[math-fun] Plouffe's 2 surprising approximations and 1 exact formula for pi.
By LatticeReduce, it appears that these are all of the form -1/240 + <algebraic>*Gamma[1/4]^8/Pi^6, e.g. for k=6, Sum[n^3/(E^(2*Pi*n/k)-1),{n,Infinity}== -1/240 + ((231 + 120 Sqrt[2] 3^(1/4) + 140 Sqrt[3] + 60 Sqrt[2] 3^(3/4)) Gamma[1/4]^8)/(5120 Pi^6) and Sum[n^3/(E^(2*Pi*n/7)-1),{n,Infinity}== 1/4 3/4 1 8 (301 + 210 Sqrt[2] 7 + 120 Sqrt[7] + 90 Sqrt[2] 7 ) Gamma[-] 1 4 -(---) + ------------------------------------------------------------------ 240 6 5120 Pi In[135]:= N[%, 33] Out[135]= 10.0000000000000001901617678886627 For k=13, the algebraic is (1/402653184000)(188822323200 + 91226112000 Sqrt[13] + 3145728000 (2128581 + 50505 Sqrt[3] + 610709 Sqrt[13] + 22176 Sqrt[39])^(1/3) + (66260343848333767461568512000000000 + 19010687557237458566381568000000000 Sqrt[13] - 31128880624384868352000000000 (50505 Sqrt[3] + 22176 Sqrt[39]))^(1/3)), which I suspect simplifies. k=163 might be, to quote Prof Watson, "Too cumbrous to be of any importance". --rwg Simon says:-> Hello everybody, these days I am working on exponential sums and I have found something of interest, like infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 13 = 119.00000000000000000000000000000009593745851025547335588584913... the precision is 31 digits. Another one is infinity ----- 3 \ n ) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 7 = 10.0000000000000001901617678886626755843593058554453334802548978434061099438\ The precision here is 15 digits. For an argument of 2*Pi*n/163, the precision is 435 digits! I don't know why I took 163... (of course). in general, these sums will be near an integer if the argument is 2*Pi*n/k and k is NOT a multiple of 2,3 or 5, strange isn't ? These are the simplest I could find, if the exponent is of the form 4m-1 then the sum is often an integer with the same conditions. This, I believe extends a little bit the known formula of Ramanujan/Berndt/etc. A good question is : does someone has a simple explanation of this ? I don't. I am preparing a paper on these results. Because, I do have another one which is EXACT, namely for pi: In general these sums with a fractional exponent are very close to integers or pi, I have a couple of series for 1/pi, 1/pi^2, etc. here is the EXACT formula with fractional arguments mixed with integer exponents. /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | 10 | ) -----------------| - 40 | ) -------------------| | / n (exp(Pi n) - 1)| | / n (exp(2 Pi n) - 1)| | ----- | | ----- | \ n = 1 / \ n = 1 / /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 10 | ) -------------------| - 10 | ) -----------------| | / n (exp(4 Pi n) - 1)| | / / Pi n \| | ----- | | ----- n |exp(----) - 1|| \ n = 1 / \ n = 1 \ 5 // /infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | + 40 | ) -------------------| - 10 | ) -------------------| | / / 2 Pi n \| | / / 4 Pi n \| | ----- n |exp(------) - 1|| | ----- n |exp(------) - 1|| \ n = 1 \ 5 // \ n = 1 \ 5 //
evalf(%); 3.14159265358979323846264338327950288419716939937510582097494459230781640628\ 62089986280348253421170679821480865132823066470938442... well, I verified up to 1000 digits and it holds.
