Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2000 USAMO Problems/Problem 1"

(→‎Solution 3)
(→‎Solution 3)
Line 38: Line 38:
 
<cmath>\text{softmax}_\tau(x,y)=\frac{xe^{x/\tau}+ye^{y/\tau}}{e^{x/\tau}+e^{y/\tau}}.</cmath> As we take <math>\tau\rightarrow 0^+</math>, observe that this converges and is strictly bounded above by <cmath>\text{max}(x,y)=\frac{x+y}{2}+\frac{|x-y|}{2}<\frac{x+y}{2}+|x-y|</cmath> so such a function does not exist.  
 
<cmath>\text{softmax}_\tau(x,y)=\frac{xe^{x/\tau}+ye^{y/\tau}}{e^{x/\tau}+e^{y/\tau}}.</cmath> As we take <math>\tau\rightarrow 0^+</math>, observe that this converges and is strictly bounded above by <cmath>\text{max}(x,y)=\frac{x+y}{2}+\frac{|x-y|}{2}<\frac{x+y}{2}+|x-y|</cmath> so such a function does not exist.  
  
−
We also have a geometric argument. The graph of softmax function is a symmetric one-to-one function with two mirrored tails reflected across the <math>x=y</math> plane (think same behavior like <math>xy=1</math> 45 degree rotated hyperbola). Bringing those tails closer together would mean the that the graph is no longer a function, being both not one-to-one yet symmetric—a contradiction (think a 45 degree rotated hyperbola with acute axes).  
+
We also have a geometric argument. The graph of softmax function is a symmetric one-to-one function with two mirrored tails reflected across the <math>x=y</math> plane (think same behavior like <math>xy=1</math> 45 degree rotated hyperbola). Bringing those tails closer together would mean the that the graph is no longer a function, being both not one-to-one yet symmetric—a contradiction (think a 45 degree rotated hyperbola with acute axes). In fact this geometry is known as a non-functional manifold with a kink (non-differentiable) at x=y.
  
 
~Lopkiloinm
 
~Lopkiloinm

Revision as of 02:56, 30 June 2025

Problem

Call a real-valued function $f$ very convex if

\[\frac {f(x) + f(y)}{2} \ge f\left(\frac {x + y}{2}\right) + |x - y|\]

holds for all real numbers $x$ and $y$. Prove that no very convex function exists.

Solution

Solution 1

Let $y \ge x$, and substitute $a = x, 2b = y-x$. Then a function is very convex if $\frac{f(a) + f(a+2b)}{2} \ge f(a + b) + 2b$, or rearranging,

\[\left[\frac{f(a+2b)-f(a+b)}{b}\right]-\left[\frac{f(a+b)-f(a)}{b}\right] \ge 4\]

Let $g(a) = \frac{f(a+b) - f(a)}{b}$, which is the slope of the secant between $(a,f(a))(a+b,f(a+b))$. Let $b$ be arbitrarily small; then it follows that $g(a+b) - g(a) > 4$, $g(a+2b) - g(a+b) > 4,\, \cdots, g(a+kb) - g(a+ [k-1]b) > 4$. Summing these inequalities yields $g(a+kb)-g(a) > 4k$. As $k \rightarrow \infty$ (but $b << \frac{1}{k}$, so $bk < \epsilon$ is still arbitrarily small), we have $\lim_{k \rightarrow \infty} g(a+kb) - g(a) = g(a + \epsilon) - g(a) > \lim_{k \rightarrow \infty} 4k = \infty$. This implies that in the vicinity of any $a$, the function becomes vertical, which contradicts the definition of a function. Hence no very convex function exists.


Solution 2

Suppose, for the sake of contradiction, that there exists a very convex function $f.$ Notice that $f(x)$ is convex if and only if $f(x) + C$ is convex, where $C$ is a constant. Thus, we may set $f(0) = 0$ for convenience.

Suppose that $f(1) = A$ and $f(-1) = B.$ By the very convex condition, \[\frac{f(0) + f\left(2^{-n}\right)}{2} \ge f\left(2^{-(n+1)}\right) + \frac{1}{2^n}.\] A straightforward induction shows that: \[f\left(2^{-n}\right) \le \frac{A - 2n}{2^n}\] for all nonnegative integers $n.$ Using a similar line of reasoning as above, \[f\left(-2^{-n}\right) \le \frac{B - 2n}{2^n}.\] Therefore, for every nonnegative integer $n,$ \[f\left(2^{-n}\right) + f\left(-2^{-n}\right) \le \frac{A+B-4n}{2^n}.\] Now, we choose $n$ large enough such that $n > \frac{A+B}{4} - 1.$ This is always possible because $A$ and $B$ are fixed for any particular function $f.$ It follows that: \[f\left(2^{-n}\right) + f\left(-2^{-n}\right) < \frac{1}{2^{n-2}}.\] However, by the very convex condition, \[f\left(2^{-n}\right) + f\left(-2^{-n}\right) \ge \frac{1}{2^{n-2}}.\] This is a contradiction. It follows that no very convex function exists.

Solution 3

The softmax function can be expressed as \[\text{softmax}_\tau(x,y)=\frac{xe^{x/\tau}+ye^{y/\tau}}{e^{x/\tau}+e^{y/\tau}}.\] As we take $\tau\rightarrow 0^+$, observe that this converges and is strictly bounded above by \[\text{max}(x,y)=\frac{x+y}{2}+\frac{|x-y|}{2}<\frac{x+y}{2}+|x-y|\] so such a function does not exist.

We also have a geometric argument. The graph of softmax function is a symmetric one-to-one function with two mirrored tails reflected across the $x=y$ plane (think same behavior like $xy=1$ 45 degree rotated hyperbola). Bringing those tails closer together would mean the that the graph is no longer a function, being both not one-to-one yet symmetric—a contradiction (think a 45 degree rotated hyperbola with acute axes). In fact this geometry is known as a non-functional manifold with a kink (non-differentiable) at x=y.

~Lopkiloinm

See Also

2000 USAMO (Problems • Resources)
Preceded by
First Question 		Followed by
Problem 2
1 • 2 • 3 • 4 • 5 • 6
All USAMO Problems and Solutions

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. AMC Logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2000_USAMO_Problems/Problem_1&oldid=252898"