"Embarrassing" open problems in maths

http://angryfaic.wordpress.com/2011/02/26/embarrassing-open-problems-in-maths

by ColinWright • 15 years ago

90 24 15 years ago

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tzs• 15 years ago

Here's an open problem that is not hard to understand (high school math is sufficient), but whose solution would be one of the most important advances in mathematics in the last 100 years, easily:

    Let H_n = 1 + 1/2 + 1/3 + ... + 1/n

So, H_1 = 1, H_2 = 1 + 1/2, and so on.

    Let divsum(n) = the sum of the divisors of n

For example, divsum(10) = 1 + 2 + 5 + 10 = 18, and divsum(8) = 1 + 2 + 4 + 8 = 15.

Here's the problem:

    Prove that for n > 1,
      divsum(n) < H_n + exp(H_n)*log(H_n)

It turns out that this is equivalent to the Riemann hypothesis. Their equivalence is shown here: http://arxiv.org/pdf/math/0008177v2

robryan• 15 years ago

The problem makes enough sense in what it's trying to prove, I can't really understand it though without knowing what it would be used for.

michael_dorfman• 15 years ago

Used for?

Many problems in math have no intended practical uses (although uses, funnily enough, tend to turn up in time.)

Knowing what it might be used for really doesn't usually offer any special insight into the nature of the problem, nor how to solve it.

grifter• 15 years ago

Interestingly enough this is most often the case with pure mathemetics: their proofs are pursued purely in the interest of understanding & attaining them - the implications & applications of which are sometimes unrealized for decades (a great example is Evariste Galois: his number theory and symmetry/group work was decades-later applied by physicists to define many of the laws physics appear to obey - check out the wiki).

mbateman• 15 years ago

A novel proof involves making novel connections, rigorously showing how different parts of mathematics can be made to work together in ways that haven't been thought of. When you look at it that way it's not hard to imagine the value in solving puzzles like this, even if it's impossible to know the specifics ahead of time (which are by nature unknown).

vecter• 15 years ago

http://en.wikipedia.org/wiki/Riemann_hypothesis#Consequences...

splicer• 15 years ago

For the first problem, he goes to the trouble of defining "graph" and "tree", as though his audience has no math background, but doesn't bother defining "labeled". For those wondering WTF an "unlabeled tree" is: http://mathworld.wolfram.com/LabeledGraph.html

splat2• 15 years ago

I'm reminded of two other "embarrassing" open problems.

A perfect number is a number whose divisors sum to return the original number. Thus, 6 is a perfect number since 1 + 2 + 3 = 6. 28 is also a perfect number since 1 + 2 + 4 + 7 + 14 = 28. But 8 is not a perfect number since 1 + 2 + 4 = 7.

1. Are there an infinite number of perfect numbers?

2. Are there any odd perfect numbers?

yequalsx• 15 years ago

A slight nitpick. Your definition should be a number whose proper divisors sum to the original number. Or it should be whose divisors sum to twice the original number.

splat2• 15 years ago

Point taken, good catch.

onemoreact• 15 years ago

These are not simple problems.

If we consider labelled trees then this question has any easy solution: there are labelled trees on n vertices. Counting the total number of graphs on n vertices is even easier. However, no closed form solution is known for unlabelled trees and this problem has been open for well over a century.

Hmm, any node can be a root node which is tricky, but not show stopper. A root node can have 1 to N-1 branches not so bad. You can order them by some sort of weight with each leaf being (depth * 2N)^N so they are strictly ordered, still looking good. But, how do you count mirror images. AKA a simple star with a center point and N edges just counts as 1.

g___• 15 years ago

Related post: Mathematical Embarassments http://rjlipton.wordpress.com/2009/12/26/mathematical-embarr...

schmittz• 15 years ago

A slight typo in the author's bit (which is fascinating), the Collatz sequence doesn't terminate in a series of 1's. By its definition, once 1 is reached, the series repeats {1,4,2,1,4....}. Thus, the number of steps required to reach 1 is called the "stopping time" of the sequence. See the Wikipedia article on the Collatz conjecture for some great visualizations of low seed value termination trajectories for the Collatz sequence:

http://en.wikipedia.org/wiki/Collatz_conjecture

hendzen• 15 years ago

Obviously e and pi are irrational. But it has not yet been proved that e + pi is irrational.

bobby07• 15 years ago

I have just found and entry on OEIS for the general formula for the first problem. http://oeis.org/A000055

G.f.: A(x) = 1 + T(x)-T^2(x)/2+T(x^2)/2, where T(x) = x + x^2 + 2*x^3 + … is g.f. for A000081

I havent checked it, I presume its wrong tho?

Someone• 15 years ago

The original article asks for a closed-form solution. "Closed-form" is not precisely defined (http://en.wikipedia.org/wiki/Closed-form_expression, http://...), but I do not think anybody would call that a closed-form solution.

"Embarrassing" open problems in maths

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