sum of inverse almost cubes
Han de Bruijn
sum(k=-infinity..+infinity) 1/(1/4 + k)^3 = 2.Pi^3
I have a truly marvelous demonstration of this proposition which this
margin is too narrow to contain. (Pierre de Fermat)
Han de Bruijn
Pubkeybreaker
Does something compel you to post nonsense??? Your claim is
trivialy false.
Robert Israel
> On Aug 18, 10:38=A0am, Han de Bruijn <umum...@gmail.com> wrote:
> > Theorem:
> >
> > sum(k=3D-infinity..+infinity) 1/(1/4 + k)^3 =3D 2.Pi^3
> >
> > I have a truly marvelous demonstration of this proposition which this
> > margin is too narrow to contain. (Pierre de Fermat)
> >
> > Han de Bruijn
>
> Does something compel you to post nonsense??? Your claim is
> trivialy false.
The equation is true, actually; I don't know about his margin.
More generally, sum_{k=-infty}^infty 1/(c + k)^3
= Pi^3*cot(Pi*c)/(1+cot(Pi*c)^2) for non-integer c. You can get
this by differentiating the well-known
sum_{k=-infty}^infty 1/(c+k)^2 = Pi^2*csc(Pi*c)^2
--
Robert Israel isr...@math.MyUniversitysInitials.ca
Department of Mathematics http://www.math.ubc.ca/~israel
University of British Columbia Vancouver, BC, Canada
Han de Bruijn
wrote:
> Pubkeybreaker <pubkeybrea...@aol.com> writes:
> > On Aug 18, 10:38=A0am, Han de Bruijn <umum...@gmail.com> wrote:
> > > Theorem:
>
> > > sum(k=3D-infinity..+infinity) 1/(1/4 + k)^3 =3D 2.Pi^3
>
> > > I have a truly marvelous demonstration of this proposition which this
> > > margin is too narrow to contain. (Pierre de Fermat)
>
> > > Han de Bruijn
>
> > Does something compel you to post nonsense??? Your claim is
> > trivialy false.
>
> The equation is true, actually; I don't know about his margin.
> More generally, sum_{k=-infty}^infty 1/(c + k)^3
> = Pi^3*cot(Pi*c)/(1+cot(Pi*c)^2) for non-integer c. You can get
> this by differentiating the well-known
> sum_{k=-infty}^infty 1/(c+k)^2 = Pi^2*csc(Pi*c)^2
With a little more algebra, I find:
1 - 1/3^3 + 1/5^3 - 1/7^3 + 1/9^3 - 1/11^3 + ..
= Pi^3/32 = sum(k=0..oo) (-1)^k/(2k+1)^3 .
But I thought it was an _original_ claim. Not so?
Han de Bruijn
quasi
Original or not, it's nice.
Which suggests a question ...
For which positive integers k does
1 - 1/3^k + 1/5^k - 1/7^k + 1/9^k - 1/11^k + ...
converge to a number of the form
a*(Pi^b)
where a is rational and b is a positive integer?
quasi
quasi
My intuition suggests that except for the known values k=1 and k=3,
there are no such positive integers k.
quasi
OwlHoot
It looks like a Zeta function, with a trivial multiplicative
property of coefficients. So maybe, like the Riemann Zeta
Function it works for any odd integer > 0.
Cheers
John Ramsden
James Waldby
> On Aug 18, 7:50 pm, quasi wrote:
...
>> Which suggests a question ...
>>
>> For which positive integers k does
>>
>> 1 - 1/3^k + 1/5^k - 1/7^k + 1/9^k - 1/11^k + ...
>>
>> converge to a number of the form
>>
>> a*(Pi^b)
>>
>> where a is rational and b is a positive integer?
> It looks like a Zeta function, with a trivial multiplicative
> property of coefficients. So maybe, like the Riemann Zeta
> Function it works for any odd integer > 0.
