𝑦 2
Argument
arg(𝑧) = 𝜃,
Polar form
𝑧 = 𝑟 ∙ (𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃 ∙ 𝑗)
Multiplication in polar form
𝑧1 · 𝑧2 = 𝑟1 𝑟2 (cos(𝜃1 + 𝜃2 ) + sin(𝜃1 + 𝜃2 ) · 𝑗
Division in polar form
𝑧1 𝑟1
= (cos(𝜃1 − 𝜃2 ) + sin(𝜃1 − 𝜃2 ) · 𝑗
𝑧2 𝑟2
𝑥, 𝑦 ∈ ℝ
−𝜋 ≤ arg(𝑧) ≤ 𝜋
or: 𝑧 = 𝑟∠𝜃
where 𝑟 = |𝑧|
Euler’s Formula
𝑒 𝑗𝜃 = 𝑐𝑜𝑠𝜃 + 𝑠𝑖𝑛𝜃 · 𝑗
Exponential form
𝑧 = 𝑟 · 𝑒 𝑗𝜃
sin/cos/hyperbolic functions
cosh(𝑗𝑥) = cos(𝑥)
sinh(𝑗𝑥) = 𝑗 sin (𝑥)
tanh(𝑗𝑥) = 𝑗 tan(𝑥)
cos(𝑗𝑥) = cosh(𝑥)
or
𝑧 = 𝑟∠𝜃
𝑒 𝑗𝜃 + 𝑒 −𝑗𝜃
2
𝑐𝑜𝑠𝜃 =
𝑠𝑖𝑛𝜃 =
sin(𝑗𝑥) = 𝑗 sinh (𝑥)
𝑒 𝑗𝜃 − 𝑒 −𝑗𝜃
2
Logarithm of complex number
ln(𝑧) = ln(|𝑧|) + arg(𝑧) · 𝑗
Power of a complex number
𝑧 𝑛 = 𝑟 𝑛 · [cos(𝑛𝜃) + sin(𝑛𝜃) · 𝑗]
Root of a complex number
1
1
𝜃 2𝜋𝑘
𝜃 2𝜋𝑘
𝑧 𝑛 = 𝑟 𝑛 · [cos ( +
) + sin ( +
) · 𝑗] , 𝑘 = 0, 1, 2, … , 𝑛 − 1
𝑛
𝑛
𝑛
𝑛
1
1
(De Moivre’s Theorem)
𝜃 2𝜋𝑘
( +
)·𝑗
𝑛
𝑧𝑛 = 𝑟𝑛 · 𝑒 𝑛
Matrix algebra:
Multiplication
Determinant for (2x2)-matrix
Matrix A of order (m,p) and matrix B of order (p,n)
𝑝
AB=C
where 𝑐𝑖𝑗 = ∑𝑘=1 𝑎𝑖𝑘 𝑏𝑘𝑗
𝑎11
𝑨 = [𝑎
21
Co-factors
𝑎12
𝑎22 ] ⇒
for {
𝑎11
det(𝑨) = |𝑎
21
𝑖 = 1, … , 𝑚
𝑗 = 1, … , 𝑛
𝑎12
𝑎22 | = 𝑎11 𝑎22 − 𝑎21 𝑎12
+ − +⋯
Sign pattern: [− + ⋯ ]
∶
The minor (the value): Determinant of remaining matrix when
removing row i and column j
Co-factor element 𝐴𝑖𝑗 ∶ Combine sign pattern and the minor
Determinant for (n x n)-matrix
det(𝑨) = ∑𝑗 𝑎𝑖𝑗 𝐴𝑖𝑗 = ∑𝑖 𝑎𝑖𝑗 𝐴𝑖𝑗
Transpose matrix 𝑨𝑇
Rows of matrix A are turned into columns (or vice versa)
Adjoint matrix
𝑎𝑑𝑗 𝑨 = (𝑨𝑐𝑜 )𝑇
Inverse matrix
