I am interested in theorems with unexpected conclusions. I don't mean an unintuitive result (like the existence of a space-filling curve), but rather a result whose conclusion seems disconnected from the hypotheses. My favorite is the following. Let $f(n)$ be the number of ways to write the nonnegative integer $n$ as a sum of powers of 2, if no power of 2 can be used more than twice. For instance, $f(6)=3$ since we can write 6 as $4+2=4+1+1=2+2+1+1$. We have $(f(0),f(1),\dots) = $ $(1,1,2,1,3,2,3,1,4,3,5,2,5,3,4,\dots)$. The conclusion is that the numbers $f(n)/f(n+1)$ run through all the reduced positive rational numbers exactly once each. See A002487 in the On-Line Encyclopedia of Integer Sequences for more information. What are other nice examples of "unexpected conclusions"?
-
4$\begingroup$ I'll send you a shirt: thenerdiestshirts.com/blog/math-shirt-cw/#more-116 $\endgroup$– Douglas ZareMar 13, 2010 at 21:48
-
2$\begingroup$ Thanks for great shirt! I guess it's difficult to make precise the difference between "unintuitive" and "unexpected conclusion." Some examples are at math.dartmouth.edu/~pw/solutions.pdf. Another nice example is the cake icing problem (demonstrations.wolfram.com/TheCakeIcingPuzzle). $\endgroup$– Richard StanleyMar 13, 2010 at 22:47
-
12$\begingroup$ Stolen from Kevin Buzzard's comment at mathoverflow.net/questions/15050/linear-algebra-problems: If A and B are real n x n matrices, A^2+B^2=AB, and AB-BA is invertible, then n is a multiple of 3. $\endgroup$– Jonas MeyerMar 13, 2010 at 23:14
33 Answers
The classical differential geometry results should definitely be mentioned here. Although it may seem not surprising for us, Gauss found his Theorema Egregium to be truly remarkable and unexpected. My favorite example is Gauss-Bonnet Theorem.
Given a $C^{\infty}$ function $f(x)$, let $\Delta_n$ be the difference between the integral of $f(x)$ and its $n$'th Riemann sum: $$\Delta_n = \int_0^1 f ~dx ~-~ \sum_{i=1}^{n} f(i/n) \frac{1}{n}$$ Clearly, $\Delta_n$ goes to $0$, but at what rate? Its easy to see that its possible for $\Delta_n$ to decay as $\Theta(1/n)$, e.g consider what happens with $f(x)=x$.
Theorem: If $f(x)$ is periodic with period $1$, then $\Delta_n$ decays faster than any polynomial in $n$.
-
$\begingroup$ It looks surprisingly! But estimations from the cited paper includes the uniform norm of $n$-th derivative, so I could not see... $\endgroup$– PetyaMar 14, 2010 at 22:32
-
$\begingroup$ Right - upper bounds on $\Delta_n$ have to depend on the function $f$ somehow. Here is a restatement of my comment in messier way that, however, makes this dependence explicit, and hopefully answers your question. For an arbitrary $C^{\infty}$ function $f$, one can prove an upper bound of $\Delta_n \leq C(f)/n$, where $C(f)$ is some functional. In fact, its obvious that $C(f) = \max_{x \in [0,1]} f'(x)$ works. The example of $f(x)=x$ shows that if $k>1$ then one cannot prove a bound of the form $\Delta_n \leq C_k(f)/n^k$. $\endgroup$– alexMar 15, 2010 at 0:02
-
$\begingroup$ ...by contrast, if $f(0)=f(1)$, then a bound of the form $\Delta_n \leq C_k(f)/n^k$ can be proven for any $k>1$. Thus in this case $\Delta_n$ decays faster than $1/n^k$ for any $k$. $\endgroup$– alexMar 15, 2010 at 0:05
-
2$\begingroup$ You need more than that $f(0)=f(1)$. For $x(1-x)$ the error is quadratic. Perhaps you want $f$ to be periodic with period $1$, and still smooth at $0$. $\endgroup$ Mar 15, 2010 at 0:57
-
1$\begingroup$ The link www-math.mit.edu/~davis/dws08.pdf given in the current revision of the answer no longer works. (And I did not find it using Wayback Machine either.) However, after all this time it seems unlikely that somebody would still know what was the linked text. $\endgroup$ Jun 12, 2022 at 9:29