Mathematical Cheatsheet

General Mathematics

Quadratic equation

Given a quadratic equation of form :

$a{x}^2+bx+c=0$

We compute the determinant :

$$\mathrm{\Delta }=b^2-4ac$$

If

$\mathrm{\Delta }>=0$ then the quadratic equation admits 2 solutions :

$$x=\frac{-b\pm \sqrt{\mathrm{\Delta }}}{2a}$$Else, if

$\mathrm{\Delta }=0$ then the quadratic equation admits only 1 solution :

$$x=\frac{-b}{2a}$$If

$\mathrm{\Delta }<0$ then the quadratic equation admits no solutions in

$\mathbb{R}$ but in

$\mathbb{C}$ we have :

$$x=\frac{-b\pm i\sqrt{-\mathrm{\Delta }}}{2a}$$

Cubic equation

Sums and Products

$\sum _{i=0}^n{i}^2=\frac{(n^2+n)(2n+1)}{6}$

$\sum _{i=0}^{n}i=\frac{n(n+1)}{2}$

• Telescopic sum.
  

  $\sum _{i=p}^q(a_{i+1}-a_i)=(a_{q+1}-a_p)$

• Telescopic products.
  

  $\prod _{i=p}^q\frac{a_{i+1}}{a_i}=\frac{a_{q+1}}{a_p}$

• Newton's Factorisation.
  for(a,b) in

  $\mathbb{R}$ and n in
  $\mathbb{N}$
  
  $(a+b{)}^n=\sum _{k=0}^n{\displaystyle (\genfrac{}{}{0ex}{}{n}{k})}{a}^k{b}^{n-k}$

• Factorisation
  

  $a^{n+1}-b^{n+1}=(a-b)\sum _{k=0}^n{a}^k-b^{n-k}$

Complex numbers

For z in

$\mathbb{C}$

$z=x+iy$

$x=Re(z)$ and

$y=Im(z)$

Additions and Multiplications.

$z+z^{\prime }=(x+x^{\prime })+i(y+y^{\prime })$

$z{z}^{\prime }=(x{x}^{\prime }-y{y}^{\prime })+i(x{y}^{\prime }+y{x}^{\prime })$

$|z|$ =

$\sqrt{Re(z{)}^2+Im(z{)}^2}$

$|z+z^{\prime }|$

$\le$

$|z|$ +

$|z^{\prime }|$

Modular Arithmetic

For a, b

$\in$

$\mathbb{N}$

$a\equiv b(\mathrm{mod}p)$ if and only if

$a-b$ can be divided by p
In this case, k

$\in$

$\mathbb{N}$ such that

$a=b+kn$

Euler's theorem

Given

$n$

$\in$

$\mathbb{Z}$ and

$a$

$\in$

$\mathbb{Z}$ with

$gcd(a,n)=1$ then

$$a^{\varphi }(n)\equiv 1(\mathrm{mod}n)$$

Fermat's little theorem

Let a

$\in$

$\mathbb{Z}$ and p a prime number

$$a^p\equiv a(\mathrm{mod}p)$$

RSA

For

$\mathrm{\forall }$p, q

$\in$

$\mathbb{N}$ (note : p and q have to be coprimes numbers)

$$n=p\ast q$$

Discrete logarithm

Linear algebra