For
example,
L(χ, s)
where χ is the Dirichlet character with label 8, 3 is given as,
1 + 1/3s – 1/5s – 1/7s + 1/9s + 1/11s – 1/13s – 1/15s + …
= 1/(1 – 1/3s) * 1/(1 + 1/5s) * 1/(1 + 1/7s)
* 1/(1 – 1/11s)
* 1/(1 + 1/13s) * …
So when on
dividing each successive number n by 8, if remainder is an even number then
χ(n) = 0; if remainder is 1 or 3, χ(n) = 1; if remainder is 5 or 7, χ(n) = – 1.
When s is a
positive odd integer, the above L-function has a value of the form kπs/√2
So for example, when s = 1,
1 + 1/3 – 1/5 – 1/7 + 1/9 + 1/11 – 1/13 – 1/15 + …
= √2π/4
It is interesting to contrast this with the Dirichlet Beta
Function (when s = 1)
i.e. 1 – 1/3 + 1/5 – 1/7 + 1/9 – 1/11
+ 1/13 – 1/15 + …
= π/4
Therefore,
(1 + 1/3 – 1/5 – 1/7 + 1/9 + …)/(1 – 1/3 + 1/5
– 1/7 + 1/9 – …) = √2
Then when s = 3,
1 + 1/33 – 1/53 – 1/73 + 1/93 + 1/113 – 1/133 – 1/153 + …
= 3π3/(64√2)
Again the above L-function, where s is a positive odd
integer, can be related to the Riemann zeta function (excluding those terms
where 2 is a factor).
(1 + 1/3s
– 1/5s – 1/7s + 1/9s
+ …)2/(1 + 1/32s + 1/52s
+ 1/72s + 1/92s + …)
= 1 + 2/3s – 2/5s – 2/7s + 2/9s + 2/11s – 2/13s – 4/15s …,
with a product over primes expression,
1/{1 – 2/(3s + 1)} *
1/{1 + 2/(5s +
1)} * 1/{1 + 2/(7s
+ 1)} * 1//{1 – 2/(11s + 1)} *
1{1 + 2/(13s
+ 1)} …
And this
will always result in a rational number.
So for
example when s = 1,
(1 + 1/3 – 1/5 – 1/7 + 1/9 + …)2/(1 + 1/32 + 1/52 + 1/72 + 1/92 + …)
= 1 + 2/3 – 2/5 – 2/7 + 2/9 + 2/11 – 2/13 – 4/15 + …,
with a product over primes expression,
1/{1 – 2/(31 + 1)} *
1/{1 + 2/(51 +
1)} * 1/{1 – 2/(71
+ 1)} * 1/{1 + 2/(111 + 1)} *
1/{1 – 2/(131
+ 1)} …
= 4/2 * 6/8
* 8/6 * 12/14 * 14/12 * …
= 1.
So this is
in its own way remarkable i.e. that the L-function,
1 + 2/3 – 2/5 – 2/7 + 2/9 + 2/11 – 2/13 – 4/15 + …
= 1
And in
brief when s = 3,
1 + 1/33 – 1/53 – 1/73 + 1/93 + …)2/(1 + 1/36 + 1/56 + 1/76 + 1/96 + …)
= 1 + 2/33 – 2/53 – 2/73 + 2/93 + 2/113 – 2/133 – 4/153…
= 135/128.
Then, L(χ,
s) where χ is the Dirichlet character with label 8, 5 is given as,
1 – 1/3s – 1/5s + 1/7s + 1/9s – 1/11s
– 1/13s + 1/15s + …
= 1/(1 + 1/3s) * 1/(1 + 1/5s) * 1/(1 – 1/7s) * 1/(1 + 1/11s) * 1/(1 +
1/13s) * …
So here
when on dividing each successive number n by 8, if remainder is an even number
then χ(n) = 0; if remainder is 1 or 7, χ(n) = 1; if remainder is 3 or 5, χ(n) =
– 1.
When s is a
positive even integer, the above L-function has a value of the form kπs/√2
So for example, when s = 2, the Dirichlet series expansion
is
1 – 1/32 – 1/52 + 1/72 + 1/92 – 1/112 – 1/132 + 1/152 + …
= π2/(8√2)
And when s = 4, the Dirichlet series expansion is
1 – 1/34 – 1/54 + 1/74 + 1/94 – 1/114 – 1/134 + 1/154 +
…
= 11π4/(768√2)
Again the above L-function, where s is a positive even
integer, can be related to the Riemann zeta function (excluding those terms
where 2 is a factor).
In this
case,
(1 – 1/3s – 1/5s + 1/7s + 1/9s – …)2/(1 + 1/32s
+ 1/52s + 1/72s + 1/92s + …)
= 1 – 2/3s – 2/5s + 2/7s + 2/9s – 2/11s – 2/13s + 4/15s + …,
with a product over primes expression,
1/{1 + 2/(3s + 1)} *
1/{1 + (2/(5s +
1)} * 1/{1 – (2/(7s + 1)} * 1/{1 + 2/(11s + 1)} *
1/{1 + 2/(13s + 1)} *
…
Furthermore,
the answer will be a rational fraction.
So when for
example s = 2,
(1 – 1/32 – 1/52 + 1/72 + 1/92 – …)2 /(1 + 1/34
+ 1/54 + 1/74 + 1/94 + …)
= 1 – 2/32 – 2/52 +
2/72 + 2/92 –
2/112 – 2/132 + 4/152 + …,
with a
product over primes expression,
1/{1 + 2/(32 + 1)} *
1/{1 + (2/(52 +
1)} * 1/{1 – (2/(72 + 1)} * 1/{1 + 2/(112 + 1)} *
1/{1 + 2/(132 + 1)} *
…
= 10/12 *
26/28 * 50/48 * 122/124 * 170/172 * …
= 3/4.
And when s
= 4,
(1 – 1/34 – 1/54 + 1/74 + 1/94 – …)2 /(1 + 1/38
+ 1/58 + 1/78 + 1/98 + …)
= 1 – 2/34 – 2/54 +
2/74 + 2/94–
2/114 – 2/134 + 4/154 + …,
with a
product over primes expression,
1/{1 + 2/(34 + 1)} *
1/{1 + (2/(54 +
1)} * 1/{1 – (2/(74 + 1)} * 1/{1 + 2/(114 + 1)} *
1/{1 + 2/(134 + 1)} *
…
= 82/84 * 626/628
* 2402/2400 * 14642/14644 * 28562/28564 * …
= 4235/4352.
No comments:
Post a Comment