Fractional Part Function

The floor function \( \lfloor x \rfloor\) is defined to be the greatest integer less than or equal to the real number \( x \). The fractional part function \( \{ x \}\) is defined to be the difference between these two:

Let \( x\) be a real number. Then \[ \{x\}= x -\lfloor x \rfloor. \]

This is the graph of the function \( y=\{x\}.\) ^[1]

For nonnegative real numbers, the fractional part is just the "part of the number after the decimal," e.g. \[ \{3.64 \} = 3.64 - \lfloor 3.64 \rfloor = 3.64 - 3 = 0.64. \] But for negative real numbers, this is no longer the case: \[ \{-3.64 \} = -3.64 - \lfloor -3.64 \rfloor = -3.64 - (-4) = 0.36. \] Note that in both cases \( \{ x \} \) is nonnegative.

\( \{ 1 \} = 1 - 1 = 0. \)

\( \{ \sqrt{2} \} = \sqrt{2}-1 = 0.4142\ldots. \)

\( \{ \pi \} = \pi - 3 = 0.14159\ldots. \)

\( \{ -17/5 \} = -17/5 - (-4) = 3/5. \)

1. \( 0 \le \{ x \} < 1 \), and \( 0 = \{ x \} \) if and only if \( x \) is an integer.
2. \( \{x \} + \{-x\} = \begin{cases} 0 & \text{if } x \text{ is an integer} \\ 1 & \text{otherwise} \end{cases} \)
3. If \( a \) and \( b \) are integers, \( b > 0 \), then \( \{ \frac{a}{b} \} = \frac{r}{b} \), where \( r \) is the remainder from dividing \( a \) by \( b \).

\( \{ 9 \} + \{-9\} = 0+0 = 0 \), but \( \{9.01\} + \{-9.01\} = 0.01+0.99 = 1 \). The fractional part is always nonnegative.

Since \( 47 = 13 \cdot 3 + 8 \), \( \left\{\frac{47}{13} \right\} = \frac{8}{13}\).

For problems involving the floor function and the fractional part function, it often helps (for ease of notation) to write \( x = n+r \) where \( n = \lfloor x \rfloor \) and \( r = \{ x\} \). Then \( n \) is an integer and \( 0 \le r < 1 \).

Let \( x \) be a positive real number such that \[ x^2+\{x\}^2 = 27. \] Find \( x\).

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Write \( x = n+r \) as suggested. Now note that \(x^2 \le x^2+\{x\}^2 < x^2+1 \), so \( 26 < x^2 \le 27\). So \(n = 5 \). Then \[ \begin{align} (5+r)^2+r^2 &= 27 \\ 25+10r+2r^2 &=27 \\ 2r^2+10r-2 &= 0 \\ \end{align} \] so \( r = \frac{-5+\sqrt{29}}2\) by the quadratic formula, and \( x = 5 + r = \frac{5+\sqrt{29}}2\).

Find the smallest real number \( m \) such that for all positive real numbers \( x \), \[ \{ x \} + \left\{ \frac1{x} \right\} < m. \]

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If \( x < 1 \), we can write \( x = 1/y, y \ge 1, \) and \( \{ x \} + \{ \frac1{x} \} = \{ y \} + \{ \frac1{y} \}. \) So we may assume \( x \ge 1 \). Then \( \{ \frac1{x} \} = \frac1{x}\). Writing \( x = n+r \), with \(n \) a positive integer, the left side becomes \[ r + \frac1{n+r}. \] For fixed \( r \), this is maximized when the denominator is minimized, i.e. \(n = 1 \).

Now \( r + \frac1{1+r} \) is an increasing function for \( 0 \le r < 1 \) (its derivative is \( 1-\frac1{(1+r)^2} \)). As \( r \to 1^- \), the sum goes to \( 1+\frac1{1+1} = 3/2 \). So the answer is \( 3/2 \).

(Exercise: if \( x \) is allowed to be negative, the answer is \( m=2\).)

There are many interesting integrals involving the fractional part function. A good way to evaluate definite integrals of this type is to break up the interval of integration into intervals on which the greatest integer function is constant; then the original integral is a sum of integrals which are easier to evaluate.

Find \[ \int_0^1 \left\{ \frac1{x} \right\} \, dx. \]

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On the interval \( (1/(n+1),1/n] \), \( \lfloor \frac1{x} \rfloor = n \). So the integral on that interval becomes \[ \begin{align} \int_{1/(n+1)}^{1/n} \left( \frac1{x}-n \right) \, dx &= \ln(1/n)-\ln(1/(n+1)) -n(1/n-1/(n+1)) \\ &= \ln(n+1)-\ln(n) - \frac1{n+1}. \end{align} \]

So the integral is \[ \sum_{n=1}^{\infty} \int_{1/(n+1)}^{1/n} \left\{ \frac1{x} \right\} \, dx = \sum_{n=1}^\infty (\ln(n+1)-\ln(n) - \frac1{n+1}) \] and the \( k \)th partial sum telescopes to \( \ln(k+1)-\left( \frac12 + \cdots + \frac1{k+1} \right) \). The limit is \( 1-\gamma \), where \( \gamma \) is the famous Euler-Mascheroni constant.

1. Schmidtke, R. Parte fraccionaria. Retrieved April 16, 2016, from https://es.wikipedia.org/wiki/Usuario:MRS~eswiki/parte_entera#/media/File:Parte_fraccionaria.png