Numerically it seems to work for k=5 and k=7; other numbers less so.
s(5)/pi^5 ~ .00325520833333333333345 and
s(7)/pi^7 ~ .000330946180555555555572
(s(k) = 100000 terms of given sum; arithmetic with 50 digits;
number of correct digits not known)
--
jiw
David Bernier
s(7) ~= (61/184320)*pi^7 ; 200,000 terms, 57 significant digits, 40 displayed.
(a := partial sum of 200,000 terms for s(7) ).
? a - (61/184320)*Pi^7
-3.0517578119659424046 E-40
David Bernier
David Bernier
According to Wikipedia, s(1), s(3), s(5), s(7) ... are the values
of Dirichlet's beta function (at 1, 3, 5 and 7 (resp.) ... );
beta_Dirichlet(5) = 5 pi^5/1536 , and so on:
< http://en.wikipedia.org/wiki/Dirichlet_beta_function > .
David Bernier
quasi
But, based on replies so far, if I correctly understand the claimed
results, it appears that with regard to what I intuited above, the
extreme opposite is true (which was my 2nd choice), namely, such an
expression is possible for _all_ odd positive integers k. Yes?
Now for even integers k, there is the known result:
1 + 1/2^2 + 1/3^2 + 1/4^2 + ... = Pi^2/6
Does this also extend? In other words, this question ...
For which positive integers k does
1 + 1/2^k + 1/3^k + 1/4^k + ...
converge to a number of the form
a*(Pi^b)
where a is rational and b is a positive integer?
Remarks:
Of course for k=1, the series diverges. But for k > 1, what's the
story? Is it just k=2? Or perhaps all even positive integers k? Or is
it all positive integers k > 1?
quasi
achille
> For which positive integers k does
>
> 1 + 1/2^k + 1/3^k + 1/4^k + ...
>
> converge to a number of the form
>
> a*(Pi^b)
>
> where a is rational and b is a positive integer?
For k = 2n even,
1 + 1/2^k + 1/3^k + 1/4^k + ...
= zeta(2n)
= ((2 pi)^(2n)(-1)^(n+1)B_{2n})/( 2 (2n)! )
where B_n are Bernoulli numbers
Han de Bruijn
Zeta function, always that Zeta function ! No wonder they can't find !
You'd better think of a generalization of: Shannon's Sampling Theorem.
That's actually where my (not Robert Israel's) proof comes from, to be
published here as soon as I find time.
Han de Bruijn
David Bernier
The set of integers k>0 with the required property
includes all even positive integers k. For k = 3,
Apery proved around 1980 that the sum is an irrational
number, and experts in irrationality proofs were
"very impressed".
zeta(s) := sum_{n=1... infty} 1/(n^s).
I think Euler thought of zeta as defined at
least for real numbers s with s > 1.
zeta(2n) = (-1)^(n+1) B_{2n} (2pi)^(2n)/(2*(2n)!) ,
where B_2n are the even index Bernoulli numbers, i.e.
B_2, B_4, B_6 ,etc.
All the Bernoulli numbers are rational, and
according to Wikipedia, there's only a variation
of notation/definition affecting B_1 (so this
has no bearing on Euler's formula for zeta(2n)).
About the Bernoulli numbers:
< http://en.wikipedia.org/wiki/Bernoulli_number > .
Test of formula for sum of inverses 30th powers [ n = 15]:
? (-1)^(16)*bernreal(30)*((2*Pi)^30)/(2*factorial(30))
1.0000000009313274324196681828717647350
? zeta(30)
1.0000000009313274324196681828717647350
David Bernier
Gerry Myerson
quasi <qu...@null.set> wrote:
> Now for even integers k, there is the known result:
>
> 1 + 1/2^2 + 1/3^2 + 1/4^2 + ... = Pi^2/6
>
> Does this also extend? In other words, this question ...
>
> For which positive integers k does
>
> 1 + 1/2^k + 1/3^k + 1/4^k + ...
