The most difficult equation in mathematics

[Demystifier]

What, in your opinion, seems to be (one of) the most difficult equation(s) in mathematics?

Here is my choice:
Find the solution $k$ of the equation
$$2^{\aleph_0}=\aleph_k$$

 

[ShayanJ]

So what about finding k for any given m in 2^{\aleph_m}=\aleph_k?
Seems a nice challenge! Is it possible to give a more difficult equation than a given one?
Anyway, I think no such thing(most difficult equation) exists in mathematics. Or at least its existence isn't a trivial fact!

 

[Demystifier]

So what about finding k for any given m in 2ℵm=ℵk 2^{\aleph_m}=\aleph_k?

It's a trivial generalization of the solution for $m=0$. :D

 

[ShayanJ]

I'm wondering which is harder! Finding k for a given m or finding m for a given k?
Has such a problem been solved in any of its cases?(or maybe in general!)(I was going to write 'considered' instead of 'solved' but I thought it surely is considered by someone! or not?)

 

[Doug Huffman]

Difficult and solved by traditional methods might be Fermat's Last solved by Andrew Wiles. Difficult and unsolved might be nonsense. How is difficulty measured, modus ponens ponendo?

 

[ShayanJ]

Difficult and solved by traditional methods might be Fermat's Last solved by Andrew Wiles. Difficult and unsolved might be nonsense. How is difficulty measured, modus ponens ponendo?

Demystifier used "most difficult equation"! Those are proofs of facts. Of course proving things can become arbitrarily difficult. (EDIT: This is not trivial too!)
And about measuring difficulty. I don't think we need to get rigorous here. If a good mathematician says that something is difficult, Its enough for me to believe that thing is difficult. (I hope you don't ask for a definition of a good mathematician!)

 

[Demystifier]

Has such a problem been solved in any of its cases?(or maybe in general!)(I was going to write 'considered' instead of 'solved' but I thought it surely is considered by someone! or not?)

The problem has been considered a lot. The most important result (by Cohen) is a proof that the problem is unsolvable by using standard axioms of set theory. Different non-standard axioms of set theory may lead to different solutions, but then the problem is how to know which axioms, if any, are the "right" ones?

For more details see
http://en.wikipedia.org/wiki/Continuum_hypothesis

 

[ShayanJ]

but then the problem is how to know which axioms, if any, are the "right" ones?

I think you should use "desired" instead of "right"!
Anyway, why just use 2? We can generalize to n^{\aleph_m}=\aleph_k. Right? Or there is some reason for that 2?

P.S.
Mathematical logic and set theory, or anything which seems that fundamental is really interesting for me but there lots of things to learn, and things with applications to physics have priority. A day will come that I'll learn those!:D

 

[Demystifier]

Or there is some reason for that 2?

Of course there is. A set with $n$ elements has $2^n$ subsets. Since the set of integers has $\aleph_0$ elements, it can be used to show that the set of reals has $2^{\aleph_0}$ elements.

 

[Demystifier]

Mathematical logic and set theory, or anything which seems that fundamental is really interesting for me but there lots of things to learn, and things with applications to physics have priority. A day will come that I'll learn those!:D

For a physicist friendly explanation of some basics in mathematical logic and set theory, including the origin of this basis $2$, see
R. Penrose, The Road to Reality

 

[ShayanJ]

Of course there is. A set with $n$ elements has $2^n$ subsets. Since the set of integers has $\aleph_0$ elements, it can be used to show that the set of reals has $2^{\aleph_0}$ elements.

Looks like I wasn't clear enough. Actually I meant Is it that we can only place 2 as the base, because of some features of aleph numbers, or its possible to put any other number as the base? Does that base have to be a natural number? An integer? Real?

 

[ShayanJ]

For a physicist friendly explanation of some basics in mathematical logic and set theory, including the origin of this basis 22, see
R. Penrose, The Road to Reality

I saw that book before. It was thick enough to make me think it contains much physics but I really didn't expect it to contain such a pure math subject! Thanks.

 

[mathman]

What, in your opinion, seems to be (one of) the most difficult equation(s) in mathematics?

Here is my choice:
Find the solution $k$ of the equation
$$2^{\aleph_0}=\aleph_k$$

The continuum hypothesis says that k = 1. The continuum hypothesis is undecidable (Gödel theorem).

 

[Demystifier]

The continuum hypothesis is undecidable (Gödel theorem).

No, that's the Cohen's theorem. And it is only valid within the standard axioms of set theory, not within all conceivable set-theory axioms.

 

[Demystifier]

Looks like I wasn't clear enough. Actually I meant Is it that we can only place 2 as the base, because of some features of aleph numbers, or its possible to put any other number as the base? Does that base have to be a natural number? An integer? Real?

