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k a July Highlights and 2025 AoPS Online Class Information
jwelsh   0
Jul 1, 2025
We are halfway through summer, so be sure to carve out some time to keep your skills sharp and explore challenging topics at AoPS Online and our AoPS Academies (including the Virtual Campus)!

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MOP (Math Olympiad Summer Program) just ended and the IMO (International Mathematical Olympiad) is right around the corner! This year’s IMO will be held in Australia, July 10th - 20th. Congratulations to all the MOP students for reaching this incredible level and best of luck to all selected to represent their countries at this year’s IMO! Did you know that, in the last 10 years, 59 USA International Math Olympiad team members have medaled and have taken over 360 AoPS Online courses. Take advantage of our Worldwide Online Olympiad Training (WOOT) courses
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0 replies
jwelsh
Jul 1, 2025
0 replies
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\(\pi_1(X,x_0)\) is abelian if and only if
EstherB   0
2 hours ago
Source: Topology - Munkres (section 52 - problem 3)
Let \( x_0 \) and \( x_1 \) be points of a path-connected space \( X \). Show that \( \pi_1(X, x_0) \) is abelian if and only if for every pair \( \alpha \) and \( \beta \) of paths from \( x_0 \) to \( x_1 \), we have \( \hat{\alpha} = \hat{\beta} \).

1. Suppose \(\pi_1(X,x_0)\) is abelian

We know that if \(X\) is path-connected, then the fundamental groups at any two basepoints are isomorphic: \(\pi_1(X,x_0) \cong \pi_1(X,x_1)\), via any path \(\alpha\) from \(x_0\) to \(x_1\). Since isomorphisms preserve algebraic properties, it follows that:
\[ \text{If } \pi_1(X, x_0) \text{ is abelian, then } \pi_1(X, x_1) \text{ is also abelian}. \]
Let \(f\) be a loop based at \(x_0\), and let \(\alpha, \beta\) be paths from \(x_0\) to \(x_1\). We want to show that:
\[ [\bar{\alpha} * f * \alpha] = [\bar{\beta} * f * \beta] \in \pi_1(X, x_1). \]
\[ [\bar{\alpha} * f * \alpha] = [\bar{\alpha} * f * (\beta * \bar{\beta}) * \alpha]. \]
Rewriting the terms:
\[ [\bar{\alpha} * f * \alpha] = [(\bar{\alpha} * f * \beta)(\bar{\beta} * \alpha)]. \]
Now, we note that:


\item \( \bar{\alpha} * f * \beta \) is a path from \(x_1 \to x_0 \to x_0 \to x_1\), i.e., a loop based at \(x_1\);
\item \( \bar{\beta} * \alpha \) also goes from \(x_1 \to x_0 \to x_1\), so it is another loop based at \(x_1\).


Thus, both belong to the group \(\pi_1(X, x_1)\), which is abelian by hypothesis. Then:
\[ [\bar{\alpha} * f * \alpha] = [\bar{\alpha} * f * \beta] \cdot [\bar{\beta} * \alpha] = [\bar{\beta} * \alpha] \cdot [\bar{\alpha} * f * \beta]. \]
Regrouping again:
\[ [\bar{\beta} * \alpha] \cdot [\bar{\alpha} * f * \beta] = [\bar{\beta} * f * \beta], \]we conclude that:
\[ \hat{\alpha}([f]) = \hat{\beta}([f]), \]therefore \(\hat{\alpha} = \hat{\beta}\).


2. Suppose that \(\hat{\alpha} = \hat{\beta}\) for every pair of paths \(\alpha, \beta : x_0 \to x_1\)

We want to show that \(\pi_1(X,x_0)\) is abelian.

Let \([f], [g] \in \pi_1(X,x_0)\), and let \(\alpha\) be a path from \(x_0\) to \(x_1\). Define:
\[ \beta := f * \alpha. \]Note that \(\beta\) is also a path from \(x_0\) to \(x_1\). By hypothesis, we have:
\[ \hat{\alpha}([g]) = \hat{\beta}([g]) \Rightarrow [\bar{\alpha} * g * \alpha] = [\bar{\beta} * g * \beta]. \]
Thus:

\begin{align*} [\bar{\alpha}] * [g] * [\alpha] &= [\bar{\beta}] * [g] * [\beta] \\ [\beta] * [\bar{\alpha}] * [g] * [\alpha] &= [g] * [\beta] \\ [f] * [g] * [\alpha] &= [g] * [\beta] \\ [f] * [g] &= [g] * [\beta] * [\bar{\alpha}] \\ [f] * [g] &= [g] * [f] \end{align*}
Thus, we conclude that the operation is indeed commutative, so \(\pi_1(X,x_0)\) is abelian.
0 replies
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EstherB
2 hours ago
0 replies
J
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Let \( p : E \to B \) be a covering map, where \( E \) is path-connected and \(
EstherB   0
6 hours ago
Source: Problem 8 of section 54 (topology book by munkres)



Let \( p : E \to B \) be a covering map, where \( E \) is path-connected and \( B \) is simply connected.
Then \( p \) is a homeomorphism.

