A pi and e equation challenge.
Danny73
(3) separate equations using pi and e where the
results are close too the value of pi
(correct too 6 decimal digits)
(((e^3/pi^2)^2)-1) = 3.141592835...
(slightly greater than pi)
(1/((e^6/pi^5)-1)) = 3.1415920835...
(slightly less than pi)
(e^6)/((pi^2 + pi)* pi^2) = 3.1415927912...
(slightly greater than pi)
The challenge --
Are more results possible that give pi
correct too 6 decimal places or more
using only pi and e and their powers in each
equation but with slightly different results
from above.
The criteria for each equation involves pi and e
and using any or all of these operators (-,+,*,/ and ^n)
giving 6 or more correct decimal places of pi in the
result.
BTW
I know (e^(pi*i)) * pi * -1 = pi
So don't try to pull any fast ones! ;-)
Also imaginary (i) is not one of the operators
that is allowed.
Dan
sttsc...@tesco.net
> (3) separate equations using pi and e where the
> results are close too the value of pi
> (correct too 6 decimal digits)
>
> (((e^3/pi^2)^2)-1) = 3.141592835...
> (slightly greater than pi)
>
> (1/((e^6/pi^5)-1)) = 3.1415920835...
> (slightly less than pi)
>
> (e^6)/((pi^2 + pi)* pi^2) = 3.1415927912...
> (slightly greater than pi)
>
> The challenge --
>
> Are more results possible that give pi
> correct too 6 decimal places or more
> using only pi and e and their powers in each
> equation but with slightly different results
> from above.
>
> The criteria for each equation involves pi and e
> and using any or all of these operators (-,+,*,/ and ^n)
> giving 6 or more correct decimal places of pi in the
> result.
You seem to be finding solutions to
say
A1*pi^5 +A2*pi^4 +A3*e^6 a.e. 0
A1*pi = -A2 + (A3/A2)(e^3/pi^2)^2)
with A1 =A2=A3 =1
would pi = (-A2/A1) +(A3/(A2*A1))e^3/pi^2)^2)
also be acceptable ?
Mike Terry
news:5410028f-c562-4749...@s9g2000yqd.googlegroups.com...
>
> (3) separate equations using pi and e where the
> results are close too the value of pi
> (correct too 6 decimal digits)
>
>
> (((e^3/pi^2)^2)-1) = 3.141592835...
> (slightly greater than pi)
>
> (1/((e^6/pi^5)-1)) = 3.1415920835...
> (slightly less than pi)
>
> (e^6)/((pi^2 + pi)* pi^2) = 3.1415927912...
> (slightly greater than pi)
>
> The challenge --
>
> Are more results possible that give pi
> correct too 6 decimal places or more
> using only pi and e and their powers in each
> equation but with slightly different results
> from above.
>
> The criteria for each equation involves pi and e
> and using any or all of these operators (-,+,*,/ and ^n)
> giving 6 or more correct decimal places of pi in the
> result.
pi = 3.141592653589793...
(If this is not acceptable because e "does not appear", replace lhs with
(pi+e-e))
Mike.
Raymond Manzoni
Your three approximations are variants of pi^4 + pi^5 ~= e^6 given
here : <http://en.wikipedia.org/wiki/Mathematical_coincidence>
Other results may be found using this link like :
e^pi - 20 + e/pi^7 = 3.1415926387...
e + (69 - e^(-8))/163 = 3.1415926538...
355/113 = 3.14159292.....
103993/33102 = 3.1415926530... :-)
Hoping this helped,
Raymond
Danny73
>
> - Show quoted text -
Thanks Raymond,
Your first equation fits the criteria for my (3) equations
with a better approximation to boot.
Your next one does not fit the criteria but is also interesting
because pi is involved indirectly with 163 along with (e).
Where e^((sqrt(163)) * pi) ~ 262537412640768744.
The next (2) do not fit the criteria.
So now there are a total of (4) equations that fit the criteria.
Are there anymore?
That link brought me someplace else.
Thanks for your input.
Dan
Danny73
>
> - Show quoted text -
Sorry I clicked on link because it was highlighted.
The actual link (wiki) was very informative.
Thanks again.
Danny73
wrote:
>
> - Show quoted text -
Sure if you have the actual integer values for a1,a2 and a3
depicted for each equation.
Raymond Manzoni
>> Your three approximations are variants of pi^4 + pi^5 ~= e^6 given
>> here : <http://en.wikipedia.org/wiki/Mathematical_coincidence>
>>
>> Other results may be found using this link like :
>>
>> e^pi - 20 + e/pi^7 = 3.1415926387...
>>
>> e + (69 - e^(-8))/163 = 3.1415926538...
>>
>> 355/113 = 3.14159292.....
>>
>> 103993/33102 = 3.1415926530... :-)
>>
>> Hoping this helped,
>> Raymond- Hide quoted text -
>>
>> - Show quoted text -
>
> Thanks Raymond,
>
> Your first equation fits the criteria for my (3) equations
> with a better approximation to boot.