𝑨−1 = |𝑨| ∙ (𝑨𝑐𝑜 )𝑇
Inverse matrix (order (2 x 2))
1
𝑎
𝑨=[
𝑐
and
(for any i or j)
𝑨 𝑨−1 = 𝑨−1 𝑨 = 𝑰
1
𝑏
𝑑
∙[
] ⇒ 𝑨−1 =
𝑑
𝑎𝑑 − 𝑏𝑐 −𝑐
−𝑏
]
𝑎
System of linear equations:
𝑨𝒙=𝒃
(a) Non-homogenous and non-singular
𝒃 ≠ 𝟎 ∧ |𝑨| ≠ 0 ⇒
𝒙 = 𝑨−1 𝒃
(unique solution)
(b) Homogenous and non-singular
𝒃 = 𝟎 ∧ |𝑨| ≠ 0 ⇒
𝒙 = 𝑨−1 𝟎 = 𝟎
(trivial solution)
(c) Non-homogenous and singular
𝒃 ≠ 𝟎 ∧ |𝑨| = 0 ⇒
{
(d) Homogenous and singular
𝒃 = 𝟎 ∧ |𝑨| = 0 ⇒
infinitely many solutions
Characteristic polynomial
𝑐(𝜆) = |𝜆𝑰 − 𝑨|
Characteristic equation
𝑐(𝜆) = 0
Eigenvalues 𝜆: 𝑨 𝒙 = 𝜆 𝒙
𝜆1 , 𝜆2 , 𝜆3 … that are solutions to 𝑐(𝜆) = 0
no solutions or
infinitely many solutions
Differentiation:
𝑓(𝑥) ′
(𝑔(𝑥)) =
Composite rule (Chain rule)
(𝑓(𝑔(𝑥))) = 𝑓 ′ (𝑔(𝑥))𝑔′(𝑥)
Inverse rule
(𝑓 −1 (𝑥)) = 𝑓′(𝑦)
or
𝑑𝑦
𝑑𝑦 𝑑𝑧
= 𝑑𝑧 𝑑𝑥
𝑑𝑥
or
𝑑𝑦
1
= 𝑑𝑥
𝑑𝑥
2
(𝑔(𝑥))
′
𝑓 ′′ (𝑥) =
For parametric representation
𝑑𝑦
𝑑𝑥
=
1
𝑑𝑦
𝑑2 𝑓
= 𝑓 (2) 𝑥
𝑑𝑥 2
𝑑𝑦
𝑑𝑡
𝑑𝑥
𝑑𝑡
𝜕𝑓
𝑑𝑦
= − 𝜕𝑥
𝜕𝑓
𝑑𝑥
𝜕𝑦
and
𝑑2𝑦
𝑑𝑥 2
𝑑𝑣
𝑣−𝑢
𝑑
𝑢
𝑑𝑥
𝑑𝑥
(
)
=
𝑑𝑥 𝑣
𝑣2
′
2nd order derivative
For implicit function
𝑓(𝑥, 𝑦) = 0
𝑑𝑢
or
𝑓′ (𝑥)𝑔(𝑥)−𝑓(𝑥)𝑔′(𝑥)
Quotient rule
=
𝑑 𝑑𝑦
( )
𝑑𝑡 𝑑𝑥
𝑑𝑥
𝑑𝑡
Taylor polynomials:
of nth degree about x = a
𝑓 ′ (𝑎)
𝑓 ′′ (𝑎)
𝑓 ′′′ (𝑎)
(𝑥 − 𝑎) +
(𝑥 − 𝑎)2 +
(𝑥 − 𝑎)3
1!
2!
3!
𝑓 (𝑛) (𝑎)
(𝑥 − 𝑎)𝑛
+⋯+
𝑛!
Note: 𝑝𝑛 (𝑥) ≈ 𝑓(𝑥) when 𝑥 ≈ 𝑎
𝑝𝑛(𝑥) = 𝑓(𝑎) +
∞
Taylor series expansion, 𝑥 ≈ 𝑎
𝑓 ′ (𝑎)
𝑓 ′′ (𝑎) 2
𝑓 (𝑛) (𝑎) 𝑛
𝑓(𝑥 + 𝑎) = 𝑓(𝑎) +
𝑥+
𝑥 +⋯= ∑
𝑥
1!
2!
𝑛!
𝑛=0
∞
Maclauren series expansion,
𝑥≈𝑎
𝑓 ′ (0)
𝑓 ′′ (0) 2
𝑓 (𝑛) (0) 𝑛
𝑓(𝑥) = 𝑓(0) +
𝑥+
𝑥 +⋯= ∑
𝑥
1!
