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k a March Highlights and 2025 AoPS Online Class Information
jlacosta   0
Mar 2, 2025
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0 replies
jlacosta
Mar 2, 2025
0 replies
A great result
steven_zhang123   7
N 5 minutes ago by Kath_revenge_krr
Show that $\lim_{n \to \infty} \sum_{k=1}^{n} (\sqrt{1+\frac{k}{n^{2} } } -1)=\frac{1}{4} $.
7 replies
steven_zhang123
Oct 31, 2024
Kath_revenge_krr
5 minutes ago
ineq-integral
Pirkuliyev Rovsen   8
N 12 minutes ago by Kath_revenge_krr
Prove that: $|\int_{0}^{2\pi }x\sin{x}dx|{\leq}\frac{ 2\pi^3  }{3}$
8 replies
Pirkuliyev Rovsen
Sep 17, 2016
Kath_revenge_krr
12 minutes ago
Weird family of sequences
AndreiVila   7
N 35 minutes ago by kamatadu
Source: Romanian District Olympiad 2025 12.3
[list=a]
[*] Let $a<b$ and $f:[a,b]\rightarrow\mathbb{R}$ be a strictly monotonous function such that $\int_a^b f(x) dx=0$. Show that $f(a)\cdot f(b)<0$.
[*] Find all convergent sequences $(a_n)_{n\geq 1}$ for which there exists a scrictly monotonous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$\int_{a_{n-1}}^{a_n} f(x)dx = \int_{a_n}^{a_{n+1}} f(x)dx,\text{ for all }n\geq 2.$$
7 replies
AndreiVila
Mar 8, 2025
kamatadu
35 minutes ago
Putnam 2014 A4
Kent Merryfield   36
N an hour ago by bjump
Suppose $X$ is a random variable that takes on only nonnegative integer values, with $E[X]=1,$ $E[X^2]=2,$ and $E[X^3]=5.$ (Here $E[Y]$ denotes the expectation of the random variable $Y.$) Determine the smallest possible value of the probability of the event $X=0.$
36 replies
Kent Merryfield
Dec 7, 2014
bjump
an hour ago
Sequence interesting problem
vickyricky   10
N an hour ago by kamatadu
Let $ a_{0} =1$ and $ b_{0} =1$ . Define $a_{n} , b_{n} $ for $ n\ge 1$ as $a_{n}=a_{n-1}+2b_{n-1} $ , $b_{n}=a_{n-1}+b_{n-1} $ . Prove that $\lim_{n\to\infty}\frac{a_n}{b_n}$ exists and find it's value .
10 replies
vickyricky
Jun 6, 2020
kamatadu
an hour ago
Very nice Darboux epsilon-delta
CatalinBordea   1
N an hour ago by QQQ43
Source: Romanian National Olympiad 2000, Grade XI, Problem 4
Let $ f:\mathbb{R}\longrightarrow\mathbb{R} $ be a function that satisfies the conditions:
$ \text{(i)}\quad \lim_{x\to\infty} (f\circ f) (x) =\infty =-\lim_{x\to -\infty} (f\circ f) (x) $
$ \text{(ii)}\quad f $ has Darboux’s property

a) Prove that the limits of $ f $ at $ \pm\infty $ exist.
b) Is possible for the limits from a) to be finite?
1 reply
CatalinBordea
Oct 2, 2018
QQQ43
an hour ago
Most Evil and Brutal Integral Ever Officially Proposed for an Integration Bee
Silver08   3
N 2 hours ago by Silver08
Source: UK University Integration Bee 2024-2025 Round 2 Relay (Singapore)
Compute: \[ \int_{1}^{2}e^{x( x+\sqrt{x^2-1} )}dx \]
3 replies
Silver08
Mar 6, 2025
Silver08
2 hours ago
IMC 2018 P4
ThE-dArK-lOrD   17
N 4 hours ago by sangsidhya
Source: IMC 2018 P4
Find all differentiable functions $f:(0,\infty) \to \mathbb{R}$ such that
$$f(b)-f(a)=(b-a)f’(\sqrt{ab}) \qquad \text{for all}\qquad a,b>0.$$
Proposed by Orif Ibrogimov, National University of Uzbekistan
17 replies
ThE-dArK-lOrD
Jul 24, 2018
sangsidhya
4 hours ago
f(x)=x-xe^(-1/x)
Sayan   6
N 4 hours ago by kamatadu
Source: ISI (BS) 2006 #6
(a) Let $f(x)=x-xe^{-\frac1x}, \ \ x>0$. Show that $f(x)$ is an increasing function on $(0,\infty)$, and $\lim_{x\to\infty} f(x)=1$.

