$pi^4 + pi^5 approx e^6$ is anything special going on here?
$begingroup$
Saw it in the news:
$$(pi^4 + pi^5)^{Largefrac16} approx 2.71828180861$$
Is this just pigeon-hole?
DISCUSSION: counterfeit $e$ using $pi$'s
Given enough integers and $pi$'s we can approximate just about any number. In formal mathematical language we say this set is dense in the real numbers:
$$ overline{ mathbb{Z}[pi]} = mathbb{R}$$
This is only part of the story since it doesn't tell us how big our integers have to be in order to approximate the constant of our choosing? Maybe we can quantify this with a notion of density?
$$ mu_N([a,b]) = frac{# |{ m + n pi: -N leq m,n leq N }cap[a,b]|}{N^2} $$
The example above works because of the constants 4, 5 and 6.
We can focus on a particular constant and ask how much effort it takes to approximate a given constant:
$$ big{ (m,n)in mathbb{Z}^2: big| m + n pi - alpha big|< epsilon big} $$
In our case we need to incorporate for square roots, cube roots and higher.
Generalization How closely can we approximate $e$ using powers of $pi$ and $n$-th roots?
$$displaystyle ( a + bpi )^{1/p} approx e $$
Here $0 leq |a|,|b|,p leq 10$
exponential-function approximation pi constants diophantine-approximation
edited Dec 27 '18 at 5:05
Martin Sleziak
44.7k10119272
asked Jun 14 '14 at 11:02
cactus314cactus314
15.5k42269
$endgroup$
|
show 6 more comments
$begingroup$
Saw it in the news:
$$(pi^4 + pi^5)^{Largefrac16} approx 2.71828180861$$
Is this just pigeon-hole?
DISCUSSION: counterfeit $e$ using $pi$'s
Given enough integers and $pi$'s we can approximate just about any number. In formal mathematical language we say this set is dense in the real numbers:
$$ overline{ mathbb{Z}[pi]} = mathbb{R}$$
This is only part of the story since it doesn't tell us how big our integers have to be in order to approximate the constant of our choosing? Maybe we can quantify this with a notion of density?
$$ mu_N([a,b]) = frac{# |{ m + n pi: -N leq m,n leq N }cap[a,b]|}{N^2} $$
The example above works because of the constants 4, 5 and 6.
We can focus on a particular constant and ask how much effort it takes to approximate a given constant:
$$ big{ (m,n)in mathbb{Z}^2: big| m + n pi - alpha big|< epsilon big} $$
In our case we need to incorporate for square roots, cube roots and higher.
Generalization How closely can we approximate $e$ using powers of $pi$ and $n$-th roots?
$$displaystyle ( a + bpi )^{1/p} approx e $$
Here $0 leq |a|,|b|,p leq 10$
exponential-function approximation pi constants diophantine-approximation
edited Dec 27 '18 at 5:05
Martin Sleziak
44.7k10119272
asked Jun 14 '14 at 11:02
cactus314cactus314
15.5k42269
$endgroup$
13
$begingroup$
A good approximation of order approx. $;10^{-7};$ ....that's all. There are literally thousands of different such approximations to all kinds of numbers.
$endgroup$
– DonAntonio
Jun 14 '14 at 11:07
4
$begingroup$
This is most likely a coincidence, as the approximation is explained well by a long-known astonishing coincidence $pi approx e^pi - 20$. For example, $log(pi^4 + pi^5) = 4log(pi)+log(1+pi)$ and applying the above with the crude estimation $exp(pi) approx 23$ in mind, one has $log(pi) approx log(exp(pi)-20) approx pi + log(1-20/exp(pi)) approx pi + log(3/23)$ and similarly $log(1+pi) approx log(exp(pi) - 19) approx pi + log(4/23)$. Adding up, we have $5pi + log(3^4/23^4cdot 4/23) approx 5.811$ which is much close to $6$.
$endgroup$
– Balarka Sen
Jun 14 '14 at 12:23
3
$begingroup$
Is there a question here?
