Can Equations Be Aesthetic Art?

In summary: I came to the conclusion that I could not decide and needed to see the back side as well. So I went around the table - the other way around than usual. I just wanted to make it short and effective, but alas, that was not meant to be. I had to walk twice around the table and as I came near the front again, I had to back off and start over. I was so focussed, I didn't even notice the other people at the table, not even that the music had stopped. The third time I came around the table, I realized that the girl had already turned away from me and I saw her boyfriend. He was looking at me with this very unfriendly look ... :uhh
  • #71
mfb said:
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.
 
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  • #72
scottdave said:
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.
 
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  • #73
Charles Link said:
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.
 
  • #74
scottdave said:
Not the last digit. Sum the digits to see if a multiple of 3.
Yes, @scottdave , @mfb 's method is clever.
 
  • #75
[itex](1+2+3+...+n)^{2}=1^{3} + 2^{3} + 3^{3} +... + n^{3}[/itex]
 
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  • #76
\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. :angel:
 
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  • #77
[tex]\begin{equation} D^{1}\left( uv\right) =uD^{1}v+vD^{1}u \end{equation}[/tex]
$$ \rm {and~ the~ binomial~ expansion~ formula} $$
[tex]\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} [/tex]
$$\rm together~ imply~ Leibniz' ~ theorem:$$
$$~D^{n}\left( uv\right) =(D^{0}uD^{1}v+D^{1}uD^0{v})^{n}~ =\sum ^{n}_{i=0}\binom {n} {i}D^nu D^{n-i}v $$
^^ 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. :oldsmile:
https://www.physicsforums.com/threads/prove-the-leibnitz-rule-of-derivatives.924400/
 
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  • #78
stoomart said:
##\overbrace{\smile}^{\theta\theta}##
I think this is the only one so far which fulfills the propositions.
 
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  • #79
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.
 
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  • #80
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}}
##
 
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  • #81
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##
 
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  • #82
Demystifier said:
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"?
 
  • #84
A Srinivasa Ramanujan formula:

[tex]\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}}}[/tex]

a beautiful combination of ##1,2,3,4,5,6## and other ...
Ssnow
 
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  • #85
fresh_42 said:
"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?
 
  • #86
Ssnow said:
a beautiful combination of 1,2,3,4,5,6 and other ...
I see a nest of golden ratios in there, (√5 ± 1 ) / 2.
 
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  • #87
Greg Bernhardt said:
We have a tie between @Orodruin and @MarkFL and someone needs to break it!
12 vs 10 at the moment.
Demystifier said:
I didn't know that there is a suggested proof. Reference?
The Wikipedia page has a link to it.
 
  • #90
@Orodruin wins! It is very elegant. Thanks all!
 
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