This is trivial, ? I do not see how. If someone has a piece of information on why this exist, I would be glad to ear from it, references, known results, etc. Me, I never saw these kind of formulas before, have a good day, Simon Plouffe
On Sun, Jan 23, 2011 at 2:30 PM, Bill Gosper <billgosper@gmail.com> wrote:
By LatticeReduce, it appears that these are all of the form
-1/240 + <algebraic>*Gamma[1/4]^8/Pi^6, e.g. for k=6, Sum[n^3/(E^(2*Pi*n/k)-1),{n,Infinity}== -1/240 + ((231 + 120 Sqrt[2] 3^(1/4) + 140 Sqrt[3] + 60 Sqrt[2] 3^(3/4)) Gamma[1/4]^8)/(5120 Pi^6)
and Sum[n^3/(E^(2*Pi*n/7)-1),{n,Infinity}==
1/4 3/4 1 8 (301 + 210 Sqrt[2] 7 + 120 Sqrt[7] + 90 Sqrt[2] 7 ) Gamma[-] 1 4 -(---) + ------------------------------------------------------------------ 240 6 5120 Pi
In[135]:= N[%, 33]
Out[135]= 10.0000000000000001901617678886627
For k=13, the algebraic is (1/402653184000)(188822323200 + 91226112000 Sqrt[13] + 3145728000 (2128581 + 50505 Sqrt[3] + 610709 Sqrt[13] + 22176 Sqrt[39])^(1/3) + (66260343848333767461568512000000000 + 19010687557237458566381568000000000 Sqrt[13] - 31128880624384868352000000000 (50505 Sqrt[3] + 22176 Sqrt[39]))^(1/3)),
which I suspect simplifies. k=163 might be, to quote Prof Watson, "Too cumbrous to be of any importance". --rwg
Simon says:->
Hello everybody,
these days I am working on exponential sums and I have found something of interest, like infinity ----- 3 \ n
) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 13
= 119.00000000000000000000000000000009593745851025547335588584913...
the precision is 31 digits.
Another one is infinity ----- 3 \ n ) ---------------
/ 2 Pi n ----- exp(------) - 1 n = 1 7
= 10.0000000000000001901617678886626755843593058554453334802548978434061099438\
The precision here is 15 digits.
For an argument of 2*Pi*n/163, the precision is 435 digits! I don't know why I took 163... (of course).
in general, these sums will be near an integer if the argument
is 2*Pi*n/k and k is NOT a multiple of 2,3 or 5, strange isn't ? These are the simplest I could find, if the exponent is of the form 4m-1 then the sum is often an integer with the same conditions. This, I believe extends
a little bit the known formula of Ramanujan/Berndt/etc.
A good question is : does someone has a simple explanation of this ? I don't. I am preparing a paper on these results.
Because, I do have another one which is EXACT, namely for pi:
In general these sums with a fractional exponent are very close to integers or pi, I have a couple of series for 1/pi, 1/pi^2, etc.
here is the EXACT formula with fractional arguments mixed with integer exponents.
By Rich's ancient number recognizer, sum(1/n/(e^(pi k n)-1)) all seem to be 3/4 ln pi -ln gamma(1/4) - k pi/24 + ln <algebraic>, e.g.
SUM(1/(N*(%E^(%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+7*LOG(2)/8-%PI/24 SUM(1/(N*(%E^(2*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+LOG(2)-%PI/12 SUM(1/(N*(%E^(3*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+7*LOG(SQRT(3)+1)/12+LOG(2-SQRT(2)*3^(1/4))/4+3*LOG(3)/8+11*LOG(2)/24-%PI/8 SUM(1/(N*(%E^(%PI*N/3)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+LOG(SQRT(3)+1)/12-LOG(2-SQRT(2)*3^(1/4))/4-LOG(3)/8+23*LOG(2)/24-%PI/72 SUM(1/(N*(%E^(4*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+11*LOG(2)/8-%PI/6 SUM(1/(N*(%E^(5*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4-LOG(5^(3/4)-5^(1/4)+2)/2+LOG(5)/2+11*LOG(2)/8-5*%PI/24 SUM(1/(N*(%E^(6*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+LOG(SQRT(3)+1)/6+3*LOG(3)/8+11*LOG(2)/12-%PI/4 which should imply your pi conjecture. --rwg
/infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \ 1 | 10 | ) -----------------| - 40 | ) -------------------|
| / n (exp(Pi n) - 1)| | / n (exp(2 Pi n) - 1)| | ----- | | ----- | \ n = 1 / \ n = 1 /
/infinity \ /infinity \
| ----- | | ----- | | \ 1 | | \ 1 | + 10 | ) -------------------| - 10 | ) -----------------|
| / n (exp(4 Pi n) - 1)| | / / Pi n \| | ----- | | ----- n |exp(----) - 1|| \ n = 1 / \ n = 1 \ 5 //
/infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \
1 | + 40 | ) -------------------| - 10 | ) -------------------| | / / 2 Pi n \| | / / 4 Pi n \| | ----- n |exp(------) - 1|| | ----- n |exp(------)
- 1|| \ n = 1 \ 5 // \ n = 1 \ 5 //
evalf(%); 3.14159265358979323846264338327950288419716939937510582097494459230781640628\
62089986280348253421170679821480865132823066470938442... well, I verified up to 1000 digits and it holds.
