Some simple but useful math reminders and tricks.

Contents

• Contents
• Divison Terminology
• Sum of Integers 1 to N
  • If N is even
  • If N is odd
  • Conclusion
• Sum of Integers 1 to N-1
  • If N is even (i.e. N-1 is odd)
  • If N is odd (i.e. N-1 is even)
  • Conclusion
• Sum of Powers of 2
  • Example N = 4
  • Conclusion
• Floor and Ceiling
  • Floor function
  • Ceiling function

Divison Terminology

How are they called again? 🤔

\[\frac{\text{dividend}}{\text{divisor}} = \frac{\text{numerator}}{\text{denominator}}\]

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.