(b) Using part (a) or otherwise, draw graphs of $y=x-1, y=x, y=x+1$, and $y=xe^{-\frac{1}{|x|}}$ for $-\infty<x<\infty$ using the same $X$ and $Y$ axes.
6 replies
Sayan
Jun 2, 2012
kamatadu
4 hours ago
Square of a rational matrix of dimension 2
loup blanc   7
N Today at 11:53 AM by ysharifi
The following exercise was posted -two months ago- on the Website StackExchange; cf.
https://math.stackexchange.com/questions/5006488/image-of-the-squaring-function-on-mathcalm-2-mathbbq
There was no solution on Stack.

-Statement of the exercise-
We consider the matrix function $f:X\in M_2(\mathbb{Q})\mapsto X^2\in M_2(\mathbb{Q})$.
Find the image of $f$.
In other words, give a method to decide whether a given matrix has or does not have at least a square root
in $M_2(\mathbb{Q})$; if the answer is yes, then give a method to calculate at least one of its roots.
7 replies
loup blanc
Feb 17, 2025
ysharifi
Today at 11:53 AM
find the isomorphism
nguyenalex   10
N Today at 11:38 AM by Royrik123456
I have the following exercise:

Let $E$ be an algebraic extension of $K$, and let $F$ be an algebraic closure of $K$ containing $E$. Prove that if $\sigma : E \to F$ is an embedding such that $\sigma(c) = c$ for all $c \in K$, then $\sigma$ extends to an automorphism of $F$.

My attempt:

Theorem (*): Suppose that $E$ is an algebraic extension of the field $K$, $F$ is an algebraically closed field, and $\sigma: K \to F$ is an embedding. Then, there exists an embedding $\tau: E \to F$ that extends $\sigma$. Moreover, if $E$ is an algebraic closure of $K$ and $F$ is an algebraic extension of $\sigma(K)$, then $\tau$ is an isomorphism.

Back to our main problem:

Since $K \subset E$ and $F$ is an algebraic extension of $K$, it follows that $F$ is an algebraic extension of $E$. Assume that there exists an embedding $\sigma : E \to F$ such that $\sigma(c) = c$ for all $c \in K$. By Theorem (*), there exists an embedding $\tau : F \to F$ that extends $\sigma$. Since $F$ is algebraically closed, $\tau(F)$ is also an algebraically closed field.

Furthermore, because $\sigma(c) = c$ for all $c \in K$ and $\tau$ is an extension of $\sigma$, we have
$$K = \sigma(K) \subset K \subset \sigma(E) \subset \tau(F) \subset F.$$
This implies that $F$ is an algebraic extension of $\tau(F)$. We conclude that $F = \tau(F)$, meaning that $\tau$ is an automorphism. (Finished!!)

