| T.R | Title | User | Personal
Name | Date | Lines |
|---|
| 1651.1 | what about other than 3? | STAR::ABBASI | I spell check | Mon Aug 10 1992 04:36 | 7 |
| What about sum(n=1..oo) 1/n^p for p>3
do we know anything about them? i.e do we have a value for this sum?
we know that for p=3, it is not known, does that mean for all p>= 3 it
is also not known?
/nasser
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| 1651.2 | more info | STAR::ABBASI | I spell check | Mon Aug 10 1992 15:52 | 2 |
| I asked about this, all even sums are known, i.e any
sum(n=1..oo) 1/n^p where p is even is known.
|
| 1651.3 | On the basis of 15 minutes of research | VMSDEV::HALLYB | Fish have no concept of fire. | Mon Aug 10 1992 20:53 | 20 |
| N sum(1/j^N) very approximately (i.e., BASIC single precision)
2 1.64473 Pi^2/6
3 1.20205 Pi^3/25.8
4 1.08232 Pi^4/90
5 1.03693 Pi^5/295
6 1.01734 Pi^6/945
7 1.00835 Pi^7/2995
8 1.00408 Pi^8/9450
9 1.00201 ...
On this tissue of evidence I conjecture that
IF R[2k] is defined by: sum(n=1..oo) 1/n^p = R[2k]*Pi^2k,
THEN R[2k+1] = SQRT(R[2k]*R[2k+2]).
John
dictated but not spell-checked :-)
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| 1651.4 | closed formula for even N's | STAR::ABBASI | I spell check | Tue Aug 11 1992 04:58 | 24 |
| > 5 1.03693 Pi^5/295
you get the 295 offcourse by doing Pi^5/1.03693, and approximate
to it.
the problem I think is that we dont know if the multiplier of
Pi^N (when N is odd is rational or irrational), I think that is
the problem here.
can you see a pattern for the odd N's multipliers?
the values for even N can all be obtained anlytically exactly
the formula is (for even N's):
2k
(2 Pi)
sum(n=1..oo) 1/n^2k = ---------- | B_2k |
2 (2K)!
the books says, this holds for all natural numbers K, where B_2k are
the Bernoulli numbers. (you can see if your even N results matches the
above formula).
will talk more on this later..
/Nasser
I spell checked (almost)
|
| 1651.6 | Riemann zeta function | CX3PT2::KOWTOW::J_MARSH | | Tue Aug 11 1992 17:06 | 10 |
| FWIW,
Sum 1/(n^k) n = 1 ... infinity
equals the Riemann zeta function with argument k.
Perhaps there exist efficient routines for calculating the Riemann zeta
function?
-- Jeff
|
| 1651.7 | | GAUSS::ROTH | Geometry is the real life! | Tue Aug 11 1992 17:16 | 15 |
| You're asking for zeta(3) which is known to be irrational, but
it is not known if it is made up of any simple relation involving
pi or not.
You can evaluate the sum accurately with Euler Maclurin summation
by basically thinking of the sum as a trapezoidal rule approx to
an integral and expressing the error comitted as an asymptotic
expansion involving the odd order derivatives at the endpoints
of summation. Maple would do this for you without your having
to even think about it, but I don't have access to it right now.
People have evaluated numerous continued fraction convergants of
this in the hopes of finding gold there...
- Jim
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| 1651.8 | We've got computers, we're tapping phone lines,... | SSAG::LARY | Laughter & hope & a sock in the eye | Tue Aug 11 1992 18:02 | 48 |
| 1651.9 | Proof of previous accuracy claim | SSAG::LARY | Laughter & hope & a sock in the eye | Tue Aug 11 1992 19:42 | 32 |
| 1651.10 | "You're no Dave Hilbert" | VMSDEV::HALLYB | Fish have no concept of fire. | Tue Aug 11 1992 21:34 | 6 |
| > Anyone have a BIGNUM package and some computer time?
I just picked up the MP package from note 26.6 and will soon be making
good use (VAX 9000-400) of Approx(K) to verify my earlier conjecture. :-)
John
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| 1651.11 | In that case... | SSAG::LARY | Laughter & hope & a sock in the eye | Wed Aug 12 1992 06:56 | 12
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