# Continuous Replicative Function

Jump to navigation
Jump to search

## Theorem

Let $f: \R \to \R$ be a real function.

Let $f$ be continuous on $\R$.

Let $f$ also be a replicative function.

Then $f$ is of the form:

- $\map f x = \paren {x - \dfrac 1 2} a$

where $a \in \R$.

## Proof

Let $f$ be a replicative function.

Then:

- $\forall n > 0: \map f {n x + 1} - \map f {n x} = \map f {x + 1} - \map f x$

If $f$ is then also continuous:

- $\forall x \in \R: \map f {x + 1} - \map f x$

and so:

- $\map g x = \map f x - c \floor x$

is both replicative and periodic.

We have:

- $\ds \int_0^1 e^{2 \pi i n x} \map g x \rd x = \dfrac 1 n \int_0^1 e^{2 \pi y} \map g y \rd y$

Expanding in a Fourier series shows:

- $\map g x = \paren {x - \dfrac 1 2} a$

for $0 < x < 1$.

Thus it follows that:

- $\map f x = \paren {x - \dfrac 1 2} a$

$\blacksquare$

## Historical Note

Donald E. Knuth in his *The Art of Computer Programming* attributes this result to Nicolaas Govert de Bruijn, but gives no details as to where this result may be found.

## Sources

- 1997: Donald E. Knuth:
*The Art of Computer Programming: Volume 1: Fundamental Algorithms*(3rd ed.) ... (previous) ... (next): $\S 1.2.4$: Integer Functions and Elementary Number Theory: Exercise $40$