>
> converge to a number of the form
>
> a*(Pi^b)
>
> where a is rational and b is a positive integer?
As others have answered, and as Euler knew,
this happens for all even k > 0, with b = k. What happens for odd k > 1
is a mystery, but the smart money is on it never happening. It has been
shown, for example, that if k = 3 gives a rational multiple of pi^3 then
the denominator of the rational is utterly enormous. Look for
"integer relation finding algorithms" and/or PSLQ.
--
Gerry Myerson (ge...@maths.mq.edi.ai) (i -> u for email)
Han de Bruijn
Yes, that fills some gaps. So alternating series of odd inverse powers
_have_ been known before yesterday. Other sources on the internet have
suggested, upon sloppy reading from my side, that they were not:
http://library.thinkquest.org/28049/Summing%20reciprocals.html
<quote> Despite this, nothing was known about sums of reciprocals of
odd powers up to 1978.Then in a tour de force of elementary though
difficult reasoning based on a curious expression for ζ(3) in terms
of binomial coefficients, R. Apery of France proved that ζ(3) is
irrational. The proof was controversial, and was referred to as
"a mixture of miracles and mysteries". Presumably, ζ(5), ζ(7), ζ(9)
... are also irrational, but it is still unknown.
The remaining question is : does ζ(3) = kπ^3 ? </quote>
But ah, _that_ is for _non_ alternating series of the kind.
http://www.hs-augsburg.de/~mueckenh/GU/GU02c.PPT#280,22,Slide 22
<quote> Reihenwert fuer ungerade Potenzen [ p ]: vergeblich gesucht
bislang noch nicht gefunden </quote>
That's again for non alternating series : sum(k=0..oo) 1/k^p = ??
Han de Bruijn
Han de Bruijn
Han de Bruijn
Related materials:
http://groups.google.nl/group/sci.math/msg/312fccff71ac71da
http://hdebruijn.soo.dto.tudelft.nl/jaar2010/dikte/document.pdf
http://hdebruijn.soo.dto.tudelft.nl/jaar2010/
The subject of our current Projects is a Comb P(x) of bell shaped
functions
p(x) on a one dimensional, infinite and equidistant grid with
discretization
(delta) on the real axis with coordinate x :
P(x) = sum(L=-oo..+oo) p(x - L.delta)
P(x) is obviously periodic with period (delta). So it can be developed
into
a Fourier series. Also the function p(x) is assumed to be symmetric
around
x = 0, meaning that P(x) = P(-x), resulting in real valued
coefficients:
a_k = 1/(delta/2) integral(-delta/2..+delta/2) P(x).cos(k.omega.x)
dx
Here omega = 2.Pi/delta . Then let the calculations continue:
= 1/(delta/2) integral(-delta/2..+delta/2)
[ sum(L=-oo..+oo) p(x - L.delta) ] cos(k.omega.x) dx
Substitute y = x - L.delta and integrate to y :
a_k = 1/(delta/2) sum(L=-oo..+oo)
integral(-delta/2 - L.delta .. +delta/2 - L.delta)
p(y) cos(k omega [y + L.delta]) dy
Where:
cos(k.omega [y + L.delta]) = cos(k.omega y + k.L.2.Pi) =
cos(k.omega.y)
Next replace y by -y and switch integration bounds:
a_k = 1/(delta/2) sum(L=-oo..+oo)
integral(L.delta - delta/2 .. L.delta + delta/2)
p(y) cos(k omega y) dy
The above integrals are precisely the adjacent pieces of another
integral
which has bounds reaching to infinity. That is, they sum up to an
infinite
integral:
a_k = 1/(delta/2) integral(-oo..+oo) p(y) cos(k omega y) dy
This proves the main result:
P(x) = sum(L=-oo..+oo) p(x - L.delta)
= a_0/2 + sum(k=0..oo) a_k cos(k omega x)
Apart from a constant, a_k is a _sampling_ of the Fourier integral of
p(x) :
A(y) = integral(-oo..+oo) p(x) cos(y x) dx ; a_k = A(k.omega)/
(delta/2)
Now let the function p(x) be _normed_ in such a way that A(0) = 1 :
A(0) = integral(-oo..+oo) p(x) dx = 1 then a_0/2 = 1/delta
Written otherwise:
P(x).delta = sum(L=-oo..+oo) p(x - L.delta).delta
= 1 + 2 sum(k=0..oo) A(k.omega) cos(k omega x)
A reason to write this is that the left hand side can be interpreted
as the
Riemann sum of the convolution integral of a constant (1) and the
shape p(x).
