# Theorems with unexpected conclusions [closed]

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"?

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## closed as no longer relevant by Felipe Voloch, Bill Johnson, Andrés E. Caicedo, Mark Sapir, Ryan BudneyJan 5 '12 at 21:50

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Very interesting question! But since it has no right answer, and you are asking for a big list, I think it should be community wiki. – Grétar Amazeen Mar 13 '10 at 21:32
I'll send you a shirt: thenerdiestshirts.com/blog/math-shirt-cw/#more-116 – Douglas Zare Mar 13 '10 at 21:48
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). – Richard Stanley Mar 13 '10 at 22:47
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. – Jonas Meyer Mar 13 '10 at 23:14

Eckmann-Hilton argument. I mean, WHY?

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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 $n$.

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It looks surprisingly! But estimations from the cited paper includes the uniform norm of $n$-th derivative, so I could not see... – Petya Mar 14 '10 at 22:32
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 $\Delta_n \leq C_k(f)/n^k$. – alex Mar 15 '10 at 0:02
...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 $k$. – alex Mar 15 '10 at 0:05
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 $0$. – Douglas Zare Mar 15 '10 at 0:57
...er, yes. Serves me right for carelessly replacing the condition "$f$ is periodic" with "$f(0)=f(1)$." – alex Mar 15 '10 at 1:15

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.

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