Chapter 6

All code for Section 6.1 is embedded in the .tex file, or else
include in the Sage script for chapter 9.

Section 6.2, "A smoothed sum of mu(n)/sigma(n)"

The Sage file boundk.sage computes

k(beta) = \prod_p (1 + p^{-(beta+1)}/((1+p^{-1})(1-p^{-beta})))

for several values of beta>0.

Output:
k(1-1/log(10^12)) is at most 1.692392
k(4/7) is at most 2.942346
k(8/15) is at most 3.243633
k(1/2) is at most 3.574861

---------------------------------------------------------

Section 6.3 and 6.4  involve no computations.

Section 6.5, "Bounds on \mathbb{h}_v"

The Sage file boundkap.sage computes

kap(beta_1, beta_2)
 = \prod_p (1 + p^{-(beta_1+beta_2)}/
                 ((1-p^(-beta_1)+p^(-1))*(1-p^(-beta_2)+p^(-1))))

for various values of beta_1 and beta_2.

Output:

kap(1,1-1/log(10^12)) is at most 1.552738
kap(1-1/log(10^12),1-1/log(10^12)) is at most 1.588493
kap(1,1/2) is at most 2.620185
kap(1-1/log(10^12),1/2) is at most 2.791671
kap(1,1/4) is at most 6.761469
kap(8/15,8/15) is at most 19.82414
kap(4/7,1/2) is at most 18.50841

./boldh_semian N v j M 

gives bounds on the quantity

\sum_{d\leq y: (d,v)=1} (\mu^2(d)/d^{(4-j)/4}) \prod_{p|d} f_j(p),

valid for all 2\leq y\leq N. Here j = 0, 1 or 2, and

f_0(p) = (p/(p-sqrt(p)+1))^2
f_1(p) = (p/(p-sqrt(p)+1)) (p/(p-p^(3/4)+1))
f_2(p) = (p/(p-p^(3/4)+1))^2

(As usual, v is either 1 or 2.)

The program uses a sieve working on blocks of size M.

The default values are
N=1000000, v=1, j=0, M=10000
If only N, v and j are specified, M is set to 2 sqrt(N) rounded up

Examples:
./boldh_semian 1000000000000 1 0
Output:
The bound is 1 + 1.41811638 log y

./boldh_semian 1000000000000 1 1
Output:
The bound is 14.19576243 y^(1/4)

./boldh_semian 1000000000000 1 2
Output:
The bound is 22.1042813 sqrt(y)

./boldh_semian 1000000000000 2 0
Output:
The bound is 1 + 0.7828167328 log y

./boldh_semian 1000000000000 2 1
Output:
The bound is 7.25779644 y^(1/4)

./boldh_semian 1000000000000 2 2
Output:
The bound is 10.28358099 sqrt(y)

Section 6.6, "Computing h_v for small arguments"

Subsection 6.6.1, "Computing h_v(y_1,y_2) for y_1, y_2 small"

./boundhvxy N r v flag

computes various bounds on
f_{r,v}(x) = \inf_{r x\leq y\leq N} \sqrt{x y} h_v(x,y)
for 1<=x<=N/r,
where r is an integer >= 1 and v = 1 or 2,
and h_v(x,y) = \sum_{d: (d,v)=1} mu(d)/sigma(d)
                            (\tilde{m}_{d v}(x/d) - zeta(2) sigma(dv)/dv)
		            (\tilde{m}_{d v}(y/d) - zeta(2) sigma(dv)/dv),
as usual.
			    
Default values: N=100000, r=1, v=1, flag=0;
	    as usual, the user may specify no values, or only N, r and v,
	    or all of them

Examples used in the text:

./boundhvxy 1000000 1 1
Output:
min: -1.542131773
minrat: -0.4681636593 at 2
integ: -0.7804306743
neginteg: -0.7812438307
mininteg: -0.8689343038

./boundhvxy 1000000 1 2
The following bounds concern f_{1,2}(x), defined as
         inf_{r x <= y <= 1000000} sqrt(x y) h_2(x,y) for r=1.
Then, for 1<=x<=1000000/r, where r=1,
f_{1,2}(x) >= -4.07843742
and f_{1,2}(x)/x >= -0.4310163289; that value may be reached near x=5
The integral of f_{1,2}/x^2 from 0 to infty is at least -1.011586799
The integral of min(0,f_{1,2})/x^2 from 0 to infty is at least -1.011586799
Define F_{1,2}(x) = max_{1<=t<=x} (-f_{1,2}(t)).
Then the integral of F_{1,2}(x)/x^2 from 0 to infty is at most  1.034883548

