Two similar irrational numbers, Coincidence or Correlation?
$begingroup$
I am trying to see if there is any relationship between two values obtained by entirely different means using $varphi=frac{1+sqrt5}2, pi, e,$ etc.
In the first equation, I was finding the base of a Kepler Triangle so that the length of the sides equaled the area. That led me to the value $2sqrt{2+sqrt5}+2$ [approximately $6.1163420545$].
That value is the base of a right triangle; that same value multiplied by the square root of $varphi$ is the side $b$; and the same value multiplied by $varphi$ is the side $c$.
The second equation is simply $pi^{e/(e-1)},$ which is approximately $6.1161695807$.
Although this could be (and probably is) completely coincidence, the figures were calculated using Microsoft Excel and are therefore very limited in their accuracy.
So given the uncertainty:
1) Is it possible that these two numbers are actually identical, which would be shown if I were to be using a program that didn't truncate values? (I find this unlikely)
2) Is there any relation between these two values that could provide a clue as to why they are so similar (or is it just complete coincidence)?
algebra-precalculus inequality convergence algorithms
edited Dec 31 '18 at 20:30
user376343
3,3583826
asked Dec 31 '18 at 19:39
mkinsonmkinson
1469
$endgroup$
|
show 7 more comments
$begingroup$
I am trying to see if there is any relationship between two values obtained by entirely different means using $varphi=frac{1+sqrt5}2, pi, e,$ etc.
In the first equation, I was finding the base of a Kepler Triangle so that the length of the sides equaled the area. That led me to the value $2sqrt{2+sqrt5}+2$ [approximately $6.1163420545$].
That value is the base of a right triangle; that same value multiplied by the square root of $varphi$ is the side $b$; and the same value multiplied by $varphi$ is the side $c$.
The second equation is simply $pi^{e/(e-1)},$ which is approximately $6.1161695807$.
Although this could be (and probably is) completely coincidence, the figures were calculated using Microsoft Excel and are therefore very limited in their accuracy.
So given the uncertainty:
1) Is it possible that these two numbers are actually identical, which would be shown if I were to be using a program that didn't truncate values? (I find this unlikely)
2) Is there any relation between these two values that could provide a clue as to why they are so similar (or is it just complete coincidence)?
algebra-precalculus inequality convergence algorithms
edited Dec 31 '18 at 20:30
user376343
3,3583826
asked Dec 31 '18 at 19:39
mkinsonmkinson
1469
$endgroup$
1
$begingroup$
143 points and no MathJax?
$endgroup$
– kelalaka
Dec 31 '18 at 19:43
$begingroup$
Give me some time to look it up. It's not a very intuitive process. I would love a general Help file somewhere that provides some suggestions like that..
$endgroup$
– mkinson
Dec 31 '18 at 19:52
1
$begingroup$
MathJax basic tutorial and quick reference
$endgroup$
– rafa11111
Dec 31 '18 at 21:01
$begingroup$
For what it's worth, my calculator gives the same values (to one less decimal place, rounded in the right direction). So I don't think they're two inaccurate versions of one number.
$endgroup$
– timtfj
Dec 31 '18 at 22:41
1
$begingroup$
I'm thinking that it must be a coincidence, but that how likely you were to get one will depend on how we characterise the set of problems being looked at. For instance it seems surprising that $π^e$ is close to $e^π$ until realise it's the same as $π^{frac1π}$ being close to $e^{frac1e}$ and look at the behaviour of $x^{frac1x}$ near $x=e$.
$endgroup$
– timtfj
Jan 2 at 15:07
|
show 7 more comments
$begingroup$
I am trying to see if there is any relationship between two values obtained by entirely different means using $varphi=frac{1+sqrt5}2, pi, e,$ etc.
In the first equation, I was finding the base of a Kepler Triangle so that the length of the sides equaled the area. That led me to the value $2sqrt{2+sqrt5}+2$ [approximately $6.1163420545$].
That value is the base of a right triangle; that same value multiplied by the square root of $varphi$ is the side $b$; and the same value multiplied by $varphi$ is the side $c$.
The second equation is simply $pi^{e/(e-1)},$ which is approximately $6.1161695807$.
Although this could be (and probably is) completely coincidence, the figures were calculated using Microsoft Excel and are therefore very limited in their accuracy.