Some bell shaped functions p(x) have a Fourier integral A(y) which
vanishes
for y outside an interval around zero. With help of MAPLE we find an
example.
Here y is in the cartesian domain and x is in the frequency domain:
> f(y) := int((1-x^2)*cos(y*x)/(2*Pi),x=-1..+1);
2 (y cos(y) - sin(y))
f(y) := - ---------------------
3
y Pi
> g(x) := int(f(y)*cos(y*x),y=-infinity..+infinity);
2 2
g(x) := -1/2 csgn(x + 1) x + 1/2 csgn(x + 1) + 1/2 csgn(x - 1) x
- 1/2 csgn(x - 1)
> int(f(y),y=-infinity..+infinity);
1
Here g(x) is a piece of a parabola above the x-axis with zeroes
(-1,+1).
The function is zero everywhere else. This is quite essential and
that's why
the theory bears great resemblance to Shannon's Sampling Theorem.
So let us define: P(x).delta = sum(L=-oo..+oo) f(x - L.delta).delta
= 1 + 2 sum(k=0..oo) g(k.omega) cos(k omega x)
If we choose now omega = 2.Pi/delta = 1, then g(k.omega) = 0 for all
(k)
so the whole Fourier series at the right hand side vanishes, except
(1):
P(x).2.Pi = sum(L=-oo..+oo) f(x - L.2.Pi).2.Pi = 1
At last, in order to preserve the third power in the denominator of
f(y),
we choose x = Pi/2 , giving:
f(Pi/2 - L.2.Pi) = 4/( Pi/2 - L.2.Pi )^3 = 1/(2.Pi^3)/(1/4 - L)^3
Conclusion:
sum(L=-oo..+oo) 1/(2.Pi^3)/(1/4 - L)^3 = 1
Some rearrangement leads to the Original Posting. The proof is
complete.
Han de Bruijn
Han de Bruijn
Google scrambled the input, so I'll try again. Mind the seventy wide.
1234567890123456789012345678901234567890123456789012345678901234567890
Related materials:
http://groups.google.nl/group/sci.math/msg/312fccff71ac71da
http://hdebruijn.soo.dto.tudelft.nl/jaar2010/dikte/document.pdf
http://hdebruijn.soo.dto.tudelft.nl/jaar2010/
The subject of our current Projects is a Comb P(x) of bell shaped
functions p(x) on a one dimensional, infinite and equidistant grid
with discretization (delta) on the real axis with coordinate x :
P(x) = sum(L=-oo..+oo) p(x - L.delta)
P(x) is obviously periodic with period (delta). So it can be developed
into a Fourier series. The function p(x) is assumed to be symmetrical
around x = 0, meaning that P(x) = P(-x), resulting in real valued
coefficients:
a_k = 1/(delta/2) integral(-delta/2..+delta/2) P(x).cos(k.omega.x) dx
Here omega = 2.Pi/delta . Then let the calculations continue:
= 1/(delta/2) integral(-delta/2..+delta/2)
[ sum(L=-oo..+oo) p(x - L.delta) ] cos(k.omega.x) dx
Substitute y = x - L.delta and integrate to y :
a_k = 1/(delta/2) sum(L=-oo..+oo)
integral(-delta/2 - L.delta .. +delta/2 - L.delta)
p(y) cos(k omega [y + L.delta]) dy
Where:
cos(k.omega [y+L.delta]) = cos(k.omega y + k.L.2.Pi) = cos(k.omega.y)
Next replace y by -y and switch integration bounds:
