The most difficult equation in mathematics

[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

 

[Dragonfall]

Yes, if N=1 :D

Or P = 0

 

[jk22]

How About finding the configuration of charges minimal that is stable in electrostatics (i just invented that one)

 

[ChrisVer]

Well I am not sure what you mean by most difficult equation in mathematics. Any differential equations that needs computers to solve (have no analytical solution) seems "difficult". An example are the coupled Boltzmann equations for the Big Bang Nucleosynthesis (it's fresh in my mind).

 

[ChrisVer]

Yes, if N=1 :D

It can as well be $N=P$ if $P$ is idempotent. Eg the projection operators when written in matrix representation satisfy that $P^2 =P$

 

[Demystifier]

1+1=2

It's not so difficult. Whitehead and Russell needed only 379 pages to prove this.
http://en.wikipedia.org/wiki/Principia_Mathematica
http://quod.lib.umich.edu/cgi/t/tex...3201.0001.001&frm=frameset&view=image&seq=401

 

[jk22]

You can also build a semi group where 1+1=1 it is just a matter of symbols we use to describe the operations.

 

[micromass]

It's not so difficult. Whitehead and Russell needed only 379 pages to prove this.
http://en.wikipedia.org/wiki/Principia_Mathematica
http://quod.lib.umich.edu/cgi/t/tex...3201.0001.001&frm=frameset&view=image&seq=401

Ah, but take a look at your post here:

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

Just like $2^{\aleph_0} = \aleph_1$ depends on the axiomatics used, so does 1+1=2

 

[micromass]

Also, whether 1+1=2 is difficult depends a lot on your definitions of 1 and 2. It is certainly possible to find a definition where 1+1=2 is trivially true by definition. For example, in the Peano system, 1+1=2 can be very easily verified. The hard part is when 1 and 2 are taken as real or complex numbers. In that case, the definitions are much more complicated and do not allow for a trivial solution. And then there are other number systems such as the hyperreals or surreals which make things even more complicated! So the difficulty lies more in the number systems used than in the actual equation.

 

[micromass]

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.

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

 

[lavinia]

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
$$2^{\aleph_0}=\aleph_k$$

What do you mean by difficult?

How about solving the Riemann hypothesis? This remains unsolved. The equation is Riemann zeta function = 0.

Solving the Laplace equation has inspired generations of research.
Solving the Heat equation has also led to generations of research.
The same holds for other PDE's e.g. Monge-Ampere equations, geometric evolution equations, the wave equation ...

The proof of Fermat's last theorem led to centuries of research. The solution shows that certain equations have no solution.

 

[phoenixthoth]

Here's one that I think is hard:

math = ?

 

[micromass]

math = math

 

[dextercioby]

Not necessarily equation, but counting problems are always terribly hard. Here's one that's really difficult: How many DISTINCT valid (within the rules of the game) endings of a standard 9x9 Sudoku game are there?