This is trivial, ? I do not see how.
If someone has a piece of information on why this exist, I would
be glad to ear from it, references, known results, etc. Me, I never saw these kind of formulas before,
have a good day, Simon Plouffe
Hello mr. Gosper, now that is interesting, you have the explicit formula for sum(1/n/(exp(Pi*n*k)-1),n=1..infinity) when k=1,2,3,4,5,6. Here with PSLQ or LLL I could get 1,2 and 4 only separatly in terms of log(Pi), Pi, log(2) and log(GAMMA(1/4)), I did not know that one could get those explicit algebrico-log-gamma expressions . What is surprising is the <cancel out> of the algebraic expression when k = 1/5, 2/5 and 4/5 and also the approximations when k = 2/13, 2/7 or 2/163, I tried to find other fractions for k and found only that explicit 1/5, 2/5 and 4/5. Nevertheles, I have found a simpler formulation of the formula for pi ; it appears on my home page at http://www.plouffe.fr/ I should switch to macsyma! best regards, Simon Plouffe
On Mon, Jan 24, 2011 at 7:58 PM, Bill Gosper <billgosper@gmail.com> wrote:
On Sun, Jan 23, 2011 at 2:30 PM, Bill Gosper <billgosper@gmail.com> wrote:
By LatticeReduce, it appears that these are all of the form
-1/240 + <algebraic>*Gamma[1/4]^8/Pi^6, e.g. for k=6, Sum[n^3/(E^(2*Pi*n/k)-1),{n,Infinity}== -1/240 + ((231 + 120 Sqrt[2] 3^(1/4) + 140 Sqrt[3] + 60 Sqrt[2] 3^(3/4)) Gamma[1/4]^8)/(5120 Pi^6)
and Sum[n^3/(E^(2*Pi*n/7)-1),{n,Infinity}==
1/4 3/4 1 8 (301 + 210 Sqrt[2] 7 + 120 Sqrt[7] + 90 Sqrt[2] 7 ) Gamma[-] 1 4 -(---) + ------------------------------------------------------------------ 240 6 5120 Pi
In[135]:= N[%, 33]
Out[135]= 10.0000000000000001901617678886627
For k=13, the algebraic is (1/402653184000)(188822323200 + 91226112000 Sqrt[13] + 3145728000 (2128581 + 50505 Sqrt[3] + 610709 Sqrt[13] + 22176 Sqrt[39])^(1/3) + (66260343848333767461568512000000000 + 19010687557237458566381568000000000 Sqrt[13] - 31128880624384868352000000000 (50505 Sqrt[3] + 22176 Sqrt[39]))^(1/3)),
which I suspect simplifies. k=163 might be, to quote Prof Watson, "Too cumbrous to be of any importance". --rwg
Simon says:->
Hello everybody,
these days I am working on exponential sums and I have found something of interest, like infinity ----- 3 \ n
) --------------- / 2 Pi n ----- exp(------) - 1 n = 1 13
= 119.00000000000000000000000000000009593745851025547335588584913...
the precision is 31 digits.
Another one is infinity ----- 3 \ n ) ---------------
/ 2 Pi n ----- exp(------) - 1 n = 1 7
= 10.0000000000000001901617678886626755843593058554453334802548978434061099438\
The precision here is 15 digits.
For an argument of 2*Pi*n/163, the precision is 435 digits! I don't know why I took 163... (of course).
in general, these sums will be near an integer if the argument
is 2*Pi*n/k and k is NOT a multiple of 2,3 or 5, strange isn't ? These are the simplest I could find, if the exponent is of the form 4m-1 then the sum is often an integer with the same conditions. This, I believe extends
a little bit the known formula of Ramanujan/Berndt/etc.
A good question is : does someone has a simple explanation of this ? I don't. I am preparing a paper on these results.
Because, I do have another one which is EXACT, namely for pi:
In general these sums with a fractional exponent are very close to integers or pi, I have a couple of series for 1/pi, 1/pi^2, etc.
here is the EXACT formula with fractional arguments mixed with integer exponents.
By Rich's ancient number recognizer, sum(1/n/(e^(pi k n)-1)) all seem to be 3/4 ln pi -ln gamma(1/4) - k pi/24 + ln <algebraic>, e.g.
