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everyone.

I would like to know are my conclusions right. The question is "what is the $n$th derivative of fractional part".

I see the plot of first derivative as a line that is parallel to the OX axe and has gaps when x is an integer, so I can write it as: $$\frac{d\{x\}}{dx} = \frac{\{x\}}{\{x\}}$$

Using this and quotient rule we can get the second derivative: $$\frac{d^2\{x\}}{dx^2} = \frac{0}{\{x\}^2}$$

Doing the same things we get the $n$th derivative formula: $$\frac{d^n\{x\}}{dx^n} = \frac{0}{\{x\}^{2^n}}$$

Bernard
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Sharlotta Neimor
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1 Answers1

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For any $x$,

$$\{x\}=x-\lfloor x\rfloor.$$

This shows that between two integers, the function is linear ($x-n$).

Taking the first derivative,

$$\{x\}'=1-\lfloor x\rfloor'.$$

As the floor function is piecewise constant, its derivative is $0$, except at the discontinuities, where it does not exist. For this reason, the first derivative of the fractional part is one between integers and undefined at integers.

The next derivatives are zero between integers (as $1'=0'=0$) and undefined at integers.

Your expressions do describe this behavior (anyway, the exponent of $\{x\}$ is irrelevant and can remain $1$ for all derivatives). From a formal point of view, they are unjustified as the computation rules that you invoke do not hold for undefined derivatives. So your conclusions are right by accident.