>
> Your next one does not fit the criteria but is also interesting
> because pi is involved indirectly with 163 along with (e).
>
> Where e^((sqrt(163)) * pi) ~ 262537412640768744.
>
> The next (2) do not fit the criteria.
>
> So now there are a total of (4) equations that fit the criteria.
>
> Are there anymore?
Since there is no 'pi' smell in that one either ;-) I shouldn't
propose it anyway :
e + (69 - e^(-8) - e^(-17))/163 = 3.1415926535877..
Of course the first one may be ameliorated too :
e^pi - 20 + e/pi^7 + e^(-2)/pi^14 = 3.141592653549... and so on...
You may too combine two results from the link to get :
e^pi-(67*pi^9/e^8-430)/12 = 3.141592678...
with the variant (e/pi)^8*((e^pi-pi)*12+430)/67 = 3.14159265498..
or combine with pi^2 + pi/24 ~= 10 to get
e^(-1863/422)-(pi^2-10)*24 = 3.14159265436...
or
e^(-12/e)-(pi^2-10)*24 = 3.14159433...
e^(-12/e)-e^6/(2*pi)^(21/2)-(pi^2-10)*24 = 3.141592653563...
and so on...
You may of course approximate pi with a predefined precision if you
accept the formula to be long enough (by adding each time a
e^(-fraction) term)!
A better 'quality criteria' could be obtained by dividing the number
of correct digits by the length of the formula
<http://en.wikipedia.org/wiki/Kolmogorov_complexity>.
Wishing you fun,
Raymond
sttsc...@tesco.net
A simple method of finding integer relations between
irrationals is to write the irrationals to a large
number of decimal places in an nx1 column vector with the identity
matrix beside it.
1 0 0 306.0196848 = pi^5 = r1
0 1 0 97.40909103 = pi^4 = r2
0 0 1 403.4287935 = e^5 = r3
As r1 >r3 -r1 = 97.4091087 >0 take row 1 from row 3
of the matrix
1 0 0 306.0196848 = r1
0 1 0 97.40909103 = r2
-1 0 1 97.4091087 = r3
As r3 is only slightly greater than r2
take row2 from row 3
1 0 0 306.0196848 = r1
0 1 0 97.40909103 = r2
-1 -1 1 0.00001717 = r3
This means that
(-1,-1,1).(pi^5,pi^4, e^6) = 0.0000172
-pi^5 -pi^4 + e^6 = 0.0000172
pi +1 -(e^6/pi^4) = -0.0000172/pi^4
pi + 0.0000172/pi^4 = (e^6/pi^4)-1
Obviously, you can reduce the values in the column
vector as close as you want to 0, but the absolute vales
in the matrix rows will grow in size correspondingly.
There is no guarantee that a1,a2,a3 will be "small"
integers.
Danny73
>
> - Show quoted text -- Hide quoted text -
>
> - Show quoted text -
Thanks again Raymond,
It is interesting on how you come up with these.
Although that last one looks most interesting --
e^(-12/e)-e^6/(2*pi)^(21/2)-(pi^2-10)*24 = 3.141592653563...
I fail to get it to come out right.
I am probably getting the order of preference wrong
for one or more of the operators.
So using my criteria for these equations it appears
there are many more equations then just (4) and
with a higher degree of accuracy also.
Danny73
>
> - Show quoted text -
Also another interesting observation.
Am I wrong in assuming that it will greatly improve
decimal digit accuracy for pi with large a1,a2 and a3?
Dan
sttsc...@tesco.net
mod 0.00001717 ( = r3), you will have at least
6 digit approximation for a1,a2,a3 of about
300/0.0002. As you continue to take row nultiples
avway from each other, you could get lucky and find
another pair or triplet of values very close
together, which means a relatively small increase
in a1,a2,a3. Even though you can approximate
pi to 100 decimal places or more, the formulas might
contain integers with 1,000,000 decimal places and that is not as
pleasing as a "surpisingly good" approximation
with "significant" small integers, such as primes or powers.
sttsc...@tesco.net
> results are close too the value of pi
> (correct too 6 decimal digits)
>
> (((e^3/pi^2)^2)-1) = 3.141592835...
> (slightly greater than pi)
>
> (1/((e^6/pi^5)-1)) = 3.1415920835...
> (slightly less than pi)
>
> (e^6)/((pi^2 + pi)* pi^2) = 3.1415927912...
> (slightly greater than pi)
>
> The challenge --
>
> Are more results possible that give pi
> correct too 6 decimal places or more
> using only pi and e and their powers in each
> equation but with slightly different results
> from above.
>
> The criteria for each equation involves pi and e
> and using any or all of these operators (-,+,*,/ and ^n)
> giving 6 or more correct decimal places of pi in the
> result.
If the sqrt(2) is acceptable
((17^2)p^2 - 95e^2)/((22^2)sqrt(2)) = 3.14159265//42..
Gives pi to 8 places and has two relatively small integer squares.