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 12:48
5
$begingroup$
@johnmangual I see a question mark but the words behind it do not make sense. Please clarify your meaning of "is this pigeonhole" as requested.
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 16:10
3
$begingroup$
$piapproxdfrac{ln(640320^3+744)}{sqrt{163}}$ , with a precision of $30$ exact decimals. Discovered independently by both Charles Hermite and Srinivasa Ramanujan. Based on the fact that $e^{pisqrt H}$ is an almost integer when H is a Heegner number.
$endgroup$
– Lucian
Jun 14 '14 at 17:23
|
show 6 more comments
$begingroup$
Saw it in the news:
$$(pi^4 + pi^5)^{Largefrac16} approx 2.71828180861$$
Is this just pigeon-hole?
DISCUSSION: counterfeit $e$ using $pi$'s
Given enough integers and $pi$'s we can approximate just about any number. In formal mathematical language we say this set is dense in the real numbers:
$$ overline{ mathbb{Z}[pi]} = mathbb{R}$$
This is only part of the story since it doesn't tell us how big our integers have to be in order to approximate the constant of our choosing? Maybe we can quantify this with a notion of density?
$$ mu_N([a,b]) = frac{# |{ m + n pi: -N leq m,n leq N }cap[a,b]|}{N^2} $$
The example above works because of the constants 4, 5 and 6.
We can focus on a particular constant and ask how much effort it takes to approximate a given constant:
$$ big{ (m,n)in mathbb{Z}^2: big| m + n pi - alpha big|< epsilon big} $$
In our case we need to incorporate for square roots, cube roots and higher.
Generalization How closely can we approximate $e$ using powers of $pi$ and $n$-th roots?
$$displaystyle ( a + bpi )^{1/p} approx e $$
Here $0 leq |a|,|b|,p leq 10$
exponential-function approximation pi constants diophantine-approximation
edited Dec 27 '18 at 5:05
Martin Sleziak
44.7k10119272
asked Jun 14 '14 at 11:02
cactus314cactus314
15.5k42269
$endgroup$
Saw it in the news:
$$(pi^4 + pi^5)^{Largefrac16} approx 2.71828180861$$
Is this just pigeon-hole?
DISCUSSION: counterfeit $e$ using $pi$'s
Given enough integers and $pi$'s we can approximate just about any number. In formal mathematical language we say this set is dense in the real numbers:
$$ overline{ mathbb{Z}[pi]} = mathbb{R}$$
This is only part of the story since it doesn't tell us how big our integers have to be in order to approximate the constant of our choosing? Maybe we can quantify this with a notion of density?
$$ mu_N([a,b]) = frac{# |{ m + n pi: -N leq m,n leq N }cap[a,b]|}{N^2} $$
The example above works because of the constants 4, 5 and 6.
We can focus on a particular constant and ask how much effort it takes to approximate a given constant:
$$ big{ (m,n)in mathbb{Z}^2: big| m + n pi - alpha big|< epsilon big} $$
In our case we need to incorporate for square roots, cube roots and higher.
Generalization How closely can we approximate $e$ using powers of $pi$ and $n$-th roots?
$$displaystyle ( a + bpi )^{1/p} approx e $$
Here $0 leq |a|,|b|,p leq 10$
exponential-function approximation pi constants diophantine-approximation
exponential-function approximation pi constants diophantine-approximation
edited Dec 27 '18 at 5:05
Martin Sleziak
44.7k10119272
asked Jun 14 '14 at 11:02
cactus314cactus314
15.5k42269
edited Dec 27 '18 at 5:05
Martin Sleziak
44.7k10119272
asked Jun 14 '14 at 11:02
cactus314cactus314
15.5k42269
edited Dec 27 '18 at 5:05
Martin Sleziak
44.7k10119272
edited Dec 27 '18 at 5:05
Martin Sleziak
44.7k10119272
edited Dec 27 '18 at 5:05
Martin Sleziak
44.7k10119272
44.7k10119272
asked Jun 14 '14 at 11:02
cactus314cactus314
15.5k42269
asked Jun 14 '14 at 11:02
cactus314cactus314
15.5k42269
asked Jun 14 '14 at 11:02
cactus314cactus314
15.5k42269
15.5k42269
13
$begingroup$
A good approximation of order approx. $;10^{-7};$ ....that's all. There are literally thousands of different such approximations to all kinds of numbers.
