Art Can Equations Be Aesthetic Art?

[Buffu]

My equation, in a separate post to allow unbiased voting:

$\nabla^2 = \Delta$

That is clever, I like it.

 

[Orodruin]

$\nabla^2 = \Delta$

I always write $\nabla^2$ in order to avoid confusion with small but finite differences ... Does that make me weird?

 

[member 587159]

I regret that I didn't submit $\sum_{k=1}^{\infty} k = \frac{-1}{12}$

 

[dextercioby]

One of the most important results in mainstream physics:

\psi = e^{i\alpha} \phi \Rightarrow \hat{\psi} = \hat{\phi}

 

[fresh_42]

I regret that I didn't submit $\sum_{k=1}^{\infty} k = \frac{-1}{12}$

And I'm glad you didn't. This weird sum shows up on PF far too often already
I regret a little that I didn't take $2^n+7^n+8^n+18^n+19^n+24^n=3^n+4^n+12^n+14^n+22^n+23^n$, the best lesson on induction I've ever seen.

 

[mfb]

I guess you can solve that equation for possible values of n...

Here is a plot

 

[fresh_42]

I guess you can solve that equation for possible values of n...

Here is a plot

Yes, but the interesting point is behind the formula: who found that, how, and even more why? Did they use a mainframe and simply tried? And how much has someone to drink before he tackles such an undertaking. Strange.

 

[mfb]

A computer should find something like that quickly. Assign prefactors of -1 (left side), 0 (don't use) or 1 (right side) and find a set that fits for n from 0 to 5.

3²⁵ = 850 billion combinations if you use 1 to 25, but most of these combinations don't need to be considered.

 

[Baluncore]

I prefer a political statement; the flow of water is power. (Sorry, no LaTex here).

H₂O = m·c²

 

[robphy]

<br /> \left( \frac{1}{2}+\frac{1}{4}+\frac{1}{8}+\frac{1}{16}+\frac{1}{32}+\frac{1}{64}+\cdots\right)=1<br />

 

[chimay]

\arctan \frac{1}{x}=\frac{\pi}{2}-\arctan x

 

[MexChemE]

The ideal gas law
PV=nRT

 

[Laurie K]

$$\frac {\lambda}{2 \pi}=\frac {\hbar}{mc}\\$$

 

[Charles Link]

I like the Planck blackbody function: $\\$ $L(\lambda,T)=\frac{2 h c^2}{\lambda^5 (e^{\frac{hc}{\lambda k_b T}}-1)}$.

 

[diogenesNY]

$E=IR$

Clean lines. Beautiful simplicity, yet breathtakingly utilitarian.
Like a cold beer on a hot day!

submitted for your approval,
diogenesNY

(I previously cited Ohm's law in a similar thread some years ago... my opinion remains unchanged... although I did have to figure out how to use latex for this one)

 

[scottdave]

$3987^{12} + 4365^{12} = 4472^{12}$

Ha. Actually pretty easy to disprove.

 

[Charles Link]

Ha. Actually pretty easy to disprove.

A numerical computation of it shows it doesn't miss by much. Fermat's last theorem says that it can't be equal, but it's much closer than I expected.

 

[mfb]

3987 and 4365 are divisible by 3, therefore their 12th powers are divisible by 3, same for the sum. 4472 is not divisible by 3, and taking the 12th power doesn't change that.

 

[Baluncore]

A numerical computation of it shows it doesn't miss by much. Fermat's last theorem says that it can't be equal, but it's much closer than I expected.

 3987^12 = x = 16134474609751291283496491970515151715346481.
 4365^12 = y = 47842181739947321332739738982639336181640625.
       x + y = 63976656349698612616236230953154487896987106.
 4472^12 = z = 63976656348486725806862358322168575784124416.
   x + y - z =  error =  1211886809373872630985912112862690.

So, it is not out by much, only by about 1.2 x 10^33.

 

[scottdave]

$E=IR$

Clean lines. Beautiful simplicity, yet breathtakingly utilitarian.
Like a cold beer on a hot day!

submitted for your approval,
diogenesNY

(I previously cited Ohm's law in a similar thread some years ago... my opinion remains unchanged... although I did have to figure out how to use latex for this one)

I like to see it with the E above, and the I and R below that.

 

[scottdave]

3987 and 4365 are divisible by 3, therefore their 12th powers are divisible by 3, same for the sum. 4472 is not divisible by 3, and taking the 12th power doesn't change that.

That is the easy method to show it.

 

[Charles Link]

That is the easy method to show it.

A check on simply the last digit does not rule out the possibility that the equality could hold. Before Fermat's last theorem was proven by Andrew Wiles, had someone come up with something like this that worked, it would have been one of the better numerical finds of the century. As I recall, as early as 1970, Fermat's theorem had already been established for exponents $n$ up to 169, so it would have been some very large numbers that would have been necessary to make such a sum.

