Solving Fizz Buzz with Cosines

Solving Fizz Buzz with Cosines

By Susam Pal on 20 Nov 2025

Fizz Buzz is a counting game that has become oddly popular in the world of computer programming as a simple test of basic programming skills. The rules of the game are straightforward. Players say the numbers aloud in order beginning with one. Whenever a number is divisible by 3, they say 'Fizz' instead. If it is divisible by 5, they say 'Buzz'. If it is divisible by both 3 and 5, the player says both 'Fizz' and 'Buzz'. Here is a typical Python program that prints this sequence:

for n in range(1, 101): if n % 15 == 0: print('FizzBuzz') elif n % 3 == 0: print('Fizz') elif n % 5 == 0: print('Buzz') else: print(n)

Here is the output: fizz-buzz.txt. Can we make the program more complicated? Perhaps we can use trigonometric functions to encode all four cases in a single closed-form expression. That is what we are going to explore in this article. By the end, we will obtain a finite Fourier series that can take any integer \( n \) and select the text to be printed.

Contents

Definitions

Before going any further, we establish a precise mathematical definition for the Fizz Buzz sequence. We begin by introducing a few functions that will help us define the Fizz Buzz sequence later.

Symbol Functions

We define a set of four functions \( \{ s_0, s_1, s_2, s_3 \} \) for integers \( n \) by: \begin{align*} s_0(n) &= n, \\ s_1(n) &= \mathtt{Fizz}, \\ s_2(n) &= \mathtt{Buzz}, \\ s_3(n) &= \mathtt{FizzBuzz}. \end{align*} We call these the symbol functions because they produce every term that appears in the Fizz Buzz sequence. The symbol function \( s_0 \) returns \( n \) itself. The functions \( s_1, \) \( s_2 \) and \( s_3 \) are constant functions that always return the literal words \( \mathtt{Fizz}, \) \( \mathtt{Buzz} \) and \( \mathtt{FizzBuzz} \) respectively, no matter what the value of \( n \) is.

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