IMC 1994 D1 P5

Art of Problem Solving

AoPS Online

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

College Math Topics in undergraduate and graduate studies

Topics in undergraduate and graduate studies

3 M G

BBookmark VNew Topic kLocked

No tags match your search

M

IMC 1994 D1 P5

G H J

Reveal topic tags

L

G H BBookmark kLocked kLocked NReply

The post below has been deleted. Click to close.

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

340 posts

j___d

#1 h Mar 6, 2017, 5:00 PM • 3 Y

Y by Adventure10, Mango247, teomihai

a) Let $f\in C[0,b]$ , $g\in C(\mathbb R)$ and let $g$ be periodic with period $b$ . Prove that $\int_0^b f(x) g(nx)\,\mathrm dx$ has a limit as $n\to\infty$ and
$\lim_{n\to\infty}\int_0^b f(x)g(nx)\,\mathrm dx=\frac 1b \int_0^b f(x)\,\mathrm dx\cdot\int_0^b g(x)\,\mathrm dx$
b) Find
$\lim_{n\to\infty}\int_0^\pi \frac{\sin x}{1+3\cos^2nx}\,\mathrm dx$

Z K Y

The post below has been deleted. Click to close.

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

2 posts

#2 h Feb 19, 2025, 1:46 PM • 2 Y

Y by teomihai, soryn

b) $I=\frac{1}{\pi} \int_0^{\pi}sinx * \int_0^{\pi}\frac{1}{1+3cos^2x}$ $\int_0^{\pi}\frac{1}{1+3cos^2x} = \int_0^{\pi}\frac{sec^{2}x}{sec^{2}x+3}$ use $sec^2x+=tan^2x+1$ and then $I' = \int_0^{\pi}\frac{dtanx}{sec^{2}x+3} = \int_0^{\pi}\frac{dtanx}{tan^{2}x+4}$ It is now a simple calculation to find the integrand has
$\frac{1}{2}arctan{\frac{tanx}{2}}$ as its primitive and observe the primitive is piecewise continuous,so we divide the interval into $\int_0^{\frac{\pi}{2}} + \int_{\frac{\pi}{2}}^{\pi}$ after a quick use of N-L formula we get $I= I'*\frac{2}{\pi}=1$

Z K Y

The post below has been deleted. Click to close.

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

2 posts

#3 h Feb 21, 2025, 5:05 AM

Y by

By the way, I found a website has more details and solutions that is https://www.imc-math.org.uk/?year=1994&item=info
I hope that will help you

Z K Y

The post below has been deleted. Click to close.

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

50 posts

#4 h Apr 5, 2025, 8:08 PM

Y by

part (a):
[ig its an overkill :read:

:read:

]
First note that:
$\int_0^b f(x)g(nx) \ dx=\int_0^{\frac{1}{n}b}f(x)g(nx) \ dx+\int_{\frac{1}{n}b}^{\frac{2}{n}b}f(x)g(nx) \ dx+\cdots + \int_{\frac{n-1}{n}b}^bf(x)g(nx) \ dx$ Also we have,
$\begin{align*} \int_{\frac{k}{n}b}^{\frac{k+1}{n}b}f(x)g(nx) \ dx &= \int_0^{\frac{b}{n}}f\left(y+\frac{k}{n}b\right)g(ny+kb) \ dy \ \ \ \ \ \ \ \ [\text{setting } y = x-\frac{k}{n}b]\\ &=\int_0^{\frac{b}{n}} f\left(y+\frac{k}{n}b\right)g(ny) \ dy \ \ \ \ \ \ \ \ \ [\text{since } g(x+kb)=g(x)]\\ &=\int_0^b \frac{1}{n}f\left(\frac{z+kb}{n}\right)g(z) \ dz \ \ \ \ \ \ \ \ \ [\text{setting } z = ny] \end{align*}$ Therefore,
$\begin{align*} \int_0^b f(x)g(nx) \ dx &= \sum_{k=0}^{n-1} \int_0^b f\left(\frac{x+kb}{n}\right)g(x) \ dx\\ &= \int_0^b g(x)\left(\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{x+kb}{n}\right)\right) \ dx \end{align*}$ Now note that since $f$ is continuous on a compact domain, it must be uniformly continuous (by Heine-Cantor theorem).
Let $M = \int_0^b |g(x)| \ dx$ .
Let $I_0 = \left[0, \frac{1}{n}b\right], \ I_1 = \left[\frac{1}{n}b, \frac{2}{n}b\right], \ldots, \ I_{n-1} = \left[\frac{n-1}{n}b, b\right]$
Note that $\frac{x+kb}{n}\in I_k$ for each $k=0,\ldots, n-1$ . Also, the length of each interval is $\frac{b}{n}$ .
Fix $\varepsilon>0$ . Note that uniform continuity of $f$ tells that there exists $\delta>0$ such that $|f(x)-f(y)|<\frac{\varepsilon}{M}$ holds wherever $0<|x-y|<\delta$ .
Now, choose $N$ large enough so that $0<\frac{b}{N}<\delta$ . Then, for all $n\ge N$ , we have $\left|f\left(\frac{x+kb}{n}\right)-f(t)\right|<\frac{\varepsilon}{M}$ for all $t\in I_k$ and $x\in (0,b)$ as $\left|\frac{x+kb}{n}-t\right|<\frac{b}{n}<\frac{b}{N}<\delta$ .
So, we have,
$\begin{align*} \left|\frac{b}{n}f\left(\frac{x+kb}{n}\right)-\int_{\frac{k}{n}b}^{\frac{k+1}{n}b}f(t) \ dt\right| &=\left|\int_{\frac{k}{n}b}^{\frac{k+1}{n}b} f\left(\frac{x+kb}{n}\right) \ dt-\int_{\frac{k}{n}b}^{\frac{k+1}{n}b}f(t) \ dt\right|\\ &\le \int_{\frac{k}{n}b}^{\frac{k+1}{n}b} \left|f\left(\frac{x+kb}{n}\right)-f(t)\right| \ dt\\ &<\int_{\frac{k}{n}b}^{\frac{k+1}{n}b} \frac{\varepsilon}{M} \ dt = \frac{b}{Mn}\varepsilon. \end{align*}$ for each $k=0, 1, \ldots, n-1$ and for all $x\in (0,b)$ . So, triangle inequality gives:
$\left|\frac{b}{n}\sum_{k=0}^{n-1}f\left(\frac{x+kb}{n}\right)-\int_0^b f(x) \ dx\right|\le \sum_{k=0}^{n-1} \left|\frac{b}{n}f\left(\frac{x+kb}{n}\right)-\int_{\frac{k}{n}b}^{\frac{k+1}{n}b}f(t) \ dt\right| < n\cdot \frac{b}{Mn}\varepsilon=\frac{b}{M}\cdot \varepsilon.$ Therefore, for all $n\ge N$ , we have
$\left|\frac{b}{n}\sum_{k=0}^{n-1}f\left(\frac{x+kb}{n}\right)-\int_0^b f(x) \ dx\right|< \frac{b}{M}\cdot \varepsilon$ $\implies \left|\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{x+kb}{n}\right)-\frac{1}{b}\int_0^b f(x) \ dx\right|< \frac{\varepsilon}{M}$
$\implies \left|\int_0^b g(x)\left(\frac{1}{n} \sum_{k=0}^{n-1}f\left(\frac{x+kb}{n}\right)\right) \ dx - \int_0^b g(x) \left(\frac{1}{b}\int_0^b f(t) \ dt\right) \ dx\right|$ $\le \int_0^b |g(x)|\left|\frac{1}{n}\sum_{k=0}^{n-1}f\left(\frac{x+kb}{n}\right)-\frac{1}{b}\int_0^b f(t) \ dt\right| \ dx$ $< \int_0^b |g(x)| (\varepsilon/M) \ dx = \varepsilon$ Now note that, $\int_0^b g(x) \left(\frac{1}{b}\int_0^b f(t) \ dt\right) \ dx = \frac{1}{b} \left(\int_0^b f(x) \ dx\right)\left(\int_0^b g(x) \ dx\right)$ and
$\int_0^b g(x)\left(\frac{1}{n} \sum_{k=0}^{n-1}f\left(\frac{x+kb}{n}\right)\right) \ dx = \int_0^b f(x)g(nx) \ dx$ So, we have,
$\left|\int_0^b f(x)g(nx) \ dx - \frac{1}{b} \left(\int_0^b f(x) \ dx\right)\left(\int_0^b g(x) \ dx\right)\right|< \varepsilon$
Hence the required limit exists and
$\boxed{\lim_{n\rightarrow \infty} \int_0^b f(x)g(nx) \ dx = \frac{1}{b} \left(\int_0^b f(x) \ dx\right)\left(\int_0^b g(x) \ dx\right).}$

https://external-preview.redd.it/Vt8un6DvXelUjrQPqHsJXaIijhQIMDRU50RjKVXe2JM.jpg?auto=webp&s=fbd4da7e4d2893906b86cebbff64976f0e1c03e2

This post has been edited 4 times. Last edited by Fibonacci_math, Apr 6, 2025, 6:27 AM

Z K Y

The post below has been deleted. Click to close.

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

50 posts

#5 h Apr 5, 2025, 8:26 PM

Y by

Part (b):

First note that $1+3\cos^2{x}$ is periodic with fundamental period $\pi$ . So using part (a) we can write
$\begin{align*} \lim_{n\rightarrow \infty} \int_0^\pi \frac{\sin{x}}{1+3\cos^2{nx}} \ dx&= \frac{1}{\pi} \left(\int_0^{\pi} \sin{x}\right)\left(\int_0^{\pi} \frac{dx}{1+3\cos^2{x}}\right)\\ &=\frac{2}{\pi}\int_0^{\pi} \frac{\csc^2{x} \ dx}{1+4\cot^2{x}}\\ &=\frac{2}{\pi} \int_{\infty}^{-\infty} \frac{-dt}{1+4t^2} \ \ \ \ \ \ [t = \cot{x}] \\ &=\frac{2}{\pi} \int_{-\infty}^{\infty} \frac{dt}{1+4t^2}\\ &=\frac{1}{\pi} [\tan^{-1}{2t}]_{-\infty}^{\infty}\\ &=\boxed{1} \end{align*}$

Z K Y

N Quick Reply

G

H

=

© 2025 AoPS Incorporated

a