2!
𝑛!
Functions of 2 or more
variables:
𝑓(𝑥, 𝑦)
Partial derivatives
𝑛=0
𝑜𝑟
𝑓(𝑥, 𝑦, 𝑧) 𝑒𝑡𝑐.
𝜕𝑓
𝑑𝑓
= [ ]
𝜕𝑥
𝑑𝑥 𝑦=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝜕𝑓
𝑑𝑓
= [ ]
𝜕𝑦
𝑑𝑦 𝑥=𝑐𝑜𝑛𝑠𝑡𝑎𝑛𝑡
𝜕𝑓
= 𝑓1′ (𝑥, 𝑦) ~ slope on the surface in x-axis direction
𝜕𝑥
𝜕𝑓
= 𝑓2′ (𝑥, 𝑦) ~ slope on the surface in y-axis direction
𝜕𝑦
Directional derivative
(Slope in any direction 𝛼)
Chain rule
𝑚𝛼 (𝑥, 𝑦) =
𝜕𝑓
𝜕𝑓
cos(𝛼) +
sin (𝛼)
𝜕𝑥
𝜕𝑦
𝑧 = 𝑓(𝑥, 𝑦)
𝑥(𝑠, 𝑡)
} ⇒
𝑦(𝑠, 𝑡)
Then:
2nd order partial derivatives
𝜕𝑧
𝜕𝑧 𝜕𝑥
𝜕𝑧 𝜕𝑦
= 𝜕𝑥 𝜕𝑠 + 𝜕𝑦 𝜕𝑠
𝜕𝑠
𝑧 = 𝐹(𝑠, 𝑡)
𝜕𝑧
𝜕𝑧 𝜕𝑥
𝜕𝑧 𝜕𝑦
= 𝜕𝑥 𝜕𝑡 + 𝜕𝑦 𝜕𝑡
𝜕𝑡
𝜕2 𝑓
𝜕 𝜕𝑓
= 𝜕𝑥 (𝜕𝑥)
𝜕𝑥 2
~ info about curvature on surface in x-axis direction
𝜕2 𝑓
𝜕 𝜕𝑓
=
( )
𝜕𝑦 2
𝜕𝑦 𝜕𝑦
~ info about curvature on surface in x-axis direction
𝜕2 𝑓
𝜕 𝜕𝑓
= 𝜕𝑥 (𝜕𝑦)
𝜕𝑥𝜕𝑦
~ increase in slope in y-direction when moving in x-axis direction
𝜕2 𝑓
𝜕 𝜕𝑓
= 𝜕𝑦 (𝜕𝑥)
𝜕𝑦𝜕𝑥
~ increase in slope in x-direction when moving in y-axis direction
𝜕2 𝑓
𝜕2 𝑓
=
𝜕𝑦𝜕𝑥
𝜕𝑥𝜕𝑦
~ the cross partials are equal (for all relevant functions)
The differential (2 variables)
𝑢 = 𝑓(𝑥, 𝑦):
∆𝑢 ≈
𝑑𝑢 =
𝜕𝑓
𝜕𝑓
∆𝑥 +
∆𝑦
𝜕𝑥
𝜕𝑦
𝑜𝑟
𝑑𝑢 =
𝜕𝑓
𝜕𝑓
𝑑𝑥 +
𝑑𝑦
𝜕𝑥
𝜕𝑦
𝜕𝑓
𝜕𝑓
∆𝑥 +
∆𝑦
𝜕𝑥
𝜕𝑦
Ordinary differential
equations:
Separation of the variables
ODE on exact form
𝑑𝑥
= ℎ(𝑡)
𝑑𝑡
ODE(1):
𝑔(𝑥)
⇒
ODE(2):
𝑑𝑥
𝑥
1
1
= 𝑓( ) ⇒ ∫
𝑑𝑦 = ∫ 𝑑𝑡
𝑑𝑡
𝑡
𝑓(𝑦) − 𝑦
𝑡
ODE(3):
𝑝(𝑡, 𝑥)
𝑑𝑥
+ 𝑞(𝑡, 𝑥) = 0
𝑑𝑡
where
∫ 𝑔(𝑥)𝑑𝑥 = ∫ ℎ(𝑡)𝑑𝑡
⇒
ODE(4):
𝑑𝑥
+ 𝑝(𝑡) ∙ 𝑥 = 𝑟(𝑡) ⇒
𝑑𝑡
𝑥
𝑡
ℎ(𝑡, 𝑥) = 𝑐
𝜕ℎ