So given the uncertainty:
1) Is it possible that these two numbers are actually identical, which would be shown if I were to be using a program that didn't truncate values? (I find this unlikely)
2) Is there any relation between these two values that could provide a clue as to why they are so similar (or is it just complete coincidence)?
algebra-precalculus inequality convergence algorithms
edited Dec 31 '18 at 20:30
user376343
3,3583826
asked Dec 31 '18 at 19:39
mkinsonmkinson
1469
$endgroup$
I am trying to see if there is any relationship between two values obtained by entirely different means using $varphi=frac{1+sqrt5}2, pi, e,$ etc.
In the first equation, I was finding the base of a Kepler Triangle so that the length of the sides equaled the area. That led me to the value $2sqrt{2+sqrt5}+2$ [approximately $6.1163420545$].
That value is the base of a right triangle; that same value multiplied by the square root of $varphi$ is the side $b$; and the same value multiplied by $varphi$ is the side $c$.
The second equation is simply $pi^{e/(e-1)},$ which is approximately $6.1161695807$.
Although this could be (and probably is) completely coincidence, the figures were calculated using Microsoft Excel and are therefore very limited in their accuracy.
So given the uncertainty:
1) Is it possible that these two numbers are actually identical, which would be shown if I were to be using a program that didn't truncate values? (I find this unlikely)
2) Is there any relation between these two values that could provide a clue as to why they are so similar (or is it just complete coincidence)?
algebra-precalculus inequality convergence algorithms
algebra-precalculus inequality convergence algorithms
edited Dec 31 '18 at 20:30
user376343
3,3583826
asked Dec 31 '18 at 19:39
mkinsonmkinson
1469
edited Dec 31 '18 at 20:30
user376343
3,3583826
asked Dec 31 '18 at 19:39
mkinsonmkinson
1469
edited Dec 31 '18 at 20:30
user376343
3,3583826
edited Dec 31 '18 at 20:30
user376343
3,3583826
edited Dec 31 '18 at 20:30
user376343
3,3583826
3,3583826
asked Dec 31 '18 at 19:39
mkinsonmkinson
1469
asked Dec 31 '18 at 19:39
mkinsonmkinson
1469
asked Dec 31 '18 at 19:39
mkinsonmkinson
1469
1469
1
$begingroup$
143 points and no MathJax?
$endgroup$
– kelalaka
Dec 31 '18 at 19:43
$begingroup$
Give me some time to look it up. It's not a very intuitive process. I would love a general Help file somewhere that provides some suggestions like that..
$endgroup$
– mkinson
Dec 31 '18 at 19:52
1
$begingroup$
MathJax basic tutorial and quick reference
$endgroup$
– rafa11111
Dec 31 '18 at 21:01
$begingroup$
For what it's worth, my calculator gives the same values (to one less decimal place, rounded in the right direction). So I don't think they're two inaccurate versions of one number.
$endgroup$
– timtfj
Dec 31 '18 at 22:41
1
$begingroup$
I'm thinking that it must be a coincidence, but that how likely you were to get one will depend on how we characterise the set of problems being looked at. For instance it seems surprising that $π^e$ is close to $e^π$ until realise it's the same as $π^{frac1π}$ being close to $e^{frac1e}$ and look at the behaviour of $x^{frac1x}$ near $x=e$.
$endgroup$
– timtfj
Jan 2 at 15:07
|
show 7 more comments
1
$begingroup$
143 points and no MathJax?
$endgroup$
– kelalaka
Dec 31 '18 at 19:43
$begingroup$
Give me some time to look it up. It's not a very intuitive process. I would love a general Help file somewhere that provides some suggestions like that..
$endgroup$
– mkinson
Dec 31 '18 at 19:52
1
$begingroup$
MathJax basic tutorial and quick reference
$endgroup$
– rafa11111
Dec 31 '18 at 21:01
$begingroup$
For what it's worth, my calculator gives the same values (to one less decimal place, rounded in the right direction). So I don't think they're two inaccurate versions of one number.
$endgroup$
– timtfj
Dec 31 '18 at 22:41
1
$begingroup$
I'm thinking that it must be a coincidence, but that how likely you were to get one will depend on how we characterise the set of problems being looked at. For instance it seems surprising that $π^e$ is close to $e^π$ until realise it's the same as $π^{frac1π}$ being close to $e^{frac1e}$ and look at the behaviour of $x^{frac1x}$ near $x=e$.
$endgroup$
– timtfj
Jan 2 at 15:07
1
1
$begingroup$
143 points and no MathJax?
