Series for pi?
By Pooya Farshim (P2572) on Sunday,
January 28, 2001 - 02:32 pm :
Prove or disprove:
By Pooya Farshim (P2572) on Monday,
January 29, 2001 - 12:58 pm :
The above sum gets close to p2/7, but not quite close enough. How can I
prove:
- the sum is convergent;
- its limit is not p2/7?
By Michael Doré (Md285) on Monday, January 29,
2001 - 11:08pm:
Hi Pooya,
This is quite interesting. I can't believe the fact that it gets so close to
p2/7 is a co-incidence yet the error never seems to go below about 0.1%.
To show it converges - first prove that in the sequence {sgn sin
n} the 1s and -1s always occur in blocks of 3 or 4, and that
between two blocks of 4 "1s" must occur a block of 4 "-1s" and
vice versa. From that try and get some sort of upper and lower
bound on the sum and hence show it converges.
By Pooya Farshim (P2572) on Tuesday,
January 30, 2001 - 01:31 pm :
Thanks Michael. I came up with this series while playing with Riemann's
re-arrangement theorem. I found it quite interesting that it goes so close
to p2/7. But how can I prove that the limit is not this value (I
calculated the sum to n=107 and it's not as close as to p2/7 as it
should be.)
By Pooya Farshim (P2572) on Tuesday,
January 30, 2001 - 04:44 pm :
Prove or disprove:
By Pooya Farshim (P2572) on Tuesday,
January 30, 2001 - 05:38 pm :
Well Michael! How about this one? If J is the value of the above integral
from 1 to n, then eJ is a rational whenever sgn sinn is positive and
is a rational multiple of p2 whenever sgn sinn is negative (clearly!)
Is there a connection between this p2 and the p2 we had above?
By Pooya Farshim (P2572) on Tuesday,
January 30, 2001 - 05:54 pm :
By the way the value of the sum to n=108 is
1.409238392475385
By Pooya Farshim (P2572) on Tuesday,
January 30, 2001 - 06:03 pm :
The convergence is obvious! The value of the sum is always
between sum to 3 and sum to 6. (i.e. between 55/30 and 73/60)
By Kerwin Hui (Kwkh2) on Tuesday,
January 30, 2001 - 06:41 pm :
In the following argument, log is taken base e.
|
= |
ó
õ
|
p
1
|
(1/x) dx - |
ó
õ
|
2p
p
|
(1/x) dx + |
ó
õ
|
3p
2p
|
- ... |
|
|
= logp- log2 + log3/2 - log4/3 + ... |
|
|
= logp- log (2*(2/3)*(4/3)*(4/5)*...) |
|
|
= logp- log( |
¥
Õ
n=1
|
(4n2/2n2 - 1)) |
|
|
= logp- log( |
¥
Õ
n=1
|
1/(1 - 1/(4n2))) |
|
From Euler's sine product [
| sinx = x |
¥
Õ
n=1
|
(1-x2/(n2p2))
|
], we have that when
| x = 1/2 p, 1 = 1/2 p |
¥
Õ
n=1
|
(1 - 1/(4n2))
|
, i.e.
|
|
ó
õ
|
¥
1
|
(sign sinx)/x dx = logp+ log(2/p) = log2 |
|
Kerwin
By Michael Doré (Md285) on Tuesday, January 30,
2001 - 11:24pm:
Yep, the key to that one is Euler's product, and I thought at first that your
first series was also a consequence of this. I'm still not entirely convinced
that your first series doesn't converge to p2/7 - perhaps it just converges
incredibly slowly after a quick start. Or maybe there is some subtlely
different series which only differs very slightly from yours which does converge
to p2/7.
The second identity is much easier to analyse than the first one
because with the first one you need to evaluate the function at
discrete points; but in the second one you can just wait till sin
x changes sign and split up the integral.
I'm not quite sure what you mean by eJ is rational,
where J is the integral from 1 to n. Is n an integer here?
By Pooya Farshim (P2572) on Wednesday,
January 31, 2001 - 01:17 pm :
Thanks Kerwin. That was pretty good. (As Expected!)
Yes! n is an integer here.
I was thinking whether g and ln(2) could play a role in the sum...
By Pooya Farshim (P2572) on
Wednesday, January 31, 2001 - 01:31 pm :
The value of the sum is not p2/7:
As sgnsin(108+1)=sgn sin(108+5)=1 and sgnsin(108+i)=-1 for i=2, 3, 4,
the value of the sum lies between sum to n=108+1 and sum to 108+4. However
p2/7 is less than the latter.
By Brad Rodgers (P1930) on Friday,
February 2, 2001 - 08:06 pm :
Could someone explain what
denotes, and why that Euler's relationship
holds?
Thanks,
Brad
By Michael Doré (Md285) on Friday, February 2,
2001 - 08:11pm:
|
|
n
Õ
i=m
|
f(i)=f(m)f(m+1)¼f(n-1)f(n) |
|
Basically it is exactly the same as
except it is a product not a sum.
I can't offer a good explanation for why Euler's sine product holds I'm afraid
- I've been trying to prove it for a while. It is a plausible
relationship because the product has roots at np where n is an integer,
as does sinx. This was how Euler first speculated that the relationship
might hold - I don't know if he proved it rigorously. I'll keep trying and get
back to you if I can prove it.
By Pooya Farshim (P2572) on Friday,
February 2, 2001 - 11:36 pm :
There is a rigorous proof of the identity in one of my
analysis books. It's long, complicated and refers back to some
previous theorems. I'd rather accept it for the moment....
By Brad Rodgers (P1930) on Saturday,
February 3, 2001 - 04:39 am :
Perhaps it's just ignorance acting at the moment, but how do we
know that
| 2×(2/3)×(4/3)¼ = |
¥
Õ
1
|
4n2/(4n2-1)
|
?
Brad
By Kerwin Hui (Kwkh2) on Saturday,
February 3, 2001 - 06:24 pm :
Brad,
Grouping the terms into blocks of 2, and a bit of arithmetic
should convince you of why it is true.
Kerwin
By Kerwin Hui (Kwkh2) on Saturday,
February 3, 2001 - 06:44 pm :
The proof of Euler's sine product
requires contour integration and various theorems concerning
complex analysis. which is probably in 2nd year
university.