Let choose $F = A$ be the field of algebraic numbers, $K=\mathbb{Q}$. Consider the embedding $\sigma: \mathbb{Q}(\sqrt{2}) \to \mathbb{Q}(\sqrt{2}) \subset A$ defined by
$$
a + b\sqrt{2} \mapsto a - b\sqrt{2}.
$$Then, according to the exercise above, $\sigma$ extends to an isomorphism
$$
\bar{\sigma}: A \to A.
$$How should we interpret $\bar{\sigma}$?
10 replies
nguyenalex
Yesterday at 3:58 PM
Royrik123456
Today at 11:38 AM
Generating functions and recursions smelling from 1000 km
Assassino9931   12
N Today at 11:22 AM by sangsidhya
Source: IMC 2022 Day 1 Problem 3
Let $p$ be a prime number. A flea is staying at point $0$ of the real line. At each minute,
the flea has three possibilities: to stay at its position, or to move by $1$ to the left or to the right.
After $p-1$ minutes, it wants to be at $0$ again. Denote by $f(p)$ the number of its strategies to do this
(for example, $f(3) = 3$: it may either stay at $0$ for the entire time, or go to the left and then to the
right, or go to the right and then to the left). Find $f(p)$ modulo $p$.
12 replies
Assassino9931
Aug 5, 2022
sangsidhya
Today at 11:22 AM
ISI 2018 #3
integrated_JRC   34
N Today at 7:08 AM by anudeep
Source: ISI 2018 B.Stat / B.Math Entrance Exam
Let $f:\mathbb{R}\to\mathbb{R}$ be a continuous function such that for all $x\in\mathbb{R}$ and for all $t\geqslant 0$, $$f(x)=f(e^tx)$$Show that $f$ is a constant function.
34 replies
integrated_JRC
May 13, 2018
anudeep
Today at 7:08 AM
Integration Bee Kaizo
Calcul8er   40
N Today at 7:07 AM by Figaro
Hey integration fans. I decided to collate some of my favourite and most evil integrals I've written into one big integration bee problem set. I've been entering integration bees since 2017 and I've been really getting hands on with the writing side of things over the last couple of years. I hope you'll enjoy!
40 replies
Calcul8er
Mar 2, 2025
Figaro
Today at 7:07 AM
Weird family of sequences
G H J
G H BBookmark kLocked kLocked NReply
Source: Romanian District Olympiad 2025 12.3
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AndreiVila
183 posts
#1
Y by
  1. Let $a<b$ and $f:[a,b]\rightarrow\mathbb{R}$ be a strictly monotonous function such that $\int_a^b f(x) dx=0$. Show that $f(a)\cdot f(b)<0$.
  2. Find all convergent sequences $(a_n)_{n\geq 1}$ for which there exists a scrictly monotonous function $f:\mathbb{R}\rightarrow\mathbb{R}$ such that $$\int_{a_{n-1}}^{a_n} f(x)dx = \int_{a_n}^{a_{n+1}} f(x)dx,\text{ for all }n\geq 2.$$
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Alphaamss
214 posts
#2
Y by
For $(a)$. Otherwise, $f(a)f(b)\geq0$, which implies $f(b)>f(a)\geq0$ or $f(a)<f(b)\leq0$. If $f(b)>f(a)\geq0$, then $f(x)>0,\forall x\in(a,b]$ and
$$0=\int_a^bf(x)dx\geq\int_{\frac{a+b}2}^bf(x)dx\geq\int_{\frac{a+b}2}^bf\left(\frac{a+b}2\right)dx=f\left(\frac{a+b}2\right)\cdot\frac{b-a}2>0,$$which is impossible.
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Levieee
118 posts
#3
Y by
for $(a.)$ another easy and cute method would be, make 3 cases
$\text{Case 1:}$ $f(a) > f(b) \geq 0$
$\text{Case 2:}$ $f(a)<f(b) \leq 0$
$\text{Case 3:}$ $f(a)<0<f(b)$
$\text{for case 1 and 2 we can get contradiction and we can show case 3 works}$

for $b.$ my guess is constant sequence but i cant get anything to prove it :( :( :(

edit: i did it :D :wow:
$\textbf{Sketch:}$ If \( f \) is a strictly monotonic function, then if
$$
\int_{a}^{b} f(x) \,dx = 0,
$$we have
$$
f(a) f(b) < 0.
$$
$$
\int_{a_i}^{a_j} f(x) \,dx = 0  \text{ for any }  a_i < a_j.
$$
Taking \( a_i < a_j < a_k \), we get
$$
f(a_i) f(a_j) < 0, \quad f(a_i) f(a_k) < 0, \quad f(a_j) f(a_k) < 0
$$which implies (\(\rightarrow \leftarrow\)).

I initially thought that only constant sequences would work, but $\textbf{@rosemarys\_baby}$ pointed out another type of sequence: sequences that are constant up to some \( n \), differ in the next term, and remain constant from there.