Since \( p \) is a covering map, it is surjective, continuous, and locally a homeomorphism.

Now we need to show that \( p \) is injective.
Suppose \( p \) is not injective.
Then there exist \( a,b \in E \) with \( a \neq b \), such that \( p(a) = p(b) \).

Since \( E \) is path-connected, there exists a path:
\[ \gamma : [0,1] \to E \quad \text{such that } \gamma(0) = a \text{ and } \gamma(1) = b \]
Consider the projection of this path in \( B \).
Let \( \alpha = p \circ \gamma : [0,1] \to B \).
Then \( \alpha \) is a path in \( B \) with
\[ \alpha(0) = p(a) = c = p(b) = \alpha(1) \]
That is, \( \alpha \) is a loop based at \( c \).
Since \( B \) is simply connected, the loop \( \alpha \) is homotopic to a point, that is, to the constant loop at \( c \) (\( \alpha \simeq k_c \)).

By the homotopy lifting theorem, we can lift the homotopy \( \alpha \simeq k_c \) to a homotopy \( \widetilde{H} \) in \( E \) starting at \( a \in E \).
Therefore, the path \( \gamma \) in \( E \) (which satisfies \( p \circ \gamma = \alpha \)) is homotopic to a constant path in \( E \) starting at \( a \).
But since \( \gamma(1) = b \), and the fixed-endpoint homotopy from \( \gamma \) to the constant path at \( a \) must end at \( a \), it follows that \( \gamma(1) = a = b \).

Thus, \( p \) is injective.

Our goal is to show that \( p^{-1}: B \to E \) is continuous.

Since \( p \) is locally a homeomorphism, for each point \( x \in E \) there exists an open set \( U_x \subseteq E \) containing \( x \) such that the restriction
\[ p|_{U_x} : U_x \to p(U_x) \]is a homeomorphism.

Since \( p \) is bijective, the family \( \{ p(U_x) \}_{x \in E} \) forms an open cover of \( B \), and the inverse function \( p^{-1} \) is well-defined on \( B \). Furthermore, on each open set \( p(U_x) \), the restriction \( p^{-1}|_{p(U_x)} \) is continuous because it is the inverse of a homeomorphism.

Since \( B \) can be covered by open sets \( \{ p(U_x) \} \) where \( p^{-1} \) is continuous, it follows that \( p^{-1} \) is continuous on all of \( B \).

Therefore, \( p \) is a homeomorphism.



0 replies
EstherB
6 hours ago
0 replies
J
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analysis
We2592   0
Today at 12:13 PM
Q) find the value of the integration $I=\int_{a}^{b} \frac{e^{-|x|}}{1+(sinhx)^2}$
0 replies
We2592
Today at 12:13 PM
0 replies
J
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about a question
Froster   0
Today at 9:31 AM
Request for Help with This Integral on Aops Online

Hey everyone on Aops Online!

I'm really stuck on computing this integral:

\int_{-a}^{a} \frac{\sqrt{a^4 + (a^2 + b^2)x^2}}{a\sqrt{a^2 + x^2}} \, dx

where a and b are constants. I've tried simplifying the integrand, using substitution (like letting x = a\tan\theta for the \sqrt{a^2 + x^2} part), and checking if it's an even function to simplify the interval (since replacing x with -x keeps the integrand the same, so it becomes 2\int_{0}^{a} \cdots dx). But I still can't get to a clear result.

Has anyone dealt with a similar integral before? Any tips on substitution, simplifying the square - root terms, or recognizing a standard integral form here? Thanks a ton for your help!
0 replies
Froster
Today at 9:31 AM
0 replies
J
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Estimating the Density
zqy648   1
N Today at 8:32 AM by watery
Source: 2024 May 谜之竞赛-6
Given non-empty subset \( I \) of the set of positive integers, a positive integer \( n \) is called good if for every prime factor \( p \) of \( n \), \( \nu_p(n) \in I \). For a positive real number \( x \), let \( S(x) \) denote the number of good numbers not exceeding \( x \).

