Determining when an integral function is eventually constant

Art of Problem Solving

AoPS Online

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

High School Olympiads Regional, national, and international math olympiads

Regional, national, and international math olympiads

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

Determining when an integral function is eventually constant

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

Source: My creation

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

43 posts

#1 h Oct 3, 2024, 5:40 PM

Y by

Let
$f\left(a\right)=\int_{0}^{1}\frac{\sin\left(2023\pi ax\right)\sin\left(2023\pi x\right)\cos\left(2024\pi x\right)\cos\left(2024\pi ax\right)}{\sin\left(\pi ax\right)\sin\left(\pi x\right)}dx$ Determine the smallest positive integer $N$ such that for all positive integers $m, n > N, f(m) = f(n).$

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

43 posts

#3 h Oct 13, 2024, 10:48 PM

Y by

bump----

Z K Y

The post below has been deleted. Click to close.

This post has been deleted. Click here to see post.

46 posts

#4 h May 18, 2025, 8:41 PM

Y by

Solution:

Let the integrand be $I(a, x) = \frac{\sin(2023\pi ax) \sin(2023\pi x) \cos(2024\pi ax) \cos(2024\pi x)}{\sin(\pi ax) \sin(\pi x)}$ .
We want to find the smallest positive integer $N$ such that for all positive integers $m, n > N$ , $f(m) = f(n)$ , where $f(a) = \int_0^1 I(a, x) \, dx$ .

Consider the case when $a$ is an even positive integer, $a = 2k$ for some positive integer $k$ .
Then $\sin(\pi ax) = \sin(2k\pi x) = 0$ .
The numerator becomes $\sin(4046k\pi x) \sin(2023\pi x) \cos(4048k\pi x) \cos(2024\pi x)$ .
Since $\sin(4046k\pi x) = 0$ , the integrand is $0$ , so $f(2k) = 0$ for all positive integers $k$ .

Consider the case when $a$ is an odd positive integer, $a = 2k+1$ for some non-negative integer $k$ .
Then $\sin(\pi ax) = \sin((2k+1)\pi x) = \pm \sin(\pi x)$ .
The integrand becomes $\frac{\sin((2k+1)2023\pi x) \sin(2023\pi x) \cos((2k+1)2024\pi x) \cos(2024\pi x)}{\pm \sin(\pi x) \sin(\pi x)}$ .

We use the identity $\frac{\sin(nx)}{\sin(x)} = 1 + 2 \cos(2x) + 2 \cos(4x) + \dots + 2 \cos((n-1)x)$ for odd $n$ .
So, $\frac{\sin(2023 \pi a x)}{\sin(\pi a x)} = 1 + 2 \sum_{j=1}^{1011} \cos(2j \pi a x)$ .

If $a=1$ , $f(1) = \int_0^1 \frac{\sin(2023\pi x) \sin(2023\pi x) \cos(2024\pi x) \cos(2024\pi x)}{\sin(\pi x) \sin(\pi x)} dx$ .

If $a$ is an odd integer, the value of $f(a)$ depends on $a$ . However, we are looking for when $f(m) = f(n)$ for large $m, n$ .
Since $f(even) = 0$ , if $m, n$ are large even integers, $f(m) = f(n) = 0$ .

The smallest positive integer $N$ such that for all positive integers $m, n > N$ , $f(m) = f(n)$ means that $f(a)$ becomes constant for $a > N$ .
Since $f(2k) = 0$ for all $k \ge 1$ , the function $f(a)$ takes the value $0$ for all even integers greater than $0$ .
Thus, if $N=1$ , for all even integers $m, n > 1$ , $f(m) = 0$ and $f(n) = 0$ , so $f(m) = f(n)$ .
The smallest positive integer $N$ is $1$ .

Final Answer: The final answer is $\boxed{1}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a