𝜕ℎ
= 𝑝(𝑡, 𝑥) and
= 𝑞(𝑡, 𝑥)
𝜕𝑥
𝜕𝑡
Test for existence of the function ℎ(𝑡, 𝑥):
First order linear ODE
, with 𝑦 =
𝜕𝑝
𝜕𝑞
=
𝜕𝑡
𝜕𝑥
𝑥(𝑡) = 𝑒 −𝑘(𝑡) ∙ ∫ 𝑒 𝑘(𝑡) ∙ 𝑟(𝑡)𝑑𝑡
where 𝑘(𝑡) = ∫ 𝑝(𝑡)𝑑𝑡
Bernoulli ODE
ODE(5):
⇒
𝑑𝑥
+ 𝑝(𝑡) ∙ 𝑥 = 𝑞(𝑡) ∙ 𝑥 𝛼
𝑑𝑡
𝑑𝑦
+ (1 − 𝛼)𝑝(𝑡)𝑦 = (1 − 𝛼)𝑞(𝑡)
𝑑𝑡
, where 𝑦 = 𝑥 1−𝛼
which is solved as an ODE(4)
Differential operator
Φ[𝑓(𝑡)] = expression where one or more derivatives of 𝑓 appear
Linear differential operator
𝐿[𝑎𝑥1 + 𝑏𝑥2 ] = 𝑎𝐿[𝑥1 ] + 𝑏𝐿[𝑥2 ]
Homogenous linear ODE
𝐿[𝑥(𝑡)] = 0
Non-homogenous linear ODE
𝐿[𝑥(𝑡)] = 𝑓(𝑡)
Linearity principle
𝑥1 and 𝑥2 solve 𝐿[𝑥(𝑡)] = 0
} ⇒ 𝑎𝑥1 + 𝑏𝑥2 is a solution
𝐿[𝑥(𝑡)] is linear
In addition:
The ODE is of order 𝑝
General solution:
} ⇒ 𝐴 𝑥 +𝐴 𝑥 + ⋯+ 𝐴 𝑥
𝑥1 , 𝑥2 , … , 𝑥𝑝 are independent solutions
1 1
2 2
𝑝 𝑝
𝑝
Independent functions
∑ 𝑘𝑖 ∙ 𝑓𝑖 (𝑡) = 0 for all 𝑡, only when 𝑘1 = 𝑘2 = ⋯ = 𝑘𝑝 = 0
𝑖=1
General solution to 𝐿[𝑥] = 𝑓(𝑡)
ODE(7):
x ∗ : Any solution to 𝐿[𝑥] = 𝑓(𝑡)
}
𝑥𝑐 : The general solution to 𝐿[𝑥] = 0
𝑑2 𝑥
𝑑𝑥
+𝑏
+ 𝑐𝑥 = 0
2
𝑑𝑡
𝑑𝑡
⇒
General solution: x ∗ + 𝑥𝑐
Linear, constant coefficient ODE
ODE(6):
Characteristic equation
𝑎𝑚2 + 𝑏𝑚 + 𝑐 = 0
General solution to ODE(6)
𝑚1 ≠ 𝑚2 ∈ ℝ ∶
𝑥(𝑡) = 𝐴𝑒 𝑚1 𝑡 + 𝐵𝑒 𝑚2 𝑡
𝑚1 = 𝑚2 = 𝑘 ∶
𝑥(𝑡) = 𝐴 ∙ 𝑡 ∙ 𝑒 𝑘𝑡 + 𝐵𝑒 𝑘𝑡
𝑚1 , 𝑚2 ∈ ℂ ∶
𝑚1 = 𝛼 + 𝛽𝑗 and 𝑚2 = 𝛼 − 𝛽𝑗 then
(The method can readily be
generalized to higher order ODE’s)
𝑎
,𝑎 ≠ 0
with solutions 𝑚1 and 𝑚2
𝑥(𝑡) = 𝑒 𝛼𝑡 ( 𝐶 · cos(𝛽𝑡) + 𝐷 · sin(𝛽𝑡))