You can use any basis you like. But if you solve the problem for basis 2 (or for any other fixed basis larger than 1), other bases should not longer be a problem.

 

[Demystifier]

Anyway, this is not a topic about continuum hypothesis. I am sure there are also other candidates for the most difficult equation in mathematics. Any suggestions?

 

[ShayanJ]

Anyway, this is not a topic about continuum hypothesis. I am sure there are also other candidates for the most difficult equation in mathematics. Any suggestions?

Solving any of the equations in this list in the most general case!(Sorry for ruining the chance for others but I want to emphasize my last sentence in post#2!)

 

[bigfooted]

These problems were and are also quite difficult to solve:
http://en.wikipedia.org/wiki/Hilbert's_problems
http://www.claymath.org/millennium-problems

A difficult problem that can be stated in simple terms: construct an algorithm that will find, if it exists, the general analytic solution of a first order ordinary differential equation y'=f(x), with f a specified (analytic) function.

Anyway, I think we can only sum up some unsolved problems in this thread. It is difficult to say 'this one is more difficult than that one, because it requires blabla theory' without a proper definition of what makes a problem 'difficult'.

 

[madness]

This is a bit outside of my area of expertise, but I think this one could be added to the list. http://en.wikipedia.org/wiki/P_versus_NP_problem

Edit: Not sure if this really counts as an equation...

 

[Demystifier]

A difficult problem that can be stated in simple terms: construct an algorithm that will find, if it exists, the general analytic solution of a first order ordinary differential equation y'=f(x), with f a specified (analytic) function.

How about $$y=\int dx\,f(x) + const$$

 

[bigfooted]

How about $$y=\int dx\,f(x) + const$$

You're a genius!
I should have used a slightly more general first order ODE of course: y'=f(x,y).

 

[ShayanJ]

How about $$y=\int dx\,f(x) + const$$

You should still note that no single algorithm exists for finding the integral of a function, even if we exclude integrals which can't be expressed in terms of elementary functions(Oops...its not applicable to such integrals, right?). So finding such an algorithm is an open problem and it seems to be very hard.

 

[Demystifier]

You're a genius!

You are sarcastic. :)

I should have used a slightly more general first order ODE of course: y'=f(x,y).

That's much harder, I admit.

 

[Demystifier]

The solutions of differential equations mentioned above can always be found numerically, to an arbitrary precision. In that sense the differential equations are not really hard. But equation in the first post is really hard, in the sense that one does not even know how to find an approximate solution.

 

[axmls]

I'm sure the standard model Lagrangian is a contender: http://nuclear.ucdavis.edu/~tgutierr/files/sml.pdf

Though someone more well-versed than me may be able to elaborate on it.

 

[ShayanJ]

I'm sure the standard model Lagrangian is a contender: http://nuclear.ucdavis.edu/~tgutierr/files/sml.pdf

Though someone more well-versed than me may be able to elaborate on it.

If you mean varying it w.r.t. to all the dynamical variables to get all the equations of motion, I should say this is only a tedious thing to do. The algorithm for doing it is very well understood(So much that I won't wonder if there is a software able to do that), its just that its complicated and time consuming and messy calculation. So it isn't even a candidate for the hardest problem in math.

 

[BiGyElLoWhAt]

Solving any of the equations in this list in the most general case!(Sorry for ruining the chance for others but I want to emphasize my last sentence in post#2!)

Haha, I don't even know what the solution is that we're looking for in most cases. Would many of these actually require multiple solutions? Some of these PDE's have multiple functions (and differentials thereof) with constraints on the equations; so I guess the first step, at least for me haha, would be to know what the solution would look like.

 

[BiGyElLoWhAt]

wow, thanks @Shyan , now I have yet another thing to eat up any possible free time I might run into (your link -> Heisenberg Ferromagnet -> spin wave -> ferromagnetism -> ? ...) ;)

 

[ShayanJ]

wow, thanks @Shyan , now I have yet another thing to eat up any possible free time I might run into (your link -> Heisenberg Ferromagnet -> spin wave -> ferromagnetism -> ? ...) ;)

When I saw the list the first time, I felt I know nothing!
Well, actually this isn't far from reality!

 

[stevendaryl]

I think you should use "desired" instead of "right"!
Anyway, why just use 2? We can generalize to n^{\aleph_m}=\aleph_k. Right? Or there is some reason for that 2?

Well, if \alpha is infinite, then there is no difference between 2^\alpha and n^\alpha.

The number x^y can be defined as the number of functions from a set with y objects into a set with x objects. If y is infinite, then this number is independent of x, as long as 0 < x < \aleph_0

 

[Adam Rifai]

i think the most difficult equation is
1 = 0.9999999...