$endgroup$
– DonAntonio
Jun 14 '14 at 11:07
4
$begingroup$
This is most likely a coincidence, as the approximation is explained well by a long-known astonishing coincidence $pi approx e^pi - 20$. For example, $log(pi^4 + pi^5) = 4log(pi)+log(1+pi)$ and applying the above with the crude estimation $exp(pi) approx 23$ in mind, one has $log(pi) approx log(exp(pi)-20) approx pi + log(1-20/exp(pi)) approx pi + log(3/23)$ and similarly $log(1+pi) approx log(exp(pi) - 19) approx pi + log(4/23)$. Adding up, we have $5pi + log(3^4/23^4cdot 4/23) approx 5.811$ which is much close to $6$.
$endgroup$
– Balarka Sen
Jun 14 '14 at 12:23
3
$begingroup$
Is there a question here?
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 12:48
5
$begingroup$
@johnmangual I see a question mark but the words behind it do not make sense. Please clarify your meaning of "is this pigeonhole" as requested.
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 16:10
3
$begingroup$
$piapproxdfrac{ln(640320^3+744)}{sqrt{163}}$ , with a precision of $30$ exact decimals. Discovered independently by both Charles Hermite and Srinivasa Ramanujan. Based on the fact that $e^{pisqrt H}$ is an almost integer when H is a Heegner number.
$endgroup$
– Lucian
Jun 14 '14 at 17:23
|
show 6 more comments
13
$begingroup$
A good approximation of order approx. $;10^{-7};$ ....that's all. There are literally thousands of different such approximations to all kinds of numbers.
$endgroup$
– DonAntonio
Jun 14 '14 at 11:07
4
$begingroup$
This is most likely a coincidence, as the approximation is explained well by a long-known astonishing coincidence $pi approx e^pi - 20$. For example, $log(pi^4 + pi^5) = 4log(pi)+log(1+pi)$ and applying the above with the crude estimation $exp(pi) approx 23$ in mind, one has $log(pi) approx log(exp(pi)-20) approx pi + log(1-20/exp(pi)) approx pi + log(3/23)$ and similarly $log(1+pi) approx log(exp(pi) - 19) approx pi + log(4/23)$. Adding up, we have $5pi + log(3^4/23^4cdot 4/23) approx 5.811$ which is much close to $6$.
$endgroup$
– Balarka Sen
Jun 14 '14 at 12:23
3
$begingroup$
Is there a question here?
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 12:48
5
$begingroup$
@johnmangual I see a question mark but the words behind it do not make sense. Please clarify your meaning of "is this pigeonhole" as requested.
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 16:10
3
$begingroup$
$piapproxdfrac{ln(640320^3+744)}{sqrt{163}}$ , with a precision of $30$ exact decimals. Discovered independently by both Charles Hermite and Srinivasa Ramanujan. Based on the fact that $e^{pisqrt H}$ is an almost integer when H is a Heegner number.
$endgroup$
– Lucian
Jun 14 '14 at 17:23
13
13
$begingroup$
A good approximation of order approx. $;10^{-7};$ ....that's all. There are literally thousands of different such approximations to all kinds of numbers.
$endgroup$
– DonAntonio
Jun 14 '14 at 11:07
$begingroup$
A good approximation of order approx. $;10^{-7};$ ....that's all. There are literally thousands of different such approximations to all kinds of numbers.