$endgroup$
– kelalaka
Dec 31 '18 at 19:43
$begingroup$
143 points and no MathJax?
$endgroup$
– kelalaka
Dec 31 '18 at 19:43
$begingroup$
Give me some time to look it up. It's not a very intuitive process. I would love a general Help file somewhere that provides some suggestions like that..
$endgroup$
– mkinson
Dec 31 '18 at 19:52
$begingroup$
Give me some time to look it up. It's not a very intuitive process. I would love a general Help file somewhere that provides some suggestions like that..
$endgroup$
– mkinson
Dec 31 '18 at 19:52
1
1
$begingroup$
MathJax basic tutorial and quick reference
$endgroup$
– rafa11111
Dec 31 '18 at 21:01
$begingroup$
MathJax basic tutorial and quick reference
$endgroup$
– rafa11111
Dec 31 '18 at 21:01
$begingroup$
For what it's worth, my calculator gives the same values (to one less decimal place, rounded in the right direction). So I don't think they're two inaccurate versions of one number.
$endgroup$
– timtfj
Dec 31 '18 at 22:41
$begingroup$
For what it's worth, my calculator gives the same values (to one less decimal place, rounded in the right direction). So I don't think they're two inaccurate versions of one number.
$endgroup$
– timtfj
Dec 31 '18 at 22:41
1
1
$begingroup$
I'm thinking that it must be a coincidence, but that how likely you were to get one will depend on how we characterise the set of problems being looked at. For instance it seems surprising that $π^e$ is close to $e^π$ until realise it's the same as $π^{frac1π}$ being close to $e^{frac1e}$ and look at the behaviour of $x^{frac1x}$ near $x=e$.
$endgroup$
– timtfj
Jan 2 at 15:07
$begingroup$
I'm thinking that it must be a coincidence, but that how likely you were to get one will depend on how we characterise the set of problems being looked at. For instance it seems surprising that $π^e$ is close to $e^π$ until realise it's the same as $π^{frac1π}$ being close to $e^{frac1e}$ and look at the behaviour of $x^{frac1x}$ near $x=e$.
$endgroup$
– timtfj
Jan 2 at 15:07
|
show 7 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Your two numbers differ at the fourth decimal, so all they have in common is 6.116....
Four digits of agreement is something we should expect to arise as a matter or random chance, unless the original expressions are picked from a pool of possible expressions with much fewer than 10,000 members.
Your expressions here look complex enough that it's easy to imagine 10,000 other "just as nice" expressions, with different numeric constants or slightly different arithmetic operations or combinations of operations.
Thus is shouldn't be a surprise to find a 4-digit coincidence.
answered Dec 31 '18 at 20:30
Henning MakholmHenning Makholm
239k17304541
$endgroup$
$begingroup$
I think this also depends on the distribution of values for the type of expression that can be produced. $4$ digits coinciding could be taken as some arbitrary number of preceding unwritten zeros also happening to be zero. Most likely the distribution is one that increases the probability of coincidences for small numbers, though.
$endgroup$
– timtfj
Jan 2 at 15:16
add a comment |
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$begingroup$
Your two numbers differ at the fourth decimal, so all they have in common is 6.116....
Four digits of agreement is something we should expect to arise as a matter or random chance, unless the original expressions are picked from a pool of possible expressions with much fewer than 10,000 members.
Your expressions here look complex enough that it's easy to imagine 10,000 other "just as nice" expressions, with different numeric constants or slightly different arithmetic operations or combinations of operations.
Thus is shouldn't be a surprise to find a 4-digit coincidence.
answered Dec 31 '18 at 20:30
Henning MakholmHenning Makholm
239k17304541
$endgroup$
$begingroup$
I think this also depends on the distribution of values for the type of expression that can be produced. $4$ digits coinciding could be taken as some arbitrary number of preceding unwritten zeros also happening to be zero. Most likely the distribution is one that increases the probability of coincidences for small numbers, though.
$endgroup$
– timtfj
Jan 2 at 15:16
add a comment |
$begingroup$
Your two numbers differ at the fourth decimal, so all they have in common is 6.116....
Four digits of agreement is something we should expect to arise as a matter or random chance, unless the original expressions are picked from a pool of possible expressions with much fewer than 10,000 members.
Your expressions here look complex enough that it's easy to imagine 10,000 other "just as nice" expressions, with different numeric constants or slightly different arithmetic operations or combinations of operations.