 

[PAllen]

Sticking to equations, find a zero of the zeta function that is not a negative even integer, and whose real part is not 1/2. Or show there is no such solution. I mean, this problem looks like it just needs the right algebraic trick, but ...

 

[MisterX]

Here's the "hard" version of that problem.
Find a method that will find an integrating factor for the equation
$$f(x,y)dx + g(x,y)dy = 0 $$
which I believe has been proven to exist but according to Tenenbaum and Pollard p. 83 no general method is known.

Edit: Oh I see I missed that this was essentially posted on the 2nd page well I'll just leave it for the reference anyway.

 

[jk22]

I stumbled about a generalization of the eigenproblem : find the numbers ai and the transformation of coordinates g such that gCg^-1=diag(ai) where C is an operator that is not forcedly linear.

 

[BiGyElLoWhAt]

I stumbled about a generalization of the eigenproblem : find the numbers ai and the transformation of coordinates g such that gCg^-1=diag(ai) where C is an operator that is not forcedly linear.

Is ai a matrix? Or just like a column vector?
Linkage?

 

[zoki85]

1+1=2

 

[jk22]

Is ai a matrix? Or just like a column vector?
Linkage?

I wrote it wrong indeed its gCg^-1(x1,...)=(...,a_i*x_i,...) where the x_i are the coordinates

 

[BiGyElLoWhAt]

Here's the "hard" version of that problem.
Find a method that will find an integrating factor for the equation
$$f(x,y)dx + g(x,y)dy = 0 $$
which I believe has been proven to exist but according to Tenenbaum and Pollard p. 83 no general method is known.

Edit: Oh I see I missed that this was essentially posted on the 2nd page well I'll just leave it for the reference anyway.

Are x and y allowed to be functions of another variable? I'm assuming not, because then the equation would be exact, and you could find a function M(x(t),y(t)) such that dM/dt = f(x,y)dx + g(x,y)dy.

I however don't see why this wouldn't be allowed in solving a differentiable equation (at least from a physics perspective, I think most variable quantities could be expressed as a function of time). Do I win?

 

[BiGyElLoWhAt]

Do I win?

No, no I don't. I think I see the flaw in my logic there haha. It's pretty straight forward for simple functions, but gets really hairy really fast.

 

[jk22]

For $2^{\aleph_0}$can we say it is simply the set of all function from N to {0,1} but such a function is simply described by 0,d1d2... Which is the coding in basis 2 of the real numbers in [0;1] hence the comtinuum and hence equals to $\aleph_1$ ?

 

[mathman]

For $2^{\aleph_0}$can we say it is simply the set of all function from N to {0,1} but such a function is simply described by 0,d1d2... Which is the coding in basis 2 of the real numbers in [0;1] hence the comtinuum and hence equals to $\aleph_1$ ?

Your statement is essentially the continuum hypothesis, that is $C=\aleph_1$.

 

[jk22]

In fact this i think proves 2^aleph0=c but we need to know if c=aleph1 the next cardinal.

But writing aleph0,1,... already makes the hypothesis that the cardinals are countable

 

[pwsnafu]

But writing aleph0,1,... already makes the hypothesis that the cardinals are countable

Err, the ordinals are uncountable, and the cardinals are indexed by the by the ordinals (under AC).
Writing aleph-0, aleph-1, ... is a transfinite sequence, not an ordinary sequence.

 

[jk22]

Well I am not really good at set theory but the cardinals are well ordered but the ch is does there exist a cardinal between N and R ? And that's why cantor created his dust set to try to construct a set inbetween you think ?

 

[jbriggs444]

Assuming the Axiom of Choice, every set has a cardinality and there exists a well ordering of the cardinalities, yes.

The Continuum Hypothesis asserts that there does not exist a cardinal between $|N|$ and $|\mathbb{R}|$.

Whatever Cantor's motivations for defining his ternary set, the fact that the Continuum Hypothesis is undecidable means that it is impossible to construct a set and prove that its cardinality is strictly between $|N|$ and $|\mathbb{R}$. If such a set could be constructed and if such a proof were possible then the Continuum Hypothesis would be decidable.

 

[jk22]

can it be proven that Ch is independent of the others axioms ? If yes then we could decide to set Ch as true or false and get two different set theories ?

In fact my question is stupid since if it were dependent we could prove it from the other axioms

 

[LCKurtz]

Keeping it "simple", how about $y' = f(x,y)$?

 

[Matterwave]

Keeping it "simple", how about $y' = f(x,y)$?

I feel like problems of this type just prompt the definition of special functions (see e.g. Bessel functions).

 

[Dragonfall]

The most difficult equation in math is \mathbf{P} \stackrel{?}{=} \mathbf{NP}

 

[Mentallic]

The most difficult equation in math is \mathbf{P} \stackrel{?}{=} \mathbf{NP}

Yes, if N=1 :D