SUM(1/(N*(%E^(%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+7*LOG(2)/8-%PI/24 SUM(1/(N*(%E^(2*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+LOG(2)-%PI/12 SUM(1/(N*(%E^(3*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+7*LOG(SQRT(3)+1)/12+LOG(2-SQRT(2)*3^(1/4))/4+3*LOG(3)/8+11*LOG(2)/24-%PI/8 SUM(1/(N*(%E^(%PI*N/3)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+LOG(SQRT(3)+1)/12-LOG(2-SQRT(2)*3^(1/4))/4-LOG(3)/8+23*LOG(2)/24-%PI/72 SUM(1/(N*(%E^(4*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+11*LOG(2)/8-%PI/6 SUM(1/(N*(%E^(5*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4-LOG(5^(3/4)-5^(1/4)+2)/2+LOG(5)/2+11*LOG(2)/8-5*%PI/24 SUM(1/(N*(%E^(6*%PI*N)-1)),N,1,INF) = -LOG(GAMMA(1/4))+3*LOG(%PI)/4+LOG(SQRT(3)+1)/6+3*LOG(3)/8+11*LOG(2)/12-%PI/4
which should imply your pi conjecture.
You'll also need Sum[1/((-1 + E^((n*Pi)/5))*n), {n, Infinity}] == -(Pi/120) - (9*Log[2])/8 - Log[3]/2 + Log[1 + Sqrt[5]] + (1/2)*Log[2 + 5^(1/4) + 5^(3/4)] + (1/2)*Log[4 + 5^(1/4) + 5^(3/4)] + (3*Log[Pi])/4-Log[Gamma[1/4]] and the conjecture Sum[1/((-1 + E^(2*n*Pi*k))*n), {n,Infinity}]- Sum[1/((-1 + E^(2*n*Pi/k))*n), {n,Infinity}]== Log[k]/2-Pi*(k-1/k)/12, which must be related to Jacobi's imaginary transformation. Yeah, this stuff is so trivial I can't prove any of it. Does anybody know how to coax a hundred or so digits out of a decently convergent Sum[...]?? Total[Table[...]] is ridiculous, and requires DIY convergence analysis. Even in 8.0. --rwg How can these sums be considered pi formulae if they prerequire precise values of e^pi?
--rwg
/infinity \ /infinity \
| ----- | | ----- | | \ 1 | | \ 1 | 10 | ) -----------------| - 40 | ) -------------------|
| / n (exp(Pi n) - 1)| | / n (exp(2 Pi n) - 1)| | ----- | | ----- | \ n = 1 / \ n = 1 /
/infinity \ /infinity \
| ----- | | ----- | | \ 1 | | \ 1 | + 10 | ) -------------------| - 10 | ) -----------------|
| / n (exp(4 Pi n) - 1)| | / / Pi n \| | ----- | | ----- n |exp(----) - 1|| \ n = 1 / \ n = 1 \ 5 //
/infinity \ /infinity \ | ----- | | ----- | | \ 1 | | \
1 | + 40 | ) -------------------| - 10 | ) -------------------| | / / 2 Pi n \| | / / 4 Pi n \| | ----- n |exp(------) - 1|| | ----- n |exp(------)
- 1|| \ n = 1 \ 5 // \ n = 1 \ 5 //
evalf(%); 3.14159265358979323846264338327950288419716939937510582097494459230781640628\
62089986280348253421170679821480865132823066470938442... well, I verified up to 1000 digits and it holds.
This is trivial, ? I do not see how.
If someone has a piece of information on why this exist, I would
be glad to ear from it, references, known results, etc. Me, I never saw these kind of formulas before,
have a good day, Simon Plouffe
On 1/25/2011 4:30 AM, Bill Gosper wrote:
Does anybody know how to coax a hundred or so digits out of a decently convergent Sum[...]?? Total[Table[...]] is ridiculous, and requires DIY convergence analysis. Even in 8.0.
A little coding does it: In[31]:= bSum[k_Real] := Module[{sum = 0, prev, n = 1, pwr, term = 1}, pwr = Exp[2 Pi k]; While[True, prev = sum; term *= pwr; sum += 1/((term - 1)*n); If[sum - prev == 0, Break[]]; n++ ]; sum] In[32]:= With[{k = 7`1000}, bSum[k] - bSum[1/k] == (Log[k]/2 - Pi*(k - 1/k)/12)] // AbsoluteTiming Out[32]= {0.1089891, True} --Sasha
participants (3)
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Bill Gosper -
Oleksandr Pavlyk -
Simon Plouffe