Thus is shouldn't be a surprise to find a 4-digit coincidence.
answered Dec 31 '18 at 20:30
Henning MakholmHenning Makholm
239k17304541
$endgroup$
$begingroup$
I think this also depends on the distribution of values for the type of expression that can be produced. $4$ digits coinciding could be taken as some arbitrary number of preceding unwritten zeros also happening to be zero. Most likely the distribution is one that increases the probability of coincidences for small numbers, though.
$endgroup$
– timtfj
Jan 2 at 15:16
add a comment |
$begingroup$
Your two numbers differ at the fourth decimal, so all they have in common is 6.116....
Four digits of agreement is something we should expect to arise as a matter or random chance, unless the original expressions are picked from a pool of possible expressions with much fewer than 10,000 members.
Your expressions here look complex enough that it's easy to imagine 10,000 other "just as nice" expressions, with different numeric constants or slightly different arithmetic operations or combinations of operations.
Thus is shouldn't be a surprise to find a 4-digit coincidence.
answered Dec 31 '18 at 20:30
Henning MakholmHenning Makholm
239k17304541
$endgroup$
Your two numbers differ at the fourth decimal, so all they have in common is 6.116....
Four digits of agreement is something we should expect to arise as a matter or random chance, unless the original expressions are picked from a pool of possible expressions with much fewer than 10,000 members.
Your expressions here look complex enough that it's easy to imagine 10,000 other "just as nice" expressions, with different numeric constants or slightly different arithmetic operations or combinations of operations.
Thus is shouldn't be a surprise to find a 4-digit coincidence.
answered Dec 31 '18 at 20:30
Henning MakholmHenning Makholm
239k17304541
answered Dec 31 '18 at 20:30
Henning MakholmHenning Makholm
239k17304541
answered Dec 31 '18 at 20:30
Henning MakholmHenning Makholm
239k17304541
answered Dec 31 '18 at 20:30
Henning MakholmHenning Makholm
239k17304541
239k17304541
$begingroup$
I think this also depends on the distribution of values for the type of expression that can be produced. $4$ digits coinciding could be taken as some arbitrary number of preceding unwritten zeros also happening to be zero. Most likely the distribution is one that increases the probability of coincidences for small numbers, though.
$endgroup$
– timtfj
Jan 2 at 15:16
add a comment |
$begingroup$
I think this also depends on the distribution of values for the type of expression that can be produced. $4$ digits coinciding could be taken as some arbitrary number of preceding unwritten zeros also happening to be zero. Most likely the distribution is one that increases the probability of coincidences for small numbers, though.
$endgroup$
– timtfj
Jan 2 at 15:16
$begingroup$
I think this also depends on the distribution of values for the type of expression that can be produced. $4$ digits coinciding could be taken as some arbitrary number of preceding unwritten zeros also happening to be zero. Most likely the distribution is one that increases the probability of coincidences for small numbers, though.
$endgroup$
– timtfj
Jan 2 at 15:16
$begingroup$
I think this also depends on the distribution of values for the type of expression that can be produced. $4$ digits coinciding could be taken as some arbitrary number of preceding unwritten zeros also happening to be zero. Most likely the distribution is one that increases the probability of coincidences for small numbers, though.
$endgroup$
– timtfj
Jan 2 at 15:16
add a comment |
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1
$begingroup$
143 points and no MathJax?
$endgroup$
– kelalaka
Dec 31 '18 at 19:43
$begingroup$
Give me some time to look it up. It's not a very intuitive process. I would love a general Help file somewhere that provides some suggestions like that..
$endgroup$
– mkinson
Dec 31 '18 at 19:52
1
$begingroup$
MathJax basic tutorial and quick reference
$endgroup$
– rafa11111
Dec 31 '18 at 21:01
$begingroup$
For what it's worth, my calculator gives the same values (to one less decimal place, rounded in the right direction). So I don't think they're two inaccurate versions of one number.
$endgroup$
– timtfj
Dec 31 '18 at 22:41
1
$begingroup$
I'm thinking that it must be a coincidence, but that how likely you were to get one will depend on how we characterise the set of problems being looked at. For instance it seems surprising that $π^e$ is close to $e^π$ until realise it's the same as $π^{frac1π}$ being close to $e^{frac1e}$ and look at the behaviour of $x^{frac1x}$ near $x=e$.
$endgroup$
– timtfj
Jan 2 at 15:07