Determine all positive real numbers \( C \) and \( \alpha \) such that \( \lim\limits_{x \to +\infty} \dfrac{S(x)}{x^\alpha} = C. \)

Proposed by Zhenqian Peng, High School Affiliated to Renmin University of China
1 reply
zqy648
Jul 13, 2025
watery
Today at 8:32 AM
J
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Great Use of Weyl Distribution
zqy648   1
N Today at 7:26 AM by EthanWYX2009
Source: 2025 March 谜之竞赛-6
Prove that for any real number \(\varepsilon > 0\), there exists a positive integer \(N\) such that for any prime \(p > N\) and any primitive root \(g\) modulo \(p\), if we define the set
\[ S = \left\{(i, j) \mid 1 \leq i < j \leq p - 1, \left\{ \frac{g^i}{p} \right\} > \left\{ \frac{g^j}{p} \right\} \right\} \]where \(\{x\}\) denotes the fractional part of the real number \(x\), then
\[ \left( \frac{1}{4} - \varepsilon \right) p^2 < |S| < \left( \frac{1}{4} + \varepsilon \right) p^2. \]Created by Zhenyu Dong and Chunji Wang
1 reply
zqy648
Jul 13, 2025
EthanWYX2009
Today at 7:26 AM
J
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NT By Probabilistic Method
EthanWYX2009   3
N Today at 7:20 AM by EthanWYX2009
Source: 2024 March 谜之竞赛-6
Given a positive integer \( k \) and a positive real number \( \varepsilon \), prove that there exist infinitely many positive integers \( n \) for which we can find pairwise coprime integers \( n_1, n_2, \cdots, n_k \) less than \( n \) satisfying
\[\gcd(\varphi(n_1), \varphi(n_2), \cdots, \varphi(n_k)) \geq n^{1-\varepsilon}.\]Proposed by Cheng Jiang from Tsinghua University
3 replies
EthanWYX2009
Jul 14, 2025
EthanWYX2009
Today at 7:20 AM
J
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infinite series
Martin.s   1
N Today at 4:38 AM by Svyatoslav
Can the infinite series

$$\sum\limits_{n=1}^{\infty}\frac{x^n}{n!}\ln(n+\alpha)$$
be expressed in terms of known functions and constants?
1 reply
Martin.s
Jul 13, 2025
Svyatoslav
Today at 4:38 AM
J
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Sequence
centslordm   10
N Today at 3:43 AM by megarnie
Determine the rest \[1,\,9,\,2,\,1,\,8,\,3,\,2,\,7,\,4,\,\ldots \]
10 replies
centslordm
Jun 17, 2025
megarnie
Today at 3:43 AM
J
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Double integral with sqrtx
mudkip42   1
N Yesterday at 9:33 PM by mudkip42
Source: MIT 18.02
Evaluate
\[ \int_0^1 \int_0^{\sqrt{x}} \frac{2xy}{1-y^4} \ dy \ dx. \]
1 reply
mudkip42
Yesterday at 9:32 PM
mudkip42
Yesterday at 9:33 PM
J
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an extremely difficult problem
szxz   1
N Yesterday at 8:30 PM by szxz
One class needs to choose n students to join a competition. The teacher hold n exams in order to choose n students. In each exam, there are no two students who have the same score. The teacher will hold one exam and pick the student whose score is the highest in that exam. Then, the teacher will hold another exam and pick the student whose score is the highest in this exam from the remaining students. In this way, the teacher will choose n students. Obviously, the order that the teacher hold the exams will affect the students will be chosen. So, how many students have the chance to be chosen at most?
For example, when n = 2 and in exam A, Alice is the first, Bob is the second, Tom is the third. In
exam B, Alice is the first, Tom is the second, Bob is the third. If the teacher choose exam A at first and then choose exam B, Alice and Tom will be chosen. If the teacher choose exam B at first and then choose exam A, Alice and Bob will be chosen. So Alice, Bob and Tom have the chance to be chosen.
1 reply
szxz
Yesterday at 8:27 PM
szxz
Yesterday at 8:30 PM
J
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Korea csat problem, so-called “Killer problem”..
darrime627   1
N Yesterday at 4:52 PM by Zennggg
Find the values of \( a \) and \( b \) such that the function \( f(x) \), which has a second derivative for all \( x \in \mathbb{R} \), satisfies the following conditions:

\[ (\gamma) \quad [f(x)]^5 + [f(x)]^3 + ax + b = \ln \left( x^2 + x + \frac{5}{2} \right) \]
\[ (\delta) \quad f(-3) f(3) < 0, \quad f'(2) > 0 \]
1 reply
darrime627
Jul 14, 2025
Zennggg
Yesterday at 4:52 PM
J
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Subset Ordered Pairs of {1, 2, ..., 10}
ahaanomegas   12
N Yesterday at 3:08 AM by mudkip42
Source: Putnam 1990 A6
If $X$ is a finite set, let $X$ denote the number of elements in $X$. Call an ordered pair $(S,T)$ of subsets of $ \{ 1, 2, \cdots, n \} $ $ \emph {admissible} $ if $ s > |T| $ for each $ s \in S $, and $ t > |S| $ for each $ t \in T $. How many admissible ordered pairs of subsets $ \{ 1, 2, \cdots, 10 \} $ are there? Prove your answer.
12 replies
ahaanomegas
Jul 12, 2013
mudkip42
Yesterday at 3:08 AM
J
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Partial sum of Taylor series of e^x
NamelyOrange   2
N Yesterday at 2:09 AM by genius_007
Source: 2022 National Taiwan University STEM Development Program Admissions Test, P1
For positive integer $n$ and positive real $x$, define $e_n(x) = \sum_{k=0}^{n}\frac{x^k}{k!}$.