$endgroup$
– DonAntonio
Jun 14 '14 at 11:07
4
4
$begingroup$
This is most likely a coincidence, as the approximation is explained well by a long-known astonishing coincidence $pi approx e^pi - 20$. For example, $log(pi^4 + pi^5) = 4log(pi)+log(1+pi)$ and applying the above with the crude estimation $exp(pi) approx 23$ in mind, one has $log(pi) approx log(exp(pi)-20) approx pi + log(1-20/exp(pi)) approx pi + log(3/23)$ and similarly $log(1+pi) approx log(exp(pi) - 19) approx pi + log(4/23)$. Adding up, we have $5pi + log(3^4/23^4cdot 4/23) approx 5.811$ which is much close to $6$.
$endgroup$
– Balarka Sen
Jun 14 '14 at 12:23
$begingroup$
This is most likely a coincidence, as the approximation is explained well by a long-known astonishing coincidence $pi approx e^pi - 20$. For example, $log(pi^4 + pi^5) = 4log(pi)+log(1+pi)$ and applying the above with the crude estimation $exp(pi) approx 23$ in mind, one has $log(pi) approx log(exp(pi)-20) approx pi + log(1-20/exp(pi)) approx pi + log(3/23)$ and similarly $log(1+pi) approx log(exp(pi) - 19) approx pi + log(4/23)$. Adding up, we have $5pi + log(3^4/23^4cdot 4/23) approx 5.811$ which is much close to $6$.
$endgroup$
– Balarka Sen
Jun 14 '14 at 12:23
3
3
$begingroup$
Is there a question here?
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 12:48
$begingroup$
Is there a question here?
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 12:48
5
5
$begingroup$
@johnmangual I see a question mark but the words behind it do not make sense. Please clarify your meaning of "is this pigeonhole" as requested.
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 16:10
$begingroup$
@johnmangual I see a question mark but the words behind it do not make sense. Please clarify your meaning of "is this pigeonhole" as requested.
$endgroup$
– Alexander Gruber♦
Jun 14 '14 at 16:10
3
3
$begingroup$
$piapproxdfrac{ln(640320^3+744)}{sqrt{163}}$ , with a precision of $30$ exact decimals. Discovered independently by both Charles Hermite and Srinivasa Ramanujan. Based on the fact that $e^{pisqrt H}$ is an almost integer when H is a Heegner number.
$endgroup$
– Lucian
Jun 14 '14 at 17:23
$begingroup$
$piapproxdfrac{ln(640320^3+744)}{sqrt{163}}$ , with a precision of $30$ exact decimals. Discovered independently by both Charles Hermite and Srinivasa Ramanujan. Based on the fact that $e^{pisqrt H}$ is an almost integer when H is a Heegner number.
$endgroup$
– Lucian
Jun 14 '14 at 17:23
|
show 6 more comments
1 Answer
1
active
oldest
votes
$begingroup$
Well known approximations for $pi$, $pi^2$ and $pi^3$ can be related to the question.
$$e^6 approx 403 = 13·31 = (3+10)·31 approx left(pi+pi^2right)pi^3= pi^4+pi^5$$
The approximations $pi approx 3$ and $pi^2 approx 10$ have similar absolute errors with opposite sign so the combination $pi+pi^2 approx 13$ is more precise. The largest root of the polynomial $x^2+x-13$ is $frac{sqrt{53}-1}{2}approx 3.140$, which approximates $pi$ with an accuracy between that of $sqrt{10}$ (one digit) and $31^frac{1}{3}$ (three digits).
answered Apr 15 '17 at 8:48
Jaume Oliver LafontJaume Oliver Lafont
3,09911033
$endgroup$
$begingroup$
this is the right idea. i think the best way is to use Poisson summation and truncate. usually one side or the other converges quickly
$endgroup$
– cactus314
Apr 15 '17 at 14:27
1
$begingroup$
Do you mean that there is a series $$left(frac{e^3}{pi^2}right)^2 = sum_{n=0}^{infty} k_npi^n$$ with $k_0=k_1=1$ and the other $k_n$ very small?