Danny73
wrote:
>
> - Show quoted text -
Interesting, and yes it should be ecceptable
where I overlooked the (sqrt) operator.
I should also include sqrt(n) as an operator as long as
(n) is an integer.
The new criteria --(-,+,*,/,(sqrt),^n) and having pi
and (e) within the equation and either pi or (e) as the result.
The other criteria is the same as above but having an integer
as the result as with the case of e^(pi*(sqrt(163))) ~
262537412640768744
or using the natural log of (e) along with pi-
((log(1.92932559972567304972024812259531566433349783047577528040155793074025543587927999223373032998386289025375167228752542895869119788592042536004022640601542769951592967417765008384707347928761185634055546471521383604751331331401391405583924e
+236))/pi)^2 ~ 29992. followed by 234 zeros
Thanks again for your interesting observations.
Raymond Manzoni
>> Since there is no 'pi' smell in that one either ;-) I shouldn't
>> propose it anyway :
>> e + (69 - e^(-8) - e^(-17))/163 = 3.1415926535877..
>>
>> Of course the first one may be ameliorated too :
>> e^pi - 20 + e/pi^7 + e^(-2)/pi^14 = 3.141592653549... and so on...
>>
>> You may too combine two results from the link to get :
>> e^pi-(67*pi^9/e^8-430)/12 = 3.141592678...
>> with the variant (e/pi)^8*((e^pi-pi)*12+430)/67 = 3.14159265498..
>>
>> or combine with pi^2 + pi/24 ~= 10 to get
>> e^(-1863/422)-(pi^2-10)*24 = 3.14159265436...
>> or
>> e^(-12/e)-(pi^2-10)*24 = 3.14159433...
>> e^(-12/e)-e^6/(2*pi)^(21/2)-(pi^2-10)*24 = 3.141592653563...
>>
>> and so on...
>>
>> You may of course approximate pi with a predefined precision if you
>> accept the formula to be long enough (by adding each time a
>> e^(-fraction) term)!
>> A better 'quality criteria' could be obtained by dividing the number
>> of correct digits by the length of the formula
>> <http://en.wikipedia.org/wiki/Kolmogorov_complexity>.
>>
>
> Thanks again Raymond,
>
> It is interesting on how you come up with these.
(entering in the kitchen... :-))
Start for example with :
pi^3 - 2 pi/10^3 ~= 31
and replace one of the terms appearing (pi, 2, pi, 10^3, 31) by
another term so that the equality 'become more exact'.
For example 10^3 should be replaced by
2*pi/(pi^3-31) = 1001.036377...
so that 1001 should be better but not good enough...
subtract 1000, take the log, invert and you get nearly 28 so that
pi^3 - 2 pi/(10^3+e^(1/28)) ~= 31 with more precision
and you got another approximation :
31/(Pi^2-2/(10^3+e^(1/28))) = 3.141592653600...
Of course I used other recipes :
- integer relations algorithms (extension of continued fractions)
<http://en.wikipedia.org/wiki/Integer_relation_algorithm>
(I use lindep() in the excellent pari/gp
<http://pari.math.u-bordeaux.fr/download.html>)
- combine two (or more) relations in one for example by dividing
the errors in the two relations and replace the result by a fraction
- add more error terms using integer relations applied to the
logarithm of the |error term|, log(e), log(pi), log(2)... (trick to get
a multiplicative answer using an additive algorithm!)
- of course many other tricks may (and should :-)) be tried!
Let's use this with pi^2 + pi/24 ~= 10 :
- search a better fraction than 24 => 2337/97
- get a new result (without 'e' terms) :
10/pi-97/2337 = 3.14159265730...
- compute the log of the error term : a= log(10/Pi-97/2337-Pi)
- use the integer relation algorithm (changing precision...)
lindep([a,1,log(pi),log(2)],8) -> [-1, 72, 14, -155]~ so that
|-log(10/Pi-Pi-97/2337)+72*log(e)+14*log(pi)-155*log(2)| << 1
- conclude that the error term was near e^72*pi^14/2^155
- continue these iterations to get :
10/pi-97/2337 = 3.14159265730...
10/pi-97/2337-e^72*pi^14/2^155 = 3.141592653589804...
10/pi-97/2337-e^72*pi^14/2^155-e*(e/pi)^114/2^24 =
3.141592653589793238465869...
10/pi-97/2337-e^72*pi^14/2^155-e*(e/pi)^114/2^24-2^84/e^31/pi^65 =
3.1415926535897932384626433800...
I'll stop here and let you play with all that too!
Enjoy!
Raymond
> Although that last one looks most interesting --
> e^(-12/e)-e^6/(2*pi)^(21/2)-(pi^2-10)*24 = 3.141592653563...
> I fail to get it to come out right.
> I am probably getting the order of preference wrong
> for one or more of the operators.
Perhaps that this output will clarify my notations :
/ 12 \ 2 exp(6)
exp| - ------ | - 24 PI - ---------- + 240
\ exp(1) / 21/2
(2 PI)
Nmacgre
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