(a) Prove that $\frac{1}{n!}\le \frac{m^{m-n-1}}{(m-1)!}$ for all integer $n\ge m >1$.

(b) Use (a) to prove that $e_m(x)\le e_n(x)\le e_{m-1}(x)+\frac{x^m}{(m-1)!(m-x)}$ for all integer $n\ge m >1$ and real $0<x<m$.

(c) Use (b) to prove that for all positive real $x$, there exists some positive $L$ such that $e_n(x)<L$ for all positive integer $n$.

(d) Prove that for positive integer $m<n$ and positive real $x,y$ such that $x+y<m$, we have $0\le e_n(x+y)-e_n(x)e_n(y)\le \frac{m^{m-n-1}(x+y)^{n+1}}{(m-1)!(m-x-y)}$.
2 replies
NamelyOrange
Jul 15, 2025
genius_007
Yesterday at 2:09 AM
J
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J
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uniformly continuous of multivariable function
keroro902   1
N May 14, 2025 by Mathzeus1024
How can I determine which of the following functions are uniformly continuous on the given domain A?

$f \left( x, y \right) = \frac{x^3 + y^2}{x^2 + y}$ , $A = \left\{ \left( x, y \right) \in \mathbb m{R}^2 \left| \right. \left| y \right| \leq \frac{x^2}{2} %Error. "nocomma" is a bad command. , x^2 + y^2 < 1 \right\}$

$g \left( x, y \right) = \frac{y^2 + 4 x^2}{y^2 - 4 x^2 - 1}$, $A = \left\{ \left( x, y \right) \in \mathbb m{R}^2 \left| 0 \leq x^2 - y^2 \leqslant 1 \right\} \right.$
1 reply
keroro902
Nov 2, 2012
Mathzeus1024
May 14, 2025
J
uniformly continuous of multivariable function
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keroro902
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keroro902
#1 h  Nov 2, 2012, 4:34 AM • 2 Y
Y by Adventure10, Mango247
How can I determine which of the following functions are uniformly continuous on the given domain A?

$f \left( x, y \right) = \frac{x^3 + y^2}{x^2 + y}$ , $A = \left\{ \left( x, y \right) \in \mathbb m{R}^2 \left| \right. \left| y \right| \leq \frac{x^2}{2} %Error. "nocomma" is a bad command. , x^2 + y^2 < 1 \right\}$

$g \left( x, y \right) = \frac{y^2 + 4 x^2}{y^2 - 4 x^2 - 1}$, $A = \left\{ \left( x, y \right) \in \mathbb m{R}^2 \left| 0 \leq x^2 - y^2 \leqslant 1 \right\} \right.$
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Mathzeus1024
1064 posts
Mathzeus1024
#2 h  May 14, 2025, 1:42 PM
Y by
For $f(x,y)$, the domain $A$ is comprised of a region that is bounded between the parabolas $-\frac{x^2}{2} \le y \le \frac{x^2}{2}$ and the circle $x^2+y^2<1$. Furthermore, $f$ is defined for all $(x,y) \in \mathbb{R}^{2} \backslash \{y=-x^2\}$. However, $y=-x^2$ intersects both $y = \pm\frac{x^2}{2}$ at $(0,0) \in A$, and $\lim_{(x,y) \rightarrow (0,0)} f(x,y) = DNE$ (does not exist). Thus, $f$ is not uniformly continuous over $A$ due to $(x,y)=(0,0)$.

For $g(x,y)$, the domain $A$ is comprised of a region that is bounded between the lines $-x \le y \le x$ and the right branch of the hyperbola $x^2-y^2 \le 1$ (which coincidentally has asymptotes $y = \pm x$). Furthermore, $g$ is defined for all $(x,y) \in \mathbb{R}^{2} \backslash \{y^2-4x^2=1\}$, which the intersection of this latter hyperbola and $A = \varnothing$. Hence, $\lim_{(x,y) \rightarrow (p,q)} g(x,y) = L$ exists for any $(x,y) = (p,q) \in A$, and $g$ is uniformly continuous over $A$.
This post has been edited 12 times. Last edited by Mathzeus1024, May 16, 2025, 4:23 PM
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