$endgroup$
– Jaume Oliver Lafont
Apr 15 '17 at 15:30
add a comment |
Your Answer
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1 Answer
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1 Answer
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active
oldest
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active
oldest
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$begingroup$
Well known approximations for $pi$, $pi^2$ and $pi^3$ can be related to the question.
$$e^6 approx 403 = 13·31 = (3+10)·31 approx left(pi+pi^2right)pi^3= pi^4+pi^5$$
The approximations $pi approx 3$ and $pi^2 approx 10$ have similar absolute errors with opposite sign so the combination $pi+pi^2 approx 13$ is more precise. The largest root of the polynomial $x^2+x-13$ is $frac{sqrt{53}-1}{2}approx 3.140$, which approximates $pi$ with an accuracy between that of $sqrt{10}$ (one digit) and $31^frac{1}{3}$ (three digits).
answered Apr 15 '17 at 8:48
Jaume Oliver LafontJaume Oliver Lafont
3,09911033
$endgroup$
$begingroup$
this is the right idea. i think the best way is to use Poisson summation and truncate. usually one side or the other converges quickly
$endgroup$
– cactus314
Apr 15 '17 at 14:27
1
$begingroup$
Do you mean that there is a series $$left(frac{e^3}{pi^2}right)^2 = sum_{n=0}^{infty} k_npi^n$$ with $k_0=k_1=1$ and the other $k_n$ very small?
$endgroup$
– Jaume Oliver Lafont
Apr 15 '17 at 15:30
add a comment |
$begingroup$
Well known approximations for $pi$, $pi^2$ and $pi^3$ can be related to the question.
$$e^6 approx 403 = 13·31 = (3+10)·31 approx left(pi+pi^2right)pi^3= pi^4+pi^5$$
The approximations $pi approx 3$ and $pi^2 approx 10$ have similar absolute errors with opposite sign so the combination $pi+pi^2 approx 13$ is more precise. The largest root of the polynomial $x^2+x-13$ is $frac{sqrt{53}-1}{2}approx 3.140$, which approximates $pi$ with an accuracy between that of $sqrt{10}$ (one digit) and $31^frac{1}{3}$ (three digits).
answered Apr 15 '17 at 8:48
Jaume Oliver LafontJaume Oliver Lafont
3,09911033
$endgroup$
$begingroup$
this is the right idea. i think the best way is to use Poisson summation and truncate. usually one side or the other converges quickly
$endgroup$
– cactus314
Apr 15 '17 at 14:27
1
$begingroup$
Do you mean that there is a series $$left(frac{e^3}{pi^2}right)^2 = sum_{n=0}^{infty} k_npi^n$$ with $k_0=k_1=1$ and the other $k_n$ very small?
$endgroup$
– Jaume Oliver Lafont
Apr 15 '17 at 15:30
add a comment |
$begingroup$
Well known approximations for $pi$, $pi^2$ and $pi^3$ can be related to the question.
$$e^6 approx 403 = 13·31 = (3+10)·31 approx left(pi+pi^2right)pi^3= pi^4+pi^5$$
The approximations $pi approx 3$ and $pi^2 approx 10$ have similar absolute errors with opposite sign so the combination $pi+pi^2 approx 13$ is more precise. The largest root of the polynomial $x^2+x-13$ is $frac{sqrt{53}-1}{2}approx 3.140$, which approximates $pi$ with an accuracy between that of $sqrt{10}$ (one digit) and $31^frac{1}{3}$ (three digits).
answered Apr 15 '17 at 8:48
Jaume Oliver LafontJaume Oliver Lafont
3,09911033
$endgroup$
Well known approximations for $pi$, $pi^2$ and $pi^3$ can be related to the question.
$$e^6 approx 403 = 13·31 = (3+10)·31 approx left(pi+pi^2right)pi^3= pi^4+pi^5$$
The approximations $pi approx 3$ and $pi^2 approx 10$ have similar absolute errors with opposite sign so the combination $pi+pi^2 approx 13$ is more precise. The largest root of the polynomial $x^2+x-13$ is $frac{sqrt{53}-1}{2}approx 3.140$, which approximates $pi$ with an accuracy between that of $sqrt{10}$ (one digit) and $31^frac{1}{3}$ (three digits).