A counterexample to the claim that only constant sequences work is:
counter example
This post has been edited 9 times. Last edited by Levieee, an hour ago
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KAME06
111 posts
#4
Y by
Idk if it's ok, otherwise, pls tell me the mistake.
WLOG, consider an increasing function. The other case is analogous.
a)Using Rolle's Theorem in $F(x)$, there exist a $k$ such $f(k)=0$. If $a<k<b$ we are done.
If $k \le a<b$, $0 \le f(a)<f(b) \Rightarrow 0 \le f(x) \Rightarrow 0<\int^a_b f(x) dx$ because is strictly monotonous and positive function. Contradiction. Analogous, $a < b \le 0$.
b)Suppose that $(a_n)_{n \ge 1}$ works. Let $b_n=\int^{a_{n}}_{a_{n-1}}f(x)dx$. As $b_2=b_3=...$, $(b_n)$ converges. Let $L_b=\lim_{n \rightarrow \infty} b_n$ and $L_a=\lim_{n \rightarrow \infty }a_n$.
Notice that $L_b=\lim_{n \rightarrow \infty} b_n =\lim_{n \rightarrow \infty} \int^{a_{n}}_{a_{n-1}}f(x)dx = \int^{L_a}_{L_a}f(x)dx =0$.
As $b_2=b_3=...$, we have that $b_n=L_b=0$ for $n>1$.
Supose that there exist $a_i, a_j, a_k$ such $a_i < a_j <a_k$ Using (a):$f(a_i)f(a_j)<0$ and $f(a_j)f(a_k)<0$. But, as $f$ is strictly monotonous function, $f(a_i)<f(a_j)<f(a_k)$. Contradiction. So $(a_n)$ can be at most two possible values. Let's call them $c, d \in \mathbb{R}$. As $a_n$ converges, WLOG, there exist a $N$ such that, for all $n \ge N$, $a_n=c$.
If $a_n$ is a sequence such that $a_n \in \{c, d\}$ and there exist a $N$ such that, for all $n \ge N$, $a_n=c$, we have got two possibilities:
-If $a_i=a_{i+1}$, $\int_{a_{i}}^{a_{i+1}} f(x)dx=0$
-If $a_i=c, a_{i+1}=d$, consider an increasing linear function $f(x)$ that pass through the midpoint of $CD$ (let's call it $m$) and $(c, f(c)); (d, f(d))$ (let's call them $C', D'$, respectively) are the intersection of the linear function and the perpendicular to the x-axis that pass through $c$ and $d$, respectively. By $ALA$, easy to see that $\triangle MCC' \cong \triangle MDD'$, so $\int_{a_{i}}^{a_{i+1}} f(x)dx=0$.
With that construction, the recurrence works.
Answer: All sequences that $a_n \in \{c, d\}$ and there exist a $N$ such that, for all $n \ge N$, $a_n=c$.
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Filipjack
814 posts
#5
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@above: in part a) you cannot use Rolle's Theorem because the function $\int\limits_c^x f(t) \mathrm{d}t$ is not necessarily differentiable. It is however continuous, so your solution for part b) is fine (this was important for the calculation of $\lim_{n \to \infty} b_n$).
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Levieee
118 posts
#6 • 1 Y
Y by KAME06
Filipjack wrote:
@above: in part a) you cannot use Rolle's Theorem because the function $\int\limits_c^x f(t) \mathrm{d}t$ is not necessarily differentiable. It is however continuous, so your solution for part b) is fine (this was important for the calculation of $\lim_{n \to \infty} b_n$).

doesn't FTOC state that if a function is continuous at $c_{1}$ then $F(x)$ must be differentiable at $c_{1}$ and $F'(c_{1})=f(c_{1})$ $c_{1} \in [a,b]$


@below my bad i didnt read that continuity wasn't given
This post has been edited 2 times. Last edited by Levieee, 5 hours ago
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Filipjack
814 posts
#7
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Levieee wrote:
doesn't FTOC state that if a function is continuous at $c_{1}$ then $F(x)$ must be differentiable at $c_{1}$ and $F'(c_{1})=f(c_{1})$ $c_{1} \in [a,b]$

The thing is: we are not given that $f$ is continuous, so you cannot apply that.

Take for example $f(x)= \begin{cases} x-1, & x \le 0 \\ x+1, & x>0 \end{cases}$ and $F(x)=\int\limits_c^x f(t) \mathrm{d}t,$ for some $c.$ For this we have $F(x)=x^2+|x|-c^2-|c|,$ which is not differentiable in $0.$ Notice also that there is no point $s$ such that $f(s)=0.$
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kamatadu
463 posts
#9
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KAME06 wrote:
Idk if it's ok, otherwise, pls tell me the mistake.

Your answer for part (b) matches with mine. So I guess it's correct. Will add mine a bit later.
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