a_k = 1/(delta/2) sum(L=-oo..+oo)
integral(L.delta - delta/2 .. L.delta + delta/2)
p(y) cos(k omega y) dy
The above integrals are precisely the adjacent pieces of another
integral which has bounds reaching to infinity. That is, they sum up
to an infinite integral:
a_k = 1/(delta/2) integral(-oo..+oo) p(y) cos(k omega y) dy
This proves the main result:
P(x) = sum(L=-oo..+oo) p(x - L.delta)
= a_0/2 + sum(k=0..oo) a_k cos(k omega x)
Apart from a constant, a_k is a _sampling_ of the Fourier integral
of p(x) :
A(y) = integral(-oo..+oo) p(x) cos(y x) dx
where a_k = A(k.omega)/(delta/2)
Now let the function p(x) be _normed_ in such a way that A(0) = 1 :
A(0) = integral(-oo..+oo) p(x) dx = 1 then a_0/2 = 1/delta
Written otherwise:
P(x).delta = sum(L=-oo..+oo) p(x - L.delta).delta
= 1 + 2 sum(k=0..oo) A(k.omega) cos(k omega x)
A reason to write this is that the left hand side can be interpreted
as the Riemann sum of the convolution integral of a constant (1) and
the shape p(x).
Some bell shaped functions p(x) have a Fourier integral A(y) which
vanishes for y outside an interval around zero. With help of MAPLE
we find an example. Here y is in the cartesian domain and x is in
the frequency domain:
> f(y) := int((1-x^2)*cos(y*x)/(2*Pi),x=-1..+1);
2 (y cos(y) - sin(y))
f(y) := - ---------------------
3
y Pi
> g(x) := int(f(y)*cos(y*x),y=-infinity..+infinity);
2 2
g(x) := -1/2 csgn(x + 1) x + 1/2 csgn(x + 1) + 1/2 csgn(x - 1) x
- 1/2 csgn(x - 1)
> int(f(y),y=-infinity..+infinity);
1
Here g(x) is a piece of a parabola above the x-axis with zeroes -1,+1.
The function is zero everywhere else. This is quite essential and thus
the theory bears great resemblance to Shannon's Sampling Theorem.
So let us define: P(x).delta = sum(L=-oo..+oo) f(x - L.delta).delta
= 1 + 2 sum(k=0..oo) g(k.omega) cos(k omega x)
If we choose now omega = 2.Pi/delta = 1, then g(k.omega) = 0 for all
(k) so the whole Fourier series at the right hand side vanishesk, with
exception of the constant = 1.
P(x).2.Pi = sum(L=-oo..+oo) f(x - L.2.Pi).2.Pi = 1
At last, in order to preserve the third power in the denominator of
f(y), we choose x = Pi/2 , giving:
f(Pi/2 - L.2.Pi) = 4/( Pi/2 - L.2.Pi )^3 = 1/(2.Pi^3)/(1/4 - L)^3
Conclusion:
sum(L=-oo..+oo) 1/(2.Pi^3)/(1/4 - L)^3 = 1
A rearrangement leads to the Original Posting. The proof is complete.
Han de Bruijn
Han de Bruijn
> On Aug 19, 1:22 pm, Han de Bruijn <umum...@gmail.com> wrote:
>
> Google scrambled the input, so I'll try again. Mind the seventy wide.
>
> 1234567890123456789012345678901234567890123456789012345678901234567890
>
> Related materials:
>
Apart from at least three typos, sorry: %s/k=0/k=1/ : vi.
Han de Bruijn