 

[scottdave]

A check on simply the last digit does not rule out the possibility that the equality could hold. Before Fermat's last theorem was proven by Andrew Wiles, had someone come up with something like this that worked, it would have been one of the better numerical finds of the century. As I recall, as early as 1970, Fermat's theorem had already been established for exponents $n$ up to 169, so it would have been some very large numbers that would have been necessary to make such a sum.

Not the last digit. Sum the digits to see if a multiple of 3.

 

[Charles Link]

Not the last digit. Sum the digits to see if a multiple of 3.

Yes, @scottdave , @mfb 's method is clever.

 

[jfizzix]

(1+2+3+...+n)^{2}=1^{3} + 2^{3} + 3^{3} +... + n^{3}

 

[hsdrop]

\begin{matrix}
. & . & . & . & . & . & . & . & . \\
. & P & P & . & . & F & F & F & . \\
. & P & . & P & . & F & .& . & . \\
. & P & P & . & . & F & F & F & . \\
. & P & . & . & . & F & . & . & . \\
. & P & . & . & . & F & . & . & . \\
. & . & . & . & . & . & . & . & . \\
\end{matrix}
I Hope everyone likes it.
I'm also hoping that it falls within the rules as well.

 

[epenguin]

\begin{equation} D^{1}\left( uv\right) =uD^{1}v+vD^{1}u \end{equation}
$$ \rm {and~ the~ binomial~ expansion~ formula} $$
\begin{equation}\left ( a^{1}b^{0}+a^{0}b^{1}\right) ^{n}=\sum ^{n}_{i=0}\binom {n} {i}a^{i}b^{n-i} \end{equation}
$$\rm together~ imply~ Leibniz' ~ theorem:$$
$$~D^{n}\left( uv\right) =(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}~ =\sum ^{n}_{i=0}\binom {n} {i}D^nu D^{n-i}v $$
^^ just fits on a page in my preview. It was not necessary for this competition that the equations be true or useful, but whether that is so can be discussed on another thread.
https://www.physicsforums.com/threads/prove-the-leibnitz-rule-of-derivatives.924400/

 

[Demystifier]

$\overbrace{\smile}^{\theta\theta}$

I think this is the only one so far which fulfills the propositions.

 

[martinbn]

What exactly is the definition of equation used for this thread? Some of the things posted, I would call formulae, some expressions and so on.

 

[Demystifier]

At least there must be some wisdom in the symbols to call it equation, like in this one:
$\widehat{\dbinom{\odot_\text{v}\odot}{\wr}}$

 

[Demystifier]

If the above is not counted as an actual equation, then I would mention one of the most difficult unsolved problems in number theory:
$a+b=c$

 

[fresh_42]

If the above is not counted as an actual equation, then I would mention one of the most difficult unsolved problems in number theory:
$a+b=c$

"Unsolved" is debatable. O.k. apparently nobody can really follow the suggested proof, but does this count for "unsolved"?

 

[Greg Bernhardt]

We have a tie between @Orodruin and @MarkFL and someone needs to break it!

 

[Ssnow]

A Srinivasa Ramanujan formula:

\frac{1}{1+\frac{e^{-2\pi\sqrt{5}}}{1+\frac{e^{-4\pi\sqrt{5}}}{1+\frac{e^{-6\pi\sqrt{5}}}{1+\cdots}}}}\,=\, \left(\frac{\sqrt{5}}{1+\sqrt[5]{5^{\frac{3}{4}}\left(\frac{\sqrt{5}-1}{2}\right)^{\frac{5}{2}}-1}}-\frac{\sqrt{5}+1}{2}\right)\cdot e^{\frac{2\pi}{\sqrt{5}}}

a beautiful combination of $1,2,3,4,5,6$ and other ...
Ssnow

 

[Demystifier]

"Unsolved" is debatable. O.k. apparently nobody can really follow the suggested proof, but does this count for "unsolved"?

I didn't know that there is a suggested proof. Reference?

 

[Baluncore]

a beautiful combination of 1,2,3,4,5,6 and other ...

I see a nest of golden ratios in there, (√5 ± 1 ) / 2.

 

[mfb]

We have a tie between @Orodruin and @MarkFL and someone needs to break it!

12 vs 10 at the moment.

I didn't know that there is a suggested proof. Reference?

The Wikipedia page has a link to it.

 

[scottdave]

We have a tie between @Orodruin and @MarkFL and someone needs to break it!

I vote for @MarkFL s equation.

 

[Charles Link]

Looks like @Ygggdrasil 's takeoff on Fermat's Last theorem has the #3 spot.

 

[Greg Bernhardt]

@Orodruin wins! It is very elegant. Thanks all!