Are primes differentiable?

Calculus Level 4

A function f p ( x ) = n + p sin x f_p(x) = \lfloor n+p \sin{x} \rfloor for x ( 0 , π ) x \in (0, \pi) , where n Z n \in \mathbb{Z} and p p is a prime number. N p N_p is the number of points where f p ( x ) f_p(x) is not differentiable. Find the last three digits of N 17497 N_{17497} .

Fun fact: 17497 is the 2014th prime.

Notation: \lfloor \cdot \rfloor denotes the floor function .


The answer is 993.

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1 solution

Chew-Seong Cheong
Aug 24, 2018

We note that f p ( x ) = n + p sin x f_p(x) = \lfloor n + p \sin x\rfloor has a point where it is not differentiable when sin x = k p \sin x = \dfrac kp , where k k is an integer and 1 k p 1 \le k \le p . k = p k=p , when x = π 2 x = \frac \pi 2 . We also note that sin x \sin x is symmetrical about π 2 \frac \pi 2 , then the points where f ( x ) f(x) is not differentiable are when:

x = sin 1 1 p , sin 1 2 p , sin 1 3 p , sin 1 p 1 p , π 2 , sin 1 p 1 p , sin 1 3 p , sin 1 2 p , sin 1 1 p x = \sin^{-1} \frac 1p,\ \sin^{-1} \frac 2p,\ \sin^{-1} \frac 3p,\ \cdots \sin^{-1} \frac {p-1}p,\ \frac \pi 2,\ \sin^{-1} \frac {p-1}p,\ \cdots \sin^{-1} \frac 3p,\ \sin^{-1} \frac 2p,\ \sin^{-1} \frac 1p

There are a total of 2 p 1 2p-1 points. For p = 17497 p=17497 , 2 p 1 = 34 993 2p-1 = 34\boxed{993} .

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