# Useful Math

Daniel Weibel
Created 4 Sep 2017
Last updated 8 Nov 2017

Some simple but useful math reminders and tricks.

# Divison Terminology

How are they called again? ðŸ¤”

$\frac{\text{dividend}}{\text{divisor}} = \frac{\text{numerator}}{\text{denominator}}$

# Sum of Integers 1 to N

Approach: pair and add up corresponding high and low elements (first + last, second-first + second-last, etc.).

## If N is even

Great, we can make $\frac{N}{2}$ pairs:

$1 + 2 + 3 + 4 + \ldots + (N-3) + (N-2) + (N-1) + N$
• $1 + N = N + 1$
• $2 + (N-1) = N + 1$
• $3 + (N-2) = N + 1$
• $4 + (N-3) = N + 1$
• $\ldots$

$\frac{N}{2}$ sums of value $N+1$:

$\frac{N}{2} \, (N+1) = \frac{N\,(N+1)}{2}$

## If N is odd

Prepend a 0 to make the number of summands even:

$0 + 1 + 2 + 3 + \ldots + (N-3) + (N-2) + (N-1) + N$
• $0 + N = N$
• $1 + (N-1) = N$
• $2 + (N-2) = N$
• $3 + (N-3) = N$
• $\ldots$

$\frac{N+1}{2}$ sums of value $N$:

$\frac{N+1}{2} \, N = \frac{N\,(N+1)}{2}$

## Conclusion

In all cases, the sum of integers 1 to $N$ is:

$\frac{N\,(N+1)}{2}$

# Sum of Integers 1 to N-1

Analogous to the sum of integers 1 to $N$.

## If N is even (i.e. N-1 is odd)

$0 + 1 + 2 + 3 + \ldots + (N-4) + (N-3) + (N-2) + (N-1)$
• $0 + (N-1) = N - 1$
• $1 + (N-2) = N - 1$
• $2 + (N-3) = N - 1$
• $3 + (N-4) = N - 1$
• $\ldots$

$\frac{N}{2}$ sums of value $N-1$:

$\frac{N}{2} \, (N-1) = \frac{N\,(N-1)}{2}$

## If N is odd (i.e. N-1 is even)

$1 + 2 + 3 + 4 + \ldots + (N-4) + (N-3) + (N-2) + (N-1)$
• $1 + (N-1) = N$
• $2 + (N-2) = N$
• $3 + (N-3) = N$
• $4 + (N-4) = N$
• $\ldots$

$\frac{N-1}{2}$ sums of value $N$:

$\frac{N-1}{2} \, N = \frac{N\,(N-1)}{2}$

## Conclusion

In all cases, the sum of integers 1 to $N-1$ is:

$\frac{N\,(N-1)}{2}$

# Sum of Powers of 2

What is the sum of the following?

$2^0 + 2^1 + 2^2 + 2^3 + \ldots + 2^N$

## Example N = 4

The solution can be easily seen by taking an example and writing the terms in binary notation:

\newcommand{\B}{\mathrm{b}} \begin{align*} 2^0 + 2^1 + 2^2 + 2^3 + 2^4 &= 1 + 2 + 4 + 8 + 16\\ &= 1_\B + 10_\B + 100_\B + 1000_\B + 10000_\B \\ &= 11111_\B \\ &= 2^5 - 1 \end{align*}

## Conclusion

The sum of the powers of 2 from 0 up to $N$ ($2^0 + 2^1 + \ldots + 2^N$), is:

$2^{N+1} - 1$

# Floor and Ceiling

## Floor function

$\mathrm{floor}(x) = \left\lfloor{x}\right\rfloor$

Go down to the nearest integer.

## Ceiling function

$\mathrm{ceiling}(x) = \left\lceil{x}\right\rceil$

Go up to the nearest integer.