answered Apr 15 '17 at 8:48
Jaume Oliver LafontJaume Oliver Lafont
3,09911033
edited Jan 29 at 16:58
answered Apr 15 '17 at 8:48
Jaume Oliver LafontJaume Oliver Lafont
3,09911033
answered Apr 15 '17 at 8:48
Jaume Oliver LafontJaume Oliver Lafont
3,09911033
answered Apr 15 '17 at 8:48
Jaume Oliver LafontJaume Oliver Lafont
3,09911033
3,09911033
$begingroup$
this is the right idea. i think the best way is to use Poisson summation and truncate. usually one side or the other converges quickly
$endgroup$
– cactus314
Apr 15 '17 at 14:27
1
$begingroup$
Do you mean that there is a series $$left(frac{e^3}{pi^2}right)^2 = sum_{n=0}^{infty} k_npi^n$$ with $k_0=k_1=1$ and the other $k_n$ very small?
$endgroup$
– Jaume Oliver Lafont
Apr 15 '17 at 15:30
add a comment |
$begingroup$
this is the right idea. i think the best way is to use Poisson summation and truncate. usually one side or the other converges quickly
$endgroup$
– cactus314
Apr 15 '17 at 14:27
1
$begingroup$
Do you mean that there is a series $$left(frac{e^3}{pi^2}right)^2 = sum_{n=0}^{infty} k_npi^n$$ with $k_0=k_1=1$ and the other $k_n$ very small?
$endgroup$
– Jaume Oliver Lafont
Apr 15 '17 at 15:30
$begingroup$
this is the right idea. i think the best way is to use Poisson summation and truncate. usually one side or the other converges quickly
$endgroup$
– cactus314
Apr 15 '17 at 14:27
$begingroup$
this is the right idea. i think the best way is to use Poisson summation and truncate. usually one side or the other converges quickly
$endgroup$
– cactus314
Apr 15 '17 at 14:27
1
1
$begingroup$
Do you mean that there is a series $$left(frac{e^3}{pi^2}right)^2 = sum_{n=0}^{infty} k_npi^n$$ with $k_0=k_1=1$ and the other $k_n$ very small?
$endgroup$
– Jaume Oliver Lafont
Apr 15 '17 at 15:30
$begingroup$
Do you mean that there is a series $$left(frac{e^3}{pi^2}right)^2 = sum_{n=0}^{infty} k_npi^n$$ with $k_0=k_1=1$ and the other $k_n$ very small?
$endgroup$
– Jaume Oliver Lafont
Apr 15 '17 at 15:30
add a comment |
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A good approximation of order approx. $;10^{-7};$ ....that's all. There are literally thousands of different such approximations to all kinds of numbers.
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– DonAntonio
Jun 14 '14 at 11:07
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This is most likely a coincidence, as the approximation is explained well by a long-known astonishing coincidence $pi approx e^pi - 20$. For example, $log(pi^4 + pi^5) = 4log(pi)+log(1+pi)$ and applying the above with the crude estimation $exp(pi) approx 23$ in mind, one has $log(pi) approx log(exp(pi)-20) approx pi + log(1-20/exp(pi)) approx pi + log(3/23)$ and similarly $log(1+pi) approx log(exp(pi) - 19) approx pi + log(4/23)$. Adding up, we have $5pi + log(3^4/23^4cdot 4/23) approx 5.811$ which is much close to $6$.
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– Balarka Sen
Jun 14 '14 at 12:23
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Is there a question here?
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– Alexander Gruber♦
Jun 14 '14 at 12:48
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@johnmangual I see a question mark but the words behind it do not make sense. Please clarify your meaning of "is this pigeonhole" as requested.
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– Alexander Gruber♦
Jun 14 '14 at 16:10
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$piapproxdfrac{ln(640320^3+744)}{sqrt{163}}$ , with a precision of $30$ exact decimals. Discovered independently by both Charles Hermite and Srinivasa Ramanujan. Based on the fact that $e^{pisqrt H}$ is an almost integer when H is a Heegner number.
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– Lucian
Jun 14 '14 at 17:23