./boundhvxy 1000000 5 1
Output:
The following bounds concern f_{1,1}(x), defined as
	 inf_{r x <= y <= 1000000} sqrt(x y) h_1(x,y) for r=1.
Then, for 1<=x<=1000000/r, where r=1,
f_{1,1}(x) >= -1.205418281
and f_{1,1}(x)/x >= -0.4453389104; that value may be reached near x=1.9999
The integral of f_{1,1}/x^2 from 0 to infty is at least -0.6418587882
The integral of min(0,f_{1,1})/x^2 from 0 to infty is at least -0.6418587882
Define F_{1,1}(x) = max_{1<=t<=x} (-f_{1,1}(t)).
Then the integral of F_{1,1}(x)/x^2 from 0 to infty is at most  0.7495649282

Subsection 6.6.2, "Computing h_v(y) for y small or moderate"

./hvofy n0 N v

computes the following quantities (for v=1 or v=2):

The integral of h_v(t)/t from 1 to N,
the integral of \dot{h}_v(t)/t from 1 to N,  (\dot{h}(t) is defined in S 5.7)
a bound on the maximum of t h_v(t) on [n0,N],
a bound on the integral of (t h_v(t))' from 1 to N,
a bound on that last integral plus N h_v(N).

We need to run:

./hvofy 1000000 1000000000 1
./hvofy 1000000 1000000000 2

give all numerical results stated in subsection 5.5.2. Running time should
be at most a couple of hours in either case.

The output of the two calls reads:
----------------------------
The integral of h_1(t)/t from 1 to 1000000000 lies in [-0.0495100146363 -0.0495100043809]
The integral of \dot{h}_1(t)/t from 1 to 1000000000 lies in [0.724970140514 0.724970252531]
The maximum of h_1(t)/t from 1000000 to 1000000000 is at most 0.04547751029
The integral of |(t h_1)'| from 1 to 1000000000 is at most: [7.012358591 7.034022551]
If we add |1000000000 h_1(1000000000)| to that, we obtain at most: 7.049495205
----------------------------
The integral of h_2(t)/t from 1 to 1000000000 lies in [2.63481249177 2.63481271383]
The integral of \dot{h}_2(t)/t from 1 to 1000000000 lies in [1.44232816757 1.44232827893]
The maximum of h_2(t)/t from 1000000 to 1000000000 is at most 0.09773372982
The integral of |(t h_2)'| from 1 to 1000000000 is at most: [17.59226478 17.60841791]
If we add |1000000000 h_2(1000000000)| to that, we obtain at most: 17.6346278
-----------------------------

(Note: It might seem clearer to output

"The integral of |(t h_1)'| from 1 to 1000000000 is at most: 7.034022551",

say, since that is the meaning of the output. As it stands, the program
outputs an interval so as to give an idea of how error terms accumulate.)

_________________________________________________________________

./hboldvofy N v flag

bounds \mathbf{h}_v(y,y) for all y\leq N.

If flag=1 (default value: 0), the program outputs values of \mathbf{h}_v(y,y)
for many values of y\leq N.

The algorithm is not particularly clever; it runs in time O(N^2) and space
O(N log N), whereas time O(N) and space O(\sqrt{N}) should be possible.

Examples:

./hboldvofy 1000000 1
Output:
Let boldh_v(y,y) = sum_{d: (d,v)=1} mu^2(d)/sigma(d)^2
                               (tildem_{dv}(y/d) - zeta(2) sigma(d v)/d v)^2 
For all y<=1000000,
boldh_v(y,y) <= 9.395565599/y
boldh_v(y,y) <= (4.112339279 + 0.858 log y)/y

./hboldvofy 1000000 2 
Output:
Let boldh_v(y,y) = sum_{d: (d,v)=1} mu^2(d)/sigma(d)^2
                               (tildem_{dv}(y/d) - zeta(2) sigma(d v)/d v)^2 
For all y<=1000000,
boldh_v(y,y) <= 18.33611067/y
boldh_v(y,y) <= (7.716944705 + 1.75 log y)/y

Section 6.7, "Excursus on the Dress-Iwaniec-Tenenbaum constant"

We compute the integral \int_1^{10^9} \dot{h}(t) dt/t using the program hvofy
(section 5.5). Other than that, this section involves no new computations,
beyond trivial bookkeeping carried out in Sage code embedded in the .tex file.
